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IntroductiontoManagementScienceModelingandCaseStudies.pdf
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Introduction to Management Science A Modeling and Case Studies Approach with Spreadsheets
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The McGraw-Hill/Irwin Series Operations and Decision Sciences
OPERATIONS MANAGEMENT
Beckman and Rosenfield, Operations, Strategy: Competing in the 21st Century, First Edition
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QUANTITATIVE METHODS AND
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Hillier and Hillier, Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets, Fifth Edition
Stevenson and Ozgur, Introduction to Management Science with Spreadsheets, First Edition
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Introduction to Management Science A Modeling and Case Studies Approach with Spreadsheets
Fifth Edition
Frederick S. Hillier Stanford University
Mark S. Hillier University of Washington
Cases developed by
Karl Schmedders University of Zurich
Molly Stephens Quinn, Emanuel, Urquhart & Sullivan, LLP
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INTRODUCTION TO MANAGEMENT SCIENCE: A MODELING AND CASE STUDIES APPROACH WITH SPREADSHEETS, FIFTH EDITION
Published by McGraw-Hill/Irwin, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY, 10020. Copyright © 2014 by The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United States of America. Previous editions © 2011, 2008, 2003. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.
Some ancillaries, including electronic and print components, may not be available to customers outside the United States.
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Library of Congress Cataloging-in-Publication Data
Hillier, Frederick S.
Introduction to management science : modeling and case studies approach with spreadsheets / Frederick S. Hillier, Stanford University, Mark S. Hillier, University of Washington ; cases developed by Karl Schmedders, University of Zurich, Molly Stephens, Quinn, Emanuel, Urquhart, Sullivan LLP.—Fifth edition. pages cm ISBN 978-0-07-802406-1 (alk. paper) 1. Management science. 2. Operations research—Data processing. 3. Electronic spreadsheets. I. Hillier, Mark S. II. Title. T56.H55 2014 005.54—dc23 2012035364
The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill, and McGraw-Hill does not guarantee the accuracy of the information presented at these sites.
www.mhhe.com
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To the memory of
Christine Phillips Hillier a beloved wife and daughter-in-law
Gerald J. Lieberman an admired mentor and one of the true giants
of our field
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About the Authors Frederick S. Hillier is professor emeritus of operations research at Stanford University. Dr. Hillier is especially known for his classic, award-winning text, Introduction to Operations Research, co-authored with the late Gerald J. Lieberman, which has been translated into well over a dozen languages and is currently in its 9th edition. The 6th edition won honorable men- tion for the 1995 Lanchester Prize (best English-language publication of any kind in the field) and Dr. Hillier also was awarded the 2004 INFORMS Expository Writing Award for the 8th edition. His other books include The Evaluation of Risky Interrelated Investments, Queueing Tables and Graphs, Introduction to Stochastic Models in Operations Research, and Introduc- tion to Mathematical Programming. He received his BS in industrial engineering and doctorate specializing in operations research and management science from Stanford University. The winner of many awards in high school and college for writing, mathematics, debate, and music, he ranked first in his undergraduate engineering class and was awarded three national fel- lowships (National Science Foundation, Tau Beta Pi, and Danforth) for graduate study. After receiving his PhD degree, he joined the faculty of Stanford University, where he earned tenure at the age of 28 and the rank of full professor at 32. Dr. Hillier’s research has extended into a variety of areas, including integer programming, queueing theory and its application, statistical quality control, and production and operations management. He also has won a major prize for research in capital budgeting. Twice elected a national officer of professional societies, he has served in many important professional and editorial capacities. For example, he served The Institute of Management Sciences as vice president for meetings, chairman of the publications committee, associate editor of Management Science, and co-general chairman of an interna- tional conference in Japan. He also is a Fellow of the Institute for Operations Research and the Management Sciences (INFORMS). He currently is continuing to serve as the founding series editor for a prominent book series, the International Series in Operations Research and Man- agement Science, for Springer Science 1 Business Media. He has had visiting appointments at Cornell University, the Graduate School of Industrial Administration of Carnegie-Mellon Uni- versity, the Technical University of Denmark, the University of Canterbury (New Zealand), and the Judge Institute of Management Studies at the University of Cambridge (England).
Mark S. Hillier, son of Fred Hillier, is associate professor of quantitative methods at the Michael G. Foster School of Business at the University of Washington. Dr. Hillier received his BS in engineering (plus a concentration in computer science) from Swarthmore College. He then received his MS with distinction in operations research and PhD in industrial engi- neering and engineering management from Stanford University. As an undergraduate, he won the McCabe Award for ranking first in his engineering class, won election to Phi Beta Kappa based on his work in mathematics, set school records on the men’s swim team, and was awarded two national fellowships (National Science Foundation and Tau Beta Pi) for gradu- ate study. During that time, he also developed a comprehensive software tutorial package, OR Courseware, for the Hillier–Lieberman textbook, Introduction to Operations Research. As a graduate student, he taught a PhD-level seminar in operations management at Stanford and won a national prize for work based on his PhD dissertation. At the University of Wash- ington, he currently teaches courses in management science and spreadsheet modeling. He has won several MBA teaching awards for the core course in management science and his elective course in spreadsheet modeling, as well as a universitywide teaching award for his work in teaching undergraduate classes in operations management. He was chosen by MBA students in 2007 as the winner of the prestigious PACCAR award for Teacher of the Year (reputed to provide the largest monetary award for MBA teaching in the nation). He also has been awarded an appointment to the Evert McCabe Endowed Faculty Fellowship. His research interests include issues in component commonality, inventory, manufacturing, and the design of production systems. A paper by Dr. Hillier on component commonality won an award for best paper of 2000–2001 in IIE Transactions. He currently is principal investigator on a grant from the Bill and Melinda Gates Foundation to lead student research projects that apply spreadsheet modeling to various issues in global health being studied by the foundation.
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About the Case Writers Karl Schmedders is professor of quantitative business administration at the University of Zurich in Switzerland and a visiting associate professor at the Kellogg School of Man- agement of Northwestern University. His research interests include management science, financial economics, and computational economics and finance. In 2003, a paper by Dr. Schmedders received a nomination for the Smith-Breeden Prize for the best paper in the Jour- nal of Finance. He received his PhD in operations research from Stanford University, where he taught both undergraduate and graduate classes in management science, including a case studies course. He received several teaching awards at Stanford, including the university- wide Walter J. Gores Teaching Award. After a post-doc at the Hoover Institution, a think tank on the Stanford campus, he became assistant professor of managerial economics and decision sciences at the Kellogg School. He was promoted to associate professor in 2001 and received tenure in 2005. In 2008 he joined the University of Zurich, where he currently teaches courses in management science, spreadsheet modeling, and computational economics and finance. At Kellogg he received several teaching awards, including the L. G. Lavengood Professor of the Year Award. Most recently he won the best professor award of the Kellogg School’s Euro- pean EMBA program (2008, 2009, and 2011) and its Miami EMBA program (2011).
Molly Stephens is a partner in the Los Angeles office of Quinn, Emanuel, Urquhart & Sullivan, LLP. She graduated from Stanford with a BS in industrial engineering and an MS in operations research. Ms. Stephens taught public speaking in Stanford’s School of Engineer- ing and served as a teaching assistant for a case studies course in management science. As a teaching assistant, she analyzed management science problems encountered in the real world and transformed these into classroom case studies. Her research was rewarded when she won an undergraduate research grant from Stanford to continue her work and was invited to speak at INFORMS to present her conclusions regarding successful classroom case studies. Follow- ing graduation, Ms. Stephens worked at Andersen Consulting as a systems integrator, expe- riencing real cases from the inside, before resuming her graduate studies to earn a JD degree with honors from the University of Texas School of Law at Austin. She is a partner in the largest law firm in the United States devoted solely to business litigation, where her practice focuses on complex financial and securities litigation.
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Preface We have long been concerned that traditional management science textbooks have not taken the best approach in introducing business students to this exciting field. Our goal when ini- tially developing this book during the late 1990s was to break out of the old mold and present new and innovative ways of teaching management science more effectively. We have been gratified by the favorable response to our efforts. Many reviewers and other users of the first four editions of the book have expressed appreciation for its various distinctive features, as well as for its clear presentation at just the right level for their business students.
Our goal for this fifth edition has been to build on the strengths of the first four editions. Co-author Mark Hillier has won several schoolwide teaching awards for his spreadsheet mod- eling and management science courses at the University of Washington while using the first four editions, and this experience has led to many improvements in the current edition. We also incorporated many user comments and suggestions. Throughout this process, we took painstaking care to enhance the quality of the preceding edition while maintaining the distinc- tive orientation of the book.
This distinctive orientation is one that closely follows the recommendations in the 1996 report of the operating subcommittee of the INFORMS Business School Education Task Force, including the following extract.
There is clear evidence that there must be a major change in the character of the (introductory management science) course in this environment. There is little patience with courses centered on algorithms. Instead, the demand is for courses that focus on business situations, include prominent non-mathematical issues, use spreadsheets, and involve model formulation and assessment more than model structuring. Such a course requires new teaching materials.
This book is designed to provide the teaching materials for such a course. In line with the recommendations of this task force, we believe that a modern introductory
management science textbook should have three key elements. As summarized in the subtitle of this book, these elements are a modeling and case studies approach with spreadsheets.
SPREADSHEETS The modern approach to the teaching of management science clearly is to use spreadsheets as a primary medium of instruction. Both business students and managers now live with spreadsheets, so they provide a comfortable and enjoyable learning environment. Modern spreadsheet software, including Microsoft Excel used in this book, now can be used to do real management science. For student-scale models (which include many practical real-world models), spreadsheets are a much better way of implementing management science models than traditional algebraic solvers. This means that the algebraic curtain that was so prevalent in traditional management science courses and textbooks now can be lifted.
However, with the new enthusiasm for spreadsheets, there is a danger of going over- board. Spreadsheets are not the only useful tool for performing management science analy- ses. Occasional modest use of algebraic and graphical analyses still have their place and we would be doing a disservice to the students by not developing their skills in these areas when appropriate. Furthermore, the book should not be mainly a spreadsheet cookbook that focuses largely on spreadsheet mechanics. Spreadsheets are a means to an end, not an end in themselves.
A MODELING APPROACH This brings us to the second key feature of the book, a modeling approach. Model formula- tion lies at the heart of management science methodology. Therefore, we heavily emphasize the art of model formulation, the role of a model, and the analysis of model results. We pri- marily (but not exclusively) use a spreadsheet format rather than algebra for formulating and presenting a model.
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Some instructors have many years of experience in teaching modeling in terms of for- mulating algebraic models (or what the INFORMS Task Force called “model structuring”). Some of these instructors feel that students should do their modeling in this way and then transfer the model to a spreadsheet simply to use the Excel Solver to solve the model. We dis- agree with this approach. Our experience (and the experience reported by many others) is that most business students find it more natural and comfortable to do their modeling directly in a spreadsheet. Furthermore, by using the best spreadsheet modeling techniques (as presented in this edition), formulating a spreadsheet model tends to be considerably more efficient and transparent than formulating an algebraic model. Another benefit is that the spreadsheet model includes all the relationships that can be expressed in an algebraic form and we often will summarize the model in this format as well.
Another break from tradition in this book (and several contemporary textbooks) is to virtually ignore the algorithms that are used to solve the models. We feel that there is no good reason why typical business students should learn the details of algorithms executed by computers. Within the time constraints of a one-term management science course, there are far more important lessons to be learned. Therefore, the focus in this book is on what we believe are these far more important lessons. High on this list is the art of modeling managerial problems on a spreadsheet.
Formulating a spreadsheet model of a real problem typically involves much more than designing the spreadsheet and entering the data. Therefore, we work through the process step by step: understand the unstructured problem, verbally develop some structure for the problem, gather the data, express the relationships in quantitative terms, and then lay out the spreadsheet model. The structured approach highlights the typical components of the model (the data, the decisions to be made, the constraints, and the measure of performance) and the different types of spreadsheet cells used for each. Consequently, the emphasis is on the mod- eling rather than spreadsheet mechanics.
A CASE STUDIES APPROACH However, all this still would be quite sterile if we simply presented a long series of brief examples with their spreadsheet formulations. This leads to the third key feature of this book—a case studies approach. In addition to examples, nearly every chapter includes one or two case studies patterned after actual applications to convey the whole process of applying management science. In a few instances, the entire chapter revolves around a case study. By drawing the student into the story, we have designed each case study to bring that chapter’s technique to life in a context that vividly illustrates the relevance of the technique for aiding managerial decision making. This storytelling, case-centered approach should make the mate- rial more enjoyable and stimulating while also conveying the practical considerations that are key factors in applying management science.
We have been pleased to have several reviewers of the first four editions express particular appreciation for our case study approach. Even though this approach has received little use in other management science textbooks, we feel that it is a real key to preparing students for the practical application of management science in all its aspects. Some of the reviewers have highlighted the effectiveness of the dialogue/scenario enactment approach used in some of the case studies. Although unconventional, this approach provides a way of demonstrating the process of managerial decision making with the help of management science. It also enables previewing some key concepts in the language of management.
Every chapter also contains full-fledged cases following the problems at the end of the chapter. These cases usually continue to employ a stimulating storytelling approach, so they can be assigned as interesting and challenging projects. Most of these cases were developed jointly by two talented case writers, Karl Schmedders (a faculty member at the University of Zurich in Switzerland) and Molly Stephens (formerly a management science consultant with Andersen Consulting). The authors also have added some cases, including several shorter ones. In addition, the University of Western Ontario Ivey School of Business (the second-largest producer of teaching cases in the world) has specially selected cases from their case collection that match the chapters in this textbook. These cases are available on
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the Ivey website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book. This website address is provided at the end of each chapter as well.
We are, of course, not the first to incorporate any of these key features into a management science textbook. However, we believe that the book currently is unique in the way that it fully incorporates all three key features together.
OTHER SPECIAL FEATURES We also should mention some additional special features of the book that are continued from the fourth edition.
• Diverse examples, problems, and cases convey the pervasive relevance of management science.
• A strong managerial perspective. • Learning objectives at the beginning of each chapter. • Numerous margin notes that clarify and highlight key points. • Excel tips interspersed among the margin notes. • Review questions at the end of each section. • A glossary at the end of each chapter. • Partial answers to selected problems in the back of the book. • Supplementary text material on the CD-ROM (as identified in the table of contents). • An Excel-based software package (MS Courseware) on the CD-ROM and website that
includes many add-ins, templates, and files (described below). • Other helpful supplements on the CD-ROM and website (described later).
A NEW SOFTWARE PACKAGE This edition continues to integrate Excel 2010 and its Solver (a product of Frontline Systems) throughout the book. However, we are excited to also add to this edition an impressive more recent product of Frontline Systems called Risk Solver Platform for Education (or RSPE for short). RSPE also is an Excel add-in and its Solver shares some of the features of the Excel Solver. However, in addition to providing all the key capabilities of the Excel Solver, RSPE adds some major new functionalities as outlined below:
• A more interactive user interface, with the model parameters always visible alongside the main spreadsheet, rather than only in the Solver dialog box.
• Parameter analysis reports that provide an easy way to see the effect of varying data in a model in a systematic way.
• A model analysis tool that reveals the characteristics of a model (e.g., whether it is linear or nonlinear, smooth or nonsmooth).
• Tools to build and solve decision trees within a spreadsheet. • The ability to build and run sophisticated Monte Carlo simulation models. • An interactive simulation mode that allows simulation results to be shown instantly when-
ever a change is made to a simulation model. • The RSPE Solver can be used in combination with computer simulation to perform simu-
lation optimization.
A CONTINUING FOCUS ON EXCEL AND ITS SOLVER As with all the preceding editions, this edition continues to focus on spreadsheet modeling in an Excel format. Although it lacks some of the functionalities of RSPE, the Excel Solver continues to provide a completely satisfactory way of solving most of the spreadsheet models encountered in this book. This edition continues to feature this use of the Excel Solver when- ever either it or the RSPE Solver could be used.
Many instructors prefer this focus because it avoids introducing other complications that might confuse their students. We agree.
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However, the key advantage of introducing RSPE in this edition is that it provides an all-in- one complement to the Excel Solver. There are some important topics in the book (including decision analysis and computer simulation) where the Excel Solver lacks the functionalities needed to deal with these kinds of problems. Multiple Excel add-ins—Solver Table, Tree- Plan, SensIt, RiskSim, Crystal Ball, and OptQuest (a module of Crystal Ball)—were intro- duced in previous editions to provide the needed functionalities. RSPE alone now replaces all of these add-ins.
OTHER SOFTWARE Each edition of this book has provided a comprehensive Excel-based software package called MS Courseware on the CD-ROM and website. RSPE replaces various Excel add-ins in this package. Otherwise, the remainder of this package is being provided again with the current edition.
This package includes Excel files that provide the live spreadsheets for all the various examples and case studies throughout the book. In addition to further investigating the exam- ples and case studies, these spreadsheets can be used by either the student or instructor as tem- plates to formulate and solve similar problems. The package also includes dozens of Excel templates for solving various models in the book.
MS Courseware includes additional software as well.
• Interactive Management Science Modules for interactively exploring certain manage- ment science techniques in depth (including techniques presented in Chapters 1, 2, 5, 10, 11, 12, and 18).
• Queueing Simulator for performing computer simulations of queueing systems (used in Chapter 12).
NEW FEATURES IN THIS EDITION We have made some important enhancements to the fifth edition.
• A Substantial Revision of Chapter 1. In addition to some updates and a new end-of- chapter case, the example at the heart of the chapter has been modernized to better attract the interest of the students. The example now deals with iWatches instead of grandfather clocks.
• A New Section Introduces Risk Solver Platform for Education (RSPE). Section 2.6 presents the basics of how to use RSPE. It is placed near the end of Chapter 2 to avoid disrupting the flow of the chapter, including the introduction of the Excel Solver.
• Parameter Analysis Reports Are Introduced and Widely Used. Parameter analysis reports are introduced in Chapter 5 for performing sensitivity analysis systematically. This key tool of RSPE also receives important use in Chapters 7, 8, and 13.
• Chapter 8 Is Revised to Better Identify the Available Solving Methods for Nonlinear Programming. The Excel Solver and the RSPE Solver share some solving methods for nonlinear programming and then the RSPE Solver adds another one. These solving meth- ods and when each one should be used are better identified now.
• A New Section on Using RSPE to Analyze a Model and Choose a Solving Method. A new Section 8.6 describes a key tool of RSPE for analyzing a model and choosing the best solving method.
• A Substantial Revision of Chapter 9 (Decision Analysis). RSPE has outstanding func- tionality for constructing and analyzing decision trees. This functionality is thoroughly exploited in the revised Chapter 9.
• A Key Revision of the First Computer Simulation Chapter. Computer simulation commonly is used to analyze complicated queueing systems, so it is natural for Chapter 12 (Computer Simulation: Basic Concepts) to refer back to Chapter 11 (Queueing Mod- els) occasionally. However, some instructors cover Chapter 12 but skip over Chapter 11. Therefore, we have revised Chapter 12 to make it as independent of Chapter 11 as possible while still covering this important kind of application of computer simulation.
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• A Major Revision of the Second Computer Simulation Chapter. Although the exam- ples remain the same, the old Chapter 13 (Computer Simulation with Crystal Ball) has been thoroughly revised to replace Crystal Ball by Risk Solver Platform for Education (RSPE). Most students already will be familiar with RSPE from preceding chapters, which should provide a gentler entry into this chapter. More importantly, this impressive, relatively new software package has some significant advantages over Crystal Ball for performing and analyzing computer simulations. However, an updated version of the old Chapter 13 still will be available on the CD-ROM (now Chapter 20) for instructors who wish to stick with Crystal Ball for the time being.
• A New Section on Decision Making with Computer Simulations. A key tool of RSPE is its use of multiple simulation runs to generate parameter analysis reports and trend charts that can provide an important guide to managerial decision making. Section 13.8 describes this approach to decision making.
• A New Section on Optimizing with Computer Simulations. Another key tool of RSPE is that its Solver can use multiple simulation runs to automatically search for an opti- mal solution for simulation models with any number of decision variables. Section 13.9 describes this approach.
• Additional Links to Articles that Describe Dramatic Real Applications. The fourth edition includes 23 application vignettes that describe in a few paragraphs how an actual application of management science had a powerful effect on a company or organization by using techniques like those being studied in that portion of the book. The current edition adds seven more vignettes based on recent applications (while deleting two old ones). We also continue the practice of adding a link to the journal articles that fully describe these applications, through a special arrangement with the Institute for Operations Research and the Management Sciences (INFORMS ® ). Thus, the instructor now can motivate his or her lectures by having the students delve into real applications that dramatically demonstrate the relevance of the material being covered in the lectures. The end-of-chapter problems also include an assignment after reading each of these articles.
We continue to be excited about this partnership with INFORMS, our field’s preeminent professional society, to provide a link to these 28 articles describing spectacular applica- tions of management science. INFORMS is a learned professional society for students, aca- demics, and practitioners in quantitative and analytical fields. Information about INFORMS journals, meetings, job bank, scholarships, awards, and teaching materials is available at www.informs.org .
• Refinements in Each Chapter. Each chapter in the fourth edition has been carefully examined and revised as needed to update and clarify the material after also taking into account the input provided by reviewers and others.
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OTHER SUPPLEMENTS The Instructor’s Edition of this book’s Online Learning Center, www.mhhe.com/hillier5e , is password-protected and a convenient place for instructors to access course supplements. Resources for professors include the complete solutions to all problems and cases, a test bank with hundreds of multiple-choice and true-false questions, and PowerPoint Presentation. The PowerPoint slides include both lecture materials for nearly every chapter and nearly all the figures (including all the spreadsheets) in the book.
The student’s CD-ROM bundled with the book provides most of the MS Courseware package. It also includes a tutorial with sample test questions (different from those in the instructor’s test bank) for self-testing quizzes on the various chapters.
The materials on the student CD-ROM can also be accessed on the Student’s Edition of the Online Learning Center, www.mhhe.com/hillier5e . The website also provides the remain- der of the MS Courseware package, as well as access to the INFORMS articles cited in the application vignettes and updates about the book, including errata. In addition, the publisher’s operations management supersite at www.mhhe.com/pom/ links to many resources on the Internet that you might find pertinent to this book.
We invite your comments, suggestions, and errata. You can contact either one of us at the e-mail addresses given below. While giving these addresses, let us also assure instructors that we will continue our policy of not providing solutions to problems and cases in the book to anyone (including your students) who contacts us. We hope that you enjoy the book.
Frederick S. Hillier Stanford University ( [email protected] )
Mark S. Hillier University of Washington ( [email protected] )
June 2012
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Acknowledgments This new edition has benefited greatly from the sage advice of many individuals. To begin, we would like to express our deep appreciation to the following individuals who provided formal reviews of the fourth edition:
Michael Cervetti University of Memphis
Jose Dula Virginia Commonwealth University
Harvey Iglarsh Georgetown University
Michael (Tony) Ratcliffe James Madison University
John Wang Montclair State University
Jinfeng Yue Middle Tennessee State University
We also are grateful for the valuable input provided by many of our students as well as various other students and instructors who contacted us via e-mail.
This book has continued to be a team effort involving far more than the two coauthors. As a third coauthor for the first edition, the late Gerald J. Lieberman provided important initial impetus for this project. We also are indebted to our case writers, Karl Schmedders and Molly Stephens, for their invaluable contributions. Ann Hillier again devoted numerous hours to sitting with a Macintosh, doing word processing and constructing figures and tables. They all were vital members of the team.
McGraw-Hill/Irwin’s editorial and production staff provided the other key members of the team, including Douglas Reiner (Publisher), Beth Baugh (Freelance Developmental Editor), and Mary Jane Lampe (Project Manager). This book is a much better product because of their guidance and hard work. It has been a real pleasure working with such a thoroughly profes- sional staff.
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1 Introduction 1
2 Linear Programming: Basic Concepts 22
3 Linear Programming: Formulation and Applications 64
4 The Art of Modeling with Spreadsheets 124
5 What-If Analysis for Linear Programming 150
6 Network Optimization Problems 194
7 Using Binary Integer Programming to Deal with Yes-or-No Decisions 232
8 Nonlinear Programming 267
9 Decision Analysis 322
10 Forecasting 384
11 Queueing Models 433
12 Computer Simulation: Basic Concepts 488
13 Computer Simulation with Risk Solver Platform 525
APPENDIXES
A Tips for Using Microsoft Excel for Modeling 599
B Partial Answers to Selected Problems 605
INDEX 609
SUPPLEMENTS ON THE CD-ROM
Supplement to Chapter 2: More about the Graphical Method for Linear Programming
Supplement to Chapter 5: Reduced Costs
Supplement to Chapter 6: Minimum Span- ning-Tree Problems
Supplement 1 to Chapter 7: Advanced For- mulation Techniques for Binary Integer Programming
Supplement 2 to Chapter 7: Some Perspec- tives on Solving Binary Integer Program- ming Problems
Supplement 1 to Chapter 9: Decision Criteria
Supplement 2 to Chapter 9: Using TreePlan Software for Decision Trees
Supplement to Chapter 11: Additional Queueing Models
Supplement to Chapter 12: The Inverse Transformation Method for Generating Random Observations
CHAPTERS ON THE CD-ROM
14 Solution Concepts for Linear Programming
15 Transportation and Assignment Problems
16 PERT/CPM Models for Project Management
17 Goal Programming
18 Inventory Management with Known Demand
19 Inventory Management with Uncertain Demand
20 Computer Simulation with Crystal Ball
Brief Contents
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Contents Chapter 1 Introduction 1
1.1 The Nature of Management Science 2 1.2 An Illustration of the Management Science
Approach: Break-Even Analysis 6 1.3 The Impact of Management Science 12 1.4 Some Special Features of This Book 14 1.5 Summary 17 Glossary 17 Learning Aids for This Chapter in Your MS Courseware 18 Solved Problem 18 Problems 18 Case 1-1 Keeping Time 20
Chapter 2 Linear Programming: Basic Concepts 22
2.1 A Case Study: The Wyndor Glass Co. Product- Mix Problem 23
2.2 Formulating the Wyndor Problem on a Spreadsheet 25
2.3 The Mathematical Model in the Spreadsheet 31 2.4 The Graphical Method for Solving Two-Variable
Problems 33 2.5 Using Excel’s Solver to Solve Linear Program-
ming Problems 38 2.6 Risk Solver Platform for Education (RSPE) 42 2.7 A Minimization Example—The Profit & Gambit
Co. Advertising-Mix Problem 46 2.8 Linear Programming from a Broader
Perspective 51 2.9 Summary 53 Glossary 53 Learning Aids for This Chapter in Your MS Courseware 54 Solved Problems 54 Problems 54 Case 2-1 Auto Assembly 60 Case 2-2 Cutting Cafeteria Costs 61 Case 2-3 Staffing a Call Center 62
Chapter 3 Linear Programming: Formulation and Applications 64
3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 65
3.2 Resource-Allocation Problems 71 3.3 Cost–Benefit–Trade-Off Problems 81
3.4 Mixed Problems 88 3.5 Transportation Problems 95 3.6 Assignment Problems 99 3.7 Model Formulation from a Broader
Perspective 102 3.8 Summary 103 Glossary 104 Learning Aids for This Chapter in Your MS Courseware 104 Solved Problems 104 Problems 105 Case 3-1 Shipping Wood to Market 114 Case 3-2 Capacity Concerns 115 Case 3-3 Fabrics and Fall Fashions 116 Case 3-4 New Frontiers 118 Case 3-5 Assigning Students to Schools 119 Case 3-6 Reclaiming Solid Wastes 120 Case 3-7 Project Pickings 121
Chapter 4 The Art of Modeling with Spreadsheets 124
4.1 A Case Study: The Everglade Golden Years Company Cash Flow Problem 125
4.2 Overview of the Process of Modeling with Spreadsheets 126
4.3 Some Guidelines for Building “Good” Spread- sheet Models 135
4.4 Debugging a Spreadsheet Model 141 4.5 Summary 144 Glossary 145 Learning Aids for This Chapter in Your MS Courseware 145 Solved Problems 145 Problems 146 Case 4-1 Prudent Provisions for Pensions 148
Chapter 5 What-If Analysis for Linear Programming 150
5.1 The Importance of What-If Analysis to Managers 151
5.2 Continuing the Wyndor Case Study 153 5.3 The Effect of Changes in One Objective Function
Coefficient 155 5.4 The Effect of Simultaneous Changes in Objective
Function Coefficients 161 5.5 The Effect of Single Changes in a
Constraint 169
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5.6 The Effect of Simultaneous Changes in the Constraints 175
5.7 Summary 179 Glossary 179 Learning Aids for This Chapter in Your MS Courseware 180 Solved Problem 180 Problems 181 Case 5-1 Selling Soap 188 Case 5-2 Controlling Air Pollution 189 Case 5-3 Farm Management 191 Case 5-4 Assigning Students to
Schools (Revisited) 193
Chapter 6 Network Optimization Problems 194
6.1 Minimum-Cost Flow Problems 195 6.2 A Case Study: The BMZ Co. Maximum Flow
Problem 202 6.3 Maximum Flow Problems 205 6.4 Shortest Path Problems 209 6.5 Summary 218 Glossary 219 Learning Aids for This Chapter in Your MS Courseware 219 Solved Problems 219 Problems 220 Case 6-1 Aiding Allies 224 Case 6-2 Money in Motion 227 Case 6-3 Airline Scheduling 229 Case 6-4 Broadcasting the Olympic
Games 230
Chapter 7 Using Binary Integer Programming to Deal with Yes-or-No Decisions 232
7.1 A Case Study: The California Manufacturing Co. Problem 233
7.2 Using BIP for Project Selection: The Tazer Corp. Problem 239
7.3 Using BIP for the Selection of Sites for Emer- gency Services Facilities: The Caliente City Problem 241
7.4 Using BIP for Crew Scheduling: The Southwest- ern Airways Problem 246
7.5 Using Mixed BIP to Deal with Setup Costs for Initiating Production: The Revised Wyndor Problem 250
7.6 Summary 254 Glossary 255 Learning Aids for This Chapter in Your MS Courseware 255
Solved Problems 255 Problems 257 Case 7-1 Assigning Art 261 Case 7-2 Stocking Sets 263 Case 7-3 Assigning Students to
Schools (Revisited) 266 Case 7-4 Broadcasting the Olympic Games
(Revisited) 266
Chapter 8 Nonlinear Programming 267
8.1 The Challenges of Nonlinear Programming 269 8.2 Nonlinear Programming with Decreasing
Marginal Returns 277 8.3 Separable Programming 287 8.4 Difficult Nonlinear Programming Problems 297 8.5 Evolutionary Solver and Genetic
Algorithms 299 8.6 Using RSPE to Analyze a Model and Choose a
Solving Method 306 8.7 Summary 310 Glossary 311 Learning Aids for This Chapter in Your MS Courseware 312 Solved Problem 312 Problems 312 Case 8-1 Continuation of the Super Grain Case
Study 317 Case 8-2 Savvy Stock Selection 318 Case 8-3 International Investments 319
Chapter 9 Decision Analysis 322
9.1 A Case Study: The Goferbroke Company Problem 323
9.2 Decision Criteria 325 9.3 Decision Trees 330 9.4 Sensitivity Analysis with Decision Trees 333 9.5 Checking Whether to Obtain More
Information 338 9.6 Using New Information to Update the
Probabilities 340 9.7 Using a Decision Tree to Analyze the Problem
with a Sequence of Decisions 344 9.8 Performing Sensitivity Analysis on the Problem
with a Sequence of Decisions 351 9.9 Using Utilities to Better Reflect the Values of
Payoffs 354 9.10 The Practical Application of Decision
Analysis 365 9.11 Summary 366 Glossary 367
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Learning Aids for This Chapter in Your MS Courseware 368 Solved Problems 368 Problems 369 Case 9-1 Who Wants to Be a Millionaire? 379 Case 9-2 University Toys and the Business Professor
Action Figures 379 Case 9-3 Brainy Business 380 Case 9-4 Smart Steering Support 382
Chapter 10 Forecasting 384
10.1 An Overview of Forecasting Techniques 385 10.2 A Case Study: The Computer Club Warehouse
(CCW) Problem 386 10.3 Applying Time-Series Forecasting Methods to
the Case Study 391 10.4 The Time-Series Forecasting Methods in
Perspective 410 10.5 Causal Forecasting with Linear Regression 413 10.6 Judgmental Forecasting Methods 418 10.7 Summary 419 Glossary 420 Summary of Key Formulas 421 Learning Aids for This Chapter in Your MS Courseware 421 Solved Problem 421 Problems 422 Case 10-1 Finagling the Forecasts 429
Chapter 11 Queueing Models 433
11.1 Elements of a Queueing Model 434 11.2 Some Examples of Queueing Systems 440 11.3 Measures of Performance for Queueing
Systems 442 11.4 A Case Study: The Dupit Corp. Problem 445 11.5 Some Single-Server Queueing Models 448 11.6 Some Multiple-Server Queueing Models 457 11.7 Priority Queueing Models 463 11.8 Some Insights about Designing Queueing
Systems 469 11.9 Economic Analysis of the Number of Servers to
Provide 473 11.10 Summary 476 Glossary 477 Key Symbols 478 Learning Aids for This Chapter in Your MS Courseware 478 Solved Problem 478 Problems 479 Case 11-1 Queueing Quandary 485 Case 11-2 Reducing In-Process Inventory 486
Chapter 12 Computer Simulation: Basic Concepts 488
12.1 The Essence of Computer Simulation 489 12.2 A Case Study: Herr Cutter’s Barber Shop
(Revisited) 501 12.3 Analysis of the Case Study 508 12.4 Outline of a Major Computer Simulation
Study 515 12.5 Summary 518 Glossary 518 Learning Aids for This Chapter in Your MS Courseware 519 Solved Problem 519 Problems 519 Case 12-1 Planning Planers 523 Case 12-2 Reducing In-Process Inventory
(Revisited) 524
Chapter 13 Computer Simulation with Risk Solver Platform 525
13.1 A Case Study: Freddie the Newsboy’s Problem 526
13.2 Bidding for a Construction Project: A Prelude to the Reliable Construction Co. Case Study 536
13.3 Project Management: Revisiting the Reliable Construction Co. Case Study 540
13.4 Cash Flow Management: Revisiting the Ever- glade Golden Years Company Case Study 546
13.5 Financial Risk Analysis: Revisiting the Think- Big Development Co. Problem 552
13.6 Revenue Management in the Travel Industry 557
13.7 Choosing the Right Distribution 562 13.8 Decision Making with Parameter Analysis
Reports and Trend Charts 575 13.9 Optimizing with Computer Simulation Using
RSPE’s Solver 583 13.10 Summary 590 Glossary 591 Learning Aids for This Chapter in Your MS Courseware 591 Solved Problem 591 Problems 592 Case 13-1 Action Adventures 596 Case 13-2 Pricing under Pressure 597
Appendix A Tips for Using Microsoft Excel for Modeling 599
Appendix B Partial Answers to Selected Problems 605
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Index 609
Supplements on the CD-ROM
Supplement to Chapter 2: More about the Graphical Method for Linear Programming
Supplement to Chapter 5: Reduced Costs Supplement to Chapter 6: Minimum Spanning-
Tree Problems Supplement 1 to Chapter 7: Advanced Formula-
tion Techniques for Binary Integer Programming Supplement 2 to Chapter 7: Some Perspectives on
Solving Binary Integer Programming Problems Supplement 1 to Chapter 9: Decision Criteria Supplement 2 to Chapter 9: Using TreePlan
Software for Decision Trees Supplement to Chapter 11: Additional Queueing
Models Supplement to Chapter 12: The Inverse Trans-
formation Method for Generating Random Observations
Chapters on the CD-ROM Chapter 14 Solution Concepts for Linear Programming
14.1 Some Key Facts about Optimal Solutions 14.2 The Role of Corner Points in Searching for an
Optimal Solution 14.3 Solution Concepts for the Simplex Method 14.4 The Simplex Method with Two Decision
Variables 14.5 The Simplex Method with Three Decision
Variables 14.6 The Role of Supplementary Variables 14.7 Some Algebraic Details for the Simplex Method 14.8 Computer Implementation of the Simplex
Method 14.9 The Interior-Point Approach to Solving Linear
Programming Problems 14.10 Summary Glossary Learning Aids for This Chapter in Your MS Courseware Problems
Chapter 15 Transportation and Assignment Problems
15.1 A Case Study: The P & T Company Distribution Problem
15.2 Characteristics of Transportation Problems 15.3 Modeling Variants of Transportation Problems 15.4 Some Other Applications of Variants of Trans-
portation Problems
15.5 A Case Study: The Texago Corp. Site Selection Problem
15.6 Characteristics of Assignment Problems 15.7 Modeling Variants of Assignment Problems 15.8 Summary Glossary Learning Aids for This Chapter in Your MS Courseware Problems Case 15-1 Continuation of the Texago Case Study
Chapter 16 PERT/CPM Models for Project Management
16.1 A Case Study: The Reliable Construction Co. Project
16.2 Using a Network to Visually Display a Project 16.3 Scheduling a Project with PERT/CPM 16.4 Dealing with Uncertain Activity Durations 16.5 Considering Time–Cost Trade-Offs 16.6 Scheduling and Controlling Project Costs 16.7 An Evaluation of PERT/CPM from a Managerial
Perspective 16.8 Summary Glossary Learning Aids for This Chapter in Your MS Courseware Problems Case 16-1 Steps to Success Case 16-2 “School’s Out Forever . . . ”
Chapter 17 Goal Programming
17.1 A Case Study: The Dewright Co. Goal-Programming Problem
17.2 Weighted Goal Programming 17.3 Preemptive Goal Programming 17.4 Summary Glossary Learning Aids for This Chapter in Your MS Courseware Problems Case 17-1 A Cure for Cuba Case 17-2 Remembering September 11
Chapter 18 Inventory Management with Known Demand
18.1 A Case Study: The Atlantic Coast Tire Corp. (ACT) Problem
18.2 Cost Components of Inventory Models 18.3 The Basic Economic Order Quantity (EOQ)
Model
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Problems Case 19-1 TNT: Tackling Newsboy’s Teachings Case 19-2 Jettisoning Surplus Stock
Chapter 20 Computer Simulation with Crystal Ball
20.1 A Case Study: Freddy the Newsboy’s Problem 20.2 Bidding for a Construction Project: A Prelude to
the Reliable Construction Co. Case Study 20.3 Project Management: Revisiting the Reliable
Construction Co. Case Study 20.4 Cash Flow Management: Revisiting the Ever-
glade Golden Years Company Case Study 20.5 Financial Risk Analysis: Revisiting the Think-
Big Development Co. Problem 20.6 Revenue Management in the Travel Industry 20.7 Choosing the Right Distribution 20.8 Decision Making with Decision Tables 20.9 Optimizing with OptQuest 20.10 Summary Glossary Learning Aids for This Chapter in Your MS Courseware Solved Problem Problems Case 20-1 Action Adventures Case 20-2 Pricing under Pressure
18.4 The Optimal Inventory Policy for the Basic EOQ Model
18.5 The EOQ Model with Planned Shortages 18.6 The EOQ Model with Quantity Discounts 18.7 The EOQ Model with Gradual Replenishment 18.8 Summary Glossary Learning Aids for This Chapter in Your MS Courseware Problems Case 18-1 Brushing Up on Inventory Control
Chapter 19 Inventory Management with Uncertain Demand
19.1 A Case Study for Perishable Products: Freddie the Newsboy’s Problem
19.2 An Inventory Model for Perishable Products 19.3 A Case Study for Stable Products: The Niko
Camera Corp. Problem 19.4 The Management Science Team’s Analysis
of the Case Study 19.5 A Continuous-Review Inventory Model for
Stable Products 19.6 Larger Inventory Systems in Practice 19.7 Summary Glossary Learning Aids for This Chapter in Your MS Courseware
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1
Chapter One
Introduction Learning Objectives
After completing this chapter, you should be able to
1. Define the term management science.
2. Describe the nature of management science.
3. Explain what a mathematical model is.
4. Use a mathematical model to perform break-even analysis.
5. Use a spreadsheet model to perform break-even analysis.
6. Identify the levels of annual savings that management science sometimes can provide to organizations.
7. Identify some special features of this book.
Welcome to the field of management science! We think that it is a particularly exciting and interesting field. Exciting because management science is having a dramatic impact on the profitability of numerous business firms around the world. Interesting because the methods used to do this are so ingenious. We are looking forward to giving you a guided tour to intro- duce you to the special features of the field.
Some students approach a course (and textbook) about management science with a cer- tain amount of anxiety and skepticism. The main source of the anxiety is the reputation of the field as being highly mathematical. This reputation then generates skepticism that such a theoretical approach can have much relevance for dealing with practical managerial prob- lems. Most traditional courses (and textbooks) about management science have only rein- forced these perceptions by emphasizing the mathematics of the field rather than its practical application.
Rest easy. This is not a traditional management science textbook. We realize that most readers of this book are aspiring to become managers, not mathematicians. Therefore, the emphasis throughout is on conveying what a future manager needs to know about manage- ment science. Yes, this means including a little mathematics here and there, because it is a major language of the field. The mathematics you do see will be at the level of high school algebra plus (in the later chapters) basic concepts of elementary probability theory. We think you will be pleasantly surprised by the new appreciation you gain for how useful and intuitive mathematics at this level can be. However, managers do not need to know any of the heavy mathematical theory that underlies the various techniques of management science. Therefore, the use of mathematics plays only a strictly secondary role in the book.
One reason we can deemphasize mathematics is that powerful spreadsheet software now is available for applying management science. Spreadsheets provide a comfortable and familiar environment for formulating and analyzing managerial problems. The spreadsheet takes care of applying the necessary mathematics automatically in the background with only a minimum of guidance by the user. This has begun to revolutionize the use of management science. In the past, technically trained management scientists were needed to carry out significant management science studies for management. Now spreadsheets are bringing many of the tools and concepts of management science within the reach of managers for conducting their own analyses. Although busy managers will continue to call upon management science teams to conduct major studies for them, they are increasingly becoming direct users themselves
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2 Chapter One Introduction
through the medium of spreadsheet software. Therefore, since this book is aimed at future managers (and management consultants), we will emphasize the use of spreadsheets for applying management science.
What does an enlightened future manager need to learn from a management science course?
1. Gain an appreciation for the relevance and power of management science. (Therefore, we include many application vignettes throughout the book that give examples of actual appli- cations of management science and the impact they had on the organizations involved.)
2. Learn to recognize when management science can (and cannot) be fruitfully applied. (Therefore, we will emphasize the kinds of problems to which the various management science techniques can be applied.)
3. Learn how to apply the major techniques of management science to analyze a variety of man- agerial problems. (Therefore, we will focus largely on how spreadsheets enable many such applications with no more background in management science than provided by this book.)
4. Develop an understanding of how to interpret the results of a management science study. (Therefore, we will present many case studies that illustrate management science studies and how their results depend on the assumptions and data that were used.)
The objectives just described are the key teaching goals of this book. We begin this process in the next two sections by introducing the nature of management
science and the impact that it is having on many organizations. (These themes will continue throughout the remaining chapters as well.) Section 1.4 then points out some of the special features of this book that you can look forward to seeing in the subsequent chapters.
1.1 THE NATURE OF MANAGEMENT SCIENCE
What is the name management science (sometimes abbreviated MS) supposed to convey? It does involve management and science or, more precisely, the science of management, but this still is too vague. Here is a more suggestive definition.
Management science is a discipline that attempts to aid managerial decision making by applying a scientific approach to managerial problems that involve quantitative factors.
Now let us see how elaborating upon each of the italicized terms in this definition conveys much more about the nature of management science.
Management Science Is a Discipline As a discipline, management science is a whole body of knowledge and techniques that are based on a scientific foundation. For example, it is analogous in some ways to the medical field. A medical doctor has been trained in a whole body of knowledge and techniques that are based on the scientific foundations of the medical field. After receiving this training and entering practice, the doctor must diagnose a patient’s illness and then choose the appropriate medical procedures to apply to the illness. The patient then makes the final decision on which medical procedures to accept. For less serious cases, the patient may choose not to consult a doctor and instead use his own basic knowledge of medical principles to treat himself. Similarly, a management scientist must receive substantial training (albeit considerably less than for a medical doctor). This training also is in a whole body of knowledge and techniques that are based on the scientific foundations of the discipline. After entering practice, the management scientist must diagnose a managerial problem and then choose the appropriate management science techniques to apply in analyzing the problem. The cognizant manager then makes the final decision as to which conclusions from this analysis to accept. For less extensive managerial problems where management science can be helpful, the manager may choose not to consult a management scientist and instead use his or her own basic knowledge of management science principles to analyze the problem.
Although it has considerably longer roots, the rapid development of the discipline began in the 1940s and 1950s. The initial impetus came early in World War II, when large numbers of scientists were called upon to apply a scientific approach to the management of the war
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1.1 The Nature of Management Science 3
effort for the allies. Another landmark event was the discovery in 1947 by George Dantzig of the simplex method for solving linear programming problems. (Linear programming is the subject of several early chapters.) Another factor that gave great impetus to the growth of the discipline was the onslaught of the computer revolution.
The traditional name given to the discipline (and the one that still is widely used today outside of business schools) is operations research . This name was applied because the teams of scientists in World War II were doing research on how to manage military opera- tions. The abbreviation OR also is widely used. This abbreviation often is combined with the one for management science (MS), thereby referring to the discipline as OR/MS. According to projections from the U.S. Bureau of Labor Statistics for the year 2013, there are approxi- mately 65,000 individuals working as operations research analysts in the United States with an average annual salary of about $79,000.
Another discipline that is closely related to management science is business analytics. Like management science, business analytics attempts to aid managerial decision making but with particular emphasis on three types of analysis: (1) descriptive analytics —the use of data (sometimes massive amounts of data) to analyze trends, (2) predictive analytics —the use of data to predict what will happen in the future (perhaps by using the forecasting techniques described in Chapter 10), and (3) prescriptive analytics —the use of data to prescribe the best course of action (frequently by using the optimization techniques described throughout this book). Broadly speaking, the techniques of the management science discipline provide the firepower for prescriptive analytics and, to a lesser extent, for predictive analytics, but not so much for descriptive analytics.
One major international professional society for the management science discipline (as well as for business analytics) is the Institute for Operations Research and the Man- agement Sciences (INFORMS). Headquartered in the United States, with over 10,000 members, this society holds major conferences in the United States each year (including an annual Conference for Business Analytics and Operations Research) plus occasional conferences elsewhere. It also publishes several prominent journals, including Manage- ment Science, Operations Research, Analytics, and Interfaces. (Articles describing actual applications of management science are featured in Interfaces, so you will see many refer- ences and links to this journal throughout the book.) In addition, a few dozen countries around the world have their own national operations research societies. (More about this in Section 1.3.)
Thus, operations research/management science (OR/MS) is a truly international discipline. (We hereafter will just use the name management science or the abbreviation MS.)
Management Science Aids Managerial Decision Making The key word here is that management science aids managerial decision making. Manage- ment scientists don’t make managerial decisions. Managers do. A management science study only provides an analysis and recommendations, based on the quantitative factors involved in the problem, as input to the cognizant managers. Managers must also take into account vari- ous intangible considerations that are outside the realm of management science and then use their best judgment to make the decision. Sometimes managers find that qualitative factors are as important as quantitative factors in making a decision.
A small informal management science study might be conducted by just a single individ- ual, who may be the cognizant manager. However, management science teams normally are used for larger studies. (We often will use the term team to cover both cases throughout the book.) Such a team often includes some members who are not management scientists but who provide other types of expertise needed for the study. Although a management science team often is entirely in-house (employees of the company), part or all of the team may instead be consultants who have been hired for just the one study. Consulting firms that partially or entirely specialize in management science currently are a growing industry.
Management Science Uses a Scientific Approach Management science is based strongly on some scientific fields, including mathematics and computer science. It also draws on the social sciences, especially economics. Since the field
operations research Management science began its rapid development dur- ing World War II with the name operations research.
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4 Chapter One Introduction
is concerned with the practical management of organizations, a management scientist should have solid training in business administration, including its various functional areas, as well.
To a considerable extent, a management science team will attempt to use the scientific method in conducting its study. This means that the team will emphasize conducting a sys- tematic investigation that includes careful data gathering, developing and testing hypotheses about the problem (typically in the form of a mathematical model), and then applying sound logic in the subsequent analysis.
When conducting this systematic investigation, the management science team typically will follow the (overlapping) steps outlined and described below.
Step 1: Defi ne the problem and gather data. In this step, the team consults with man- agement to clearly identify the problem of concern and ascertain the appropriate objec- tives for the study. The team then typically spends a surprisingly large amount of time gathering relevant data about the problem with the assistance of other key individuals in the organization. A common frustration is that some key data are either very rough or completely unavailable. This may necessitate installing a new computer-based manage- ment information system.
Another increasingly common problem is that there may be too much data available to be easily analyzed. Dramatic advances in computerized data capture, processing power, data transmission, and storage capabilities are enabling organizations to integrate their various databases into massive data warehouses. This has led to the development of data- mining software for extracting hidden predictive information, correlations, and patterns from large databases.
Fortunately, the rapid development of the information technology (IT) fi eld in recent years is leading to a dramatic improvement in the quantity and quality of data that may be available to the management science (MS) team. Corporate IT now is often able to provide the computational resources and databases, as well as any helpful data mining, that are needed by the MS team. Thus, the MS team often will collaborate closely with the IT group.
Step 2: Formulate a model (typically a mathematical model) to represent the problem. Models, or approximate representations, are an integral part of everyday life. Common examples include model airplanes, portraits, globes, and so on. Similarly, mod- els play an important role in science and business, as illustrated by models of the atom, models of genetic structure, mathematical equations describing physical laws of motion or chemical reactions, graphs, organization charts, and industrial accounting systems. Such models are invaluable for abstracting the essence of the subject of inquiry, showing interrelationships, and facilitating analysis.
Mathematical models are also approximate representations, but they are expressed in terms of mathematical symbols and expressions. Such laws of physics as F 5 ma and E 5 mc 2 are familiar examples. Similarly, the mathematical model of a business problem is the system of equations and related mathematical expressions that describes the es- sence of the problem.
With the emergence of powerful spreadsheet technology, spreadsheet models now are widely used to analyze managerial problems. A spreadsheet model lays out the rel- evant data, measures of performance, interrelationships, and so forth, on a spreadsheet in an organized way that facilitates fruitful analysis of the problem. It also frequently incor- porates an underlying mathematical model to assist in the analysis, but the mathematics is kept in the background so the user can concentrate on the analysis.
The modeling process is a creative one. When dealing with real managerial problems (as opposed to some cut-and-dried textbook problems), there normally is no single “cor- rect” model but rather a number of alternative ways to approach the problem. The mod- eling process also is typically an evolutionary process that begins with a simple “verbal model” to defi ne the essence of the problem and then gradually evolves into increasingly more complete mathematical models (perhaps in a spreadsheet format).
We further describe and illustrate such mathematical models in the next section.
Step 3: Develop a computer-based procedure for deriving solutions to the problem from the model. The beauty of a well-designed mathematical model is that it enables the
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1.1 The Nature of Management Science 5
use of mathematical procedures to fi nd good solutions to the problem. These procedures usually are run on a computer because the calculations are too extensive to be done by hand. In some cases, the management science team will need to develop the procedure. In others, a standard software package already will be available for solving the model. When the mathematical model is incorporated into a spreadsheet, the spreadsheet software nor- mally includes a Solver that usually will solve the model.
Step 4: Test the model and refi ne it as needed. Now that the model can be solved, the team needs to thoroughly check and test the model to make sure that it provides a suffi ciently accurate representation of the real problem. A number of questions should be addressed, perhaps with the help of others who are particularly familiar with the problem. Have all the relevant factors and interrelationships in the problem been accurately incor- porated into the model? Does the model seem to provide reasonable solutions? When it is applied to a past situation, does the solution improve upon what was actually done? When assumptions about costs and revenues are changed, do the solutions change in a plausible manner?
Step 5: Apply the model to analyze the problem and develop recommendations for management. The management science team now is ready to solve the model, perhaps under a variety of assumptions, in order to analyze the problem. The resulting recommen- dations then are presented to the managers who must make the decisions about how to deal with the problem.
If the model is to be applied repeatedly to help guide decisions on an ongoing basis, the team might also develop a decision support system. This is an interactive computer-based system that aids managerial decision making. The system draws current data from databases or management information systems and then solves the various versions of the model specifi ed by the manager.
Step 6: Help to implement the team’s recommendations that are adopted by management. Once management makes its decisions, the management science team nor- mally is asked to help oversee the implementation of the new procedures. This includes providing some information to the operating management and personnel involved on the rationale for the changes that are being made. The team also makes sure that the new operating system is consistent with its recommendations as they have been modifi ed and approved by management. If successful, the new system may be used for years to come. With this in mind, the team monitors the initial experience with the system and seeks to identify any modifi cations that should be made in the future.
Management Science Considers Quantitative Factors Many managerial problems revolve around such quantitative factors as production quantities, revenues, costs, the amounts available of needed resources, and so on. By incorporating these quantitative factors into a mathematical model and then applying mathematical procedures to solve the model, management science provides a uniquely powerful way of analyzing such managerial problems. Although management science is concerned with the practical manage- ment of organizations, including taking into account relevant qualitative factors, its special contribution lies in this unique ability to deal with the quantitative factors.
The Special Products Company example discussed below will illustrate how management science considers quantitative factors.
1. When did the rapid development of the management science discipline begin? 2. What is the traditional name given to this discipline that still is widely used outside of business
schools? 3. What does a management science study provide to managers to aid their decision making? 4. Upon which scientific fields and social sciences is management science especially based? 5. What is a decision support system? 6. What are some common quantitative factors around which many managerial problems
revolve?
Review Questions
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6 Chapter One Introduction
1.2 AN ILLUSTRATION OF THE MANAGEMENT SCIENCE APPROACH: BREAK-EVEN ANALYSIS
The Special Products Company produces expensive and unusual gifts to be sold in stores that cater to affluent customers who already have everything. The latest new-product pro- posal to management from the company’s Research Department is a first-of-its-kind iWatch. This iWatch would combine the features of a top-of-the-line atomic wristwatch and a next- generation smartphone, including the ability to respond to voice commands or questions with voice responses. It also would connect to the Internet wirelessly to provide weather, sports scores, stock quotes, and more. An extensive research-and-development project would be needed to develop the iWatch. The proposal is to provide a generous budget of $10 million for this project in order to provide as many desirable features as possible within this budget. It is clear that the production costs for the iWatch would be very large because of the extreme miniaturization that would be required, so the selling price would need to be far beyond the reach of middle-class customers. Therefore, the marketing of the iWatch would be aimed at wealthy customers who want the most advanced products regardless of cost.
Management needs to decide whether to develop and market this new product and, if so, how many of these watches to produce. Before making these decisions, a sales forecast will be obtained to estimate how many watches can be sold. Since most of these sales would occur quickly during the relatively brief time before the “next big thing” arrives to take over the market, there would be only one production run for the iWatch and the number produced would be set equal to the sales forecast. Following the production run, the iWatch would be marketed as aggressively as needed to sell this entire inventory if possible. Management now needs a management science study to be conducted to determine how large this sales potential needs to be to make the iWatch profitable after considering all the prospective revenues and costs, so let’s next look at the estimates of these financial figures.
If the company goes ahead with this product, the research-and-development cost of $10 million is referred to as a fixed cost because it remains the same regardless of how many watches are produced and sold. (However, note that this cost would not be incurred if management decides not to introduce the product since the research-and-development project then would not be undertaken.)
In addition to this fixed cost, there is a production cost that varies with the number of watches produced. This variable cost is $1,000 per watch produced, which adds up to $1,000 times the number of watches produced. (The cost for each additional unit produced, $1,000, is referred to as the marginal cost. ) Each watch sold would generate a unit revenue of $2,000 for the company.
Spreadsheet Modeling of the Problem You will see throughout this book that spreadsheets provide a very convenient way of using a management science approach for modeling and analyzing a wide variety of managerial problems. This certainly is true for the Special Products Company problem as well, as we now will demonstrate.
Figure 1.1 shows a spreadsheet formulation of this problem after obtaining a sales forecast that indicates 30,000 watches can be sold. The data have been entered into cells C4 to C7. Cell C9 is used to record a trial value for the decision as to how many watches to produce. As one of the many possibilities that eventually might be tried, Figure 1.1 shows the specific trial value of 20,000.
Cells F4 to F7 give the resulting total revenue, total costs, and profit (loss) by using the Excel equations shown under the spreadsheet in Figure 1.1 . The Excel equations could have been written using cell references (e.g., F6 5 C6*C9). However, the spreadsheet model is made clearer by giving “range names” to key cells or blocks of cells. (A range name is a descriptive name given to a cell or range of cells that immediately identifies what is there. Appendix A provides details about how to incorporate range names into a spreadsheet model.) To define a name for a selected cell (or range of cells), click on the name box (on the left of the formula bar above the spreadsheet) and type a name. These cell names then can be used in other formulas to create an equation that is easy to decipher (e.g., TotalVariable- Cost 5 MarginalCost*ProductionQuantity rather than the more cryptic F6 5 C6*C9). Note
A cost that remains the same regardless of the pro- duction volume is referred to as a fixed cost, whereas a cost that varies with the production volume is called a variable cost.
Excel Tip: To update formulas throughout the spreadsheet to incorporate a newly defined range name, choose Apply Names from the Define Name menu on the Formulas tab
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1.2 An Illustration of the Management Science Approach: Break-Even Analysis 7
that spaces are not allowed in range names. When a range name has more than one word, we have used capital letters to distinguish the start of each new word (e.g., ProductionQuantity).
The lower left-hand corner of Figure 1.1 lists the names of the quantities in the spreadsheet in alphabetical order and then gives cell references where the quantities are found. Although this isn’t particularly necessary for such a small spreadsheet, you should find it helpful for the larger spreadsheets found later in the book.
This same spreadsheet is provided for you live in your MS Courseware on the CD-ROM. (All the spreadsheets in the book are included in your MS Courseware.) As you can see for yourself by bringing up and playing with the spreadsheet, it provides a straightforward way of performing what-if analysis on the problem. What-if analysis involves addressing such ques- tions as what happens if the sales forecast should have been considerably lower? What hap- pens if some of the cost and revenue estimates are wrong? Simply enter a variety of new values for these quantities in the spreadsheet and see what happens to the profit shown in cell F7.
The lower right-hand corner of Figure 1.1 introduces two useful Excel functions, the MIN( a, b ) function and the IF( a, b, c ) function. The equation for cell F4 uses the MIN( a, b ) function, which gives the minimum of a and b. In this case, the estimated number of watches that will be sold is the minimum of the sales forecast and the production quantity, so
F4 5 UnitRevenue*MIN(SalesForecast, ProductionQuantity)
enters the unit revenue (from cell C4) times the minimum of the sales forecast (from C7) and the production quantity (from C9) into cell F4.
Also note that the equation for cell F5 uses the IF( a, b, c ) function, which does the follow- ing: If statement a is true, it uses b; otherwise, it uses c. Therefore,
F5 5 IF(ProductionQuantity . 0, FixedCost, 0)
says to enter the fixed cost (C5) into cell F5 if the production quantity (C9) is greater than zero, but otherwise enter 0 (the fixed cost is avoided if production is not initiated).
The spreadsheet in Figure 1.1 , along with its equations for the results in column F, con- stitutes a spreadsheet model for the Special Products Company problem. You will see many examples of such spreadsheet models throughout the book.
Excel Tip: A list of all the defined names and their corresponding cell refer- ences can be pasted into a spreadsheet by choosing Paste Names from the Use in Formula menu on the Formulas tab, and then clicking on Paste List.
A spreadsheet is a conve- nient tool for performing what-if analysis.
The Excel function MIN (a, b) gives the minimum of the numbers in the cells whose addresses are a and b.
The Excel function IF (a, b, c) tests if a is true. If so, it uses b; otherwise it uses c.
1 Special Products Co. Break-Even Analysis A B C D E F
2
3
4
5
6
7
8
9
Data Results Unit Revenue
Fixed Cost
Marginal Cost
Sales Forecast
$40,000,000
$10,000,000
$20,000,000
Total Revenue
Total Fixed Cost
Total Variable Cost
Profit (Loss)
Production Quantity
3
E F
4
5
6
7
Results
Total Revenue
Total Fixed Cost
Total Variable Cost
Profit (Loss)
=UnitRevenue * MIN(SalesForecast, ProductionQuantity)
=IF(ProductionQuantity > 0, FixedCost, 0)
=MarginalCost * ProductionQuantity
=TotalRevenue – (TotalFixedCost + TotalVariableCost)
Range Name
FixedCost
MarginalCost
ProductionQuantity
Profit
SalesForecast
TotalFixedCost
TotalRevenue
TotalVariableCost
UnitRevenue
Cell
C5
C6
C9
F7
C7
F5
F4
F6
C4
$2,000
$10,000,000
$1,000
30,000
20,000
$10,000,000
FIGURE 1.1 A spreadsheet formulation of the Special Products Company problem.
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8 Chapter One Introduction
This particular spreadsheet model is based on an underlying mathematical model that uses algebra to spell out the equations in cells F4:F7 and then to derive some additional useful information. Let us take a look at this mathematical model next.
Expressing the Problem Mathematically The issue facing management is to make the following decision.
Decision to be made: Number of watches to produce (if any).
Since this number is not yet known, we introduce an algebraic variable Q to represent this quantity. Thus,
Q 5 Number of watches to produce,
where Q is referred to as a decision variable. Naturally, the value chosen for Q should not exceed the sales forecast for the number of watches that can be sold. Choosing a value of 0 for Q would correspond to deciding not to introduce the product, in which case none of the costs or revenues described in the preceding paragraph would be incurred.
The objective is to choose the value of Q that maximizes the company’s profit from this new product. The management science approach is to formulate a mathematical model to represent this problem by developing an equation that expresses the profit in terms of the decision variable Q. To get there, it is necessary first to develop equations in terms of Q for the total cost and revenue generated by the watches.
If Q 5 0, no cost is incurred. However, if Q > 0, there is both a fixed cost and a variable cost.
Fixed cost 5 $10 million (if Q . 0)
Therefore, the total cost would be
Total cost 5 e0 if Q 5 0 $10 million 1 $1,000Q if Q . 0
Since each watch sold would generate a revenue of $2,000 for the company, the total rev- enue from selling Q watches would be
Total revenue 5 $2,000Q
Consequently, the profit from producing and selling Q watches would be
Profit 5 Total revenue 2 Total cost
5 e0 if Q 5 0 $2,000Q 2 ($10 million 1 $1,000Q) if Q . 0
Thus, since $2,000 Q 2 $1,000 Q 5 $1,000 Q
Analysis of the Problem This last equation shows that the attractiveness of the proposed new product depends greatly on the value of Q, that is, on the number of watches that can be produced and sold. A small value of Q means a loss (negative profit) for the company, whereas a sufficiently large value would generate a positive profit for the company. For example, look at the difference between Q 5 2,000 and Q 5 20,000.
Profit 5 2$10 million 1 $1,000 (2,000) 5 2$8 million if Q 5 20
Figure 1.2 plots both the company’s total cost and total revenue for the various values of Q. Note that the cost line and the revenue line intersect at Q 5 10,000. For any value of Q < 10,000, cost exceeds revenue, so the gap between the two lines represents the loss to the company. For any Q > 10,000, revenue exceeds cost, so the gap between the two lines now
Variable cost 5 $1,000 Q
Profit 5 2$10 million 1 $1,000Q if Q . 0
Profit 5 2$10 million 1 $1,000 (20,000) 5 $10 million if Q 5 200
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1.2 An Illustration of the Management Science Approach: Break-Even Analysis 9
shows positive profit. At Q 5 10,000, the profit is 0. Since 10,000 units is the production and sales volume at which the company would break even on the proposed new product, this vol- ume is referred to as the break-even point. This is the point that must be exceeded to make it worthwhile to introduce the product. Therefore, the crucial question is whether the sales forecast for how many watches can be sold is above or below the break-even point.
Figure 1.2 illustrates the graphical procedure for finding the break-even point. Another alternative is to use an algebraic procedure to solve for the point. Because the profit is 0 at this point, the procedure consists of solving the following equation for the unknown Q.
Profit 5 2$10 million 1 $1,000Q 5 0
Thus,
$1,000Q 5 $10 million
A Complete Mathematical Model for the Problem The preceding analysis of the problem made use of a basic mathematical model that consisted of the equation for profit expressed in terms of Q. However, implicit in this analysis were some additional factors that can be incorporated into a complete mathematical model for the problem.
Two of these factors concern restrictions on the values of Q that can be considered. One of these is that the number of watches produced cannot be less than 0. Therefore,
Q $ 0 is one of the constraints for the complete mathematical model. Another restriction on the value of Q is that it should not exceed the number of watches that can be sold. A sales forecast has not yet been obtained, so let the symbol s represent this currently unknown value.
s 5 Sales forecast (not yet available) of the number of watches that can be sold
Consequently,
Q 5 $10 million
$1,000
Q 5 10,000
constraints A constraint in a mathemat- ical model is an inequality or equation that expresses some restrictions on the values that can be assigned to the decision variables.
FIGURE 1.2 Break-even analysis for the Special Products Company shows that the cost line and revenue line intersect at Q 5 10,000 watches, so this is the break-even point for the proposed new product.
Revenue/Cost in $millions
Production Quantity (thousands)
Total revenue = $2000Q
Total cost = $10 million + $1000Q if Q > 0
Break-even point = 10,000 watches
Profit
Fixed cost = $10 million if Q > 0
Q
50
40
30
20
10
0 5 10 15 20 25
Loss
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10 Chapter One Introduction
Q # s
is another constraint, where s is a parameter of the model whose value has not yet been chosen. The final factor that should be made explicit in the model is the fact that management’s
objective is to make the decision that maximizes the company’s profit from this new product. Therefore, the complete mathematical model for this problem is to find the value of the deci- sion variable Q so as to
Maximize profit 5 b0 if Q 5 0 2 $10 million 1 $1,000Q if Q . 0
subject to
Q # s
Q $ 0
where the algebraic expression given for Profit is called the objective function for the model. The value of Q that solves this model depends on the value that will be assigned to the parameter s (the future forecast of the number of units that can be sold). Because the break- even point is 10,000 here is how the solution for Q depends on s.
Solution for Mathematical Model
Break-even point 5 Fixed cost
Unit revenue 2 Marginal cost 5
$10 million $2,000 2 $1,000
5 10,000
If s # 10,000, then set Q 5 0 If s . 10,000, then set Q 5 s
Therefore, the company should introduce the product and produce the number of units that can be sold only if this production and sales volume exceeds the break-even point.
What-if Analysis of the Mathematical Model A mathematical model is intended to be only an approximate representation of the problem. For example, some of the numbers in the model inevitably are only estimates of quantities that cannot be determined precisely at this time.
The above mathematical model is based on four numbers that are only estimates—the fixed cost of $10 million, the marginal cost of $1,000, the unit revenue of $2,000, and the sales forecast (after it is obtained). A management science study usually devotes consider- able time to investigating what happens to the recommendations of the model if any of the estimates turn out to considerably miss their targets. This is referred to as what-if analysis.
To assist you in performing what-if analysis on this kind of model in a straightforward and enjoyable way, we have provided a Break-Even Analysis module in the Interactive Man- agement Science Modules at www.mhhe.com/hillier5e . (All of the modules in this software package also are included on your CD-ROM.) By following the simple directions given there, you can drag either the cost line or the revenue line to change the fixed cost, the marginal cost, or the unit revenue. This immediately enables you to see the effect on the break-even point if any of these cost or revenue numbers should turn out to have values that are some- what different than their estimates in the model.
Incorporating the Break-Even Point into the Spreadsheet Model A key finding of the above mathematical model is its formula for the break-even point,
Break-even point 5 Fixed cost
Unit revenue 2 Marginal cost
Therefore, once both the quantities in this formula and the sales forecast have been carefully estimated, the solution for the mathematical model specifies what the production quantity should be.
By contrast, although the spreadsheet in Figure 1.1 enables trying a variety of trial values for the production quantity, it does not directly indicate what the production quantity should be. Figure 1.3 shows how this spreadsheet can be expanded to provide this additional guidance.
parameter The constants in a math- ematical model are referred to as the parameters of the model.
objective function The objective function for a mathematical model is a mathematical expression that gives the measure of performance for the prob- lem in terms of the decision variables.
what-if analysis Since estimates can be wrong, what-if analysis is used to check the effect on the recommendations of a model if the estimates turn out to be wrong.
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1.2 An Illustration of the Management Science Approach: Break-Even Analysis 11
Suppose that a sales forecast of 30,000 has been obtained, as shown in cell C7. As indicated by its equation at the bottom of the figure, cell F9 calculates the break-even point by dividing the fixed cost ($10 million) by the net profit per watch sold ($1,000), where this net profit is the unit revenue ($2,000) minus the marginal cost ($1,000). Since the sales forecast of 30,000 exceeds the break-even point of 10,000 this forecast has been entered into cell C9.
If desired, the complete mathematical model for break-even analysis can be fully incorpo- rated into the spreadsheet by requiring that the model solution for the production quantity be entered into cell C9. This would be done by using the equation
C9 5 IF(SalesForecast . BreakEvenPoint, SalesForecast, 0)
However, the disadvantage of introducing this equation is that it would eliminate the pos- sibility of trying other production quantities that might still be of interest. For example, if management does not have much confidence in the sales forecast and wants to minimize the danger of producing more watches than can be sold, consideration would be given to production quantities smaller than the forecast. For example, the trial value shown in cell C9 of Figure 1.1 might be chosen instead. As in any application of management science, a math- ematical model can provide useful guidance but management needs to make the final decision after considering factors that may not be included in the model.
1 Special Products Co. Break-Even Analysis A B C D E F
2
3
4
5
6
7
8
9
Data
30,000
Results Unit Revenue
Fixed Cost
Marginal Cost
Sales Forecast
$60,000,000
$10,000,000
$30,000,000
Total Revenue
Total Fixed Cost
Total Variable Cost
Profit (Loss)
Break-Even PointProduction Quantity
3
E F
4
5
6
7
8
9
Results
Total Revenue
Total Fixed Cost
Total Variable Cost
Profit (Loss)
=UnitRevenue * MIN(SalesForecast, ProductionQuantity)
=IF(ProductionQuantity > 0, FixedCost, 0)
=MarginalCost * ProductionQuantity
=TotalRevenue – (TotalFixedCost + TotalVariableCost)
Break-Even Point =FixedCost/(UnitRevenue – MarginalCost)
Range Name
BreakEvenPoint
FixedCost
MarginalCost
ProductionQuantity
Profit
SalesForecast
TotalFixedCost
TotalRevenue
TotalVariableCost
UnitRevenue
Cell
F9
C5
C6
C9
F7
C7
F5
F4
F6
C4
$2,000
$10,000,000
$1,000
30,000 $20,000,000
10,000
FIGURE 1.3 An expansion of the spreadsheet in Figure 1.1 that uses the solution for the mathematical model to calculate the break-even point.
1. How do the production and sales volume of a new product need to compare to its break-even point to make it worthwhile to introduce the product?
2. What are the factors included in the complete mathematical model for the Special Products Company problem, in addition to an equation for profit?
3. What is the purpose of what-if analysis? 4. How can a spreadsheet be used to perform what-if analysis? 5. What does the MIN( a, b ) Excel function do? 6. What does the IF( a, b, c ) Excel function do?
Review Questions
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12
1.3 THE IMPACT OF MANAGEMENT SCIENCE
Management science (or operations research as it is commonly called by practitioners) has had an impressive impact on improving the efficiency of numerous organizations around the world. In the process, management science has made a significant contribution to increasing the productivity of the economies of various countries. There now are a few dozen member countries in the International Federation of Operational Research Societies (IFORS), with each country having a national operations research society. Both Europe and Asia have fed- erations of such societies to coordinate holding international conferences and publishing international journals in those continents. In addition, we described in Section 1.1 how the Institute for Operations Research and the Management Sciences (INFORMS) is a particularly prominent international society in this area. Among its various journals is one called Inter- faces that regularly publishes articles describing major management science studies and the impact they had on their organizations.
Management science (MS) has had numerous applications of various types in business and industry, sometimes resulting in annual savings of millions, or even hundreds of millions, of dollars. As an example, many hundreds of management scientists work on such airline prob- lems as how to most effectively assign airplanes and crews to flights and how to develop fare structures that maximize revenue. For decades, financial services firms have used portfolio selection techniques that were developed by management scientists who won the Nobel Prize in Economics for their work. Management science models have become a core component of the marketing discipline. Multinational corporations rely on MS for guiding the management of their supply chains. There are numerous other examples of MS applications that are having a dramatic impact on the companies involved.
Management science also is widely used outside business and industry. For example, it is having an increasing impact in the health care area, with applications involving improved management of health care delivery and operations, disease modeling, clinical diagnosis and decision making, radiation therapy, and so on. Applications of MS also abound at various levels of government, ranging from dealing with national security issues at the federal level to managing the delivery of emergency services at the municipal level. Other key governmental applications involve the use of MS modeling to help guide energy, environmental, and global warming policies. Some of the earliest MS applications were military applications, including logistical planning and war gaming, and these continue today.
These are just a sampling of the numerous applications of management science that are having a major impact on the organizations involved. The list goes on and on.
The most important appli- cations of management science in business and industry have resulted in annual savings in the hun- dreds of millions of dollars.
Management science also has had a major impact in the health care area, in guiding key governmental policies, and in military applications.
Federal Express (FedEx) is the world’s largest express trans- portation company. Every working day, it delivers more than 6.5 million documents, packages, and other items throughout the United States and more than 220 countries and territories around the world. In some cases, these ship- ments can be guaranteed overnight delivery by 10:30 AM the next morning.
The logistical challenges involved in providing this ser- vice are staggering. These millions of daily shipments must be individually sorted and routed to the correct general location (usually by aircraft) and then delivered to the exact destination (usually by motorized vehicle) in an amazingly short period of time. How is this possible?
Management science (which usually is referred to as oper- ations research within FedEx) is the technological engine that drives this company. Ever since the company’s found- ing in 1973, management science (MS) has helped make its major business decisions, including equipment investment,
route structure, scheduling, finances, and location of facili- ties. After MS was credited with literally saving the company during its early years, it became the custom to have MS rep- resented at the weekly senior management meetings and, indeed, several of the senior corporate vice presidents have come up from the outstanding FedEx MS group.
FedEx has come to be acknowledged as a world-class company. It routinely ranks among the top companies on Fortune magazine’s annual listing of the “World’s Most Admired Companies.” It also was the first winner (in 1991) of the prestigious prize now known as the INFORMS Prize, which is awarded annually for the effective and repeated integration of management science into organizational deci- sion making in pioneering, varied, novel, and lasting ways.
Source: R. O. Mason, J. L. McKenney, W. Carlson, and D. Copeland, “Absolutely, Positively Operations Research: The Federal Express Story,” Interfaces 27, no. 2 (March–April 1997), pp. 17–36. (A link to this article is provided on our Website, www.mhhe.com/hillier5e.)
An Application Vignette
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1.3 The Impact of Management Science 13
To give you a better notion of the wide applicability of management science, we list some actual applications in Table 1.1 . Note the diversity of organizations and applications in the first two columns. The third column identifies the section where an “application vignette” devotes several paragraphs to describing the application and also references an article that provides full details. (You can see the first of these application vignettes in this section.) The last column indicates that these applications typically resulted in annual savings in the many millions of dollars. Furthermore, additional benefits not recorded in the table (e.g., improved service to customers and better managerial control) sometimes were considered to be even more important than these financial benefits. (You will have an opportunity to investigate these less tangible benefits further in Problems 1.9 and 1.10.)
A link to the articles in Interfaces that describes these applications in detail is included on our Website, www.mhhe.com/hillier5e . We are grateful to INFORMS for our special partnership to make these articles available to you through this link. We think you will find these articles interesting and enlightening in illustrating the dramatic impact that management science sometimes can have on the success of a variety of organizations.
You also will see a great variety of applications of management science throughout the book in the form of case studies, examples, and end-of-chapter cases. Some of the applica- tions are similar to ones described in application vignettes, but many others are quite differ- ent. However, they all generally fall into one of three broad categories, namely, applications
Organization Area of Application Section Annual Savings
Federal Express Logistical planning of shipments 1.3 Not estimated Swift & Company Improve sales and manufacturing performance 2.1 $12 million Samsung Electronics Reduce manufacturing times and inventory levels 2.6 $200 million
more revenue INDEVAL Settle all securities transactions in Mexico 3.2 $150 million United Airlines Plan employee work schedules at airports and reservations offices 3.3 $6 million Procter & Gamble Redesign the production and distribution system 3.5 $200 million Welch’s Optimize use and movement of raw materials 4.3 $150,000 Pacific Lumber Company Long-term forest ecosystem management 5.4 $398 million NPV Hewlett-Packard Product portfolio management 6.1 $180 million Norwegian companies Maximize flow of natural gas through offshore pipeline network 6.3 $140 million Canadian Pacific Railway Plan routing of rail freight 6.4 $100 million Waste Management Develop a route-management system for trash collection and disposal 7.1 $100 million MISO Administer the transmission of electricity in 13 states 7.2 $700 million Netherlands Railways Optimize operation of a railway network 7.4 $105 million Continental Airlines Reassign crews to flights when schedule disruptions occur 7.5 $40 million Bank Hapoalim Group Develop a decision-support system for investment advisors 8.2 $31 million more
revenue DHL Optimize the use of marketing resources 8.4 $22 million Workers’ Compensation Board
Manage high-risk disability claims and rehabilitation 9.3 $4 million
Westinghouse Evaluate research and development projects 9.7 Not estimated Conoco-Phillips Evaluate petroleum exploration projects 9.9 Not estimated L. L. Bean Forecast staffing needs at call centers 10.2 $300,000 Taco Bell Forecast the level of business throughout the day 10.3 $13 million CSAV Optimize global shipping 10.6 $81 million General Motors Improve the throughput of its production lines 11.5 $200 million KeyCorp Improve efficiency of bank teller service 11.6 $20 million Federal Aviation Administration
Manage air traffic flows in severe weather 12.2 $200 million
Sasol Improve the efficiency of its production processes 12.4 $23 million Merrill Lynch Pricing analysis for providing financial services 13.5 $50 million
more revenue
TABLE 1.1 Applications of Management Science to Be Described in Application Vignettes
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14 Chapter One Introduction
in the areas of operations management, finance, and marketing. Tables 1.2 , 1.3 , and 1.4 list these applications in these three respective areas, where the first column identifies where the application is described. In the second column of each table, note the many different ways in which management science can have a real impact in helping improve managerial decisions.
Even these long lists of applications in Tables 1.1 to 1.4 are just a sample of the numer- ous important ways in which management science is applied in organizations around the world. We do not have enough space to provide a more comprehensive compilation of the important applications. (Other applications are included in the supplementary chapters on the CD-ROM.) A hallmark of management science is its great flexibility in dealing with new managerial problems as they arise.
1.4 SOME SPECIAL FEATURES OF THIS BOOK
The focus of this book is on teaching what an enlightened future manager needs to learn from a management science course. It is not on trying to train technical analysts. This focus has led us to include a number of special features that we hope you enjoy.
Location Type of Application
Sec. 2.1 onward A case study: What is the most profitable mix of products? Case 2-1 Which mix of car models should be produced? Case 2-2 Which mix of ingredients should go into the casserole in a university cafeteria? Case 2-3 Which mix of customer–service agents should be hired to staff a call center? Sec. 3.3 Personnel scheduling of customer–service agents Sec. 3.5 Minimize the cost of shipping a product from factories to customers Sec. 3.7 Optimize the assignment of personnel to tasks Case 3-1 How should a product be shipped to market? Case 3-3 Which mix of women’s clothing should be produced for next season? Cases 3-5, 5-4, 7-3 Develop a plan for assigning students to schools so as to minimize busing costs Case 3-6 Which mixes of solid waste materials should be amalgamated into different grades
of a salable product? Case 3-7 How should qualified managers be assigned to new R&D projects? Case 5-2 Develop and analyze a steel company’s plan for pollution abatement Case 5-3 Plan the mix of livestock and crops on a farm with unpredictable weather Sec. 6.1 Minimize the cost of operating a distribution network Secs. 6.2, 6.3 A case study: Maximize the flow of goods through a distribution network Sec. 6.4 Find the shortest path from an origin to a destination Case 6-1 Logistical planning for a military campaign Case 6-3 Develop the most profitable flight schedules for an airline Cases 6-4, 7-4 Operate and expand a private computer network Sec. 7.2 Choose the best combination of R&D projects to pursue Sec. 7.3 Select the best sites for emergency services facilities Sec. 7.4 Airline crew scheduling Sec. 7.5 Production planning when setup costs are involved Case 7-2 Make inventory decisions for a retailer’s warehouse Sec. 8.3 Production planning when overtime is needed Sec. 8.5 Find the shortest route to visit all the American League ballparks Sec. 11.2 Many examples of commercial service systems, internal service systems, and trans-
portation service systems that can be analyzed with queueing models Sec. 11.4 onward A case study: An analysis of competing proposals for more quickly providing main-
tenance services to customers Cases 11-2, 12-2 Analysis of proposals for reducing in-process inventory Sec. 12.1 Comparison of whether corrective maintenance or preventive maintenance is
better Secs. 12.2, 12.3 A case study: Would it be profitable for the owner of a small business to add an
associate? Case 12-1 Analysis of proposals for relieving a production bottleneck Sec. 13.1 A case study: How much of a perishable product should be added to a retailer’s
inventory? Sec. 13.3 Plan a complex project to ensure a strong likelihood of meeting the project
deadline
TABLE 1.2 Case Studies and Examples in the Area of Operations Management
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1.4 Some Special Features of this Book 15
One special feature is that the entire book revolves around modeling as an aid to mana- gerial decision making. This is what is particularly relevant to a manager. Although they may not use this term, all managers often engage in at least informal modeling (abstracting the essence of a problem to better analyze it), so learning more about the art of modeling is important. Since managers instigate larger management science studies done by others, they also need to be able to recognize the kinds of managerial problems where such a study might be helpful. Thus, a future manager should acquire the ability both to recognize when a man- agement science model might be applicable and to properly interpret the results from analyz- ing the model. Therefore, rather than spending substantial time in this book on mathematical theory, the mechanics of solution procedures, or the manipulation of spreadsheets, the focus is on the art of model formulation, the role of a model, and the analysis of model results. A wide range of model types is considered.
Location Type of Application
Sec. 1.2 Break-even analysis Case 1-1 Break-even analysis and what-if analysis Sec. 3.2 An airline choosing which airplanes to purchase Sec. 3.2 Capital budgeting of real-estate development projects Case 3-2 Develop a schedule for investing in a company’s computer equipment Sec. 4.1 onward A case study: Develop a financial plan for meeting future cash flow needs Case 4-1 Develop an investment and cash flow plan for a pension fund Sec. 6.4 Minimize the cost of car ownership Case 6-2 Find the most cost-effective method of converting various foreign currencies
into dollars Sec. 7.1 A case study: Determine the most profitable combination of investments Case 7-1 Develop an investment plan for purchasing art Sec. 8.2 Portfolio selection that balances expected return and risk Sec. 8.5 Select a portfolio to beat the market as frequently as possible Case 8-2 Determine an optimal investment portfolio of stocks Case 8-3 Develop a long-range plan to purchase and sell international bonds Sec. 9.1 onward A case study: Choose whether to drill for oil or sell the land instead Case 9-1 Choose a strategy for the game show, “Who Wants to Be a Millionaire?” Sec. 12.1 Analysis of a new gambling game Sec. 13.2 Choose the bid to submit in a competitive bidding process Sec. 13.4 Develop a financial plan when future cash flows are somewhat unpredictable Sec. 13.5 Risk analysis when assessing financial investments Sec. 13.6 How much overbooking should be done in the travel industry? Case 13-1 Analysis of how a company’s cash flows might evolve over the next year Case 13-2 Calculate the value of a European call option
TABLE 1.3 Case Studies and Examples in the Area of Finance
Location Type of Application
Secs. 2.6, 3.3 Determine the best mix of advertising media Case 2-1 Evaluate whether an advertising campaign would be worthwhile Secs. 3.1, 3.4 A case study: Which advertising plan best achieves managerial goals? Case 3-4 Develop a representative marketing survey Case 5-1 Analysis of the trade-off between advertising costs and the resulting increase
in sales of several products Sec. 6.4 Balance the speed of bringing a new product to market and the associated
costs Sec. 8.3 Deal with nonlinear marketing costs Case 8-1 Refine the advertising plan developed in the case study presented in Sections
3.1 and 3.4 Case 9-2 Should a company immediately launch a new product or test-market it first? Case 9-3 Should a company buy additional marketing research before deciding whether
to launch a new product? Case 9-4 Plan a sequence of decisions for a possible new product Sec. 10.2 onward A case study: Manage a call center for marketing goods over the telephone Case 10-1 Improve forecasts of demand for a call center Case 11-1 Estimate customer waiting times for calling into a call center
TABLE 1.4 Case Studies and Examples in the Area of Marketing
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16 Chapter One Introduction
Another special feature is a heavy emphasis on case studies to better convey these ideas in an interesting way in the context of applications. Every subsequent chapter includes at least one case study that introduces and illustrates the application of that chapter’s techniques in a realistic setting. In a few instances, the entire chapter revolves around a case study. Although consider- ably smaller and simpler than most real studies (to maintain clarity), these case studies are pat- terned after actual applications requiring a major management science study. Consequently, they convey the whole process of such a study, some of the pitfalls involved, and the complementary roles of the management science team and the manager responsible for the decisions to be made.
To complement these case studies, every chapter also includes major cases at the end. These realistic cases can be used for individual assignments, team projects, or case studies in class. In addition, the University of Western Ontario Ivey School of Business (the second larg- est producer of teaching cases in the world) also has specially selected cases from its case col- lection that match the chapters in this textbook. These cases are available on the Ivey Website, www.cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
The book also places heavy emphasis on conveying the impressive impact that manage- ment science is having on improving the efficiency of numerous organizations around the world. Therefore, you will see many examples of actual applications throughout the book in the form of boxed application vignettes, such as the one already shown in Section 1.3. You then will have the opportunity to learn more about these actual applications by reading the articles fully describing them that are accessed by a link on our Website. As indicated in Table 1.1 , these applications sometimes resulted in annual savings of millions, tens of mil- lions, or even hundreds of millions of dollars.
In addition, we try to provide you with a broad perspective about the nature of the real world of management science in practice. It is easy to lose sight of this world when cranking through textbook exercises to master the mechanics of a series of techniques. Therefore, we shift some emphasis from mastering these mechanics to seeing the big picture. The case stud- ies, cases, and descriptions of actual applications are part of this effort.
Another feature is the inclusion of one or more solved problems for each chapter to help you get started on your homework for that chapter. The statement of each solved problem is given just above the Problems section of the chapter, and then the complete solution is given on both the CD-ROM and the Website for the book.
The last, but certainly not the least, of the special features of this book is the accompany- ing software. We will describe and illustrate how to use today’s premier spreadsheet package, Microsoft Excel, to formulate many management science models in a spreadsheet format. Excel 2007 implemented a significant overhaul of the user interface. Excel 2010 brought other significant changes and improvements. Many of the models considered in this book can be solved using standard Excel. Some Excel add-ins also are available to solve other models. Appendix A provides a primer on the use of Excel.
Included with the book is an extensive collection of software that we collectively refer to as MS Courseware. This collection includes spreadsheet files, Frontline Systems’ Risk Solver Platform for Education, and a package of Interactive Management Science Modules. Each of these is briefly described below.
MS Courseware includes numerous spreadsheet files for every chapter in this book. Each time a spreadsheet example is presented in the book, a live spreadsheet that shows the for- mulation and solution for the example also is available in MS Courseware. This provides a convenient reference, or even useful templates, when you set up spreadsheets to solve similar problems. Also, for many models in the book, template spreadsheet files are provided that already include all the equations necessary to solve the model. You simply enter the data for the model and the solution is immediately calculated.
Included with standard Excel is an add-in, called Solver, which is used to solve most of the optimization models considered in the first half of this book. Solver is a product of Frontline Systems, Inc. New with this edition of the textbook is a very powerful software package from Frontline Systems, Inc., called Risk Solver Platform for Education (RSPE). Some special fea- tures of RSPE are a significantly enhanced version of the basic Solver included with Excel, the ability to build decision trees within Excel, as covered in Chapter 9, and tools to build computer simulation models within Excel, as covered in Chapter 13.
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Chapter 1 Glossary 17
As mentioned in Section 1.2, another learning aid accompanying the book is the pack- age of Interactive Management Science Modules provided at www.mhhe.com/hillier5e . This innovative tool includes several modules that enable you to interactively explore several man- agement science techniques in depth. For your convenience, an offline version of this package also is included in your MS Courseware on the CD-ROM.
Most of the software used in this book is compatible with both Excel for Windows PCs and Excel for Macintosh computers (Macs). Some software (e.g., Risk Solver Platform for Education) is not directly compatible with Macs, although it works well on any recent (Intel) Mac with Boot Camp or virtualization software. For the most up-to-date information on soft- ware compatibility and relevant differences between Windows PC versions and Mac ver- sions, please refer to the Software Compatibility link at www.mhhe.com/hillier5e .
We should point out that Excel is not designed for dealing with the really large manage- ment science models that occasionally arise in practice. More powerful software packages that are not based on spreadsheets generally are used to solve such models instead. How- ever, management science teams, not managers, primarily use these sophisticated packages (including using modeling languages to help input the large models). Since this book is aimed mainly at future managers rather than future management scientists, we will not have you use these packages.
To alert you to relevant material in MS Courseware, the end of each chapter has a list entitled “Learning Aids for This Chapter in Your MS Courseware.”
1.5 Summary Management science is a discipline area that attempts to aid managerial decision making by applying a scientific approach to managerial problems that involve quantitative factors. The rapid development of this discipline began in the 1940s and 1950s. The onslaught of the computer revolution has since continued to give great impetus to its growth. Further impetus now is being provided by the widespread use of spread- sheet software, which greatly facilitates the application of management science by managers and others.
A major management science study involves conducting a systematic investigation that includes careful data gathering, developing and testing hypotheses about the problem (typically in the form of a mathematical model), and applying sound logic in the subsequent analysis. The management science team then presents its recommendations to the managers who must make the decisions about how to resolve the problem. Smaller studies might be done by managers themselves with the aid of spreadsheets.
A major part of a typical management science study involves incorporating the quantitative factors into a mathematical model (perhaps incorporated into a spreadsheet) and then applying mathematical proce- dures to solve the model. Such a model uses decision variables to represent the quantifiable decisions to be made. An objective function expresses the appropriate measure of performance in terms of these decision variables. The constraints of the model express the restrictions on the values that can be assigned to the decision variables. The parameters of the model are the constants that appear in the objective function and the constraints. An example involving break-even analysis was used to illustrate a mathematical model.
Management science has had an impressive impact on improving the efficiency of numerous organi- zations around the world. In fact, many award-winning applications have resulted in annual savings in the millions, tens of millions, or even hundreds of millions of dollars.
The focus of this book is on emphasizing what an enlightened future manager needs to learn from a management science course. Therefore, the book revolves around modeling as an aid to managerial decision making. Many case studies (within the chapters) and cases (at the end of chapters) are used to better convey these ideas.
Glossary break-even point The production and sales volume for a product that must be exceeded to achieve a profit. (Section 1.2), 9 business analytics A discipline closely related to management science that makes extensive use of data to analyze trends, make forecasts, and apply optimization techniques. (Section 1.1), 3
constraint An inequality or equation in a math- ematical model that expresses some restrictions on the values that can be assigned to the decision variables. (Section 1.2), 9 decision support system An interactive com- puter-based system that aids managerial decision making. (Section 1.1), 5
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18 Chapter One Introduction
Chapter 1 Excel Files:
Special Products Co. Example
Interactive Management Science Modules:
Module for Break-Even Analysis
Learning Aids for This Chapter in Your MS Courseware
Solved Problem (See the CD-ROM or Website for the Solution)
1.S1. Make or Buy? Power Notebooks, Inc., plans to manufacture a new line of notebook computers. Management is trying to decide whether to purchase the LCD screens for the computers from an outside supplier or to manufacture the screens in-house. The screens cost $100 each from the outside supplier. To set up the assem- bly process required to produce the screens in-house would cost $100,000. The company could then produce each screen for $75. The number of notebooks that eventually will be produced ( Q ) is unknown at this point.
a. Set up a spreadsheet that will display the total cost of both op- tions for any value of Q. Use trial and error with the spread- sheet to determine the range of production volumes for which each alternative is best.
b. Use a graphical procedure to determine the break-even point for Q (i.e., the quantity at which both options yield the same cost).
c. Use an algebraic procedure to determine the break-even point for Q.
Problems
1.1. The manager of a small firm is considering whether to pro- duce a new product that would require leasing some special equip- ment at a cost of $20,000 per month. In addition to this leasing cost, a production cost of $10 would be incurred for each unit of the product produced. Each unit sold would generate $20 in revenue.
Develop a mathematical expression for the monthly profit that would be generated by this product in terms of the number of units produced and sold per month. Then determine how large this number needs to be each month to make it profitable to pro- duce the product. 1.2. Refer to Problem 1.1. A sales forecast has been obtained that indicates that 4,000 units of the new product could be sold. This forecast is considered to be quite reliable, but there is consid- erable uncertainty about the accuracy of the estimates given for the leasing cost, the marginal production cost, and the unit revenue.
Use the Break-Even Analysis module in the Interactive Man- agement Science Modules to perform what-if analysis on these estimates.
a. How large can the leasing cost be before this new product ceases to be profitable?
b. How large can the marginal production cost be before this new product ceases to be profitable?
c. How small can the unit revenue be before this new product ceases to be profitable?
1.3. Management of the Toys R4U Company needs to decide whether to introduce a certain new novelty toy for the upcom- ing Christmas season, after which it would be discontinued. The total cost required to produce and market this toy would be $500,000 plus $15 per toy produced. The company would receive revenue of $35 for each toy sold.
a. Assuming that every unit of this toy that is pro- duced is sold, write an expression for the profit in terms of the number produced and sold. Then find the break-even point that this number must exceed to make it worthwhile to introduce this toy.
decision variable An algebraic variable that represents a quantifiable decision to be made. (Section 1.2), 8 mathematical model An approximate repre- sentation of, for example, a business problem that is expressed in terms of mathematical symbols and expressions. (Section 1.1), 4 model An approximate representation of some- thing. (Section 1.1), 4 MS Courseware The name of the software package that is shrinkwrapped with the book or is on its Website. (Section 1.4), 16 objective function A mathematical expression in a model that gives the measure of performance for a problem in terms of the decision variables. (Section 1.2), 10
operations research The traditional name for management science that still is widely used out- side of business schools. (Section 1.1), 3 parameter One of the constants in a mathemat- ical model. (Section 1.2), 10 range name A descriptive name given to a cell or range of cells that immediately identifies what is there. (Section 1.2), 6 spreadsheet model An approximate represen- tation of, for example, a business problem that is laid out on a spreadsheet in a way that facilitates analysis of the problem. (Section 1.1), 4 what-if analysis Analysis of how the recom- mendations of a model might change if any of the estimates providing the numbers in the model eventually need to be corrected. (Section 1.2), 10
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Chapter 1 Problems 19
b. Now assume that the number that can be sold might be less than the number produced. Write an expres- sion for the profit in terms of these two numbers.
c. Formulate a spreadsheet that will give the profit in part b for any values of the two numbers.
d. Write a mathematical expression for the constraint that the number produced should not exceed the number that can be sold.
1.4. A reliable sales forecast has been obtained indicating that the Special Products Company (see Section 1.2) would be able to sell 30,000 iWatches, which appears to be enough to justify introducing this new product. However, management is concerned that this conclusion might change if more accurate estimates were available for the research-and-development cost, the marginal production cost, and the unit revenue. Therefore, before a final decision is made, management wants what-if anal- ysis done on these estimates.
Use the spreadsheet from Figure 1.3 (see this chapter’s Excel files) and trial-and-error to perform what-if analysis by indepen- dently investigating each of the following questions.
a. How large can the research-and-development cost be before the watches cease to be profitable?
b. How large can the marginal production cost be before the watches cease to be profitable?
c. How small can the unit revenue be before the watches cease to be profitable?
1.5. Reconsider the problem facing the management of the Special Products Company as presented in Section 1.2.
A more detailed investigation now has provided better esti- mates of the data for the problem. The research-and-development cost still is estimated to be $10 million, but the new estimate of the marginal production cost is $1,300. The revenue from each watch sold now is estimated to be $1,700.
a. Use a graphical procedure to find the new break- even point.
b. Use an algebraic procedure to find the new break- even point.
c. State the mathematical model for this problem with the new data.
d. Incorporate this mathematical model into a spread- sheet with a sales forecast of 30,000. Use this spreadsheet model to find the new break-even point, and then determine the production quantity and the estimated total profit indicated by the model.
e. Suppose that management fears that the sales fore- cast may be overly optimistic and so does not want to consider producing more than 20,000 watches. Use the spreadsheet from part d to determine what the production quantity should be and the estimated total profit that would result.
1.6. The Best-for-Less Corp. supplies its two retail outlets from its two plants. Plant A will be supplying 30 shipments next month. Plant B has not yet set its production schedule for next month but has the capacity to produce and ship any amount up to a maximum of 50 shipments. Retail outlet 1 has submitted its order for 40 shipments for next month. Retail outlet 2 needs a minimum of 25 shipments next month but would be happy to receive more. The production costs are the same at the two
plants but the shipping costs differ. The shipping cost per ship- ment from each plant to each retail outlet is given below, along with a summary of the other data.
The distribution manager, Jennifer Lopez, now needs to develop a plan for how many shipments to send from each plant to each of the retail outlets next month. Her objective is to mini- mize the total shipping cost.
Unit Shipping Cost
Retail Outlet 1 Retail Outlet 2 Supply
Plant A $700 $400 5 30 shipments Plant B $800 $600 ≤ 50 shipments
Needed 5 40 shipments $ 25 shipments
a. Identify the individual decisions that Jennifer needs to make. For each of these decisions, define a deci- sion variable to represent the decision.
b. Write a mathematical expression for the total ship- ping cost in terms of the decision variables.
c. Write a mathematical expression for each of the constraints on what the values of the decision vari- ables can be.
d. State a complete mathematical model for Jennifer’s problem.
e. What do you think Jennifer’s shipping plan should be? Explain your reasoning. Then express your shipping plan in terms of the decision variables.
1.7. The Water Sports Company soon will be producing and marketing a new model line of motor boats. The production manager, Michael Jensen, now is facing a make-or-buy decision regarding the outboard motor to be installed on each of these boats. Based on the total cost involved, should the motors be produced internally or purchased from a vendor? Producing them internally would require an investment of $1 million in new facilities as well as a production cost of $1,600 for each motor produced. If purchased from a vendor instead, the price would be $2,000 per motor.
Michael has obtained a preliminary forecast from the com- pany’s marketing division that 3,000 boats in this model line will be sold.
a. Use spreadsheets to display and analyze Michael’s two options. Which option should be chosen?
b. Michael realizes from past experience that prelimi- nary sales forecasts are quite unreliable, so he wants to check on whether his decision might change if a more careful forecast differed significantly from the preliminary forecast. Determine a break-even point for the production and sales volume below which the buy option is better and above which the make option is better.
1.8. Reconsider the Special Products Company problem pre- sented in Section 1.2.
Although the company is well qualified to do most of the work in producing the iWatch, it currently lacks much expertise in one key area, namely, developing and producing a miniature camera to be embedded into the iWatch. Therefore, management
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20 Chapter One Introduction
is negative below the break-even point and positive above the break-even point. Using this expression as the objective function, state the overall mathe- matical model (including constraints) for the prob- lem of determining whether to choose the make option and, if so, how many units of the LCD dis- play (one per watch) to produce.
d. Use a graphical procedure to find the break-even point described in part c.
e. Use an algebraic procedure to find the break-even point described in part c.
f. Use a spreadsheet model to find the break-even point described in part c. What is the conclusion about what the company should do?
1.9. Select one of the applications of management science listed in Table 1.1 . Read the article that is referenced in the application vignette presented in the section shown in the third column. (A link to all these articles is provided on our Website, www.mhhe.com/hillier5e . ) Write a two-page summary of the application and the benefits (including nonfinancial benefits) it provided.
1.10. Select three of the applications of management science listed in Table 1.1 . For each one, read the article that is refer- enced in the application vignette presented in the section shown in the third column. (A link to all these articles is provided on our Website, www.mhhe.com/hillier5e . ) For each one, write a one-page summary of the application and the benefits (including nonfinancial benefits) it provided. 1.11. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 1.3. Summarize the five major reasons given for why operations research/management science (OR/MS) was so successful.
now is considering contracting out this part of the job to another company that has this expertise. If this were done, the Special Products Company would reduce its research-and-development cost to $5 million, as well as reduce its marginal production cost to $750. However, the Special Products Company also would pay this other company $500 for each miniature camera and so would incur a total marginal cost of $1,250 (including its payment to the other company) while still obtaining revenue of $2,000 for each watch produced and sold. However, if the company does all the production itself, all the data presented in Section 1.2 still apply. After obtaining an analysis of the sales potential, management believes that 30,000 watches can be sold.
Management now wants to determine whether the make option (do all the development and production internally) or the buy option (contract out the development and production of the miniature cameras) is better. a. Use a spreadsheet to display and analyze the buy
option. Show the relevant data and financial output, including the total profit that would be obtained by producing and selling 30,000 watches.
b. Figure 1.3 shows the analysis for the make option. Compare these results with those from part a to deter- mine which option (make or buy) appears to be better.
c. Another way to compare these two options is to find a break-even point for the production and sales volume, below which the buy option is better and above which the make option is better. Begin this process by developing an expression for the differ- ence in profit between the make and buy options in terms of the number of grandfather clocks to pro- duce for sale. Thus, this expression should give the incremental profit from choosing the make option rather than the buy option, where this incremental profit is 0 if 0 watches are produced but otherwise
Case 1-1
Keeping Time
Founded nearly 50 years ago by Alfred Lester-Smith, Beautiful Clocks specializes in developing and marketing a diverse line of large ornamental clocks for the finest homes. Tastes have changed over the years, but the company has prospered by continually updating its product line to satisfy its affluent clientele. The Lester- Smith family continues to own a majority share of the company and the grandchildren of Alfred Lester-Smith now hold several of the top managerial positions. One of these grandchildren is Mer- edith Lester-Smith, the new CEO of the company.
Meredith feels a great responsibility to maintain the family heritage with the company. She realizes that the company needs to continue to develop and market exciting new products. Since the 50 th anniversary of the founding of the company is rapidly approaching, she has decided to select a particularly special new product to launch with great fanfare on this anniversary. But what should it be? As she ponders this crucial decision, Mer- edith’s thoughts go back to the magnificent grandfather clock that her grandparents had in their home many years ago. She had
admired the majesty of that clock as a child. How about launch- ing a modern version of this clock?
This is a difficult decision. Meredith realizes that grand- father clocks now are largely out of style. However, if she is so nostalgic about the memory of the grandfather clock in her grandparents’ home, wouldn’t there be a considerable number of other relatively wealthy couples with similar memories who would welcome the prestige of adding the grandeur of a beauti- fully designed limited-edition grandfather clock in their home? Maybe. This also would highlight the heritage and continuity of the company. It all depends on whether there would be enough sales potential to make this a profitable product.
Meredith had an excellent management science course as part of her MBA program in college, so she realizes that break-even analysis is needed to help make this decision. With this in mind, she instructs several staff members to investigate this prospective product further, including developing estimates of the related costs and revenues as well as forecasting the potential sales.
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Additional Case 21
One month later, the preliminary estimates of the relevant financial figures come back. The cost of designing the grand- father clock and then setting up the production facilities to produce this product would be approximately $250,000. There would be only one production run for this limited- edition grandfather clock. The additional cost for each clock produced would be roughly $2,000. The marketing depart- ment estimates that their price for selling the clocks can be successfully set at about $4,500 apiece, but a firm forecast of how many clocks can be sold at this price has not yet been obtained. However, it is believed that the sales likely would reach into three digits.
Meredith wants all these numbers pinned down considerably further. However, she feels that some analysis can be done now to draw preliminary conclusions.
a. Assuming that all clocks produced are sold, develop a spread- sheet model for estimating the profit or loss from producing any particular number of clocks.
b. Use this spreadsheet to find the break-even point by trial and error.
c. Develop the corresponding mathematical expression for the estimated profit in terms of the number of clocks produced.
d. Use a graphical procedure to find the break-even point. e. Use the algebraic procedure to find the break-even point.
A fairly reliable forecast now has been obtained indicat- ing that the company would be able to sell 300 of the limited- edition grandfather clocks, which appears to be enough to justify introducing this new product. However, Meredith is concerned
that this conclusion might change if more accurate estimates were available for the various costs and revenues. Therefore, she wants sensitivity analysis done on these estimates. Use the Break-Even Analysis module in the Interactive Management Science Modules to perform sensitivity analysis by indepen- dently investigating each of the following questions.
f. How large can the cost of designing this product and setting up the production facilities be before the grandfather clocks cease to be profitable?
g. How large can the production cost for each additional clock be before the grandfather clocks cease to be profitable?
h. If both of the costs identified in parts f and g were 50% larger than their initial estimates, would producing and selling the grandfather clocks still be profitable?
i. How small can the price for selling each clock be before the grandfather clocks cease to be profitable?
Now suppose that 300 grandfather clocks are produced but only 200 are sold.
j. Would it still be profitable to produce and sell the grand- father clocks under this circumstance?
Additional Case An additional case for this chapter is also available at the Uni- versity of Western Ontario Ivey School of Business Website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
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22
Chapter Two
Linear Programming: Basic Concepts Learning Objectives
After completing this chapter, you should be able to
1. Explain what linear programming is.
2. Identify the three key questions to be addressed in formulating any spreadsheet model.
3. Name and identify the purpose of the four kinds of cells used in linear programming spreadsheet models.
4. Formulate a basic linear programming model in a spreadsheet from a description of the problem.
5. Present the algebraic form of a linear programming model from its formulation on a spreadsheet.
6. Apply the graphical method to solve a two-variable linear programming problem.
7. Use Excel to solve a linear programming spreadsheet model.
The management of any organization regularly must make decisions about how to allocate its resources to various activities to best meet organizational objectives. Linear programming is a powerful problem-solving tool that aids management in making such decisions. It is appli- cable to both profit-making and not-for-profit organizations, as well as governmental agen- cies. The resources being allocated to activities can be, for example, money, different kinds of personnel, and different kinds of machinery and equipment. In many cases, a wide variety of resources must be allocated simultaneously. The activities needing these resources might be various production activities (e.g., producing different products), marketing activities (e.g., advertising in different media), financial activities (e.g., making capital investments), or some other activities. Some problems might even involve activities of all these types (and perhaps others), because they are competing for the same resources.
You will see as we progress that even this description of the scope of linear program- ming is not sufficiently broad. Some of its applications go beyond the allocation of resources. However, activities always are involved. Thus, a recurring theme in linear programming is the need to find the best mix of activities—which ones to pursue and at what levels.
Like the other management science techniques, linear programming uses a mathematical model to represent the problem being studied. The word linear in the name refers to the form of the mathematical expressions in this model. Programming does not refer to computer pro- gramming; rather, it is essentially a synonym for planning. Thus, linear programming means the planning of activities represented by a linear mathematical model.
Because it comprises a major part of management science, linear programming takes up several chapters of this book. Furthermore, many of the lessons learned about how to apply linear programming also will carry over to the application of other management science techniques.
This chapter focuses on the basic concepts of linear programming.
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2.1 A Case Study: The Wyndor Glass Co. Product-Mix Problem 23
2.1 A CASE STUDY: THE WYNDOR GLASS CO. PRODUCT-MIX PROBLEM
Jim Baker has had an excellent track record during his seven years as manager of new product development for the Wyndor Glass Company. Although the company is a small one, it has been experiencing considerable growth largely because of the innovative new products devel- oped by Jim’s group. Wyndor’s president, John Hill, has often acknowledged publicly the key role that Jim has played in the recent success of the company.
Therefore, John felt considerable confidence six months ago in asking Jim’s group to develop the following new products:
• An 8-foot glass door with aluminum framing. • A 4-foot 3 6-foot double-hung, wood-framed window.
Although several other companies already had products meeting these specifications, John felt that Jim would be able to work his usual magic in introducing exciting new features that would establish new industry standards.
Background The Wyndor Glass Co. produces high-quality glass products, including windows and glass doors that feature handcrafting and the finest workmanship. Although the products are expen- sive, they fill a market niche by providing the highest quality available in the industry for the most discriminating buyers. The company has three plants that simultaneously produce the components of its products.
Plant 1 produces aluminum frames and hardware. Plant 2 produces wood frames. Plant 3 produces the glass and assembles the windows and doors.
Because of declining sales for certain products, top management has decided to revamp the company’s product line. Unprofitable products are being discontinued, releasing produc- tion capacity to launch the two new products developed by Jim Baker’s group if management approves their release.
The 8-foot glass door requires some of the production capacity in Plants 1 and 3, but not Plant 2. The 4-foot 3 6-foot double-hung window needs only Plants 2 and 3.
Management now needs to address two issues:
1. Should the company go ahead with launching these two new products? 2. If so, what should be the product mix —the number of units of each produced per week—
for the two new products?
Management’s Discussion of the Issues Having received Jim Baker’s memorandum describing the two new products, John Hill now has called a meeting to discuss the current issues. In addition to John and Jim, the meeting includes Bill Tasto, vice president for manufacturing, and Ann Lester, vice president for marketing.
Let’s eavesdrop on the meeting. John Hill (president): Bill, we will want to rev up to start production of these products as soon as we can. About how much production output do you think we can achieve? Bill Tasto (vice president for manufacturing): We do have a little available production capacity, because of the products we are discontinuing, but not a lot. We should be able to achieve a production rate of a few units per week for each of these two products. John: Is that all? Bill: Yes. These are complicated products requiring careful crafting. And, as I said, we don’t have much production capacity available. John: Ann, will we be able to sell several of each per week? Ann Lester (vice president for marketing): Easily.
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24
John: Good. Now there’s one more issue to resolve. With this limited production capacity, we need to decide how to split it between the two products. Do we want to produce the same number of both products? Or mostly one of them? Or even just produce as much as we can of one and postpone launching the other one for a little while? Jim Baker (manager of new product development): It would be dangerous to hold one of the products back and give our competition a chance to scoop us. Ann: I agree. Furthermore, launching them together has some advantages from a mar- keting standpoint. Since they share a lot of the same special features, we can combine the advertising for the two products. This is going to make a big splash. John: OK. But which mixture of the two products is going to be most profitable for the company? Bill: I have a suggestion. John: What’s that? Bill: A couple times in the past, our Management Science Group has helped us with these same kinds of product-mix decisions, and they’ve done a good job. They ferret out all the relevant data and then dig into some detailed analysis of the issue. I’ve found their input very helpful. And this is right down their alley. John: Yes, you’re right. That’s a good idea. Let’s get our Management Science Group working on this issue. Bill, will you coordinate with them?
The meeting ends.
The Management Science Group Begins Its Work At the outset, the Management Science Group spends considerable time with Bill Tasto to clarify the general problem and specific issues that management wants addressed. A par- ticular concern is to ascertain the appropriate objective for the problem from management’s viewpoint. Bill points out that John Hill posed the issue as determining which mixture of the two products is going to be most profitable for the company.
Therefore, with Bill’s concurrence, the group defines the key issue to be addressed as follows.
Question: Which combination of production rates (the number of units produced per week) for the two new products would maximize the total profit from both of them?
The group also concludes that it should consider all possible combinations of production rates of both new products permitted by the available production capacities in the three plants. For example, one alternative (despite Jim Baker’s and Ann Lester’s objections) is to forgo pro- ducing one of the products for now (thereby setting its production rate equal to zero) in order
The issue is to find the most profitable mix of the two new products.
Swift & Company is a diversified protein-producing busi- ness based in Greeley, Colorado. With annual sales of over $8 billion, beef and related products are by far the largest portion of the company’s business.
To improve the company’s sales and manufacturing per- formance, upper management concluded that it needed to achieve three major objectives. One was to enable the company’s customer service representatives to talk to their more than 8,000 customers with accurate information about the availability of current and future inventory while considering requested delivery dates and maximum prod- uct age upon delivery. A second was to produce an efficient shift-level schedule for each plant over a 28-day horizon. A third was to accurately determine whether a plant can ship a requested order-line-item quantity on the requested date
and time given the availability of cattle and constraints on the plant’s capacity.
To meet these three challenges, a management science team developed an integrated system of 45 linear program- ming models based on three model formulations to dynami- cally schedule its beef-fabrication operations at five plants in real time as it receives orders. The total audited benefits realized in the first year of operation of this system were $12.74 million, including $12 million due to optimizing the product mix. Other benefits include a reduction in orders lost, a reduction in price discounting, and better on-time delivery.
Source: A. Bixby, B. Downs, and M. Self, “A Scheduling and Capable-to-Promise Application for Swift & Company, Interfaces 36, no. 1 (January–February 2006), pp. 69–86. (A link to this article is provided on our website, www.mhhe.com/hillier5e . )
An Application Vignette
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2.2 Formulating the Wyndor Problem on a Spreadsheet 25
to produce as much as possible of the other product. (We must not neglect the possibility that maximum profit from both products might be attained by producing none of one and as much as possible of the other.)
The Management Science Group next identifies the information it needs to gather to con- duct this study:
1. Available production capacity in each of the plants. 2. How much of the production capacity in each plant would be needed by each product. 3. Profitability of each product.
Concrete data are not available for any of these quantities, so estimates have to be made. Estimating these quantities requires enlisting the help of key personnel in other units of the company.
Bill Tasto’s staff develops the estimates that involve production capacities. Specifically, the staff estimates that the production facilities in Plant 1 needed for the new kind of doors will be available approximately four hours per week. (The rest of the time Plant 1 will con- tinue with current products.) The production facilities in Plant 2 will be available for the new kind of windows about 12 hours per week. The facilities needed for both products in Plant 3 will be available approximately 18 hours per week.
The amount of each plant’s production capacity actually used by each product depends on its production rate. It is estimated that each door will require one hour of production time in Plant 1 and three hours in Plant 3. For each window, about two hours will be needed in Plant 2 and two hours in Plant 3.
By analyzing the cost data and the pricing decision, the Accounting Department estimates the profit from the two products. The projection is that the profit per unit will be $300 for the doors and $500 for the windows.
Table 2.1 summarizes the data now gathered. The Management Science Group recognizes this as being a classic product-mix problem .
Therefore, the next step is to develop a mathematical model —that is, a linear programming model —to represent the problem so that it can be solved mathematically. The next four sections focus on how to develop this model and then how to solve it to find the most profitable mix between the two products, assuming the estimates in Table 2.1 are accurate.
Production Time Used for Each Unit Produced
Plant Doors Windows Available per Week 1 1 hour 0 4 hours 2 0 2 hours 12 hours 3 3 hours 2 hours 18 hours
Unit profit $300 $500
TABLE 2.1 Data for the Wyndor Glass Co. Product-Mix Problem
2.2 FORMULATING THE WYNDOR PROBLEM ON A SPREADSHEET
Spreadsheets provide a powerful and intuitive tool for displaying and analyzing many man- agement problems. We now will focus on how to do this for the Wyndor problem with the popular spreadsheet package Microsoft Excel. 1
1 Other spreadsheet packages with similar capabilities also are available, and the basic ideas presented here are still applicable.
1. What were the two issues addressed by management? 2. The Management Science Group was asked to help analyze which of these issues? 3. How did this group define the key issue to be addressed? 4. What information did the group need to gather to conduct its study?
Review Questions
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26 Chapter Two Linear Programming: Basic Concepts
Formulating a Spreadsheet Model for the Wyndor Problem Figure 2.1 displays the Wyndor problem by transferring the data in Table 2.1 onto a spread- sheet. (Columns E and F are being reserved for later entries described below.) We will refer to the cells showing the data as data cells . To distinguish the data cells from other cells in the spreadsheet, they are shaded light blue. (In the textbook figures, the light blue shading appears as light gray.) The spreadsheet is made easier to interpret by using range names. (As mentioned in Section 1.2, a range name is simply a descriptive name given to a cell or range of cells that immediately identifies what is there. Excel allows you to use range names instead of the corresponding cell addresses in Excel equations, since this usually makes the equations much easier to interpret at a glance.) The data cells in the Wyndor Glass Co. prob- lem are given the range names UnitProfit (C4:D4), HoursUsedPerUnitProduced (C7:D9), and Hours Available (G7:G9). To enter a range name, first select the range of cells, then click in the name box on the left of the formula bar above the spreadsheet and type a name. (See Appendix A for further details about defining and using range names.)
Three questions need to be answered to begin the process of using the spreadsheet to for- mulate a mathematical model (in this case, a linear programming model ) for the problem.
1. What are the decisions to be made? 2. What are the constraints on these decisions? 3. What is the overall measure of performance for these decisions?
The preceding section described how Wyndor’s Management Science Group spent consider- able time with Bill Tasto, vice president for manufacturing, to clarify management’s view of their problem. These discussions provided the following answers to these questions.
1. The decisions to be made are the production rates (number of units produced per week) for the two new products.
2. The constraints on these decisions are that the number of hours of production time used per week by the two products in the respective plants cannot exceed the number of hours available.
3. The overall measure of performance for these decisions is the total profit per week from the two products.
Figure 2.2 shows how these answers can be incorporated into the spreadsheet. Based on the first answer, the production rates of the two products are placed in cells C12 and D12 to locate them in the columns for these products just under the data cells. Since we don’t know yet what these production rates should be, they are just entered as zeroes in Figure 2.2 . (Actually, any trial solution can be entered, although negative production rates should be excluded since they are impossible.) Later, these numbers will be changed while seeking the best mix of production rates. Therefore, these cells containing the decisions to be made are called changing cells . To highlight the changing cells, they are shaded bright yellow with a light border. (In the textbook figures, the bright yellow appears as gray.) The changing cells are given the range name UnitsProduced (C12:D12).
Excel Tip: Cell shading and borders can be added by using the borders but- ton and the fill color button in the Font Group of the Home tab.
Excel Tip: See the margin notes in Section 1.2 for tips on adding range names.
These are the three key questions to be addressed in formulating any spread- sheet model.
Some students find it helpful to organize their thoughts by answering these three key questions before beginning to formu- late the spreadsheet model.
The changing cells contain the decisions to be made.
FIGURE 2.1 The initial spreadsheet for the Wyndor problem after transferring the data in Table 2.1 into data cells.
1
A B C D E F G
2
3
4
5
6
7
8
9
Wyndor Glass Co. Product-Mix Problem
Doors Windows
Unit Profit
Plant 1
Plant 2
Plant 3
Hours Used per Unit Produced
Hours
Available
$300 $500
1
0
3
0
2
2
4
12
18
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2.2 Formulating the Wyndor Problem on a Spreadsheet 27
Using the second answer, the total number of hours of production time used per week by the two products in the respective plants is entered in cells E7, E8, and E9, just to the right of the corresponding data cells. The total number of production hours depends on the production rates of the two products, so this total is zero when the production rates are zero. With posi- tive production rates, the total number of production hours used per week in a plant is the sum of the production hours used per week by the respective products. The production hours used by a product is the number of hours needed for each unit of the product times the number of units being produced. Therefore, when positive numbers are entered in cells C12 and D12 for the number of doors and windows to produce per week, the data in cells C7:D9 are used to calculate the total production hours per week as follows:
Production hours in Plant 1 5 1(# of doors) 1 0(# of windows)
Production hours in Plant 2 5 0(# of doors) 1 2(# of windows)
Production hours in Plant 3 5 3(# of doors) 1 2(# of windows)
Consequently, the Excel equations for the three cells in column E are
E7 5 C7*C12 1 D7*D12
E8 5 C8*C12 1 D8*D12
E9 5 C9*C12 1 D9*D12
where each asterisk denotes multiplication. Since each of these cells provides output that depends on the changing cells (C12 and D12), they are called output cells .
Notice that each of the equations for the output cells involves the sum of two products. There is a function in Excel called SUMPRODUCT that will sum up the product of each of the individual terms in two different ranges of cells when the two ranges have the same number of rows and the same number of columns. Each product being summed is the prod- uct of a term in the first range and the term in the corresponding location in the second range. For example, consider the two ranges, C7:D7 and C12:D12, so that each range has one row and two columns. In this case, SUMPRODUCT (C7:D7, C12:D12) takes each of the individual terms in the range C7:D7, multiplies them by the corresponding term in the range C12:D12, and then sums up these individual products, just as shown in the first equa- tion above. Applying the range name for UnitsProduced (C12:D12), the formula becomes SUMPRODUCT(C7:D7, UnitsProduced). Although optional with such short equations, this function is especially handy as a shortcut for entering longer equations.
The formulas in the output cells E7:E9 are very similar. Rather than typing each of these formulas separately into the three cells, it is quicker (and less prone to typos) to type the formula just once in E7 and then copy the formula down into cells E8 and E9. To do this, first enter the formula 5 SUMPRODUCT(C7:D7, UnitsProduced) in cell E7. Then select cell
The colon in C7:D9 is Excel shorthand for the range from C7 to D9; that is, the entire block of cells in column C or D and in row 7, 8, or 9.
Output cells show quanti- ties that are calculated from the changing cells.
The SUMPRODUCT func- tion is used extensively in linear programming spread- sheet models.
FIGURE 2.2 The complete spreadsheet for the wyndor problem with an initial trial solu- tion (both production rates equal to zero) entered into the changing cells (C12 and D12).
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
0
0
0
≤
≤
≤
Windows
Doors
0 0
Windows
$0
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$300 $500
1
0
3
0
2
2
4
12
18
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28 Chapter Two Linear Programming: Basic Concepts
E7 and drag the fill handle (the small box on the lower right corner of the cell cursor) down through cells E8 and E9.
When copying formulas, it is important to understand the difference between relative and absolute references. In the formula in cell E7, the reference to cells C7:D7 is based upon the relative position to the cell containing the formula. In this case, this means the two cells in the same row and immediately to the left. This is known as a relative reference . When this for- mula is copied to new cells using the fill handle, the reference is automatically adjusted to refer to the new cell(s) at the same relative location (the two cells in the same row and immediately to the left). The formula in E8 becomes 5 SUMPRODUCT(C8:D8, UnitsProduced) and the formula in E9 becomes 5 SUMPRODUCT(C9:D9, UnitsProduced). This is exactly what we want, since we always want the hours used at a given plant to be based upon the hours used per unit produced at that same plant (the two cells in the same row and immediately to the left).
In contrast, the reference to the UnitsProduced in E7 is called an absolute reference . These references do not change when they are filled into other cells but instead always refer to the same absolute cell locations.
To make a relative reference, simply enter the cell address (e.g., C7:D7). References referred to by a range name are treated as absolute references. Another way to make an abso- lute reference to a range of cells is to put $ signs in front of the letter and number of the cell reference (e.g., $C$12:$D$12). See Appendix A for more details about relative and absolute referencing and copying formulas.
Next, # signs are entered in cells F7, F8, and F9 to indicate that each total value to their left cannot be allowed to exceed the corresponding number in column G. (On the computer # (or $ ) is often represented as , 5 (or . 5 ), since there is no # (or $ ) key on the keyboard.) The spreadsheet still will allow you to enter trial solutions that violate the # signs. However, these # signs serve as a reminder that such trial solutions need to be rejected if no changes are made in the numbers in column G.
Finally, since the answer to the third question is that the overall measure of performance is the total profit from the two products, this profit (per week) is entered in cell G12. Much like the numbers in column E, it is the sum of products. Since cells C4 and D4 give the profit from each door and window produced, the total profit per week from these products is
Profit 5 $300(# of doors) 1 $500(# of windows)
Hence, the equation for cell G12 is
G12 5 SUMPRODUCT(C4:D4, C12:D12)
Utilizing range names of TotalProfit (G12), UnitProfit (C4:D4), and UnitsProduced (C12:D12), this equation becomes
TotalProfit 5 SUMPRODUCT(UnitProfit, UnitsProduced)
This is a good example of the benefit of using range names for making the resulting equation easier to interpret.
TotalProfit (G12) is a special kind of output cell. It is our objective to make this cell as large as possible when making decisions regarding production rates. Therefore, TotalProfit (G12) is referred to as the objective cell . This cell is shaded orange with a heavy border. (In the textbook figures, the orange appears as gray and is distinguished from the changing cells by its darker shading and heavy border.)
The bottom of Figure 2.3 summarizes all the formulas that need to be entered in the Hours Used column and in the Total Profit cell. Also shown is a summary of the range names (in alphabetical order) and the corresponding cell addresses.
This completes the formulation of the spreadsheet model for the Wyndor problem. With this formulation, it becomes easy to analyze any trial solution for the production
rates. Each time production rates are entered in cells C12 and D12, Excel immediately cal- culates the output cells for hours used and total profit. For example, Figure 2.4 shows the spreadsheet when the production rates are set at four doors per week and three windows per week. Cell G12 shows that this yields a total profit of $2,700 per week. Also note that E7 5 G7, E8 , G8, and E9 5 G9, so the # signs in column F are all satisfied. Thus, this
You can make the column absolute and the row rela- tive (or vice versa) by put- ting a $ sign in front of only the letter (or number) of the cell reference.
Excel Tip: After entering a cell reference, repeatedly pressing the F4 key (or command-T on a Mac) will rotate among the four possi- bilities of relative and abso- lute references (e.g., C12, $C$12, C$12, $C12).
One easy way to enter a # (or $) in a spreadsheet is to type , (or .) with underlining turned on.
The objective cell contains the overall measure of per- formance for the decisions in the changing cells.
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2.2 Formulating the Wyndor Problem on a Spreadsheet 29
trial solution is feasible. However, it would not be feasible to further increase both production rates, since this would cause E7 . G7 and E9 . G9.
Does this trial solution provide the best mix of production rates? Not necessarily. It might be possible to further increase the total profit by simultaneously increasing one production rate and decreasing the other. However, it is not necessary to continue using trial and error to explore such possibilities. We shall describe in Section 2.5 how the Excel Solver can be used to quickly find the best (optimal) solution.
This Spreadsheet Model Is a Linear Programming Model The spreadsheet model displayed in Figure 2.3 is an example of a linear programming model. The reason is that it possesses all the following characteristics.
FIGURE 2.3 The spreadsheet model for the Wyndor problem, including the formulas for the objective cell Total- Profit (G12) and the other output cells in column E, where the goal is to maxi- mize the objective cell.
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
0
0
0
≤
≤
≤
Windows
Doors
0 0
Windows
$0
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
5
E
Hours
Used6
11
12
7
8
9
=SUMPRODUCT(C7:D7, UnitsProduced)
G
Total Profit
=SUMPRODUCT(UnitProfit, UnitsProduced)
=SUMPRODUCT(C8:D8, UnitsProduced)
=SUMPRODUCT(C9:D9, UnitsProduced)
Range Name
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
TotalProfit
UnitProfit
UnitsProduced
Cell
G7:G9
E7:E9
C7:D9
G12
C4:D4
C12:D12
$300 $500
1
0
3
0
2
2
4
12
18
FIGURE 2.4 The spreadsheet for the Wyndor problem with a new trial solution entered into the chang- ing cells, UnitsProduced (C12:D12).
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
4
6
18
≤
≤
≤
Windows
Doors
4 3
Windows
$2,700
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$300 $500
1
0
3
0
2
2
4
12
18
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30 Chapter Two Linear Programming: Basic Concepts
Characteristics of a Linear Programming Model on a Spreadsheet
1. Decisions need to be made on the levels of a number of activities, so changing cells are used to display these levels. (The two activities for the Wyndor problem are the production of the two new products, so the changing cells display the number of units produced per week for each of these products.)
2. These activity levels can have any value (including fractional values) that satisfy a number of constraints. (The production rates for Wyndor’s new products are restricted only by the constraints on the number of hours of production time available in the three plants.)
3. Each constraint describes a restriction on the feasible values for the levels of the activi- ties, where a constraint commonly is displayed by having an output cell on the left, a mathematical sign (#, $ , or 5 ) in the middle, and a data cell on the right. (Wyndor’s three constraints involving hours available in the plants are displayed in Figures 2.2 – 2.4 by hav- ing output cells in column E, # signs in column F, and data cells in column G.)
4. The decisions on activity levels are to be based on an overall measure of performance, which is entered in the objective cell. The goal is to either maximize the objective cell or minimize the objective cell, depending on the nature of the measure of performance. (Wyndor’s overall measure of performance is the total profit per week from the two new products, so this measure has been entered in the objective cell G12, where the goal is to maximize this objective cell.)
5. The Excel equation for each output cell (including the objective cell) can be expressed as a SUMPRODUCT function, 2 where each term in the sum is the product of a data cell and a changing cell. (The bottom of Figure 2.3 shows how a SUMPRODUCT function is used for each output cell for the Wyndor problem.)
Linear programming models are not the only models that can have characteristics 1, 3, and 4. However, characteristics 2 and 5 are key assumptions of linear programming. Therefore, these are the two key characteristics that together differentiate a linear programming model from other kinds of mathematical models that can be formulated on a spreadsheet.
Characteristic 2 rules out situations where the activity levels need to have integer values. For example, such a situation would arise in the Wyndor problem if the decisions to be made were the total numbers of doors and windows to produce (which must be integers) rather than the numbers per week (which can have fractional values since a door or window can be started in one week and completed in the next week). When the activity levels do need to have integer values, a similar kind of model (called an integer programming model) is used instead by making a small adjustment on the spreadsheet, as will be illustrated in Section 3.2.
Characteristic 5 describes the so-called proportionality assumption of linear programming, which states that each term in an output cell must be proportional to a particular changing cell. This prohibits those cases where the Excel equation for an output cell cannot be expressed as a SUMPRODUCT function. To illustrate such a case, suppose that the weekly profit from producing Wyndor’s new windows can be more than doubled by doubling the production rate because of economies in marketing larger amounts. This would mean that this weekly profit is not simply proportional to this production rate, so the Excel equation for the objective cell would need to be more complicated than a SUMPRODUCT function. Consideration of how to formulate such models will be deferred to Chapter 8.
Summary of the Formulation Procedure The procedure used to formulate a linear programming model on a spreadsheet for the Wyndor problem can be adapted to many other problems as well. Here is a summary of the steps involved in the procedure.
1. Gather the data for the problem (such as summarized in Table 2.1 for the Wyndor problem). 2. Enter the data into data cells on a spreadsheet. 3. Identify the decisions to be made on the levels of activities and designate changing cells
for displaying these decisions.
2 There also are some special situations where a SUM function can be used instead because all the numbers that would have gone into the corresponding data cells are 1’s.
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2.3 The Mathematical Model in the Spreadsheet 31
4. Identify the constraints on these decisions and introduce output cells as needed to specify these constraints.
5. Choose the overall measure of performance to be entered into the objective cell. 6. Use a SUMPRODUCT function to enter the appropriate value into each output cell (includ-
ing the objective cell).
This procedure does not spell out the details of how to set up the spreadsheet. There gen- erally are alternative ways of doing this rather than a single “right” way. One of the great strengths of spreadsheets is their flexibility for dealing with a wide variety of problems.
2.3 THE MATHEMATICAL MODEL IN THE SPREADSHEET
There are two widely used methods for formulating a linear programming model. One is to formulate it directly on a spreadsheet, as described in the preceding section. The other is to use algebra to present the model. The two versions of the model are equivalent. The only dif- ference is whether the language of spreadsheets or the language of algebra is used to describe the model. Both versions have their advantages, and it can be helpful to be bilingual. For example, the two versions lead to different, but complementary, ways of analyzing problems like the Wyndor problem (as discussed in the next two sections). Since this book emphasizes the spreadsheet approach, we will only briefly describe the algebraic approach.
Formulating the Wyndor Model Algebraically The reasoning for the algebraic approach is similar to that for the spreadsheet approach. In fact, except for making entries on a spreadsheet, the initial steps are just as described in the preceding section for the Wyndor problem.
1. Gather the relevant data ( Table 2.1 in Section 2.1). 2. Identify the decisions to be made (the production rates for the two new products). 3. Identify the constraints on these decisions (the production time used in the respective
plants cannot exceed the amount available). 4. Identify the overall measure of performance for these decisions (the total profit from the
two products). 5. Convert the verbal description of the constraints and measure of performance into quanti-
tative expressions in terms of the data and decisions (see below).
To start performing step 5, note that Table 2.1 indicates that the number of hours of pro- duction time available per week for the two new products in the respective plants are 4, 12, and 18. Using the data in this table for the number of hours used per door or window produced then leads to the following quantitative expressions for the constraints:
Plant 1: (# of doors) # 4 Plant 2: 2(# of windows) # 12 Plant 3: 3(# of doors) 1 2(# of windows) # 18
In addition, negative production rates are impossible, so two other constraints on the deci- sions are
(# of doors) $ 0 (# of windows) $ 0
The overall measure of performance has been identified as the total profit from the two products. Since Table 2.1 gives the unit profits for doors and windows as $300 and $500,
A linear programming model can be formulated either as a spreadsheet model or as an algebraic model.
1. What are the three questions that need to be answered to begin the process of formulating a linear programming model on a spreadsheet?
2. What are the roles for the data cells, the changing cells, the output cells, and the objective cell when formulating such a model?
3. What is the form of the Excel equation for each output cell (including the objective cell) when formulating such a model?
Review Questions
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32 Chapter Two Linear Programming: Basic Concepts
respectively, the expression obtained in the preceding section for the total profit per week from these products is
Profit 5 $300(# of doors) 1 $500(# of windows)
The goal is to make the decisions (number of doors and number of windows) so as to maxi- mize this profit, subject to satisfying all the constraints identified above.
To state this objective in a compact algebraic model, we introduce algebraic symbols to represent the measure of performance and the decisions. Let
P 5 Profit (total profit per week from the two products, in dollars) D 5 # of doors (number of the special new doors to be produced per week) W 5 # of windows (number of the special new windows to be produced per week)
Substituting these symbols into the above expressions for the constraints and the measure of performance (and dropping the dollar signs in the latter expression), the linear programming model for the Wyndor problem now can be written in algebraic form as shown below.
Algebraic Model
Choose the values of D and W so as to maximize
P 5 300D 1 500W
subject to satisfying all the following constraints:
D # 4
2W # 12 3D 1 2W # 18
and
D $ 0 W $ 0
Terminology for Linear Programming Models Much of the terminology of algebraic models also is sometimes used with spreadsheet mod- els. Here are the key terms for both kinds of models in the context of the Wyndor problem.
1. D and W (or C12 and D12 in Figure 2.3 ) are the decision variables . 2. 300 D 1 500 W [or SUMPRODUCT (UnitProfit, UnitsProduced)] is the objective
function . 3. P (or G12) is the value of the objective function (or objective value for short). 4. D $ 0 and W $ 0 (or C12 $ 0 and D12 $ 0) are called the nonnegativity constraints
(or nonnegativity conditions ). 5. The other constraints are referred to as functional constraints (or structural constraints ). 6. The parameters of the model are the constants in the algebraic model (the numbers in the
data cells). 7. Any choice of values for the decision variables (regardless of how desirable or undesirable
the choice) is called a solution for the model. 8. A feasible solution is one that satisfies all the constraints, whereas an infeasible
solution violates at least one constraint. 9. The best feasible solution, the one that maximizes P (or G12), is called the optimal
solution . (It is possible to have a tie for the best feasible solution, in which case all the tied solutions are called optimal solutions.)
Comparisons So what are the relative advantages of algebraic models and spreadsheet models? An alge- braic model provides a very concise and explicit statement of the problem. Sophisticated software packages that can solve huge problems generally are based on algebraic models because of both their compactness and their ease of use in rescaling the size of a problem.
Management scientists often use algebraic models, but managers generally pre- fer spreadsheet models.
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2.4 The Graphical Method for Solving Two-Variable Problems 33
Management science practitioners with an extensive mathematical background find algebraic models very useful. For others, however, spreadsheet models are far more intuitive. Both managers and business students training to be managers generally live with spreadsheets, not algebraic models. Therefore, the emphasis throughout this book is on spreadsheet models.
2.4 THE GRAPHICAL METHOD FOR SOLVING TWO-VARIABLE PROBLEMS
Linear programming problems having only two decision variables, like the Wyndor problem, can be solved by a graphical method .
Although this method cannot be used to solve problems with more than two decision vari- ables (and most linear programming problems have far more than two), it still is well worth learning. The procedure provides geometric intuition about linear programming and what it is trying to achieve. This intuition is helpful in analyzing larger problems that cannot be solved directly by the graphical method.
It is more convenient to apply the graphical method to the algebraic version of the lin- ear programming model rather than the spreadsheet version. We shall briefly illustrate the method by using the algebraic model obtained for the Wyndor problem in the preceding sec- tion. (A far more detailed description of the graphical method, including its application to the Wyndor problem, is provided in the supplement to this chapter on the CD-ROM.) For this purpose, keep in mind that
D 5 Production rate for the special new doors (the number in changing cell C12 of the spreadsheet) W 5 Production rate for the special new windows (the number in changing cell D12 of the spreadsheet)
The key to the graphical method is the fact that possible solutions can be displayed as points on a two-dimensional graph that has a horizontal axis giving the value of D and a verti- cal axis giving the value of W. Figure 2.5 shows some sample points.
Notation: Either ( D, W ) 5 (2, 3) or just (2, 3) refers to the solution where D 5 2 and W 5 3, as well as to the corresponding point in the graph. Similarly, ( D, W ) 5 (4, 6) means D 5 4 and W 5 6, whereas the origin (0, 0) means D 5 0 and W 5 0.
To find the optimal solution (the best feasible solution), we first need to display graphi- cally where the feasible solutions are. To do this, we must consider each constraint, identify the solutions graphically that are permitted by that constraint, and then combine this informa- tion to identify the solutions permitted by all the constraints. The solutions permitted by all the constraints are the feasible solutions and the portion of the two-dimensional graph where the feasible solutions lie is referred to as the feasible region .
The shaded region in Figure 2.6 shows the feasible region for the Wyndor problem. We now will outline how this feasible region was identified by considering the five constraints one at a time.
To begin, the constraint D $ 0 implies that consideration must be limited to points that lie on or to the right of the W axis. Similarly, the constraint W $ 0 restricts consideration to the points on or above the D axis.
Next, consider the first functional constraint, D # 4, which limits the usage of Plant 1 for producing the special new doors to a maximum of four hours per week. The solutions permit- ted by this constraint are those that lie on, or to the left of, the vertical line that intercepts the D axis at D 5 4, as indicated by the arrows pointing to the left from this line in Figure 2.6 .
graphical method The graphical method pro- vides helpful intuition about linear programming.
feasible region The points in the feasible region are those that satisfy every constraint.
1. When formulating a linear programming model, what are the initial steps that are the same with either a spreadsheet formulation or an algebraic formulation?
2. When formulating a linear programming model algebraically, algebraic symbols need to be introduced to represent which kinds of quantities in the model?
3. What are decision variables for a linear programming model? The objective function? Non- negativity constraints? Functional constraints?
4. What is meant by a feasible solution for the model? An optimal solution?
Review Questions
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FIGURE 2.5 Graph showing the points ( D, W ) 5 (2, 3) and ( D, W ) 5 (4, 6) for the Wyndor Glass Co. product-mix problem.
W
D
8
7
6
5
4
3
2
1
1 2 3 4 5 6 7 8 –2 –1
A product mix of D = 2 and W = 3
A product mix of D = 4 and W = 6
Production Rate (units per week) for Doors
Origin
P ro
d u
ct io
n R
at e
(u n
it s
p er
w ee
k )
fo r
W in
d ow
s
0
–1
–2
(2, 3)
(4, 6)
FIGURE 2.6 Graph showing how the feasible region is formed by the constraint bound- ary lines, where the arrows indicate which side of each line is permit- ted by the corresponding constraint.
W
D
10
8
6
4
2
2 4 6 8
Production Rate for Doors
P ro
d u
ct io
n R
at e
fo r
W in
d ow
s
0
3D + 2W = 18
D = 4
2W = 12
Feasible region
34 Chapter Two Linear Programming: Basic Concepts
The second functional constraint, 2 W # 12, has a similar effect, except now the bound- ary of its permissible region is given by a horizontal line with the equation, 2 W 5 12 (or W 5 6), as indicated by the arrows pointing downward from this line in Figure 2.6 . The line forming the boundary of what is permitted by a constraint is sometimes referred to as
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2.4 The Graphical Method for Solving Two-Variable Problems 35
a constraint boundary line , and its equation may be called a constraint boundary equation . Frequently, a constraint boundary line is identified by its equation.
For each of the first two functional constraints, D # 4 and 2 W # 12, note that the equation for the constraint boundary line ( D 5 4 and 2 W 5 12, respectively) is obtained by replacing the inequality sign with an equality sign. For any constraint with an inequality sign (whether a functional constraint or a nonnegativity constraint), the general rule for obtaining its con- straint boundary equation is to substitute an equality sign for the inequality sign.
We now need to consider one more functional constraint, 3 D 1 2 W # 18. Its constraint boundary equation
3D 1 2W 5 18
includes both variables, so the boundary line it represents is neither a vertical line nor a hori- zontal line. Therefore, the boundary line must intercept (cross through) both axes somewhere. But where?
When a constraint boundary line is neither a vertical line nor a horizontal line, the line intercepts the D axis at the point on the line where W 5 0. Similarly, the line intercepts the W axis at the point on the line where D 5 0.
Hence, the constraint boundary line 3 D 1 2 W 5 18 intercepts the D axis at the point where W 5 0.
When W 5 0, 3D 1 2W 5 18 so the intercept with the D axis is at
becomes 3D 5 18
D 5 6
Similarly, the line intercepts the W axis where D 5 0.
When D 5 0, 3D 1 2W 5 18 so the intercept with the D axis is at
becomes 2W 5 18
W 5 9
Consequently, the constraint boundary line is the line that passes through these two intercept points, as shown in Figure 2.6 .
As indicated by the arrows emanating from this line in Figure 2.6 , the solutions permitted by the constraint 3 D 1 2 W # 18 are those that lie on the origin side of the constraint bound- ary line 3 D 1 2 W 5 18. The easiest way to verify this is to check whether the origin itself, ( D, W ) 5 (0, 0), satisfies the constraint. 3 If it does, then the permissible region lies on the side of the constraint boundary line where the origin is. Otherwise, it lies on the other side. In this case,
3(0) 1 2(0) 5 0 so ( D, W ) 5 (0, 0) satisfies
3D 1 2W # 18
(In fact, the origin satisfies any constraint with a # sign and a positive right-hand side.) A feasible solution for a linear programming problem must satisfy all the constraints
simultaneously. The arrows in Figure 2.6 indicate that the nonnegative solutions permitted by each of these constraints lie on the side of the constraint boundary line where the origin is (or on the line itself). Therefore, the feasible solutions are those that lie nearer to the origin than all three constraint boundary lines (or on the line nearest the origin).
Having identified the feasible region, the final step is to find which of these feasible solu- tions is the best one—the optimal solution. For the Wyndor problem, the objective happens to be to maximize the total profit per week from the two products (denoted by P ). Therefore, we want to find the feasible solution ( D, W ) that makes the value of the objective function
P 5 300D 1 500W
as large as possible. To accomplish this, we need to be able to locate all the points ( D, W ) on the graph that give
a specified value of the objective function. For example, consider a value of P 5 1,500 for the objective function. Which points ( D, W ) give 300 D 1 500 W 5 1,500?
For any constraint with an inequality sign, its con- straint boundary equation is obtained by replacing the inequality sign by an equal- ity sign.
The location of a slanting constraint boundary line is found by identifying where it intercepts each of the two axes.
Checking whether (0, 0) satisfies a constraint indi- cates which side of the constraint boundary line satisfies the constraint.
3 The one case where using the origin to help determine the permissible region does not work is if the con- straint boundary line passes through the origin. In this case, any other point not lying on this line can be used just like the origin.
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FIGURE 2.7 Graph showing three objective function lines for the Wyndor Glass Co. product-mix problem, where the top one passes through the optimal solution.
W
D
8
6
4
2
2 4 6 8 10
Production Rate for Doors
Production Rate for Windows
0
Feasible regionP = 1,500 = 300D + 500W
Optimal solution
P = 3,000 = 300D + 500W
P = 3,600 = 300D + 500W
(2, 6)
36 Chapter Two Linear Programming: Basic Concepts
This equation is the equation of a line. Just as when plotting constraint boundary lines, the location of this line is found by identifying its intercepts with the two axes. When W 5 0, this equation yields D 5 5, and similarly, W 5 3 when D 5 0, so these are the two intercepts, as shown by the bottom slanting line passing through the feasible region in Figure 2.7 .
P 5 1,500 is just one sample value of the objective function. For any other specified value of P, the points (D, W) that give this value of P also lie on a line called an objective function line.
An objective function line is a line whose points all have the same value of the objective function.
For the bottom objective function line in Figure 2.7 , the points on this line that lie in the feasible region provide alternate ways of achieving an objective function value of P 5 1,500. Can we do better? Let us try doubling the value of P to P 5 3,000. The corresponding objec- tive function line
300D 1 500W 5 3,000
is shown as the middle line in Figure 2.7 . (Ignore the top line for the moment.) Once again, this line includes points in the feasible region, so P 5 3,000 is achievable.
Let us pause to note two interesting features of these objective function lines for P 5 1,500 and P 5 3,000. First, these lines are parallel. Second, doubling the value of P from 1,500 to 3,000 also doubles the value of W at which the line intercepts the W axis from W 5 3 to W 5 6. These features are no coincidence, as indicated by the following properties.
Key Properties of Objective Function Lines: All objective function lines for the same prob- lem are parallel. Furthermore, the value of W at which an objective function line intercepts the W axis is proportional to the value of P.
These key properties of objective function lines suggest the strategy to follow to find the optimal solution. We already have tried P 5 1,500 and P 5 3,000 in Figure 2.7 and found that their objective function lines include points in the feasible region. Increasing P again will generate another parallel objective function line farther from the origin. The objective func- tion line of special interest is the one farthest from the origin that still includes a point in the feasible region. This is the third objective function line in Figure 2.7 . The point on this line
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2.4 The Graphical Method for Solving Two-Variable Problems 37
that is in the feasible region, ( D, W ) 5 (2, 6), is the optimal solution since no other feasible solution has a larger value of P.
Optimal Solution
D 5 2 (Produce 2 special new doors per week) W 5 6 (Produce 6 special new windows per week)
These values of D and W can be substituted into the objective function to find the value of P.
P 5 300D 1 500W 5 300(2) 1 500(6) 5 3,600
This has been a fairly quick description of the graphical method. You can go to the supple- ment to this chapter on the CD-ROM if you would like to see a fuller description of how to use the graphical method to solve the Wyndor problem.
The Interactive Management Science Modules (available at www.mhhe.com/hillier5e or in your CD-ROM) also includes a module that is designed to help increase your understand- ing of the graphical method. This module, called Graphical Linear Programming and Sen- sitivity Analysis, enables you to immediately see the constraint boundary lines and objective function lines that result from any linear programming model with two decision variables. You also can see how the objective function lines lead you to the optimal solution. Another key feature of the module is the ease with which you can use the graphical method to perform what-if analysis to see what happens if any changes occur in the data for the problem. (We will focus on what-if analysis for linear programming, including the role of the graphical method for this kind of analysis, in Chapter 5.)
This module also can be used to check out how unusual situations can arise when solv- ing linear programming problems. For example, it is possible for a problem to have multiple solutions that tie for being an optimal solution. (Hint: See what happens if the unit profit for the windows in the Wyndor problem were reduced to $200.) It is possible for a problem to have no optimal solutions because it has no feasible solutions. (Hint: See what happens if the Wyndor management decides to require that the total number of doors and windows produced per week must be at least 10 to justify introducing these new products.) Still another remote possibility is that a linear programming model has no optimal solution because the constraints do not prevent increasing (when maximizing) the objective function value indefinitely. (Hint: See what happens when the functional constraints for Plants 2 and 3 in the Wyndor problem are inadvertently left out of the model.) Try it. (See Problem 2.17.)
Summary of the Graphical Method The graphical method can be used to solve any linear programming problem having only two decision variables. The method uses the following steps:
1. Draw the constraint boundary line for each functional constraint. Use the origin (or any point not on the line) to determine which side of the line is permitted by the constraint.
2. Find the feasible region by determining where all constraints are satisfied simultaneously. 3. Determine the slope of one objective function line. All other objective function lines will
have the same slope. 4. Move a straight edge with this slope through the feasible region in the direction of improv-
ing values of the objective function. Stop at the last instant that the straight edge still passes through a point in the feasible region. This line given by the straight edge is the optimal objective function line.
5. A feasible point on the optimal objective function line is an optimal solution.
Check out this module in the Interactive Management Science Modules to learn more about the graphical method.
1. The graphical method can be used to solve linear programming problems with how many deci- sion variables?
2. What do the axes represent when applying the graphical method to the Wyndor problem? 3. What is a constraint boundary line? A constraint boundary equation? 4. What is the easiest way of determining which side of a constraint boundary line is permitted by
the constraint?
Review Questions
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38 Chapter Two Linear Programming: Basic Concepts
2.5 USING EXCEL’S SOLVER TO SOLVE LINEAR PROGRAMMING PROBLEMS
The graphical method is very useful for gaining geometric intuition about linear program- ming, but its practical use is severely limited by only being able to solve tiny problems with two decision variables. Another procedure that will solve linear programming problems of any reasonable size is needed. Fortunately, Excel includes a tool called Solver that will do this once the spreadsheet model has been formulated as described in Section 2.2. (Section 2.6 will show how Risk Solver Platform, which includes a more advanced version of Solver, can be used to solve this same problem.) To access Solver the first time, you need to install it. Click the Office Button, choose Excel Options, then click on Add-Ins on the left side of the window, select Manage Excel Add-Ins at the bottom of the window, and then press the Go button. Make sure Solver is selected in the Add-Ins dialog box, and then it should appear on the Data tab. For Excel 2011 (for the Mac), go to www.solver.com/mac to download and install the Solver application.
Figure 2.3 in Section 2.2 shows the spreadsheet model for the Wyndor problem. The val- ues of the decision variables (the production rates for the two products) are in the changing cells, UnitsProduced (C12:D12), and the value of the objective function (the total profit per week from the two products) is in the objective cell TotalProfit (G12). To get started, an arbi- trary trial solution has been entered by placing zeroes in the changing cells. Solver will then change these to the optimal values after solving the problem.
This procedure is started by choosing Solver on the Data tab (for Excel 2007 or 2010 on a PC), or choosing Solver in the Tools menu (for Excel 2011 on a Mac). Figure 2.8 shows the Solver dialog box that is used to tell Solver where each component of the model is located on the spreadsheet.
You have the choice of typing the range names, typing the cell addresses, or clicking on the cells in the spreadsheet. Figure 2.8 shows the result of using the first choice, so TotalProfit (rather than G12) has been entered for the objective cell. Since the goal is to maximize the objective cell, Max also has been selected. The next entry in the Solver dialog box identifies the changing cells, which are UnitsProduced (C12:D12) for the Wyndor problem.
Next, the cells containing the functional constraints need to be specified. This is done by clicking on the Add button on the Solver dialog box. This brings up the Add Constraint dialog box shown in Figure 2.9 . The # signs in cells F7, F8, and F9 of Figure 2.3 are a reminder that the cells in HoursUsed (E7:E9) all need to be less than or equal to the corresponding cells in HoursAvailable (G7:G9). These constraints are specified for Solver by entering HoursUsed (or E7:E9) on the left-hand side of the Add Constraint dialog box and HoursAvailable (or G7:G9) on the right-hand side. For the sign between these two sides, there is a menu to choose between , 5 , 5 , or . 5 , so , 5 has been chosen. This choice is needed even though # signs were previously entered in column F of the spreadsheet because the Solver only uses the constraints that are specified with the Add Constraint dialog box.
If there were more functional constraints to add, you would click on Add to bring up a new Add Constraint dialog box. However, since there are no more in this example, the next step is to click on OK to go back to the Solver dialog box.
Before asking Solver to solve the model, two more steps need to be taken. We need to tell Solver that nonnegativity constraints are needed for the changing cells to reject negative production rates. We also need to specify that this is a linear programming problem so the simplex method (the standard method used by Solver to solve linear programming problems) can be used. This is demonstrated in Figure 2.10 , where the Make Unconstrained Variables Non-Negative option has been checked and the Solving Method chosen is Simplex LP (rather than GRG Nonlinear or Evolutionary, which are used for solving nonlinear problems). The Solver dialog box shown in this figure now summarizes the complete model.
Now you are ready to click on Solve in the Solver dialog box, which will start the solving of the problem in the background. After a few seconds (for a small problem), Solver will then indicate the results. Typically, it will indicate that it has found an optimal solution, as specified in the Solver Results dialog box shown in Figure 2.11 . If the model has no feasible solutions or no optimal solution, the dialog box will indicate that instead by stating that
Excel Tip: If you select cells by clicking on them, they will first appear in the dialog box with their cell addresses and with dollar signs (e.g., C9:D9). You can ignore the dollar signs. Solver eventu- ally will replace both the cell addresses and the dollar signs with the correspond- ing range name (if a range name has been defined for the given cell addresses), but only after either adding a constraint or closing and reopening the Solver dialog box.
Solver Tip: To select changing cells, click and drag across the range of cells. If the changing cells are not contiguous, you can type a comma and then select another range of cells. Up to 200 changing cells can be selected with the basic version of Solver that comes with Excel.
The Add Constraint dialog box is used to specify all the functional constraints.
When solving a linear pro- gramming problem, be sure to specify that nonnegativ- ity constraints are needed and that the model is linear by choosing Simplex LP.
Solver Tip: The message “Solver could not find a feasible solution” means that there are no solu- tions that satisfy all the constraints. The message “The Objective Cell values do not converge” means that Solver could not find a best solution, because better solutions always are available (e.g., if the con- straints do not prevent infi- nite profit). The message “The linearity conditions required by this LP Solver are not satisfied” means Simplex LP was chosen as the Solving Method, but the model is not linear.
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2.5 Using Excel’s Solver to Solve Linear Programming Problems 39
“Solver could not find a feasible solution” or that “The Objective Cell values do not con- verge.” (Section 14.1 will describe how these possibilities can occur.) The dialog box also presents the option of generating various reports. One of these (the Sensitivity Report) will be discussed in detail in Chapter 5.
FIGURE 2.9 The Add Constraint dia- log box after specifying that cells E7, E8, and E9 in Figure 2.3 are required to be less than or equal to cells G7, G8, and G9, respectively.
FIGURE 2.8 The Solver dialog box after specifying the first components of the model for the Wyndor prob- lem. TotalProfit (G12) is being maximized by changing UnitsProduced (C12:D12). Figure 2.9 will demonstrate the addition of constraints and then Figure 2.10 will demonstrate the changes needed to specify that the problem being considered is a linear programming problem. (The default Solving Method shown here— GRG Nonlinear — is not applicable to linear programming problems.)
The Incomplete Solver Dialog Box
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40 Chapter Two Linear Programming: Basic Concepts
After solving the model and clicking OK in the Solver Results dialog box, Solver replaces the original numbers in the changing cells with the optimal numbers, as shown in Figure 2.12 . Thus, the optimal solution is to produce two doors per week and six windows per week, just as was found by the graphical method in the preceding section. The spreadsheet also indicates the corresponding number in the objective cell (a total profit of $3,600 per week), as well as the numbers in the output cells HoursUsed (E7:E9).
The entries needed for the Solver dialog box are summarized in the Solver Parameters box shown on the bottom left of Figure 2.12 . This more compact summary of the Solver Parameters will be shown for all of the many models that involve the Solver throughout the book.
At this point, you might want to check what would happen to the optimal solution if any of the numbers in the data cells were to be changed to other possible values. This is easy to do because Solver saves all the addresses for the objective cell, changing cells, constraints, and so on when you save the file. All you need to do is make the changes you want in the data cells and then click on Solve in the Solver dialog box again. (Chapter 5 will focus on this kind of what-if analysis, including how to use the Solver’s Sensitivity Report to expedite the analysis.)
FIGURE 2.10 The completed Solver dia- log box after specifying the entire model in terms of the spreadsheet.
The Completed Solver Dialog Box
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FIGURE 2.12 The spreadsheet obtained after solving the Wyndor problem.
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
2
12
18
Windows
Doors
2 6
Windows
$3,600
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
5
E
Hours
Used6
11
12
7
8
9
=SUMPRODUCT(C7:D7, UnitsProduced)
G
Total Profit
=SUMPRODUCT(UnitProfit, UnitsProduced)
=SUMPRODUCT(C8:D8, UnitsProduced)
=SUMPRODUCT(C9:D9, UnitsProduced)
Range Name
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
TotalProfit
UnitProfit
UnitsProduced
Cell
G7:G9
E7:E9
C7:D9
G12
C4:D4
C12:D12
$300 $500
1
0
3
0
2
2
4
12
18
≤
≤
≤
Solver Parameters Set Objective Cell: Total Profit To: Max By Changing Variable Cells: UnitsProduced Subject to the Constraints: HoursUsed <= HoursAvailable Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
2.5 Using Excel’s Solver to Solve Linear Programming Problems 41
FIGURE 2.11 The Solver Results dialog box that indicates that an optimal solution has been found.
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42 Chapter Two Linear Programming: Basic Concepts
To assist you with experimenting with these kinds of changes, your MS Courseware includes Excel files for this chapter (as for others) that provide a complete formulation and solution of the examples here (the Wyndor problem and the one in Section 2.7) in a spread- sheet format. We encourage you to “play” with these examples to see what happens with dif- ferent data, different solutions, and so forth. You might also find these spreadsheets useful as templates for homework problems.
2.6 RISK SOLVER PLATFORM FOR EDUCATION (RSPE)
Frontline Systems, the original developer of the standard Solver included with Excel (hereaf- ter referred to as Excel’s Solver), also has developed Premium versions of Solver that provide greatly enhanced functionality. The company now features a particularly powerful Premium Solver called Risk Solver Platform. New with this edition, we are excited to provide access to the Excel add-in, Risk Solver Platform for Education (RSPE) from Frontline Systems. Instructions for installing this software are on a supplementary insert included with the book and also on the book’s website, www.mhhe.com/hillier5e.
While Excel’s Solver is sufficient for most of the problems considered in this book, RSPE includes a number of important features not available with Excel’s Solver. Where either Excel’s Solver or RSPE can be used, the book will often use the term Solver generically to mean either Excel’s Solver or RSPE. Where there are differences, the book will include instructions for both Excel’s Solver and RSPE. The enhanced features of RSPE will be high- lighted as they come up throughout the book. However, if you and your instructor prefer to focus on only using Excel’s Solver, you will find that there is plenty of material to cover in the book that does not require the use of RSPE.
When RSPE is installed, a new tab is available on the Excel ribbon called Risk Solver Platform. Choosing this tab will reveal the ribbon shown in Figure 2.13 . The buttons on this ribbon will be used to interact with RSPE. This same figure also reveals a nice feature of RSPE—the Solver Options and Model Specifications pane (showing the objective cell, changing cells, constraints, etc.)—that can be seen alongside your main spreadsheet, with both visible simultaneously. This pane can be toggled on (to see the model) or off (to hide the model and leave more room for the spreadsheet) by clicking on the Model button on the far left of the Risk Solver Platform ribbon. Also, since the model was already set up with Excel’s Solver in Section 2.5, it is already set up in the RSPE Model pane, with the objective speci- fied as TotalProfit (G12) with changing cells UnitsProduced (C12:D12) and the constraints HoursUsed (E7:E9) , 5 HoursAvailable (G7:G9). The data for Excel’s Solver and RSPE are compatible with each other. Making a change with one makes the same change in the other. Thus, you can work with either Excel’s Solver or RSPE, and then go back and forth, without losing any Solver data.
If the model had not been previously set up with Excel’s Solver, the steps for doing so with RSPE are analogous to the steps used with Excel’s Solver as covered in Section 2.5. In both cases, we need to specify the location of the objective cell, the changing cells, and the functional constraints, and then click to solve the model. However, the user interface is somewhat different. RSPE uses the buttons on the Risk Solver Platform ribbon instead of the Solver dialog box. We will now walk you through the steps to set up the Wyndor problem in RSPE.
To specify TotalProfit (G12) as the objective cell, select the cell in the spreadsheet and then click on the Objective button on the Risk Solver Platform ribbon. As shown in Figure 2.14 , this will drop down a menu where you can choose to minimize (Min) or maximize (Max) the objective cell. Within the options of Min or Max are further options (Normal, Expected, VaR, etc.). For now, we will always choose the Normal option.
1. Which dialog box is used to enter the addresses for the objective cell and the changing cells? 2. Which dialog box is used to specify the functional constraints for the model? 3. Which options normally need to be chosen to solve a linear programming model?
Review Questions
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2.6 Risk Solver Platform for Education (RSPE) 43
FIGURE 2.13 The screenshot for the Wyndor problem that shows both the ribbon and the Solver Options and Model Specifications pane that are revealed after choosing the tab called Risk Solver Platform on the Excel ribbon.
FIGURE 2.14 The screenshot for the Wyndor problem that shows the drop-down menu generated by clicking on the Objec- tive button on the Risk Solver Platform ribbon after choosing TotalProfit (G12) as the objective cell.
To specify UnitsProduced (C12:D12) as the changing cells, select these cells in the spread- sheet and then click on the Decisions button on the Risk Solver Platform ribbon. As shown in Figure 2.15 , this will drop down a menu where you can choose various options (Plot, Normal, Recourse). For linear programming, we will always choose the Normal option.
Next the functional constraints need to be specified. For the Wyndor problem, the functional constraints are HoursUsed (E7:E9) , 5 HoursAvailable (G7:G9). To enter these constraints in RSPE, select the cells representing the left-hand side of these constraints (HoursUsed, or E7:E9) and click the Constraints button on the Risk Solver Platform ribbon. As shown in Figure 2.16 , this drops down a menu for various kinds of constraints. For linear program- ming functional constraints, choose Normal Constraint and then the type of constraint desired (either , 5 , 5 , or . 5 ). For the Wyndor problem, choosing , 5 would then bring up the Add Constraint dialog box shown in Figure 2.17 . This is much like the Add Constraint dialog box for Excel’s Solver (see Figure 2.9 ). HoursUsed and , 5 already are filled in when the Add Constraint dialog box is brought up (because the HoursUsed cells were selected and , 5 was chosen under the Constraints button menu). The Add Constraint dialog box then can be used to fill in the remaining right-hand side of the constraint—HoursAvailable (G7:G9)—by clicking in the box labeled Constraint and choosing these cells on the spreadsheet. Figure 2.17 shows the dialog box after this has been done.
RSPE Tip: Another way to add an objective, chang- ing cells, or constraints using RSPE is to click on the big green plus ( 1 ) on the Mode pane and choose Add Objective, Add Vari- able, or Add Constraint, respectively.
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44 Chapter Two Linear Programming: Basic Concepts
Changes to the model can easily be made within the Model pane shown in Figure 2.13 . For example, to delete an element of the model (e.g., the objective, changing cells, or constraints), select that part of the model and then click on the red X near the top of the Model pane. To change an element of the model, click on that element in the Model pane. The bottom of the Model pane will then show information about that element. For example, clicking on the HoursUsed , 5 HoursAvailable constraint in the Model pane will then show the informa- tion seen in Figure 2.18 . Clicking on any piece of the information will allow you to change it (e.g., you can change , 5 to . 5 , or you can change the cell references for either side of the constraint).
Selecting the Engine tab at the top of the Model pane will show information about the algo- rithm that will be used to solve the problem as well as a variety of options for that algorithm.
RSPE Tip: Double-click- ing on any element of the model (e.g., the objective, any of the changing cells, or any of the constraints) will bring up a dialog box allowing you to make changes to that part of the model.
FIGURE 2.16 The screenshot of the Wyndor problem that shows the drop-down menu generated by click- ing on the Constraints button on the Risk Solver Platform ribbon after choosing HoursUsed (E7:E9) as the left-hand side of the functional constraints.
FIGURE 2.17 The Add Constraint dialog box that is brought up after choosing , 5 in Figure 2.16 . This dialog box also shows HoursAvailable (G7:G9) as the right-hand side of the functional constraints after clicking in the Constraint box and choosing these cells on the spreadsheet.
FIGURE 2.15 The screenshot for the Wyndor problem that shows the drop-down menu generated by click- ing on the Decision but- ton on the Risk Solver Platform ribbon after choosing UnitsProduced (C12:D12) as the chang- ing variable cells.
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2.6 Risk Solver Platform for Education (RSPE) 45
FIGURE 2.18 This figure shows the result of clicking on the HoursUsed , 5 Hours Available constraint in the Model pane prior to con- sidering possible changes in the constraint.
The drop-down menu at the top will allow you to choose the algorithm. For a linear pro- gramming model (such as the Wyndor problem), you will want to choose the Standard LP/ Quadratic Engine. This is equivalent to the Simplex LP option in Excel’s Solver. To make unconstrained variables nonnegative (as we did in Figure 2.10 with Excel’s Solver), be sure that the Assume Nonnegative option is set to true. Figure 2.19 shows the model pane after making these selections.
Once the model is all set up in RSPE, the model would be solved by clicking on the Optimize button on the Risk Solver Platform ribbon. Just like Excel’s Solver, this will then display the results of solving the model on the spreadsheet, as shown in Figure 2.20 . As seen in this figure, the Output tab of the Model pane also will show a summary of the solution process, including the message (similar to Figure 2.11 ) that “Solver found a solution. All constraints and optimality conditions are satisfied.”
1. Which button on the Risk Solver Platform ribbon should be pressed to specify the objective cell?
2. Which button on the Risk Solver Platform ribbon should be pressed to specify the changing cells?
3. Which button on the Risk Solver Platform ribbon should be pressed to enter the constraints? 4. Which button on the Risk Solver Platform ribbon should be pressed to solve the model?
Review Questions
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46 Chapter Two Linear Programming: Basic Concepts
FIGURE 2.20 A screen shot showing the final solution of the Wyndor problem and the Output tab of the model pane showing a summary of the solution process.
FIGURE 2.19 The model pane after selecting the Standard LP/Quadratic Engine and setting the Assume Non- negative option to True.
2.7 A MINIMIZATION EXAMPLE—THE PROFIT & GAMBIT CO. ADVERTISING-MIX PROBLEM
The analysis of the Wyndor Glass Co. case study in Sections 2.2, 2.5, and 2.6 illustrated how to formulate and solve one type of linear programming model on a spreadsheet. The same general approach can be applied to many other problems as well. The great flexibility of linear programming and spreadsheets provides a variety of options for how to adapt the
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2.7 A Minimization Example—The Profit & Gambit Co. Advertising-Mix Problem 47
formulation of the spreadsheet model to fit each new problem. Our next example illustrates some options not used for the Wyndor problem.
Planning an Advertising Campaign The Profit & Gambit Co. produces cleaning products for home use. This is a highly com- petitive market, and the company continually struggles to increase its small market share. Management has decided to undertake a major new advertising campaign that will focus on the following three key products:
• A spray prewash stain remover. • A liquid laundry detergent. • A powder laundry detergent.
This campaign will use both television and the print media. A commercial has been devel- oped to run on national television that will feature the liquid detergent. The advertisement for the print media will promote all three products and will include cents-off coupons that consumers can use to purchase the products at reduced prices. The general goal is to increase the sales of each of these products (but especially the liquid detergent) over the next year by a significant percentage over the past year. Specifically, management has set the following goals for the campaign:
• Sales of the stain remover should increase by at least 3 percent. • Sales of the liquid detergent should increase by at least 18 percent. • Sales of the powder detergent should increase by at least 4 percent.
Table 2.2 shows the estimated increase in sales for each unit of advertising in the respective outlets. 4 (A unit is a standard block of advertising that Profit & Gambit commonly purchases, but other amounts also are allowed.) The reason for 2 1 percent for the powder detergent in the Television column is that the TV commercial featuring the new liquid detergent will take away some sales from the powder detergent. The bottom row of the table shows the cost per unit of advertising for each of the two outlets.
Management’s objective is to determine how much to advertise in each medium to meet the sales goals at a minimum total cost.
Formulating a Spreadsheet Model for This Problem The procedure summarized at the end of Section 2.2 can be used to formulate the spreadsheet model for this problem. Each step of the procedure is repeated below, followed by a descrip- tion of how it is performed here.
1. Gather the data for the problem. This has been done as presented in Table 2.2 . 2. Enter the data into data cells on a spreadsheet. The top half of Figure 2.21 shows this
spreadsheet. The data cells are in columns C and D (rows 4 and 8 to 10), as well as in cells
4 A simplifying assumption is being made that each additional unit of advertising in a particular outlet will yield the same increase in sales regardless of how much advertising already is being done. This becomes a poor assumption when the levels of advertising under consideration can reach a saturation level (as in Case 8.1), but is a reasonable approximation for the small levels of advertising being considered in this problem.
Increase in Sales per Unit of Advertising
Product Television Print Media Minimum
Required Increase Stain remover 0% 1% 3% Liquid detergent 3 2 18 Powder detergent 21 4 4
Unit cost $1 million $2 million
TABLE 2.2 Data for the Profit & Gambit Co. Advertising- Mix Problem
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FIGURE 2.21 The spreadsheet model for the Profit & Gambit problem, including the formulas for the objec- tive cell TotalCost (G14) and the other output cells in column E, as well as the specifications needed to set up Solver. The changing cells, Adver- tisingUnits (C14:D14), show the optimal solution obtained by Solver.
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
13
14
Profit & Gambit Co. Advertising-Mix Problem
Television
3%
18%
8%
≥ ≥
≥
Print Media
Television
4 3
Print Media
10
Total Cost
($millions)
Unit Cost ($millions)
Stain Remover
Liquid Detergent
Powder Detergent
Advertising Units
Increase in Sales per Unit of Advertising
Minimum
Increase
Increased
Sales
6
E
Increased
Sales7
12
13
14
8
9
10
=SUMPRODUCT(C8:D8, AdvertisingUnits)
G
Total Cost
($millions)
=SUMPRODUCT(UnitCost, AdvertisingUnits)
=SUMPRODUCT(C9:D9, AdvertisingUnits)
=SUMPRODUCT(C10:D10, AdvertisingUnits)
Range Name
AdvertisingUnits
IncreasedSales
IncreasedSalesPerUnitAdvertising
MinimumIncrease
TotalCost
UnitCost
Cells
C14: D14
E8: E10
C8: D10
G8: G10
G14
C4: D4
1 2
0%
3%
-1%
1%
2%
4%
3%
18%
4%
Solver Parameters Set Objective Cell: TotalCost To: Min By Changing (Variable) Cells: AdvertisingUnits Subject to the Constraints: IncreasedSales >= MinimumIncrease Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
48 Chapter Two Linear Programming: Basic Concepts
G8:G10. Note how this particular formatting of the spreadsheet has facilitated a direct transfer of the data from Table 2.2 .
3. Identify the decisions to be made on the levels of activities and designate changing cells for making these decisions. In this case, the activities of concern are advertising on televi- sion and advertising in the print media, so the levels of these activities refer to the amount of advertising in these media. Therefore, the decisions to be made are
Decision 1: TV 5 Number of units of advertising on television
Decision 2: PM 5 Number of units of advertising in the print media
The two gray cells with light borders in Figure 2.21 —C14 and D14—have been desig- nated as the changing cells to hold these numbers:
TV S cell C14 PM S cell D14 with AdvertisingUnits as the range name for these cells. (See the bottom of Figure 2.21 for
a list of all the range names.) These are natural locations for the changing cells, since each one is in the column for the corresponding advertising medium. To get started, an arbitrary
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2.7 A Minimization Example—The Profit & Gambit Co. Advertising-Mix Problem 49
trial solution (such as all zeroes) is entered into these cells. ( Figure 2.21 shows the optimal solution after having already applied Solver.)
4. Identify the constraints on these decisions and introduce output cells as needed to spec- ify these constraints. The three constraints imposed by management are the goals for the increased sales for the respective products, as shown in the rightmost column of Table 2.2 . These constraints are
Stain remover: Total increase in sales $ 3%
Liquid detergent: Total increase in sales $ 18%
Powder detergent: Total increase in sales $ 4%
The second and third columns of Table 2.2 indicate that the total increases in sales from both forms of advertising are
Total for stain remover 5 1% of PM
Total for liquid detergent 5 3% of TV 1 2% of PM
Total for powder detergent 5 21% of TV 1 4% of PM
Consequently, since rows 8, 9, and 10 in the spreadsheet are being used to provide infor- mation about the three products, cells E8, E9, and E10 are introduced as output cells to show the total increase in sales for the respective products. In addition, $ signs have been entered in column F to remind us that the increased sales need to be at least as large as the numbers in column G. (The use of $ signs here rather than # signs is one key difference from the spreadsheet model for the Wyndor problem in Figure 2.3 .)
5. Choose the overall measure of performance to be entered into the objective cell. Manage- ment’s stated objective is to determine how much to advertise in each medium to meet the sales goals at a minimum total cost. Therefore, the total cost of the advertising is entered in the objective cell TotalCost (G14). G14 is a natural location for this cell since it is in the same row as the changing cells. The bottom row of Table 2.2 indicates that the number going into this cell is
Cost 5 ($1 million) TV 1 ($2 million) PM S cell G14
6. Use a SUMPRODUCT function to enter the appropriate value into each output cell (includ- ing the objective cell). Based on the above expressions for cost and total increases in sales, the SUMPRODUCT functions needed here for the output cells are those shown under the right side of the spreadsheet in Figure 2.21 . Note that each of these functions involves the relevant data cells and the changing cells, AdvertisingUnits (C14:D14).
This spreadsheet model is a linear programming model, since it possesses all the charac- teristics of such models enumerated in Section 2.2.
Applying Solver to This Model The procedure for using Solver to obtain an optimal solution for this model is basically the same as described in Section 2.5 (for Excel’s Solver) or Section 2.6 (for RSPE). The Solver parameters are shown below the left-hand side of the spreadsheet in Figure 2.21 . In addi- tion to specifying the objective cell and changing cells, the constraints that IncreasedSales $ MinimumIncrease have been specified in this box by using the Add Constraint dialog box. Since the objective is to minimize total cost, Min also has been selected. (This is in contrast to the choice of Max for the Wyndor problem.)
Two options are also specified at the bottom of the Solver Parameters box on the lower left-hand side of Figure 2.21 . The changing cells need nonnegativity constraints (specified in the main Solver dialog box in Excel’s Solver, or on the Engine tab of the Model pane in RSPE) because negative values of advertising levels are not possible alternatives. Choosing the Simplex LP solving method in Excel’s Solver (or the Standard LP/Quadratic Engine in RSPE) specifies that this is a linear programming model.
After running Solver, the optimal solution shown in the changing cells of the spreadsheet in Figure 2.21 is obtained.
Unlike the Wyndor prob- lem, we need to use $ signs for these constraints.
Unlike the Wyndor prob- lem, the objective now is to minimize the objective cell.
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50
Samsung Electronics Corp., Ltd. (SEC), is a leading mer- chant of dynamic and static random access memory devices and other advanced digital integrated circuits. Its site at Kiheung, South Korea (probably the largest semiconductor fabrication site in the world), fabricates more than 300,000 silicon wafers per month and employs over 10,000 people.
Cycle time is the industry’s term for the elapsed time from the release of a batch of blank silicon wafers into the fabrication process until completion of the devices that are fabricated on those wafers. Reducing cycle times is an ongoing goal since it both decreases costs and enables offering shorter lead times to potential customers, a real key to maintaining or increasing market share in a very competitive industry.
Three factors present particularly major challenges when striving to reduce cycle times. One is that the prod- uct mix changes continually. Another is that the company often needs to make substantial changes in the fab-out schedule inside the target cycle time as it revises forecasts
of customer demand. The third is that the machines of a general type are not homogeneous so only a small number of machines are qualified to perform each device step.
A management science team developed a huge linear programming model with tens of thousands of decision variables and functional constraints to cope with these challenges. The objective function involved minimizing back-orders and finished-goods inventory.
The ongoing implementation of this model enabled the company to reduce manufacturing cycle times to fabricate dynamic random access memory devices from more than 80 days to less than 30 days. This tremendous improvement and the resulting reduction in both manufacturing costs and sale prices enabled Samsung to capture an additional $200 million in annual sales revenue.
Source: R. C. Leachman, J. Kang, and Y. Lin, “SLIM: Short Cycle Time and Low Inventory in Manufacturing at Samsung Electronics,” Interfaces 32, no. 1 (January–February 2002), pp. 61–77. (A link to this article is provided on our website, www.mhhe.com/hillier5e. )
An Application Vignette
Optimal Solution
C14 5 4 (Undertake 4 units of advertising on television) C14 5 3 (Undertake 3 units of advertising in the print media)
The objective cell indicates that the total cost of this advertising plan would be $10 million.
The Mathematical Model in the Spreadsheet When performing step 5 of the procedure for formulating a spreadsheet model, the total cost of advertising was determined to be
Cost 5 TV 1 2 PM (in millions of dollars)
where the goal is to choose the values of TV (number of units of advertising on television) and PM (number of units of advertising in the print media) so as to minimize this cost. Step 4 identified three functional constraints:
Stain remover: 1% of PM $ 3%
Liquid detergent: 3% of TV 1 2% of PM $ 18%
Powder detergent: 21% of TV 1 4% of PM $ 4%
Choosing the Make Variables Non-Negative option with Solver recognized that TV and PM cannot be negative. Therefore, after dropping the percentage signs from the functional con- straints, the complete mathematical model in the spreadsheet can be stated in the following succinct form.
Minimize Cost 5 TV 1 2 PM (in millions of dollars) subject to
Stain remover increased sales: PM $ 3
Liquid detergent increased sales: 3 TV 1 2 PM $ 18
Powder detergent increased sales: 2TV 1 4 PM $ 4 and
TV $ 0 PM $ 0
Implicit in this statement is “Choose the values of TV and PM so as to. . . .” The term “subject to” is shorthand for “Choose these values subject to the requirement that the values satisfy all the following constraints.”
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FIGURE 2.22 Graph showing two objec- tive function lines for the Profit & Gambit Co. advertising-mix problem, where the bottom one passes through the opti- mal solution.
PM
TV
Feasible region
10
4
5
optimal solution
(4,3)
10
(4,3)
150
Amount of TV advertising
Cost = 15 = TV + 2 PM
Cost = 10 = TV + 2 PM
2.8 Linear Programming from a Broader Perspective 51
This model is the algebraic version of the linear programming model in the spreadsheet. Note how the parameters (constants) of this algebraic model come directly from the numbers in Table 2.2 . In fact, the entire model could have been formulated directly from this table.
The differences between this algebraic model and the one obtained for the Wyndor prob- lem in Section 2.3 lead to some interesting changes in how the graphical method is applied to solve the model. To further expand your geometric intuition about linear programming, we briefly describe this application of the graphical method next.
Since this linear programming model has only two decision variables, it can be solved by the graphical method described in Section 2.4. The method needs to be adapted in two ways to fit this particular problem. First, because all the functional constraints now have a $ sign with a positive right-hand side, after obtaining the constraint boundary lines in the usual way, the arrows indicating which side of each line satisfies that constraint now all point away from the origin. Second, the method is adapted to minimization by moving the objective function lines in the direction that reduces Cost and then stopping at the last instant that an objective function line still passes through a point in the feasible region, where such a point then is an optimal solution. The supplement to this chapter includes a description of how the graphical method is applied to the Profit & Gambit problem in this way.
Figure 2.22 shows the resulting final graph that identifies the optimal solution as
TV 5 4 (use 4 units of TV advertising)
PM 5 3 (use 3 units of print media advertising)
2.8 LINEAR PROGRAMMING FROM A BROADER PERSPECTIVE
Linear programming is an invaluable aid to managerial decision making in all kinds of com- panies throughout the world. The emergence of powerful spreadsheet packages has helped to further spread the use of this technique. The ease of formulating and solving small linear programming models on a spreadsheet now enables some managers with a very modest back- ground in management science to do this themselves on their own desktop.
1. What kind of product is produced by the Profit & Gambit Co.? 2. Which advertising media are being considered for the three products under consideration? 3. What is management’s objective for the problem being addressed? 4. What was the rationale for the placement of the objective cell and the changing cells in the
spreadsheet model? 5. The algebraic form of the linear programming model for this problem differs from that for the
Wyndor Glass Co. problem in which two major ways?
Review Questions
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52 Chapter Two Linear Programming: Basic Concepts
Many linear programming studies are major projects involving decisions on the levels of many hundreds or thousands of activities. For such studies, sophisticated software packages that go beyond spreadsheets generally are used for both the formulation and solution pro- cesses. These studies normally are conducted by technically trained teams of management scientists, sometimes called operations research analysts, at the instigation of management. Management needs to keep in touch with the management science team to ensure that the study reflects management’s objectives and needs. However, management generally does not get involved with the technical details of the study.
Consequently, there is little reason for a manager to know the details of how linear pro- gramming models are solved beyond the rudiments of using Solver. (Even most management science teams will use commercial software packages for solving their models on a computer rather than developing their own software.) Similarly, a manager does not need to know the technical details of how to formulate complex models, how to validate such a model, how to interact with the computer when formulating and solving a large model, how to efficiently perform what-if analysis with such a model, and so forth. Therefore, these technical details are de-emphasized in this book. A student who becomes interested in conducting technical analyses as part of a management science team should plan to take additional, more techni- cally oriented courses in management science.
So what does an enlightened manager need to know about linear programming? A man- ager needs to have a good intuitive feeling for what linear programming is. One objective of this chapter is to begin to develop that intuition. That’s the purpose of studying the graphical method for solving two-variable problems. It is rare to have a real linear programming prob- lem with as few as two decision variables. Therefore, the graphical method has essentially no practical value for solving real problems. However, it has great value for conveying the basic notion that linear programming involves pushing up against constraint boundaries and moving objective function values in a favorable direction as far as possible. Chapter 14 on the CD-ROM also demonstrates that this approach provides considerable geometric insight into how to analyze larger models by other methods.
A manager must also have an appreciation for the relevance and power of linear program- ming to encourage its use where appropriate. For future managers using this book, this appre- ciation is being promoted by using application vignettes and their linked articles to describe real applications of linear programming and the resulting impact, as well as by including (in miniature form) various realistic examples and case studies that illustrate what can be done.
Certainly a manager must be able to recognize situations where linear programming is applicable. We focus on developing this skill in Chapter 3, where you will learn how to rec- ognize the identifying features for each of the major types of linear programming problems (and their mixtures).
In addition, a manager should recognize situations where linear programming should not be applied. Chapter 8 will help to develop this skill by examining certain underlying assump- tions of linear programming and the circumstances that violate these assumptions. That chap- ter also describes other approaches that can be applied where linear programming should not.
A manager needs to be able to distinguish between competent and shoddy studies using lin- ear programming (or any other management science technique). Therefore, another goal of the upcoming chapters is to demystify the overall process involved in conducting a management science study, all the way from first studying a problem to final implementation of the manage- rial decisions based on the study. This is one purpose of the case studies throughout the book.
Finally, a manager must understand how to interpret the results of a linear programming study. He or she especially needs to understand what kinds of information can be obtained through what-if analysis, as well as the implications of such information for managerial deci- sion making. Chapter 5 focuses on these issues.
1. Does management generally get heavily involved with the technical details of a linear program- ming study?
2. What is the purpose of studying the graphical method for solving problems with two decision variables when essentially all real linear programming problems have more than two?
3. List the things that an enlightened manager should know about linear programming.
Review Questions
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Chapter 2 Glossary 53
Linear programming is a powerful technique for aiding managerial decision making for certain kinds of problems. The basic approach is to formulate a mathematical model called a linear programming model to represent the problem and then to analyze this model. Any linear programming model includes decision variables to represent the decisions to be made, constraints to represent the restrictions on the feasible values of these decision variables, and an objective function that expresses the overall measure of performance for the problem.
Spreadsheets provide a flexible and intuitive way of formulating and solving a linear programming model. The data are entered into data cells. Changing cells display the values of the decision variables, and an objective cell shows the value of the objective function. Output cells are used to help specify the constraints. After formulating the model on the spreadsheet, Solver is used to quickly find an optimal solution. Risk Solver Platform for Education also provides a more powerful method for finding an opti- mal solution.
The graphical method can be used to solve a linear programming model having just two decision variables. This method provides considerable insight into the nature of linear programming models and optimal solutions.
2.9 Summary
Glossary absolute reference A reference to a cell (or a column or a row) with a fixed address, as indi- cated either by using a range name or by placing a $ sign in front of the letter and number of the cell reference. (Section 2.2), 28 changing cells The cells in the spreadsheet that show the values of the decision variables. (Sec- tion 2.2), 26 constraint A restriction on the feasible val- ues of the decision variables. (Sections 2.2 and 2.3), 30 constraint boundary equation The equation for the constraint boundary line. (Section 2.4), 35 constraint boundary line For linear program- ming problems with two decision variables, the line forming the boundary of the solutions that are permitted by the constraint. (Section 2.4), 35 data cells The cells in the spreadsheet that show the data of the problem. (Section 2.2), 26 decision variable An algebraic variable that represents a decision regarding the level of a par- ticular activity. The value of the decision variable appears in a changing cell on the spreadsheet. (Section 2.3), 32 feasible region The geometric region that con- sists of all the feasible solutions. (Section 2.4), 33 feasible solution A solution that simultane- ously satisfies all the constraints in the linear programming model. (Section 2.3), 32 functional constraint A constraint with a func- tion of the decision variables on the left-hand side. All constraints in a linear programming model that are not nonnegativity constraints are called functional constraints. (Section 2.3), 32 graphical method A method for solving lin- ear programming problems with two decision variables on a two-dimensional graph. (Section 2.4), 33 infeasible solution A solution that violates at least one of the constraints in the linear program- ming model. (Section 2.3), 32
linear programming model The mathematical model that represents a linear programming prob- lem. (Sections 2.2 and 2.3), 26
nonnegativity constraint A constraint that expresses the restriction that a particular decision variable must be nonnegative (greater than or equal to zero). (Section 2.3), 32
objective cell The cell in the spreadsheet that shows the overall measure of performance of the decisions. (Section 2.2), 28
objective function The part of a linear pro- gramming model that expresses what needs to be either maximized or minimized, depending on the objective for the problem. The value of the objec- tive function appears in the objective cell on the spreadsheet. (Section 2.3), 32
objective function line For a linear program- ming problem with two decision variables, a line whose points all have the same value of the objective function. (Section 2.4), 36
optimal solution The best feasible solution according to the objective function. (Section 2.3), 32 output cells The cells in the spreadsheet that provide output that depends on the changing cells. These cells frequently are used to help specify constraints. (Section 2.2), 27 parameter The parameters of a linear program- ming model are the constants (coefficients or right-hand sides) in the functional constraints and the objective function. Each parameter rep- resents a quantity (e.g., the amount available of a resource) that is of importance for the analysis of the problem. (Section 2.3), 32 product-mix problem A type of linear pro- gramming problem where the objective is to find the most profitable mix of production levels for the products under consideration. (Section 2.1), 25 range name A descriptive name given to a cell or range of cells that immediately identifies what is there. (Section 2.2), 26
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54 Chapter Two Linear Programming: Basic Concepts
relative reference A reference to a cell whose address is based upon its position rela- tive to the cell containing the formula. (Section 2.2), 28 solution Any single assignment of values to the decision variables, regardless of whether the
assignment is a good one or even a feasible one. (Section 2.3), 32 Solver The spreadsheet tool that is used to specify the model in the spreadsheet and then to obtain an optimal solution for that model. (Section 2.5), 38
Chapter 2 Excel Files:
Wyndor Example
Profit & Gambit Example
Interactive Management Science Modules:
Module for Graphical Linear Programming and Sensitivity Analysis
Excel Add-in:
Risk Solver Pl atform for Education (RSPE)
Supplement to Chapter 2 on the CD-ROM:
More About the Graphical Method for Linear Programming
Learning Aids for This Chapter in Your MS Courseware
Solved Problems (See the CD-ROM or Website for the Solution) 2.S1. Back Savers Production Problem Back Savers is a company that produces backpacks primarily for students. They are considering offering some combination of two different models—the Collegiate and the Mini. Both are made out of the same rip-resistant nylon fabric. Back Savers has a long-term contract with a supplier of the nylon and receives a 5,000-square- foot shipment of the material each week. Each Collegiate requires 3 square feet while each Mini requires 2 square feet. The sales forecasts indicate that at most 1,000 Collegiates and 1,200 Minis can be sold per week. Each Collegiate requires 45 minutes of labor to produce and generates a unit profit of $32. Each Mini requires 40 minutes of labor and generates a unit profit of $24. Back Sav- ers has 35 laborers that each provides 40 hours of labor per week. Management wishes to know what quantity of each type of back- pack to produce per week.
a. Formulate and solve a linear programming model for this problem on a spreadsheet.
b. Formulate this same model algebraically. c. Use the graphical method by hand to solve this model.
2.S2. Conducting a Marketing Survey The marketing group for a cell phone manufacturer plans to conduct a telephone survey to determine consumer attitudes toward a new cell phone that is currently under development.
In order to have a sufficient sample size to conduct the anal- ysis, they need to contact at least 100 young males (under age 40), 150 older males (over age 40), 120 young females (under age 40), and 200 older females (over age 40). It costs $1 to make a daytime phone call and $1.50 to make an evening phone call (because of higher labor costs). This cost is incurred whether or not anyone answers the phone. The table below shows the likelihood of a given customer type answering each phone call. Assume the survey is conducted with whoever first answers the phone. Also, because of limited evening staffing, at most one-third of phone calls placed can be evening phone calls. How should the marketing group conduct the telephone survey so as to meet the sample size requirements at the lowest possible cost?
a. Formulate and solve a linear programming model for this problem on a spreadsheet.
b. Formulate this same model algebraically.
Who Answers? Daytime Calls Evening Calls
Young male 10% 20% Older male 15% 30% Young female 20% 20% Older female 35% 25% No answer 20% 5%
2.1. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 2.1. Briefly describe how linear programming was applied in this study. Then list the vari- ous financial and nonfinancial benefits that resulted from this study.
We have inserted the symbol E* (for Excel) to the left of each problem or part where Excel should be used. The symbol R* is used instead if Risk Solver Platform for Education should be used instead of the Excel Solver to solve the problem. An aster- isk on the problem number indicates that at least a partial answer is given in the back of the book.
Problems
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Chapter 2 Problems 55
2.2. Reconsider the Wyndor Glass Co. case study intro- duced in Section 2.1. Suppose that the estimates of the unit prof- its for the two new products now have been revised to $600 for the doors and $300 for the windows. E* a. Formulate and solve the revised linear program-
ming model for this problem on a spreadsheet. b. Formulate this same model algebraically. c. Use the graphical method to solve this revised
model. 2.3. Reconsider the Wyndor Glass Co. case study intro- duced in Section 2.1. Suppose that Bill Tasto (Wyndor’s vice president for manufacturing) now has found a way to provide a little additional production time in Plant 2 to the new products. a. Use the graphical method to find the new optimal
solution and the resulting total profit if one addi- tional hour per week is provided.
b. Repeat part a if two additional hours per week are provided instead.
c. Repeat part a if three additional hours per week are provided instead.
d. Use these results to determine how much each additional hour per week would be worth in terms of increasing the total profit from the two new products.
E*2.4. Use Solver to do Problem 2.3. 2.5. The following table summarizes the key facts about two products, A and B, and the resources, Q, R, and S, required to produce them.
Resource Usage per Unit Produced
Resource Product
A Product
B
Amount of Resource Available
Q 2 1 2 R 1 2 2 S 3 3 4
Profit/unit $3,000 $2,000
All the assumptions of linear programming hold. E* a. Formulate and solve a linear programming model
for this problem on a spreadsheet. b. Formulate this same model algebraically.
2.6.* This is your lucky day. You have just won a $20,000 prize. You are setting aside $8,000 for taxes and partying expenses, but you have decided to invest the other $12,000. Upon hearing this news, two different friends have offered you an opportunity to become a partner in two different entrepreneurial ventures, one planned by each friend. In both cases, this investment would involve expending some of your time next summer as well as putting up cash. Becoming a full partner in the first friend’s venture would require an invest- ment of $10,000 and 400 hours, and your estimated profit (ignoring the value of your time) would be $9,000. The cor- responding figures for the second friend’s venture are $8,000 and 500 hours, with an estimated profit to you of $9,000. However, both friends are flexible and would allow you to
come in at any fraction of a full partnership you would like. If you choose a fraction of a full partnership, all the above figures given for a full partnership (money investment, time investment, and your profit) would be multiplied by this same fraction.
Because you were looking for an interesting summer job anyway (maximum of 600 hours), you have decided to partici- pate in one or both friends’ ventures in whichever combination would maximize your total estimated profit. You now need to solve the problem of finding the best combination. a. Describe the analogy between this problem and the
Wyndor Glass Co. problem discussed in Section 2.1. Then construct and fill in a table like Table 2.1 for this problem, identifying both the activities and the resources.
b. Identify verbally the decisions to be made, the con- straints on these decisions, and the overall measure of performance for the decisions.
c. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions.
E* d. Formulate a spreadsheet model for this problem. Identify the data cells, the changing cells, and the objective cell. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function. Then use Solver to solve this model.
e. Indicate why this spreadsheet model is a linear pro- gramming model.
f. Formulate this same model algebraically. g. Identify the decision variables, objective function,
nonnegativity constraints, functional constraints, and parameters in both the algebraic version and spreadsheet version of the model.
h. Use the graphical method by hand to solve this model. What is your total estimated profit?
i. Use the Graphical Linear Programming and Sensitivity Analysis module in your Interactive Management Science Modules to apply the graphi- cal method to this model.
2.7. You are given the following linear programming model in algebraic form, where x 1 and x 2 are the decision variables and Z is the value of the overall measure of performance.
Maximize Z 5 x1 1 2x2
subject to Constraint on resource 1: x 1 1 x 2 # 5 (amount available) Constraint on resource 2: x 1 1 3 x 2 # 9 (amount available) and
x1 $ 0 x2 $ 0
a. Identify the objective function, the functional con- straints, and the nonnegativity constraints in this model.
E* b. Incorporate this model into a spreadsheet. c. Is ( x 1 , x 2 ) 5 (3, 1) a feasible solution? d. Is ( x 1 , x 2 ) 5 (1, 3) a feasible solution? E* e. Use Solver to solve this model.
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56 Chapter Two Linear Programming: Basic Concepts
2.8. You are given the following linear programming model in algebraic form, where x 1 and x 2 are the decision variables and Z is the value of the overall measure of performance.
Maximize Z 5 3x1 1 2x2
subject to
Constraint on resource 1: 3 x 1 1 x 2 # 9 (amount available) Constraint on resource 2: x 1 1 2 x 2 # 8 (amount available)
and
x1 $ 0 x2 $ 0
a. Identify the objective function, the functional con- straints, and the nonnegativity constraints in this model.
E* b. Incorporate this model into a spreadsheet. c. Is ( x 1 , x 2 ) 5 (2, 1) a feasible solution? d. Is ( x 1 , x 2 ) 5 (2, 3) a feasible solution? e. Is ( x 1 , x 2 ) 5 (0, 5) a feasible solution? E* f. Use Solver to solve this model.
2.9. The Whitt Window Company is a company with only three employees that makes two different kinds of handcrafted windows: a wood-framed and an aluminum framed window. They earn $60 profit for each wood-framed window and $30 profit for each aluminum-framed window. Doug makes the wood frames and can make 6 per day. Linda makes the alumi- num frames and can make 4 per day. Bob forms and cuts the glass and can make 48 square feet of glass per day. Each wood- framed window uses 6 square feet of glass and each aluminum- framed window uses 8 square feet of glass.
The company wishes to determine how many windows of each type to produce per day to maximize total profit.
a. Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Section 2.1. Then construct and fill in a table like Table 2.1 for this problem, identifying both the activities and the resources.
b. Identify verbally the decisions to be made, the con- straints on these decisions, and the overall measure of performance for the decisions.
c. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions.
E* d. Formulate a spreadsheet model for this problem. Identify the data cells, the changing cells, and the objective cell. Also show the Excel equation for each output cell expressed as a SUMPROD- UCT function. Then use Solver to solve this model.
e. Indicate why this spreadsheet model is a linear pro- gramming model.
f. Formulate this same model algebraically.
g. Identify the decision variables, objective function, nonnegativity constraints, functional constraints, and parameters in both the algebraic version and spreadsheet version of the model.
h. Use the graphical method to solve this model.
i. A new competitor in town has started making wood-framed windows as well. This may force the company to lower the price it charges and so lower the profit made for each wood-framed window. How would the optimal solution change (if at all) if the profit per wood-framed window decreases from $60 to $40? From $60 to $20?
j. Doug is considering lowering his working hours, which would decrease the number of wood frames he makes per day. How would the optimal solution change if he only makes 5 wood frames per day?
2.10. The Apex Television Company has to decide on the number of 27" and 20" sets to be produced at one of its factories. Market research indicates that at most 40 of the 27" sets and 10 of the 20" sets can be sold per month. The maximum number of work-hours available is 500 per month. A 27" set requires 20 work-hours and a 20" set requires 10 work-hours. Each 27" set sold produces a profit of $120 and each 20" set produces a profit of $80. A wholesaler has agreed to purchase all the television sets produced if the numbers do not exceed the maxima indi- cated by the market research. E* a. Formulate and solve a linear programming model
for this problem on a spreadsheet. b. Formulate this same model algebraically. c. Solve this model by using the Graphical Linear Pro-
gramming and Sensitivity Analysis module in your Interactive Management Science Modules to apply the graphical method.
2.11. The WorldLight Company produces two light fixtures (Products 1 and 2) that require both metal frame parts and elec- trical components. Management wants to determine how many units of each product to produce so as to maximize profit. For each unit of Product 1, one unit of frame parts and two units of electrical components are required. For each unit of Product 2, three units of frame parts and two units of electrical components are required. The company has 200 units of frame parts and 300 units of electrical components. Each unit of Product 1 gives a profit of $1, and each unit of Product 2, up to 60 units, gives a profit of $2. Any excess over 60 units of Product 2 brings no profit, so such an excess has been ruled out. a. Identify verbally the decisions to be made, the con-
straints on these decisions, and the overall measure of performance for the decisions.
b. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions.
E* c. Formulate and solve a linear programming model for this problem on a spreadsheet.
d. Formulate this same model algebraically. e. Solve this model by using the Graphical Linear
Programming and Sensitivity Analysis module in your Interactive Management Science Modules to apply the graphical method. What is the resulting total profit?
2.12. The Primo Insurance Company is introducing two new product lines: special risk insurance and mortgages. The expected profit is $5 per unit on special risk insurance and $2 per unit on mortgages.
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Chapter 2 Problems 57
Management wishes to establish sales quotas for the new product lines to maximize total expected profit. The work requirements are shown below: a. Identify verbally the decisions to be made, the con-
straints on these decisions, and the overall measure of performance for the decisions.
b. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions.
Work-Hours per Unit
Department Special
Risk Mortgage Work-Hours
Available
Underwriting 3 2 2,400 Administration 0 1 800 Claims 2 0 1,200
E* c. Formulate and solve a linear programming model for this problem on a spreadsheet.
d. Formulate this same model algebraically.
2.13.* You are given the following linear programming model in algebraic form, with x 1 and x 2 as the decision variables and constraints on the usage of four resources:
Maximize Profit 5 2x1 1 x2
subject to
x2 # 10 (resource 1) 2x1 1 5x2 # 60 (resource 2) x1 1 x2 # 18 (resource 3)
3x1 1 x2 # 44 (resource 4)
and
x1 $ 0 x2 $ 0
a. Use the graphical method to solve this model. E* b. Incorporate this model into a spreadsheet and then
use Solver to solve this model. R*2.14. Use Risk Solver Platform for Education to formulate and solve the model shown in Problem 2.13 in a spreadsheet. 2.15. Because of your knowledge of management science, your boss has asked you to analyze a product mix problem involving two products and two resources. The model is shown below in algebraic form, where x 1 and x 2 are the production rates for the two products and P is the total profit.
Maximize P 5 3x1 1 2x2
subject to
x1 1 x2 # 8 (resource 1)
2x1 1 x2 # 10 (resource 2)
and
x1 $ 0 x2 $ 0
a. Use the graphical method to solve this model. E* b. Incorporate this model into a spreadsheet and then
use Solver to solve this model.
R*2.16. Use Risk Solver Platform for Education to formulate and solve the model shown in Problem 2.15 in a spreadsheet. 2.17. Independently consider each of the following changes in the Wyndor problem. In each case, apply the graphical method by hand to this new version of the problem, describe your con- clusion, and then explain how and why the nature of this conclu- sion is different from the original Wyndor problem. a. The unit profit for the windows now is $200. b. To justify introducing these two new products,
Wyndor management now requires that the total number of doors and windows produced per week must be at least 10.
c. The functional constraints for Plants 2 and 3 now have been inadvertently deleted from the model.
2.18. Do Problem 2.17 by using the Graphical Linear Pro- gramming and Sensitivity Analysis module in your Interactive Management Science Modules. 2.19. Weenies and Buns is a food processing plant that manufactures hot dogs and hot dog buns. They grind their own flour for the hot dog buns at a maximum rate of 200 pounds per week. Each hot dog bun requires 0.1 pound of flour. They currently have a contract with Pigland, Inc., which specifies that a delivery of 800 pounds of pork product is delivered every Monday. Each hot dog requires 1/4 pound of pork product. All the other ingredients in the hot dogs and hot dog buns are in plentiful supply. Finally, the labor force at Weenies and Buns consists of five employees working full time (40 hours per week each). Each hot dog requires three minutes of labor, and each hot dog bun requires two minutes of labor. Each hot dog yields a profit of $0.20, and each bun yields a profit of $0.10.
Weenies and Buns would like to know how many hot dogs and how many hot dog buns they should produce each week so as to achieve the highest possible profit. a. Identify verbally the decisions to be made, the con-
straints on these decisions, and the overall measure of performance for the decisions.
b. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions.
E* c. Formulate and solve a linear programming model for this problem on a spreadsheet.
d. Formulate this same model algebraically. e. Use the graphical method to solve this model.
Decide yourself whether you would prefer to do this by hand or by using the Graphical Linear Program- ming and Sensitivity Analysis module in your Inter- active Management Science Modules.
2.20. The Oak Works is a family-owned business that makes handcrafted dining room tables and chairs. They obtain the oak from a local tree farm, which ships them 2,500 pounds of oak each month. Each table uses 50 pounds of oak while each chair uses 25 pounds of oak. The family builds all the furniture itself and has 480 hours of labor available each month. Each table or chair requires six hours of labor. Each table nets Oak Works $400 in profit, while each chair nets $100 in profit. Since chairs are often sold with the tables, they want to produce at least twice as many chairs as tables.
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58 Chapter Two Linear Programming: Basic Concepts
Item Calories (per oz.)
Calories from Fat (per oz.)
Vitamin A (IU per oz.)
Vitamin C (mg per oz.)
Protein (g, per oz.)
Cost (per oz.)
Beef tips 54 19 0 0 8 40¢ Gravy 20 15 0 1 0 35¢ Peas 15 0 15 3 1 15¢ Carrots 8 0 350 1 1 18¢ Dinner roll 40 10 0 0 1 10¢
The Oak Works would like to decide how many tables and chairs to produce so as to maximize profit. a. Formulate and solve a linear programming model
for this problem on a spreadsheet. by using the Excel Solver.
R* b. Use Risk Solver Platform for Education to formu- late and solve this model in a spreadsheet.
c. Formulate this same model algebraically.
2.21. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 2.6. Briefly describe how linear program- ming was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 2.22. Nutri-Jenny is a weight-management center. It pro- duces a wide variety of frozen entrées for consumption by its clients. The entrées are strictly monitored for nutritional content to ensure that the clients are eating a balanced diet. One new entrée will be a beef sirloin tips dinner. It will consist of beef tips and gravy, plus some combination of peas, carrots, and a dinner roll. Nutri-Jenny would like to determine what quantity of each item to include in the entrée to meet the nutritional requirements, while costing as little as possible. The nutritional information for each item and its cost are given in the following table.
The nutritional requirements for the entrée are as follows: (1) it must have between 280 and 320 calories, (2) calories from fat should be no more than 30 percent of the total number of calories, and (3) it must have at least 600 IUs of vitamin A, 10 milligrams of vitamin C, and 30 grams of protein. Further- more, for practical reasons, it must include at least 2 ounces of beef, and it must have at least half an ounce of gravy per ounce of beef.
E* a. Formulate and solve a linear programming model for this problem on a spreadsheet. by using the Excel Solver.
R* b. Use Risk Solver Platform for Education to formu- late and solve this model in a spreadsheet.
c. Formulate this same model algebraically.
2.23. Ralph Edmund loves steaks and potatoes. There- fore, he has decided to go on a steady diet of only these two foods (plus some liquids and vitamin supplements) for all his meals. Ralph realizes that this isn’t the healthiest diet, so he wants to make sure that he eats the right quantities of the two foods to satisfy some key nutritional require- ments. He has obtained the following nutritional and cost information.
Grams of Ingredient per Serving
Ingredient Steak Potatoes
Daily Requirement
(grams)
Carbohydrates 5 15 $ 50 Protein 20 5 $ 40 Fat 15 2 # 60
Cost per serving $4 $2
Ralph wishes to determine the number of daily servings (may be fractional) of steak and potatoes that will meet these require- ments at a minimum cost.
a. Identify verbally the decisions to be made, the con- straints on these decisions, and the overall measure of performance for the decisions.
b. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions.
E* c. Formulate and solve a linear programming model for this problem on a spreadsheet.
d. Formulate this same model algebraically.
e. Use the graphical method by hand to solve this model.
f. Use the Graphical Linear Programming and Sen- sitivity Analysis module in your Interactive Man- agement Science Modules to apply the graphical method to this model.
2.24. Dwight is an elementary school teacher who also raises pigs for supplemental income. He is trying to decide what to feed his pigs. He is considering using a combination of pig feeds available from local suppliers. He would like to feed the pigs at minimum cost while also making sure each pig receives an ade- quate supply of calories and vitamins. The cost, calorie content, and vitamin content of each feed is given in the table below.
Contents Feed Type A Feed Type B
Calories (per pound) 800 1,000 Vitamins (per pound) 140 units 70 units Cost (per pound) $0.40 $0.80
Each pig requires at least 8,000 calories per day and at least 700 units of vitamins. A further constraint is that no more than 1/3 of the diet (by weight) can consist of Feed Type A, since it contains an ingredient that is toxic if consumed in too large a quantity.
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Chapter 2 Problems 59
Increase in Sales per Unit of Advertising
Product Television Print Media
Minimum Required Increase
Stain remover 0% 1.5% 3% Liquid detergent 3 4 18 Powder detergent 21 2 4 Unit cost $1 million $2 million
Food Item Calories from Fat
Total Calories
Vitamin C (mg) Fiber (g) Cost (¢)
Bread (1 slice) 15 80 0 4 6 Peanut butter (1 tbsp) 80 100 0 0 5 Jelly (1 tbsp) 0 70 4 3 8 Apple 0 90 6 10 35 Milk (1 cup) 60 120 2 0 20 Cranberry juice (1 cup) 0 110 80 1 40
a. Identify verbally the decisions to be made, the con- straints on these decisions, and the overall measure of performance for the decisions.
b. Convert these verbal descriptions of the con- straints and the measure of performance into quantitative expressions in terms of the data and decisions.
E* c. Formulate and solve a linear programming model for this problem on a spreadsheet.
d. Formulate this same model algebraically.
2.25. Reconsider the Profit & Gambit Co. problem described in Section 2.6. Suppose that the estimated data given in Table 2.2 now have been changed as shown in the table that accompanies this problem. E* a. Formulate and solve a linear programming model
on a spreadsheet for this revised version of the problem.
b. Formulate this same model algebraically. c. Use the graphical method to solve this model. d. What were the key changes in the data that caused
your answer for the optimal solution to change from the one for the original version of the problem?
e. Write a paragraph to the management of the Profit & Gambit Co. presenting your conclusions from the above parts. Include the potential effect of further refining the key data in the below table. Also point out the leverage that your results might provide to management in negotiating a decrease in the unit cost for either of the advertising media.
2.26. You are given the following linear programming model in algebraic form, with x 1 and x 2 as the decision variables:
Minimize Cost 5 40x1 1 50x2
subject to Constraint 1: 2x1 1 3x2 $ 30
Constraint 2: x1 1 x2 $ 12
Constraint 3: 2x1 1 x2 $ 20
and x1 $ 0 x2 $ 0
a. Use the graphical method to solve this model.
b. How does the optimal solution change if the objec- tive function is changed to Cost 5 40 x 1 1 70 x 2 ?
c. How does the optimal solution change if the third functional constraint is changed to 2 x 1 1 x 2 $ 15?
E* d. Now incorporate the original model into a spread- sheet and use Solver to solve this model.
E* e. Use Excel to do parts b and c.
2.27. The Learning Center runs a day camp for 6-10 year olds during the summer. Its manager, Elizabeth Reed, is trying to reduce the center’s operating costs to avoid having to raise the tuition fee. Elizabeth is currently planning what to feed the chil- dren for lunch. She would like to keep costs to a minimum, but also wants to make sure she is meeting the nutritional requirements of the children. She has already decided to go with peanut butter and jelly sandwiches, and some combination of apples, milk, and/ or cranberry juice. The nutritional content of each food choice and its cost are given in the table that accompanies this problem.
The nutritional requirements are as follows. Each child should receive between 300 and 500 calories, but no more than 30 percent of these calories should come from fat. Each child should receive at least 60 milligrams (mg) of vitamin C and at least 10 grams (g) of fiber.
To ensure tasty sandwiches, Elizabeth wants each child to have a minimum of 2 slices of bread, 1 tablespoon (tbsp) of pea- nut butter, and 1 tbsp of jelly, along with at least 1 cup of liquid (milk and/or cranberry juice).
Elizabeth would like to select the food choices that would minimize cost while meeting all these requirements. E* a. Formulate and solve a linear programming model
for this problem on a spreadsheet. b. Formulate this same model algebraically.
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60 Chapter Two Linear Programming: Basic Concepts
Case 2-1
Auto Assembly
Automobile Alliance, a large automobile manufacturing com- pany, organizes the vehicles it manufactures into three families: a family of trucks, a family of small cars, and a family of midsized and luxury cars. One plant outside Detroit, Michigan, assembles two models from the family of midsized and luxury cars. The first model, the Family Thrillseeker, is a four-door sedan with vinyl seats, plastic interior, standard features, and excellent gas mileage. It is marketed as a smart buy for middle-class families with tight budgets, and each Family Thrillseeker sold generates a modest profit of $3,600 for the company. The second model, the Classy Cruiser, is a two-door luxury sedan with leather seats, wooden interior, custom features, and navigational capabilities. It is marketed as a privilege of affluence for upper-middle-class families, and each Classy Cruiser sold generates a healthy profit of $5,400 for the company.
Rachel Rosencrantz, the manager of the assembly plant, is currently deciding the production schedule for the next month. Specifically, she must decide how many Family Thrillseek- ers and how many Classy Cruisers to assemble in the plant to maximize profit for the company. She knows that the plant pos- sesses a capacity of 48,000 labor-hours during the month. She also knows that it takes six labor-hours to assemble one Fam- ily Thrillseeker and 10.5 labor-hours to assemble one Classy Cruiser.
Because the plant is simply an assembly plant, the parts required to assemble the two models are not produced at the plant. Instead, they are shipped from other plants around the Michigan area to the assembly plant. For example, tires, steer- ing wheels, windows, seats, and doors all arrive from various supplier plants. For the next month, Rachel knows that she will only be able to obtain 20,000 doors from the door supplier. A recent labor strike forced the shutdown of that particular sup- plier plant for several days, and that plant will not be able to meet its production schedule for the next month. Both the Fam- ily Thrillseeker and the Classy Cruiser use the same door part.
In addition, a recent company forecast of the monthly demands for different automobile models suggests that the demand for the Classy Cruiser is limited to 3,500 cars. There is no limit on the demand for the Family Thrillseeker within the capacity limits of the assembly plant.
a. Formulate and solve a linear programming model to deter- mine the number of Family Thrillseekers and the number of Classy Cruisers that should be assembled.
Before she makes her final production decisions, Rachel plans to explore the following questions independently, except where otherwise indicated.
b. The marketing department knows that it can pursue a targeted $500,000 advertising campaign that will raise the demand for the Classy Cruiser next month by 20 percent. Should the campaign be undertaken?
c. Rachel knows that she can increase next month’s plant capac- ity by using overtime labor. She can increase the plant’s labor-hour capacity by 25 percent. With the new assembly plant capacity, how many Family Thrillseekers and how many Classy Cruisers should be assembled?
d. Rachel knows that overtime labor does not come without an extra cost. What is the maximum amount she should be willing to pay for all overtime labor beyond the cost of this labor at regular-time rates? Express your answer as a lump sum.
e. Rachel explores the option of using both the targeted advertising campaign and the overtime labor-hours. The advertising campaign raises the demand for the Classy Cruiser by 20 percent, and the overtime labor increases the plant’s labor-hour capacity by 25 percent. How many Family Thrillseekers and how many Classy Cruisers should be assembled using the advertising campaign and overtime labor-hours if the profit from each Classy Cruiser sold continues to be 50 percent more than for each Family Thrillseeker sold?
f. Knowing that the advertising campaign costs $500,000 and the maximum usage of overtime labor-hours costs $1,600,000 beyond regular time rates, is the solution found in part e a wise decision compared to the solution found in part a?
g. Automobile Alliance has determined that dealerships are actually heavily discounting the price of the Family Thrillseekers to move them off the lot. Because of a profit- sharing agreement with its dealers, the company is not mak- ing a profit of $3,600 on the Family Thrillseeker but instead is making a profit of $2,800. Determine the number of Fam- ily Thrillseekers and the number of Classy Cruisers that should be assembled given this new discounted profit.
h. The company has discovered quality problems with the Family Thrillseeker by randomly testing Thrillseekers at the end of the assembly line. Inspectors have discovered that in over 60 percent of the cases, two of the four doors on a Thrillseeker do not seal properly. Because the percentage of defective Thrillseekers determined by the random testing is so high, the floor foreman has decided to perform qual- ity control tests on every Thrillseeker at the end of the line. Because of the added tests, the time it takes to assemble one Family Thrillseeker has increased from 6 hours to 7.5 hours. Determine the number of units of each model that should be assembled given the new assembly time for the Family Thrillseeker.
i. The board of directors of Automobile Alliance wishes to capture a larger share of the luxury sedan market and there- fore would like to meet the full demand for Classy Cruisers. They ask Rachel to determine by how much the profit of her assembly plant would decrease as compared to the profit found in part a. They then ask her to meet the full demand for Classy Cruisers if the decrease in profit is not more than $2,000,000.
j. Rachel now makes her final decision by combining all the new considerations described in parts f, g, and h. What are her final decisions on whether to undertake the advertising campaign, whether to use overtime labor, the number of Family Thrillseekers to assemble, and the number of Classy Cruisers to assemble?
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Case 2-2 Cutting Cafeteria Costs 61
Case 2-2
Cutting Cafeteria Costs
A cafeteria at All-State University has one special dish it serves like clockwork every Thursday at noon. This supposedly tasty dish is a casserole that contains sautéed onions, boiled sliced potatoes, green beans, and cream of mushroom soup. Unfortu- nately, students fail to see the special quality of this dish, and they loathingly refer to it as the Killer Casserole. The students reluctantly eat the casserole, however, because the cafeteria pro- vides only a limited selection of dishes for Thursday’s lunch (namely, the casserole).
Maria Gonzalez, the cafeteria manager, is looking to cut costs for the coming year, and she believes that one sure way to cut costs is to buy less expensive and perhaps lower quality ingredients. Because the casserole is a weekly staple of the caf- eteria menu, she concludes that if she can cut costs on the ingre- dients purchased for the casserole, she can significantly reduce overall cafeteria operating costs. She therefore decides to invest time in determining how to minimize the costs of the casserole while maintaining nutritional and taste requirements.
Maria focuses on reducing the costs of the two main ingre- dients in the casserole, the potatoes and green beans. These two ingredients are responsible for the greatest costs, nutritional con- tent, and taste of the dish.
Maria buys the potatoes and green beans from a wholesaler each week. Potatoes cost $0.40 per pound (lb), and green beans cost $1.00 per lb.
All-State University has established nutritional requirements that each main dish of the cafeteria must meet. Specifically, the dish must contain 180 grams (g) of protein, 80 milligrams (mg) of iron, and 1,050 mg of vitamin C. (There are 454 g in one lb and 1,000 mg in one g.) For simplicity when planning, Maria assumes that only the potatoes and green beans contribute to the nutritional content of the casserole.
Because Maria works at a cutting-edge technological uni- versity, she has been exposed to the numerous resources on the World Wide Web. She decides to surf the Web to find the nutritional content of potatoes and green beans. Her research yields the following nutritional information about the two ingredients.
Potatoes Green Beans
Protein 1.5 g per 100 g 5.67 g per 10 ounces Iron 0.3 mg per 100 g 3.402 mg per 10 ounces Vitamin C 12 mg per 100 g 28.35 mg per 10 ounces
(There are 28.35 g in one ounce.) Edson Branner, the cafeteria cook who is surprisingly con-
cerned about taste, informs Maria that an edible casserole must contain at least a six-to-five ratio in the weight of potatoes to green beans.
Given the number of students who eat in the cafeteria, Maria knows that she must purchase enough potatoes and green beans to prepare a minimum of 10 kilograms (kg) of casserole each week. (There are 1,000 g in one kg.) Again, for simplicity in planning, she assumes that only the potatoes and green beans determine the amount of casserole that can be prepared. Maria does not establish an upper limit on the amount of casserole to
prepare since she knows all leftovers can be served for many days thereafter or can be used creatively in preparing other dishes.
a. Determine the amount of potatoes and green beans Maria should purchase each week for the casserole to minimize the ingredient costs while meeting nutritional, taste, and demand requirements. Before she makes her final decision, Maria plans to explore
the following questions independently, except where otherwise indicated.
b. Maria is not very concerned about the taste of the casserole; she is only concerned about meeting nutritional requirements and cutting costs. She therefore forces Edson to change the recipe to allow only for at least a one-to-two ratio in the weight of potatoes to green beans. Given the new recipe, determine the amount of potatoes and green beans Maria should purchase each week.
c. Maria decides to lower the iron requirement to 65 mg since she determines that the other ingredients, such as the onions and cream of mushroom soup, also provide iron. Determine the amount of potatoes and green beans Maria should pur- chase each week given this new iron requirement.
d. Maria learns that the wholesaler has a surplus of green beans and is therefore selling the green beans for a lower price of $0.50 per lb. Using the same iron requirement from part c and the new price of green beans, determine the amount of potatoes and green beans Maria should purchase each week.
e. Maria decides that she wants to purchase lima beans instead of green beans since lima beans are less expensive and provide a greater amount of protein and iron than green beans. Maria again wields her absolute power and forces Edson to change the recipe to include lima beans instead of green beans. Maria knows she can purchase lima beans for $0.60 per lb from the wholesaler. She also knows that lima beans contain 22.68 g of protein and 6.804 mg of iron per 10 ounces of lima beans and no vitamin C. Using the new cost and nutritional content of lima beans, determine the amount of potatoes and lima beans Maria should purchase each week to minimize the ingredient costs while meeting nutritional, taste, and demand require- ments. The nutritional requirements include the reduced iron requirement from part c.
f. Will Edson be happy with the solution in part e? Why or why not?
g. An All-State student task force meets during Body Aware- ness Week and determines that All-State University’s nutri- tional requirements for iron are too lax and that those for vitamin C are too stringent. The task force urges the univer- sity to adopt a policy that requires each serving of an entrée to contain at least 120 mg of iron and at least 500 mg of vita- min C. Using potatoes and lima beans as the ingredients for the dish and using the new nutritional requirements, deter- mine the amount of potatoes and lima beans Maria should purchase each week.
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62 Chapter Two Linear Programming: Basic Concepts
Case 2-3
Staffing a Call Center
California Children’s Hospital has been receiving numerous customer complaints because of its confusing, decentralized appointment and registration process. When customers want to make appointments or register child patients, they must contact the clinic or department they plan to visit. Several problems exist with this current strategy. Parents do not always know the most appropriate clinic or department they must visit to address their children’s ailments. They therefore spend a significant amount of time on the phone being transferred from clinic to clinic until they reach the most appropriate clinic for their needs. The hos- pital also does not publish the phone numbers of all clinics and departments, and parents must therefore invest a large amount of time in detective work to track down the correct phone number. Finally, the various clinics and departments do not communicate with each other. For example, when a doctor schedules a refer- ral with a colleague located in another department or clinic, that department or clinic almost never receives word of the referral. The parent must contact the correct department or clinic and provide the needed referral information.
In efforts to reengineer and improve its appointment and reg- istration process, the children’s hospital has decided to centralize the process by establishing one call center devoted exclusively to appointments and registration. The hospital is currently in the middle of the planning stages for the call center. Lenny Davis, the hospital manager, plans to operate the call center from 7 am to 9 pm during the weekdays.
Several months ago, the hospital hired an ambitious manage- ment consulting firm, Creative Chaos Consultants, to forecast the number of calls the call center would receive each hour of the day. Since all appointment and registration-related calls would be received by the call center, the consultants decided that they could forecast the calls at the call center by totaling the number of appointment and registration-related calls received by all clinics and departments. The team members visited all the clinics and departments, where they diligently recorded every call relating to appointments and registration. They then totaled these calls and altered the totals to account for calls missed dur- ing data collection. They also altered totals to account for repeat calls that occurred when the same parent called the hospital many times because of the confusion surrounding the decentral- ized process. Creative Chaos Consultants determined the aver- age number of calls the call center should expect during each hour of a weekday. The following table provides the forecasts.
Work Shift Average Number of Calls
7 AM to 9 AM 40 calls per hour 9 AM to 11 AM 85 calls per hour 11 AM to 1 PM 70 calls per hour 1 PM to 3 PM 95 calls per hour 3 PM to 5 PM 80 calls per hour 5 PM to 7 PM 35 calls per hour 7 PM to 9 PM 10 calls per hour
After the consultants submitted these forecasts, Lenny became interested in the percentage of calls from Spanish speakers since the hospital services many Spanish patients. Lenny knows that
he has to hire some operators who speak Spanish to handle these calls. The consultants performed further data collection and determined that, on average, 20 percent of the calls were from Spanish speakers.
Given these call forecasts, Lenny must now decide how to staff the call center during each two-hour shift of a week- day. During the forecasting project, Creative Chaos Consul- tants closely observed the operators working at the individual clinics and departments and determined the number of calls operators process per hour. The consultants informed Lenny that an operator is able to process an average of six calls per hour. Lenny also knows that he has both full-time and part- time workers available to staff the call center. A full-time employee works eight hours per day, but because of paper- work that must also be completed, the employee spends only four hours per day on the phone. To balance the schedule, the employee alternates the two-hour shifts between answering phones and completing paperwork. Full-time employees can start their day either by answering phones or by completing paperwork on the first shift. The full-time employees speak either Spanish or English, but none of them are bilingual. Both Spanish-speaking and English-speaking employees are paid $10 per hour for work before 5 pm and $12 per hour for work after 5 pm. The full-time employees can begin work at the beginning of the 7 am to 9 am shift, 9 am to 11 am shift, 11 am to 1 pm shift, or 1 pm to 3 pm shift. The part-time employ- ees work for four hours, only answer calls, and only speak En glish. They can start work at the beginning of the 3 pm to 5 pm shift or the 5 pm to 7 p m shift, and, like the full-time employees, they are paid $10 per hour for work before 5 pm and $12 per hour for work after 5 pm.
For the following analysis, consider only the labor cost for the time employees spend answering phones. The cost for paper- work time is charged to other cost centers.
a. How many Spanish-speaking operators and how many English-speaking operators does the hospital need to staff the call center during each two-hour shift of the day in order to answer all calls? Please provide an integer number since half a human operator makes no sense.
b. Lenny needs to determine how many full-time employees who speak Spanish, full-time employees who speak English, and part-time employees he should hire to begin on each shift. Creative Chaos Consultants advises him that linear pro- gramming can be used to do this in such a way as to mini- mize operating costs while answering all calls. Formulate a linear programming model of this problem.
c. Obtain an optimal solution for the linear programming model formulated in part b to guide Lenny’s decision.
d. Because many full-time workers do not want to work late into the evening, Lenny can find only one qualified English- speaking operator willing to begin work at 1 pm. Given this new constraint, how many full-time English-speaking opera- tors, full-time Spanish-speaking operators, and part-time operators should Lenny hire for each shift to minimize oper- ating costs while answering all calls?
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Additional Cases 63
e. Lenny now has decided to investigate the option of hiring bilingual operators instead of monolingual operators. If all the operators are bilingual, how many operators should be working during each two-hour shift to answer all phone calls? As in part a, please provide an integer answer.
f. If all employees are bilingual, how many full-time and part- time employees should Lenny hire to begin on each shift to minimize operating costs while answering all calls? As in part b, formulate a linear programming model to guide Len- ny’s decision.
g. What is the maximum percentage increase in the hourly wage rate that Lenny can pay bilingual employees over monolin- gual employees without increasing the total operating costs?
h. What other features of the call center should Lenny explore to improve service or minimize operating costs?
Source: This case is based on an actual project completed by a team of master’s students in what is now the Department of Management Science and Engineering at Stanford University.
Additional Cases Additional cases for this chapter are also available at the University of Western Ontario Ivey School of Business web- site, cases.ivey.uwo.ca/cases, in the segment of the Case- Mate area designated for this book.
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64
Chapter Three
Linear Programming: Formulation and Applications Learning Objectives
After completing this chapter, you should be able to
1. Recognize various kinds of managerial problems to which linear programming can be applied.
2. Describe the five major categories of linear programming problems, including their identifying features.
3. Formulate a linear programming model from a description of a problem in any of these categories.
4. Describe the difference between resource constraints and benefit constraints, includ- ing the difference in how they arise.
5. Describe fixed-requirement constraints and where they arise.
6. Identify the kinds of Excel functions that linear programming spreadsheet models use for the output cells, including the objective cell.
7. Identify the four components of any linear programming model and the kind of spreadsheet cells used for each component.
8. Recognize managerial problems that can be formulated and analyzed as linear pro- gramming problems.
9. Understand the flexibility that managers have in prescribing key considerations that can be incorporated into a linear programming model.
Linear programming problems come in many guises. And their models take various forms. This diversity can be confusing to both students and managers, making it difficult to recog- nize when linear programming can be applied to address a managerial problem. Since manag- ers instigate management science studies, the ability to recognize the applicability of linear programming is an important managerial skill. This chapter focuses largely on developing this skill.
The usual textbook approach to trying to teach this skill is to present a series of diverse examples of linear programming applications. The weakness of this approach is that it empha- sizes differences rather than the common threads between these applications. Our approach will be to emphasize these common threads—the identifying features —that tie together linear programming problems even when they arise in very different contexts. We will describe some broad categories of linear programming problems and the identifying features that characterize them. Then we will use diverse examples, but with the purpose of illustrating and emphasizing the common threads among them.
We will focus on five key categories of linear programming problems: resource-allocation problems, cost–benefit–trade-off problems, mixed problems, transportation problems, and
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3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 65
assignment problems. In each case, an important identifying feature is the nature of the restric- tions on what decisions can be made, and thus the nature of the resulting functional constraints in the linear programming model. For each category, you will see how the basic data for a problem lead directly to a linear programming model with a certain distinctive form. Thus, model formulation becomes a by-product of proper problem formulation.
The chapter begins with a case study that initially involves a resource-allocation problem. We then return to the case study in Section 3.4, where additional managerial considerations turn the problem into a mixed problem.
Sections 3.2 to 3.6 focus on the five categories of linear programming problems in turn. Section 3.7 then takes a broader look at the formulation of linear programming models from a managerial perspective. That section (along with Section 3.4) highlights the importance of having the model accurately reflect the managerial view of the problem. These (and other) sections also describe the flexibility available to managers for having the model structured to best fit their view of the important considerations.
3.1 A CASE STUDY: THE SUPER GRAIN CORP. ADVERTISING-MIX PROBLEM
Claire Syverson, vice president for marketing of the Super Grain Corporation, is facing a daunting challenge: how to break into an already overly crowded breakfast cereal market in a big way. Fortunately, the company’s new breakfast cereal— Crunchy Start —has a lot going for it: Great taste. Nutritious. Crunchy from start to finish. She can recite the litany in her sleep now. It has the makings of a winning promotional campaign.
However, Claire knows that she has to avoid the mistakes she made in her last campaign for a breakfast cereal. That had been her first big assignment since she won this promotion, and what a disaster! She thought she had developed a really good campaign. But somehow it had failed to connect with the most crucial segments of the market—young children and par- ents of young children. She also has concluded that it was a mistake not to include cents-off coupons in the magazine and newspaper advertising. Oh well. Live and learn.
But she had better get it right this time, especially after the big stumble last time. The com- pany’s president, David Sloan, already has impressed on her how important the success of Crunchy Start is to the future of the company. She remembers exactly how David concluded the conversation. “The company’s shareholders are not happy. We need to get those earnings headed in the right direction again.” Claire had heard this tune before, but she saw in David’s eyes how deadly serious he was this time.
Claire often uses spreadsheets to help organize her planning. Her management science course in business school impressed upon her how valuable spreadsheet modeling can be. She regrets that she did not rely more heavily on spreadsheet modeling for the last campaign. That was a mistake that she is determined not to repeat.
Now it is time for Claire to carefully review and formulate the problem in preparation for formulating a spreadsheet model.
The Problem Claire already has employed a leading advertising firm, Giacomi & Jackowitz, to help design a nationwide promotional campaign that will achieve the largest possible exposure for Crunchy Start. Super Grain will pay this firm a fee based on services performed (not to exceed $1 million) and has allocated an additional $4 million for advertising expenses.
Giacomi & Jackowitz has identified the three most effective advertising media for this product:
Medium 1: Television commercials on Saturday morning programs for children. Medium 2: Advertisements in food and family-oriented magazines. Medium 3: Advertisements in Sunday supplements of major newspapers.
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66 Chapter Three Linear Programming: Formulation and Applications
The problem now is to determine which levels should be chosen for these advertising activi- ties to obtain the most effective advertising mix.
To determine the best mix of activity levels for this particular advertising problem, it is necessary (as always) to identify the overall measure of performance for the problem and then the contribution of each activity toward this measure. An ultimate goal for Super Grain is to maximize its profits, but it is difficult to make a direct connection between advertising exposure and profits. Therefore, as a rough surrogate for profit, Claire decides to use expected number of exposures as the overall measure of performance, where each viewing of an adver- tisement by some individual counts as one exposure.
Giacomi & Jackowitz has made preliminary plans for advertisements in the three media. The firm also has estimated the expected number of exposures for each advertisement in each medium, as given in the bottom row of Table 3.1 .
The number of advertisements that can be run in the different media are restricted by both the advertising budget (a limit of $4 million) and the planning budget (a limit of $1 million for the fee to Giacomi & Jackowitz). Another restriction is that there are only five commercial spots available for running different commercials (one commercial per spot) on children’s television programs Saturday morning (medium 1) during the time of the promotional cam- paign. (The other two media have an ample number of spots available.)
Consequently, the three resources for this problem are:
Resource 1: Advertising budget ($4 million). Resource 2: Planning budget ($1 million). Resource 3: TV commercial spots available (5).
Table 3.1 shows how much of the advertising budget and the planning budget would be used by each advertisement in the respective media.
• The first row gives the cost per advertisement in each medium. • The second row shows Giacomi & Jackowitz’s estimates of its total cost (including
overhead and profit) for designing and developing each advertisement for the respective media. 1 (This cost represents the billable fee from Super Grain.)
• The last row then gives the expected number of exposures per advertisement.
Analysis of the Problem Claire decides to formulate and solve a linear programming model for this problem on a spreadsheet. The formulation procedure summarized at the end of Section 2.2 guides this pro- cess. Like any linear programming model, this model will have four components:
1. The data 2. The decisions 3. The constraints 4. The measure of performance
Costs
Cost Category Each TV
Commercial Each Magazine Ad Each Sunday Ad Ad budget $300,000 $150,000 $100,000 Planning budget 90,000 30,000 40,000
Expected number of exposures 1,300,000 600,000 500,000
TABLE 3.1 Cost and Exposure Data for the Super Grain Corp. Advertising-Mix Problem
1 When presenting its estimates in this form, the firm is making two simplifying assumptions. One is that its cost for designing and developing each additional advertisement in a medium is roughly the same as for the first advertisement in that medium. The second is that its cost when working with one medium is unaffected by how much work it is doing (if any) with the other media.
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3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 67
The spreadsheet needs to be formatted to provide the following kinds of cells for these components:
Data → data cells Decisions → changing cells Constraints → output cells Measure of performance → objective cell
Figure 3.1 shows the spreadsheet model formulated by Claire. Let us see how she did this by considering each of the components of the model individually.
Four kinds of cells are needed for these four com- ponents of a spreadsheet model.
=SUMPRODUCT(ExposuresPerAd,NumberOfAds)
1
A B C D FE G H
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Super Grain Corp. Advertising-Mix Problem
TV Spots
300
90
150
30
100
40
4,000
1,000
4,000
1,000
Magazine Ads SS Ads
SS AdsTV Spots
5
0 20 10
Magazine Ads
17,000
Total Exposures
(thousands)
Exposures per Ad
(thousands)
Ad Budget
Planning Budget
Number of Ads
Max TV Spots
Cost per Ad ($thousands)
Budget
Available
Budget
Spent
6
F
Budget
Spent7
11
12
13
8
9
=SUMPRODUCT(C8:E8,NumberOfAds)
H
Total Exposures
(thousands)
=SUMPRODUCT(C9:E9,NumberOfAds)
Range Name
BudgetAvailable
BudgetSpent
CostPerAd
ExposuresPerAd
MaxTVSpots
NumberOfAds
TotalExposures
TVSpots
Cells
H8: H9
F8: F9
C8: E9
C4: E4
C15
C13: E13
H13
C13
1,300 600 500
≤ ≤
≤
Solver Parameters Set Objective Cell: TotalExposures To: Max By Changing Variable Cells: NumberOfAds
Subject to the Constraints: BudgetSpent <= BudgetAvailable
TVSpots <= MaxTVSpots
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 3.1 The spreadsheet model for the Super Grain problem (Section 3.1), including the objective cell TotalExposures (H13) and the other output cells BudgetSpent (F8:F9), as well as the specifications needed to set up Solver. The changing cells NumberOfAds (C13:E13) show the optimal solution obtained by Solver.
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68 Chapter Three Linear Programming: Formulation and Applications
The Data One important kind of data is the information given earlier about the amounts available of the three resources for the problem (the advertising budget, the planning budget, and the commercial spots available). Table 3.1 provides the other key data for the problem. Using units of thousands of dollars, these data have been transferred directly into data cells in the spreadsheet in Figure 3.1 and given these range names: ExposuresPerAd (C4:E4), CostPerAd (C8:E9), BudgetAvailable (H8:H9), and MaxTVSpots (C15).
The Decisions The problem has been defined as determining the most effective advertising mix among the three media selected by Giacomi & Jackowitz. Therefore, there are three decisions:
Decision 1: TV 5 Number of commercials for separate spots on television.
Decision 2: M 5 Number of advertisements in magazines.
Decision 3: SS 5 Number of advertisements in Sunday supplements.
The changing cells to hold these numbers have been placed in row 13 in the columns for these media:
TV S cell C13 M S cell D13 SS S cell E13
These changing cells are collectively referred to by the range name NumberOfAds (C13:E13).
The Constraints These changing cells need to be nonnegative. In addition, constraints are needed for the three resources. The first two resources are the ad budget and planning budget. The amounts avail- able for these two budgets are shown in the range BudgetAvailable (H8:H9). As suggested by the # signs entered into column G, the corresponding constraints are
Total spending on advertising # 4,000 (Ad budget in $1,000s)
Total cost of planning # 1,000 (Planning budget in $1,000s)
Using the data in columns C, D, and E for the resources, these totals are
Total spending on advertising 5 300 TV 1 150 M 1 100 SS
Total cost of planning 5 90 TV 1 30 M 1 40 SS
These sums of products on the right-hand side are entered into the output cells BudgetSpent (F8:F9) by using the SUMPRODUCT functions shown in the lower right-hand side of Figure 3.1 . Although the # signs entered in column G are only cosmetic (trial solutions still can be entered in the changing cells that violate these inequalities), they will serve as a reminder later to use these same # signs when entering the constraints in Solver.
The third resource is TV spots for different commercials. Five such spots are available for purchase. The number of spots used is one of the changing cells (C13). Since this cell will be used in a constraint, we assign the cell its own range name: TVSpots (C13). The maximum number of TV spots available is in the data cell MaxTVSpots (C15). Thus, the required con- straint is TVSpots # MaxTVSpots.
The Measure of Performance Claire Syverson is using expected number of exposures as the overall measure of perfor- mance, so let
Exposure 5 Expected number of exposures (in thousands) from all the advertising
The data cells ExposuresPerAd (C4:E4) provide the expected number of exposures (in thou- sands) per advertisement in the respective media and the changing cells NumberOfAds (C13:E13) give the number of each type of advertisement. Therefore,
Excel Tip: Range names may overlap. For instance, we have used NumberOfAds to refer to the whole range of changing cells, C13:E13, and TVSpots to refer to the single cell, C13.
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3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 69
Exposure 5 1,300TV 1 600M 1 500SS
5 SUMPRODUCT (ExposuresPerAd, NumberOfAds)
is the formula that needs to be entered into the objective cell, TotalExposures (H13).
Summary of the Formulation The above analysis of the four components of the model has formulated the following linear programming model (in algebraic form) on the spreadsheet:
Maximize Exposure 5 1,300TV 1 600M 1 500SS
subject to
Ad spending: 300 TV 1 150 M 1 100 SS # 4,000
Planning costs: 90 TV 1 30 M 1 40 SS # 1,000
Number of television spots: TV # 5
and
TV $ 0 M $ 0 SS $ 0
The difficult work of defining the problem and gathering all the relevant data in Table 3.1 leads directly to this formulation.
Solving the Model To solve the spreadsheet model formulated above, some key information needs to be entered into Solver. The lower left-hand side of Figure 3.1 shows the needed entries: the objective cell (TotalExposures), the changing cells (NumberOfAds), the goal of maximizing the objective cell, and the constraints BudgetSpent # BudgetAvailable and TVSpots # MaxTVSpots. Two options are also specified at the bottom of the Solver Parameters box on the lower left-hand side of Figure 3.1 . The changing cells need non- negativity constraints because negative values of advertising are not possible. Choose the Simplex LP (Excel’s Solver) or Standard LP/Quadratic Engine (RSPE) solving method, because this is a linear programming model. Running Solver then finds an optimal solu- tion for the model and display it in the changing cells.
The optimal solution given in row 13 of the spreadsheet provides the following plan for the promotional campaign:
Do not run any television commercials. Run 20 advertisements in magazines. Run 10 advertisements in Sunday supplements.
Since TotalExposures (H13) gives the expected number of exposures in thousands, this plan would be expected to provide 17,000,000 exposures.
Evaluation of the Adequacy of the Model When she chose to use a linear programming model to represent this advertising-mix prob- lem, Claire recognized that this kind of model does not provide a perfect match to this prob- lem. However, a mathematical model is intended to be only an approximate representation of the real problem. Approximations and simplifying assumptions generally are required to have a workable model. All that is really needed is that there be a reasonably high correlation between the prediction of the model and what would actually happen in the real problem. The team now needs to check whether this criterion is satisfied.
One assumption of linear programming is that fractional solutions are allowed. For the current problem, this means that a fractional number (e.g., 3½) of television commercials (or of ads in magazines or Sunday supplements) should be allowed. This is technically true, since a commercial can be aired for less than a normal run, or an ad can be run in just a fraction of the usual magazines or Sunday supplements. However, one defect of the model is that it
Excel Tip: With Excel’s Solver, the Solver dialog box is used to tell Solver the location on the spread- sheet of several of the elements of the model: the changing cells, the objective cell, and the constraints. With RSPE, the Decisions, Constraints, and Objective menu on the RSPE ribbon are used along with the Model pane.
Linear programming models allow fractional solutions.
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70 Chapter Three Linear Programming: Formulation and Applications
assumes that Giacomi & Jackowitz’s cost for planning and developing a commercial or ad that receives only a fraction of its usual run is only that fraction of its usual cost, even though the actual cost would be the same as for a full run. Fortunately, the optimal solution obtained was an integer solution (0 television commercials, 20 ads in magazines, and 10 ads in Sunday supplements), so the assumption that fractional solutions are allowed was not even needed.
Although it is possible to have a fractional number of a normal run of commercials or ads, a normal run tends to be much more effective than a fractional run. Therefore, it would have been reasonable for Claire to drop the assumption that fractional solutions are allowed. If Claire had done this and the optimal solution for the linear programming model had not turned out to be integer, constraints can be added to require the changing cells to be integer. (The TBA Airlines example in the next section provides an illustration of this type of con- straint.) After adding such constraints, the model is called an integer programming model instead of a linear programming model, but it still can be readily solved by Solver.
Another key assumption of linear programming is that the appropriate equation for each of the output cells, including the objective cell, is one that can be expressed as a SUMPRODUCT of data cells and changing cells (or occasionally just a SUM of changing cells). For the objec- tive cell (cell H13) in Figure 3.1 , this implies that the expected number of exposures to be obtained from each advertising medium is proportional to the number of advertisements in that medium. This proportionality seems true, since each viewing of the advertisements by some individual counts as another exposure. Another implication of using a SUMPRODUCT func- tion is that the expected number of exposures to be obtained from an advertising medium is unaffected by the number of advertisements in the other media. Again, this implication seems valid, since viewings of advertisements in different media count as separate exposures.
Although a SUMPRODUCT function is appropriate for calculating the expected number of exposures, the choice of this number for the overall measure of performance is somewhat questionable. Management’s real objective is to maximize the profit generated as a result of the advertising campaign, but this is difficult to measure so expected number of exposures was selected to be a surrogate for profit. This would be valid if profit were proportional to the expected number of exposures. However, proportionality is only an approximation in this case because too many exposures for the same individual reach a saturation level where the impact (potential profit) from one more exposure is substantially less than for the first exposure. (When proportionality is not a reasonable approximation, Chapter 8 will describe nonlinear models that can be used instead.)
To check how reasonable it is to use expected number of exposures as a surrogate for profit, Claire meets with Sid Jackowitz, one of the senior partners of Giacomi & Jackowitz. Sid indicates that the contemplated promotional campaign (20 advertisements in magazines and 10 in Sunday supplements) is a relatively modest one well below saturation levels. Most readers will only notice these ads once or twice, and a second notice is very helpful for rein- forcing the first one. Furthermore, the readership of magazines and Sunday supplements is sufficiently different that the interaction of the advertising impact in these two media should be small. Consequently, Claire concludes that using expected number of exposures for the objective cell in Figure 3.1 provides a reasonable approximation. (A continuation of this case study in Case 8-1 will delve into the more complicated analysis that is required in order to use profit directly as the measure of performance to be recorded in the objective cell instead of making this approximation.)
Next, Claire quizzes Sid about his firm’s costs for planning and developing advertisements in these media. Is it reasonable to assume that the cost in a given medium is proportional to the number of advertisements in that medium? Is it reasonable to assume that the cost of developing advertisements in one medium would not be substantially reduced if the firm had just finished developing advertisements in another medium that might have similar themes? Sid acknowledges that there is some carryover in ad planning from one medium to another, especially if both are print media (e.g., magazines and Sunday supplements), but that the carryover is quite limited because of the distinct differences in these media. Furthermore, he feels that the proportionality assumption is quite reasonable for any given medium since the amount of work involved in planning and developing each additional advertisement in the medium is nearly the same as for the first one in the medium. The total fee that Super
Linear programming mod- els should use SUM or SUMPRODUCT functions for the output cells, includ- ing the objective cell.
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3.2 Resource-Allocation Problems 71
Grain will pay Giacomi & Jackowitz will eventually be based on a detailed accounting of the amount of work done by the firm. Nevertheless, Sid feels that the cost estimates previously provided by the firm (as entered in cells C9, D9, and E9 in units of thousands of dollars) give a reasonable basis for roughly projecting what the fee will be for any given plan (the entries in the changing cells) for the promotional campaign.
Based on this information, Claire concludes that using a SUMPRODUCT function for cell F9 provides a reasonable approximation. Doing the same for cell F8 is clearly justified. Given her earlier conclusions as well, Claire decides that the linear programming model incorpo- rated into Figure 3.1 (plus any expansions of the model needed later for the detailed planning) is a sufficiently accurate representation of the real advertising-mix problem. It will not be necessary to refine the results from this model by turning next to a more complicated kind of mathematical model (such as those to be described in Chapter 8).
Therefore, Claire sends a memorandum to the company’s president, David Sloan, describ- ing a promotional campaign that corresponds to the optimal solution from the linear program- ming model (no TV commercials, 20 ads in magazines, and 10 ads in Sunday supplements). She also requests a meeting to evaluate this plan and discuss whether some modifications should be made.
We will pick up this story again in Section 3.4.
1. What is the problem being addressed in this case study? 2. What overall measure of performance is being used? 3. What are the assumptions of linear programming that need to be checked to evaluate the
adequacy of using a linear programming model to represent the problem under consideration?
Review Questions
3.2 RESOURCE-ALLOCATION PROBLEMS
In the opening paragraph of Chapter 2, we described managerial problems involving the allo- cation of an organization’s resources to its various productive activities. Those were resource- allocation problems.
Resource-allocation problems are linear programming problems involving the allocation of resources to activities. The identifying feature for any such problem is that each functional con- straint in the linear programming model is a resource constraint , which has the form
Amount of resource used # Amount of resource available
for one of the resources.
The amount of a resource used depends on which activities are undertaken, the levels of those activities, and how heavily those activities need to use the resource. Thus, the resource constraints place limits on the levels of the activities. The objective is to choose the levels of the activities so as to maximize some overall measure of performance (such as total profit) from the activities while satisfying all the resource constraints.
Beginning with the case study and then the Wyndor Glass Co. product-mix problem, we will look at four examples that illustrate the characteristics of resource-allocation problems. These examples also demonstrate how this type of problem can arise in a variety of contexts.
The Super Grain Corp. Advertising-Mix Problem The linear programming model formulated in Section 3.1 for the Super Grain case study is one example of a resource-allocation problem. The three activities under consideration are the advertising in the three types of media chosen by Giacomi & Jackowitz.
Activity 1: TV commercials Activity 2: Magazine ads Activity 3: Sunday ads
The decisions being made are the levels of these activities, that is, the number of TV com- mercials, magazine ads, and Sunday ads to run.
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72 Chapter Three Linear Programming: Formulation and Applications
The resources to be allocated to these activities are
Resource 1: Advertising budget ($4 million). Resource 2: Planning budget ($1 million). Resource 3: TV spots available for different commercials (5).
where the amounts available of these resources are given in parentheses. Thus, this problem has three resource constraints:
1. Advertising budget used # $4 million 2. Planning budget used # $1 million 3. TV spots used # 5
Rows 8–9 and cells C13:C15 in Figure 3.1 show these constraints in a spreadsheet. Cells C8:E9 give the amount of the advertising budget and the planning budget used by each unit of each activ- ity, that is, the amount used by one TV spot, one magazine ad, and one Sunday ad, respectively.
Cells C4, D4, and E4 on this spreadsheet give the contribution per unit of each activity to the overall measure of performance (expected number of exposures).
Characteristics of Resource-Allocation Problems Other resource-allocation problems have the same kinds of characteristics as the Super Grain problem. In each case, there are activities where the decisions to be made are the levels of these activities. The contribution of each activity to the overall measure of performance is proportional to the level of that activity. Commonly, this measure of performance is the total profit from the activities, but occasionally it is something else (as in the Super Grain problem).
Every problem of this type has a resource constraint for each resource. The amounts of the resources used depend on the levels of the activities. For each resource, the amount used by each activity is proportional to the level of that activity.
The manager or management science team studying a resource allocation problem needs to gather (with considerable help) three kinds of data:
1. The amount available of each resource. 2. The amount of each resource needed by each activity. Specifically, for each combination
of resource and activity, the amount of the resource used per unit of the activity must be estimated.
3. The contribution per unit of each activity to the overall measure of performance.
Generally there is considerable work involved in developing these data. A substantial amount of digging and consultation is needed to obtain the best estimates available in a timely fashion. This step is critical. Well-informed estimates are needed to obtain a valid linear pro- gramming model for guiding managerial decisions. The dangers involved in inaccurate esti- mates are one reason why what-if analysis (Chapter 5) is such an important part of most linear programming studies.
The Wyndor Glass Co. Product-Mix Problem The product-mix problem facing the management of the Wyndor Glass Co. in Section 2.1 is to determine the most profitable mix of production rates for the two new products, consider- ing the limited availability of spare production capacity in the company’s three plants. This is a resource-allocation problem.
The activities under consideration are
Activity 1: Produce the special new doors. Activity 2: Produce the special new windows.
The decisions being made are the levels of these activities, that is, the production rates for the doors and windows. Production rate is being measured as the number of units (doors or win- dows) produced per week. Management’s objective is to maximize the total profit generated by the two new products, so the overall measure of performance is total profit. The contribu- tion of each product to profit is proportional to the production rate for that product.
An initial step in formulat- ing any resource-allocation problem is to identify the activities and the resources.
For each proposed activ- ity, a decision needs to be made as to how much of the activity to do. In other words, what should the level of the activity be?
These three kinds of data are needed for any resource-allocation problem.
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3.2 Resource-Allocation Problems 73
The resources to be allocated to these activities are
Resource 1: Production capacity in Plant 1. Resource 2: Production capacity in Plant 2. Resource 3: Production capacity in Plant 3.
Each of the three functional constraints in the linear programming model formulated in Section 2.2 (see rows 7–9 of the spreadsheet in Figure 2.3 or 2.4) is a resource constraint for one of these three resources. Column E shows the amount of production capacity used in each plant and column G gives the amount available.
Table 2.1 in Section 2.1 provides the data for the Wyndor problem. You already have seen how the numbers in Table 2.1 become the parameters in the linear programming model in either its spreadsheet formulation (Section 2.2) or its algebraic form (Section 2.3).
The TBA Airlines Problem TBA Airlines is a small regional company that specializes in short flights in small passenger air- planes. The company has been doing well and management has decided to expand its operations.
The Problem The basic issue facing management now is whether to purchase more small airplanes to add some new short flights or to start moving into the national market by purchasing some large airplanes for new cross-country flights (or both). Many factors will go into management’s final decision, but the most important one is which strategy is likely to be most profitable.
The first row of Table 3.2 shows the estimated net annual profit (inclusive of capital recov- ery costs) from each type of airplane purchased. The second row gives the purchase cost per airplane and also notes that the total amount of capital available for airplane purchases is $250 million. The third row records the fact that management does not want to purchase more than five small airplanes because of limited possibilities for adding lucrative short flights, whereas they have not specified a maximum number for large airplanes (other than that imposed by the limited capital available).
How many airplanes of each type should be purchased to maximize the total net annual profit?
Formulation This is a resource-allocation problem. The activities under consideration are
Activity 1: Purchase small airplanes. Activity 2: Purchase large airplanes.
The decisions to be made are the levels of these activities, that is,
S 5 Number of small airplanes to purchase L 5 Number of large airplanes to purchase
The one resource to be allocated to these activities is
Resource: Investment capital ($250 million).
Thus, there is a single resource constraint:
Investment capital spent # $250 million
In addition, management has specified one side constraint:
Number of small airplanes purchased # 5
Figure 3.2 shows the formulation of a spreadsheet model for this problem, where the data in Table 3.2 have been transferred into the data cells—UnitProfit (C4:D4), CapitalPerUnitPurchased
Small Airplane Large Airplane Capital Available
Net annual profit per airplane $7 million $22 million Purchase cost per airplane $25 million $75 million $250 million Maximum purchase quantity 5 No maximum
TABLE 3.2 Data for the TBA Airlines Problem
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74 Chapter Three Linear Programming: Formulation and Applications
(C8:D8), CapitalAvailable (G8), and MaxSmallAirplanes (C14). The resource constraint then appears in cells C8:G8 while C12:C14 shows the side constraint. The objective for this problem is to maximize the total net annual profit, so the equation for the objective cell is
TotalProfit (G12) 5 SUMPRODUCT (UnitProfit, UnitsPurchased)
Since the TBA Airlines problem is a resource-allocation problem, this spreadsheet model has essentially the same form as the Super Grain and Wyndor problems except for one small difference. The changing cells in this case must have integer values since it is not feasible for the company to purchase and operate a fraction of an airplane. Therefore, constraints that the changing cells need to be integer are added. With Excel’s Solver, use the Add Constraint dialog box to choose the range of these cells (C12:D12) as the left-hand side and then choose int from the pop-up menu between the left-hand and right-hand side. In RSPE, choose the changing cells (C12:D12), and then under the Constraint menu on the RSPE ribbon, choose Integer under the Variable Type/Bound submenu.
These changing cells in Figure 3.2 show the optimal solution, ( S, L ) 5 (1, 3), obtained after running Solver.
1
2
3
4
5
6
7
8
9
1 0
1 1
1 2
1 3
1 4
A B C D E F G
TBA Airlines Airplane Purchasing Problem
Small Airplane Large Airplane
Unit Profit ($millions) 7 22
Capital Capital
Spent Available
Capital ($millions) 25 75 250 <= 250
T otal Profit
Small Airplane Large Airplane ($millions)
Number Purchased 1 3 73
<=
Maximum Small Airplanes 5
Capital per Unit Purchased
6
7
8
E
Capital
Spent
=SUMPRODUCT(CapitalPerUnitPurchased,NumberPurchased)
10
11
12
G
Total Profit
($millions)
=SUMPRODUCT(UnitProfit,NumberPurchased)
Range Name Cells
Capital Available G8
CapitalPerUnitPurchased C8:D8
CapitalSpent E8
MaxSmallAirplanes C14
NumberPurchased C12:D12
SmallAirplanes C12
TotalProfit G12
UnitProfit C4:D4
Solver Parameters Set Objective Cell: TotalProfit To: Max By Changing Variable Cells: NumberPurchased
Subject to the Constraints: CapitalSpent <= CapitalAvailable
NumberPurchased = integer
SmallAirplanes <= MaxSmallAirplanes
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 3.2 A spreadsheet model for the TBA Airlines integer programming problem where the changing cells, UnitsProduced (C12:D12), show the optimal airplane purchases obtained by Solver, and the objective cell, TotalProfit (G12), gives the resulting total profit in mil- lions of dollars.
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3.2 Resource-Allocation Problems 75
One of the assumptions of linear programming is that the changing cells are allowed to have any values, including fractional values, that satisfy the functional and nonnegativity constraints. Therefore, technically speaking, the TBA problem is not a linear programming problem because of adding the constraints
NumberPurchased 5 integer
that are displayed in the Solver Parameters box in Figure 3.2 . Such a problem that fits lin- ear programming except for adding such constraints is called an integer programming problem . The method used by Solver to solve integer programming problems is quite differ- ent from that for solving linear programming problems. In fact, integer programming prob- lems tend to be much more difficult to solve than linear programming problems so there is considerably more limitation on the size of the problem. However, this doesn’t matter to a spreadsheet modeler dealing with small problems. From his or her viewpoint, there is virtu- ally no distinction between linear programming and integer programming problems. They are formulated in exactly the same way. Then, at the very end, a decision needs to be made as to whether any of the changing cells need to be restricted to integer values. If so, those con- straints are added as described above. Keep this option in mind as we continue to discuss the formulation of various types of linear programming problems throughout the chapter.
Summary of the Formulation The above formulation of a model with one resource constraint and one side constraint for the TBA Airlines problem now can be summarized (in algebraic form) as follows:
Maximize Profit 5 7S 1 22L
subject to
25S 1 75L # 250
S # 5 and
S $ 0 L $ 0
Capital Budgeting Financial planning is one of the most important areas of application for resource-allocation problems. The resources being allocated in this area are quite different from those for applica- tions in the production planning area (such as the Wyndor Glass Co. product-mix problem), where the resources tend to be production facilities of various kinds. For financial planning, the resources tend to be financial assets such as cash, securities, accounts receivable, lines of credit, and so forth. Our specific example involves capital budgeting, where the resources are amounts of investment capital available at different points in time.
The Problem The Think-Big Development Co. is a major investor in commercial real-estate development projects. It currently has the opportunity to share in three large construction projects:
Project 1: Construct a high-rise office building. Project 2: Construct a hotel. Project 3: Construct a shopping center.
Each project requires each partner to make investments at four different points in time: a down payment now, and additional capital after one, two, and three years. Table 3.3 shows for each project the total amount of investment capital required from all the partners at these four points in time. Thus, a partner taking a certain percentage share of a project is obligated to invest that percentage of each of the amounts shown in the table for the project.
All three projects are expected to be very profitable in the long run. So the management of Think-Big wants to invest as much as possible in some or all of them. Management is willing to commit all the company’s investment capital currently available, as well as all additional
Excel Tip: To constrain a range of changing cells to be integer in Excel’s Solver, choose the range of cells in the left-hand side of the Add Constraint dialog box and choose int from the pop-up menu. Clicking OK then enters the constraint that these cells 5 integer in the Solver dialog box. In RSPE, select the range of cells to be constrained integer, and then under the Constraint menu on the RSPE ribbon, choose Integer under the Variable Type/Bound submenu.
Excel Tip: Even when a changing cell is constrained to be integer, rounding errors occasionally will cause Excel to return a non- integer value very close to an integer (e.g., 1.23E-10, meaning 0.000000000123). To make the spreadsheet cleaner, you may replace these “ugly” representations by their proper integer val- ues in the changing cells.
Excel Tip: In Excel’s Solver Options, the Integer Optimality (%) setting (1 percent by default) causes Solver to stop solv- ing an integer programming problem when it finds a fea- sible solution whose objec- tive function value is within the specified percentage of being optimal. In RSPE, the equivalent setting is Integer Tolerance under the Engine tab of the Solver model (see Figure 2.19). This is useful to speed up solving large problems. For smaller prob- lems (e.g., homework prob- lems), this option should be set to 0 to guarantee finding an optimal solution.
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76
investment capital expected to become available over the next three years. The objective is to determine the investment mix that will be most profitable, based on current estimates of profitability.
Since it will be several years before each project begins to generate income, which will continue for many years thereafter, we need to take into account the time value of money in evaluating how profitable it might be. This is done by discounting future cash outflows (capital invested) and cash inflows (income), and then adding discounted net cash flows, to calculate a project’s net present value.
Based on current estimates of future cash flows (not included here except for outflows), the estimated net present value for each project is shown in the bottom row of Table 3.3 . All the investors, including Think-Big, then will split this net present value in proportion to their share of the total investment.
For each project, participation shares are being sold to major investors, such as Think-Big, who become the partners for the project by investing their proportional shares at the four spec- ified points in time. For example, if Think-Big takes a 10 percent share of the office building, it will need to provide $4 million now, and then $6 million, $9 million, and $1 million in 1 year, 2 years, and 3 years, respectively.
A key part of a country’s financial infrastructure is its secu- rities markets. By allowing a variety of financial institutions and their clients to trade stocks, bonds, and other finan- cial securities, they help fund both public and private ini- tiatives. Therefore, the efficient operation of its securities markets plays a crucial role in providing a platform for the economic growth of the country.
Each central securities depository and its system for quickly settling security transactions are part of the opera- tional backbone of securities markets and a key component of financial system stability. In Mexico, an institution called INDEVAL provides both the central securities depository and its security settlement system for the entire country. This security settlement system uses electronic book entries, modifying cash and securities balances, for the various par- ties in the transactions.
The total value of the securities transactions the INDEVAL settles averages over $250 billion daily. This makes INDEVAL the main liquidity conduit for Mexico’s entire financial sec- tor. Therefore, it is extremely important that INDEVAL’s sys- tem for clearing securities transactions be an exceptionally efficient one that maximizes the amount of cash that can be delivered almost instantaneously after the transactions. Because of past dissatisfaction with this system, INDEVAL’s board of directors ordered a major study in 2005 to com- pletely redesign the system.
Following more than 12,000 man-hours devoted to this redesign, the new system was successfully launched in November 2008. The core of the new system is a large linear programming model that is applied many times daily to choose which pending transactions should be set- tled immediately with the depositor’s available balances. Linear programming is ideally suited for this application because it can maximize the value of the transactions settled while taking into account the various relevant constraints.
This application of linear programming has substan- tially enhanced and strengthened the Mexican financial infrastructure by reducing its daily liquidity requirements by $130 billion. It also reduces the intraday financing costs for market participants by more than $150 million annu- ally. This application also led to INDEVAL winning the pres- tigious first prize in the 2010 international competition for the Franz Edelman Award for Achievement in Operations Research and the Management Sciences.
Source: D. Muñoz, M. de Lascurain, O. Romeo-Hernandez, F. Solis, L. de los Santoz, A. Palacios-Brun, F. Herrería, and J. Villaseñor, “INDEVAL Develops a New Operating and Settlement System Using Operations Research,” Interfaces 41, no. 1 (January–February 2011), pp. 8–17. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
Investment Capital Requirements
Year Office Building Hotel Shopping Center
0 $40 million $80 million $90 million 1 60 million 80 million 50 million 2 90 million 80 million 20 million 3 10 million 70 million 60 million
Net present value $45 million $70 million $50 million
TABLE 3.3 Financial Data for the Projects Being Considered for Partial Investment by the Think- Big Development Co.
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3.2 Resource-Allocation Problems 77
The company currently has $25 million available for capital investment. Projections are that another $20 million will become available after one year, $20 million more after two years, and another $15 million after three years. What share should Think-Big take in the respective projects to maximize the total net present value of these investments?
Formulation This is a resource-allocation problem. The activities under consideration are
Activity 1: Invest in the construction of an office building. Activity 2: Invest in the construction of a hotel. Activity 3: Invest in the construction of a shopping center.
Thus, the decisions to be made are the levels of these activities, that is, what participation share to take in investing in each of these projects. A participation share can be expressed as either a fraction or a percentage of the entire project, so the entire project is considered to be one “unit” of that activity.
The resources to be allocated to these activities are the funds available at the four invest- ment points. Funds not used at one point are available at the next point. (For simplicity, we will ignore any interest earned on these funds.) Therefore, the resource constraint for each point must reflect the cumulative funds to that point.
Resource 1: Total investment capital available now. Resource 2: Cumulative investment capital available by the end of one year. Resource 3: Cumulative investment capital available by the end of two years. Resource 4: Cumulative investment capital available by the end of three years.
Since the amount of investment capital available is $25 million now, another $20 million in one year, another $20 million in two years, and another $15 million in three years, the amounts available of the resources are the following:
Amount of resource 1 available 5 $25 million Amount of resource 2 available 5 $(25 1 20) million 5 $45 million Amount of resource 3 available 5 $(25 1 20 1 20) million 5 $65 million Amount of resource 4 available 5 $(25 1 20 1 20 1 15) million 5 $80 million
Table 3.4 shows all the data involving these resources. The rightmost column gives the amounts of resources available calculated above. The middle columns show the cumulative amounts of the investment capital requirements listed in Table 3.3 . For example, in the Office Building column of Table 3.4 , the second number ($100 million) is obtained by adding the first two numbers ($40 million and $60 million) in the Office Building column of Table 3.3 . The Data As with any resource-allocation problem, three kinds of data need to be gathered. One is the amounts available of the resources, as given in the rightmost column of Table 3.4 . A second is the amount of each resource needed by each project, which is given in the middle columns of this table. A third is the contribution of each project to the overall measure of performance (net present value), as given in the bottom row of Table 3.3 .
The first step in formulating the spreadsheet model is to enter these data into data cells in the spreadsheet. In Figure 3.3 , the data cells (and their range names) are NetPresentValue
Cumulative Investment Capital Required for an Entire Project
Resource Office
Building Hotel Shopping
Center Amount of
Resource Available
1 (Now) $ 40 million $ 80 million $ 90 million $25 million 2 (End of year 1) 100 million 160 million 140 million 45 million 3 (End of year 2) 190 million 240 million 160 million 65 million 4 (End of year 3) 200 million 310 million 220 million 80 million
TABLE 3.4 Resource Data for the Think-Big Development Co. Investment-Mix Problem
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78 Chapter Three Linear Programming: Formulation and Applications
(C5:E5), CapitalRequired (C9:E12), and CapitalAvailable (H9:H12). To save space on the spreadsheet, these numbers are entered in units of millions of dollars.
The Decisions With three activities under consideration, there are three decisions to be made.
Decision 1: OB 5 Participation share in the office building. Decision 2: H 5 Participation share in the hotel. Decision 3: SC 5 Participation share in the shopping center.
For example, if Think-Big management were to decide to take a one-tenth participation share (i.e., a 10 percent participation share) in each of these projects, then
OB 5 0.1 5 10% H 5 0.1 5 10% SC 5 0.1 5 10%
However, it may not be desirable to take the same participation share (expressed as either a fraction or a percentage) in each of the projects, so the idea is to choose the best combination
=SUMPRODUCT(NetPresentValue,ParticipationShare)
Cumulative
Capital
Spent
Cumulative
Capital
Available
1
A B C D FE G H
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Think-Big Development Co. Capital Budgeting Program
Office
Building Hotel
Shopping
Center
Office
Building
0.00% 16.50%
Hotel
13.11%
Shopping
Center
18.11
Total NPV
($millions)
Net Present Value
($millions)
Now
End of Year 1
End of Year 2
25
44.76
60.58
80
≤ ≤ ≤ ≤End of Year 3
Participation Share
Cumulative Capital Required ($millions)
6
F
Cumulative
Capital
Spent
7
14
15
16
8
9
10
11
12
=SUMPRODUCT(C9:E9,ParticipationShare)
H
Total NPV
($millions)
=SUMPRODUCT(C10:E10,ParticipationShare)
=SUMPRODUCT(C11:E11,ParticipationShare)
=SUMPRODUCT C12:E12,ParticipationShare)
Range Name
CapitalAvailable
CapitalRequired
CapitalSpent
ParticipationShare
NetPresentValue
TotalNPV
Cells
H9:H12
C9:E12
F9:F12
C16:E16
C5:E5
H16
45 70 50
40
100
80
160
90
140
190 240 160
200 310 220
25
45
65
80
Solver Parameters Set Objective Cell: TotalNPV To: Max By Changing Variable Cells: ParticipationShare
Subject to the Constraints: CapitalSpent <= CapitalAvailable
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 3.3 The spreadsheet model for the Think-Big problem, including the formulas for the objective cell TotalNPV (H16) and the other output cells CapitalSpent (F9:F12), as well as the specifications needed to set up Solver. The changing cells ParticipationShare (C16:E16) show the optimal solution obtained by Solver.
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3.2 Resource-Allocation Problems 79
of values of OB, H, and SC. In Figure 3.3 , the participation shares (expressed as percentages) have been placed in changing cells under the data cells (row 16) in the columns for the three projects, so
OB S cell C16 H S D16 SC S cell E16
where these cells are collectively referred to by the range name ParticipationShare (C16:E16). The Constraints The numbers in these changing cells make sense only if they are nonnega- tive, so the Make Variables Nonnegative option will need to be selected in the Excel’s Solver dialog box (or equivalently in RSPE, set the Assume Non-Negative option to True in the Engine tab of the Model pane). In addition, the four resources require resource constraints:
Total invested now # 25 (millions of dollars available) Total invested within 1 year # 45 (millions of dollars available) Total invested within 2 years # 65 (millions of dollars available) Total invested within 3 years # 80 (millions of dollars available)
The data in columns C, D, and E indicate that (in millions of dollars)
Total invested now 5 40 OB 1 80 H 1 90 SC Total invested within 1 year 5 100 OB 1 160 H 1 140 SC Total invested within 2 years 5 190 OB 1 240 H 1 160 SC Total invested within 3 years 5 200 OB 1 310 H 1 220 SC
These totals are calculated in the output cells CapitalSpent (F9:F12) using the SUMPRODUCT function, as shown below the spreadsheet in Figure 3.3 . Finally, # signs are entered into col- umn G to indicate the resource constraints that will need to be entered in Solver. The Measure of Performance The objective is to
Maximize NPV 5 total net present value of the investments
NetPresentValue (C5:E5) shows the net present value of each entire project, while Partici- pationShare (C16:E16) shows the participation share for each of the projects. Therefore, the total net present value of all the participation shares purchased in all three projects is (in mil- lions of dollars)
NPV 5 45 OB 1 70 H 1 50 SC
5 SUMPRODUCT (NetPresentValue, ParticipationShare)
S cell H16
Summary of the Formulation This completes the formulation of the linear programming model on the speadsheet, as summarized below (in algebraic form).
Maximize NPV 5 45 OB 1 70 H 1 50 SC
subject to
Total invested now: 40 OB 1 80 H 1 90 SC # 25
Total invested within 1 year: 100 OB 1 160 H 1 140 SC # 45
Total invested within 2 years: 190 OB 1 240 H 1 160 SC # 65
Total invested within 3 years: 200 OB 1 310 H 1 220 SC # 80
and
OB $ 0 H $ 0 SC $ 0
where all these numbers are in units of millions of dollars. Note that this model possesses the key identifying feature for resource-allocation prob-
lems, namely, each functional constraint is a resource constraint that has the form
Amount of resource used # Amount of resource available
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80 Chapter Three Linear Programming: Formulation and Applications
Solving the Model The lower left-hand side of Figure 3.3 shows the entries needed in Solver to specify the model, along with the selection of the usual two options. The spreadsheet shows the resulting optimal solution in row 16, namely,
Invest nothing in the office building. Invest in 16.50 percent of the hotel. Invest in 13.11 percent of the shopping center.
TotalNPV (H16) indicates that this investment program would provide a total net present value of $18.11 million.
This amount actually is only an estimate of what the total net present value would turn out to be, depending on the accuracy of the financial data given in Table 3.3 . There is some uncer- tainty about the construction costs for the three real estate projects, so the actual investment capital requirements for years 1, 2, and 3 may deviate somewhat from the amounts specified in this table. Because of the risk involved in these projects, the net present value for each one also might deviate from the amounts given at the bottom of the table. Chapter 5 describes one approach to analyzing the effect of such deviations. Chapters 12 and 13 will present another technique, called computer simulation, for systematically taking future uncertainties into account. Section 13.5 will focus on further analysis of this same example.
Another Look at Resource Constraints These examples of resource-allocation problems illustrate a variety of resources: financial allocations for advertising and planning purposes, TV commercial spots available for pur- chase, available production capacities of different plants, the total amount of capital avail- able for investment, and cumulative investment capital available by certain times. However, these illustrations only scratch the surface of the realm of possible resources that need to be allocated to activities in resource-allocation problems. In fact, by interpreting resource suf- ficiently broadly, any restriction on the decisions to be made that has the form
Amount used # Amount available
can be thought of as a resource constraint, where the thing whose amount is being measured is the corresponding “resource.” Since any functional constraint with a # sign in a linear pro- gramming model (including the side constraint in the TBA Airlines example) can be verbal- ized in this form, any such constraint can be thought of as a resource constraint.
Hereafter, we will use resource constraint to refer to any functional constraint with a # sign in a linear programming model. The constant on the right-hand side represents the amount avail- able of a resource. Therefore, the left-hand side represents the amount used of this resource. In the algebraic form of the constraint, the coefficient (positive or negative) of each decision vari- able is the resource usage per unit of the corresponding activity.
Summary of the Formulation Procedure for Resource-Allocation Problems The four examples illustrate that the following steps are used for any resource-allocation problem to define the specific problem, gather the relevant data, and then formulate the linear programming model.
1. Since any linear programming problem involves finding the best mix of levels of various activities, identify these activities for the problem at hand. The decisions to be made are the levels of these activities.
2. From the viewpoint of management, identify an appropriate overall measure of perfor- mance (commonly profit, or a surrogate for profit) for solutions of the problem.
3. For each activity, estimate the contribution per unit of the activity to this overall measure of performance.
4. Identify the resources that must be allocated to the activities. 5. For each resource, identify the amount available and then the amount used per unit of each
activity.
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3.3 Cost–Benefit–Trade-Off Problems 81
6. Enter the data gathered in steps 3 and 5 into data cells in a spreadsheet. A convenient for- mat is to have the data associated with each activity in a separate column, the data for the unit profit and each constraint in a separate row, and to leave two blank columns between the activity columns and the amount of resource available column. Figure 3.4 shows a template of the overall format of a spreadsheet model for resource-allocation problems.
7. Designate changing cells for displaying the decisions on activity levels. 8. For the two blank columns created in step 6, use the left one as a Totals column for output
cells and enter # signs into the right one for all the resources. In the row for each resource, use the SUMPRODUCT function to enter the total amount used in the Totals column.
9. Designate an objective cell for displaying the overall measure of performance. Use a SUMPRODUCT function to enter this measure of performance.
All the functional constraints in this linear programming model in a spreadsheet are resource constraints, that is, constraints with a # sign. This is the identifying feature that classifies the problem as being a resource-allocation problem.
Activities
Unit Profit Profit per unit of activity Resources
Used Resources Available
SUMPRODUCT
(resource used per unit,
changing cells)
Resource used per unit of activity
Total Profit
Level of Activity Changing cells SUMPRODUCT(profit per unit, changing cells)
C on
st ra
in ts
≤
FIGURE 3.4 A template of a spreadsheet model for pure resource-allocation problems.
1. What is the identifying feature for a resource-allocation problem? 2. What is the form of a resource constraint? 3. What are the three kinds of data that need to be gathered for a resource-allocation problem? 4. Compare the types of activities for the four examples of resource-allocation problems. 5. Compare the types of resources for the four examples of resource-allocation problems.
Review Questions
3.3 COST–BENEFIT–TRADE-OFF PROBLEMS
Cost–benefit–trade-off problems have a form that is very different from resource-allocation problems. The difference arises from managerial objectives that are very different for the two kinds of problems.
For resource-allocation problems, limits are set on the use of various resources (including financial resources), and then the objective is to make the most effective use (according to some overall measure of performance) of these given resources.
For cost–benefit–trade-off problems, management takes a more aggressive stance, pre- scribing what benefits must be achieved by the activities under consideration (regardless of the resulting resource usage), and then the objective is to achieve all these benefits with mini- mum cost. By prescribing a minimum acceptable level for each kind of benefit, and then mini- mizing the cost needed to achieve these levels, management hopes to obtain an appropriate trade-off between cost and benefits. (You will see in Chapter 5 that what-if analysis plays a key role in providing the additional information needed for management to choose the best trade-off between cost and benefits.)
A cost–benefit–trade-off formulation enables man- agement to specify mini- mum goals for the benefits that need to be achieved by the activities.
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82 Chapter Three Linear Programming: Formulation and Applications
Cost–benefit–trade-off problems are linear programming problems where the mix of levels of various activities is chosen to achieve minimum acceptable levels for various benefits at a minimum cost. The identifying feature is that each functional constraint is a benefit constraint , which has the form
Level achieved $ Minimum acceptable level
for one of the benefits.
Interpreting benefit broadly, we can think of any functional constraint with a $ sign as a benefit constraint. In most cases, the minimum acceptable level will be prescribed by manage- ment as a policy decision, but occasionally this number will be dictated by other circumstances.
For any cost–benefit–trade-off problem, a major part of the study involves identifying all the activities and benefits that should be considered and then gathering the data relevant to these activities and benefits.
Three kinds of data are needed:
1. The minimum acceptable level for each benefit (a managerial policy decision). 2. For each benefit, the contribution of each activity to that benefit (per unit of the activity). 3. The cost per unit of each activity.
Let’s examine two examples of cost–benefit–trade-off problems.
The Profit & Gambit Co. Advertising-Mix Problem As described in Section 2.7, the Profit & Gambit Co. will be undertaking a major new adver- tising campaign focusing on three cleaning products. The two kinds of advertising to be used are television and the print media. Management has established minimum goals—the mini- mum acceptable increase in sales for each product—to be gained by the campaign.
The problem is to determine how much to advertise in each medium to meet all the sales goals at a minimum total cost.
The activities in this cost–benefit–trade-off problem are
Activity 1: Advertise on television. Activity 2: Advertise in the print media.
The benefits being sought from these activities are
Benefit 1: Increased sales for a spray prewash stain remover. Benefit 2: Increased sales for a liquid laundry detergent. Benefit 3: Increased sales for a powder laundry detergent.
Management wants these increased sales to be at least 3 percent, 18 percent, and 4 percent, respectively. As shown in Section 2.7, each benefit leads to a benefit constraint that incor- porates the managerial goal for the minimum acceptable level of increase in the sales for the corresponding product, namely,
Level of benefit 1 achieved $ 3% Level of benefit 2 achieved $ 18% Level of benefit 3 achieved $ 4%
The data for this problem are given in Table 2.2 (Section 2.7). Section 2.7 describes how the linear programming model is formulated directly from the numbers in this table.
This example provides an interesting contrast with the Super Grain Corp. case study in Section 3.1, which led to a formulation as a resource-allocation problem. Both are advertis- ing-mix problems, yet they lead to entirely different linear programming models. They differ because of the differences in the managerial view of the key issues in each case:
• As the vice president for marketing of Super Grain, Claire Syverson focused first on how much to spend on the advertising campaign and then set limits (an advertising budget of $4 million and a planning budget of $1 million) that led to resource constraints.
These three kinds of data are needed for any cost– benefit–trade-off problem.
An initial step in formulat- ing any cost–benefit–trade- off problem is to identify the activities and the benefits.
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• The management of Profit & Gambit instead focused on what it wanted the advertising campaign to accomplish and then set goals (minimum required increases in sales) that led to benefit constraints.
From this comparison, we see that it is not the nature of the application that determines the classification of the resulting linear programming formulation. Rather, it is the nature of the restrictions imposed on the decisions regarding the mix of activity levels. If the restrictions involve limits on the usage of resources, that identifies a resource-allocation problem. If the restrictions involve goals on the levels of benefits, that characterizes a cost–benefit–trade-off problem. Frequently, the nature of the restrictions arise from the way management frames the problem.
However, we don’t want you to get the idea that every linear programming problem falls entirely and neatly into either one type or the other. In the preceding section and this one, we are looking at pure resource-allocation problems and pure cost–benefit–trade-off problems. Although many real problems tend to be either one type or the other, it is fairly common to have both resource constraints and benefit constraints, even though one may predominate. (In the next section, you will see an example of how both types of constraints can arise in the same problem when the management of the Super Grain Corp. introduces additional con- siderations into the analysis of their advertising-mix problem.) Furthermore, we still need to consider additional categories of linear programming problems in the remaining sections of this chapter.
Now, another example of a pure cost–benefit–trade-off problem.
Personnel Scheduling One of the common applications of cost–benefit–trade-off analysis involves personnel sched- uling for a company that provides some kind of service, where the objective is to schedule the work times of the company’s employees so as to minimize the cost of providing the level of service specified by management. The following example illustrates how this can be done.
The Problem Union Airways is adding more flights to and from its hub airport and so needs to hire addi- tional customer service agents. However, it is not clear just how many more should be hired. Management recognizes the need for cost control while also consistently providing a satisfac- tory level of service to the company’s customers, so a desirable trade-off between these two factors is being sought. Therefore, a management science team is studying how to schedule the agents to provide satisfactory service with the smallest personnel cost.
Cost control is essential for survival in the airline industry. Therefore, upper management of United Airlines initiated a management science study to improve the utilization of personnel at the airline’s reservations offices and air- ports by matching work schedules to customer needs more closely. The number of employees needed at each location to provide the required level of service varies greatly dur- ing the 24-hour day and might fluctuate considerably from one half hour to the next.
Trying to design the work schedules for all the employ- ees at a given location to meet these service requirements most efficiently is a nightmare of combinatorial consider- ations. Once an employee arrives, he or she will be there continuously for the entire shift (2 to 10 hours, depending on the employee), except for either a meal break or short rest breaks every two hours. Given the minimum number of employees needed on duty for each half-hour interval
over a 24-hour day (this minimum changes from day to day over a seven-day week), how many employees of each shift length should begin work at what start time over each 24-hour day of a seven-day week? Fortunately, linear pro- gramming thrives on such combinatorial nightmares. The linear programming model for some of the locations sched- uled involves over 20,000 decisions!
This application of linear programming was credited with saving United Airlines more than $6 million annually in just direct salary and benefit costs. Other benefits included improved customer service and reduced workloads for sup- port staff.
Source: T. J. Holloran and J. E. Bryne, “United Airlines Station Manpower Planning System,” Interfaces 16, no. 1 (January–February 1986), pp. 39–50. (A link to this article is provided on our website, www.mhhe.com/hillier5e. )
An Application Vignette
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84 Chapter Three Linear Programming: Formulation and Applications
Based on the new schedule of flights, an analysis has been made of the minimum number of customer service agents that need to be on duty at different times of the day to provide a satisfactory level of service. (The queueing models presented in Chapter 11 can be used to determine the minimum numbers of agents needed to keep customer waiting times reason- able.) These numbers are shown in the last column of Table 3.5 for the time periods given in the first column. The other entries in this table reflect one of the provisions in the company’s current contract with the union that represents the customer service agents. The provision is that each agent works an eight-hour shift. The authorized shifts are
Shift 1: 6:00 am to 2:00 pm. Shift 2: 8:00 am to 4:00 pm. Shift 3: Noon to 8:00 pm. Shift 4: 4:00 pm to midnight. Shift 5: 10:00 pm to 6:00 am.
Check marks in the main body of Table 3.5 show the time periods covered by the respective shifts. Because some shifts are less desirable than others, the wages specified in the contract differ by shift. For each shift, the daily compensation (including benefits) for each agent is shown in the bottom row. The problem is to determine how many agents should be assigned to the respective shifts each day to minimize the total personnel cost for agents, based on this bottom row, while meeting (or surpassing) the service requirements given in the last column.
Formulation This problem is, in fact, a pure cost–benefit–trade-off problem. To formulate the problem, we need to identify the activities and benefits involved.
Activities correspond to shifts. The level of each activity is the number of agents assigned to that shift. A unit of each activity is one agent assigned to that shift.
Thus, the general description of a linear programming problem as finding the best mix of activ- ity levels can be expressed for this specific application as finding the best mix of shift sizes.
Benefits correspond to time periods. For each time period, the benefit provided by the activities is the service that agents pro- vide customers during that period. The level of a benefit is measured by the number of agents on duty during that time period.
Once again, a careful formulation of the problem, including gathering all the relevant data, leads rather directly to a spreadsheet model. This model is shown in Figure 3.5 , and we out- line its formulation below.
Time Periods Covered by Shift
Time Period 1 2 3 4 5 Minimum Number of Agents Needed
6:00 AM to 8:00 AM ✔ 48 8:00 AM to 10:00 AM ✔ ✔ 79 10:00 AM to noon ✔ ✔ 65 Noon to 2:00 PM ✔ ✔ ✔ 87 2:00 PM to 4:00 PM ✔ ✔ 64 4:00 PM to 6:00 PM ✔ ✔ 73 6:00 PM to 8:00 PM ✔ ✔ 82 8:00 PM to 10:00 PM ✔ 43 10:00 PM to midnight ✔ ✔ 52 Midnight to 6:00 AM ✔ 15
Daily cost per agent $170 $160 $175 $180 $195
TABLE 3.5 Data for the Union Airways Personnel Scheduling Problem
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3.3 Cost–Benefit–Trade-Off Problems 85
The Data As indicated in this figure, all the data in Table 3.5 have been entered directly into the data cells CostPerShift (C5:G5), ShiftWorksTimePeriod (C8:G17), and MinimumNeeded (J8:J17). For the ShiftWorksTimePeriod (C8:G17) data, an entry of 1 indicates that the corre- sponding shift includes that time period whereas 0 indicates not. Like any cost–benefit–trade- off problem, these numbers indicate the contribution of each activity to each benefit. Each agent working a shift contributes either 0 or 1 toward the minimum number of agents needed in a time period.
Shift Works Time Period? (1=yes, 0=no)
1
A B C FED G IH
2
3
4
6AM–2PM
Shift
5
6
7
8
9
10
13
14
15
16
17
18
19
20
21
11
12
Union Airways Personnel Scheduling Problem
Cost per Shift
8AM–4PM
Shift
Noon–8PM
Shift
4PM–Midnight
Shift
10PM–6AM
Shift
6AM–2PM
Shift
8AM–4PM
Shift
Noon–8PM
Shift
4PM–Midnight
Shift
10PM–6AM
Shift Total Cost
$30,610
Time Period
Number Working
6AM–8AM
8AM–10AM
10AM–12PM
12PM–2PM
2PM–4PM
4PM–6PM
6PM–8PM
8PM–10PM
10PM–12AM
12AM–6AM
J
Minimum
Needed
Total
Working
48
79
79
118
70
82
82
43
58
15
6
H
Total
Working7
20
21
8
9
10
11
12
13
14
15
17
16
=SUMPRODUCT(C8:G8,NumberWorking)
=SUMPRODUCT(C9:G9,NumberWorking)
=SUMPRODUCT(C10:G10,NumberWorking)
=SUMPRODUCT(C11:G11,NumberWorking)
=SUMPRODUCT(C12:G12,NumberWorking)
=SUMPRODUCT(C13:G13,NumberWorking)
=SUMPRODUCT(C14:G14,NumberWorking)
=SUMPRODUCT(C15:G15,NumberWorking)
=SUMPRODUCT(C16:G16,NumberWorking)
=SUMPRODUCT(C17:G17,NumberWorking)
J
Total Cost
=SUMPRODUCT(CostPerShift,NumberWorking)
Range Name
CostPerShift
MinimumNeeded
NumberWorking
ShiftWorksTimePeriod
TotalCost
TotalWorking
Cells
C5:G5
J8:J17
C21:G21
C8:G17
J21
H8:H17
$170 $160 $175 $180 $195
1
1
1
1
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
1
1
48 31 39 43 15
48
79
65
87
64
73
82
43
52
15
≥ ≥
≥ ≥
≥
≥
≥
≥
≥ ≥
Solver Parameters Set Objective Cell: TotalCost To: Min By Changing Variable Cells: NumberWorking
Subject to the Constraints: NumberWorking = integer
TotalWorking >= MinimumNeeded
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 3.5 The spreadsheet model for the Union Airways problem, including the formulas for the objective cell TotalCost (J21) and the other output cells TotalWorking (H8:H17), as well as the specifications needed to set up Solver. The changing cells NumberWorking (C21:G21) show the optimal solution obtained by Solver.
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86 Chapter Three Linear Programming: Formulation and Applications
The Decisions Since the activities in this case correspond to the five shifts, the decisions to be made are
S 1 5 Number of agents to assign to Shift 1 (starts at 6 am )
S 2 5 Number of agents to assign to Shift 2 (starts at 8 am )
S 3 5 Number of agents to assign to Shift 3 (starts at noon)
S 4 5 Number of agents to assign to Shift 4 (starts at 4 pm )
S 5 5 Number of agents to assign to Shift 5 (starts at 10 pm )
The changing cells to hold these numbers have been placed in the activity columns in row 21, so
S1 S cell C21 S2 S cell D21 c S5 S cell G21
where these cells are collectively referred to by the range name NumberWorking (C21:G21).
The Constraints These changing cells need to be nonnegative. In addition, we need 10 benefit constraints, where each one specifies that the total number of agents serving in the corresponding time period listed in column B must be no less than the minimum acceptable number given in column J. Thus, these constraints are
Total number of agents serving 628 am $ 48 (min. acceptable) Total number of agents serving 8210 am $ 79 (min. acceptable)
# # #
Total number of agents serving midnight26 am $ 15 (min. acceptable)
Since columns C to G indicate which of the shifts serve each of the time periods, these totals are
Total number of agents serving 628 am 5 S1 Total number of agents serving 8210 am 5 S1 1 S2
# # #
Total number of agents serving midnight 26 am 5 S5
These totals are calculated in the output cells TotalWorking (H8:H17) using the SUMPROD- UCT functions shown below the spreadsheet in Figure 3.5 .
One other type of constraint is that the number of agents assigned to each shift must have an integer value. These constraints for the five shifts should be added in the same way as described for the TBA Airlines problem in Section 3.2. In particular, with Excel’s Solver they are added in the Add Constraint dialog box by entering NumberWorking on the left-hand side and then choosing int from the pop-up menu between the left-hand side and the right-hand side. The set of constraints, NumberWorking 5 integer, then appears in the Solver Parameters, as shown in Figure 3.5 . In RSPE, select the range of cells to be constrained integer, and then under the Constraint menu on the RSPE ribbon, choose Integer under the Variable Type/Bound submenu.
The Measure of Performance The objective is to
Minimize Cost 5 Total daily personnel cost for all agents
Since CostPerShift (C5:G5) gives the daily cost per agent on each shift and NumberWorking (C21:G21) gives the number of agents working each shift,
Cost 5 170S1 1 160S2 1 175S3 1 180S4 1 195S5 (in dollars)
5 SUMPRODUCT (CostPerShift, NumberWorking)
S cell J21 Summary of the Formulation The above steps provide the complete formulation of the linear programming model on a spreadsheet, as summarized below (in algebraic form).
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3.3 Cost–Benefit–Trade-Off Problems 87
Minimize Cost 5 170S1 1 160S2 1 175S3 1 180S4 1 195S5 (in dollars)
subject to
Total agents 628 am: S1 $ 48 Total agents 8210 am: S1 1 S2 $ 79 # # # Total agents midnight26 am: S5 $ 15
and
S1 $ 0 S2 $ 0 S3 $ 0 S4 $ 0 S5 $ 0
Solving the Model The lower left-hand corner of Figure 3.5 shows the entries needed in Solver, along with the selection of the usual two options. After solving, NumberWorking (C21:G21) in the spreadsheet shows the resulting optimal solution for the number of agents that should be assigned to each shift. TotalCost (J21) indicates that this plan would cost $30,610 per day.
Summary of the Formulation Procedure for Cost–Benefit–Trade- Off Problems The nine steps in formulating any cost–benefit–trade-off problem follow the same pattern as presented at the end of the preceding section for resource-allocation problems, so we will not repeat them here. The main differences are that the overall measure of performance now is the total cost of the activities (or some surrogate of total cost chosen by management) in steps 2 and 3, benefits now replace resources in steps 4 and 5, and $ signs now are entered to the right of the output cells for benefits in step 8. Figure 3.6 shows a template of the format of a spreadsheet model for cost–benefit–trade-off problems.
All the functional constraints in the resulting model are benefit constraints, that is, con- straints with a $ sign. This is the identifying feature of a pure cost–benefit–trade-off problem.
Activities
Unit Cost Cost per unit of activity
SUMPRODUCT
(benefit per unit,
changing cells)
Benefit achieved per unit of activity
Total Cost
Level of Activity Changing cells SUMPRODUCT(cost per unit, changing cells)
C on
st ra
in ts
Benefit Achieved
Benefit Needed
≥
FIGURE 3.6 A template of a spreadsheet model for pure cost–benefit–trade-off problems.
1. What is the difference in managerial objectives between resource-allocation problems and cost–benefit–trade-off problems?
2. What is the identifying feature of a cost–benefit–trade-off problem? 3. What is the form of a benefit constraint? 4. What are the three kinds of data that need to be gathered for a cost–benefit–trade-off problem? 5. Compare the types of activities for the two examples of cost–benefit–trade-off problems. 6. Compare the types of benefits for the two examples of cost–benefit–trade-off problems.
Review Questions
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88 Chapter Three Linear Programming: Formulation and Applications
3.4 MIXED PROBLEMS
Sections 3.2 and 3.3 each described a broad category of linear programming problems— resource-allocation and cost–benefit–trade-off problems. As summarized in Table 3.6 , each features one of the first two types of functional constraints shown there. In fact, the identi- fying feature of a pure resource-allocation problem is that all its functional constraints are resource constraints. The identifying feature of a pure cost–benefit–trade-off problem is that all its functional constraints are benefit constraints. (Keep in mind that the functional con- straints include all the constraints of a problem except its nonnegativity constraints.)
The bottom row of Table 3.6 shows the last of the three types of functional constraints, namely, fixed-requirement constraints , which require that the left-hand side of each such constraint must exactly equal some fixed amount. Thus, since the left-hand side rep- resents the amount provided of some quantity, the form of a fixed-requirement constraint is
Amount provided 5 Required amount
The identifying feature of a pure fixed-requirements problem is that it is a linear pro- gramming problem where all its functional constraints are fixed-requirement constraints. The next two sections will describe two particularly prominent types of fixed-requirement prob- lems called transportation problems and assignment problems.
However, before turning to these types of problems, we first will use a continuation of the Super Grain case study from Section 3.1 to illustrate how many linear programming problems fall into another broad category called mixed problems.
Many linear programming problems do not fit completely into any of the previously discussed categories (pure resource-allocation problems, cost–benefit–trade-off problems, and fixed- requirement problems) because the problem’s functional constraints include more than one of the types shown in Table 3.6 . Such problems are called mixed problems .
Now let us see how a more careful analysis of the Super Grain case study turns this resource- allocation problem into a mixed problem that includes all three types of functional constraints shown in Table 3.6 .
Super Grain Management Discusses Its Advertising-Mix Problem The description of the Super Grain case study in Section 3.1 ends with Clair Syverson (Super Grain’s vice president for marketing) sending a memorandum to the company’s president, David Sloan, requesting a meeting to evaluate her proposed promotional campaign for the company’s new breakfast cereal.
Soon thereafter, Claire Syverson and David Sloan meet to discuss plans for the campaign. David Sloan (president): Thanks for your memo, Claire. The plan you outline for the promotional campaign looks like a reasonable one. However, I am surprised that it does not make any use of TV commercials. Why is that? Claire Syverson (vice president for marketing): Well, as I described in my memo, I used a spreadsheet model to see how to maximize the number of exposures from the
Type Form* Typical Interpretation Main Usage
Resource constraint LHS # RHS For some resource, Amount used # Amount available
Resource-allocation problems and mixed problems
Benefit constraint LHS $ RHS For some benefit, Level achieved $ Minimum acceptable level
Cost–benefit–trade-off problems and mixed problems
Fixed-requirement constraint LHS 5 RHS For some quantity, Amount provided 5 Required amount
Fixed-requirements problems and mixed problems
*LHS 5 Left-hand side (a SUMPRODUCT function). RHS 5 Right-hand side (a constant).
TABLE 3.6 Types of Functional Constraints
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3.4 Mixed Problems 89
campaign and this turned out to be the plan that does this. I also was surprised that it did not include TV commercials, but the model indicated that introducing commercials would pro- vide less exposures on a dollar-for-dollar basis than magazine ads and Sunday supplement ads. Don’t you think it makes sense to use the plan that maximizes the number of exposures? David: Not necessarily. Some exposures are a lot less important than others. For example, we know that middle-aged adults are not big consumers of our cereals, so we don’t care very much how many of those people see our ads. On the other hand, young children are big con- sumers. Having TV commercials on the Saturday morning programs for children is our pri- mary method of reaching young children. You know how important it will be to get young children to ask their parents for Crunchy Start. That is our best way of generating first-time sales. Those commercials also get seen by a lot of parents who are watching the programs with their kids. What we need is a commercial that is appealing to both parents and kids, and that gets the kids immediately bugging their parents to go buy Crunchy Start. I think that is a real key to a successful campaign. Claire: Yes, that makes a lot of sense. In fact, I already have set some goals regarding the number of young children and the number of parents of young children that need to be reached by this promotional campaign. David: Good. Did you include those goals in your spreadsheet model? Claire: No, I didn’t. David: Well, I suggest that you incorporate them directly into your model. I suspect that maximizing exposures while also meeting your goals will give us a high impact plan that includes some TV commercials. Claire: Good idea. I’ll try it. David: Are there any other factors that the plan in your memo doesn’t take into account as well as you would like? Claire: Well, yes, one. The plan doesn’t take into account my budget for cents-off cou- pons in magazines and newspapers. David: You should be able to add that to your model as well. Why don’t you go back and see what happens when you incorporate these additional considerations? Claire: OK, will do. You seem to have had a lot of experience with spreadsheet modeling. David: Yes. It is a great tool as long as you maintain some healthy skepticism about what comes out of the model. No model can fully take into account everything that we must con- sider when dealing with managerial problems. This is especially true the first time or two you run the model. You need to keep asking, what are the missing quantitative consider- ations that I still should add to the model? Then, after you have made the model as complete as possible and obtained a solution, you still need to use your best managerial judgment to weigh intangible considerations that cannot be incorporated into the model.
Incorporating Additional Managerial Considerations into the Super Grain Model Therefore, David and Claire conclude that the spreadsheet model needs to be expanded to incorporate some additional considerations. In particular, since the promotional campaign is for a breakfast cereal that should have special appeal to young children, they feel that two audiences should be targeted— young children and parents of young children. (This is why one of the three advertising media recommended by Giacomi & Jackowitz is commercials on children’s television programs Saturday morning.) Consequently, Claire now has set two new goals for the campaign.
Goal 1: The advertising should be seen by at least five million young children. Goal 2: The advertising should be seen by at least five million parents of young children.
In effect, these two goals are minimum acceptable levels for two special benefits to be achieved by the advertising activities.
Benefit 1: Promoting the new breakfast cereal to young children. Benefit 2: Promoting the new breakfast cereal to parents of young children.
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Because of the way the goals have been articulated, the level of each of these benefits is measured by the number of people in the specified category that are reached by the advertising.
To enable constructing the corresponding benefit constraints (as described in Section 3.3), Claire asks Giacomi & Jackowitz to estimate how much each advertisement in each of the media will contribute to each benefit, as measured by the number of people reached in the specified category. These estimates are given in Table 3.7 .
It is interesting to observe that management wants special consideration given to these two kinds of benefits even though the original spreadsheet model ( Figure 3.1 ) already takes them into account to some extent. As described in Section 3.1, the expected number of exposures is the overall measure of performance to be maximized. This measure counts up all the times that an advertisement is seen by any individual, including all those individuals in the target audiences. However, maximizing this general measure of performance does not ensure that the two specific goals prescribed by management (Claire Syverson) will be achieved. Claire feels that achieving these goals is essential to a successful promotional campaign. Therefore, she complements the general objective with specific benefit constraints that do ensure that the goals will be achieved. Having benefit constraints added to incorporate managerial goals into the model is a prerogative of management.
Claire has one more consideration she wants to incorporate into the model. She is a strong believer in the promotional value of cents-off coupons (coupons that shoppers can clip from printed advertisements to obtain a refund of a designated amount when purchasing the adver- tised item). Consequently, she always earmarks a major portion of her annual marketing bud- get for the redemption of these coupons. She still has $1,490,000 left from this year’s allotment for coupon redemptions. Because of the importance of Crunchy Start to the company, she has decided to use this entire remaining allotment in the campaign promoting this cereal.
This fixed amount for coupon redemptions is a fixed requirement that needs to be expressed as a fixed-requirement constraint. As described at the beginning of this section, the form of a fixed-requirement constraint is that, for some type of quantity,
Amount provided 5 Required amount
In this case, the quantity involved is the amount of money provided for the redemption of cents-off coupons. To specify this constraint in the spreadsheet, we need to estimate how much each advertisement in each of the media will contribute toward fulfilling the required amount for the quantity. Both medium 2 (advertisements in food and family-oriented maga- zines) and medium 3 (advertisements in Sunday supplements of major newspapers) will fea- ture cents-off coupons. The estimates of the amount of coupon redemption per advertisement in each of these media is given in Table 3.8 .
Benefit constraints are use- ful for incorporating mana- gerial goals into the model.
Number Reached in Target Category (in millions)
Target Category Each TV
Commercial Each
Magazine Ad Each
Sunday Ad Minimum
Acceptable Level
Young children 1.2 0.1 0 5
Parents of young children 0.5 0.2 0.2 5
TABLE 3.7 Benefit Data for the Revised Super Grain Corp. Advertising-Mix Problem
Contribution toward Required Amount
Requirement Each
TV Spot Each
Magazine Ad Each
Sunday Ad Required Amount
Coupon redemption 0 $40,000 $120,000 $1,490,000
TABLE 3.8 Data for the Fixed-Requirement Constraint for the Revised Super Grain Corp. Advertising-Mix Problem
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3.4 Mixed Problems 91
TotalExposures
($thousands)
=SUMPRODUCT(ExposuresPerAd,NumberOfAds)
1
A B C D FE G H
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Super Grain Corp. Advertising-Mix Problem
TV Spots
3,775
1,000
Magazine Ads SS Ads
SS AdsTV Spots
3 14 7.75
Magazine Ads
16,175
Exposures per Ad
(thousands)
Ad Budget
Planning Budget
Number of Ads
SS AdsTV Spots Magazine Ads
Coupon Redemption
per Ad ($thousands)
Maximum TV Spots
Cost per Ad ($thousands) Budget AvailableBudget Spent
5
5.85
Young Children
Parents of Young Children
Number Reached per Ad (millions) Minimum AcceptableTotal Reached
1,490
Total Redeemed Required Amount
Total Exposures
(thousands)
6
F
Budget Spent
7
17
18
19
8
9
=SUMPRODUCT(C7:E7,NumberOfAds)
H
Total Exposures
(thousands)
=SUMPRODUCT(C8:E8,NumberOfAds)
Total Reached10
11
12
13
=SUMPRODUCT(C11:E11,NumberOfAds)
=SUMPRODUCT(C12:E12,NumberOfAds)
14
15
Total Redeemed
=SUMPRODUCT(CouponRedemptionPerAd,
NumberOfAds)
Range Name
BudgetAvailable
BudgetSpent
CostPerAd
CouponRedemptionPerAd
ExposuresPerAd
MaxTVSpots
MinimumAcceptable
NumberOfAds
NumberReachedPerAd
RequiredAmount
TotalExposures
TotalReached
TotalRedeemed
TVSpots
Cells
H7:H8
F7:F8
C7:E8
C15:E15
C4:E4
C21
H11:H12
C19:E19
C11:E12
H15
H19
F11:F12
F15
C19
1,300 600 500
300
90
150
30
100
40
4,000
1,000
0 40 120 1,490
5
1.2
0.5
0.1
0.2
0
0.2
5
5
≤
≤
≤
=
≥ ≥
Solver Parameters
Set Objective Cell: TotalExposures To: Max By Changing Variable Cells: NumberOfAds
Subject to the Constraints: BudgetSpent <= Budget Available
TVSpots <= MaxTVSpots
TotalReached >= MinimumAcceptable
TotalRedeemed = RequiredAmount
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 3.7 The spreadsheet model for the revised Super Grain problem, including the formulas for the objective cell TotalExposures (H19) and the other output cells in column F, as well as the specifications needed to set up Solver. The changing cells NumberOfAds (C19:E19) show the optimal solution obtained by Solver.
Formulation of the Revised Spreadsheet Model Figure 3.7 shows one way of formatting the spreadsheet to expand the original spreadsheet model in Figure 3.1 to incorporate the additional managerial considerations. We then outline the four components of the revised model next.
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92 Chapter Three Linear Programming: Formulation and Applications
The Data Additional data cells in NumberReachedPerAd (C11:E12), MinimumAcceptable (H11:H12), CouponRedemptionPerAd (C15:E15), and RequiredAmount (H15) give the data in Tables 3.7 and 3.8 .
The Decisions Recall that, as before, the decisions to be made are
TV 5 Number of commercials on television
M 5 Number of advertisements in magazines
SS 5 Number of advertisements in Sunday supplements
The changing cells to hold these numbers continue to be in NumberOfAds (C19:E19).
The Constraints In addition to the original constraints, we now have two benefit constraints and one fixed- requirement constraint. As specified in rows 11 and 12, columns F to H, the benefit con- straints are
Total number of young children reached $ 5 (goal 1 in millions)
Total number of parents reached $ 5 (goal 2 in millions)
Using the data in columns C to E of these rows,
Total number of young children reached 5 1.2TV 1 0.1M 1 0SS 5 SUMPRODUCT (C11:E11, NumberOfAds) S cell F11
Total number of parents reached 5 0.5TV 1 0.2M 1 0.2SS 5 SUMPRODUCT (C12:E12, NumberOfAds) S cell F12
These output cells are given the range name TotalReached (F11:F12). The fixed-requirement constraint indicated in row 15 is that
Total coupon redemption 5 1,490 (allotment in $1,000s)
CouponRedemptionPerAd (C15:E15) gives the number of coupons redeemed per ad, so
Total coupon redemption 5 0TV 1 40M 1 120SS
5 SUMPRODUCT (CouponRedemptionPerAd, NumberOfAds)
S cell F15
These same constraints are specified in Solver, along with the original constraints, in Figure 3.7 .
The Measure of Performance The measure of performance continues to be
Exposure 5 1,300TV 1 600M 1 500SS
5 SUMPRODUCT (ExposuresPerAd, NumberOfAds)
S cell H19
so the objective cell is again TotalExposures (H19).
Summary of the Formulation The above steps have resulted in formulating the following linear programming model (in algebraic form) on a spreadsheet.
Maximize Exposure 5 1,300TV 1 600M 1 500SS
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3.4 Mixed Problems 93
subject to the following constraints:
1. Resource constraints:
300TV 1 150M 1 100SS # 4,000 (ad budget in $1,000s) 90TV 1 30M 1 40SS # 1,000 (planning budget in $1,000s)
TV # 5 (television spots available)
2. Benefit constraints:
1.2TV 1 0.1M $ 5 (millions of young children)
0.5TV 1 0.2M 1 0.2SS $ 5 (millions of parents)
3. Fixed-requirement constraint:
40M 1 120SS 5 1,490 (coupon budget in $1,000s)
4. Nonnegativity constraints:
TV $ 0 M $ 0 SS $ 0
Solving the Model The lower left-hand corner of Figure 3.7 shows the entries needed in Solver, along with the selection of the usual two options. Solver then finds the optimal solution given in row 19. This optimal solution provides the following plan for the promotional campaign:
Run 3 television commercials. Run 14 advertisements in magazines. Run 7.75 advertisements in Sunday supplements (so the eighth advertisement would appear in only 75 percent of the newspapers).
Although the expected number of exposures with this plan is only 16,175,000, versus the 17,000,000 with the first plan shown in Figure 3.1 , both Claire Syverson and David Sloan feel that the new plan does a much better job of meeting all of management’s goals for this campaign. They decide to adopt the new plan.
This case study illustrates a common theme in real applications of linear programming— the continuing evolution of the linear programming model. It is common to make later adjust- ments in the initial version of the model, perhaps even many times, as experience is gained in using the model. Frequently, these adjustments are made to more adequately reflect some important managerial considerations. This may result in a mixed problem because the new functional constraints needed to incorporate the managerial considerations may be of a differ- ent type from those in the original model.
Other Examples This case study provides a relatively simple example of a small mixed problem. Most of the mixed problems that arise in practice are much larger, sometimes involving hundreds or thousands of activities and hundreds or thousands of constraints. At first glance, these larger problems may seem considerably more complicated than the case study. However, the impor- tant thing to remember is that any linear programming problem can have only three types of functional constraints—resource constraints, benefit constraints, and fixed-requirement constraints—where each type is formulated just as illustrated above for the case study.
There are numerous kinds of managerial problems to which linear programming can be applied. We don’t have nearly enough space available to give examples of all the most impor- tant kinds of applications. However, if you would like to explore this further, we suggest that you go through the five solved problems that are summarized in front of the Problems sec- tion for this chapter. Reading the seven cases that follow the Problems section, as well as the application vignettes in both this chapter and the preceding chapter, also will further illustrate the unusually wide applicability of linear programming.
Meanwhile, we soon will turn to two more categories of linear programming problems in the next two sections.
A model may need to be modified a number of times before it adequately incor- porates all the important considerations.
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Summary of the Formulation Procedure for Mixed Linear Programming Problems The procedure for formulating mixed problems is similar to the one outlined at the end of Section 3.2 for resource-allocation problems. However, pure resource-allocation prob- lems only have resource constraints whereas mixed problems can include all three types of functional constraints (resource constraints, benefit constraints, and fixed-requirement constraints). Therefore, the following summary for formulating mixed problems includes separate steps for dealing with these different types of constraints. Also see Figure 3.8 for a template of the format for a spreadsheet model of mixed problems. (This format works well for most mixed problems, including those encountered in this chapter, but more flexibility is occasionally needed, as will be illustrated in the next chapter.)
1. Since any linear programming problem involves finding the best mix of levels of various activities, identify these activities for the problem at hand. The decisions to be made are the levels of these activities.
2. From the viewpoint of management, identify an appropriate overall measure of perfor- mance for solutions of the problem.
3. For each activity, estimate the contribution per unit of the activity to this overall measure of performance.
4. Identify any resources that must be allocated to the activities (as described in Section 3.2). For each one, identify the amount available and then the amount used per unit of each activity.
5. Identify any benefits to be obtained from the activities (as described in Section 3.3). For each one, identify the minimum acceptable level prescribed by management and then the benefit contribution per unit of each activity.
6. Identify any fixed requirements that, for some type of quantity, the amount provided must equal a required amount (as described in Section 3.4). For each fixed requirement, identify the required amount and then the contribution toward this required amount per unit of each activity.
Activities
Unit Profit or Cost Profit/cost per unit of activity Resources
Used Resources Available
Resource used per unit of activity SUMPRODUCT
(resource used per unit,
changing cells)
Benefit Achieved
Benefit Needed
Benefit achieved per unit of activity
SUMPRODUCT
(benefit per unit,
changing cells)
Total Profit or Cost Level of Activity Changing cells
C on
st ra
in ts
SUMPRODUCT(profit/cost per unit, changing cells)
=
≤
≥
FIGURE 3.8 A template of a spreadsheet model for mixed problems.
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3.5 Transportation Problems 95
7. Enter the data gathered in steps 3–6 into data cells in a spreadsheet. 8. Designate changing cells for displaying the decisions on activity levels. 9. Use output cells to specify the constraints on resources, benefits, and fixed requirements.
10. Designate an objective cell for displaying the overall measure of performance.
1. What types of functional constraints can appear in a mixed linear programming problem? 2. What managerial goals needed to be incorporated into the expanded linear programming
model for the Super Grain Corp. problem? 3. Which categories of functional constraints are included in the new linear programming model? 4. Why did management adopt the new plan even though it provides a smaller expected number
of exposures than the original plan recommended by the original linear programming model?
Review Questions
3.5 TRANSPORTATION PROBLEMS
One of the most common applications of linear programming involves optimizing a shipping plan for transporting goods. In a typical application, a company has several plants producing a certain product that needs to be shipped to the company’s customers (or perhaps to distribu- tion centers). How much should each plant ship to each customer in order to minimize the total cost? Linear programming can provide the answer. This type of linear programming problem is called a transportation problem .
This kind of application normally needs two kinds of functional constraints. One kind specifies that the amount of the product produced at each plant must equal the total amount shipped to customers. The other kind specifies that the total amount received from the plants by each customer must equal the amount ordered. These are fixed-requirement constraints, which makes the problem a fixed-requirements problem. However, there also are variations of this problem where resource constraints or benefit constraints are needed.
Transportation problems and assignment problems (described in the next section) are such important types of linear programming problems that the entire Chapter 15 on the CD-ROM is devoted to further describing these two related types of problems and providing examples of a wide variety of applications.
We provide below an example of a typical transportation problem.
The Big M Company Transportation Problem The Big M Company produces a variety of heavy duty machines at two factories. One of its products is a large turret lathe. Orders have been received from three customers to pur- chase some of these turret lathes next month. These lathes will be shipped individually, and Table 3.9 shows what the cost will be for shipping each lathe from each factory to each cus- tomer. This table also shows how many lathes have been ordered by each customer and how many will be produced by each factory. The company’s distribution manager now wants to determine how many machines to ship from each factory to each customer to minimize the total shipping cost.
Figure 3.9 depicts the distribution network for this problem. This network ignores the geo- graphical layout of the factories and customers and instead lines up the two factories in one column on the left and the three customers in one column on the right. Each arrow shows one of the shipping lanes through this distribution network.
Shipping Cost for Each Lathe
To Customer 1 Customer 2 Customer 3 Output
From Factory 1 $700 $900 $800 12 lathes Factory 2 800 900 700 15 lathes
Order size 10 lathes 8 lathes 9 lathes
TABLE 3.9 Some Data for the Big M Company Distribution- Network Problem
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Formulation of the Problem in Linear Programming Terms We need to identify the activities and requirements of this transportation problem to for- mulate it as a linear programming problem. In this case, two kinds of activities have been mentioned—the production of the turret lathes at the two factories and the shipping of these lathes along the various shipping lanes. However, we know the specific amounts to be pro- duced at each factory, so no decisions need to be made about the production activities. The decisions to be made concern the levels of the shipping activities —how many lathes to ship through each shipping lane. Therefore, we need to focus on the shipping activities for the linear programming formulation.
The activities correspond to shipping lanes, depicted by arrows in Figure 3.9 . The level of each activity is the number of lathes shipped through the corresponding ship- ping lane.
Just as any linear programming problem can be described as finding the best mix of activ- ity levels, this one involves finding the best mix of shipping amounts for the various shipping lanes. The decisions to be made are
S F1-C1 5 Number of lathes shipped from Factory 1 to Customer 1
S F1-C2 5 Number of lathes shipped from Factory 1 to Customer 2
S F1-C3 5 Number of lathes shipped from Factory 1 to Customer 3
S F2-C1 5 Number of lathes shipped from Factory 2 to Customer 1
S F2-C2 5 Number of lathes shipped from Factory 2 to Customer 2
S F2-C3 5 Number of lathes shipped from Factory 2 to Customer 3
so six changing cells will be needed in the spreadsheet. The objective is to
Minimize Cost 5 Total cost for shipping the lathes
Using the shipping costs given in Table 3.9 ,
Cost 5 700SF1-C1 1 900SF1-C2 1 800SF1-C3 1 800SF1-C1 1 900SF2-C2 1 700SF2-C3
is the quantity in dollars to be entered into the objective cell. (We will use a SUMPRODUCT function to do this a little later.)
12 lathes
produced
15 lathes
produced
10 lathes
needed
8 lathes
needed
9 lathes
needed
F1
F2
C1
C2
C3
$70 0/l
ath e
$700/lathe
$900/lathe
$90 0/la
the
$800/lathe
$8 00
/la th
e
FIGURE 3.9 The distribution network for the Big M Company problem.
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The spreadsheet model also will need five constraints involving fixed requirements. Both Table 3.9 and Figure 3.9 show these requirements.
Requirement 1: Factory 1 must ship 12 lathes. Requirement 2: Factory 2 must ship 15 lathes. Requirement 3: Customer 1 must receive 10 lathes. Requirement 4: Customer 2 must receive 8 lathes. Requirement 5: Customer 3 must receive 9 lathes.
Thus, there is a specific requirement associated with each of the five locations in the distribu- tion network shown in Figure 3.9 .
All five of these requirements can be expressed in constraint form as
Amount provided 5 Required amount
For example, Requirement 1 can be expressed algebraically as
SF1-C1 1 SF1-C2 1 SF1-C3 5 12
where the left-hand side gives the total number of lathes shipped from Factory 1, and 12 is the required amount to be shipped from Factory 1. Therefore, this constraint restricts S F1-C1 , S F1-C2 , and S F1-C3 to values that sum to the required amount of 12. In contrast to the # form for resource constraints and the $ form for benefit constraints, the constraints express fixed requirements that must hold with equality, so this transportation problem falls into the category of fixed-requirements problems introduced in the preceding section. However, Chapter 15 (on the CD-ROM) gives several examples that illustrate how variants of transportation problems can have resource constraints or benefit constraints as well. For example, if 12 lathes represent the manufacturing capacity of Factory 1 (the maximum number that can be shipped) rather than a requirement for how many must be shipped, the constraint just given for Requirement 1 would become a # resource constraint instead. Such variations can be incorporated readily into the spreadsheet model.
Formulation of the Spreadsheet Model In preparation for formulating the model, the problem has been formulated above by iden- tifying the decisions to be made, the constraints on these decisions, and the overall mea- sure of performance, as well as gathering all the important data displayed in Table 3.9 . All this information leads to the spreadsheet model shown in Figure 3.10 . The data cells include ShippingCost (C5:E6), Output (H11:H12), and OrderSize (C15:E15), incorporating all the data from Table 3.9 . The changing cells are UnitsShipped (C11:E12), which give the deci- sions on the amounts to be shipped through the respective shipping lanes. The output cells are TotalShippedOut (F11:F12) and TotalToCustomer (C13:E13), where the SUM functions
Careful problem formula- tion needs to precede model formulation.
Procter & Gamble (P & G) makes and markets over 300 brands of consumer goods worldwide. The company has grown continuously over its long history tracing back to the 1830s. To maintain and accelerate that growth, a major management science study was undertaken to strengthen P & G’s global effectiveness. Prior to the study, the com- pany’s supply chain consisted of hundreds of suppliers, over 50 product categories, over 60 plants, 15 distribution cen- ters, and over 1,000 customer zones. However, as the com- pany moved toward global brands, management realized that it needed to consolidate plants to reduce manufactur- ing expenses, improve speed to market, and reduce capital investment. Therefore, the study focused on redesigning the company’s production and distribution system for its North American operations. The result was a reduction in
the number of North American plants by almost 20 percent, saving over $200 million in pretax costs per year.
A major part of the study revolved around formulating and solving transportation problems for individual prod- uct categories. For each option regarding keeping certain plants open, and so forth, solving the corresponding trans- portation problem for a product category showed what the distribution cost would be for shipping the product category from those plants to the distribution centers and customer zones.
Source: J. D. Camm, T. E. Chorman, F. A. Dill, J. R. Evans, D. J. Sweeney, and G. W. Wegryn, “Blending OR/MS, Judgment, and GIS: Restructuring P & G’s Supply Chain,” Interfaces 27, no. 1 (January–February 1997), pp. 128–142. (A link to this article is pro- vided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
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98 Chapter Three Linear Programming: Formulation and Applications
entered into these cells are shown below the spreadsheet in Figure 3.10 . The constraints are that TotalShippedOut is required to equal Output and TotalToCustomer is required to equal OrderSize. These constraints have been specified on the spreadsheet and entered into Solver. The objective cell is TotalCost (H15), where its SUMPRODUCT function gives the total shipping cost. The lower left-hand corner of Figure 3.10 shows the entries needed in Solver, along with the selection of the usual two options.
The layout of the spreadsheet is different than for all the prior linear programming exam- ples in the book. Rather than a separate column for each activity and a separate row for each constraint, the cost data and changing cells are laid out in a table format. This format provides a more natural and compact way of displaying the constraints and results.
UnitsShipped (C11:E12) in the spreadsheet in Figure 3.10 shows the result of applying Solver to obtain an optimal solution for the number of lathes to ship through each shipping lane. TotalCost (H15) indicates that the total shipping cost for this shipping plan is $20,500.
Since any transportation problem is a special type of linear programming problem, it makes the standard assumption that fractional solutions are allowed. However, we actually don’t want this assumption for this particular application since only integer numbers of lathes
Here is an example where SUM functions are used for output cells instead of SUMPRODUCT functions.
=SUMPRODUCT(ShippingCost,UnitsShipped)
1
A B C D FE G H
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Big M Company Distribution Problem
Customer 1 Customer 2
Factory 1
Factory 2
Factory 1
Factory 2
Customer 3
Customer 1
10 2
Customer 2
0
0 6 9
12
15
=
=
Customer 3
10 8 9
= = =
$20,500
Total Cost
Shipping Cost
(per Lathe)
Units Shipped
Total to Customer
Order Size
Output
Total
Shipped
Out
8
F
Total
Shipped9
14
15
10
11
Out
H
Total Cost
13
B C
Total to Customer =SUM(C11:C12)
D
=SUM(D11:D12)
E
=SUM(E11:E12)
=SUM(C11:E11)
12 =SUM(C12:E12)
Range Name
OrderSize
Output
ShippingCost
TotalCost
TotalShippedOut
TotalToCustomer
UnitsShipped
Cells
C15:E15
H11:H12
C5:E6
H15
F11:F12
C13:E13
C11:E12
$700 $900 $800
$800 $900 $700
12
15
10 8 9
Solver Parameters
Set Objective Cell: TotalCost To: Min By Changing Variable Cells: UnitsShipped
Subject to the Constraints: TotalShippedOut = Output
TotalToCustomer = OrderSize
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 3.10 The spreadsheet model for the Big M Company problem, including the formulas for the objective cell TotalCost (H15) and the other output cells TotalShippedOut (F11:F12) and TotalToCustomer (C13:E13), as well as the specifications needed to set up Solver. The changing cells UnitsShipped (C11:E12) show the optimal solution obtained by Solver.
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3.6 Assignment Problems 99
3.6 ASSIGNMENT PROBLEMS
We now turn to another special type of linear programming problem called assignment problems . As the name suggests, this kind of problem involves making assignments. Frequently, these are assignments of people to jobs. Thus, many applications of the assignment problem involve aiding managers in matching up their personnel with tasks to be performed. Other applications might instead involve assigning machines, vehicles, or plants to tasks.
Here is a typical example.
An Example: The Sellmore Company Problem The marketing manager of the Sellmore Company will be holding the company’s annual sales conference soon for sales regional managers and personnel. To assist in the administra- tion of the conference, he is hiring four temporary employees (Ann, Ian, Joan, and Sean), where each will handle one of the following four tasks:
1. Word processing of written presentations. 2. Computer graphics for both oral and written presentations. 3. Preparation of conference packets, including copying and organizing written materials. 4. Handling of advance and on-site registrations for the conference.
He now needs to decide which person to assign to each task.
can be shipped from a factory to a customer. Fortunately, even while making the standard assumption, the numbers in the optimal solution shown in UnitsShipped (C11:E12) only have integer values. This is no coincidence. Because of the form of its model, almost any trans- portation problem (including this one) is guaranteed in advance to have an optimal solution that has only integer values despite the fact that fractional solutions also are allowed. In par- ticular, as long as the data for the problem includes only integer values for all the supplies and demands (which are the outputs and order sizes in the Big M Company problem), any transportation problem with feasible solutions is guaranteed to have an optimal solution with integer values for all its decision variables. Therefore, it is not necessary to add constraints to the model that require these variables to have only integer values.
To summarize, here is the algebraic form of the linear programming model that has been formulated in the spreadsheet:
Minimize Cost 5 700SF1-C1 1 900SF1-C2 1 800SF1-C3 1 800SF2-C1
1 900SF2-C2 1 700SF2-C3
subject to the following constraints:
1. Fixed-requirement constraints:
SF1-C1 1 SF1-C2 1 SF1-C3 5 12 (Factory 1) SF2-C1 1 SF2-C2 1 SF2-C3 5 15 (Factory 2)
SF1-C1 1 SF2-C1 5 10 (Customer 1) SF1-C2 1 SF2-C2 5 8 (Customer 2)
SF1-C3 1 SF2-C3 5 9 (Customer 3)
2. Nonnegativity constraints:
SF1-C1 $ 0 SF1-C2 $ 0 SF1-C3 $ 0 SF2-C1 $ 0 SF2-C2 $ 0 SF2-C3 $ 0
1. Why are transportation problems given this name? 2. What is an identifying feature of transportation problems? 3. How does the form of a fixed-requirement constraint differ from that of a resource constraint?
A benefit constraint? 4. What are the quantities with fixed requirements in the Big M Company problem?
Review Questions
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100 Chapter Three Linear Programming: Formulation and Applications
Although each temporary employee has at least the minimal background necessary to per- form any of the four tasks, they differ considerably in how efficiently they can handle the dif- ferent types of work. Table 3.10 shows how many hours each would need for each task. The rightmost column gives the hourly wage based on the background of each employee.
Formulation of a Spreadsheet Model Figure 3.11 shows a spreadsheet model for this problem. Table 3.10 is entered at the top. Com- bining these required times and wages gives the cost (cells D15:G18) for each possible assign- ment of a temporary employee to a task, using equations shown at the bottom of Figure 3.11 . This cost table is just the way that any assignment problem is displayed. The objective is to determine which assignments should be made to minimize the sum of the associated costs.
The values of 1 in Supply (J24:J27) indicate that each person (assignee) listed in column C must perform exactly one task. The values of 1 in Demand (D30:G30) indicate that each task must be performed by exactly one person. These requirements then are specified in the constraints given in Solver.
Each of the changing cells Assignment (D24:G27) is given a value of 1 when the corre- sponding assignment is being made, and a value of 0 otherwise. Therefore, the Excel equation for the objective cell, TotalCost 5 SUMPRODUCT(Cost, Assignment), gives the total cost for the assignments being made. The Solver Parameters box specifies that the goal is to mini- mize this objective cell.
The changing cells in Figure 3.11 show the optimal solution obtained after running Solver. This solution is
Assign Ann to prepare conference packets. Assign Ian to do the computer graphics. Assign Joan to handle registrations. Assign Sean to do the word processing.
The total cost given in cell J30 is $1,957.
Characteristics of Assignment Problems Note that all the functional constraints of the Sellmore Co. problem (as shown in cells H24:J27 and D28:G30 of Figure 3.11 ) are fixed-requirement constraints which require each person to perform exactly one task and require each task to be performed by exactly one person. Thus, like the Big M Company transportation problem, the Sellmore Co. is a fixed-requirements problem. This is a characteristic of all pure assignment problems. However, Chapter 15 (on the CD-ROM) gives some examples of variants of assignment problems where this is not the case.
Like the changing cells Assignment (D24:G27) in Figure 3.11 , the changing cells in the spreadsheet model for any pure assignment problem gives a value of 1 when the corre- sponding assignment is being made, and a value of 0 otherwise. Since the fixed-requirement constraints require only each row or column of changing cells to add up to 1 (which could happen, e.g., if two of the changing cells in the same row or column had a value of 0.5 and the rest 0), this would seem to necessitate adding the constraints that each of the changing cells must be integer. After choosing the Solver option to make the changing cells nonnegative, this then would force each of the changing cells to be 0 or 1. However, it turned out to be unnecessary to add the constraints that require the changing cells to have values of 0 or 1 in
Decisions need to be made regarding which person to assign to each task.
Using Cost (D15:G18), the objective is to mini- mize the total cost of the assignments.
A value of 1 in a chang- ing cell indicates that the corresponding assignment is being made, whereas 0 means that the assignment is not being made.
Excel Tip: When solving an assignment problem, rounding errors occasion- ally will cause Excel to return a noninteger value very close to 0 (e.g., 1.23 E-10, meaning 0.000000000123) or very close to 1 (e.g., 0.9999912). To make the spreadsheet cleaner, you may replace these “ugly” representations by their proper value of 0 or 1 in the changing cells.
Required Time per Task (Hours)
Temporary Employee
Word Processing Graphics Packets Registrations
Hourly Wage
Ann 35 41 27 40 $14 Ian 47 45 32 51 12 Joan 39 56 36 43 13 Sean 32 51 25 46 15
TABLE 3.10 Data for the Sellmore Co. Problem
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3.6 Assignment Problems 101
13
B DC E F G
Cost
=D6*I6
14
=D7*I7
15
Assignee
=D8*I8
16
17
18 =D9*I9
=E6*I6
=E7*I7
=E8*I8
=E9*I9
=G6*I6
=G7*I7
=G8*I8
=G9*I9
=F6*I6
Word
Processing Graphics RegistrationsPackets
=F7*I7
=F8*I8
=F9*I9
Task
Task
A B C D E F G H I J
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Sellmore Co. Assignment Problem
Required Time
(Hours)
Cost
Word
Processing
Ann
Ian
Joan
Assignee
35
47
39
Graphics
41
45
56
Packets
27
32
36
40
51
43
Sean 32 51 25 46
Registrations
Assignment Word
Processing
Ann
Ian
Joan
Assignee
0
0
0
Graphics
0
1
0
Packets
1
0
0
0
0
1
Sean
Ann
Ian
Joan
Sean
1 0 0 0
Registrations
$14
$12
$13
$15
Hourly
Wage
Total Assigned
Total
Assignments
1
1
1
1
Supply
1
1
1
1
$1,957
=
=
=
=
1 1 1 1
= = = = Total Cost
Demand 1 1 1 1
Range Name
Assignment
Cost
Demand
HourlyWage
RequiredTime
Supply
TotalAssigned
TotalAssignments
TotalCost
Cells
D24:G27
D15:G18
D30:G30
I6:I9
D6:G9
J24:J27
D28:G28
H24:H27
J30
22
H
23
25
24
Total
Assignments
=SUM(D25:G25)
26 =SUM(D26:G26)
27 =SUM(D27:G27)
=SUM(D24:G24)
29
J
30
Total Cost
=SUMPRODUCT(Cost,Assignment)
28
C D E F G
Total Assigned =SUM(D24:D27) =SUM(F24:F27) =SUM(G24:G27)=SUM(E24:E27)
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Task
Word
Processing
Ann
Ian
Joan
Assignee
$490
$564
$507
Graphics
$574
$540
$728
Packets
$378
$384
$468
$560
$612
$559
Sean $480 $765 $375 $690
Registrations
Solver Parameters
Set Objective Cell: TotalCost To: Min By Changing Variable Cells: Assignment
Subject to the Constraints: TotalAssigned = Demand
TotalAssignments = Supply
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 3.11 A spreadsheet formulation of the Sellmore Co. problem as an assignment problem, including the objective cell TotalCost (J30) and the other output cells Cost (D15:G18), TotalAssignments (H24:H27), and TotalAssigned (D28:G28), as well as the specifications needed to set up the model. The values of 1 in the changing cells Assignment (D24:G27) show the optimal plan obtained by Solver for assigning the people to the tasks.
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102 Chapter Three Linear Programming: Formulation and Applications
Figure 3.11 because Solver gave an optimal solution that had only values of 0 or 1 anyway. In fact, a general characteristic of pure assignment problems is that Solver always provides such an optimal solution without needing to add these additional constraints.
As described further in Chapter 15, another interesting characteristic of any pure assign- ment problem is that it can be viewed as a special type of pure transportation problem. In particular, every fixed-requirement constraint in the corresponding transportation problem would require that either a row or column of changing cells add up to 1. This would result in Solver giving an optimal solution where every changing cell has a value of either 0 or 1, just as for the original assignment problem.
1. Why are assignment problems given this name? 2. Pure assignment problems have what type of functional constraints? 3. What is the interpretation of the changing cells in the spreadsheet model of a pure assignment
problem?
Review Questions
3.7 MODEL FORMULATION FROM A BROADER PERSPECTIVE
Formulating and analyzing a linear programming model provides information to help manag- ers make their decisions. That means the model must accurately reflect the managerial view of the problem:
• The overall measure of performance must capture what management wants accomplished. • When management limits the amounts of resources that will be made available to the activ-
ities under consideration, these limitations should be expressed as resource constraints. • When management establishes minimum acceptable levels for benefits to be gained from
the activities, these managerial goals should be incorporated into the model as benefit constraints.
• If management has fixed requirements for certain quantities, then fixed-requirement con- straints are needed.
With the help of spreadsheets, some managers now are able to formulate and solve small linear programming models themselves. However, larger linear programming models may be formulated by management science teams, not managers. When this is done, the management science team must thoroughly understand the managerial view of the problem. This requires clear communication with management from the very beginning of the study and maintaining effective communication as new issues requiring managerial guidance are identified. Man- agement needs to clearly convey its view of the problem and the important issues involved. A manager cannot expect to obtain a helpful linear programming study without making clear just what help is wanted.
As is necessary in any textbook, the examples in this chapter are far smaller, simpler, and more clearly spelled out than is typical of real applications. Many real studies require formu- lating complicated linear programming models involving hundreds or thousands of decisions and constraints. In these cases, there usually are many ambiguities about just what should be incorporated into the model. Strong managerial input and support are vital to the success of a linear programming study for such complex problems.
When dealing with huge real problems, there is no such thing as “the” correct linear pro- gramming model for the problem. The model continually evolves throughout the course of the study. Early in the study, various techniques are used to test initial versions of the model to identify the errors and omissions that inevitably occur when constructing such a large model. This testing process is referred to as model validation .
Once the basic formulation has been validated, there are many reasonable variations of the model that might be used. Which variation to use depends on such factors as the assumptions about the problem that seem most reasonable, the estimates of the parameters of the model that seem most reliable, and the degree of detail desired in the model.
In large linear programming studies, a good approach is to begin with a relatively simple version of the model and then use the experience gained with this model to evolve toward
Both the measure of perfor- mance and the constraints in a model need to reflect the managerial view of the problem.
Linear programming stud- ies need strong managerial input and support.
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more elaborate models that more nearly reflect the complexity of the real problem. This pro- cess of model enrichment continues only as long as the model remains reasonably easy to solve. It must be curtailed when the study’s results are needed by management. Managers often need to curb the natural instinct of management science teams to continue adding “bells and whistles” to the model rather than winding up the study in a timely fashion with a less elegant but adequate model.
When managers study the output of the current model, they often detect some undesirable characteristics that point toward needed model enrichments. These enrichments frequently take the form of new benefit constraints to satisfy some managerial goals not previously articulated. (Recall that this is what happened in the Super Grain case study.)
Even though many reasonable variations of the model could be used, an optimal solution can be solved for only with respect to one specific version of the model at a time. This is why what- if analysis is such an important part of a linear programming study. After obtaining an optimal solution with respect to one specific model, management will have many what-if questions:
• What if the estimates of the parameters in the model are incorrect? • How do the conclusions change if different plausible assumptions are made about the
problem? • What happens when certain managerial options are pursued that are not incorporated into
the current model?
Chapter 5 is devoted primarily to describing how what-if analysis addresses these and related issues, as well as how managers use this information.
Because managers instigate management science studies, they need to know enough about linear programming models and their formulation to be able to recognize managerial prob- lems to which linear programming can be applied. Furthermore, since managerial input is so important for linear programming studies, managers need to understand the kinds of manage- rial concerns that can be incorporated into the model. Developing these two skills have been the most important goals of this chapter.
What-if analysis addresses some key questions that remain after formulating and solving a model.
1. A linear programming model needs to reflect accurately whose view of the problem? 2. What is meant by model validation? 3. What is meant by the process of model enrichment? 4. Why is what-if analysis an important part of a linear programming study?
Review Questions
Functional constraints with a # sign are called resource constraints, because they require that the amount used of some resource must be less than or equal to the amount available of that resource. The identifying feature of resource-allocation problems is that all their functional constraints are resource constraints.
Functional constraints with a $ sign are called benefit constraints, since their form is that the level achieved for some benefit must be greater than or equal to the minimum acceptable level for that benefit. Frequently, benefit constraints express goals prescribed by management. If every functional constraint is a benefit constraint, then the problem is a cost–benefit–trade-off problem.
Functional constraints with an 5 sign are called fixed-requirement constraints, because they express the fixed requirement that, for some quantity, the amount provided must be equal to the required amount. The identifying feature of fixed-requirements problems is that their functional constraints are fixed- requirement constraints. One prominent type of fixed-requirements problem is transportation problems, which typically involve finding a shipping plan that minimizes the total cost of transporting a product from a number of plants to a number of customers. Another prominent type is assignment problems, which typi- cally involves assigning people to tasks so as to minimize the total cost of performing these tasks.
Linear programming problems that do not fit into any of these three categories are called mixed problems.
In many real applications, management science teams formulate and analyze large linear program- ming models to help guide managerial decision making. Such teams need strong managerial input and support to help ensure that their work really meets management’s needs.
3.8 Summary
3.8 Summary 103
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104 Chapter Three Linear Programming: Formulation and Applications
Glossary assignment problem A type of linear program- ming problem that typically involves assigning people to tasks so as to minimize the total cost of performing these tasks. (Section 3.6), 99 benefit constraint A functional constraint with a $ sign. The left-hand side is interpreted as the level of some benefit that is achieved by the activities under consideration, and the right-hand side is the minimum acceptable level for that ben- efit. (Section 3.3), 82 cost–benefit–trade-off problem A type of linear programming problem involving the trade-off between the total cost of the activi- ties under consideration and the benefits to be achieved by these activities. Its identifying fea- ture is that each functional constraint in the lin- ear programming model is a benefit constraint. (Section 3.3), 82 fixed-requirement constraint A functional constraint with an 5 sign. The left-hand side rep- resents the amount provided of some type of quan- tity, and the right-hand side represents the required amount for that quantity. (Section 3.4), 88 fixed-requirements problem A type of linear programming problem concerned with optimiz- ing how to meet a number of fixed requirements. Its identifying feature is that each functional constraint in its model is a fixed-requirement constraint. (Section 3.4), 88 identifying feature A feature of a model that identifies the category of linear programming problem it represents. (Chapter introduction), 64
integer programming problem A variation of a linear programming problem that has the additional restriction that some or all of the decision variables must have integer values. (Section 3.2), 75 mixed problem Any linear programming prob- lem that includes at least two of the three types of functional constraints (resource constraints, benefit constraints, and fixed-requirement con- straints). (Section 3.4), 88 model enrichment The process of using expe- rience with a model to identify and add important details that will provide a better representation of the real problem. (Section 3.7), 103 model validation The process of checking and testing a model to develop a valid model. (Section 3.7), 102 resource-allocation problem A type of linear programming problem concerned with allocating resources to activities. Its identifying feature is that each functional constraint in its model is a resource constraint. (Section 3.2), 71 resource constraint A functional constraint with a # sign. The left-hand side represents the amount of some resource that is used by the activities under consideration, and the right- hand side represents the amount available of that resource. (Section 3.2), 71 transportation problem A type of linear pro- gramming problem that typically involves finding a shipping plan that minimizes the total cost of transporting a product from a number of plants to a number of customers. (Section 3.5), 95
Chapter 3 Excel Files:
Super Grain Example
TBA Airlines Example
Think-Big Example
Union Airways Example
Big M Example Revised Super Grain Example Sellmore Example
Excel Add-in:
Risk Solver Platform for Education (RSPE)
Learning Aids for This Chapter in Your MS Courseware
Solved Problems (See the CD-ROM or Website for the Solutions) 3.S1. Farm Management Dwight and Hattie have run the family farm for over 30 years. They are currently planning the mix of crops to plant on their 120-acre farm for the upcoming season. The table gives the labor-hours and fertilizer required per acre, as well as the total expected profit per acre for each of the potential crops under consideration. Dwight, Hattie, and their children can work at most 6,500 total hours during the upcoming season. They have 200 tons of fertilizer available. What mix of crops should be planted to maximize the family’s total profit? a. Formulate and solve a linear programming model for this
problem in a spreadsheet. b. Formulate this same model algebraically.
Crop
Labor Required
(hours per acre)
Fertilizer Required (tons per
acre)
Expected Profit (per
acre)
Oats 50 1.5 $500 Wheat 60 2 $600 Corn 105 4 $950
3.S2. Diet Problem The kitchen manager for Sing Sing prison is trying to decide what to feed its prisoners. She would like to offer some com- bination of milk, beans, and oranges. The goal is to minimize
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Chapter 3 Problems 105
cost, subject to meeting the minimum nutritional requirements imposed by law. The cost and nutritional content of each food, along with the minimum nutritional requirements, are shown below. What diet should be fed to each prisoner?
a. Formulate and solve a linear programming model for this problem in a spreadsheet.
b. Formulate this same model algebraically.
Milk (gallons)
Navy Beans (cups)
Oranges (large Calif.
Valencia)
Minimum Daily
Requirement
Niacin (mg) 3.2 4.9 0.8 13.0 Thiamin (mg) 1.12 1.3 0.19 1.5 Vitamin C (mg) 32.0 0.0 93.0 45.0 Cost ($) 2.00 0.20 0.25
3.S3. Cutting Stock Problem Decora Accessories manufactures a variety of bathroom acces- sories, including decorative towel rods and shower curtain rods. Each of the accessories includes a rod made out of stainless steel. However, many different lengths are needed: 12, 18, 24, 40, and 60 inches. Decora purchases 60-inch rods from an out- side supplier and then cuts the rods as needed for their products. Each 60-inch rod can be used to make a number of smaller rods. For example, a 60-inch rod could be used to make a 40-inch and an 18-inch rod (with 2 inches of waste), or five 12-inch rods (with no waste). For the next production period, Decora needs twenty-five 12-inch rods, fifty-two 18-inch rods, forty- five 24-inch rods, thirty 40-inch rods, and twelve 60-inch rods. What is the fewest number of 60-inch rods that can be purchased to meet their production needs? Formulate and solve an integer programming model in a spreadsheet.
3.S4. Producing and Distributing AEDs at Heart Start Heart Start produces automated external defibrillators in each of two different plants (A and B). The unit production costs and monthly production capacity of the two plants are indicated in the table below. The automated external defibrillators are sold
through three wholesalers. The shipping cost from each plant to the warehouse of each wholesaler along with the monthly demand from each wholesaler are also indicated in the table. The management of Heart Start now has asked their top man- agement scientist (you) to address the following two questions. How many automated external defibrillators should be produced in each plant, and how should they be distributed to each of the three wholesaler warehouses so as to minimize the combined cost of production and shipping? Formulate and solve a linear programming model in a spreadsheet.
3.S5. Bidding for Classes In the MBA program at a prestigious university in the Pacific Northwest, students bid for electives in the second year of their program. Each student has 100 points to bid (total) and must take two electives. There are four electives available: Manage- ment Science (MS), Finance (Fin), Operations Management (OM), and Marketing (Mkt). Each class is limited to 5 students. The bids submitted for each of the 10 students are shown in the table below.
Student Bids for Classes
Student MS Fin OM Mkt
George 60 10 10 20 Fred 20 20 40 20 Ann 45 45 5 5 Eric 50 20 5 25 Susan 30 30 30 10 Liz 50 50 0 0 Ed 70 20 10 0 David 25 25 35 15 Tony 35 15 35 15 Jennifer 60 10 10 20
a. Formulate and solve a spreadsheet model to determine an as- signment of students to classes so as to maximize the total bid points of the assignments.
b. Does the resulting solution seem like a fair assignment? c. Which alternative objectives might lead to a fairer
assignment?
Unit Shipping Cost Unit
Production Cost
Monthly Production Capacity
Warehouse 1
Warehouse 2
Warehouse 3
Plant A $22 $14 $30 $600 100 Plant B $16 $20 $24 $625 120
Monthly Demand 80 60 70
3.1. Reconsider the Super Grain Corp. case study as presented in Section 3.1. The advertising firm, Giacomi & Jackowitz, now has suggested a fourth promising advertising medium—radio commercials—to promote the company’s new breakfast cereal, Crunchy Start. Young children are potentially major consumers of this cereal, but parents of young children (the major potential
We have inserted the symbol E* to the left of each problem (or its parts) where Excel should be used (unless your instructor gives you contrary instructions). Either the Excel’s Solver or RSPE may be used to solve such problems. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
Problems
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purchasers) often are too busy to do much reading (so may miss the company’s advertisements in magazines and Sunday supple- ments) or even to watch the Saturday morning programs for children where the company’s television commercials are aired. However, these parents do tend to listen to the radio during the commute to and from work. Therefore, to better reach these parents, Giacomi & Jackowitz suggests giving consideration to running commercials for Crunchy Start on nationally syndicated radio programs that appeal to young adults during typical com- muting hours.
Giacomi & Jackowitz estimates that the cost of develop- ing each new radio commercial would be $50,000, and that the expected number of exposures per commercial would be 900,000. The firm has determined that 10 spots are available for different radio commercials, and each one would cost $200,000 for a normal run.
E* a. Formulate and solve a spreadsheet model for the revised advertising-mix problem that includes this fourth advertising medium. Identify the data cells, the changing cells, and the objective cell. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function.
b. Indicate why this spreadsheet model is a linear pro- gramming model.
c. Express this model in algebraic form.
3.2 Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 3.2. Briefly describe how linear program- ming was applied in this study. Then list the various benefits that resulted from this study. 3.3.* Consider a resource-allocation problem having the fol- lowing data.
Resource Usage per Unit of Each
Activity
Resource 1 2 Amount of
Resource Available
1 2 1 10 2 3 3 20 3 2 4 20
Contribution per unit
$20 $30
Contribution per unit 5 profit per unit of the activity.
E* a. Formulate a linear programming model for this problem on a spreadsheet.
E* b. Use the spreadsheet to check the following solu- tions: ( x 1 , x 2 ) 5 (2, 2), (3, 3), (2, 4), (4, 2), (3, 4), (4, 3). Which of these solutions are feasible? Which of these feasible solutions has the best value of the objective function?
E* c. Use Solver to find an optimal solution. d. Express this model in algebraic form. e. Use the graphical method to solve this model. 3.4. Consider a resource-allocation problem having the fol- lowing data.
Resource Usage per Unit of Each
Activity
Resource 1 2 3
Amount of Resource Available
A 30 20 0 500 B 0 10 40 600 C 20 20 30 1,000
Contribution per unit $50 $40 $70
Contribution per unit 5 profit per unit of the activity.
E* a. Formulate and solve a linear programming model for this problem on a spreadsheet.
b. Express this model in algebraic form.
E*3.5. Consider a resource-allocation problem having the fol- lowing data.
Resource Usage per Unit of Each Activity
Resource 1 2 3 4
Amount of Resource Available
P 3 5 22 4 400 Q 4 21 3 2 300 R 6 3 2 21 400 S 22 2 5 3 300
Contribution per unit
$11 $9 $8 $9
Contribution per unit 5 profit per unit of the activity.
a. Formulate a linear programming model for this problem on a spreadsheet.
b. Make five guesses of your own choosing for the optimal solution. Use the spreadsheet to check each one for feasibility and, if feasible, for the value of the objective function. Which feasible guess has the best objective function value?
c. Use Solver to find an optimal solution.
3.6.* The Omega Manufacturing Company has discontinued the production of a certain unprofitable product line. This act created considerable excess production capacity. Management is considering devoting this excess capacity to one or more of three products, Products 1, 2, and 3. The available capacity of the machines that might limit output is summarized in the fol- lowing table.
Machine Type Available Time
(in Machine-Hours per Week)
Milling machine 500 Lathe 350 Grinder 150
The number of machine-hours required for each unit of the respective products are shown in the next table.
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Productivity Coefficient (in Machine-Hours per Unit)
Machine Type Product 1 Product 2 Product 3
Milling machine 9 3 5 Lathe 5 4 0 Grinder 3 0 2
The Sales Department indicates that the sales potential for Products 1 and 2 exceeds the maximum production rate and that the sales potential for product 3 is 20 units per week. The unit profit would be $50, $20, and $25, respectively, for Products 1, 2, and 3. The objective is to determine how much of each prod- uct Omega should produce to maximize profit. a. Indicate why this is a resource-allocation problem
by identifying both the activities and the limited resources to be allocated to these activities.
b. Identify verbally the decisions to be made, the con- straints on these decisions, and the overall measure of performance for the decisions.
c. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions.
E* d. Formulate a spreadsheet model for this problem. Iden- tify the data cells, the changing cells, the objective cell, and the other output cells. Also show the Excel equa- tion for each output cell expressed as a SUMPROD- UCT function. Then use Solver to solve the model.
e. Summarize the model in algebraic form. 3.7. Ed Butler is the production manager for the Bilco Cor- poration, which produces three types of spare parts for automo- biles. The manufacture of each part requires processing on each of two machines, with the following processing times (in hours).
Part
Machine A B C
1 0.02 0.03 0.05 2 0.05 0.02 0.04
Each machine is available 40 hours per month. Each part manufactured will yield a unit profit as follows:
Part
A B C
Profit $50 $40 $30
Ed wants to determine the mix of spare parts to produce to maximize total profit. a. Identify both the activities and the resources for this
resource-allocation problem. E* b. Formulate a linear programming model for this
problem on a spreadsheet. E* c. Make three guesses of your own choosing for the
optimal solution. Use the spreadsheet to check each one for feasibility and, if feasible, for the value of the objective function. Which feasible guess has the best objective function value?
E* d. Use Solver to find an optimal solution. e. Express the model in algebraic form. E*3.8. Consider the following algebraic formulation of a resource-allocation problem with three resources, where the decisions to be made are the levels of three activities ( A 1 , A 2 , and A 3 ).
Maximize Profit 5 20A1 1 40A2 1 30A3
subject to Resource 1: 3 A 1 1 5 A 2 1 4 A 3 # 400 (amount available) Resource 2: A 1 1 A 2 1 A 3 # 100 (amount available) Resource 3: A 1 1 3 A 2 1 2 A 3 # 200 (amount available)
and
A1 $ 0 A2 $ 0 A3 $ 0
Formulate and solve the spreadsheet model for this problem. 3.9. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 3.3. Briefly describe how linear program- ming was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 3.10. Consider a cost–benefit–trade-off problem having the following data.
Benefit Contribution per Unit of Each
Activity
Benefit 1 2
Minimum Acceptable
Level
1 5 3 60 2 2 2 30 3 7 9 126
Unit cost $60 $50
E* a. Formulate a linear programming model for this problem on a spreadsheet.
E* b. Use the spreadsheet to check the following solu- tions: ( x 1 , x 2 ) 5 (7, 7), (7, 8), (8, 7), (8, 8), (8, 9), (9, 8). Which of these solutions are feasible? Which of these feasible solutions has the best value of the objective function?
E* c. Use Solver to find an optimal solution. d. Express the model in algebraic form. e. Use the graphical method to solve this model. E*3.11. Consider a cost–benefit–trade-off problem having the following data.
Benefit Contribution per Unit of Each Activity
Benefit 1 2 3 4
Minimum Acceptable
Level
P 2 21 4 3 80 Q 1 4 21 2 60 R 3 5 4 21 110
Unit cost $400 $600 $500 $300
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a. Formulate a linear programming model for this problem on a spreadsheet.
b. Make five guesses of your own choosing for the optimal solution. Use the spreadsheet to check each one for feasibility and, if feasible, for the value of the objective function. Which feasible guess has the best objective function value?
c. Use Solver to find an optimal solution. 3.12.* Fred Jonasson manages a family-owned farm. To sup- plement several food products grown on the farm, Fred also raises pigs for market. He now wishes to determine the quantities of the available types of feed (corn, tankage, and alfalfa) that should be given to each pig. Since pigs will eat any mix of these feed types, the objective is to determine which mix will meet cer- tain nutritional requirements at a minimum cost. The number of units of each type of basic nutritional ingredient contained within a kilogram of each feed type is given in the following table, along with the daily nutritional requirements and feed costs.
Nutritional Ingredient
Kilogram of Corn
Kilogram of Tankage
Kilogram of Alfalfa
Minimum Daily
Requirement
Carbohydrates 90 20 40 200 Protein 30 80 60 180 Vitamins 10 20 60 150
Cost (¢) 84 72 60
E* a. Formulate a linear programming model for this problem on a spreadsheet.
E* b. Use the spreadsheet to check if ( x 1 , x 2 , x 3 ) 5 (1, 2, 2) is a feasible solution and, if so, what the daily cost would be for this diet. How many units of each nutritional ingredient would this diet provide daily?
E* c. Take a few minutes to use a trial-and-error approach with the spreadsheet to develop your best guess for the optimal solution. What is the daily cost for your solution?
E* d. Use Solver to find an optimal solution. e. Express the model in algebraic form. 3.13. Maureen Laird is the chief financial officer for the Alva Electric Co., a major public utility in the Midwest. The com- pany has scheduled the construction of new hydroelectric plants 5, 10, and 20 years from now to meet the needs of the growing population in the region served by the company. To cover the construction costs, Maureen needs to invest some of the compa- ny’s money now to meet these future cash flow needs. Maureen may purchase only three kinds of financial assets, each of which costs $1 million per unit. Fractional units may be purchased. The assets produce income 5, 10, and 20 years from now, and that income is needed to cover minimum cash flow requirements in those years, as shown in the following table.
Income per Unit of Asset
Year Asset 1 Asset 2 Asset 3
Minimum Cash Flow Required
5 $2 million $1 million $0.5 million $400 million 10 0.5 million 0.5 million 1 million 100 million 20 0 1.5 million 2 million 300 million
Maureen wishes to determine the mix of investments in these assets that will cover the cash flow requirements while minimiz- ing the total amount invested. E* a. Formulate a linear programming model for this
problem on a spreadsheet. E* b. Use the spreadsheet to check the possibility of
purchasing 100 units of asset 1, 100 units of asset 2, and 200 units of asset 3. How much cash flow would this mix of investments generate 5, 10, and 20 years from now? What would be the total amount invested?
E* c. Take a few minutes to use a trial-and-error approach with the spreadsheet to develop your best guess for the optimal solution. What is the total amount invested for your solution?
E* d. Use Solver to find an optimal solution. e. Summarize the model in algebraic form.
3.14. Web Mercantile sells many household products through an online catalog. The company needs substantial warehouse space for storing its goods. Plans now are being made for leas- ing warehouse storage space over the next five months. Just how much space will be required in each of these months is known. However, since these space requirements are quite different, it may be most economical to lease only the amount needed each month on a month-by-month basis. On the other hand, the addi- tional cost for leasing space for additional months is much less than for the first month, so it may be less expensive to lease the maximum amount needed for the entire five months. Another option is the intermediate approach of changing the total amount of space leased (by adding a new lease and/or having an old lease expire) at least once but not every month.
The space requirement and the leasing costs for the various leasing periods are as follows.
Month Required Space (Square Feet)
1 30,000 2 20,000 3 40,000 4 10,000 5 50,000
Leasing Period (Months)
Cost per Sq. Ft. Leased
1 $ 65 2 100 3 135 4 160 5 190
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Chapter 3 Problems 109
The objective is to minimize the total leasing cost for meet- ing the space requirements.
a. Indicate why this is a cost–benefit–trade-off prob- lem by identifying both the activities and the ben- efits being sought from these activities.
b. Identify verbally the decisions to be made, the con- straints on these decisions, and the overall measure of performance for the decisions.
c. Convert these verbal descriptions of the con- straints and the measure of performance into quantitative expressions in terms of the data and decisions.
E* d. Formulate a spreadsheet model for this problem. Identify the data cells, the changing cells, the objec- tive cell, and the other output cells. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function. Then use Solver to solve the model.
e. Summarize the model in algebraic form.
E*3.15. Consider the following algebraic formulation of a cost– benefit–trade-off problem involving three benefits, where the decisions to be made are the levels of four activities ( A 1 , A 2 , A 3 , and A 4 ):
Minimize Cost 5 2 A1 1 A2 2 A3 1 3 A4
subject to Benefit 1: 3 A 1 1 2 A 2 2 2 A 3 1 5 A 4 $ 80 (minimum
acceptable level)
Benefit 2: A 1 2 A 2 1 A 4 $ 10 (minimum acceptable
level)
Benefit 3: A 1 1 A 2 2 A 3 1 2 A 4 $ 30 (minimum acceptable
level)
and
A1 $ 0 A2 $ 0 A3 $ 0 A4 $ 0
Formulate and solve the spreadsheet model for this problem.
Unit Shipping Cost
To From Customer 1 Customer 2 Customer 3 Output
Factory 1 $600 $800 $700 400 units Factory 2 400 900 600 500 units
Order size 300 units 200 units 400 units
3.16. Larry Edison is the director of the Computer Center for Buckly College. He now needs to schedule the staffing of the center. It is open from 8 am until midnight. Larry has monitored the usage of the center at various times of the day and deter- mined that the following number of computer consultants are required.
3.18. The Fagersta Steelworks currently is working two mines to obtain its iron ore. This iron ore is shipped to either of two storage facilities. When needed, it then is shipped on to the company’s steel plant. The diagram below depicts this dis- tribution network, where M1 and M2 are the two mines, S1 and S2 are the two storage facilities, and P is the steel plant. The
Time of Day
Minimum Number of Consultants Required
to Be on Duty
8 AM–noon 6 Noon–4 PM 8 4 PM–8 PM 12 8 PM–midnight 6
Two types of computer consultants can be hired: full-time and part-time. The full-time consultants work for eight consecu- tive hours in any of the following shifts: morning (8 am –4 pm ), afternoon (noon–8 pm ), and evening (4 pm –midnight). Full-time consultants are paid $14 per hour.
Part-time consultants can be hired to work any of the four shifts listed in the table. Part-time consultants are paid $12 per hour.
An additional requirement is that during every time period, there must be at least two full-time consultants on duty for every part-time consultant on duty.
Larry would like to determine how many full-time and part-time consultants should work each shift to meet the above requirements at the minimum possible cost. a. Which category of linear programming problem
does this problem fit? Why? E* b. Formulate and solve a linear programming model
for this problem on a spreadsheet. c. Summarize the model in algebraic form. 3.17.* The Medequip Company produces precision medical diagnostic equipment at two factories. Three medical centers have placed orders for this month’s production output. The fol- lowing table shows what the cost would be for shipping each unit from each factory to each of these customers. Also shown are the number of units that will be produced at each factory and the number of units ordered by each customer.
A decision now needs to be made about the shipping plan for how many units to ship from each factory to each customer. a. Which category of linear programming problem
does this problem fit? Why? E* b. Formulate and solve a linear programming model
for this problem on a spreadsheet. c. Summarize this formulation in algebraic form.
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3.20. The Metalco Company desires to blend a new alloy of 40 percent tin, 35 percent zinc, and 25 percent lead from several available alloys having the following properties.
Alloy
Property 1 2 3 4 5
Percentage of tin 60 25 45 20 50 Percentage of zinc 10 15 45 50 40 Percentage of lead 30 60 10 30 10 Cost ($/lb) 22 20 25 24 27
The objective is to determine the proportions of these alloys that should be blended to produce the new alloy at a minimum cost. a. Identify all the requirements that will need to be
expressed in fixed-requirement constraints. E* b. Formulate and solve a linear programming model
for this problem on a spreadsheet. c. Express this model in algebraic form. 3.21. The Weigelt Corporation has three branch plants with excess production capacity. Fortunately, the corporation has a new product ready to begin production, and all three plants have this capability, so some of the excess capacity can be used in this way. This product can be made in three sizes—large, medium, and small—that yield a net unit profit of $420, $360, and $300, respectively. Plants 1, 2, and 3 have the excess capacity to pro- duce 750, 900, and 450 units per day of this product, respec- tively, regardless of the size or combination of sizes involved.
The amount of available in-process storage space also imposes a limitation on the production rates of the new product. Plants 1, 2, and 3 have 13,000, 12,000, and 5,000 square feet, respectively, of in-process storage space available for a day’s production of this product. Each unit of the large, medium, and small sizes produced per day requires 20, 15, and 12 square feet, respectively.
Sales forecasts indicate that if available, 900, 1,200, and 750 units of the large, medium, and small sizes, respectively, would be sold per day.
At each plant, some employees will need to be laid off unless most of the plant’s excess production capacity can be used to produce the new product. To avoid layoffs if possible, manage- ment has decided that the plants should use the same percentage of their excess capacity to produce the new product.
Management wishes to know how much of each of the sizes should be produced by each of the plants to maximize profit. E* a. Formulate and solve a linear programming model
for this mixed problem on a spreadsheet. b. Express the model in algebraic form. 3.22.* A cargo plane has three compartments for storing cargo: front, center, and back. These compartments have capac- ity limits on both weight and space, as summarized below.
Compartment
Weight Capacity
(Tons)
Space Capacity
(Cubic Feet)
Front 12 7,000 Center 18 9,000 Back 10 5,000
diagram also shows the monthly amounts produced at the mines and needed at the plant, as well as the shipping cost and the maximum amount that can be shipped per month through each shipping lane.
S1
P
S2
$2,000/ton
30 tons max. $400/ton 70 tons max.
40 tons produced
60 tons produced
$1,700/ton
30 tons m ax.
M1
$1,100/ton
50 tons max. M2
$1 ,6
00 /to
n
50 to
ns m
ax .
$8 00
/to n
70 to
ns ma
x.
100 tons needed
Management now wants to determine the most economical plan for shipping the iron ore from the mines through the distri- bution network to the steel plant.
a. Identify all the requirements that will need to be expressed in fixed-requirement constraints.
E* b. Formulate and solve a linear programming model for this problem on a spreadsheet.
c. Express this model in algebraic form.
3.19.* Al Ferris has $60,000 that he wishes to invest now in order to use the accumulation for purchasing a retirement annu- ity in five years. After consulting with his financial advisor, he has been offered four types of fixed-income investments, which we will label as investments A, B, C, and D.
Investments A and B are available at the beginning of each of the next five years (call them years 1 to 5). Each dollar invested in A at the beginning of a year returns $1.40 (a profit of $0.40) two years later (in time for immediate reinvestment). Each dol- lar invested in B at the beginning of a year returns $1.70 three years later.
Investments C and D will each be available at one time in the future. Each dollar invested in C at the beginning of year 2 returns $1.90 at the end of year 5. Each dollar invested in D at the beginning of year 5 returns $1.30 at the end of year 5.
Al wishes to know which investment plan maximizes the amount of money that can be accumulated by the beginning of year 6. a. For this problem, all its functional constraints can
be expressed as fixed-requirement constraints. To do this, let A t , B t , C t , and D t be the amounts invested in investments A, B, C, and D, respectively, at the beginning of year t for each t where the investment is available and will mature by the end of year 5. Also let R t be the number of available dollars not invested at the beginning of year t (and so available for investment in a later year). Thus, the amount invested at the beginning of year t plus R t must equal the number of dollars available for invest- ment at that time. Write such an equation in terms of the relevant variables above for the beginning of each of the five years to obtain the five fixed- requirement constraints for this problem.
b. Formulate a complete linear programming model for this problem in algebraic form.
E* c. Formulate and solve this model on a spreadsheet.
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to employ. They would like to maximize their net profit —their gross profit from sales minus their labor costs. E* a. Formulate and solve a linear programming model
for this problem on a spreadsheet. b. Summarize this formulation in algebraic form. E*3.24. Oxbridge University maintains a powerful mainframe computer for research use by its faculty, Ph.D. students, and research associates. During all working hours, an operator must be available to operate and maintain the computer, as well as to perform some programming services. Beryl Ingram, the director of the computer facility, oversees the operation.
It is now the beginning of the fall semester and Beryl is con- fronted with the problem of assigning different working hours to her operators. Because all the operators are currently enrolled in the university, they are available to work only a limited number of hours each day.
There are six operators (four undergraduate students and two graduate students). They all have different wage rates because of differences in their experience with computers and in their pro- gramming ability. The following table shows their wage rates, along with the maximum number of hours that each can work each day.
Maximum Hours of Availability
Operators Wage Rate Mon. Tue. Wed. Thurs. Fri.
K. C. $10.00/hour 6 0 6 0 6 D. H. $10.10/hour 0 6 0 6 0 H. B. $9.90/hour 4 8 4 0 4 S. C. $9.80/hour 5 5 5 0 5 K. S. $10.80/hour 3 0 3 8 0 N. K. $11.30/hour 0 0 0 6 2
Each operator is guaranteed a certain minimum number of hours per week that will maintain an adequate knowledge of the operation. This level is set arbitrarily at 8 hours per week for the undergraduate students (K. C., D. H., H. B., and S. C.) and 7 hours per week for the graduate students (K. S. and N. K.).
The computer facility is to be open for operation from 8 am to 10 pm Monday through Friday with exactly one operator on duty during these hours. On Saturdays and Sundays, the com- puter is to be operated by other staff.
Because of a tight budget, Beryl has to minimize cost. She wishes to determine the number of hours she should assign to each operator on each day. Formulate and solve a spreadsheet model for this problem. 3.25. Slim-Down Manufacturing makes a line of nutritionally complete, weight-reduction beverages. One of its products is a strawberry shake that is designed to be a complete meal. The strawberry shake consists of several ingredients. Some informa- tion about each of these ingredients is given next.
Furthermore, the weight of the cargo in the respective com- partments must be the same proportion of that compartment’s weight capacity to maintain the balance of the airplane.
The following four cargoes have been offered for shipment on an upcoming flight as space is available.
Cargo Weight (Tons)
Volume (Cubic Feet/Ton)
Profit ($/Ton)
1 20 500 320 2 16 700 400 3 25 600 360 4 13 400 290
Any portion of these cargoes can be accepted. The objec- tive is to determine how much (if any) of each cargo should be accepted and how to distribute each among the compartments to maximize the total profit for the flight. E* a. Formulate and solve a linear programming model
for this mixed problem on a spreadsheet. b. Express the model in algebraic form. 3.23. Comfortable Hands is a company that features a prod- uct line of winter gloves for the entire family—men, women, and children. They are trying to decide what mix of these three types of gloves to produce.
Comfortable Hands’s manufacturing labor force is union- ized. Each full-time employee works a 40-hour week. In addi- tion, by union contract, the number of full-time employees can never drop below 20. Nonunion, part-time workers also can be hired with the following union-imposed restrictions: (1) each part-time worker works 20 hours per week and (2) there must be at least two full-time employees for each part-time employee.
All three types of gloves are made out of the same 100 per- cent genuine cowhide leather. Comfortable Hands has a long- term contract with a supplier of the leather and receives a 5,000-square-foot shipment of the material each week. The mate- rial requirements and labor requirements, along with the gross profit per glove sold (not considering labor costs), are given in the following table.
Glove
Material Required
(Square Feet)
Labor Required (Minutes)
Gross Profit
(per Pair)
Men’s 2 30 $ 8 Women’s 1.5 45 10 Children’s 1 40 6
Each full-time employee earns $13 per hour, while each part-time employee earns $10 per hour. Management wishes to know what mix of each of the three types of gloves to produce per week, as well as how many full-time and part-time workers
Ingredient
Calories from Fat
(per tbsp.)
Total Calories
(per tbsp.)
Vitamin Content
(mg/tbsp.) Thickeners (mg/tbsp.)
Cost (¢/tbsp.)
Strawberry flavoring 1 50 20 3 10 Cream 75 100 0 8 8 Vitamin supplement 0 0 50 1 25 Artificial sweetener 0 120 0 2 15 Thickening agent 30 80 2 25 6
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E*3.28. The Cost-Less Corp. supplies its four retail outlets from its four plants. The shipping cost per shipment from each plant to each retail outlet is given below.
Unit Shipping Cost
Retail Outlet: 1 2 3 4
Plant 1 $500 $600 $400 $200 2 200 900 100 300 3 300 400 200 100 4 200 100 300 200
Plants 1, 2, 3, and 4 make 10, 20, 20, and 10 shipments per month, respectively. Retail outlets 1, 2, 3, and 4 need to receive 20, 10, 10, and 20 shipments per month, respectively.
The distribution manager, Randy Smith, now wants to deter- mine the best plan for how many shipments to send from each plant to the respective retail outlets each month. Randy’s objec- tive is to minimize the total shipping cost.
Formulate this problem as a transportation problem on a spreadsheet and then use Solver to obtain an optimal solution. E*3.29. The Childfair Company has three plants producing child push chairs that are to be shipped to four distribution
The nutritional requirements are as follows. The beverage must total between 380 and 420 calories (inclusive). No more than 20 percent of the total calories should come from fat. There must be at least 50 milligrams (mg) of vitamin content. For taste reasons, there must be at least two tablespoons (tbsp.) of strawberry flavoring for each tbsp. of artificial sweetener. Finally, to maintain proper thickness, there must be exactly 15 mg of thickeners in the beverage.
Management would like to select the quantity of each ingre- dient for the beverage that would minimize cost while meeting the above requirements. a. Identify the requirements that lead to resource con-
straints, to benefit constraints, and to fixed-require- ment constraints.
E* b. Formulate and solve a linear programming model for this problem on a spreadsheet.
c. Summarize this formulation in algebraic form. 3.26. Joyce and Marvin run a day care for preschoolers. They are trying to decide what to feed the children for lunches. They would like to keep their costs down, but they also need to meet the nutritional requirements of the children. They have already decided to go with peanut butter and jelly sandwiches, and some combination of graham crackers, milk, and orange juice. The nutritional content of each food choice and its cost are given in the table below.
Food Item Calories from Fat
Total Calories
Vitamin C (mg)
Protein (g)
Cost (¢)
Bread (1 slice) 10 70 0 3 5 Peanut butter (1 tbsp.) 75 100 0 4 4 Strawberry jelly (1 tbsp.) 0 50 3 0 7 Graham cracker (1 cracker) 20 60 0 1 8 Milk (1 cup) 70 150 2 8 15 Juice (1 cup) 0 100 120 1 35
The nutritional requirements are as follows. Each child should receive between 400 and 600 calories. No more than 30 percent of the total calories should come from fat. Each child should consume at least 60 milligrams (mg) of vitamin C and 12 grams (g) of protein. Furthermore, for practical reasons, each child needs exactly 2 slices of bread (to make the sandwich), at least twice as much peanut butter as jelly, and at least 1 cup of liquid (milk and/or juice).
Joyce and Marvin would like to select the food choices for each child that minimize cost while meeting the above requirements. a. Identify the requirements that lead to resource con-
straints, to benefit constraints, and to fixed-require- ment constraints.
E* b. Formulate and solve a linear programming model for this problem on a spreadsheet.
c. Express the model in algebraic form. 3.27. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 3.5. Briefly describe how the model for the transportation problem was applied in this study. Then list the vari- ous financial and nonfinancial benefits that resulted from this study.
centers. Plants 1, 2, and 3 produce 12, 17, and 11 shipments per month, respectively. Each distribution center needs to receive 10 shipments per month. The distance from each plant to the respective distribution centers is given below.
Distance to Distribution Center (Miles)
1 2 3 4
Plant 1 800 1,300 400 700 2 1,100 1,400 600 1,000 3 600 1,200 800 900
The freight cost for each shipment is $100 plus 50 cents/mile. How much should be shipped from each plant to each of the
distribution centers to minimize the total shipping cost? Formulate this problem as a transportation problem on a
spreadsheet and then use Solver to obtain an optimal solution. E*3.30. The Onenote Co. produces a single product at three plants for four customers. The three plants will produce 60, 80, and 40 units, respectively, during the next week. The firm has
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Formulate this problem on a spreadsheet and then use Solver to obtain the optimal solution identified above. 3.34. Four cargo ships will be used for shipping goods from one port to four other ports (labeled 1, 2, 3, 4). Any ship can be used for making any one of these four trips. However, because of differences in the ships and cargoes, the total cost of loading, transporting, and unloading the goods for the different ship–port combinations varies considerably, as shown in the following table.
Port
1 2 3 4
Ship 1 $500 $400 $600 $700 2 600 600 700 500 3 700 500 700 600 4 500 400 600 600
The objective is to assign the four ships to four different ports in such a way as to minimize the total cost for all four shipments. a. Describe how this problem fits into the format for
an assignment problem. E* b. Formulate and solve this problem on a spreadsheet. E*3.35. Reconsider Problem 3.10. Now distribution centers 1, 2, and 3 must receive exactly 10, 20, and 30 units per week, respectively. For administrative convenience, management has decided that each distribution center will be supplied totally by a single plant, so that one plant will supply one distribution center and the other plant will supply the other two distribution centers. The choice of these assignments of plants to distribu- tion centers is to be made solely on the basis of minimizing total shipping cost.
Formulate and solve a spreadsheet model for this problem. 3.36. Vincent Cardoza is the owner and manager of a machine shop that does custom order work. This Wednesday afternoon, he has received calls from two customers who would like to place rush orders. One is a trailer hitch com- pany that would like some custom-made heavy-duty tow bars. The other is a mini-car-carrier company that needs some cus- tomized stabilizer bars. Both customers would like as many as possible by the end of the week (two working days). Since both products would require the use of the same two machines, Vincent needs to decide and inform the customers this after- noon about how many of each product he will agree to make over the next two days.
Each tow bar requires 3.2 hours on machine 1 and 2 hours on machine 2. Each stabilizer bar requires 2.4 hours on machine 1 and 3 hours on machine 2. Machine 1 will be available for 16 hours over the next two days and machine 2 will be avail- able for 15 hours. The profit for each tow bar produced would be $130 and the profit for each stabilizer bar produced would be $150.
Vincent now wants to determine the mix of these production quantities that will maximize the total profit.
a. Formulate an integer programming model in alge- braic form for this problem.
E* b. Formulate and solve the model on a spreadsheet.
made a commitment to sell 40 units to customer 1, 60 units to customer 2, and at least 20 units to customer 3. Both customers 3 and 4 also want to buy as many of the remaining units as pos- sible. The net profit associated with shipping a unit from plant i for sale to customer j is given by the following table.
Customer
1 2 3 4
Plant 1 $800 $700 $500 $200 2 500 200 100 300 3 600 400 300 500
Management wishes to know how many units to sell to cus- tomers 3 and 4 and how many units to ship from each of the plants to each of the customers to maximize profit. Formulate and solve a spreadsheet model for this problem. E*3.31. The Move-It Company has two plants building forklift trucks that then are shipped to three distribution centers. The production costs are the same at the two plants, and the cost of shipping each truck is shown below for each combination of plant and distribution center.
Distribution Center
1 2 3
Plant A $800 $700 $400 B 600 800 500
A total of 60 forklift trucks are produced and shipped per week. Each plant can produce and ship any amount up to a maxi- mum of 50 trucks per week, so there is considerable flexibility on how to divide the total production between the two plants so as to reduce shipping costs. However, each distribution center must receive exactly 20 trucks per week.
Management’s objective is to determine how many forklift trucks should be produced at each plant, and then what the over- all shipping pattern should be to minimize total shipping cost. Formulate and solve a spreadsheet model for this problem. E*3.32. Redo Problem 3.31 when any distribution center may receive any quantity between 10 and 30 forklift trucks per week in order to further reduce total shipping cost, provided only that the total shipped to all three distribution centers must still equal 60 trucks per week. E*3.33. Consider the assignment problem having the following cost table.
Job
1 2 3
Person A $5 $7 $4 B 3 6 5 C 2 3 4
The optimal solution is A-3, B-1, C-2, with a total cost of $10.
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purchases. Regardless of which airplanes are purchased, air travel of all distances is expected to be sufficiently large that these planes would be utilized at essentially maximum capac- ity. It is estimated that the net annual profit (after capital recov- ery costs are subtracted) would be $4.2 million per long-range plane, $3 million per medium-range plane, and $2.3 million per short-range plane.
It is predicted that enough trained pilots will be available to the company to crew 30 new airplanes. If only short-range planes were purchased, the maintenance facilities would be able to handle 40 new planes. However, each medium-range plane is equivalent to 11/3 short-range planes, and each long-range plane is equivalent to 12/3 short-range planes in terms of their use of the maintenance facilities.
The information given here was obtained by a preliminary analysis of the problem. A more detailed analysis will be con- ducted subsequently. However, using the preceding data as a first approximation, management wishes to know how many planes of each type should be purchased to maximize profit. E* a. Formulate and solve a spreadsheet model for this
problem. b. Formulate this model in algebraic form.
3.37. Pawtucket University is planning to buy new copier machines for its library. Three members of its Management Science Department are analyzing what to buy. They are con- sidering two different models: Model A, a high-speed copier, and Model B, a lower speed but less expensive copier. Model A can handle 20,000 copies a day and costs $6,000. Model B can handle 10,000 copies a day but only costs $4,000. They would like to have at least six copiers so that they can spread them throughout the library. They also would like to have at least one high-speed copier. Finally, the copiers need to be able to handle a capacity of at least 75,000 copies per day. The objective is to determine the mix of these two copiers that will handle all these requirements at minimum cost. E* a. Formulate and solve a spreadsheet model for this
problem. b. Formulate this same model in algebraic form. 3.38. Northeastern Airlines is considering the purchase of new long-, medium-, and short-range jet passenger airplanes. The purchase price would be $67 million for each long-range plane, $50 million for each medium-range plane, and $35 million for each short-range plane. The board of directors has autho- rized a maximum commitment of $1.5 billion for these
Case 3-1
Shipping Wood to Market Alabama Atlantic is a lumber company that has three sources of wood and five markets to be supplied. The annual availability of wood at sources 1, 2, and 3 is 15, 20, and 15 million board feet, respectively. The amount that can be sold annually at mar- kets 1, 2, 3, 4, and 5 is 11, 12, 9, 10, and 8 million board feet, respectively.
In the past, the company has shipped the wood by train. However, because shipping costs have been increasing, the alternative of using ships to make some of the deliveries is being investigated. This alternative would require the company to invest in some ships. Except for these investment costs, the ship- ping costs in thousands of dollars per million board feet by rail and by water (when feasible) would be the following for each route.
Unit Cost by Rail ($1,000s) to Market Unit Cost by Ship ($1,000s) to Market
Source 1 2 3 4 5 1 2 3 4 5
1 61 72 45 55 66 31 38 24 — 35 2 69 78 60 49 56 36 43 28 24 31 3 59 66 63 61 47 — 33 36 32 26
The capital investment (in thousands of dollars) in ships required for each million board feet to be transported annually by ship along each route is given next.
Unit Investment for Ships ($1,000s) to Market
Source 1 2 3 4 5
1 275 303 238 — 285 2 293 318 270 250 265 3 — 283 275 268 240
Considering the expected useful life of the ships and the time value of money, the equivalent uniform annual cost of these investments is one-tenth the amount given in the table. The objec- tive is to determine the overall shipping plan that minimizes the total equivalent uniform annual cost (including shipping costs).
You are the head of the management science team that has been assigned the task of determining this shipping plan for each of the three options listed next.
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Case 3-2 Capacity Concerns 115
Finally, consider the fact that these results are based on cur- rent shipping and investment costs, so that the decision on the option to adopt now should take into account management’s projection of how these costs are likely to change in the future. For each option, describe a scenario of future cost changes that would justify adopting that option now.
Option 1: Continue shipping exclusively by rail. Option 2: Switch to shipping exclusively by water (except where only rail is feasible). Option 3: Ship by either rail or water, depending on which is less expensive for the particular route.
Present your results for each option. Compare.
Case 3-2
Capacity Concerns
Bentley Hamilton throws the business section of The New York Times onto the conference room table and watches as his associ- ates jolt upright in their overstuffed chairs.
Mr. Hamilton wants to make a point. He throws the front page of the The Wall Street Journal on
top of The New York Times and watches as his associates widen their eyes once heavy with boredom.
Mr. Hamilton wants to make a big point. He then throws the front page of the Financial Times on top
of the newspaper pile and watches as his associates dab the fine beads of sweat off their brows.
Mr. Hamilton wants his point indelibly etched into his asso- ciates’ minds.
“I have just presented you with three leading financial news- papers carrying today’s top business story,” Mr. Hamilton declares in a tight, angry voice. “My dear associates, our com- pany is going to hell in a hand basket! Shall I read you the head- lines? From The New York Times, ‘ CommuniCorp stock drops to lowest in 52 weeks.’ From The Wall Street Journal, ‘Commu- niCorp loses 25 percent of the pager market in only one year.’ Oh, and my favorite, from the Financial Times, ‘CommuniCorp cannot CommuniCate: CommuniCorp stock drops because of internal communications disarray.’ How did our company fall into such dire straits?”
Mr. Hamilton throws a transparency showing a line sloping slightly upward onto the overhead projector. “This is a graph of our productivity over the last 12 months. As you can see from the graph, productivity in our pager production facility has increased steadily over the last year. Clearly, productivity is not the cause of our problem.”
Mr. Hamilton throws a second transparency showing a line sloping steeply upward onto the overhead projector. “This is a graph of our missed or late orders over the last 12 months.” Mr. Hamilton hears an audible gasp from his associates. “As
you can see from the graph, our missed or late orders have increased steadily and significantly over the past 12 months. I think this trend explains why we have been losing market share, causing our stock to drop to its lowest level in 52 weeks. We have angered and lost the business of retailers, our customers who depend upon on-time deliveries to meet the demand of consumers.”
“Why have we missed our delivery dates when our produc- tivity level should have allowed us to fill all orders?” Mr. Ham- ilton asks. “I called several departments to ask this question.”
“It turns out that we have been producing pagers for the hell of it!” Mr. Hamilton says in disbelief. “The marketing and sales departments do not communicate with the manufacturing department, so manufacturing executives do not know what pag- ers to produce to fill orders. The manufacturing executives want to keep the plant running, so they produce pagers regardless of whether the pagers have been ordered. Finished pagers are sent to the warehouse, but marketing and sales executives do not know the number and styles of pagers in the warehouse. They try to communicate with warehouse executives to determine if the pagers in inventory can fill the orders, but they rarely receive answers to their questions.”
Mr. Hamilton pauses and looks directly at his associates. “Ladies and gentlemen, it seems to me that we have a serious internal communications problem. I intend to correct this prob- lem immediately. I want to begin by installing a companywide computer network to ensure that all departments have access to critical documents and are able to easily communicate with each other through e-mail. Because this intranet will represent a large change from the current communications infrastructure, I expect some bugs in the system and some resistance from employees. I therefore want to phase in the installation of the intranet.”
Mr. Hamilton passes the following time line and require- ments chart to his associates (IN 5 intranet).
Month 1 Month 2 Month 3 Month 4 Month 5
IN education Install IN in sales
Install IN in manufacturing
Install IN in warehouse
Install IN in marketing
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“There are several factors that complicate your decision,” Mr. Hamilton continues. “Two server manufacturers are willing to offer discounts to CommuniCorp. SGI is willing to give you a discount of 10 percent off each server purchased, but only if you purchase servers in the first or second month. Sun is will- ing to give you a 25 percent discount off all servers purchased in the first two months. You are also limited in the amount of money you can spend during the first month. CommuniCorp has already allocated much of the budget for the next two months, so you only have a total of $9,500 available to purchase servers in months 1 and 2. Finally, the manufacturing department requires at least one of the three more powerful servers. Have your deci- sion on my desk at the end of the week.”
a. Emily first decides to evaluate the number and type of serv- ers to purchase on a month-to-month basis. For each month, formulate a spreadsheet model to determine which servers Emily should purchase in that month to minimize costs in that month and support the new users given your results for the preceding months. How many and which types of servers should she purchase in each month? How much is the total cost of the plan?
b. Emily realizes that she could perhaps achieve savings if she bought a larger server in the initial months to support users in the final months. She therefore decides to evaluate the num- ber and type of servers to purchase over the entire planning period. Formulate a spreadsheet model to determine which servers Emily should purchase in which months to minimize total cost and support all new users. How many and which
Department Number of Employees
Sales 60 Manufacturing 200 Warehouse 30 Marketing 75
Mr. Hamilton proceeds to explain the time line and require- ments chart. “In the first month, I do not want to bring any department onto the intranet; I simply want to disseminate infor- mation about it and get buy-in from employees. In the second month, I want to bring the sales department onto the intranet since the sales department receives all critical information from customers. In the third month, I want to bring the manufacturing department onto the intranet. In the fourth month, I want to install the intranet at the warehouse, and in the fifth and final month, I want to bring the marketing department onto the intranet. The requirements chart above lists the number of employees requir- ing access to the intranet in each department.”
Mr. Hamilton turns to Emily Jones, the head of Corporate Information Management. “I need your help in planning for the installation of the intranet. Specifically, the company needs to purchase servers for the internal network. Employees will con- nect to company servers and download information to their own desktop computers.”
Mr. Hamilton passes Emily the following chart detailing the types of servers available, the number of employees each server supports, and the cost of each server.
Type of Server Number of Employees Server Supports Cost of Server
Standard Intel Pentium PC Up to 30 employees $ 2,500 Enhanced Intel Pentium PC Up to 80 employees 5,000 SGI Workstation Up to 200 employees 10,000 Sun Workstation Up to 2,000 employees 25,000
“Emily, I need you to decide what servers to purchase and when to purchase them to minimize cost and to ensure that the company possesses enough server capacity to follow the intranet implementation timeline,” Mr. Hamilton says. “For example, you may decide to buy one large server during the first month to support all employees, or buy several small servers during the first month to support all employees, or buy one small server each month to support each new group of employees gaining access to the intranet.”
types of servers should she purchase in each month? How much is the total cost of the plan?
c. Why is the answer using the first method different from that using the second method?
d. Are there other costs for which Emily is not accounting in her problem formulation? If so, what are they?
e. What further concerns might the various departments of CommuniCorp have regarding the intranet?
Case 3-3
Fabrics and Fall Fashions From the 10th floor of her office building, Katherine Rally watches the swarms of New Yorkers fight their way through the streets infested with yellow cabs and the sidewalks littered with hot dog stands. On this sweltering July day, she pays particular attention to the fashions worn by the various women and won- ders what they will choose to wear in the fall. Her thoughts are not simply random musings; they are critical to her work since she owns and manages TrendLines, an elite women’s clothing company.
Today is an especially important day because she must meet with Ted Lawson, the production manager, to decide upon next month’s production plan for the fall line. Specifically, she must determine the quantity of each clothing item she should produce given the plant’s production capacity, limited resources, and demand forecasts. Accurate planning for next month’s production is critical to fall sales since the items produced next month will appear in stores during September and women generally buy the majority of the fall fashions when they first appear in September.
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She turns back to her sprawling glass desk and looks at the numerous papers covering it. Her eyes roam across the clothing patterns designed almost six months ago, the lists of material requirements for each pattern, and the lists of demand forecasts for each pattern determined by customer surveys at fashion shows. She remembers the hectic and sometimes nightmarish days of designing the fall line and presenting it at fashion shows in New York, Milan, and Paris. Ultimately, she paid her team of six designers a total of $860,000 for their work on her fall line. With the cost of hiring runway models, hair stylists, and make-up artists; sewing and fitting clothes; building the set; choreographing and rehearsing the show; and renting the confer- ence hall, each of the three fashion shows cost her an additional $2,700,000.
She studies the clothing patterns and material requirements. Her fall line consists of both professional and casual fashions. She determined the price for each clothing item by taking into account the quality and cost of material, the cost of labor and machining, the demand for the item, and the prestige of the TrendLines brand name.
The fall professional fashions include.
Clothing Item Material Requirements Price Labor and Machine Cost
Tailored wool slacks 3 yards of wool $300 $160 2 yards of acetate for lining
Cashmere sweater 1.5 yards of cashmere 450 150 Silk blouse 1.5 yards of silk 180 100 Silk camisole 0.5 yard of silk 120 60 Tailored skirt 2 yards of rayon 270 120
1.5 yards of acetate for lining Wool blazer 2.5 yards of wool 320 140
1.5 yards of acetate for lining
Clothing Item Material Requirements Price Labor and Machine Cost
Velvet pants 3 yards of velvet $350 $175 2 yards of acetate for lining
Cotton sweater 1.5 yards of cotton 130 60 Cotton miniskirt 0.5 yard of cotton 75 40 Velvet shirt 1.5 yards of velvet 200 160 Button-down blouse 1.5 yards of rayon 120 90
The fall casual fashions include.
She knows that for the next month, she has ordered 45,000 yards of wool, 28,000 yards of acetate, 9,000 yards of cashmere, 18,000 yards of silk, 30,000 yards of rayon, 20,000 yards of vel- vet, and 30,000 yards of cotton for production. The prices of the materials are listed below.
Material Price per Yard
Wool $ 9.00 Acetate 1.50 Cashmere 60.00 Silk 13.00 Rayon 2.25 Velvet 12.00 Cotton 2.50
Any material that is not used in production can be sent back to the textile wholesaler for a full refund, although scrap material cannot be sent back to the wholesaler.
She knows that the production of both the silk blouse and cot- ton sweater leaves leftover scraps of material. Specifically, for the production of one silk blouse or one cotton sweater, 2 yards of silk and cotton, respectively, are needed. From these 2 yards, 1.5 yards are used for the silk blouse or the cotton sweater and 0.5 yard is left as scrap material. She does not want to waste the material, so she plans to use the rectangular scrap of silk or cot- ton to produce a silk camisole or cotton miniskirt, respectively. Therefore, whenever a silk blouse is produced, a silk camisole is also produced. Likewise, whenever a cotton sweater is pro- duced, a cotton miniskirt is also produced. Note that it is pos- sible to produce a silk camisole without producing a silk blouse and a cotton miniskirt without producing a cotton sweater.
The demand forecasts indicate that some items have lim- ited demand. Specifically, because the velvet pants and velvet shirts are fashion fads, TrendLines has forecasted that it can sell only 5,500 pairs of velvet pants and 6,000 velvet shirts. TrendLines does not want to produce more than the forecasted
demand because once the pants and shirts go out of style, the company cannot sell them. TrendLines can produce less than the forecasted demand, however, since the company is not required to meet the demand. The cashmere sweater also has limited demand because it is quite expensive, and TrendLines knows it can sell at most 4,000 cashmere sweaters. The silk blouses and camisoles have limited demand because many women think silk is too hard to care for, and TrendLines projects that it can sell at most 12,000 silk blouses and 15,000 silk camisoles.
The demand forecasts also indicate that the wool slacks, tailored skirts, and wool blazers have a great demand because they are basic items needed in every professional wardrobe. Specifically, the demand is 7,000 pairs of wool slacks and 5,000 wool blazers. Katherine wants to meet at least 60 percent of the demand for these two items to maintain her loyal customer base
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therefore get no refund for the velvet. How does this fact change the production plan?
d. What is an intuitive economic explanation for the difference between the solutions found in parts b and c?
e. The sewing staff encounters difficulties sewing the arms and lining into the wool blazer since the blazer pattern has an awkward shape and the heavy wool material is difficult to cut and sew. The increased labor time to sew a wool blazer increases the labor and machine cost for each blazer by $80. Given this new cost, how many of each clothing item should TrendLines produce to maximize profit?
f. The textile wholesaler informs Katherine that since another textile customer canceled his order, she can obtain an extra 10,000 yards of acetate. How many of each clothing item should TrendLines now produce to maximize profit?
g. TrendLines assumes that it can sell every item that was not sold during September and October in a big sale in Novem- ber at 60 percent of the original price. Therefore, it can sell all items in unlimited quantity during the November sale. (The previously mentioned upper limits on demand only con- cern the sales during September and October.) What should the new production plan be to maximize profit?
and not lose business in the future. Although the demand for tai- lored skirts could not be estimated, Katherine feels she should make at least 2,800 of them.
a. Ted is trying to convince Katherine not to produce any velvet shirts since the demand for this fashion fad is quite low. He argues that this fashion fad alone accounts for $500,000 of the fixed design and other costs. The net contribution (price of clothing item – materials cost – labor cost) from selling the fashion fad should cover these fixed costs. Each velvet shirt generates a net contribution of $22. He argues that given the net contribution, even satisfying the maximum demand will not yield a profit. What do you think of Ted’s argument?
b. Formulate and solve a linear programming problem to maxi- mize profit given the production, resource, and demand constraints.
Before she makes her final decision, Katherine plans to explore the following questions independently, except where otherwise indicated.
c. The textile wholesaler informs Katherine that the velvet can- not be sent back because the demand forecasts show that the demand for velvet will decrease in the future. Katherine can
Case 3-4
New Frontiers
Rob Richman, president of AmeriBank, takes off his glasses, rubs his eyes in exhaustion, and squints at the clock in his study. It reads 3 am. For the last several hours, Rob has been por- ing over AmeriBank’s financial statements from the last three quarters of operation. AmeriBank, a medium-sized bank with branches throughout the United States, is headed for dire eco- nomic straits. The bank, which provides transaction, savings, investment, and loan services, has been experiencing a steady decline in its net income over the past year, and trends show that the decline will continue. The bank is simply losing customers to nonbank and foreign bank competitors.
AmeriBank is not alone in its struggle to stay out of the red. From his daily industry readings, Rob knows that many American banks have been suffering significant losses because of increasing competition from nonbank and foreign bank competitors offering services typically in the domain of American banks. Because the nonbank and foreign bank competitors specialize in particular ser- vices, they are able to better capture the market for those services by offering less expensive, more efficient, more convenient ser- vices. For example, large corporations now turn to foreign banks and commercial paper offerings for loans, and affluent Americans now turn to money-market funds for investment. Banks face the daunting challenge of distinguishing themselves from nonbank and foreign bank competitors.
Rob has concluded that one strategy for distinguishing AmeriBank from its competitors is to improve services that nonbank and foreign bank competitors do not readily pro- vide: transaction services. He has decided that a more conve- nient transaction method must logically succeed the automatic teller machine, and he believes that electronic banking over the
Internet allows this convenient transaction method. Over the Internet, customers are able to perform transactions on their desktop computers either at home or work. The explosion of the Internet means that many potential customers understand and use the World Wide Web. He therefore feels that if AmeriBank offers Web banking (as the practice of Internet banking is com- monly called), the bank will attract many new customers.
Before Rob undertakes the project to make Web banking possible, however, he needs to understand the market for Web banking and the services AmeriBank should provide over the Internet. For example, should the bank only allow customers to access account balances and historical transaction information over the Internet, or should the bank develop a strategy to allow customers to make deposits and withdrawals over the Internet? Should the bank try to recapture a portion of the investment mar- ket by continuously running stock prices and allowing customers to make stock transactions over the Internet for a minimal fee?
Because AmeriBank is not in the business of performing surveys, Rob has decided to outsource the survey project to a professional survey company. He has opened the project up for bidding by several survey companies and will award the proj- ect to the company that is willing to perform the survey for the least cost. Rob provided each survey company with a list of sur- vey requirements to ensure that AmeriBank receives the needed information for planning the Web banking project.
Because different age groups require different services, AmeriBank is interested in surveying four different age groups. The first group encompasses customers who are 18 to 25 years old. The bank assumes that this age group has limited yearly income and performs minimal transactions. The second group
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Case 3-5 Assigning Students to Schools 119
c. After submitting its bid, Sophisticated Surveys is informed that it has the lowest cost but that AmeriBank does not like the solution. Specifically, Rob feels that the selected survey population is not representative enough of the banking cus- tomer population. Rob wants at least 50 people of each age group surveyed in each region. What is the new bid made by Sophisticated Surveys?
d. Rob feels that Sophisticated Surveys oversampled the 18-to- 25-year-old population and the Silicon Valley population. He imposes a new constraint that no more than 600 individuals can be surveyed from the 18-to-25-year-old population and no more than 650 individuals can be surveyed from the Sili- con Valley population. What is the new bid?
e. When Sophisticated Surveys calculated the cost of reaching and surveying particular individuals, the company thought that reaching individuals in young populations would be easiest. In a recently completed survey, however, Sophisti- cated Surveys learned that this assumption was wrong. The new costs for surveying the 18-to-25-year-old population are listed below.
Region Cost per Person
Silicon Valley $6.50 Big cities 6.75 Small towns 7.00
Given the new costs, what is the new bid? f. To ensure the desired sampling of individuals, Rob imposes
even stricter requirements. He fixes the exact percentage of people that should be surveyed from each population. The requirements are listed next.
Population Percentage of People Surveyed
18 to 25 25% 26 to 40 35 41 to 50 20 51 and over 20 Silicon Valley 20 Big cities 50 Small towns 30
By how much would these new requirements increase the cost of surveying for Sophisticated Surveys? Given the 15 percent profit margin, what would Sophisticated Surveys bid?
encompasses customers who are 26 to 40 years old. This age group has significant sources of income, performs many trans- actions, requires numerous loans for new houses and cars, and invests in various securities. The third group encompasses cus- tomers who are 41 to 50 years old. These customers typically have the same level of income and perform the same number of transactions as the second age group, but the bank assumes that these customers are less likely to use Web banking since they have not become as comfortable with the explosion of com- puters or the Internet. Finally, the fourth group encompasses customers who are 51 years of age and over. These customers commonly crave security and require continuous information on retirement funds. The bank believes that it is highly unlikely that customers in this age group will use Web banking, but the bank desires to learn the needs of this age group for the future. AmeriBank wants to interview 2,000 customers with at least 20 percent from the first age group, at least 27.5 percent from the second age group, at least 15 percent from the third age group, and at least 15 percent from the fourth age group.
Rob understands that some customers are uncomfortable with using the Internet. He therefore wants to ensure that the survey includes a mix of customers who know the Internet well and those that have less exposure to the Internet. To ensure that AmeriBank obtains the correct mix, he wants to interview at least 15 percent of customers from the Silicon Valley where Internet use is high, at least 35 percent of customers from big cities where Internet use is medium, and at least 20 percent of customers from small towns where Internet use is low.
Sophisticated Surveys is one of three survey companies competing for the project. It has performed an initial analysis of these survey requirements to determine the cost of surveying different populations. The costs per person surveyed are listed in the following table.
Age Group
Region 18 to 25 26 to 40 41 to 50 51 and over
Silicon Valley $4.75 $6.50 $6.50 $5.00 Big cities 5.25 5.75 6.25 6.25 Small towns 6.50 7.50 7.50 7.25
Sophisticated Surveys explores the following options cumulatively.
a. Formulate a linear programming model to minimize costs while meeting all survey constraints imposed by AmeriBank.
b. If the profit margin for Sophisticated Surveys is 15 percent of cost, what bid will it submit?
Case 3-5
Assigning Students to Schools The Springfield School Board has made the decision to close one of its middle schools (sixth, seventh, and eighth grades) at the end of this school year and reassign all of next year’s middle
school students to the three remaining middle schools. The school district provides busing for all middle school students who must travel more than approximately a mile, so the school
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board wants a plan for reassigning the students that will mini- mize the total busing cost. The annual cost per student for busing from each of the six residential areas of the city to each of the schools is shown in the following table (along with other basic data for next year), where 0 indicates that busing is not needed and a dash indicates an infeasible assignment.
Busing Cost per Student
Area Number of Students
Percentage in 6th Grade
Percentage in 7th Grade
Percentage in 8th Grade School 1 School 2 School 3
1 450 32 38 30 $300 $ 0 $700 2 600 37 28 35 — 400 500 3 550 30 32 38 600 300 200 4 350 28 40 32 200 500 — 5 500 39 34 27 0 — 400 6 450 34 28 38 500 300 0
School capacity: 900 1,100 1,000
The school board also has imposed the restriction that each grade must constitute between 30 and 36 percent of each school’s population. The above table shows the percentage of each area’s middle school population for next year that falls into each of the three grades. The school attendance zone boundaries can be drawn so as to split any given area among more than one school, but assume that the percentages shown in the table will continue to hold for any partial assignment of an area to a school.
You have been hired as a management science consultant to assist the school board in determining how many students in each area should be assigned to each school.
a. Formulate and solve a linear programming model for this problem.
b. What is your resulting recommendation to the school board?
After seeing your recommendation, the school board expresses concern about all the splitting of residential areas among multiple schools. They indicate that they “would like to keep each neighborhood together.”
c. Adjust your recommendation as well as you can to enable each area to be assigned to just one school. (Adding this restriction may force you to fudge on some other constraints.)
How much does this increase the total busing cost? (This line of analysis will be pursued more rigorously in Case 7-3.)
The school board is considering eliminating some busing to reduce costs. Option 1 is to only eliminate busing for students trav- eling 1 to 1.5 miles, where the cost per student is given in the table
as $200. Option 2 is to also eliminate busing for students traveling 1.5 to 2 miles, where the estimated cost per student is $300.
d. Revise the model from part a to fit Option 1, and solve. Com- pare these results with those from part b, including the reduc- tion in total busing cost.
e. Repeat part d for Option 2.
The school board now needs to choose among the three alter- native busing plans (the current one or Option 1 or Option 2). One important factor is busing costs. However, the school board also wants to place equal weight on a second factor: the inconve- nience and safety problems caused by forcing students to travel by foot or bicycle a substantial distance (more than a mile, and especially more than 1.5 miles). Therefore, they want to choose a plan that provides the best trade-off between these two factors.
f. Use your results from parts b, d, and e to summarize the key information related to these two factors that the school board needs to make this decision.
g. Which decision do you think should be made? Why?
Note: This case will be continued in later chapters (Cases 5-4 and 7-3), so we suggest that you save your analysis, including your basic spreadsheet model.
Case 3-6
Reclaiming Solid Wastes
The Save-It Company operates a reclamation center that collects four types of solid waste materials and then treats them so that they can be amalgamated (treating and amalgamating are separate pro- cesses) into a salable product. Three different grades of this prod- uct can be made, depending on the mix of the materials used. (See the first table.) Although there is some flexibility in the mix for each grade, quality standards specify the minimum or maximum amount of the materials allowed in that product grade. (This mini- mum or maximum amount is the weight of the material expressed
as a percentage of the total weight for that product grade.) For each of the two higher grades, a fixed percentage is specified for one of the materials. These specifications are given in the first table along with the cost of amalgamation and the selling price for each grade.
The reclamation center collects its solid waste materials from some regular sources and so is normally able to maintain a steady rate for treating them. The second table gives the quanti- ties available for collection and treatment each week, as well as the cost of treatment, for each type of material.
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Case 3-7 Project Pickings 121
maximize the total weekly profit (total sales income minus total amalgamation cost).
a. Formulate this problem in linear programming terms by identifying all the activities, resources, benefits, and fixed requirements that lurk within it.
b. Formulate and solve a spreadsheet model for this linear pro- gramming problem.
c. Express this linear programming model in the spreadsheet in algebraic form.
The Save-It Company is solely owned by Green Earth, an organization that is devoted to dealing with environmen- tal issues; Save-It’s profits are all used to help support Green Earth’s activities. Green Earth has raised contributions and grants, amounting to $30,000 per week, to be used exclusively to cover the entire treatment cost for the solid waste materials. The board of directors of Green Earth has instructed the management of Save-It to divide this money among the materials in such a way that at least half of the amount available of each material is actually collected and treated. These additional restrictions are listed in the second table.
Within the restrictions specified in the two tables, manage- ment wants to allocate the materials to product grades so as to
Grade Specification Amalgamation Cost per Pound
Selling Price per Pound
Material 1: Not more than 30% of the total A Material 2: Not less than 40% of the total $3.00 $8.50
Material 3: Not more than 50% of the total Material 4: Exactly 20% of the total
Material 1: Not more than 50% of the total B Material 2: Not less than 10% of the total 2.50 7.00
Material 4: Exactly 10% of the total
C Material 1: Not more than 70% of the total 2.00 5.50
Material Available
Pounds/Week Treatment Cost
per Pound Additional Restrictions
1 3,000 $3.00 1. For each material, at least half of the pounds/week available should be collected and treated.
2. $30,000 per week should be used to treat these materials.
2 2,000 6.00 3 4,000 4.00 4 1,000 5.00
Case 3-7
Project Pickings
Tazer, a pharmaceutical manufacturing company, entered the pharmaceutical market 15 years ago with the introduction of six new drugs. Five of the six drugs were simply permutations of existing drugs and therefore did not sell very heavily. The sixth drug, however, addressed hypertension and was a huge success. Since Tazer had a patent on the hypertension drug, it experienced no competition, and profits from the hypertension drug alone kept Tazer in business. Pharmaceutical patents remain in force for 20 years, so this one has five more years before it expires.
During the past 15 years, Tazer continued a moderate amount of research and development, but it never stumbled upon a drug as successful as the hypertension drug. One reason is that the company never had the motivation to invest heavily in innova- tive research and development. The company was riding the profit wave generated by its hypertension drug and did not feel the need to commit significant resources to finding new drug breakthroughs.
Now Tazer is beginning to fear the pressure of competition. Tazer knows that once the patent expires in five years, generic drug manufacturing companies will swarm into the market like vultures. Historical trends show that generic drugs decrease sales of branded drugs by 75 percent.
Tazer is therefore looking to invest significant amounts of money in research and development this year to begin the search for a new breakthrough drug that will offer the company the same success as the hypertension drug. Tazer believes that if the company begins extensive research and development now, the probability of finding a successful drug shortly after the expiration of the hypertension patent will be high.
As head of research and development at Tazer, you are responsible for choosing potential projects and assigning project directors to lead each of the projects. After researching the needs of the market, analyzing the shortcomings of current drugs,
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and interviewing numerous scientists concerning the promising areas of medical research, you have decided that your depart- ment will pursue five separate projects, which are listed below:
Project Up: Develop a more effective antidepressant that does not cause serious mood swings. Project Stable: Develop a drug that addresses manic-depression. Project Choice: Develop a less intrusive birth control method for women. Project Hope: Develop a vaccine to prevent HIV infection. Project Release: Develop a more effective drug to lower blood pressure.
For each of the five projects, you are only able to specify the medical ailment the research should address since you do not know what compounds will exist and be effective without research.
You also have five senior scientists to lead the five projects. You know that scientists are very temperamental people and will only work well if they are challenged and motivated by the proj- ect. To ensure that the senior scientists are assigned to projects they find motivating, you have established a bidding system for the projects. You have given each of the five scientists 1,000 bid points. They assign bids to each project, giving a higher number of bid points to projects they most prefer to lead.
The following table provides the bids from the five senior scientists for the five individual projects.
Project Dr. Kvaal Dr. Zuner Dr. Tsai Dr. Mickey Dr. Rollins
Project Up 100 0 100 267 100 Project Stable 400 200 100 153 33 Project Choice 200 800 100 99 33 Project Hope 200 0 100 451 34 Project Release 100 0 600 30 800
You decide to evaluate a variety of scenarios you think are likely.
a. Given the bids, you need to assign one senior scientist to each of the five projects to maximize the preferences of the scientists. What are the assignments?
b. Dr. Rollins is being courted by Harvard Medical School to accept a teaching position. You are fighting desperately to keep her at Tazer, but the prestige of Harvard may lure her away. If this were to happen, the company would give up the project with the least enthusiasm. Which project would not be done?
c. You do not want to sacrifice any project since researching only four projects decreases the probability of finding a breakthrough new drug. You decide that either Dr. Zuner or Dr. Mickey could lead two projects. Under these new con- ditions with just four senior scientists, which scientists will lead which projects to maximize preferences?
d. After Dr. Zuner was informed that she and Dr. Mickey are being considered for two projects, she decided to change her bids. Dr. Zuner’s new bids for each of the projects are shown next.
Project Up: 20 Project Stable: 450 Project Choice: 451 Project Hope: 39 Project Release: 40
Under these new conditions with just four senior scien- tists, which scientists will lead which projects to maximize preferences?
e. Do you support the assignments found in part d? Why or why not?
f. Now you again consider all five scientists. You decide, how- ever, that several scientists cannot lead certain projects. In par- ticular, Dr. Mickey does not have experience with research on the immune system, so he cannot lead Project Hope. His fam- ily also has a history of manic-depression, and you feel that he would be too personally involved in Project Stable to serve as an effective project leader. Dr. Mickey therefore cannot lead Project Stable. Dr. Kvaal also does not have experience with research on the immune system and cannot lead Project Hope. In addition, Dr. Kvaal cannot lead Project Release because he does not have experience with research on the cardiovascular system. Finally, Dr. Rollins cannot lead Project Up because her family has a history of depression and you feel she would be too personally involved in the project to serve as an effec- tive leader. Because Dr. Mickey and Dr. Kvaal cannot lead two of the five projects, they each have only 600 bid points.
Dr. Rollins has only 800 bid points because she cannot lead one of the five projects. The following table provides the new bids of Dr. Mickey, Dr. Kvaal, and Dr. Rollins.
Project Dr. Mickey Dr. Kvaal Dr. Rollins
Project Up 300 86 Can’t lead Project Stable Can’t lead 343 50 Project Choice 125 171 50 Project Hope Can’t lead Can’t lead 100 Project Release 175 Can’t lead 600
Which scientists should lead which projects to maximize preferences?
g. You decide that Project Hope and Project Release are too complex to be led by only one scientist. Therefore, each of these projects will be assigned two scientists as project lead- ers. You decide to hire two more scientists in order to staff all projects: Dr. Arriaga and Dr. Santos. Because of religious reasons, neither of them want to lead Project Choice and so they assign 0 bid points to this project. The next table lists all projects, scientists, and their bids.
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Additional Cases 123
Additional Cases Additional cases for this chapter are also available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
Which scientists should lead which projects to maximize preferences?
h. Do you think it is wise to base your decision in part g only on an optimal solution for a variant of an assignment problem?
Project Dr. Kvaal Dr. Zuner Dr. Tsai Dr. Mickey Dr. Rollins Dr. Arriaga Dr. Santos
Project Up 86 0 100 300 Can’t lead 250 111 Project Stable 343 200 100 Can’t lead 50 250 1 Project Choice 171 800 100 125 50 0 0 Project Hope Can’t lead 0 100 Can’t lead 100 250 333 Project Release Can’t lead 0 600 175 600 250 555
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Chapter Four
The Art of Modeling with Spreadsheets Learning Objectives
After completing this chapter, you should be able to
1. Describe the general process for modeling in spreadsheets.
2. Describe some guidelines for building good spreadsheet models.
3. Apply both the general process for modeling in spreadsheets and the guidelines in this chapter to develop your own spreadsheet model from a description of the problem.
4. Identify some deficiencies in a poorly formulated spreadsheet model.
5. Apply a variety of techniques for debugging a spreadsheet model.
Nearly all managers now make extensive use of spreadsheets to analyze business problems. What they are doing is modeling with spreadsheets.
Spreadsheet modeling is a major emphasis throughout this book. Section 1.2 in Chapter 1 introduced a spreadsheet model for performing break-even analysis. Section 2.2 in Chapter 2 described how to use spreadsheets to formulate linear programming models. Chapter 3 focused on spreadsheet models for five key categories of linear programming problems: resource- allocation problems, cost–benefit–trade-off problems, mixed problems, transportation problems, and assignment problems. Many kinds of spreadsheet models are discussed in subsequent chapters as well. However, those presentations focus mostly on the charac- teristics of spreadsheet models that fit the management science techniques (such as linear programming) being covered in those chapters. We devote this chapter instead to the general process of building models with spreadsheets.
Modeling in spreadsheets is more an art than a science. There is no systematic procedure that invariably will lead to a single correct spreadsheet model. For example, if two people were given exactly the same business problem to analyze with a spreadsheet, their spread- sheet models would likely look quite different. There is no one right way of modeling any given problem. However, some models will be better than others.
Although no completely systematic procedure is available for modeling in spreadsheets, there is a general process that should be followed. This process has four major steps: (1) plan the spreadsheet model, (2) build the model, (3) test the model, and (4) analyze the model and its results. After introducing a case study in Section 4.1, the next section will describe this plan-build-test-analyze process in some detail and illustrate the process in the context of the case study. Section 4.2 also will discuss some ways of overcoming common stumbling blocks in the modeling process.
Unfortunately, despite its logical approach, there is no guarantee that the plan-build-test- analyze process will lead to a “good” spreadsheet model. A good spreadsheet model is easy to understand, easy to debug, and easy to modify. Section 4.3 presents some guidelines for building such models. This section also uses the case study in Section 4.1 to illustrate the dif- ference between appropriate formulations and poor formulations of a model.
Even with an appropriate formulation, the initial versions of large spreadsheet models commonly will include some small but troublesome errors, such as inaccurate references to cell addresses or typographical errors when entering equations into cells. Indeed, some
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4.1 A Case Study: The Everglade Golden Years Company Cash Flow Problem 125
studies have shown that a surprisingly large number of errors typically occur in the first version of such models. These errors often can be difficult to track down. Section 4.4 pres- ents some helpful ways to debug a spreadsheet model and root out such errors.
The overriding goal of this chapter is to provide a solid foundation for becoming a successful spreadsheet modeler. However, this chapter by itself will not turn you into a highly skilled mod- eler. Ultimately, to reach this point you also will need to study various examples of good spread- sheet models in the different areas of management science and then have lots of practice in formulating your own models. This process will continue throughout the remainder of this book.
4.1 A CASE STUDY: THE EVERGLADE GOLDEN YEARS COMPANY CASH FLOW PROBLEM
The Everglade Golden Years Company operates upscale retirement communities in certain parts of southern Florida. The company was founded in 1946 by Alfred Lee, who was in the right place at the right time to enjoy many successful years during the boom in the Florida economy as many wealthy retirees flooded into the area. Today, the company continues to be run by the Lee family, with Alfred’s grandson, Sheldon Lee, as the CEO.
The past few years have been difficult ones for Everglade. The demand for retirement community housing has been light and Everglade has been unable to maintain full occupancy. However, this market has picked up recently and the future is looking brighter. Everglade has recently broken ground for the construction of a new retirement community and has more new construction planned over the next 10 years (2014 through 2023).
Julie Lee is the chief financial officer (CFO) at Everglade. She has spent the last week in front of her computer trying to come to grips with the company’s imminent cash flow problem. Julie has projected Everglade’s net cash flows over the next 10 years as shown in Table 4.1 . With less money currently coming in than would be provided by full occupancy and with all the construc- tion costs for the new retirement community, Everglade will have negative cash flow for the next few years. With only $1 million in cash reserves, it appears that Everglade will need to take out some loans in order to meet its financial obligations. Also, to protect against uncertainty, company policy dictates maintaining a balance of at least $500,000 in cash reserves at all times.
The company’s bank has offered two types of loans to Everglade. The first is a 10-year loan with interest-only payments made annually and then the entire principal repaid in a single balloon payment after 10 years. The interest rate on this long-term loan is a favorable 5 percent per year. The second option is a series of one-year loans. These loans can be taken out each year as needed, but each must be repaid (with interest) the following year. Each new loan can be used to help repay the loan for the preceding year if needed. The interest rate for these short-term loans currently is projected to be 7 percent per year.
Armed with her cash flow projections and the loan options from the bank, Julie schedules a meeting with the CEO, Sheldon Lee. Their discussion is as follows:
Julie: Well, we really seem to be in a pickle. There is no way to meet our cash flow prob- lems without borrowing money.
With only $1 million in cash reserves and negative cash flows looming soon, loans will be needed to observe the company policy of maintaining a balance of at least $500,000 at all times.
Year Projected Net Cash Flow
(millions of dollars)
2014 28 2015 22 2016 24 2017 3 2018 6 2019 3 2020 24 2021 7 2022 22 2023 10
TABLE 4.1 Projected Net Cash Flows for the Everglade Golden Years Company over the Next Ten Years
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Sheldon: I was afraid of that. What are our options? Julie: I’ve talked to the bank, and we can take out a 10-year loan with an interest rate of 5 percent, or a series of one-year loans at a projected rate of 7 percent. Sheldon: Wow. That 5 percent rate sounds good. Can we just borrow all that we need using the 10-year loan? Julie: That was my initial reaction as well. However, after looking at the cash flow pro- jections, I’m not sure the answer is so clear-cut. While we have negative cash flow for the next few years, the situation looks much brighter down the road. With a 10-year loan, we are obligated to keep the loan and make the interest payments for 10 years. The one-year loans are more flexible. We can borrow the money only in the years we need it. This way we can save on interest payments in the future. Sheldon: Okay. I can see how the flexibility of the one-year loans could save us some money. Those loans also will look better if interest rates come down in future years. Julie: Or they could go higher instead. There’s no way to predict future interest rates, so we might as well just plan on the basis of the current projection of 7 percent per year. Sheldon: Yes, you’re right. So which do you recommend, a 10-year loan or a series of one-year loans? Julie: Well, there’s actually another option as well. We could consider a combination of the two types of loans. We could borrow some money long-term to get the lower interest rate and borrow some money short-term to retain flexibility. Sheldon: That sounds complicated. What we want is a plan that will keep us solvent throughout the 10 years and then leave us with as large a cash balance as possible at the end of the 10 years after paying off all the loans. Could you set this up on a spreadsheet to figure out the best plan? Julie: You bet. I’ll try that and get back to you. Sheldon: Great. Let’s plan to meet again next week when you have your report ready.
You’ll see in the next two sections how Julie carefully develops her spreadsheet model for this cash flow problem.
The objective is to develop a financial plan that will keep the company solvent and then maximize the cash balance in 2024, after all the loans are paid off.
1. What is the advantage of the long-term loan for Everglade? 2. What is the advantage of the series of short-term loans for Everglade? 3. What is the objective for the financial plan that needs to be developed?
Review Questions
4.2 OVERVIEW OF THE PROCESS OF MODELING WITH SPREADSHEETS
You will see later that a linear programming model can be incorporated into a spreadsheet to solve this problem. However, you also will see that the format of this spreadsheet model does not fit readily into any of the categories of linear programming models described in Chapter 3. Even the template given in Figure 3.8 that shows the format for a spreadsheet model of mixed problems (the broadest category of linear programming problems) does not help in formulating the model for the current problem. The reason is that the Everglade cash flow management problem is an example of a more complicated type of linear programming problem (a dynamic problem with many time periods) that requires starting from scratch in carefully formulating the spreadsheet model. Therefore, this example will nicely illustrate the process of modeling with spreadsheets when dealing with complicated problems of any type, including those discussed later in the book that do not fit linear programming.
When presented with a problem like the Everglade problem, the temptation is to jump right in, launch Excel, and start entering a model. Resist this urge. Developing a spreadsheet model without proper planning inevitably leads to a model that is poorly organized and filled with “spaghetti code.”
Part of the challenge of planning and developing a spreadsheet model is that there is no standard procedure to follow. It is more an art than a science. However, to provide you with
Spaghetti code is a term from computer program- ming. It refers to computer code that is not logically organized and thus jumps all over the place, so it is jumbled like a plate of spaghetti.
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some structure as you begin learning this art, we suggest that you follow the modeling process depicted in Figure 4.1 .
As suggested by the figure, the four major steps in this process are to (1) plan, (2) build, (3) test, and (4) analyze the spreadsheet model. The process mainly flows in this order. How- ever, the two-headed arrows between Build and Test indicate a back-and-forth process where testing frequently results in returning to the Build step to fix some problems discovered dur- ing the Test step. This back and forth movement between Build and Test may occur several times until the modeler is satisfied with the model. At the same time that this back and forth movement is occurring, the modeler may be involved with further building of the model. One strategy is to begin with a small version of the model to establish its basic logic and then, after testing verifies its accuracy, to expand to a full-scale model. Even after completing the testing and then the analyzing of the model, the process may return to the Build step or even the Plan step if the Analysis step reveals inadequacies in the model.
Each of these four major steps may also include some detailed steps. For example, Figure 4.1 lists three detailed steps within the Plan step. Initially, when dealing with a fairly complicated problem, it is helpful to take some time to perform each of these detailed steps manually one at a time. However, as you become more experienced with modeling in spread- sheets, you may find yourself merging some of the detailed steps and quickly performing them mentally. An experienced modeler often is able to do some of these steps mentally, without working them out explicitly on paper. However, if you find yourself getting stuck, it is likely that you are missing a key element from one of the previous detailed steps. You then should go back a step or two and make sure that you have thoroughly completed those preceding steps.
We now describe the various components of the modeling process in the context of the Everglade cash flow problem. At the same time, we also point out some common stumbling blocks encountered while building a spreadsheet model and how these can be overcome.
Plan: Visualize Where You Want to Finish One common stumbling block in the modeling process occurs right at the very beginning. Given a complicated situation like the one facing Julie at Everglade, sometimes it can be difficult to decide how to even get started. At this point, it can be helpful to think about where you want to end up. For example, what information should Julie provide in her report to Sheldon? What should the “answer” look like when presenting the recommended approach to the problem?
A modeler might go back and forth between the Build and Test steps several times.
Plan
Build
Test
Analyze
• Visualize where you want to finish
• Do some calculations by hand
• Sketch out a spreadsheet
Start with a
small-scale model
Try different trial solutions to check the logic
Evaluate proposed solutions and/or optimize
with Solver
Expand the model
to full scale
If the solution reveals inadequacies
in the model, return to Plan or Build
FIGURE 4.1 A flow diagram for the general plan-build-test- analyze process for modeling with spreadsheets.
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What kinds of numbers need to be included in the recommendation? The answers to these ques- tions can quickly lead you to the heart of the problem and help get the modeling process started.
The question that Julie is addressing is which loan, or combination of loans, to use and in what amounts. The long-term loan is taken in a single lump sum. Therefore, the “answer” should include a single number indicating how much money to borrow now at the long-term rate. The short-term loan can be taken in any or all of the 10 years, so the “answer” should include 10 numbers indicating how much to borrow at the short-term rate in each given year. These will be the changing cells in the spreadsheet model.
What other numbers should Julie include in her report to Sheldon? The key numbers would be the projected cash balance each year, the amount of the interest payments, and when loan payments are due. These will be output cells in the spreadsheet model.
It is important to distinguish between the numbers that represent decisions (changing cells) and those that represent results (output cells). For instance, it may be tempting to include the cash balances as changing cells. These cells clearly change depending on the decisions made. However, the cash balances are a result of how much is borrowed, how much is paid, and all of the other cash flows. They cannot be chosen independently but instead are a function of the other numbers in the spreadsheet. The distinguishing characteristic of changing cells (the loan amounts) is that they do not depend on anything else. They represent the independent decisions being made. They impact the other numbers, but not vice versa.
At this stage in the process, you should have a clear idea of what the answer will look like, including what and how many changing cells are needed, and what kind of results (output cells) should be obtained.
Plan: Do Some Calculations by Hand When building a model, another common stumbling block can arise when trying to enter a formula in one of the output cells. For example, just how does Julie keep track of the cash balances in the Everglade cash flow problem? What formulas need to be entered? There are a lot of factors that enter into this calculation, so it is easy to get overwhelmed.
If you are getting stuck at this point, it can be a very useful exercise to do some calcula- tions by hand. Just pick some numbers for the changing cells and determine with a calculator or pencil and paper what the results should be. For example, pick some loan amounts for Everglade and then calculate the company’s resulting cash balance at the end of the first couple of years. Let’s say Everglade takes a long-term loan of $6 million and then adds short- term loans of $2 million in 2014 and $5 million in 2015. How much cash would the company have left at the end of 2014 and at the end of 2015?
These two quantities can be calculated by hand as follows. In 2014, Everglade has some initial money in the bank ($1 million), a negative cash flow from its business operations ( 2 $8 million), and a cash inflow from the long-term and short-term loans ($6 million and $2 million, respectively). Thus, the ending balance for 2014 would be:
Ending balance (2014) 5 Starting balance $1 million
1 Cash flow (2014) 2$8 million
1 LT loan (2014) 1$6 million
1 ST loan (2014) 1$2 million $1 million
The calculations for the year 2015 are a little more complicated. In addition to the starting balance left over from 2014 ($1 million), negative cash flow from business operations for 2015 ( 2 $2 million), and a new short-term loan for 2015 ($5 million), the company will need to make interest payments on its 2014 loans as well as pay back the short-term loan from 2014. The ending balance for 2015 is therefore:
Ending balance (2015) 5 Starting balance (from end of 2014) $1 million
1Cash flow (2015) 2$2 million
1ST loan (2015) 1$5 million
2LT interest payment 2(5%)($6 million)
At this point, you should know what changing cells and output cells are needed.
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4.2 Overview of the Process of Modeling with Spreadsheets 129
2ST interest payment 2(7%)($2 million)
2ST loan payback (2014) 2$2 million $1.56 million
Doing calculations by hand can help in a couple of ways. First, it can help clarify what formula should be entered for an output cell. For instance, looking at the by-hand calculations above, it appears that the formula for the ending balance for a particular year should be
Ending balance 5 Starting balance 1 Cash flow 1 Loans 2Interest payments 2 Loan paybacks
It now will be a simple exercise to enter the proper cell references in the formula for the ending balance in the spreadsheet model. Second, hand calculations can help to verify the spreadsheet model. By plugging in a long-term loan of $6 million, along with short-term loans of $2 million in 2014 and $5 million in 2015, into a completed spreadsheet, the ending balances should be the same as calculated above. If they’re not, this suggests an error in the spreadsheet model (assuming the hand calculations are correct).
Plan: Sketch Out a Spreadsheet Any model typically has a large number of different elements that need to be included on the spreadsheet. For the Everglade problem, these would include some data cells (interest rates, starting balance, minimum balances, and cash flows), some changing cells (loan amounts), and a number of output cells (interest payments, loan paybacks, and ending balances). There- fore, a potential stumbling block can arise when trying to organize and lay out the spreadsheet model. Where should all the pieces fit on the spreadsheet? How do you begin putting together the spreadsheet?
Before firing up Excel and blindly entering the various elements, it can be helpful to sketch a layout of the spreadsheet. Is there a logical way to arrange the elements? A little planning at this stage can go a long way toward building a spreadsheet that is well organized. Don’t bother with numbers at this point. Simply sketch out blocks on a piece of paper for the various data cells, changing cells, and output cells, and label them. Concentrate on the layout. Should a block of numbers be laid out in a row or a column, or as a two-dimensional table? Are there common row or column headings for different blocks of cells? If so, try to arrange the blocks in consistent rows or columns so they can utilize a single set of headings. Try to arrange the spreadsheet so that it starts with the data at the top and progresses logically toward the objec- tive cell at the bottom. This will be easier to understand and follow than if the data cells, changing cells, output cells, and objective cell are all scattered throughout the spreadsheet.
A sketch of a potential spreadsheet layout for the Everglade problem is shown in Figure 4.2 . The data cells for the interest rates, starting balance, and minimum cash balance are at the
Hand calculations can clarify what formulas are needed for the output cells.
Plan where the various blocks of data cells, chang- ing cells, and output cells should go on the spread- sheet by sketching your layout ideas on paper.
LT Rate
ST Rate
Start Balance
Minimum Cash
Cash
Flow
LT
Loan
ST
Loan
LT
Interest
ST
Interest
LT
Payback
ST
Payback
Ending
Balance
Minimum
Balance
2014
2015
.
.
.
.
2023
2024
≥
FIGURE 4.2 Sketch of the spreadsheet for Everglade’s cash flow problem.
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130 Chapter Four The Art of Modeling with Spreadsheets
top of the spreadsheet. All of the remaining elements in the spreadsheet then follow the same structure. The rows represent the different years (from 2014 through 2024). All the various cash inflows and outflows are then broken out in the columns, starting with the projected cash flow from the business operations (with data for each of the 10 years), continuing with the loan inflows, interest payments, and loan paybacks, and culminating with the ending balance (calculated for each year). The long-term loan is a one-time loan (in 2014), so it is sketched as a single cell. The short-term loan can occur in any of the 10 years (2014 through 2023), so it is sketched as a block of cells. The interest payments start one year after the loans. The long-term loan is paid back 10 years later (2024).
Organizing the elements with a consistent structure, like in Figure 4.2 , not only saves hav- ing to retype the year labels for each element, but also makes the model easier to understand. Everything that happens in a given year is arranged together in a single row.
It is generally easiest to start sketching the layout with the data. The structure of the rest of the model should then follow the structure of the data cells. For example, once the projected cash flows data are sketched as a vertical column (with each year in a row), then it follows that the other cash flows should be structured the same way.
There is also a logical progression to the spreadsheet. The data for the problem are located at the top and left of the spreadsheet. Then, since the cash flow, loan amounts, interest pay- ments, and loan paybacks are all part of the calculation for the ending balance, the columns are arranged this way, with the ending balance directly to the right of all these other elements. Since Sheldon has indicated that the objective is to maximize the ending balance in 2024, this cell is designated to be the objective cell.
Each year, the balance must be greater than the minimum required balance ($500,000). Since this will be a constraint in the model, it is logical to arrange the balance and minimum balance blocks of numbers adjacent to each other in the spreadsheet. You can put the $ signs on the sketch to remind yourself that these will be constraints.
Build: Start with a Small Version of the Spreadsheet Once you’ve thought about a logical layout for the spreadsheet, it is finally time to open a new worksheet in Excel and start building the model. If it is a complicated model, you may want to start by building a small, readily manageable version of the model. The idea is to first make sure that you’ve got the logic of the model worked out correctly for the small version before expanding the model to full scale.
For example, in the Everglade problem, we could get started by building a model for just the first two years (2014 and 2015), like the spreadsheet shown in Figure 4.3 .
This spreadsheet is set up to follow the layout suggested in the sketch of Figure 4.2 . The loan amounts are in columns D and E. Since the interest payments are not due until the fol- lowing year, the formulas in columns F and G refer to the loan amounts from the preceding year (LTLoan, or D11, for the long-term loan, and E11 for the short-term loan). The loan pay- ments are calculated in columns H and I. Column H is blank because the long-term loan does not need to be repaid until 2024. The short-term loan is repaid one year later, so the formula in cell I12 refers to the short-term loan taken the preceding year (cell E11). The ending balance in 2014 is the starting balance plus the sum of all the various cash flows that occur in 2014 (cells C11:I11). The ending balance in 2015 is the ending balance in 2014 (cell J11) plus the sum of all the various cash flows that occur in 2015 (cells C12:I12). All these formulas are summarized below the spreadsheet in Figure 4.3 .
Building a small version of the spreadsheet works very well for spreadsheets that have a time dimension. For example, instead of jumping right into a 10-year planning problem, we can start with the simpler problem of just looking at a couple of years. Once this smaller model is working correctly, you then can expand the model to 10 years.
Even if a spreadsheet model does not have a time dimension, the same concept of starting small can be applied. For example, if certain constraints considerably complicate a problem, start by working on a simpler problem without the difficult constraints. Get the simple model working and then move on to tackle the difficult constraints. If a model has many sets of out- put cells, you can build up a model piece by piece by working on one set of output cells at a time, making sure each set works correctly before moving on to the next.
The sketch of a spreadsheet in Figure 4.2 has a logical progression, starting with the data on the top left and then moving through the calculations toward the objective cell on the bottom right.
Work out the logic for a small version of the spreadsheet model before expanding to full size.
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4.2 Overview of the Process of Modeling with Spreadsheets 131
Test: Test the Small Version of the Model If you do start with a small version of the model first, be sure to test this version thoroughly to make sure that all the logic is correct. It is far better to fix a problem early, while the spread- sheet is still a manageable size, rather than later after an error has been propagated throughout a much larger spreadsheet.
To test the spreadsheet, try entering values in the changing cells for which you know what the values of the output cells should be, and then see if the spreadsheet gives the results that you expect. For example, in Figure 4.3 , if zeroes are entered for the loan amounts, then the interest payments and loan payback quantities also should be zero. If $1 million is borrowed for both the long-term loan and the short-term loan, then the interest payments the following year should be $50,000 and $70,000, respectively. (Recall that the interest rates are 5 percent and 7 percent, respectively.) If Everglade takes out a $6 million long-term loan and a $2 million short-term loan in 2014, plus a $5 million short-term loan in 2015, then the ending balances should be $1 million for 2014 and $1.56 million for 2015 (based on the calculations done earlier by hand). All these tests work correctly for the spreadsheet in Figure 4.3 , so we can be fairly certain that it is correct.
If the output cells are not giving the results that you expect, then carefully look through the formulas to see if you can determine and fix the problem. Section 4.4 will give further guid- ance on some ways to debug a spreadsheet model.
Build: Expand the Model to Full-Scale Size Once a small version of the spreadsheet has been tested to make sure all the formulas are correct and everything is working properly, the model can be expanded to full-scale size. Excel’s fill commands often can be used to quickly copy the formulas into the remainder of
Try entering numbers in the changing cells for which you know what the values of the output cells should be.
(all cash figures in millions of dollars)
1
A B C D IHGFE J K
2
3
4
5
6
7
8
9
10
11
12
Everglade Cash Flow Management Problem (Years 2014 and 2015)
LT Rate
ST Rate
1
0.5
Start Balance
Minimum Cash
2014
2015
Year
9
10
11
12
F G
LT ST
H
LT
J K L
Ending
Balance
Minimum
Balance
Range Name
LTLoan
LTRate
MinimumCash
StartBalance
STRate
Cell
D11
C3
C7
C6
C4
Cash
Flow
LT
Loan
LT
Interest
–0.30
ST
Interest
LT
Payback
ST
Payback
L
Ending
Balance
Minimum
Balance
–0.14 –2.00
2
5
1.00
1.56
≥
≥
0.50
0.50
ST
Loan
= –LTRate*LTLoan = –STRate*E11 = –E11
Payback
I
ST
PaybackInterest Interest
=StartBalance+SUM(C11:I11)
=J11+SUM(C12:I12)
≥ ≥
=MinimumCash
=MinimumCash
5%
7%
–8
–2
6
FIGURE 4.3 A small version (years 2014 and 2015 only) of the spreadsheet for the Everglade cash flow management problem.
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132 Chapter Four The Art of Modeling with Spreadsheets
the model. For Figure 4.3 , the formulas in columns F, G, I, J, and L can be copied using the Fill Down command in the Editing Group of the Home tab to obtain all the formulas shown in Figure 4.4 . For example, selecting cells G12:G21 and choosing Fill Down will take the formula in cell G12 and copy it (after adjusting the cell address in Column E for the formula) into cells G13 through G21.
Before using the fill commands to copy formulas, be sure that the relative and absolute references have been used appropriately. (Appendix A provides details about relative and absolute references.) For example, in G12 ( 5 2 STRate* E11) using a range name for STRate makes this an absolute reference. When copied into cells G13:G21, the short-term loan inter- est rate used will always be the value in STRate (C4). The reference to E11 (the loan amount from the previous year) is a relative reference. E11 is two cells to the left and one cell up. When the formula is copied from G12 into G13:G21, the reference in each of these cells will continue to be two cells to the left and one cell up. This is exactly what we want, since we always want the interest payment to be based on the short-term loan that was taken one year ago (two cells to the left and one cell up).
After using the Fill Down command to copy the formulas in columns F, G, I, J, and L and entering the LT loan payback into cell H21, the complete model appears as shown in Figure 4.4 .
Test: Test the Full-Scale Version of the Model Just as it was important to test the small version of the model, it needs to be tested again after it is expanded to full-scale size. The procedure is the same one followed for testing the small version, including the ideas that will be presented in Section 4.4 for debugging a spreadsheet model.
Analyze: Analyze the Model Before using Solver, the spreadsheet in Figure 4.4 is merely an evaluative model for Everglade. It can be used to evaluate any proposed solution, including quickly determining what interest and loan payments will be required and what the resulting balances will be at the end of each year. For example, LTLoan (D11) and STLoan (E11:E20) in Figure 4.4 show one possible plan, which turns out to be unacceptable because EndingBalance (J11:J21) indicates that a negative ending balance would result in three of the years.
To optimize the model, Solver is used as shown in Figure 4.5 to specify the objective cell, the changing cells, and the constraints. Everglade management wants to find a combination of loans that will keep the company solvent throughout the next 10 years (2014–2023) and then will leave as large a cash balance as possible in 2024 after paying off all the loans. Therefore, the objective cell to be maximized is EndBalance (J21) and the changing cells are the loan amounts LTLoan (D11) and STLoan (E11:E20). To assure that Everglade maintains a mini- mum balance of at least $500,000 at the end of each year, the constraints for the model are EndingBalance (J11:J21) $ MinimumBalance (L11:L21).
After running Solver, the optimal solution is shown in Figure 4.5 . The changing cells, LTLoan (D11) and STLoan (E11:E20) give the loan amounts in the various years. The objec- tive cell EndBalance (J21) indicates that the ending balance in 2024 will be $5.39 million.
Conclusion of the Case Study The spreadsheet model developed by Everglade’s CFO, Julie Lee, is the one shown in Figure 4.5 . Her next step is to submit a report to her CEO, Sheldon Lee, that recommends the plan obtained by this model.
Soon thereafter, Sheldon and Julie meet to discuss her report.
Sheldon: Thanks for your report, Julie. Excellent job. Your spreadsheet really lays every- thing out in a very understandable way. Julie: Thanks. It took a little while to get the spreadsheet organized properly and to make sure it was operating correctly, but I think the time spent was worthwhile. Sheldon: Yes, it was. You can’t rush those things. But one thing is still bothering me. Julie: What’s that? Sheldon: It has to do with our forecasts for the company’s future cash flows. We have been assuming that the cash flows in the coming years will be the ones shown in column C
Excel Tip: A shortcut for filling down or filling across to the right is to select the cell you want to copy, click on the fill handle (the small box in the lower right-hand corner of the selection rectangle), and drag through the cells you want to fill.
Excel Tip: A shortcut for changing a cell reference from relative to absolute is to press the F4 key on a PC or command-T on a Mac.
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4.2 Overview of the Process of Modeling with Spreadsheets 133
(all cash figures in millions of dollars)
1
A B C D IHGFE J K
2
3
4
5
6
7
8
9
10
13
14
15
16
17
18
19
20
21
11
12
Everglade Cash Flow Management Problem
LT Rate
ST Rate
Start Balance
Minimum Cash
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
Year
–8
–2
–4
3
6
3
–4
7
–2
10
Cash
Flow
LT
Loan
LT
Interest
ST
Interest
LT
Payback
ST
Payback
L
Ending
Balance
Minimum
Balance
≥
≥
≥
≥ ≥
≥
≥
≥
≥ ≥
≥
ST
Loan
–0.30
–0.30
–0.30
–0.30
–0.30
–0.30
–0.30
–0.30
–0.30
–0.30
1.00
1.56
–8.09
–5.39
0.31
3.01
–1.29
5.41
3.11
12.81
6.51
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
–2
–5
0
0
0
0
0
0
0
0
–0.14
–0.35
0
0
0
0
0
0
0
0 –6
IHGF J K
9
10
14
15
16
17
18
19
20
21
11
12
13
LT
Interest
ST
Interest
LT
Payback
ST
Payback
L
Ending
Balance
Minimum
Balance
≥
≥
≥
≥ ≥
≥
≥
≥
≥ ≥
≥
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=StartBalance+SUM(C11:I11) =MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=J11+SUM(C12:I12)
=J12+SUM(C13:I13)
=J13+SUM(C14:I14)
=J14+SUM(C15:I15)
=J15+SUM(C16:I16)
=J16+SUM(C17:I17)
=J17+SUM(C18:I18)
=J18+SUM(C19:I19)
=J19+SUM(C20:I20)
=–E11
=–E12
=–E13
=–E14
=–E15
=–E16
=–E17
=–E18
=–E19
=–E20
=–STRate*E11
=–STRate*E12
=–STRate*E13
=–STRate*E14
=–STRate*E15
=–STRate*E16
=–STRate*E17
=–STRate*E18
=–STRate*E19
=–STRate*E20 =–LTLoan
Range Name
CashFlow
EndBalance
EndingBalance
LTLoan
LTRate
MinimumBalance
MinimumCash
StartBalance
STLoan
STRate
Cells
C11:C20
J21
J11:J21
D11
C3
L11:L21
C7
C6
E11:E20
C4
5%
7%
1
0.5
6
=J20+SUM(C21:I21)
5
0
0
0
0
0
0
0
2
0
FIGURE 4.4 A complete spreadsheet model for the Everglade cash flow management problem, including the equations entered into the objective cell EndBalance (J21) and all the other output cells, to be used before calling on Solver. The entries in the changing cells, LTLoan (D11) and STLoan (E11:E20), are only a trial solution at this stage.
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134 Chapter Four The Art of Modeling with Spreadsheets
(all cash figures in millions of dollars)
1
A B C D IHGFE J K
2
3
4
5
6
7
8
9
10
13
14
15
16
17
18
19
20
21
11
12
Everglade Cash Flow Management Problem
LT Rate
ST Rate
Start Balance
Minimum Cash
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
Year
–8
–2
–4
3
6
3
–4
7
–2
Cash
Flow
LT
Loan
LT
Interest
ST
Interest
LT
Payback
ST
Payback
L
Ending
Balance
Minimum
Balance
ST
Loan
2.85
5.28
9.88
7.81
2.59
0
4.23
0
0
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
0.50
0.50
0.50
0.50
0.50
0.50
0.50
2.74
0.51
10.27
5.39
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
–2.85
–5.28
–9.88
–7.81
–2.59
0
– 4.23
0
0
0
–0.20
–0.37
–0.69
–0.55
–0.18
0
–0.30
0
0
0 – 4.65
Range Name
CashFlow
EndBalance
EndingBalance
LTLoan
LTRate
MinimumBalance
MinimumCash
StartBalance
STLoan
STRate
Cells
C11:C20
J21
J11:J21
D11
C3
L11:L21
C7
C6
E11:E20
C4
IHGF J K
9
10
14
15
16
17
18
19
20
21
11
12
13
LT
Interest
ST
Interest
LT
Payback
ST
Payback
L
Ending
Balance
Minimum
Balance
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
= StartBalance+SUM(C11:I11) =MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=J11+SUM(C12:I12)
=J12+SUM(C13:I13)
=J13+SUM(C14:I14)
=J14+SUM(C15:I15)
=J15+SUM(C16:I16)
=J16+SUM(C17:I17)
=J17+SUM(C18:I18)
=J18+SUM(C19:I19)
=J19+SUM(C20:I20)
=–E11
=–E12
=–E13
=–E14
=–E15
=–E16
=–E17
=–E18
=–E19
=–E20
=–STRate*E11
=–STRate*E12
=–STRate*E13
=–STRate*E14
=–STRate*E15
=–STRate*E16
=–STRate*E17
=–STRate*E18
=–STRate*E19
=–STRate*E20 =–LTLoan
5%
7%
1
0.5
10
4.65
0
=J20+SUM(C21:I21)
≥
≥
≥ ≥ ≥
≥
≥
≥
≥ ≥
≥
≥
≥
≥
≥ ≥
≥
≥
≥
≥ ≥
≥
Solver Parameters Set Objective Cell: EndBalance To: Max By Changing Variable Cells: LTLoan, STLoan
Subject to the Constraints: EndingBalance >= MinimumBalance
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 4.5 A complete spreadsheet model for the Everglade cash flow management problem after calling on Solver to obtain the optimal solu- tion shown in the changing cells, LTLoan (D11) and STLoan (E11:E20). The objective cell EndBalance (J21) indicates that the resulting cash balance in 2024 will be $5.39 million if all the data cells prove to be accurate.
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4.3 Some Guidelines for Building “Good” Spreadsheet Models 135
of your spreadsheet. Those are good estimates, but we both know that they are only esti- mates. A lot of changes that we can’t foresee now are likely to occur over the next 10 years. When there is a shift in the economy, or when other unexpected developments occur that impact the company, those cash flows can change a lot. How do we know if your recom- mended plan will still be a good one if those kinds of changes occur? Julie: A very good question. To answer it, we should do some what-if analysis to see what would happen if those kinds of changes occur. Now that the spreadsheet is set up properly, it will be very easy to do that by simply changing some of the cash flows in column C and seeing what would happen with the current plan. You can try out any change or changes you want and immediately see the effect. Each time you change a future cash flow, you also have the option of trying out changes on short-term loan amounts to see what kind of adjustments would be needed to maintain a balance of at least $500,000 in every year.
OK, are you ready? Shall we do some what-if analysis now? Sheldon: Let’s do.
Fortunately, Julie had set up the spreadsheet properly (providing a data cell for the cash flow in each of the next 10 years) to enable performing what-if analysis immediately by simply trying different numbers in some of these data cells. (The next chapter will focus on describing the importance of what-if analysis and alternative ways of performing this kind of analysis.) After spending half an hour trying different numbers, Sheldon and Julie conclude that the plan in Figure 4.5 will be a sound initial financial plan for the next 10 years, even if future cash flows deviate somewhat from current forecasts. If deviations do occur, adjust- ments will of course need to be made in the short-term loan amounts. At any point, Julie also will have the option of returning to the company’s bank to try to arrange another long-term loan for the remainder of the 10 years at a lower interest rate than that offered for short-term loans. If so, essentially the same spreadsheet model as in Figure 4.5 can be used, along with Solver, to find the optimal adjusted financial plan for the remainder of the 10 years.
A management science technique called computer simulation provides another effective way of taking the uncertainty of future cash flows into account. Chapters 12 and 13 will describe this technique and Section 13.4 will be devoted to continuing the analysis of this same case study.
Providing a data cell for each piece of data makes it easy to check what would happen if the correct value for a piece of data differs from its initial estimate.
1. What is a good way to get started with a spreadsheet model if you don’t even know where to begin?
2. What are two ways in which doing calculations by hand can help you? 3. Describe a useful way to get started organizing and laying out a spreadsheet. 4. What types of values should be put into the changing cells to test the model? 5. What is the difference between an absolute cell reference and a relative cell reference?
Review Questions
4.3 SOME GUIDELINES FOR BUILDING “GOOD” SPREADSHEET MODELS
There are many ways to set up a model on a spreadsheet. While one of the benefits of spread- sheets is the flexibility they offer, this flexibility also can be dangerous. Although Excel pro- vides many features (such as range names, shading, borders, etc.) that allow you to create “good” spreadsheet models that are easy to understand, easy to debug, and easy to modify, it is also easy to create “bad” spreadsheet models that are difficult to understand, difficult to debug, and difficult to modify. The goal of this section is to provide some guidelines that will help you to create “good” spreadsheet models.
Enter the Data First Any spreadsheet model is driven by the data in the spreadsheet. The form of the entire model is built around the structure of the data. Therefore, it is always a good idea to enter and care- fully lay out all the data before you begin to set up the rest of the model. The model structure then can conform to the layout of the data as closely as possible.
All the data should be laid out on the spreadsheet before beginning to formu- late the rest of the spread- sheet model.
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Often, it is easier to set up the rest of the model when the data are already on the spread- sheet. In the Everglade problem (see Figure 4.5 ), the data for the cash flows have been laid out in the first columns of the spreadsheet (B and C), with the year labels in column B and the data in cells C11:C20. Once the data are in place, the layout for the rest of the model quickly falls into place around the structure of the data. It is only logical to lay out the changing cells and output cells using the same structure, with each of the various cash flows in columns that utilize the same row labels from column B.
Now reconsider the spreadsheet model developed in Section 2.2 for the Wyndor Glass Co. problem. The spreadsheet is repeated here in Figure 4.6 . The data for the Hours Used per Unit Produced have been laid out in the center of the spreadsheet in cells C7:D9. The output cells, HoursUsed (E7:E9), then have been placed immediately to the right of these data and to the left of the data on HoursAvailable(G7:G9), where the row labels for these output cells are the same as for all these data. This makes it easy to interpret the three constraints being laid out in rows 7–9 of the spreadsheet model. Next, the changing cells and objective cell have been placed together in row 12 below the data, where the column labels for the changing cells are the same as for the columns of data above.
The locations of the data occasionally will need to be shifted somewhat to better accom- modate the overall model. However, with this caveat, the model structure generally should conform to the data as closely as possible.
Organize and Clearly Identify the Data Related data should be grouped together in a convenient format and entered into the spread- sheet with labels that clearly identify the data. For data that are laid out in tabular form, the table should have a heading that provides a general description of the data and then each row and column should have a label that will identify each entry in the table. The units of the data also should be identified. Different types of data should be well separated in the spreadsheet. However, if two tables need to use the same labels for either their rows or columns, then be consistent in making them either rows in both tables or columns in both tables.
In the Wyndor Glass Co. problem ( Figure 4.6 ), the three sets of data have been grouped into tables and clearly labeled Unit Profit, Hours Used per Unit Produced, and Hours Avail- able. The units of the data are identified (dollar signs are included in the unit profit data and hours are indicated in the labels of the time data). Finally, all three data tables make consistent use of rows and columns. Since the Unit Profit data have their product labels (Doors and Win- dows) in columns C and D, the Hours Used per Unit Produced data use this same structure.
Provide labels in the spreadsheet that clearly identify all the data.
Welch’s, Inc., is the world’s largest processor of Concord and Niagara grapes with annual sales surpassing $550 million per year. Such products as Welch’s grape jelly and Welch’s grape juice have been enjoyed by generations of American consumers.
Every September, growers begin delivering grapes to processing plants that then press the raw grapes into juice. Time must pass before the grape juice is ready for conver- sion into finished jams, jellies, juices, and concentrates.
Deciding how to use the grape crop is a complex task given changing demand and uncertain crop quality and quantity. Typical decisions include what recipes to use for major product groups, the transfer of grape juice between plants, and the mode of transportation for these transfers.
Because Welch’s lacked a formal system for optimizing raw material movement and the recipes used for produc- tion, a management science team developed a preliminary linear programming model. This was a large model with 8,000 decision variables that focused on the component level of detail. Small-scale testing proved that the model worked.
To make the model more useful, the team then revised it by aggregating demand by product group rather than by component. This reduced its size to 324 decision variables and 361 functional constraints. The model then was incor- porated into a spreadsheet.
The company has run the continually updated version of this spreadsheet model each month since 1994 to pro- vide senior management with information on the optimal logistics plan generated by Solver. The savings from using and optimizing this model were approximately $150,000 in the first year alone. A major advantage of incorporating the linear programming model into a spreadsheet has been the ease of explaining the model to managers with dif- fering levels of mathematical understanding. This has led to a widespread appreciation of the management science approach for both this application and others.
Source: E. W. Schuster and S. J. Allen, “Raw Material Management at Welch’s, Inc.,” Interfaces 28, no. 5 (September–October 1998), pp. 13–24. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
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4.3 Some Guidelines for Building “Good” Spreadsheet Models 137
This structure also is carried through to the changing cells (Units Produced). Similarly, the data for each plant (rows 7–9) are in the rows for both the Hours Used per Unit Produced data and the Hours Available data. Keeping the data oriented the same way is not only less confusing, but it also makes it possible to use the SUMPRODUCT function. Recall that the SUMPRODUCT function assumes that the two ranges are exactly the same shape (i.e., the same number of rows and columns). If the Unit Profit data and the Units Produced data had not been oriented the same way (e.g., one in a column and the other in a row), it would not have been possible to use the SUMPRODUCT function in the Total Profit calculation.
Similarly, for the Everglade problem ( Figure 4.5 ), the five sets of data have been grouped into cells and tables and clearly labeled ST Rate, LT Rate, Start Balance, Cash Flow, and Minimum Cash. The units of the data are identified (cells F5:I5 specify that all cash figures are in millions of dollars), and all the tables make consistent use of rows and columns (years in the rows).
Enter Each Piece of Data into One Cell Only If a piece of data is needed in more than one formula, then refer to the original data cell rather than repeating the data in additional places. This makes the model much easier to modify. If the value of that piece of data changes, it only needs to be changed in one place. You do not need to search through the entire model to find all the places where the data value appears.
For example, in the Everglade problem ( Figure 4.5 ), there is a company policy of maintaining a cash balance of at least $500,000 at all times. This translates into a constraint for the minimum
Every formula using the same piece of data should refer to the same single data cell.
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
2
12
18
Windows
Doors
2 6
Windows
$3,600
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
5
E
Hours
Used6
11
12
7
8
9
=SUMPRODUCT(C7:D7,UnitsProduced)
G
Total Profit
=SUMPRODUCT(Unit Profit,UnitsProduced)
=SUMPRODUCT(C8:D8,UnitsProduced)
=SUMPRODUCT(C9:D9,UnitsProduced)
Range Name
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
TotalProfit
UnitProfit
UnitsProduced
Cells
G7:G9
E7:E9
C7:D9
G12
C4:D4
C12:D12
$300 $500
1
0
3
0
2
2
4
12
18
≤ ≤ ≤
Solver Parameters Set Objective Cell: TotalProfit To: Max By Changing Variable Cells: UnitsProduced
Subject to the Constraints: HoursUsed <= HoursAvailable
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 4.6 The spreadsheet model formulated in Section 2.2 for the Wyndor Glass Co. product-mix problem.
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138 Chapter Four The Art of Modeling with Spreadsheets
balance of $500,000 at the end of each year. Rather than entering the minimum cash position of 0.5 (in millions of dollars) into all the cells in column L, it is entered once in MinimumCash (C7) and then referred to by the cells in MinimumBalance (L11:L21). Then, if this policy were to change to, say, a minimum of $200,000 cash, the number would need to be changed in only one place.
Separate Data from Formulas Avoid using numbers directly in formulas. Instead, enter any needed numbers into data cells and then refer to the data cells as needed. For example, in the Everglade problem ( Figure 4.5 ), all the data (the interest rates, starting balance, minimum cash, and projected cash flows) are entered into separate data cells on the spreadsheet. When these numbers are needed to cal- culate the interest charges (in columns F and G), loan payments (in column H and I), ending balances (column J), and minimum balances (column L), the data cells are referred to rather than entering these numbers directly in the formulas.
Separating the data from the formulas has a couple advantages. First, all the data are vis- ible on the spreadsheet rather than buried in formulas. Seeing all the data makes the model easier to interpret. Second, the model is easier to modify since changing data only requires modifying the corresponding data cells. You don’t need to modify any formulas. This proves to be very important when it comes time to perform what-if analysis to see what the effect would be if some of the estimates in the data cells were to take on other plausible values.
Keep It Simple Avoid the use of powerful Excel functions when simpler functions are available that are eas- ier to interpret. As much as possible, stick to SUMPRODUCT or SUM functions. This makes the model easier to understand and also helps to ensure that the model will be linear. (Linear models are considerably easier to solve than others.) Try to keep formulas short and simple. If a complicated formula is required, break it out into intermediate calculations with subtotals. For example, in the Everglade spreadsheet, each element of the loan payments is broken out explicitly: LT Interest, ST Interest, LT Payback, and ST Payback. Some of these columns could have been combined (e.g., into two columns with LT Payments and ST Payments, or even into one column for all Loan Payments). However, this makes the formulas more com- plicated and also makes the model harder to test and debug. As laid out, the individual formu- las for the loan payments are so simple that their values can be predicted easily without even looking at the formula. This simplifies the testing and debugging of the model.
Use Range Names One way to refer to a block of related cells (or even a single cell) in a spreadsheet formula is to use its cell address (e.g., L11:L21 or C3). However, when reading the formula, this requires looking at that part of the spreadsheet to see what kind of information is given there. As mentioned previously in Sections 1.2 and 2.2, a better alternative is to assign a descriptive range name to the block of cells that immediately identifies what is there. (This is done by selecting the block of cells, clicking on the name box on the left of the formula bar above the spreadsheet, and then typing a name.) This is especially helpful when writing a formula for an output cell. Writing the formula in terms of range names instead of cell addresses makes the formula much easier to interpret. Range names also make the description of the model in Solver much easier to understand.
Figure 4.5 illustrates the use of range names for the Everglade spreadsheet model. For example, consider the formula for long-term interest in cell F12. Since the long-term rate is given in cell C3 and the long-term loan amount is in cell D11, the formula for the long-term interest could have been written as 5 2 C3*D11. However, by using the range name LTRate for cell C3 and the range name LTLoan for cell D11, the formula instead becomes 5 2 LTRate*LTLoan, which is much easier to interpret at a glance.
On the other hand, be aware that it is easy to get carried away with defining range names. Defining too many range names can be more trouble than it is worth. For example, when related data are grouped together in a table, we recommend giving a range name only for the entire table rather than for the individual rows and columns. In general, we suggest defining range names only for each group of data cells, the changing cells, the objective cell, and both sides of each group of constraints (the left-hand side and the right-hand side).
Formulas should refer to data cells for any needed numbers.
Make the spreadsheet as easy to interpret as possible.
Range names make formulas much easier to interpret.
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4.3 Some Guidelines for Building “Good” Spreadsheet Models 139
Care also should be taken to assure that it is easy to quickly identify which cells are referred to by a particular range name. Use a name that corresponds exactly to the label on the spreadsheet. For example, in Figure 4.5 , columns J and L are labeled Ending Balance and Minimum Balance on the spreadsheet, so we use the range names EndingBalance and MinimumBalance. Using exactly the same name as the label on the spreadsheet makes it quick and easy to find the cells that are referred to by a range name.
When desired, a list of all the range names and their corresponding cell addresses can be pasted directly into the spreadsheet by choosing Paste from the Use in Formula menu on the Formulas tab, and then clicking Paste List. Such a list (after reformatting) is included below essentially all the spreadsheets displayed in this text.
When modifying an existing model that utilizes range names, care should be taken to assure that the range names continue to refer to the correct range of cells. When inserting a row or column into a spreadsheet model, it is helpful to insert the row or column into the middle of a range rather than at the end. For example, to add another product to a product- mix model with four products, add a column between Products 2 and 3 rather than after Product 4. This will automatically extend the relevant range names to span across all five columns since these range names will continue to refer to everything between Product 1 and Product 4, including the newly inserted column for the fifth product. Similarly, delet- ing a row or column from the middle of a range will contract the span of the relevant range names appropriately. You can double-check the cells that are referred to by a range name by choosing that range name from the name box (on the left of the formula bar above the spreadsheet). This will highlight the cells that are referred to by the chosen range name.
Use Relative and Absolute References to Simplify Copying Formulas Whenever multiple related formulas will be needed, try to enter the formula just once and then use Excel’s fill commands to replicate the formula. Not only is this quicker than retyping the formula, but it is also less prone to error.
We saw a good example of this when discussing the expansion of the model to full-scale size in the preceding section. Starting with the two-year spreadsheet in Figure 4.3 , fill com- mands were used to copy the formulas in columns F, G, I, J, and L for the remaining years to create the full-scale, 10-year spreadsheet in Figure 4.4 .
Using relative and absolute references for related formulas not only aids in building a model but also makes it easier to modify an existing model or template. For example, suppose that you have formulated a spreadsheet model for a product-mix problem but now wish to modify the model to add another resource. This requires inserting a row into the spreadsheet. If the output cells are written with proper relative and absolute references, then it is simple to copy the existing formulas into the inserted row.
Use Borders, Shading, and Colors to Distinguish between Cell Types It is important to be able to easily distinguish between the data cells, changing cells, output cells, and objective cell in a spreadsheet. One way to do this is to use different borders and cell shading for each of these different types of cells. In the text, data cells appear lightly shaded, changing cells are shaded a medium amount with a light border, output cells appear with no shading, and the objective cell is shaded darkly with a heavy border.
In the spreadsheet files in MS Courseware, data cells are light blue, changing cells are yel- low, and the objective cell is orange. Obviously, you may use any scheme that you like. The important thing is to be consistent, so that you can quickly recognize the types of cells. Then, when you want to examine the cells of a certain type, the color will immediately guide you there.
Show the Entire Model on the Spreadsheet Solver uses a combination of the spreadsheet and the Solver dialog box (or the model pane in RSPE) to specify the model to be solved. Therefore, it is possible to include certain elements of the model (such as the #, 5 , or $ signs and/or the right-hand sides of the constraints) in
Spaces are not allowed in range names. When a range name has more than one word, we have used capital letters to dis- tinguish the start of each new word in a range name (e.g., MinimumBalance). Another way is to use the underscore character (e.g., Minimum_Balance).
Excel’s fill commands provide a quick and reliable way to replicate a formula into multiple cells.
Make it easy to spot all the cells of the same type.
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140 Chapter Four The Art of Modeling with Spreadsheets
Solver without displaying them in the spreadsheet. However, we strongly recommend that every element of the model be displayed on the spreadsheet. Every person using or adapt- ing the model, or referring back to it later, needs to be able to interpret the model. This is much easier to do by viewing the model on the spreadsheet than by trying to decipher it from Solver. Furthermore, a printout of the spreadsheet does not include information from Solver.
In particular, all the elements of a constraint should be displayed on the spreadsheet. For each constraint, three adjacent cells should be used for the total of the left-hand side, the #, 5 , or $ sign in the middle, and the right-hand side. (Note in Figure 4.5 that this was done in columns J, K, and L of the spreadsheet for the Everglade problem.) As mentioned earlier, the changing cells and objective cell should be highlighted in some manner (e.g., with borders and/or cell shading). A good test is that you should not need to go to Solver to determine any element of the model. You should be able to identify the changing cells, the objective cell, and all the constraints in the model just by looking at the spreadsheet.
A Poor Spreadsheet Model It is certainly possible to set up a linear programming spreadsheet model without utilizing any of these ideas. Figure 4.7 shows an alternative spreadsheet formulation for the Everglade problem that violates nearly every one of these guidelines. This formulation can still be solved using Solver, which in fact yields the same optimal solution as in Figure 4.5 . However, the
Display every element of the model on the spread- sheet rather than relying on only Solver to include certain elements.
1
A B C D FE
2
3
4
5
LT ST Ending
Loan
4.65
6
7
8
9
10
11
12
13
14
15
Year
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
Loan
2.85
5.28
9.88
7.81
2.59
0
4.23
0
0
0
2024
Balance
0.50
0.50
0.50
0.50
0.50
0.50
0.50
2.74
0.51
10.27
5.39
A Poor Formulation of the Everglade Cash Flow Problem
3
E
Ending
Balance4
5
6
=1–8+C5+D5
=E5–2+D6–$C$5*(0.05)–D5*(1.07)
7 =E6–4+D7–$C$5*(0.05)–D6*(1.07)
8 =E7+3+D8–$C$5*(0.05)–D7*(1.07)
9 =E8+6+D9–$C$5*(0.05)–D8*(1.07)
10 =E9+3+D10–$C$5*(0.05)–D9*(1.07)
11 =E10–4+D11–$C$5*(0.05)–D10*(1.07)
12 =E11+7+D12–$C$5*(0.05)–D11*(1.07)
13 =E12–2+D13–$C$5*(0.05)–D12*(1.07)
14 =E13+10+D14–$C$5*(0.05)–D13*(1.07)
15 =E14+D15–$C$5*(1.05)–D14*(1.07)
Solver Parameters Set Objective Cell: E15 To: Max By Changing Variable Cells: C5, D5:D14
Subject to the Constraints: E5:E15 >= 0.5
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 4.7 A poor formulation of the spreadsheet model for the Everglade cash flow man- agement problem.
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4.4 Debugging a Spreadsheet Model 141
formulation has many problems. It is not clear which cells yield the solution (borders and/ or shading are not used to highlight the changing cells and objective cell). Without going to Solver, the constraints in the model cannot be identified (the spreadsheet does not show the entire model). The spreadsheet also does not show most of the data. For example, to deter- mine the data used for the projected cash flows, the interest rates, or the starting balance, you need to dig into the formulas in column E (the data are not separate from the formulas). If any of these data change, the actual formulas need to be modified rather than simply changing a number on the spreadsheet. Furthermore, the formulas and the model in Solver are difficult to interpret (range names are not utilized).
Compare Figures 4.5 and 4.7 . Applying the guidelines for good spreadsheet models (as is done for Figure 4.5 ) results in a model that is easier to understand, easier to debug, and easier to modify. This is especially important for models that will have a long life span. If this model is going to be reused months later, the “good” model of Figure 4.5 immediately can be understood, modified, and reapplied as needed, whereas deciphering the spreadsheet model of Figure 4.7 again would be a great challenge.
1. Which part of the model should be entered first on the spreadsheet? 2. Should numbers be included in formulas or entered separately in data cells? 3. How do range names make formulas and the model in Solver easier to interpret? How should
range names be chosen? 4. What are some ways to distinguish data cells, changing cells, output cells, and objective cells
on a spreadsheet? 5. How many cells are needed to completely specify a constraint on a spreadsheet?
Review Questions
4.4 DEBUGGING A SPREADSHEET MODEL
No matter how carefully it is planned and built, even a moderately complicated model usually will not be error-free the first time it is run. Often the mistakes are immediately obvious and quickly corrected. However, sometimes an error is harder to root out. Following the guide- lines in Section 4.3 for developing a good spreadsheet model can make the model much easier to debug. Even so, much like debugging a computer program, debugging a spreadsheet model can be a difficult task. This section presents some tips and a variety of Excel features that can make debugging easier.
As a first step in debugging a spreadsheet model, test the model using the principles dis- cussed in the first subsection on testing in Section 4.2. In particular, try different values for the changing cells for which you can predict the correct result in the output cells and see if they calculate as expected. Values of 0 are good ones to try initially because usually it is then obvious what should be in the output cells. Try other simple values, such as all 1s, where the correct results in the output cells are reasonably obvious. For more complicated values, break out a calculator and do some manual calculations to check the various output cells. Include some very large values for the changing cells to ensure that the calculations are behaving reasonably for these extreme cases.
If you have defined range names, be sure that they still refer to the correct cells. Some- times they can become disjointed when you add rows or columns to the spreadsheet. To test the range names, you can either select the various range names in the name box, which will highlight the selected range in the spreadsheet, or paste the entire list of range names and their references into the spreadsheet.
Carefully study each formula to be sure it is entered correctly. A very useful feature in Excel for checking formulas is the toggle to switch back and forth between viewing the for- mulas in the worksheet and viewing the resulting values in the output cells. By default, Excel shows the values that are calculated by the various output cells in the model. Typing control-~ switches the current worksheet to instead display the formulas in the output cells, as shown in Figure 4.8 . Typing control-~ again switches back to the standard view of displaying the values in the output cells (like Figure 4.5 ).
Debugging a spreadsheet model sometimes is as challenging as debugging a computer program.
If you have added rows or columns to the spreadsheet, make sure that each of the range names still refers to the correct cells.
Excel Tip: Pressing control-~ on a PC (or command-~ on a Mac) toggles the worksheet between viewing values and viewing formulas in all the output cells.
toggle The toggle feature in Excel is a great way to check the formulas for the output cells.
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C onfirm
ing P ages
1 4 2 C
h a p
te r Fo
u r T
he A rt of M
odeling w ith Spreadsheets
(all cash figures in millions of dollars)
1
A B C D IHGFE J K
2
3
4
5
6
7
8
9
10
13
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15
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21
11
12
Everglade Cash Flow Management Problem
LT Rate
ST Rate
Start Balance
Minimum Cash
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
Year
Cash
Flow
LT
Loan
LT
Interest
ST
Interest
LT
Payback
ST
Payback
L
Ending
Balance
Minimum
Balance
ST
Loan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=StartBalance+SUM(C11:I11) =MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=J11+SUM(C12:I12)
=J12+SUM(C13:I13)
=J13+SUM(C14:I14)
=J14+SUM(C15:I15)
=J15+SUM(C16:I16)
=J16+SUM(C17:I17)
=J17+SUM(C18:I18)
=J18+SUM(C19:I19)
=J19+SUM(C20:I20)
=J20+SUM(C21:I21)
=–E11
=–E12
=–E13
=–E14
=–E15
=–E16
=–E17
=–E18
=–E19
=–E20
=–STRate*E11
=–STRate*E12
=–STRate*E13
=–STRate*E14
=–STRate*E15
=–STRate*E16
=–STRate*E17
=–STRate*E18
=–STRate*E19
=–STRate*E20 =–LTLoan
0.05
0.07
1
0.5
–8
–2
–4
3
6
3
–4
7
–2
10
4.65124 2.84759
5.28073
9.88295
7.80732
2.58639
0
4.23256
0
0
0
≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥
FIGURE 4.8 The spreadsheet obtained by toggling the spreadsheet in Figure 4.5 once to replace the values in the output cells by the formulas entered into those cells. Using the toggle feature in Excel once more will restore the view of the spreadsheet shown in Figure 4.5.
h il2
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4.4 Debugging a Spreadsheet Model 143
Another useful set of features built into Excel are the auditing tools . The auditing tools are available in the Formula Auditing group of the Formulas Tab.
The auditing tools can be used to graphically display which cells make direct links to a given cell. For example, selecting LTLoan (D11) in Figure 4.5 and then Trace Dependents generates the arrows on the spreadsheet shown in Figure 4.9 .
You now can immediately see that LTLoan (D11) is used in the calculation of LT Inter- est for every year in column F, in the calculation of LTPayback (H21), and in the calculation of the ending balance in 2014 (J11). This can be very illuminating. Think about what output cells LTLoan should impact directly. There should be an arrow to each of these cells. If, for example, LTLoan is missing from any of the formulas in column F, the error will be immedi- ately revealed by the missing arrow. Similarly, if LTLoan is mistakenly entered in any of the short-term loan output cells, this will show up as extra arrows.
You also can trace backward to see which cells provide the data for any given cell. These can be displayed graphically by choosing Trace Precedents. For example, choosing Trace Precedents for the ST Interest cell for 2015 (G12) displays the arrows shown in Figure 4.10 . These arrows indicate that the ST Interest cell for 2015 (G12) refers to the ST Loan in 2014 (E11) and to STRate (C4).
When you are done, choose Remove Arrows.
Excel’s auditing tools enable you to either trace forward or backward to see the linkages between cells.
(all cash figures in millions of dollars)
1
A B C D IHGFE J K
2
3
4
5
6
7
8
9
10
13
14
15
16
17
18
19
20
21
11
12
Everglade Cash Flow Management Problem
LT Rate
ST Rate
1
0.5
Start Balance
Minimum Cash
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
Year
–8
–2
–4
3
6
3
–4
7
–2
10
Cash
Flow
LT
Loan
LT
Interest
ST
Interest
LT
Payback
ST
Payback
L
Ending
Balance
Minimum
Balance
ST
Loan
2.85
5.28
9.88
7.81
2.59
0
4.23
0
0
0
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
0.50
0.50
0.50
0.50
0.50
0.50
0.50
2.74
0.51
10.27
5.39
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
–2.85
–5.28
–9.88
–7.81
–2.59
0
–4.23
0
0
0
–0.20
–0.37
–0.69
–0.55
–0.18
0
–0.30
0
0
0 –4.65
5%
7%
≥
≥
≥
≥
≥
≥
≥
≥
≥
≥
≥
4.65
FIGURE 4.9 The spreadsheet obtained by using the Excel auditing tools to trace the dependents of the LT Loan value in cell D11 of the spread- sheet in Figure 4.5.
1. What is a good first step for debugging a spreadsheet model? 2. How do you toggle between viewing formulas and viewing values in output cells? 3. Which Excel tool can be used to trace the dependents or precedents for a given cell?
Review Questions
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144 Chapter Four The Art of Modeling with Spreadsheets
1
A B C D IHGFE J K
2
3
4
5
6
7
8
9
10
13
14
15
16
17
18
19
20
21
11
12
Everglade Cash Flow Management Problem
LT Rate
ST Rate
1
0.5
Start Balance
Minimum Cash
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
Year
–8 4.65
–2
–4
3
6
3
–4
7
–2
10
Cash
Flow
LT
Loan
LT
Interest
ST
Interest
LT
Payback
ST
Payback
L
Ending
Balance
Minimum
Balance
ST
Loan
2.85
5.28
9.88
7.81
2.59
0
4.23
0
0
0
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
–0.23
0.50
0.50
0.50
0.50
0.50
0.50
0.50
2.74
0.51
10.27
5.39
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
–2.85
–5.28
–9.88
–7.81
–2.59
0
–4.23
0
0
0
–0.20
–0.37
–0.69
–0.55
–0.18
0
–0.30
0
0
0 –6.65
(all cash figures in millions of dollars)
5%
7%
≥
≥
≥
≥
≥
≥ ≥
≥
≥ ≥
≥
FIGURE 4.10 The spreadsheet obtained by using the Excel auditing tools to trace the precedents of the ST Interest (2015) calculation in cell G12 of the spreadsheet in Figure 4.5.
There is a considerable art to modeling well with spreadsheets. This chapter focuses on providing a foundation for learning this art.
The general process of modeling in spreadsheets has four major steps: (1) plan the spreadsheet model, (2) build the model, (3) test the model, and (4) analyze the model and its results. During the planning step, it is helpful to begin by visualizing where you want to finish and then doing some cal- culations by hand to clarify the needed computations before starting to sketch out a logical layout for the spreadsheet. Then, when you are ready to undertake the building step, it is a good idea to start by building a small, readily manageable version of the model before expanding the model to full-scale size. This enables you to test the small version first to get all the logic straightened out correctly before expanding to a full-scale model and undertaking a final test. After completing all of this, you are ready for the analysis step, which involves applying the model to evaluate proposed solutions and perhaps using Solver to optimize the model.
Using this plan-build-test-analyze process should yield a spreadsheet model, but it doesn’t guarantee that you will obtain a good one. Section 4.3 describes in detail the following guidelines for building “good” spreadsheet models:
• Enter the data first. • Organize and clearly identify the data. • Enter each piece of data into one cell only. • Separate data from formulas. • Keep it simple. • Use range names. • Use relative and absolute references to simplify copying formulas. • Use borders, shading, and colors to distinguish between cell types. • Show the entire model on the spreadsheet.
4.5 Summary
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Chapter 4 Solved Problems 145
Even if all these guidelines are followed, a thorough debugging process may be needed to eliminate the errors that lurk within the initial version of the model. It is important to check whether the output cells are giving correct results for various values of the changing cells. Other items to check include whether range names refer to the appropriate cells and whether formulas have been entered into output cells correctly. Excel provides a number of useful features to aid in the debugging process. One is the ability to toggle the worksheet between viewing the results in the output cells and the formulas entered into those output cells. Several other helpful features are available with Excel’s auditing tools.
Glossary auditing tools A set of tools provided by Excel to aid in debugging a spreadsheet model. (Section 4.4), 143 range name A descriptive name given to a range of cells that immediately identifies what is there. (Section 4.3), 138
toggle The act of switching back and forth between viewing the results in the output cells and viewing the formulas entered into those out- put cells. (Section 4.4), 141
Chapter 4 Excel Files:
Everglade Case Study
Wyndor Example
Everglade Problem 4.12
Everglade Problem 4.13
Excel Add-in:
Risk Solver Platform for Education (RSPE)
Learning Aids for This Chapter in Your MS Courseware
Solved Problems (See the CD-ROM or Website for the Solutions)
4.S1. Production and Inventory Planning Model Surfs Up produces high-end surfboards. A challenge faced by Surfs Up is that their demand is highly seasonal. Demand exceeds production capacity during the warm summer months, but is very low in the winter months. To meet the high demand during the summer, Surfs Up typically pro- duces more surfboards than are needed in the winter months and then carries inventory into the summer months. Their production facility can produce at most 50 boards per month using regular labor at a cost of $125 each. Up to 10 addi- tional boards can be produced by utilizing overtime labor at a cost of $135 each. The boards are sold for $200. Because of storage cost and the opportunity cost of capital, each board held in inventory from one month to the next incurs a cost of $5 per board. Since demand is uncertain, Surfs Up would like to maintain an ending inventory (safety stock) of at least 10 boards during the warm months (May–September) and at least 5 boards during the other months (October–April). It is now the start of January and Surfs Up has 5 boards in inventory. The forecast of demand over the next 12 months is shown in the table below. Formulate and solve a linear programming model in a spreadsheet to determine how many surfboards should be produced each month to maximize total profit.
4.S2. Aggregate Planning: Manpower Hiring/ Firing/Training Cool Power produces air-conditioning units for large commercial properties. Because of the low cost and efficiency of its products, the company has been growing from year to year. Also, seasonality in construction and weather conditions create production require- ments that vary from month to month. Cool Power currently has 10 fully trained employees working in manufacturing. Each trained employee can work 160 hours per month and is paid a monthly wage of $4,000. New trainees can be hired at the beginning of any month. Because of their lack of initial skills and required training, a new trainee provides only 100 hours of useful labor in the first month, but is still paid a full monthly wage of $4,000. Furthermore, because of required interviewing and training, there is a $2,500 hiring cost for each employee hired. After one month, a trainee is considered fully trained. An employee can be fired at the begin- ning of any month, but must be paid two weeks of severance pay ($2,000). Over the next 12 months, Cool Power forecasts the labor requirements shown in the table on the next page. Since manage- ment anticipates higher requirements next year, Cool Power would like to end the year with at least 12 fully trained employees. How many trainees should be hired and/or workers fired in each month to meet the labor requirements at the minimum possible cost? For- mulate and solve a linear programming spreadsheet model.
Forecasted Demand
Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.
10 14 15 20 45 65 85 85 40 30 15 15
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146 Chapter Four The Art of Modeling with Spreadsheets
Labor Requirements (hours)
Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.
1,600 2,000 2,000 2,000 2,800 3,200 3,600 3,200 1,600 1,200 800 800
a. Visualize where you want to finish. What numbers will top management need? What are the decisions that need to be made? What should the objective be?
b. Suppose that Reboot were to produce 5,000 pairs of boots in each of the first two quarters. Calculate by hand the ending inventory, profit from sales, and inventory costs for quarters 1 and 2.
c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell.
E* d. Build a spreadsheet model for quarters 1 and 2, and then thoroughly test the model.
E* e. Expand the model to full scale and then solve it. E*4.4.* The Fairwinds Development Corporation is consider- ing taking part in one or more of three different development projects—A, B, and C—that are about to be launched. Each project requires a significant investment over the next few years and then would be sold upon completion. The projected cash flows (in millions of dollars) associated with each project are shown in the table below.
Year Project A Project B Project C
1 24 28 210 2 26 28 27 3 26 24 27 4 24 24 25 5 0 30 23 6 0 0 44
Fairwinds has $10 million available now and expects to receive $6 million from other projects by the end of each year (1 through 6) that would be available for the ongoing investments the follow- ing year in projects A, B, and C. By acting now, the company may participate in each project either fully, fractionally (with other development partners), or not at all. If Fairwinds participates at less than 100 percent, then all the cash flows associated with that project are reduced proportionally. Company policy requires end- ing each year with a cash balance of at least $1 million. a. Visualize where you want to finish. What numbers
are needed? What are the decisions that need to be made? What should the objective be?
b. Suppose that Fairwinds were to participate in Proj- ect A fully and in Project C at 50 percent. Calculate by hand what the ending cash positions would be after year 1 and year 2.
c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell.
E* d. Build a spreadsheet model for years 1 and 2, and then thoroughly test the model.
E* e. Expand the model to full scale, and then solve it.
We have inserted the symbol E* (for Excel) to the left of each problem or part where Excel should be used. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
E*4.1. Consider the Everglade cash flow problem discussed in this chapter. Suppose that extra cash is kept in an interest- bearing savings account. Assume that any cash left at the end of a year earns 3 percent interest the following year. Make any necessary modifications to the spreadsheet and re-solve. (The original spreadsheet for this problem is available on the CD-ROM.) 4.2.* The Pine Furniture Company makes fine country furni- ture. The company’s current product lines consist of end tables, coffee tables, and dining room tables. The production of each of these tables requires 8, 15, and 80 pounds of pine wood, respec- tively. The tables are handmade and require one hour, two hours, and four hours, respectively. Each table sold generates $50, $100, and $220 profit, respectively. The company has 3,000 pounds of pine wood and 200 hours of labor available for the coming week’s production. The chief operating officer (COO) has asked you to do some spreadsheet modeling with these data to analyze what the product mix should be for the coming week and make a recommendation. a. Visualize where you want to finish. What num-
bers will the COO need? What are the decisions that need to be made? What should the objective be?
b. Suppose that Pine Furniture were to produce three end tables and three dining room tables. Calculate by hand the amount of pine wood and labor that would be required, as well as the profit generated from sales.
c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell.
E* d. Build a spreadsheet model and then solve it.
4.3. Reboot, Inc., is a manufacturer of hiking boots. Demand for boots is highly seasonal. In particular, the demand in the next year is expected to be 3,000, 4,000, 8,000, and 7,000 pairs of boots in quarters 1, 2, 3, and 4, respectively. With its current production facility, the company can produce at most 6,000 pairs of boots in any quarter. Reboot would like to meet all the expected demand, so it will need to carry inventory to meet demand in the later quarters. Each pair of boots sold generates a profit of $20 per pair. Each pair of boots in inventory at the end of a quarter incurs $8 in storage and capital recovery costs. Reboot has 1,000 pairs of boots in inventory at the start of quar- ter 1. Reboot’s top management has given you the assignment of doing some spreadsheet modeling to analyze what the produc- tion schedule should be for the next four quarters and make a recommendation.
Problems
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Chapter 4 Problems 147
a. Visualize where you want to finish. What numbers will Allen require? What are the decisions that need to be made? What should the objective be?
b. Suppose one apprentice is hired in January. Calcu- late by hand how many labor-hours would be avail- able in January and February. Calculate by hand the total costs in January and February.
c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell.
E* d. Build a spreadsheet model for January and February and thoroughly test the model.
E* e. Build and solve a linear programming spreadsheet model to maximize the profit over all 12 months.
4.8. Refer to the scenario described in Problem 3.14 (Chapter 3), but ignore the instructions given there. Focus instead on using spreadsheet modeling to address Web Mercan- tile’s problem by doing the following. a. Visualize where you want to finish. What numbers
will Web Mercantile require? What are the decisions that need to be made? What should the objective be?
b. Suppose that Web Mercantile were to lease 30,000 square feet for all five months and then 20,000 addi- tional square feet for the last three months. Calcu- late the total costs by hand.
c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell.
E* d. Build a spreadsheet model for months 1 and 2, and then thoroughly test the model.
E* e. Expand the model to full scale, and then solve it. 4.9.* Refer to the scenario described in Problem 3.16 (Chapter 3), but ignore the instructions given there. Focus instead on using spreadsheet modeling to address Larry Edison’s problem by doing the following. a. Visualize where you want to finish. What numbers
will Larry require? What are the decisions that need to be made? What should the objective be?
b. Suppose that Larry were to hire three full-time workers for the morning shift, two for the afternoon shift, and four for the evening shift, as well as three part-time workers for each of the four shifts. Cal- culate by hand how many workers would be work- ing at each time of the day and what the total cost would be for the entire day.
c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell.
E* d. Build a spreadsheet model and then solve it. 4.10. Refer to the scenario described in Problem 3.19 (Chapter 3), but ignore the instructions given there. Focus instead on using spreadsheet modeling to address Al Ferris’s problem by doing the following.
4.5. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 4.3. Briefly describe how spread- sheet modeling was applied in this study. Then list the vari- ous financial and nonfinancial benefits that resulted from this study.
4.6. Decorum, Inc., manufactures high-end ceiling fans. Their sales are seasonal with higher demand in the warmer sum- mer months. Typically, sales average 400 units per month. How- ever, in the hot summer months (June, July, and August), sales spike up to 600 units per month. Decorum can produce up to 500 units per month at a cost of $300 each. By bringing in temporary workers, Decorum can produce up to an additional 75 units at a cost of $350 each. Decorum sells the ceiling fans for $500 each. Decorum can carry inventory from one month to the next, but at a cost of $20 per ceiling fan per month. Decorum has 25 units in inventory at the start of January. Assuming Decorum must produce enough ceiling fans to meet demand, how many ceiling fans should Decorum produce each month (using their regular labor force and/or temporary workers) over the course of the next year so as to maximize their total profit?
a. Visualize where you want to finish. What numbers will Decorum require? What are the decisions that need to be made? What should the objective be?
b. Suppose Decorum builds 450 ceiling fans in January and 550 (utilizing temporary workers) in February. Calculate by hand the total costs for January and February.
c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell.
E* d. Build a spreadsheet model for January and February and thoroughly test the model.
E* e. Build and solve a linear programming spreadsheet model to maximize the profit over all 12 months.
4.7. Allen Furniture is a manufacturer of hand-crafted furniture. At the start of January, Allen employs 20 trained craftspeople. They have forecasted their labor needs over the next 12 months as shown in the table below. Each trained craftsperson provides 200 labor-hours per month and is paid a wage of $3,000 per month. Hiring new craftspeople requires advertising, interviewing, and then training at a cost of $2,500 per hire. New hires are called apprentices for their first month. Apprentices spend their first month observing and learning. They are paid $2,000 for the month, but provide no labor. In their second month, apprentices are reclassified as trained craftspeople. The union contract allows for firing a craftsper- son at the beginning of a month, but $1,500 must be given in severance pay. Moreover, at most 10% of the trained crafts- people can be fired in any month. Allen would like to start next year with at least 25 trained craftspeople. How many apprentices should be hired and how many craftspeople should be fired in each month to meet the labor requirements at the minimum possible cost?
Labor Requirements (hours)
Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.
3,400 4,000 4,200 4,200 3,000 2,800 3,000 4,000 4,500 5,000 5,200 4,800
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148 Chapter Four The Art of Modeling with Spreadsheets
E*4.12. Refer to the spreadsheet file named “Everglade Problem 4.12” contained on the CD-ROM. This file contains a formula- tion of the Everglade problem considered in this chapter. How- ever, three errors are included in this formulation. Use the ideas presented in Section 4.4 for debugging a spreadsheet model to find the errors. In particular, try different trial values for which you can predict the correct results, use the toggle to examine all the formu- las, and use the auditing tools to check precedence and dependence relationships among the various changing cells, data cells, and out- put cells. Describe the errors found and how you found them. E*4.13. Refer to the spreadsheet file named “Everglade Problem 4.13” contained on the CD-ROM. This file contains a formula- tion of the Everglade problem considered in this chapter. How- ever, three errors are included in this formulation. Use the ideas presented in Section 4.4 for debugging a spreadsheet model to find the errors. In particular, try different trial values for which you can predict the correct results, use the toggle to examine all the formulas, and use the auditing tools to check precedence and dependence relationships among the various changing cells, data cells, and output cells. Describe the errors found and how you found them.
a. Visualize where you want to finish. What numbers will Al require? What are the decisions that need to be made? What should the objective be?
b. Suppose that Al were to invest $20,000 each in investment A (year 1), investment B (year 2), and investment C (year 2). Calculate by hand what the ending cash position would be after each year.
c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell.
E* d. Build a spreadsheet model for years 1 through 3, and then thoroughly test the model.
E* e. Expand the model to full scale, and then solve it. 4.11. In contrast to the spreadsheet model for the Wyndor Glass Co. product-mix problem shown in Figure 4.6 , the spreadsheet given below is an example of a poorly formulated spreadsheet model for this same problem. Referring to Section 4.3, identify the guidelines violated by the model below. Then, explain how each guideline has been violated and why the model in Figure 4.6 is a better alternative.
1
A B C D
2
3
4
5
6
7
8
Doors Produced
Windows Produced
Hours Used (Plant 1)
Hours Used (Plant 2)
Hours Used (Plant 3)
Total Profit
2
6
2
12
18
$3,600
Wyndor Glass Co. (Poor Formulation)
5
B C
6
7
8
=1*C3+0*C4Hours Used (Plant 1)
=0*C3+2*C4Hours Used (Plant 2)
=3*C3+2*C4Hours Used (Plant 3)
=300*C3+500*C4Total Profit
Solver Parameters
Set Objective Cell: C8 To: Max By Changing Variable Cells:
C3:C4
Subject to the Constraints: C5 <= 4
C6 <= 12
C7 <= 18
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
Case 4-1
Prudent Provisions for Pensions Among its many financial products, the Prudent Financial Ser- vices Corporation (normally referred to as PFS) manages a well-regarded pension fund that is used by a number of com- panies to provide pensions for their employees. PFS’s manage- ment takes pride in the rigorous professional standards used in operating the fund. Since the near-collapse of the financial
markets during the protracted Great Recession that began in late 2007, PFS has redoubled its efforts to provide prudent manage- ment of the fund.
It is now December 2013. The total pension payments that will need to be made by the fund over the next 10 years are shown in the following table.
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Additional Case 149
payments for the year occur at the beginning of the year imme- diately after these interest payments (including a year’s interest from the money market fund) are received. The entire face value of a bond also will be received on its maturity date. Since the current price of each bond is less than its face value, the actual yield of the bond exceeds its coupon rate. Bond 3 is a zero-cou- pon bond, so it pays no interest but instead pays a face value on the maturity date that greatly exceeds the purchase price.
PFS would like to make the smallest possible investment (including any deposit into the money market fund) on January 1, 2014, to cover all its required pension payments through 2023. Some spreadsheet modeling needs to be done to see how to do this.
a. Visualize where you want to finish. What numbers are needed by PFS management? What are the decisions that need to be made? What should the objective be?
b. Suppose that PFS were to invest $30 million in the money market fund and purchase 10,000 units each of bond 1 and bond 2 on January 1, 2014. Calculate by hand the payments received from bonds 1 and 2 on January 1 of 2015 and 2016. Also calculate the resulting balance in the money market fund on January 1 of 2014, 2015, and 2016 after receiving these payments, making the pension payments for the year, and depositing any excess into the money market fund.
c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objec- tive cell.
d. Build a spreadsheet model for years 2014 through 2016, and then thoroughly test the model.
e. Expand the model to consider all years through 2023, and then solve it.
Additional Case Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book.
Year Pension Payments ($ millions)
2014 8 2015 12 2016 13 2017 14 2018 16 2019 17 2020 20 2021 21 2022 22 2023 24
By using interest as well, PFS currently has enough liquid assets to meet all these pension payments. Therefore, to safe- guard the pension fund, PFS would like to make a number of investments whose payouts would match the pension payments over the next 10 years. The only investments that PFS trusts for the pension fund are a money market fund and bonds. The money market fund pays an annual interest rate of 2 percent. The characteristics of each unit of the four bonds under consideration are shown in the table below.
Current Price
Coupon Rate
Maturity Date
Face Value
Bond 1 $980 4% Jan. 1, 2015 $1,000 Bond 2 920 2 Jan. 1, 2017 1,000 Bond 3 750 0 Jan. 1, 2019 1,000 Bond 4 800 3 Jan. 1, 2022 1,000
All of these bonds will be available for purchase on January 1, 2014, in as many units as desired. The coupon rate is the per- centage of the face value that will be paid in interest on January 1 of each year, starting one year after purchase and continuing until (and including) the maturity date. Thus, these interest pay- ments on January 1 of each year are in time to be used toward the pension payments for that year. Any excess interest payments will be deposited into the money market fund. To be conserva- tive in its financial planning, PFS assumes that all the pension
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Chapter Five
What-If Analysis for Linear Programming Learning Objectives
After completing this chapter, you should be able to
1. Explain what is meant by what-if analysis.
2. Summarize the benefits of what-if analysis.
3. Enumerate the different kinds of changes in the model that can be considered by what-if analysis.
4. Describe how the spreadsheet formulation of the problem can be used to perform any of these kinds of what-if analysis.
5. Use Parameters with Risk Solver Platform for Education (RSPE) to systematically investigate the effect of changing either one or two data cells to various other trial values.
6. Find how much any single coefficient in the objective function can change without changing the optimal solution.
7. Evaluate simultaneous changes in objective function coefficients to determine whether the changes are small enough that the original optimal solution must still be optimal.
8. Predict how the value in the objective cell would change if a small change were to be made in the right-hand side of one or more of the functional constraints.
9. Find how much the right-hand side of a single functional constraint can change before this prediction becomes no longer valid.
10. Evaluate simultaneous changes in right-hand sides to determine whether the changes are small enough that this prediction must still be valid.
Chapters 2 to 4 have described and illustrated how to formulate a linear programming model on a spreadsheet to represent a variety of managerial problems, and then how to use Solver to find an optimal solution for this model. You might think that this would finish our story about linear programming: Once the manager learns the optimal solution, she would immediately implement this solution and then turn her attention to other matters. However, this is not the case. The enlightened manager demands much more from linear programming, and linear programming has much more to offer her—as you will discover in this chapter.
An optimal solution is only optimal with respect to a particular mathematical model that provides only a rough representation of the real problem. A manager is interested in much more than just finding such a solution. The purpose of a linear programming study is to help guide management’s final decision by providing insights into the likely consequences of pur- suing various managerial options under a variety of assumptions about future conditions. Most of the important insights are gained while conducting analysis after finding an optimal solution for the original version of the basic model. This analysis is commonly referred to as what-if analysis because it involves addressing some questions about what would happen to the optimal solution if different assumptions were made about future conditions. Spread- sheets play a central role in addressing these what-if questions.
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5.1 The Importance of What-If Analysis to Managers 151
This chapter focuses on the types of information provided by what-if analysis and why it is valuable to managers. The first section provides an overview. Section 5.2 returns to the Wyndor Glass Co. product-mix case study (Section 2.1) to describe the what-if analysis that is needed in this situation. The subsequent sections then perform this what-if analysis, using a variety of procedures that are applicable to any linear programming problem.
5.1 THE IMPORTANCE OF WHAT-IF ANALYSIS TO MANAGERS
The examples and problems in the preceding chapters on linear programming have provided the data needed to determine precisely all the numbers that should go into the data cells for the spreadsheet formulation of the linear programming model. (Recall that these num- bers are referred to as the parameters of the model . ) Real applications seldom are this straightforward. Substantial time and effort often are needed to track down the needed data. Even then, it may be possible to develop only rough estimates of the parameters of the model.
For example, in the Wyndor case study, two key parameters of the model are the coeffi- cients in the objective function that represent the unit profits of the two new products. These parameters were estimated to be $300 for the doors and $500 for the windows. However, these unit profits depend on many factors—the costs of raw materials, production, shipping, advertising, and so on, as well as such things as the market reception to the new products and the amount of competition encountered. Some of these factors cannot be estimated with real accuracy until long after the linear programming study has been completed and the new prod- ucts have been on the market for some time.
Therefore, before Wyndor’s management makes a decision on the product mix, it will want to know what the effect would be if the unit profits turn out to differ significantly from the estimates. For example, would the optimal solution change if the unit profit for the doors turned out to be $200 instead of the estimate of $300? How inaccurate can the estimate be in either direction before the optimal solution changes?
Such questions are addressed in Section 5.3 when only one estimate is inaccurate. Section 5.4 will address similar questions when multiple estimates are inaccurate.
If the optimal solution will remain the same over a wide range of values for a particular coefficient in the objective function, then management will be content with a fairly rough estimate for this coefficient. On the other hand, if even a small error in the estimate would change the optimal solution, then management will want to take special care to refine this estimate. Management sometimes will get involved directly in adjusting such estimates to its satisfaction.
Here then is a summary of the first benefit of what-if analysis:
1. Typically, many of the parameters of a linear programming model are only estimates of quan- tities (e.g., unit profits) that cannot be determined precisely at this time. What-if analysis reveals how close each of these estimates needs to be to avoid obtaining an erroneous optimal solution, and therefore pinpoints the sensitive parameters (those parameters where extra care is needed to refine their estimates because even small changes in their values can change the optimal solution).
Several sections describe how what-if analysis provides this benefit for the most important parameters. Sections 5.3 and 5.4 do this for the coefficients in the objective function (these numbers typically appear in the spreadsheet in the row for the unit contribution of each activ- ity toward the overall measure of performance). Sections 5.5 and 5.6 do the same for the right-hand sides of the functional constraints (these are the numbers that typically are in the right-hand column of the spreadsheet just to the right of the # , $ , or 5 signs).
Businesses operate in a dynamic environment. Even when management is satisfied with the current estimates and implements the corresponding optimal solution, conditions may change later. For example, suppose that Wyndor’s management is satisfied with $300 as the estimate of the unit profit for the doors, but increased competition later forces a price reduc- tion that reduces this unit profit. Does this change the optimal product mix? The what-if analysis shown in Section 5.3 immediately indicates in advance which new unit profits would leave the optimal product mix unchanged, which can help guide management in its new
In real applications, many of the numbers in the model may be only rough estimates.
What happens to the opti- mal solution if an error is made in estimating a parameter of the model?
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152 Chapter Five What-If Analysis for Linear Programming
pricing decision. Furthermore, if the optimal product mix is unchanged, then there is no need to solve the model again with the new coefficient. Avoiding solving the model again is no big deal for the tiny two-variable Wyndor problem, but it is extremely welcome for real applica- tions that may have hundreds or thousands of constraints and variables. In fact, for such large models, it may not even be practical to re-solve the model repeatedly to consider the many possible changes of interest.
Thus, here is the second benefit of what-if analysis:
2. If conditions change after the study has been completed (a common occurrence), what-if analysis leaves signposts that indicate (without solving the model again) whether a resulting change in a parameter of the model changes the optimal solution.
Again, several subsequent sections describe how what-if analysis does this. These sections focus on studying how changes in the parameters of a linear programming
model affect the optimal solution. This type of what-if analysis commonly is referred to as sensitivity analysis , because it involves checking how sensitive the optimal solution is to the value of each parameter. Sensitivity analysis is a vital part of what-if analysis.
However, rather than being content with the passive sensitivity analysis approach of check- ing the effect of parameter estimates being inaccurate, what-if analysis often goes further to take a proactive approach. An analysis may be made of various possible managerial actions that would result in changes to the model.
A prime example of this proactive approach arises when certain parameters of the model represent managerial policy decisions rather than quantities that are largely outside the con- trol of management. For example, for the Wyndor product-mix problem, the right-hand sides of the three functional constraints (4, 12, 18) represent the number of hours of production time in the three respective plants being made available per week for the production of the two new products. Management can change these three resource amounts by altering the production levels for the old products in these plants. Therefore, after learning the optimal solution, management will want to know the impact on the profit from the new products if these resource amounts are changed in certain ways. One key question is how much this profit can be increased by increasing the available production time for the new products in just one of the plants. Another is how much this profit can be increased by simultaneously making helpful changes in the available production times in all the plants. If the profit from the new products can be increased enough to more than compensate for the profit lost by decreasing the production levels for certain old products, management probably will want to make the change.
We now can summarize the third benefit of what-if analysis:
3. When certain parameters of the model represent managerial policy decisions, what-if analysis provides valuable guidance to management regarding the impact of altering these policy decisions.
Sections 5.5 and 5.6 will explore this benefit further. What-if analysis sometimes goes even further in providing helpful guidance to manage-
ment, such as when analyzing alternate scenarios for how business conditions might evolve. However, this chapter will focus on the three benefits summarized above.
What happens to the opti- mal solution if conditions change in the future?
What happens if managerial policy decisions change?
1. What are the parameters of a linear programming model? 2. How can inaccuracies arise in the parameters of a model? 3. What does what-if analysis reveal about the parameters of a model that are only estimates? 4. Is it always inappropriate to make only a fairly rough estimate for a parameter of a model?
Why? 5. How is it possible for the parameters of a model to be accurate initially and then become inac-
curate at a later date? 6. How does what-if analysis help management prepare for changing conditions? 7. What is meant by sensitivity analysis? 8. For what kinds of managerial policy decisions does what-if analysis provide guidance?
Review Questions
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5.2 Continuing the Wyndor Case Study 153
5.2 CONTINUING THE WYNDOR CASE STUDY
We now return to the case study introduced in Section 2.1 involving the Wyndor Glass Co. product-mix problem.
To review briefly, recall that the company is preparing to introduce two new products:
• An 8-foot glass door with aluminum framing. • A 4-foot 3 6-foot double-hung wood-framed window.
To analyze which mix of the two products would be most profitable, the company’s Manage- ment Science Group introduced two decision variables:
D 5 Production rate of this new kind of door
W 5 Production rate of this new kind of window
where this rate measures the number of units produced per week. Three plants will be involved in the production of these products. Based on managerial decisions regarding how much these plants will continue to be used to produce current products, the number of hours of production time per week being made available in Plants 1, 2, and 3 for the new products is 4, 12, and 18, respectively. After obtaining rough estimates that the profit per unit will be $300 for the doors and $500 for the windows, the Management Science Group then formu- lated the linear programming model shown in Figure 2.13 and repeated here in Figure 5.1 , where the objective is to choose the values of D and W in the changing cells UnitsProduced (C12:D12) so as to maximize the total profit (per week) given in the objective cell TotalProfit (G12). Applying Solver to this model yielded the optimal solution shown on this spreadsheet and summarized as follows.
Optimal Solution
D 5 2 (Produce 2 doors per week.)
W 5 6 (Produce 6 windows per week.)
Profit 5 3,600 (The estimated total weekly profit is $3,600.)
However, this optimal solution assumes that all the estimates that provide the parameters of the model (as shown in the UnitProfit (C4:D4), HoursUsedPerUnitProduced (C7:D9), and HoursAvailable (G7:G9) data cells) are accurate.
The head of the Management Science Group, Lisa Taylor, now is ready to meet with man- agement to discuss the group’s recommendation that the above product mix be used.
Management’s Discussion of the Recommended Product Mix Lisa Taylor (head of Management Science Group): I asked for this meeting so we could explore what questions the two of you would like us to pursue further. In particular, I am especially concerned that we weren’t able to better pin down just what the numbers should be to go into our model. Which estimates do you think are the shakiest? Bill Tasto (vice president for manufacturing): Without question, the estimates of the unit profits for the two products. Since the products haven’t gone into production yet, all we could do is analyze the data from similar current products and then try to project what the changes would be for these new products. We have some numbers, but they are pretty rough. We would need to do a lot more work to pin down the numbers better. John Hill (president): We may need to do that. Lisa, do you have a way of checking how far off one of these estimates can be without changing the optimal product mix? Lisa: Yes, we do. We can quickly find what we call the allowable range for each unit profit. As long as the true value of the unit profit is within this allowable range, and the other unit profit is correct, the optimal product mix will not change. If this range is pretty wide, you don’t need to worry about refining the estimate of the unit profit. However, if the range is quite narrow, then it is important to pin down the estimate more closely.
The allowable range for a unit profit indicates how far its estimate can be off with- out affecting the optimal product mix.
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154 Chapter Five What-If Analysis for Linear Programming
John: What happens if both estimates are off? Lisa: We can provide a way of checking whether the optimal product mix might change for any new combination of unit profits you think might be the true one. John: Great. That’s what we need. There’s also one more thing. Bill gave you the num- bers for how many hours of production time we’re making available per week in the three plants for these new products. I noticed you used these numbers on your spreadsheet. Lisa: Yes. They’re the right-hand sides of our constraints. Is something wrong with these numbers? John: No, not at all. I just wanted to let you know that we haven’t made a final decision on whether these are the numbers we want to use. We would like your group to provide us with some analysis of what the effect would be if we change any of those numbers. How much more profit could we get from the new products for each additional hour of production time per week we provide in one of the plants? That sort of thing. Lisa: Yes, we can get that analysis to you right away also. John: We might also be interested in changing the available production hours for two or three of the plants. Lisa: No problem. We’ll give you information about that as well.
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Doors
2
12
18
Windows
Doors
2 6
Windows
$3,600
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
5
E
Hours
Used6
11
12
7
8
9
=SUMPRODUCT(C7:D7, UnitsProduced)
G
Total Profit
=SUMPRODUCT(UnitProfit, UnitsProduced)
=SUMPRODUCT(C8:D8, UnitsProduced)
=SUMPRODUCT(C9:D9, UnitsProduced)
Range Name
DoorsProduced
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
TotalProfit
UnitProfit
UnitsProduced
WindowsProduced
Cells
C12
G7:G9
E7:E9
C7:D9
G12
C4:D4
C12:D12
D12
$300 $500
1
0
3
0
2
2
4
12
18
Wyndor Glass Co. Product-Mix Problem
≤
≤ ≤
Solver Parameters
Set Objective Cell: TotalProfit To: Max By Changing Variable Cells:
UnitsProduced
Subject to the Constraints: HoursUsed <= HoursAvailable
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
FIGURE 5.1 The spreadsheet model and its optimal solution for the original Wyndor problem before beginning what-if analysis.
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5.3 The Effect of Changes in One Objective Function Coefficient 155
Summary of Management’s What-If Questions Here is a summary of John Hill’s what-if questions that Lisa and her group will be addressing in the coming sections.
1. What happens if the estimate of the unit profit of one of Wyndor’s new products is inac- curate? (Section 5.3)
2. What happens if the estimates of the unit profits of both of Wyndor’s new products are inaccurate? (Section 5.4)
3. What happens if a change is made in the number of hours of production time per week being made available to Wyndor’s new products in one of the plants? (Section 5.5)
4. What happens if simultaneous changes are made in the number of hours of production time per week being made available to Wyndor’s new products in all the plants? (Section 5.6)
1. Which estimates of the parameters in the linear programming model for the Wyndor problem are most questionable?
2. Which numbers in this model represent tentative managerial decisions that management might want to change after receiving the Management Science Group’s analysis?
Review Questions
5.3 THE EFFECT OF CHANGES IN ONE OBJECTIVE FUNCTION COEFFICIENT
Section 5.1 began by discussing the fact that many of the parameters of a linear programming model typically are only estimates of quantities that cannot be determined precisely at the time. What-if analysis (or sensitivity analysis in particular) reveals how close each of these estimates needs to be to avoid obtaining an erroneous optimal solution.
We focus in this section on how sensitivity analysis does this when the parameters involved are coefficients in the objective function. (Recall that each of these coefficients gives the unit contribution of one of the activities toward the overall measure of performance.) In the pro- cess, we will address the first of the what-if questions posed by Wyndor management in the preceding section.
Question 1: What happens if the estimate of the unit profit of one of Wyndor’s new products is inaccurate?
To start this process, first consider the question of what happens if the estimate of $300 for the unit profit for Wyndor’s new kind of door is inaccurate. To address this question, let
P D 5 Unit profit for the new kind of door
5 Cell C4 in the spreadsheet (see Figure 5.1 )
Although P D 5 $300 in the current version of Wyndor’s linear programming model, we now want to explore how much larger or how much smaller P D can be and still have ( D, W ) 5 (2, 6) as the optimal solution. In other words, how much can the estimate of $300 for the unit profit for these doors be off before the model will give an erroneous optimal solution?
Using the Spreadsheet to Do Sensitivity Analysis One of the great strengths of a spreadsheet is the ease with which it can be used interactively to perform various kinds of what-if analysis, including the sensitivity analysis being con- sidered in this section. Once Solver has been set up to obtain an optimal solution, you can immediately find out what would happen if one of the parameters of the model were to be changed to some other value. All you have to do is make this change on the spreadsheet and then run Solver again.
To illustrate, Figure 5.2 shows what would happen if the unit profit for doors were to be decreased from P D 5 $300 to P D 5 $200. Comparing with Figure 5.1 , there is no change at all in the optimal solution. In fact, the only changes in the new spreadsheet are the new value of P D in cell C4 and a decrease of $200 in the total profit shown in cell G12 (because each
Run Solver again and the spreadsheet immediately reveals the effect of chang- ing any values in the data cells.
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156 Chapter Five What-If Analysis for Linear Programming
of the two doors produced per week provides $100 less profit). Because the optimal solution does not change, we now know that the original estimate of P D 5 $300 can be considerably too high without invalidating the model’s optimal solution.
But what happens if this estimate is too low instead? Figure 5.3 shows what would happen if P D were to be increased to P D 5 $500. Again, there is no change in the optimal solution.
Because the original value of P D 5 $300 can be changed considerably in either direction without changing the optimal solution, P D is said to be not a sensitive parameter. It is not necessary to pin down this estimate with great accuracy to have confidence that the model is providing the correct optimal solution.
This may be all the information that is needed about P D . However, if there is a good pos- sibility that the true value of P D will turn out to be outside this broad range from $200 to $500, further investigation would be desirable. How much higher or lower can P D be before the optimal solution would change?
Figure 5.4 demonstrates that the optimal solution would indeed change if P D were increased all the way up to P D 5 $1,000. Thus, we now know that this change occurs some- where between $500 and $1,000 during the process of increasing P D .
Using a Parameter Analysis Report (RSPE) to Do Sensitivity Analysis Systematically To pin down just when the optimal solution will change, we could continue selecting new values of P D at random. However, a better approach is to systematically consider a range of values of P D .
FIGURE 5.2 The revised Wyndor prob- lem where the estimate of the unit profit for doors has been decreased from PD 5 $300 to PD 5 $200, but no change occurs in the optimal solution.
1
A B C D E F G
2
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5
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7
8
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10
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Wyndor Glass Co. Product-Mix Problem
Doors
2
12
18
Windows
Doors
2 6
Windows
$3,400
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$200 $500
1
0
3
0
2
2
4
12
18
≤ ≤
≤
FIGURE 5.3 The revised Wyndor prob- lem where the estimate of the unit profit for doors has been increased from PD 5 $300 to PD 5 $500, but no change occurs in the optimal solution.
1
A B C D E F G
2
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Wyndor Glass Co. Product-Mix Problem
Doors
2
12
18
Windows
Doors
2 6
Windows
$4,000
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$500 $500
1
0
3
0
2
2
4
12
18
≤
≤ ≤
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5.3 The Effect of Changes in One Objective Function Coefficient 157
The Risk Solver Platform for Education (RSPE), first introduced in Section 2.6, can gener- ate a parameter analysis report that is designed to do just this sort of analysis. Instructions for installing RSPE are on a supplementary insert included with the book and also on the book’s website, www.mhhe.com/hillier5e .
The data cell containing a parameter that will be systematically varied (C4 in this case) is referred to as a parameter cell . A parameter analysis report is used to show the results in the changing cells and/or the objective cell for various trial values in the parameter cell. For each trial value, these results are obtained by using Solver to re-solve the problem.
To generate a parameter analysis report, the first step is to define the parameter cell. In this case, select cell C4 (the unit profit for doors) and choose Optimization under the Parameters menu on the RSPE ribbon. In the parameter cell dialog box, shown in Figure 5.5 , enter the range of trial values for the parameter cell. The entries shown specify that P D will be system- atically varied from $100 to $1,000. If desired, additional parameter cells could be defined in this same way, but we will not do so at this point.
Next choose Optimization . Parameter Analysis under the Reports menu on the RSPE ribbon. This brings up the dialog box shown in Figure 5.6 that allows you to specify which parameter cells to vary and which results to show. The choice of which parameter cells to vary is made under Parameters in the bottom half of the dialog box. Clicking on (..) will select all of the parameter cells defined so far (moving them to the box on the right). In the Wyndor example, only one parameter has been defined, so this causes the single parameter cell (UnitProfitPerDoor) to appear on the right. If more parameter cells had been defined, particular parameter cells can be chosen for immediate analysis by clicking on the 1 next
FIGURE 5.4 The revised Wyndor problem where the esti- mate of the unit profit for doors has been increased from PD 5 $300 to PD 5 $1,000, which results in a change in the optimal solution.
1
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Wyndor Glass Co. Product-Mix Problem
Doors
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Windows
Doors
4 3
Windows
$5,500
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$1,000 $500
1
0
3
0
2
2
4
12
18
≤
≤ ≤
FIGURE 5.5 The parameter cell dialog box for PD specifies here that this parameter cell for the Wyndor problem will be systematically varied from $100 to $1,000.
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158 Chapter Five What-If Analysis for Linear Programming
to Wyndor to reveal the list of parameter cells that have been defined in the Wyndor spread- sheet. Clicking on (.) then moves individual parameter cells to the list on the right.
The choice of which results to show as the parameter cell is varied is made in the upper half of the dialog box. Clicking on (..) will cause all of the changing cells (DoorsProduced and WindowsProduced) and the objective cell (Total Profit) to appear in the list on the right. To instead choose a subset of these cells, click on the small 1 next to Variables (or Objective) to reveal a list of all the changing cells (or objective cell) and then click on . to move that changing cell (or objective cell) to the right.
Finally, enter the number of Major Axis Points to specify how many different values of the parameter cell will be shown in the parameter analysis report. The values will be spread evenly between the lower and upper values specified in the parameters cell dialog box in Figure 5.5 . With 10 major axis points, a lower value of $100, and an upper value of $1000, the parameter analysis report will show results for P D of $100, $200, $300,. . . , $1000.
Clicking on the OK button generates the parameter analysis report shown in Figure 5.7 . One at a time, the trial values listed in the first column of the table are put into the parameter cell (UnitProfitPerDoor) and then Solver is called on to re-solve the problem. The optimal results for that particular trial value of the parameter cell are then shown in the remaining columns— DoorsProduced, WindowsProduced, and TotalProfit. This is repeated automatically for each remaining trial value of the parameter cell. The end result (which happens very quickly for small problems) is the completely-filled-in parameter analysis report shown in Figure 5.7 .
The parameter analysis report reveals that the optimal solution remains the same all the way from P D 5 $100 (and perhaps lower) to P D 5 $700, but that a change occurs somewhere between $700 and $800. We next could systematically consider values of P D between $700 and $800 to determine more closely where the optimal solution changes. However, here is a shortcut. The range of values of P D over which ( D, W ) 5 (2, 6) remains as the optimal solu- tion is referred to as the allowable range for an objective function coefficient , or
The allowable range for a coefficient in the objective function is the range of val- ues for this coefficient over which the optimal solu- tion for the original model remains optimal.
FIGURE 5.6 The dialog box for the parameter analysis report specifies here for the Wyndor problem that the UnitProfitPerDoor parameter cell will be varied and that results from all the changing cells (DoorsProduced and WindowsProduced) and the objective cell (Total- Profit) will be shown.
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5.3 The Effect of Changes in One Objective Function Coefficient 159
just the allowable range for short. Upon request, Solver will provide a report called the sensitivity report that reveals exactly what this allowable range is.
Using the Sensitivity Report to Find the Allowable Range As was shown in Figure 2.12, when Solver gives the message that it has found a solution, it also gives on the right a list of three reports that can be provided. By selecting the second one (labeled Sensitivity), you will obtain the sensitivity report.
Figure 5.8 shows the relevant part of this report for the Wyndor problem. The Final Value column indicates the optimal solution. The next column gives the reduced costs, which can provide some useful information when any of the changing cells equal zero in the optimal solution, which is not the case here. (For a zero-valued changing cell, the corresponding reduced cost can be used to determine what the effect would be of either increasing that changing cell or making a change in its coefficient in the objective function. Because of the relatively technical nature of these interpretations of reduced costs, we will not discuss them further here, but will provide a full explanation in the supplement to this chapter on the CD-ROM.) The next three columns provide the information needed to identify the allow- able range for each coefficient in the objective function. The Objective Coefficient column gives the current value of each coefficient, and then the next two columns give the allowable increase and the allowable decrease from this value to remain within the allowable range.
For example, consider P D , the coefficient of D in the objective function. Since D is the production rate for these special doors, the Doors row in the table provides the following information (without the dollar sign) about P D :
Current value of P D : 300
Allowable increase in P D : 450 So P D # 300 1 450 5 750
Allowable decrease in P D : 300 So P D $ 300 2 300 5 0
Allowable range for P D : 0 # P D # 750
Reduced costs are described in the supplement to this chapter on the CD-ROM. Don’t worry about this relatively technical subject (unless your instructor assigns this supplement).
The sensitivity report gen- erated by Solver reveals the allowable range for each coefficient in the objective function.
FIGURE 5.7 The parameter analysis report that shows the effect of systematically varying the estimate of the unit profit for doors for the Wyndor problem.
1
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UnitProfitPerDoor
A B C D
DoorsProduced WindowsProduced TotalProfit
$100
$200
$300
$400
$500
$600
$700
$800
$900
$1,000
$3,200
$3,400
$3,600
$3,800
$4,000
$4,200
$4,400
$4,700
$5,100
$5,500
2
2
2
2
2
2
2
4
4
4
6
6
6
6
6
6
6
3
3
3
Variable Cells
Cell Name Final Value
Reduced Cost
Objective Coefficient
Allowable Increase
Allowable Decrease
$C$12 DoorsProduced 2 0 300 450 300 $D$12 WindowsProduced 6 0 500 1E 1 30 300
FIGURE 5.8 Part of the sensitivity report generated by Solver for the original Wyndor problem (Figure 5.1), where the last three columns enable identifying the allowable ranges for the unit profits for doors and windows.
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160 Chapter Five What-If Analysis for Linear Programming
Therefore, if P D is changed from its current value (without making any other change in the model), the current solution ( D, W ) 5 (2, 6) will remain optimal so long as the new value of P D is within this allowable range.
Figure 5.9 provides graphical insight into this allowable range. For the original value of P D 5 300, the solid line in the figure shows the slope of the objective function line passing through (2, 6). At the lower end of the allowable range, when P D 5 0, the objective function line that passes through (2, 6) now is line B in the figure, so every point on the line segment between (0, 6) and (2, 6) is an optimal solution. For any value of P D , 0, the objective func- tion line will have rotated even further so that (0, 6) becomes the only optimal solution. At the upper end of the allowable range, when P D 5 750, the objective function line that passes through (2, 6) becomes line C, so every point on the line segment between (2, 6) and (4, 3) becomes an optimal solution. For any value of P D . 750, the objective function line is even steeper than line C, so (4, 3) becomes the only optimal solution.
The module called Graphical Linear Programming and Sensitivity Analysis in the Interac- tive Management Science Modules (available at www.mhhe.com/hillier5e/ or in your CD- ROM) is designed to help you perform this kind of graphical analysis. After you enter the model for the original Wyndor problem, the module provides you with the graph shown in Figure 5.9 (without the dashed lines). You then can simply drag one end of the objective func- tion line up or down to see how far you can increase or decrease P D before ( D, W ) 5 (2, 6) will no longer be optimal.
Conclusion: The allowable range for P D is 0 # P D # 750, because ( D, W ) 5 (2, 6) remains optimal over this range but not beyond. (When P D 5 0 or P D 5 750, there are multiple optimal solutions, but ( D, W ) 5 (2, 6) still is one of them.) With the range this wide around the original estimate of $300 ( P D 5 300) for the unit profit for doors, we can be quite confident of obtain- ing the correct optimal solution for the true unit profit even though the discussion in Section 5.2 indicates that this estimate is fairly rough.
The sensitivity report also can be used to find the allowable range for the unit profit for Wyndor’s other new product. In particular, let
P W 5 Unit profit for Wyndor’s new kind of window
5 Cell D4 in the spreadsheet
Check out this module in the Interactive Management Science Modules to gain graphical insight into the allowable range.
FIGURE 5.9 The two dashed lines that pass through solid constraint boundary lines are the objective function lines when PD (the unit profit for doors) is at an endpoint of its allowable range, 0 # PD # 750, since either line or any objective function line in between still yields (D, W ) 5 (2, 6) as an optimal solution for the Wyndor problem.
Production rate for windows
D
W
8
6
4
2
2 4 6
Production rate for doors
0
(2, 6) is optimal for 0 PD 750
PD = 0 (Profit = 0D + 500W)
PD = 300 (Profit = 300D + 500W)
PD = 750 (Profit = 750D + 500W)
Line A
Line C
Line B
Feasible region
≤≤
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5.4 The Effect of Simultaneous Changes in Objective Function Coefficients 161
Referring to the Windows row of the sensitivity report ( Figure 5.8 ), this row indicates that the allowable decrease in P W is 300 (so P W $ 500 2 300 5 200) and the allowable increase is 1E 1 30. What is meant by 1E 1 30? This is shorthand in Excel for 10 30 (1 with 30 zeroes after it). This tremendously huge number is used by Excel to represent infinity. Therefore, the allowable range of P W is obtained from the sensitivity report as follows:
Current value of P W : 500
Allowable increase in P W : Unlimited So P W has no upper limit
Allowable decrease in P W : 300 So P W $ 500 2 300 5 200
Allowable range: P W $ 200
The allowable range is quite wide for both objective function coefficients. Thus, even though P D 5 $300 and P W 5 $500 were only rough estimates of the true unit profit for the doors and windows, respectively, we can still be confident that we have obtained the correct optimal solution.
We are not always so lucky. For some linear programming problems, even a small change in the value of certain coefficients in the objective function can change the optimal solution. Such coefficients are referred to as sensitive parameters. The sensitivity report will immedi- ately indicate which of the objective function coefficients (if any) are sensitive parameters. These are parameters that have a small allowable increase and/or a small allowable decrease. Hence, extra care should be taken to refine these estimates.
Once this has been done and the final version of the model has been solved, the allowable ranges continue to serve an important purpose. As indicated in Section 5.1, the second benefit of what-if analysis is that if conditions change after the study has been completed (a common occurrence), what-if analysis leaves signposts that indicate (without solving the model again) whether a resulting change in a parameter of the model changes the optimal solution. Thus, if weeks, months, or even years later, the unit profit for one of Wyndor’s new products changes substantially, its allowable range indicates immediately whether the old optimal product mix still is the appropriate one to use. Being able to draw an affirmative conclusion without reconstructing and solving the revised model is extremely helpful for any linear programming problem, but especially so when the model is a large one.
A parameter is consid- ered sensitive if even a small change in its value can change the optimal solution.
1. What is meant by the allowable range for a coefficient in the objective function? 2. What is the significance if the true value for a coefficient in the objective function turns out to
be so different from its estimate that it lies outside its allowable range? 3. In Solver’s sensitivity report, what is the interpretation of the Objective Coefficient column?
The Allowable Increase column? The Allowable Decrease column?
Review Questions
5.4 THE EFFECT OF SIMULTANEOUS CHANGES IN OBJECTIVE FUNCTION COEFFICIENTS
The coefficients in the objective function typically represent quantities (e.g., unit profits) that can only be estimated because of considerable uncertainty about what their true values will turn out to be. The allowable ranges described in the preceding section deal with this uncertainty by focusing on just one coefficient at a time. In effect, the allowable range for a particular coef- ficient assumes that the original estimates for all the other coefficients are completely accurate so that this coefficient is the only one whose true value may differ from its original estimate.
In actuality, the estimates for all the coefficients (or at least more than one of them) may be inaccurate simultaneously. The crucial question is whether this is likely to result in obtain- ing the wrong optimal solution. If so, greater care should be taken to refine these estimates as much as possible, at least for the more crucial coefficients. On the other hand, if what-if analysis reveals that the anticipated errors in estimating the coefficients are unlikely to affect the optimal solution, then management can be reassured that the current linear programming model and its results are providing appropriate guidance.
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162
The Pacific Lumber Company (PALCO) is a large timber- holding company with headquarters in Scotia, California. The company has over 200,000 acres of highly productive forest lands that support five mills located in Humboldt County in northern California. The lands include some of the most spectacular redwood groves in the world that have been given or sold at low cost to be preserved as parks. PALCO manages the remaining lands intensively for sustained timber production, subject to strong forest prac- tice laws. Since PALCO’s forests are home to many species of wildlife, including endangered species such as spotted owls and marbled murrelets, the provisions of the federal Endangered Species Act also need to be carefully observed.
To obtain a sustained yield plan for the entire land- holding, PALCO management contracted with a team of management science consultants to develop a 120-year, 12-period, long-term forest ecosystem management plan. The management science team performed this task by for- mulating and applying a linear programming model to
optimize the company’s overall timberland operations and profitability after satisfying the various constraints. The model was a huge one with approximately 8,500 functional constraints and 353,000 decision variables.
A major challenge in applying the linear programming model was the many uncertainties in estimating what the parameters of the model should be. The major factors caus- ing these uncertainties were the continuing fluctuations in market supply and demand, logging costs, and environ- mental regulations. Therefore, the management science team made extensive use of detailed sensitivity analysis. The resulting sustained yield plan increased the company’s present net worth by over $398 million while also generat- ing a better mix of wildlife habitat acres.
Source: L. R. Fletcher, H. Alden, S. P. Holmen, D. P. Angelis, and M. J. Etzenhouser, “Long-Term Forest Ecosystem Planning at Pacific Lumber,” Interfaces 29, no. 1 (January–February 1999), pp. 90–112. (A link to this article is provided on our website, www.mhhe.com/hillier5e .)
An Application Vignette
This section focuses on how to determine, without solving the problem again, whether the optimal solution might change if certain changes occur simultaneously in the coefficients of the objective function (due to their true values differing from their estimates). In the process, we will address the second of Wyndor management’s what-if questions.
Question 2: What happens if the estimates of the unit profits of both of Wyndor’s new products are inaccurate?
Using the Spreadsheet for This Analysis Once again, a quick-and-easy way to address this kind of question is to simply try out differ- ent estimates on the spreadsheet formulation of the model and see what happens each time after running Solver again.
In this case, the optimal product mix indicated by the model is heavily weighted toward producing the windows (6 per week) rather than the doors (only 2 per week). Since there is equal enthusiasm for both new products, management is concerned about this imbalance. Therefore, management has raised a what-if question. What would happen if the estimate of the unit profit for the doors ($300) were too low and the corresponding estimate for the windows ($500) were too high? Management feels that the estimates could easily be off in these directions. If this were the case, would this lead to a more balanced product mix being the most profitable one?
This question can be answered in a matter of seconds simply by substituting new estimates of the unit profits in the original spreadsheet in Figure 5.1 and running Solver again. Figure 5.10 shows that new estimates of $450 for doors and $400 for windows causes no change at all in the solution for the optimal product mix. (The total profit does change, but this occurs only because of the changes in the unit profits.) Would even larger changes in the estimates of unit profits finally lead to a change in the optimal product mix? Figure 5.11 shows that this does happen, yielding a relatively balanced product mix of ( D, W ) 5 (4, 3), when estimates of $600 for doors and $300 for windows are used.
Using a Two-Way Parameter Analysis Report (RSPE) for This Analysis Using RPSE, a two-way parameter analysis report provides a way of systematically investi- gating the effect if the estimates of both unit profits are inaccurate. This kind of parameter analysis table shows the results in a single output cell for various trial values in two parameter
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5.4 The Effect of Simultaneous Changes in Objective Function Coefficients 163
cells. Therefore, for example, it can be used to show how TotalProfit (G12) in Figure 5.1 varies over a range of trial values in the two parameter cells, UnitProfitPerDoor (C4) and UnitProfitPerWindow (D4). For each pair of trial values in these data cells, Solver is called on to re-solve the problem.
To create such a two-way parameter analysis report for the Wyndor problem, both Unit- ProfitPerDoor (C4) and UnitProfitPerWindow (D4) need to be defined as parameter cells. In turn, select cell C4 and D4, then choose Optimization under the Parameters menu on the RSPE ribbon, and then enter the range of trial values for each parameter cell (as was done in Figure 5.5 in the previous section). For this example, UnitProfitPerDoor (C4) will be varied from $300 to $600 while UnitProfitPerWindow (D4) will be varied from $100 to $500.
Next, choose Optimization . Parameter Analysis under the reports menu on the RSPE ribbon to bring up the dialog box shown in Figure 5.12 . For a two-way parameter analysis report, two parameter cells are chosen, but only a single result can be shown. Under Param- eters, clicking on (..) chooses both of the defined parameter cells, UnitProfitPerDoor and UnitProfitPerWindow. Under Results, click on (,,) to clear out the list of cells on the right, click on the 1 next to Objective to reveal the objective cell (TotalProfit), select TotalProfit, and then click on . to move this cell to the right.
The next step is to turn on the option called Vary Parameters Independently. This will allow both parameter cells to be varied independently over their entire ranges. The number of different values of the first parameter cell and the second parameter cell to be shown in the parameter analysis report are entered in Major Axis Points and Minor Axis Points, respec- tively. These values will be spread evenly over the range of values specified in the parameter
FIGURE 5.11 The revised Wyndor prob- lem where the estimates of the unit profits for doors and windows have been changed to $600 and $300, respectively, which results in a change in the optimal solution.
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
4
6
18
Windows
Doors
4 3
Windows
$3,300
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$600 $300
1
0
3
0
2
2
4
12
18
≤
≤
≤
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
2
12
18
Windows
Doors
2 6
Windows
$3,300
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$450 $400
1
0
3
0
2
2
4
12
18
≤ ≤
≤
FIGURE 5.10 The revised Wyndor problem where the esti- mates of the unit profits for doors and windows have been changed to PD 5 $450 and PW 5 $400, respectively, but no change occurs in the optimal solution.
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164 Chapter Five What-If Analysis for Linear Programming
dialog box for each parameter cell. Therefore, choosing 4 and 5 for the respective number of values, as shown in Figure 5.12 , will vary UnitProfitPerDoor over the four values of $300, $400, $500, and $600 while simultaneously varying UnitProfitPerWindow over the five val- ues of $100, $200, $300, $400, and $500.
Clicking on the OK button generates the parameter analysis report shown in Figure 5.13 . The trial values for the respective parameter cells are listed in the first column and first row of the table. For each combination of a trial value from the first column and from the first row, Solver has solved for the value of the output cell of interest (the objective cell for this example) and entered it into the corresponding column and row of the table.
It also is possible to choose either DoorsProduced or WindowsProduced instead of Total- Profit as the result to show in the dialog box in Figure 5.12 . A similar parameter analysis report then could have been generated to show either the optimal number of doors to produce or the optimal number of windows to produce for each combination of values for the unit profits. These two parameter analysis reports are shown in Figure 5.14 . The upper right-hand corner (cell F3) of both reports, taken together, gives the optimal solution of ( D, W ) 5 (2, 6) when using the original unit-profit estimates of $300 for doors and $500 for windows.
FIGURE 5.12 The dialog box for the parameter analysis report specifies here that the UnitProfitPerDoor and UnitProfitPerWindow parameter cells will be varied and results from the objective cell (Total- Profit) will be shown for the Wyndor problem.
FIGURE 5.13 The parameter analysis report that shows how the optimal total profit changes when systemati- cally varying the estimate of both the unit profit for doors and the unit profit for windows for the Wyndor problem.
1
2
3
4
5
6
UnitProfitPerDoor
$300
$400
$500
$600
$100
$1,500
$1,900
$2,300
$2,700
$200
$1,800
$2,200
$2,600
$3,000
$300
$2,400
$2,600
$2,900
$3,300
$400
$3,000
$3,200
$3,400
$3,600
$500
$3,600
$3,800
$4,000
$4,200
TotalProfit
A B C D E F
UnitProfitPerWindow
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5.4 The Effect of Simultaneous Changes in Objective Function Coefficients 165
Moving down from this cell corresponds to increasing this estimate for doors, while moving to the left amounts to decreasing the estimate for windows. (The cells when moving up or to the right of H26 are not shown because these changes would only increase the attractiveness of ( D, W ) 5 (2, 6) as the optimal solution.) Note that ( D, W ) 5 (2, 6) continues to be the optimal solution for all the cells near H26. This indicates that the original estimates of unit profit would need to be very inaccurate indeed before the optimal product mix would change. Although the estimates are fairly rough, management is confident that they are not that inac- curate. Therefore, there is no need to expend the considerable effort that would be needed to refine the estimates.
At this point, it continues to appear that ( D, W ) 5 (2, 6) is the best product mix for initiat- ing the production of the two new products (although additional what-if questions remain to be addressed in subsequent sections). However, we also now know from Figure 5.14 that as conditions change in the future, if the unit profits for both products change enough, it may be advisable to change the product mix later. We still need to leave clear signposts behind to signal when a future change in the product mix should be considered, as described next.
Gleaning Additional Information from the Sensitivity Report The preceding section described how the data in the sensitivity report enable finding the allowable range for an individual coefficient in the objective function when that coefficient is the only one that changes from its original value. These same data (the allowable increase and allowable decrease in each coefficient) also can be used to analyze the effect of simultaneous changes in these coefficients. Here is how.
The 100 Percent Rule for Simultaneous Changes in Objective Function Coefficients: If simulta- neous changes are made in the coefficients of the objective function, calculate for each change the percentage of the allowable change (increase or decrease) for that coefficient to remain within its allowable range. If the sum of the percentage changes does not exceed 100 percent, the original optimal solution definitely will still be optimal. (If the sum does exceed 100 per- cent, then we cannot be sure.)
This rule does not spell out what happens if the sum of the percentage changes does exceed 100 percent. The consequence depends on the directions of the changes in the coefficients. Exceeding 100 percent may or may not change the optimal solution, but so long as 100 per- cent is not exceeded, the original optimal solution definitely will still be optimal.
Keep in mind that we can safely use the entire allowable increase or decrease in a sin- gle objective function coefficient only if none of the other coefficients have changed at all. With simultaneous changes in the coefficients, we focus on the percentage of the allowable increase or decrease that is being used for each coefficient.
What-if analysis shows that there is no need to refine Wyndor’s estimates of the unit profits for doors and windows.
A sum # 100 percent guar- antees that the original opti- mal solution is still optimal.
FIGURE 5.14 The pair of parameter analysis reports that show how the optimal number of doors to produce (top report) and the optimal number of windows to produce (bottom report) change when systemati- cally varying the estimate of both the unit profit for doors and the unit profit for windows for the Wyn- dor problem.
1
2
3
4
5
6
UnitProfitPerDoor
$300
$400
$500
$600
$100
3
3
3
3
$200
6
3
3
3
$300
6
6
3
3
$400
6
6
6
6
$500
6
6
6
6
WindowsProduced
A B C D E F
UnitProfitPerWindow
1
2
3
4
5
6
UnitProfitPerDoor
$300
$400
$500
$600
$100
4
4
4
4
$200
2
4
4
4
$300
2
2
4
4
$400
2
2
2
2
$500
2
2
2
2
DoorsProduced
A B C D E F
UnitProfitPerWindow
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166 Chapter Five What-If Analysis for Linear Programming
To illustrate, consider the Wyndor problem again, along with the information provided by the sensitivity report in Figure 5.8 . Suppose conditions have changed after the initial study, and the unit profit for doors ( P D ) has increased from $300 to $450 while the unit profit for windows ( P W ) has decreased from $500 to $400. The calculations for the 100 percent rule then are
P D : $300 → $450
Percentage of allowable increase 5 100 a450 2 300 450
b% 5 33 1@3% P W : $500 → $400
Percentage of allowable decrease 5 100 a500 2 400 300
b% 5 33 1@3% Sum 5 66
2@3%
Since the sum of the percentages does not exceed 100 percent, the original optimal solution ( D, W ) 5 (2, 6) definitely is still optimal, just as we found earlier in Figure 5.10 .
Now suppose conditions have changed even further, so P D has increased from $300 to $600 while P W has decreased from $500 to $300. The calculations for the 100 percent rule now are
P D : $300 → $600
Percentage of allowable increase 5 100 a600 2 300 450
b% 5 66 2@3% P W : $500 → $300
Percentage of allowbale decrease 5 100 a500 2 300 300
b% 5 66 2@3% Sum 5 133 1@3%
Since the sum of the percentages now exceeds 100 percent, the 100 percent rule says that we can no longer guarantee that ( D, W ) 5 (2, 6) is still optimal. In fact, we found earlier in both Figures 5.11 and 5.14 that the optimal solution has changed to ( D, W ) 5 (4, 3).
These results suggest how to find just where the optimal solution changes while P D is being increased and P W is being decreased in this way. Since 100 percent is midway between 66⅔ percent and 1331/3 percent, the sum of the percentage changes will equal 100 percent when the values of P D and P W are midway between their values in the above cases. In particu- lar, P D 5 $525 is midway between $450 and $600 and P W 5 $350 is midway between $400 and $300. The corresponding calculations for the 100 percent rule are
P D : $300 → $525
Percentage of allowable increase 5 100 a525 2 300 450
b% 5 50% P W : $500 → $350
Percentage of allowable decrease 5 100 a500 2 350 300
b% 5 50% Sum 5 100%
Although the sum of the percentages equals 100 percent, the fact that it does not exceed 100 percent guarantees that ( D, W ) 5 (2, 6) is still optimal. Figure 5.15 shows graphically that both (2, 6) and (4, 3) are now optimal, as well as all the points on the line segment connecting these two points. However, if P D and P W were to be changed any further from their original val- ues (so that the sum of the percentages exceeds 100 percent), the objective function line would be rotated so far toward the vertical that ( D, W ) 5 (4, 3) would become the only optimal solution.
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5.4 The Effect of Simultaneous Changes in Objective Function Coefficients 167
At the same time, keep in mind that having the sum of the percentages of allowable changes exceed 100 percent does not automatically mean that the optimal solution will change. For example, suppose that the estimates of both unit profits are halved. The resulting calculations for the 100 percent rule are
P D : $300 → $150
Percentage of allowable decrease 5 100 a300 2 150 300
b% 5 50% P W : $500 → $250
Percentage of allowable decrease 5 100 a500 2 250 300
b% 5 83% Sum 5 133%
Even though this sum exceeds 100 percent, Figure 5.16 shows that the original optimal solu- tion is still optimal. In fact, the objective function line has the same slope as the original objec- tive function line (the solid line in Figure 5.9 ). This happens whenever proportional changes are made to all the unit profits, which will automatically lead to the same optimal solution.
Comparisons You now have seen three approaches to investigating what happens if simultaneous changes occur in the coefficients of the objective function: (1) try out changes directly on a spread- sheet, (2) use a two-way parameter analysis report and (3) apply the 100 percent rule.
The spreadsheet approach is a good place to start, especially for less experienced model- ers, because it is simple and quick. If you are only interested in checking one specific set of changes in the coefficients, you can immediately see what happens after making the changes in the spreadsheet.
Here is an example where the original optimal solu- tion is still optimal even though the sum exceeds 100 percent.
FIGURE 5.15 When the estimates of the unit profits for doors and windows change to PD 5 $525 and PW 5 $350, which lies at the edge of what is allowed by the 100 percent rule, the graphical method shows that (D, W ) 5 (2, 6) still is an optimal solu- tion, but now every other point on the line segment between this solution and (4, 3) also is optimal.
W
D
10
8
6
4
2
2 4 6 8
Production rate for doors
Production rate for windows
0
(4, 3)
(2, 6)
Objective function line now is Profit = $3,150 = 525D + 350W since PD = $525 and PW = $350
Entire line segment is optimal
Feasible region
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168 Chapter Five What-If Analysis for Linear Programming
More often, there will be numerous possibilities for what the true values of the coefficients will turn out to be, because of uncertainty in the original estimates of these coefficients. The parameter analysis report is useful for systematically checking a variety of possible changes in one or two objective function coefficients. Trying out representative possibilities on the spreadsheet may provide all the insight that is needed. Perhaps the optimal solution for the original model will remain optimal over nearly all these possibilities, so this solution can be confidently used. Or perhaps it will become clear that the original estimates need to be refined before selecting a solution.
When the spreadsheet approach and/or parameter analysis report does not provide a clear conclusion, the 100 percent rule can usefully complement this approach in the following ways:
• The 100 percent rule can be used to determine just how large the changes in the objective function coefficients need to be before the original optimal solution may no longer be optimal.
• When the model has a large number of decision variables (as is common for real prob- lems), it may become impractical to use the spreadsheet approach to systematically try out a variety of simultaneous changes in many or all of the coefficients in the objective func- tion because of the huge number of representative possibilities. The parameter analysis report can only be used to systematically check possible changes in—at most—two coef- ficients at a time. However, by dividing each coefficient’s allowable increase or allowable decrease by the number of decision variables, the 100 percent rule immediately indicates how much each coefficient can be safely changed without invalidating the current optimal solution.
• After completing the study, if conditions change in the future that cause some or all of the coefficients in the objective function to change, the 100 percent rule quickly indi- cates whether the original optimal solution must remain optimal. If the answer is affirma- tive, there is no need to take all the time that may be required to reconstruct the (revised) spreadsheet model. The time saved can be very substantial for large models.
FIGURE 5.16 When the estimates of the unit profits for doors and windows change to PD 5 $150 and PW 5 $250 (half their original values), the graphical method shows that the optimal solution still is (D, W ) 5 (2, 6), even though the 100 percent rule says that the optimal solution might change.
W
D
8
6
4
2
2 4 6 8
Production rate for doors
Production rate for windows
0
(2, 6)
Feasible region
Profit = $1,800 = 150D + 250W Optimal solution
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5.5 The Effect of Single Changes in a Constraint 169
1. In the 100 percent rule for simultaneous changes in objective function coefficients, what are the percentage changes that are being considered?
2. In this 100 percent rule, if the sum of the percentage changes does not exceed 100 percent, what does this say about the original optimal solution?
3. In this 100 percent rule, if the sum of the percentage changes exceeds 100 percent, does this mean that the original optimal solution is no longer optimal?
Review Questions
5.5 THE EFFECT OF SINGLE CHANGES IN A CONSTRAINT
We now turn our focus from the coefficients in the objective function to the effect of chang- ing the functional constraints. The changes might occur either in the coefficients on the left- hand sides of the constraints or in the values of the right-hand sides.
We might be interested in the effect of such changes for the same reason we are interested in this effect for objective function coefficients, namely, that these parameters of the model are only estimates of quantities that cannot be determined precisely at this time so we want to determine the effect if these estimates are inaccurate.
However, a more common reason for this interest is the one discussed at the end of Section 5.1, namely, that the right-hand sides of the functional constraints may well represent managerial policy decisions rather than quantities that are largely outside the control of management. There- fore, after the model has been solved, management will want to analyze the effect of altering these policy decisions in a variety of ways to see if these decisions can be improved. What-if analysis provides valuable guidance to management in determining the effect of altering these policy decisions. (Recall that this was cited as the third benefit of what-if analysis in Section 5.1.)
This section describes how to perform what-if analysis when making changes in just one spot (a coefficient or a right-hand side) of a single constraint. The next section then will deal with simultaneous changes in the constraints.
The procedure for determining the effect if a single change is made in a constraint is the same regardless of whether the change is in a coefficient on the left-hand side or in the value on the right-hand side. (The one exception is that the Solver sensitivity report provides infor- mation about changes in the right-hand side but does not do so for the left-hand side.) There- fore, we will illustrate the procedure by making changes in a right-hand side.
In particular, we return to the Wyndor case study to address the third what-if question posed by Wyndor management in Section 5.2.
Question 3: What happens if a change is made in the number of hours of production time per week being made available to Wyndor’s new products in one of the plants?
The number of hours available in each plant is the value of the right-hand side for the corre- sponding constraint, so we want to investigate the effect of changing this right-hand side for one of the plants. With the original optimal solution, ( D, W ) 5 (2, 6), only 2 of the 4 avail- able hours in Plant 1 are used, so changing this number of available hours (barring a large decrease) would have no effect on either the optimal solution or the resulting total profit from the two new products. However, it is unclear what would happen if the number of available hours in either Plant 2 or Plant 3 were to be changed. Let’s start with Plant 2.
Using the Spreadsheet for This Analysis Referring back to Section 5.2, Figure 5.1 shows the spreadsheet model for the original Wyn- dor problem before beginning what-if analysis. The optimal solution is ( D, W ) 5 (2, 6) with a total profit of $3,600 per week from the two new products. Cell G8 shows that 12 hours of production time per week are being made available for the new products in Plant 2.
To see what happens if a specific change is made in this number of hours, all you need to do is substitute the new number in cell G8 and run Solver again. For example, Figure 5.17 shows the result if the number of hours is increased from 12 to 13. The corresponding optimal solu- tion in C12:D12 gives a total profit of $3,750. Thus, the resulting change in profit would be
Incremental profit 5 $3,750 2 $3,600 5 $150
When the right-hand sides represent managerial policy decisions, what-if analysis provides guidance regard- ing the effect of altering these decisions.
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170 Chapter Five What-If Analysis for Linear Programming
Since this increase in profit is obtained by adding just one more hour in Plant 2, it would be interesting to see the effect of adding several more hours. Figure 5.18 shows the effect of adding five more hours. Comparing Figure 5.18 to Figure 5.17 , the additional profit from providing five more hours would be
Incremental profit 5 $4,500 2 $3,750 5 $750 from adding 5 hours 5 $150 per hour added
Would adding even more hours increase profit even further? Figure 5.19 shows what would happen if a total of 20 hours per week were made available to the new products in Plant 2. Both the optimal solution and the total profit are the same as in Figure 5.18 , so increasing from 18 to 20 hours would not help. (The reason is that the 18 hours available in Plant 3 prevent producing more than 9 windows per week, so only 18 hours can be used in Plant 2.) Thus, it appears that 18 hours is the maximum that should be considered for Plant 2.
However, the fact that the total profit from the two new products can be increased sub- stantially by increasing the number of hours per week made available to the new products from 12 to 18 does not mean that these additional hours should be provided automatically. The production time made available for these two new products can be increased only if it is decreased for other products. Therefore, management will need to assess the disadvan- tages of decreasing the production time for any other products (including both lost profit
So far, each additional hour provided in Plant 2 adds $150 to profit.
Now management needs to consider the trade-off between adding production time for the new products and decreasing it for other products.
FIGURE 5.17 The revised Wyndor problem where the hours available in Plant 2 per week have been increased from 12 (as in Figure 5.1) to 13, which results in an increase of $150 in the total profit per week from the two new products.
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
1.66667
13
18
Windows
Doors
1.667 6.5
Windows
$3,750
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$300 $500
1
0
3
0
2
2
4
13
18
≤
≤
≤
FIGURE 5.18 A further revision of the Wyndor problem in Figure 5.17 to further increase the hours avail- able in Plant 2 from 13 to 18, which results in a further increase in total profit of $750 (which is the $150 per hour added in Plant 2).
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
0
18
18
Windows
Doors
0 9
Windows
$4,500
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$300 $500
1
0
3
0
2
2
4
18
18
≤ ≤
≤
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5.5 The Effect of Single Changes in a Constraint 171
and less tangible disadvantages) before deciding whether to increase the production time for the new products. This analysis also might lead to decreasing the production time made available to the two new products in one or more of the plants.
Using a Parameter Analysis Report (RSPE) for This Analysis Using RPSE, a parameter analysis report can be used to systematically determine the effect of making various changes in one of the parameters in a constraint. For example, Figure 5.20 displays a parameter analysis report (obtained by following the same procedure used to gener- ate a parameter analysis report in Section 5.3) to show how the changing cells and total profit change as the number of available hours in Plant 2 range between 4 and 20.
An interesting pattern is apparent in the incremental profit column. Starting at 12 hours available at Plant 2 (the current allotment), each additional hour allocated yields an additional $150 in profit (up to making 18 hours available). Similarly, if hours are taken away from
FIGURE 5.19 A further revision of the Wyndor problem in Figure 5.18 to further increase the hours avail- able in Plant 2 from 18 to 20, which results in no change in total profit because the optimal solu- tion cannot make use of these additional hours.
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
0
18
18
Windows
Doors
0 9
Windows
$4,500
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$300 $500
1
0
3
0
2
2
4
20
18
≤
≤
≤
FIGURE 5.20 The parameter analysis report that shows the effect of varying the number of hours of production time being made available per week in Plant 2 for Wyndor’s new products.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
$2,200
$2,450
$2,700
$2,850
$3,000
$3,150
$3,300
$3,450
$3,600
$3,750
$3,900
$4,050
$4,200
$4,350
$4,500
$4,500
$4,500
$250
$250
$150
$150
$150
$150
$150
$150
$150
$150
$150
$150
$150
$150
$0
$0
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9
9
4
4
4
3.666666667
3.333333333
3
2.666666667
2.333333333
2
1.666666667
1.333333333
1
0.666666667
0.333333333
0
0
0
DoorsProduced WindowsProduced TotalProfit Incremental ProfitHoursAvailableInPlant2
A B C D E
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172 Chapter Five What-If Analysis for Linear Programming
Plant 2, each hour lost causes a loss of $150 profit (down to making six hours available). This rate of change in the profit for increases or decreases in the right-hand side of a constraint is known as the shadow price.
Given an optimal solution and the corresponding value of the objective function for a linear programming model, the shadow price for a functional constraint is the rate at which the value of the objective function can be increased by increasing the right-hand side of the constraint by a small amount.
However, the shadow price of $150 for the Plant 2 constraint is valid only within a range of values near 12 (in particular, between 6 hours and 18 hours). If the number of available hours is increased beyond 18 hours, then the incremental profit drops to zero. If the available hours are reduced below six hours, then profit drops at a faster rate of $250 per hour. There- fore, letting RHS denote the value of the right-hand side, the shadow price of $150 is valid for
6 # RHS # 18
This range is known as the allowable range for the right-hand side (or just allowable range for short).
The allowable range for the right-hand side of a functional constraint is the range of values for this right-hand side over which this constraint’s shadow price remains valid.
Unlike the allowable range for objective function coefficients described in Section 5.3, a change that is within the allowable range for the right-hand side does not mean the original solution is still optimal. In fact, any time the shadow price is not zero, a change to a right-hand side leads to a change in the optimal solution. The shadow price indicates how much the value of the objective function will change as the optimal solution changes.
Using the Sensitivity Report to Obtain the Key Information As illustrated earlier, it is straightforward to use a parameter analysis report to calculate the shadow price for a functional constraint, as well as to find (or at least closely approximate) the allowable range for the right-hand side of this constraint over which the shadow price remains valid. However, this same information also can be obtained immediately from Solv- er’s sensitivity report for all the functional constraints. Figure 5.21 shows the full sensitivity report provided by Solver for the original Wyndor problem after obtaining the optimal solu- tion given in Figure 5.1 . The top half is the part already shown in Figure 5.8 for finding allow- able ranges for the objective function coefficients. The bottom half focuses on the functional constraints, including providing the shadow prices for these constraints in the fourth column. The first three columns remind us that (1) the output cells for these constraints in Figure 5.1 are cells E7 to E9, (2) these cells give the number of production hours used per week in the three plants, and (3) the final values in these cells are 2, 12, and 18 (as shown in column E of Figure 5.1 ). (We will discuss the last three columns a little later.)
The shadow price given in the fourth column for each constraint tells us how much the value of the objective function [objective cell (G12) in Figure 5.1 ] would increase if the
In general, the shadow price for a constraint reveals the rate at which the objective cell can be increased by increasing the right-hand side of that constraint. This remains valid as long as the right-hand side is within its allowable range.
This allowable range focuses on a right-hand side and the corresponding shadow price. In contrast to the allowable range for objective function coeffi- cients described in Section 5.3, it does not indicate whether the original solu- tion is still optimal, just whether the shadow price remains valid.
Constraints
Cell Name Final Value
Shadow Price
Constraint R. H. Side
Allowable Increase
Allowable Decrease
$E$7 Plant 1 Used 2 0 4 1E 1 30 2 $E$8 Plant 2 Used 12 150 12 6 6 $E$9 Plant 3 Used 18 100 18 6 6
Variable Cells
Cell Name Final Value
Reduced Cost
Objective Coefficient
Allowable Increase
Allowable Decrease
$C$12 DoorsProduced 2 0 300 450 300 $D$12 WindowsProduced 6 0 500 1E 1 30 300
FIGURE 5.21 The complete sensitivity report generated by Solver for the original Wyndor problem as formulated in Figure 5.1.
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5.5 The Effect of Single Changes in a Constraint 173
right-hand side of that constraint (cell G7, G8, or G9) were to be increased by 1. Conversely, it also tells us how much the value of the objective function would decrease if the right-hand side were to be decreased by 1. The shadow price for the Plant 1 constraint is 0, because this plant already is using less hours (2) than are available (4) so there would be no benefit to mak- ing an additional hour available. However, Plants 2 and 3 are using all the hours available to them for the two new products (with the product mix given by the changing cells). Thus, it is not surprising that the shadow prices indicate that the objective cell would increase if the hours available in either Plant 2 or Plant 3 were to be increased.
To express this information in the language of management, the value of the objective function for this problem [objective cell (G12) in Figure 5.1 ] represents the total profit in dollars per week from the two new products under consideration. The right-hand side of each functional constraint represents the number of hours of production time being made avail- able per week for these products in the plant that corresponds to this constraint. Therefore, the shadow price for a functional constraint informs management as to how much the total profit from the two new products could be increased for each additional hour of production time made available to these products per week in the corresponding plant. Conversely, the shadow price indicates how much this profit would decrease for each reduction of an hour of production time in that plant. This interpretation of the shadow price remains valid as long as the change in the number of hours of production time is not very large.
Specifically, this interpretation of the shadow price remains valid as long as the number of hours of production time remains within its allowable range. Solver’s sensitivity report pro- vides all the data needed to identify the allowable range of each functional constraint. Refer back to the bottom of this report given in Figure 5.21 . The final three columns enable calcu- lation of this range. The “Constraint R. H. Side” column indicates the original value of the right-hand side before any change is made. Adding the number in the “Allowable Increase” column to this original value then gives the upper endpoint of the allowable range. Similarly, subtracting the number in the “Allowable Decrease” column from this original value gives the lower endpoint. Using the fact that 1E 1 30 represents infinity ( ̀ ), these calculations of the allowable ranges are shown below, where a subscript has been added to each RHS to identify the constraint involved.
Plant 1 constraint: 4 2 2 # RHS 1 # 4 1 ̀ , so 2 # RHS 1 (no upper limit)
Plant 2 constraint: 12 2 6 # RHS 2 # 12 1 6, so 6 # RHS 2 # 18
Plant 3 constraint: 18 2 6 # RHS 3 # 18 1 6, so 12 # RHS 3 # 24
In the case of the Plant 2 constraint, Figure 5.22 provides graphical insight into why 6 # RHS # 18 is the range of validity for the shadow price. The optimal solution for the original problem, ( D, W ) 5 (2, 6), lies at the intersection of line B and line C. The equation for line B is 2 W 5 12 because this is the constraint boundary line for the Plant 2 constraint (2 W # 12). However, if the value of this right-hand side (RHS 2 5 12) is changed, line B will either shift upward (for a larger value of RHS 2 ) or downward (for a smaller value of RHS 2 ). As line B shifts, the boundary of the feasible region shifts accordingly and the optimal solu- tion continues to lie at the intersection of the shifted line B and line C—provided the shift in line B is not so large that this intersection is no longer feasible. Each time RHS 2 is increased (or decreased) by 1, this intersection shifts enough to increase (or decrease) Profit by the amount of the shadow price ($150). Figure 5.22 indicates that this intersection remains fea- sible (and so optimal) as RHS 2 increases from 12 to 18, because the feasible region expands upward as line B shifts upward. However, for values of RHS 2 larger than 18, this intersection is no longer feasible because it gives a negative value of D (the production rate for doors). Thus, each increase of 1 above 18 no longer increases Profit by the amount of the shadow price. Similarly, as RHS 2 decreases from 12 to 6, this intersection remains feasible (and so optimal) as line B shifts down accordingly. However, for values of RHS 2 less than 6, this intersection is no longer feasible because it violates the Plant 1 constraint ( D # 4) whose boundary line is line A. Hence, each decrease of 1 below 6 no longer decreases Profit by the amount of the shadow price. Consequently, 6 # RHS # 18 is the allowable range over which the shadow price is valid.
The shadow prices reveal the relationship between profit and the amount of production time made avail- able in the plants.
Here is how to find the allowable ranges for the right-hand sides from the sensitivity report.
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174 Chapter Five What-If Analysis for Linear Programming
Summary Recall again that the right-hand side of each functional constraint for the Wyndor problem represents the number of hours of production time per week in the corresponding plant that is being made available to the two new products. The shadow price for each constraint reveals how much the total profit from these new products would increase for each additional hour of production time made available in the corresponding plant for these products. This inter- pretation of the shadow price remains valid as long as the number of hours remains within its allowable range. Therefore, each shadow price can be applied by management to evaluate a change in its original decision regarding the number of hours as long as the new number is within the corresponding allowable range. This evaluation also would need to take into account how the change in the number of hours made available to the new products would impact the production rates and profits for the company’s other products.
FIGURE 5.22 A graphical interpretation of the allowable range, 6 # RHS2 # 18, for the right-hand side of Wyn- dor’s Plant 2 constraint.
W
D
10
8
6
4
2
2 4 6 8
Production rate for doors
Production rate for windows
0
(4, 3)
(2, 6)
(0, 9)
Line A (D = 4)
Line C (3D + 2W = 18)
2W = 6 Profit = 300(4) + 500(3) = $2,700
2W = 18 Profit = 300(0) + 500(9) = $4,500
2W = 12 Profit = 300(2) + 500(6) = $3,600 Line B
Feasible
region for
original
problem
1. Why might it be of interest to investigate the effect of making changes in a functional constraint?
2. Why might it be possible to alter the right-hand side of a functional constraint? 3. What is meant by a shadow price? 4. How can a shadow price be found by using the spreadsheet? By using a parameter analysis
report? By using Solver’s sensitivity report? 5. Why are shadow prices of interest to managers? 6. Can shadow prices be used to determine the effect of decreasing rather than increasing the
right-hand side of a functional constraint? 7. What does a shadow price of 0 tell a manager? 8. Which columns of Solver’s sensitivity report are used to find the allowable range for the right-
hand side of a functional constraint? 9. Why are these allowable ranges of interest to managers?
Review Questions
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5.6 The Effect of Simultaneous Changes in the Constraints 175
5.6 THE EFFECT OF SIMULTANEOUS CHANGES IN THE CONSTRAINTS
The preceding section described how to perform what-if analysis to investigate the effect of changes in a single spot of a constraint. We now turn our consideration to the effect of simul- taneous changes in the constraints.
The need to consider these simultaneous changes arises frequently. There may be con- siderable uncertainty about the estimates for a number of the parameters in the functional constraints, so questions will arise as to the effect if the true values of the parameters simulta- neously deviate significantly from the estimates. Since the right-hand sides of the constraints often represent managerial policy decisions, questions will arise about what would happen if some of these decisions were to be changed. These decisions frequently are interrelated and so need to be considered simultaneously.
We outline next how the usual three methods for performing what-if analysis can be applied to considering simultaneous changes in the constraints. The third one (using Solver’s sensitivity report) is only helpful for changing right-hand sides. For the first two (using the spreadsheet and using a parameter analysis report), the procedure is the same regardless of whether the changes are in the coefficients on the left-hand sides or in the right-hand sides of the constraints (or both). Since changing the right-hand sides is the more important case, we will focus on this case.
In particular, we now will deal with the last of Wyndor management’s what-if questions.
Question 4: What happens if simultaneous changes are made in the number of hours of produc- tion time per week being made available to Wyndor’s new products in all the plants?
In particular, after seeing that the Plant 2 constraint has the largest shadow price (150), versus a shadow price of 100 for the Plant 3 constraint, management now is interested in exploring a specific type of simultaneous change in these production hours. By shifting the production of one of the company’s current products from Plant 2 to Plant 3, it is possible to increase the number of production hours available to the new products in Plant 2 by decreasing the num- ber of production hours available in Plant 3 by the same amount. Management wonders what would happen if these simultaneous changes in production hours were made.
Using the Spreadsheet for This Analysis According to the shadow prices, the effect of shifting one hour of production time per week from Plant 3 to Plant 2 would be as follows.
RHS 2 : 12 → 13 Change in total profit 5 Shadow price 5 $150 RHS 3 : 18 → 17 Change in total profit 5 2 Shadow price 5 2 100
Net increase in total profit 5 $50
However, we don’t know if these shadow prices remain valid if both right-hand sides are changed by this amount.
A quick way to check this is to substitute the new right-hand sides into the original spread- sheet in Figure 5.1 and run Solver again. The resulting spreadsheet in Figure 5.23 shows that the net increase in total profit (from $3,600 to $3,650) is indeed $50, so the shadow prices are valid for these particular simultaneous changes in right-hand sides.
How long will these shadow prices remain valid if we continue shifting production hours from Plant 3 to Plant 2? We could continue checking this by substituting other combinations of right-hand sides into the spreadsheet and re-solving each time. However, a more systematic way of doing this is to use a parameter analysis report, as described below.
Using a Parameter Analysis Report for This Analysis Since it can become tedious, or even impractical, to use the spreadsheet to investigate a large number of simultaneous changes in the right-hand sides, let us see how a parameter analysis report can be used to do this analysis more systematically.
We could use a two-way parameter analysis report to investigate how the profit and optimal production rates vary for different combinations of the number of hours available in Plant 2 and Plant 3. However, in this case we aren’t interested in all combinations of hours in the two plants,
Managerial policy decisions involving right-hand sides frequently are interrelated, so changes in these deci- sions should be considered simultaneously.
We now are checking to see whether the shadow prices remain valid for evaluat- ing specific simultaneous changes in the right-hand sides.
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176 Chapter Five What-If Analysis for Linear Programming
but rather only those combinations that involve a simple shifting of available hours from Plant 3 to Plant 2. For this analysis, we will see that a one-way parameter analysis table is sufficient.
For each hour reduced in Plant 3, an additional hour is made available in Plant 2. Thus, the number of available hours in Plant 2 is a function of the number of available hours in Plant 3. In particular, since there are 30 total hours available at the two plants (RHS 2 1 RHS 3 5 30), the number of available hours in Plant 2 (RHS 2 ) is
RHS2 5 30 2 RHS3
Figure 5.24 shows the Wyndor Glass Co. spreadsheet with the data cell for the number of available hours in Plant 2 replaced by the above formula. Because of this formula, whenever the number of available hours in Plant 3 is reduced, the number of available hours in Plant 2 will automatically increase by the same amount. Now a one-way parameter analysis report can be used to investigate various numbers of available hours in Plant 3 (with the correspond- ing automatic adjustment made to the available hours at Plant 2). HoursAvailableInPlant3 (G9) is specified as a parameter cell with a range of trial values from 18 down to 12. A parameter analysis table is then generated in Figure 5.25 to show how DoorsProduced (C12), WindowsProduced (D12), and TotalProfit (G12) vary as HoursAvailableInPlant3 (G9) varies from 18 down to 12. A column was added to the left of the parameter analysis table to show how HoursAvailableInPlant2 varies correspondingly with the different values of HoursAvail- ableInPlant3. Also, we have calculated the incremental profit (in column F) for each hour shifted from Plant 3 to Plant 2.
Again there is a pattern to the incremental profit. For each hour shifted from Plant 3 to Plant 2 (up to 3 hours), an additional profit of $50 is achieved. However, if more than 3 hours are shifted, the incremental profit becomes 2 $250. Thus, it appears worthwhile to shift up to 3 available hours from Plant 3 to Plant 2, but no more.
Although a one-way parameter analysis report is limited to enumerating trial values for only one data cell, you have just seen how such a parameter analysis report still can system- atically investigate a large number of simultaneous changes in two data cells by entering a formula for the second data cell in terms of the first one. The two data cells considered above happened to be right-hand sides of constraints, but either or both could have been coefficients on the left-hand side instead. It is even possible to enter formulas for multiple data cells in terms of the one whose trial values are being enumerated. Furthermore, by using a two-way parameter analysis report, trial values can be enumerated simultaneously for two data cells, with the possibility of entering formulas for additional data cells in terms of these two.
Gleaning Additional Information from the Sensitivity Report Despite the versatility of a parameter analysis report, it cannot handle a number of impor- tant cases. The most important one is where management wants to explore various possi- bilities for changing its policy decisions that correspond to changing several right-hand sides
By entering a formula into one data cell in terms of another one, a one-way parameter analysis report is able to investigate inter- related trial values in both data cells.
FIGURE 5.23 The revised Wyndor problem where column G in Figure 5.1 has been changed by shifting one of the hours available in Plant 3 to Plant 2 and then re-solving.
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
1.33333
13
17
Windows
Doors
1.333 6.5
Windows
$3,650
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
$300 $500
1
0
3
0
2
2
4
13
17
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5.6 The Effect of Simultaneous Changes in the Constraints 177
simultaneously in a variety of ways. Although the spreadsheet can be used to see the effect of any combination of simultaneous changes, it can take an exorbitant amount of time to systematically investigate a large number of simultaneous changes in right-hand sides in this way. Fortunately, Solver’s sensitivity report provides valuable information for guiding such an investigation. In particular, there is a 100 percent rule (analogous to the one presented in Section 5.4) that uses this information to conduct this kind of investigation.
Recall that the 100 percent rule described in Section 5.4 is used to investigate simultaneous changes in objective function coefficients. The new 100 percent rule presented next investi- gates simultaneous changes in right-hand sides in a similar way.
The data needed to apply the new 100 percent rule for the Wyndor problem are given by the last three columns in the bottom part of the sensitivity report in Figure 5.21 . Keep in mind that we can safely use the entire allowable decrease or increase from the current value of a right-hand side only if none of the other right-hand sides are changed at all. With simultane- ous changes in the right-hand sides, we focus for each change on the percentage of the allow- able decrease or increase that is being used. As detailed next, the 100 percent rule basically says that we can safely make the simultaneous changes only if the sum of these percentages does not exceed 100 percent.
This 100 percent rule reveals whether the simulta- neous changes in the right- hand sides are small enough to guarantee that the shadow prices are still valid.
FIGURE 5.24 By inserting a formula into cell G8 that keeps the total number of hours available in Plant 2 and Plant 3 equal to 30, it will be pos- sible to generate a one-way parameter analysis report (see Figure 5.25) that shows the effect of shifting more and more of the hours available from Plant 3 to Plant 2.
5
G H
6
8
9
7
Hours
Available
=H8-G9 30
18
4 Total (Plants 2 & 3)
1
A B C D E F G H
2
3
4
5
6
7
8
9
10
11
12
Wyndor Glass Co. Product-Mix Problem
Doors
2
12
18
Windows
Doors
2 6
Windows
$3,600
Total Profit
Unit Profit
Plant 1
Plant 2
Plant 3
Units Produced
Hours Used per Unit Produced
Hours
Available
Hours
Used
Total (Plants 2 & 3)
$300 $500
1
0
3
0
2
2
4
12
18
30
FIGURE 5.25 The parameter analysis report that shows the effect of shifting more and more of the hours available from Plant 3 to Plant 2 for the Wyndor problem.
A 1 2
3
4 5
6
7
8
12
13
14 15
16
17
18
18
17
16 15
14
13
12
2
1.333
0.667 0
0
0
0
6
6.5
7 7.5
7
6.5
6
$3,600
$3,650
$3,700 $3,750
$3,500
$3,250
$3,000
$50
$50 $50
-$250
-$250
-$250
B C D E F HoursAvailableInPlant2 HoursAvailableInPlant3 DoorsProduced WindowsProduced TotalProfit Incremental Profit
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178 Chapter Five What-If Analysis for Linear Programming
The 100 Percent Rule for Simultaneous Changes in Right-Hand Sides: The shadow prices remain valid for predicting the effect of simultaneously changing the right-hand sides of some of the functional constraints as long as the changes are not too large. To check whether the changes are small enough, calculate for each change the percentage of the allowable change (decrease or increase) for that right-hand side to remain within its allowable range. If the sum of the percentage changes does not exceed 100 percent, the shadow prices definitely will still be valid. (If the sum does exceed 100 percent, then we cannot be sure.)
To illustrate this rule, consider again the simultaneous changes (shifting one hour of pro- duction time per week from Plant 3 to Plant 2) that led to Figure 5.23 . The calculations for the 100 percent rule in this case are
RHS 2 : 12 → 13
Percentage of allowable increase 5 100 a13 2 12 6
b 5 16 2@3% RHS 3 : 18 → 17
Percentage of allowable decrease 5 100 a18 2 17 6
b 5 16 2@3% Sum 5 33 1@3%
Since the sum of 331/3 percent is less than 100 percent, the shadow prices definitely are valid for predicting the effect of these changes, as was illustrated with Figure 5.23 .
The fact that 331/3 percent is one-third of 100 percent suggests that the changes can be three times as large as above without invalidating the shadow prices. To check this, let us apply the 100 percent rule with these larger changes.
RHS 2 : 12 → 15
Percentage of allowable increase 5 a15 2 12 6
b% 5 50% RHS 3 : 18 → 15
Percentage of allowable decrease 5 a18 2 15 6
b% 5 50% Sum 5 100%
Because the sum does not exceed 100 percent, the shadow prices are still valid, but these are the largest changes in the right-hand sides that can provide this guarantee. In fact, Figure 5.25 demonstrated that the shadow prices become invalid for larger changes.
1. Why might it be of interest to investigate the effect of making simultaneous changes in the functional constraints?
2. How can the spreadsheet be used to investigate simultaneous changes in the functional constraints?
3. What are the capabilities of a parameter analysis report for investigating simultaneous changes in the functional constraints?
4. Why might a manager be interested in considering simultaneous changes in right-hand sides? 5. What is the 100 percent rule for simultaneous changes in right-hand sides? 6. What are the data needed to apply the 100 percent rule for simultaneous changes in right-
hand sides? 7. What is guaranteed if the sum of the percentages of allowable changes in the right-hand sides
does not exceed 100 percent? 8. What is the conclusion if the sum of the percentages of allowable changes in the right-hand
sides does exceed 100 percent?
Review Questions
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Chapter 5 Glossary 179
What-if analysis is analysis done after finding an optimal solution for the original version of the basic model. This analysis provides important insights to help guide managerial decision making. This chap- ter describes how this is done when the basic model is a linear programming model. The spreadsheet for the model, the parameter analysis report available with RSPE, and the sensitivity report generated by Solver all play a central role in this process.
The coefficients in the objective function typically represent quantities that can only be roughly estimated when the model is formulated. Will the optimal solution obtained from the model be the correct one if the true value of one of these coefficients is significantly different from the estimate used in the model? The spreadsheet can be used to quickly check specific changes in the coeffi- cient. The parameter analysis report enables the systematic investigation of many trial values for this coefficient. For a broader investigation, the allowable range for each coefficient identifies the interval within which the true value must lie in order for this solution to still be the correct optimal solution. These ranges are easily calculated from the data in the sensitivity report provided by Solver.
What happens if there are significant inaccuracies in the estimates of two or more coefficients in the objective function? Specific simultaneous changes can be checked with the spreadsheet. A two-way parameter analysis report can systematically investigate various simultaneous changes in two coeffi- cients. To go further, the 100 percent rule for simultaneous changes in objective function coefficients provides a convenient way of checking whole ranges of simultaneous changes, again by using the data in Solver’s sensitivity report.
What-if analysis usually extends to considering the effect of changes in the functional constraints as well. Occasionally, changes in the coefficients of these constraints will be considered because of the uncertainty in their original estimates. More frequently, the changes considered will be in the right-hand sides of the constraints. The right-hand sides frequently represent managerial policy decisions. In such cases, shadow prices provide valuable guidance to management about the potential effects of altering these policy decisions. The shadow price for each constraint is easily found by using the spreadsheet, a parameter analysis report, or the sensitivity report.
Shadow price analysis can be validly applied to investigate possible changes in right-hand sides as long as these changes are not too large. The allowable range for each right-hand side indicates just how far it can be changed, assuming no other changes are made. If other right-hand sides are changed as well, the 100 percent rule for simultaneous changes in right-hand sides enables checking whether the changes definitely are not too large. Solver’s sensitivity report provides the key infor- mation needed to find each allowable range or to apply this 100 percent rule. Both the spreadsheet and the parameter analysis report also can sometimes be used to help investigate these simultaneous changes.
5.7 Summary
Glossary allowable range for an objective function coefficient The range of values for a particular coefficient in the objective function over which the optimal solution for the original model remains optimal. (Section 5.3), 158 allowable range for the right-hand side The range of values for the right-hand side of a functional constraint over which this con- straint’s shadow price remains valid. (Section 5.5), 172 parameter cell A data cell containing a param- eter that will be systematically varied when generating a parameter analysis report. (Section 5.3), 157 parameters of the model The parameters of a linear programming model are the constants (coefficients or right-hand sides) in the functional constraints and the objective function. (Section 5.1), 151
sensitive parameter A parameter is consid- ered sensitive if even a small change in its value can change the optimal solution. (Section 5.1), 151 sensitivity analysis The part of what-if analy- sis that focuses on individual parameters of the model. It involves checking how sensitive the optimal solution is to the value of each param- eter. (Section 5.1), 152 shadow price The shadow price for a func- tional constraint is the rate at which the optimal value of the objective function can be increased by increasing the right-hand side of the constraint by a small amount. (Section 5.5), 172 what-if analysis Analysis that addresses ques- tions about what would happen to the optimal solution if different assumptions were made about future conditions. (Chapter introduction), 150
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180 Chapter Five What-If Analysis for Linear Programming
Chapter 5 Excel Files: Wyndor Example Profit & Gambit Example
Excel Add-in: Risk Solver Platform for Education (RSPE)
Interactive Management Science Modules:
Module for Graphical Linear Programming and Sensitivity Analysis
Supplement to Chapter 5 on the CD-ROM: Reduced Costs
Learning Aids for This Chapter in Your MS Courseware
Solved Problem (See the CD-ROM or Website for the Solution) 5.S1. Sensitivity Analysis at Stickley Furniture Stickley Furniture is a manufacturer of fine hand-crafted furni- ture. During the next production period, management is consid- ering producing dining room tables, dining room chairs, and/or bookcases. The time required for each item to go through the two stages of production (assembly and finishing), the amount of wood required (fine cherry wood), and the corresponding unit profits are given in the following table, along with the amount of each resource available in the upcoming production period.
Tables Chairs Bookcases Available
Assembly (minutes)
80 40 50 8,100
Finishing) (minutes)
30 20 30 4,500
Wood (pounds)
80 10 50 9,000
Unit profit $360 $125 $300
After formulating a linear programming model to determine the production levels that would maximize profit, the solved model and the corresponding sensitivity report are shown below.
a. Suppose the profit per table increases by $100. Will this change the optimal production quantities? What can be said about the change in total profit?
b. Suppose the profit per chair increases by $100. Will this change the optimal production quantities? What can be said about the change in total profit?
c. Suppose the profit per table increases by $90 and the profit per bookcase decreases by $50. Will this change the optimal produc- tion quantities? What can be said about the change in total profit?
d. Suppose a worker in the assembly department calls in sick, so eight fewer hours now are available in the assembly de- partment. How much would this affect total profit? Would it change the optimal production quantities?
e. Explain why the shadow price for the wood constraint is zero. f. A new worker has been hired who is trained to do both as-
sembly and finishing. She will split her time between the two areas, so there now are four additional hours available in both assembly and finishing. How much would this affect total profit? Would this change the optimal production quantities?
g. Based on the sensitivity report, is it wise to have the new worker in part f split her time equally between assembly and finishing, or would some other plan be better?
B C D E F G H
3
4
5
6
7
8
9
10
11
12
Unit Profit $360 $125 $300
Assembly (minutes)
Resources Required per Unit Available
Finishing (minutes)
Woods (pounds)
80
30
80
40
20
10
50
30
50
8,100
4,500
9,000
Used
8,100
4,500
8,100
Production 20 0 130
Tables Chairs Bookcases
Tables Chairs Bookcases
$46,200
Total Profit
Variable Cells
Cell Name Final Value
Reduced Cost
Objective Coefficient
Allowable Increase
Allowable Decrease
$C$12 Production Tables 20 0 360 120 60 $D$12 Production Chairs 0 288.333 125 88.333 1E 1 30 $E$12 Production Bookcases 130 0 300 60 75
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Chapter 5 Problems 181
Constraints
Cell Name Final Value
Shadow Price
Constraint R. H. Side
Allowable Increase
Allowable Decrease
$F$7 Assembly (minutes) Used 8,100 2 8,100 900 600 $F$8 Finishing (minutes) Used 4,500 6.67 4,500 360 1,462.5 $F$9 Wood (pounds) Used 8,100 0 9,000 1E 1 30 900
h. Use a parameter analysis report to determine how the op- timal production quantities and total profit will change de- pending on how the new worker in part f allocates her time between assembly and finishing. In particular, assume 0,
1, 2, . . . , or 8 hours are added to assembly, with a corre- sponding 8, 7, 6, . . . , or 0 hours added to finishing. (The original spreadsheet is contained on the CD included with the textbook.)
each of these estimates can be off before the optimal solution would change. Begin exploring this ques- tion for the first activity (producing toys) by using the spreadsheet and Solver to manually generate a table that gives the optimal solution and total profit as the unit profit for this activity increases in 50¢ increments from $2.00 to $4.00. What conclusion can be drawn about how much the estimate of this unit profit can differ in each direction from its original value of $3.00 before the optimal solution would change?
E* c. Repeat part b for the second activity (producing subassemblies) by generating a table as the unit profit for this activity increases in 50¢ increments from 2 $3.50 to 2 $1.50 (with the unit profit for the first activity fixed at $3).
E* d. Use a parameter analysis report to systematically generate all the data requested in parts b and c, except use 25¢ increments instead of 50¢ incre- ments. Use these data to refine your conclusions in parts b and c.
E* e. Use Solver’s sensitivity report to find the allowable range for the unit profit of each activity.
E* f. Use a two-way parameter analysis report to system- atically generate the total profit as the unit profits of the two activities are changed simultaneously as described in parts b and c.
g. Use the information provided by Solver’s sensitiv- ity report to describe how far the unit profits of the two activities can change simultaneously before the optimal solution might change.
5.2. Consider a resource-allocation problem having the fol- lowing data:
Resource Usage per Unit of Each Activity
Resource 1 2
Amount of Resource Available
1 1 2 10 2 1 3 12
Unit profit $2 $5
The objective is to determine the number of units of each activ- ity to undertake so as to maximize the total profit.
We have inserted the symbol E* to the left of each problem (or its parts) where Excel should be used (unless your instructor gives you contrary instructions). An asterisk on the problem number indi- cates that at least a partial answer is given in the back of the book.
5.1.* One of the products of the G. A. Tanner Company is a special kind of toy that provides an estimated unit profit of $3. Because of a large demand for this toy, management would like to increase its production rate from the current level of 1,000 per day. However, a limited supply of two subassemblies (A and B) from vendors makes this difficult. Each toy requires two subassemblies of type A, but the vendor providing these subas- semblies would only be able to increase its supply rate from the current 2,000 per day to a maximum of 3,000 per day. Each toy requires only one subassembly of type B, but the vendor provid- ing these subassemblies would be unable to increase its supply rate above the current level of 1,000 per day.
Because no other vendors currently are available to provide these subassemblies, management is considering initiating a new production process internally that would simultaneously produce an equal number of subassemblies of the two types to supplement the supply from the two vendors. It is estimated that the company’s cost for producing one subassembly of each type would be $2.50 more than the cost of purchasing these subas- semblies from the two vendors. Management wants to determine both the production rate of the toy and the production rate of each pair of subassemblies (one A and one B) that would maxi- mize the total profit.
Viewing this problem as a resource-allocation problem, one of the company’s managers has organized its data as follows:
Resource Usage per Unit of Each Activity
Resource Produce
Toys Produce
Subassemblies
Amount of Resource Available
Subassembly A 2 21 3,000 Subassembly B 1 21 1,000
Unit profit $3 2$2.50
E* a. Formulate and solve a spreadsheet model for this problem.
E* b. Since the stated unit profits for the two activities are only estimates, management wants to know how much
Problems
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182 Chapter Five What-If Analysis for Linear Programming
d. If the estimates change for more than one of the unit shipping costs, how can you use the sensitiv- ity report to determine whether the optimal solution might change?
E*5.4.* Consider the Union Airways problem presented in Sec- tion 3.3, including the spreadsheet in Figure 3.5 showing its for- mulation and optimal solution.
Management is about to begin negotiations on a new contract with the union that represents the company’s customer service agents. This might result in some small changes in the daily costs per agent given in Table 3.5 for the various shifts. Sev- eral possible changes listed below are being considered sepa- rately. In each case, management would like to know whether the change might result in the solution in Figure 3.5 no longer being optimal. Answer this question in parts a to e by using the spreadsheet and Solver directly. If the optimal solution changes, record the new solution.
a. The daily cost per agent for shift 2 changes from $160 to $165.
b. The daily cost per agent for shift 4 changes from $180 to $170.
c. The changes in parts a and b both occur.
d. The daily cost per agent increases by $4 for shifts 2, 4, and 5, but decreases by $4 for shifts 1 and 3.
e. The daily cost per agent increases by 2 percent for each shift.
f. Use Solver to generate the sensitivity report for this problem. Suppose that the above changes are being considered later without having the spreadsheet model immediately available on a computer. Show in each case how the sensitivity report can be used to check whether the original optimal solution must still be optimal.
g. For each of the five shifts in turn, use a parameter analysis report to systematically generate the opti- mal solution and total cost when the only change is that the daily cost per agent on that shift increases in $3 increments from $15 less than the current cost up to $15 more than the current cost.
E*5.5. Consider the Think-Big Development Co. problem pre- sented in Section 3.2, including the spreadsheet in Figure 3.3 showing its formulation and optimal solution. In parts a-g, use the spreadsheet and Solver to check whether the optimal solu- tion would change and, if so, what the new optimal solution would be, if the estimates in Table 3.3 of the net present values of the projects were to be changed in each of the following ways. (Consider each part by itself.) a. The net present value of project 1 (a high-rise office
building) increases by $200,000. b. The net present value of project 2 (a hotel) increases
by $200,000. c. The net present value of project 1 decreases by
$5 million. d. The net present value of project 3 (a shopping cen-
ter) decreases by $200,000. e. All three changes in parts b, c, and d occur
simultaneously.
While doing what-if analysis, you learn that the estimates of the unit profits are accurate only to within 6 50 percent. In other words, the ranges of likely values for these unit profits are $1 to $3 for activity 1 and $2.50 to $7.50 for activity 2. E* a. Formulate a spreadsheet model for this problem
based on the original estimates of the unit profits. Then use Solver to find an optimal solution and to generate the sensitivity report.
E* b. Use the spreadsheet and Solver to check whether this optimal solution remains optimal if the unit profit for activity 1 changes from $2 to $1. From $2 to $3.
E* c. Also check whether the optimal solution remains optimal if the unit profit for activity 1 still is $2 but the unit profit for activity 2 changes from $5 to $2.50. From $5 to $7.50.
E* d. Use a parameter analysis report to systematically generate the optimal solution and total profit as the unit profit of activity 1 increases in 20¢ increments from $1 to $3 (without changing the unit profit of activity 2). Then do the same as the unit profit of activity 2 increases in 50¢ increments from $2.50 to $7.50 (without changing the unit profit of activity 1). Use these results to estimate the allowable range for the unit profit of each activity.
e. Use the Graphical Linear Programming and Sensi- tivity Analysis module in your Interactive Manage- ment Science Modules to estimate the allowable range for the unit profit of each activity.
E* f. Use the sensitivity report to find the allowable range for the unit profit of each activity. Then use these ranges to check your results in parts b-e.
E* g. Use a two-way parameter analysis report to system- atically generate the optimal total profit as the unit profits of the two activities are changed simultane- ously as described in part d.
h. Use the Graphical Linear Programming and Sensi- tivity Analysis module to interpret the results in part g graphically.
E*5.3. Consider the Big M Co. problem presented in Section 3.5, including the spreadsheet in Figure 3.10 showing its formu- lation and optimal solution.
There is some uncertainty about what the unit costs will be for shipping through the various shipping lanes. Therefore, before adopting the optimal solution in Figure 3.10, manage- ment wants additional information about the effect of inaccura- cies in estimating these unit costs.
Use Solver to generate the sensitivity report preparatory to addressing the following questions. a. Which of the unit shipping costs given in Table 3.9
has the smallest margin for error without invali- dating the optimal solution given in Figure 3.10? Where should the greatest effort be placed in esti- mating the unit shipping costs?
b. What is the allowable range for each of the unit shipping costs?
c. How should the allowable range be interpreted to management?
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Chapter 5 Problems 183
the two stages of production (molding and finishing), the mate- rial required (clay), and the corresponding unit profits are given in the following table, along with the amount of each resource available in the upcoming production period.
Plates Mugs Steins Available
Molding (minutes) 4 6 3 2,400 Finishing (minutes) 8 14 12 7,200 Clay (ounces) 5 4 3 3,000
Unit Profit $3.10 $4.75 $4.00
A linear programming model has been formulated in a spreadsheet to determine the production levels that would maxi- mize profit. The solved spreadsheet model and corresponding sensitivity report are shown below.
For each of the following parts, answer the question as spe- cifically and completely as is possible without re-solving the problem with Solver. Note: Each part is independent (i.e., any change made in one part does not apply to any other parts). a. Suppose the profit per plate decreases from $3.10
to $2.80. Will this change the optimal production quantities? What can be said about the change in total profit?
b. Suppose the profit per stein increases by $0.30 and the profit per plate decreases by $0.25. Will this change the optimal production quantities? What can be said about the change in total profit?
f. The net present values of projects 1, 2, and 3 change to $46 million, $69 million, and $49 million, respectively.
g. The net present values of projects 1, 2, and 3 change to $54 million, $84 million, and $60 million, respectively.
h. Use Solver to generate the sensitivity report for this problem. For each of the preceding parts, sup- pose that the change occurs later without having the spreadsheet model immediately available on a com- puter. Show in each case how the sensitivity report can be used to check whether the original optimal solution must still be optimal.
i. For each of the three projects in turn, use a param- eter analysis report to systematically generate the optimal solution and the total net present value when the only change is that the net present value of that project increases in $1 million increments from $5 million less than the current value up to $5 mil- lion more than the current value.
5.6. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 5.4. Briefly describe how what-if analysis was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 5.7. University Ceramics manufactures plates, mugs, and steins that include the campus name and logo for sale in cam- pus bookstores. The time required for each item to go through
1
2
3
4
5
6
7
8
9
10
A B C D E F G
Plates Mugs Steins
Unit Profit $3.10 $4.75 $4.00
Resource Required per Unit Used Available
Molding (minutes) 4 2,400 <= 2,400
Finishing (minutes) 8 7,200 <= 7,200
Clay (ounces) 5
6
14
4
3
12
3 2,700 <= 3,000
Plates Mugs Steins Total Profit
Production 300 0 400 $2,530
Variable Cells
Cell Name Final Value
Reduced Cost
Objective Coefficient
Allowable Increase
Allowable Decrease
$B$10 Production Plates 300 0 3.10 2.23 0.37 $C$10 Production Mugs 0 20.46 4.75 0.46 $D$10 Production Steins 400 0 4.00 0.65 1.37
Constraints
Cell Name Final Value
Shadow Price
Constraint R. H. Side
Allowable Increase
Allowable Decrease
$E$5 Molding (minutes) Used 2,400 0.22 2400 200 600 $E$6 Finishing (minutes) Used 7,200 0.28 7200 2400 2400 $E$7 Cream (ounces) Used 2,700 0 3000 1E 1 30
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184 Chapter Five What-If Analysis for Linear Programming
company has only 200 gallons of milk, 150 pounds of sugar, and 60 gallons of cream left in its inventory. The linear programming formulation for this problem is shown below in algebraic form.
Let
C 5 Gallons of chocolate ice cream produced
V 5 Gallons of vanilla ice cream produced
B 5 Gallons of banana ice cream produced
Maximize Profit 5 1.00 C 1 0.90 V 1 0.95 B
subject to
Milk: 0.45 C 1 0.50 V 1 0.40 B # 200 gallons
Sugar: 0.50 C 1 0.40 V 1 0.40 B # 150 pounds
Cream: 0.10 C 1 0.15 V 1 0.20 B # 60 gallons
and
C $ 0 V $ 0 B $ 0
This problem was solved using Solver. The spreadsheet (already solved) and the sensitivity report are shown below. (Note: The numbers in the sensitivity report for the milk con- straint are missing on purpose, since you will be asked to fill in these numbers in part f. )
For each of the following parts, answer the question as spe- cifically and completely as possible without solving the problem again with Solver. Note: Each part is independent (i.e., any change made to the model in one part does not apply to any other parts).
c. Suppose a worker in the molding department calls in sick. Now eight fewer hours are available that day in the molding department. How much would this affect total profit? Would it change the optimal production quantities?
d. Suppose one of the workers in the molding depart- ment is also trained to do finishing. Would it be a good idea to have this worker shift some of her time from the molding department to the finishing department? Indicate the rate at which this would increase or decrease total profit per minute shifted. How many minutes can be shifted before this rate might change?
e. The allowable decrease for the mugs’ objective coefficient and for the available clay constraint are both missing from the sensitivity report. What num- bers should be there? Explain how you were able to deduce each number.
5.8. Ken and Larry, Inc., supplies its ice cream parlors with three flavors of ice cream: chocolate, vanilla, and banana. Due to extremely hot weather and a high demand for its products, the company has run short of its supply of ingredients: milk, sugar, and cream. Hence, they will not be able to fill all the orders received from their retail outlets, the ice cream parlors. Due to these circumstances, the company has decided to choose the amount of each flavor to produce that will maximize total profit, given the constraints on the supply of the basic ingredients.
The chocolate, vanilla, and banana flavors generate, respec- tively, $1.00, $0.90, and $0.95 of profit per gallon sold. The
Variable Cells
Cell Name Final Value
Reduced Cost
Objective Coefficient
Allowable Increase
Allowable Decrease
$B$10 Gallons Produced Chocolate 0 20.0375 1 0.0375 1E 1 30 $C$10 Gallons Produced Vanilla 300 0 0.9 0.05 0.0125 $D$10 Gallons Produced Banana 75 0 0.95 0.0214 0.05
Constraints
Cell Name Final Value
Shadow Price
Constraint R. H. Side
Allowable Increase
Allowable Decrease
$E$5 Milk Used $E$6 Sugar Used 150 1.875 150 10 30 $E$7 Cream Used 60 1 60 15 3.75
A B C D E F G
1
2
3
4
5
6
7
8
9
10
Unit Profit $1.00 $0.90 $0.95
Resource Resources Used per Gallon Produced Available
Milk
Sugar
Cream
0.45
0.5
0.1
0.5
0.4
0.15
0.4
0.4
0.2
200
150
60
Used
180
150
60
Gallons Produced
Chocolate Vanilla Banana
0 300 75
Chocolate Vanilla Banana
$341.25
Total Profit
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Chapter 5 Problems 185
Rocking Chair
Dining Room Table Armoire Available
Assembly (minutes)
100 180 120 3,600
Finishing (minutes)
60 80 80 2,000
Wood (pounds)
30 180 120 4,000
Unit Profit $240 $720 $600
A linear programming model has been formulated in a spreadsheet to determine the production levels that would maxi- mize profit. The solved spreadsheet model and corresponding sensitivity report are shown below.
For each of the following parts, answer the question as spe- cifically and completely as is possible without re-solving the problem with Solver. Note: Each part is independent (i.e., any change made in one problem part does not apply to any other parts). a. Suppose the profit per armoire decreases by $50.
Will this change the optimal production quantities? What can be said about the change in total profit?
b. Suppose the profit per table decreases by $60 and the profit per armoire increases by $90. Will this change the optimal production quantities? What can be said about the change in total profit?
a. What is the optimal solution and total profit? b. Suppose the profit per gallon of banana changes to
$1.00. Will the optimal solution change and what can be said about the effect on total profit?
c. Suppose the profit per gallon of banana changes to 92¢. Will the optimal solution change and what can be said about the effect on total profit?
d. Suppose the company discovers that three gallons of cream have gone sour and so must be thrown out. Will the optimal solution change and what can be said about the effect on total profit?
e. Suppose the company has the opportunity to buy an additional 15 pounds of sugar at a total cost of $15. Should it do so? Explain.
f. Fill in all the sensitivity report information for the milk constraint, given just the optimal solution for the problem. Explain how you were able to deduce each number.
5.9. Colonial Furniture produces hand-crafted colonial style furniture. Plans are now being made for the production of rocking chairs, dining room tables, and/or armoires over the next week. These products go through two stages of production (assembly and finishing). The following table gives the time required for each item to go through these two stages, the amount of wood required (fine cherry wood), and the corresponding unit profits, along with the amount of each resource available next week.
1
2
3
4
5
6
7
8
9
10
11
12
A B C D E F G
Rocking Dining Room
Chair Table Armoire
Unit Profit $240 $720 $600
Resource Required per Unit Used Available
Assembly (minutes) 100 120 3,600 <= 3,600
Finishing (minutes) 60 80 2,000 <= 2,000
Wood (pounds) 30
180
80
180 120 3,600 <= 4,000
Rocking Dining Room
Chair Table Armoire Total Profit
Production 0 10 15 $16,200
Variable Cells
Cell Name Final Value
Reduced Cost
Objective Coefficient
Allowable Increase
Allowable Decrease
$B$12 Production Chair 0 2230 240 230 1E 1 30 $C$12 Production Table 10 0 720 180 120 $D$12 Production Armoire 15 0 600 120 120
Constraints
Cell Name Final Value
Shadow Price
Constraint R. H. Side
Allowable Increase
Allowable Decrease
$E$6 Assembly (minutes) Used 3,600 2.00 3,600 400 600 $E$7 Finishing (minutes) Used 2,000 4.50 2,000 400 400 $E$8 Wood (pounds) Used 3,600 4,000
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186 Chapter Five What-If Analysis for Linear Programming
E* d. Use Solver to check the effect of the changes speci- fied in part b.
E* e. Use a parameter analysis report to systematically generate the optimal solution and total profit as the unit profit for grandfather clocks is increased in $20 increments from $150 to $450 (with no change in the unit profit for wall clocks). Then do the same as the unit profit for wall clocks is increased in $20 increments from $50 to $350 (with no change in the unit profit for grandfather clocks). Use this infor- mation to estimate the allowable range for the unit profit of each type of clock.
E* f. Use a two-way parameter analysis report to sys- tematically generate the optimal total profit as the unit profits for the two types of clocks are changed simultaneously as specified in part e, except use $50 increments instead of $20 increments.
E* g. For each of the three partners in turn, use Solver to determine the effect on the optimal solution and the total profit if that partner alone were to increase his or her maximum number of work hours available per week by 5 hours.
E* h. Use a parameter analysis report to systemati- cally generate the optimal solution and the total profit when the only change is that David’s maxi- mum number of hours available to work per week changes to each of the following values: 35, 37, 39, 41, 43, 45. Then do the same when the only change is that LaDeana’s maximum number of hours avail- able to work per week changes in the same way. Then do the same when the only change is that Lydia’s maximum number of hours available to work per week changes to each of the following values: 15, 17, 19, 21, 23, 25.
E* i. Generate a sensitivity report and use it to determine the allowable range for the unit profit for each type of clock and the allowable range for the maximum number of hours each partner is available to work per week.
j. To increase the total profit, the three partners have agreed that one of them will slightly increase the maximum number of hours available to work per week. The choice of which one will be based on which one would increase the total profit the most. Use the sensitivity report to make this choice. (Assume no change in the original estimates of the unit profits.)
k. Explain why one of the shadow prices is equal to zero.
l. Can the shadow prices in the sensitivity report be validly used to determine the effect if Lydia were to change her maximum number of hours available to work per week from 20 to 25? If so, what would be the increase in the total profit?
m. Repeat part l if, in addition to the change for Lydia, David also were to change his maximum number of hours available to work per week from 40 to 35.
n. Use graphical analysis to verify your answer in part m.
c. Suppose a part-time worker in the assembly depart- ment calls in sick, so that now four fewer hours are available that day in the assembly department. How much would this affect total profit? Would it change the optimal production quantities?
d. Suppose one of the workers in the assembly depart- ment is also trained to do finishing. Would it be a good idea to have this worker shift some of his time from the assembly department to the finishing department? Indicate the rate at which this would increase or decrease total profit per minute shifted. How many minutes can be shifted before this rate might change?
e. The shadow price and allowable range for the wood constraint are missing from the sensitivity report. What numbers should be there? Explain how you were able to deduce each number.
5.10. David, LaDeana, and Lydia are the sole partners and workers in a company that produces fine clocks. David and LaDeana are each available to work a maximum of 40 hours per week at the company, while Lydia is available to work a maxi- mum of 20 hours per week.
The company makes two different types of clocks: a grandfa- ther clock and a wall clock. To make a clock, David (a mechani- cal engineer) assembles the inside mechanical parts of the clock while LaDeana (a woodworker) produces the hand-carved wood casings. Lydia is responsible for taking orders and shipping the clocks. The amount of time required for each of these tasks is shown next.
Time Required
Task Grandfather
Clock Wall Clock
Assemble clock mechanism
6 hours 4 hours
Carve wood casing
8 hours 4 hours
Shipping 3 hours 3 hours
Each grandfather clock built and shipped yields a profit of $300, while each wall clock yields a profit of $200.
The three partners now want to determine how many clocks of each type should be produced per week to maximize the total profit. a. Formulate a linear programming model in algebraic
form for this problem. b. Use the Graphical Linear Programming and Sensi-
tivity Analysis module in your Interactive Manage- ment Science Modules to solve the model. Then use this module to check if the optimal solution would change if the unit profit for grandfather clocks were changed from $300 to $375 (with no other changes in the model). Then check if the optimal solution would change if, in addition to this change in the unit profit for grandfather clocks, the estimated unit profit for wall clocks also changed from $200 to $175.
E* c. Formulate and solve the original version of this model on a spreadsheet.
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Chapter 5 Problems 187
d. Repeat part b for the second functional constraint where its right-hand side is incremented by 1 from 6 to 18.
e. Use Solver’s sensitivity report to determine the shadow price for each functional constraint and the allowable range for the right-hand side of each of these constraints.
5.13. Consider a resource-allocation problem having the fol- lowing data:
Resource Usage per Unit of Each Activity
Resource 1 2
Amount of Resource Available
1 1 3 8 2 1 1 4
Unit profit $1 $2
The objective is to determine the number of units of each activity to undertake so as to maximize the total profit. a. Use the graphical method to solve this model. b. Use graphical analysis to determine the shadow
price for each of these resources by solving again after increasing the amount of the resource avail- able by one.
E* c. Use the spreadsheet model and Solver instead to do parts a and b.
E* d. For each resource in turn, use a parameter analysis report to systematically generate the optimal solu- tion and the total profit when the only change is that the amount of that resource available increases in increments of 1 from 4 less than the original value up to 6 more than the original value. Use these results to estimate the allowable range for the amount available for each resource.
E* e. Use Solver’s sensitivity report to obtain the shadow prices. Also use this report to find the range for the amount of each resource available over which the corresponding shadow price remains valid.
f. Describe why these shadow prices are useful when management has the flexibility to change the amounts of the resources being made available.
5.14. Follow the instructions of Problem 5.13 for a resource- allocation problem that again has the objective of maximizing total profit and that has the following data:
Resource Usage per Unit of Each Activity
Resource 1 2
Amount of Resource Available
1 1 0 4 2 1 3 15 3 2 1 10
Unit profit $3 $2
E*5.11.* Reconsider Problem 5.1. After further negotiations with each vendor, management of the G. A. Tanner Company has learned that either of them would be willing to consider increasing their supply of their respective subassemblies over the previously stated maxima (3,000 subassemblies of type A per day and 1,000 of type B per day) if the company would pay a small premium over the regular price for the extra subassem- blies. The size of the premium for each type of subassembly remains to be negotiated. The demand for the toy being pro- duced is sufficiently high that 2,500 per day could be sold if the supply of subassemblies could be increased enough to support this production rate. Assume that the original estimates of unit profits given in Problem 5.1 are accurate.
a. Formulate and solve a spreadsheet model for this problem with the original maximum supply lev- els and the additional constraint that no more than 2,500 toys should be produced per day.
b. Without considering the premium, use the spread- sheet and Solver to determine the shadow price for the subassembly A constraint by solving the model again after increasing the maximum supply by one. Use this shadow price to determine the maximum premium that the company should be willing to pay for each subassembly of this type.
c. Repeat part b for the subassembly B constraint.
d. Estimate how much the maximum supply of subas- semblies of type A could be increased before the shadow price (and the corresponding premium) found in part b would no longer be valid by using a parameter analysis report to generate the optimal solution and total profit (excluding the premium) as the maximum supply increases in increments of 100 from 3,000 to 4,000.
e. Repeat part d for subassemblies of type B by using a parameter analysis report as the maximum supply increases in increments of 100 from 1,000 to 2,000.
f. Use Solver’s sensitivity report to determine the shadow price for each of the subassembly con- straints and the allowable range for the right-hand side of each of these constraints.
E*5.12. Reconsider the model given in Problem 5.2. While doing what-if analysis, you learn that the estimates of the right- hand sides of the two functional constraints are accurate only to within 6 50 percent. In other words, the ranges of likely values for these parameters are 5 to 15 for the first right-hand side and 6 to 18 for the second right-hand side. a. After solving the original spreadsheet model, deter-
mine the shadow price for the first functional con- straint by increasing its right-hand side by one and solving again.
b. Use a parameter analysis report to generate the opti- mal solution and total profit as the right-hand side of the first functional constraint is incremented by 1 from 5 to 15. Use this table to estimate the allow- able range for this right-hand side, that is, the range over which the shadow price obtained in part a is valid.
c. Repeat part a for the second functional constraint.
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188 Chapter Five What-If Analysis for Linear Programming
E*5.17. Consider the Union Airways problem presented in Sec- tion 3.3, including the spreadsheet in Figure 3.5 showing its for- mulation and optimal solution.
Management now is considering increasing the level of ser- vice provided to customers by increasing one or more of the numbers in the rightmost column of Table 3.5 for the minimum number of agents needed in the various time periods. To guide them in making this decision, they would like to know what impact this change would have on total cost.
Use Solver to generate the sensitivity report in preparation for addressing the following questions. a. Which of the numbers in the rightmost column of
Table 3.5 can be increased without increasing total cost? In each case, indicate how much it can be increased (if it is the only one being changed) with- out increasing total cost.
b. For each of the other numbers, how much would the total cost increase per increase of 1 in the num- ber? For each answer, indicate how much the num- ber can be increased (if it is the only one being changed) before the answer is no longer valid.
c. Do your answers in part b definitely remain valid if all the numbers considered in part b are simultane- ously increased by 1?
d. Do your answers in part b definitely remain valid if all 10 numbers are simultaneously increased by 1?
e. How far can all 10 numbers be simultaneously increased by the same amount before your answers in part b may no longer be valid?
E*5.15.* Consider the Super Grain Corp. case study as pre- sented in Section 3.1, including the spreadsheet in Figure 3.1 showing its formulation and optimal solution. Use Solver to gen- erate the sensitivity report. Then use this report to independently address each of the following questions.
a. How much could the total expected number of exposures be increased for each additional $1,000 added to the advertising budget?
b. Your answer in part a would remain valid for how large of an increase in the advertising budget?
c. How much could the total expected number of exposures be increased for each additional $1,000 added to the planning budget?
d. Your answer in part c would remain valid for how large of an increase in the planning budget?
e. Would your answers in parts a and c definitely remain valid if both the advertising budget and planning budget were increased by $100,000 each?
f. If only $100,000 can be added to either the advertis- ing budget or the planning budget, where should it be added to do the most good?
g. If $100,000 must be removed from either the adver- tising budget or the planning budget, from which budget should it be removed to do the least harm?
E*5.16. Follow the instructions of Problem 5.15 for the continu- ation of the Super Grain Corp. case study as presented in Section 3.4 including the spreadsheet in Figure 3.7 showing its formula- tion and optimal solution.
Case 5-1
Selling Soap
Reconsider the Profit & Gambit Co. advertising-mix problem presented in Section 2.7. Recall that a major advertising cam- paign is being planned that will focus on three key products: a stain remover, a liquid detergent, and a powder detergent. Man- agement has made the following policy decisions about what needs to be achieved by this campaign.
• Sales of the stain remover should increase by at least 3 percent. • Sales of the liquid detergent should increase by at least
18 percent.
• Sales of the powder detergent should increase by at least 4 percent.
The spreadsheet in Figure 2.21 shows the linear program- ming model that was formulated for this problem. The minimum required increases in the sales of the three products are given in the data cells Minimum Increase (G8:G10). The changing cells Advertising Units (C14:D14) indicate that an optimal solu- tion for the model is to undertake four units of advertising on television and three units of advertising in the print media. The objective cell TotalCost (G14) shows that the total cost for this advertising campaign would be $10 million.
After receiving this information, Profit & Gambit manage- ment now wants to analyze the trade-off between the total adver- tising cost and the resulting benefits achieved by increasing the sales of the three products. Therefore, a management science team (you) has been given the assignment of developing the information that management will need to analyze this trade-off and decide whether it should change any of its policy decisions regarding the required minimum increases in the sales of the three products. In particular, management needs some detailed information about how the total advertising cost would change if it were to change any or all of these policy decisions.
a. For each of the three products in turn, use graphical analy- sis to determine how much the total advertising cost would change if the required minimum increase in the sales of that product were to be increased by 1 percent (without changing the required minimum increases for the other two products).
b. Use the spreadsheet shown in Figure 2.21 (available on the CD-ROM) to obtain the information requested in part a.
c. For each of the three products in turn, use a parameter anal- ysis report to determine how the optimal solution for the model and the resulting total advertising cost would change
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Case 5-2 Controlling Air Pollution 189
e. Suppose that all the original numbers in Minimum Increase (G8:G10) were to be increased simultaneously by the same amount. How large can this amount be before the shadow prices provided by the sensitivity report may no longer be valid?
f. Below is the beginning of a memorandum from the man- agement science team to Profit & Gambit management that is intended to provide management with the information it needs to perform its trade-off analysis. Write the rest of this memorandum based on a summary of the results obtained in the preceding parts. Present your information in clear, simple terms that use the language of management. Avoid technical terms such as shadow prices, allowable ranges, and so forth.
if the required minimum increase in the sales of that prod- uct were to be systematically varied over a range of values (without changing the required minimum increases for the other two products). In each case, start the range of values at 0 percent and increase by 1 percent increments up to double the original minimum required increase.
d. Use Solver to generate the sensitivity report and indicate how the report is able to provide the information requested in part a. Also use the report to obtain the allowable range for the required minimum increase in the sales of each product. Interpret how each of these allowable ranges relates to the results obtained in part c.
MEMORANDUM
To: Profit & Gambit management From: The Management Science Team Subject: The trade-off between advertising expenditures and increased sales
As instructed, we have been continuing our analysis of the plans for the major new advertis- ing campaign that will focus on our spray prewash stain remover, our liquid formulation laundry detergent, and our powder laundry detergent.
Our recent report presented our preliminary conclusions on how much advertising to do in the different media to meet the sales goals at a minimum total cost:
Allocate $4 million to advertising on television.
Allocate $6 million to advertising in the print media.
Total advertising cost: $10 million.
We estimate that the resulting increases in sales will be
Stain remover: 3 percent increase in sales
Liquid detergent: 18 percent increase in sales
Powder detergent: 8 percent increase in sales
You had specified that these increases should be at least 3 percent, 18 percent, and 4 percent, respectively, so we have met the minimum levels for the first two products and substantially exceeded it for the third.
However, you also indicated that your decisions on these minimum required increases in sales (3 percent, 18 percent, and 4 percent) had been tentative ones. Now that we have more specific information on what the advertising costs and the resulting increases in sales will be, you plan to reevaluate these decisions to see if small changes might improve the trade-off between advertis- ing cost and increased sales.
To assist you in reevaluating your decisions, we now have analyzed this trade-off for each of the three products. Our best estimates are the following.
Case 5-2
Controlling Air Pollution
The Nori & Leets Co. is one of the major producers of steel in its part of the world. It is located in the city of Steeltown and is the only large employer there. Steeltown has grown and prospered along with the company, which now employs nearly 50,000 residents. Therefore, the attitude of the townspeople
always has been, “What’s good for Nori & Leets is good for the town.” However, this attitude is now changing; uncon- trolled air pollution from the company’s furnaces is ruining the appearance of the city and endangering the health of its residents.
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190 Chapter Five What-If Analysis for Linear Programming
A recent stockholders’ revolt resulted in the election of a new enlightened board of directors for the company. These directors are determined to follow socially responsible policies, and they have been discussing with Steeltown city officials and citizens’ groups what to do about the air pollution problem. Together they have worked out stringent air quality standards for the Steeltown airshed.
The three main types of pollutants in this airshed are particu- late matter, sulfur oxides, and hydrocarbons. The new standards require that the company reduce its annual emission of these pollutants by the amounts shown in the following table.
Pollutant
Required Reduction in Annual Emission Rate
(million pounds)
Particulates 60 Sulfur oxides 150 Hydrocarbons 125
The board of directors has instructed management to have the engineering staff determine how to achieve these reductions in the most economical way.
The steelworks have two primary sources of pollution, namely, the blast furnaces for making pig iron and the open- hearth furnaces for changing iron into steel. In both cases, the engineers have decided that the most effective abatement meth- ods are (1) increasing the height of the smokestacks, 1 (2) using filter devices (including gas traps) in the smokestacks, and (3) including cleaner, high-grade materials among the fuels for the furnaces. Each of these methods has a technological limit on how heavily it can be used (e.g., a maximum feasible increase in the height of the smokestacks), but there also is considerable flex- ibility for using the method at a fraction of its technological limit.
The next table shows how much emissions (in millions of pounds per year) can be eliminated from each type of furnace by fully using any abatement method to its technological limit.
1 Subsequent to this study, this particular abatement method has become a controversial one. Because its effect is to reduce ground- level pollution by spreading emissions over a greater distance, environmental groups contend that this creates more acid rain by keeping sulfur oxides in the air longer. Consequently, the U.S. Envi- ronmental Protection Agency adopted new rules to remove incen- tives for using tall smokestacks.
Reduction in Emission Rate from the Maximum Feasible Use of an Abatement Method
Taller Smokestacks Filters Better Fuels
Pollutant Blast
Furnaces Open-Hearth
Furnaces Blast
Furnaces Open-Hearth
Furnaces Blast
Furnaces Open-Hearth
Furnaces
Particulates 12 9 25 20 17 13 Sulfur oxides 35 42 18 31 56 49 Hydrocarbons 37 53 28 24 29 20
For purposes of analysis, it is assumed that each method also can be less fully used to achieve any fraction of the abatement capacities shown in this table. Furthermore, the fractions can be different for blast furnaces and open-hearth furnaces. For either type of furnace, the emission reduction achieved by each method
is not substantially affected by whether or not the other methods also are used.
After these data were developed, it became clear that no sin- gle method by itself could achieve all the required reductions. On the other hand, combining all three methods at full capacity on both types of furnaces (which would be prohibitively expensive if the company’s products are to remain competitively priced) is much more than adequate. Therefore, the engineers concluded that they would have to use some combination of the methods, perhaps with fractional capacities, based on their relative costs. Furthermore, because of the differences between the blast and the open-hearth furnaces, the two types probably should not use the same combination.
An analysis was conducted to estimate the total annual cost that would be incurred by each abatement method. A method’s annual cost includes increased operating and maintenance expenses, as well as reduced revenue due to any loss in the effi- ciency of the production process caused by using the method. The other major cost is the start-up cost (the initial capital out- lay) required to install the method. To make this one-time cost commensurable with the ongoing annual costs, the time value of money was used to calculate the annual expenditure that would be equivalent in value to this start-up cost.
This analysis led to the total annual cost estimates given in the next table for using the methods at their full abatement capacities.
Total Annual Cost from the Maximum Feasible Use of an Abatement Method
Abatement Method Blast Furnaces
Open-Hearth Furnaces
Taller smokestacks $8 million $10 million Filters 7 million 6 million Better fuels 11 million 9 million
It also was determined that the cost of a method being used at a lower level is roughly proportional to the fraction of the abate- ment capacity (given in the preceding table) that is achieved. Thus, for any given fraction achieved, the total annual cost would be roughly that fraction of the corresponding quantity in the cost table.
The stage now is set to develop the general framework of the company’s plan for pollution abatement. This plan needs to specify which types of abatement methods will be used and at what fractions of their abatement capacities for (1) the blast fur- naces and (2) the open-hearth furnaces.
You have been asked to head a management science team to analyze this problem. Management wants you to begin by
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Case 5-3 Farm Management 191
policy standards (increasing one while decreasing another) are available without increasing the total cost.
d. Identify the parameters of the linear programming model that should be classified as sensitive parameters. Make a result- ing recommendation about which parameters should be esti- mated more closely, if possible.
e. Analyze the effect of an inaccuracy in estimating each cost parameter given in the third table. If the true value were 10 percent less than the estimated value, would this change the optimal solution? Would it change if the true value were 10 percent more than the estimated value? Make a resulting recommendation about where to focus further work in esti- mating the cost parameters more closely.
f. For each pollutant, specify the rate at which the total cost of an optimal solution would change with any small change in the required reduction in the annual emission rate of the pol- lutant. Also specify how much this required reduction can be changed (up or down) without affecting the rate of change in the total cost.
g. For each unit change in the policy standard for particulates given in the first table, determine the change in the opposite direction for sulfur oxides that would keep the total cost of an optimal solution unchanged. Repeat this for hydrocarbons instead of sulfur oxides. Then do it for a simultaneous and equal change for both sulfur oxides and hydrocarbons in the opposite direction from particulates.
h. Letting u denote the percentage increase in all the policy standards given in the first table, use a parameter analysis report to systematically find an optimal solution and the total cost for the revised linear programming problem for each u 5 10, 20, 30, 40, 50. Considering the tax incentive offered by the city, use these results to determine which value of u (including the option of u 5 0) should be chosen by the com- pany to minimize its total cost of both pollution abatement and taxes.
i. For the value of u chosen in part h, generate the sensitivity report and repeat parts f and g so that the decision makers can make a final decision on the relative values of the policy standards for the three pollutants.
determining which plan would minimize the total annual cost of achieving the required reductions in annual emission rates for the three pollutants.
a. Identify verbally the components of a linear programming model for this problem.
b. Display the model on a spreadsheet. c. Obtain an optimal solution and generate the sensitivity report.
Management now wants to conduct some what-if analysis with your help. Since the company does not have much prior experience with the pollution abatement methods under con- sideration, the cost estimates given in the third table are fairly rough, and each one could easily be off by as much as 10 percent in either direction. There also is some uncertainty about the val- ues given in the second table, but less so than for the third table. By contrast, the values in the first table are policy standards and so are prescribed constants.
However, there still is considerable debate about where to set these policy standards on the required reductions in the emission rates of the various pollutants. The numbers in the first table actually are preliminary values tentatively agreed upon before learning what the total cost would be to meet these stan- dards. Both the city and company officials agree that the final decision on these policy standards should be based on the trade- off between costs and benefits. With this in mind, the city has concluded that each 10 percent increase in the policy standards over the current values (all the numbers in the first table) would be worth $3.5 million to the city. Therefore, the city has agreed to reduce the company’s tax payments to the city by $3.5 million for each 10 percent increase in the policy standards (up to 50 percent) that is accepted by the company.
Finally, there has been some debate about the relative values of the policy standards for the three pollutants. As indicated in the first table, the required reduction for particulates now is less than half of that for either sulfur oxides or hydrocarbons. Some have argued for decreasing this disparity. Others contend that an even greater disparity is justified because sulfur oxides and hydrocarbons cause considerably more damage than particu- lates. Agreement has been reached that this issue will be reex- amined after information is obtained about which trade-offs in
Case 5-3
Farm Management
The Ploughman family owns and operates a 640-acre farm that has been in the family for several generations. The Ploughmans always have had to work hard to make a decent living from the farm and have had to endure some occasional difficult years. Stories about earlier generations overcom- ing hardships due to droughts, floods, and so forth, are an important part of the family history. However, the Plough- mans enjoy their self-reliant lifestyle and gain considerable satisfaction from continuing the family tradition of success- fully living off the land during an era when many family farms are being abandoned or taken over by large agricultural corporations.
John Ploughman is the current manager of the farm, while his wife Eunice runs the house and manages the farm’s finances. John’s father, Grandpa Ploughman, lives with them and still puts in many hours working on the farm. John and Eunice’s older children, Frank, Phyllis, and Carl, also are given heavy chores before and after school.
The entire family can produce a total of 4,000 person-hours’ worth of labor during the winter and spring months and 4,500 person-hours during the summer and fall. If any of these person- hours are not needed, Frank, Phyllis, and Carl will use them to work on a neighboring farm for $5/hour during the winter and spring months and $5.50/hour during the summer and fall.
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192 Chapter Five What-If Analysis for Linear Programming
The above estimates of the net value per acre planted in each of the three crops assumes good weather conditions. Adverse weather conditions would harm the crops and greatly reduce the resulting value. The scenarios particularly feared by the family are a drought, a flood, an early frost, both a drought and an early frost, and both a flood and an early frost. The estimated net val- ues for the year under these scenarios are shown next.
Net Value per Acre Planted
Scenario Soybeans Corn Wheat
Drought 2$10 2$15 0 Flood 15 20 $10 Early frost 50 40 30 Drought and early frost 215 220 210 Flood and early frost 10 10 5
e. Find an optimal solution under each scenario after making the necessary adjustments to the linear programming model formulated in part b. In each case, what is the prediction regarding the family’s monetary worth at the end of the year?
f. For the optimal solution obtained under each of the six sce- narios (including the good weather scenario considered in parts a-d ), calculate what the family’s monetary worth would be at the end of the year if each of the other five scenarios occurs instead. In your judgment, which solution provides the best balance between yielding a large monetary worth under good weather conditions and avoiding an overly small monetary worth under adverse weather conditions?
Grandpa has researched what the weather conditions were in past years as far back as weather records have been kept and obtained the data shown on the next page. With these data, the family has decided to use the following approach to making its planting and livestock decisions. Rather than the optimistic approach of assuming that good weather conditions will prevail (as done in parts a-d ), the average net value under all weather conditions will be used for each crop (weighting the net values under the various scenarios by the frequencies in the above table).
Scenario Frequency
Good weather 40% Drought 20 Flood 10 Early frost 15 Drought and early frost 10 Flood and early frost 5
g. Modify the linear programming model formulated in part b to fit this new approach.
h. Repeat part c for this modified model. i. Use a shadow price obtained in part h to analyze whether it
would be worthwhile for the family to obtain a bank loan with a 10 percent interest rate to purchase more livestock now beyond what can be obtained with the $20,000 from the investment fund.
j. For each of the three crops, use the sensitivity report obtained in part h to identify how much latitude for error is available in estimating the net value per acre planted for that crop without changing the optimal solution. Which two net values
The farm supports two types of livestock, dairy cows and laying hens, as well as three crops: soybeans, corn, and wheat. (All three are cash crops, but the corn also is a feed crop for the cows and the wheat also is used for chicken feed.) The crops are harvested during the late summer and fall. During the winter months, John, Eunice, and Grandpa make a decision about the mix of livestock and crops for the coming year.
Currently, the family has just completed a particularly suc- cessful harvest that has provided an investment fund of $20,000 that can be used to purchase more livestock. (Other money is available for ongoing expenses, including the next planting of crops.) The family currently has 30 cows valued at $35,000 and 2,000 hens valued at $5,000. They wish to keep all this livestock and perhaps purchase more. Each new cow would cost $1,500, and each new hen would cost $3.
Over a year’s time, the value of a herd of cows will decrease by about 10 percent and the value of a flock of hens will decrease by about 25 percent due to aging.
Each cow will require two acres of land for grazing and 10 person-hours of work per month, while producing a net annual cash income of $850 for the family. The corresponding figures for each hen are no significant acreage, 0.05 person-hours per month, and an annual net cash income of $4.25. The chicken house can accommodate a maximum of 5,000 hens, and the size of the barn limits the herd to a maximum of 42 cows.
For each acre planted in each of the three crops, the next table gives the number of person-hours of work that will be required during the first and second halves of the year, as well as a rough estimate of the crop’s net value (in either income or savings in purchasing feed for the livestock).
To provide much of the feed for the livestock, John wants to plant at least one acre of corn for each cow in the coming year’s herd and at least 0.05 acre of wheat for each hen in the coming year’s flock.
John, Eunice, and Grandpa now are discussing how much acreage should be planted in each of the crops and how many cows and hens to have for the coming year. Their objective is to maximize the family’s monetary worth at the end of the coming year (the sum of the net income from the livestock for the coming year plus the net value of the crops for the coming year plus what remains from the investment fund plus the value of the livestock at the end of the coming year plus income from working on a neighboring farm minus living expenses of $40,000 for the year).
Data per Acre Planted
Soybeans Corn Wheat
Winter and spring, person-hours
1.0 0.9 0.6
Summer and fall, person-hours
1.4 1.2 0.7
Net value $70 $60 $40
a. Identify verbally the components of a linear programming model for this problem.
b. Display the model on a spreadsheet. c. Obtain an optimal solution and generate the sensitivity
report. What does the model predict regarding the family’s monetary worth at the end of the coming year?
d. Find the allowable range for the net value per acre planted for each of the three crops.
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Additional Cases 193
in general terms, an organization faces an uncertain future where any one of a number of scenarios may unfold. Which one will occur depends on conditions that are outside the control of the organization. The organization needs to choose the levels of various activities, but the unit contribution of each activity to the overall measure of performance is greatly affected by which scenario unfolds. Under these circumstances, what is the best mix of activities?
k. Think about specific situations outside of farm management that fit this description. Describe one.
need to be estimated most carefully? If both estimates are incorrect simultaneously, how close do the estimates need to be to guarantee that the optimal solution will not change? Use a two way parameter analysis report to systematically generate the optimal monetary worth as these two net values are varied simultaneously over ranges that go up to twice as far from the estimates as needed to guarantee that the optimal solution will not change.
This problem illustrates a kind of situation that is frequently faced by various kinds of organizations. To describe the situation
Case 5-4
Assigning Students to Schools (Revisited)
Reconsider Case 3-5. The Springfield School Board still has the policy of providing busing for all middle school students who must travel more than approximately a mile. Another cur- rent policy is to allow splitting residential areas among multiple schools if this will reduce the total busing cost. (This latter pol- icy will be reversed in Case 7-3.) However, before adopting a busing plan based on part a of Case 3-5, the school board now wants to conduct some what-if analysis.
a. If you have not already done so for part a of Case 3-5, formu- late and solve a linear programming model for this problem on a spreadsheet.
b. Use Solver to generate the sensitivity report.
One concern of the school board is the ongoing road con- struction in area 6. These construction projects have been delaying traffic considerably and are likely to affect the cost of busing students from area 6, perhaps increasing costs as much as 10 percent.
c. Use the sensitivity report to check how much the busing cost from area 6 to school 1 can increase (assuming no change in the costs for the other schools) before the current opti- mal solution would no longer be optimal. If the allowable increase is less than 10 percent, use Solver to find the new optimal solution with a 10 percent increase.
d. Repeat part c for school 2 (assuming no change in the costs for the other schools).
e. Now assume that the busing cost from area 6 would increase by the same percentage for all the schools. Use the sensitivity report to determine how large this percentage can be before the current optimal solution might no longer be optimal. If the allowable increase is less than 10 percent, use Solver to find the new optimal solution with a 10 percent increase.
The school board has the option of adding portable class- rooms to increase the capacity of one or more of the middle schools for a few years. However, this is a costly move that the board would only consider if it would significantly decrease
busing costs. Each portable classroom holds 20 students and has a leasing cost of $2,500 per year. To analyze this option, the school board decides to assume that the road construction in area 6 will wind down without significantly increasing the busing costs from that area.
f. For each school, use the corresponding shadow price from the sensitivity report to determine whether it would be worth- while to add any portable classrooms.
g. For each school where it is worthwhile to add any portable classrooms, use the sensitivity report to determine how many could be added before the shadow price would no longer be valid (assuming this is the only school receiving portable classrooms).
h. If it would be worthwhile to add portable classrooms to more than one school, use the sensitivity report to determine the combinations of the number to add for which the shadow prices definitely would still be valid. Then use the shadow prices to determine which of these combinations is best in terms of minimizing the total cost of busing students and leasing portable classrooms. Use Solver for finding the corre- sponding optimal solution for assigning students to schools.
i. If part h was applicable, modify the best combination of portable classrooms found there by adding one more to the school with the most favorable shadow price. Use Solver to find the corresponding optimal solution for assigning stu- dents to schools and to generate the corresponding sensitiv- ity report. Use this information to assess whether the plan developed in part h is the best one available for minimizing the total cost of busing students and leasing portables. If not, find the best plan.
Additional Cases Additional cases for this chapter also are available at the Univer- sity of Western Ontario Ivey School of Business website, cases .ivey.uwo.ca/cases , in the segment of the CaseMate area desig- nated for this book.
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Chapter Six
Network Optimization Problems Learning Objectives
After completing this chapter, you should be able to
1. Formulate network models for various types of network optimization problems.
2. Describe the characteristics of minimum-cost flow problems, maximum flow prob- lems, and shortest path problems.
3. Identify some areas of application for these types of problems.
4. Identify several categories of network optimization problems that are special types of minimum-cost flow problems.
5. Formulate and solve a spreadsheet model for a minimum-cost flow problem, a maxi- mum flow problem, or a shortest path problem from a description of the problem.
Networks arise in numerous settings and in a variety of guises. Transportation, electrical, and communication networks pervade our daily lives. Network representations also are widely used for problems in such diverse areas as production, distribution, project planning, facilities location, resource management, and financial planning—to name just a few examples. In fact, a network representation provides such a powerful visual and conceptual aid for portraying the relationships between the components of systems that it is used in virtually every field of scientific, social, and economic endeavor.
One of the most exciting developments in management science in recent decades has been the unusually rapid advance in both the methodology and application of network opti- mization problems. A number of algorithmic breakthroughs have had a major impact, as have ideas from computer science concerning data structures and efficient data manipu- lation. Consequently, algorithms and software now are available and are being used to solve huge problems on a routine basis that would have been completely intractable a few decades ago.
This chapter presents the network optimization problems that have been particularly helpful in dealing with managerial issues. We focus on the nature of these problems and their applications rather than on the technical details and the algorithms used to solve the problems.
You already have seen some examples of network optimization problems in Chapter 3. In particular, transportation problems (described in Section 3.5) have a network representation, as illustrated in Figure 3.9, and assignment problems (Section 3.6) have a similar network representation (as described in Chapter 15 on the CD-ROM). Therefore, both transportation problems and assignment problems are simple types of network optimization problems.
Like transportation problems and assignment problems, many other network optimization problems (including all the types considered in this chapter) also are special types of linear programming problems. Consequently, after formulating a spreadsheet model for these prob- lems, they can be readily solved by Solver.
Section 6.1 discusses an especially important type of network optimization problem called a minimum-cost flow problem. A typical application involves minimizing the cost of shipping goods through a distribution network.
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6.1 Minimum-Cost Flow Problems 195
Section 6.3 presents maximum flow problems, which are concerned with such issues as how to maximize the flow of goods through a distribution network. Section 6.2 lays the groundwork by introducing a case study of a maximum flow problem.
Section 6.4 considers shortest path problems. In their simplest form, the objective is to find the shortest route between two locations.
A supplement to this chapter on the CD-ROM discusses minimum spanning-tree prob- lems, which are concerned with minimizing the cost of providing connections between all users of a system. This is the only network optimization problem considered in this book that is not, in fact, a special type of linear programming problem.
6.1 MINIMUM-COST FLOW PROBLEMS
Before describing the general characteristics of minimum-cost flow problems, let us first look at a typical example.
An Example: The Distribution Unlimited Co. Problem The Distribution Unlimited Co. has two factories producing a product that needs to be shipped to two warehouses. Here are some details.
Factory 1 is producing 80 units. Factory 2 is producing 70 units. Warehouse 1 needs 60 units. Warehouse 2 needs 90 units.
(Each unit corresponds to a full truckload of the product.) Figure 6.1 shows the distribution network available for shipping this product, where F1 and
F2 are the two factories, W1 and W2 are the two warehouses, and DC is a distribution center. The arrows show feasible shipping lanes. In particular, there is a rail link from Factory 1 to Warehouse 1 and another from Factory 2 to Warehouse 2. (Any amounts can be shipped along these rail links.) In addition, independent truckers are available to ship up to 50 units from each factory to the distribution center, and then to ship up to 50 units from the distribution center to each warehouse. (Whatever is shipped to the distribution center must subsequently be shipped on to the warehouses.) Management’s objective is to determine the shipping plan (how many units to ship along each shipping lane) that will minimize the total shipping cost.
The shipping costs differ considerably among these shipping lanes. The cost per unit shipped through each lane is shown above the corresponding arrow in the network in Figure 6.2 .
To make the network less crowded, the problem usually is presented even more com- pactly, as shown in Figure 6.3 . The number in square brackets next to the location of each
The objective is to mini- mize the total shipping cost through the distribution network.
FIGURE 6.1 The distribution network for the Distribution Unlimited Co. problem, where each feasible ship- ping lane is represented by an arrow.
W2
W1
90 units needed
60 units needed
70 units produced
80 units produced
F2
F1
DC
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196 Chapter Six Network Optimization Problems
facility indicates the net number of units (outflow minus inflow) generated there. Thus, the number of units terminating at each warehouse is shown as a negative number. The number at the distribution center is 0 since the number of units leaving minus the number of units arriv- ing must equal 0. The number on top of each arrow shows the unit shipping cost along that shipping lane. Any number in square brackets underneath an arrow gives the maximum num- ber of units that can be shipped along that shipping lane. (The absence of a number in square brackets underneath an arrow implies that there is no limit on the shipping amount there.) This network provides a complete representation of the problem, including all the necessary data, so it constitutes a network model for this minimum-cost flow problem.
Since this is such a tiny problem, you probably can see what the optimal solution must be. (Try it.) This solution is shown in Figure 6.4 , where the shipping amount along each shipping lane is given in parentheses. (To avoid confusion, we delete the unit shipping costs and ship- ping capacities in this figure.) Combining these shipping amounts with the unit shipping costs given in Figures 6.2 and 6.3 , the total shipping cost for this solution (when starting by listing the costs from F1, then from F2, and then from DC) is
Total shipping cost 5 30($700) 1 50($300) 1 30($500) 1 40($1,000)
1 30($200) 1 50($400)
5 $117,000
Figure 6.3 illustrates how a minimum-cost flow problem can be completely depicted by a network.
FIGURE 6.2 The data for the distribu- tion network for the Dis- tribution Unlimited Co. problem.
W2
W1 $700/unit
$1,000/unit 90 units needed
60 units needed
70 units produced
80 units produced
$5 00
/u ni
t
[5 0
un its
m ax
.]
$2 00
/u ni
t
[5 0
un its
m ax
.]
$300/unit
[50 units m ax.]
$400/unit
[50 units m ax.]
F2
F1
DC
FIGURE 6.3 A network model for the Distribution Unlimited Co. problem as a minimum-cost flow problem.
W2
W1 $700
[0]
$1,000
[–60]
[–90]
[80]
[70]
$5 00
[5 0]
$2 00
[5 0]
$300[50]
$400[50]
F2
F1
DC
FIGURE 6.4 The optimal solution for the Distribution Unlim- ited Co. problem, where the shipping amounts are shown in parentheses over the arrows.
W2
W1 (30)
[0]
(40)
[–60]
[–90]
[80]
[70]
(3 0)
(3 0)
(50)
(50)
F2
F1
DC
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General Characteristics This example possesses all the general characteristics of any minimum-cost flow problem. Before summarizing these characteristics, here is the terminology you will need.
Terminology 1. The model for any minimum-cost flow problem is represented by a network with flow
passing through it. 2. The circles in the network are called nodes . 3. Each node where the net amount of flow generated (outflow minus inflow) is a fixed posi-
tive number is a supply node . (Thus, F1 and F2 are the supply nodes in Figure 6.3 .) 4. Each node where the net amount of flow generated is a fixed negative number is a demand
node . (Consequently, W1 and W2 are the demand nodes in the example.) 5. Any node where the net amount of flow generated is fixed at zero is a transshipment
node . (Thus, DC is the transshipment node in the example.) Having the amount of flow out of the node equal the amount of flow into the node is referred to as conservation of flow .
6. The arrows in the network are called arcs . 7. The maximum amount of flow allowed through an arc is referred to as the capacity of
that arc.
Using this terminology, the general characteristics of minimum-cost flow problems (the model for this type of problem) can be described in terms of the following assumptions.
Assumptions of a Minimum-Cost Flow Problem 1. At least one of the nodes is a supply node. 2. At least one of the other nodes is a demand node. 3. All the remaining nodes are transshipment nodes. 4. Flow through an arc is only allowed in the direction indicated by the arrowhead, where the
maximum amount of flow is given by the capacity of that arc. (If flow can occur in both directions, this would be represented by a pair of arcs pointing in opposite directions.)
5. The network has enough arcs with sufficient capacity to enable all the flow generated at the supply nodes to reach all the demand nodes.
A supply node has net flow going out whereas a demand node has net flow coming in.
Since the arrowhead on an arc indicates the direction in which flow is allowed, a pair of arcs pointing in opposite directions is used if flow can occur in both directions.
Hewlett-Packard (HP) offers many innovative products to meet the diverse needs of more than 1 billion customers. The breadth of its product offering has helped the com- pany achieve unparalleled market reach. However, offering multiple similar products also can cause serious problems— including confusing sales representatives and customers— that can adversely affect the revenue and costs for any par- ticular product. Therefore, it is important to find the right balance between too much product variety and too little.
With this in mind, HP top management made managing product variety a strategic business priority. HP has been a leader in applying management science to its important business problems for decades, so it was only natural that many of the company’s top management scientists were called on to address this problem as well.
The heart of the methodology that was developed to address this problem involved formulating and apply- ing a network optimization model. After excluding pro- posed products that do not have a sufficiently high return on investment, the remaining proposed products can be
envisioned as flows through a network that can help fill some of the projected orders on the right-hand side of the network. The resulting model is a special type of minimum cost flow problem (related to the special type discussed in the next two sections).
Following its implementation by the beginning of 2005, this application of a minimum cost flow problem had a dra- matic impact in enabling HP businesses to increase oper- ational focus on their most critical products. This yielded companywide profit improvements of over $500 million between 2005 and 2008, and then about $180 million annually thereafter. It also yielded a variety of important qualitative benefits for HP.
These dramatic results led to HP winning the prestigious first prize in the 2009 Franz Edelman Award for Achieve- ment in Operations Research and the Management Sciences.
Source: J. Ward and 20 co-authors, “HP Transforms Product Port- folio Management with Operations Research,” Interfaces 40, no. 1 (January–February 2010), pp. 17–32. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
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198 Chapter Six Network Optimization Problems
6. The cost of the flow through each arc is proportional to the amount of that flow, where the cost per unit flow is known.
7. The objective is to minimize the total cost of sending the available supply through the net- work to satisfy the given demand. (An alternative objective is to maximize the total profit from doing this.)
A solution for this kind of problem needs to specify how much flow is going through each arc. To be a feasible solution, the amount of flow through each arc cannot exceed the capacity of that arc and the net amount of flow generated at each node must equal the specified amount for that node. The following property indicates when the problem will have feasible solutions.
Feasible Solutions Property: Under the above assumptions, a minimum-cost flow problem will have feasible solutions if and only if the sum of the supplies from its supply nodes equals the sum of the demands at its demand nodes.
Note that this property holds for the Distribution Unlimited Co. problem, because the sum of its supplies is 80 1 70 5 150 and the sum of its demands is 60 1 90 5 150.
For many applications of minimum-cost flow problems, management desires a solution with integer values for all the flow quantities (e.g., integer numbers of full truckloads along each shipping lane). The model does not include any constraints that require this for feasible solutions. Fortunately, such constraints are not needed because of the following property.
Integer Solutions Property: As long as all its supplies, demands, and arc capacities have inte- ger values, any minimum-cost flow problem with feasible solutions is guaranteed to have an optimal solution with integer values for all its flow quantities.
See in Figure 6.3 that all the assumptions needed for this property to holds are satisfied for the Distribution Unlimited Co. problem. In particular, all the supplies (80 and 70), demands (60 and 90), and arc capacities (50) have integer values. Therefore, all the flow quantities in the optimal solution given in Figure 6.4 (30 three times, 50 two times, and 40) have integer values. This ensures that only full truckloads will be shipped into and out of the distribution center. (Remember that each unit corresponds to a full truckload of the product.)
Now let us see how to obtain an optimal solution for the Distribution Unlimited Co. prob- lem by formulating a spreadsheet model and then applying Solver.
Using Excel to Formulate and Solve Minimum-Cost Flow Problems Figure 6.5 shows a spreadsheet model that is based directly on the network representation of the problem in Figure 6.3 . The arcs are listed in columns B and C, along with their capacities (unless unlimited) in column F and their costs per unit flow in column G. The changing cells Ship (D4:D9) show the flow amounts through these arcs and the objective cell TotalCost (D11) provides the total cost of this flow by using the equation
D11 5 SUMPRODUCT(Ship, UnitCost)
The first set of constraints in the Solver Parameters box, D5:D8 # Capacity (F5:F8), ensures that the arc capacities are not exceeded.
Similarly, Column I lists the nodes, column J calculates the actual net flow generated at each node (given the flows in the changing cells), and column L specifies the net amount of flow that needs to be generated at each node. Thus, the second set of constraints in the Solver Parameters box is NetFlow (J4:J8) 5 SupplyDemand (L4:L8), requiring that the actual net amount of flow generated at each node must equal the specified amount.
The equations entered into NetFlow (J4:J8) use the difference of two SUMIF functions to calculate the net flow (outflow minus inflow) generated at each node. In each case, the first SUMIF function calculates the flow leaving the node and the second one calculates the flow entering the node. For example, consider the F1 node (I4). SUMIF(From,I4,Ship) sums each individual entry in Ship (D4:D9) if that entry is in a row where the entry in From (B4:B9) is the same as in I4. Since I4 5 F1 and the only rows that have F1 in the From column are rows 4 and 5, the sum in the Ship column is only over these same rows, so this sum is D4 1 D5. Similarly, SUMIF(To,I4,Ship) sums each individual entry in Ship (D4:D9) if
The objective is to mini- mize the total cost of sup- plying the demand nodes.
Capacity constraints like these are needed in any minimum-cost flow prob- lem that has any arcs with limited capacity.
Any minimum-cost flow problem needs net flow constraints like this for every node.
Excel Tip: SUMIF(A, B, C) adds up each entry in the range C for which the cor- responding entry in range A equals B. This function is especially useful in network problems for calculating the net flow generated at a node.
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6.1 Minimum-Cost Flow Problems 199
that entry is in a row where the entry in To (C4:C9) is the same as in I4. However, F1 never appears in the To column, so this sum is 0. Therefore, the overall equation for J4 yields J4 5 D4 1 D5 5 30 1 50 5 80, which is the net flow generated at the F1 node.
While it appears more complicated to use the SUMIF function rather than just entering J4 5 D4 1 D5, J5 5 D8 1 D9, J6 5 D6 1 D7 2 D5 2 D8, and so on, it is actually simpler. The SUMIF formula only needs to be entered once (in cell J4). It can then be copied down into the remaining cells in NetFlow (J5:J8). For a problem with many nodes, this is much quicker and (perhaps more significantly) less prone to error. In a large problem, it is all too easy to miss an arc when determining which cells in the Ship column to add and subtract to calculate the net flow for a given node.
The first Solver option specifies that the flow amounts cannot be negative. The second acknowledges that this is still a linear programming problem.
Running Solver gives the optimal solution shown in Ship (D4:D9). This is the same solu- tion as displayed in Figure 6.4 .
Solving Large Minimum-Cost Flow Problems More Efficiently Because minimum-cost flow problems are a special type of linear programming problem, and the simplex method can solve any linear programming problem, it also can solve any mini- mum-cost flow problem in the standard way. For example, Solver uses the simplex method to solve this type (or any other type) of linear programming problem. This works fine for small problems, like the Distribution Unlimited Co. problem, and for considerably larger
FIGURE 6.5 A spreadsheet model for the Distribution Unlimited Co. minimum-cost flow problem, including the objective cell TotalCost (D11) and the other output cells NetFlow (J4:J8), as well as the equations entered into these cells and the other specifications needed to set up the model. The changing cells Ship (D4:D9) show the optimal shipping quantities through the distribution network obtained by Solver.
1
A B C D E F G H I J
2
3
4
5
6
7
8
9
10
11
Distribution Unlimited Co. Minimum Cost Flow Problem
From
F1
F1
DC
DC
F2
Nodes
F1
F2
DC
W1
W2
Net Flow
80
70
0
-60
-90
Supply/Demand
K L
=
=
=
=
=
F2
Total Cost
To
W1
DC
W1
W2
DC
W2
Ship
30
50
30
50
30
Capacity
40
Unit Cost
$117,000
Range Name
Capacity
From
NetFlow
Nodes
Ship
SupplyDemand
To
TotalCost
UnitCost
Cells
F5:F8
B4:B9
J4:J8
I4:I8
D4:D9
L4:L8
C4:C9
D11
G4:G9
3
J
4
6
5
Net Flow
=SUMIF(From,I4,Ship)-SUMIF(To,I4,Ship)
=SUMIF(From,I6,Ship)-SUMIF(To,I6,Ship)
=SUMIF(From,I5,Ship)-SUMIF(To,I5,Ship)
8
7
=SUMIF(From,I8,Ship)-SUMIF(To,I8,Ship)
=SUMIF(From,I7,Ship)-SUMIF(To,I7,Ship)
C D
Total Cost =SUMPRODUCT(Ship,UnitCost)11
50
50
50
50
$700
$300
$200
$400
$500
$1,000
80
70
0
-60
-90
≤
≤
≤
≤
Solver Parameters
Set Objective Cell: TotalCost To: Min By Changing Variable Cells:
Ship
Subject to the Constraints: D5:D8 <= Capacity
NetFlow = SupplyDemand
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
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200 Chapter Six Network Optimization Problems
ones as well. Therefore, the approach illustrated in Figure 6.5 will serve you well for any minimum-cost flow problem encountered in this book and for many that you will encounter subsequently.
However, we should mention that a different approach is sometimes needed in practice to solve really big problems. Because of the special form of minimum-cost flow problems, it is possible to greatly streamline the simplex method to solve them far more quickly. In particu- lar, rather than going through all the algebra of the simplex method, it is possible to execute the same steps far more quickly by working directly with the network for the problem.
This streamlined version of the simplex method is called the network simplex method . The network simplex method can solve some huge problems that are much too large for the simplex method.
Like the simplex method, the network simplex method not only finds an optimal solu- tion but also can be a valuable aid to managers in conducting the kinds of what-if analyses described in Chapter 5.
Many companies now use the network simplex method to solve their minimum-cost flow problems. Some of these problems are huge, with many tens of thousands of nodes and arcs. Occasionally, the number of arcs will even be far larger, perhaps into the millions.
Although Solver does not, other commercial software packages for linear programming commonly include the network simplex method.
An important advance in recent years has been the development of excellent graphical interfaces for modeling minimum-cost flow problems. These interfaces make the design of the model and the interpretation of the output of the network simplex method completely visual and intuitive with no mathematics involved. This is very helpful for managerial deci- sion making.
Some Applications Probably the most important kind of application of minimum-cost flow problems is to the operation of a distribution network, such as the one depicted in Figures 6.1 – 6.4 for the Dis- tribution Unlimited Co. problem. As summarized in the first row of Table 6.1 , this kind of application involves determining a plan for shipping goods from their sources (factories, etc.) to intermediate storage facilities (as needed) and then on to the customers.
For some applications of minimum-cost flow problems, all the transshipment nodes are processing facilities rather than intermediate storage facilities. This is the case for solid waste management, as indicated in Table 6.1 . Here, the flow of materials through the network begins at the sources of the solid waste, then goes to the facilities for processing these waste materials into a form suitable for landfill, and then sends them on to the various landfill loca- tions. However, the objective still is to determine the flow plan that minimizes the total cost, where the cost now is for both shipping and processing.
In other applications, the demand nodes might be processing facilities. For example, in the third row of Table 6.1 , the objective is to find the minimum-cost plan for obtaining supplies from various possible vendors, storing these goods in warehouses (as needed), and then ship- ping the supplies to the company’s processing facilities (factories, etc.).
The next kind of application in Table 6.1 (coordinating product mixes at plants) illustrates that arcs can represent something other than a shipping lane for a physical flow of materials.
network simplex method The network simplex method can solve much larger minimum-cost flow problems (sometimes with millions of nodes and arcs) than can the simplex method used by Solver.
Kind of Application Supply Nodes Transshipment Nodes Demand Nodes
Operation of a distribution network Sources of goods Intermediate storage facilities Customers Solid waste management Sources of solid waste Processing facilities Landfill locations Operation of a supply network Vendors Intermediate warehouses Processing facilities Coordinating product mixes at plants Plants Production of a specific product Market for a specific
product
Cash flow management Sources of cash at a specific time
Short-term investment options Needs for cash at a specific time
TABLE 6.1 Typical Kinds of Applications of Minimum-Cost Flow Problems
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6.1 Minimum-Cost Flow Problems 201
This application involves a company with several plants (the supply nodes) that can produce the same products but at different costs. Each arc from a supply node represents the produc- tion of one of the possible products at that plant, where this arc leads to the transshipment node that corresponds to this product. Thus, this transshipment node has an arc coming in from each plant capable of producing this product, and then the arcs leading out of this node go to the respective customers (the demand nodes) for this product. The objective is to deter- mine how to divide each plant’s production capacity among the products so as to minimize the total cost of meeting the demand for the various products.
The last application in Table 6.1 (cash flow management) illustrates that different nodes can represent some event that occurs at different times. In this case, each supply node repre- sents a specific time (or time period) when some cash will become available to the company (through maturing accounts, notes receivable, sales of securities, borrowing, etc.). The supply at each of these nodes is the amount of cash that will become available then. Similarly, each demand node represents a specific time (or time period) when the company will need to draw on its cash reserves. The demand at each such node is the amount of cash that will be needed then. The objective is to maximize the company’s income from investing the cash between each time it becomes available and when it will be used. Therefore, each transshipment node represents the choice of a specific short-term investment option (e.g., purchasing a certificate of deposit from a bank) over a specific time interval. The resulting network will have a suc- cession of flows representing a schedule for cash becoming available, being invested, and then being used after the maturing of the investment.
Special Types of Minimum-Cost Flow Problems There are five important categories of network problems that turn out to be special types of minimum-cost flow problems.
One is the transportation problems discussed in Section 3.5. Figure 3.9 shows the net- work representation of a typical transportation problem. In our current terminology, the sources and destinations of a transportation problem are the supply nodes and demand nodes, respectively. Thus, a transportation problem is just a minimum-cost flow problem without any transshipment nodes and without any capacity constraints on the arcs (all of which go directly from a supply node to a demand node).
A second category is the assignment problems discussed in Section 3.6. Recall that this kind of problem involves assigning a group of people (or other operational units) to a group of tasks where each person is to perform a single task. An assignment problem can be viewed as a special type of transportation problem whose sources are the assignees and whose destinations are the tasks. This then makes the assignment problem also a special type of minimum-cost flow problem with the characteristics described in the preceding paragraph. In addition, each person is a supply node with a supply of 1 and each task is a demand node with a demand of 1.
A third special type of minimum-cost flow problem is transshipment problems . This kind of problem is just like a transportation problem except for the additional feature that the shipments from the sources (supply nodes) to the destinations (demand nodes) might also pass through intermediate transfer points (transshipment nodes) such as distribution centers. Like a transportation problem, there are no capacity constraints on the arcs. Conse- quently, any minimum-cost flow problem where each arc can carry any desired amount of flow is a transshipment problem. For example, if the data in Figure 6.2 were altered so that any amounts (within the ranges of the supplies and demands) could be shipped into and out of the distribution center, the Distribution Unlimited Co. would become just a transshipment problem. 1
Because of their close relationship to a general minimum-cost flow problem, we will not discuss transshipment problems further.
The other two important special types of minimum-cost flow problems are maximum flow problems and shortest path problems, which will be described in Sections 6.3 and 6.4 after presenting a case study of a maximum flow problem in the next section.
transshipment problem A transshipment problem is just a minimum-cost flow problem that has unlimited capacities for all its arcs.
1 Be aware that a minimum-cost flow problem that does have capacity constraints on the arcs is sometimes referred to as a capacitated transshipment problem. We will not use this terminology.
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202 Chapter Six Network Optimization Problems
In case you are wondering why we are bothering to point out that these five kinds of problems are special types of minimum-cost flow problems, here is one very important rea- son. It means that the network simplex method can be used to solve large problems of any of these types that might be difficult or impossible for the simplex method to solve. It is true that other efficient special-purpose algorithms also are available for each of these kinds of problems. However, recent implementations of the network simplex method have become so powerful that it now provides an excellent alternative to these other algorithms in most cases. This is especially valuable when the available software package includes the network simplex method but not another relevant special-purpose algorithm. Furthermore, even after finding an optimal solution, the network simplex method can continue to be helpful in aiding mana- gerial what-if sessions along the lines discussed in Chapter 5.
The network simplex method can be used to solve huge problems of any of these five special types.
1. Name and describe the three kinds of nodes in a minimum-cost flow problem. 2. What is meant by the capacity of an arc? 3. What is the usual objective for a minimum-cost flow problem? 4. What property is necessary for a minimum-cost flow problem to have feasible solutions? 5. What is the integer solutions property for minimum-cost flow problems? 6. What is the name of the streamlined version of the simplex method that is designed to solve
minimum-cost flow problems very efficiently? 7. What are a few typical kinds of applications of minimum-cost flow problems? 8. Name five important categories of network optimization problems that turn out to be special
types of minimum-cost flow problems.
Review Questions
6.2 A CASE STUDY: THE BMZ CO. MAXIMUM FLOW PROBLEM
What a day! First being called into his boss’s office and then receiving an urgent telephone call from the company president himself. Fortunately, he was able to reassure them that he has the situation under control.
Although his official title is Supply Chain Manager for the BMZ Company, Karl Schmidt often tells his friends that he really is the company’s crisis manager. One crisis after another. The supplies needed to keep the production lines going haven’t arrived yet. Or the supplies have arrived but are unusable because they are the wrong size. Or an urgent shipment to a key customer has been delayed. This current crisis is typical. One of the company’s most important distribution centers—the one in Los Angeles—urgently needs an increased flow of shipments from the company.
Karl was chosen for this key position because he is considered a rising young star. Hav- ing just received his MBA degree from a top American business school four years ago, he is the youngest member of upper-level management in the entire company. His business school training in the latest management science techniques has proven invaluable in improving sup- ply chain management throughout the company. The crises still occur, but the frequent chaos of past years has been eliminated.
Karl has a plan for dealing with the current crisis. This will mean calling on management science once again.
Background The BMZ Company is a European manufacturer of luxury automobiles. Although its cars sell well in all the developed countries, its exports to the United States are particularly impor- tant to the company.
BMZ has a well-deserved reputation for providing excellent service. One key to main- taining this reputation is having a plentiful supply of automobile replacement parts readily available to the company’s numerous dealerships and authorized repair shops. These parts are mainly stored in the company’s distribution centers and then delivered promptly when needed. One of Karl Schmidt’s top priorities is avoiding shortages at these distribution centers.
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6.2 A Case Study: The BMZ Co. Maximum Flow Problem 203
The company has several distribution centers in the United States. However, the clos- est one to the Los Angeles center is over 1,000 miles away in Seattle. Since BMZ cars are becoming especially popular in California, it is particularly important to keep the Los Ange- les center well supplied. Therefore, the fact that supplies there are currently dwindling is a matter of real concern to BMZ top management—as Karl learned forcefully today.
Most of the automobile replacement parts are produced at the company’s main factory in Stuttgart, Germany, along with the production of new cars. It is this factory that has been sup- plying the Los Angeles center with spare parts. Some of these parts are bulky, and very large numbers of certain parts are needed, so the total volume of the supplies has been relatively massive—over 300,000 cubic feet of goods arriving monthly. Now a much larger amount will be needed over the next month to replenish the dwindling inventory.
The Problem Karl needs to execute a plan quickly for shipping as much as possible from the main factory to the distribution center in Los Angeles over the next month. He already has recognized that this is a maximum flow problem —a problem of maximizing the flow of replacement parts from the factory to this distribution center.
The factory is producing far more than can be shipped to this one distribution center. Therefore, the limiting factor on how much can be shipped is the limited capacity of the com- pany’s distribution network.
This distribution network is depicted in Figure 6.6 , where the nodes labeled ST and LA are the factory in Stuttgart and the distribution center in Los Angeles, respectively. There is a rail head at the factory, so shipments first go by rail to one of three European ports: Rotter- dam (node RO), Bordeaux (node BO), and Lisbon (node LI). They then go by ship to ports in the United States, either New York (node NY) or New Orleans (node NO). Finally, they are shipped by truck from these ports to the distribution center in Los Angeles.
The organizations operating these railroads, ships, and trucks are independently owned companies that ship goods for numerous firms. Because of prior commitments to their regular customers, these companies are unable to drastically increase the allocation of space to any sin- gle customer on short notice. Therefore, the BMZ Co. is only able to secure a limited amount of shipping space along each shipping lane over the next month. The amounts available are given in Figure 6.6 , using units of hundreds of cubic meters. (Since each unit of 100 cubic meters is a little over 3,500 cubic feet, these are large volumes of goods that need to be moved.)
The problem is to maximize the flow of automobile replacement parts from the factory in Stuttgart, Germany, to the distribu- tion center in Los Angeles.
FIGURE 6.6 The BMZ Co. distribu- tion network from its main factory in Stuttgart, Germany, to a distribution center in Los Angeles.
LA
NY
NO
RO
BO ST
LI
Los Angeles
New York
New Orleans
[70 units max.]
[8 0 u
ni ts
m ax
.]
[60 un
its ma
x.]
[40 units max.]
[5 0 u
ni ts
m ax
.]
[30 uni
ts m ax.]
Lisbon
[40 un
its ma
x.] [70 units max.]
[50 units max.]
Stuttgart
Rotterdam
Bordeaux
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204 Chapter Six Network Optimization Problems
Model Formulation Figure 6.7 shows the network model for this maximum flow problem. Rather than showing the geographical layout of the distribution network, this network simply lines up the nodes (representing the cities) in evenly spaced columns. The arcs represent the shipping lanes, where the capacity of each arc (given in square brackets under the arc) is the amount of ship- ping space available along that shipping lane. The objective is to determine how much flow to send through each arc (how many units to ship through each shipping lane) to maximize the total number of units flowing from the factory in Stuttgart to the distribution center in Los Angeles.
Figure 6.8 shows the corresponding spreadsheet model for this problem when using the format introduced in Figure 6.5 . The main difference from the model in Figure 6.5 is the change in the objective. Since we are no longer minimizing the total cost of the flow through the network, column G in Figure 6.5 can be deleted in Figure 6.8 . The objective cell MaxFlow (D14) in Figure 6.8 now needs to give the total number of units flowing from Stuttgart to Los Angeles. Thus, the equations at the bottom of the figure include D14 5 I4, where I4 gives the net flow leaving Stuttgart to go to Los Angeles. As in Figure 6.5 , the equations in Figure 6.8 entered into NetFlow (I4:I10) again use the difference of two SUMIF functions to calculate the net flow generated at each node. Since the objective is to maximize the flow shown in MaxFlow (D14), the Solver Parameters box specifies that this objective cell is to be maximized. After running Solver, the optimal solution shown in the changing cells Ship (D4:D12) is obtained for the amount that BMZ should ship through each shipping lane.
However, Karl is not completely satisfied with this solution. He has an idea for doing even better. This will require formulating and solving another maximum flow problem. (This story continues in the middle of the next section.)
In contrast to the spread- sheet model in Figure 6.5 , which minimizes TotalCost (D11), the spreadsheet model in Figure 6.8 maxi- mizes the objective cell MaxFlow (D14).
FIGURE 6.7 A network model for the BMZ Co. problem as a maximum flow problem, where the number in square brackets below each arc is the capacity of that arc.
RO
STLA BO
LI
NY
NO
[8 0]
[6 0]
[5 0]
[4 0]
[70]
[30] [40]
[50]
[70]
1. What is the current crisis facing the BMZ Co.? 2. When formulating this problem in network terms, what is flowing through BMZ’s distribution
network? From where to where? 3. What is the objective of the resulting maximum flow problem?
Review Questions
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6.3 Maximum Flow Problems 205
6.3 MAXIMUM FLOW PROBLEMS
Like a minimum-cost flow problem, a maximum flow problem is concerned with flow through a network. However, the objective now is different. Rather than minimizing the cost of the flow, the objective now is to find a flow plan that maximizes the amount flowing through the network. This is how Karl Schmidt was able to find a flow plan that maximizes the number of units of automobile replacement parts flowing through BMZ’s distribution network from its factory in Stuttgart to the distribution center in Los Angeles.
General Characteristics Except for the difference in objective (maximize flow versus minimize cost), the character- istics of the maximum flow problem are quite similar to those for the minimum-cost flow problem. However, there are some minor differences, as we will discuss after summarizing the assumptions.
Assumptions of a Maximum Flow Problem
1. All flow through the network originates at one node, called the source , and terminates at one other node, called the sink . (The source and sink in the BMZ problem are the factory and the distribution center, respectively.)
FIGURE 6.8 A spreadsheet model for the BMZ Co. maximum flow problem, including the equations entered into the objective cell MaxFlow (D14) and the other output cells NetFlow (I4:I10), as well as the other specifications needed to set up the model. The changing cells Ship (D4:D12) show the optimal shipping quantities through the distribution network obtained by Solver.
1
A B C D E F G H I
2
3
4
5
6
7
8
9
10
11
BMZ Co. Maximum Flow Problem
From
Stuttgart
Stuttgart
Stuttgart
Rotterdam
Bordeaux
Nodes
Stuttgart
Rotterdam
Bordeaux
Lisbon
New York
New Orleans
Los Angeles
Net Flow
150
0
0
0
0
0
-150
Supply/Demand
J K
=
=
=
=
=Bordeaux
Lisbon
New York
New Orleans
Rotterdam
Bordeaux
Lisbon
New York
New York
New Orleans
New Orleans
Los Angeles
Los Angeles
Maximum Flow
To Ship Capacity
70
50
40
30
80
70
30
50
30
150
12
13
14
Range Name
Capacity
From
MaxFlow
NetFlow
Nodes
Ship
SupplyDemand
To
Cells
F4:F12
B4:B12
D14
I4:I10
H4:H10
D4:D12
K5:K9
C4:C12
3
I
4
6
5
Net Flow
=SUMIF(From,H4,Ship)-SUMIF(To,H4,Ship)
=SUMIF(From,H6,Ship)-SUMIF(To,H6,Ship)
=SUMIF(From,H5,Ship)-SUMIF(To,H5,Ship)
8
7
=SUMIF(From,H8,Ship)-SUMIF(To,H8,Ship)
=SUMIF(From,H7,Ship)-SUMIF(To,H7,Ship)
10
9
=SUMIF(From,H10,Ship)-SUMIF(To,H10,Ship)
=SUMIF(From,H9,Ship)-SUMIF(To,H9,Ship)
C D
Maximum Flow =I414
70
50
50
30
80
70
40
60
40
0
0
0
0
0
≤ ≤
≤
≤
≤
≤ ≤
≤
≤
Solver Parameters Set Objective Cell: MaxFlow To: Max By Changing Variable Cells:
Ship Subject to the Constraints:
I5:I9 = SupplyDemand Ship <= Capacity
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
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206
2. All the remaining nodes are transshipment nodes. (These are nodes RO, BO, LI, NY, and NO in the BMZ problem.)
3. Flow through an arc is only allowed in the direction indicated by the arrowhead, where the maximum amount of flow is given by the capacity of that arc. At the source, all arcs point away from the node. At the sink, all arcs point into the node.
4. The objective is to maximize the total amount of flow from the source to the sink. This amount is measured in either of two equivalent ways, namely, either the amount leaving the source or the amount entering the sink. (Cells D14 and I4 in Figure 6.8 use the amount leaving the source.)
The source and sink of a maximum flow problem are analogous to the supply nodes and demand nodes of a minimum-cost flow problem. These are the only nodes in both problems that do not have conservation of flow (flow out equals flow in). Like the supply nodes, the source generates flow. Like the demand nodes, the sink absorbs flow.
However, there are two differences between these nodes in a minimum-cost flow problem and the corresponding nodes in a maximum flow problem.
One difference is that, whereas supply nodes have fixed supplies and demand nodes have fixed demands, the source and sink do not. The reason is that the objective is to maximize the flow leaving the source and entering the sink rather than fixing this amount.
The second difference is that, whereas the number of supply nodes and the number of demand nodes in a minimum-cost flow problem may be more than one, there can be only one source and only one sink in a maximum flow problem. However, variants of maximum flow problems that have multiple sources and sinks can still be solved by Solver, as you now will see illustrated by the BMZ case study introduced in the preceding section.
Continuing the Case Study with Multiple Supply Points and Multiple Demand Points Here is Karl Schmidt’s idea for how to improve upon the flow plan obtained at the end of Sec- tion 6.2 (as given in column D of Figure 6.8 ).
The company has a second, smaller factory in Berlin, north of its Stuttgart factory, for pro- ducing automobile parts. Although this factory normally is used to help supply distribution centers in northern Europe, Canada, and the northern United States (including one in Seattle), it also is able to ship to the distribution center in Los Angeles. Furthermore, the distribution center in Seattle has the capability of supplying parts to the customers of the distribution cen- ter in Los Angeles when shortages occur at the latter center.
The objective is to find a flow plan that maximizes the flow from the source to the sink.
Although a maximum flow problem has only a single source and a single sink, variants with multiple sources and sinks also can be solved, as illustrated in the next subsection.
The network for transport of natural gas on the Norwegian Continental Shelf, with approximately 5,000 miles of subsea pipelines, is the world’s largest offshore pipeline network. Gassco is a company entirely owned by the Norwegian state, which operates this network. Another company that is largely state owned, StatoilHydro, is the main Norwegian supplier of natural gas to markets throughout Europe and elsewhere.
Gassco and StatoilHydro together use management science techniques to optimize both the configuration of the network and the routing of the natural gas. The main model used for this routing is a multicommodity network- flow model in which the different hydrocarbons and con- taminants in natural gas constitute the commodities. The objective function for the model is to maximize the total flow of the natural gas from the supply points (the off- shore drilling platforms) to the demand points (typically
import terminals). However, in addition to the usual sup- ply-and-demand constraints, the model also includes con- straints involving pressure-flow relationships, maximum delivery pressures, and technical pressure bounds on pipe- lines. Therefore, this model is a generalization of the model for the maximum flow problem described in this section.
This key application of management science, along with a few others, has had a dramatic impact on the effi- ciency of the operation of this offshore pipeline network. The resulting accumulated savings were estimated to be approximately $2 billion in the period 1995–2008.
Source: F. Rømo, A. Tomasgard, L. Hellemo, M. Fodstad, B. H. Eidesen, and B. Pedersen, “Optimizing the Norwegian Natural Gas Production and Transport,” Interfaces 39, no. 1 (January–February 2009), pp. 46–56. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
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6.3 Maximum Flow Problems 207
In this light, Karl now has developed a better plan for addressing the current inventory shortages in Los Angeles. Rather than simply maximizing shipments from the Stuttgart fac- tory to Los Angeles, he has decided to maximize the total shipments from both factories to the distribution centers in both Los Angeles and Seattle.
Figure 6.9 shows the network model representing the expanded distribution network that encompasses both factories and both distribution centers. In addition to the nodes shown in Figures 6.6 and 6.7 , node BE is the second, smaller factory in Berlin; nodes HA and BN are additional ports used by this factory in Hamburg and Boston, respectively; and node SE is the distribution center in Seattle. As before, the arcs represent the shipping lanes, where the num- ber in square brackets below each arc is the capacity of that arc, that is, the maximum number of units that can be shipped through that shipping lane over the next month.
The corresponding spreadsheet model is displayed in Figure 6.10 . The format is the same as in Figure 6.8 . However, the objective cell MaxFlow (D21) now gives the total flow from Stuttgart and Berlin, so D21 5 I4 1 I5 (as shown by the equation for this objective cell given at the bottom of the figure).
The changing cells Ship (D4:D19) in this figure show the optimal solution obtained for the number of units to ship through each shipping lane over the next month. Comparing this solution with the one in Figure 6.8 shows the impact of Karl Schmidt’s decision to expand the distribution network to include the second factory and the distribution center in Seattle. As indicated in column I of the two figures, the number of units going to Los Angeles directly has been increased from 150 to 160, in addition to the 60 units going to Seattle as a backup for the inventory shortage in Los Angeles. This plan solved the crisis in Los Angeles and won Karl commendations from top management.
Some Applications The applications of maximum flow problems and their variants are somewhat similar to those for minimum-cost flow problems described in the preceding section when management’s objective is to maximize flow rather than to minimize cost. Here are some typical kinds of applications.
1. Maximize the flow through a distribution network, as for the BMZ Co. problem. 2. Maximize the flow through a company’s supply network from its vendors to its processing
facilities. 3. Maximize the flow of oil through a system of pipelines. 4. Maximize the flow of water through a system of aqueducts. 5. Maximize the flow of vehicles through a transportation network.
FIGURE 6.9 A network model for the expanded BMZ Co. prob- lem as a variant of a maxi- mum flow problem, where the number in square brackets below each arc is the capacity of that arc.
NY
NO
BN
ROSE
BOLA
LI
HA
[20 ]
[3 0]
[40 ]
[40]
[1 0]
[80 ]
[50 ]
[70]
[30]
[40]
[60] [50]
[70]
[20]
[4 0]
BE
ST
[60]
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208 Chapter Six Network Optimization Problems
FIGURE 6.10 A spreadsheet model for the expanded BMZ Co. problem as a variant of a maximum flow problem with sources in both Stuttgart and Berlin and sinks in both Los Angeles and Seattle. Using the objective cell MaxFlow (D21) to maximize the total flow from the two sources to the two sinks, Solver yields the optimal shipping plan shown in the changing cells Ship (D4:D19).
1
A B C D E F G H I
2
3
4
5
6
7
8
9
10
11
BMZ Co. Expanded Maximum Flow Problem
From
Stuttgart
Stuttgart
Stuttgart
Berlin
Berlin
Nodes
Stuttgart
Berlin
Hamburg
Rotterdam
Bordeaux
Lisbon
Boston
New York
New Orleans
Los Angeles
Seattle
Net Flow
140
80
0
0
0
0
0
0
0
-160
-60
Supply/Demand
J K
=
=
=
=
=
=
=
Rotterdam
Bordeaux
Bordeaux
Lisbon
Rotterdam
Bordeaux
Lisbon
Rotterdam
Hamburg
New York
New York
New Orleans
New Orleans
Maximum Flow
To Ship Capacity
70
40
60
30
40
30
30
20
60
Hamburg
Hamburg
New Orleans
New York
New York
Boston
Boston
New York
Boston
Los Angeles
Los Angeles
Seattle
Los Angeles
Seattle
80
40
10
20
30
30
70
220
12
13
14
15
16
17
18
19
20
21
Range Name
Capacity
From
MaxFlow
NetFlow
Nodes
Ship
SupplyDemand
To
Cells
F4:F19
B4:B19
D21
I4:I14
H4:H14
D4:D19
K6:K12
C4:C19
3
I
4
6
5
Net Flow
=SUMIF(From,H4,Ship)-SUMIF(To,H4,Ship)
=SUMIF(From,H6,Ship)-SUMIF(To,H6,Ship)
=SUMIF(From,H5,Ship)-SUMIF(To,H5,Ship)
8
7
=SUMIF(From,H8,Ship)-SUMIF(To,H8,Ship)
=SUMIF(From,H7,Ship)-SUMIF(To,H7,Ship)
10
9
=SUMIF(From,H10,Ship)-SUMIF(To,H10,Ship)
=SUMIF(From,H9,Ship)-SUMIF(To,H9,Ship)
12
11
=SUMIF(From,H12,Ship)-SUMIF(To,H12,Ship)
=SUMIF(From,H11,Ship)-SUMIF(To,H11,Ship)
14
13
=SUMIF(From,H14,Ship)-SUMIF(To,H14,Ship)
=SUMIF(From,H13,Ship)-SUMIF(To,H13,Ship)
C D
Maximum Flow =I4+I521
70
50
60
40
50
30
40
20
60
80
40
10
20
30
40
70
0
0
0
0
0
0
0
≤
≤
≤
≤
≤
≤
≤
≤
≤
≤
≤
≤
≤
≤
≤
≤
Solver Parameters
Set Objective Cell: MaxFlow To: Max By Changing Variable Cells:
Ship
Subject to the Constraints: I6:I12 = Sup plyDemand
Ship <= Capacity
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
Solving Very Large Problems The expanded BMZ network in Figure 6.9 has 11 nodes and 16 arcs. However, the networks for most real applications are considerably larger, and occasionally vastly larger. As the num- ber of nodes and arcs grows into the hundreds or thousands, the formulation and solution approach illustrated in Figures 6.8 and 6.10 quickly becomes impractical.
Fortunately, management scientists have other techniques available for formulating and solving huge problems with many tens of thousands of nodes and arcs. One technique is to
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6.4 Shortest Path Problems 209
reformulate a variant of a maximum flow problem so that an extremely efficient special- purpose algorithm for maximum flow problems still can be applied. Another is to reformulate the problem to fit the format for a minimum-cost flow problem so that the network simplex method can be applied. These special algorithms are available in some software packages, but not in Solver. Thus, if you should ever encounter a maximum flow problem or a variant that is beyond the scope of Solver (which won’t happen in this book), rest assured that it probably can be formulated and solved in another way.
1. How does the objective of a maximum flow problem differ from that for a minimum-cost flow problem?
2. What are the source and the sink for a maximum flow problem? For each, in what direction do all their arcs point?
3. What are the two equivalent ways in which the total amount of flow from the source to the sink can be measured?
4. The source and sink of a maximum flow problem are different from the supply nodes and demand nodes of a minimum-cost flow problem in what two ways?
5. What are a few typical kinds of applications of maximum flow problems?
Review Questions
6.4 SHORTEST PATH PROBLEMS
The most common applications of shortest path problems are for what the name suggests— finding the shortest path between two points. Here is an example.
An Example: The Littletown Fire Department Problem Littletown is a small town in a rural area. Its fire department serves a relatively large geo- graphical area that includes many farming communities. Since there are numerous roads throughout the area, many possible routes may be available for traveling to any given farming community from the fire station. Since time is of the essence in reaching a fire, the fire chief wishes to determine in advance the shortest path from the fire station to each of the farming communities.
Figure 6.11 shows the road system connecting the fire station to one of the farming com- munities, including the mileage along each road. Can you find which route from the fire sta- tion to the farming community minimizes the total number of miles?
Model Formulation for the Littletown Problem Figure 6.12 gives the network representation of this problem, which ignores the geographical layout and the curves in the roads. This network model is the usual way of representing a shortest path problem. The junctions now are nodes of the network, where the fire station and farming community are two additional nodes labeled as O (for origin ) and T (for destinat- ion ), respectively. Since travel (flow) can go in either direction between the nodes, the lines
The objective is to find the shortest route from the fire station to the farming community.
FIGURE 6.11 The road system between the Littletown Fire Sta- tion and a certain farming community, where A, B, . . . , H are junctions and the number next to each road shows its dis- tance in miles. 3
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Fire Station
Farming Community
C
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8
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connecting the nodes now are referred to as links 2 instead of arcs. A link between a pair of nodes allows travel in either direction, whereas an arc allows travel in only the direction indi- cated by an arrowhead, so the lines in Figure 6.12 need to be links instead of arcs. (Notice that the links do not have an arrowhead at either end.)
Have you found the shortest path from the origin to the destination yet? (Try it now before reading further.) It is
O S A S B S E S F S T
with a total distance of 19 miles. This problem (like any shortest path problem) can be thought of as a special kind of min-
imum-cost flow problem (Section 6.1) where the miles traveled now are interpreted to be the cost of flow through the network. A trip from the fire station to the farming community is inter- preted to be a flow of 1 on the chosen path through the network, so minimizing the cost of this flow is equivalent to minimizing the number of miles traveled. The fire station is considered to be the one supply node, with a supply of 1 to represent the start of this trip. The farming com- munity is the one demand node, with a demand of 1 to represent the completion of this trip. All the other nodes in Figure 6.12 are transshipment nodes, so the net flow generated at each is 0.
Figure 6.13 shows the spreadsheet model that results from this interpretation. The format is basically the same as for the minimum-cost flow problem formulated in Figure 6.5 , except
2 Another name sometimes used is undirected arc, but we will not use this terminology.
links In a shortest path problem, travel goes from the origin to the destination through a series of links (such as roads) that connect pairs of nodes (junctions) in the network.
Incorporated in 1881, Canadian Pacific Railway (CPR) was North America’s first transcontinental railway. CPR trans- ports rail freight over a 14,000-mile network extending from Montreal to Vancouver and throughout the U.S. Northwest and Midwest. Alliances with other carriers extend CPR’s market reach into the major business centers of Mexico as well.
Every day CPR receives approximately 7,000 new ship- ments from its customers going to destinations across North America and for export. It must route and move these ship- ments in railcars over the network of track, where a railcar may be switched a number of times from one locomotive engine to another before reaching its destination. CPR must coordinate the shipments with its operational plans for 1,600 locomotives, 65,000 railcars, over 5,000 train crew members, and 250 train yards.
CPR management turned to a management science con- sulting firm, MultiModal Applied Systems, to work with CPR employees in developing a management science approach to this problem. A variety of management science tech- niques were used to create a new operating strategy. How- ever, the foundation of the approach was to represent the
flow of blocks of railcars as flow through a network where each node corresponds to both a location and a point in time. This representation then enabled the application of network optimization techniques. For example, numerous shortest path problems are solved each day as part of the overall approach.
This application of management science is saving CPR roughly US$100 million per year. Labor productivity, loco- motive productivity, fuel consumption, and railcar veloc- ity have improved very substantially. In addition, CPR now provides its customers with reliable delivery times and has received many awards for its improvement in service. This application of network optimization techniques also led to CPR winning the prestigious First Prize in the 2003 inter- national competition for the Franz Edelman Award for Achievement in Operations Research and the Management Sciences.
Source: P. Ireland, R. Case, J. Fallis, C. Van Dyke, J. Kuehn, and M. Meketon, “The Canadian Pacific Railway Transforms Opera- tions by Using Models to Develop Its Operating Plans,” Interfaces 34, no. 1 (January–February 2004), pp. 5–14. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
FIGURE 6.12 The network representa- tion of Figure 6.11 as a shortest path problem.
A
B
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D
O
C
F
G T
H
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1
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5 3
4 2 7 4
2 7
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(Origin) (Destination) 5
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6.4 Shortest Path Problems 211
FIGURE 6.13 A spreadsheet model for the Littletown Fire Department shortest path problem, including the equations entered into the objective cell TotalDistance (D29) and the other output cells SupplyDemand (K4:K13). The values of 1 in the changing cells OnRoute (D4:D27) reveal the optimal solution obtained by Solver for the shortest path (19 miles) from the fire station to the farming community.
1
A B C D E F G H I
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Littletown Fire Department Shortest Path Problem
From
Fire St.
Fire St.
Fire St.
A
A
Nodes
Fire St.
A
B
C
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G
H
Farm Comm.
Net Flow
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Supply/Demand
J K
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Total Distance
To On Route Distance
0
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Farm Comm.
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Farm Comm.
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Net Flow
=SUMIF(From,H4,OnRoute)-SUMIF(To,H4,OnRoute)
=SUMIF(From,H6,OnRoute)-SUMIF(To,H6,OnRoute)
=SUMIF(From,H5,OnRoute)-SUMIF(To,H5,OnRoute)
8
7
=SUMIF(From,H8,OnRoute)-SUMIF(To,H8,OnRoute)
=SUMIF(From,H7,OnRoute)-SUMIF(To,H7,OnRoute)
10
9
=SUMIF(From,H10,OnRoute)-SUMIF(To,H10,OnRoute)
=SUMIF(From,H9,OnRoute)-SUMIF(To,H9,OnRoute)
12
11
=SUMIF(From,H12,OnRoute)-SUMIF(To,H12,OnRoute)
=SUMIF(From,H11,OnRoute)-SUMIF(To,H11,OnRoute)
13 =SUMIF(From,H13,OnRoute)-SUMIF(To,H13,OnRoute)
C D
Total Distance =SUMPRODUCT(OnRoute,Distance)29
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Solver Parameters
Set Objective Cell: TotalDistance To: Min By Changing Variable Cells:
OnRoute
Subject to the Constraints: NetFlow = SupplyDemand
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP Range Name
Distance
From
NetFlow
Nodes
OnRoute
SupplyDemand
To
TotalDistance
Cells
F4:F27
B4:B27
I4:I13
H4:H13
D4:D27
K4:K13
C4:C27
D29
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212 Chapter Six Network Optimization Problems
now there are no arc capacity constraints and the unit cost column is replaced by a column of distances in miles. The flow quantities given by the changing cells OnRoute (D4:D27) are 1 for each arc that is on the chosen path from the fire station to the farming community and 0 other- wise. The objective cell TotalDistance (D29) gives the total distance of this path in miles. (See the equation for this cell at the bottom of the figure.) Columns B and C together list all the verti- cal links in Figure 6.12 twice, once as a downward arc and once as an upward arc, since either direction might be on the chosen path. The other links are only listed as left-to-right arcs, since this is the only direction of interest for choosing a shortest path from the origin to the destination.
Column K shows the net flow that needs to be generated at each of the nodes. Using the equations at the bottom of the figure, each column I cell then calculates the actual net flow at that node by adding the flow out and subtracting the flow in. The corresponding constraints, Nodes (H4:H13) 5 SupplyDemand (K4:K13), are specified in the Solver Parameters box.
The solution shown in OnRoute (D4:D27) is the optimal solution obtained after running Solver. It is exactly the same as the shortest path given earlier.
Just as for minimum-cost flow problems and maximum flow problems, special algorithms are available for solving large shortest path problems very efficiently, but these algorithms are not included in Solver. Using a spreadsheet formulation and Solver is fine for problems of the size of the Littletown problem and somewhat larger, but you should be aware that vastly larger problems can still be solved by other means.
General Characteristics Except for more complicated variations beyond the scope of this book, all shortest path problems share the characteristics illustrated by the Littletown problem. Here are the basic assumptions.
Assumptions of a Shortest Path Problem
1. You need to choose a path through the network that starts at a certain node, called the origin , and ends at another certain node, called the destination .
2. The lines connecting certain pairs of nodes commonly are links (which allow travel in either direction), although arcs (which only permit travel in one direction) also are allowed.
3. Associated with each link (or arc) is a nonnegative number called its length. (Be aware that the drawing of each link in the network typically makes no effort to show its true length other than giving the correct number next to the link.)
4. The objective is to find the shortest path (the path with the minimum total length) from the origin to the destination.
Some Applications Not all applications of shortest path problems involve minimizing the distance traveled from the origin to the destination. In fact, they might not even involve travel at all. The links (or arcs) might instead represent activities of some other kind, so choosing a path through the network corresponds to selecting the best sequence of activities. The numbers giving the “lengths” of the links might then be, for example, the costs of the activities, in which case the objective would be to determine which sequence of activities minimizes the total cost.
Here are three categories of applications.
1. Minimize the total distance traveled, as in the Littletown example. 2. Minimize the total cost of a sequence of activities, as in the example that follows in the
subsection below. 3. Minimize the total time of a sequence of activities, as in the example involving the Quick
Company at the end of this section.
An Example of Minimizing Total Cost Sarah has just graduated from high school. As a graduation present, her parents have given her a car fund of $21,000 to help purchase and maintain a certain three-year-old used car for college. Since operating and maintenance costs go up rapidly as the car ages, Sarah’s parents tell her that she will be welcome to trade in her car on another three-year-old car one or more times during the next three summers if she determines that this would minimize her total net
This spreadsheet model is like one for a minimum- cost flow problem with no arc capacity constraints except that distances replace unit costs and travel on a chosen path is interpreted as a flow of 1 through this path.
The objective is to find the shortest path from the ori- gin to the destination.
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6.4 Shortest Path Problems 213
cost. They also inform her that they will give her a new car in four years as a college gradu- ation present, so she should definitely plan to trade in her car then. (These are pretty nice parents!)
Table 6.2 gives the relevant data for each time Sarah purchases a three-year-old car. For example, if she trades in her car after two years, the next car will be in ownership year 1 dur- ing her junior year, and so forth.
When should Sarah trade in her car (if at all) during the next three summers to minimize her total net cost of purchasing, operating, and maintaining the car(s) over her four years of college?
Figure 6.14 shows the network formulation of this problem as a shortest path problem. Nodes 1, 2, 3, and 4 are the end of Sarah’s first, second, third, and fourth years of college, respectively. Node 0 is now, before starting college. Each arc from one node to a second node corresponds to the activity of purchasing a car at the time indicated by the first of these two nodes and then trading it in at the time indicated by the second node. Sarah begins by purchas- ing a car now, and she ends by trading in a car at the end of year 4, so node 0 is the origin and node 4 is the destination.
The number of arcs on the path chosen from the origin to the destination indicates how many times Sarah will purchase and trade in a car. For example, consider the path
0 1 3 4
This corresponds to purchasing a car now, then trading it in at the end of year 1 to purchase a second car, then trading in the second car at the end of year 3 to purchase a third car, and then trading in this third car at the end of year 4.
Since Sarah wants to minimize her total net cost from now (node 0) to the end of year 4 (node 4), each arc length needs to measure the net cost of that arc’s cycle of purchasing, main- taining, and trading in a car. Therefore,
Arc length 5 Purchase price 1 Operating and maintenance costs 2 Trade-in value
For example, consider the arc from node 1 to node 3. This arc corresponds to purchasing a car at the end of year 1, operating and maintaining it during ownership years 1 and 2, and then trading it in at the end of ownership year 2. Consequently,
Length of arc from 1 to 3 5 12,000 1 2,000 1 3,000 2 6,500
5 10,500 (in dollars)
Sarah needs a schedule for trading in her car that will minimize her total net cost.
Operating and Maintenance Costs for Ownership Year
Trade-in Value at End of Ownership Year
Purchase Price 1 2 3 4 1 2 3 4
$12,000 $2,000 $3,000 $4,500 $6,500 $8,500 $6,500 $4,500 $3,000
TABLE 6.2 Sarah’s Data Each Time She Purchases a Three- Year-Old Car
FIGURE 6.14 Formulation of the problem of when Sarah should trade in her car as a shortest path problem. The node labels measure the number of years from now. Each arc represents purchasing a car and then trading it in later.
(Origin) (Destination) 5,500 5,500 5,5005,500
10,500
17,000
25,000
10,500
10,500
17,000
0 1 2 3 4
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214 Chapter Six Network Optimization Problems
The arc lengths calculated in this way are shown next to the arcs in Figure 6.14 . Adding up the lengths of the arcs on any path from node 0 to node 4 then gives the total net cost for that particular plan for trading in cars over the next four years. Therefore, finding the shortest path from the origin to the destination identifies the plan that will minimize Sarah’s total net cost.
Figure 6.15 shows the corresponding spreadsheet model, formulated in just the same way as for Figure 6.13 except that distances are now costs. Thus, the objective cell TotalCost (D23) now gives the total cost that is to be minimized. The changing cells OnRoute (D12:D21) in the figure display the optimal solution obtained after running Solver. Since values of 1 indi- cate the path being followed, the shortest path turns out to be
0 2 4
Trade in the first car at the end of year 2. Trade in the second car at the end of year 4.
The length of this path is 10,500 1 10,500 5 21,000, so Sarah’s total net cost is $21,000, as given by the objective cell. Recall that this is exactly the amount in Sarah’s car fund provided by her parents. (These are really nice parents!)
An Example of Minimizing Total Time The Quick Company has learned that a competitor is planning to come out with a new kind of product with great sales potential. Quick has been working on a similar product that had been scheduled to come to market in 20 months. However, research is nearly complete and Quick’s management now wishes to rush the product out to meet the competition.
There are four nonoverlapping phases left to be accomplished, including the remaining research (the first phase) that currently is being conducted at a normal pace. However, each phase can instead be conducted at a priority or crash level to expedite completion. These are the only levels that will be considered for the last three phases, whereas both the normal level and these two levels will be considered for the first phase. The times required at these levels are shown in Table 6.3 .
Management now has allocated $30 million for these four phases. The cost of each phase at the levels under consideration is shown in Table 6.4 .
Management wishes to determine at which level to conduct each of the four phases to minimize the total time until the product can be marketed, subject to the budget restriction of $30 million.
Figure 6.16 shows the network formulation of this problem as a shortest path problem. Each node indicates the situation at that point in time. Except for the destination, a node is identified by two numbers:
1. The number of phases completed. 2. The number of millions of dollars left for the remaining phases.
The origin is now, when 0 phases have been completed and the entire budget of $30 million is left. Each arc represents the choice of a particular level of effort (identified in parentheses below the arc) for that phase. [There are no crash arcs emanating from the (2, 12) and (3, 3) nodes because this level of effort would require exceeding the budget of $30 million for the four phases.] The time (in months) required to perform the phase with this level of effort then is the length of the arc (shown above the arc). Time is chosen as the measure of arc length because the objective is to minimize the total time for all four phases. Summing the arc lengths for any particular path through the network gives the total time for the plan cor- responding to that path. Therefore, the shortest path through the network identifies the plan that minimizes total time.
The sum of the arc lengths on any path through this network gives the total net cost of the corresponding plan for trading in cars.
The objective cell now is TotalCost instead of TotalDistance.
The objective is to mini- mize the total time for the project.
The sum of the arc lengths on any path through this network gives the total time of the corresponding plan for preparing the new product.
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6.4 Shortest Path Problems 215
All four phases have been completed as soon as any one of the four nodes with a first label of 4 has been reached. So why doesn’t the network just end with these four nodes rather than hav- ing an arc coming out of each one? The reason is that a shortest path problem is required to have only a single destination. Consequently, a dummy destination is added at the right-hand side.
FIGURE 6.15 A spreadsheet model that formulates Sarah’s problem as a shortest path problem where the objec- tive is to minimize the total cost instead of the total distance. The bottom of the figure shows the equations entered in the objective cell TotalCost (D23) and the other out- put cells Cost (E12:E21) and NetFlow (H12:H16). After applying Solver, the values of 1 in the changing cells OnRoute (D12:D21) identify the shortest (least expen- sive) path for scheduling trade-ins.
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Sarah's Car Purchasing Problem
From
Year 0
Year 0
Year 0
Year 0
Year 1
Year 2
Year 3
Year 4
Year 1
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Year 0
Year 1
Year 2
Year 3
Year 4
Net Flow
1
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-1
Supply/Demand
H I J
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Year 1
Year 1
Year 2
Year 2
Year 3
Year 1
Year 2
Year 3
Year 4
Year 2
Year 3
Year 4
Year 3
Year 4
Year 4
Total Cost
To On Route Cost
$5,500
$10,500
$17,000
$25,000
$5,500
$10,500
$17,000
$5,500
$10,500
$5,500
$21,000
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Operating &
Maint. Cost
Trade-in
Value at End
of Year
Purchase
Price
$2,000
$3,000
$4,500
$6,500
$8,500
$6,500
$4,500
$3,000
$12,000
1
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Range Name
Cost
From
NetFlow
Nodes
OnRoute
OpMaint1
OpMaint2
OpMaint3
OpMaint4
PurchasePrice
SupplyDemand
To
TotalCost
TradeIn1
TradeIn2
TradeIn3
TradeIn4
Cells
E12:E21
B12:B21
H12:H16
G12:G16
D12:D21
C5
C6
C7
C8
E5
J12:J16
C12:C21
D23
D5
D6
D7
D8 11
H
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Net Flow
=SUMIF(From,G12,OnRoute)-SUMIF(To,G12,OnRoute)
=SUMIF(From,G14,OnRoute)-SUMIF(To,G14,OnRoute)
=SUMIF(From,G13,OnRoute)-SUMIF(To,G13,OnRoute)
16
15
=SUMIF(From,G16,OnRoute)-SUMIF(To,G16,OnRoute)
=SUMIF(From,G15,OnRoute)-SUMIF(To,G15,OnRoute)
11
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Cost
=PurchasePrice+OpMaint1-TradeIn1
=PurchasePrice+OpMaint1+OpMaint2+OpMaint3-TradeIn3
=PurchasePrice+OpMaint1+OpMaint2-TradeIn2
16
15
=PurchasePrice+OpMaint1-TradeIn1
=PurchasePrice+OpMaint1+OpMaint2+OpMaint3+OpMaint4-TradeIn4
17
19
18
=PurchasePrice+OpMaint1+OpMaint2-TradeIn2
=PurchasePrice+OpMaint1-TradeIn1
=PurchasePrice+OpMaint1+OpMaint2+OpMaint3-TradeIn3
21
20
=PurchasePrice+OpMaint1-TradeIn1
=PurchasePrice+OpMaint1+OpMaint2-TradeIn2
C D
Total Cost =SUMPRODUCT(OnRoute,Cost)23
Solver Parameters
Set Objective Cell: TotalCost To: Min By Changing Variable Cells:
OnRoute
Subject to the Constraints: NetFlow = SupplyDemand
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
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216 Chapter Six Network Optimization Problems
Level Remaining Research Development
Design of Manufacturing System
Initiate Production and Distribution
Normal 5 months — — — Priority 4 months 3 months 5 months 2 months Crash 2 months 2 months 3 months 1 month
Level Remaining Research Development
Design of Manufacturing System
Initiate Production and Distribution
Normal $3 million — — — Priority 6 million $6 million $ 9 million $3 million Crash 9 million 9 million 12 million 6 million
TABLE 6.4 Cost for the Phases of Preparing Quick Co.’s New Product
FIGURE 6.16 Formulation of the Quick Co. problem as a shortest path problem. Except for the dummy destination, the arc labels indicate, first, the number of phases completed and, second, the amount of money left (in millions of dollars) for the remaining phases. Each arc length gives the time (in months) to per- form that phase.
2 (Priority)
2 (Priority)
2 (Priority)
1(Crash)
1(Crash)
2 (Priority)
1(Crash)
5 (Priority)
5 (Priority)
5 (Priority)
3(Crash)
3(Crash)
5 (Priority)
3(Crash)
4, 9
4, 6
4, 3
4, 0
1, 27
1, 240, 30 T
1, 21
4 (Priority)
5
(N or
ma l)
2(Crash)
2, 21
2, 18
2, 15
2, 12
3, 12
3, 9
3, 6
3, 3
3
(Pri orit
y)
3
(Pri orit
y)
2 (Crash)
3
(Pri orit
y)
2 (Crash)
2 (Crash)
(Destination)(Origin)
0
0
0
0
TABLE 6.3 Time Required for the Phases of Prepar- ing Quick Co.’s New Product
When real travel through a network can end at more than one node, an arc with length 0 is inserted from each of these nodes to a dummy destination so that the network will have just a single destination.
Since each of the arcs into the dummy destination has length 0, this addition to the network does not affect the total length of a path from the origin to its ending point.
Figure 6.17 displays the spreadsheet model for this problem. Once again, the format is the same as in Figures 6.13 and 6.15 , except now the quantity of concern in column F and the objective cell TotalTime (D32) is time rather than distance or cost. Since Solver has already been run, the changing cells OnRoute (D4:D30) indicate which arcs lie on the path that mini- mizes the total time. Thus, the shortest path is
0, 30 1, 21 2, 15 3, 3 4, 0 T
with a total length of 2 1 3 1 3 1 2 1 0 5 10 months, as given by TotalTime (D32). The resulting plan for the four phases is shown in Table 6.5 . Although this plan does consume the
The objective cell now is TotalTime instead of TotalDistance.
Phase Level Time Cost Remaining research Crash 2 months $ 9 million Development Priority 3 months 6 million Design of manufacturing system Crash 3 months 12 million Initiate production and distribution Priority 2 months 3 million
Total 10 months $30 million
TABLE 6.5 The Optimal Solution Obtained by Solver for Quick Co.’s Shortest Path Problem
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6.4 Shortest Path Problems 217
FIGURE 6.17 A spreadsheet model that formulates the Quick Co. problem as a shortest path problem where the objective is to minimize the total time instead of the total distance, so the objective cell is TotalTime (D32). The other output cells are NetFlow (I4:I20). The values of 1 in the changing cells OnRoute (D4:D30) reveal the shortest (quickest) path obtained by Solver.
1
A B C D E F G H I
2
3
4
5
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Quick Co. Product Development Scheduling Problem
From
(0, 30)
(0, 30)
(0, 30)
(1, 27)
(1, 27)
Nodes
(0, 30)
(1, 27)
(1, 24)
(1, 21)
(2, 21)
(2, 18)
(2, 15)
(2, 12)
(3, 12)
(3, 9)
(3, 6)
(3, 3)
(4, 9)
(4, 6)
(4, 3)
(4, 0)
(T)
Net Flow
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-1
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
Supply/Demand
J K
(1, 24)
(1, 24)
(1, 21)
(1, 21)
(1, 27)
(1, 24)
(1, 21)
(2, 21)
(2, 18)
(2, 18)
(2, 15)
(2, 15)
(2, 12)
To On Route Time
0
0
0
0
1
0
1
0
0
(2, 21)
(2, 21)
(2, 18)
(2, 18)
(2, 15)
(2, 15)
(2, 12)
(3, 12)
(3, 9)
(3, 9)
(3, 6)
(3, 6)
(3, 3)
(3, 3)
0
0
1
0
0
0
0
(3, 12)
(3, 12)
(3, 9)
(3, 9)
(3, 6)
(3, 6)
(3, 3)
(4, 9)
(4, 6)
(4, 6)
(4, 3)
(4, 3)
(4, 0)
(4, 0)
0
0
0
1
(4, 9)
(4, 6)
(4, 3)
(4, 0)
(T)
(T)
(T)
(T)
0
0
0
1
0
0
0
Total Time 10
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
4
5
3
2
3
2
2
3
2
3
5
3
5
5
3
5
1
2
1
2
2
1
2
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-1
(continued)
entire budget of $30 million, it reduces the time until the product can be brought to market from the originally planned 20 months down to just 10 months.
Given this information, Quick’s management now must decide whether this plan provides the best trade-off between time and cost. What would be the effect on total time of spending a few million more dollars? What would be the effect of reducing the spending somewhat instead? It is easy to provide management with this information as well by quickly solving some shortest path problems that correspond to budgets different from $30 million. The ulti- mate decision regarding which plan provides the best time–cost trade-off then is a judgment decision that only management can make.
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218 Chapter Six Network Optimization Problems
Range Name
From
NetFlow
Nodes
OnRoute
SupplyDemand
Time
To
TotalTime
Cells
B4:B30
I4:I20
H4:H20
D4:D30
K4:K20
F4:F30
C4:C30
D32
3
I
4
6
5
Net Flow
=SUMIF(From,H4,OnRoute)-SUMIF(To,H4,OnRoute)
=SUMIF(From,H6,OnRoute)-SUMIF(To,H6,OnRoute)
=SUMIF(From,H5,OnRoute)-SUMIF(To,H5,OnRoute)
8
7
=SUMIF(From,H8,OnRoute)-SUMIF(To,H8,OnRoute)
=SUMIF(From,H7, OnRoute)-SUMIF(To,H7,OnRoute)
10
9
=SUMIF(From,H10,OnRoute)-SUMIF(To,H10,OnRoute)
=SUMIF(From,H9,OnRoute)-SUMIF(To,H9,OnRoute)
12
11
=SUMIF(From,H12,OnRoute)-SUMIF(To,H12,OnRoute)
=SUMIF(From,H11,OnRoute)-SUMIF(To,H11,OnRoute)
13 =SUMIF(From,H13,OnRoute)-SUMIF(To,H13,OnRoute)
15
14
=SUMIF(From,H15,OnRoute)-SUMIF(To,H15,OnRoute)
=SUMIF(From,H14,OnRoute)-SUMIF(To,H14,OnRoute)
17
16
=SUMIF(From,H17,OnRoute)-SUMIF(To,H17,OnRoute)
=SUMIF(From,H16,OnRoute)-SUMIF(To,H16,OnRoute)
19
18
=SUMIF(From,H19,OnRoute)-SUMIF(To,H19,OnRoute)
=SUMIF(From,H18,OnRoute)-SUMIF(To,H18,OnRoute)
20 =SUMIF(From,H20,OnRoute)-SUMIF(To,H20,OnRoute)
C D
Total Time =SUMPRODUCT(OnRoute,Time)32
Solver Parameters
Set Objective Cell: TotalTime To: Min By Changing Variable Cells:
OnRoute
Subject to the Constraints: NetFlow = SupplyDemand
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
FIGURE 6.17 (continued)
Networks of some type arise in a wide variety of contexts. Network representations are very useful for portraying the relationships and connections between the components of systems. Each component is represented by a point in the network called a node, and then the connections between components (nodes) are represented by lines called arcs (for one-way travel) or links (for two-way travel).
Frequently, a flow of some type must be sent through a network, so a decision needs to be made about the best way to do this. The kinds of network optimization models introduced in this chapter pro- vide a powerful tool for making such decisions.
The model for minimum-cost flow problems plays a central role among these network optimization models, both because it is so broadly applicable and because it can be readily solved. Solver solves spreadsheet formulations of reasonable size, and the network simplex method can be used to solve larger problems, including huge problems with tens of thousands of nodes and arcs. A minimum-cost flow problem typically is concerned with optimizing the flow of goods through a network from their points of origin (the supply nodes ) to where they are needed (the demand nodes ). The objective is to minimize the total cost of sending the available supply through the network to satisfy the given demand. One typical application (among several) is to optimize the operation of a distribution network.
Special types of minimum-cost flow problems include transportation problems and assignment prob- lems (discussed in Chapter 3) as well as two prominent types introduced in this chapter: maximum flow problems and shortest path problems.
6.5 Summary
1. What are the origin and the destination in the Littletown Fire Department example? 2. What is the distinction between an arc and a link? 3. What are the supply node and the demand node when a shortest path problem is interpreted
as a minimum-cost flow problem? With what supply and demand? 4. What are three measures of the length of a link (or arc) that lead to three categories of applica-
tions of shortest path problems? 5. What is the objective for Sarah’s shortest path problem? 6. When does a dummy destination need to be added to the formulation of a shortest path problem? 7. What kind of trade-off does the management of the Quick Co. need to consider in making its
final decision about how to expedite its new product to market?
Review Questions
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Chapter 6 Solved Problems 219
Given the limited capacities of the arcs in the network, the objective of a maximum flow problem is to maximize the total amount of flow from a particular point of origin (the source ) to a particular termi- nal point (the sink ). For example, this might involve maximizing the flow of goods through a company’s supply network from its vendors to its processing facilities.
A shortest path problem also has a beginning point (the origin ) and an ending point (the destination ), but now the objective is to find a path from the origin to the destination that has the minimum total length. For some applications, length refers to distance, so the objective is to minimize the total distance traveled. However, some applications instead involve minimizing either the total cost or the total time of a sequence of activities.
Glossary arc A channel through which flow may occur from one node to another, shown as an arrow between the nodes pointing in the direction in which flow is allowed. (Section 6.1), 197 capacity of an arc The maximum amount of flow allowed through the arc. (Section 6.1), 197 conservation of flow Having the amount of flow out of a node equal the amount of flow into that node. (Section 6.1), 197 demand node A node where the net amount of flow generated (outflow minus inflow) is a fixed negative number, so that flow is absorbed there. (Section 6.1), 197 destination The node at which travel through the network is assumed to end for a shortest path problem. (Section 6.4), 212 dummy destination A fictitious destination introduced into the formulation of a shortest path problem with multiple possible termination points to satisfy the requirement that there be just a single destination. (Section 6.4), 216 length of a link or arc The number (typically a distance, a cost, or a time) associated with includ- ing the link or arc in the selected path for a short- est path problem. (Section 6.4), 212 link A channel through which flow may occur in either direction between a pair of
nodes, shown as a line between the nodes. (Section 6.4), 210 network simplex method A streamlined version of the simplex method for solving minimum-cost flow problems very efficiently. (Section 6.1), 200 node A junction point of a network, shown as a labeled circle. (Section 6.1), 197 origin The node at which travel through the network is assumed to start for a shortest path problem. (Section 6.4), 212 sink The node for a maximum flow problem at which all flow through the network terminates. (Section 6.3), 205 source The node for a maximum flow problem at which all flow through the network originates. (Section 6.3), 205 supply node A node where the net amount of flow generated (outflow minus inflow) is a fixed positive number. (Section 6.1), 197 transshipment node A node where the amount of flow out equals the amount of flow in. (Section 6.1), 197 transshipment problem A special type of minimum-cost flow problem where there are no capacity constraints on the arcs. (Section 6.1), 201
Chapter 6 Excel Files:
Distribution Unlimited Example
BMZ Example
Expanded BMZ Example
Littletown Fire Department Example
Sarah Example
Quick Example
Supplement to Chapter 6 on the CD-ROM:
Minimum Spanning-Tree Problems
Learning Aids for This Chapter in Your MS Courseware
Solved Problems (See the CD-ROM or Website for the Solutions) 6.S1. Distribution at Heart Beats Heart Beats is a manufacturer of medical equipment. The com- pany’s primary product is a device used to monitor the heart during medical procedures. This device is produced in two
factories and shipped to two warehouses. The product is then shipped on demand to four third-party wholesalers. All ship- ping is done by truck. The product distribution network is shown below. The annual production capacity at factories 1 and
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220 Chapter Six Network Optimization Problems
6.S2. Assessing the Capacity of a Pipeline Network Exxo 76 is an oil company that operates the pipeline network shown below, where each pipeline is labeled with its maximum flow rate in million cubic feet (MMcf) per day. A new oil well has been constructed near A. They would like to transport oil from the well near A to their refinery at G. Formulate and solve a network optimization model to determine the maximum flow rate from A to G.
A
B
D
F
E
G
C
4
6
16 15
14
12
15
20
20
2
8
23
6.S3. Driving to the Mile-High City Sarah and Jennifer have just graduated from college at the Uni- versity of Washington in Seattle and want to go on a road trip. They have always wanted to see the mile-high city of Denver. Their road atlas shows the driving time (in hours) between vari- ous city pairs, as shown below. Formulate and solve a network optimization model to find the quickest route from Seattle to Denver.
2 is 400 and 250, respectively. The annual demand at wholesal- ers 1, 2, 3, and 4 is 200, 100, 150, and 200, respectively. The cost of shipping one unit in each shipping lane is shown on the arcs. Because of limited truck capacity, at most 250 units can be shipped from Factory 1 to Warehouse 1 each year. Formu- late and solve a network optimization model in a spreadsheet to determine how to distribute the product at the lowest possible annual cost.
Problems We have inserted the symbol E* to the left of each problem (or its parts) where Excel should be used (unless your instruc- tor gives you contrary instructions). An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
6.1. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 6.1. Briefly describe how the model for a special type of minimum-cost flow problem was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 6.2.* Consider the transportation problem having the follow- ing data.
Destination
1 2 3 Supply Source
1 6 7 4 40 2 5 8 6 60
Demand 30 40 30
a. Formulate a network model for this problem as a minimum-cost flow problem by drawing a network similar to Figure 6.3 .
E* b. Formulate and solve a spreadsheet model for this problem.
[400]
[−200]
[−100]
[−150]
[−200]
[250]
[250]
$25
$35
$40
$60
$35
$55
$50
$65
F1
F2
WH 1
WH 2
WS 2
WS
WS 2
3
WS 4
Butte
Seattle
Billings 3
Portland
Boise
Salt Lake
City Grand
Junction
Denver4
1
7
4
12
14
7
7
10
7
6
9
5
7 Cheyenne
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Chapter 6 Problems 221
6.3. The Makonsel Company is a fully integrated company that both produces goods and sells them at its retail outlets. After production, the goods are stored in the company’s two warehouses until needed by the retail outlets. Trucks are used to transport the goods from the two plants to the warehouses, and then from the warehouses to the three retail outlets.
Using units of full truckloads, the first table below shows each plant’s monthly output, its shipping cost per truckload sent to each warehouse, and the maximum amount that it can ship per month to each warehouse.
For each retail outlet (RO), the second table below shows its monthly demand, its shipping cost per truckload from each warehouse, and the maximum amount that can be shipped per month from each warehouse.
Management now wants to determine a distribution plan (number of truckloads shipped per month from each plant to each warehouse and from each warehouse to each retail outlet) that will minimize the total shipping cost. a. Draw a network that depicts the company’s distri-
bution network. Identify the supply nodes, trans- shipment nodes, and demand nodes in this network.
b. Formulate a network model for this problem as a minimum-cost flow problem by inserting all the necessary data into the network drawn in part a. (Use the format depicted in Figure 6.3 to display these data.)
E* c. Formulate and solve a spreadsheet model for this problem.
6.4. The Audiofile Company produces boomboxes. How- ever, management has decided to subcontract out the production of the speakers needed for the boomboxes. Three vendors are available to supply the speakers. Their price for each shipment of 1,000 speakers is shown below.
Vendor Price
1 $22,500 2 22,700 3 22,300
Each shipment would go to one of the company’s two ware- houses. In addition to the price for each shipment, each vendor would charge a shipping cost for which it has its own formula
based on the mileage to the warehouse. These formulas and the mileage data are shown below.
Vendor Charge
per Shipment Warehouse 1 Warehouse 2
1 $300 1 40¢/mile 1,600 miles 400 miles 2 $200 1 50¢/mile 500 miles 600 miles 3 $500 1 20¢/mile 2,000 miles 1,000 miles
Whenever one of the company’s two factories needs a ship- ment of speakers to assemble into the boomboxes, the company hires a trucker to bring the shipment in from one of the ware- houses. The cost per shipment is given next, along with the num- ber of shipments needed per month at each factory.
Unit Shipping Cost
Factory 1 Factory 2
Warehouse 1 $200 $700 Warehouse 2 400 500
Monthly demand 10 6
Each vendor is able to supply as many as 10 shipments per month. However, because of shipping limitations, each vendor is only able to send a maximum of six shipments per month to each warehouse. Similarly, each warehouse is only able to send a maximum of six shipments per month to each factory.
Management now wants to develop a plan for each month regarding how many shipments (if any) to order from each ven- dor, how many of those shipments should go to each warehouse, and then how many shipments each warehouse should send to each factory. The objective is to minimize the sum of the pur- chase costs (including the shipping charge) and the shipping costs from the warehouses to the factories. a. Draw a network that depicts the company’s supply
network. Identify the supply nodes, transshipment nodes, and demand nodes in this network.
b. This problem is only a variant of a minimum-cost flow problem because the supply from each vendor is a maximum of 10 rather than a fixed amount of 10. However, it can be converted to a full-fledged minimum-cost flow problem by adding a dummy demand node that receives (at zero cost) all the
Unit Shipping Cost Shipping Capacity
To
From RO1 RO2 RO3 RO1 RO2 RO3
Warehouse 1 $470 $505 $490 100 150 100 Warehouse 2 390 410 440 125 150 75
Demand 150 200 150 150 200 150
Unit Shipping Cost Shipping Capacity
To
From Warehouse 1 Warehouse 2 Warehouse 1 Warehouse 2 Output
Plant 1 $425 $560 125 150 200 Plant 2 510 600 175 200 300
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222 Chapter Six Network Optimization Problems
6.6. Reconsider Problem 6.5. Suppose now that, for admin- istrative convenience, management has decided that all 130 units per month needed at the distribution center in Los Angeles must come from the Stuttgart factory (node ST) and all 50 units per month needed at the distribution center in Seattle must come from the Berlin factory (node BE). For each of these distribution centers, Karl Schmidt wants to determine the shipping plan that will minimize the total shipping cost. a. For the distribution center in Los Angeles, for-
mulate a network model for this problem as a minimum-cost flow problem by inserting all the necessary data into the distribution network shown in Figure 6.6 . (Use the format depicted in Figure 6.3 to display these data.)
E* b. Formulate and solve a spreadsheet model for the problem formulated in part a.
c. For the distribution center in Seattle, draw its distri- bution network emanating from the Berlin factory at node BE.
d. Repeat part a for the distribution center in Seattle by using the network drawn in part c.
E* e. Formulate and solve a spreadsheet model for the problem formulated in part d.
f. Add the total shipping costs obtained in parts b and e. Compare this sum with the total shipping cost obtained in part c of Problem 6.5 (as given in the back of the book).
6.7. Consider the maximum flow problem formulated in Figures 6.7 and 6.8 for the BMZ case study. Redraw Figure 6.7 and insert the optimal shipping quantities (cells
unused supply capacity at the vendors. Formu- late a network model for this minimum-cost flow problem by inserting all the necessary data into the network drawn in part a supplemented by this dummy demand node. (Use the format depicted in Figure 6.3 to display these data.)
E* c. Formulate and solve a spreadsheet model for the company’s problem.
6.5.* Consider Figure 6.9 (in Section 6.3), which depicts the BMZ Co. distribution network from its factories in Stuttgart and Berlin to the distribution centers in both Los Angeles and Seat- tle. This figure also gives in brackets the maximum amount that can be shipped through each shipping lane.
In the weeks following the crisis described in Section 6.2, the distribution center in Los Angeles has successfully replen- ished its inventory. Therefore, Karl Schmidt (the supply chain manager for the BMZ Co.) has concluded that it will be suf- ficient hereafter to ship 130 units per month to Los Angeles and 50 units per month to Seattle. (One unit is a hundred cubic meters of automobile replacement parts.) The Stuttgart factory (node ST in the figure) will allocate 130 units per month and the Berlin factory (node BE) will allocate 50 units per month out of their total production to cover these shipments. However, rather than resuming the past practice of supplying the Los Angeles distribution center from only the Stuttgart factory and supplying the Seattle distribution center from only the Berlin factory, Karl has decided to allow either factory to supply either distribution center. He feels that this additional flexibility is likely to reduce the total shipping cost.
The following table gives the shipping cost per unit through each of these shipping lanes.
Unit Shipping Cost to Node
To
From LI BO RO HA NO NY BN LA SE
Node
ST $3,200 $2,500 $2,900 — — — — — — BE — — $2,400 $2,000 — — — — — LI — — — — $6,100 — — — — BO — — — — $6,800 $5,400 — — — RO — — — — — $5,900 — — — HA — — — — — $6,300 $5,700 — — NO — — — — — — — $3,100 — NY — — — — — — — $4,200 $4,000 BN — — — — — — — $3,400 $3,000
Karl wants to determine the shipping plan that will minimize the total shipping cost. a. Formulate a network model for this problem as a
minimum-cost flow problem by inserting all the necessary data into the distribution network shown in Figure 6.9 . (Use the format depicted in Figure 6.3 to display these data.)
E* b. Formulate and solve a spreadsheet model for this problem.
c. What is the total shipping cost for this optimal solution?
D4:D12 in Figure 6.8 ) in parentheses above the respective arcs. Examine the capacities of these arcs. Explain why these arc capacities ensure that the shipping quantities in parenthe- ses must be an optimal solution because the maximum flow cannot exceed 150.
6.8. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 6.3. Briefly describe how a generalization of the model for the maximum flow problem was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study.
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Chapter 6 Problems 223
E*6.9. Formulate and solve a spreadsheet model for the maxi- mum flow problem shown at the top of the next column, where node A is the source, node F is the sink, and the arc capacities are the numbers in square brackets shown next to the arcs.
B
A F
C
D
E
[7]
[9] [2]
[4]
[6]
[7]
[6]
[3]
[9]
6.10. The diagram depicts a system of aqueducts that origi- nate at three rivers (nodes R1, R2, and R3) and terminate at a major city (node T), where the other nodes are junction points in the system.
R1
A
B
C F
E
T
D
R3
R2
Using units of thousands of acre feet, the following tables show the maximum amount of water that can be pumped through each aqueduct per day.
To
From A B C
R1 75 65 — R2 40 50 60 R3 — 80 70
To To
From D E F From T
A 60 45 — D 120 B 70 55 45 E 190 C — 70 90 F 130
The city water manager wants to determine a flow plan that will maximize the flow of water to the city. a. Formulate this problem as a maximum flow prob-
lem by identifying a source, a sink, and the trans- shipment nodes, and then drawing the complete network that shows the capacity of each arc.
E* b. Formulate and solve a spreadsheet model for this problem.
6.11. The Texago Corporation has four oil fields, four refin- eries, and four distribution centers in the locations identified in the next tables. A major strike involving the transportation
industries now has sharply curtailed Texago’s capacity to ship oil from the four oil fields to the four refineries and to ship petro- leum products from the refineries to the distribution centers. Using units of thousands of barrels of crude oil (and its equiva- lent in refined products), the following tables show the maxi- mum number of units that can be shipped per day from each oil field to each refinery and from each refinery to each distribution center.
Refinery
Oil Field New
Orleans Charleston Seattle St.
Louis
Texas 11 7 2 8 California 5 4 8 7 Alaska 7 3 12 6 Middle East 8 9 4 15
Distribution Center
Refinery Pittsburgh Atlanta Kansas
City San
Francisco
New Orleans 5 9 6 4 Charleston 8 7 9 5 Seattle 4 6 7 8 St. Louis 12 11 9 7
The Texago management now wants to determine a plan for how many units to ship from each oil field to each refinery and from each refinery to each distribution center that will maximize the total number of units reaching the distribution centers. a. Draw a rough map that shows the location of Tex-
ago’s oil fields, refineries, and distribution cen- ters. Add arrows to show the flow of crude oil and then petroleum products through this distribution network.
b. Redraw this distribution network by lining up all the nodes representing oil fields in one column, all the nodes representing refineries in a second column, and all the nodes representing distribution centers in a third column. Then add arcs to show the possible flow.
c. Use the distribution network from part b to formu- late a network model for Texago’s problem as a variant of a maximum flow problem.
E* d. Formulate and solve a spreadsheet model for this problem.
6.12. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 6.4. Briefly describe how network opti- mization models (including for shortest path problems) were applied in this study. Then list the various financial and nonfi- nancial benefits that resulted from this study. E*6.13. Reconsider the Littletown Fire Department problem presented in Section 6.4 and depicted in Figure 6.11 . Due to maintenance work on the one-mile road between nodes A and B, a detour currently must be taken that extends the trip between these nodes to four miles.
Formulate and solve a spreadsheet model for this revised problem to find the new shortest path from the fire station to the farming community.
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224 Chapter Six Network Optimization Problems
j
1 2 3
i 0 $8,000 $18,000 $31,000 1 10,000 21,000 2 12,000
Management wishes to determine at what times (if any) the trac- tor should be replaced to minimize the total cost for the tractor(s) over three years. a. Formulate a network model for this problem as a
shortest path problem. E* b. Formulate and solve a spreadsheet model for this
problem. 6.16. One of Speedy Airlines’s flights is about to take off from Seattle for a nonstop flight to London. There is some flex- ibility in choosing the precise route to be taken, depending upon weather conditions. The following network depicts the possible routes under consideration, where SE and LN are Seattle and London, respectively, and the other nodes represent various intermediate locations.
A
SE B
D
E 4.7
4. 6
3.4
3.6
3.2
3.5
3.4
C F
3.3
3.5
3.4
3. 8
4.2
LN 3.6
The winds along each arc greatly affect the flying time (and so the fuel consumption). Based on current meteorologi- cal reports, the flying times (in hours) for this particular flight are shown next to the arcs. Because the fuel consumed is so expensive, the management of Speedy Airlines has established a policy of choosing the route that minimizes the total flight time. a. What plays the role of distances in interpreting this
problem to be a shortest path problem? E* b. Formulate and solve a spreadsheet model for this
problem.
6.14. You need to take a trip by car to another town that you have never visited before. Therefore, you are studying a map to determine the shortest route to your destination. Depending on which route you choose, there are five other towns (call them A, B, C, D, E) through which you might pass on the way. The map shows the mileage along each road that directly connects two towns without any intervening towns. These numbers are sum- marized in the following table, where a dash indicates that there is no road directly connecting these two towns without going through any other towns.
Miles between Adjacent Towns
Town A B C D E Destination
Origin 40 60 50 — — — A 10 — 70 — — B 20 55 40 — C — 50 — D 10 60 E 80
a. Formulate a network model for this problem as a shortest path problem by drawing a network where nodes represent towns, links represent roads, and numbers indicate the length of each link in miles.
E* b. Formulate and solve a spreadsheet model for this problem.
c. Use part b to identify your shortest route. d. If each number in the table represented your cost
(in dollars) for driving your car from one town to the next, would the answer in part c now give your minimum-cost route?
e. If each number in the table represented your time (in minutes) for driving your car from one town to the next, would the answer in part c now give your minimum-time route?
6.15.* At a small but growing airport, the local airline com- pany is purchasing a new tractor for a tractor-trailer train to bring luggage to and from the airplanes. A new mechanized luggage system will be installed in three years, so the tractor will not be needed after that. However, because it will receive heavy use, so that the running and maintenance costs will increase rapidly as it ages, it may still be more economical to replace the trac- tor after one or two years. The next table gives the total net discounted cost associated with purchasing a tractor (purchase price minus trade-in allowance, plus running and maintenance costs) at the end of year i and trading it in at the end of year j (where year 0 is now).
Case 6-1
Aiding Allies Commander Votachev steps into the cold October night and deeply inhales the smoke from his cigarette, savoring its warmth. He surveys the destruction surrounding him— shattered windows, burning buildings, torn roads—and smiles.
His two years of work training revolutionaries east of the Ural Mountains has proven successful; his troops now occupy seven strategically important cities in the Russian Federa- tion: Kazan, Perm, Yekaterinburg, Ufa, Samara, Saratov, and
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Case 6-1 Aiding Allies 225
Orenburg. His siege is not yet over, however. He looks to the west. Given the political and economic confusion in the Rus- sian Federation at this time, he knows that his troops will be able to conquer Saint Petersburg and Moscow shortly. Com- mander Votachev will then be able to rule with the wisdom and control exhibited by his communist predecessors Lenin and Stalin.
Across the Pacific Ocean, a meeting of the top security and foreign policy advisors of the United States is in progress at the White House. The president has recently been briefed about the communist revolution masterminded by Commander Votachev and is determining a plan of action. The president reflects upon a similar October long ago in 1917, and he fears the possibility of a new age of radical Communist rule accom- panied by chaos, bloodshed, escalating tensions, and possibly nuclear war. He therefore decides that the United States needs to respond and to respond quickly. Moscow has requested assis- tance from the United States military, and the president plans to send troops and supplies immediately.
The president turns to General Lankletter and asks him to describe the preparations being taken in the United States to send the necessary troops and supplies to the Russian Federation.
General Lankletter informs the president that along with troops, weapons, ammunition, fuel, and supplies, aircraft, ships, and vehicles are being assembled at two port cities with airfields: Boston and Jacksonville. The aircraft and ships will transfer all troops and cargo across the Atlantic Ocean to the Eurasian continent. The general hands the president a list of the types of aircraft, ships, and vehicles being assembled along with a description of each type. The list is shown next.
Transportation Type Name Capacity Speed
Aircraft C-141 Starlifter 150 tons 400 miles per hour Ship Transport 240 tons 35 miles per hour Vehicle Palletized Load 16,000 kilograms 60 miles per hour
System Truck
All aircraft, ships, and vehicles are able to carry both troops and cargo. Once an aircraft or ship arrives in Europe, it stays there to support the armed forces.
The president then turns to Tabitha Neal, who has been nego- tiating with the NATO countries for the last several hours to use their ports and airfields as stops to refuel and resupply before heading to the Russian Federation. She informs the president that the following ports and airfields in the NATO countries will be made available to the U.S. military.
Ports Airfields
Napoli London Hamburg Berlin Rotterdam Istanbul
The president stands and walks to the map of the world pro- jected on a large screen in the middle of the room. He maps the progress of troops and cargo from the United States to three strategic cities in the Russian Federation that have not yet been seized by Commander Votachev. The three cities are Saint Petersburg, Moscow, and Rostov. He explains that the troops and cargo will be used both to defend the Russian cities and to
launch a counter attack against Votachev to recapture the cities he currently occupies. (The map is shown at the end of the case.)
The president also explains that all Starlifters and transports leave Boston or Jacksonville. All transports that have traveled across the Atlantic must dock at one of the NATO ports to unload. Palletized load system trucks brought over in the trans- ports will then carry all troops and materials unloaded from the ships at the NATO ports to the three strategic Russian cities not yet seized by Votachev. All Starlifters that have traveled across the Atlantic must land at one of the NATO airfields for refuel- ing. The planes will then carry all troops and cargo from the NATO airfields to the three Russian cities.
a. Draw a network showing the different routes troops and supplies may take to reach the Russian Federation from the United States.
b. Moscow and Washington do not know when Commander Votachev will launch his next attack. Leaders from the two countries therefore have agreed that troops should reach each of the three strategic Russian cities as quickly as possible. The president has determined that the situation is so dire that cost is no object—as many Starlifters, transports, and trucks as are necessary will be used to transfer troops and cargo from the United States to Saint Petersburg, Moscow, and Rostov. Therefore, no limitations exist on the number of troops and amount of cargo that can be transferred between any cities.
The president has been given the information in the next table about the length of the available routes between cities.
Given the distance and the speed of the transportation used between each pair of cities, how can the president most
quickly move troops from the United States to each of the three strategic Russian cities? Highlight the path(s) on the network. How long will it take troops and supplies to reach Saint Petersburg? Moscow? Rostov?
c. The president encounters only one problem with his first plan: He has to sell the military deployment to Congress. Under the War Powers Act, the president is required to con- sult with Congress before introducing troops into hostilities or situations where hostilities will occur. If Congress does not give authorization to the president for such use of troops, the president must withdraw troops after 60 days. Congress also has the power to decrease the 60-day time period by passing a concurrent resolution.
The president knows that Congress will not authorize sig- nificant spending for another country’s war, especially when voters have paid so much attention to decreasing the national debt. He therefore decides that he needs to find a way to get the needed troops and supplies to Saint Petersburg, Moscow, and Rostov at the minimum cost.
Each Russian city has contacted Washington to commu- nicate the number of troops and supplies the city needs at
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226 Chapter Six Network Optimization Problems
used between the United States and the Russian Federation on the network.
d. Once the president releases the number of planes, ships, and trucks that will travel between the United States and the Russian Federation, Tabitha Neal contacts each of the American cities and NATO countries to indicate the number of planes to expect at the airfields, the number of ships to expect at the docks, and the number of trucks to expect traveling across the roads. Unfortunately, Tabitha learns that several additional restrictions exist that cannot be immediately eliminated. Because of air- field congestion and unalterable flight schedules, only a limited number of planes may be sent between any two cities. These plane limitations are given below.
From To Maximum Number of
Airplanes
Boston Berlin 300 Boston Istanbul 500 Boston London 500 Jacksonville Berlin 500 Jacksonville Istanbul 700 Jacksonville London 600 Berlin Saint Petersburg 500 Istanbul Saint Petersburg 0 London Saint Petersburg 1,000 Berlin Moscow 300 Istanbul Moscow 100 London Moscow 200 Berlin Rostov 0 Istanbul Rostov 900 London Rostov 100
a minimum for reinforcement. After analyzing the requests, General Lankletter has converted the requests from numbers of troops, gallons of gasoline, and so on, to tons of cargo for easier planning. The requirements are listed below.
City Requirements
Saint Petersburg 320,000 tons Moscow 440,000 tons Rostov 240,000 tons
Both in Boston and Jacksonville, there are 500,000 tons of the necessary cargo available. When the United States decides to send a plane, ship, or truck between two cities, several costs occur: fuel costs, labor costs, maintenance costs, and appropriate port or airfield taxes and tariffs. These costs are listed next.
The president faces a number of restrictions when trying to satisfy the requirements. Early winter weather in northern Russia has brought a deep freeze with much snow. Therefore, General Lankletter is opposed to sending truck convoys in the area. He convinces the president to supply Saint Petersburg only through the air. Moreover, the truck routes into Rostov are quite limited, so that from each port, at most 2,500 trucks can be sent to Rostov. The Ukrainian government is very sen- sitive about American airplanes flying through its air space. It restricts the U.S. military to at most 200 flights from Berlin to Rostov and to at most 200 flights from London to Rostov. (The U.S. military does not want to fly around the Ukraine and is thus restricted by the Ukrainian limitations.)
How does the president satisfy each Russian city’s mili- tary requirements at minimum cost? Highlight the path to be
From To Cost
Boston Berlin $50,000 per Starlifter Boston Hamburg $30,000 per transport Boston Istanbul $55,000 per Starlifter Boston London $45,000 per Starlifter Boston Rotterdam $30,000 per transport Boston Napoli $32,000 per transport Jacksonville Berlin $57,000 per Starlifter Jacksonville Hamburg $48,000 per transport Jacksonville Istanbul $61,000 per Starlifter Jacksonville London $49,000 per Starlifter Jacksonville Rotterdam $44,000 per transport Jacksonville Napoli $56,000 per transport Berlin Saint Petersburg $24,000 per Starlifter Hamburg Saint Petersburg $3,000 per truck Istanbul Saint Petersburg $28,000 per Starlifter London Saint Petersburg $22,000 per Starlifter Rotterdam Saint Petersburg $3,000 per truck Napoli Saint Petersburg $5,000 per truck Berlin Moscow $22,000 per Starlifter Hamburg Moscow $4,000 per truck Istanbul Moscow $25,000 per Starlifter London Moscow $19,000 per Starlifter Rotterdam Moscow $5,000 per truck Napoli Moscow $5,000 per truck Berlin Rostov $23,000 per Starlifter Hamburg Rostov $7,000 per truck Istanbul Rostov $2,000 per Starlifter London Rostov $4,000 per Starlifter Rotterdam Rostov $8,000 per truck Napoli Rostov $9,000 per truck
From To (Kilometers)
Boston Berlin 7,250 km Boston Hamburg 8,250 Boston Istanbul 8,300 Boston London 6,200 Boston Rotterdam 6,900 Boston Napoli 7,950 Jacksonville Berlin 9,200 Jacksonville Hamburg 9,800 Jacksonville Istanbul 10,100 Jacksonville London 7,900 Jacksonville Rotterdam 8,900 Jacksonville Napoli 9,400 Berlin Saint Petersburg 1,280 Hamburg Saint Petersburg 1,880 Istanbul Saint Petersburg 2,040 London Saint Petersburg 1,980 Rotterdam Saint Petersburg 2,200 Napoli Saint Petersburg 2,970 Berlin Moscow 1,600 Hamburg Moscow 2,120 Istanbul Moscow 1,700 London Moscow 2,300 Rotterdam Moscow 2,450 Napoli Moscow 2,890 Berlin Rostov 1,730 Hamburg Rostov 2,470 Istanbul Rostov 990 London Rostov 2,860 Rotterdam Rostov 2,760 Napoli Rostov 2,800
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Case 6-2 Money in Motion 227
In addition, because some countries fear that citizens will become alarmed if too many military trucks travel the pub- lic highways, they object to a large number of trucks traveling through their countries. These objections mean that a limited number of trucks are able to travel between certain ports and Russian cities. These limitations are listed below.
From To Maximum Number of Trucks
Rotterdam Moscow 600 Rotterdam Rostov 750 Hamburg Moscow 700 Hamburg Rostov 500 Napoli Moscow 1,500 Napoli Rostov 1,400
Tabitha learns that all shipping lanes have no capacity limits due to the American control of the Atlantic Ocean.
The president realizes that due to all the restrictions, he will not be able to satisfy all the reinforcement requirements of the three Russian cities. He decides to disregard the cost issue and instead to maximize the total amount of cargo he can get to the Russian cities. How does the president maxi- mize the total amount of cargo that reaches the Russian Fed- eration? Highlight the path(s) used between the United States and the Russian Federation on the network.
Case 6-2
Money in Motion
Jake Nguyen runs a nervous hand through his once finely combed hair. He loosens his once perfectly knotted silk tie. And he rubs his sweaty hands across his once immaculately pressed trousers. Today has certainly not been a good day.
Over the past few months, Jake had heard whispers circu- lating from Wall Street—whispers from the lips of investment bankers and stockbrokers famous for their outspokenness. They had whispered about a coming Japanese economic collapse— whispered because they had believed that publicly vocalizing their fears would hasten the collapse.
And, today, their very fears have come true. Jake and his colleagues gather around a small television dedicated exclu- sively to the Bloomberg channel. Jake stares in disbelief as he listens to the horrors taking place in the Japanese market. And the Japanese market is taking the financial markets in all other East Asian countries with it on its tailspin. He goes numb. As manager of Asian foreign investment for Grant Hill Associates, a small West Coast investment boutique specializing in currency trading, Jake bears personal responsibility for any negative impacts of the collapse. And Grant Hill Associates will experi- ence negative impacts.
Jake had not heeded the whispered warnings of a Japanese collapse. Instead, he had greatly increased the stake Grant Hill
Associates held in the Japanese market. Because the Japanese market had performed better than expected over the past year, Jake had increased investments in Japan from $2.5 million to $15 million only one month ago. At that time, one dollar was worth 80 yen.
No longer. Jake realizes that today’s devaluation of the yen means that one dollar is worth 125 yen. He will be able to liqui- date these investments without any loss in yen, but now the dol- lar loss when converting back into U.S. currency would be huge. He takes a deep breath, closes his eyes, and mentally prepares himself for serious damage control.
Jake’s meditation is interrupted by a booming voice calling for him from a large, corner office. Grant Hill, the president of Grant Hill Associates, yells, “Nguyen, get the hell in here!”
Jake jumps and looks reluctantly toward the corner office hiding the furious Grant Hill. He smooths his hair, tightens his tie, and walks briskly into the office.
Grant Hill meets Jake’s eyes upon his entrance and continues yelling, “I don’t want one word out of you, Nguyen! No excuses; just fix this debacle! Get all of our money out of Japan! My gut tells me this is only the beginning! Get the money into safe U.S. bonds! NOW! And don’t forget to get our cash positions out of Indonesia and Malaysia ASAP with it!”
Yekaterinburg
St. Petersburg
Perm
Ufa Moscow
Boston
Jacksonville
BerlinLondon
Napoli Istanbul
Rostov
Saratov Orenburg
Samara
Kazan
Hamburg
Rotterdam
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228 Chapter Six Network Optimization Problems
Jake has enough common sense to say nothing. He nods his head, turns on his heels, and practically runs out of the office.
Safely back at his desk, Jake begins formulating a plan to move the investments out of Japan, Indonesia, and Malaysia. His experiences investing in foreign markets have taught him that when playing with millions of dollars, how he gets money out of a foreign market is almost as important as when he gets money out of the market. The banking partners of Grant Hill Associates charge different transaction fees for converting one currency into another one and wiring large sums of money around the globe.
And now, to make matters worse, the governments in East Asia have imposed very tight limits on the amount of money an individual or a company can exchange from the domestic cur- rency into a particular foreign currency and withdraw it from the country. The goal of this dramatic measure is to reduce the outflow of foreign investments out of those countries to prevent a complete collapse of the economies in the region. Because of Grant Hill Associates’ cash holdings of 10.5 billion Indonesian rupiahs and 28 million Malaysian ringgits, along with the hold- ings in yen, it is not clear how these holdings should be con- verted back into dollars.
Jake wants to find the most cost-effective method to con- vert these holdings into dollars. On his company’s website, he
always can find on-the-minute exchange rates for most curren- cies in the world (see Table 1 ).
The table states that, for example, 1 Japanese yen equals 0.008 U.S. dollars. By making a few phone calls, he discovers the transaction costs his company must pay for large currency transactions during these critical times (see Table 2 ).
Jake notes that exchanging one currency for another one results in the same transaction cost as a reverse conversion. Finally, Jake finds out the maximum amounts of domestic cur- rencies his company is allowed to convert into other currencies in Japan, Indonesia, and Malaysia (see Table 3 ).
a. Formulate Jake’s problem as a minimum-cost flow problem, and draw the network for his problem. Identify the supply and demand nodes for the network.
b. Which currency transactions must Jake perform to con- vert the investments from yens, rupiahs, and ringgits into U.S. dollars to ensure that Grant Hill Associates has the maximum dollar amount after all transactions have occurred? How much money does Jake have to invest in U.S. bonds?
c. The World Trade Organization forbids transaction limits because they promote protectionism. If no transaction limits
TABLE 1 Currency Exchange Rates
To From Yen Rupiah Ringgit U.S. Dollar Canadian Dollar Euro Pound Peso
Japanese yen 1 50 0.04 0.008 0.01 0.0064 0.0048 0.0768 Indonesian rupiah 1 0.0008 0.00016 0.0002 0.000128 0.000096 0.001536 Malaysian ringgit 1 0.2 0.25 0.16 0.12 1.92 U.S. dollar 1 1.25 0.8 0.6 9.6 Canadian dollar 1 0.64 0.48 7.68 European euro 1 0.75 12 English pound 1 16 Mexican peso 1
TABLE 2 Transaction Cost (Percent)
To From Yen Rupiah Ringgit U.S. Dollar Canadian Dollar Euro Pound Peso
Yen — 0.5 0.5 0.4 0.4 0.4 0.25 0.5 Rupiah — 0.7 0.5 0.3 0.3 0.75 0.75 Ringgit — 0.7 0.7 0.4 0.45 0.5 U.S. dollar — 0.05 0.1 0.1 0.1 Canadian dollar — 0.2 0.1 0.1 Euro — 0.05 0.5 Pound — 0.5 Peso —
TABLE 3 Transaction Limits in Equivalent of 1,000 Dollars
To From Yen Rupiah Ringgit U.S. Dollar Canadian Dollar Euro Pound Peso
Yen — 5,000 5,000 2,000 2,000 2,000 2,000 4,000 Rupiah 5,000 — 2,000 200 200 1,000 500 200 Ringgit 3,000 4,500 — 1,500 1,500 2,500 1,000 1,000
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Case 6-3 Airline Scheduling 229
exist, what method should Jake use to convert the Asian holdings from the respective currencies into dollars?
d. In response to the World Trade Organization’s mandate forbidding transaction limits, the Indonesian government introduces a new tax to protect its currency that leads to a 500 percent increase in transaction costs for transactions of rupiahs. Given these new transaction costs but no transaction
limits, what currency transactions should Jake perform to convert the Asian holdings from the respective currencies into dollars?
e. Jake realizes that his analysis is incomplete because he has not included all aspects that might influence his planned currency exchanges. Describe other factors that Jake should examine before he makes his final decision.
Case 6-3
Airline Scheduling
Richard Cook is very concerned. Until recently, he has always had the golden touch, having successfully launched two start- up companies that made him a very wealthy man. However, the timing could not have been worse for his latest start-up— a regional airline called Northwest Commuter that operates on the west coast of the United States. All had been well at the beginning. Four airplanes had been leased and the company had become fairly well established as a no-frills airline provid- ing low-cost commuter flights between the west coast cities of
(including some new ones) for the coming year that could feasi- bly be flown by the four airplanes.
A little over a decade ago, Richard had been an honor gradu- ate of a leading MBA program. He had enjoyed the management science course he took then and he has decided to apply spread- sheet modeling to analyze his problem.
The leasing cost for each airplane is $30,000 per day. At the end of the day, an airplane might remain in the city where it landed on its last flight. Another option is to fly empty overnight
Flight Number From To Depart Arrive
Expected Revenue ($000)
1257 Seattle San Francisco 8:00 AM 10:00 AM 37 2576 Seattle Portland 9:30 AM 10:30 AM 20 8312 Seattle San Francisco 9:30 AM 11:30 AM 25 1109 Seattle San Francisco 12:00 PM 2:00 PM 27 3752 Seattle San Francisco 2:30 PM 4:30 PM 23 2498 Seattle Portland 3:00 PM 4:00 PM 18 8787 Seattle San Francisco 5:00 PM 7:00 PM 29 8423 Seattle Portland 6:30 PM 7:30 PM 27 7922 Portland Seattle 9:00 AM 10:00 AM 20 5623 Portland San Francisco 9:30 AM 11:00 AM 23 2448 Portland San Francisco 11:00 AM 12:30 PM 19 1842 Portland Seattle 12:00 PM 1:00 PM 21 3487 Portland Seattle 2:00 PM 3:00 PM 22 4361 Portland San Francisco 4:00 PM 5:30 PM 29 4299 Portland Seattle 6:00 PM 7:00 PM 27 1288 San Francisco Seattle 8:00 AM 10:00 AM 32 3335 San Francisco Portland 8:30 AM 10:00 AM 26 9348 San Francisco Seattle 10:30 AM 12:30 PM 24 7400 San Francisco Seattle 12:00 PM 2:00 PM 27 7328 San Francisco Portland 12:00 PM 1:30 PM 24 6386 San Francisco Portland 4:00 PM 5:30 PM 28 6923 San Francisco Seattle 5:00 PM 7:00 PM 32
Seattle, Portland, and San Francisco. Achieving fast turnaround times between flights had given Northwest Commuter an impor- tant competitive advantage. Then the cost of jet fuel began spi- raling upward and the company began going heavily into the red (like so many other airlines at the time). Although some of the flights were still profitable, others were losing a lot of money. Fortunately, jet fuel costs now are starting to come down, but it has become clear to Richard that he needs to find new ways for Northwest Commuter to become a more efficient airline. In particular, he wants to start by dropping unprofitable flights and then identifying the most profitable combination of flights
to another city to be ready to start a flight from there the next morning. The cost of this latter option is $5,000.
The table above shows the 22 possible flights that are being considered for the coming year. The last column gives the esti- mated net revenue (in thousands of dollars) for each flight, given the average number of passengers anticipated for that flight.
a. To simplify the analysis, assume for now that there is vir- tually no turnaround time between flights so the next flight can begin as soon as the current flight ends. (If an immediate next flight is not available, the airplane would wait until the
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230 Chapter Six Network Optimization Problems
next scheduled flight from that city.) Develop a network that displays some of the feasible routings of the flights. ( Hint: Include separate nodes for each half hour between 8:00 am and 7:30 pm in each city.) Then develop and apply the corre- sponding spreadsheet model that finds the feasible combina- tion of flights that maximizes the total profit.
b. Richard is considering leasing additional airplanes to achieve economies of scale. The leasing cost of each one again would be $30,000 per day. Perform what-if analysis to determine whether it would be worthwhile to have 5, 6, or 7 airplanes instead of 4.
c. Now repeat part a under the more realistic assumption that there is a minimum turnaround time of 30 minutes on the
ground for unloading and loading passengers between the arrival of a flight and the departure of the next flight by the same airplane. (Most airlines use a considerably longer turn- around time.) Does this change the number of flights that can be flown?
d. Richard now is considering having each of the four airplanes carry freight instead of flying empty if it flies overnight to another city. Instead of a cost of $5,000, this would result in net revenue of $5,000. Adapt the spreadsheet model used in part c to find the feasible combination of flights that maxi- mizes the total profit. Does this change the number of air- planes that fly overnight to another city?
Case 6-4
Broadcasting the Olympic Games
The management of the WBC television network has been celebrating for days. What a coup! After several unsuccessful attempts in recent decades, they finally have hit the big jackpot. They have won the bidding war to gain the rights to broadcast the next Summer Olympic Games!
The price was enormous. However, the advertising income also will be huge. Even if the network loses some money in the process, the gain in prestige should make it all worthwhile. After all, the entire world follows these games closely every four years. Now the entire world receiving the feed of the broadcast from the WBC network will learn what a preeminent network it is.
each link in the network is shown in the diagram below (in GB/s). WBC can divide the transmission and route it through multiple paths of the network from A to G, so long as the total bandwidth required on each link does not exceed the capacity of that link.
a. By utilizing the entire computer network, what is the maxi- mum bandwidth available (in GB/s) for transmission from the general site of the Olympic Games (node A) to the home studios (node G)? Set up and solve a linear programming spreadsheet model.
A
B
C
D
E
F
G
12
9
7
5
10
9
8
13
6
3
4
6
However, reality also is setting in for WBC management. Telecasting the entire Olympic Games will be an enormously complex task. Many different sporting events will be occurring simultaneously in far-flung venues. An unprecedented amount of live television and live-on-the-Internet coverage of the vari- ous sporting events needs to be planned.
Due to the high amount of bandwidth that will be required to transmit the coverage of the games back to its home studios, WBC needs to upgrade its computer network. It operates a pri- vate computer network as shown in the network diagram in the right-hand column. The games will be held near node A. WBC’s home studios are located at node G. At peak times, coverage of the games will require 35 GB/s (GB per second) to be sent through the network from node A to node G. The capacity of
b. WBC would like to expand the capacity of the network so it can handle the peak requirement of 35 GB/s from the Olym- pics site (A) to the home studios (G). WBC can increase the capacity of each link of the computer network by install- ing additional fiber optic cables. The table on the next page shows the existing capacity of each network segment (in GB/s), the maximum additional capacity that can be added (in GB/s), and the cost to increase the capacity (in millions of dollars per unit GB/s added). Make a copy of the spreadsheet model used to solve part a and make any revisions necessary to solve this new problem.
Note: This case will be continued in the next chapter (Case 7-4), so we suggest that you save your spreadsheet model from part b.
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Additional Cases 231
Network Segment Existing
Capacity (GB/s) Maximum Additional
Capacity (GB/s)
Cost per GB/s of Additional Capacity
($million)From To
A B 13 6 2.8 A C 6 4 2.5 A D 10 3 2.8 B D 9 4 2.5 B E 5 5 3.1 B F 7 3 1.6 C D 8 5 3.9 D E 3 2 2.8 D G 12 5 1.6 E F 4 2 4.6 E G 6 4 2.9 F G 9 5 1.8
Additional Cases Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
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232
Chapter Seven
Using Binary Integer Programming to Deal with Yes-or-No Decisions Learning Objectives
After completing this chapter, you should be able to
1. Describe how binary decision variables are used to represent yes-or-no decisions.
2. Use binary decision variables to formulate constraints for mutually exclusive alterna- tives and contingent decisions.
3. Formulate a binary integer programming model for the selection of projects.
4. Formulate a binary integer programming model for the selection of sites for facilities.
5. Formulate a binary integer programming model for crew scheduling in the travel industry.
6. Formulate other basic binary integer programming models from a description of the problems.
7. Use mixed binary integer programming to deal with setup costs for initiating the pro- duction of a product.
The preceding chapters have considered various kinds of problems where decisions need to be made about how much to do of various activities. Thus, the decision variables in the result- ing model represent the level of the corresponding activities.
We turn now to a common type of problem where, instead of how-much decisions, the decisions to be made are yes-or-no decisions . A yes-or-no decision arises when a particu- lar option is being considered and the only possible choices are yes, go ahead with this option, or no, decline this option.
The natural choice of a decision variable for a yes-or-no decision is a binary variable. Binary variables are variables whose only possible values are 0 and 1. Thus, when repre- senting a yes-or-no decision, a binary decision variable is assigned a value of 1 for choos- ing yes and a value of 0 for choosing no.
Models that fit linear programming except that they use binary decision variables are called binary integer programming (BIP) models. (We hereafter will use the BIP abbreviation.) A pure BIP model is one where all the variables are binary variables, whereas a mixed BIP model is one where only some of the variables are binary variables.
A BIP model can be considered to be a special type of integer programming model. A gen- eral integer programming model is simply a linear programming model except for also having constraints that some or all of the decision variables must have integer values (0, 1, 2, . . .). A BIP model further restricts these integer values to be only 0 or 1.
However, BIP problems are quite different from general integer programming problems because of the difference in the nature of the decisions involved. Like linear programming problems, general integer programming problems involve how-much decisions, but where these decisions make sense only if they have integer values. For example, the TBA Airlines
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7.1 A Case Study: The California Manufacturing Co. Problem 233
problem presented in Section 3.2 is a general integer programming problem because it is essen- tially a linear programming problem except that its how-much decisions (how many small air- planes and how many large airplanes to purchase) only make sense if they have integer values. By contrast, BIP problems involve yes-or-no decisions instead of how-much decisions.
The preceding chapters already have focused on problems involving how-much decisions and how such techniques as linear programming or integer programming can be used to ana- lyze these problems. Therefore, this chapter will be devoted instead to problems involving yes- or-no decisions and how BIP models can be used to analyze this special category of problems.
BIP problems arise with considerable frequency in a wide variety of applications. To illus- trate this, we begin with a case study and then present some more examples in the subsequent sections. One of the supplements to this chapter on the CD-ROM also provides additional formulation examples for BIP problems.
You will see throughout this chapter that BIP problems can be formulated on a spreadsheet just as readily as linear programming problems. The Solver also can solve BIP problems of modest size. You normally will have no problem solving the small BIP problems found in this book, but Solver may fail on somewhat larger problems. To provide some perspective on this issue, we include another supplement on the CD-ROM that is entitled Some Perspec- tives on Solving Binary Integer Programming Problems. The algorithms available for solving BIP problems (including the one used by Solver) are not nearly as efficient as those for linear programming, so this supplement discusses some of the difficulties and pitfalls involved in solving large BIP problems. One option with any large problem that fits linear programming except that it has decision variables that are restricted to integer values (but not necessarily just 0 and 1) is to ignore the integer constraints and then to round the solution obtained to integer values. This is a reasonable option in some cases but not in others. The supplement emphasizes that this is a particularly dangerous shortcut with BIP problems.
7.1 A CASE STUDY: THE CALIFORNIA MANUFACTURING CO. PROBLEM
The top management of the California Manufacturing Company wants to develop a plan for the expansion of the company. Therefore, a management science study will be conducted to help guide the decisions that need to be made. The president of the company, Armando Ortega, is about to meet with the company’s top management scientist, Steve Chan, to discuss the study that management wants done. Let’s eavesdrop on this meeting.
Armando Ortega (president): OK, Steve, here is the situation. With our growing busi- ness, we are strongly considering building a new factory. Maybe even two. The factory needs to be close to a large, skilled labor force, so we are looking at Los Angeles and San Francisco as the potential sites. We also are considering building one new warehouse. Not more than one. This warehouse would make sense in saving shipping costs only if it is in the same city as a new factory. Either Los Angeles or San Francisco. If we decide not to build a new factory at all, we definitely don’t want the warehouse either. Is this clear, so far? Steve Chan (management scientist): Yes, Armando, I understand, What are your criteria for making these decisions? Armando Ortega: Well, all the other members of top management have joined me in addressing this issue. We have concluded that these two potential sites are very comparable on nonfinancial grounds. Therefore, we feel that these decisions should be based mainly on financial considerations. We have $10 million of capital available for this expansion and we want it to go as far as possible in improving our bottom line. Which feasible combination of investments in factories and warehouses in which locations will be most profitable for the company in the long run? In your language, we want to maximize the total net present value of these investments. Steve Chan: That’s very clear. It sounds like a classical management science problem. Armando Ortega: That’s why I called you in, Steve. I would like you to conduct a quick management science study to determine the most profitable combination of investments. I also would like you to take a look at the amount of capital being made available and its
What is the most profit- able combination of investments?
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234
With headquarters in Houston, Texas, Waste Management, Inc. (a Fortune 100 company), is the leading provider of com- prehensive waste-management services in North America. Its network of operations includes 293 active landfill disposal sites, 16 waste-to-energy plants, 72 landfill gas-to-energy facilities, 146 recycling plants, 346 transfer stations, and 435 collection operations (depots) to provide services to nearly 20 million residential customers and 2 million commercial customers throughout the United States and Canada.
The company’s collection-and-transfer vehicles need to follow nearly 20,000 daily routes. With an annual operating cost of nearly $120,000 per vehicle, management wanted to have a comprehensive route-management system that would make every route as profitable and efficient as possi- ble. Therefore, a management science team that included a number of consultants was formed to attack this problem.
The heart of the route-management system developed by this team is a huge mixed BIP model that optimizes the routes assigned to the respective collection-and-transfer
vehicles. Although the objective function takes several fac- tors into account, the primary goal is the minimization of total travel time. The main decision variables are binary vari- ables that equal 1 if the route assigned to a particular vehicle includes a particular possible leg and that equal 0 otherwise. A geographical information system (GIS) provides the data about the distance and time required to go between any two points. All of this is imbedded within a Web-based Java appli- cation that is integrated with the company’s other systems.
It is estimated that the recent implementation of this comprehensive route-management system will increase the company’s cash flow by $648 million over a five-year period, largely because of savings of $498 million in opera- tional expenses over this same period. It also is providing better customer service.
Source: S. Sahoo, S. Kim, B.-I. Kim, B. Krass, and A. Popov, Jr., “Routing Optimization for Waste Management,” Interfaces 35, no. 1 (January–February 2005), pp. 24–36. (A link to this article is provided on our website, www.mhhe.com/hillier5e . )
An Application Vignette
effect on how much profit we can get from these investments. The decision to make $10 million available is only a tentative one. That amount is stretching us, because we now are investigating some other interesting project proposals that would require quite a bit of capital, so we would prefer to use less than $10 million on these particular investments if the last few million don’t buy us much. On the other hand, this expansion into either Los Ange- les or San Francisco, or maybe both of these key cities, is our number one priority. It will have a real positive impact on the future of this company. So we are willing to go out and raise some more capital if it would give us a lot of bang for the buck. Therefore, we would like you to do some what-if analysis to tell us what the effect would be if we were to change the amount of capital being made available to anything between $5 million and $15 million. Steve Chan: Sure, Armando, we do that kind of what-if analysis all the time. We refer to it as sensitivity analysis because it involves checking how sensitive the outcome is to the amount of capital being made available. Armando Ortega: Good. Now, Steve, I need your input within the next couple weeks. Can you do it? Steve Chan: Well, Armando, as usual, the one question is whether we can gather all the necessary data that quickly. We’ll need to get good estimates of the net present value of each of the possible investments. I’ll need a lot of help in digging out that information. Armando Ortega: I thought you would say that. I already have my staff working hard on developing those estimates. I can get you together with them this afternoon. Steve Chan: Great. I’ll get right on it.
Background The California Manufacturing Company is a diversified company with several facto- ries and warehouses throughout California, but none yet in Los Angeles or San Francisco. Because the company is enjoying increasing sales and earnings, management feels that the time may be ripe to expand into one or both of those prime locations. A basic issue is whether to build a new factory in either Los Angeles or San Francisco, or perhaps even in both cities. Management also is considering building at most one new warehouse, but will restrict the choice of location to a city where a new factory is being built.
The decisions to be made are listed in the second column of Table 7.1 in the form of yes- or-no questions. In each case, giving an answer of yes to the question corresponds to the decision to make the investment to build the indicated facility (a factory or a warehouse) in
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7.1 A Case Study: The California Manufacturing Co. Problem 235
Decision Number
Yes-or-No Question
Decision Variable
Net Present Value (Millions)
Capital Required (Millions)
1 Build a factory in Los Angeles? x1 $8 $6 2 Build a factory in San Francisco? x2 5 3 3 Build a warehouse in Los Angeles? x3 6 5 4 Build a warehouse in San Francisco? x4 4 2
Capital available: $10 million
TABLE 7.1 Data for the California Manufacturing Co. Problem
Decision Number
Decision Variable
Possible Value
Interpretation of a Value of 1
Interpretation of a Value of 0
1 x1 0 or 1 Build a factory in Los Angeles Do not build this factory 2 x2 0 or 1 Build a factory in San Francisco Do not build this factory 3 x3 0 or 1 Build a warehouse in Los Angeles Do not build this warehouse 4 x4 0 or 1 Build a warehouse in San Francisco Do not build this warehouse
TABLE 7.2 Binary Decision Variables for the California Manufacturing Co. Problem
the indicated location (Los Angeles or San Francisco). The capital required for the investment is given in the rightmost column, where management has made the tentative decision that the total amount of capital being made available for all the investments is $10 million. (Note that this amount is inadequate for some of the combinations of investments.) The fourth col- umn shows the estimated net present value (net long-run profit considering the time value of money) if the corresponding investment is made. (The net present value is 0 if the investment is not made.) Much of the work of Steve Chan’s management science study (with substantial help from the president’s staff) goes into developing these estimates of the net present values. As specified by the company’s president, Armando Ortega, the objective now is to find the feasible combination of investments that maximizes the total net present value.
Introducing Binary Decision Variables for the Yes-or-No Decisions As summarized in the second column of Table 7.1 , the problem facing management is to make four interrelated yes-or-no decisions. To formulate a mathematical model for this prob- lem, Steve Chan needs to introduce a decision variable for each of these decisions. Since each decision has just two alternatives, choose yes or choose no, the corresponding decision variable only needs to have two values (one for each alternative). Therefore, Steve uses a binary variable, whose only possible values are 0 and 1, where 1 corresponds to the decision to choose yes and 0 corresponds to choosing no.
These decision variables are shown in the second column of Table 7.2 . The final two col- umns give the interpretation of a value of 1 and 0, respectively.
Dealing with Interrelationships between the Decisions Recall that management wants no more than one new warehouse to be built. In terms of the corresponding decision variables, x 3 and x 4 , this means that no more than one of these vari- ables is allowed to have the value 1. Therefore, these variables must satisfy the constraint
x3 1 x4 # 1
as part of the mathematical model for the problem. These two alternatives (build a warehouse in Los Angeles or build a warehouse in San
Francisco) are referred to as mutually exclusive alternatives because choosing one of these alternatives excludes choosing the other. Groups of two or more mutually exclusive alternatives arise commonly in BIP problems. For each such group where at most one of the alternatives can be chosen, the constraint on the corresponding binary decision variables has the form shown above, namely, the sum of these variables must be less than or equal to 1. For some groups of mutually exclusive alternatives, management will exclude the possibility
With a group of mutually exclusive alternatives, only one of the corresponding binary decision variables can equal 1.
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236 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
of choosing none of the alternatives, in which case the constraint will set the sum of the cor- responding binary decision variables equal to 1.
The California Manufacturing Co. problem also has another important kind of restriction. Management will allow a warehouse to be built in a particular city only if a factory also is being built in that city. For example, consider the situation for Los Angeles (LA).
If decide no, do not build a factory in LA (i.e., if choose x 1 5 0), then cannot build a warehouse in LA (i.e., must choose x 3 5 0).
If decide yes, do build a factory in LA (i.e., if choose x 1 5 1), then can either build a warehouse in LA or not (i.e., can choose either x 3 5 1 or 0).
How can these interrelationships between the factory and warehouse decisions for LA be expressed in a constraint for a mathematical model? The key is to note that, for either value of x 1 , the permissible value or values of x 3 are less than or equal to x 1 . Since x 1 and x 3 are binary variables, the constraint
x3 # x1
forces x 3 to take on a permissible value given the value of x 1 . Exactly the same reasoning leads to
x4 # x2
as the corresponding constraint for San Francisco. Just as for Los Angeles, this constraint forces having no warehouse in San Francisco ( x 4 5 0) if a factory will not be built there ( x 2 5 0), whereas going ahead with the factory there ( x 2 5 1) leaves open the decision to build the warehouse there ( x 4 5 0 or 1).
For either city, the warehouse decision is referred to as a contingent decision , because the decision depends on a prior decision regarding whether to build a factory there. In general, one yes-or-no decision is said to be contingent on another yes-or-no decision if it is allowed to be yes only if the other is yes. As above, the mathematical constraint expressing this relation- ship requires that the binary variable for the former decision must be less than or equal to the binary variable for the latter decision.
The rightmost column of Table 7.1 reveals one more interrelationship between the four decisions, namely, that the amount of capital expended on the four facilities under consid- eration cannot exceed the amount available ($10 million). Therefore, the model needs to include a constraint that requires
Capital expended # $10 million
How can the amount of capital expended be expressed in terms of the four binary decision vari- ables? To start this process, consider the first yes-or-no decision (build a factory in Los Angeles?). Combining the information in the rightmost column of Table 7.1 and the first row of Table 7.2 ,
Capital expended on factory in Los Angeles 5 b$6 million if x1 5 1 0 if x1 5 0
5 $6 million times x1
By the same reasoning, the amount of capital expended on the other three investment oppor- tunities (in units of millions of dollars) is 3 x 2 , 5 x 3 , and 2 x 4 , respectively. Consequently,
Capital expended 5 6x1 1 3x2 1 5x3 1 2x4 (in millions of dollars)
Therefore, the constraint becomes
6x1 1 3x2 1 5x3 1 2x4 # 10
The BIP Model As indicated by Armando Ortega in his conversation with Steve Chan, management’s objec- tive is to find the feasible combination of investments that maximizes the total net present value of these investments. Thus, the value of the objective function should be
NPV 5 Total net present value
One yes-or-no decision is contingent on another yes- or-no decision if the first one is allowed to be yes only if the other one is yes.
Excel Tip: Beware that rounding errors can occur with Excel. Therefore, even when you add a constraint that a changing cell has to be binary, Excel occasion- ally will return a noninteger value very close to an inte- ger (e.g., 1.23E-10, meaning 0.000000000123). When this happens, you can replace the noninteger value with the proper integer value.
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7.1 A Case Study: The California Manufacturing Co. Problem 237
If the investment is made to build a particular facility (so that the corresponding decision variable has a value of 1), the estimated net present value from that investment is given in the fourth column of Table 7.1 . If the investment is not made (so the decision variable equals 0), the net present value is 0. Therefore, continuing to use units of millions of dollars,
NPV 5 8x1 1 5x2 1 6x3 1 4x4
is the quantity to enter into the objective cell to be maximized. Incorporating the constraints developed in the preceding subsection, the complete BIP model
then is shown in Figure 7.1 . The format is basically the same as for linear programming models.
Excel Tip: In the Solver Options, the Integer Opti- mality (%) setting (1 percent by default) causes Solver to stop solving an integer programming problem when it finds a feasible solution whose objective function value is within the specified percentage of being opti- mal. (In RSPE, this option is called Integer Tolerance and is found in the Engine tab of the Model pane.) This is useful for very large BIP problems since it may enable finding a near-optimal solu- tion when finding an optimal solution in a reasonable period of time is not pos- sible. For smaller problems (e.g., all the problems in this book), this option should be set to 0 to guarantee finding an optimal solution.
FIGURE 7.1 A spreadsheet formula- tion of the BIP model for the California Manu- facturing Co. case study where the changing cells BuildFactory? (C18:D18) and BuildWarehouse? (C16:D16) give the opti- mal solution obtained by Solver.
Capital
Spent
Capital
Available
Range Name
Build?
BuildWarehouse?
BuildFactory?
CapitalAvailable
CapitalRequired
CapitalSpent
MaxWarehouses
NPV
TotalNPV
TotalWarehouses
Cells
C16:D18
C16:D16
C18:D18
G12
C10:D12
E12
G16
C4:D6
D20
E16
20
C D
Total NPV ($millions) =SUMPRODUCT(NPV,Build?)
10
E
11
12
13
14
15
16
Capital
Spent
= SUMPRODUCT(CapitalRequired,Build?)
Total
Warehouses
= SUM(BuildWarehouse?)
A B C D E F G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
California Manufacturing Co. Facility Location Problem
Warehouse
Factory
NPV ($millions)
Capital Required
($millions)
LA SF
Warehouse
Factory
Build? LA
0
SF
0
1 1
Warehouse
Factory
LA SF
9
Total
Warehouses
0
Maximum
Warehouses15
16
17
18
19
20 Total NPV ($millions) $13
6 4
8 5
5 2
6 3 10
1
Solver Parameters Set Objective Cell: TotalNPV To: Max By Changing Variable Cells:
BuildWarehouse?, BuildFactory? Subject to the Constraints:
BuildFactory? = binary BuildWarehouse? = binary BuildWarehouse? <= BuildFactory? CapitalSpent <= CapitalAvailable TotalWarehouses <= MaxWarehouses
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
? ?
?
?
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238 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
The one key difference arises when using Solver. Each of the decision variables (cells C18:D18 and C16:D16) is constrained to be binary. In Excel’s Solver, this is accomplished in the Add Constraint dialog box by choosing each range of changing cells as the left-hand side and then choosing bin from the pop-up menu. In RSPE, this is accomplished by selecting each range of changing cells, and then under the Constraints menu on the RSPE ribbon, choose Binary in the Variable Type/Bound submenu. The other constraints shown in Solver (see the lower left-hand side of Figure 7.1 ) have been made quite intuitive by using the suggestive range names given in the lower right-hand side of the figure. For convenience, the equations entered into the output cells in E12 and D20 use a SUMPRODUCT function that includes C17:D17 and either C11:D11 or C5:D5 because the blanks or # signs in these rows are interpreted as zeroes by Solver.
Solver gives the optimal solution shown in C18:D18 and C16:D16 of the spreadsheet, namely, build factories in both Los Angeles and San Francisco, but do not build any ware- houses. The objective cell (D20) indicates that the total net present value from building these two factories is estimated to be $13 million.
Performing Sensitivity Analysis Now that Steve Chan has used the BIP model to determine what should be done when the amount of capital being made available to these investments is $10 million, his next task is to perform what-if analysis on this amount. Recall that Armando Ortega wants him to determine what the effect would be if this amount were changed to anything else between $5 million and $15 million.
In Chapter 5, we described three different methods of performing what-if analysis on a linear programming spreadsheet model when there is a change in a constraint: using trial and error with the spreadsheet, generating a parameter analysis report, or referring to Solver’s sensitivity report. The first two of these can be used on integer programming problems in exactly the same way as for linear programming problems. The third method, however, does not work. The sensitivity report is not available for integer programming problems. This is because the concept of a shadow price and allowable range no longer applies. In contrast to linear programming, the objective function values for an integer programming problem do not change in a predictable manner when the right-hand side of a constraint is changed.
It is straightforward to determine the impact of changing the amount of available capital by trial and error. Simply try different values in the data cell CapitalAvailable (G12) and re-solve with Solver. However, a more systematic way to perform this analysis is to gener- ate a parameter analysis report using RSPE. The parameter analysis report works for integer programming models in exactly the same way as it does for linear programming models (as described in Section 5.3 in the subsection entitled Using a Parameter Analysis Report (RSPE) to Do Sensitivity Analysis Systematically ).
After defining CapitalAvailable (G12) as a parameter cell with values ranging from 5 to 15 ($millions), the parameter analysis report shown in Figure 7.2 was generated by executing the series of steps outlined in Section 5.3. Note how Figure 7.2 shows the effect on the opti- mal solution and the resulting total net present value of varying the amount of capital being made available.
What-if analysis also could be performed on any of the other data cells—NPV (C4:D6), CapitalRequired (C10:D12), and MaxWarehouses (G16)—in a similar way with a parameter analysis report (or by using trial and error with the spreadsheet). However, a careful job was done in developing good estimates of the net present value of each of the possible invest- ments, and there is little uncertainty in the values entered in the other data cells, so Steve Chan decides that further what-if analysis is not needed.
Management’s Conclusion Steve Chan’s report is delivered to Armando Ortega within the two-week deadline. The report recommends the plan presented in Figure 7.1 (build a factory in both Los Angeles and San Francisco but no warehouses) if management decides to stick with its tentative decision to make $10 million of capital available for these investments. One advantage of this plan is that it only uses $9 million of this capital, which frees up $1 million of capital for other project proposals currently being investigated. The report also highlights the results shown in Fig- ure 7.2 while emphasizing two points. One is that a heavy penalty would be paid (a reduction
Note how helpful the range names are for interpreting this BIP spreadsheet model.
Solver’s sensitivity report is not available for integer programming problems.
Trial-and-error and/or a parameter analysis report can be used to perform sen- sitivity analysis for integer programming problems.
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7.2 Using BIP for Project Selection: The Tazer Corp. Problem 239
in the total net present value from $13 million to $9 million) if the amount of capital being made available were to be reduced below $9 million. The other is that increasing the amount of capital being made available by just $1 million (from $10 million to $11 million) would enable a substantial increase of $4 million in the total net present value (from $13 million to $17 million). However, a much larger further increase in the amount of capital being made available (from $11 million to $14 million) would be needed to enable a considerably smaller further increase in the total net present value (from $17 million to $19 million).
Armando Ortega deliberates with other members of top management before making a decision. It is quickly concluded that increasing the amount of capital being made available all the way up to $14 million would be stretching the company’s financial resources too dangerously to justify the relatively small payoff. However, there is considerable discussion of the pros and cons of the two options of using either $9 million or $11 million of capital. Because of the large payoff from the latter option (an additional $4 million in total net pres- ent value), management finally decides to adopt the plan presented in row 8 of Figure 7.2 . Thus, the company will build new factories in both Los Angeles and San Francisco as well as a new warehouse in San Francisco, with an estimated total net present value of $17 million. However, because of the large capital requirements of this plan, management also decides to defer building the warehouse until the two factories are completed so that their profits can help finance the construction of the warehouse.
1. What are the four interrelated decisions that need to be made by the management of the California Manufacturing Co.?
2. Why are binary decision variables appropriate to represent these decisions? 3. What is the objective specified by management for this problem? 4. What are the mutually exclusive alternatives in this problem? What is the form of the resulting
constraint in the BIP model? 5. What are the contingent decisions in this problem? For each one, what is the form of the
resulting constraint in the BIP model? 6. What is the tentative managerial decision on which sensitivity analysis needs to be performed?
Review Questions
7.2 USING BIP FOR PROJECT SELECTION: THE TAZER CORP. PROBLEM
The California Manufacturing Co. case study focused on four proposed projects: (1) build a factory in Los Angeles, (2) build a factory in San Francisco, (3) build a warehouse in Los Angeles, and (4) build a warehouse in San Francisco. Management needed to make yes-or-no decisions about which of these projects to select. This is typical of many applications of BIP. However, the nature of the projects may vary considerably from one application to the next. Instead of the proposed construction projects in the case study, our next example involves the selection of research and development projects.
This example is adapted from both Case 3-7 and its continuation in the supplement to Chapter 13, but all the relevant information is repeated below.
FIGURE 7.2 The parameter analysis report generated by RSPE that shows the effect on the optimal solution and the resulting total net present value of sys- tematically varying the amount of capital being made available for these investments.
A
5 1 2 3
0 1 1 1 1
1 9 9 9 9 13 13 17 17 17 19 19
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1
0 0 0
0 0 0 0
0 0
0 0
0 0
0 0 0 1 1
6 7 8 9
10 11 12 13 14 15
4 5 6 7 8 9 10 11 12
CapitalAvailable BuildWarehouseLA? BuildWarehouseSF? BuildFactoryLA? BuildFactorySF? TotalINPV B C D E F
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240 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
The Tazer Corp. Problem Tazer Corp., a pharmaceutical manufacturing company, is beginning the search for a new breakthrough drug. The following five potential research and development projects have been identified for attempting to develop such a drug.
Project Up: Develop a more effective antidepressant that does not cause serious mood swings. Project Stable: Develop a drug that addresses manic depression. Project Choice: Develop a less intrusive birth control method for women. Project Hope: Develop a vaccine to prevent HIV infection. Project Release: Develop a more effective drug to lower blood pressure.
In contrast to Case 3-7, Tazer management now has concluded that the company cannot devote enough money to research and development to undertake all of these projects. Only $1.2 billion is available, which will be enough for only two or three of the projects. The first row of Table 7.3 shows the amount needed (in millions of dollars) for each of these projects. The second row estimates each project’s probability of being successful. If a project is suc- cessful, it is estimated that the resulting drug would generate the revenue shown in the third row. Thus, the expected revenue (in the statistical sense) from a potential drug is the product of its numbers in the second and third rows, whereas its expected profit is this expected rev- enue minus the investment given in the first row. These expected profits are shown in the bottom row of Table 7.3 .
Tazer management now wants to determine which of these projects should be undertaken to maximize their expected total profit.
Formulation with Binary Variables Because the decision for each of the five proposed research and development projects is a yes-or-no decision, the corresponding decision variables are binary variables. Thus, the deci- sion variable for each project has the following interpretation.
Decision Variable 5 b1, if approve the project 0, if reject the project
Let x 1 , x 2 , x 3 , x 4 , and x 5 denote the decision variables for the respective projects in the order in which they are listed in Table 7.3 .
If a project is rejected, there is neither any profit nor any loss, whereas the expected profit if a project is approved is given in the bottom row of Table 7.3 . Therefore, when using units of millions of dollars, the expected total profit is
P 5 300x1 1 120x2 1 170x3 1 100x4 1 70x5
The objective is to select the projects to approve that will maximize this expected total profit while satisfying the budget constraint.
Other than requiring the decision variables to be binary, the budget constraint limit- ing the total investment to no more than $1.2 billion is the only constraint that has been imposed by Tazer management on the selection of these research and development projects.
The objective is to choose the projects that will maxi- mize the expected profit while satisfying the budget constraint.
Project
1 (Up) 2 (Stable) 3 (Choice) 4 (Hope) 5 (Release)
R&D investment ($million) 400 300 600 500 200 Success rate 50% 35% 35% 20% 45% Revenue if successful ($million) 1,400 1,200 2,200 3,000 600 Expected profit ($million) 300 120 170 100 70
TABLE 7.3 Data for the Tazer Project Selection Problem
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241
7.3 USING BIP FOR THE SELECTION OF SITES FOR EMERGENCY SERVICES FACILITIES: THE CALIENTE CITY PROBLEM
Although the problem encountered in the California Manufacturing Co. case study can be described as a project selection problem (as was done at the beginning of the preceding sec- tion), it could just as well have been called a site selection problem. Recall that the company’s management needed to select a site (Los Angeles or San Francisco) for its new factory as
Referring to the first row of Table 7.3 , this constraint can be expressed in terms of the deci- sion variables as
400x1 1 300x2 1 600x3 1 500x4 1 200x5 # 1,200
With this background, the stage now is set for formulating a BIP spreadsheet model for this problem.
A BIP Spreadsheet Model for the Tazer Problem Figure 7.3 shows a BIP spreadsheet model for this problem. The data in Table 7.3 have been trans- ferred into cells C5:G8. The changing cells are DoProject? (C10:G10) and the objective cell is TotalExpectedProfit (H8). The one functional constraint is depicted in cells H5:J5. In addition, the changing cells DoProject? are constrained to be binary, as shown in the Solver Parameters box.
The changing cells DoProject? (C10:G10) in Figure 7.3 show the optimal solution that has been obtained by Solver, namely,
Choose Project Up, Project Choice, and Project Release.
The objective cell indicates that the resulting total expected profit is $540 million.
1. How are binary variables used to represent managerial decisions on which projects from a group of proposed projects should be selected for approval?
2. What types of projects are under consideration in the Tazer Corp. problem? 3. What is the objective for this problem?
Review Questions
The Midwest Independent Transmission Operator, Inc. (MISO) is a nonprofit organization formed in 1998 to administer the generation and transmission of electricity throughout the midwestern United States. It serves over 40 million customers (both individuals and businesses) through its control of nearly 60,000 miles of high-voltage transmission lines and more than 1,000 power plants capa- ble of generating 146,000 megawatts of electricity. This infrastructure spans 13 midwestern states plus the Cana- dian province of Manitoba.
The key mission of any regional transmission organi- zation is to reliably and efficiently provide the electricity needed by its customers. MISO transformed the way this was done by using mixed binary integer programming to minimize the total cost of providing the needed electricity. Each main binary variable in the model represents a yes-or- no decision about whether a particular power plant should be on during a particular time period. After solving this model, the results are then fed into a linear programming
model to set electricity output levels and establish prices for electricity trades.
The mixed BIP model is a massive one with about 3,300,000 continuous variables, 450,000 binary variables, and 3,900,000 functional constraints. A special technique (Lagrangian relaxation) is used to solve such a huge model.
This innovative application of management science yielded savings of approximately $2.5 billion over the four years from 2007 to 2010, with an additional savings of about $7 billion expected through 2020. These dramatic results led to MISO winning the prestigious first prize in the 2011 international competition for the Franz Edelman Award for Achievement in Operations Research and the Management Sciences.
Source: B. Carlson and 12 co-authors, “MISO Unlocks Billions in Savings through the Application of Operations Research for Energy and Ancillary Services Markets,” Interfaces 42, no. 1 (January–February 2012), pp. 58–73. (A link to this article is provided on our website, www.mhhe.com/hillier5e . )
An Application Vignette
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242 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
FIGURE 7.3 A spreadsheet formulation of the BIP model for the Tazer Corp. project selection problem where the changing cells DoProject? (C10:G10) give the optimal solution obtained by Solver.
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A B C D E F G H I J
Tazer Corp. Project Selection Problem
Stable Choice Hope Release Total Budget
R&D Investment ($million) 300 600 500 200 1,200 <= 1,200
Success Rate 35% 35% 20% 45%
Revenue If Successful ($million) 1,200 2,200 3,000 600
Expected Profit ($million)
Up
400
50%
1,400
300 120 170 100 70 540
Do Project? 1 0 1 0 1
8
B C D E F G
Expected Profit ($million) =C7*C6-C5 =D7*D6-D5 =E7*E6-E5 =F7*F6-F5 =G7*G6-G5
Range Name Cells Budget J5
DoProject? C10:G10
ExpectedProfit C8:G8
RandDInvestment C5:G5
Revenue C7:G7
SuccessRate C6:G6
TotalExpectedProfit H8
TotalRandD H5
4
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H
Total
=SUMPRODUCT(RandDInvestment,DoProject?)
=SUMPRODUCT(ExpectedProfit,DoProject?)
Solver Parameters Set Objective Cell: TotalExpectedProfit To: Max By Changing Variable Cells:
DoProject?
Subject to the Constraints: DoProject? = binary
TotalRandD <= Budget
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
well as for its possible new warehouse. For either of the possible sites for the new factory (or the warehouse), there is a yes-or-no decision for whether that site should be selected, so it becomes natural to represent each such decision by a binary decision variable.
Various kinds of site selection problems are one of the most common types of applications of BIP. The kinds of facilities for which sites need to be selected can be of any type. In some cases, several sites are to be selected for several facilities of a particular type, whereas only a single site is to be selected in other cases.
We will focus here on the selection of sites for emergency services facilities. These facili- ties might be fire stations, police stations, ambulance centers, and so forth. In any of these cases, the overriding concern commonly is to provide facilities close enough to each part of the area being served that the response time to an emergency anywhere in the area will be suf- ficiently small. The form of the BIP model then will be basically the same regardless of the specific type of emergency services being considered.
To illustrate, let us consider an example where sites are being selected for fire stations. For simplicity, this example will divide the area being served into only eight tracts instead of the many dozens or hundreds that would be typical in real applications.
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7.3 Using BIP for the Selection of Sites for Emergency Services Facilities: The Caliente City Problem 243
The Caliente City Problem Caliente City is located in a particularly warm and arid part of the United States, so it is especially prone to the occurrence of fires. The city has become a popular place for senior citizens to move to after retirement, so it has been growing rapidly and spreading well beyond its original borders. However, the city still has only one fire station, located in the congested center of the original town site. The result has been some long delays in fire trucks reaching fires in the outer parts of the city, causing much more damage than would have occurred with a prompt response. The city’s residents are very unhappy about this, so the city council has directed the city manager to develop a plan for locating multiple fire stations throughout the city (including perhaps moving the current fire station) that would greatly reduce the response time to any fire. In particular, the city council has adopted the following policy about the maximum acceptable response time for fire trucks to reach a fire anywhere in the city after being notified about the fire.
Response time # 10 minutes
Having had a management science course in college, the city manager recognizes that BIP provides her with a powerful tool for analyzing this problem. To get started, she divides the city into eight tracts and then gathers data on the estimated response time for a fire in each tract from a potential fire station in each of the eight tracts. These data are shown in Table 7.4 . For example, if a decision were to be made to locate a fire station in tract 1 and if that fire station were to be used to respond to a fire in any of the tracts, the second column of Table 7.4 shows what the (estimated) response time would be. (Since the response time would exceed 10 minutes for a fire in tracts 3, 5, 6, 7, or 8, a fire station actually would need to be located nearer to each of these tracts to satisfy the city council’s new policy.) The bot- tom row of Table 7.4 shows what the cost would be of acquiring the land and constructing a fire station in any of the eight tracts. (The cost is far less for tract 5 because the current fire station already is there so only a modest renovation is needed if the decision is made to retain a fire station there.)
The objective now is to determine which tracts should receive a fire station to minimize the total cost of the stations while ensuring that each tract has at least one station close enough to respond to a fire in no more than 10 minutes.
Formulation with Binary Variables For each of the eight tracts, there is a yes-or-no decision as to whether that tract should receive a fire station. Therefore, we let x 1 , x 2 , . . . , x 8 denote the corresponding binary deci- sion variables, where
xj 5 b1, if tract j is selected to receive a fire station0, if not for j 5 1, 2, . . . , 8.
The objective is to mini- mize the total cost of ensur- ing a response time of no more than 10 minutes.
Fire Station in Tract
1 2 3 4 5 6 7 8
1 2 8 18 9 23 22 16 28 Response times (minutes) for a fire in tract
2 9 3 10 12 16 14 21 25 3 17 8 4 20 21 8 22 17 4 10 13 19 2 18 21 6 12 5 21 12 16 13 5 11 9 12 6 25 15 7 21 15 3 14 8 7 14 22 18 7 13 15 2 9 8 30 24 15 14 17 9 8 3
Cost of station ($thousands)
350 250 450 300 50 400 300 200
TABLE 7.4 Response Time and Cost Data for the Caliente City Problem
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244 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
Since the objective is to minimize the total cost of the fire stations that will satisfy the city council’s new policy on response times, the total cost needs to be expressed in terms of these decision variables. Using units of thousands of dollars while referring to the bottom row of Table 7.4 , the total cost is
C 5 350x1 1 250x2 1 450x3 1 300x4 1 50x5 1 400x6 1 300x7 1 200x8
We also need to formulate constraints in terms of these decision variables that will ensure that no response times exceed 10 minutes. For example, consider tract 1. When a fire occurs there, the row for tract 1 in Table 7.4 indicates that the only tracts close enough that a fire station would provide a response time not exceeding 10 minutes are tract 1 itself, tract 2, and tract 4. Thus, at least one of these three tracts needs to have a fire station. This requirement is expressed in the constraint
x1 1 x2 1 x4 $ 1
Incidentally, this constraint is called a set covering constraint because it covers the requirement of having a fire station located in at least one member of the set of tracts (tracts 1, 2, and 4) that are within 10 minutes of tract 1. In general, any constraint where a sum of binary variables is required to be greater-than-or-equal-to one is referred to as a set covering constraint.
Applying the above reasoning for tract 1 to all the tracts leads to the following constraints.
Tract 1: x1 1 x2 1 x4 $ 1 Tract 2: x1 1 x2 1 x3 $ 1 Tract 3: x2 1 x3 1 x6 $ 1 Tract 4: x1 1 x4 1 x7 $ 1 Tract 5: 1 x5 1 x7 $ 1 Tract 6: x3 1 x6 1 x8 $ 1 Tract 7: x4 1 x7 1 x8 $ 1 Tract 8: x6 1 x7 1 x8 $ 1
These set covering constraints (along with requiring the variables to be binary) are all that is needed to ensure that each tract has at least one fire station close enough to respond to a fire in no more than 10 minutes.
This type of BIP model (minimizing total cost where all the functional constraints are set covering constraints) is called a set covering problem . Such problems arise fairly fre- quently. In fact, you will see another example of a set covering problem in Section 7.4.
Having identified the nature of the constraints for the Caliente City problem, it now is fairly straightforward to formulate its BIP spreadsheet model.
A BIP Spreadsheet Model for the Caliente City Problem Figure 7.4 shows a BIP spreadsheet model for this problem. The data cells ResponseTime (D5:K12) show all the response times given in Table 7.4 and CostOfStation (D14:K14) provides the cost data from the bottom row of this table. There is a yes-or-no decision for each tract as to whether a fire station should be located there, so the changing cells are StationInTract? (D29:K29). The objective is to minimize total cost, so the objective cell is TotalCost (N29). The set covering constraints are displayed in cells L17:N24. The Station- InTract? changing cells have been constrained to be binary, as can be seen in the Solver Parameters box.
After running Solver, the optimal solution shown in the changing cells StationInTract? (D29:K29) in Figure 7.4 is obtained, namely,
Select tracts 2, 7, and 8 as the sites for fire stations.
The objective cell TotalCost (N29) indicates that the resulting total cost is $750,000.
This constraint ensures that the response time for a fire in tract 1 will be no more than 10 minutes.
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7.3 Using BIP for the Selection of Sites for Emergency Services Facilities: The Caliente City Problem 245
FIGURE 7.4 A spreadsheet formulation of the BIP model for the Caliente City site selection problem where the changing cells StationInTract? (D29:K29) show the optimal solution obtained by Solver.
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A B C D E F G H I J K L M N
Caliente City Fire Station Location Problem
Fire Station in Tract
1 2 3 4 5 6 7 8
1 2 8 18 9 23 22 16 28
Response 2 9 3 10 12 16 14 21 25
Times 3 17 8 4 20 21 8 22 17
(minutes) 4 10 13 19 2 18 21 6 12
for a Fire 5 21 12 16 13 5 11 9 12
in Tract 6 25 15 7 21 15 3 14 8
7 14 22 18 7 13 15 2 9
8 30 24 15 14 17 9 8 3
Cost of Station 350 250 450 300 50 400 300 200
($thousands) Number
Covering
1 1 1 0 1 0 0 0 0 1 >= 1
Response 2 1 1 1 0 0 0 0 0 1 >= 1
Time 3 0 1 1 0 0 1 0 0 1 >= 1
<= 4 1 0 0 1 0 0 1 0 1 >= 1
10 5 0 0 0 0 1 0 1 0 1 >= 1
Minutes? 6 0 0 1 0 0 1 0 1 1 >= 1
7 0 0 0 1 0 0 1 1 2 >= 1
8 0 0 0 0 0 1 1 1 2 >= 1
Total
Fire Station in Tract Cost
1 2 3 4 5 6 7 8 ($thousands)
Station in Tract? 0 1 0 0 0 0 1 1 750
15
16
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18
19
J K L
Number
Covering
=IF(J5<=MaxResponseTime,1,0) =IF(K5<=MaxResponseTime,1,0) =SUMPRODUCT(D17:K17,StationInTract?)
=IF(J6<=MaxResponseTime,1,0) =IF(K6<=MaxResponseTime,1,0) =SUMPRODUCT(D18:K18,StationInTract?)
=IF(J7<=MaxResponseTime,1,0) =IF(K7<=MaxResponseTime,1,0) =SUMPRODUCT(D19:K19,StationInTract?)
Range Name Cells
CostOfStation D14:K14
MaxResponseTime B21
NumberCovering L17:L24
One N17:N24
ResponseTime D5:K12
StationInTract? D29:K29
TotalCost N29
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N
Total
Cost
($thousands)
=SUMPRODUCT(CostOfStation,StationInTract?)
Solver Parameters Set Objective Cell: TotalCost To: Min By Changing Variable Cells:
StationInTract?
Subject to the Constraints: StationInTract? = binary
NumberCovering >= One
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
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246 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
FIGURE 7.5 The arrows show the 11 Southwestern Airways flights that need to be cov- ered by the three crews based in San Francisco.
Seattle (SEA)
San Francisco (SFO)
Los Angeles (LAX)
Denver (DEN)
Chicago (ORD)
1. How are binary variables used to represent managerial decisions regarding which site or sites should be selected for new facilities?
2. What are some types of emergency services facilities for which sites may need to be selected? 3. What was the objective for the Caliente City problem? 4. What is a set covering constraint and what is a set covering problem?
Review Questions
7.4 USING BIP FOR CREW SCHEDULING: THE SOUTHWESTERN AIRWAYS PROBLEM
Throughout the travel industry (airlines, rail travel, cruise ships, tour companies, etc.), one of the most challenging problems in maintaining an efficient operation is the scheduling of its crews who serve customers during their travels. Given many feasible overlapping sequences of trips for a crew, to which ones should a crew be assigned so as to cover all the trips at a minimum cost? Thus, for each feasible sequence of trips, there is a yes-or-no decision as to whether a crew should be assigned to that sequence, so a binary decision variable can be used to represent that decision.
For many years, airline companies have been using BIP models to determine how to do their crew scheduling in the most cost-efficient way. Some airlines have saved many mil- lions of dollars annually through this application of BIP. Consequently, other segments of the travel industry now are also using BIP in this way. For example, the application vignette in this section describes how Netherlands Railways achieved a dramatic increase in profits by applying BIP (and related techniques) in a variety of ways, including crew scheduling.
To illustrate the approach, consider the following miniature example of airline crew scheduling.
The Southwestern Airways Problem Southwestern Airways needs to assign its crews to cover all its upcoming flights. We will focus on the problem of assigning three crews based in San Francisco (SFO) to the 11 flights shown in Figure 7.5 . These same flights are listed in the first column of Table 7.5 . The other 12 columns show the 12 feasible sequences of flights for a crew. (The numbers in each column indicate the order of the flights.) At most, three of the sequences need to be chosen (one per crew) in such a way that every flight is covered. (It is permissible to have more than one crew on a flight, where the extra crews would fly as passengers, but union contracts require that the extra crews still be paid for their time as if they were working.) The cost of assigning a crew to a particular sequence of flights is given (in thousands of dollars) in the bottom row of the table. The objective is to minimize the total cost of the crew assignments that cover all the flights.
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247
Netherlands Railways (Nederlandse Spoorwegen Reizigers) is the main Dutch railway operator of passenger trains. In this densely populated country, about 5,500 passenger trains currently transport approximately 1.1 million passen- gers on an average workday. The company’s operating rev- enues are approximately 1.5 billion euros (approximately $2 billion) per year.
The amount of passenger transport on the Dutch rail- way network has steadily increased over the years, so a national study in 2002 concluded that three major infra- structure extensions should be undertaken. As a result, a new national timetable for the Dutch railway system, specify ing the planned departure and arrival times of every train at every station, would need to be developed. There- fore, the management of Netherlands Railways directed that an extensive management science study should be conducted over the next few years to develop an optimal overall plan for both the new timetable and the usage of the available resources (rolling-stock units and train crews) for meeting this timetable. A task force consisting of sev- eral members of the company’s Department of Logistics and several prominent management science scholars from European universities or a software company was formed to conduct this study.
The new timetable was launched in December 2006, along with a new system for scheduling the allocation of rolling-stock units (various kinds of passenger cars and other train units) to the trains meeting this timetable. A new system also was implemented for scheduling the assignment of crews (with a driver and a number of con- ductors in each crew) to the trains. Binary integer program- ming and related techniques were used to do all of this. For example, the BIP model used for crew scheduling closely resembles (except for its vastly larger size) the one shown in this section for the Southwestern Airlines problem.
This application of management science immediately resulted in an additional annual profit of approximately $60 million for the company and this additional profit is expected to increase to $105 million annually in the coming years. These dramatic results led to Netherlands Railways winning the prestigious First Prize in the 2008 international competition for the Franz Edelman Award for Achievement in Operations Research and the Management Sciences.
Source: L. Kroon, D. Huisman, E. Abbink, P.-J. Fioole, M. Fischetti, G. Maróti, A. Schrijver, A. Steenbeck, and R. Ybema, “The New Dutch Timetable: The OR Revolution,” Interfaces 39, no. 1 (January–February 2009), pp. 6–17. (A link to this article is provided on our website, www.mhhe.com/hillier5e . )
An Application Vignette
Feasible Sequence of Flights
Flight 1 2 3 4 5 6 7 8 9 10 11 12
1. San Francisco to Los Angeles (SFO–LAX)
1 1 1 1
2. San Francisco to Denver (SFO–DEN)
1 1 1 1
3. San Francisco to Seattle (SFO–SEA)
1 1 1 1
4. Los Angeles to Chicago (LAX–ORD)
2 2 3 2 3
5. Los Angeles to San Francisco (LAX–SFO)
2 3 5 5
6. Chicago to Denver (ORD–DEN)
3 3 4
7. Chicago to Seattle (ORD–SEA)
3 3 3 3 4
8. Denver to San Francisco (DEN–SFO)
2 4 4 5
9. Denver to Chicago (DEN–ORD)
2 2 2
10. Seattle to San Francisco (SEA–SFO)
2 4 4 5
11. Seattle to Los Angeles (SEA–LAX)
2 2 4 4 2
Cost, $1,000s 2 3 4 6 7 5 7 8 9 9 8 9
TABLE 7.5 Data for the Southwestern Airways Problem
Formulation with Binary Variables With 12 feasible sequences of flights, we have 12 yes-or-no decisions:
Should sequence j be assigned to a crew? (j 5 1, 2, c, 12)
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248 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
Therefore, we use 12 binary variables to represent these respective decisions:
xj 5 b1, if sequence j is assigned to a crew0, otherwise Since the objective is to minimize the total cost of the three crew assignments, we now
need to express the total cost in terms of these binary decision variables. Referring to the bot- tom row of Table 7.5 , this total cost (in units of thousands of dollars) is
C 5 2x1 1 3x2 1 4x3 1 6x4 1 7x5 1 5x6 1 7x7 1 8x8 1 9x9 1 9x10 1 8x11 1 9x12
With only three crews available to cover the flights, we also need the constraint
x1 1 x2 1 c1 x12 # 3
The most interesting part of this formulation is the nature of each constraint that ensures that a corresponding flight is covered. For example, consider the last flight in Table 7.5 (Seat- tle to Los Angeles). Five sequences (namely, sequences 6, 9, 10, 11, and 12) include this flight. Therefore, at least one of these five sequences must be chosen. The resulting constraint is
x6 1 x9 1 x10 1 x11 1 x12 $ 1
For each of the 11 flights, the constraint that ensures that the flight is covered is con- structed in the same way from Table 7.5 by requiring that at least one of the flight sequences that includes that flight is assigned to a crew. Thus, 11 constraints of the following form are needed.
Flight 1: x1 1 x4 1 x7 1 x10 $ 1 Flight 2: x2 1 x5 1 x8 1 x11 $ 1
# # #
Flight 11: x6 1 x9 1 x10 1 x11 1 x12 $ 1
Note that these constraints have the same form as the constraints for the Caliente City problem in Section 7.3 (a sum of certain binary variables $ 1), so these too are set covering constraints. Therefore, this crew scheduling problem is another example of a set covering problem (where this particular set covering problem also includes the side constraint that x 1 1 x 2 1 ? ? ? 1 x 12 # 3).
Having identified the nature of the constraints, the stage now is set for formulating a BIP spreadsheet model for this problem.
A BIP Spreadsheet Model for the Southwestern Airways Problem Figure 7.6 shows a spreadsheet formulation of the complete BIP model for this problem. The changing cells FlySequence? (C22:N22) contain the values of the 12 binary decision variables. The data in IncludesSegment? (C8:N18) and Cost (C5:N5) come directly from Table 7.5 . The last three columns of the spreadsheet are used to show the set covering con- straints, Total $ AtLeastOne, and the side constraint, TotalSequences # NumberOfCrews. Finally, the changing cells FlySequence? have been constrained to be binary, as shown in the Solver Parameters box.
Solver provides the optimal solution shown in FlySequence? (C22:N22). In terms of the x j variables, this solution is
x3 5 1 (assign sequence 3 to a crew)
x4 5 1 (assign sequence 4 to a crew)
x11 5 1 (assign sequence 11 to a crew)
and all other x j 5 0, for a total cost of $18,000 as given by TotalCost (Q24). (Another optimal solution is x 1 5 1, x 5 5 1, x 12 5 1, and all other x j 5 0.)
These are set covering con- straints, just like the con- straints in the Caliente City problem in Section 7.3.
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7.4 Using BIP for Crew Scheduling: The Southwestern Airways Problem 249
FIGURE 7.6 A spreadsheet formulation of the BIP model for the Southwestern Airways crew scheduling problem, where FlySequence (C22:N22) shows the optimal solution obtained by Solver. The list of flight sequences under consideration is given in cells A25:D37.
Flight Sequence Key
Flight Sequence
1
A B C
2
3
4 1
D
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I
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N O P Q
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SFO-DEN-SFO
SFO-SEA-SFO
SFO-LAX-ORD-DEN-SFO
SFO-DEN-ORD-DEN-SFO
SFO-SEA-LAX-SFO
SFO-LAX-ORD-SEA-SFO
SFO-DEN-ORD-SEA-SFO
SFO-SEA-LAX-ORD-DEN-SFO
SFO-LAX-ORD-SEA-LAX-SFO
SFO-DEN-ORD-SEA-LAX-SFO
SFO-SEA-LAX-ORD-SEA-SFO
Southwestern Airways Crew Scheduling Problem
Total Cost ($thousands)24
P Q
=SUMPRODUCT(Cost,FlySequence?)
7
O
Total
8
9
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=SUMPRODUCT(C8:N8,FlySequence?)
=SUMPRODUCT(C9:N9,FlySequence?)
=SUMPRODUCT(C10:N10,FlySequence?)
11
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=SUMPRODUCT(C11:N11,FlySequence?)
=SUMPRODUCT(C12:N12,FlySequence?)
=SUMPRODUCT(C13:N13,FlySequence?)
14
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=SUMPRODUCT(C14:N14,FlySequence?)
=SUMPRODUCT(C15:N15,FlySequence?)
=SUMPRODUCT(C16:N16,FlySequence?)
17
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=SUMPRODUCT(C17:N17,FlySequence?)
=SUMPRODUCT(C18:N18,FlySequence?)
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Total
Sequences
=SUM(FlySequence?)
Range Name
AtLeastOne
Cost
FlySequence?
IncludesSegment?
NumberOfCrews
Total
TotalCost
TotalSequences
Cells
Q8:Q18
C5:N5
C22:N22
C8:N18
Q22
O8:O18
Q24
O22
Cost ($thousands)
1 2 3 4 5 6 7 8 9 10 11 12
0 0 1 1 0 0 0 0 0 0 1 0
Total
Sequences
3
Number
of Crews
18Total Cost ($thousands)
Fly Sequence?
Includes Segment?
SFO–LAX
SFO–DEN
SFO–SEA
LAX–ORD
LAX–SFO
ORD–DEN
ORD–SEA
DEN–SFO
DEN–ORD
SEA–SFO
SEA–LAX
Total
At
Least
One
2 3 4 6 7 5 7 8 9 9 8 9
1
1 0 0 1 0 0 1 0 0 1 0 0
0 1 0 0 1 0 0 1 0 0 1 0
0 0 1 0 0 1 0 0 1 0 0 1
0 0 0 1 0 0 1 0 1 1 0 1
1 0 0 0 0 1 0 0 0 1 1 0
0 0 0 1 0 0 0 1 0 0 0
0 0 0 0 0 0 1 1 0 1 1 1
0 1 0 1 1 0 0 0 1 0 0 0
0 0 0 0 1 0 0 1 0 0 1 0
0 0 1 0 0 0 1 1 0 0 0 1
0 0 0 0 0 1 0 0 1 1 1 1
1
1
1
1
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1
1
1
1
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3≤
≥ ≥
≥
≥
≥
≥ ≥
≥
≥
≥
≥
Solver Parameters Set Objective Cell: TotalCost
To: Min By Changing Variable Cells:
FlySequence?
Subject to the Constraints: FlySequence? = binary
Total >= AtLeastOne
TotalSequences <= NumberOfCrews
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
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250 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
We should point out that this BIP model is a tiny one compared to the ones typically used in actual practice. Airline crew scheduling problems involving thousands of possible flight sequences now are being solved by using models similar to the one shown above but with thousands of binary variables rather than just a dozen.
Many airlines are solving huge BIP models of this kind.
1. What is the crew scheduling problem that is encountered by companies in the travel industry? 2. What are the yes-or-no decisions that need to be made when addressing a crew scheduling
problem? 3. For the Southwestern Airways problem, there is a constraint for each flight to ensure that this
flight is covered by a crew. Describe the mathematical form of this constraint. Then explain in words what this constraint is saying.
Review Questions
7.5 USING MIXED BIP TO DEAL WITH SETUP COSTS FOR INITIATING PRODUCTION: THE REVISED WYNDOR PROBLEM
All of the examples considered thus far in this chapter have been pure BIP problems (prob- lems where all the decision variables are binary variables). However, mixed BIP problems (problems where only some of the decision variables are binary variables) also arise quite frequently because only some of the decisions to be made are yes-or-no decisions and the rest are how-much decisions.
One important example of this type is the product-mix problem introduced in Chapter 2, but now with the added complication that a setup cost must be incurred to initiate the pro- duction of each product. Therefore, in addition to the how-much decisions of how much to produce of each product, there also is a prior yes-or-no decision for each product of whether to perform a setup to enable initiating its production.
To illustrate this type of problem, we will consider a revised version of the Wyndor Glass Co. product-mix problem that was described in Section 2.1 and analyzed throughout most of Chapter 2.
The Revised Wyndor Problem with Setup Costs Suppose now that the Wyndor Glass Co. will only devote one week each month to the pro- duction of the special doors and windows described in Section 2.1, so the question now is how many doors and windows to produce during each of these week-long production runs. Since the decisions to be made are no longer the rates of production for doors and windows, but rather how many doors and windows to produce in individual production runs, these quanti- ties now are required to be integer.
Each time Wyndor’s plants convert from the production of other products to the produc- tion of these doors and windows for a week, the following setup costs would be incurred to initiate this production.
Setup cost to produce doors 5 $700 Setup cost to produce windows 5 $1,300
Otherwise, all the original data given in Table 2.2 still apply, including a unit profit of $300 for doors and $500 for windows when disregarding these setup costs.
Table 7.6 shows the resulting net profit from producing any feasible quantity for either product. Note that the large setup cost for either product makes it unprofitable to produce less than three units of that product.
The dots in Figure 7.7 show the feasible solutions for this problem. By adding the appro- priate entries in Table 7.6 , the figure also shows the calculation of the total net profit P for each of the corner points. The optimal solution turns out to be
(D, W ) 5 (0, 6) with P 5 1,700
By contrast, the original solution
(D, W ) 5 (2, 6) with P 5 1,600
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7.5 Using Mixed BIP to Deal with Setup Costs for Initiating Production: The Revised Wyndor Problem 251
now gives a smaller value of P. The reason that this original solution (which gave P 5 3,600 for the original problem) is no longer optimal is that the setup costs reduce the total net profit so much:
P 5 3,600 2 700 2 1,300 5 1,600
Therefore, the graphical method for linear programming can no longer be used to find the optimal solution for this new problem with setup costs.
How can we formulate a model for this problem so that it fits a standard kind of model that can be solved by Solver? Table 7.6 shows that the net profit for either product is no longer directly proportional to the number of units produced. Therefore, as it stands, the problem no longer fits either linear programming or BIP. Before, for the original problem without setup costs, the objective function was simply P 5 300 D 1 500 W. Now we need to subtract from this expression each setup cost if the corresponding product will be produced, but we should not subtract the setup cost if the product will not be produced. This is where binary variables come to the rescue.
Formulation with Binary Variables For each product, there is a yes-or-no decision regarding whether to perform the setup that would enable initiating the production of the product, so the setup cost is incurred only if the decision is yes. Therefore, we can introduce a binary variable for each setup cost and
Net Profit ($)
Number of Units Produced Doors Windows
0 0 (300) 2 0 5 0 0 (500) 2 0 5 0 1 1 (300) 2 700 5 2400 1 (500) 2 1,300 5 2800 2 2 (300) 2 700 5 2100 2 (500) 2 1,300 5 2300 3 3 (300) 2 700 5 200 3 (500) 2 1,300 5 200 4 4 (300) 2 700 5 500 4 (500) 2 1,300 5 700 5 Not feasible 5 (500) 2 1,300 5 1,200 6 Not feasible 6 (500) 2 1,300 5 1,700
TABLE 7.6 Net Profit ($) for the Revised Wyndor Problem
FIGURE 7.7 The dots are the feasible solutions for the revised Wyndor problem. Also shown is the calculation of the total net profit P (in dollars) for each corner point from the net profits given in Table 7.6 .
W
D
8
6
4
2
2 4 6 8
Production quantity for doors
Production quantity for windows
0
(0, 6) gives P = 1,700
(2, 6) gives P = –100 + 1,700 = 1,600
(4, 3) gives P = 500 + 200 = 700
(4, 0) gives P = 500 (0, 0) gives P = 0
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252
associate each value of the binary variable with one of the two possibilities for the setup cost. In particular, let
y1 5 b1, if perform the setup to produce doors0, if not y2 5 b1, if perform the setup to produce windows0, if not
Therefore, the objective function now can be written as
P 5 300D 1 500W 2 700y1 2 1,300y2
which fits the format for mixed BIP. Since a setup is required to produce the corresponding product, these binary variables can
be related directly to the production quantities as follows:
y1 5 b1, if D . 0 can hold (can produce doors)0, if D 5 0 must hold (cannot produce doors) y2 5 b1, if W . 0 can hold (can produce windows)0, if W 5 0 must hold (cannot produce windows)
We need to include constraints in the model that will ensure that these relationships will hold. (An algorithm solving the model only recognizes the objective function and the con- straints, not the definitions of the variables.)
So what are the constraints of the model? We still need all the constraints of the original model. We also need constraints that D and W are integers, and that y 1 and y 2 are binary. In addition, we need some ordinary linear programming constraints that will ensure the follow- ing relationships:
If y1 5 0, then D 5 0.
If y2 5 0, then W 5 0.
These binary variables enable subtracting each setup cost only if the setup is performed.
Prior to its merger with United Airlines in 2010, Continental Airlines was a major United States air carrier that trans- ported passengers, cargo, and mail. It operated more than 2,000 daily departures to well over 100 domestic destina- tions and nearly 100 foreign destinations.
Airlines like Continental face schedule disruptions daily because of unexpected events, including inclement weather, aircraft mechanical problems, and crew unavailability. Another consequence of such disruptions is that crews may not be in positions to service their remaining scheduled flights. Airlines must reassign crews quickly to cover open flights and to return them to their original schedules in a cost-effective manner while honoring all government regula- tions, contractual obligations, and quality-of-life requirements.
To address such problems, a management science team at Continental Airlines developed a detailed mixed BIP model for reassigning crews to flights as soon as such emergencies arise. Because the airline has thousands of crews and daily flights, the model needed to be huge to consider all possible pairings of crews with flights. There- fore, the model has millions of decision variables and many thousands of constraints. (Most of these decision variables
are binary variables and the rest are general integer vari- ables). In its first year of use (mainly in 2001), the model was applied four times to recover from major schedule disruptions (two snowstorms, a flood, and the September 11 terrorist attacks). This led to savings of approximately $40 million. Subsequent applications extended to many daily minor disruptions as well.
Although other airlines subsequently scrambled to apply management science in a similar way, this initial advantage over other airlines in being able to recover more quickly from schedule disruptions with fewer delays and cancelled flights left Continental Airlines in a relatively strong posi- tion as the airline industry struggled through a difficult period during the initial years of the 21st century. This ini- tiative led to Continental winning the prestigious First Prize in the 2002 international competition for the Franz Edel- man Award for Achievement in Operations Research and the Management Sciences.
Source: G. Yu, M. Argüello, C. Song, S. M. McGowan, and A. White, “A New Era for Crew Recovery at Continental Airlines,” Interfaces 33, no. 1 (January–February 2003), pp. 5–22. (A link to this article is provided on our website, www.mhhe.com/hillier5e . )
An Application Vignette
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7.5 Using Mixed BIP to Deal with Setup Costs for Initiating Production: The Revised Wyndor Problem 253
(If y 1 5 1, no restrictions are placed on D other than those already imposed by the other con- straints, and the same thing applies to W if y 2 5 0.)
It is possible with Excel to use the IF function to represent this relationship between y 1 and D and between y 2 and W. Unfortunately, the IF function does not fit the assumptions of linear programming. Consequently, Solver has difficulty solving spreadsheet models that use this function. This is why another formulation with ordinary linear programming constraints is needed instead to express these relationships.
Since the other constraints impose bounds on D and W of 0 # D # 4 and 0 # W # 6, here are some ordinary linear programming constraints that ensure these relationships:
D # 4y1 W # 6y2
Note that setting y 1 5 0 gives D # 0, which forces the nonnegative D to be D 5 0, whereas setting y 1 5 1 gives D # 4, which allows all the values of D already allowed by the other constraints. Then check that the same conclusions apply for W when setting y 2 5 0 and y 2 5 1.
It was not necessary to choose 4 and 6 for the respective coefficients of y 1 and y 2 in these two constraints. Any coefficients larger than 4 and 6 would have the same effect. You just need to avoid smaller coefficients, since this would impose undesired restrictions on D and W when y 1 5 1 and y 2 5 1.
On larger problems, it is sometimes difficult to determine the smallest acceptable coef- ficients for these binary variables. Therefore, it is common to formulate the model by just using a reasonably large number (say, 99 in this case) that is safely larger than the smallest acceptable coefficient.
With this background, we now are ready to formulate a mixed BIP spreadsheet model for this problem that uses the number 99 in these constraints.
A Mixed BIP Spreadsheet Model for the Revised Wyndor Problem Figure 7.8 shows one way of formulating this model. The format for the first 14 rows is the same as for the original Wyndor problem, so the difference arises in rows 15–17 of the spreadsheet. The values of the binary variables, y 1 and y 2 , appear in the new chang- ing cells, Setup? (C17:D17). The bottom of the figure identifies the equations entered into the output cells in row 16, C16 5 99*C17 and D16 5 99*D17. Consequently, the con- straints, UnitsProduced (C14:D14) # OnlyIfSetup (C16:D16), impose the relationships that D # 99 y 1 and W # 99 y 2 .
The changing cells in this spreadsheet show the optimal solution obtained after apply- ing Solver. Thus, this solution is to not produce any doors ( y 1 5 0 and D 5 0) but to per- form the setup to enable producing 6 windows ( y 2 5 1 and W 5 6) to obtain a net profit of $1,700.
Note that this optimal solution does indeed satisfy the requirements that D 5 0 must hold when y 1 5 0 and that W . 0 can hold when y 2 5 1. The constraints do permit performing a setup to produce a product and then not producing any units ( y 1 5 1 with D 5 0 or y 2 5 1 with W 5 0), but the objective function causes an optimal solution automatically to avoid this foolish option of incurring the setup cost for no purpose.
These constraints force the model to refuse production if the corresponding setup is not performed.
If a how-much decision x can only be done (i.e., x . 0) if a corresponding yes-no deci- sion y is done (i.e., y 5 1), this can be enforced with a big-number constraint such as x # 99y. If y 5 0, this becomes x # 0. If y 5 1, then this becomes x # 99. The big number (99 in this example) is chosen to be big enough so that it is safely larger than x could ever become.
1. How does a mixed BIP problem differ from a pure BIP problem? 2. Why is a linear programming formulation no longer valid for a product-mix problem when
there are setup costs for initiating production? 3. How can a binary variable be defined in terms of whether a setup is performed to initiate the
production of a certain product? 4. What caused the optimal solution for the revised Wyndor problem to differ from that for the
original Wyndor problem?
Review Questions
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254 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
Managers frequently must make yes-or-no decisions, where the only two possible choices are yes, go ahead with a particular option, or no, decline this option. A binary integer programming (BIP) model considers many options simultaneously, with a binary decision variable for each option. Mixed BIP models include some continuous decision variables as well.
The California Manufacturing Co. case study involves yes-or-no decisions on whether a new factory should be built in certain cities and then whether a new warehouse also should be built in certain cities. This case study also introduced the modeling of mutually exclusive alternatives and contingent deci- sions, as well as the performance of sensitivity analysis for BIP models.
7.6 Summary
FIGURE 7.8 A spreadsheet model for the revised Wyndor problem, where Solver gives the optimal solu- tion shown in the chang- ing cells, UnitsProduced (C14:D14) and Setup? (C17:D17).
1
A B C D E F G H
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Wyndor Glass Co. Product-Mix with Setup Costs
Doors
0
12
12
Windows
Doors
0 6
0 99
0 1 Total Profit
−Total Setup Cost
Production Profit
$1,700
$1,300
$3,000
Windows
Unit Profit
Setup Cost
Plant 1
Plant 2
Plant 3
Units Prod'd
Only if Set Up
Setup?
Hours Used per Unit Produced
Hours
Available
Hours
Used
Range Name
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
OnlyIfSetup
ProductionProfit
Setup?
SetupCost
TotalProfit
TotalSetupCost
UnitProfit
UnitsProduced
Cells
G9:G11
E9:E11
C9:D11
C16:D16
H15
C17:D17
C5:D5
H17
H16
C4:D4
C14:D14
15
G H
16
17
Production Profit
−Total Setup Cost
Total Profit
=SUMPRODUCT(UnitProfit,UnitsProduced)
=SUMPRODUCT(SetupCost,Setup?)
=ProductionProfit − TotalSetupCost
16
B
Only if Set Up
C
=99*C17
D
=99*D17
7
E
8
9
Hours
Used
=SUMPRODUCT(C9:D9,UnitsProduced)
=SUMPRODUCT(C10:D10,UnitsProduced)10
11 =SUMPRODUCT(C11:D11,UnitsProduced)
$300 $500
$700 $1,300
1
0
3
0
2
2
4
12
18
Solver Parameters Set Objective Cell: TotalProfit To: Max By Changing Variable Cells:
UnitsProduced, Setup?
Subject to the Constraints: Setup? = binary
UnitsProduced = integer
HoursUsed <= HoursAvailable
UnitsProduced <= OnlyIfSetup
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
≤ ≤
≤
≤ ≤
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Chapter 7 Solved Problems 255
Many companies have saved millions of dollars by formulating and solving BIP models for a wide variety of applications. We have described and illustrated some of the most important types, including the selection of projects (e.g., research and development projects), the selection of sites for facilities (e.g., emergency services facilities such as fire stations), and crew scheduling in the travel industry (e.g., airlines). We also have discussed how to use mixed BIP to deal with setup costs for initiating production when addressing product-mix problems.
Glossary binary decision variable A binary variable that represents a yes-or-no decision by assigning a value of 1 for choosing yes and a value of 0 for choosing no. (Introduction), 232 binary integer programming A type of prob- lem or model that fits linear programming except that it uses binary decision variables. (Introduction), 232 binary variable A variable whose only pos- sible values are 0 and 1. (Introduction), 232 BIP Abbreviation for binary integer program- ming. (Introduction), 232 contingent decision A yes-or-no decision is a contingent decision if it can be yes only if a certain other yes-or-no decision is yes. (Section 7.1), 236 mixed BIP model A BIP model where only some of the variables are restricted to be binary variables. (Introduction), 232
mutually exclusive alternatives A group of alternatives where choosing any one alterna- tive excludes choosing any of the others. (Section 7.1), 235 pure BIP model A BIP model where all the variables are restricted to be binary variables. (Introduction), 232 set covering constraint A constraint that requires the sum of certain binary variables to be greater than or equal to 1. (Section 7.3), 244 set covering problem A type of BIP model where the objective is to minimize some quantity such as total cost and all the functional constraints are set covering constraints. (Section 7.3), 244 yes-or-no decision A decision whose only possible choices are (1) yes, go ahead with a cer- tain option or (2) no, decline this option. (Introduction), 232
Chapter 7 Excel Files:
California Mfg. Case Study
Tazer Corp. Example
Caliente City Example
Southwestern Airways Example
Revised Wyndor Example
Excel Add-in:
Risk Solver Platform for Education (RSPE) Supplements to This Chapter on the CD-ROM:
Advanced Formulation Techniques for Binary Integer Programming
Some Perspectives on Solving Binary Integer Programming Problems
Learning Aids for This Chapter in Your MS Courseware
Solved Problems (See the CD-ROM or Website for the Solutions) 7.S1. Capital Budgeting with Contingency Constraints A company is planning its capital budget over the next several years. There are eight potential projects under consideration. A calculation has been made of the expected net present value of each project, along with the cash outflow that would be required over the next four years. These data, along with the cash that is available each year, are shown in the next table. There also are the following contingency constraints: (a) at least one of project 1, 2, or 3 must be done, (b) projects 6 and 7 cannot both be done, and (c) project 5 can only be done if project 6 is done.
Formulate and solve a BIP model in a spreadsheet to determine which projects should be pursued to maximize the total expected net present value.
7.S2. Locating Search-and-Rescue Teams The Washington State legislature is trying to decide on loca- tions at which to base search-and-rescue teams. The teams are expensive, so the legislature would like as few as possible while still providing the desired level of service. In particular, since response time is critical, the legislature would like every county to either have a team located in that county or in an adjacent
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256 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
Cash Outflow Required ($million) Cash
Available
($million) Project
1 2 3 4 5 6 7 8
Year 1 1 3 0 3 3 7 2 5 20 Year 2 2 2 2 2 2 3 3 4 20 Year 3 2 3 4 2 3 3 6 2 20 Year 4 2 1 0 5 4 2 1 2 20
NPV ($mil)
10 12 11 15 24 17 16 18
Counties 1. Clallum
2. Jefferson
3. Grays Harbor
4. Pacific
5. Wahkiakum
6. Kitsap
7. Mason
8. Thurston
9. Whatcom
10. Skagit
11. Snohomish
12. King
13. Pierce
14. Lewis
15. Cowlitz
16. Clark
17. Skamania
18. Okanogan
19. Chelan
20. Douglas
21. Kittitas
22. Grant
23. Yakima
24. Klickitat
25. Benton
26. Ferry
27. Stevens
28. Pend Oreille
29. Lincoln
30. Spokane
31. Adams
32. Whitman
33. Franklin
34. Walla Walla
35. Columbia
36. Garfield
37. Asotin
1
2
3
4
5 15
14
16
17
24
25
22
34 35
36
32
3029
28 27
2618
9
10
11
12
138
7
6 20 19
21 31
33
37
23
county. (The locations and names of the counties are shown above.) Formulate and solve a BIP model in a spreadsheet to determine where the teams should be located.
7.S3. Warehouse Site Selection Consider a small company that produces a single product in two plants and serves customers in five different regions.
The company has been using a make-to-order policy of pro- ducing the product only in the quantities needed to fill the orders that have come in from the various regions. However, because of the problems caused by the sporatic production schedule, management has decided to smooth out the production rate and ship the product to one or more storage warehouses, which then will use inventory to fill the incoming regional orders. Manage- ment now needs to decide where to locate the company’s new warehouse(s). There are three locations under consideration. For each location, there is a fixed monthly cost associated with leas- ing and operating the warehouse there. Furthermore, each poten- tial warehouse location has a maximum capacity for monthly shipments restricted primarily by the number of trucking docks
at the site. The product costs $400 to produce at Plant 1 and $300 to produce at Plant 2. The shipping cost from each plant to each potential warehouse location is shown in the first table below. The fixed leasing and operating cost (if open), the ship- ping costs, and the capacity (maximum monthly shipments) of each potential warehouse location are shown in the second table below. The monthly demand in each of the customer regions is expected to be 200, 225, 100, 150, and 175 units, respectively. Formulate and solve a BIP model in a spreadsheet to determine which warehouse(s) should be used and how the product should be distributed from plant to warehouse(s) to customer.
Shipping Costs and Capacity of the Plants
Shipping Cost (per unit) Capacity (units/month)WH 1 WH 2 WH 3
Plant 1 $25 $50 $75 500 Plant 2 $50 $75 $25 400
Fixed Cost, Shipping Costs, and Capacity of the Warehouses
Fixed Cost (per month)
Shipping Cost (per unit) Capacity
(units/month)Region 1 Region 2 Region 3 Region 4 Region 5
WH 1 $50,000 $30 $70 $75 $55 $40 700 WH 2 $30,000 $55 $30 $45 $45 $70 500 WH 3 $70,000 $70 $30 $50 $60 $55 1,000
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Chapter 7 Problems 257
as well as the amount of investment required to undertake the project.
Development Project
1 2 3 4 5
Estimated profit (millions) $1 $ 1.8 $ 1.6 $0.8 $1.4 Capital required (millions) 6 12 10 4 8
The owners of the firm, Dave Peterson and Ron Johnson, have raised $20 million of investment capital for these projects. Dave and Ron now want to select the combination of projects that will maximize their total estimated long-run profit (net present value) without investing more than $20 million. a. Formulate a BIP model in algebraic form for this
problem. E* b. Formulate and solve this model on a spreadsheet. E* c. Perform sensitivity analysis on the amount of
investment capital made available for the devel- opment projects by generating a parameter analy- sis report with RSPE to solve the model with the following amounts of investment capital (in mil- lions of dollars): 16, 18, 20, 22, 24, 26, 28, and 30. Include both the changing cells and the objective cell as output cells in the parameter analysis report.
E*7.6. The board of directors of General Wheels Co. is con- sidering seven large capital investments. Each investment can be made only once. These investments differ in the estimated long-run profit (net present value) that they will generate as well as in the amount of capital required, as shown by the fol- lowing table.
Investment Opportunity
Estimated Profit (millions)
Capital Required (millions)
1 $17 $43 2 10 28 3 15 34 4 19 48 5 7 17 6 13 32 7 9 23
The total amount of capital available for these investments is $100 million. Investment opportunities 1 and 2 are mutually exclusive, and so are 3 and 4. Furthermore, neither 3 nor 4 can be undertaken unless one of the first two opportunities is under- taken. There are no such restrictions on investment opportunities 5, 6, and 7. The objective is to select the combination of capi- tal investments that will maximize the total estimated long-run profit (net present value). a. Formulate and solve a BIP model on a spreadsheet
for this problem. b. Perform sensitivity analysis on the amount of
capital made available for the investment oppor- tunities by generating a parameter analysis report
To the left of the problems (or their parts), we have inserted an E* whenever Excel should be used (unless your instructor gives you contrary instructions). An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
7.1. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 7.1. Briefly describe how mixed BIP was applied in this study. Then list the various financial and nonfi- nancial benefits that resulted from this study. 7.2. Reconsider the California Manufacturing Co. case study presented in Section 7.1. The mayor of San Diego now has contacted the company’s president, Armando Ortega, to try to persuade him to build a factory and perhaps a warehouse in that city. With the tax incentives being offered the company, Armando’s staff estimates that the net present value of building a factory in San Diego would be $7 million and the amount of capital required to do this would be $4 million. The net present value of building a warehouse there would be $5 million and the capital required would be $3 million. (This option will only be considered if a factory also is being built there.)
Armando has asked Steve Chan to revise his previous man- agement science study to incorporate these new alternatives into the overall problem. The objective still is to find the feasible combination of investments that maximizes the total net present value, given that the amount of capital available for these invest- ments is $10 million. a. Formulate a BIP model in algebraic form for this
problem. E* b. Formulate and solve this model on a spreadsheet. 7.3.* A young couple, Eve and Steven, want to divide their main household chores (marketing, cooking, dishwashing, and laundering) between them so that each has two tasks but the total time they spend on household duties is kept to a minimum. Their efficiencies on these tasks differ, where the time each would need to perform the task is given by the following table.
Time Needed per Week (Hours)
Marketing Cooking Dish Washing Laundry
Eve 4.5 7.8 3.6 2.9 Steven 4.9 7.2 4.3 3.1
a. Formulate a BIP model in algebraic form for this problem.
E* b. Formulate and solve this model on a spreadsheet. 7.4. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 7.2. Briefly describe how mixed binary inte- ger programming was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 7.5. A real-estate development firm, Peterson and Johnson, is considering five possible development projects. Using units of millions of dollars, the following table shows the estimated long- run profit (net present value) that each project would generate,
Problems
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258 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
columns and the only paths considered always move forward one column at a time.
3
5
6
(Destination)(Origin)
3
2
3
6
4
C
O
A
B D
T
The numbers along the links represent distances (in miles), and the objective is to find the shortest path from the origin to the destination.
This problem also can be formulated as a BIP model involv- ing both mutually exclusive alternatives and contingent deci- sions. Formulate and solve this BIP model on a spreadsheet. Identify the constraints for (1) mutually exclusive alternatives and (2) contingent decisions. 7.9. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 7.4. Briefly describe how BIP and related techniques were applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 7.10. Speedy Delivery provides two-day delivery service of large parcels across the United States. Each morning at each col- lection center, the parcels that have arrived overnight are loaded onto several trucks for delivery throughout the area. Since the competitive battlefield in this business is speed of delivery, the parcels are divided among the trucks according to their geo- graphical destinations to minimize the average time needed to make the deliveries.
On this particular morning, the dispatcher for the Blue River Valley Collection Center, Sharon Lofton, is hard at work. Her three drivers will be arriving in less than an hour to make the day’s deliveries. There are nine parcels to be delivered, all at locations many miles apart. As usual, Sharon has loaded these locations into her computer. She is using her company’s special software package, a decision support system called Dispatcher. The first thing Dispatcher does is use these locations to generate a con- siderable number of attractive possible routes for the individual delivery trucks. These routes are shown in the table below (where the numbers in each column indicate the order of the deliveries), along with the estimated time required to traverse the route.
Dispatcher is an interactive system that shows these routes to Sharon for her approval or modification. (For example, the
with RSPE to solve the model with the following amounts of capital (in millions of dollars): 80, 90, 100, 110, . . . , and 200. Include both the chang- ing cells and the objective cell as output cells in the parameter analysis report.
E*7.7. The Fly-Right Airplane Company builds small jet air- planes to sell to corporations for use by their executives. To meet the needs of these executives, the company’s customers sometimes order a custom design of the airplanes being pur- chased. When this occurs, a substantial start-up cost is incurred to initiate the production of these airplanes.
Fly-Right has recently received purchase requests from three customers with short deadlines. However, because the com- pany’s production facilities already are almost completely tied up filling previous orders, it will not be able to accept all three orders. Therefore, a decision now needs to be made on the num- ber of airplanes the company will agree to produce (if any) for each of the three customers.
The relevant data are given in the next table. The first row gives the start-up cost required to initiate the production of the airplanes for each customer. Once production is under way, the marginal net revenue (which is the purchase price minus the marginal production cost) from each airplane produced is shown in the second row. The third row gives the percentage of the available production capacity that would be used for each airplane produced. The last row indicates the maximum number of airplanes requested by each customer (but less will be accepted).
Customer
1 2 3
Start-up cost $3 million $2 million 0 Marginal net revenue $2 million $3 million $0.8 million Capacity used per plane 20% 40% 20% Maximum order 3 planes 2 planes 5 planes
Fly-Right now wants to determine how many airplanes to produce for each customer (if any) to maximize the company’s total profit (total net revenue minus start-up costs). Formulate and solve a spreadsheet model with both integer variables and binary variables for this problem. E*7.8. Consider the following special type of shortest path problem (discussed in Section 6.4) where the nodes are in
Attractive Possible Route
Delivery Location 1 2 3 4 5 6 7 8 9 10
A 1 1 1 B 2 1 2 2 2 C 3 3 3 3 D 2 1 1 E 2 2 3 F 1 2 G 3 1 2 3 H 1 3 1 I 3 4 2
Time (in hours) 6 4 7 5 4 6 5 3 7 6
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station to minimize the total cost of stations while ensuring that each tract has at least one station close enough to respond to a medical emergency in no more than 15 minutes. In con- trast to the original problem, note that the total number of paramedic stations is no longer fixed. Furthermore, if a tract without a station has more than one station within 15 minutes, it is no longer necessary to assign this tract to just one of these stations. a. Formulate the algebraic form of a pure BIP model
with five binary variables for this problem. E* b. Display and solve this model on a spreadsheet. 7.13. Reconsider the Southwestern Airways crew scheduling problem presented in Section 7.4. Because of a blizzard in the Chicago area, all the flights into and out of Chicago (including flights 4, 6, 7, and 9 in Table 7.5 ) have been canceled for the time being, so a new crew scheduling plan needs to be devel- oped to cover the seven remaining flights in Table 7.5 .
The 12 feasible sequences of flights still are the ones shown in Table 7.5 after deleting the canceled flights. When flights into and out of Chicago had originally been part of a sequence, a crew now would fly as passengers on a Southwest- ern Airways flight to the next city in the sequence to cover the remaining flights in the sequence. For example, flight sequence 4 now would be San Francisco to Los Angeles to Denver to San Francisco, where a crew would fly as passen- gers on a flight from Los Angeles to Denver (not shown in the table) to enable serving as the crew from Denver to San Francisco. (Since the original sequence 5 included a round- trip from Denver to Chicago and back, a crew assigned to this sequence now would simply lay over in Denver to await the flight from Denver to San Francisco.) The cost of assigning a crew to any sequence still would be the same as shown in the bottom row of Table 7.5 .
The objective still is to minimize the total cost of the crew assignments that cover all the flights. The fact that only 7 flights
computer may not know that flooding has made a particular route infeasible.) After Sharon approves these routes as attrac- tive possibilities with reasonable time estimates, Dispatcher next formulates and solves a BIP model for selecting three routes that minimize their total time while including each delivery location on exactly one route. E* a. Using the data in the table, demonstrate how Dis-
patcher can formulate and solve this BIP model on a spreadsheet.
b. Describe how the problem addressed in part a is analogous to the crew scheduling problem described in Section 7.4.
E*7.11. An increasing number of Americans are moving to a warmer climate when they retire. To take advantage of this trend, Sunny Skies Unlimited is undertaking a major real-estate development project. The project is to develop a completely new retirement community (to be called Pilgrim Haven) that will cover several square miles. One of the decisions to be made is where to locate the two paramedic stations that have been allocated to the community to respond to medical emergen- cies. For planning purposes, Pilgrim Haven has been divided into five tracts, with no more than one paramedic station to be located in any given tract. Each station is to respond to all the medical emergencies that occur in the tract in which it is located as well as in the other tracts that are assigned to this station. Thus, the decisions to be made consist of (1) the tracts to receive a paramedic station and (2) the assignment of each of the other tracts to one of the paramedic stations. The objective is to minimize the overall average of the response times to medical emergencies.
The following table gives the average response time to a medical emergency in each tract (the rows) if that tract is served by a station in a given tract (the columns). The last column gives the forecasted average number of medical emergencies that will occur in each of the tracts per day.
Fire Station in Tract Average Frequency of Medical Emergencies
per Day1 2 3 4 5
Response 1 5 20 15 25 10 2 times (min.) 2 12 4 20 15 25 1 to a medical 3 30 15 6 25 15 3 emergency 4 20 10 15 4 12 1 in tract 5 15 25 12 10 5 3
Formulate and solve a BIP model on a spreadsheet for this problem. Identify any constraints that correspond to mutually exclusive alternatives or contingent decisions.
7.12. Reconsider Problem 7.11. The management of Sunny Skies Unlimited now has decided that the decision regarding the locations of the paramedic stations should be based mainly on costs.
The cost of locating a paramedic station in a tract is $200,000 for tract 1, $250,000 for tract 2, $400,000 for tract 3, $300,000 for tract 4, and $500,000 for tract 5. Management’s objective now is to determine which tracts should receive a
now need to be covered instead of 11 increases the chance that fewer than three crews will need to be assigned to a flight sequence this time. (The flights where these crews fly as passen- gers do not need to be covered since they already are assigned to crews that are not based in San Francisco.) a. Formulate a BIP model in algebraic form for this
problem. E* b. Formulate and solve this problem on a spreadsheet. 7.14. Yakima Construction Corporation (YCC) is consid- ering a number of different development projects. The cash outflows that would be required to complete each project are
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a. Formulate a BIP model in algebraic form for this problem.
E* b. Formulate and solve this model on a spreadsheet. 7.17. The school board for the Bellevue School District has made the decision to purchase 1,350 additional Macintosh com- puters for computer laboratories in all its schools. Based on past experience, the school board also has directed that these computers should be purchased from some combination of three companies— Educomp, Macwin, and McElectronics. In all three cases, the companies charge a discounted variable cost per computer and a fixed delivery and installation cost for these large sales to school districts. The table below shows these charges as well as the capacity (the maximum number of computers that can be sold from the limited inventory) for each of the companies.
Educomp Macwin McElectronics
Capacity 700 700 1,000 Fixed cost $45,000 $35,000 $50,000 Variable cost $750 $775 $700
The school board wants to determine the minimum-cost plan for meeting its computer needs. a. Formulate a BIP model in algebraic form for this
problem. E* b. Formulate and solve this model on a spreadsheet. E* c. Now suppose that Macwin has not submitted its
final bid yet, so the per computer cost is not known with certainty. Generate a parameter analysis report with RSPE to show the optimal order quantities and total cost of the optimal solution when the cost per computer for Macwin is $680, $690, $700, $710, . . . , $790, or $800.
E*7.18. Noble Amazon sells books online. Management is try- ing to determine the best sites for the company’s warehouses. The five potential sites under consideration are listed in the first column of the table at the bottom of the page. Most of the sales come from customers in the United States. The average weekly demand from each region of the country, the average shipping
indicated in the table below, along with the expected net present value of each project (all values in millions of dollars).
Project
1 2 3 4 5
Year 1 $ 8 $10 $12 $4 $14 Year 2 6 8 6 3 6 Year 3 3 7 6 2 5 Year 4 0 5 6 0 7
NPV $12 $15 $20 $9 $23
Each project must be done in full (with the corresponding cash flows for all four years) or not done at all. Furthermore, there are the following additional considerations. Project 1 cannot be done unless Project 2 is also undertaken, and projects 3 and 4 would compete with each other, so they should not both be cho- sen. YCC expects to have the following cash available to invest in these projects: $40 million for year 1, $25 million for year 2, $16 million for year 3, and $12 million for year 4. Any available money not spent in a given year is then available to spend the following year. YCC’s policy is to choose their projects so as to maximize their total expected NPV. a. Formulate a BIP model in algebraic form for this
problem. E* b. Formulate and solve this model on a spreadsheet. 7.15. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 7.5. Briefly describe how mixed BIP was applied in this study. Then list the various financial and nonfi- nancial benefits that resulted from this study. 7.16. An electrical utility needs to generate 6,500 megawatts of electricity today. It has five generators. If any electricity is generated by a given generator, that generator must be started up and a fixed start-up cost is incurred. There is an additional cost for each megawatt generated by a generator. These costs, as well as the maximum capacity of each generator, are shown in the following table. The objective is to determine the minimum cost plan that meets the electrical needs for today.
Generator
A B C D E
Fixed start-up cost $3,000 $2,000 $2,500 $1,500 $1,000 Cost per megawatt /day generated $5 $4 $6 $6 $7 Maximum capacity (MW/day) 2,100 1,800 2,500 1,500 3,000
Average Shipping Cost ($/book) Fixed Cost
(per week)
Warehouse Capacity
(books/week)Warehouse Site Northwest Southwest Midwest Southeast Northeast
Spokane, WA $2.40 $3.50 $4.80 $6.80 $5.75 $40,000 20,000 Reno, NV $3.25 $2.30 $3.40 $5.25 $6.00 $30,000 20,000 Omaha, NE $4.05 $3.25 $2.85 $4.30 $4.75 $25,000 15,000 Harrisburg, PA $5.25 $6.05 $4.30 $3.25 $2.75 $40,000 25,000 Jacksonville, FL $6.95 $5.85 $4.80 $2.10 $3.50 $30,000 15,000
Customer demand (per week)
8,000 12,000 9,000 14,000 17,000
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Case 7-1 Assigning Art 261
cost from each warehouse site to each region of the country, the fixed cost per week of each warehouse if it is operated, and the maximum capacity of each warehouse (if it is operated) are shown in the table. Formulate and solve a mixed BIP model in a spreadsheet to determine which warehouse sites Noble Amazon should operate and how books should be distributed from each warehouse to each region of the country to minimize total cost. E*7.19. Aberdeen Computer Corp. (ACC) is located in Aber- deen, Washington. The company has developed the WebSurfer, a low-cost e-mail and Web-surfing appliance. This product is manufactured at four plants, located in Atlanta, Kansas City, Aberdeen, and Austin. After production, the WebSurfers are shipped to three warehouses, located in Nashville, San Jose, and Houston. ACC sells the WebSurfers through the retail channel. In particular, five different retailers currently sell the WebSurfer—Sears, Best Buy, Fry’s, Comp USA, and Office Max. ACC makes weekly shipments to the main warehouses of these five retailers. The shipping cost from each plant to each warehouse, along with the production cost and weekly produc- tion capacity at each plant, are given in the table below.
Shipping Cost ($/unit) Production
Cost ($/unit) Capacity
(units/week)Plant Nashville San Jose Houston
Atlanta $30 $40 $50 $208 200 Kansas City $25 $45 $40 $214 300 Aberdeen $45 $30 $55 $215 300 Austin $30 $50 $30 $210 400
The shipping cost from each warehouse to each customer, the variable cost (cost per unit moved through the warehouse), the capacity (maximum number of units that can be moved through the warehouse per week) for each warehouse, and the weekly demand for each customer are given in the table below.
Shipping Cost ($/unit) Variable
Cost ($/unit)
Capacity (units/week)Warehouse Sears
Best Buy Fry’s
Comp USA
Office Max
Nashville $40 $45 $30 $25 $20 $4 300 San Jose $15 $50 $25 $15 $40 $5 500 Houston $50 $35 $15 $40 $50 $5 500
Customer demand (per week)
100 50 75 300 150
production and distribution of the WebSurfer from the various plants, through the warehouses, to the customers that will minimize total costs.
Plant Fixed Cost ($/week)
Atlanta $8,000 Kansas City $9,000 Aberdeen $9,000 Austin $10,000
Warehouse Fixed Cost ($/week)
Nashville $4,000 San Jose $5,000 Houston $5,000
b. Now suppose that ACC is considering saving money by closing some of its production facilities
and/or warehouses. Suppose there is a fixed cost to operate each plant and each warehouse as indi- cated in the tables above. Add binary variables to your model in part a to incorporate the decision of which plants and warehouses to keep open so as to
Case 7-1
Assigning Art
It had been a dream come true for Ash Briggs, a struggling art- ist living in the San Francisco Bay area. He had made a trip to the corner grocery store late one Friday afternoon to buy some milk, and, on impulse, he had also purchased a California lottery ticket. One week later, he was a multimillionaire.
Ash did not want to squander his winnings on materialistic, triv- ial items. Instead he wanted to use his money to support his true pas- sion: art. Ash knew all too well the difficulties of gaining recognition as an artist in this post-industrial, technological society where artistic appreciation is rare and financial support even rarer. He therefore
a. Formulate and solve a linear programming model in a spreadsheet to determine the plan for weekly
minimize total cost (including the fixed costs for any plant or warehouse that is operated).
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262 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
wire mesh sculpture to the delicate lines in a two-dimensional computer-generated drawing. He therefore wants at least one wire-mesh sculpture displayed if a computer-generated draw- ing is displayed. Alternatively, he wants at least one computer- generated drawing displayed if a wire-mesh sculpture is dis- played. Furthermore, Ash wants to expose viewers to all painting styles, but he wants to limit the number of paintings displayed to achieve a balance in the exhibit between paintings and other art forms. He therefore decides to include at least one photo-realistic painting, at least one cubist painting, at least one expressionist painting, at least one watercolor painting, and at least one oil painting. At the same time, he wants the number of paintings to be no greater than twice the number of other art forms.
Ash wants all his own paintings included in the exhibit since he is sponsoring the exhibit and since his paintings celebrate the San Francisco Bay area, the home of the exhibit.
Ash possesses personal biases for and against some artists. Ash is currently having a steamy affair with Candy Tate, and he wants both of her paintings displayed. Ash counts both David Lyman and Rick Rawls as his best friends, and he does not want
decided to use the money to fund an exhibit of up-and-coming mod- ern artists at the San Francisco Museum of Modern Art.
Ash approached the museum directors with his idea, and the directors became excited immediately after he informed them that he would fund the entire exhibit in addition to donating $1 million to the museum. Celeste McKenzie, a museum director, was assigned to work with Ash in planning the exhibit. The exhibit was slated to open one year from the time Ash met with the directors, and the exhibit pieces would remain on display for two months.
Ash began the project by combing the modern art commu- nity for potential artists and pieces. He presented a list (shown below) of artists, their pieces, and the price of displaying each piece 1 to Celeste.
Ash possesses certain requirements for the exhibit. He believes the majority of Americans lack adequate knowledge of art and artistic styles, and he wants the exhibit to educate Ameri- cans. Ash wants visitors to become aware of the collage as an art form, but he believes collages require little talent. He there- fore decides to include only one collage. Additionally, Ash wants viewers to compare the delicate lines in a three-dimensional
1 The display price includes the cost of paying the artist for loaning the piece to the museum, transporting the piece to San Francisco, con- structing the display for the piece, insuring the piece while it is on display, and transporting the piece back to its origin.
Artist Piece Description of Piece Price
Colin Zweibell Perfection A wire-mesh sculpture of the human body $300,000 Burden A wire-mesh sculpture of a mule 250,000 The Great Equalizer A wire-mesh sculpture of a gun 125,000
Rita Losky Chaos Reigns A series of computer-generated drawings 400,000 Who Has Control? A computer-generated drawing intermeshed with lines
of computer code 500,000
Domestication A pen-and-ink drawing of a house 400,000 Innocence A pen-and-ink drawing of a child 550,000
Norm Marson Aging Earth A sculpture of trash covering a larger globe 700,000 Wasted Resources A collage of various packaging materials 575,000
Candy Tate Serenity An all-blue watercolor painting 200,000 Calm before the Storm A painting with an all-blue watercolor background and a black
watercolor center 225,000
Robert Bayer Void An all-black oil painting 150,000 Sun An all-yellow oil painting 150,000
David Lyman Storefront Window A photo-realistic painting of a jewelry store display window 850,000 Harley A photo-realistic painting of a Harley-Davidson motorcycle 750,000
Angie Oldman Consumerism A collage of magazine advertisements 400,000 Reflection A mirror (considered a sculpture) 175,000 Trojan Victory A wooden sculpture of a condom 450,000
Rick Rawls Rick A photo-realistic self-portrait (painting) 500,000 Rick II A cubist self-portrait (painting) 500,000 Rick III An expressionist self-portrait (painting) 500,000
Bill Reynolds Beyond A science fiction oil painting depicting Mars colonization 650,000 Pioneers An oil painting of three astronauts aboard the space shuttle 650,000
Bear Canton Wisdom A pen-and-ink drawing of an Apache chieftain 250,000 Superior Powers A pen-and-ink drawing of a traditional Native American rain dance 350,000 Living Land An oil painting of the Grand Canyon 450,000
Helen Row Study of a Violin A cubist painting of a violin 400,000 Study of a Fruit Bowl A cubist painting of a bowl of fruit 400,000
Ziggy Lite My Namesake A collage of Ziggy cartoons 300,000 Narcissism A collage of photographs of Ziggy Lite 300,000
Ash Briggs All That Glitters A watercolor painting of the Golden Gate Bridge 50,000* The Rock A watercolor painting of Alcatraz 50,000* Winding Road A watercolor painting of Lombard Street 50,000* Dreams Come True A watercolor painting of the San Francisco Museum of Modern Art 50,000*
*Ash does not require personal compensation, and the cost for moving his pieces to the museum from his home in San Francisco is minimal. The cost of displaying his pieces therefore only includes the cost of constructing the display and insuring the pieces.
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Case 7-2 Stocking Sets 263
Please explore the following questions independently except where otherwise indicated.
a. Ash decides to allocate $4 million to fund the exhibit. Given the pieces available and the specific requirements from Ash and Celeste, formulate and solve a binary integer program- ming problem to maximize the number of pieces displayed in the exhibit without exceeding the budget. How many pieces are displayed? Which pieces are displayed?
b. To ensure that the exhibit draws the attention of the pub- lic, Celeste decides that it must include at least 20 pieces. Formulate and solve a binary integer programming prob- lem to minimize the cost of the exhibit while displaying at least 20 pieces and meeting the requirements set by Ash and Celeste. How much does the exhibit cost? Which pieces are displayed?
c. An influential patron of Rita Losky’s work who chairs the museum’s board of directors learns that Celeste requires at least 20 pieces in the exhibit. He offers to pay the minimum amount required on top of Ash’s $4 million to ensure that exactly 20 pieces are displayed in the exhibit and that all of Rita’s pieces are displayed. How much does the patron have to pay? Which pieces are displayed?
to play favorites among these two artists. He therefore decides to display as many pieces from David Lyman as from Rick Rawls and to display at least one piece from each of them. Although Ziggy Lite is very popular within art circles, Ash believes Ziggy makes a mockery of art. Ash will therefore only accept one dis- play piece from Ziggy, if any at all.
Celeste also possesses her own agenda for the exhibit. As a museum director, she is interested in representing a diverse pop- ulation of artists, appealing to a wide audience, and creating a politically correct exhibit. To advance feminism, she decides to include at least one piece from a female artist for every two pieces included from a male artist. To advance environmentalism, she decides to include either one or both of the pieces Aging Earth and Wasted Resources. To advance Native American rights, she decides to include at least one piece by Bear Canton. To advance science, she decides to include at least one of the following pieces: Chaos Reigns, Who Has Control?, Beyond, and Pioneers.
Celeste also understands that space is limited at the museum. The museum only has enough floor space for four sculptures and enough wall space for 20 paintings, collages, and drawings.
Finally, Celeste decides that if Narcissism is displayed, Reflection should also be displayed since Reflection also sug- gests narcissism.
Case 7-2
Stocking Sets Daniel Holbrook, an expediter at the local warehouse for Furniture City, sighed as he moved boxes and boxes of inventory to the side to reach the shelf where the particular item he needed was located. He dropped to his hands and knees and squinted at the inven- tory numbers lining the bottom row of the shelf. He did not find the number he needed. He worked his way up the shelf until he found the number matching the number on the order slip. Just his luck! The item was on the top row of the shelf! Daniel walked back through the warehouse to find a ladder, stumbling over boxes of inventory littering his path. When he finally climbed the ladder to reach the top shelf, his face crinkled in frustration. Not again! The item he needed was not in stock! All he saw above the inventory number was an empty space covered with dust!
Daniel trudged back through the warehouse to make the dreaded phone call. He dialed the number of Brenda Sims, the saleswoman on the kitchen showroom floor of Furniture City, and informed her that the particular light fixture the customer had requested was not in stock. He then asked her if she wanted him to look for the rest of the items in the kitchen set. Brenda told him that she would talk to the customer and call him back.
Brenda hung up the phone and frowned. Mr. Davidson, her cus- tomer, would not be happy. Ordering and receiving the correct light fixture from the regional warehouse would take at least two weeks.
Brenda then paused to reflect upon business during the last month and realized that over 80 percent of the orders for kitchen sets could not be filled because items needed to complete the sets were not in stock at the local warehouse. She also realized that Furniture City was losing customer goodwill and business because of stockouts. The furniture megastore was gaining a reputation for slow service and delayed deliveries, causing cus- tomers to turn to small competitors that sold furniture directly from the showroom floor.
Brenda decided to investigate the inventory situation at the local warehouse. She walked the short distance to the building next door and gasped when she stepped inside the warehouse. What she saw could only be described as chaos. Spaces allo- cated for some items were overflowing into the aisles of the warehouse while other spaces were completely bare. She walked over to one of the spaces overflowing with inventory to deter- mine what item was overstocked. She could not believe her eyes! The warehouse had at least 30 rolls of pea-green wallpa- per! No customer had ordered pea-green wallpaper since 1973!
Brenda marched over to Daniel demanding an explanation. Daniel said that the warehouse had been in such a chaotic state since his arrival one year ago. He said the inventory problems occurred because management had a policy of stocking every furniture item on the showroom floor in the local warehouse. Management only replenished inventory every three months, and when inventory was replenished, management ordered every item regardless of whether it had been sold. Daniel also said that he had tried to make management aware of the problems with overstock- ing unpopular items and understocking popular items, but manage- ment would not listen to him because he was simply an expediter.
Brenda understood that Furniture City required a new inven- tory policy. Not only was the megastore losing money by making customers unhappy with delivery delays, but it was also losing money by wasting warehouse space. By changing the inventory policy to stock only popular items and replenish them immedi- ately when sold, Furniture City would ensure that the majority of customers would receive their furniture immediately and that the valuable warehouse space would be utilized effectively.
Brenda needed to sell her inventory policy to management. Using her extensive sales experience, she decided that the most effective sales strategy would be to use her kitchen department
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as a model for the new inventory policy. She would identify all kitchen sets comprising 85 percent of customer orders. Given the fixed amount of warehouse space allocated to the kitchen department, she would identify the items Furniture City should stock to satisfy the greatest number of customer orders. She would then calculate the revenue from satisfying customer orders under the new inventory policy, using the bottom line to persuade management to accept her policy.
Brenda analyzed her records over the past three years and determined that 20 kitchen sets were responsible for 85 percent of the customer orders. These 20 kitchen sets were composed of up to eight features in a variety of styles. Brenda listed each feature and its popular styles in the tables below.
Brenda then created a table (given on the next page) showing the 20 kitchen sets and the particular features composing each set. To simplify the table, she used the codes shown in paren- theses below to represent the particular feature and style. For example, kitchen set 1 consists of floor tile T2, wallpaper W2, light fixture L4, cabinet C2, countertop O2, dishwasher D2, sink S2, and range R2. Notice that sets 14 through 20 do not contain dishwashers.
Floor Tile Wallpaper Light Fixtures Cabinets
(T1) White textured tile (W1) Plain ivory paper (L1) One large rectangular frosted fixture (C1) Light solid wood cabinets (T2) Ivory textured tile (W2) Ivory paper with
dark brown pinstripes (L2) Three small square frosted fixtures (C2) Dark solid wood cabinets
(T3) White checkered tile with blue trim
(W3) Blue paper with marble texture
(L3) One large oval frosted fixture (C3) Light-wood cabinets with glass doors
(T4) White checkered tile with light yellow trim
(W4) Light yellow paper with marble texture
(L4) Three small frosted globe fixtures (C4) Dark-wood cabinets with glass doors
Countertops Dishwashers Sinks Ranges
(O1) Plain light-wood countertops
(D1) White energy-saving dishwasher (S1) Sink with separate hot and cold water taps
(R1) White electric oven
(O2) Stained light-wood countertops
(D2) Ivory energy-saving dishwasher (S2) Divided sink with separate hot and cold water taps and garbage disposal
(R2) Ivory electric oven
(O3) White lacquer-coated countertops
(S3) Sink with one hot and cold water tap
(R3) White gas oven
(O4) Ivory lacquer-coated countertops
(S4) Divided sink with one hot and cold water tap and gar bage disposal
(R4) Ivory gas oven
Brenda knew she had only a limited amount of warehouse space allocated to the kitchen department. The warehouse could hold 50 square feet of tile and 12 rolls of wallpaper in the inven- tory bins. The inventory shelves could hold two light fixtures, two cabinets, three countertops, and two sinks. Dishwashers and ranges are similar in size, so Furniture City stored them in similar locations. The warehouse floor could hold a total of four dishwashers and ranges.
Every kitchen set always includes exactly 20 square feet of tile and exactly five rolls of wallpaper. Therefore, 20 square feet of a particular style of tile and five rolls of a particular style of wallpaper are required for the styles to be in stock.
a. Formulate and solve a binary integer programming problem to maximize the total number of kitchen sets (and thus the number of customer orders) Furniture City stocks in the local warehouse. Assume that when a customer orders a kitchen set, all the particular items composing that kitchen set are replenished at the local warehouse immediately.
b. How many of each feature and style should Furniture City stock in the local warehouse? How many different kitchen sets are in stock?
c. Furniture City decides to discontinue carrying nursery sets, and the warehouse space previously allocated to the nurs- ery department is divided between the existing departments at Furniture City. The kitchen department receives enough additional space to allow it to stock both styles of dishwash- ers and three of the four styles of ranges. How does the opti- mal inventory policy for the kitchen department change with this additional warehouse space?
d. Brenda convinces management that the kitchen department should serve as a testing ground for future inventory policies. To provide adequate space for testing, management decides
to allocate all the space freed by the nursery department to the kitchen department. The extra space means that the kitchen department can store not only the dishwashers and ranges from part c, but also all sinks, all countertops, three of the four light fixtures, and three of the four cabinets. How much does the additional space help?
e. How would the inventory policy be affected if the items composing a kitchen set could not be replenished immedi- ately? Under what conditions is the assumption of immediate replenishment nevertheless justified?
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Case 7-3
Assigning Students to Schools (Revisited)
Reconsider Case 3-5. The Springfield School Board now has made the decision to prohibit the splitting of residential areas among multiple schools. Thus, each of the six areas must be assigned to a single school.
a. Formulate and solve a BIP model for this problem under the current policy of providing busing for all middle school stu- dents who must travel more than approximately a mile.
b. Referring to part c of Case 3-5, determine how much the total busing cost increases because of the decision to prohibit the splitting of residential areas among multiple schools.
c, d, e, f. Repeat parts d, e, f, g of Case 3-5 under the new school board decision to prohibit splitting of residential areas among multiple schools.
Case 7-4
Broadcasting the Olympic Games (Revisited)
Reconsider part b of Case 6-4. Use the spreadsheet model devel- oped there to incorporate the following consideration.
An additional concern not considered in Case 6-4 is that the routers at nodes C and F are each already maxed out. If any additional capacity is built into or out of node C or node F, then the routers at these stations must be upgraded. This would cost a total of $2 million at node C or a total of $3 mil- lion at node F. Which network segments should be increased in capacity and by how much so as to increase the total capacity of the network enough to transmit the peak requirement of 35
GB/s from the Olympics site (A) to the home studios (G) at the lowest possible cost?
Additional Cases An additional case for this chapter also is available at the Uni- versity of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
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Chapter Eight
Nonlinear Programming Learning Objectives
After completing this chapter, you should be able to
1. Describe how a nonlinear programming model differs from a linear programming model.
2. Recognize when a nonlinear programming model is needed to represent a problem.
3. Formulate a nonlinear programming model from a description of the problem.
4. Construct nonlinear formulas needed for nonlinear programming models.
5. Distinguish between nonlinear programming problems that should be easy to solve and those that may be difficult (if not impossible) to solve.
6. Use the Nonlinear Solver to solve simple types of nonlinear programming problems.
7. Use the multistart feature of Solver to attempt to solve some more difficult nonlin- ear programming problems.
8. Use Evolutionary Solver to attempt to solve some difficult nonlinear programming problems.
9. Recognize when the separable programming technique is applicable to enable using linear programming with a nonlinear objective function.
10. Apply the separable programming technique when applicable.
11. Use RSPE to analyze a model and choose the most appropriate solving method.
The previous chapters have introduced you to a wide variety of management science models, including various types of linear programming and integer programming models. However, one characteristic shared by all these linear programming and integer programming models is that they all are linear models, that is, models where all the functions (mathematical relation- ships) involved are linear.
When formulating a linear model in a spreadsheet, this means that the Excel functions being used to express the formulas in output cells include only sums (e.g., C1 1 C2, or SUM(C1:C2), or C1 2 C2) or products of a number (or data cell) and a changing cell (e.g., 2*C4 or the SUMPRODUCT of data cells with changing cells). If any output cell includes the multiplication or division of changing cells (e.g., C4*C5 or C3/C6 or C4^2) or uses almost any Excel function other than SUM or SUMPRODUCT (such as ROUND, ABS, IF, MAX, MIN, SQRT, etc.), then the resulting model will typically not be linear.
Table 8.1 gives various examples of formulas that could be entered into output cells when the data cells are in column D and the changing cells are in column C. The formulas on the left are all linear while those on the right are not. The first four examples in each column are quite similar. Can you see why the formulas on the left are linear while those on the right are not? The key to seeing this distinction is that a linear formula permits any calculations that involve only the data cells but restricts each changing cell to having only the most basic arithmetic operations: addition or subtraction and multiplication or division by a constant. By contrast, note how each of the formulas on the right side of Table 8.1 involves more compli- cated operations on changing cells, including such Excel operations as ROUND, MAX, MIN, and ABS, as well as IF and SUMIF operations where the IF portion involves changing cells.
Despite the versatility of linear models, managers occasionally encounter problems where such a model does not quite fit because at least one of the formulas that needs to be entered
A formula automatically becomes nonlinear if it ever multiplies or divides a changing cell by another changing cell or if it assigns an exponent (other than 1) to any changing cell.
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into output cells is not linear. In most cases, this occurs because the formula for the objective cell needs to be nonlinear, although cases also arise occasionally where a nonlinear formula is needed for one or more output cells representing the left-hand side of a functional constraint. If the model is a linear programming model except for having at least one nonlinear formula for an output cell (such as the objective cell), then it is called a nonlinear programming model.
For simplicity, we will focus in this chapter on the common case where the only nonlin- ear formula is for the objective cell. However, do be aware that this approach also can be extended to the other output cells when nonlinear constraints are needed in the model.
Formulating and solving nonlinear programming models often is considerably more chal- lenging than formulating and solving linear programming models. However, these challenges frequently can be overcome, sometimes in relatively straightforward ways. Rest easy. This chapter focuses on the relatively straightforward types of nonlinear programming which require only reasonably routine spreadsheet modeling and the application of the Nonlinear Solver. This is all a manager (or future manager) needs to know about nonlinear program- ming. A management science specialist should be called on to deal with more difficult types of nonlinear programming.
Because of the close relationship between linear and nonlinear programming, it is some- times unclear which technique should be used to analyze a managerial problem. This occurs for problems where the appropriate formula for the objective cell is nonlinear but is reason- ably close to being linear. In this case, one alternative is to use a linear approximation for the formula so that linear programming can be applied. The advantage is greater ease of formulat- ing and solving the model. Since a model is intended to be only an idealized representation of the real problem, this alternative is reasonable if the linear approximation is a good one. However, the major advantage of using nonlinear programming instead is the greater preci- sion it provides in seeking the best solution for the real problem. When the appropriate non- linear programming model is not an overly difficult one to formulate and solve, it makes good sense to use this model. If desired, a linear programming model still can be used to perform some quick preliminary analysis, including some what-if analysis, but the greater precision of nonlinear programming should not be foregone lightly for the final analysis.
Section 8.1 discusses the challenges encountered when using nonlinear programming and introduces the Nonlinear Solver. Fortunately, there are some “easy” types of nonlinear pro- gramming problems that arise fairly frequently. Two such types are presented in Sections 8.2 and 8.3. Section 8.4 then describes how some “difficult” nonlinear programming problems still can be solved by applying the Nonlinear Solver multiple times with different starting solutions. However, the Nonlinear Solver is unable to solve some other nonlinear program- ming problems. Fortunately, both Excel’s Solver (for Excel 2010 and later) and RSPE provide an additional procedure called Evolutionary Solver for coping with such problems. Evolution- ary Solver is described in Section 8.5. Finally, after formulating a model of the problem under consideration, Section 8.6 describes how RSPE can be used to analyze the model and choose the most appropriate solving method.
Nonlinear programming often provides greater precision than linear pro- gramming for analyzing managerial problems.
Linear Formulas Nonlinear Formulas
SUMPRODUCT(D4:D6, C4:C6) SUMPRODUCT(C4:C6, C1:C3) [(D1 1 D2)/D3]* C4 [(C1 1 C2)/C3]* D4 IF(D2 . 5 2, 2*C3, 3*C4) IF(C2 . 5 2, 2*C3, 3*C4) SUMIF(D1:D6, 4, C1:C6) SUMIF(C1:C6, 4, D1:D6) SUM(D4:D6) ROUND(C1) 2*C1 1 3*C4 1 C6 MAX(C1, 3) C1 1 C2 1 C3 MIN(C1, C2)
ABS(C1) SQRT(C1) C1* C2 C1 / C2 C1 ^ 2
Note: Data cells are in D1:D6; changing cells are in C1:C6.
TABLE 8.1 Examples of Linear and Nonlinear Formulas in a Spreadsheet When the Data Cells Are in Col- umn D and the Changing Cells Are in Column C
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8.1 THE CHALLENGES OF NONLINEAR PROGRAMMING
In almost every respect, a nonlinear programming model is indistinguishable from a linear programming model. In both cases, decisions need to be made regarding the levels of a num- ber of activities, where these activity levels can have any value (including a fractional value) that satisfies a number of constraints. The decisions regarding activity levels are to be based on an overall measure of performance. When the model is formulated in a spreadsheet, the changing cells display the activity levels, output cells help to represent the constraints, and the objective cell shows the overall measure of performance.
The only way to distinguish a nonlinear programming model from a linear programming model is to examine the formulas entered into the output cells. It is a nonlinear programming model if one or more of these formulas is nonlinear instead of linear. In many applications such a model has only one nonlinear formula and it is the one entered into the objective cell. (This is the case we focus on in this chapter.)
Despite such a small difference in the appearance of the two kinds of models, their appli- cation differs in three major ways.
1. Nonlinear programming is used to model nonproportional relationships between activity levels and the overall measure of performance, whereas linear programming assumes a proportional relationship.
2. Constructing the nonlinear formula(s) needed for a nonlinear programming model is considerably more difficult than developing the linear formulas used in linear programming.
3. Solving a nonlinear programming model is often much more difficult (if it is possible at all) than solving a linear programming model.
As these comparisons indicate, using nonlinear programming instead of linear programming raises some new challenges. Let us examine these challenges a little more closely.
The Challenge of Nonproportional Relationships When either a linear programming model or a nonlinear programming model is formulated in a spreadsheet, the objective cell needs to show the overall measure of performance that results from the activity levels that are displayed in the changing cells. However, nonlinear programming uses a more complicated relationship between the activity levels and the overall measure of performance than does linear programming.
In the case of linear programming, this relationship is assumed to be a particularly simple one. To illustrate, consider again the Wyndor Glass Co. problem introduced in Section 2.1 and formulated in Section 2.2. The activities for this problem are the production of the special new doors and the special new windows, where the levels of these activities are
D 5 Number of doors to be produced per week W 5 Number of windows to be produced per week
The overall measure of performance is the total weekly profit obtained from the production and sale of these doors and windows. The unit profit has been estimated to be $300 for each door and $500 for each window. The graphs in Figure 8.1 show the resulting relationship between the level of each activity ( D and W ) and the contribution of that activity to the overall measure of performance. The straight line in each graph shows a proportional relationship because the weekly profit from each product is proportional to the production rate for that product. These straight lines also indicate that the objective function
Profit 5 $300D 1 $500W
is linear. The fact that this formula being entered into the objective cell is linear helps to make the overall model a linear programming model.
As illustrated by the Wyndor Glass Co. problem, every linear programming problem assumes a proportional relationship between each activity and the overall measure of perfor- mance. A summary of this key assumption is given just below Figure 8.1.
A nonlinear programming model has the same appear- ance as a linear program- ming model except for having a nonlinear formula in at least one output cell (commonly the objective cell).
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Proportionality Assumption of Linear Programming: The contribution of each activity to the value of the objective function is proportional to the level of the activity. 1 In other words, the term in the objective function involving this activity consists of a coefficient times a decision variable, where the coefficient is the contribution per unit of this activity and the decision variable is the level of this activity. (For example, for each product in the Wyndor Glass Co. problem, the coefficient is the product’s unit profit and the decision variable is the production rate for the product.)
Nonlinear programming problems arise when this assumption is violated. This occurs whenever any activity has a nonproportional relationship with the overall measure of performance because the contribution of the activity to this measure of performance is not proportional to the level of the activity.
Figure 8.2 shows four examples of different types of nonproportional relationships. (For definiteness, these graphs assume that the overall measure of performance is profit, but any other measure to be maximized also could be used.)
The first of these examples, shown in Figure 8.2(a) , illustrates a profit graph with decreas- ing marginal returns.
Consider any activity where a graph of its profit versus the level of the activity is plotted. Sup- pose that the slope (steepness) of the graph never increases but sometimes decreases as the level of the activity increases. Then the activity is said to have decreasing marginal returns .
Similarly, in problems where the objective is to minimize the total cost of the activities, an activity is said to have decreasing marginal returns if the slope of its cost graph never decreases but sometimes increases as the level of the activity increases. 2
Since frequently it is difficult to continue increasing profit at the same rate as the level of an activity keeps getting pushed higher and higher, activities with decreasing marginal returns are fairly common. For example, it may be necessary to lower the price of a product to increase its sales. Alternatively, if the price is held constant, the marketing costs may need to go up more than proportionally to attain increases in the level of sales. (The next section begins with an example in which marketing costs behave in this way.) Decreasing marginal
Nonlinear programming problems arise when the proportionality assumption of linear programming is violated.
Many activities have decreasing marginal returns.
300
0 2
Production rate for doors Production rate for windows
4
600
900
1,200
Weekly
profit
($)
D
500
1,000
1,500
2,000
2,500
3,000
0 2 4 6
Weekly
profit
($)
W
FIGURE 8.1 Profit graphs for the Wyndor Glass Co. that show the weekly profit from each product versus the production rate for that product.
1 The same assumption also is made about the contribution of each activity to the left-hand side of each functional constraint, but we are focusing in this chapter on how to deal with a lack of proportionality in the objective function. 2 Using mathematical terminology, a profit graph with decreasing marginal returns is said to be a concave function, whereas a cost graph with decreasing marginal returns is said to be a convex function. We are using the more suggestive economic term, decreasing marginal returns, to cover both cases (including for functions of multiple decision variables).
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returns also occur when less efficient facilities and personnel need to be used to increase the level of an activity.
Figure 8.2(b) illustrates a profit graph that is piecewise linear because it consists of a sequence of connected line segments. As the level of the activity increases, the slope of the profit graph remains the same within each line segment but then decreases at the kink where the next line segment begins. Since the slope never increases as the level of the activ- ity increases but does decrease at the kinks, this profit graph also fits the definition of having decreasing marginal returns. This kind of graph might occur, for example, because overtime needs to be used to increase the level of the activity beyond the first kink, and then even more expensive weekend overtime is needed to increase the level beyond the second kink.
Figure 8.2(c) provides an example of a nonproportional relationship that does not quite have decreasing marginal returns. The reason it does not is that there are spots called discontinuities where the profit graph is disconnected because it suddenly jumps up or down. Such discontinui- ties could occur, for example, because quantity discounts for purchasing a component of a prod- uct become available when the production level for the product rises above certain thresholds.
Having activities with decreasing marginal returns is not the only way in which the pro- portionality assumption can be violated. For example, another way is to have activities with increasing marginal returns, as illustrated by Figure 8.2(d) . In this case, the slope of the profit graph never decreases but sometimes increases as the level of the activity increases. (Simi- larly, a cost graph exhibits increasing marginal returns if its slope never increases but some- times decreases as the level of the activity increases.) This can occur because of the greater efficiencies sometimes achieved at higher levels of an activity.
Profit graphs are used when the overall objective is to maximize the total profit from all the activities. However, cost graphs are needed instead when the overall objective is to mini- mize the total cost of all the activities. An activity can violate the proportionality assumption in the same ways that are illustrated in Figure 8.2 if its cost graph has any of the shapes shown
An activity has increasing marginal returns if its effi- ciency increases as the level of the activity is increased.
Profit
Level of an activity
(a)
Profit
Level of an activity
(b)
Profit
Level of an activity
(c)
Profit
Level of an activity
(d)
FIGURE 8.2 Examples of profit graphs with nonproportional relationships: (a) decreas- ing marginal returns; (b) piecewise linear with decreasing marginal returns; (c) decreasing marginal returns except for discontinuities; and (d) increasing marginal returns.
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in Figure 8.3 . For each case, note how this cost graph bends in the opposite way from the corresponding profit graph in Figure 8.2 . Thus, an increasing slope in the cost graph reflects decreasing marginal returns, whereas a decreasing slope reflects increasing marginal returns. (The same conclusion applies to graphs where the objective is to minimize some overall mea- sure of performance other than total cost.)
Figures 8.2 and 8.3 illustrate only some of the possible nonproportional relationships. For example, an activity might have neither decreasing marginal returns nor increasing marginal returns because the slope of its graph sometimes decreases and sometimes increases as the level of the activity increases.
In addition, sometimes there are interactions between activities that cause (or help to cause) the objective function to be nonlinear. To illustrate, consider the Wyndor Glass Co. problem again. Suppose now that a major advertising campaign will be required to market either new product if it is produced by itself, but that the same single campaign can be used to effectively promote both products if both are produced. Because a major cost is saved for the second prod- uct, their joint profit is somewhat more than the sum of their individual profits when each is produced by itself. In particular, the appropriate objective function might be, say,
Profit 5 $300D 1 $500W 1 $100DW
where DW denotes the product of D and W. Because of the cross-product term, $100 DW, this objective function is nonlinear even though, when either D or W is fixed at some value, the proportionality assumption still holds for the other product.
When there are interactions between activities, the total profit from all the activities some- times will still have decreasing marginal returns. (The common technical term for this is that the objective function is concave. ) The intuitive interpretation of decreasing marginal returns (a profit graph that never bends up but sometimes bends down) continues to apply here. We will not bother with the complex technical definition that is needed in this case.
Even when the proportion- ality assumption is satis- fied, interactions between activities still can lead to a nonlinear programming model.
Cost
Level of an activity
(a)
Cost
Level of an activity
(b)
Cost
Level of an activity
(c)
Cost
Level of an activity
(d)
FIGURE 8.3 Examples of cost graphs with nonproportional relationships: (a) decreas- ing marginal returns; (b) piecewise linear with decreasing marginal returns; (c) decreasing marginal returns except for discontinuities; and (d) increasing marginal returns. Each cost graph bends in the opposite way from the corresponding profit graph in Figure 8.2.
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The Challenge of Constructing Nonlinear Formulas For a linear programming model, it is relatively simple to construct the formula that needs to be entered into the objective cell by using a SUMPRODUCT function. For example, when the objective cell gives the total profit from all the activities (as for the Wyndor Glass Co. problem), each product being summed is simply the product of the unit profit for an activity (as given in a data cell) and the level of that activity (as given in a changing cell).
Considerably more work is needed for a nonlinear programming problem. Even when there are no interactions between activities, it is necessary to construct a nonlinear formula for each activity that represents the contribution of that activity to the objective function that needs to be entered into the objective cell. For example, when the objective is to maximize total profit, the nonlinear formula for each activity needs to correspond to the profit graph for that activity.
One useful method for fitting a nonlinear formula to a graph begins by assuming a general form for the formula. For a profit graph with decreasing marginal returns, it is common to assume a quadratic form, such as
Profit from an activity 5 ax2 1 bx 1 c
where x is the level of the activity, a is a negative constant, b is a positive constant, and c is any kind of constant. Another possibility is to assume a logarithmic form, like
Profit from an activity 5 a ln(x) 1 b
where ln( x ) is called the natural logarithm of x. In either case, the next step is to find the appropriate values of the parameters (e.g., a, b,
and c ). Excel has a built-in curve fitting method to find the values of the parameters that best fit the data. For example, suppose that prior data (or at least estimates) are available on the profit that would be achieved at several levels of the activity, as shown in the spreadsheet in Figure 8.4 .
The first step in applying the curve fitting method in Excel is to graph the profit data (profit versus level of the activity), using an X–Y scatter chart. Next, right-click on one of the data points and choose “Add Trendline” from the menu that pops up. This brings up the dialog box shown in Figure 8.5 . Use this dialog box to choose the form of the equation that you want Excel to fit to the data. For example, to fit a quadratic equation to the data, choose Polynomial with Order 2.
Next choose the option to “Display Equation on chart” and click on OK. Excel then chooses the parameters for the equation of the chosen form that most closely fits the graphed data. For example, the quadratic equation that most closely matches the data in Figure 8.4 is
Profit 5 20.3002 x2 1 5.661 x 1 6.1477
This quadratic form for a profit graph (or a cost graph) is widely used.
The order of a polynomial is the highest exponent used in the polynomial. For a quadratic equation, the order is 2.
1
A B C
Level of Activity Profit
D E F G H
2
3
4
5
6
7
8
9
10
11
12
13
Constructing a Nonlinear Formula
$35
$30
$25
$20
$15
$10
$5
$0
0 5
Level of Activity (x)
Profit vs. Level of Activity
10
P ro
fi t
2
4
5
7
10
$16
$24
$28
$30
$33
FIGURE 8.4 An example of an activ- ity for which prior data are available on the profit versus the level of the activity, so Excel’s curve fitting method can be applied.
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This equation is shown and plotted directly on the graph of profit versus activity level, as shown in Figure 8.6 .
The quadratic form provides at least a reasonable approximation for many profit graphs, so it is used often. However, it is prudent to check whether the approximation actually is a reasonable one for any particular activity. This is done by estimating the profit that would be obtained by the activity at several different levels in addition to the data being used by the curve fitting method, and then checking whether the profit at these other levels is reasonably close to what is given by the formula. If it is not, one alternative is to collect more data and reapply the curve fitting method to seek a better overall fit. Another alternative is to adopt a different form for the formula (e.g., logarithmic or a polynomial with an order higher than 2) and then to apply the corresponding curve fitting method.
The Challenge of Solving Nonlinear Programming Models It is easy to solve linear programming models with either Solver or a variety of other software packages. Very large problems are solved routinely every day. In fact, the most advanced software packages now are successfully solving amazingly huge problems. Furthermore, the solution obtained is guaranteed to be optimal.
A variety of algorithms also have been developed for solving, or at least attempting to solve, particular kinds of nonlinear programming models. For example, the Nonlinear Solver is a
FIGURE 8.5 The Format Trendline dialog box that is used to perform the curve fitting method in Excel. For this example, the Polynomial of Order 2 option (a qua- dratic equation) is chosen for the type of regression. The option to display the equation on the chart is selected.
1
A B C
Level of Activity Profit
D E F G H
2
3
4
5
6
7
8
9
10
11
12
13
Constructing a Nonlinear Formula
$35
$30
$25
$20
$15
$10
$5
$0
0 5
Level of Activity (x)
Profit vs. Level of Activity
10
P ro
fi t
y = −0.3002x2 + 5.661x + 6.1477
2
4
5
7
10
$16
$24
$28
$30
$33
FIGURE 8.6 The quadratic equation found by Excel that most closely matches the profit versus level of activity data for the example intro- duced in Figure 8.4.
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solving method provided by Solver that uses one of the best of these algorithms to deal with some nonlinear programming models. (The Nonlinear Solver is called by choosing “GRG Nonlinear” as the solving method in either Excel’s Solver dialog box or on the Engine tab of RSPE’s Model pane.)
However, even when using the Nonlinear Solver, dealing with nonlinear programming models can be a real challenge. They often are much more difficult to solve than linear pro- gramming models. Furthermore, even when a solution is obtained, it sometimes cannot be guaranteed to be optimal.
Fortunately, some types of nonlinear programming models are relatively easy to solve. Cases ( a ) and ( b ) in Figure 8.2 (when maximizing) or in Figure 8.3 (when minimizing) are examples of “easy” types of nonlinear programming models, namely, types where the activi- ties have decreasing marginal returns. As long as all the activities fit either case (except for any that still satisfy the proportionality assumption), formulating the model in a spreadsheet is not especially difficult and Solver [while using the Nonlinear Solver for case (a)] can readily solve the model if it is not unusually large. The next section focuses on case ( a ) and Section 8.3 considers case ( b ).
Unfortunately, other types of nonlinear programming tend to be more difficult. For exam- ple, even though case ( c ) in Figures 8.2 and 8.3 has decreasing marginal returns except at the discontinuities in the graph, the presence of such discontinuities for any of the activi- ties makes it uncertain that the Nonlinear Solver will successfully solve the model. Having increasing marginal returns, as in case ( d ), also can create serious complications.
Far more complicated nonlinear programming models can be constructed than any of those suggested by Figures 8.2 and 8.3 . For example, consider the following model in algebraic form.
Maximize Profit 5 0.5x5 2 6x4 1 24.5x3 2 39x2 1 20x
subject to
x # 5
x $ 0
In this case, there is only a single activity, where x represents the level of this activity. Fur- thermore, there is only a single functional constraint ( x # 5) in addition to the nonnegativity constraint. Nevertheless, Figure 8.7 demonstrates what a difficult time the Nonlinear Solver has in attempting to cope with this problem. The model is straightforward to formulate in a spreadsheet, with x (C5) as the changing cell and Profit (C8) as the objective cell, where the formula for Profit is entered as shown in the lower left-hand corner of Figure 8.7 . Note that in the Solver Parameters box shown in the lower right-hand corner of Figure 8.7 , GRG Nonlinear is chosen as the Solving Method (in the Solver dialog box for Excel’s Solver or on the Engine tab of the Model pane for RSPE). Choosing GRG Nonlinear instructs Solver to use the Nonlinear Solver to attempt to solve the model. Trying to use the Simplex LP solving method (or Standard LP/Quadratic Engine for RSPE) instead will result in an error message if, like this example, the model is not linear.
When x 5 0 is entered as the initial value in the changing cell, the left spreadsheet in Fig- ure 8.7 shows that the Nonlinear Solver then indicates that x 5 0.371 is the optimal solution with Profit 5 $3.19. However, if x 5 3 is entered as the initial value instead, as in the middle spreadsheet in Figure 8.7 , the Nonlinear Solver obtains x 5 3.126 as the optimal solution with Profit 5 $6.13. Trying still another initial value of x 5 4.7 in the right spreadsheet, the Non- linear Solver now indicates an optimal solution of x 5 5 with Profit $0. What is going on here?
Plotting the profit graph for such a complicated objective function is a difficult task, but doing so in Figure 8.8 does help to explain the Nonlinear Solver’s difficulties with this problem. Start- ing at x 5 0, the profit graph does indeed climb to a peak at x 5 0.371, as reported in the left spreadsheet of Figure 8.7 . Starting at x 5 3 instead, the graph climbs to a peak at x 5 3.126, which is the solution found in the middle spreadsheet. Using the right spreadsheet’s starting solu- tion of x 5 4.7, the graph climbs until it reaches the boundary imposed by the x # 5 constraint, so x 5 5 is the peak in that direction. These three peaks are referred to as the local maxima (or local optima ) because each one is a maximum of the graph within a local neighborhood of that point. However, only the largest of these local maxima is the global maximum , that
Although some nonlinear programming models can be very difficult to solve, those that have decreasing marginal returns generally are relatively easy.
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is, the highest point on the entire graph. Thus, the middle spreadsheet in Figure 8.7 did succeed in finding the optimal solution at x 5 3.126 with Profit 5 $6.13.
The algorithm used by the Nonlinear Solver can be thought of as a mountain climbing pro- cedure. It starts at the initial solution entered into the changing cells and then begins climbing that mountain until it reaches the peak (or is blocked from climbing further by reaching the boundary imposed by the constraints). The procedure terminates when it reaches this peak (or boundary) and reports this solution. It has no way of detecting whether there is a taller moun- tain somewhere else on the profit graph.
When the objective cell is to be minimized instead of maximized, this algorithm reverses direction and climbs down until it reaches the lowest point in that valley (or is blocked by a boundary). Once again, it has no way of detecting whether there is a lower valley somewhere else on the cost graph.
The reason that having decreasing marginal returns for all the activities (except any with a proportional relationship) is an easy type of nonlinear programming problem is that the profit graph (when maximizing) has only one mountain. Therefore, a local maximum at the peak of the mountain (or a boundary) also is a global maximum, so the solution obtained by
When maximizing, the Nonlinear Solver only climbs to a local maximum and stops. This local maxi- mum may or may not be the global maximum.
A nonlinear program- ming problem needs to have decreasing marginal returns to guarantee that the solution obtained by the Nonlinear Solver actually is optimal.
1
A B C D E
Maximum
5
$3.19
?x =
Profit = 0.5x5-6x4+24.5x3-39x2+20x
Profit = 0.5x5-6x4+24.5x3-39x2+20x Maximum x Profit
E5
C5
C8 = =0.5*x^5-6*x^4+24.5*x^3-39*x^2+20*x
=
2
3
4
5
6
7
8
Solver Solution
(Starting with x = 0)
Range Name Cell
7
B C
8
1
A B C D E
Maximum
5
$6.13
?x =
Profit = 0.5x5-6x4+24.5x3-39x2+20x
=
2
3
4
5
6
7
8
Solver Solution
(Starting with x = 3)
1
A B C D E
Maximum
5
$0.00
?x =
Profit = 0.5x5-6x4+24.5x3-39x2+20x
=
2
3
4
5
6
7
8
Solver Solution
(Starting with x = 4.7)
0.371 3.126 5.000
Solver Parameters Set Objective Cell: Profit To: Max By Changing Variable Cells:
x
Subject to the Constraints: x <= Maximum
Solver Options: Make Variables Nonnegative
Solving Method: GRG Nonlinear
FIGURE 8.7 An example of a complicated nonlinear programming model where the Nonlinear Solver obtains three different final solutions when it starts with three different initial solutions.
2
-2
-4
-6
2
4
6
4 x
Profit ($)FIGURE 8.8 The profit graph for the example considered in Figure 8.7.
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8.2 Nonlinear Programming with Decreasing Marginal Returns 277
the Nonlinear Solver is guaranteed to be optimal. For example, the profit graph based on a quadratic form in Figure 8.6 has decreasing marginal returns, so it has only a single moun- tain and its peak (which happens to be at x 5 9.43) is the global maximum. Similarly, when minimizing a cost graph with decreasing marginal returns, there is only one valley so the local minimum at the bottom (or a boundary) also is a global minimum.
Figure 8.7 suggests that one way of dealing with more difficult problems that may have multiple local optima is to apply the Nonlinear Solver repeatedly with a variety of starting solutions and then adopt the best of the final solutions obtained. Although this will not guar- antee finding a globally optimal solution, it often will provide a good chance of finding at least a very good solution. Therefore, this is a reasonable approach for some relatively small problems, particularly when a systematic procedure is used to provide a comprehensive cross- section of starting solutions. Solver provides an option for doing this in an automatic way (as will be described in Section 8.4).
However, this kind of approach is not very practical for problems with a large number of decision variables, since a huge number of starting solutions would be required to provide a comprehensive cross-section for such problems. What is needed is an algorithm that occa- sionally will “jump” from the current mountain to another more promising mountain on the profit graph so that the algorithm is likely to eventually reach the tallest mountain on its own regardless of which starting solution is entered into the changing cells. Solver provides such an algorithm called Evolutionary Solver. Although Evolutionary Solver has its limitations as well, it provides an excellent complement to the Nonlinear Solver for attempting to cope with many nonlinear programming problems. Evolutionary Solver and its use are described in Section 8.5.
1. What are the features of linear programming models that are shared by nonlinear program- ming models?
2. How does the appearance of a nonlinear programming model differ from that of a linear programming model?
3. In what three major ways does the application of nonlinear programming models differ from that of linear programming models?
4. What is the proportionality assumption of linear programming that is violated by nonlinear programming problems?
5. When an activity has decreasing marginal returns, how does the slope of its profit graph behave?
6. What could cause the profit graph of an activity to be piecewise linear with decreasing mar- ginal returns?
7. What is a common assumption about the form of the formula for the profit of an activity when applying a curve fitting method?
8. What are the types of nonlinear programming models that are relatively easy to solve? 9. When it is given a starting solution, how does the Nonlinear Solver then proceed to attempt
to solve a maximization problem with multiple local maxima? 10. What can be done to give the Nonlinear Solver a better chance of obtaining an optimal solu-
tion (or at least a very good solution) for a maximization problem with multiple local maxima?
Review Questions
8.2 NONLINEAR PROGRAMMING WITH DECREASING MARGINAL RETURNS
In this section, we will focus on nonlinear programming problems having the following characteristics.
1. The same constraints as for a linear programming model. 2. A nonlinear objective function. 3. Each activity that violates the proportionality assumption of linear programming has
decreasing marginal returns (as defined in the preceding section and illustrated in Figures 8.2[a] and 8.3[a] ).
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278 Chapter Eight Nonlinear Programming
This is a particularly simple type of nonlinear programming problem. The Nonlinear Solver can readily solve such problems if they are not unusually large. Furthermore, the solution obtained is guaranteed to be optimal for this type of problem.
For some problems of this type, the objective function will include cross-product terms involving the product of two or more decision variables. In this case, whenever all but one of the decision variables are fixed at particular values, the effect on the value of the objec- tive function of increasing the one decision variable must still satisfy either proportionality or decreasing marginal returns for the third characteristic to hold. (The precise mathematical description of the third characteristic is that an objective function being maximized is required to be concave whereas an objective function being minimized is required to be convex. )
As discussed in the preceding section, it is fairly common for an activity to provide less and less return as the level of the activity is increased, so the activity has decreasing marginal returns. Consequently, nonlinear programming problems with decreasing marginal returns arise fairly frequently. We will go through two examples in some detail to illustrate how this happens and then describe how to formulate and solve such a problem.
In some cases, when the nonlinear objective function is reasonably close to being linear, a linear programming model will be used as an approximation to perform the preliminary analysis and then a more precise nonlinear programming model will be used to do the detailed analysis. This is what is happening below as the story of the Wyndor Glass Co. case study continues to unfold.
Continuation of the Wyndor Glass Co. Case Study As described in Section 2.1, the Wyndor Glass Co. produces high-quality glass products, where different parts of the production are performed in three plants. It now is launching two new products (a special kind of door and a special kind of window), where the anticipated profit has been estimated to be $300 per door and $500 per window. Section 2.2 discusses how these estimates of the unit profits, along with information regarding constraints, have led to the formulation of a linear programming model whose objective function to be maximized is Profit 5 $300 D 1 $500 W, where D and W are the number of doors and windows to be produced per week, respectively.
To refresh your memory, Figure 8.9 shows the spreadsheet model that was formulated in Section 2.2 for this problem. Having run Solver, the changing cells UnitsProduced (C12:D12) give the optimal solution, ( D, W ) 5 (2, 6), and the objective cell TotalProfit (G12) indicates that this will yield a weekly profit of $3,600, according to the model.
This model assumes that the profit from either of these new products would be propor- tional to the production rate for the product. However, this is a questionable assumption. Therefore, before making a final decision on the production rates, Wyndor management wants a more precise analysis to be done, as described in the following conversation between two members of management.
John Hill (Wyndor president): How are your marketing plans coming along for the launch of our two new products, Ann? Will it be very expensive? Ann Lester (Wyndor vice president for marketing): That depends on what sales vol- ume we need to generate. Our market research indicates that we would be able to sell small numbers of the new doors and windows with virtually no advertising. However, it also tells us that we would need an extensive advertising campaign if we produce close to what our plants can handle. Has a final decision been made yet on the production rates? John: No, it hasn’t. In fact, that’s why I asked you to come see me. We would like to ask for your help. Ann: Sure. What can I do? John: Well, basically what we want is your updated input on what the marketing costs per week would need to be to sell each product if the production rate were to be set at each of several alternative values. Ann: Sure, I can do that. When they started their analysis before, I was asked to estimate the marketing cost per door and per window. I told them $75 per door and $200 per window. Those looked like good estimates at the time.
The Nonlinear Solver can readily solve such prob- lems because the solution it obtains is guaranteed to be optimal for this type of problem.
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8.2 Nonlinear Programming with Decreasing Marginal Returns 279
John: Yes. Those cost estimates were factored in when they developed their estimated profits of $300 per door and $500 per window. Do your cost estimates still look pretty close? Ann: No, not really. I don’t think it makes sense any more to figure our marketing costs on a per door or per window basis. As I was saying before, our costs would be very small with low production rates, but would need to be very substantial with high production rates. Therefore, figuring $75 per door and $200 per window is much too large with low production rates, about right at medium production rates, and much too small at high production rates. John: Yes, that’s what I suspected. That’s why we want you to forget about doing it now on a per door or per window basis and instead estimate your weekly marketing costs for each product if the production rate were to be set at each of several alternative values. This will enable the Management Science Group to perform a more precise analysis of what the production rates should be. Ann: That makes sense. I’ll pull these new estimates together right away.
After receiving these estimates, the Management Science Group plotted the weekly mar- keting cost for each product versus the production rate of the product. Each of these plots showed that the marketing cost increases roughly with the square of the production rate as this rate is increased. Therefore, a quadratic form was assumed to apply Excel’s curve fitting procedure to each of these plots.
A linear formula is no longer adequate for estimat- ing the marketing costs.
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
TotalProfit
UnitProfit
UnitsProduced
G7:G9
E7:E9
C7:D9
G12
C4:D4
C12:D12
Range Name Cells
1
A B C D E F G
2
3 Doors Windows
4 Unit Profit
5
6 Hours Used per Unit Produced
7
8
9
Hours
Used
2
12
18
≤
≤ ≤
Hours
Available
Plant 1
Plant 2
Plant 3
10
11
12 Units Produced
Doors Windows
Wyndor Glass Co. Product-Mix Problem
2 6
Total Profit
Hours5
E
6
7
8
9
=SUMPRODUCT(C7:D7,UnitsProduced)
=SUMPRODUCT(C8:D8,UnitsProduced)
=SUMPRODUCT(C9:D9,UnitsProduced)
Used
Total Profit11
G
12 =SUMPRODUCT(UnitProfit,UnitsProduced)
$300 $500
1
0
3
0
2
2
4
12
18
$3,600
Solver Parameters Set Objective Cell: Profit To: Max By Changing Variable Cells: UnitsProduced
Subject to the Constraints: HoursUsed <= HoursAvailable
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 8.9 The spreadsheet model that was formulated in Section 2.2 for the original Wyndor problem introduced in Section 2.1.
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280 Chapter Eight Nonlinear Programming
This curve fitting procedure estimated that the weekly marketing costs required to sustain a production rate of D doors per week would be roughly
Marketing cost for doors 5 $25D2
for any fractional or integer value of D permitted by the production constraints. Excluding marketing costs, the gross profit per door sold is about $375. Therefore, the weekly net profit would be roughly
Net profit for doors 5 $375D 2 $25D2
The corresponding estimates per week for windows are
Marketing cost for windows 5 $66⅔W2
Gross profit for windows 5 $700W
Net profit for windows 5 $700W 2 $66⅔W2
Figure 8.10 shows the resulting profit graphs for both products. Note that both curves show decreasing marginal returns, where this becomes particularly pronounced for larger values of W.
Combining the net profit for doors and for windows, the new objective function to be maximized for this problem is
Profit = $375D 2 $25D2 1 $700W 2 $66⅔W2
subject to the same constraints as before. Because the terms involving D 2 and W 2 have expo- nents different from 1 for these decision variables, this objective function is a nonlinear function. Therefore, the overall problem is a nonlinear programming problem. Furthermore, because this objective function has a quadratic form (and the problem has all three character- istics listed at the beginning of this section), the overall problem is a special type of nonlinear programming problem called a quadratic programming problem. This is a common type of nonlinear programming problem and also a particularly convenient type to formulate and solve. Special algorithms have been developed just to solve quadratic programming problems very efficiently, so commercial management science software packages often include such an algorithm to enable solving huge problems of this type. RSPE includes a special algorithm for quadratic programming problems whereas Excel’s Nonlinear Solver only uses a general algorithm for solving any nonlinear programming problem with decreasing marginal returns.
A Spreadsheet Formulation Figure 8.11 shows the formulation of a spreadsheet model for this problem. It is interest- ing to compare this model with the one for the original Wyndor problem in Figure 8.9 . At
The new estimates of mar- keting costs cause both the doors and windows to have decreasing marginal returns.
A quadratic programming problem has linear con- straints and an objective function that has both a quadratic form and decreas- ing marginal returns.
D
1,200
1,000
800
600
400
200
0 2 4 Production rate for doors
Weekly profit ($)
W
1,800
1,600
1,400
1,200
1,000
800
600
400
200
0 2 4 6 Production rate for windows
Weekly profit ($)
FIGURE 8.10 The smooth curves are the profit graphs for Wyn- dor’s doors and windows for the version of its problem where nonlinear marketing costs must be considered.
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Hours5
E
6
7
8
9
=SUMPRODUCT(C7:D7,UnitsProduced)
=SUMPRODUCT(C8:D8,UnitsProduced)
=SUMPRODUCT(C9:D9,UnitsProduced)
Used
DoorsProduced
GrossProfitFromSales
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
MarketingCost
TotalMarketingCost
TotalProfit
UnitProfit
UnitsProduced
WindowsProduced
C12
H12
G7:G9
E7:E9
C7:D9
C14:D14
H14
H16
C4:D4
C12:D12
D12
Range Name Cells
1
A B C D E F G H
2
3 Doors Windows
4 Unit Profit (Gross)
5
6 Hours Used per Unit Produced
7
8
9
Hours
Used
3.214
8.357
18
≤
≤ ≤
Hours
Available
Plant 1
Plant 2
Plant 3
10
11
12
13
14 Marketing Cost $258 $1,164 $1,422
15
16
Units Produced
Doors Windows
Wyndor Problem with Nonlinear Marketing Costs
3.214 4.179 $4,130
Total Profit
Total Marketing Cost
Gross Profit from Sales
Gross Profit from Sales
$2,708
12
G H
13
=SUMPRODUCT(UnitProfit,UnitsProduced)
Total Marketing Cost14
15
=SUM(MarketingCost)
Total Profit16 =GrossProfitFromSales−TotalMarketingCost
Marketing Cost14
B C D
=25*(DoorsProduced^2) =66.667*(WindowsProduced^2)
$375 $700
4
12
18
1
0
3
0
2
2
Solver Parameters Set Objective Cell: Profit To: Max By Changing (Variable) Cells:
UnitsProduced
Subject to the Constraints: HoursUsed <= HoursAvailable
Solver Options: Make Variables Nonnegative
Solving Method: GRG Nonlinear
or Quadratic (RSPE)
8.2 Nonlinear Programming with Decreasing Marginal Returns 281
FIGURE 8.11 A spreadsheet model for the Wyndor nonlinear pro- gramming problem with nonlinear marketing costs, where the changing cells UnitsProduced (C12:D12) show the optimal produc- tion rates and the objec- tive cell TotalProfit (H16) gives the resulting total profit per week.
first glance, they appear to be nearly the same. A closer examination reveals four significant differences.
First, the unit profits in row 4 of Figure 8.9 have been replaced here by the gross unit prof- its, which exclude the marketing costs.
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W
D
6
5
4
3
2
1
0 1 2 3 4 5
Production rate for doors
P ro
d u
ct io
n r
at e
fo r
w in
d ow
s
Profit = $2,800 Profit = $2,708 Profit = $2,600 Profit = $2,500
(33/14, 45/28) = optimal solution
FIGURE 8.12 Graphical display of the nonlinear programming formulation of the Wyn- dor problem with nonlin- ear marketing costs. The curves are objective func- tion curves for some sam- ple values of Profit and the one (Profit 5 $2,708) that passes through the optimal solution, (D, W ) 5 (33/14, 4 5/28).
282 Chapter Eight Nonlinear Programming
Second, to take the marketing costs into account in calculating the objective cell Total- Profit (H16), the spreadsheet in Figure 8.11 has added several output cells: GrossProfitFrom- Sales (H12), MarketingCost (C14:D14), and TotalMarketingCost (H14).
Third, a fundamental difference lies in the equations entered into certain output cells. In Figure 8.9 , the formula for TotalProfit (G12) is expressed in terms of the SUMPRODUCT function that is characteristic of linear programming when each product is the product of a data cell and a changing cell. In Figure 8.11 , something else is needed for calculating the marketing cost portion of total profit because that portion of the objective function is nonlin- ear. For example, consider the term involving D 2 in the objective function. Because the value of D appears in DoorsProduced (C12), Excel expresses D 2 as DoorsProduced^2, where the symbol ^ indicates that the number following this symbol (2) is the exponent of the number in DoorsProduced (C12). The same approach is used for expressing W 2 . Therefore, the formula for total marketing cost is
Total Marketing Cost 5 SUM(MarketingCost)
5 25 * (DoorsProduced¿2) 1 66.667 * (WindowsProduced¿2) The formula for the objective cell then becomes
TotalProfit (H16) 5 GrossProfitFromSales (H12) 2 TotalMarketingCost (H14)
The fourth difference arises in the selection of the Solver options at the bottom of Figures 8.9 and 8.11 . In contrast to Figure 8.9 , GRG Nonlinear (i.e., the Nonlinear Solver) is chosen as the solving method (or alternatively, with RSPE, Standard LP/Quadratic Engine can be chosen). These settings (or the Evolutionary Solver method discussed in Section 8.5) are required when attempting to solve any nonlinear programming model.
For this particular model, running Solver provides the optimal solution shown in UnitsProduced (C12:D12), namely,
D 5 3.214 (produce an average of 3.214 doors per week)
W 5 4.179 (produce an average of 4.179 windows per week)
where TotalProfit (H16) shows a resulting weekly profit of $2,708. These strange values of D and W certainly are not intuitive. Figure 8.12 conveys some graphical intuition into why this answer was obtained. The feasible region is the same as for the original Wyndor problem in Chapter 2. However, instead of having objective function lines with which to search for an optimal solution, plotting the points that give any constant value for our nonlinear objective function now gives an objective function curve instead. Thus, when using the objective func- tion to calculate Profit for various feasible and infeasible values of ( D, W ), each of the four
Excel Tip: When a chang- ing cell needs to be raised to some power in a formula, the symbol ^ is placed between the changing cell and the exponent.
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8.2 Nonlinear Programming with Decreasing Marginal Returns 283
curves in the figure shows all the values of ( D, W ) that give the fixed value of Profit indicated for that curve. (Plotting these points is a tedious and difficult process, so we will not bother with the details of how it is done.) The figure shows that increasing Profit moves the objective function curve to the right. The largest value of Profit such that the objective function curve still passes through any points in the feasible region is Profit 5 $2,708. Therefore, using frac- tions, the one feasible point that the Profit 5 $2,708 curve passes through,
(D, W ) 5 (3 3@14, 4 5@28)
is the optimal solution. Since these are not particularly convenient fractions with which to plan production sched-
ules, they should be adjusted slightly. The curve for Profit 5 $2,708 in Figure 8.12 indicates that any point on the slanting line at the boundary of the feasible region that is close to the optimal solution will provide a weekly profit very close to $2,708. For example,
(D, W ) 5 (3 1@3, 4)
gives a weekly profit of $2,706, so management prefers this more convenient production schedule.
By contrast, consider the linear programming solution in Figure 8.9 , ( D, W ) 5 (2, 6), that does not take the nonlinearities in the marketing costs into account. When using the current objective function that incorporates these nonlinearities, ( D, W ) 5 (2, 6) provides a weekly profit of only $2,450. This illustrates the kind of improvement that can be obtained by replac- ing an approximate linear programming model by a more precise nonlinear programming model.
Applying Nonlinear Programming to Portfolio Selection It now is common practice for professional managers of large stock portfolios to use com- puter models based partially on nonlinear programming to guide them. Because investors are concerned about both the expected return (gain) and the risk associated with their invest- ments, nonlinear programming is used to determine a portfolio that, under certain assump- tions, provides an optimal trade-off between these two factors. This approach is based largely on path-breaking research done by Harry Markowitz and William Sharpe that helped them win the 1990 Nobel Prize in economics.
One way of formulating their approach is as a nonlinear version of the cost–benefit trade- off problems discussed in Section 3.3. In this case, the cost involved is the risk associated with the investments. The benefit is the expected return from the portfolio of investments. There- fore, the general form of the model is
Minimize Risk
subject to
Expected return $ Minimum acceptable level
The measure of risk used here is a basic quantity from probability theory called the vari- ance of the return. Using standard formulas from probability theory, the objective function then can be expressed as a nonlinear function of the decision variables (the fractions of the total investment to invest in the respective stocks) that yields decreasing marginal returns for the stocks. By adding the constraint on expected return, as well as nonnegativity constraints and a constraint that the fractions of the total investment invested in the respective stocks sum to 1, we thereby obtain a simple type of nonlinear programming model for optimizing the selection of the portfolio.
To illustrate the approach, we now will focus on a small numerical example where just three stocks (securities) are being considered for inclusion in the portfolio. Thus, the decision variables are
S1 5 Fraction of total investment invested in Stock 1
S2 5 Fraction of total investment invested in Stock 2
S3 5 Fraction of total investment invested in Stock 3
This model focuses on the trade-off between risk and expected return from the portfolio of investments.
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284
Since these fractions need to sum to 1,
S1 1 S2 1 S3 5 1
will be included as one of the constraints of the model. Table 8.2 gives the needed data for these three stocks. The second column provides the
expected return for each of these stocks, so the expected return for the overall portfolio is
Expected return 5 (21S1 1 30S2 1 8S3)%
The investor’s current choice of the minimum acceptable level for this quantity is
Minimum acceptable expected return 5 18%
Since the expected returns of Stocks 1 and 2 exceed 18%, this minimum acceptable level will be achieved if these stocks comprise a sufficiently large portion of the portfolio.
However, Stocks 1 and 2 are much riskier than Stock 3. There is no certainty that the expected returns shown in Table 8.2 actually will be achieved, but there is much more uncertainty for Stocks 1 and 2 than for Stock 3. Each stock has an underlying probability distribution of what its return will turn out to be. In each case, the standard deviation (i.e., the square root of the vari- ance) of this distribution provides a measure of how spread out this distribution is, since there is roughly a two-thirds probability that the return will turn out to be within one standard deviation of the expected return. This measure of the risk of a stock is given in the third column of Table 8.2 .
However, the risk for the portfolio cannot be obtained solely from the third column, since this column only gives the risk for each individual stock considered in isolation. The risk for the portfolio also is affected by whether the particular stocks tend to move up and down
The challenge is to find the right balance between the high return but high risk from Stocks 1 and 2 and the low risk but low return from Stock 3.
The Bank Hapoalim Group is Israel’s largest banking group, providing services within Israel through a network of 327 branches, 9 regional business centers, and various domes- tic subsidiaries. It also operates worldwide through 37 branches, offices, and subsidiaries in major financial centers in North and South America and Europe.
A major part of Bank Hapoalim’s business involves pro- viding investment advisors for its customers. To stay ahead of its competitors, management embarked on a restruc- turing program to provide these investment advisors with state-of-the-art methodology and technology. A manage- ment science team was formed to do this.
The team concluded that it needed to develop a flex- ible decision-support system for the investment advisors that could be tailored to meet the diverse needs of every customer. Each customer would be asked to provide exten- sive information about his or her needs, including choosing among various alternatives regarding his or her investment objectives, investment horizon, choice of an index to strive to exceed, preference with regard to liquidity and currency, and so forth. A series of questions also would be asked to ascertain the customer’s risk-taking classification.
The natural choice of the model to drive the resulting decision-support system (called the Opti-Money System) was the classical nonlinear programming model for port- folio selection described in this section of the book, with modifications to incorporate all the information about the needs of the individual customer. This model generates an optimal weighting of 60 possible asset classes of equities and bonds in the portfolio, and the investment advisor then works with the customer to choose the specific equi- ties and bonds within these classes.
In one recent year, the bank’s investment advisors held some 133,000 consultation sessions with 63,000 customers while using this decision-support system. The annual earnings over benchmarks to customers who fol- low the investment advice provided by the system total approximately US$244 million, while adding more than US$31 million to the bank’s annual income.
Source: M. Avriel, H. Pri-Zan, R. Meiri, and A. Peretz, “Opti-Money at Bank Hapoalim: A Model-Based Investment Decision-Support System for Individual Customers,” Interfaces 34, no. 1 (January– February 2004), pp. 39–50. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
Stock Expected Return Risk
(Standard Deviation) Pair of Stocks
Joint Risk per Stock
(Covariance)
1 21% 25% 1 and 2 0.040 2 30 45 1 and 3 20.005 3 8 5 2 and 3 20.010
TABLE 8.2 Data for the Stocks of the Portfolio Selection Example
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8.2 Nonlinear Programming with Decreasing Marginal Returns 285
together (increased risk) or tend to move in opposite directions (decreased risk). In the right- most column of Table 8.2 , the positive joint risk for Stocks 1 and 2 indicates that these two stocks have some tendency to move in the same direction. However, the negative joint risk for the other two pairs of stocks shows that Stock 3 tends to go up when either Stock 1 or 2 goes down, and vice versa. (In the terminology of probability theory, the joint risk for each of two stocks is the covariance of their returns, as given in the rightmost column of Table 8.2 , so the total joint risk for two stocks is two times this covariance.)
The data in Table 8.2 typically are obtained by taking samples of the returns of the stocks from a number of previous years and then calculating the averages, standard deviations, and covariances for these samples. Adjustments in the resulting estimate of at least the expected return of a stock also may be made if it appears that the current prospects for the stock are some- what different than in previous years. Using the formula from probability theory for calculating the overall variance from individual variances and covariances, the risk for the entire portfolio is
Risk 5 (0.25S1) 2
1 (0.45S2) 2
1 (0.05S3) 2
1 2(0.04)S1S2 1 2(20.005)S1S3 1 2(20.01)S2S3
Therefore, the algebraic form of the nonlinear programming model for this example is
Minimize Risk 5 (0.25S1) 2
1 (0.45S2) 2
1 (0.05S3) 2
1 2(0.04)S1S2 1 2(20.005)S1S3 1 2(20.01)S2S3
subject to
21S1 1 30S2 1 8S3 $ 18
S1 1 S2 1 S3 5 1
and
S1 $ 0 S2 $ 0 S3 $ 0
Fortunately, the objective function for this model has decreasing marginal returns. (This is not obvious, but it has been verified that Risk, measured by the variance of the return for the entire portfolio, always has decreasing marginal returns for any portfolio.) Furthermore, this is a quadratic programming model since the objective function is quadratic (terms consisting of a coefficient times the product of two variables are allowed in a quadratic function) and the model has all three characteristics listed at the beginning of this section. Therefore, this is a particularly simple type of nonlinear programming model to solve.
Figure 8.13 shows the corresponding spreadsheet model after having applied Solver. For ease of interpretation, the changing cells Portfolio (C14:E14) give the values of S 1 , S 2 , and S 3 as percentages rather than fractions. These cells indicate that the optimal solution is
S 1 5 40.2%: Allocate 40.2% of the portfolio to Stock 1 S 2 5 21.7%: Allocate 21.7% of the portfolio to Stock 2 S 3 5 38.1%: Allocate 38.1% of the portfolio to Stock 3
Thus, despite its relatively low return, including a substantial amount of Stock 3 in the portfo- lio is worthwhile to counteract the high risk associated with Stocks 1 and 2. Expected-Return (C19) indicates that this portfolio still achieves an expected return of 18 percent, which equals the minimum acceptable level. The objective cell Variance (C21) gives the risk for the port- folio, namely, the variance of the return for the entire portfolio, as 0.0238. To help interpret this quantity, StandDev (C23) calculates the corresponding standard deviation of the return for the portfolio as "0.0238 5 0.154 5 15.4 percent. The fact that this standard deviation is less than the expected return is encouraging, because this indicates that it is fairly unlikely that the actual return that eventually is achieved by the portfolio will turn out to be negative. The standard deviation is this small, despite the much larger standard deviations of the returns for stocks 1 and 2 given in StockStandDev (C6:E6), because of the very small standard devia- tion for stock 3 and the negative values in Covar13 (E9) and Covar23 (E10).
This is an example of a cost–benefit trade-off problem since it involves finding the best trade-off between cost (risk) and a benefit (expected return). Except for the form of the objec- tive function, it is analogous to the cost–benefit trade-off problems discussed in Section 3.3.
This kind of quadratic pro- gramming model is widely used by portfolio managers.
There is a good chance that the return for the portfolio will not deviate from the expected return by more than the standard deviation of the return.
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286 Chapter Eight Nonlinear Programming
As discussed further in Chapter 5, analysis of such a problem seldom ends with finding an optimal solution for the original version of the model. The minimum acceptable level stated in the model for the benefit (or benefits) involved is a tentative policy decision. After learning the resulting cost, further analysis is needed to find the best trade-off between costs and ben- efits. This analysis involves varying the minimum acceptable level for the benefit and seeing what the effect is on the cost. If a lot more benefit can be obtained for relatively little cost, this probably should be done. On the other hand, if decreasing the benefit a little would save a lot of cost, the minimum acceptable level probably should be decreased.
One way of applying this approach to the current example is to generate a parameter analy- sis report with RSPE as described in Chapter 5 to generate a table that gives the expected return and risk provided by an optimal solution for the model for a range of values of the minimum acceptable expected return. Figure 8.14 shows such a table. In the parlance of the world of finance, the pairs of values in columns F and G are referred to as points on the effi- cient frontier. In fact, the right-hand side of Figure 8.14 shows a plot of this efficient frontier. After examining enough such points, the investor then can make a personal decision about which one provides the best trade-off between expected return and risk.
An investor needs the kind of table and graph shown in Figure 8.14 to decide on which portfolio provides the best trade-off between expected return and risk.
Covar12
Covar13
Covar23
Covariance
ExpectedReturn
MinExpectedReturn
OneHundredPercent
Portfolio
SD1
SD2
SD3
StandDev
Stock1
Stock2
Stock3
StockExpectedReturn
StockStandDev
Total
Variance
D9
E9
E10
C9:E11
C19
E19
H14
C14:E14
C6
D6
E6
C23
C14
D14
E14
C4:E4
C6:E6
F14
C21
Range Name Cells
1
2
3
4
5
6
7
8
9
10
11
12
13
14 Portfolio 40.2% 21.7% 38.1% 15
16
17
18
Minimum Expected Return
19
20
21
22
23
Expected Return
Risk (Variance)
Risk (Stand. Dev.)
Portfolio 18% ≥ 18%
0.0238
15.4%
A B C D E F G H
Stock 1 Stock 2 Stock 3
Expected Return 21% 30% 8%
25% 45% 5%Risk (Stand. Dev.)
Stock 1 Stock 2 Stock 3
Stock 1 Stock 2 Stock 3
100% = 100%
Total
Joint Risk (Covar.) -0.005
-0.010
0.040Stock 1
Stock 2
Stock 3
Portfolio Selection Problem (Nonlinear Programming)
19 =SUMPRODUCT(StockExpectedReturn,Portfolio)
20
21
22
23
=((SD1*Stock1)^2)+((SD2*Stock2)^2)+((SD3*Stock3)^2)+2*Covar12*Stock1*Stock2+2*Covar13*Stock1*Stock3+2*Covar23*Stock2*Stock3
=SQRT(Variance)
Risk (Variance)
Risk (Stand. Dev.)
B C
Expected Return
Total13 F
14 =SUM(Portfolio)
Solver Parameters Set Objective Cell: Variance To: Min By Changing Variable Cells:
Portfolio
Subject to the Constraints: ExpectedReturn >= MinExpectedReturn
Total = OneHundredPercent
Solver Options: Make Variables Nonnegative
Solving Method: GRG Nonlinear
or Quadratic (RSPE)
FIGURE 8.13 A spreadsheet model for the portfolio selection example of nonlinear programming, where the changing cells Portfolio (C14:E14) give the optimal portfolio and the objective cell Variance (C21) shows the resulting risk.
1. What are the three characteristics of a simple type of nonlinear programming problem that can be readily solved by the Nonlinear Solver?
2. For this simple type of nonlinear programming problem, how does the graphical display for a two-variable problem differ from that for a two-variable linear programming problem?
3. What additional characteristic must this type of nonlinear programming problem have in order to be a quadratic programming problem?
4. When applying nonlinear programming to portfolio selection, a trade-off is being sought between which two factors?
Review Questions
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8.3 Separable Programming 287
8.3 SEPARABLE PROGRAMMING
Section 8.1 described several types of nonproportional relationships between an activity and the overall measure of performance for a problem. One such relationship is decreasing mar- ginal returns and Section 8.2 has just focused on nonlinear programming problems where this type of relationship holds for all the activities. We now turn our attention to a related kind of nonproportional relationship where the activities again have decreasing marginal returns. However, the difference is that the profit or cost graph for each activity now is piecewise linear because it consists of a sequence of connected line segments. Figure 8.2(b) in Section 8.1 illus- trated such a profit graph (or the graph for any other measure of performance to be maximized) and Figure 8.3(c) did the same for a cost graph (or any related graph with minimization).
There is a special technique called separable programming that is designed to deal with this kind of nonlinear programming problem. Thus, the total profit (or cost) is simply the sum of the profits (or costs) obtained directly from these piecewise linear profit (or cost) graphs for the individual activities. (No cross-product terms are allowed and each graph must have decreasing marginal returns.) Because of the line segments in each profit or cost graph, this technique converts the formulation of the model into a linear programming model. This enables solving the model extremely efficiently and then applying the powerful tools of what- if analysis for linear programming.
The next episode in the saga of the Wyndor Glass Co. problem illustrates this technique.
The Wyndor Glass Co. Problem When Overtime Is Needed The company now is ready to begin production of its special new doors and windows, based on the planning described in Chapter 2, Chapter 5, and Section 8.2. Because of the nonlinear marketing costs discussed in Section 8.2, the current plan is to use production rates of
(D, W ) 5 (3⅓, 4)
where D and W are the number of doors and windows to be produced per week, respectively. However, there now is a new development that might alter this production plan for the first
four months. In particular, the company has accepted a special order for hand-crafted goods to be made
in Plants 1 and 2 throughout the next four months. Filling this order will require borrowing certain employees from the work crews for the regular products, so the remaining workers will need to work overtime to utilize the full production capacity of each plant’s machinery and equipment for these products.
For nonlinear programming problems with decreasing marginal returns where the profit or cost graphs also are piecewise linear, the separable programming technique converts the problem into an equiva- lent linear programming problem.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
10.0% 8.1% 4.3% 87.6% 0.0015 3.9% 10.0%
12.0%
14.0%
16.0%
18.0%
20.0%
22.0%
24.0%
26.0%
28.0%
30.0%
5.6%
8.6%
12.0%
15.4%
18.9%
22.5%
26.1%
30.8%
37.3%
45.0%
0.0032
0.0074
0.0143
0.0238
0.0359
0.0506
0.0680
0.0946
0.1394
0.2025
75.2%
62.8%
50.5%
38.1%
25.7%
13.4%
1.0%
0.0%
0.0%
0.0%
8.6%
13.0%
17.3%
21.7%
26.1%
30.4%
34.8%
55.6%
77.8%
100.0%
16.2%
24.2%
32.2%
40.2%
48.2%
56.2%
64.2%
44.4%
22.2%
0.0%
12.0%
14.0%
16.0%
18.0%
20.0%
22.0%
24.0%
26.0%
28.0%
30.0%
B C D E F G H I J K L M N
MinExpected Return VarianceStock 1 Stock 2 Stock 3 St. Deviation Return
0.0% 0.0% 50.0%10.0% 20.0% 30.0% 40.0%
5.0%
10.0%
15.0%
20.0%
25.0%
30.0%
35.0%
FIGURE 8.14 The parameter analysis report that shows the trade-off between expected return and risk when the model of Figure 8.13 is modified by varying the minimum acceptable expected return as a parameter cell.
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288 Chapter Eight Nonlinear Programming
Because of this new development, management has asked the Management Science Group to quickly update its model and check whether the current production plan still would be the most profitable one to use during the first four months. To get a quick handle on the problem, the group decides to ignore the nonlinearities in the marketing costs for now and simply modify the original spreadsheet model shown in Figure 8.9 (in the preced- ing section) to take overtime into account. (After considering this simpler version of the problem, the nonlinearities in the marketing costs will be brought back into the analysis later in this section.)
The constraints in this original model, HoursUsed (E7:E9) # HoursAvailable (G7:G9), are still valid, where overtime would be used to fill some of the hours of production time avail- able in Plants 1 and 2 as given by cells G7 and G8. However, the objective function no longer is valid because the additional cost of using overtime work reduces the profit obtained from each unit of product produced in this way.
For the portion of the work done in Plants 1 and 2, Table 8.3 shows the maximum num- ber of units of each product that can be produced per week on regular time and on overtime. Plant 3 does not need to use overtime, so its unchanged constraint is given in parentheses at the bottom. The fourth column is the sum of the second and third columns, where these sums are implied by the original constraints for plants 1 and 2 ( D # 4 and 2 W # 12, so W # 6). The final two columns give the estimated profit for each unit produced on regular time and on overtime (in Plants 1 and 2), based on the original estimates of marketing costs rather than those developed in Section 8.2.
Figure 8.15 plots the weekly profit from each product versus its production rate. Note that the slope (steepness) of each profit graph decreases when the production rate is increased sufficiently to require overtime, because the profit per unit produced shown in Table 8.3 is less on overtime than on regular time. Thus, these two products have decreasing marginal returns.
Management had considered hiring some temporary workers to avoid the extra expense of using overtime. However, this would mean incurring some training costs, as well as inef- ficiencies from using inexperienced workers. Therefore, because this is a temporary situation where regular production can resume in four months, management has decided to go ahead and use overtime.
However, management does insist that the work crew for each product be fully utilized on regular time before any overtime is used. Furthermore, it feels that the current plans for the production rates should be changed temporarily if this would improve overall profitability.
Applying Separable Programming to This Problem Since each profit graph in Figure 8.15 is not a straight line, the profit from each product is not proportional to its production rate. Consequently, the proportionality assumption of linear programming (discussed in Section 8.1) is violated. However, each profit graph does consist of two straight lines (line segments) that are connected together at the point where the slope changes. Thus, within each line segment, the profit graph looks like the proportionality assumption still holds. This suggests the following key idea.
The Separable Programming Technique: For each activity that violates the proportionality assumption, separate its profit graph into parts, with a line segment in each part. Then, instead of using a single decision variable to represent the level of each such activity, introduce a sepa- rate new decision variable for each line segment on that activity’s profit graph. Since the pro- portionality assumption holds for these new decision variables, formulate a linear programming model in terms of these variables.
Without worrying about the new estimates of nonlinear marketing costs yet, how should the original Wyndor model be modified to con- sider overtime?
The model needs to provide a solution that uses over- time for a product only if all available regular time for that product has been fully utilized.
The key idea is to have a separate decision variable for each line segment in a profit graph (or cost graph).
Maximum Weekly Production Profit per Unit Produced
Product Regular Time Overtime Total Regular Time Overtime
Doors 3 1 4 $300 $200 Windows 3 3 6 500 100
(and 3D 1 2W # 18)
TABLE 8.3 Data for the Original Wyndor Problem When Overtime Is Needed
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8.3 Separable Programming 289
For the Wyndor problem, these new decision variables are
DR 5 Number of doors produced per week on regular time
DO 5 Number of doors produced per week on overtime
WR 5 Number of windows produced per week on regular time
WO 5 Number of windows produced per week on overtime
The unit profits associated with these variables are given in the final two columns of Table 8.3 , so these numbers become the coefficients in the objective function. The second and third columns give the maximum values of these variables, so corresponding constraints are intro- duced into the model. The three functional constraints in the model for the original Wyndor problem also need to hold, but with D replaced by ( D R 1 D O ) and W replaced by ( W R 1 W O ). Section 2.3 shows the algebraic form of the model for the original Wyndor problem (where P denotes the total profit per week from the two products, in dollars). Now that we are sepa- rately considering the production rates of the two products on regular time and on overtime, this model now expands to choosing the values of the four new decision variables so as to
Maximize P 5 300DR 1 200DO 1 500WR 1 100WO,
subject to
DR # 3, DO # 1, WR # 3, WD # 3
DR 1 DO # 4
2(WR 1 WO) # 12
3(DR 1 DO) 1 2(WR 1 WO) # 18
and
DR $ 0 DO $ 0 WR $ 0 WD $ 0
(The second and third rows of constraints could be deleted because they are ensured by the first row of constraints.) The key to the separable programming technique is that this model is a linear programming model that fits what was originally a nonlinear programming problem.
The resulting spreadsheet model for this linear programming formulation of the problem is shown in Figure 8.16 . The changing cells UnitsProduced (C14:D15) include separate cells for each of the four decision variables. The new constraints, UnitsProduced (C14:D15) # Maximum (F14:G15), enforce the upper bounds on these decision variables indicated by the second and third columns of Table 8.3 . The new output cells TotalProduced (C16:D16) sum the production quantities on regular time and overtime for each of the products. This then enables calculating
D
1,100
900
0 3 4 Production rate for doors
Weekly profit ($)
W
1,800
1,500
0 3 6 Production rate for windows
Weekly profit ($)
FIGURE 8.15 Profit graphs for the Wyn- dor Glass Co. that show the total weekly profit from each product versus the production rate for that product when over- time is needed to exceed a production rate of three units per week. At this point, these profit graphs are based on the original estimates of marketing costs rather than the esti- mates of nonlinear mar- keting costs developed in Section 8.2.
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290 Chapter Eight Nonlinear Programming
the hours used with the equation, HoursUsed (E8:E10) 5 SUMPRODUCT (HoursUsedPer UnitProduced, TotalProduced). Otherwise, the model is basically the same as the original linear programming model in Figure 8.9 . Note that Simplex LP has been chosen as the solving method, because the new model also has been formulated to become a linear programming model. The proportionality assumption now is satisfied for the new decision variables. Therefore, the model can be solved very efficiently. This ability to reformulate the original model to make it fit linear programming is what makes separable programming a valuable technique.
However, there is one important factor that is not taken into account explicitly in this for- mulation. Recall that management insists that regular time production be fully utilized before using any overtime on each product. There are no constraints in the model that enforce this
Hours6
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7
8
9
10
=SUMPRODUCT(C8:D8,TotalProduced)
=SUMPRODUCT(C9:D9,TotalProduced)
=SUMPRODUCT(C10:D10,TotalProduced)
Used
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
Maximum
TotalProduced
TotalProfit
UnitProfit
UnitsProduced
G8:G10
E8:E10
C8:D10
F14:G15
C16:D16
D18
C4:D5
C14:D15
Range Name Cells
1
A B C D E F G
2
3 Doors Windows
4 Regular
Overtime
Regular
Overtime 1 0
Total Produced
Total Profit
4 3
Unit Profit
5
6
Hours Used per Unit Produced
Units Produced
7
8
9
Hours
Used
4
6
18
Hours
Available
≤
≤ ≤
Plant 1
Plant 2
Plant 310
11
12
13
14
15
16
17
18
Doors Windows
Maximum
Doors Windows
Wyndor Problem with Overtime (Separable Programming)
3 3
≤ ≤
$2,600
Total Produced16
B C D
=SUM(C14:C15) =SUM(D14:D15)
Total Profit18
C D
=SUMPRODUCT(UnitProfit,UnitsProduced)
$300 $500
$200 $100
4
12
18
1
0
3
0
2
2
1
3
3
3
Solver Parameters Set Objective Cell: TotalProfit To: Max By Changing Variable Cells:
UnitsProduced
Subject to the Constraints: HoursUsed <= HoursAvailable
UnitsProduced <= Maximum
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
FIGURE 8.16 A spreadsheet model for the Wyndor separable programming problem when overtime is needed, where the changing cells UnitsProduced (C14:D15) give the optimal produc- tion rates obtained by Solver and the objective cell TotalProfit (D18) shows the resulting total profit per week. This model is based on the profit graphs in Fig- ure 8.15 and so does not incorporate the nonlinear marketing costs developed in Section 8.2.
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8.3 Separable Programming 291
restriction. Consequently, it actually is feasible in the model to have D O . 0 when D R , 3, or to have W O . 0 when W R , 3.
Fortunately, even though such a solution is feasible in the model, it cannot be optimal. The reason is that the activities (producing the two products) have decreasing marginal returns, since the unit profit on overtime is less than on regular time for each product. Therefore, to maximize the total profit, an optimal solution automatically will use up all regular time for a product before starting on overtime.
The key is to have decreasing marginal returns. Without it, the linear programming model with this approach may not provide a legitimate optimal solution. This is the reason that sepa- rable programming is only applicable when the activities have decreasing marginal returns (except for those activities that satisfy the proportionality assumption).
Figure 8.16 shows the changing cells UnitsProduced (C14:D15) after using Solver to obtain an optimal solution. This optimal solution is
DR 5 3, DO 5 1: Produce 4 doors per week
WR 5 3, WO 5 0: Produce 3 windows per week
for a total profit of $2,600 per week given by the objective cell TotalProfit (D18). This com- pares with a total profit of $2,567 per week for the previous plan (produce 3⅓ doors and 4 windows per week) that had been adopted before the need to use overtime arose.
Applying Separable Programming with Smooth Profit Graphs In some applications of separable programming, the profit graphs will be curves rather than a series of line segments. This occurs when the marginal return from an activity decreases on a continuous basis rather than just at certain points.
For example, the solid curve in Figure 8.17 shows such a profit graph for an activity. To apply separable programming, this curve can then be approximated by a series of line seg- ments, such as the dashed-line segments in the figure. By introducing a new decision variable for each of the line segments (and repeating this for other activities with such profit graphs), the approach just illustrated by the Wyndor example can again be used to convert the overall problem into a linear programming problem.
This is not the only way to solve problems where the activities have profit graphs with shapes similar to the one shown in Figure 8.17 . Section 8.2 discusses problems of just this same type. The Nonlinear Solver can readily solve such problems by using a nonlinear pro- gramming model that employs the formulas for the profit graphs. The advantage is that no approximation is needed, whereas separable programming uses the kind of approximation illustrated in Figure 8.17 .
However, the separable programming approach also has certain advantages. One is that converting the problem into a linear programming problem tends to make it quicker to solve, which can be very helpful for large problems. Another advantage is that a linear program- ming formulation makes available Solver’s Sensitivity Report, which is a great aid to what-if
Decreasing marginal returns are needed to use the separable programming technique.
The approximation in Figure 8.17 only requires estimating the profit at the three dots rather than esti- mating a formula for the entire profit graph.
Level of activity
Profit
Profit graph
Approximation
FIGURE 8.17 The solid curve shows a profit graph for an activ- ity whose marginal return decreases on a continuous basis. The dashed-line segments display the kind of approximation used by separable programming.
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292 Chapter Eight Nonlinear Programming
analysis, whereas the sensitivity information provided when using a nonlinear programming model is not nearly as useful. A third important advantage is that the separable programming approach only requires estimating the profit from each activity at a few points, such as the dots in Figure 8.17 . Therefore, it is not necessary to use a curve fitting method to estimate the formula for the profit graph, where this estimation would have introduced an approximation into the process.
The end of the Wyndor story (below) illustrates the application of both approaches.
The Wyndor Problem with Both Overtime Costs and Nonlinear Marketing Costs The spreadsheet model in Figure 8.16 provides a good quick estimate of approximately what the production rates should be for the new products over the next four months. This is useful for planning purposes, but the model is a somewhat rough one because it does not take into account the new estimates of nonlinear marketing costs that were developed in Section 8.2. Therefore, the next step for Wyndor’s Management Science Group is to enhance the model by incorporating these new estimates.
Recall that Ann Lester, Wyndor’s vice president for marketing, now is estimating that the marketing costs will need to be $25 D 2 and $66⅔ W 2 to sustain sales of D doors and W win- dows per week. These costs then would need to be subtracted from the gross profit for each product (the profit excluding marketing costs) to obtain that product’s profit. Since the origi- nal estimates of marketing costs had been $75 per door and $200 per window when estimating the unit profits given in Table 8.3 , the group now needs to use the data shown in Table 8.4 .
Based on these data, the fourth column of Table 8.5 shows the weekly profit that would be obtained by producing D doors per week for various values of D. This profit is calculated by subtracting the marketing costs in the third column from the gross profit in the second column. The rightmost column gives the incremental profit from the last increase of 1 in the value of D. Thus, the incremental profit is calculated by taking the profit in the same row and subtracting the profit in the preceding row. Note the large drop in the incremental profit at D 5 4 because overtime must be used to increase D above 3.
Table 8.6 provides the corresponding calculations for windows. In this case, the incremen- tal profit at W 5 4, W 5 5, and W 5 6 actually is negative because of the large extra costs of the overtime that is needed to increase W above 3.
The solid curves in Figure 8.18 show the entire profit graphs for the doors and windows. The slope of each graph always is decreasing as the production rate increases, so both activities have decreasing marginal returns. This decrease in the slope is almost imperceptible at small production rates and then becomes more pronounced at larger rates. There also is a kink in each graph at D 5 3 or W 5 3 because overtime is required to increase the production rate further.
Maximum Weekly Production Gross Unit Profit
Product Regular
Time Overtime Total Regular
Time Overtime Marketing
Costs
Doors 3 1 4 $375 $275 $25D2 Windows 3 3 6 700 300 66⅔W2
TABLE 8.4 Data for the Wyndor Problem with Both Overtime Costs and Nonlinear Marketing Costs
D Gross Profit Marketing Costs Profit Incremental
Profit
0 0 0 0 — 1 $ 375 $ 25 $ 350 $350 2 750 100 650 300 3 1,125 225 900 250 4 1,400 400 1,000 100
TABLE 8.5 Calculations of Wyn- dor’s Weekly Profit from Producing D Doors per Week
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8.3 Separable Programming 293
The Management Science Group now wants to use separable programming to determine what the production rates should be to maximize total profit. For this purpose, the group uses the dashed-line segments in Figure 8.18 to obtain piecewise linear graphs that closely approximate the actual profit graphs. The one place where the approximation is not really close is when the profit graph for windows goes from $1,500, at W 5 3, to $600, at W 5 6 (an average decrease of
It is very reasonable to use separable programming when the piecewise linear graphs approximate the actual profit graphs this closely.
W Gross Profit Marketing Costs Profit Incremental
Profit
0 0 0 0 — 1 $ 700 $ 66⅔ $ 633⅓ $ 633⅓ 2 1,400 266⅔ 1,133⅓ 500 3 2,100 600 1,500 366⅓ 4 2,400 1,066⅔ 1,333⅓ 2166⅓ 5 2,700 1,666⅔ 1,033⅓ 2300
6 3,000 2,400 600 2433⅓
TABLE 8.6 Calculations of Wyn- dor’s Weekly Profit from Producing W Windows per Week
200
400
600
800
1,000
0 1 2 3 4
Production rate for doors Production rate for windows
Weekly
Profit
($)
D
200
400
600
800
1,000
1,200
1,400
1,600
0 1 2 3 4 5 6
Weekly
Profit
($)
W
FIGURE 8.18 The solid curves show the profit graphs for Wyn- dor’s doors and windows when both overtime costs and nonlinear marketing costs are incorporated into the problem. The dashed- line segments display the approximation used by the separable programming model in Figure 8.19.
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294 Chapter Eight Nonlinear Programming
$300 per unit of W ). Since the profit decreases when W is increased above W 5 3, it seems unde- sirable to increase W this much. Therefore, a particularly close approximation is not needed in this part of the graph, so only a single line segment is used between W 5 3 and W 5 6.
Figure 8.19 shows the separable programming spreadsheet model that is based on the piecewise linear profit graphs in Figure 8.18 . This model is very similar to the separable
Hours9
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=SUMPRODUCT(C11:D11,TotalProduced)
=SUMPRODUCT(C12:D12,TotalProduced)
=SUMPRODUCT(C13:D13,TotalProduced)
Used
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
Maximum
TotalProduced
TotalProfit
UnitProfit
UnitsProduced
G11:G13
E11:E13
C11:D13
F17:G20
C21:D21
D23
C4:D7
C17:D20
Range Name Cells
1
A B C D E F G
2
3 Doors Windows
4 Regular (0–1)
Regular (1–2)
Regular (2–3)
Regular (0–1)
Regular (1–2)
Regular (2–3)
Overtime
Overtime 1 0
Total Produced
Total Profit
4 3
Unit Profit
$350.00 $633.33
$300.00 $500.00
$250.00 $366.67
$100.00 -$300.00
5
6
Hours Used per Unit Produced
Units Produced
7
8
9 Hours
Used
4
6
18
Hours
Available
4
12
18
≤
≤ ≤
1
0
3
0
2
2
Plant 1
Plant 2
Plant 3
10
11
12
13
14
15
16
17
18
19
1
1
1
1
20
21
22
23
Doors Windows
Maximum
Doors Windows
Wyndor with Overtime and Marketing Costs (Separable)
1
1
1
1
1
1
1
3
1
1
≤ ≤
≤ ≤
$2,500
Total Produced21
B C D
=SUM(C17:C20) =SUM(D17:D20)
Total Profit23
C D
=SUMPRODUCT(UnitProfit,UnitsProduced)
Solver Parameters Set Objective Cell: TotalProfit To: Max By Changing Variable Cells: UnitsProduced
Subject to the Constraints: HoursUsed <= HoursAva ilable
UnitsProduced <= Maximum
Solver Options: Make Variables Nonnegative Solving Method: Simplex LP
FIGURE 8.19 A spreadsheet model for the separable program- ming formulation of the Wyndor problem when overtime is needed and nonlinear marketing costs also are incorporated into the problem. By summing the columns of the chang- ing cells UnitsProduced (C17:D20), Total Produced (C21:D21) gives the optimal produc- tion rates obtained by Solver. The objective cell TotalProfit (D23) shows a resulting weekly profit of $2,500.
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8.3 Separable Programming 295
programming spreadsheet model in Figure 8.16 that does not incorporate the new estimates of nonlinear marketing costs. The latter model is based on the piecewise linear profit graphs in Figure 8.15 , each of which has only two line segments. Therefore, each of the sets of cells, UnitProfit (C4:D5) and UnitsProduced (C14:D15), has only two rows. Because each of the piecewise linear profit graphs in Figure 8.18 has four line segments, each of the cor- responding sets of cells in Figure 8.19 , UnitProfit (C4:D7) and UnitsProduced (C17:D20), has four rows. The numbers in UnitProfit (C4:D7) are the slopes of the corresponding line segments in Figure 8.18 . These slopes come directly from the incremental profits given in Tables 8.5 and 8.6 , except for cell D7. This cell is based on the line segment from W 5 3 to W 5 6 in Figure 8.18 , which has a slope of 2 $300 since the profit is decreasing at the rate of $300 per unit increase in W. This slope of 2 $300 is the average of the last three incre- mental profits in Table 8.6 . All the other line segments in Figure 8.18 run over only one unit of D or W, so the slope of each of these line segments equals the corresponding incremental profit in Table 8.5 or 8.6 .
The changing cells in Figure 8.19 , UnitsProduced (C17:D20), give the optimal solution obtained by Solver. TotalProduced (C21:D21), which 5 SUM(UnitsProduced), gives the cor- responding total production rates, namely,
D 5 4: Produce 4 doors per week, including 1 on overtime
W 5 3: Produce 3 windows per week
The objective cell TotalProfit (D23) indicates that the resulting weekly profit would be $2,500.
To check these results, the Management Science Group also formulates and runs the corresponding nonlinear programming model that employs the formulas for the smooth profit graphs in Figure 8.18 . This spreadsheet model is shown in Figure 8.20 . It is nearly the same as the one in Figure 8.11 that does not include overtime costs. The one differ- ence is that the single row for UnitProfit (C4:D4) and for UnitsProduced (C12:D12) in Figure 8.11 now is split into two rows each, (C4:D5) and (C15:D16), to differentiate between regular time and overtime. Because of the kink in the profit graphs at D 5 3 and W 5 3, the pairs of rows are needed to provide separate formulas for the two parts of each profit graph on either side of the kink. TotalProduced (C17:D17) provides the same opti- mal solution, ( D, W ) 5 (4, 3), as the separable programming model in Figure 8.19 , with a total profit of $2,500.
Based on these results, Wyndor management now adopts production rates of ( D, W ) 5 (4, 3) for the next four months while overtime is needed. Following that period, the plan is to switch to ( D, W ) 5 (3⅓, 4) because of the results obtained in Section 8.2.
This completes the Wyndor case study. One key lesson is that a management science study may involve developing more than just a single model to represent a problem. As the study goes on and additional relevant considerations come to light, the original model may evolve through a series of enhancements into a rather different kind of model. For example, what started as a linear programming model may end up needing to be a nonlinear programming model.
Separable programming calculates unit profits for different parts of a profit graph by using the slopes of the line segments in the piecewise linear approxi- mation of the profit graph.
Figures 8.19 and 8.20 pro- vide a comparison between a separable programming model and the kind of non- linear programming model discussed in Section 8.2.
Here is one key lesson from the Wyndor case study.
1. For each activity that violates the proportionality assumption, what must be the shape of its profit graph (or at least an approximation of the profit graph) in order to apply separable programming?
2. What kind of mathematical model is eventually formulated when applying the separable pro- gramming technique?
3. For problems where the activities have profit graphs with shapes similar to the one shown in Figure 8.17 , what are some advantages of using the kind of approximation displayed in this figure to enable applying separable programming?
4. For these same problems, what is an advantage of instead using a nonlinear programming model that directly employs the formulas for the profit graphs?
Review Questions
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296 Chapter Eight Nonlinear Programming
Hours7
E
8
9
10
11
=SUMPRODUCT(C9:D9,TotalProduced)
=SUMPRODUCT(C10:D10,TotalProduced)
=SUMPRODUCT(C11:D11,TotalProduced)
Used
DoorsProduced
GrossProfitFromSales
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
MarketingCost
Maximum
TotalMarketingCost
TotalProduced
TotalProfit
UnitProfit
UnitsProduced
WindowsProduced
C17
H18
G9:G11
E9:E11
C9:D11
C19:D19
F15:G16
H19
C17:D17
H20
C4:D5
C15:D16
D17
Range Name Cells
1
A B C D E F G H
2
3 Doors Windows
4 Regular
Overtime
Regular
Overtime 1 0
Total Produced
Marketing Cost
Total Profit
Total Marketing Cost
Gross Profit from Sales
$2,500
$1,000
$3,500
4 3
Unit Profit (Gross)
$375 $700
$275 $3005
6
Hours Used per Unit Produced
7
8
9
Hours
Used
4
6
18
Hours
Available
4
12
18
≤
≤ ≤
1
0
3
0
2
2
Plant 1
Plant 2
Plant 3
10
11
12
13
14
15
16
17
18
19
20
DoorsUnits Produced Windows
Maximum
Doors Windows
Wyndor With Overtime and Marketing Costs (Nonlinear Programming)
3
1
3
3
33
≤ ≤
$600$400
Total Produced17
18
19
B C D
=SUM(C15:C16) =SUM(D15:D16)
Marketing Cost =25*(DoorsProduced^2) =66.667*(WindowsProduced^2)
Gross Profit from Sales
Total Marketing Cost
Total Profit
18
19
20
G H
=SUMPRODUCT(UnitProfit,UnitsProduced)
=SUM(MarketingCost)
=GrossProfitFromSales–TotalMarketingCost
Solver Parameters Set Objective Cell: TotalProfit To: Max By Changing Variable Cells:
UnitsProduced
Subject to the Constraints: HoursUsed <= HoursAvailable
UnitsProduced <= Maximum
Solver Options: Make Variables Nonnegative
Solving Method: GRG Nonlinear
or Quadratic (RSPE)
FIGURE 8.20 A spreadsheet model for the nonlinear program- ming formulation of the Wyndor problem when overtime is needed and nonlinear marketing costs also are incorporated into the problem. Total Produced (C17:D17) gives the optimal produc- tion rates obtained by Solver and the objective cell TotalProfit (H20) shows the resulting weekly profit. Note that this nonlinear program- ming formulation gives the same results as the separable program- ming formulation in Figure 8.19.
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8.4 Difficult Nonlinear Programming Problems 297
8.4 DIFFICULT NONLINEAR PROGRAMMING PROBLEMS
We saw in Section 8.2 that even if a model has a nonlinear objective function, so long as the model has certain properties (e.g., linear constraints and maximizing an objective function with decreasing marginal returns), the Nonlinear Solver can easily find an optimal solution. Furthermore, we saw in Section 8.3 that in some cases separable programming can be used to model (or approximate) a nonlinear problem in such a way that linear programming can be used to efficiently find an optimal solution.
However, nonlinear programming problems come in many guises and forms. For example, for problems where the objective is to maximize total profit, some might have increasing marginal returns for the profit from certain activities. Some might have nonlin- ear functions in the constraints. Some might have profit graphs with several disconnected curves. These other kinds of nonlinear programming problems often are much more dif- ficult, if not impossible, to solve. The reason is that there may be many locally optimal solutions that are not globally optimal. We saw in Figures 8.7 and 8.8 how the Nonlinear Solver can get stuck at these locally optimal solutions without ever finding the globally optimal solution.
One approach for attempting to solve problems that may have multiple local optima is to run the Nonlinear Solver many times, each time starting with a different initial solution entered into the changing cells on the spreadsheet. For each run, the Nonlinear Solver will start its search at the given initial solution (the starting point) and move in a direction that improves the objective function until it finds a local optimum. By trying many starting points, the goal is to find most or all of the local optima. We then pick the best solution found from all the trials. At a minimum, we are likely to end up with a solution that is better than if we just take the first local optimum that the Nonlinear Solver finds. With luck, one of the starting points will yield the globally optimal solution.
For example, consider the model in Figure 8.7 with the corresponding profit graph in Fig- ure 8.8 . For any starting point x less than 1.5, the objective function increases by moving toward the local maximum at x 5 0.371 (Profit 5 $3.19). Thus, for any starting point x less than 1.5 (including the starting point x 5 0 tried in the leftmost spreadsheet in Figure 8.7 ), the Nonlinear Solver’s search will move toward and eventually converge to this local maxi- mum. Similarly, for any starting point x between 1.5 and 4.6 (such as x 5 3 tried in the center spreadsheet in Figure 8.7 ), the Nonlinear Solver will converge to the local (and global) maxi- mum at x 5 3.126 (Profit 5 $6.13). Finally, for any starting point x greater than 4.6 (such as x 5 4.7 tried in the rightmost spreadsheet in Figure 8.7 ), the Nonlinear Solver will converge to the local maximum at x 5 5 (Profit 5 $0). By trying several starting points, three differ- ent local optima are found. The best of these local optima is x 5 3.126 with a corresponding profit of $6.13.
The Nonlinear Solver includes an automated way of trying multiple starting points. In Excel’s Solver, clicking on the Options button in Solver and then choosing the GRG Non- linear tab brings up the Nonlinear Solver Options dialog box shown on the left side of Figure 8.21 . Selecting the Use Multistart option causes the Nonlinear Solver to randomly select 100 different starting points. (The number of starting points can be varied by chang- ing the Population Size option.) In RSPE, the Multistart option (and Population Size) are set on the Engine tab in the Model pane, as shown on the right side of Figure 8.21 . When Multistart is enabled, the Nonlinear Solver then provides the best solution found after solv- ing with each of the different starting points.
Unfortunately, there is no guarantee in general of finding a globally optimal solution, no matter how many different starting points are tried. Also, if the profit graphs are not smooth (e.g., if they have discontinuities or kinks), as is typically the case if functions like IF, ABS, MAX, or ROUND are used, then the Nonlinear Solver may not even be able to find local optima. Fortunately, there is another approach available to attempt to solve these difficult nonlinear programming problems. We explore this new approach in the next section.
The Nonlinear Solver often has difficulty solving non- linear programming models if the constraints are non- linear or if the profit graphs for any activities are either not smooth or have increas- ing marginal returns.
If there are multiple local optima, running the Non- linear Solver many times with different starting points can sometimes find the global optimum.
This approach has some major limitations.
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298
Deutsche Post DHL is the largest logistics service provider worldwide. It employs over half a million people in more than 220 countries while delivering 3 million items and over 70 million letters each day with over 150,000 vehicles. The dramatic story of how DHL quickly achieved this lofty status is one that combines enlightened managerial lead- ership, an innovative marketing campaign, and the appli- cation of nonlinear programming to optimize the use of marketing resources.
Starting as just a German postal service, senior manage- ment developed a visionary plan to begin the 21st century by transforming the company into a truly global logistics business. The first step was to acquire and integrate a num- ber of similar companies that already had a strong pres- ence in various other parts of the world. Because customers who operate on a global scale expect to deal with just one provider, the next step was to develop an aggressive mar- keting program based on extensive marketing research to re-brand DHL as a superior truly global company that could fully meet the needs of these customers. These marketing
activities were pursued vigorously in more than 20 of the largest countries on four continents.
This kind of marketing program is very expensive, so it is important to use the limited marketing resources as effec- tively as possible. Therefore, management scientists devel- oped a brand choice model with an objective function that measures this effectiveness. Nonlinear programming then was used to maximize this objective function budget with- out exceeding the total marketing budget.
This innovative use of marketing theory and non- linear programming led to a substantial increase in the global brand value of DHL that enabled it to catapult into a market-leading position. This increase from 2003 to 2008 was estimated to be $1.32 billion (a 32 percent increase). The corresponding return on investment was 38 percent.
Source: M. Fischer, W. Giehl, and T. Freundt, “Managing Global Brand Investments at DHL,” Interfaces 41, no. 1 (January–February 2011), pp. 35–50. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
FIGURE 8.21 On the left, Excel’s Nonlinear Solver Options dialog box provides several parameters for the Nonlinear Solver. The Multistart option causes the Nonlinear Solver to try many random starting points. (The number of starting points can be adjusted by changing the Popu- lation Size.) On the right, the same options are available in RSPE under the Engine tab on the Model pane under Global Optimization.
1. The Nonlinear Solver has difficulty solving nonlinear programming problems with certain prop- erties. List three of these properties.
2. What is a method for attempting to solve problems with multiple locally optimal solutions?
Review Questions
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8.5 Evolutionary Solver and Genetic Algorithms 299
8.5 EVOLUTIONARY SOLVER AND GENETIC ALGORITHMS
A major new feature of Excel 2010 (and also available in RSPE) is a new and improved Solver. In addition to retaining the solution methods (called Simplex LP and GRG Nonlinear in Excel 2010) that were included with Solver in older versions of Excel, this new Solver added a new search procedure called Evolutionary Solver .
Evolutionary Solver uses an entirely different approach than the Nonlinear Solver to search for an optimal solution for a nonlinear programming model. The philosophy of Evolutionary Solver is based on genetics, evolution, and the survival of the fittest. Hence, this type of algo- rithm is sometimes called a genetic algorithm .
When dealing with a nonlinear programming problem, the Nonlinear Solver starts with a single solution (the starting point), and then moves in directions that will improve this solu- tion. At any point in time, the Nonlinear Solver is only keeping track of a single solution (the best one found so far). In contrast, Evolutionary Solver begins by randomly generating a large set of candidate solutions, called the population . Throughout the solution process, Evolu- tionary Solver keeps track of the whole population of candidate solutions. Much like trying different starting points with the Nonlinear Solver, this attention to many candidate solutions can help avoid being trapped at a local optimum.
After generating the population, Evolutionary Solver next creates a new generation of the population. The existing population of candidate solutions is paired off to create “off- spring” for the next generation. Borrowing from the principles of genetics, the offspring com- bine some elements from each parent. For example, an offspring could combine some of the changing cell values from one parent and some from the other, while other changing cells might be averaged between the two parents.
Among the population of solutions in any generation, some solutions will be good (or “fit”) and some will be bad (or “unfit”). The level of fitness is determined by evaluating the objective function at each of the candidate solutions in the population. A penalty is subtracted for any solution that does not satisfy one or more of the constraints. Then, borrowing from the princi- ples of evolution and the survival of the fittest, the “fit” members of the population are allowed to reproduce frequently (create many offspring), while the “unfit” members are not allowed to reproduce. In this way, the population eventually evolves to become more and more fit.
Another key feature of genetic algorithms is mutation . Like gene mutation in biology, Evolutionary Solver will occasionally make a random change in a member of the population. For example, the value of one changing cell might be replaced with a new random value. This mutation can create offspring that are far removed from the rest of the population. This is important, since it can help the algorithm get unstuck if it is getting trapped near a local optimum.
Evolutionary Solver keeps creating new generations of solutions until there have been no improvements in several consecutive generations. The algorithm then terminates and the best solution found so far is reported.
Now let us look at an example where Evolutionary Solver is needed to solve the problem.
Selecting a Portfolio to Beat the Market In Section 8.2, we developed a model for finding a portfolio of stocks that minimizes the risk (variance of the return from the portfolio) subject to achieving at least some desired minimum expected return. The Nonlinear Solver (or the Quadratic Solver in RSPE) could be used for that problem because the constraints were linear whereas the objective function had a smooth quadratic form and had decreasing marginal returns.
Now consider another common goal of portfolio managers—to beat the market. Figure 8.22 shows a spreadsheet model for pursuing this goal when choosing a portfolio from five large stocks traded on the New York Stock Exchange (NYSE): Disney (DIS), Boeing (BA), General Electric (GE), Procter & Gamble (PG), and McDonald’s (MCD). The quarterly performance (return) of each of these stocks over a six-year period (2006–2011) is shown in StockData (D4:H27). The performance of the market as a whole, as measured by the NYSE Composite Index, is shown in column K.
Evolutionary Solver uses the principles of genetics, evolution, and the survival of the fittest.
Each pair of parents creates offspring that resemble the parents.
Only the fit parents are allowed to create many offspring.
Random mutations occa- sionally occur in the offspring.
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If we assume that past performance is somewhat of an indicator of the future, then pick- ing a portfolio that beat the market most often during these six years might yield a portfolio that will more than likely beat the market in the future. Thus, the model in Figure 8.22 uses the objective of choosing the portfolio that beat the market for the largest number of quarters during this period.
The changing cells in this model are Portfolio (D31:H31), representing the percentage of the portfolio to invest in each individual stock. The return of the given portfolio for each quar- ter is calculated in column I. Column J then compares the return of the portfolio to the return of the market and determines whether the portfolio beat the market using the IF functions shown below the spreadsheet in Figure 8.22 . The number of quarters in which the portfolio
The objective is to find the portfolio that beat the mar- ket most frequently.
BeatMarket? Market NumberBeatingTheMarket OneHundredPercent OneHundredPercent2 Portfolio Return StockData Sum ZeroPercent
J4:J27 K4:K27 J36 D33:H33 K31 D31:H31 I4:I27 D4:H27 I31 D29:H29
Range Name Cells
2
3
4 =SUMPRODUCT(Portfolio,D4:H4)
=SUMPRODUCT(Portfolio,D5:H5)
Beat
Market?Return
=IF(Return>Market,"Yes","No")
5
=SUMPRODUCT(Portfolio,D6:H6)
=IF(Return>Market,"Yes","No")
6
=SUMPRODUCT(Portfolio,D7:H7)
=IF(Return>Market,"Yes","No")
=IF(Return>Market,"Yes","No")7
8
9 : :
: :
I J
30
31
Sum
=SUM(Portfolio)
I
34
35
36 =COUNTIF(BeatMarket?,"Yes")
Beating the Market
Number of Quarters
J
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
A B C D E F G H I J K
Beating the Market (Evolutionary Solver) Beat Market
Quarter Year DIS BA GE PG MCD Return Market? (NYSE) Q4 Q3 Q2 Q1 Q4 Q3 Q2 Q1 Q4 Q3 Q2 Q1 Q4 Q3 Q2 Q1 Q4 Q3 Q2 Q1 Q4 Q3 Q2 Q1
0% 0% 0% 0% 0% <= <= <= <= <= Sum
Portfolio 20.0% 20.0% 20.0% 20.0% 20.0% 100% = 100% <= <= <= <= <=
100% 100% 100% 100% 100% Number of Quarters Beating the Market
15
26.43% -22.76% -17.60%
0.53% 13.95% -1.33% 6.69%
-13.08% 35.05% 0.85%
28.62% 20.64%
-15.78% -25.03% -12.21% -11.22% -14.54% -16.40%
9.56% 8.55% 0.46%
13.07% -3.36% 5.46%
11.41%
-9.39% 14.88% 14.57% 5.09%
-9.78% 8.26%
18.78% 17.69% 28.45%
-19.94% -24.97% -1.62% -0.57% -2.81% -5.12% 0.75%
0.47% 11.87% 3.05% 7.56%
16.34%
-0.87%
21.99% 18.81% -18.54% -5.17% 10.39% 13.47% 13.53%
-20.28% 21.08% -7.30% 41.00% 16.79%
-35.72% -35.26% -3.19%
-27.08% 0.77%
-9.70% 8.89% 9.01%
-4.21% 6.19% 7.88%
-4.54% -0.04%
0.20% 6.46%
4.06% -3.54% 8.09% 0.76%
-4.46% 5.10% 5.48%
14.25% 9.46%
-23.30% -10.72% 15.31%
-12.72% -4.08% 4.90%
15.60% -2.59% -1.25% 4.22%
12.08% -2.98% 0.04%
4.86% 11.64% -0.07% 3.83%
13.96% -0.45% 7.77%
10.34% 0.15% 6.22%
-11.43% 1.69%
10.41% 1.46%
-4.69% 11.01% 7.29%
12.69% 1.61%
16.10% 16.42% -2.20% 1.89%
15.10% 17.76% -10.77%
0.33% 7.12% 7.73% 8.01%
-9.61% 15.45% 5.63%
20.34% 16.31%
-21.24% -18.86%
1.74% -10.03% -5.07% -3.06% 8.42% 5.36%
-0.58% 10.29% 7.22% 0.66% 5.93% No
No No
No
No
No No
No No
Yes Yes Yes
Yes
Yes
Yes Yes
Yes Yes Yes Yes
Yes Yes Yes Yes 10.09%
-18.36% -1.02% 5.54% 9.38%
12.54% -13.13%
3.66% 3.97%
17.03% 18.60%
-13.52% -23.57% -13.02% -1.56% -9.68% -2.98% 1.68% 6.60% 1.34% 7.90% 3.68%
-0.78% 6.18%
2011 2011 2011 2011 2010 2010 2010 2010 2009 2009 2009 2009 2008 2008 2008 2008 2007 2007 2007 2007 2006 2006 2006 2006
FIGURE 8.22 A spreadsheet model (prior to using Solver) for selecting a portfolio that beat the market most frequently in recent quarters. A start- ing solution has been entered in the changing cells Portfolio (D31:H31). The objective cell is NumberBeatingTheMarket (J36).
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8.5 Evolutionary Solver and Genetic Algorithms 301
beat the market is then calculated in the objective cell, NumberBeatingTheMarket (J36). As seen in the figure, a portfolio that was evenly split among the five stocks (20 percent in each) would have beaten the market in 15 of the 24 quarters during this six-year period.
The Nonlinear Solver would have little to no chance of solving this model. The objective function is not smooth since changes in the Portfolio can cause instantaneous (nonsmooth) jumps in the objective cell (the number of quarters that the portfolio beats the market). However, the objective cell remains constant for small changes in the changing cells until the change is signifi- cant enough to cause a quarter in column J to switch from Yes to No (or No to Yes). An unfor- tunate consequence of this is that nearly every solution is a local maximum, since very small changes in the portfolio will lead to no improvement in the objective cell. Thus, the Nonlinear Solver typically will stop its search immediately and report the initial solution as a local maxi- mum. Since the Nonlinear Solver cannot solve this model, we will try the Evolutionary Solver.
Applying Evolutionary Solver to Portfolio Selection to Beat the Market Figure 8.23 shows the entries for Excel’s Solver dialog box that are needed to solve the model shown in Figure 8.22 . The objective cell is NumberBeatingTheMarket (cell J36) and the chang- ing cells are Portfolio (D31:H31). It also must include that (1) the portfolio needs to sum to 100 percent and (2) each individual stock must represent between 0 and 100 percent of the portfolio.
With Excel’s Solver, there is a dropdown menu that gives a choice of which Solving Method to employ. Evolutionary Solver is selected in Figure 8.23 for this model. In RSPE, the solving method is chosen in the Engine tab of the Model pane, as shown on the right side of Figure 8.24 .
When using Excel’s Solver, clicking the Options button and then the Evolutionary tab yields the Evolutionary Solver Options dialog box shown on the left side of Figure 8.24 .
The Nonlinear Solver can’t handle this kind of problem.
FIGURE 8.23 Excel’s Solver dialog box that is used to complete the spreadsheet model introduced in Figure 8.22. Selecting Evolutionary from the dropdown menu specifies that Evolution- ary Solver will be used to solve the problem.
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This allows changing various parameters of the search, such as the maximum time without improvement to allow the search to continue, the size of the population, and the mutation rate. The same parameters are available in RSPE in the Engine tab of the Model pane, as shown on the right side of Figure 8.24 . The default values for the parameters displayed in Figure 8.24 are reasonable ones for most small applications. However, feel free to experiment with these parameters. For example, increasing the population size or mutation rate may help with searches that are getting stuck.
The Require Bounds on Variables option is selected by default. This forces all changing cells to be constrained with lower and upper limits. We strongly recommend that you use this option to place bounds on the changing cells whenever possible. This greatly narrows the area over which Evolutionary Solver needs to search and can increase the chances of finding a good solution.
Running Solver then causes Evolutionary Solver to begin its search. Within a minute or so, the solution shown in Figure 8.25 was found. This represents a portfolio that beat the market in 19 of the 24 quarters over this six-year period. Is this solution optimal? Perhaps not. Unfor- tunately, there is no way to guarantee that we have found an optimal solution. However, it likely is a good solution (close to optimal).
Applying Evolutionary Solver to a Traveling Salesman Problem Becky Thomas has just completed her MBA degree from the University of Washington Busi- ness School in Seattle and would like to celebrate during the summer with a driving tour of the United States, including visits to see a major league baseball game in every American League city. She would then return home to Seattle to start a job in the fall. She would like to plan her route to minimize the travel distance.
This is an example of a well-known problem in management science called the traveling- salesman problem. In the generic version of this problem, a salesperson needs to plan a sales trip to a certain set of cities in some order. Starting from a given location (home) and then returning home at the end, the objective is to find the route that minimizes the total travel distance (or time).
Figure 8.26 displays a map of the United States showing each of the 14 American League cities. Becky will start and end her tour in Seattle. After listing the cities to be visited in
Use the Require Bounds on Variables option whenever possible.
This solution found by Evolutionary Solver is not guaranteed to be optimal, but it probably is at least close to optimal.
FIGURE 8.24 On the left, Excel’s Evolutionary Solver Options dialog box provides several param- eters for Evolutionary Solver. The default values shown here are reasonable choices for most small applications. On the right, the same options are available in RSPE under the Engine tab on the Model pane.
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8.5 Evolutionary Solver and Genetic Algorithms 303
10.09% -18.36%
-1.02% 5.54% 9.38%
12.54% -13.13%
3.66% 3.97%
17.03% 18.60%
-13.52% -23.57% -13.02%
-1.56% -9.68% -2.98% 1.68% 6.60% 1.34% 7.90% 3.68%
-0.78% 6.18%
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
A B C D E F G H I J K
Beating the Market (Evolutionary Solver) Beat Market
Quarter Year DIS BA GE PG MCD Return Market? (NYSE) Q4 2011 Q3 2011 Q2 2011 Q1 2011 Q4 2010 Q3 2010 Q2 2010 Q1 2010 Q4 2009 Q3 2009 Q2 2009 Q1 2009 Q4 2008 Q3 2008 Q2 2008 Q1 2008 Q4 2007 Q3 2007 Q2 2007 Q1 2007 Q4 2006 Q3 2006 Q2 2006 Q1 2006
0% 0% 0% 0% 0% <= <= <= <= <= Sum
Portfolio 33.1% 7.6% 27.3% 0.2% 31.7% 100% = 100% <= <= <= <= <=
100% 100% 100% 100% 100% Number of Quarters Beating the Market
19
26.43% -22.76% -17.60%
0.53% 13.95% -1.33% 6.69%
-13.08% 35.05%
0.85% 28.62% 20.64%
-15.78% -25.03% -12.21% -11.22% -14.54% -16.40%
9.56% 8.55% 0.46%
13.07% -3.36% 5.46%
11.41%
-9.39% 14.88% 14.57%
5.09% -9.78% 8.26%
18.78% 17.69% 28.45%
-19.94% -24.97%
-1.62% -0.57% -2.81% -5.12% 0.75%
0.47% 11.87%
3.05% 7.56%
16.34%
-0.87%
21.99% 18.81% -18.54% -5.17% 10.39% 13.47% 13.53%
-20.28% 21.08% -7.30% 41.00% 16.79%
-35.72% -35.26%
-3.19% -27.08%
0.77% -9.70% 8.89% 9.01%
-4.21% 6.19% 7.88%
-4.54% -0.04%
0.20% 6.46%
4.06% -3.54% 8.09% 0.76%
-4.46% 5.10% 5.48%
14.25% 9.46%
-23.30% -10.72% 15.31%
-12.72% -4.08% 4.90%
15.60% -2.59% -1.25% 4.22%
12.08% -2.98% 0.04%
4.86% 11.64% -0.07% 3.83%
13.96% -0.45% 7.77%
10.34% 0.15% 6.22%
-11.43% 1.69%
10.41% 1.46%
-4.69% 11.01%
7.29% 12.69%
1.61% 16.10% 16.42% -2.20% 1.89%
15.10% 20.37% -12.40%
-0.78% 8.80% 9.64%
10.32% -9.93% 13.64%
7.58% 19.32% 17.58%
-21.25% -19.30%
1.00% -8.01% -3.32% -2.09% 5.75% 6.84%
-0.45% 11.73%
8.15% 0.97% 6.87%
No
No No
No
No
Yes Yes Yes Yes Yes
Yes Yes Yes Yes
Yes Yes
Yes Yes Yes Yes
Yes Yes Yes Yes
FIGURE 8.25 After running Solver, Evolutionary Solver found the solution shown in the changing cells Portfolio (D31:H31) for the model formu- lated in Figures 8.22 and 8.23. The objective cell NumberBeatingTheMarket (J36) indicates that this portfolio beat the market in 19 of these 24 quarters. Running Solver again probably would lead to at least a slightly different solution for the portfolio.
Seattle
Oakland
Anaheim
Texas
Kansas City
Chicago
Minnestota Toronto
Detroit
Cleveland
Baltimore
New York
Boston
Tampa Bay
FIGURE 8.26 A map of the United States showing the 14 cities (including Toronto, Canada) with American League ballparks.
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304 Chapter Eight Nonlinear Programming
alphabetical order, each city has been labeled by both a number (an integer between 1 and 13) and a letter code, as given in cells B6:C18 and E3:Q4 of Figure 8.27 . The data cells are Dis- tance (D5:Q18), giving the travel distance between each pair of cities. A decision needs to be made about the order in which to visit the other cities before returning to Seattle. Therefore, the respective changing cells Route (D22:P22) indicate which city (referenced by its number label) is visited at each stage of the route. In other words, the first city visited after Seattle will have its number label in D22, the second in E22, and so on. The spreadsheet in Figure 8.27 shows the route if the cities are visited in alphabetical order (Anaheim 1, Baltimore 2, Boston 3, etc.). The total distance of this route is 18,962 miles.
Row 23 displays the letter code for each city based upon the number label in row 22, using the INDEX function of Excel. Row 24 uses the INDEX function to look up the distance to travel to each city from the preceding city in the route. The objective cell TotalMilesTraveled (Q26) adds up the total miles driven in the route.
Because each city needs to be visited exactly once, the one constraint needed in this model is that the changing cells all must be unique integers from 1 to 13. Both Excel’s Solver (Excel 2010 and later) and RSPE include a type of constraint, called alldifferent, that does just what we need. When n changing cells are choosing integers from 1 to n, constraining these chang- ing cells to be alldifferent forces them to be unique integers from 1 to n. To implement the alldifferent constraint in Excel’s Solver, press the Add button in Solver to bring up the Add Constraint dialog box. Select the changing cells (Route) for the left-hand side of the dialog box and then choose dif in the pull-down menu in the center of the dialog box, as shown in Figure 8.28 . In RSPE, select the changing cells (Route) and then choose alldifferent in the Variable Type/Bound submenu of the Constraints menu.
The resulting model is not linear because of the INDEX function used to calculate dis- tances and the alldifferent constraint. However, the Evolutionary Solver can be used to try to find a good route. After solving with Evolutionary Solver, the resulting solution is shown in
INDEX(Range, i) returns the ith element in Range, where Range is a block of cells. INDEX(Range, a, b) returns the element in row a and column b of Range.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
A B C D E F G H I J K L M N O P Q
Tour of the American League Ballparks
Distance 1 2 3 4 5 6 7 8 9 10 11 12 13
(miles) SEA ANA BAL BOS CHI CLE DET KC MIN NY OAK TB TEX TOR
SEA 0 1134 2708 3016 2052 2391 2327 1858 1653 2841 810 3077 2131 2564
1 ANA 1134 0 2647 3017 2048 2382 2288 1577 1857 2794 387 2490 1399 2523
2 BAL 2708 2647 0 427 717 358 514 1070 1113 199 2623 950 1357 457
3 BOS 3016 3017 427 0 994 657 799 1435 1390 222 3128 1293 1753 609
4 CHI 2052 2048 717 994 0 348 279 542 410 809 2173 1160 921 515 5 CLE 2391 2382 358 657 348 0 172 819 758 471 2483 1108 1189 296 6 DET 2327 2288 514 799 279 172 0 769 685 649 2399 1184 1156 240 7 KC 1858 1577 1070 1435 542 819 769 0 443 1233 1861 1171 505 1006
8 MIN 1653 1857 1113 1390 410 758 685 443 0 1217 1979 1573 949 906 9 NY 2841 2794 199 222 809 471 649 1233 1217 0 2930 1150 1559 516
10 OAK 810 387 2623 3128 2173 2483 2399 1861 1979 2930 0 2823 1752 2627 11 TB 3077 2490 950 1293 1160 1108 1184 1171 1573 1150 2823 0 1079 1348
12 TEX 2131 1399 1357 1753 921 1189 1156 505 949 1559 1752 1079 0 1435
13 TOR 2564 2523 457 609 515 296 240 1006 906 516 2627 1348 1435 0
Tour Start 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th 11th 12th 13th End
Route 1 2 3 4 5 6 7 8 9 10 11 12 13
City SEA ANA BAL BOS CHI CLE DET KC MIN NY OAK TB TEX TOR SEA
Miles Traveled 1134 2647 427 994 348 172 769 443 1217 2930 2823 1079 1435 2564
Total Miles Traveled 18,982
23
24
B C D E P Q
City SEA =INDEX($C$6:$C$18,D22) =INDEX($C$6:$C$18,E22) =INDEX($C$6:$C$18,P22) SEA
Miles Traveled =INDEX(D6:D18,D22) =INDEX($E$6:$Q$18,D22,E22) =INDEX($E$6:$Q$18,O22,P22) =INDEX(D6:D18,P22)
26
P Q
Total Miles Traveled =SUM(D24:Q24)
FIGURE 8.27 A spreadsheet model for determining the minimum-distance route from Seattle to the other 13 American League cities and back to Seattle. The changing cells are Route (D22:P22) and the objective cell is TotalMilesTraveled (Q26).
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8.5 Evolutionary Solver and Genetic Algorithms 305
both D22:P22 and D23:P23 in Figure 8.29 . This route is much improved over the one shown in Figure 8.27 , with a total distance of 9,040 miles instead of 18,962 miles. In this case, Evo- lutionary Solver has succeeded in finding the optimal solution.
Advantages and Disadvantages of Evolutionary Solver Evolutionary Solver has two significant advantages over the Nonlinear Solver for solving dif- ficult nonlinear programming problems. First, the complexity of the objective function does not impact Evolutionary Solver. As long as the function can be evaluated for a given candi- date solution (in order to determine the level of fitness), it does not matter if the function has kinks or discontinuities or many local optima. Second, by evaluating whole populations of candidate solutions that aren’t necessarily in the same neighborhood as the current best solu- tion, Evolutionary Solver keeps from getting trapped at a local optimum. Furthermore, even if the whole population eventually evolves toward a solution that is only locally optimal, muta- tion allows for the possibility of getting unstuck. In fact, because of the random mutations, Evolutionary Solver is guaranteed to eventually find an optimal solution for any optimization problem if it runs forever, but this is impractical of course.
Evolutionary Solver can handle problems with com- plicated objective functions and many local optima.
FIGURE 8.28 The Add Constraint dia- log box for Excel’s Solver showing the alldifferent constraint.
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
A B C D E F G H I J K L M N O P Q Tour of the American League Ballparks
Distance 1 2 3 4 5 6 7 8 9 10 11 12 13 (miles) SEA ANA BAL BOS CHI CLE DET KC MIN NY OAK TB TEX TOR
SEA 0 1134 2708 3016 2052 2391 2327 1858 1653 2841 810 3077 2131 2564 1 ANA 1134 0 2647 3017 2048 2382 2288 1577 1857 2794 387 2490 1399 2523 2 BAL 2708 2647 0 427 717 358 514 1070 1113 199 2623 950 1357 457 3 BOS 3016 3017 427 0 994 657 799 1435 1390 222 3128 1293 1753 609 4 CHI 2052 2048 717 994 0 348 279 542 410 809 2173 1160 921 515 5 CLE 2391 2382 358 657 348 0 172 819 758 471 2483 1108 1189 296 6 DET 2327 2288 514 799 279 172 0 769 685 649 2399 1184 1156 240 7 KC 1858 1577 1070 1435 542 819 769 0 443 1233 1861 1171 505 1006 8 MIN 1653 1857 1113 1390 410 758 685 443 0 1217 1979 1573 949 906 9 NY 2841 2794 199 222 809 471 649 1233 1217 0 2930 1150 1559 516
10 OAK 810 387 2623 3128 2173 2483 2399 1861 1979 2930 0 2823 1752 2627 11 TB 3077 2490 950 1293 1160 1108 1184 1171 1573 1150 2823 0 1079 1348 12 TEX 2131 1399 1357 1753 921 1189 1156 505 949 1559 1752 1079 0 1435 13 TOR 2564 2523 457 609 515 296 240 1006 906 516 2627 1348 1435 0
Tour Start 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th 11th 12th 13th End Route 10 1 12 11 2 9 3 13 5 6 4 7 8
City SEA OAK ANA TEX TB BAL NY BOS TOR CLE DET CHI KC MIN SEA Miles Traveled 810 387 1399 1079 950 199 222 609 296 172 279 542 443 1653
Total Miles Traveled 9,040
Range Name Cells CityName C5:C18 Distance D5:Q18 MilesTraveled D24:Q24 Route D22:P22 TotalMilesTraveled Q26
Solver Parameters Set Objective Cell: TotalMilesDriven To: Min By Changing Variable Cells: Route Subject to the Constraints:
Route = alldifferent Solver Options: Solving Method: Evolutionary
FIGURE 8.29 After adding the alldifferent constraint, Route (D22:P22) and C23:Q23 show the route from Seattle to the other 13 American League cities and back to Seattle that has been obtained by Evolutionary Solver.
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On the other hand, it must be pointed out that Evolutionary Solver is not a panacea. First, it can take much longer than the Nonlinear Solver to find a final solution. With certain choices of options, the search for better solutions can continue for hours or even days. Second, Evo- lutionary Solver does not perform well on models that have many constraints. For instance, it would not perform very well on many of the models considered in Chapters 2–6, even though the Linear Solver can solve these models almost instantaneously. Third, Evolutionary Solver is a random process. Running Evolutionary Solver again on the same model usually will yield a different final solution. Finally, the best solution found typically is not optimal (although it may be very close). Evolutionary Solver is not an optimizer in the same sense that the other solving methods in Solver are. It does not continuously move toward better solutions until it reaches a local optimum. Rather it is more like an intelligent search engine, trying out dif- ferent random solutions. Thus, while it is quite likely to end up with a solution that is very close to optimal, it almost never returns the exact optimal solution on most types of nonlinear programming problems. (However, its chances of finding an optimal solution are much better on problems like the beat-the-market example where the objective cell only takes on integer values.) Consequently, it often can be beneficial to run the Nonlinear Solver after the Evolu- tionary Solver, starting with the final solution obtained by the Evolutionary Solver, to see if this solution can be improved by searching around its neighborhood.
However, Evolutionary Solver also has several limitations.
1. Why is the algorithm used by Evolutionary Solver called a genetic algorithm? 2. What criterion does Evolutionary Solver use to choose which members of a generation are fit
and which are unfit? 3. How does mutation help Evolutionary Solver? 4. What are two advantages that Evolutionary Solver has over other solving methods in Solver for
solving difficult nonlinear programming problems? 5. What are three disadvantages of Evolutionary Solver compared to the other solving methods in
Solver?
Review Questions
8.6 USING RSPE TO ANALYZE A MODEL AND CHOOSE A SOLVING METHOD
Solver is a powerful tool for solving certain kinds of management science problems, because it provides a choice of solving methods to best fit the problem under consideration. You now have seen all of these solving methods in action, namely,
1. Simplex LP (the Linear Solver) that has been used to solve linear programming and BIP problems in the preceding chapters, as well as in Section 8.3.
2. GRG Nonlinear (the Nonlinear Solver) that has been used to solve, or attempt to solve, certain nonlinear programming problems in Sections 8.1, 8.2, and 8.4.
3. Quadratic (the Quadratic Solver available only in RSPE’s Solver) that is used to solve quadratic programming problems such as the first example in Section 8.2.
4. Evolutionary (the Evolutionary Solver) that has been used to solve certain relatively dif- ficult nonlinear programming problems in Section 8.5.
Each of these solving methods can be very effective when it is a good fit for the problem under consideration. When the spreadsheet model is a linear programming problem, the Lin- ear Solver has some nice features. It is very fast at finding an optimal solution (considerably faster than the Nonlinear Solver and orders of magnitude faster than the Evolutionary Solver on problems of comparable size). When desired, it also provides a very informative Sensitiv- ity Report for doing what-if analysis, as discussed in Chapter 5. The Linear Solver can even be used to solve special kinds of nonlinear programming problems where the separable pro- gramming technique can be used to convert the problem into an equivalent linear program- ming problem, as was demonstrated in Section 8.3.
Unfortunately, the very fast Linear Solver cannot solve other kinds of nonlinear program- ming problems. However, if the problem has certain nice characteristics (e.g., the objective function is smooth and has decreasing marginal returns), then the Nonlinear Solver (or even
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8.6 Using RSPE to Analyze a Model and Choose a Solving Method 307
the Quadratic Solver if the objective function has a quadratic form) can usually solve the problem very effectively. If the objective function is more complicated (e.g., it has many peaks like the example shown in Figure 8.8 ), the MultiStart feature of the Nonlinear Solver still may allow (but doesn’t guarantee) finding an optimal solution. If the profit or cost graphs are not smooth (e.g., if they have discontinuities or kinks)—as may happen if functions like IF, ABS, MAX, or ROUND are used—only the Evolutionary Solver typically will be effec- tive at finding good solutions.
Using RSPE to Choose a Solving Method These general guidelines for the applicability of the various solving methods may make it obvious which one should be used for the particular model under consideration. However, this sometimes is far from obvious, especially when dealing with larger models where it is difficult to identify the characteristics of the objective function. Fortunately, RSPE includes a feature that can analyze your model to determine its type so that you can choose the solv- ing method that is most appropriate for that model type. If you wish, it even includes a fea- ture that will automatically select the solving method that is most appropriate for the model. Finally, if you are trying to create a certain type of model, but the analysis indicates that the model is not of this type, a report can be generated that pinpoints which part of the model does not fit the intended type. (These special features provided by RSPE for analyzing your model are not available with Excel’s standard Solver.)
To demonstrate these features of RSPE, reconsider the portfolio selection example that has been formulated in Figure 8.13 . Under the Optimize menu on the RSPE ribbon, choose Analyze without Solving (or press the Analyze without Solving button—the button with a spreadsheet icon with a checkmark—on the upper-right side of the Model tab in the RSPE Model pane). This causes RSPE to analyze the model and provide a report under the Model Diagnosis section in the Model pane, as shown in Figure 8.30 .
The Model Diagnosis lists the model type as QP Convex. The QP means that the model is a quadratic programming problem (a special kind of nonlinear programming problem
RSPE allows you to analyze your model to determine its type so an appropriate solv- ing method can be chosen. It can even automatically choose the right solving method to apply to your model. If a model does not fit an intended type, RSPE can also generate a report to tell you why. These fea- tures are not available with Excel’s standard Solver.
FIGURE 8.30 After clicking on Analyze Model without Solving, the Model Diagnosis sec- tion of the Model tab in the Model pane shows information about the model. For the portfolio selection example for- mulated in Figure 8.13, it indicates that the model is a convex quadratic pro- gramming model.
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308 Chapter Eight Nonlinear Programming
described in Section 8.2). Convex means that the objective function has decreasing marginal returns (as illustrated in graph (a) of Figure 8.3 for minimization problems like this one) and is therefore the kind of model that the Nonlinear Solver (or with RSPE, the Quadratic Solver) can solve easily. (The Quadratic Solver is designed to solve quadratic programming problems even more efficiently than the Nonlinear Solver can.) If the Model Diagnosis had listed the model type as NLP Convex instead, then the Nonlinear Solver should be used but the Qua- dratic Solver would not be applicable.
Under the Variables–Functions–Dependencies section of the Model Diagnosis in Figure 8.30 , the analysis of the model is presented in even more detail. Under Vars, the analysis lists how many decision variables there are (3), how many of them occur in a smooth way in the objective function (3), and how many are linear within the objective function (0). The Fcns column gives the same information about all the functions in the model—there are three in all, one of which is smooth (but not linear) and two of which are linear. This is because the objective function (calculating the variance of the portfolio) is smooth but nonlinear, while the other two functions (the expected return and the sum of the portfolio weights) are linear. The Dpns column counts every time each variable is used among all the functions—there are nine dependencies in all, since all three variables are used in all three functions, where all three variables are smooth (but not linear) in the objective function and the remaining six dependencies are linear in the other two functions.
If the same analysis were applied to the original Wyndor problem ( Figure 8.9 ) or the trav- eling salesman problem formulated in Figure 8.28 , then the model diagnosis would be as shown on the left and right side of Figure 8.31 , respectively. The left side of Figure 8.31 indicates that the Wyndor problem is LP Convex (a linear programming model) with all linear variables and functions and should therefore be solved by the Linear Solver. The right side of this figure indicates that the traveling salesman problem is NSP (a nonsmooth problem ) and can therefore be solved effectively by only the Evolutionary Solver.
Under the Engine tab of the RSPE Model pane (see Figure 8.21 ), there is an option to Automatically Select Engine. Turning this option on will cause RSPE to use the Solver engine to select the solving method that is most appropriate for the type of model being solved. If RSPE analyzes the model and determines it to be linear, it will use the Linear Solver. If RSPE finds that the model is nonlinear but convex, it will use either the Nonlinear Solver or (if the objective function has a quadratic form) the Quadratic Solver. If RSPE determines that the model is not smooth, it will use the Evolutionary Solver.
Using RSPE to Further Analyze the Model For some applications, it may appear that the appropriate model for the problem under con- sideration should at least come close to fitting a desired model type (e.g., linear program- ming) for which a very efficient solving method is available. However, after completing an initial formulation of the model, the kind of model diagnosis illustrated by Figures 8.30 and 8.31 may indicate that the model instead fits a less desirable type (e.g., NSP, a nonsmooth problem). This may occur because of a mistake in the formulation. Alternatively, there may be just a small spot in the model where a simple reformulation or approximation could be made to make the model fit the intended model type. Another possible finding is that the
Turning on the Automati- cally Select Engine option causes RSPE to auto- matically select the solving method (Linear, Nonlinear, Quadratic, or Evolutionary) that is most appropriate for your model.
(a) (b)
FIGURE 8.31 The Model Diagnosis section for (a) the original Wyndor problem shown in Figure 8.9 and (b) the traveling salesman problem for- mulated in Figure 8.28. These indicate that these models are linear and nonsmooth, respectively.
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8.6 Using RSPE to Analyze a Model and Choose a Solving Method 309
model actually needs to not fit the intended type in order to adequately represent the problem under consideration. In all these cases, it would be very helpful to be able to pinpoint what in the model is causing it to not fit the intended type.
Rather than spending a lot of time trying to pinpoint these trouble spots through a visual inspection of the model, RSPE provides a way of doing this very quickly and automatically. To illustrate how this is done, we will assume that linear programming has been chosen as the intended model type.
After completing the initial formulation of the model, suppose that Solver gives the error message “The linearity conditions required by this Solver engine are not satisfied.” This indi- cates that the model does not yet satisfy the linearity assumptions of linear programming. Perhaps it is possible to quickly spot what went wrong, but if not, let us look at an example of how RSPE would do this job for you.
Reconsider the portfolio selection example that has been formulated in Figure 8.13 . We know that this model is not linear because of the Risk (Variance) calculation in Variance (C31). However, if we did not recognize this fact, RSPE can provide a structure report to indi- cate why the model is not linear. First click the Option button on the RSPE ribbon and select the Optimization tab. This brings up the dialog box shown in Figure 8.32 . Under Intended Model Type, choose Linear to indicate the type of model you are trying to create. Next choose Analyze Model without Solving under the Optimize menu. As we saw in Figure 8.30 , this analysis indicates that the model is not linear because it is QP Convex (a quadratic pro- gramming problem) instead. Finally, under the Reports menu on the RSPE ribbon, choose Structure under Optimization. This generates the Structure Report shown in Figure 8.33 . In addition to showing how many variables, functions, and dependencies are linear (as we already saw in the Model Diagnosis section of Figure 8.30 ), this report also indicates which functions and which variables are exceptions to the intended model type (linear programming in this example). In particular, the bottom left-hand side of the report reveals that the reason the model is not linear is because of the function in the objective cell Variance (C21). The
FIGURE 8.32 The Optimization tab of the Options dialog box is used here to specify what type of model is desired. For this example, the Intended Model Type is Linear.
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310 Chapter Eight Nonlinear Programming
bottom right-hand side further spells out that there are three exceptions to linearity within this function—the changing cells Stock1 (C14), Stock2 (D14), and Stock3 (E14) all appear in this function in a nonlinear way. Therefore, the RSPE structure report has completely pinpointed what in the model is causing it to not be linear.
For this example, the appropriate conclusion would be that the model actually should not be made to fit linear programming because the function in the objective cell needs to have a quadratic form. However, in other applications, the conclusion might be that the model should be changed to the linear programming type, either because a mistake was made in the initial formulation or because an appropriate reformulation or approximation can be made to fit linear programming.
Using a structure report, RSPE can pinpoint where a model is not linear.
A B C D E F G H I
7
8
All 3 93
3 93
2 60
Smooth
Linear
Variance Variance Variance Variance VarianceStock1 Stock2 Stock3
Exception 1Objective Cell (Min) Exception 2 Exception 3
9
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J K L
13
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Statistics
Variables Functions Dependents
Cell Name Variable Formula Variable Formula Variable Formula
FIGURE 8.33 The Structure Report for the portfolio selection problem formulated in Figure 8.13 when the intended model type is linear. The report indicates that the objective cell (Variance) is not linear and that the three changing cells—Stock1, Stock2, and Stock3—all appear in this function in a nonlinear way.
1. When Excel functions like IF, MAX, ABS, or ROUND are used, the resulting model tends to be of what type?
2. If RSPE analyzes the model and determines it is QP Convex, which solving method should be used?
3. If RSPE analyzes the model and determines it is NLP Convex, which solving method should be used?
4. If RSPE analyzes the model and determines it is NSP (nonsmooth), which solving method should be used?
5. If the intended model type is linear (linear programming) but Solver gives an error message indicating that the model is not linear, what report in RSPE can be useful to pinpoint where the model is not linear?
Review Questions
A nonlinear programming model has the same characteristics as a linear programming model with one key exception. All the mathematical expressions (including the objective function) are linear in a linear programming model, but at least one of these expressions (often just the objective function) is nonlinear in a nonlinear programming model. When formulating the model in a spreadsheet, this means that the model becomes a nonlinear programming model if a nonlinear formula needs to be entered into the objective cell (and perhaps into some other output cells). This chapter focuses on the common case where only the objective cell needs a nonlinear formula. This cell needs a nonlinear formula whenever there is a nonpro- portional relationship between the level of any of the activities and the overall measure of performance for the problem. This kind of relationship violates the proportionality assumption of linear programming.
Formulating and solving a nonlinear programming model tends to be more difficult than formulat- ing and solving a linear programming model. For example, some nonlinear programming models have a number of locally optimal solutions where only one of these solutions is globally optimal and most of the others are greatly inferior. Unfortunately, after a starting solution is entered into the spreadsheet, the Nonlinear Solver will find only one of these locally optimal solutions, with no indication of whether this is the one that also is globally optimal or perhaps one that is far from optimal. Which locally optimal solution is found depends on the choice of the starting solution.
However, when a nonlinear programming model has decreasing marginal returns, it generally is easy to solve. For this type of problem, a locally optimal solution automatically is also a globally optimal
8.7 Summary
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solution. Therefore, the solution found by Solver is guaranteed to be the best solution for the model (or at least one of those tied for being the best).
Some nonlinear programming problems with decreasing marginal returns can be solved in an even easier way. This occurs when the profit graphs (or cost graphs) for the activities are piecewise linear (or at least can be closely approximated by piecewise linear graphs). In this case, the separable program- ming technique can be applied to convert the problem into a linear programming problem, which is the easiest type of problem for Solver.
For more difficult nonlinear programming problems that might have a number of locally optimal solutions, one approach is to run Solver many times, each time starting with a different initial solution entered into the changing cells on the spreadsheet. Solver includes an option to try many random starting points automatically.
However, this method of dealing with difficult problems has two major limitations. First, it is impractical for problems with more than a few changing cells. Second, it does not work for problems with such complicated objective functions that Solver is not even able to find local optima. If such problems don’t have many constraints, another search procedure called Evolutionary Solver is likely to perform well. Using concepts from genetics, evolution, and the survival of the fittest, this procedure is able to gradually move toward the best of the local optima. Given enough search time (which may be a very long time), it frequently succeeds in finding a solution that is very close to optimal.
Because Solver provides a variety of solving methods for solving different types of nonlinear pro- gramming problems, it sometimes is difficult to determine which solving method is the appropriate one for a particular problem. After formulating a spreadsheet model of the problem, RSPE includes a feature that can analyze the model to determine its type, which then would imply which solving method should be used. If desired, RSPE also can automatically select the solving method. If you intend in advance to formulate a specific type of model (including perhaps a linear programming model) but RSPE tells you it is of a different type (perhaps because you made a formulation mistake), RSPE also can pinpoint what in the model is causing it to not fit the intended type.
Glossary curve fitting method A method for using the known values in a profit or cost graph to find the equation for the graph that best fits these data. (Section 8.1), 273 decreasing marginal returns An activity with a profit graph has decreasing marginal returns if the slope (steepness) of this profit graph never increases but sometimes decreases as the level of the activity increases. (Section 8.1), 270 discontinuity A spot in a graph where it is dis- connected because it suddenly jumps up or down. (Section 8.1), 271 Evolutionary Solver A search procedure provided in Solver that uses concepts from genet- ics, evolution, and the survival of the fittest. (Section 8.5), 299 generation A new set of candidate solutions that is created by Evolutionary Solver by pairing off members of the existing population of candi- date solutions to create “offspring” for the next generation. (Section 8.5), 299 genetic algorithm A type of algorithm that uses concepts from genetics. (Section 8.5), 299 global maximum The highest point on an entire graph. (Section 8.1), 275 local maximum A point at which a graph reaches its maximum within a local neighborhood of that point. (Section 8.1), 275 mutation Similar to gene mutation in biology, this is a random change that Evolutionary Solver occasionally makes in a member of the current population. (Section 8.5), 299
Nonlinear Solver A solving method provided by Solver that is used to solve, or attempt to solve, some nonlinear programming problems. (Section 8.1) 274
nonproportional relationship An activity has a nonproportional relationship with the overall measure of performance for a problem if the contribution of the activity to this measure is not proportional to the level of the activity. (Section 8.1), 270
piecewise linear A graph is piecewise linear if it consists of a sequence of connected line seg- ments. (Section 8.1), 271
population The large set of candidate solu- tions that is randomly generated by Evolutionary Solver. (Section 8.5), 299
proportional relationship An activity has a proportional relationship with the overall measure of performance for a problem if the contribution of the activity to this measure is proportional to the level of the activity. (Section 8.1), 269
proportionality assumption A basic assump- tion of linear programming that requires the contribution of each activity to the value of the objective function to be proportional to the level of that activity. (Section 8.1), 270
quadratic programming A special type of nonlinear programming where the objective func- tion has both a quadratic form and decreasing marginal returns and all the constraints are linear. (Section 8.2), 280
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Chapter 8 Excel Files: Constructing a Nonlinear Formula An Example with Multiple Local Maxima Original Wyndor Problem Wyndor Problem with Nonlinear Marketing Costs Portfolio Selection Example Wyndor Problem with Overtime Wyndor Problem with Overtime and Marketing Costs (Separable Programming)
Wyndor Problem with Overtime and Marketing Costs (Nonlinear Programming) Beating the Market Example Tour of American League Ballparks
Excel Add-in: Risk Solver Platform for Education (RSPE)
Learning Aids for This Chapter in Your MS Courseware
Solved Problem (See the CD-ROM or Website for the Solution) 8.S1. Airline Ticket Pricing Model Business travelers tend to be less price sensitive than leisure travelers. Knowing this, airlines have discovered that extra profit can be generated by using separate pricing for these two types of customers. For example, airlines often charge more for a midweek flight (mostly business travelers) than for travel that includes a Saturday night stay (mostly leisure travelers).
Suppose an airline has estimated demand versus price for mid- week travel (mostly business travelers) and for travel that includes a Saturday night stay (mostly leisure travelers) as shown in the table below. This flight is served by a Boeing 777 with capacity for 300 travelers. The fixed cost of operating the flight is $30,000. The variable cost per passenger (for food and fuel) is $30.
Demand
Price Midweek Saturday Night Stay Total
$200 150 465 615 $300 105 210 315 $400 82 127 209 $500 63 82 145 $600 49 60 109 $700 35 45 80 $800 27 37 64
a. One function that can used to estimate demand ( D ) as a function of price ( P ) is a linear demand function, where
D 5 a 2 bP. For positive values of a and b, this will give lower demand when the price is higher. However, a nonlin- ear demand function usually can provide a better fit to the data. For example, one such function is a constant elasticity demand function, where D 5 aP b . For positive values of a and negative values of b, this also will give lower demand when the price is higher. Graph the above data and use the Add Trendline feature of Excel to find the constant elastic- ity demand function that best fits the data in the above table for midweek demand, Saturday night stay demand, and total demand.
b. For this part, assume that the airline charges a single price to all customers. Using the demand function for total demand determined in part a, formulate and solve a nonlinear pro- gramming model in a spreadsheet to determine what the price should be so as to achieve the highest profit for the airline.
c. Now assume that the airline charges separate prices for mid- week and Saturday night stay tickets. Using the two demand functions for midweek and Saturday night stay tickets deter- mined in part a, formulate and solve a nonlinear program- ming spreadsheet model to determine what the prices of the two types of tickets should be so as to maximize the profit for the airline.
d. How much extra profit can the airline achieve by charging higher prices for midweek tickets than for Saturday night stay tickets?
Production Rate Profit
200 $ 9,500 500 22,500 800 34,000 1,000 40,000
a. Draw a profit graph for this product by plotting the profits for the four production rates and then drawing a smooth curve through the four points by hand. (Start the graph with a profit of 0 at a production rate of 0.)
To the left of each of the following problems (or their parts), we have inserted an E* whenever Excel should be used (unless your instructor gives you contrary instructions). An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
8.1. The J. P. Atkins Company will soon be introducing a new product. Estimates have been made of the monthly profit that would be generated by this product for each of four alter- native values of the monthly production rate, as shown to the right.
Problems
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Sales Profit
(millions of dollars)
0 0 100,000 15 200,000 18 300,000 13 400,000 4 500,000 1 600,000 6 700,000 30 800,000 70
a. Draw a profit graph for this microchip by plotting the profits for the various sales levels and then drawing a smooth curve that passes through (or very near) these points.
b. Does the microchip have decreasing marginal returns, increasing marginal returns, or neither?
E* c. Use Excel’s curve fitting method to (1) obtain a nonlinear formula with a quadratic form (a polyno- mial of order 2) for the profit graph and then (2) construct the graph.
E* d. Repeat part c when using the Excel option of a polynomial of order 3 instead of order 2.
e. Which of the Excel options used in parts c and d does a better job of fitting the profit graph to the data?
8.4. The following table shows the estimated daily profit from a new product for several of the alternative choices for the production rate.
Production Rate (R) Profit per Day (P)
0 0 1 $ 95 2 184 3 255 4 320
Because the profit goes up less than proportionally with the production rate (decreasing marginal returns), the management science team analyzing what this production rate (and the pro- duction rates of some other products) should be has decided to approximate the profit ( P ) by a simple nonlinear function of the production rate R. a. One such approximation is P 5 $100 R 2 $5 R 2 .
How closely does this nonlinear function approxi- mate the five values of P given in the table?
b. Repeat part a for the approximation, P 5 $104 R 2 $6 R 2 .
c. Which of these two nonlinear functions provides the better fit to all the data?
E* d. Use Excel’s curve fitting method to (1) obtain a nonlinear formula with a quadratic form for the profit graph and then (2) construct the graph.
8.5. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 8.2. Briefly describe how nonlinear pro- gramming was applied in this study. Then list the various finan- cial and nonfinancial benefits that resulted from this study.
b. Does the proportionality assumption of linear pro- gramming seem to be satisfied reasonably well for this product?
c. To the extent that profit is not strictly proportional to the production rate, does this product have decreasing marginal returns, increasing marginal returns, or neither?
E* d. Use Excel’s curve fitting method to (1) obtain a nonlinear formula with a quadratic form for the profit graph and then (2) construct the graph.
8.2. Consider the following three cases for how the profit from an activity varies with the level of the activity.
Profit ($)
Level of Activity Case 1 Case 2 Case 3
0 0 0 0 1 9 6 5 2 16 14 6 3 21 24 3 4 24 36 4 5 25 50 7
a. For each case, draw the profit graph by plotting the profits for the various levels of the activity and then drawing a smooth curve through the points by hand.
b. For each case, indicate whether the activity has decreasing marginal returns, increasing marginal returns, or neither.
c. How would your answers in part b change if the graphs plotted in part a were cost graphs instead of profit graphs?
E* d. For each case, use Excel’s curve fitting method to (1) obtain a nonlinear formula with a quadratic form for the profit graph and then (2) construct the graph. For any case where the activity has neither decreasing marginal returns nor increasing marginal returns, comment on how good a fit is provided by using a quadratic form.
8.3. The Chiplet Corporation is about to launch the produc- tion and marketing of a new microchip that is more powerful than anything that is currently on the market. Not surprisingly, the profitability of this microchip will depend greatly on its reception in this highly competitive and fast-moving market. If the sales are fairly low, the company will still be able to make a respectable profit because it will have enough available produc- tion capacity to produce the microchip with its current facili- ties. However, if sales are somewhat higher, the company will need to expand its production facilities, which will have the effect of depressing the profit from the microchip if sales only reach a moderate level. (Fully meeting this demand would still be worthwhile because one of top management’s prime goals is to continue increasing the company’s market share as it points toward future generations of microchips already under devel- opment.) Fortunately, if sales reach a relatively high level, the profit from the microchip will become very substantial. The following table shows the estimated profit for various levels of sales over the short lifetime of this microchip.
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deviation of the probability distribution of the profit from the corresponding portfolio. Ann uses the notation m and s for the mean and standard devia- tion. She recalls that, for typical probability distri- butions, the probability is fairly high (about 0.8 or 0.9) that the return will exceed m 2 s , and the prob- ability is extremely high (often close to 0.999) that the profit will exceed m 2 3 s . Calculate m 2 s and m 2 3 s for the four portfolios obtained in part c. Which portfolio will give Ann the highest m among those that also give m 2 s $ 0?
8.8. Reconsider the portfolio selection example given in Section 8.2. A fourth stock (Stock 4) now has been found that gives a good balance between expected return and risk. Using the same units as in Table 8.2 , its expected return is 17% and its risk is 18%. Its joint risk per stock with Stocks 1, 2, and 3 is 2 0.015, 2 0.025, and 0.003, respectively. a. Still using a minimum acceptable expected return of
18%, formulate the revised quadratic programming model in algebraic form for this problem.
E* b. Display and solve this model on a spreadsheet. E* c. Develop a revision of the parameter analysis report
in Figure 8.14 for this revised problem. 8.9. The management of the Albert Hanson Company is trying to determine the best product mix for two new products. Because these products would share the same production facili- ties, the total number of units produced of the two products com- bined cannot exceed two per hour. Because of uncertainty about how well these products will sell, the profit from producing each product provides decreasing marginal returns as the production rate is increased. In particular, with a production rate of R 1 units per hour, it is estimated that Product 1 would provide a profit per hour of $200R1 2 $100R1
2. If the production rate of product 2 is R 2 units per hour, its estimated profit per hour would be $300R2 2 $100R2
2. a. Formulate a quadratic programming model in alge-
braic form for determining the product mix that maximizes the total profit per hour.
E* b. Formulate this model on a spreadsheet. c. Use RSPE’s Analyze without Solving feature to
confirm that the model is QP Convex. d. Solve the model using the appropriate solving method. 8.10. The B. J. Jensen Company specializes in the production of power saws and power drills for home use. Sales are rela- tively stable throughout the year except for a jump upward dur- ing the Christmas season. Since the production work requires considerable work and experience, the company maintains a stable employment level and then uses overtime to increase pro- duction in November. The workers also welcome this opportu- nity to earn extra money for the holidays.
B. J. Jensen, Jr., the current president of the company, is overseeing the production plans being made for the upcoming November. He has obtained the data at the top of the next page.
However, Mr. Jensen now has learned that, in addition to the limited number of labor hours available, two other factors will limit the production levels that can be achieved this Novem- ber. One is that the company’s vendor for power supply units will only be able to provide 10,000 of these units for Novem- ber (2,000 more than his usual monthly shipment). Each power saw and each power drill requires one of these units. Second, the
E*8.6. Reconsider the portfolio selection example, including its spreadsheet model in Figure 8.13 , given in Section 8.2. Note in Table 8.2 that Stock 2 has the highest expected return and stock 3 has by far the lowest. Nevertheless, the changing cells Portfolio (C14:E14) provide an optimal solution that calls for purchasing far more of Stock 3 than of Stock 2. Although pur- chasing so much of Stock 3 greatly reduces the risk of the portfo- lio, an aggressive investor may be unwilling to own so much of a stock with such a low expected return.
For the sake of such an investor, add a constraint to the model that specifies that the percentage of Stock 3 in the portfolio can- not exceed the amount specified by the investor. Then compare the expected return and risk (standard deviation of the return) of the optimal portfolio with that in Figure 8.13 when the upper bound on the percentage of Stock 3 allowed in the portfolio is set at the following values. a. 20% b. 0% c. Generate a parameter analysis report using RSPE to
systematically try all the percentages at 5% inter- vals from 0% to 50%.
8.7.* A stockbroker, Richard Smith, has just received a call from his most important client, Ann Hardy. Ann has $50,000 to invest and wants to use it to purchase two stocks. Stock 1 is a solid blue-chip security with a respectable growth poten- tial and little risk involved. Stock 2 is much more speculative. It is being touted in two investment newsletters as having out- standing growth potential, but also is considered very risky. Ann would like a large return on her investment, but also has con- siderable aversion to risk. Therefore, she has instructed Richard to analyze what mix of investments in the two stocks would be appropriate for her. She also informs him that her plan is to hold the stock being purchased now for three years before selling it.
After doing some research on the historical performances of the two stocks and on the current prospects for the companies involved, Richard is able to make the following estimates. If the entire $50,000 were to be invested in Stock 1 now, the profit when sold in three years would have an expected value of $12,500 and a standard deviation of $5,000. If the entire $50,000 were to be invested in Stock 2 now, the profit when sold in three years would have an expected value of $20,000 and a standard deviation of $30,000. The two stocks behave independently in different sec- tors of the market so Richard’s calculation from historical data is that the covariance of the profits from the two stocks is 0.
Richard now is ready to use a spreadsheet model to determine how to allocate the $50,000 to the two stocks so as to minimize Ann’s risk while providing an expected profit that is at least as large as her minimum acceptable value. He asks Ann to decide what her minimum acceptable value is. a. Without yet assigning a specific numerical value to
the minimum acceptable expected profit, formulate a quadratic programming model in algebraic form for this problem.
E* b. Display this model on a spreadsheet. E* c. Solve this model for four cases: Minimum accept-
able expected profit 5 $13,000, $15,000, $17,000, and $19,000.
d. Ann was a statistics major in college and so under- stands well that the expected return and risk in this model represent estimates of the mean and standard
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vendor who supplies a key part for the gear assemblies will only be able to provide 15,000 for November (4,000 more than for other months). Each power saw requires two of these parts and each power drill requires one.
Mr. Jensen now wants to determine how many power saws and how many power drills to produce in November to maxi- mize the company’s total profit. a. Draw the profit graph for each of these two products. E* b. Use separable programming to formulate a lin-
ear programming model on a spreadsheet for this problem. Then solve the model. What does this say about how many power saws and how many power drills to produce in November?
8.11.* The Dorwyn Company has two new products (special kinds of doors and windows) that will compete with the two new products for the Wyndor Glass Co. (described in Section 2.1). Using units of hundreds of dollars for the objective function, the linear programming model in algebraic form shown below has been formulated to determine the most profitable product mix.
Maximize Profit 5 4D 1 6W
subject to
D 1 3W # 8
5D 1 2W # 14
and
D $ 0 W $ 0
However, because of the strong competition from Wyndor, Dorwyn management now realizes that the company will need to make a strong marketing effort to generate substantial sales of these products. In particular, it is estimated that achieving a pro- duction and sales rate of D doors per week will require weekly marketing costs of D 3 hundred dollars (so $100 for D 5 1, $800 for D 5 2, $2,700 for D 5 3, etc.). The corresponding market- ing costs for windows are estimated to be 2 W 2 hundred dollars. Thus, the objective function in the model should be
Profit 5 4D 1 6W 2 D3 2 2W2
Dorwyn management now would like to use the revised model to determine the most profitable product mix. E* a. Formulate and solve this nonlinear programming
model on a spreadsheet. b. Construct tables to show the profit data for each
product when the production rate is 0, 1, 2, 3. c. Draw a figure that plots the weekly profit points
for each product when the production rate is 0, 1, 2, 3. Connect the pairs of consecutive points with (dashed) line segments.
E* d. Use separable programming based on this figure to formulate an approximate linear programming
Maximum Monthly Production*
Profit per Unit Produced
Regular Time Overtime Regular Time Overtime
Power saws 3,000 2,000 $150 $50 Power drills 5,000 3,000 100 75
* Assuming adequate supplies of materials from the company’s vendors.
model on a spreadsheet for this problem. Then solve the model. What does this say to Dorwyn manage- ment about which product mix to use?
e. Compare the solution based on a separable pro- gramming approximation in part d with the solution obtained in part a for the exact nonlinear program- ming model.
8.12. The MFG Corporation is planning to produce and mar- ket three different products. Let x 1 , x 2 , and x 3 denote the number of units of the three respective products to be produced. The pre- liminary estimates of their potential profitability are as follows.
For the first 15 units produced of Product 1, the unit profit would be approximately $360. The unit profit would be only $30 for any additional units of Product 1. For the first 20 units pro- duced of Product 2, the unit profit is estimated at $240. The unit profit would be $120 for each of the next 20 units and $90 for any additional units. For the first 10 units of Product 3, the unit profit would be $450. The unit profit would be $300 for each of the next 5 units and $180 for any additional units.
Certain limitations on the use of needed resources impose the following constraints on the production of the three products:
x1 1 x2 1 x3 # 60
3x1 1 2x2 # 200
x1 1 2x3 # 70
Management wants to know what values of x 1 , x 2 , and x 3 should be chosen to maximize the total profit. a. Plot the profit graph for each of the three products. E* b. Use separable programming to formulate a linear
programming model on a spreadsheet for this prob- lem. Then solve the model. What is the resulting recommendation to management about the values of x 1 , x 2 , and x 3 to use?
8.13. Suppose that separable programming has been applied to a certain problem (the “original problem”) to convert it to the fol- lowing equivalent linear programming model in algebraic form:
Maximize Profit 5 5x11 1 4x12 1 2x13 1 4x21 1 x22
subject to
3x11 1 3x12 1 3x13 1 2x21 1 2x22 # 25
2x11 1 2x12 1 2x13 2 x21 2 x22 # 10
and 0 # x11 # 2 0 # x21 # 3 0 # x12 # 3 0 # x22 # 1
0 # x13
What was the mathematical model for the original problem? Answer this by plotting the profit graph for each of the original activities and then writing the constraints for the original prob- lem in terms of the original decision variables.
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E* f. Use Excel and Solver to formulate and solve the original nonlinear programming model directly. Compare with the answers obtained after complet- ing part e.
g. Use calculus to find the value of x 1 that maximizes 3 x 1 2 ( x 1 2 1)
2 , the net profit from the first prod- uct. Also use calculus to find the value of x 2 that maximizes 3 x 2 2 ( x 2 2 2)
2 , the net profit from the second product. Show that these values satisfy the constraints for the nonlinear programming model. Then compare these values with the answers obtained in parts e and f.
8.15. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 8.4. Briefly describe how nonlinear pro- gramming was applied in this study. Then list the various finan- cial and nonfinancial benefits that resulted from this study. E*8.16. Consider the following nonlinear programming problem.
Maximize Profit 5 x5 2 13x4 1 59x3 2 107x2 1 61x
subject to
0 # x # 5
a. Formulate this problem in a spreadsheet and then use the Nonlinear Solver and the Multistart feature to solve this problem.
b. Use RSPE’s Analyze without Solving feature to determine what type of model this is.
c. Use Evolutionary Solver to solve this problem.
E*8.17. Consider the following nonlinear programming problem.
Maximize Profit 5 100x6 2 1,359x5 1 6,836x4 2 15,670x3 1 15,870x2 2 5,095x
subject to
0 # x # 5
a. Formulate this problem in a spreadsheet and then use the Nonlinear Solver and the Multistart feature to solve this problem.
b. Use RSPE’s Analyze without Solving feature to determine what type of model this is.
c. Use Evolutionary Solver to solve this problem.
E*8.18. Because of population growth, the state of Washing- ton has been given an additional seat in the House of Repre- sentatives, making a total of ten. The state legislature, which is currently controlled by the Republicans, needs to develop a plan for redistricting the state. There are 18 major cities in the state of Washington that need to be assigned to 1 of the 10 congressional districts. The table on the next page gives the numbers of registered Democrats and registered Republicans in each city. Each district must contain between 150,000 and 350,000 of these registered voters. Use Evolutionary Solver to assign each city to 1 of the 10 congressional districts in order to maximize the number of districts that have more reg- istered Republicans than registered Democrats. (Hint: Use the SUMIF function.)
8.14. Jim Matthews, vice president for marketing of the J. R. Nickel Company, is planning advertising campaigns for two unrelated products. These two campaigns need to use some of the same resources. Therefore, Jim knows that his decisions on the levels of the two campaigns need to be made jointly after considering these resource constraints. In particular, letting x 1 and x 2 denote the levels of campaigns 1 and 2, respectively, these constraints are 4 x 1 1 x 2 # 20 and x 1 1 4 x 2 # 20.
In facing these decisions, Jim is well aware that there is a point of diminishing returns when raising the level of an adver- tising campaign too far. At that point, the cost of additional advertising becomes larger than the increase in net revenue (excluding advertising costs) generated by the advertising. After careful analysis, he and his staff estimate that the net profit from the first product (including advertising costs) when conducting the first campaign at level x 1 would be 3 x 1 2 ( x 1 2 1)
2 in mil- lions of dollars. The corresponding estimate for the second prod- uct is 3 x 2 2 ( x 2 2 2)
2 . Letting P be total net profit, this analysis led to the following
nonlinear programming model for determining the levels of the two advertising campaigns:
Maximize P 5 3x1 2 (x1 2 1) 2
1 3x2 2 (x2 2 2) 2
subject to
4x1 1 x2 # 20
x1 1 4x2 # 20
and x1 $ 0 x2 $ 0
a. Construct tables to show the profit data for each product when the level of its advertising campaign is x 1 5 0, 1, 2, 2.5, 3, 4, 5 (for the first product) or x 2 5 0, 1, 2, 3, 3.5, 4, 5 (for the second product).
b. Use these profit data to draw rough-hand a smooth profit graph for each product. (Note that these profit graphs start at negative values when x 1 5 0 or x 2 5 0 because the products would lose money if there is no advertising to support them.)
c. On the profit graph for the first product, draw an approximation of this profit graph by inserting a dashed-line segment between the profit at x 1 5 0 and x 1 5 2, between the profit at x 1 5 2 and x 1 5 4, and between the profit at x 1 5 4 and x 1 5 5. Then do the same on the profit graph for the second product with x 2 5 0, 2, 4, 5.
E* d. Use separable programming with the approxima- tion of the profit graphs obtained in part c to for- mulate an approximate linear programming model on a spreadsheet for Jim Matthews’s problem. Then solve this model. What does this solution say the levels of the advertising campaigns should be? What would the total net profit from the two prod- ucts be?
E* e. Repeat parts c and d except using x 1 5 0, 2, 2.5, 3, 5 and x 2 5 0, 3, 3.5, 4, 5 for the approximations of the profit graphs in part c. (These particular approxima- tions actually lead to the exact optimal solution for Jim Matthews’s problem.)
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Case 8-1 Continuation of the Super Grain Case Study 317
optimize the portfolio again when considering the data for the first three years only (Q1 2006 through Q4 2008).
b. For how many quarters does this same portfolio beat the market for the next three years (Q1 2009 through Q4 2011)?
c. Comment on the results from parts a and b. E*8.20. Reconsider the portfolio optimization problem consid- ered in Section 8.5, where the goal was to select the portfolio that beat the market for the largest number of quarters over the last six years. a. Use Evolutionary Solver to instead find a portfolio
that did not lose money in the largest number of quarters.
b. Use Evolutionary Solver to instead find a portfolio that yielded a return of at least 10 percent for the largest number of quarters.
8.21. Reconsider the Wyndor Glass Co. problem introduced in Section 2.1.
E* a. Solve this problem using the Linear Solver. E* b. Starting with an initial solution of producing 0 doors
and 0 windows, solve this problem using Evolution- ary Solver.
c. Comment on the performance of the two approaches.
City Democrat
(Thousands) Republican (Thousands)
1 152 62 2 81 59 3 75 83 4 34 52 5 62 87 6 38 87 7 48 69 8 74 49 9 98 62 10 66 72 11 83 75 12 86 82 13 72 83 14 28 53 15 112 98 16 45 82 17 93 68 18 72 98
8.19. Reconsider the portfolio optimization problem consid- ered in Section 8.5, where the goal was to select the portfolio that beat the market for the largest number of quarters over the last six years. E* a. Using the naive solution (20 percent in each stock)
as a starting point, apply Evolutionary Solver to
Case 8-1
Continuation of the Super Grain Case Study
Reconsider the Super Grain case study introduced in Section 3.1 and continued in Section 3.4. Recall that Claire Syverson, vice president for marketing of the Super Grain Corporation, is planning an advertising campaign for the company’s new break- fast cereal (Crunchy Start) with the help of a leading advertising firm, Giacomi & Jackowitz. The campaign will use three adver- tising media: television commercials on Saturday morning pro- grams for children, advertisements in food- and family-oriented magazines, and advertisements in Sunday supplements of major newspapers. The problem being addressed is to determine the best mix of these advertising media.
The spreadsheet in Figure 3.7 shows the revised linear pro- gramming model that was used to address this problem. This model includes constraints on advertising expenditures, plan- ning expenditures, and the use of cents-off coupons, as well as on managerial goals involving the numbers of young children and their parents that should be reached by the advertising. The changing cells NumberOfAds (C19:E19) show the optimal num- ber of advertisements to place in each of the three media accord- ing to this model. The objective cell TotalExposures (H19) gives an estimate of the resulting total number of exposures, where each viewing of an advertisement by some individual counts as one exposure.
The ultimate goal of the advertising campaign is to maximize the company’s profits that are attained because of the resulting
sales. However, it is difficult to make a direct connection between advertising exposure and profits. Therefore, the total number of exposures was chosen to be a rough surrogate for profit. This is why the objective cell in Figure 3.7 (and in Figure 3.1) gives the total number of exposures instead of the total profit.
Claire is uneasy with having done this. She realizes that her assumption—that the total profit from the introduction of Crunchy Start is proportional to the total number of exposures from the advertising campaign—is only a rough approxima- tion. The most important reason is that running too many adver- tisements in an advertising medium reaches a saturation level, where the impact of one more advertisement is substantially less than that for the first advertisement in that medium. Neverthe- less, when the objective cell gives the total number of exposures, having an individual see the advertisement one more time after being saturated counts the same (one more exposure) as seeing the advertisement for the first time.
To check the results in Figure 3.7, Claire decides to try using profit directly as the measure of performance to be recorded in the objective cell. She carefully defines profit as the total profit obtained from first-time sales of Crunchy Start that occur because of the advertising campaign. Excluded from consider- ation are profits from impulse purchases of Crunchy Start by customers who have seen no advertisements but are attracted to this new cereal with its shiny box trumpeting its virtues sitting
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318 Chapter Eight Nonlinear Programming
these points. (Fractional advertisements are allowed by using only a portion of the available outlets.)
b. For each of the advertising media, use Excel’s curve fitting method to (1) obtain a nonlinear formula for the sales graph and then (2) construct the graph. In each case, try three Excel options for the form of the graph—a polynomial of order 2 (the quadratic form), a polynomial of order 3, and the loga- rithmic form—and then choose the option that you feel pro- vides the best fit.
on a store shelf, since these sales have no relevance for evalu- ating the advertising campaign. Repeat purchases of Crunchy Start also are excluded from consideration because these depend mainly on the reaction to the cereal from the first purchase instead of the advertising campaign.
Claire asks Sid Jackowitz, one of the senior partners of Gia- comi & Jackowitz, to develop estimates of the number of first- time purchases of Crunchy Start that should result from various numbers of advertisements in each medium. His estimates are shown in the following tables.
Number of TV Spots
Number of Sales
1 1,000,000 2 1,750,000 3 2,450,000 4 2,800,000 5 3,000,000
Number of Ads in Sunday Supplements
Number of Sales
2 1,200,000 4 2,200,000 6 3,000,000 8 3,500,000 10 3,750,000
Number of Magazine Ads
Number of Sales
5 700,000 10 1,200,000 15 1,550,000 20 1,800,000 25 2,000,000
Sid also reports that it is reasonable to assume that sales resulting from advertising in one of the media are not sub- stantially affected by the amount of advertising in the other media since the audiences for the different media are some- what different.
It is estimated that the company’s gross profits from Crunchy Start will be 75¢ per sale. However, this gross profit excludes the advertising costs and planning costs for the advertising campaign. Therefore, Claire wants to include these costs in her definition of the total profit that should be considered for deter- mining the best advertising mix.
a. For each of the three advertising media, draw a graph of the number of sales versus the number of advertisements by plot- ting the sales for the five points provided by Sid Jackowitz and then drawing a smooth curve through (or very near)
c. Using your results from part b, write an expression for the total profit (as defined by Claire) in terms of the number of advertisements of each type.
d. Using your result from part c, revise the spreadsheet model in Figure 3.7 (available on the CD-ROM) so that it maximizes total profit instead of the total number of exposures, and then solve.
e. Use the sales tables provided by Sid Jackowitz to apply sepa- rable programming to this problem when maximizing total profit.
f. Compare your results in parts d and e with those in Figure 3.7 and then give your recommendation (with a brief explanation) for the best advertising mix. Do you feel it was worthwhile to introduce a nonlinear profit function into the model in order to refine the linear programming model used in Figure 3.7?
Case 8-2
Savvy Stock Selection
Ever since the day she took her first economics class in high school, Lydia wondered about the financial practices of her par- ents. They worked very hard to earn enough money to live a comfortable middle-class life, but they never made their money work for them. They simply deposited their hard-earned pay- checks in savings accounts earning a nominal amount of interest. (Fortunately, there always was enough money available when it came time to pay her college bills.) She promised herself that when she became an adult, she would not follow the same finan- cially conservative practices as her parents.
And Lydia kept this promise. She took every available finance course in her business program at college. Having landed a coveted job on Wall Street upon graduation, she now begins every morning by watching the CNN financial reports. She plays investment games on the World Wide Web, finding portfolios that maximize her return while minimizing her risk. And she reads The Wall Street Journal and Financial Times.
Lydia also reads the investment advice columns of the finan- cial magazines. She decides to follow the current advice given by her two favorite columnists. In his monthly column, edi- tor Jonathan Taylor recommends three stocks that he believes will rise far above market average. In addition, the well-known mutual fund guru Donna Carter advocates the purchase of three additional stocks that she thinks will outperform the market over the next year.
Bigbell (ticker symbol on the stock exchange: BB), one of the nation’s largest telecommunications companies, trades at a price–earnings ratio well below market average. Huge investments over the last eight months have depressed earn- ings considerably. However, with its new cutting-edge tech- nology, the company is expected to significantly raise its profit margins. Taylor predicts that the stock will rise from its current price of $60 per share to $72 per share within the next year.
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Case 8-3 International Investments 319
Automobile Alliance (AUA) is a leading car manufacturer from the Detroit area that just recently introduced two new mod- els. These models show very strong initial sales, and therefore the company’s stock is predicted to rise from $20 to $26 over the next year.
On the World Wide Web, Lydia found data about the risk involved in the stocks of these companies. The historical vari- ances of return of the six stocks and their covariances are shown in the following table.
Lotsofplace (LOP) is one of the leading hard drive manu- facturers in the world. The industry recently underwent major consolidation, as fierce price wars over the last few years were followed by many competitors going bankrupt or being bought by Lotsofplace and its competitors. Due to reduced competition in the hard drive market, revenues and earnings are expected to rise considerably over the next year. Taylor predicts a one-year increase of 42 percent in the stock of Lotsofplace from the cur- rent price of $127 per share.
Internetlife (ILI) has survived the many ups and downs of Internet companies. With the next Internet frenzy just around the
Company BB LOP ILI HEAL QUI AUA
Variance 0.032 0.1 0.333 0.125 0.065 0.08
Covariances LOP ILI HEAL QUI AUA
BB 0.005 0.03 20.031 20.027 0.01 LOP 0.085 20.07 20.05 0.02 ILI 20.11 20.02 0.042 HEAL 0.05 20.06 QUI 20.02
corner, Taylor expects a doubling of this company’s stock price from $4 to $8 within a year.
Healthtomorrow (HEAL) is a leading biotechnology com- pany that is about to get approval for several new drugs from the Food and Drug Administration, which will help earnings to grow 20 percent over the next few years. In particular, a new drug to significantly reduce the risk of heart attacks is supposed to reap huge profits. Also, due to several new great-tasting medications for children, the company has been able to build an excellent image in the media. This public relations coup will surely have a positive effect on the sale of its over-the-counter medications. Carter is convinced that the stock will rise from $50 to $75 per share within a year.
Quicky (QUI) is a fast-food chain that has been vastly expanding its network of restaurants all over the United States. Carter has followed this company closely since it went public some 15 years ago when it had only a few dozen restaurants on the West Coast of the United States. Since then the company has expanded, and it now has restaurants in every state. Due to its emphasis on healthy foods, it is capturing a growing mar- ket share. Carter believes that the stock will continue to perform well above market average for an increase of 46 percent in one year from its current stock price of $150.
a. At first, Lydia wants to ignore the risk of all the investments. Given this strategy, what is her optimal investment portfo- lio; that is, what fraction of her money should she invest in each of the six different stocks? What is the total risk of her portfolio?
b. Lydia decides that she doesn’t want to invest more than 40 percent in any individual stock. While still ignoring risk, what is her new optimal investment portfolio? What is the total risk of her new portfolio?
c. Now Lydia wants to take into account the risk of her invest- ment opportunities. For use in the following parts, formulate a quadratic programming model that will minimize her risk (measured by the variance of the return from her portfolio) while ensuring that her expected return is at least as large as her choice of a minimum acceptable value.
d. Lydia wants to ensure that she receives an expected return of at least 35 percent. She wants to reach this goal at minimum risk. What investment portfolio allows her to do that?
e. What is the minimum risk Lydia can achieve if she wants an expected return of at least 25 percent? Of at least 40 percent?
f. Do you see any problems or disadvantages with Lydia’s approach to her investment strategy?
Case 8-3
International Investments Charles Rosen relaxes in a plush, overstuffed recliner by the fire, enjoying the final vestiges of his week-long winter vaca- tion. As a financial analyst working for a large investment firm in Germany, Charles has very few occasions to enjoy these private moments since he is generally catching red-eye flights around the world to evaluate various investment opportuni- ties. Charles pats the loyal golden retriever lying at his feet
and takes a swig of brandy, enjoying the warmth of the liq- uid. He sighs and realizes that he must begin attending to his own financial matters while he still has the time during the holiday. He opens a folder placed conspicuously on the top of a large stack of papers. The folder contains information about an investment Charles made when he graduated from college four years ago.
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320 Chapter Eight Nonlinear Programming
affects his potential return on the B bonds, but he also knows that most likely a strategy exists for maximizing his return on the bonds. He might be able to decrease the tax he has to pay on interest income by selling portions of his bonds in dif- ferent years. Charles considers his strategy viable because the government requires investors to pay taxes on interest income only when they sell their B bonds. For example, if Charles were to sell one-third of his B bonds on December 31 of the sixth year, he would have to pay taxes on the interest income of DM (6,351 2 6,100).
Charles asks himself several questions. Should he keep all the bonds until the end of the seventh year? If so, he would earn 0.7823 times DM 30,000 in interest income, but he would have to pay very substantial taxes for that year. Considering these tax payments, Charles wonders if he should sell a portion of the bonds at the end of this year (the fifth year) and at the end of next year.
If Charles sells his bonds, his alternative investment oppor- tunities are limited. He could purchase a certificate of deposit (CD) paying 4.0 percent interest, so he investigates this alterna- tive. He meets with an investment advisor from the local branch of a bank, and the advisor tells him to keep the B bonds until the end of the seventh year. She argues that even if he had to pay 30 percent in taxes on the 9.00 percent rate of interest that the B bonds would be paying in their last year (see Table 1 ), this strategy would still result in a net rate of 6.30 percent inter- est, which is much better than the 4.0 percent interest he could obtain on a CD.
Charles concludes that he will make all his transactions on December 31, regardless of the year. Also, since he intends to attend business school in the United States in the fall of the sev- enth year and plans to pay his tuition for his second, third, and fourth semester with his investment, he does not plan to keep his money in Germany beyond December 31 of the seventh year.
(For the first three parts, assume that if Charles sells a portion of his bonds, he will put the money under his mattress earning zero percent interest. For the subsequent parts, assume that he could invest the proceeds of the bonds in the certificate of deposit.)
a. Formulate a separable programming model to be used in the following parts.
b. What is the optimal investment strategy for Charles? c. What is fundamentally wrong with the advice Charles got
from the investment advisor at the bank? d. Now that Charles is considering investing in the certificate of
deposit, what is his optimal investment strategy? e. What would his optimal investment strategy for the fifth,
sixth, and seventh years have been if he had originally invested DM 50,000?
f. Charles and his fiancée have been planning to get married after his first year in business school. However, Charles learns that for married couples, the tax-free amount of inter- est earnings each year is DM 12,200. How much money could Charles save on his DM 30,000 investment by getting married this year (the fifth year for his investment)?
g. Due to a recession in Germany, interest rates are low and are expected to remain low. However, since the American economy is booming, interest rates are expected to rise in the United States. A rise in interest rates would lead to a rise of the dollar in comparison to the mark. Analysts at Charles’s
Charles remembers his graduation day fondly. He obtained a degree in business administration and was full of investment ideas that were born while he had been daydreaming in his numerous finance classes. Charles maintained a well-paying job throughout college, and he was able to save a large portion of the college fund that his parents had invested for him.
Upon graduation, Charles decided that he should transfer the college funds to a more lucrative investment opportunity. Since he had signed on to work in Germany, he evaluated investment opportunities in that country. Ultimately, he decided to invest 30,000 German marks (DM) in so-called B bonds that would mature in seven years. Charles purchased the bonds just four years ago last week (in early January of what will be called the “first year” in this discussion). He considered the bonds an excellent investment opportunity since they offered high interest rates (see Table 1 ) that would rise over the subsequent seven years and because he could sell the bonds whenever he wanted after the first year. He calculated the amount that he would be paid if he sold bonds originally worth DM 100 on the last day of any of the seven years (see Table 2 ). The amount paid included the principal plus the interest. For example, if he sold bonds originally worth DM 100 on December 31 of the sixth year, he would be paid DM 163.51 (the principal is DM 100 and the interest is DM 63.51).
Charles did not sell any of the bonds during the first four years. Last year, however, the German federal government intro- duced a capital gains tax on interest income. The German gov- ernment designated that the first DM 6,100 a single individual earns in interest per year would be tax-free. Any interest income beyond DM 6,100 would be taxed at a rate of 30 percent. For example, if Charles earned interest income of DM 10,100, he would be required to pay 30 percent of DM 4,000 (DM 10,100 – DM 6,100) in taxes, or DM 1,200. His after-tax income would therefore be DM 8,900.
Because of the new tax implemented last year, Charles has decided to reevaluate the investment. He knows that the new tax
Year Interest Rate Annual Percentage Yield
1 7.50% 7.50% 2 8.50 8.00 3 8.50 8.17 4 8.75 8.31 5 9.00 8.45 6 9.00 8.54 7 9.00 8.61
TABLE 1 Interest Rates over the Seven Years
Year DM
1 107.50 2 116.64 3 126.55 4 137.62 5 150.01 6 163.51 7 178.23
TABLE 2 Total Return on 100 DM
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Additional Cases 321
Additional Cases Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
investment bank expect the dollar to remain at the current exchange rate of DM 1.50 per dollar for the fifth year and then to rise to DM 1.80 per dollar by the end of the sev- enth year. Therefore, Charles is considering investing at the beginning of the sixth year in a two-year American munici- pal bond paying 3.6 percent tax-exempt interest to help pay tuition. How much money should he plan to convert into dol- lars by selling B bonds for this investment?
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322
Chapter Nine
Decision Analysis Learning Objectives
After completing this chapter, you should be able to
1. Identify the kind of decision-making environment for which decision analysis is needed.
2. Describe the logical way in which decision analysis organizes a problem.
3. Formulate a payoff table from a description of the problem.
4. Describe and evaluate several alternative criteria for making a decision based on a payoff table.
5. Apply Bayes’ decision rule to solve a decision analysis problem.
6. Formulate and solve a decision tree for dealing with a sequence of decisions.
7. Use RSPE to construct and solve a decision tree.
8. Perform sensitivity analysis with Bayes’ decision rule.
9. Determine whether it is worthwhile to obtain more information before making a decision.
10. Use new information to update the probabilities of the states of nature.
11. Use data tables to perform sensitivity analysis when dealing with a sequence of decisions.
12. Use utilities to better reflect the values of payoffs.
13. Describe some common features in the practical application of decision analysis.
The previous chapters have focused mainly on managerial decision making when the con- sequences of alternative decisions are known with a reasonable degree of certainty. This decision-making environment enabled formulating helpful mathematical models (linear pro- gramming, integer programming, nonlinear programming, etc.) with objective functions that specify the estimated consequences of any combination of decisions. Although these conse- quences usually cannot be predicted with complete certainty, they could at least be estimated with enough accuracy to justify using such models (along with what-if analysis, etc.).
However, managers often must make decisions in environments that are fraught with much more uncertainty. Here are a few examples.
1. A manufacturer introducing a new product into the marketplace. What will be the reac- tion of potential customers? How much should be produced? Should the product be test marketed in a small region before deciding upon full distribution? How much advertising is needed to launch the product successfully?
2. A financial firm investing in securities. Which are the market sectors and individual secu- rities with the best prospects? Where is the economy headed? How about interest rates? How should these factors affect the investment decisions?
3. A government contractor bidding on a new contract. What will be the actual costs of the project? Which other companies might be bidding? What are their likely bids?
4. An agricultural firm selecting the mix of crops and livestock for the upcoming season. What will be the weather conditions? Where are prices headed? What will costs be?
5. An oil company deciding whether to drill for oil in a particular location. How likely is there to be oil in that location? How much? How deep will they need to drill? Should geologists investigate the site further before drilling?
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9.1 A Case Study: The Goferbroke Company Problem 323
This is the kind of decision making in the face of great uncertainty that decision analysis is designed to address. Decision analysis provides a framework and methodology for rational decision making when the outcomes are uncertain.
The first section introduces a case study that will be carried throughout the chapter to illus- trate the various phases involved in applying decision analysis. Section 9.2 focuses on choos- ing an appropriate decision criterion and then the next section describes how decision trees can be used to structure and analyze a decision analysis problem. Section 9.4 discusses how sensi- tivity analysis can be performed efficiently with the help of decision trees. The subsequent four sections deal with whether it would be worthwhile to obtain more information and, if so, how to use this information for making a sequence of decisions. Section 9.9 then describes how to analyze the problem while calibrating the possible outcomes to reflect their true value to the decision maker. Finally, Section 9.10 discusses the practical application of decision analysis.
In addition, a supplement to this chapter on the CD-ROM presents a detailed description and evaluation of various decision criteria.
9.1 A CASE STUDY: THE GOFERBROKE COMPANY PROBLEM
Max Flyer is the founder and sole owner of the Goferbroke Company, which develops oil wells in unproven territory. Max’s friends refer to him affectionately as a wildcatter. How- ever, he prefers to think of himself as an entrepreneur. He has poured his life’s savings into the company in the hope of making it big with a large strike of oil.
Now his chance possibly has come. His company has purchased various tracts of land that larger oil companies have spurned as unpromising even though they are near some large oil fields. Now Max has received an exciting report about one of these tracts. A consulting geolo- gist has just informed Max that he believes there is one chance in four of oil there.
Max has learned from bitter experience to be skeptical about the chances of oil reported by consulting geologists. Drilling for oil on this tract would require an investment of about $100,000. If the land turns out to be dry (no oil), the entire investment would be lost. Since his company does not have much capital left, this loss would be quite serious.
On the other hand, if the tract does contain oil, the consulting geologist estimates that there would be enough there to generate a net revenue of approximately $800,000, leaving an approximate profit of
Profit if find oil 5 Revenue if find oil 2 Drilling cost 5 $800,000 2 $100,000 5 $700,000
Although this wouldn’t be quite the big strike for which Max has been waiting, it would pro- vide a very welcome infusion of capital into the company to keep it going until he hopefully can hit the really big gusher.
There is another option. Another oil company has gotten wind of the consulting geolo- gist’s report and so has offered to purchase the tract of land from Max for $90,000. This is very tempting. This too would provide a welcome infusion of capital into the company, but without incurring the large risk of a very substantial loss of $100,000.
Table 9.1 summarizes the decision alternatives and prospective payoffs that face Max. So Max is in a quandary about what to do. Fortunately, help is at hand. Max’s daughter
Jennifer has recently earned her degree from a fine business school and now has come to
Should Max sell the land instead of drilling for oil there?
TABLE 9.1 Prospective Profits for the Goferbroke Company
Profit Status of Land
Oil DryAlternative
Drill for oil $700,000 2$100,000 Sell the land 90,000 90,000
Chance of status 1 in 4 3 in 4
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324 Chapter Nine Decision Analysis
work for her proud dad. He asks her to apply her business training to help him analyze the problem. Having studied management science in college, she recommends applying decision analysis. Having paid for her fine education, he agrees to give it a try.
Jennifer begins by interviewing her dad about the problem.
Jennifer: How much faith do you put in the consulting geologist’s assessment that there is one chance in four of oil on this tract? Max: Not too much. These guys sometimes seem to pull numbers out of the air. He has convinced me that there is some chance of oil there. But it could just as well be one chance in three, or one chance in five. They don’t really know. Jennifer: Is there a way of getting more information to pin these odds down better? This is an important option with the decision analysis approach. Max: Yes. We could arrange for a detailed seismic survey of the land. That would pin down the odds somewhat better. But you don’t really find out until you drill. Furthermore, these seismic surveys cost you an arm and a leg. I got a quote for this tract. 30,000 bucks! Then it might say oil is likely, so we drill and we might not find anything. Then I’m out another 100,000 bucks! Losing $130,000 would almost put us out of business. Jennifer: OK. Let’s put the seismic survey on the back burner for now. Here is another key consideration. It sounds like we need to go beyond dollars and cents to look at the con- sequences of the possible outcomes. Losing $130,000 would hurt a lot more than gaining $130,000 would help. Max: That’s for sure! Jennifer: Well, decision analysis has a way of taking this into account by using what are called utilities. The utility of an outcome measures the true value to you of that outcome rather than just the monetary value. Max: Sounds good. Jennifer: Now this is what I suggest we do. We’ll start out simple, without considering the option of the seismic survey and without getting into utilities. I’ll introduce you to how decision analysis organizes our problem and to the options it provides for the criterion to use for making your decision. You’ll be able to choose the criterion that feels right to you. Then we’ll look at whether it might be worthwhile to do the seismic survey and, if so, how to best use its information. After that, we’ll get into the nitty gritty of carefully analyzing the prob- lem, including incorporating utilities. I think when we finish the process and you make your decision, you’ll feel quite comfortable that you are making the best one. Max: Good. Let’s get started.
Here is the tutorial that Jennifer provided her dad about the logical way in which decision analysis organizes a problem.
Decision Analysis Terminology Decision analysis has a few special terms.
The decision maker is the individual or group responsible for making the decision (or sequence of decisions) under consideration. For the Goferbroke Co. problem, the decision maker is Max. Jennifer (the management scientist) can help perform the analyses, but the objective is to assist the decision maker in identifying the best possible decision from the decision maker’s perspective.
The alternatives are the options for the decision to be made by the decision maker. Max’s alternatives at this point are to drill for oil or to sell the tract of land.
The outcome of the decision to be made will be affected by random factors that are outside the control of the decision maker. These random factors determine the situation that will be found when the decision is executed. Each of these possible situations is referred to as a pos- sible state of nature . For the Goferbroke Co. problem, the possible states of nature are that the tract contains oil or that it is dry (no oil).
The decision maker generally will have some information about the relative likelihood of the possible states of nature. This information may be in the form of just subjective estimates based on the experience or intuition of an individual, or there may be some degree of hard
The utility of an outcome measures the true value to the decision maker of that outcome.
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9.2 Decision Criteria 325
evidence involved (such as is contained in the consulting geologist’s report). When these estimates are expressed in the form of probabilities, they are referred to as the prior prob- abilities of the respective states of nature. For the Goferbroke Co. problem, the consulting geologist has provided the prior probabilities given in Table 9.2 . Although these are unlikely to be the true probabilities based on more information (such as through a seismic survey), they are the best available estimates of the probabilities prior to obtaining more information. (Later in the chapter, we will analyze whether it would be worthwhile to conduct a seismic survey, so the current problem of what to do without a seismic survey will be referred to here- after as the first Goferbroke Co. problem.)
Each combination of a decision alternative and a state of nature results in some outcome. The payoff is a quantitative measure of the value to the decision maker of the consequences of the outcome. In most cases, the payoff is expressed as a monetary value, such as the profit. As indicated in Table 9.1 , the payoff for the Goferbroke Co. at this stage is profit. (In Section 9.9, the company’s payoffs will be reexpressed in terms of utilities.)
The Payoff Table When formulating the problem, it is important to identify all the relevant decision alternatives and the possible states of nature. After identifying the appropriate measure for the payoff from the perspective of the decision maker, the next step is to estimate the payoff for each combination of a decision alternative and a state of nature. These payoffs then are displayed in a payoff table .
Table 9.3 shows the payoff table for the first Goferbroke Co. problem. The payoffs are given in units of thousands of dollars of profit. Note that the bottom row also shows the prior probabilities of the states of nature, as given earlier in Table 9.2 .
State of Nature Prior Probability
The tract of land contains oil 0.25 The tract of land is dry (no oil) 0.75
TABLE 9.2 Prior Probabilities for the First Goferbroke Co. Problem
State of Nature
Alternative Oil Dry
Drill for oil 700 2100 Sell the land 90 90
Prior probability 0.25 0.75
TABLE 9.3 Payoff Table (Profit in $1,000s) for the First Goferbroke Co. Problem
1. What are the decision alternatives being considered by Max? 2. What is the consulting geologist’s assessment of the chances of oil on the tract of land? 3. How much faith does Max put in the consulting geologist’s assessment of the chances of oil? 4. What option is available for obtaining more information about the chances of oil? 5. What is meant by the possible states of nature? 6. What is meant by prior probabilities? 7. What do the payoffs represent in a payoff table?
Review Questions
9.2 DECISION CRITERIA
Given the payoff table for the first Goferbroke Co. problem shown in Table 9.3 , what criterion should be used in deciding whether to drill for oil or sell the land? There is no single correct answer for this question that is appropriate for every decision maker. The choice of a decision criterion depends considerably on the decision maker’s own temperament and attitude toward decision making, as well as the circumstances of the decision to be made. Ultimately, Max Flyer, as the owner of the Goferbroke Co., needs to decide which decision criterion is most appropriate for this situation from his personal viewpoint.
Over a period of many decades (and even centuries), a considerable number of criteria have been suggested for how to make a decision when given the kind of information provided by a payoff table. All these criteria consider the payoffs in some way and some also take into account
There is no single decision criterion that is best for every situation.
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the prior probabilities of the states of nature, but other criteria do not use probabilities in any way. Each criterion has some rationale as well as some drawbacks. However, in recent decades, a substantial majority of management scientists has concluded that one of these criteria (Bayes’ decision rule) is a particularly appropriate criterion for most decision makers in most situations. Therefore, after describing and discussing Bayes’ decision rule in this section, the rest of the chapter will focus on how to apply this particular criterion in a variety of contexts.
However, before turning to Bayes’ decision rule, we briefly introduce three alternative decision criteria below. All of these alternative criteria are particularly simple and intuitive. At the same time, each criterion is quite superficial in the sense that it focuses on only one piece of information provided by the payoff table and ignores the rest (including the pieces considered by the other two criteria). Nevertheless, many individuals informally apply one or more of these criteria at various times in their lives. The first two make no use of prior prob- abilities, which can be quite reasonable when it is difficult or impossible to obtain relatively reliable values for these probabilities. Bayes’ decision rule is quite different from these alter- native criteria in that it makes full use of all the information in the payoff table by applying a more structured approach to decision making.
The CD-ROM includes a supplement to this chapter entitled Decision Criteria that provides a much more detailed discussion and critique of these three alternative decision criteria as well as three others that are somewhat more complicated. (These three others included in the supplement are the equally likely criterion that assigns equal probabilities to all the states of nature, the minimax regret criterion that minimizes the regret that can be felt afterward if the decision does not turn out well, and the realism criterion that uses the decision maker’s pessimism–optimism index.)
Decision Making without Probabilities: The Maximax Criterion The maximax criterion is the decision criterion for the eternal optimist. It says to focus only on the best that can happen to us. Here is how this criterion works:
1. Identify the maximum payoff from any state of nature for each decision alternative. 2. Find the maximum of these maximum payoffs and choose the corresponding decision
alternative.
The rationale for this criterion is that it gives an opportunity for the best possible outcome (the largest payoff in the entire payoff table) to occur. All that is needed is for the right state of nature to occur, which the eternal optimist believes is likely.
Table 9.4 shows the application of this criterion to the first Goferbroke problem. It begins with the payoff table ( Table 9.3 ) without the prior probabilities (since these probabilities are ignored by this criterion). An extra column on the right then shows the maximum payoff for each decision alternative. Since the maximum of these maxima (700) must be the largest payoff in the entire payoff table, the corresponding decision alternative (drill for oil) is selected by this criterion.
The biggest drawback of this criterion is that it completely ignores the prior probabilities. For example, it always would say that Goferbroke should drill for oil even if the chance of finding oil were minuscule. Another drawback is that it ignores all the payoffs except the largest one. For example, it again would say that Goferbroke should drill for oil even if the payoff from selling the land were 699 ($699,000).
Decision Making without Probabilities: The Maximin Criterion The maximin criterion is the criterion for the total pessimist. In contrast to the maximax cri- terion, it says to focus only on the worst that can happen to us. Here is how this criterion works:
1. Identify the minimum payoff from any state of nature for each decision alternative. 2. Find the maximum of these minimum payoffs and choose the corresponding decision alternative.
Bayes’ decision rule is the recommended decision cri- terion for most situations.
The maximax criterion always chooses the decision alternative that can give the largest possible payoff.
This criterion ignores the prior probabilities.
The maximin criterion always chooses the decision alternative that provides the best guarantee for its mini- mum possible payoff.
State of Nature
Alternative Oil Dry Maximum in Row Drill for oil 700 2100 700 ← Maximax Sell the land 90 90 90
TABLE 9.4 Application of the Maximax Criterion to the First Goferbroke Co. Problem
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9.2 Decision Criteria 327
The rationale for this criterion is that it provides the best possible protection against being unlucky. Even if each possible choice of a decision alternative were to lead to its worst state of nature occurring, which the total pessimist thinks is likely, the choice indicated by this criterion gives the best possible payoff under these circumstances.
The application of this criterion to the first Goferbroke problem is shown in Table 9.5 . The basic difference from Table 9.4 is that the numbers in the right-hand column now are the minimum rather than the maximum in each row. Since 90 is the maximum of these two num- bers, the alternative to be chosen is to sell the land.
The drawbacks of this criterion are similar to those for the maximax criterion. Because it completely ignores prior probabilities, it always would say that Goferbroke should sell the land even if it were almost certain to find oil if it drilled. Because it ignores all the payoffs except the maximin payoff, it again would say that Goferbroke should sell the land even if the payoff from drilling successfully for oil were 10,000 ($10 million).
Decision Making with Probabilities: The Maximum Likelihood Criterion The maximum likelihood criterion says to focus on the most likely state of nature as follows.
1. Identify the state of nature with the largest prior probability. 2. Choose the decision alternative that has the largest payoff for this state of nature.
The rationale for this criterion is that by basing our decision on the assumption that the most likely state of nature will occur, we are giving ourselves a better chance of a favorable outcome than by assuming any other state of nature.
Table 9.6 shows the application of this criterion to the first Goferbroke Co. problem. This table is identical to the payoff table given in Table 9.3 except for also showing step 1 (select the dry state of nature) and step 2 (select the sell the land alternative) of the criterion. Since dry is the state of nature with the larger prior probability, we only consider the payoffs in its column ( 2 100 and 90). The larger of these two payoffs is 90, so we choose the corresponding alternative, sell the land.
This criterion has a number of drawbacks. One is that with a considerable number of states of nature, the most likely state can have a fairly low prior probability, in which case it would make little sense to base the decision solely on this one state. Another more serious drawback is that it completely ignores all the payoffs (including any extremely large payoffs and any disastrous payoffs) throughout the payoff table except those for the single most likely state of nature. For example, no matter how large the payoff for finding oil, it automatically would say that Goferbroke should sell the land instead of drilling for oil whenever the dry state has a little larger prior probability than the oil state.
This criterion also ignores the prior probabilities.
The maximum likelihood criterion assumes that the most likely state of nature will occur and chooses accordingly.
This criterion ignores all the payoffs except for the most likely state of nature.
State of Nature
Alternative Oil Dry Minimum in Row
Drill for oil 700 2100 2100 Sell the land 90 90 90 ← Maximin
TABLE 9.5 Application of the Maximin Criterion to the First Goferbroke Co. Problem
State of Nature
Alternative Oil Dry
Drill for oil 700 2100 Step 2:
Sell the land 90 90 ← Maximum
Prior probability 0.25 0.75 ↑
Step 1: Maximum
TABLE 9.6 Application of the Maximum Likelihood Criterion to the First Goferbroke Co. Problem
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Decision Making with Probabilities: Bayes’ Decision Rule Bayes’ decision rule directly uses the prior probabilities of the possible states of nature as summarized below.
1. For each decision alternative, calculate the weighted average of its payoffs by multiplying each payoff by the prior probability of the corresponding state of nature and then sum- ming these products. Using statistical terminology, refer to this weighted average as the expected payoff (EP) for this decision alternative.
2. Bayes’ decision rule says to choose the alternative with the largest expected payoff.
The spreadsheet in Figure 9.1 shows the application of this criterion to the first Goferbroke Co. problem. Columns B, C, and D display the payoff table first given in Table 9.3 . Cells F5 and F6 then execute step 1 of the procedure by using the equations entered into these cells, namely,
F5 5 SUMPRODUCT(PriorProbability, DrillPayoff) 5 0.25(700) 1 0.75( 2 100) 5 100 F6 5 SUMPRODUCT(PriorProbability, SellPayoff) 5 0.25(90) 1 0.75(90) 5 90
Since expected payoff 5 100 for the drill alternative (cell F5), versus a smaller value of expected payoff 5 90 for the sell the land alternative (cell F6), this criterion says to drill for oil.
Like all the others, this criterion cannot guarantee that the selected alternative will turn out to have been the best one after learning the true state of nature. However, it does provide another guarantee described below.
The expected payoff for a particular decision alternative can be interpreted as what the average payoff would become if the same situation were to be repeated numerous times. Therefore, on the average, repeatedly using Bayes’ decision rule to make decisions will lead to larger payoffs in the long run than any other criterion (assuming the prior probabilities are valid).
Thus, if the Goferbroke Co. owned many tracts of land with this same payoff table, drilling for oil on all of them would provide an average payoff of about 100 ($100,000), versus only 90 ($90,000) for selling. As the following calculations indicate, this is the average payoff from
On the average, Bayes’ decision rule provides larger payoffs in the long run than any other criterion.
DrillPayoff
ExpectedPayoff
PriorProbability
SellPayoff
C5:D5
F5:F6
C8:D8
C6:D6
Range Name Cells
1
2
3 Payoff Table State of Nature
4
5
6
Expected
Payoff
7
8
Oil DryAlternative
Drill
Sell
Prior Probability
A B C D E F
Bayes' Decision Rule for the Goferbroke Co.
3
4
5
6
Expected
Payoff
=SUMPRODUCT(PriorProbability,DrillPayoff)
=SUMPRODUCT(PriorProbability,SellPayoff)
F
700
90
–100
90
0.25 0.75
100
90
FIGURE 9.1 This spreadsheet shows the application of Bayes’ decision rule to the first Goferbroke Co. problem, where a comparison of the expected payoffs in cells F5:F6 indicates that the Drill alternative should be chosen because it has the largest expected payoff.
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9.2 Decision Criteria 329
drilling that results from having oil in an average of one tract out of every four (as indicated by the prior probabilities).
Oil found in one tract: Payoff 5 700 Three tracts are dry: Payoff 5 3(2100) 5 2300
Total payoff 5 400
Average payoff 5 400
4 5 100
However, achieving this average payoff might require going through a long stretch of dry tracts until the “law of averages” can prevail to reach 25 percent of the tracts having oil. Surviving a long stretch of bad luck may not be feasible if the company does not have adequate financing.
This criterion also has its share of critics. Here are the main criticisms.
1. There usually is considerable uncertainty involved in assigning values to prior probabili- ties, so treating these values as true probabilities will not reveal the true range of possible outcomes. (Section 9.4 discusses how sensitivity analysis can address this concern.)
2. Prior probabilities inherently are at least largely subjective in nature, whereas sound deci- sion making should be based on objective data and procedures. (Section 9.6 describes how new information sometimes can be obtained to improve prior probabilities and make them more objective.)
3. By focusing on average outcomes, expected (monetary) payoffs ignore the effect that the amount of variability in the possible outcomes should have on the decision making. For example, since Goferbroke does not have the financing to sustain a large loss, selling the land to assure a payoff of 90 ($90,000) may be preferable to an expected payoff of 100 ($100,000) from drilling. Selling would avoid the risk of a large loss from drilling when the land is dry. (Section 9.9 will discuss how utilities can be used to better reflect the value of payoffs.)
So why is this criterion commonly referred to as Bayes’ decision rule? The reason is that it is often credited to the Reverend Thomas Bayes, a nonconforming 18th century English minister who won renown as a philosopher and mathematician, although the same basic idea has even longer roots in the field of economics. Bayes’ philosophy of decision making still is very influential today, and some management scientists even refer to themselves as Bayesians because of their devotion to this philosophy.
More recently, it has become somewhat popular to also call this criterion the expected monetary value (EMV) criterion . The reason for this alternative name is that the payoffs in the payoff table often represent monetary values (such as the number of dollars of profit), in which case the expected payoff for each decision alternative is its expected monetary value. However, the name is a misnomer for those cases where the measure of the payoff is some- thing other than monetary value (as in Section 9.9). Therefore, we will consistently use the single name, Bayes’ decision rule, to refer to this criterion in all situations.
Because of its popularity, the rest of the chapter focuses on procedures that are based on this criterion.
Max’s Reaction Max: So where does this leave us? Jennifer: Well, now you need to decide which criterion seems most appropriate to you in this situation. Max: Well, I can’t say that I am very excited about any of the criteria. But it sounded like this last one is a popular one. Jennifer: Yes, it is. Max: Why? Jennifer: Really, two reasons. First, this is the criterion that uses all the available informa- tion. The prior probabilities may not be as accurate as we would like, but they do give us valuable information about roughly how likely each of the possible states of nature is. Many management scientists feel that using this key information should lead to better decisions.
By considering only expected payoffs, Bayes’ decision rule fails to give special consideration to the possibility of disastrously large losses.
Bayes’ decision rule uses all the information provided by the payoff table.
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Max: I’m not ready to accept that yet. But what is the second reason? Jennifer: Remember that this is the criterion that focuses on what the average payoff would be if the same situation were repeated numerous times. We called this the expected payoff. Consistently selecting the decision alternative that provides the best expected payoff would provide the most payoff to the company in the long run. Doing what is best in the long run seems like rational decision making for a manager.
1. How does the maximax criterion select a decision alternative? What kind of person might find this criterion appealing?
2. What are some criticisms of the maximax criterion? 3. How does the maximin criterion select a decision alternative? What kind of person might find
this criterion appealing? 4. What are some criticisms of the maximin criterion? 5. Which state of nature does the maximum likelihood criterion focus on? 6. What are some criticisms of the maximum likelihood criterion? 7. How does Bayes’ decision rule select a decision alternative? 8. How is the expected payoff for a decision alternative calculated? 9. What are some criticisms of Bayes’ decision rule?
Review Questions
9.3 DECISION TREES
The spreadsheet in Figure 9.1 illustrates one useful way of performing decision analysis with Bayes’ decision rule. Another enlightening way to apply this decision rule is to use a decision tree to display and analyze the problem graphically. The decision tree for the first Goferbroke Co. problem is shown in Figure 9.2 . Starting on the left side and moving to the right side shows the progression of events. First, a decision is made as to whether to drill for oil or sell the land. If the decision is to drill, the next event is to learn whether the state of nature is that the land contains oil or is dry. Finally, the payoff is obtained that results from these events.
In the terminology of decision trees, the junction points are called nodes (or forks) and the lines emanating from the nodes are referred to as branches . A distinction is then made between the following two types of nodes.
A decision node , represented by a square, indicates that a decision needs to be made at that point in the process. An event node (or chance node), represented by a circle, indicates that a random event occurs at that point.
Thus, node A in Figure 9.2 is a decision node since the decision on whether to drill or sell occurs there. Node B is an event node since a random event, the occurrence of one of the pos- sible states of nature, takes place there. Each of the two branches emanating from this node corresponds to one of the possible random events, where the number in parentheses along the branch gives the probability that this event will occur.
A decision tree can be very helpful for visualizing and analyzing a problem. When the problem is as small as the one in Figure 9.2 , using the decision tree in the analysis process is optional. However, one nice feature of decision trees is that they also can be used for more complicated problems where a sequence of decisions needs to be made. You will see this illustrated for the full Goferbroke Co. problem in Sections 9.7 and 9.9 when a decision on whether to conduct a seismic survey is made before deciding whether to drill or sell.
Spreadsheet Software for Decision Trees We will describe and illustrate how to use Risk Solver Platform for Education (RSPE) to construct and analyze decision trees in Excel. Instructions for installing this software are on a supplementary insert included with the book and also on the book’s website, www.mhhe .com/hillier5e . If you are a Mac user (RSPE is not compatible with Mac versions of Excel) or you or your instructor simply prefer to use different software, a supplement to this chapter on the CD-ROM and website contains instructions for TreePlan, another popular Excel add-in for constructing and analyzing decision trees in Excel.
To begin creating a decision tree using RSPE, select Add Node from the Decision Tree/ Node menu. This brings up the dialog box shown in Figure 9.3 . Here you can choose the type
B
700
Payoff
–100
Oi l (0
.25 )
Dry (0.75)A
Sell 90
Dr ill
FIGURE 9.2 The decision tree for the first Goferbroke Co. problem as presented in Table 9.3.
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The Workers’ Compensation Board (WCB) of British Columbia, Canada, is responsible for the occupational health and safety, rehabilitation, and compensation interests of the province’s workers and employers. The WCB serves more than 165,000 employers who employ about 1.8 million workers in British Columbia. It spends approximately US$1 billion annually on compensation and rehabilitation.
A key factor in controlling WCB costs is to identify those short-term disability claims that pose a potentially high financial risk of converting into a far more expensive long- term disability claim unless there is intensive early claim- management intervention to provide the needed medical treatment and rehabilitation. The question was how to accurately identify these high-risk claims so as to minimize the expected total cost of claim compensation and claim- management intervention.
A management science team was formed to study this problem by applying decision analysis. For each of numerous
categories of injury claims based on the nature of the injury, the gender and age of the worker, and so on, a decision tree was used to evaluate whether that category should be classified as low risk (not requiring intervention) or high risk (requiring intervention), depending on the severity of the injury. For each category, a calculation was made of the cutoff point on the critical number of short-term disabil- ity claim days paid that would trigger claim-management intervention so as to minimize the expected cost of claim payments and intervention.
This application of decision analysis with decision trees is saving WCB approximately US$4 million per year while also enabling some injured workers to return to work sooner.
Source: E. Urbanovich, E. E. Young, M. L. Puterman, and S. O. Fattedad, “Early Detection of High-Risk Claims at the Workers’ Compensation Board of British Columbia,” Interfaces 33, no. 4 (July–August 2003), pp. 15–26. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
331
of node (Decision or Event), give names to each of the branches, and specify a value for each branch (the partial payoff associated with that branch). The default names for the branches of a decision node in RSPE are Decision 1 and Decision 2. These can be changed (or more branches added) by double-clicking on the branch name (or in the next blank row to add a branch) and typing in a new name. The initial node in the first Goferbroke problem (node A in Figure 9.2 ) is a decision node with two branches: Drill and Sell. The payoff associated with drilling is –100 (the $100,000 cost of drilling) and the payoff associated with selling is 90 (the $90,000 selling price). After making all of these entries as shown in Figure 9.3 , clicking OK then yields the decision tree shown in Figure 9.4 .
If the decision is to drill, the next event is to learn whether or not the land contains oil. To create an event node, click on the cell containing the triangle terminal node at the end of
RSPE Tip: The type of node, the branch names, and the values (or partial payoffs) of each branch can be entered in the RSPE Decision Tree dialog box. Alternatively, the names and values can be entered (or changed) after the fact by typing them directly into the spreadsheet.
FIGURE 9.3 The Decision Tree dialog box used to specify that the initial node of the first Goferbroke problem is a decision node with two branches, Drill and Sell, with values (partial payoffs) of 2100 and 90, respectively.
1
2 Drill
Sell
3
4
5 2
90
-100
90
-100
90
-100
90
6
7
9
8
A B C D E F G
FIGURE 9.4 The initial, partial decision tree created by RSPE by selecting Add Node from the Decision Tree/Node menu on the RSPE rib- bon and specifying a Decision node with two branches named Drill and Sell, with partial payoffs of 2100 and 90, respectively.
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the drill branch (cell F3 in Figure 9.4 ) and choose Add Node from the Decision Tree/Node menu on the RSPE ribbon to bring up the dialog box shown in Figure 9.5 . The node is an event node with two branches, Oil and Dry, with probabilities 0.25 and 0.75, respectively, and values (partial payoffs) of 800 and 0, respectively, as entered into the dialog box in Figure 9.5 . After clicking OK, the final decision tree is shown in Figure 9.6 . (Note that RSPE, by default, shows all probabilities as a percentage, with 25% and 75% in H1 and H6, rather than 0.25 and 0.75.)
At any time, you also can click on any existing node and make changes using various choices under the Decision Tree menu on the RSPE ribbon. For example, under the Node submenu, you can choose Add Node, Change Node, Delete Node, Copy Node, or Paste Node. Under the Branch submenu, you can Add Branch, Change Branch, or Delete Branch.
At each stage in constructing a decision tree, RSPE automatically solves for the optimal policy with the current tree when using Bayes’ decision rule. The number inside each deci- sion node indicates which branch should be chosen (assuming the branches emanating from that node are numbered consecutively from top to bottom). Thus, for the final decision tree in Figure 9.6 , the number 1 in cell B9 specifies that the first branch (the Drill alternative) should be chosen. The number on both sides of each terminal node is the payoff if that node is reached. The number 100 in cells A10 and E6 is the expected payoff (the measure of perfor- mance for Bayes’ decision rule) at those stages in the process.
This description of the decision analysis tools of RSPE may seem somewhat complicated. However, we think that you will find the procedure quite intuitive when you execute it on a computer. If you spend considerable time with RSPE, you also will find that it has many help- ful features that haven’t been described in this brief introduction.
Max’s Reaction Max: I like this decision tree thing. It puts everything into perspective. Jennifer: Good. Max: But one thing still really bothers me. Jennifer: I think I can guess.
RSPE Tip: To create a new node at the end of a tree, select the cell contain- ing the terminal node and choose Add Node from the Decision Tree/Node menu on the RSPE ribbon. Select the type of node and enter names, values, and (if an event node) probabilities for each branch.
RSPE Tip: To make changes to a node, select the node and choose an appropriate option under the Decision Tree menu on the RSPE ribbon.
RSPE always identifies the optimal policy for the cur- rent decision tree according to Bayes’ decision rule.
FIGURE 9.5 The Decision Tree dialog box used to specify that the second node of the first Goferbroke problem is an event node with two branches, Oil and Dry, with values (partial payoffs) of 800 and 0, and with probabilities of 0.25 and 0.75, respectively.
1
2
Drill
Sell
Oil
Dry
3
4
5
1
100
-100 100 75%
90 90
90
800
0
700
-100
700
-100
6
7
9
8
10
11
12
13
14
A B C D E F G H I J K
25%
FIGURE 9.6 The decision tree constructed and solved by RSPE for the first Goferbroke Co. problem as presented in Table 9.3, where the 1 in cell B9 indicates that the top branch (the Drill alternative) should be chosen.
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Max: Yes. I’ve made it pretty plain that I don’t want to make my decision based on believing the consulting geologist’s numbers. One chance in four of oil. Hah! It’s just an educated guess. Jennifer: Well, let me ask this. What is the key factor in deciding whether to drill for oil or sell the land? Max: How likely it is that there is oil there. Jennifer: Doesn’t the consulting geologist help in determining this? Max: Definitely. I hardly ever drill without his input. Jennifer: So shouldn’t your criterion for deciding whether to drill be based directly on this input? Max: Yes, it should. Jennifer: But then I don’t understand why you keep objecting to using the consulting geologist’s numbers. Max: I’m not objecting to using his input. This input is vital to my decision. What I object to is using his numbers, one chance in four of oil, as being the gospel truth. That is what this Bayes’ decision rule seems to do. We both saw what a close decision this was, 100 versus 90. What happens if his numbers are off some, as they probably are? This is too important a decision to be based on some numbers that are largely pulled out of the air. Jennifer: OK, I see. Now he says that there is one chance in four of oil, a 25 percent chance. Do you think that is the right ballpark at least? If not 25 percent, how much lower might it be? Or how much higher? Max: I usually add and subtract 10 percent from whatever the consulting geologist says. So I suppose the chance of oil is likely to be somewhere between 15 percent and 35 percent. Jennifer: Good. Now we’re getting somewhere. I think I know exactly what we should do next. Max: What’s that? Jennifer: There is a management science technique that is designed for just this kind of situation. It is called sensitivity analysis. It will allow us to investigate what happens if the consulting geologist’s numbers are off. Max: Great! Let’s do it.
1. What is a decision tree? 2. What is a decision node in a decision tree? An event node? 3. What symbols are used to represent decision nodes and event nodes?
Review Questions
9.4 SENSITIVITY ANALYSIS WITH DECISION TREES
Sensitivity analysis (an important type of what-if analysis introduced in Section 5.1) com- monly is used with various applications of management science to study the effect if some of the numbers included in the mathematical model are not correct. In this case, the mathemati- cal model is represented by the decision tree shown in Figure 9.6 . The numbers in this tree that are most questionable are the prior probabilities in cells H1 and H6, so we will initially focus the sensitivity analysis on these numbers.
It is helpful to start this process by consolidating the data and results on the spreadsheet below the decision tree, as in Figure 9.7 . As indicated by the formulas at the bottom of the fig- ure, the cells giving the results make reference to the corresponding output cells on the deci- sion tree. Similarly, the data cells on the decision tree now reference the corresponding data cells below the tree. Consequently, the user can experiment with various alternative values in the data cells below and the results will simultaneously change in both the decision tree and the results section below the tree to reflect the new data.
Consolidating the data and results offers a couple of advantages. First, it assures that each piece of data is in only one place. Each time that piece of data is needed in the decision tree, a reference is made to the single data cell below. This greatly simplifies sensitivity analysis. To change a piece of data, it needs to be changed in only one place rather than searching through
Excel Tip: Consolidating the data and results on the spreadsheet makes it easier to do sensitivity analysis and also makes the model and results easier to interpret.
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the entire tree to find and change all occurrences of that piece of data. 1 A second advantage of consolidating the data and results is that it makes it easy for anyone to interpret the model. It is not necessary to understand RSPE or how to read a decision tree in order to see what data were used in the model or what the suggested plan of action and expected payoff are.
1 In this very simple decision tree, this advantage does not become evident since each piece of data is only used once in the tree anyway. However, in later sections, when the possibility of seismic testing is consid- ered, some data will be repeated many times in the tree and this advantage will become more clear.
1
2
Drill
Sell
Oil
Dry
3
4
5
1
100
-100 100
90 90
90
800
25%
75%
0
700
-100
700
-100
6
7
9
8
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11
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14
15
16
17 Data
18
19
20
21
22
23
24
25
26
Cost of Drilling
Revenue if Oil
Revenue if Sell
Revenue if Dry
Probability of Oil
Action
Expected Payoff
A
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Drill
=–CostOfDrilling
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1
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=ProbabilityOfOil
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Sell
=RevenueIfSell
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Expected Payoff
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=IF(B9=1,"Drill","Sell")
=A10
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=RevenueIfOil
=RevenueIfDry
Dry
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Action
CostOfDrilling
ExpectedPayoff
ProbabilityOfOil
RevenueIfDry
RevenueIfOil
RevenueIfSell
E24
E18
E26
E22
E21
E19
E20
Range Name Cell
FIGURE 9.7 In preparation for per- forming sensitivity analy- sis on the first Goferbroke Co. problem, the data and results have been consoli- dated on the spreadsheet below the decision tree.
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9.4 Sensitivity Analysis with Decision Trees 335
The sum of the two prior probabilities must equal one, so increasing one of these prob- abilities automatically decreases the other one by the same amount, and vice versa. This is enforced on the decision tree in Figure 9.7 by the equation in cell H6—the probability of a dry site 5 H6 5 1 2 ProbabilityOfOil (E22). Max has concluded that the true chances of having oil on the tract of land are likely to lie somewhere between 15 and 35 percent. In other words, the true prior probability of having oil is likely to be in the range from 0.15 to 0.35, so the cor- responding prior probability of the land being dry would range from 0.85 to 0.65.
We can begin sensitivity analysis by simply trying different trial values for the prior prob- ability of oil. This is done in Figure 9.8 , first with this probability at the lower end of the range (0.15) and next with this probability at the upper end (0.35). When the prior probability of oil is only 0.15, the decision swings over to selling the land by a wide margin (an expected payoff of 90 versus only 20 for drilling). However, when this probability is 0.35, the decision is to drill by a wide margin (expected payoff 5 180 versus only 90 for selling). Thus, the decision is very sensitive to the prior probability of oil. This sensitivity analysis has revealed that it is important to do more, if possible, to pin down just what the true value of the probability of oil is.
Using a Data Table to Do Sensitivity Analysis Systematically To pin down just where the suggested course of action changes, we could continue selecting new trial values of the prior probability of oil at random. However, a better approach is to systematically consider a range of values. A feature built into Excel, called a data table, is designed to perform just this sort of analysis. Data tables are used to show the results of cer- tain output cells for various trial values of a data cell.
To use data tables, first make a table on the spreadsheet with headings as shown in col- umns I, J, and K in Figure 9.9 . In the first column of the table (I19:I29), list the trial values for the data cell (the prior probability of oil), except leave the first row blank. The headings of the next columns specify which output will be evaluated. For each of these columns, use the first row of the table (cells J18:K18) to write an equation that refers to the relevant output cell. In this case, the cells of interest are Action (E24) and ExpectedPayoff (E26), so the equations for J18:K18 are those shown below the spreadsheet in Figure 9.9 .
Next, select the entire table (I18:K29) and then choose Data Table from the What-If Anal- ysis menu of the Data tab. In the Data Table dialog box (as shown at the bottom left of Figure 9.9 ), indicate the column input cell (E22), which refers to the data cell that is being changed in the first column of the table. Nothing is entered for the row input cell because no row is being used to list the trial values of a data cell in this case.
Clicking OK then generates the data table shown in Figure 9.10 . For each trial value for the data cell listed in the first column of the table, the corresponding output cell values are calculated and displayed in the other columns of the table. (The entries in the first row of the table come from the original solution in the spreadsheet.)
Figure 9.10 reveals that the best course of action switches from Sell to Drill for a prior probability of oil somewhere between 0.23 and 0.25. Trial and error (or algebra) can be used to pin this number down more precisely. It turns out to be 0.2375.
For a problem with more than two possible states of nature, the most straightforward approach is to focus the sensitivity analysis on only two states at a time as described above. This again would involve investigating what happens when the prior probability of one state increases as the prior probability of the other state decreases by the same amount, holding fixed the prior probabilities of the remaining states. This procedure then can be repeated for as many other pairs of states as desired.
Max’s Reaction Max: That data table paints a pretty clear picture. I think I’m getting a much better handle on the problem. Jennifer: Good. Max: Less than a 23¾ percent chance of oil, I should sell. If it’s more, I should drill. It confirms what I suspected all along. This is a close decision, and it all boils down to picking the right number for the chances of oil. I sure wish I had more to go on than the consulting geologist’s numbers.
A data table displays the results of selected output cells for various trial values of a data cell.
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336 Chapter Nine Decision Analysis
1
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Drill
Sell
Oil
Dry
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Revenue if Oil
Revenue if Sell
Revenue if Dry
Probability of Oil
Action
Expected Payoff
A B C D E F G H I J K
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800
90
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0.15
Sell
90
1
2
Drill
Sell
Oil
Dry
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Revenue if Oil
Revenue if Sell
Revenue if Dry
Probability of Oil
Action
Expected Payoff
A B C D E F G H I J K
100
800
90
0
0.35
Drill
180
FIGURE 9.8 Performing sensitivity analysis for the first Goferbroke Co. problem by trying alternative val- ues (0.15 and 0.35) of the prior probability of oil.
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9.4 Sensitivity Analysis with Decision Trees 337
Jennifer: You talked earlier about the possibility of paying $30,000 to get a detailed seis- mic survey of the land. Max: Yes, I might have to do that. But 30,000 bucks! I’m still not sure that it’s worth that much dough. Jennifer: I have a quick way of checking that. It’s another technique I learned in my man- agement science course. It’s called finding the expected value of perfect information (EVPI) . The expected value of perfect information is the increase in the expected payoff you would get if the seismic survey could tell you for sure if there is oil there. Max: But it can’t tell you for sure.
The next section describes how to find and use the expected value of perfect information.
16
A B C D E F G H I J K L M
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Probability
of Oil Action
Expected
Payoff
Action
ExpectedPayoff
E24
E26
Range Name Cell
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J K
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=Action
Expected
Payoff
=ExpectedPayoff
Data
Cost of Drilling
Revenue if Oil
Revenue if Sell
Revenue if Dry
Probability of Oil
Action
Expected Payoff
100
800
90
0
0.25
Drill
100
Select these
cells
(I18:K29),
before
choosing
Data Table
from the
What-If
Analysis is
menu of the
Data tab
15%
Drill 100
17%
19%
21%
23%
25%
27%
29%
31%
33%
35%
FIGURE 9.9 Expansion of the spread- sheet in Figure 9.7 to pre- pare for generating a data table, where the choice of E22 for the column input cell in the Data Table dialog box indicates that this is the data cell that is being changed in the first column of the data table.
16
I J K
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Drill
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of Oil
15%
Action
Drill
Expected
Payoff
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17%
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23%
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27%
29%
31%
33%
35%
FIGURE 9.10 After the preparation displayed in Figure 9.9, clicking OK generates this data table that shows the optimal action and expected payoff for various trial values of the prior probability of oil.
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338 Chapter Nine Decision Analysis
Jennifer: Yes, I know. But finding out for sure if oil is there is what we refer to as perfect information. So the increase in the expected payoff if you find out for sure is the expected value of perfect information. We know that’s better than you actually can do with a seismic survey. Max: Right. Jennifer: OK, suppose we find that the expected value of perfect information is less than $30,000. Since that is better than we can do with a seismic survey, that tells us right off the bat that it wouldn’t pay to do the seismic survey. Max: OK, I get it. But what if this expected value of perfect information is more than $30,000? Jennifer: Then you don’t know for sure whether the seismic survey is worth it until you do some more analysis. This analysis takes some time, whereas it is very quick to calculate the expected value of perfect information. So it is well worth simply checking whether the expected value of perfect information is less than $30,000 and, if so, saving a lot of additional work. Max: OK. Let’s do it.
1. Why might it be helpful to use sensitivity analysis with Bayes’ decision rule? 2. When preparing to perform sensitivity analysis, what are a couple of advantages of consolidat-
ing the data and results on the spreadsheet that contains the decision tree? 3. What is shown by a data table when it is used to perform sensitivity analysis? 4. What conclusion was drawn for the first Goferbroke Co. problem regarding how the decision
should depend on the prior probability of oil?
Review Questions
9.5 CHECKING WHETHER TO OBTAIN MORE INFORMATION
Prior probabilities may provide somewhat inaccurate estimates of the true probabilities of the states of nature. Might it be worthwhile for Max to spend some money for a seismic survey to obtain better estimates? The quickest way to check this is to pretend that it is possible for the same amount of money to actually determine which state is the true state of nature (“perfect information”) and then determine whether obtaining this information would make this expen- diture worthwhile. If having perfect information would not be worthwhile, then it definitely would not be worthwhile to spend this money just to learn more about the probabilities of the states of nature.
The key quantities for performing this analysis are
EP (without more info) 5 Expected payoff from applying Bayes’ decision rule with the original prior probabilities
5 100 (as given in Figure 9.7)
EP (with perfect info) 5 Expected payoff if the decision could be made after learning the true state of nature
EVPI 5 Expected value of perfect information
C 5 Cost of obtaining more information 5 30 (cost of the seismic survey in thousands of dollars)
The expected value of perfect information is calculated as
EVPI 5 EP (with perfect info) 2 EP (without more info)
After calculating EP (with perfect info) and then EVPI, the last step is to compare EVPI with C.
If C . EVPI, then it is not worthwhile to obtain more information. If C # EVPI, then it might be worthwhile to obtain more information.
To calculate EP (with perfect info), we pretend that the decision can be made after learning the true state of nature. Given the true state of nature, we then would automatically choose the alternative with the maximum payoff for that state. Thus, we drill if we know there is oil,
Definitely identifying the true state of nature is referred to as perfect infor- mation. This represents the best outcome of seeking more information.
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9.5 Checking Whether to Obtain More Information 339
whereas we sell if we know the site is dry. The prior probabilities still give our best estimate of the probability that each state of nature will turn out to be the true one. EP (with perfect info) is therefore the weighted average of the maximum payoff for each state, multiplying each maximum payoff by the prior probability of the corresponding state of nature. Thus,
EP (with perfect info) 5 (0.25)(700) 1 (0.75)(90) 5 242.5
RSPE also can be used to calculate EP (with perfect info) by constructing and solving the decision tree shown in Figure 9.11 . The clever idea here is to start the decision tree with an event node whose branches are the various states of nature (oil and dry in this case). Since a decision node follows each of these branches, the decision is being made with perfect infor- mation about the true state of nature. Therefore, the expected payoff of 242.5 obtained by RSPE in cell A11 is the expected payoff with perfect information.
Since EP (with perfect info) 5 242½, we now can calculate the expected value of perfect information as
EVPI 5 EP (with perfect info) 2 EP (without more info) 5 242.5 2 100 5 142.5
Conclusion: EVPI . C, since 142.5 . 30. Therefore, it might be worthwhile to do the seismic survey.
Max’s Reaction Max: So you’re telling me that if the seismic survey could really be definitive in deter- mining whether oil is there, doing the survey would increase my average payoff by about $142,500? Jennifer: That’s right. Max: So after subtracting the $30,000 cost of the survey, I would be ahead $112,500. Well, too bad the surveys aren’t that good. In fact, they’re not all that reliable. Jennifer: Tell me more. How reliable are they?
Starting the decision tree with an event node whose branches are the various states of nature corresponds to starting with perfect information about the true state of nature.
1
2
Oil
Dry
Drill
Sell
Drill
Sell
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A B C D E F G H I J K
2
FIGURE 9.11 By starting with an event node involving the states of nature, RSPE uses this decision tree to obtain the expected payoff with perfect information for the first Goferbroke Co. problem.
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340 Chapter Nine Decision Analysis
Max: Well, they come back with seismic soundings. If the seismic soundings are favor- able, then oil is fairly likely. If they are unfavorable, then oil is pretty unlikely. But you can’t tell for sure. Jennifer: OK. Suppose oil is there. How often would you get favorable seismic soundings? Max: I can’t give you an exact number. Maybe 60 percent. Jennifer: OK, good. Now suppose that the land is dry. How often would you still get favorable seismic soundings? Max: Too often! I’ve lost a lot of money drilling when the seismic survey said to and then nothing was there. That’s why I don’t like to spend the 30,000 bucks. Jennifer: Sure. So it tells you to drill when you shouldn’t close to half the time? Max: No. It’s not that bad. But fairly often. Jennifer: Can you give me a percentage? Max: OK. Maybe 20 percent. Jennifer: Good. Thanks. Now I think we can do some analysis to determine whether it is really worthwhile to do the seismic survey. Max: How do you do the analysis? Jennifer: Well, I’ll describe the process in detail pretty soon. But here is the general idea. We’ll do some calculations to determine what the chances of oil would be if the seismic soundings turn out to be favorable. Then we’ll calculate the chances if the soundings are unfavorable. We called the consulting geologist’s numbers prior probabilities because they were prior to obtaining more information. The improved numbers are referred to as posterior probabilities . Max: OK. Jennifer: Then we’ll use these posterior probabilities to determine the average payoff, after subtracting the $30,000 cost, if we do the seismic survey. If this payoff is better than we would do without the seismic survey, then we should do it. Otherwise, not. Max: That makes sense.
Posterior probabilities are the revised probabilities of the states of nature after doing a test or sur- vey to improve the prior probabilities.
1. What is meant by perfect information regarding the states of nature? 2. How can the expected payoff with perfect information be calculated from the payoff table? 3. How should a decision tree be constructed to obtain the expected payoff with perfect informa-
tion by solving the tree? 4. What is the formula for calculating the expected value of perfect information? 5. What is the conclusion if the cost of obtaining more information is more than the expected
value of perfect information? 6. What is the conclusion if the cost of obtaining more information is less than the expected value
of perfect information? 7. Which of these two cases occurs in the Goferbroke Co. problem?
Review Questions
9.6 USING NEW INFORMATION TO UPDATE THE PROBABILITIES
The prior probabilities of the possible states of nature often are quite subjective in nature, so they may be only very rough estimates of the true probabilities. Fortunately, it frequently is possible to do some additional testing or surveying (at some expense) to improve these esti- mates. These improved estimates are called posterior probabilities.
In the case of the Goferbroke Co., these improved estimates can be obtained at a cost of $30,000 by conducting a detailed seismic survey of the land. The possible findings from such a survey are summarized below.
Possible Findings from a Seismic Survey FSS: Favorable seismic soundings; oil is fairly likely. USS: Unfavorable seismic soundings; oil is quite unlikely.
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9.6 Using New Information to Update the Probabilities 341
To use either finding to calculate the posterior probability of oil (or of being dry), it is nec- essary to estimate the probability of obtaining this finding for each state of nature. During the conversation at the end of the preceding section, Jennifer elicited these estimates from Max, as summarized in Table 9.7 . (Max actually only estimated the probability of favorable seismic soundings, but subtracting this number from one gives the probability of unfavorable seismic soundings.) The symbol used in the table for each of these estimated probabilities is
P(finding 0 state) 5 Probability that the indicated finding will occur, given that the state of nature is the indicated one
This kind of probability is referred to as a conditional probability, because it is conditioned on being given the state of nature.
Recall that the prior probabilities are
P(Oil) 5 0.25 P(Dry) 5 0.75
The next step is to use these probabilities and the probabilities in Table 9.7 to obtain a combined probability called a joint probability. Each combination of a state of nature and a finding from the seismic survey will have a joint probability that is determined by the following formula.
P(state and finding) 5 P(state) P(finding 0 state) For example, the joint probability that the state of nature is Oil and the finding from the seis- mic survey is favorable (FSS) is
P(Oil and FSS) 5 P(Oil) P(FSS 0 Oil) 5 0.25(0.6) 5 0.15
The calculation of all these joint probabilities is shown in the third column of the probability tree diagram given in Figure 9.12 . The case involved is identified underneath each branch of the tree and the probability is given over the branch. The first column gives the prior probabilities and then the probabilities from Table 9.7 are shown in the second column. Multiplying each probability in the first column by a probability in the second column gives the corresponding joint probability in the third column.
Having found each joint probability of both a particular state of nature and a particular finding from the seismic survey, the next step is to use these probabilities to find each prob- ability of just a particular finding without specifying the state of nature. Since any finding can be obtained with any state of nature, the formula for calculating the probability of just a particular finding is
P(finding) 5 P(Oil and finding) 1 P(Dry and finding)
For example, the probability of a favorable finding (FSS) is
P(FSS) 5 P(Oil and FSS) 1 P(Dry and FSS) 5 0.15 1 0.15 5 0.3
where the two joint probabilities on the right-hand side of this equation are found on the first and third branches of the third column of the probability tree diagram. The calculation of both P (FSS) and P (USS) is shown underneath the diagram. (These are referred to as unconditional probabilities to differentiate them from the conditional probabilities of a finding given the state of nature, shown in the second column.)
Each joint probability in the third column of the prob- ability tree diagram is the product of the probabilities in the first two columns.
The probability of a finding is the sum of the corre- sponding joint probabilities in the third column of the probability tree diagram.
P (finding | state)
State of Nature Favorable (FSS) Unfavorable (USS)
Oil P(FSS | Oil) 5 0.6 P(USS | Oil) 5 0.4 Dry P(FSS | Dry) 5 0.2 P(USS | Dry) 5 0.8
TABLE 9.7 Probabilities of the Possible Findings from the Seismic Survey, Given the State of Nature, for the Goferbroke Co. Problem
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342 Chapter Nine Decision Analysis
Finally, we now are ready to calculate each posterior probability of a particular state of nature given a particular finding from the seismic survey. The formula involves combining the joint probabilities in the third column with the unconditional probabilities underneath the diagram as follows.
P(state 0 finding) 5 P(state and finding) P(finding)
For example, the posterior probability that the true state of nature is oil, given a favorable finding (FSS) from the seismic survey, is
P(Oil 0 FSS) 5 P(Oil and FSS) P(FSS)
5 0.15 0.3
5 0.5
The fourth column of the probability tree diagram shows the calculation of all the posterior probabilities. The arrows indicate how each numerator comes from the corresponding joint probability in the third column and the denominator comes from the corresponding uncondi- tional probability below the diagram.
By using the formulas given earlier for the joint probabilities and unconditional prob- abilities, each posterior probability also can be calculated directly from the prior probabilities (first column) and the conditional probabilities (second column) as follows.
P(state 0 finding) 5 P(state) P(finding 0 state) P(Oil) P(finding 0 Oil) 1 P(Dry) P(finding 0 Dry)
The arrows in the prob- ability tree diagram show where the numbers come from for calculating the posterior probabilities.
Prior Probabilities
P (state)
Conditional Probabilities
P (finding | state)
Joint Probabilities
P (state and finding)
Posterior Probabilities
P (state | finding)
0.2 5
O il
0.6
FS S, g
ive n O
il
0.25(0.6) = 0.15 Oil and FSS
0.25(0.4) = 0.1 Oil and USS
0.75(0.2) = 0.15 Dry and FSS
0.75(0.8) = 0.6 Dry and USS
0.75Dry
0.4USS, given Oil
0.2
FS S, g
ive n D
ry
0.8USS, given Dry
0.15 0.3
= 0.5
0.1 0.7
= 0.14
0.15 0.3
= 0.5
0.6 0.7
= 0.86
Dry, given USS
Dry, given FSS
Oil, given USS
Oil, given FSS
Unconditional probabilities: P(FSS) = 0.15 + 0.15 = 0.3
P(finding) P(USS) = 0.1 + 0.6 = 0.7
FIGURE 9.12 Probability tree diagram for the Goferbroke Co. problem showing all the probabilities leading to the calculation of each posterior probability of the state of nature given the finding of the seismic survey.
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9.6 Using New Information to Update the Probabilities 343
For example, the posterior probability of oil, given a favorable finding (FSS), is
P(Oil 0 FSS) 5 P(Oil) P(FSS 0 Oil) P(Oil) P(FSS 0 Oil) 1 P(Dry) P(FSS 0 Dry)
5 0.25(0.6)
0.25(0.6) 1 0.75(0.2)
5 0.5
This formula for a posterior probability is known as Bayes’ theorem , in honor of its discov- ery by the Reverend Bayes. The clever Reverend Bayes found that any posterior probability can be found in this way for any decision analysis problem, regardless of how many states of nature it has. The denominator in the formula would contain one such term for each of the states of nature. Note that the probability tree diagram also is applying Bayes’ theorem, but in smaller steps rather than a single long formula.
Table 9.8 summarizes all the posterior probabilities calculated in Figure 9.12 . After you learn the logic of calculating posterior probabilities, we suggest that you use
the computer to perform these rather lengthy calculations. We have provided an Excel tem- plate (labeled Posterior Probabilities) for this purpose in this chapter’s Excel files in your MS Courseware. Figure 9.13 illustrates the use of this template for the Goferbroke Co. problem. All you do is enter the prior probabilities and the conditional probabilities from the first two columns of Figure 9.12 into the top half of the template. The posterior probabilities then immediately appear in the bottom half. (The equations entered into the cells in columns E through H are similar to those for column D shown at the bottom of the figure.)
Max’s Reaction Max: So this is saying that even with favorable seismic soundings, I still only have one chance in two of finding oil. No wonder I’ve been disappointed so often in the past when I’ve drilled after receiving a favorable seismic survey. I thought those surveys were sup- posed to be more reliable than that. So now I’m even more unenthusiastic about paying 30,000 bucks to get a survey done. Jennifer: But one chance in two of oil. Those are good odds. Max: Yes, they are. But I’m likely to lay out 30,000 bucks and then just get an unfavor- able survey back. Jennifer: My calculations indicate that you have about a 70 percent chance of that happening. Max: See what I mean? Jennifer: But even an unfavorable survey tells you a lot. Just one chance in seven of oil then. That might rule out drilling. So a seismic survey really does pin down the odds of oil a lot better. Either one chance in two or one chance in seven instead of the ballpark estimate of one chance in four from the consulting geologist. Max: Yes, I suppose that’s right. I really would like to improve the consulting geologist’s numbers. It sounds like you’re recommending that we do the seismic survey. Jennifer: Well, actually, I’m not quite sure yet. What we’ll do is sketch out a decision tree, showing the decision on whether to do the seismic survey and then the decision on whether to drill or sell. Then we’ll work out the average payoffs for these decisions on the decision tree. Max: OK, let’s do it. I want to make a decision soon.
P (state | finding)
Finding Oil Dry
Favorable (FSS) P(Oil | FSS) 5 1/2 P(Dry | FSS) 5 1/2 Unfavorable (USS) P(Oil | USS) 5 1/7 P(Dry | USS) 5 6/7
TABLE 9.8 Posterior Probabilities of the States of Nature, Given the Finding from the Seismic Survey, for the Goferbroke Co. Problem
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344 Chapter Nine Decision Analysis
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P(Finding I State)
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A B C D E F G H
Template for Posterior Probabilities
Posterior
Probabilities:
P(State I Finding)
State of Nature
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B C D
Posterior
Probabilities:
Finding =B6
=D5
=E5
P(Finding)
P(State I Finding)
State of Nature
=F5
=G5
=H5
=SUMPRODUCT(C6:C10,D6:D10)
=SUMPRODUCT(C6:C10,E6:E10)
=SUMPRODUCT(C6:C10,F6:F10)
=SUMPRODUCT(C6:C10,G6:G10)
=SUMPRODUCT(C6:C10,H6:H10)
=C6*D6/SUMPRODUCT(C6:C10,D6:D10)
=C6*E6/SUMPRODUCT(C6:C10,E6:E10)
=C6*F6/SUMPRODUCT(C6:C10,F6:F10)
=C6*G6/SUMPRODUCT(C6:C10,G6:G10)
=C6*H6/SUMPRODUCT(C6:C10,H6:H10)
Oil
Dry
0.25
0.75
Finding P(Finding)
FSS
USS
0.3
0.7
Oil
0.5
0.14286
Dry
0.5
0.85714
FSS
0.6
0.2
USS
0.4
0.8
Finding
FIGURE 9.13 The Posterior Prob- abilities template in your MS Courseware enables efficient calculation of posterior probabilities, as illustrated here for the Goferbroke Co. problem.
1. What are posterior probabilities of the states of nature? 2. What are the possible findings from a seismic survey for the Goferbroke Co.? 3. What probabilities need to be estimated in addition to prior probabilities in order to begin
calculating posterior probabilities? 4. What five kinds of probabilities are considered in a probability tree diagram? 5. What is the formula for calculating P (state and finding)? 6. What is the formula for calculating P (finding)? 7. What is the formula for calculating a posterior probability, P (state | finding), from P (state and
finding) and P (finding)? 8. What is the name of the famous theorem for how to calculate posterior probabilities?
Review Questions
9.7 USING A DECISION TREE TO ANALYZE THE PROBLEM WITH A SEQUENCE OF DECISIONS
We now turn our attention to analyzing the full Goferbroke Co. problem with the help of a decision tree. For the full problem, there is a sequence of two decisions to be made. First, should a seismic survey be conducted? Second, after obtaining the results of the seismic sur- vey (if it is conducted), should the company drill for oil or sell the land?
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9.7 Using a Decision Tree to Analyze the Problem with a Sequence of Decisions 345
As described in Section 9.3, a decision tree provides a graphical display of the progression of decisions and random events for the problem. Figure 9.2 in that section shows the decision tree for the first Goferbroke problem where the only decision under consideration is whether to drill for oil or sell the land. Figure 9.6 then shows the same decision tree as it would be constructed and solved with RSPE.
Constructing the Decision Tree Now that a prior decision needs to be made on whether to conduct a seismic survey, this same decision tree needs to be expanded as shown in Figure 9.14 (before including any numbers). Recall that each square in the tree represents a decision node, where a decision needs to be made, and each circle represents an event node, where a random event will occur.
Thus, the first decision (should we have a seismic survey done?) is represented by deci- sion node a in Figure 9.14 . The two branches leading out of this node correspond to the two alternatives for this decision. Node b is an event node representing the random event of the outcome of the seismic survey. The two branches emanating from node b represent the two possible outcomes of the survey. Next comes the second decision (nodes c, d, and e ) with its two possible choices. If the decision is to drill for oil, then we come to another event node (nodes f, g, and h ), where its two branches correspond to the two possible states of nature.
The next step is to insert numbers into the decision tree as shown in Figure 9.15 . The num- bers under or over the branches that are not in parentheses are the cash flows (in thousands of dollars) that occur at those branches. For each path through the tree from node a to a final branch, these same numbers then are added to obtain the resulting total payoff shown in bold- face to the right of that branch. The last set of numbers is the probabilities of random events. In particular, since each branch emanating from an event node represents a possible random event, the probability of this event occurring from this node has been inserted in parentheses along this branch. From event node h, the probabilities are the prior probabilities of these states of nature, since no seismic survey has been conducted to obtain more information in this case. However, event nodes f and g lead out of a decision to do the seismic survey (and then to drill). Therefore, the probabilities from these event nodes are the posterior probabili- ties of the states of nature, given the outcome of the seismic survey, where these numbers are obtained from Table 9.8 or from cells D15:E16 in Figure 9.13 . Finally, we have the two
The numbers in parentheses are probabilities.
b g
f
h
Un fav
or ab
le
Favorable
a
e
d
c
D o
se ism
ic su
rv ey
No seismic survey Dr
ill
Sell
Oil
Dry
Dr ill
Sell
Oil
Dry
Dr ill
Sell
Oil
Dry
FIGURE 9.14 The decision tree for the full Goferbroke Co. problem (before including any numbers) when first deciding whether to con- duct a seismic survey.
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346
The Westinghouse Science and Technology Center is the Westinghouse Electric Corporation’s main research and development (R&D) arm to develop new technology. The process of evaluating R&D projects to decide which ones should be initiated and then which ones should be con- tinued as progress is made (or not made) is particularly challenging for management because of the great uncer- tainties and very long time horizons involved. The actual launch date for an embryonic technology may be years, even decades, removed from its inception as a modest R&D proposal to investigate the technology’s potential.
As the center came under increasing pressure to reduce costs and deliver high-impact technology quickly, the cen- ter’s controller funded a management science project to improve this evaluation process. The management science team developed a decision tree approach to analyzing any R&D proposal while considering its complete sequence of key decision points. The first decision point is whether to
fund the proposed embryonic project for the first year or so. If its early technical milestones are reached, the next decision point is whether to continue funding the proj- ect for some period. This may then be repeated one or more times. If the late technical milestones are reached, the next decision point is whether to prelaunch because the innovation still meets strategic business objectives. If a strategic fit is achieved, the final decision point is whether to commercialize the innovation now or to delay its launch or to abandon it altogether. A decision tree with a pro- gression of decision nodes and intervening event nodes provides a natural way of depicting and analyzing such an R&D project.
Source: R. K. Perdue, W. J. McAllister, P. V. King, and B. G. Berkey, “Valuation of R and D Projects Using Options Pricing and Decision Analysis Models,” Interfaces 29, no. 6 (November–December 1999), pp. 57–74. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
b g
f
h
Un fav
or ab
le (0
.7 )
0
0Favorable (0.3)
a
e
d
c
D o
se ism
ic su
rv ey
–3 0
0 No seismic survey
Dr ill
–1 00
90
Sell
Oil (0 .25)
800 0
Dry (0.75)
Dr ill
–1 00
90
Sell
Oil (0 .5)
800
0
Dry (0.5)
Dr ill
–1 00
90
Sell
Oil (0. 143)
800
0 Dry (0.857)
Payoff
670
–130
60
670
–130
60
700
–100
90
FIGURE 9.15 The decision tree in Figure 9.14 after adding both the probabilities of random events and the payoffs.
branches emanating from event node b. The numbers here are the probabilities of these find- ings from the seismic survey, Favorable (FSS) or Unfavorable (USS), as given underneath the probability tree diagram in Figure 9.12 or in cells C15:C16 of Figure 9.13 .
Performing the Analysis Having constructed the decision tree, including its numbers, we now are ready to analyze the problem by using the following procedure.
1. Start at the right side of the decision tree and move left one column at a time. For each column, perform either step 2 or step 3 depending on whether the nodes in that column are event nodes or decision nodes.
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9.7 Using a Decision Tree to Analyze the Problem with a Sequence of Decisions 347
2. For each event node, calculate its expected payoff by multiplying the expected payoff of each branch (shown in boldface to the right of the branch) by the probability of that branch and then summing these products. Record this expected payoff for each event node in boldface next to the node, and designate this quantity as also being the expected payoff for the branch leading to this node.
3. For each decision node, compare the expected payoffs of its branches and choose the alter- native whose branch has the largest expected payoff. In each case, record the choice on the decision tree.
To begin the procedure, consider the rightmost column of nodes, namely, event nodes f, g, and h. Applying step 2, their expected payoffs (EP) are calculated as
EP 5 1 7
(670) 1 6 7
(2130) 5 215.7 for node f
EP 5 1 2
(670) 1 1 2
(2130) 5 270 for node g
EP 5 1 4
(700) 1 3 4
(2100) 5 100 for node h
These expected payoffs then are placed above these nodes, as shown in Figure 9.16 . Next, we move one column to the left, which consists of decision nodes c, d, and e. The
expected payoff for a branch that leads to an event node now is recorded in boldface over that event node. Therefore, step 3 can be applied as follows:
Node c: Drill alternative has EP 5 2 15.7 Sell alternative has EP 5 60 60 . 2 15.7, so choose the Sell alternative Node d: Drill alternative has EP 5 270 Sell alternative has EP 5 60 270 . 60, so choose the Drill alternative
The expected payoff needs to be calculated for each event node.
b g
f
h
Un fav
or ab
le (0
.7 )
0
0Favorable (0.3)
a
e
d
c
D o
se ism
ic su
rv ey
–3 0
0 No seismic survey
Dr ill
–1 00
90
Sell
Oil (0 .25)
800 0
Dry (0.75)
Dr ill
–1 00
90
Sell
Oil (0 .5)
800
0
Dry (0.5)
Dr ill
–1 00
90
Sell
Oil (0. 143)
800
0 Dry (0.857)
Payoff
670
–130
60
670
–130
60
700
–100
90
123
123
270
100
60
100
270
–15.7
FIGURE 9.16 The final decision tree that records the analysis for the full Goferbroke Co. problem when using monetary payoffs.
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348 Chapter Nine Decision Analysis
Node e: Drill alternative has EP 5 100 Sell alternative has EP 5 90 100 . 90, so choose the Drill alternative
The expected payoff for each chosen alternative now would be recorded in boldface over its decision node, as shown in Figure 9.16 . The chosen alternative also is indicated by inserting a double dash as a barrier through each rejected branch.
Next, moving one more column to the left brings us to node b. Since this is an event node, step 2 of the procedure needs to be applied. The expected payoff for each of its branches is recorded over the following decision node. Therefore, the expected payoff is
EP 5 0.7(60) 1 0.3(270) 5 123 for node b
as recorded over this node in Figure 9.16 . Finally, we move left to node a, a decision node. Applying step 3 yields
Node a: Do seismic survey has EP 5 123 No seismic survey has EP 5 100 123 . 100, so choose Do seismic survey.
This expected payoff of 123 now would be recorded over the node, and a double dash inserted to indicate the rejected branch, as already shown in Figure 9.16 .
This procedure has moved from right to left for analysis purposes. However, having com- pleted the decision tree in this way, the decision maker now can read the tree from left to right to see the actual progression of events. The double dashes have closed off the undesirable paths. Therefore, given the payoffs for the final outcomes shown on the right side, Bayes’ decision rule says to follow only the open paths from left to right to achieve the largest pos- sible expected payoff.
Following the open paths from left to right in Figure 9.16 yields the following optimal policy, according to Bayes’ decision rule.
Optimal Policy Do the seismic survey. If the result is unfavorable, sell the land. If the result is favorable, drill for oil. The expected payoff (including the cost of the seismic survey) is 123 ($123,000).
Expected Value of Sample Information We have assumed so far that the cost of the seismic survey for the full Goferbroke Co. prob- lem is known in advance to be $30,000. However, suppose that there is uncertainty about this cost. How would this change the analysis described above?
In this case, the analysis would begin by identifying two key quantities,
EP (with more info) 5 Expected payoff (excluding the cost of the survey) when the survey is done
EP (without more info) 5 Expected payoff when the survey is not done
where Bayes’ decision rule is applied to find both quantities. EP (with more info) is obtained by using the top half of the decision tree in Figure 9.17 except that the (unknown) cost of the seismic survey is not included, so all the payoffs and expected payoffs would be 30 larger than shown there. Therefore, cell E19 indicates that
EP (with more info) 5 123 1 30 5 153
EP (without more info) is described at the beginning of Section 9.5, and is obtained here from the bottom half of the decision tree in Figure 9.17 without any change, so cell E42 shows that
EP (without more info) 5 100
A double dash indicates a rejected decision.
The open paths (no double dashes) provide the optimal decision at each decision node.
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9.7 Using a Decision Tree to Analyze the Problem with a Sequence of Decisions 349
1
2
Do Survey
No Survey
Unfavorable
Favorable
Drill
Sell
Drill
Sell
Drill
Sell
Dry
Oil
3
4
5
1
1
1
2
70%
30%
-100
-100
0
100
90 90
270
0 60
25%
75%
800 700
90 60
-100 270
90 60
-100 -15.714
0
1000
Dry
Oil
-130
50%
50%
800 670
0
Dry
Oil
-130
14.286%
85.714%
800 670
-130
60
60
700
-100
90
670
-130
670
0
6
7
9
8
10
11
12
13
14
15
16
17
18
19 -30 123
20
21
22
23
24
25
26
27
28
29
30 123
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
A
Decision Tree for Goferbroke Co. Problem (with Survey)
B C D E F G H I J K L M N O P Q R S
FIGURE 9.17 The decision tree constructed and solved by RSPE for the full Goferbroke Co. problem that also considers whether to do a seismic survey.
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350 Chapter Nine Decision Analysis
Now we can calculate the expected value of sample information (EVSI) (where “sam- ple information” refers to the information from the seismic survey in this case) as
EVSI 5 EP (with more info) 2 EP (without more info) 5 153 2 100 5 53
Let
C 5 Best available estimate of the cost of the seismic survey (in thousands of dollars)
The final step in the analysis is to compare C and EVSI.
If C , EVSI, then perform the seismic survey. If C $ EVSI, then do not perform the seismic survey.
Using RSPE Using the procedures described in Section 9.3, the Decision Tree tools of RSPE can be used to construct and solve this same decision tree on a spreadsheet. Figure 9.17 shows the decision tree obtained with RSPE. Although the form is somewhat different, note that this decision tree is completely equivalent to the one in Figure 9.16 . Besides the convenience of constructing the tree directly on a spreadsheet, RSPE also provides the key advantage of automatically solving the decision tree. Rather than relying on hand calculations as in Figure 9.16 , RSPE instantaneously calculates all the expected payoffs at each stage of the tree, as shown below and to the left of each node, as soon as the decision tree is constructed. Instead of using dou- ble dashes, RSPE puts a number inside each decision node indicating which branch should be chosen (assuming the branches emanating from that node are numbered consecutively from top to bottom).
Max’s Reaction Max: I see that this decision tree gives me some numbers to compare alternatives. But how reliable are those numbers? Jennifer: Well, you have to remember that these average payoffs for the alternatives at the decision nodes are based on both the payoffs on the right and the probabilities at the event nodes. These probabilities are based in turn on the consulting geologist’s numbers and the numbers you gave me on how frequently you get favorable seismic soundings when you have oil or when the land is dry. Max: That doesn’t sound so good. You know what I think about the consulting geologist’s numbers. And the numbers I gave you were pretty rough estimates. Jennifer: True. So the average payoffs shown in the decision tree are only approxima- tions. This is when some sensitivity analysis can be helpful, like we did earlier before we considered doing the seismic survey. Max: OK. So let’s do it.
1. What does a decision tree display? 2. What is happening at a decision node? 3. What is happening at an event node? 4. What kinds of numbers need to be inserted into a decision tree before beginning the
analysis? 5. When performing the analysis, where do you begin on the decision tree and in which direction
do you move for dealing with the nodes? 6. What calculation needs to be performed at each event node? 7. What comparison needs to be made at each decision node? 8. What is meant by the expected value of sample information and how might it be used?
Review Questions
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9.8 Performing Sensitivity Analysis on the Problem with a Sequence of Decisions 351
9.8 PERFORMING SENSITIVITY ANALYSIS ON THE PROBLEM WITH A SEQUENCE OF DECISIONS
Section 9.4 describes how the decision tree created with RSPE ( Figures 9.6 and 9.7 ) was used to perform sensitivity analysis for the first Goferbroke problem where the only decision being made was whether to drill for oil or sell the land (without conducting the seismic survey). The focus was on one particularly critical piece of data, the prior probability of oil, so the analysis involved checking whether the decision would change if the original value of this prior prob- ability (0.25) were changed to various other trial values. New trial values first were consid- ered in a trial-and-error manner ( Figure 9.8 ) and then were investigated more systematically by constructing a data table ( Figure 9.10 ).
Since Max Flyer wants to consider whether to have a seismic survey conducted before deciding whether to drill or sell, the relevant decision tree now is the one in Figure 9.17 instead of the one in Figure 9.6 . With this sequence of decisions and the resulting need to obtain and apply posterior probabilities, conducting sensitivity analysis becomes somewhat more involved. Let’s see how it is done.
Organizing the Spreadsheet As was done in Section 9.4, it is helpful to begin by consolidating the data and results into one section of the spreadsheet, as shown in Figure 9.18 . The data cells in the decision tree now make reference to the consolidated data cells to the right of the decision tree (cells V4:V11). Similarly, the summarized results to the right of the decision tree make reference to the output cells within the decision tree (the decision nodes in cells B29, F41, J11, and J26, as well as the expected payoff in cell A30).
The probability data in the decision tree are complicated by the fact that the posterior probabilities will need to be updated any time a change is made in any of the prior prob- ability data. Fortunately, the template for calculating posterior probabilities (as shown in Figure 9.13 ) can be used to do these calculations. The relevant portion of this template (B3:H19) has been copied (using the Copy and Paste commands in the Edit menu) to the spreadsheet in Figure 9.18 (now appearing in U30:AA46). The data for the template refer to the probability data in the data cells PriorProbabilityOfOil (V9), ProbFSSGivenOil (V10), and ProbUSSGivenDry (V11), as shown in the formulas for cells V33:X34 at the bottom of Figure 9.18 . The template automatically calculates the probability of each finding and the posterior probabilities (in cells V42:X43) based on these data. The decision tree then refers to these calculated probabilities when they are needed, as shown in the formulas for cells P3:P11 in Figure 9.18 .
While it takes some time and effort to consolidate the data and results, including all the necessary cross-referencing, this step is truly essential for performing sensitivity analysis. Many pieces of data are used in several places on the decision tree. For example, the reve- nue if Goferbroke finds oil appears in cells P6, P21, and L36. Performing sensitivity analy- sis on this piece of data now requires changing its value in only one place (cell V6) rather than three (cells P6, P21, and L36). The benefits of consolidation are even more important for the probability data. Changing any prior probability may cause all the posterior prob- abilities to change. By including the posterior probability template, the prior probability can be changed in one place and then all the other probabilities are calculated and updated appropriately.
After making any change in the cost data, revenue data, or probability data in Figure 9.18 , the spreadsheet nicely summarizes the new results after the actual work to obtain these results is instantly done by the posterior probability template and the decision tree. Therefore, exper- imenting with alternative data values in a trial-and-error manner is one useful way of per- forming sensitivity analysis.
Now let’s see how this sensitivity analysis can be done more systematically by using a data table.
Consolidating the data and results on the spreadsheet is important for sensitivity analysis.
Organize the spreadsheet so that any piece of data needs to be changed in only one place.
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352 Chapter Nine Decision Analysis
1
2
Do Survey
No Survey
Unfavorable
Favorable
Drill
Sell
Drill
Sell
Drill
Sell
Dry
Oil
3
4
5
1
1
1
2
70%
30%
-100
-100
0
100
270
0 60
25%
75%
800 700
90 60
-100 270
90 60
-100 -15.714
0
1000
Dry
Oil
-130
50%
50%
800 670
0
Dry
Oil
-130
14.3%
85.7%
800 670
-130
60
60
700
-100
90
670
-130
670
0
6
7
9
8
10
11
12
13
14
15
16
17
18
19 -30 123
20
21
22
23
24
25
26
27
28
29
30 123
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
A
Decision Tree for Goferbroke Co. Problem (with Survey)
B C D E F G H I J K L M N O P Q R S T U V W X Y Z AA
Data
Action
Data:
Expected Payoff ($thousands)
Prior Probability of Oil
Cost of Survey
Cost of Drilling
Revenue if Oil
Revenue if Sell
Revenue if Dry
Do Survey?
If No If Yes
Drill
State of
Nature
Prior
Probability
P(Finding I State)
Finding
Posterior
Probabilities:
Finding P(Finding )
P(State I Finding)
State of Nature
Drill
Sell
If Favorable
If Unfavorable
P(FSS I Oil)
P(USS I Dry)
0
0.25
0.8
Yes
123
0.6
30
100
800
90
U V W X
14
15 Do Survey?
If No If Yes
=IF(B29=1,"Yes","No")
=IF(J26=1,"Drill","Sell")
=IF(J11=1,"Drill","Sell")
=IF(F41=1,"Drill","Sell")
16
17
18
U V W X
30
31
32
33
34
19
20
P
3
4
5
6
7
8
9
10
11
=ProbOilGivenUSS
Oil
=RevenueIfOil
V
23
24
25
Expected
Payoff
($thousands)
26 =A30
=ProbDryGivenUSS
Dry
=RevenueIfDry
CostOfDrilling
CostOfSurvey
PriorProbabilityOfOil
ProbDryGivenFSS
ProbDryGivenUSS
ProbFSS
ProbFSSGivenOil
ProbOilGivenFSS
ProbOilGivenUSS
ProbUSS
ProbUSSGivenDry
RevenueIfDry
RevenueIfOil
RevenueIfSell
V5
V4
V9
X42
X43
V42
V10
W42
W43
V43
V11
V8
V6
V7
Range Name Cell
If Favorable
If Unfavorable
Data:
State of
Nature
Oil
Dry
Prior
Probability
=PriorProbabilityOfOil
=1–PriorProbabilityOfOil
=ProbFSSGivenOil
=1–ProbUSSGivenDry
=1–ProbFSSGivenOil
=ProbUSSGivenDry
P(Finding I State)
FSS
Finding
USS
Oil
Dry
0.25
0.75
FSS
0.6
0.2
USS
0.4
0.8
FSS
USS
0.3
0.7
Oil
0.5
0.143
Dry
0.5
0.857
Action
46 9090
FIGURE 9.18 In preparation for performing sensitivity analysis on the full Goferbroke Co. problem, the data and results have been consolidated on the spreadsheet to the right of the decision tree.
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9.8 Performing Sensitivity Analysis on the Problem with a Sequence of Decisions 353
Using a Data Table to Do Sensitivity Analysis Systematically To systematically determine how the decisions and expected payoffs change as the prior prob- ability of oil (or any other data) changes, a data table can be generated with Excel by using the same procedure described in Section 9.4. First make a table on the spreadsheet with headings as shown in columns Y through AD in Figure 9.19 . In the first column of the table (Y5:Y15), list the trial values for the data cell (the prior probability of oil), except leave the first row blank. The headings of the next columns specify which output will be evaluated. For each of these columns, use the first row of the table (cells Y4:AD4) to write an equation that refers to the relevant output cell. In this case, the cells of interest are (1) the decision of whether to do the survey (V15), (2) if so, whether to drill if the survey is favorable or unfavorable (W19 and W20), (3) if not, whether to drill (U19), and (4) the value of ExpectedPayoff (V26). The equa- tions for Y4:AD4 referring to these output cells are shown below the spreadsheet in Figure 9.19 .
Next, select the entire table (Y4:AD15) and then choose Data Table from the What-If Analysis menu of the Data tab. In the Data Table dialog box (as shown at the bottom left of Figure 9.19 ), indicate the column input cell (V9), which refers to the data cell that is being changed in the first column of the table.
Clicking OK then generates the table shown in Figure 9.19 . For each trial value for the data cell listed in the first column of the table, the corresponding output cell values are calcu- lated and displayed in the other columns of the table. Some of the output in the data table is not relevant. For example, when the decision is to not do the survey in column Z, the results in columns AA and AB (what to do given favorable or unfavorable survey results) are not relevant. Similarly, when the decision is to do the survey in column Z, the results in column AC (what to do if you don’t do the survey) are not relevant. The relevant output has been formatted in boldface to make it stand out compared to the irrelevant output.
Figure 9.19 reveals that the optimal initial decision switches from Sell without a survey to doing the survey somewhere between 0.1 and 0.2 for the prior probability of oil and then switches again to Drill without a survey somewhere between 0.3 and 0.4. Using the spread- sheet in Figure 9.18 , trial-and-error analysis soon leads to the following conclusions about how the optimal policy depends on this probability.
Optimal Policy
Let p 5 Prior probability of oil. If p # 0.168, then sell the land (no seismic survey). If 0.169 # p # 0.308, then do the survey: drill if favorable and sell if not. If p $ 0.309, then drill for oil (no seismic survey).
Max’s Reaction Max: Very interesting. I especially liked the way we were able to use that sensitivity analysis spreadsheet to see immediately what would happen when we change some of the numbers. And there was one thing that I found particularly encouraging. Jennifer: What was that? Max: When we changed that prior probability of oil to nearly every other plausible value, it kept coming back with the same answer. Do the seismic survey and then drill only if the survey is favorable. Otherwise, sell. So the consulting geologist’s numbers can be off by quite a bit and we still would be doing the right thing. Jennifer: Yes, that was a key finding, wasn’t it? OK. Does this mean that you are com- fortable now with a decision to proceed with the seismic survey and then either drill or sell depending on the outcome of the survey? Max: Not quite. There is still one thing that bothers me. Jennifer: What’s that? Max: Suppose the seismic survey gives us a favorable seismic sounding, so we drill. If the land turns out to be dry, then I’m out 130,000 bucks! As I said at the beginning, that would nearly put us out of business. That scares me. I currently am shorter of working capital than I normally am. Therefore, losing $130,000 now would hurt more than it normally does. It doesn’t look like this approach is really taking that into account.
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Jennifer: No, you’re right. It really doesn’t. This approach just looks at average monetary values. That isn’t good enough when you’re dealing with such large amounts. You wouldn’t be willing to flip a coin to determine whether you win or lose $130,000, right? Max: No, I sure wouldn’t. Jennifer: OK, that’s the tipoff. As I mentioned the first time we talked about this problem, I think the circumstances here indicate that we need to go beyond dollars and cents to look at the consequences of the possible outcomes. Fortunately, decision analysis has a way of doing this by introducing utilities. The basic idea is that the utility of an outcome measures the true value to you of that outcome rather than just the monetary value. So by expressing payoffs in terms of utilities, the decision tree analysis would find the average utility at each node instead of the average monetary value. So now the decisions would be based on giving you the highest possible average utility.
Considering average mon- etary values isn’t good enough when uncomfort- ably large losses can occur.
FIGURE 9.19 The data table that shows the optimal policy and expected payoff for various trial values of the prior probability of oil.
1
2
3
4
5
6
7
9
8
10
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of Oil
Do If No
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Payoff
($thousands)
If Survey If Survey
UnfavorableFavorableSurvey?
Yes
Yes
Yes
No
No
No
No
No
No
No
No
No
0
0.1
0.2
0.3
0.4
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1
Drill
Drill
Drill
Drill
Drill Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill
Drill 123
90
90
102.8
143.2
220
300
380
460
540
620
700
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Sell Sell
Sell
Sell
Sell
Sell
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Probability
Y Z AA AB AC AD
1
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of Oil
Do If No
Expected
=ExpectedPayoff
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If Survey If Survey
UnfavorableFavorableSurvey?
=V15 =W19 =W20 =U19
Survey
Probability
Y Z AA AB AC AD
1. When preparing to perform sensitivity analysis, how should one begin organizing the spread- sheet that contains the decision tree?
2. Performing sensitivity analysis on a certain piece of data should require changing its value in how many places on the spreadsheet?
3. What conclusion was drawn for the full Goferbroke problem regarding how the decision should depend on the prior probability of oil?
Review Questions
9.9 USING UTILITIES TO BETTER REFLECT THE VALUES OF PAYOFFS
Thus far, when applying Bayes’ decision rule, we have assumed that the expected payoff in mon- etary terms is the appropriate measure of the consequences of taking an action. However, in many situations where very large amounts of money are involved, this assumption is inappropriate.
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For example, suppose that an individual is offered the choice of (1) accepting a 50–50 chance of winning $100,000 or (2) receiving $40,000 with certainty. Many people would pre- fer the $40,000 even though the expected payoff on the 50–50 chance of winning $100,000 is $50,000. A company may be unwilling to invest a large sum of money in a new product, even when the expected profit is substantial, if there is a risk of losing its investment and thereby becoming bankrupt. People buy insurance even though it is a poor investment from the view- point of the expected payoff.
Do these examples invalidate Bayes’ decision rule? Fortunately, the answer is no, because there is a way of transforming monetary values to an appropriate scale that reflects the deci- sion maker’s preferences. This scale is called the utility function for money.
Utility Functions for Money Figure 9.20 shows a typical utility function U(M) for money M . The intuitive interpretation is that it indicates that an individual having this utility function would value obtaining $30,000 twice as much as $10,000 and would value obtaining $100,000 twice as much as $30,000. This reflects the fact that the person’s highest-priority needs would be met by the first $10,000. Hav- ing this decreasing slope of the function as the amount of money increases is referred to as having a decreasing marginal utility for money. Such an individual is referred to as being risk averse .
However, not all individuals have a decreasing marginal utility for money. Some people are risk seekers instead of risk averse, and they go through life looking for the “big score.” The slope of their utility function increases as the amount of money increases, so they have an increasing marginal utility for money.
Figure 9.21 compares the shape of the utility function for money for risk-averse and risk- seeking individuals. Also shown is the intermediate case of a risk-neutral individual , who prizes money at its face value. Such an individual’s utility for money is simply proportional to the amount of money involved. Although some people appear to be risk neutral when only small amounts of money are involved, it is unusual to be truly risk neutral with very large amounts.
It also is possible to exhibit a mixture of these kinds of behavior. For example, an individ- ual might be essentially risk neutral with small amounts of money, then become a risk seeker with moderate amounts, and then turn risk averse with large amounts. In addition, one’s atti- tude toward risk can shift over time depending on circumstances.
Managers of a business firm need to consider the company’s circumstances and the collec- tive philosophy of top management in determining the appropriate attitude toward risk when making managerial decisions.
Two different individuals can have very different util- ity functions for money.
U(M)
M
1
0.75
0.5
0.25
$10,000 $30,000 $60,000 $100,000 0
FIGURE 9.20 A typical utility function for money, where U(M) is the utility of obtaining an amount of money M.
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The fact that different people have different utility functions for money has an important implication for decision making in the face of uncertainty.
When a utility function for money is incorporated into a decision analysis approach to a problem, this utility function must be constructed to fit the current preferences and values of the decision maker involved. (The decision maker can be either a single individual or a group of people.)
The key to constructing the utility function for money to fit the decision maker is the fol- lowing fundamental property of utility functions.
Fundamental Property: Under the assumptions of utility theory, the decision maker’s utility function for money has the property that the decision maker is indifferent between two alterna- tive courses of action if the two alternatives have the same expected utility.
To illustrate, suppose that the decision maker has the utility function shown in Figure 9.20 . Further suppose that the decision maker is offered the following opportunity.
Offer: An opportunity to obtain either $100,000 (utility 5 1) with probability p or nothing (utility 5 0) with probability (1 2 p ).
Thus, by weighting the two possible utilities (1 and 0) by their probabilities, the expected utility is
E(utility) 5 p 1 0(1 2 p) 5 p for this offer
Therefore, for each of the following three pairs of alternatives, the above fundamental prop- erty indicates that the decision maker is indifferent between the first and second alternatives.
1. First alternative: The offer with p 5 0.25, so E (utility) 5 0.25. Second alternative: Definitely obtain $10,000, so utility 5 0.25.
2. First alternative: The offer with p 5 0.5, so E (utility) 5 0.5. Second alternative: Definitely obtain $30,000, so utility 5 0.5.
3. First alternative: The offer with p 5 0.75, so E (utility) 5 0.75. Second alternative: Definitely obtain $60,000, so utility 5 0.75.
This example also illustrates one way in which the decision maker’s utility function for money can be constructed in the first place. The decision maker would be made the same hypothetical offer to obtain a large amount of money (e.g., $100,000) with probability p, or nothing (utility 5 0) otherwise. Then, for each of a few smaller amounts of money (e.g., $10,000, $30,000, and $60,000), the decision maker would be asked to choose a value of p that would make him or her indifferent between the offer and definitely obtaining that amount of money. The utility of the smaller amount of money then is p times the utility of the large amount. When the utility of the large amount has been set equal to 1, as in Figure 9.20 , this conveniently makes the utility of the smaller amount simply equal to p. The utility values in Figure 9.20 imply that the decision maker has chosen p 5 0.25 when M 5 $10,000, p 5 0.5 when M 5 $30,000, and p 5 0.75 when M 5 $60,000. (Constructing the utility function in this way is an example of the equivalent lottery method described later in this section.)
The scale of the utility function is irrelevant. In other words, it doesn’t matter whether the values of U ( M ) at the dashed lines in Figure 9.20 are 0.25, 0.5, 0.75, 1 (as shown) or 10,000,
In all three of these cases, a decision maker with the utility function in Figure 9.20 would be indifferent between the two alternatives because they have the same expected utility.
U(M)
M (a) Risk averse
U(M)
M (b) Risk seeking
U(M)
M (c) Risk neutral
FIGURE 9.21 The shape of the utility function for money for (a) risk-averse, (b) risk- seeking, and (c) risk- neutral individuals.
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9.9 Using Utilities to Better Reflect the Values of Payoffs 357
20,000, 30,000, 40,000, or whatever. All the utilities can be multiplied by any positive con- stant without affecting which decision alternative will have the largest expected utility. It also is possible to add the same constant (positive or negative) to all the utilities without affecting which decision alternative will have the largest expected utility.
For these reasons, we have the liberty to set the value of U ( M ) arbitrarily for two values of M, so long as the higher monetary value has the higher utility. It is particularly convenient to set U ( M ) 5 0 for the smallest value of M under consideration and to set U ( M ) 5 1 for the largest M, as was done in Figure 9.20 . By assigning a utility of 0 to the worst outcome and a utility of 1 to the best outcome, and then determining the utilities of the other outcomes accordingly, it becomes easy to see the relative utility of each outcome along the scale from worst to best.
Now we are ready to summarize the basic role of utility functions in decision analysis.
When the decision maker’s utility function for money is used to measure the relative worth of the various possible monetary outcomes, Bayes’ decision rule replaces monetary payoffs by the corresponding utilities. Therefore, the optimal decision (or series of decisions) is the one that maximizes the expected utility.
Only utility functions for money have been discussed here. However, we should mention that utility functions can sometimes still be constructed when some or all of the important consequences of the decision alternatives are not monetary in nature. (For example, the con- sequences of a doctor’s decision alternatives in treating a patient involve the future health of the patient.) This is not necessarily easy, since it may require making value judgments about the relative desirability of rather intangible consequences. Nevertheless, under these circum- stances, it is important to incorporate such value judgments into the decision process.
Dealing with the Goferbroke Co. Problem Recall that the Goferbroke Co. is operating without much capital, so a loss of $100,000 would be quite serious. As the owner of the company, Max already has gone heavily into debt to keep going. The worst-case scenario would be to come up with $30,000 for a seismic survey and then still lose $100,000 by drilling when there is no oil. This scenario would not bankrupt the company at this point but definitely would leave it in a precarious financial position.
On the other hand, striking oil is an exciting prospect, since earning $700,000 finally would put the company on fairly solid financial footing.
Max is the decision maker for this problem. Therefore, to prepare for using utilities to ana- lyze the problem, it is necessary to construct Max’s utility function for money, U ( M ), where we will express the amount of money M in units of thousands of dollars.
We start by assigning utilities of 0 and 1, respectively, to the smallest and largest possible payoffs. Since the smallest possible payoff is M 5 2 130 (a loss of $130,000) and the largest is M 5 700 (a gain of $700,000), this gives U ( 2 130) 5 0 and U(700) 5 1.
To determine the utilities for other possible monetary payoffs, it is necessary to probe Max’s attitude toward risk. Especially important are his feelings about the consequences of the worst possible loss ($130,000) and the best possible gain ($700,000), as well as how he compares these consequences. Let us eavesdrop as Jennifer probes these feelings with Max.
Interviewing Max Jennifer: Well now, these utilities are intended to reflect your feelings about the true value to you of these various possible payoffs. Therefore, to pin down what your utilities are, we need to talk some about how you feel about these payoffs and their consequences for the company. Max: Fine. Jennifer: A good place to begin would be the best and worst possible cases. The possibil- ity of gaining $700,000 or losing $130,000. Max: Those are the big ones all right. Jennifer: OK, suppose you drill without paying for a seismic survey and then you find oil, so your profit is about $700,000. What would that do for the company? Max: A lot. That would finally give me the capital I need to become more of a major player in this business. I then could take a shot at finding a big oil field. That big strike I’ve talked about.
The objective now is to maximize the expected util- ity rather than the expected payoff in monetary terms.
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Jennifer: OK, good. Now let’s talk about the consequences if you were to get that biggest possible loss instead. Suppose you pay for a seismic survey, then you drill and the land is dry. So you’re out about $130,000. How bad would that be? What kind of future would the company have? Max: Well, let me put it this way. It would put the company in a pretty uncomfortable financial position. I would need to work hard on getting some more financing. Then we would need to cautiously work our way out of the hole by forming some partnerships for some low-risk, low-gain drilling. But I think we could do it. I’ve been in that position a couple times before and come out of it. We’d be OK. Jennifer: It sounds like you wouldn’t be overly worried about such a loss as long as you have reasonable odds for a big payoff to justify this risk. Max: That’s right. Jennifer: OK, now let’s talk about those odds. What I’m going to do is set up a simpler hypo- thetical situation. Suppose these are your alternatives. One is to drill. If you find oil, you clear $700,000. If the land is dry, you’re out $130,000. The only other alternative is to sell the land for $90,000. I know this isn’t your actual situation since $700,000 does not include the cost of a survey whereas the loss of $130,000 does, but let’s pretend that these are your alternatives. Max: I don’t understand why you want to talk about a situation that is different from what we are facing. Jennifer: Trust me. Considering these kinds of hypothetical situations are going to enable us to determine your utilities. Max: OK. Jennifer: Now presumably if you had a 50–50 chance of either clearing $700,000 or los- ing $130,000, you would drill. Max: Sure. Jennifer: If you had a smaller chance, say one-in-four of gaining $700,000, versus a three- in-four chance of losing $130,000, would you choose to drill or sell the land for $90,000? Max: Well, that’s almost the original decision we were trying to make, before we considered the seismic survey. However, there is one big difference. Now you’re asking me to suppose that the loss if there is no oil is $130,000 rather than $100,000. The higher loss would be quite a bit more painful. I wouldn’t be willing to take this risk with just a one-in-four chance of gaining $700,000. Jennifer: OK, so now we know that the point at which you would be indifferent between going ahead or not is somewhere between having a one-in-four chance and a 50–50 chance of gaining $700,000 rather than losing $130,000. Let’s see if we can pin down just where your point of indifference is within this range from one-in-four and 50–50. Let’s try a one-in- three chance. Would you go ahead and drill with a one-in-three chance of gaining $700,000 ver- sus a two-in-three chance of losing $130,000, or would you choose to sell the land for $90,000? Max: Hmm. That’s not so clear. What would be the average payoff in this case? Jennifer: Almost $147,000. Max: Not bad. Hmm, one chance in three of gaining $700,000. That’s tempting. But two chances in three of losing $130,000 with all the problems involved with that. I don’t know. $90,000 would be a sure thing. That’s a tough one. Jennifer: OK, let’s try this. Suppose your chances of gaining $700,000 were a little better than one-in-three. Would you do it? Max: Yes, I think so. Jennifer: And if your chances were a little under one-in-three? Max: Then I don’t think I would do it. Jennifer: OK. You’ve convinced me that your point of indifference is one-in-three. That’s exactly what I needed to know.
Finding U (90) Max has indeed given Jennifer just the information she needs to determine U (90), Max’s utility for a payoff of 90 (a gain of $90,000). Recall that U ( 2 130) already has been set at U ( 2 130) 5 0 and that U (700) already has been set at U (700) 5 1. Here is the procedure that Jennifer is using to find U (90).
The point of indifference is the point where the deci- sion maker is indifferent between two hypothetical alternatives.
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9.9 Using Utilities to Better Reflect the Values of Payoffs 359
The decision maker (Max) is offered two alternatives, A 1 and A 2 .
A 1 : Obtain a payoff of 700 with probability p. Obtain a payoff of 2 130 with probability (1 2 p ).
A 2 : Definitely obtain a payoff of 90.
Question to the decision maker: What value of p makes you indifferent between these two alternatives? Recall that Max has chosen p 5 1/3.
For a given choice of p, the expected utility for A 1 is
E(utility for A1) 5 pU(700) 1 (1 2 p)U(2130) 5 p(1) 1 (1 2 p)(0) 5 p
If the decision maker is indifferent between the two alternatives, the fundamental property of utility functions says that the two alternatives must have the same expected utility. Therefore, the utility for A 2 must also be p. Since Max chose a point of indifference of p 5 1/3, the utility for A 2 must be 1/3, so U (90) 5 1/3.
The Equivalent Lottery Method for Determining Utilities The above procedure for finding U (90) illustrates that the key to finding the utility for any payoff M is having the decision maker select a point of indifference between two alternatives, where one of them ( A 1 ) involves a lottery between the largest payoff and the smallest payoff and the other alternative ( A 2 ) is to receive a sure payoff of M. At the point of indifference, the lottery is equivalent to the sure payoff in the sense that they have the same expected utility, so the procedure is referred to as the equivalent lottery method . Here is an outline of the procedure.
Equivalent Lottery Method
1. Determine the largest potential payoff, M 5 maximum, and assign it a utility of 1:
U(maximum) 5 1
2. Determine the smallest potential payoff, and assign it a utility of 0:
U(minimum) 5 0
3. To determine the utility of another potential payoff M, the decision maker is offered the following two hypothetical alternatives:
A 1 : Obtain a payoff of maximum with probability p. Obtain a payoff of minimum with probability 1 2 p.
A 2 : Definitely obtain a payoff of M.
Question to the decision maker: What value of p makes you indifferent between these two alternatives? Then U ( M ) 5 p.
Constructing Max’s Utility Function for Money We now have found utilities for three possible payoffs (–130, 90, and 700) for Goferbroke. Plotting these values on a graph of the utility function U(M) versus the monetary payoff M and then drawing a smooth curve through these points gives the curve shown in Figure 9.22 .
This curve is an estimate of Max’s utility function for money. To find the utility val- ues for the other possible payoffs ( 2 100, 60, and 670), Max could repeat step 3 of the equivalent lottery method for M 5 2 100, M 5 60, and M 5 670. However, since 2 100 is so close to 2 130, 60 is so close to 90, and 670 is so close to 700, an alternative is to estimate these utilities as the values on the curve in Figure 9.22 at M 5 2 100, M 5 60, and M 5 670. Following the corresponding dotted lines in the figure leads to U ( 2 100) 5 0.05, U (60) 5 0.30, and U (670) 5 0.97. Table 9.9 gives the complete list of possible payoffs and their utilities.
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Monetary Payoff, M Utility, U(M)
2130 0 2100 0.05 60 0.30 90 0.333 670 0.97 700 1
TABLE 9.9 Utilities for the Goferbroke Co. Problem
For comparative purposes, the dashed line in Figure 9.22 shows the utility function that would result if Max were completely risk neutral. By nature, Max is inclined to be a risk seeker. However, the difficult financial circumstances of his company that he badly wants to keep solvent has forced him to adopt a moderately risk-averse stance in addressing his current decisions.
Using a Decision Tree to Analyze the Problem with Utilities Now that Max’s utility function for money has been constructed in Table 9.9 (and Figure 9.22 ), this information can be used with a decision tree as summarized next.
The procedure for using a decision tree to analyze the problem now is identical to that described in Section 9.7 except for substituting utilities for monetary payoffs. Therefore, the value obtained to evaluate each node of the tree now is the expected utility there rather than the expected (monetary) payoff. Consequently, the optimal decision selected by Bayes’ decision rule maximizes the expected utility for the overall problem.
Thus, using RSPE once again, our final decision tree with utilities shown in Figure 9.23 closely resembles the one in Figure 9.17 given in Section 9.7. The nodes and branches are exactly the same, as are the probabilities for the branches emanating from the event nodes. However, the key difference from Figure 9.17 is that the monetary payoff at each terminal node now has been replaced by the corresponding utility from Table 9.9 . (This was accom- plished with RSPE by entering this same utility as the “cash flow” at the terminal branch and then entering “cash flows” of 0 at all the preceding branches.) It is these utilities that have been used by RSPE to compute the expected utilities given next to all the nodes.
These expected utilities lead to the same decisions as in Figure 9.17 at all decision nodes except the bottom one in cell F41. The decision at this node now switches to sell instead of drill. However, the solution procedure still leaves this node on a closed path, as indicated by the 1 in cell B29. Therefore, the overall optimal policy remains the same as that obtained in Figure 9.17 (do the seismic survey; sell if the result is unfavorable; drill if the result is favorable).
At each terminal branch, enter the utility of that outcome as the “cash flow” there and then do not change the default value of 0 for the “cash flow” at the preceding branches.
U(M)
M
1.00
0.75
0.50
0.25
−100 100 200 300 400 500 600 700 Thousands of dollars
M ax
’s u
til ity
fu nc
tio n
R is
k- ne
ut ra
l u til
ity fu
nc tio
n
FIGURE 9.22 Max’s utility function for money as the owner of Goferbroke Co.
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A
Decision Tree for Goferbroke Co. Problem (with Max's Utility Function)
B C D E F G H I J K L M N O P Q R S
FIGURE 9.23 The final decision tree constructed and solved by RSPE for the full Goferbroke Co. problem when using Max’s utility function for money to maximize expected utility.
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The approach of maximizing the expected monetary payoff used in the preceding sections was equivalent to assuming that the decision maker is neutral toward risk. By using utility theory with an appropriate utility function, the optimal solution now reflects the decision maker’s attitude about risk. Because Max adopted only a moderately risk-averse stance, the optimal policy did not change from before. For a somewhat more risk-averse decision maker, the optimal solution would switch to the more conservative approach of immediately selling the land (no seismic survey).
Jennifer and Max are to be commended for incorporating utilities into a decision analysis approach to his problem. Utilities help to provide a rational approach to decision making in the face of uncertainty. However, many managers are not sufficiently comfortable with the relatively abstract notion of utilities, or with working with probabilities to construct a utility function, to be willing to use this approach. Consequently, utilities are not used nearly as widely in practice as some of the other techniques of decision analysis described in this chap- ter, including Bayes’ decision rule (with monetary payoffs) and decision trees.
Another Approach for Estimating U ( M ) The procedure described earlier for constructing U ( M ) asks the decision maker to repeatedly apply the equivalent lottery method, which requires him (or her) each time to make a difficult decision about which probability would make him indifferent between two alternatives. Many managers would be uncomfortable with making this kind of decision. Therefore, an alterna- tive approach is sometimes used instead to estimate the utility function for money.
This approach assumes that the utility function has a certain mathematical form and then adjusts this form to fit the decision maker’s attitude toward risk as closely as possible. For example, one particularly popular form to assume (because of its relative simplicity) is the exponential utility function ,
U(M ) 5 1 2 e2 M R
where R is the decision maker’s risk tolerance. This utility function has the kind of shape shown in Figure 9.21 ( a ), so it is designed to fit a risk-averse individual. A great aversion to risk corresponds to a small value of R (which would cause the curve in this figure to bend sharply), whereas a small aversion to risk corresponds to a large value of R (which gives a much more gradual bend in the curve).
A decision maker’s risk tolerance tends to vary over time as his (or her) wealth changes. He tends to have a higher risk tolerance when he is relatively wealthy than when he is not. How- ever, given his current degree of wealth, the exponential utility function assumes that he has a constant risk tolerance over the entire range of potential outcomes that may be realized after the decision is made. This often is a reasonable assumption. Unfortunately, it is a questionable assumption in Max’s situation because he is unusually risk averse about the worst possible outcome (a loss of $130,000) but has a very high risk tolerance when comparing large potential gains. This is why Jennifer never raised the possibility of using an exponential utility function.
In other situations where the consequences of the potential losses are not as severe, assuming an exponential utility function may provide a reasonable approximation. In such a case, here is an easy way of estimating the appropriate value of R. The decision maker would be asked to choose the number R that would make him indifferent between the following two alternatives.
A 1 : A 50–50 gamble where he would gain R dollars with probability 0.5 and lose R /2 dollars with probability 0.5.
A 2 : Neither gain nor lose anything.
For example, if the decision maker were indifferent between doing nothing or taking a 50–50 gamble where he would gain $1,000 with probability 0.5 and lose $500 with probability 0.5, then R 5 1,000.
Using RSPE with an Exponential Utility Function RSPE includes the option of using the exponential utility function. Clicking on the Options button on the RSPE ribbon and choosing Tree reveals the options shown in Figure 9.24 . The Certainty Equivalents section gives two choices—Expected Values or Exponential Utility
The previous approach of maximizing the expected monetary payoff assumes a risk-neutral decision maker.
Since R measures the deci- sion maker’s risk toler- ance, the aversion to risk decreases as R increases.
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Function. Choosing the latter revises the decision tree to incorporate the exponential util- ity function. The Decision Node EV/CE section has two choices, Maximize or Minimize, which enable you to specify whether the objective is to maximize the measure of performance (expected utility in this case) or to minimize that measure. (Maximize is the default choice, as has been used throughout the chapter.) The Risk Tolerance box is used to enter a value to be used for R when calculating the exponential utility function. (These same options are also available on the Platform tab of the Model pane in RSPE.)
To illustrate, suppose that the exponential utility function with a risk tolerance of R 5 1,000 were to be used as a rough approximation for analyzing the full Goferbroke Co. problem. (Since this problem expresses payoffs in units of thousands of dollars, R 5 1,000 here is equivalent to using R 5 1,000,000 when payoffs are in units of dollars.) The resulting deci- sion tree is shown in Figure 9.25 . There are now two values calculated below and to the left of each node. The lower number represents the expected utility value at that stage in the deci- sion tree. The upper number represents the certain payoff that is equivalent to this expected utility value. For example, cell A31 indicates that the expected value of the exponential utility
RSPE Tip: The Decision Tree section of the Platform tab on the Model pane lets you specify whether to use expected monetary values or the exponential utility func- tion for applying Bayes’ decision rule. The entry in Decision Node EV/CE (either Maximize or Mini- mize) also lets you specify whether the objective is to maximize the measure of performance (expected pay- off or expected utility) or to minimize that measure.
FIGURE 9.24 The Tree tab of the RSPE Options dialog box allows you to set several options for how the decision tree is solved. Here the options are set to use the expo- nential utility function, to maximize profit, and to use an R value of 1,000.
Following the merger of Conoco Inc. and the Phillips Petro- leum Company in 2002, ConocoPhillips became the third- largest integrated energy company in the United States with $160 billion in assets and 38,000 employees. Like any company in this industry, the management of Conoco- Phillips must grapple continually with decisions about the allocation of limited investment capital across a set of risky petroleum exploration projects. These decisions have a great impact on the profitability of the company.
In the early 1990s, the then Phillips Petroleum Company became an industry leader in the application of sophisti- cated management science methodology to aid these deci- sions by developing a decision analysis software package called DISCOVERY. The user interface allows a geologist or engineer to model the uncertainties associated with a project and then the software interprets the inputs and constructs a decision tree that shows all the decision nodes
(including opportunities to obtain additional seismic infor- mation) and the intervening event nodes. A key feature of the software is the use of an exponential utility function to incorporate management’s attitudes about financial risk. An intuitive questionnaire is used to measure corporate risk preferences in order to determine an appropriate value of the risk tolerance parameter for this utility function.
Management uses the software to (1) evaluate petro- leum exploration projects with a consistent risk-taking pol- icy across the company, (2) rank projects in terms of overall preference, (3) identify the firm’s appropriate level of par- ticipation in these projects, and (4) stay within budget.
Source: M. R. Walls, G. T. Morahan, and J. S. Dyer, “Decision Analysis of Exploration Opportunities in the Onshore U.S. at Phillips Petroleum Company,” Interfaces 25, no. 6 (November– December 1995), pp. 39–56. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
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1
2
Do Survey
No Survey
Unfavorable
Favorable
Drill
Sell
Drill
Sell
Drill
Sell
Dry
Oil
3
4
5
1
2
1
2
70%
30%
-100
-0.105
-100
0
48.1147
0.0470
192.047
0.175
0 60
0.0582
25%
75%
800 700
0.503
90 60
-100 192.047
0.175
90 60
0.0582
0.0582
-100
-0.0492
0
90
0.08607
0
Dry
Oil
-130
50%
50%
800 670
0
Dry
Oil
-130
-0.139
14.3%
85.7%
800 670
0.488
-0.139
0.488
-130
60
60
700
-100
90
90 90
0.0861
670
-130
670
0
6
7
9
8
10
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16
17
18
19 -30 97.8160
0.093220
21
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30 98
0.093231
32
33
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35
36
37
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45
46
47
Decision Tree for Goferbroke Co. (with an Exponential Utility Function)
A B C D E F G H I J K L M N O P Q R S
RiskTolerance E48
Range Name Cells
48 Risk Tolerance 1,000
-47.981
FIGURE 9.25 The final decision tree constructed and solved by RSPE for the full Goferbroke Co. problem when using an exponential utility func- tion with R 5 1,000.
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9.10 The Practical Application of Decision Analysis 365
function for this decision would be 0.0932. This expected utility is equivalent to a certain payoff of $98,000, as indicated in cell A30.
The exponential utility function leads to the same decisions as in Figure 9.23 . The overall optimal policy remains to do the seismic survey; sell if the result is unfavorable; drill if the result is favorable. However, the optimal policy changes when the value of R is decreased far enough. For values of R less than 728, the optimal policy switches to not doing the survey and selling the land. Thus, a more risk-averse decision maker would make the safer decision for Goferbroke—sell the land and receive $90,000 for sure.
9.10 THE PRACTICAL APPLICATION OF DECISION ANALYSIS
In one sense, the Goferbroke Co. problem is a very typical application of decision analysis. Like other applications, Max needed to make his decisions (Do a seismic survey? Drill for oil or sell the land?) in the face of great uncertainty. The decisions were difficult because their payoffs were so unpredictable. The outcome depended on factors that were outside Max’s control (does the land contain oil or is it dry?). He needed a framework and methodology for rational decision making in this uncertain environment. These are the usual characteristics of applications of decision analysis.
However, in other ways, the Goferbroke problem is not such a typical application. It was oversimplified to include only two possible states of nature (oil and dry), whereas there actu- ally would be a considerable number of distinct possibilities. For example, the actual state might be dry, a small amount of oil, a moderate amount, a large amount, and a huge amount, plus different possibilities concerning the depth of the oil and soil conditions that impact the cost of drilling to reach the oil. Max also was considering only two alternatives for each of two decisions. Real applications commonly involve more decisions, more alternatives to be considered for each one, and many possible states of nature.
Problems as tiny as the Goferbroke problem can be readily analyzed and solved by hand. However, real applications typically involve large decision trees, whose construction and analysis require the use of a software package (such as RSPE introduced in this chapter). In some cases, the decision tree can explode in size with many thousand terminal branches. Special algebraic tech- niques are being developed and incorporated into the solvers for dealing with such large problems.
Other kinds of graphical techniques also are available to complement the decision tree in representing and solving decision analysis problems. One that has become quite popular is called the influence diagram , which provides another helpful way of showing the inter- relationships among a decision maker’s alternatives, uncertainties, and values.
Although the Goferbroke problem only involved a single decision maker (Max) assisted by a single analyst (Jennifer), many strategic business decisions are made collectively by management. One technique for group decision making is called decision conferencing . This is a process where the group comes together for discussions in a decision conference with the help of an analyst and a group facilitator. The facilitator works directly with the group to help it structure and focus discussions, think creatively about the problem, bring assumptions to the surface, and address the full range of issues involved. The analyst uses decision analysis to assist the group in exploring the implications of the various decision alternatives. With the assistance of a computerized group decision support system, the analyst
The Goferbroke Co. prob- lem could have included many more states of nature.
1. What are utilities intended to reflect? 2. What is the shape of the utility function for money for a risk-averse individual? A risk-seeking
individual? A risk-neutral individual? 3. What is the fundamental property of utility functions? 4. What is the lottery when using the equivalent lottery method? 5. Given two hypothetical alternatives where one of them involves a probability p, what is meant
by the point of indifference between these two alternatives? 6. When using utilities with a decision tree, what kind of value is obtained to evaluate each node
of the tree? 7. What decisions did Max make regarding the full Goferbroke Co. problem?
Review Questions
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builds and solves models on the spot, and then performs sensitivity analysis to respond to what-if questions from the group.
Applications of decision analysis commonly involve a partnership between the manage- rial decision maker (whether an individual or a group) and an analyst (whether an individual or a team) with training in management science. Some managers are not as fortunate as Max in having a staff member (let alone a daughter) like Jennifer who is qualified to serve as the analyst. Therefore, a considerable number of management consulting firms specializing in decision analysis have been formed to fill this role.
If you would like to do more reading about the practical application of decision analysis, a good place to begin would be the leadoff article 2 in the first issue of the journal Decision Analysis that was founded in 2004 to focus on applied research in decision analysis. This leadoff article provides a detailed discussion of various publications that present applications of decision analysis.
2 D. L. Keefer, C. W. Kirkwood, and J. L. Corner, “Perspective on Decision Analysis Applications,” Decision Analysis 1 (2004), pp. 4–22.
1. How does the Goferbroke Co. problem compare with typical applications of decision analysis? 2. What is the purpose of an influence diagram? 3. Who are the typical participants in a decision-conferencing process? 4. Where can a manager go for expert help in applying decision analysis if a qualified analyst is
not available on staff?
Review Questions
Decision analysis is a valuable technique for decision making in the face of great uncertainty. It pro- vides a framework and methodology for rational decision making when the outcomes are uncertain.
In a typical application, a decision maker needs to make either a single decision or a short sequence of decisions (with additional information perhaps becoming available between decisions). A number of alternatives are available for each decision. Uncontrollable random factors affect the payoff that would be obtained from a decision alternative. The possible outcomes of the random factors are referred to as the possible states of nature.
Which state of nature actually occurs will be learned only after making the decisions. However, prior to the decisions, it often is possible to estimate prior probabilities of the respective states of nature.
Various alternative decision criteria are available for making the decisions. A particularly popular one is Bayes’ decision rule, which uses the prior probabilities to determine the expected payoff for each deci- sion alternative and then chooses the one with the largest expected payoff. This is the criterion (accom- panied by sensitivity analysis) that is mostly used in practice, so it is the focus of much of the chapter.
Sensitivity analysis is very helpful for evaluating the effect of having inaccurate estimates of the data for the problem, including the probabilities, revenues, and costs. Data tables can be used to systemati- cally vary the data and see how it affects the optimal decisions or expected payoffs.
It sometimes is possible to pay for a test or survey to obtain additional information about the prob- abilities of the various states of nature. Calculating the expected value of perfect information provides a quick way of checking whether doing this might be worthwhile.
When more information is obtained, the updated probabilities are referred to as posterior probabili- ties. A probability tree diagram is helpful for calculating these new probabilities.
For problems involving a sequence of decisions (including perhaps a decision on whether to obtain more information), a decision tree commonly is used to graphically display the progression of decisions and random events. The calculations for applying Bayes’ decision rule then can be performed directly on the decision tree one event node or decision node at a time. Spreadsheet packages, such as RSPE, are very helpful for constructing and solving decision trees.
When the problem involves the possibility of uncomfortably large losses, utilities provide a way of incorporating the decision maker’s attitude toward risk into the analysis. Bayes’ decision rule then is applied by expressing payoffs in terms of utilities rather than monetary values.
Decision analysis is widely used. Versatile software packages for personal computers have become an integral part of the practical application of decision analysis.
9.11 Summary
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Chapter 9 Glossary 367
Glossary alternatives The options available to the deci- sion maker for the decision under consideration. (Section 9.1), 324 Bayes’ decision rule A popular criterion for decision making that uses probabilities to calcu- late the expected payoff for each decision alter- native and then chooses the one with the largest expected payoff. (Section 9.2), 328 Bayes’ theorem A formula for calculating a posterior probability of a state of nature. (Section 9.6), 343 branch A line emanating from a node in a deci- sion tree. (Section 9.3), 330 decision conferencing A process used for group decision making. (Section 9.10), 365 decision maker The individual or group responsible for making the decision under consid- eration. (Section 9.1), 324 decision node A point in a decision tree where a decision needs to be made. (Section 9.3), 330 decision tree A graphical display of the pro- gression of decisions and random events to be considered. (Sections 9.3 and 9.7), 330, 345 equivalent lottery method The procedure for finding the decision maker’s utility for a specific amount of money by comparing two hypothetical alternatives where one involves a gamble. (Section 9.9), 359 event node A point in a decision tree where a random event will occur. (Section 9.3), 330 expected monetary value (EMV) criterion An alternative name for Bayes’ decision rule when the payoffs have monetary values. (Section 9.2), 329 expected payoff (EP) For a decision alterna- tive, it is the weighted average of the payoffs, using the probabilities of the states of nature as the weights. (Section 9.2), 328 expected value of perfect information (EVPI) The increase in the expected payoff that could be obtained if it were possible to learn the true state of nature before making the deci- sion. (Sections 9.4 and 9.5), 337, 338 expected value of sample information (EVSI) The increase in the expected payoff that could be obtained by performing a test to obtain more information, excluding the cost of the test. (Section 9.7), 350 exponential utility function A utility func- tion designed to fit some risk-averse individuals. (Section 9.9), 362 influence diagram A diagram that comple- ments the decision tree for representing and analyzing decision analysis problems. (Section 9.10), 365 maximax criterion A very optimistic decision criterion that does not use prior probabilities and
simply chooses the decision alternative that could give the largest possible payoff. (Section 9.2), 326 maximin criterion A very pessimistic decision criterion that does not use prior probabilities and simply chooses the decision alternative that pro- vides the best guarantee for its minimum possible payoff. (Section 9.2), 326 maximum likelihood criterion A criterion for decision making with probabilities that focuses on the most likely state of nature. (Section 9.2), 327 node A junction point in a decision tree. (Section 9.3), 330 payoff A quantitative measure of the outcome from a decision alternative and a state of nature. (Section 9.1), 325 payoff table A table giving the payoff for each combination of a decision alternative and a state of nature. (Section 9.1), 325 point of indifference The point where the decision maker is indifferent between the two hypothetical alternatives in the equivalent lottery method. (Section 9.9), 358 posterior probabilities Revised probabilities of the states of nature after doing a test or survey to improve the prior probabilities. (Sections 9.5 and 9.6), 340 prior probabilities The estimated probabilities of the states of nature prior to obtaining addi- tional information through a test or survey. (Section 9.1), 325 probability tree diagram A diagram that is helpful for calculating the posterior probabilities of the states of nature. (Section 9.6), 341 risk-averse individual An individual whose util- ity function for money has a decreasing slope as the amount of money increases.(Section 9.9), 355 risk-neutral individual An individual whose utility for money is proportional to the amount of money involved. (Section 9.9), 355 risk seeker An individual whose utility func- tion for money has an increasing slope as the amount of money increases. (Section 9.9), 355 sensitivity analysis The study of how other plausible values for the probabilities of the states of nature (or for the payoffs) would affect the recommended decision alternative. (Sections 9.4 and 9.8), 333, 351 states of nature The possible outcomes of the random factors that affect the payoff that would be obtained from a decision alternative. (Section 9.1), 324 utility The utility of an outcome measures the instrinsic value to the decision maker of that out- come. (Sections 9.1 and 9.9), 324 utility function for money, U ( M ) A plot of utility versus the amount of money M being received. (Section 9.9), 355
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Chapter 9 Excel Files:
Bayes’ Decision Rule for First Goferbroke Problem
Decision Tree for First Goferbroke Problem
Data Table for First Goferbroke Problem
EP with Perfect Info for First Goferbroke Problem
Decision Tree for EVPI for First Goferbroke Problem
Template for Posterior Probabilities
Decision Tree for Full Goferbroke Problem (with Data Table)
Decision Tree for Full Goferbroke Problem with Max’s Utility Function
Decision Tree for Full Goferbroke Problem with Exponential Utility Function
Excel Add-in:
Risk Solver Platform for Education (RSPE)
Supplement to This Chapter on the CD-ROM:
Decision Criteria Using TreePlan Software for Decision Trees
“Ch. 9 Supplement” Excel Files:
Template for Maximax Criterion
Template for Maximin Criterion
Template for Realism Criterion
Template for Minimax Regret Criterion
Template for Maximum Likelihood Criterion
Template for Equally Likely Criterion
Learning Aids for This Chapter in Your MS Courseware
Solved Problems (See the CD-ROM or Website for the Solutions) 9.S1. New Vehicle Introduction The General Ford Motors Corporation (GFMC) is planning the introduction of a brand new SUV—the Vector. There are two options for production. One is to build the Vector at the com- pany’s existing plant in Indiana, sharing production time with its line of minivans that are currently being produced there. If sales of the Vector are just moderate, this will work out well as there is sufficient capacity to produce both types of vehicles at the same plant. However, if sales of the Vector are strong, this option would require the operation of a third shift, which would lead to significantly higher costs.
A second option is to open a new plant in Georgia. This plant would have sufficient capacity to meet even the largest projec- tions for sales of the Vector. However, if sales are only moder- ate, the plant would be underutilized and therefore less efficient.
This is a new design, so sales are hard to predict. However, GFMC predicts that there would be about a 60 percent chance of strong sales (annual sales of 100,000), and a 40 percent chance of moderate sales (annual sales of 50,000). The average revenue per Vector sold is $30,000. Production costs per vehicle for the two pro- duction options depend upon sales, as indicated in the table below.
Expected Production Cost per Vehicle for the Vector ($thousands)
Moderate Sales
Strong Sales
Shared plant in Indiana 16 24
Dedicated plant in Georgia 22 20
The amortized annual cost of plant construction and other associated fixed costs for the Georgia plant would total $400 million per year (regardless of sales volume). The fixed costs for adding Vector production to the plant in Indiana would total $200 million per year (regardless of sales volume).
a. Construct a decision tree to determine which production op- tion maximizes the expected annual profit, considering fixed costs, production costs, and sales revenues.
b. Because of the uncertainty in expected sales for the Vec- tor, GFMC is considering conducting a marketing survey to determine customer attitudes toward the Vector to bet- ter predict the likelihood of strong sales. The marketing survey would give one of two results—a positive attitude or a negative attitude toward the design. GFMC has used this marketing survey for other vehicles. For vehicles that eventually had strong sales, the marketing survey indicated positive attitudes toward the design 70 percent of the time and negative attitudes 30 percent of the time. For vehicles that eventually had moderate sales, the marketing survey indicated positive attitudes toward the design 20 percent of the time and negative attitudes 80 percent of the time. As- suming GFMC conducts such a survey, construct a decision tree to to determine how the company should proceed and what the expected annual profit would be (ignoring the cost of the survey).
c. What is the expected value of sample information in part b? What does this say about how large the cost of the marketing survey can be before it would no longer be worthwhile to conduct the survey?
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Chapter 9 Problems 369
9.S2. Settle or Go to Trial Meredith Delgado owns a small firm that has developed soft- ware for organizing and playing music on a computer. Her soft- ware contains a number of unique features that she has patented, so her company’s future has looked bright.
However, there now has been an ominous development. It appears that a number of her patented features were copied in similar software developed by MusicMan Software, a huge soft- ware company with annual sales revenue in excess of $1 billion. Meredith is distressed. MusicMan Software has stolen her ideas and that company’s marketing power is likely to enable it to cap- ture the market and drive Meredith out of business.
In response, Meredith has sued MusicMan Software for pat- ent infringement. With attorney fees and other expenses, the cost of going to trial (win or lose) is expected to be $1 million. She feels that she has a 60 percent chance of winning the case, in which case she would receive $5 million in damages. If she
loses the case, she gets nothing. Moreover, if she loses the case, there is a 50 percent chance that the judge would also order Meredith to pay for court expenses and lawyer fees for Music- Man (an additional $1 million cost). MusicMan Software has offered Meredith $1.5 million to settle this case out of court.
a. Construct and use a decision tree to determine whether Meredith should go to court or accept the settlement offer, assuming she wants to maximize her expected payoff.
b. To implement the equivalent lottery method to determine appropriate utility values for all the possible payoffs in this problem, what questions would need to be asked of Meredith?
c. Suppose that Meredith’s attitude toward risk is such that she would be indifferent between doing nothing and a gamble where she would win $1 million with 50 percent probability and lose $500 thousand with 50 percent probability. Use the expo- nential utility function to re-solve the decision tree from part a.
wants. However, because these strawberries already are very ripe, she will need to sell them tomorrow and then discard any that remain unsold. Jean estimates that she will be able to sell 10, 11, 12, or 13 cases tomorrow. She can purchase the strawberries for $3 per case and sell them for $8 per case. Jean now needs to decide how many cases to purchase.
Jean has checked the store’s records on daily sales of straw- berries. On this basis, she estimates that the prior probabilities are 0.2, 0.4, 0.3, and 0.1 for being able to sell 10, 11, 12, and 13 cases of strawberries tomorrow. a. Develop a decision analysis formulation of this
problem by identifying the decision alternatives, the states of nature, and the payoff table.
b. If Jean is dubious about the accuracy of these prior probabilities and so chooses to ignore them and use the maximax criterion, how many cases of straw- berries should she purchase?
c. How many cases should be purchased if she uses the maximin criterion?
d. How many cases should be purchased if she uses the maximum likelihood criterion?
e. How many cases should be purchased according to Bayes’ decision rule?
f. Jean thinks she has the prior probabilities just about right for selling 10 cases and selling 13 cases, but is uncertain about how to split the prior probabilities for 11 cases and 12 cases. Reapply Bayes’ decision rule when the prior probabilities of 11 and 12 cases are ( i ) 0.2 and 0.5, ( ii ) 0.3 and 0.4, and ( iii ) 0.5 and 0.2.
9.4.* Warren Buffy is an enormously wealthy investor who has built his fortune through his legendary investing acumen. He currently has been offered three major investments and he would like to choose one. The first one is a conservative investment that would perform very well in an improving economy and only suffer a small loss in a worsening economy. The second is a speculative investment that would perform extremely well in an improving economy but would do very badly in a worsening
To the left of the following problems (or their parts), we have inserted the symbol R whenever RSPE can be used. The symbol T indicates that the Excel template for posterior probabilities can be helpful. Nearly all the problems can be conveniently formu- lated in a spreadsheet format, so no special symbol is used to designate this. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
9.1. You are given the following payoff table (in units of thousands of dollars) for a decision analysis problem without probabilities.
State of Nature
Alternative S1 S2 S3
A1 6 2 4 A2 3 4 3 A3 8 1 5
a. Which alternative should be chosen under the maxi- max criterion?
b. Which alternative should be chosen under the maxi- min criterion?
9.2. Follow the instructions of Problem 9.1 with the follow- ing payoff table.
State of Nature
Alternative S1 S2 S3 S4
A1 25 30 20 24 A2 17 14 31 21 A3 22 22 22 22 A4 29 21 26 27
9.3. Jean Clark is the manager of the Midtown Saveway Grocery Store. She now needs to replenish her supply of straw- berries. Her regular supplier can provide as many cases as she
Problems
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the prior probability of an improving economy fixed at 0.1). Plot these expected profits on a single graph that has expected profit as the vertical axis and the prior probability of a stable economy as the horizontal axis. For each of the three investments, draw a line segment connecting its two points on this graph to show how its expected profit would vary with the prior probability of a stable econ- omy. Use this graph to describe how the choice of the investment depends on the prior probability of a stable economy.
economy. The third is a countercyclical investment that would lose some money in an improving economy but would perform well in a worsening economy.
Warren believes that there are three possible scenarios over the lives of these potential investments: (1) an improving econ- omy, (2) a stable economy, and (3) a worsening economy. He is pessimistic about where the economy is headed, and so has assigned prior probabilities of 0.1, 0.5, and 0.4, respectively, to these three scenarios. He also estimates that his profits under these respective scenarios are those given by the following table.
Improving Economy
Stable Economy
Worsening Economy
Conservative investment $ 30 million $ 5 million $210 million Speculative investment 40 million 10 million 230 million Countercyclical investment 210 million 0 15 million
Prior probability 0.1 0.5 0.4
Which investment should Warren make under each of the following criteria? a. Maximax criterion. b. Maximin criterion. c. Maximum likelihood criterion. d. Bayes’ decision rule. 9.5. Reconsider Problem 9.4. Warren Buffy decides that Bayes’ decision rule is his most reliable decision criterion. He believes that 0.1 is just about right as the prior probability of an improving economy, but is quite uncertain about how to split the remaining probabilities between a stable economy and a worsen- ing economy. Therefore, he now wishes to do sensitivity analy- sis with respect to these latter two prior probabilities. a. Reapply Bayes’ decision rule when the prior prob-
ability of a stable economy is 0.3 and the prior prob- ability of a worsening economy is 0.6.
b. Reapply Bayes’ decision rule when the prior prob- ability of a stable economy is 0.7 and the prior prob- ability of a worsening economy is 0.2.
c. Construct a decision tree by hand for this problem with the original prior probabilities.
R d. Use RSPE to construct and solve a decision tree for this problem with the original prior probabilities.
R e. In preparation for performing sensitivity analysis, consolidate the data and results on the same spread- sheet as the decision tree constructed in part d (as was done in Figure 9.7 for the case study).
R f. Use the spreadsheet (including the decision tree) obtained in parts d and e to do parts a and b.
R g. Expanding the spreadsheet as needed, generate a data table that shows which investment Warren should make and the resulting expected profit for the following prior probabilities of a stable econ- omy: 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9.
h. For each of the three investments, find the expected profit when the prior probability of a stable economy is 0 and then when it is 0.9 (with
9.6. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 9.3. Briefly describe how decision analy- sis was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 9.7.* Consider a decision analysis problem whose payoffs (in units of thousands of dollars) are given by the following payoff table.
State of Nature
Alternative S1 S2
A1 80 25 A2 30 50 A3 60 40
Prior probability 0.4 0.6
a. Which alternative should be chosen under the maxi- max criterion?
b. Which alternative should be chosen under the maxi- min criterion?
c. Which alternative should be chosen under the maxi- mum likelihood criterion?
d. Which alternative should be chosen under Bayes’ decision rule?
R e. Use RSPE to construct and solve a decision tree for this problem.
R f. Expanding the spreadsheet containing this decision tree as needed, perform sensitivity analysis with the decision tree by re-solving when the prior probabil- ity of S 1 is 0.2 and again when it is 0.6.
R g. Now perform this sensitivity analysis systematically by generating a data table that shows the best alter- native (according to Bayes’ decision rule) and the resulting expected payoff as the prior probability of S 1 increases in increments of 0.04 from 0.2 to 0.6.
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After referring to historical meteorological records, Dwight also has estimated the following prior probabilities for the weather during the growing season:
Dry 0.3
Moderate 0.5
Damp 0.2
a. Develop a decision analysis formulation of this problem by identifying the decision alternatives, the states of nature, and the payoff table.
R b. Construct a decision tree for this problem and use Bayes’ decision rule to determine which crop to grow.
R c. Using Bayes’ decision rule, do sensitivity analysis with respect to the prior probabilities of moder- ate weather and damp weather (without changing the prior probability of dry weather) by re-solving when the prior probability of moderate weather is 0.2, 0.3, 0.4, and 0.6.
9.10. Barbara Miller makes decisions according to Bayes’ decision rule. For her current problem, Barbara has constructed the following payoff table (in units of hundreds of dollars) and she now wishes to maximize the expected payoff.
State of Nature
Alternative S1 S2 S3
A1 2x 50 10 A2 25 40 90 A3 35 3x 30
Prior probability 0.4 0.2 0.4
The value of x currently is 50, but there is an opportunity to increase x by spending some money now.
What is the maximum amount Barbara should spend to increase x to 75? 9.11. You are given the following payoff table (in units of thousands of dollars) for a decision analysis problem.
State of Nature
Alternative S1 S2 S3
A1 4 0 0 A2 0 2 0 A3 3 0 1
Prior probability 0.2 0.5 0.3
9.8. You are given the following payoff table (in units of thousands of dollars) for a decision analysis problem.
State of Nature
Alternative S1 S2 S3
A1 220 170 110 A2 200 180 150
Prior probability 0.6 0.3 0.1
a. Which alternative should be chosen under the maxi- max criterion?
b. Which alternative should be chosen under the maxi- min criterion?
c. Which alternative should be chosen under the maxi- mum likelihood criterion?
d. Which alternative should be chosen under Bayes’ decision rule?
e. Construct a decision tree by hand for this problem.
R f. Use RSPE to construct and solve a decision tree for this problem.
R g. Perform sensitivity analysis with this decision tree by generating a data table that shows what hap- pens when the prior probability of S 1 increases in increments of 0.05 from 0.3 to 0.7 while the prior probability of S 3 remains fixed at its original value. Then use trial and error to estimate the value of the prior probability of S 1 at which the best alternative changes as this prior probability increases.
R h. Repeat part g when it is the prior probability of S 2 that remains fixed at its original value.
R i. Repeat part g when it is the prior probability of S 1 that remains fixed at its original value while the prior probability of S 2 increases in increments of 0.05 from 0 to 0.4.
j. If you feel that the true probabilities of the states of nature should be within 10 percent of the given prior probabilities, which alternative would you choose?
9.9. Dwight Moody is the manager of a large farm with 1,000 acres of arable land. For greater efficiency, Dwight always devotes the farm to growing one crop at a time. He now needs to make a decision on which one of four crops to grow during the upcoming growing season. For each of these crops, Dwight has obtained the following estimates of crop yields and net incomes per bushel under various weather conditions.
Expected Yield, Bushels/Acre
Weather Crop 1 Crop 2 Crop 3 Crop 4
Dry 20 15 30 40 Moderate 35 20 25 40 Damp 40 30 25 40
Net income per bushel $1.00 $1.50 $1.00 $0.50
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a. According to Bayes’ decision rule, which alterna- tive should be chosen?
b. Find the expected value of perfect information. R c. Check your answer in part b by recalculating it with
the help of a decision tree. d. You are given the opportunity to spend $1,000 to
obtain more information about which state of nature is likely to occur. Given your answer to part b, might it be worthwhile to spend this money?
9.12.* Betsy Pitzer makes decisions according to Bayes’ deci- sion rule. For her current problem, Betsy has constructed the fol- lowing payoff table (in units of dollars).
State of Nature
Alternative S1 S2 S3
A1 50 100 2100 A2 0 10 210 A3 20 40 240
Prior probability 0.5 0.3 0.2
a. Which alternative should Betsy choose? b. Find the expected value of perfect information. R c. Check your answer in part b by recalculating it with
the help of a decision tree. d. What is the most that Betsy should consider pay-
ing to obtain more information about which state of nature will occur?
9.13. Using Bayes’ decision rule, consider the decision anal- ysis problem having the following payoff table (in units of thou- sands of dollars).
State of Nature
Alternative S1 S2 S3
A1 2100 10 100 A2 210 20 50 A3 10 10 60
Prior probability 0.2 0.3 0.5
a. Which alternative should be chosen? What is the resulting expected payoff?
b. You are offered the opportunity to obtain infor- mation that will tell you with certainty whether the first state of nature S 1 will occur. What is the maximum amount you should pay for the informa- tion? Assuming you will obtain the information, how should this information be used to choose an alternative? What is the resulting expected payoff (excluding the payment)?
c. Now repeat part b if the information offered con- cerns S 2 instead of S 1 .
d. Now repeat part b if the information offered con- cerns S 3 instead of S 1 .
R e. Now suppose that the opportunity is offered to pro- vide information that will tell you with certainty which state of nature will occur (perfect informa- tion). What is the maximum amount you should
pay for the information? Assuming you will obtain the information, how should this information be used to choose an alternative? What is the resulting expected payoff (excluding the payment)?
f. If you have the opportunity to do some testing that will give you partial additional information (not perfect information) about the state of nature, what is the maximum amount you should consider pay- ing for this information?
9.14. Reconsider the Goferbroke Co. case study, including its analysis in Sections 9.6 and 9.7. With the help of the consulting geologist, Jennifer Flyer now has obtained some historical data that provides more precise information than Max could supply on the likelihood of obtaining favorable seismic soundings on similar tracts of land. Specifically, when the land contains oil, favorable seismic soundings are obtained 80 percent of the time. This percentage changes to 40 percent when the land is dry. a. Revise Figure 9.12 to find the new posterior
probabilities. T b. Use the corresponding Excel template to check your
answers in part a. c. Revise Figure 9.16 to find the new decision tree.
What is the resulting optimal policy? R d. Use RSPE to construct and solve this new decision
tree. 9.15. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 9.7. Briefly describe how decision analy- sis was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 9.16.* Vincent Cuomo is the credit manager for the Fine Fab- rics Mill. He is currently faced with the question of whether to extend $100,000 of credit to a potential new customer, a dress manufacturer. Vincent has three categories for the creditworthi- ness of a company—poor risk, average risk, and good risk—but he does not know which category fits this potential customer. Experience indicates that 20 percent of companies similar to this dress manufacturer are poor risks, 50 percent are average risks, and 30 percent are good risks. If credit is extended, the expected profit for poor risks is 2 $15,000, for average risks $10,000, and for good risks $20,000. If credit is not extended, the dress manufacturer will turn to another mill. Vincent is able to con- sult a credit-rating organization for a fee of $5,000 per company evaluated. For companies whose actual credit records with the mill turn out to fall into each of the three categories, the follow- ing table shows the percentages that were given each of the three possible credit evaluations by the credit-rating organization.
Actual Credit Record
Credit Evaluation Poor Average Good
Poor 50% 40% 20% Average 40 50 40 Good 10 10 40
a. Develop a decision analysis formulation of this problem by identifying the decision alternatives, the states of nature, and the payoff table when the credit-rating organization is not used.
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T f. Use the corresponding Excel template to obtain the answers for part e.
g. Given that the research predicts S 1 , use Bayes’ deci- sion rule to determine which decision alternative should be chosen and the resulting expected payoff.
h. Repeat part g when the research predicts S 2 .
i. Given that research is done, what is the expected payoff when using Bayes’ decision rule?
j. Use the preceding results to determine the optimal policy regarding whether to do the research and the choice of the decision alternative.
R k. Construct and solve the decision tree to show the anal- ysis for the entire problem. (Using RSPE is optional.)
9.18. An athletic league does drug testing of its athletes, 10 percent of whom use drugs. The test, however, is only 95 percent reliable. That is, a drug user will test positive with probability 0.95 and negative with probability 0.05, and a non- user will test negative with probability 0.95 and positive with probability 0.05.
Develop a probability tree diagram to determine the posterior probability of each of the following outcomes of testing an athlete. a. The athlete is a drug user, given that the test is
positive. b. The athlete is not a drug user, given that the test is
positive. c. The athlete is a drug user, given that the test is
negative. d. The athlete is not a drug user, given that the test is
negative. T e. Use the corresponding Excel template to check your
answers in the preceding parts.
9.19. Management of the Telemore Company is consider- ing developing and marketing a new product. It is estimated to be twice as likely that the product would prove to be success- ful as unsuccessful. If it were successful, the expected profit would be $1,500,000. If unsuccessful, the expected loss would be $1,800,000. A marketing survey can be conducted at a cost of $100,000 to predict whether the product would be success- ful. Past experience with such surveys indicates that successful products have been predicted to be successful 80 percent of the time, whereas unsuccessful products have been predicted to be unsuccessful 70 percent of the time. a. Develop a decision analysis formulation of this
problem by identifying the decision alternatives, the states of nature, and the payoff table when the mar- ket survey is not conducted.
b. Assuming the market survey is not conducted, use Bayes’ decision rule to determine which decision alternative should be chosen.
c. Find the expected value of perfect information. Does this answer indicate that consideration should be given to conducting the market survey?
T d. Assume now that the market survey is conducted. Find the posterior probabilities of the respective states of nature for each of the two possible predic- tions from the market survey.
R e. Use RSPE to construct and solve the decision tree for this entire problem.
b. Assuming the credit-rating organization is not used, use Bayes’ decision rule to determine which deci- sion alternative should be chosen.
c. Find the expected value of perfect information. Does this answer indicate that consideration should be given to using the credit-rating organization?
d. Assume now that the credit-rating organization is used. Develop a probability tree diagram to find the posterior probabilities of the respective states of nature for each of the three possible credit evalua- tions of this potential customer.
T e. Use the corresponding Excel template to obtain the answers for part d.
f. Draw the decision tree for this entire problem by hand. Use this decision tree to determine Vincent’s optimal policy.
R g. Use RSPE to construct and solve this decision tree. R h. Find the expected value of sample information. If
the fee for using the credit-rating organization is open to negotiation, how large can the fee be and use of this organization still be worthwhile?
9.17. You are given the following payoff table (in units of dollars).
State of Nature
Alternative S1 S2
A1 400 2100 A1 0 100
Prior probability 0.4 0.6
You have the option of paying $100 to have research done to better predict which state of nature will occur. When the true state of nature is S 1 , the research will accurately predict S 1 60 percent of the time (but will inaccurately predict S 2 40 percent of the time). When the true state of nature is S 2 , the research will accurately predict S 2 80 percent of the time (but will inaccurately predict S 1 20 percent of the time). a. Given that the research is not done, use Bayes’ deci-
sion rule to determine which decision alternative should be chosen.
R b. Use a decision tree to help find the expected value of perfect information. Does this answer indicate that it might be worthwhile to do the research?
c. Given that the research is done, find the joint prob- ability of each of the following pairs of outcomes: ( i ) the state of nature is S 1 and the research predicts S 1 , ( ii ) the state of nature is S 1 and the research pre- dicts S 2 , ( iii ) the state of nature is S 2 and the research predicts S 1 , and ( iv ) the state of nature is S 2 and the research predicts S 2 .
d. Find the unconditional probability that the research predicts S 1 . Also find the unconditional probability that the research predicts S 2 .
e. Given that the research is done, use your answers in parts c and d to determine the posterior probabilities of the states of nature for each of the two possible predictions of the research.
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this decision tree. According to this analysis, which decision alternative should be chosen?
9.22.* Reconsider Problem 9.21. Management of Silicon Dynamics now is considering doing full-fledged market research at an estimated cost of $1 million to predict which of the two levels of demand is likely to occur. Previous experience indi- cates that such market research is correct two-thirds of the time. a. Find the expected value of perfect information for
this problem. b. Does the answer in part a indicate that it might be
worthwhile to perform this market research? c. Develop a probability tree diagram to obtain the
posterior probabilities of the two levels of demand for each of the two possible outcomes of the market research.
T d. Use the corresponding Excel template to check your answers in part c.
R9.23.* Reconsider Problem 9.22. The management of Silicon Dynamics now wants to see a decision tree displaying the entire problem. a. Use RSPE to construct and solve this decision tree. b. Find the expected value of sample information.
How large can the cost of doing full-fledged market research be and still be worthwhile?
c. Assume now that the estimate of $1 million for the cost of doing full-fledged market research is correct but that there is some uncertainty in the financial data ($15 million, $6 million, and $600) stated in Problem 9.20. Each could vary from its base value by as much as 10 percent. For each one, perform sensitivity analysis to find what would happen if its value were at either end of this range of variabil- ity (without any change in the other two pieces of data). Then do the same for the eight cases where all these pieces of data are at one end or the other of their ranges of variability.
9.24. You are given the following decision tree, where the numbers in parentheses are probabilities and the numbers on the right are payoffs at these terminal points.
2,500
–700
900
800
750
(0.8)
(0.6)
(0.2)
(0.4)
9.20. The Hit-and-Miss Manufacturing Company produces items that have a probability p of being defective. These items are produced in lots of 150. Past experience indicates that p for an entire lot is either 0.05 or 0.25. Furthermore, in 80 percent of the lots produced, p equals 0.05 (so p equals 0.25 in 20 percent of the lots). These items are then used in an assembly, and ulti- mately their quality is determined before the final assembly leaves the plant. Initially the company can either screen each item in a lot at a cost of $10 per item and replace defective items or use the items directly without screening. If the latter action is chosen, the cost of rework is ultimately $100 per defective item. Because screening requires scheduling of inspectors and equip- ment, the decision to screen or not screen must be made two days before the screening is to take place. However, one item can be taken from the lot and sent to a laboratory for inspection, and its quality (defective or nondefective) can be reported before the screen/no-screen decision must be made. The cost of this ini- tial inspection is $125.
a. Develop a decision analysis formulation of this problem by identifying the decision alternatives, the states of nature, and the payoff table if the single item is not inspected in advance.
b. Assuming the single item is not inspected in advance, use Bayes’ decision rule to determine which decision alternative should be chosen.
c. Find the expected value of perfect information. Does this answer indicate that consideration should be given to inspecting the single item in advance?
T d. Assume now that the single item is inspected in advance. Find the posterior probabilities of the respective states of nature for each of the two pos- sible outcomes of this inspection.
R e. Construct and solve the decision tree for this entire problem.
R f. Find the expected value of sample information. If the cost of using the laboratory to inspect the single item in advance is open to negotiation, how large can the cost of using the laboratory be and still be worthwhile?
9.21.* Silicon Dynamics has developed a new computer chip that will enable it to begin producing and marketing a personal computer if it so desires. Alternatively, it can sell the rights to the computer chip for $15 million. If the company chooses to build computers, the profitability of the venture depends on the company’s ability to market the computer during the first year. It has sufficient access to retail outlets that it can guarantee sales of 10,000 computers. On the other hand, if this computer catches on, the company can sell 100,000 machines. For analysis pur- poses, these two levels of sales are taken to be the two possible outcomes of marketing the computer, but it is unclear what their prior probabilities are. The cost of setting up the assembly line is $6 million. The difference between the selling price and the variable cost of each computer is $600. a. Develop a decision analysis formulation of this
problem by identifying the decision alternatives, the states of nature, and the payoff table.
b. Construct a decision tree for this problem by hand. R c. Assuming the prior probabilities of the two levels of
sales are both 0.5, use RSPE to construct and solve
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If William predicts a losing season instead, what is the posterior probability of a winning season? Of a losing season? Show how these answers are obtained from a probability tree diagram.
T e. Use the corresponding Excel template to obtain the answers requested in part d.
f. Draw the decision tree for this entire problem by hand. Analyze this decision tree to determine the optimal policy regarding whether to hire William and whether to undertake the campaign.
R g. Use RSPE to construct and solve this decision tree. R h. Find the expected value of sample information. If
the fee for hiring William Walsh is open to nego- tiation, how large can William’s fee be and still be worthwhile?
9.27. The comptroller of the Macrosoft Corporation has $100 million of excess funds to invest. She has been instructed to invest the entire amount for one year in either stocks or bonds (but not both) and then to reinvest the entire fund in either stocks or bonds (but not both) for one year more. The objective is to maximize the expected monetary value of the fund at the end of the second year.
The annual rates of return on these investments depend on the economic environment, as shown in the following table.
Rate of Return
Economic Environment Stocks Bonds
Growth 20% 5% Recession 210 10 Depression 250 20
The probabilities of growth, recession, and depression for the first year are 0.7, 0.3, and 0, respectively. If growth occurs in the first year, these probabilities remain the same for the sec- ond year. However, if a recession occurs in the first year, these probabilities change to 0.2, 0.7, and 0.1, respectively, for the second year. a. Construct by hand the decision tree for this problem
and then analyze the decision tree to identify the optimal policy.
R b. Use RSPE to construct and solve the decision tree. 9.28. On Monday, a certain stock closed at $10 per share. On Tuesday, you expect the stock to close at $9, $10, or $11 per share, with respective probabilities 0.3, 0.3, and 0.4. On Wednesday, you expect the stock to close 10 percent lower, unchanged, or 10 percent higher than Tuesday’s close, with the following probabilities.
Today’s Close 10 Percent
Lower Unchanged 10 Percent
Higher
$ 9 0.4 0.3 0.3 10 0.2 0.2 0.6 11 0.1 0.2 0.7
On Tuesday, you are directed to buy 100 shares of the stock before Thursday. All purchases are made at the end of the day, at the known closing price for that day, so your only options are to
a. Analyze this decision tree to obtain the optimal policy.
R b. Use RSPE to construct and solve the same decision tree.
9.25. You are given the following decision tree, with the probabilities at event nodes shown in parentheses and with the payoffs at terminal points shown on the right. Analyze this deci- sion tree to obtain the optimal policy.
–5
–10
30
0
10
40
–10
10
(0.5) (0.5)
(0.5)
(0.3)
(0.7)
(0.5)
(0.4)
(0.6)
9.26.* The Athletic Department of Leland University is con- sidering whether to hold an extensive campaign next year to raise funds for a new athletic field. The response to the cam- paign depends heavily on the success of the football team this fall. In the past, the football team has had winning seasons 60 percent of the time. If the football team has a winning season (W) this fall, then many of the alumni will contribute and the cam- paign will raise $3 million. If the team has a losing season (L), few will contribute and the campaign will lose $2 million. If no campaign is undertaken, no costs are incurred. On September 1, just before the football season begins, the Athletic Department needs to make its decision about whether to hold the campaign next year. a. Develop a decision analysis formulation of this
problem by identifying the decision alternatives, the states of nature, and the payoff table.
b. According to Bayes’ decision rule, should the cam- paign be undertaken?
c. What is the expected value of perfect information? d. A famous football guru, William Walsh, has offered
his services to help evaluate whether the team will have a winning season. For $100,000, he will care- fully evaluate the team throughout spring practice and then throughout preseason workouts. William then will provide his prediction on September 1 regarding what kind of season, W or L, the team will have. In similar situations in the past when evaluating teams that have winning seasons 50 per- cent of the time, his predictions have been correct 75 percent of the time. Considering that this team has more of a winning tradition, if William predicts a winning season, what is the posterior probability that the team actually will have a winning season? What is the posterior probability of a losing season?
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Then do the same for the four cases where both financial figures are at one end or the other of their ranges of variability.
R9.31. Chelsea Bush is an emerging candidate for her party’s nomination for president of the United States. She now is con- sidering whether to run in the high-stakes Super Tuesday pri- maries. If she enters the Super Tuesday (S.T.) primaries, she and her advisers believe that she will either do well (finish first or second) or do poorly (finish third or worse) with probabili- ties 0.4 and 0.6, respectively. Doing well on Super Tuesday will net the candidate’s campaign approximately $16 million in new contributions, whereas a poor showing will mean a loss of $10 million after numerous TV ads are paid for. Alterna- tively, she may choose not to run at all on Super Tuesday and incur no costs.
Chelsea’s advisors realize that her chances of success on Super Tuesday may be affected by the outcome of the smaller New Hampshire (N.H.) primary occurring three weeks before Super Tuesday. Political analysts feel that the results of New Hampshire’s primary are correct two-thirds of the time in pre- dicting the results of the Super Tuesday primaries. Among Chelsea’s advisers is a decision analysis expert who uses this information to calculate the following probabilities:
P (Chelsea does well in S.T. primaries, given she does well in N.H.) 5 4/7
P (Chelsea does well in S.T. primaries, given she does poorly in N.H.) 5 ¼
P (Chelsea does well in N.H. primary) 5 7/15
The cost of entering and campaigning in the New Hampshire primary is estimated to be $1.6 million.
Chelsea feels that her chance of winning the nomination depends largely on having substantial funds available after the Super Tuesday primaries to carry on a vigorous campaign the rest of the way. Therefore, she wants to choose the strategy (whether to run in the New Hampshire primary and then whether to run in the Super Tuesday primaries) that will maximize her expected funds after these primaries.
a. Construct and solve the decision tree for this problem.
b. There is some uncertainty in the estimates of a gain of $16 million or a loss of $10 million depending on the showing on Super Tuesday. Either amount could differ from this estimate by as much as 25 percent in either direction. For each of these two financial figures, perform sensitivity analysis to check how the results in part a would change if the value of the financial figure were at either end of this range of variability (without any change in the value of the other financial figure). Then do the same for the four cases where both financial figures are at one end or the other of their ranges of variability.
R9.32. The executive search being conducted for Western Bank by Headhunters Inc. may finally be bearing fruit. The posi- tion to be filled is a key one—vice president for Information Processing—because this person will have responsibility for developing a state-of-the-art management information system that will link together Western’s many branch banks. However, Headhunters feels it has found just the right person, Matthew
buy at the end of Tuesday or at the end of Wednesday. You wish to determine the optimal strategy for whether to buy on Tuesday or defer the purchase until Wednesday, given the Tuesday clos- ing price, to minimize the expected purchase price. a. Develop and evaluate a decision tree by hand for
determining the optimal strategy. R b. Use RSPE to construct and solve the decision tree. R9.29. Jose Morales manages a large outdoor fruit stand in one of the less affluent neighborhoods of San Jose, California. To replenish his supply, Jose buys boxes of fruit early each morning from a grower south of San Jose. About 90 percent of the boxes of fruit turn out to be of satisfactory quality, but the other 10 percent are unsatisfactory. A satisfactory box contains 80 percent excellent fruit and will earn $200 profit for Jose. An unsatisfactory box contains 30 percent excellent fruit and will pro- duce a loss of $1,000. Before Jose decides to accept a box, he is given the opportunity to sample one piece of fruit to test whether it is excellent. Based on that sample, he then has the option of rejecting the box without paying for it. Jose wonders (1) whether he should continue buying from this grower, (2) if so, whether it is worthwhile sampling just one piece of fruit from a box, and (3) if so, whether he should be accepting or rejecting the box based on the outcome of this sampling.
Use RSPE (and the Excel template for posterior probabili- ties) to construct and solve the decision tree for this problem. 9.30.* The Morton Ward Company is considering the intro- duction of a new product that is believed to have a 50–50 chance of being successful. One option is to try out the product in a test market, at an estimated cost of $2 million, before making the introduction decision. Past experience shows that ultimately successful products are approved in the test market 80 percent of the time, whereas ultimately unsuccessful products are approved in the test market only 25 percent of the time. If the product is successful, the net profit to the company will be $40 million; if unsuccessful, the net loss will be $15 million. a. Discarding the test market option, develop a deci-
sion analysis formulation of the problem by identi- fying the decision alternatives, states of nature, and payoff table. Then apply Bayes’ decision rule to determine the optimal decision alternative.
b. Find the expected value of perfect information. R c. Now including the option of trying out the product
in a test market, use RSPE (and the Excel template for posterior probabilities) to construct and solve the decision tree for this problem.
R d. Find the expected value of sample information. How large can the cost of trying out the product in a test market be and still be worthwhile to do?
R e. Assume now that the estimate of $2 million for the cost of trying out the product in a test market is correct. However, there is some uncertainty in the stated profit and loss figures ($40 million and $15 million). Either could vary from its base by as much as 25 percent in either direction. For each of these two financial figures, perform sensitivity analysis to check how the results in part c would change if the value of the financial figure were at either end of this range of variability (without any change in the value of the other financial figure).
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taking the earthquake insurance. Should you take the insurance?
9.36. For your graduation present from college, your parents are offering you your choice of two alternatives. The first alter- native is to give you a money gift of $19,000. The second alter- native is to make an investment in your name. This investment will quickly have the following two possible outcomes.
Outcome Probability
Receive $10,000 0.3 Receive $30,000 0.7
Your utility for receiving M thousand dollars is given by the util- ity function U(M) 5 "M 1 6. Which choice should you make to maximize expected utility? 9.37. Reconsider Problem 9.36. You now are uncertain about what your true utility function for receiving money is, so you are in the process of constructing this utility function by using the equivalent lottery method and units of thousands of dollars. You have concluded that you are indifferent between the two alterna- tives offered to you by your parents. Use this information to find U (19) after setting U (10) 5 0 and U (30) 5 1. 9.38. You wish to construct your personal utility function U ( M ) for receiving M thousand dollars. After setting U (0) 5 0, you next set U (10) 5 1 as your utility for receiving $10,000. You next want to find U (1) and then U (5). a. You offer yourself the following two hypothetical
alternatives:
A 1 : Obtain $10,000 with probability p. Obtain 0 with probability (1 2 p ).
A 2 : Definitely obtain $1,000.
You then ask yourself the question: What value of p makes you indifferent between these two alter- natives? Your answer is p 5 0.125. Find U (1) by using the equivalent lottery method.
b. You next repeat part a except for changing the sec- ond alternative to definitely receiving $5,000. The value of p that makes you indifferent between these two alternatives now is p 5 0.5625. Find U (5).
c. Repeat parts a and b, but now use your personal choices for p.
9.39. You are given the following payoff table.
State of Nature
Alternative S1 S2
A1 25 36 A2 100 0 A3 0 49
Prior probability p 1 2 p
a. Assume that your utility function for the payoffs is U(x) 5 "x. Plot the expected utility of each deci- sion alternative versus the value of p on the same graph. For each decision alternative, find the range of values of p over which this alternative maximizes the expected utility.
Fenton, who has an excellent record in a similar position for a midsized bank in New York.
After a round of interviews, Western’s president believes that Matthew has a probability of 0.7 of designing the management information system successfully. If Matthew is successful, the company will realize a profit of $2 million (net of Matthew’s salary, training, recruiting costs, and expenses). If he is not suc- cessful, the company will realize a net loss of $600,000.
For an additional fee of $40,000, Headhunters will provide a detailed investigative process (including an extensive back- ground check, a battery of academic and psychological tests, etc.) that will further pinpoint Matthew’s potential for success. This process has been found to be 90 percent reliable, that is, a candidate who would successfully design the management information system will pass the test with probability 0.9, and a candidate who would not successfully design the system will fail the test with probability 0.9.
Western’s top management needs to decide whether to hire Matthew and whether to have Headhunters conduct the detailed investigative process before making this decision. a. Construct and solve the decision tree for this prob-
lem to identify the optimal policy. b. Now suppose that Headhunters’s fee for administer-
ing its detailed investigative process is negotiable. What is the maximum amount that Western Bank should pay?
9.33. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 9.9. Briefly describe how decision analy- sis was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 9.34. Reconsider the Goferbroke Co. case study, including the application of utilities in Section 9.9. Max Flyer now has decided that, given the company’s precarious financial situation, he needs to take a much more risk-averse approach to the prob- lem. Therefore, he has revised the utilities given in Table 9.9 as follows: U ( 2 130) 5 0, U ( 2 100) 5 0.07, U (60) 5 0.40, U (90) 5 0.45, U (670) 5 0.99, and U (700) 5 1. a. Analyze the revised decision tree corresponding
to Figure 9.23 by hand to obtain the new optimal policy.
R b. Use RSPE to construct and solve this revised deci- sion tree.
9.35.* You live in an area that has a possibility of incurring a massive earthquake, so you are considering buying earthquake insurance on your home at an annual cost of $180. The prob- ability of an earthquake damaging your home during one year is 0.001. If this happens, you estimate that the cost of the damage (fully covered by earthquake insurance) will be $160,000. Your total assets (including your home) are worth $250,000. a. Apply Bayes’ decision rule to determine which
alternative (take the insurance or not) maximizes your expected assets after one year.
b. You now have constructed a utility function that measures how much you value having total assets worth x dollars ( x $ 0). This utility function is U(x) 5 "x. Compare the utility of reducing your total assets next year by the cost of the earthquake insurance with the expected utility next year of not
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9.41. Consider the following decision tree, where the prob- abilities for each event node are shown in parentheses.
Payoff
$6,859
–$1,331
–$125
0
Utility
x1/3
19
–11
–5
0
$ x $ x
+$7,600
(0.60)
–$590
(0.40)
$16
+$141
(0.50)
$0
–$600
(0.50)
A1
A2
–$141
The dollar amount given next to each branch is the cash flow generated along that branch, where these intermediate cash flows add up to the total net cash flow shown to the right of each terminal branch. (The unknown amount for the top branch is represented by the variable x. ) The decision maker has a util- ity function U ( y ) 5 y 1/3 where y is the total net cash flow after a terminal branch. The resulting utilities for the various terminal branches are shown to the right of the decision tree.
Use these utilities to analyze the decision tree. Then deter- mine the value of x for which the decision maker is indifferent between decision alternatives A 1 and A 2 .
R b. Now assume that your utility function is the expo- nential utility function with a risk tolerance of R 5 50. Use RSPE to construct and solve the result- ing decision tree in turn for p 5 0.25, p 5 0.5, and p 5 0.75.
R9.40. Dr. Switzer has a seriously ill patient but has had trou- ble diagnosing the specific cause of the illness. The doctor now has narrowed the cause down to two alternatives: disease A or disease B. Based on the evidence so far, she feels that the two alternatives are equally likely.
Beyond the testing already done, there is no test available to determine if the cause is disease B. One test is available for disease A, but it has two major problems. First, it is very expen- sive. Second, it is somewhat unreliable, giving an accurate result only 80 percent of the time. Thus, it will give a positive result (indicating disease A) for only 80 percent of patients who have disease A, whereas it will give a positive result for 20 percent of patients who actually have disease B instead.
Disease B is a very serious disease with no known treatment. It is sometimes fatal, and those who survive remain in poor health with a poor quality of life thereafter. The prognosis is similar for victims of disease A if it is left untreated. However, there is a fairly expensive treatment available that eliminates the danger for those with disease A, and it may return them to good health. Unfortunately, it is a relatively radical treatment that always leads to death if the patient actually has disease B instead.
The probability distribution for the prognosis for this patient is given for each case in the following table, where the column headings (after the first one) indicate the disease for the patient.
Outcome Probabilities
No Treatment Receive Treatment
for Disease A
Outcome A B A B
Die 0.2 0.5 0 1.0 Survive with poor health 0.8 0.5 0.5 0 Return to good health 0 0 0.5 0
The patient has assigned the following utilities to the possible outcomes.
Outcome Utility
Die 0 Survive with poor health 10 Return to good health 30
In addition, these utilities should be incremented by 2 2 if the patient incurs the cost of the test for disease A and by 2 1 if the patient (or the patient’s estate) incurs the cost of the treatment for disease A.
Use decision analysis with a complete decision tree to deter- mine if the patient should undergo the test for disease A and then how to proceed (receive the treatment for disease A?) to maximize the patient’s expected utility.
R9.42. Reconsider the Goferbroke Co. case study when using utilities, as presented in Section 9.9.
a. Beginning with the decision tree shown in Figure 9.23 (available in one of this chapter’s Excel files), prepare to perform sensitivity analysis by expand- ing and organizing the spreadsheet to (1) con- solidate the data and results in one section and (2) incorporate the Excel template for posterior proba- bilities in another section (similar to what was done in Figure 9.18 ).
b. Perform sensitivity analysis by re-solving the deci- sion tree (after using the Excel template for poste- rior probabilities to revise these probabilities) when the prior probability of oil is changed in turn to 0.15, 0.2, 0.3, and 0.35.
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Case 9-2 University Toys and the Business Professor Action Figures 379
Case 9-1
Who Wants to Be a Millionaire?
You are a contestant on “Who Wants to Be a Millionaire?” You already have answered the $250,000 question correctly and now must decide if you would like to answer the $500,000 question. You can choose to walk away at this point with $250,000 in win- nings or you may decide to answer the $500,000 question. If you answer the $500,000 question correctly, you can then choose to walk away with $500,000 in winnings or go on and try to answer the $1,000,000 question. If you answer the $1,000,000 question correctly, the game is over and you win $1,000,000. If you answer either the $500,000 or the $1,000,000 question incorrectly, the game is over immediately and you take home “only” $32,000.
A feature of the game “Who Wants to Be a Millionaire?” is that you have three “lifelines”—namely “50–50,” “ask the audi- ence,” and “phone a friend.” At this point (after answering the $250,000 question), you already have used two of these lifelines, but you have the “phone a friend” lifeline remaining. With this option, you may phone a friend to obtain advice on the correct answer to a question before giving your answer. You may use this option only once (i.e., you can use it on either the $500,000
question or the $1,000,000 question, but not both). Since some of your friends are smarter than you are, “phone a friend” sig- nificantly improves your odds for answering a question correctly. Without “phone a friend,” if you choose to answer the $500,000 question you have a 65 percent chance of answering correctly, and if you choose to answer the $1,000,000 question you have a 50 percent chance of answering correctly (the questions get progres- sively more difficult). With “phone a friend,” you have an 80 per- cent chance of answering the $500,000 question correctly and a 65 percent chance of answering the $1,000,000 question correctly.
a. Use RSPE to construct and solve a decision tree to decide what to do. What is the best course of action, assuming your goal is to maximize your expected winnings?
b. Use the equivalent lottery method to determine your personal utility function (in particular, your utility values for all of the possible payoffs in the game).
c. Re-solve the decision tree, replacing the payoffs with your utility values, to maximize your expected utility. Does the best course of action change?
Case 9-2
University Toys and the Business Professor Action Figures
University Toys has developed a brand new product line—a series of Business Professor Action Figures (BPAFs) featuring likenesses of popular professors at the local business school. Management needs to decide how to market the dolls.
One option is to immediately ramp up production and simul- taneously launch an ad campaign in the university newspaper. This option would cost $1,000. Based on past experience, new action figures either take off and do well or fail miserably. Hence, the prediction is for one of two possible outcomes— total sales of 2,500 units or total sales of only 250 units. Uni- versity Toys receives revenue of $2 per unit sold. Management currently thinks that there is about a 50 percent chance that the product will do well (sell 2,500 units) and a 50 percent chance that it will do poorly (sell 250 units).
Another option is to test-market the product. The company could build a few units, put up a display in the campus book- store, and see how they sell without any further advertising. This would require less capital for the production run and no money for advertising. Again, the prediction is for one of two possible outcomes for this test market, namely, the product will either do well (sell 200 units) or do poorly (sell 20 units). The cost for this option is estimated to be $100. University Toys receives revenue of $2 per unit sold for the test market as well. The company has often test marketed toys in this manner. Products that sell well when fully marketed have also sold well in the test market 80 percent of the time. Products that sell poorly when fully mar- keted also sell poorly in the test market 60 percent of the time.
There is a complication with the test market option, however. A rival toy manufacturer is rumored to be considering the devel- opment of Law School Professor Action Figures (LSPAF). After doing the test marketing, if University Toys decides to go ahead and ramp up production and fully market the BPAF, the cost of doing so would still be $1,000. However, the sales prospects depend upon whether LSPAF has been introduced into the mar- ket or not. If LSPAF has not entered the market, then the sales prospects will be the same as described above (i.e., 2,500 units if BPAF does well, or 250 units if BPAF does poorly, on top of any units sold in the test market). However, if LSPAF has been introduced, the increased competition will diminish sales of BPAF. In particular, management expects in this case to sell 1,000 if BPAF does well or 100 units if it does poorly (on top of any units sold in the test market). Note that the probability of BPAF doing well or doing poorly is not affected by LSPAF, just the final sales totals of each possibility. The probability that LSPAF will enter the market before the end of the test market is 20 percent. On the other hand, if University Toys markets BPAF immediately, they are guaranteed to beat the LSPAF to market (thus making LSPAF a nonfactor).
a. Suppose that the test marketing is done. Use the Posterior Probabilities template to determine the likelihood that the BPAF would sell well if fully marketed, given that it sells well in the test market and then given that it sells poorly in the test market.
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d. Generate a data table that shows how the expected payoff and the test marketing decision changes as the probability that the LSPAFs will enter the market varies from 0 percent to 100 percent (at 10 percent increments).
e. At what probability does the test marketing decision change?
b. Use RSPE to develop and solve a decision tree to help Uni- versity Toys decide the best course of action and the expected payoff.
c. Now suppose that University Toys is uncertain of the prob- ability that the LSPAFs will enter the market before the test marketing would be completed (if it were done). How would you expect the expected payoff to vary as the probability that the LSPAFs will enter the market changes?
Case 9-3
Brainy Business
While El Niño is pouring its rain on northern California, Charlotte Rothstein, CEO, major shareholder, and founder of Cerebrosoft, sits in her office, contemplating the decision she faces regarding her company’s newest proposed product, Brainet. This has been a particularly difficult decision. Brainet might catch on and sell very well. However, Charlotte is con- cerned about the risk involved. In this competitive market, marketing Brainet also could lead to substantial losses. Should she go ahead anyway and start the marketing campaign? Or just abandon the product? Or perhaps buy additional market- ing research information from a local market research company before deciding whether to launch the product? She has to make a decision very soon and so, as she slowly drinks from her glass of high-protein-power multivitamin juice, she reflects on the events of the past few years.
Cerebrosoft was founded by Charlotte and two friends after they had graduated from business school. The company is located in the heart of Silicon Valley. Charlotte and her friends managed to make money in their second year in business and have continued to do so every year since. Cerebrosoft was one of the first companies to sell software over the World Wide Web and to develop PC-based software tools for the multimedia sec- tor. Two of the products generate 80 percent of the company’s revenues: Audiatur and Videatur. Each product has sold more than 100,000 units during the past year. Business is done over the Web: Customers can download a trial version of the soft- ware, test it, and if they are satisfied with what they see, they can purchase the product (by using a password that enables them to disable the time counter in the trial version). Both products are priced at $75.95 and are sold exclusively over the Web.
Although the World Wide Web is a network of computers of different types, running different kinds of software, a stan- dardized protocol between the computers enables them to com- municate. Users can surf the Web and visit computers many thousands of miles away, accessing information available at the site. Users also can make files available on the Web, and this is how Cerebrosoft generates its sales. Selling software over the Web eliminates many of the traditional cost factors of consumer products: packaging, storage, distribution, sales force, and so on. Instead, potential customers can download a trial version, take a look at it (that is, use the product) before its trial period expires, and then decide whether to buy it. Furthermore, Cere- brosoft can always make the most recent files available to the
customer, avoiding the problem of having outdated software in the distribution pipeline.
Charlotte is interrupted in her thoughts by the arrival of Jean- nie Korn. Jeannie is in charge of marketing for online products and Brainet has had her particular attention from the beginning. She is more than ready to provide the advice that Charlotte has requested. “Charlotte, I think we should really go ahead with Brainet. The software engineers have convinced me that the current version is robust and we want to be on the market with this as soon as possible! From the data for our product launches during the past two years, we can get a rather reliable estimate of how the market will respond to the new product, don’t you think? And look!” She pulls out some presentation slides. “Dur- ing that time period we launched 12 new products altogether and 4 of them sold more than 30,000 units during the first six months alone! Even better: The last two we launched even sold more than 40,000 copies during the first two quarters!” Charlotte knows these numbers as well as Jeannie does. After all, two of these launches have been products she herself helped to develop. But she feels uneasy about this particular product launch. The company has grown rapidly during the past three years and its financial capabilities are already rather stretched. A poor prod- uct launch for Brainet would cost the company a lot of money, something that isn’t available right now due to the investments Cerebrosoft has recently made.
Later in the afternoon, Charlotte meets with Reggie Ruffin, a jack of all trades and the production manager. Reggie has a solid track record in his field and Charlotte wants his opinion on the Brainet project.
“Well, Charlotte, quite frankly, I think that there are three main factors that are relevant to the success of this project: com- petition, units sold, and cost—ah, and, of course, our pricing. Have you decided on the price yet?”
“I am still considering which of the three strategies would be most beneficial to us. Selling for $50.00 and trying to maximize revenues—or selling for $30.00 and trying to maximize market share. Of course, there is still your third alternative; we could sell for $40.00 and try to do both.”
At this point, Reggie focuses on the sheet of paper in front of him. “And I still believe that the $40.00 alternative is the best one. Concerning the costs, I checked the records; basically we have to amortize the development costs we incurred for Brainet. So far we have spent $800,000 and we expect to spend another
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80 percent of the time, while 15 percent of the time they predicted moderate competition in that setting. Given that the competition turned out to be moderate, they predicted severe competition 15 percent of the time and moderate competition 80 percent of the time. Finally, for the case of weak competition, the numbers were 90 percent of the time a correct prediction, 7 percent of the time a ‘moderate’ prediction and 3 percent of the time a ‘severe’ prediction.”
Charlotte feels that all these numbers are too much for her. “Don’t we have a simple estimate of how the market will react?”
“Some prior probabilities, you mean? Sure, from our past experience, the likelihood of facing severe competition is 20 percent, whereas it is 70 percent for moderate competition
The next morning, Charlotte is sipping from another power drink. Jeannie and Reggie will be in her office any moment now and, with their help, she will have to decide what to do with Brainet. Should they launch the product? If so, at what price?
When Jeannie and Reggie enter the office, Jeannie immedi- ately bursts out: “Guys, I just spoke to our marketing research company. They say that they could do a study for us about the competitive situation for the introduction of Brainet and deliver the results within a week.”
“How much do they want for the study?” “I knew you’d ask that, Reggie. They want $10,000, and I
think it’s a fair deal.” At this point, Charlotte steps into the conversation. “Do
we have any data on the quality of the work of this marketing research company?”
“Yes, I do have some reports here. After analyzing them, I have come to the conclusion that the predictions of the mar- keting research company are pretty good: Given that the com- petition turned out to be severe, they predicted it correctly
$50,000 per year for support and shipping the CDs to those who want a hardcopy on top of their downloaded software.” Reggie next hands a report to Charlotte. “Here we have some data on the industry. I just received that yesterday, hot off the press. Let’s see what we can learn about the industry here.” He shows Charlotte some of the highlights. Reggie then agrees to compile the most relevant information contained in the report and have it ready for Charlotte the following morning. It takes him long into the night to gather the data from the pages of the report, but in the end he produces three tables, one for each of the three alternative pric- ing strategies. Each table shows the corresponding probability of various amounts of sales given the level of competition (severe, moderate, or weak) that develops from other companies.
Level of Competition
Sales Severe Moderate Weak
50,000 units 0.2 0.25 0.3 30,000 units 0.25 0.3 0.35 20,000 units 0.55 0.45 0.35
TABLE 1 Probability Distribution of Unit Sales, Given a High Price ($50)
Level of Competition
Sales Severe Moderate Weak
50,000 units 0.25 0.30 0.40 30,000 units 0.35 0.40 0.50 20,000 units 0.40 0.30 0.10
TABLE 2 Probability Distribution of Unit Sales, Given a Medium Price ($40)
Level of Competition
Sales Severe Moderate Weak
50,000 units 0.35 0.40 0.50 30,000 units 0.40 0.50 0.45 20,000 units 0.25 0.10 0.05
TABLE 3 Probability Distribution of Unit Sales, Given a Low Price ($30)
and 10 percent for weak competition,” says Jeannie, her num- bers always ready when needed.
All that is left to do now is to sit down and make sense of all this . . .
a. For the initial analysis, ignore the opportunity of obtaining more information by hiring the marketing research company. Identify the decision alternatives and the states of nature. Construct the payoff table. Then formulate the decision prob- lem in a decision tree. Clearly distinguish between decision and event nodes and include all the relevant data.
b. What is Charlotte’s decision if she uses the maximum likeli- hood criterion?
c. What is Charlotte’s decision if she uses Bayes’ decision rule? d. Now consider the possibility of doing the market research.
Develop the corresponding decision tree. Calculate the rele- vant probabilities and analyze the decision tree. Should Cere- brosoft pay the $10,000 for the marketing research? What is the overall optimal policy?
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Case 9-4
Smart Steering Support
On a sunny May morning, Marc Binton, CEO of Bay Area Automobile Gadgets (BAAG), enters the conference room on the 40th floor of the Gates building in San Francisco, where BAAG’s offices are located. The other executive officers of the company have already gathered. The meeting has only one item on its agenda: planning a research and development project to develop a new driver support system (DSS). Brian Huang, manager of Research and Development, is walking around ner- vously. He has to inform the group about the R&D strategy he has developed for the DSS. Marc has identified DSS as the stra- tegic new product for the company. Julie Aker, vice president of Marketing, will speak after Brian. She will give detailed infor- mation about the target segment, expected sales, and marketing costs associated with the introduction of the DSS.
BAAG builds electronic nonaudio equipment for luxury cars. Founded by a group of Stanford graduates, the company sold its first product—a car routing system relying on a tech- nology called Global Positioning Satellites (GPS)—a few years ago. Such routing systems help drivers to find directions to their desired destinations using satellites to determine the exact position of the car. To keep up with technology and to meet the wishes of its customers, the company has added a number of new features to its router during the last few years. The DSS will be a completely new product, incorporating recent developments in GPS as well as voice recognition and display technologies. Marc strongly supports this product, as it will give BAAG a competi- tive advantage over its Asian and European competitors.
Driver support systems have been a field of intense research for more than a decade. These systems provide the driver with a wide range of information, such as directions, road condi- tions, traffic updates, and so forth. The information exchange can take place verbally or via projection of text onto the wind- screen. Other features help the driver avoid obstacles that have been identified by cars ahead on the road (these cars transmit the information to the following vehicles). Marc wants to incorpo- rate all these features and other technologies into one support system that would then be sold to BAAG’s customers in the automobile industry.
After all the attendees have taken their seats, Brian starts his presentation: “Marc asked me to inform you about our efforts with the driver support system, particularly the road scanning device. We have reached a stage where we basically have to make a go or no-go decision concerning the research for this device, which, as you all know by now, is a key feature in the DSS. We have already integrated the other devices, such as the GPS-based positioning and direction system. The question with which we have to deal is whether to fund basic research into the road scanning device. If this research is successful, we then will have to decide if we want to develop a product based on these results—or if we just want to sell the technology without developing a product. If we do decide to develop the product ourselves, there is a chance that the product development pro- cess might not be successful. In that case, we could still sell the technology. In the case of successful product development, we would have to decide whether to market the product. If we decide not to market the developed product, we could at least
sell the product concept that was the result of our successful research and development efforts. Doing so would earn more than just selling the technology prematurely. If, on the other hand, we decide to market the driver support system, then we are faced with the uncertainty of how the product will be received by our customers.”
“You completely lost me,” snipes Marc. Max, Julie’s assistant, just shakes his head and murmurs,
“those techno-nerds. . . .” Brian starts to explain: “Sorry for the confusion. Let’s just go
through it again, step by step.” “Good idea—and perhaps make smaller steps!” Julie obvi-
ously dislikes Brian’s style of presentation. “OK, the first decision we are facing is whether to invest in
research for the road scanning device.” “How much would that cost us?” asks Marc. “Our estimated budget for this is $300,000. Once we invest
that money, the outcome of the research effort is somewhat uncertain. Our engineers assess the probability of successful research at 80 percent.”
“That’s a pretty optimistic success rate, don’t you think?” Julie remarks sarcastically. She still remembers the disaster with Brian’s last project, the fingerprint-based car-security-system. After spending half a million dollars, the development engineers concluded that it would be impossible to produce the security system at an attractive price.
Brian senses Julie’s hostility and shoots back: “In engineer- ing, we are quite accustomed to these success rates—something we can’t say about marketing. . . .”
“What would be the next step?” intervenes Marc. “Hm, sorry. If the research is not successful, then we can
only sell the DSS in its current form.” “The profit estimate for that scenario is $2 million,” Julie
throws in. “If, however, the research effort is successful, then we will
have to make another decision, namely, whether to go on to the development stage.”
“If we wouldn’t want to develop a product at that point, would that mean that we would have to sell the DSS as it is now?” asks Max.
“Yes, Max. Except that additionally we would earn some $200,000 from selling our research results to GM. Their research division is very interested in our work and they have offered that money for our findings.”
“Ah, now that’s good news,” remarks Julie. Brian continues, “If, however, after successfully complet-
ing the research stage, we decide to develop a new product, then we’ll have to spend another $800,000 for that task, at a 35 percent chance of not being successful.”
“So you are telling us we’ll have to spend $800,000 for a ticket in a lottery where we have a 35 percent chance of not win- ning anything?” asks Julie.
“Julie, don’t focus on the losses but on the potential gains! The chance of winning in this lottery, as you call it, is 65 percent. I believe that that’s much more than with a normal lottery ticket,” says Marc.
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“What are the chances of ending up with a high, medium, or low acceptance of the new DSS?” asks Brian.
“We can see those numbers at the bottom of the slide,” says Julie, while she is turning toward the projection behind her. There is a 30 percent chance of high market acceptance and a 20 percent chance of low market acceptance.
At this point, Marc moves in his seat and asks: “Given all these numbers and bits of information, what are you suggesting that we do?”
a. Organize the available data on cost and profit estimates in a table.
b. Formulate the problem in a decision tree. Clearly distinguish between decision and event nodes.
c. Calculate the expected payoffs for each node in the decision tree.
d. What is BAAG’s optimal policy according to Bayes’ deci- sion rule?
e. What would be the expected value of perfect information on the outcome of the research effort?
f. What would be the expected value of perfect information on the outcome of the development effort?
g. Marc is a risk-averse decision maker. In a number of inter- views, it has been determined that he would just barely be willing to consider taking a 50-50 gamble where he either makes $1.2 million or loses $600 thousand. Based on Marc’s level of risk aversion, use the exponential utility function to determine BAAG’s optimal policy.
Additional Cases Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
“Thanks, Marc,” says Brian. “Once we invest that money in development, we have two possible outcomes: either we will be successful in developing the road scanning device or we won’t. If we fail, then once again we’ll sell the DSS in its current form and cash in the $200,000 from GM for the research results. If the development process is successful, then we have to decide whether to market the new product.”
“Why wouldn’t we want to market it after successfully developing it?” asks Marc.
“That’s a good question. Basically what I mean is that we could decide not to sell the product ourselves but instead give the right to sell it to somebody else, to GM for example. They would pay us $1 million for it.”
“I like those numbers!” remarks Julie. “Once we decide to build the product and market it, we will
face the market uncertainties and I’m sure that Julie has those numbers ready for us. Thanks.”
At this point, Brian sits down and Julie comes forward to give her presentation. Immediately some colorful slides are pro- jected on the wall behind her as Max operates the computer.
“Thanks, Brian. Well, here’s the data we have been able to gather from some marketing research. The acceptance of our new product in the market can be high, medium, or low.” Julie is pointing to some figures projected on the wall behind her. “Our estimates indicate that high acceptance would result in profits of $8.0 million and that medium acceptance would give us $4.0 million. In the unfortunate case of a poor reception by our customers, we still expect $2.2 million in profit. I should mention that these profits do not include the additional costs of marketing or R&D expenses.”
“So, you are saying that in the worst case we’ll make barely more money than with the current product?” asks Brian.
“Yes, that’s what I am saying.” “What budget would you need for the marketing of our DSS
with the road scanner?” asks Marc. “For that we would need an additional $200,000 on top of
what has already been included in the profit estimates,” Julie replies.
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Chapter Ten
Forecasting Learning Objectives
After completing this chapter, you should be able to
1. Describe some important types of forecasting applications.
2. Identify two common measures of the accuracy of forecasting methods.
3. Adjust forecasting data to consider seasonal patterns.
4. Describe several forecasting methods that use the pattern of historical data to fore- cast a future value.
5. Apply these methods either by hand or with the software provided.
6. Compare these methods to identify the conditions when each is particularly suitable.
7. Describe and apply an approach to forecasting that relates the quantity of interest to one or more other quantities.
8. Describe several forecasting methods that use expert judgment.
How much will the economy grow over the next year? Where is the stock market headed? What about interest rates? How will consumer tastes be changing? What will be the hot new products?
Forecasters have answers to all these questions. Unfortunately, these answers will more than likely be wrong. Nobody can accurately predict the future every time.
Nevertheless, the future success of any business depends heavily on how savvy its man- agement is in spotting trends and developing appropriate strategies. The leaders of the best companies often seem to have a sixth sense for when to change direction to stay a step ahead of the competition, but this sixth sense actually is guided by frequent use of the best forecast- ing techniques. These companies seldom get into trouble by badly misestimating what the demand will be for their products. Many other companies do. The ability to forecast well makes the difference.
When historical sales data are available, some proven statistical forecasting methods have been developed for using these data to forecast future demand. Such methods assume that historical trends will continue, so management then needs to make adjustments to reflect current changes in the marketplace.
Several judgmental forecasting methods that solely use expert judgment also are available. These methods are especially valuable when little or no historical sales data are available or when major changes in the marketplace make these data unreliable for forecast- ing purposes.
Forecasting product demand is just one important application of these forecasting meth- ods. For other applications, forecasts might be needed for such diverse quantities as the need for spare parts, production yields, and staffing needs. Forecasting techniques also are heavily used for forecasting economic trends on a regional, national, or even international level.
We begin the chapter with an overview of forecasting techniques in their simplest form. Section 10.2 then introduces a typical case study that will be carried through much of the chapter to illustrate how these techniques commonly need to be adapted to incorporate such practical considerations as seasonal demand patterns. Sections 10.3–10.5 focus on statistical forecasting methods and Section 10.6 on judgmental forecasting methods. Excel templates for various forecasting methods and a forecasting module that is part of your Interactive Man- agement Science Modules are included in your MS Courseware.
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10.1 An Overview of Forecasting Techniques 385
10.1 AN OVERVIEW OF FORECASTING TECHNIQUES
To illustrate various forecasting techniques, consider the following problem.
A Forecasting Problem Fastchips is a leading producer of microprocessors. Six months ago, it launched the sales of its latest microprocessor. The month-by-month sales (in thousands) of the microprocessor over the initial six months have been
17 25 24 26 30 28
In this highly competitive market, sales can shift rather quickly, depending largely on when competitors launch the latest version of their microprocessors. Therefore, it always is impor- tant to have a forecast of the next month’s sales to guide what the production level should be.
Let us look at some alternative ways of obtaining this forecast.
Some Forecasting Techniques The most straightforward technique is the last-value forecasting method (sometimes also called the naive method ), which says simply to use the last month’s sales as the forecast for the next month. For Fastchips, this yields
Forecast 5 28
This is a reasonable forecasting method when conditions tend to change so quickly that sales before the last month’s are not a reliable indicator of future sales.
The averaging forecasting method says to use the average of all the monthly sales to date as the forecast for the next month. This gives
Forecast 5 17 1 25 1 24 1 26 1 30 1 28
6 5 25
for Fastchips. This is a reasonable forecasting method when conditions tend to remain so stable that even the earliest sales are a reliable indicator of future sales (a dubious assumption for Fastchips).
The moving-average forecasting method provides a middle ground between the last-value and averaging method by using the average of the monthly sales for only the most recent months as the forecast for the next month. The number of months being used must be specified. For example, a three-month moving average forecast for Fastchips is
Forecast 5 26 1 30 1 28
3 5 28
This is a reasonable forecasting method when conditions tend to change occasionally but not extremely rapidly.
The exponential smoothing forecasting method provides a more sophisticated ver- sion of the moving-average method in the way that it gives primary consideration to sales in only the most recent months. In particular, rather than giving equal weight to the sales in the most recent months, the exponential smoothing method gives the greatest weight to the last month and then progressively smaller weights to the older months. (The formula for this method will be given in Section 10.3.) This forecasting method is a reasonable one under the same conditions as described for the moving-average method.
Additional sophistication is added to the exponential smoothing forecasting method by using exponential smoothing with trend . This latter method adjusts exponential smoothing by also directly considering any current upward or downward trend in the sales. (Formulas are given in Section 10.3.)
If the sales data show a relatively constant trend in some direction, then linear regression provides a reasonable forecasting method. This method uses a two-dimensional graph with sales measured along the vertical axis and time measured along the horizontal axis. After plot- ting the sales data month by month, this method finds a line passing through the midst of the data as closely as possible. The extension of the line into future months provides the forecast of sales in these future months.
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386 Chapter Ten Forecasting
Section 10.5 fully presents the linear regression method. The other forecasting methods mentioned above are described in detail in Section 10.3. The discussion in both sections is in the context of the case study introduced in Section 10.2.
Which of these forecasting techniques should Fastchips use? On the basis of the sales data so far, it appears that either the moving-average forecasting method or the exponen- tial smoothing forecasting method would be a reasonable choice. However, as time goes on, further analysis should be done to see which of the forecasting methods provides the small- est forecasting errors (the difference between the actual sales and forecasted sales). After determining the forecasting error in each of a number of months for any forecasting method, one common measure of the accuracy of that method is the average of these forecasting errors. (This average is referred to as the mean absolute deviation , which is abbreviated as MAD . ) Because large forecasting errors are far more serious than small ones, another popular measure of the accuracy of a forecasting method is the average of the square of its forecasting errors. (This measure is referred to as the mean square error , which is abbrevi- ated as MSE . ) Throughout this chapter, MAD and MSE values are used to help analyze which forecasting method should be used in the case study.
For some types of products, the sales to be anticipated in a particular month are influenced by the time of the year. For example, a product which is popular for Christmas presents could well have December sales that are twice as large as January sales. For any product influenced by seasonal factors, it is important to incorporate these seasonal factors into the forecasts. This plays a fundamental role in the analysis of the case study throughout the chapter.
Although we have described the various forecasting techniques in terms of forecasts of sales month by month for the Fastchips problem, other forecasting applications can be somewhat different. The quantity being forecasted might be something other than sales and the periods involved might be something like quarters or years instead of months. For example, this chap- ter’s case study involves forecasts of the number of calls to a call center on a quarterly basis.
When using any of these forecasting techniques, it is also important to look behind the numbers to try to understand what is driving the quantity being forecasted in order to adjust the forecast provided by the forecasting technique in an appropriate way. This is a key lesson provided by the analysis of the case study. When there are factors that are driving changes in the quantity being fore- casted, the judgmental forecasting methods described in Section 10.6 also can play a useful role.
A summary of the formulas for all the forecasting techniques described above (and elabo- rated upon throughout the chapter) is provided at the end of the chapter.
1. What is the last-value forecasting method and when might it be a reasonable method to use? 2. What is the averaging forecasting method and when might it be a reasonable method to use? 3. What is the moving-average forecasting method and when might it be a reasonable method
to use? 4. How does the exponential smoothing forecasting method differ from the moving-average
forecasting method? 5. How does exponential smoothing with trend differ from the exponential smoothing forecast-
ing method? 6. How does the linear regression forecasting method obtain forecasts? 7. What are the two main measures of the accuracy of a forecasting method?
Review Questions
10.2 A CASE STUDY: THE COMPUTER CLUB WAREHOUSE (CCW) PROBLEM
The Computer Club Warehouse (commonly referred to as CCW ) sells various computer prod- ucts at bargain prices by taking telephone orders (as well as website and fax orders) directly from customers. Its products include desktop and laptop computers, peripherals, hardware accesso- ries, supplies, software (including games), and computer-related furniture. The company mails catalogs to its customers and numerous prospective customers several times per year, as well as publishing minicatalogs in computer magazines. These catalogs prominently display the 800 toll- free telephone number to call to place an order. These calls come into the company’s call center.
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10.2 A Case Study: The Computer Club Warehouse (CCW) Problem 387
The CCW Call Center The call center is never closed. During busy hours, it is staffed by dozens of agents. Their sole job is to take and process customer orders over the telephone. (A second, much smaller call center uses another 800 number for callers making inquiries or reporting problems. This case study focuses on just the main call center.)
New agents receive a week’s training before beginning work. This training emphasizes how to efficiently and courteously process an order. An agent is expected not to average more than five minutes per call. Records are kept and an agent who does not meet this target by the end of the probationary period will not be retained. Although the agents are well-paid, the tedium and time pressure associated with the job leads to a fairly high turnover rate.
A large number of telephone trunks are provided for incoming calls. If an agent is not free when the call arrives, it is placed on hold with a recorded message and background music. If all the trunks are in use (referred to as saturation ), an incoming call receives a busy signal instead.
Although some customers who receive a busy signal, or who hang up after being on hold too long, will try again later until they get through, many do not. Therefore, it is very impor- tant to have enough agents on duty to minimize these problems. On the other hand, because of the high labor costs for the agents, CCW tries to avoid having so many agents on duty that they have significant idle time.
Consequently, obtaining forecasts of the demand for the agents is crucial to the company.
The Call Center Manager, Lydia Weigelt The current manager of the call center is Lydia Weigelt. As the top student in her graduating class from business school, she was wooed by several top companies before choosing CCW. Extremely bright and hard driving, she is being groomed to enter top management at CCW in the coming years.
When Lydia was hired a little over three years ago, she was assigned to her current posi- tion in order to learn the business from the ground up. The call center is considered to be the nerve center of the entire CCW operation.
Before Lydia’s arrival, the company had suffered from serious management problems with the call center. Orders were not being processed efficiently. A few were even misdirected. Staffing levels never seemed to be right. Management directives to adjust the levels kept overcompensating in the opposite direction. Data needed to get a handle on the staffing level problem hadn’t been kept. Morale was low.
All that changed when Lydia arrived. One of her first moves was to install procedures for gathering the data needed to make decisions on staffing levels. The key data included a detailed record of call volume and how much of the volume was being handled by each agent. Efficiency improved substantially. Despite running a tight ship, Lydia took great pains to praise and reward good work. Morale increased dramatically.
Although gratified by the great improvement in the operation of the call center, Lydia still has one major frustration. At the end of each quarter, when she knows how many agents are not being retained at the end of their probationary period, she makes a decision on how many new agents to hire to go through the next training session (held at the beginning of each quarter). She has developed an excellent procedure for estimating the staffing level needed to cover any particular call volume. However, each time she has used this procedure to set the staffing level for the upcoming quarter, based on her forecast of the call volume, the forecast usually has turned out to be considerably off. Therefore, she still isn’t getting the right staff- ing levels.
Lydia has concluded that her next project should be to develop a better forecasting method to replace the current one.
Lydia’s Current Forecasting Method Thanks to Lydia’s data-gathering procedures installed shortly after her arrival, reliable data on call volume now are available for the past three years. Figure 10.1 shows the average num- ber of calls received per day in each of the four quarters of these years. The right side also dis- plays these same data to show the pattern graphically. This graph was generated by selecting the data in D4:D15 and then either selecting a Line chart from the Insert tab (for Excel 2010) or Chart tab (for Excel 2011).
Better forecasts of call vol- ume are needed.
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388
Note that the sales in Quarter 4 jump up each year due to Christmas purchases. When Lydia first joined CCW, the president told her about the “25 percent rule” that the company had traditionally used to forecast call volume (and sales).
The 25 Percent Rule: Since sales are relatively stable through the year except for a substantial increase during the Christmas season, assume that each quarter’s call volume will be the same as for the preceding quarter, except for adding 25 percent for Quarter 4. Thus,
Forecast for Quarter 2 5 Call volume for Quarter 1 Forecast for Quarter 3 5 Call volume for Quarter 2 Forecast for Quarter 4 5 1.25(Call volume for Quarter 3)
The forecast for the next year’s Quarter 1 then would be obtained from the current year’s Quarter 4 by
Forecast for next Quarter 1 5 Call volume for Quarter 4
1.25
Forecasts need to take into account the seasonal pat- tern of increased sales in Quarter 4 due to Christmas purchases.
1
A B C
Year Quarter
D E F G H I
2
3
4
5
6
7
8
1
1
1
1
2
1
2
3
4
1
9
10
11
12
13
14
15
2
2
2
3
3
3
3
2
3
4
1
2
3
4
Call Volume
CCW's Average Daily Call Volume
6,000 1
7,000
8,000
9,000
10,000
2 3 4 1 2 3 4 1 2 3 4
Year 1
Quarter
Year 2 Year 3
A ve
ra ge
D ai
ly C
al l V
ol u
m e
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
FIGURE 10.1 The average number of calls received per day at the CCW call center in each of the four quarters of the past three years.
L.L. Bean, Inc., is a widely known retailer of high-quality outdoor goods and apparel, with an annual sales volume close to $2 billion. The company markets its products pri- marily through the mailing of millions of copies of various catalogs each year. Thus, in addition to online and mail orders, much of its sales volume is generated through orders taken at the company’s call center, which is open seven days per week. The sales volume is seasonal, with an especially large spike during the Christmas season. The sales within each week tend to slowly decrease from Monday through Sunday, except for a sharp drop on the day of a holiday and a sharp rise immediately after the arrival of a catalog.
Staffing the call center at an appropriate level day by day is critical for the company. Understaffing results in lost sales from customers who don’t get through to the call center and then give up. Overstaffing results in excessive labor costs. Therefore, accurate forecasts of daily call volumes are needed.
Because previous subjective forecasting methods had proven to be unsatisfactory, L.L. Bean management hired a team of management science consultants to improve the forecasting procedures for orders coming into the call
center. After L.L. Bean call-center managers compiled an exhaustive list of 35 possible factors that could logically affect call volumes, this team developed and helped imple- ment a highly sophisticated time-series forecasting method (the Box and Jenkins autoregressive/integrated/moving- average method). This methodology incorporates all the important factors, including seasonal patterns, the effect of holidays, and the effect of the arrival of catalogs. Each week, forecasts are obtained for the daily call volumes for the next three weeks. The forecasts for the last of the three weeks are then used to determine the Monday-to-Sunday work schedule at the call center two weeks in advance.
The improved precision of this forecasting methodol- ogy is estimated to have saved L.L. Bean $300,000 annually through enhanced scheduling efficiency. The computeriza- tion of the methodology also greatly reduced the labor costs required to prepare the forecast every week.
Source: B. H. Andrews and S. M. Cunningham, “L.L. Bean Improves Call-Center Forecasting,” Interfaces 25, no. 6 (November– December 1995), pp. 1–13. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
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10.2 A Case Study: The Computer Club Warehouse (CCW) Problem 389
This is the forecasting method that Lydia has been using. Figure 10.2 shows the forecasts that Lydia obtained with this method. Column F gives the
forecasting error (the deviation of the forecast from what then turned out to be the true value of the call volume) in each case. Since the total of the 11 forecasting errors is 4,662, the average is
Average forecasting error 5 4,662
11 5 424
As mentioned in Section 10.1, the average forecasting error is commonly called MAD, which stands for mean absolute deviation. Its formula is
MAD 5 Sum of forecasting errors
Number of forecasts
MAD is simply the average of the forecasting errors.
1
A B C
Year Quarter
D E F G H I J K L
2
3
4
5
6
7
8
1
1
1
1
2
1
3
4
19
10
11
12
13
14
15
16
17
18
19
20
2
2
2
2
3
3
3
3
4
4
4
4
2
3
4
1
2
3
4
1
2
3
4
Data Forecast Error
Forecasting
344
104
55
644
193
720
1,006
13
170
1,127
286
Lydia's Current Forecasting Method for CCW's Average Daily Call Volume
Mean Absolute Deviation
MAD =
Mean Square Error
MSE =
Data
Forecast
ForecastingError
D5:D20
E5:E20
F5:F20
Range Name Cells
Forecast Error
Forecasting3
4
5
6
7
8
9
10
11
12
=D5
=D6
=1.25*D7
=D8/1.25
=D9
:
:
=ABS(D6-E6)
=ABS(D7-E7)
=ABS(D8-E8)
=ABS(D9-E9)
=ABS(D10-E10)
:
:
E F
=AVERAGE(ForecastingError)MAD =
MSE =
5
H I
=SUMSQ(ForecastingError)/ COUNT(ForecastingError)8
H I
424
317,815
6,809
6,465
8,211
6,613
7,257
7,064
9,730
6,979
6,992
6,822
9,936
7,720
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
6,000 1
7,000
8,000
9,000
10,000
2 3 4 1 2 3 4 1 2 3 4 1
Year 1
Quarter
True Value Forecast
Year 2 Year 3 Year 4
A ve
ra ge
D ai
ly C
al l
V ol
u m
e
FIGURE 10.2 This spreadsheet records the results of applying the 25 percent rule over the past three years to forecast the average daily call volume for the upcoming quarter.
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390 Chapter Ten Forecasting
Thus, in this case, cell I5 gives
MAD 5 424
To put this value of MAD 5 424 into perspective, note that 424 is over 5 percent of the aver- age daily call volume in most quarters. With forecasting errors ranging as high as 1,127, two of the errors are well over 10 percent. Although errors of this size are common in typical applications of forecasting, greater accuracy is needed for this particular application. Errors of 5 and 10 percent make it impossible to properly set the staffing level for a quarter. No wonder Lydia is mad about the poor job that the 25 percent rule is doing. A better forecasting method is needed.
Another popular measure of the accuracy of forecasting methods is called the mean square error and is abbreviated as MSE. Its formula is
MSE 5 Sum of square of forecasting errors
Number of forecasts
Thus, in Figure 10.2 ,
MSE 5 (344)2 1 (104)2 1 c1 (286)2
11 5 317,815
as given in cell I8. The advantage of squaring the forecasting errors is that it increases the weight of the large errors relative to the weight given to the small errors. Small errors are only to be expected with even the best forecasting methods and, since such errors have no seri- ous consequences, decreasing their weight is desirable. It is the large forecasting errors that have serious consequences. Therefore, it is good to give a relatively large penalty to a fore- casting method that occasionally allows large forecasting errors while rewarding a forecast- ing method that consistently keeps the errors reasonably small. When comparing two such methods, this can result in the former kind of method receiving the larger value of MSE even though it has the smaller MAD value. Thus, MSE provides a useful complement to MAD by providing additional information about how consistently a forecasting method avoids seriously large errors. However, the disadvantage of MSE compared to MAD is a greater difficulty in interpreting the significance of its value for an individual forecasting method. Consequently, Lydia (who is familiar with both measures) will focus most of her attention on MAD values while keeping her eye on MSE values as well.
The Plan to Find a Better Forecasting Method Lydia remembers taking a management science course in college. She recalls that one of the topics in the course was forecasting, so she decides to review her textbook and class notes on this topic.
This review reminds her that she is dealing with what is called a time series.
A time series is a series of observations over time of some quantity of interest. For example, the series of observations of average daily call volume for the most recent 12 quarters, as given in Figure 10.1 , constitute a time series.
She also is reminded that a variety of statistical methods are available for using the historical data from a time series to forecast a future observation in the series. These types of statistical forecasting methods are referred to as time-series forecasting methods . Her task now is to review these methods, along with other statistical forecasting methods, and assess which one is best suited for her particular forecasting problem.
To assist her in this task for a few weeks, Lydia gains approval from the CCW president to contract for the services of a consultant (a former classmate) from a management science consulting firm that specializes largely in forecasting.
The next section describes their approach to the problem.
MSE is the average of the square of the forecasting errors.
A bad forecasting error greatly increases the value of the MSE.
1. How does the Computer Club Warehouse (CCW) operate? 2. What are the consequences of not having enough agents on duty in the CCW call center? Of
having too many? 3. What is Lydia’s current major frustration?
Review Questions
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10.3 Applying Time-Series Forecasting Methods to the Case Study 391
4. What is CCW’s 25 percent rule? 5. What is MAD? 6. What is MSE? 7. What is a time series?
10.3 APPLYING TIME-SERIES FORECASTING METHODS TO THE CASE STUDY
As mentioned at the end of the preceding section, time-series forecasting methods are statis- tical forecasting methods that use a time series (a series of observations over time of some quantity of interest) to forecast a future observation in the series based on its previous values. Several of these methods were introduced in Section 10.1. We focus in this section on further describing these methods and applying them to the case study, where the quantity of interest to be forecasted is the average daily call volume for the next quarter.
Figure 10.1 in the preceding section highlights the seasonal pattern of CCW’s call vol- umes, with a large jump up each fourth quarter due to Christmas shopping. Therefore, before considering specific forecasting methods, Lydia and the consultant begin by addressing how to deal with this seasonal pattern.
Considering Seasonal Effects For many years, the folklore at CCW has been that the call volume (and sales) will be pretty stable over the first three quarters of a year and then will jump up by about 25 percent in Quarter 4. This has been the basis for the 25 percent rule.
To check how close this folklore still is to reality, the consultant uses the data previously given in Figure 10.1 to calculate the average daily call volume for each quarter over the past three years. For example, the average for Quarter 1 is
Average (Quarter1) 5 6,809 1 7,257 1 6,992
3 5 7,019
These averages for all four quarters are shown in the second column of Table 10.1 . Underneath this column, the overall average over all four quarters is calculated to be 7,529. Dividing the aver- age for each quarter by this overall average gives the seasonal factor shown in the third column.
In general, the seasonal factor for any period of a year (a quarter, a month, etc.) measures how that period compares to the overall average for an entire year. Specifically, using historical data, the seasonal factor is calculated to be
Seasonal factor 5 Average for the period
Overall average
This Excel template cal- culates seasonal factors on either a monthly or quar- terly basis.
Quarter Three-Year
Average Seasonal
Factor
1 7,019 7,019 7,529
5 0.93
2 6,784 6,784 7,529
5 0.90
3 7,434 7,434 7,529
5 0.99
4 8,880 8,880 7,529
5 1.18
Total 5 30,117
Average 5 30,117
4 5 7,529
TABLE 10.1 Calculation of the Seasonal Factors for the CCW Problem
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392 Chapter Ten Forecasting
Your MS Courseware includes an Excel template for calculating these seasonal factors. Figure 10.3 shows this template applied to the CCW problem.
Note the significant differences in the seasonal factors for the first three quarters, with Quarter 3 considerably above the other two. This makes sense to Lydia, who has long sus- pected that back-to-school buying should give a small boost to sales in Quarter 3.
In contrast to the 25 percent rule, the seasonal factor for Quarter 4 is only 19 percent higher than that for Quarter 3. (However, the Quarter 4 factor is about 25 percent above 0.94, which is the average of the seasonal factors for the first three quarters.)
Although data on call volumes are not available prior to the most recent three years, reli- able sales data have been kept. Upon checking these data several years back, Lydia finds the same seasonal patterns occurring.
Conclusion: The seasonal factors given in Table 10.1 appear to accurately reflect subtle but important differences in all the seasons. Therefore, these factors now will be used, instead of the 25 percent rule, to indicate seasonal patterns until such time as future data indicate a shift in these patterns.
The Seasonally Adjusted Time Series It is much easier to analyze sales data and detect new trends if the data are first adjusted to remove the effect of seasonal patterns. To remove the seasonal effects from the time series shown in Figure 10.1 , each of these average daily call volumes needs to be divided by the cor- responding seasonal factor given in Table 10.1 and Figure 10.3 . Thus, the formula is
Seasonally adjusted call volume 5 Actual call volume
Seasonal factor
Applying this formula to all 12 call volumes in Figure 10.1 gives the seasonally adjusted call volumes shown in column F of the Excel template in Figure 10.4 .
1
A B C D E F G
2
3
4
5
6
7
8
9
10
11
12
13
Year
1
1
1
1
2
2
2
2
Quarter
True
Value Type of Seasonality
3
Quarter
1
2
3
4
Estimate for Seasonal Factor
14 3
15 3
16 3
Estimating Seasonal Factors for CCW
SeasonalFactor
TrueValue
TypeOfSeasonality
G10:G21
D5:D41
F5
Range Name Cells
1
2
3
4
1
2
3
4
1
2
3
4
Quarterly
Estimate for
Seasonal Factor
=AVERAGE(D5,D9,D13)/AVERAGE(TrueValue)
=AVERAGE(D6,D10,D14)/AVERAGE(TrueValue)
=AVERAGE(D7,D11,D15)/AVERAGE(TrueValue)
=AVERAGE(D8,D12,D16)/AVERAGE(TrueValue)
8
9
10
11
12
13
G
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
0.9323
0.9010
0.9873
1.1794
FIGURE 10.3 The Excel template in your MS Courseware for calculating seasonal fac- tors is applied here to the CCW problem.
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10.3 Applying Time-Series Forecasting Methods to the Case Study 393
In effect, these seasonally adjusted call volumes show what the call volumes would have been if the calls that occur because of the time of the year (Christmas shopping, back-to- school shopping, etc.) had been spread evenly throughout the year instead. Compare the plots in Figures 10.4 and 10.1 . After considering the smaller vertical scale in Figure 10.4 , note how much less fluctuation this figure has than Figure 10.1 because of removing seasonal effects. However, this figure still is far from completely flat because fluctuations in call volume occur for other reasons besides just seasonal effects. For example, hot new products attract a flurry of calls. A jump also occurs just after the mailing of a catalog. Some random fluctuations occur without any apparent explanation. Figure 10.4 enables seeing and analyzing these fluc- tuations in sales volumes that are not caused by seasonal effects.
The pattern in these remaining fluctuations in the seasonally adjusted time series (especially the pattern for the most recent data points) is particularly helpful for forecast- ing where the next data point will fall. Thus, in Figure 10.4 , the data points fall in the range between 6,635 and 8,178, with an average of 7,529. However, the last few data points are trending upward above this average, and the last point is the highest in the entire time series. This suggests that the next data point for the upcoming quarter prob- ably will be above the 7,529 average and may well be near or even above the last data point of 8,178.
The various time-series forecasting methods use different approaches to projecting forward the pattern in the seasonally adjusted time series to forecast the next data point. The main methods will be presented in this section.
Removing seasonal effects provides a much clearer picture of trends.
1
A B C
Year Quarter
D E F G H I J
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
1
1
1
2
2
2
2
3
3
3
3
1
2
3
4
1
2
3
4
1
2
3
4
Factor Call Volume
Seasonal Actual
Call Volume
Seasonally Adjusted
Call Volume
Seasonally Adjusted
Seasonally Adjusted Time Series for CCW
0.93
0.90
0.99
1.18
0.93
0.90
0.99
1.18
0.93
0.90
0.99
1.18
3
4
5
6
7
8
9
10
=E5/D5
=E6/D6
=E7/D7
=E8/D8
:
:
F
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
6,000
1
6,500
7,000
7,500
8,000
8,500
9,000
2 3 4 1 2 3 4 1 2 3 4 Year 1
Quarter Year 2 Year 3A
ve ra
ge D
ai ly
C al
l V ol
u m
e
7,322
7,183
6,635
7,005
7,803
7,849
7,863
7,393
7,518
7,580
8,029
8,178
FIGURE 10.4 The seasonally adjusted time series for the CCW problem obtained by dividing each actual average daily call volume in Figure 10.1 by the corresponding seasonal factor obtained in Figure 10.3.
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394 Chapter Ten Forecasting
After obtaining a forecast for the seasonally adjusted time series, all these methods then convert this forecast to a forecast of the actual call volume (without seasonal adjustments), as outlined below.
Outline for Forecasting Call Volume 1. Select a time-series forecasting method. 2. Apply this method to the seasonally adjusted time series to obtain a forecast of the season-
ally adjusted call volume for the next quarter. 1 3. Multiply this forecast by the corresponding seasonal factor in Table 10.1 to obtain a fore-
cast of the actual call volume (without seasonal adjustment).
The following descriptions of forecasting methods focus on how to perform step 2, that is, how to forecast the next data point for a given time series. We also include a spreadsheet in each case that applies steps 2 and 3 throughout the past three years and then calculates both MAD (the average forecasting error) and MSE (the average of the square of the forecasting errors). Lydia and the consultant are paying particular attention to the MAD values to assess which method seems best suited for forecasting CCW call volumes.
The Last-Value Forecasting Method The last-value forecasting method ignores all the data points in a time series except the last one. It then uses this last value as the forecast of what the next data point will turn out to be, so the formula is simply
Forecast 5 Last value
Figure 10.5 shows what would have happened if this method had been applied to the CCW problem over the past three years. (We are supposing that the seasonal factors given in Table 10.1 already were being used then.) Column E gives the true values of the seasonally adjusted call volumes from column F of Figure 10.4 . Each of these values then becomes the seasonally adjusted forecast for the next quarter, as shown in column F.
Rows 22–33 show separate plots of these values in columns E and F. Note how the plot of the seasonally adjusted forecasts follows exactly the same path as the plot of the seasonally adjusted call volumes but shifted to the right by one quarter. Therefore, each time there is a large shift up or down in the call volume, the forecasts are one quarter late in catching up with the shift.
Multiplying each seasonally adjusted forecast in column F by the corresponding seasonal factor in column K gives the forecast of the actual call volume (without seasonal adjustment) presented in column G. The difference between this forecast and the actual call volume in column D gives the forecasting error in column H.
Thus, column G is using the following formula:
Actual forecast 5 Seasonal factor 3 Seasonally adjusted forecast
as indicated by the equations at the bottom of the figure. For example, since cell K9 gives 0.93 as the seasonal factor for Quarter 1, the forecast of the actual call volume for Year 2, Quarter 1 given in cell G10 is
Actual forecast 5 (0.93)(7,005) 5 6,515
Since the true value of this call volume turned out to be 7,257, the forecasting error calculated in cell H10 for this quarter is
Forecasting error 5 7,257 2 6,515 5 742
Summing these forecasting errors over all 11 quarters of forecasts gives a total of 3,246, so the average forecasting error given in cell K23 is
MAD 5 3,246
11 5 295
If seasonal adjustments are not needed, you can obtain your forecast directly from the original time series and then skip step 3.
1 This forecast also can be projected ahead to subsequent quarters, but we are focusing on just the next quarter.
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10.3 Applying Time-Series Forecasting Methods to the Case Study 395
1
A B C
Year Quarter
D E F G H I J K
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
1
1
1
1
2
2
2
2
3
3
3
3
4
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
Quarter
Type of Seasonality
1
2
3
4
True
Value
Seasonally
Adjusted
Value
Seasonally
Adjusted
Forecast
Actual
Forecast
Forecasting
Error
7,322
7,183
6,635
7,005
7,803
7,849
7,863
7,393
7,518
7,580
8,029
8,178
124
543
436
742
41
14
554
116
56
445
175
Last-Value Forecasting Method with Seasonality for CCW
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
7,322
7,183
6,635
7,005
7,803
7,849
7,863
7,393
7,518
7,580
8,029
8,178
Seasonal Factor
0.93
0.90
0.99
1.18
Mean Absolute Deviation
MAD = 295
Mean Square Error
MSE = 145,909
ActualForecast
ForecastingError
MAD
MSE
SeasonalFactor
SeasonallyAdjustedForecast
SeasonallyAdjustedValue
TrueValue
TypeOfSeasonality
G6:G30
H6:H30
K23
K26
K9:K20
F6:F30
E6:E30
D6:D30
K6
Range Name Cells Seasonally
Adjusted
Value
Seasonally
Adjusted
Forecast
Actual
Forecast
Forecasting
Error
3
4
5
6
7
8
9
10
11
12
=D6/K9
=D7/K10
=D8/K11
=D9/K12
=D10/K9
=D11/K10
=K10*F7
=K11*F8
=K12*F9
=K9*F10
=K10*F11
=ABS(D7-G7)
=ABS(D8-G8)
=ABS(D9-G9)
=ABS(D10-G10)
=ABS(D11-G11)
:
=E6
=E7
=E8
=E9
=E10
: : :
E F G H
=AVERAGE(ForecastingError)MAD = MSE =23
J K
=SUMSQ(ForecastingError)/COUNT(ForecastingError)26
J K
6,000
1 2 3 4 1 2 3 4 1 2 3 4 1
6,500
7,000
7,500
8,000
8,500
9,000
Quarter
Year 2Year 1 Year 3 Year 4
S ea
so n
al ly
A d
ju st
ed A
ve ra
ge D
ai ly
C al
l V ol
u m
e
Seasonally
adjusted
value
Seasonally
adjusted
forecast
6,589
7,112
7,830
6,515
7,023
7,770
9,278
6,876
6,766
7,504
9,475
7,606
Quarterly
13 : : : :
FIGURE 10.5 The Excel template in your MS Courseware for the last-value method with seasonal adjustments is applied here to the CCW problem.
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396 Chapter Ten Forecasting
This compares with MAD 5 424 for the 25 percent rule that Lydia has been using (as described in the preceding section).
Similarly, the average of the square of these forecasting errors is calculated in cell K26 as
MSE 5 (124)2 1 (543)2 1 c1 (175)2
11 5 145,909
This value also is considerably less than the corresponding value, MSE 5 317,815, shown in Figure 10.2 for the 25 percent rule.
Except for its graph, Figure 10.5 displays one of the templates in this chapter’s Excel files. In fact, your MS Courseware includes two Excel templates for each of the forecast- ing methods presented in this section. One template performs all the calculations for you for the case where no seasonal adjustments are needed. The second template does the same when seasonal adjustments are included, as illustrated by this figure. With all templates of the second type, you have complete flexibility for what to enter as the seasonal factors. One option is to calculate these factors based on historical data (as was done with another Excel template in Figure 10.3 ). Another is to estimate them based on historical experience, as with the 25 percent rule.
The 25 percent rule actually is a last-value forecasting method as well, but with different seasonal factors. Since this rule holds that the call volume in the fourth quarter will average 25 percent more than each of the first three quarters, its seasonal factors are essentially 0.94 for Quarters 1, 2, and 3 and 1.18 (25 percent more than 0.94) for Quarter 4. Thus, the lower value of MAD in Figure 10.5 is entirely due to refining the seasonal factors in Table 10.1 .
Lydia is enthusiastic to see the substantial improvement obtained by simply refining the seasonal factors. However, the consultant quickly adds a note of caution. The forecasts obtained in Figure 10.5 are using the same data that were used to calculate these refined sea- sonal factors, which creates some bias for these factors to tend to perform better than on new data (future call volumes). Fortunately, Lydia also has checked older sales data to confirm that these seasonal factors seem quite accurate. The consultant agrees that it appears that these factors should provide a significant improvement over the 25 percent rule.
The last-value forecasting method sometimes is called the naive method , because statis- ticians consider it naive to use just a sample size of one when additional relevant data are avail- able. However, when conditions are changing rapidly, it may be that the last value is the only relevant data point for forecasting the next value under current conditions. Therefore, manag- ers who are anything but naive do occasionally use this method under such circumstances.
The Averaging Forecasting Method The averaging forecasting method goes to the other extreme. Rather than using just a sample size of one, this method uses all the data points in the time series and simply averages these points. Thus, the forecast of what the next data point will turn out to be is
Forecast 5 Average of all data to date
Using the corresponding Excel template to apply this method to the CCW problem over the past three years gives the seasonally adjusted forecasts shown in column F of Figure 10.6 . At the bottom of the figure, the equation entered into each of the column F cells is just the average of the column E cells in the preceding rows. The middle of the figure shows a plot of these seasonally adjusted forecasts for all three years next to the true values of the seasonally adjusted call volumes. Note how each forecast lies at the average of all the preceding call vol- umes. Therefore, each time there is a large shift in the call volume, the subsequent forecasts are very slow in catching up with the shift.
Multiplying all the seasonally adjusted forecasts in column F by the corresponding sea- sonal factors in column K then gives the forecasts of the actual call volumes shown in column G. Based on the resulting forecasting errors given in column H, the average forecasting error in this case (cell K23) is
MAD 5 400
This method is a good one to use when conditions are changing rapidly.
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10.3 Applying Time-Series Forecasting Methods to the Case Study 397
1
A B C
Year Quarter
D E F G H I J K
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
1
1
1
1
2
2
2
2
3
3
3
3
4
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
Quarter
Type of Seasonality
1
2
3
4
True
Value
Seasonally
Adjusted
Value
Seasonally
Adjusted
Forecast
Actual
Forecast
Forecasting
Error
7,322
7,252
7,047
7,036
7,190
7,300
7,380
7,382
7,397
7,415
7,471
7,530
124
611
49
713
593
557
16
127
165
608
834
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
7,322
7,183
6,635
7,005
7,803
7,849
7,863
7,393
7,518
7,580
8,029
8,178
Quarterly
Seasonal Factor
0.93
0.90
0.99
1.18
Mean Absolute Deviation
MAD = 400
Mean Square Error
MSE = 242,876
ActualForecast
ForecastingError
MAD
MSE
SeasonalFactor
SeasonallyAdjustedForecast
SeasonallyAdjustedValue
TrueValue
TypeOfSeasonality
G6:G30
H6:H30
K23
K26
K9:K20
F6:F30
E6:E30
D6:D30
K6
Range Name Cells Seasonally
Adjusted
Value
Seasonally
Adjusted
Forecast
Actual
Forecast
Forecasting
Error
3
4
5
6
7
8
9
10
11
12
=D6/K9
=D7/K10
=D8/K11
=D9/K12
=D10/K9
=D11/K10
=K10*F7
=K11*F8
=K12*F9
=K9*F10
=K10*F11
=ABS(D7-G7)
=ABS(D8-G8)
=ABS(D9-G9)
=ABS(D10-G10)
=ABS(D11-G11)
:
=AVERAGE(E$6:E6)
=AVERAGE(E$6:E7)
=AVERAGE(E$6:E8)
=AVERAGE(E$6:E9)
=AVERAGE(E$6:E10)
: : :
E F G H
=AVERAGE(ForecastingError)MAD = MSE =23
J K
=SUMSQ(ForecastingError)/COUNT(ForecastingError)26
J K
Averaging Forecasting Method with Seasonality for CCW
6,000
1 2 3 4 1 2 3 4 1 2 3 4 1
6,500
7,000
7,500
8,000
8,500
9,000
Quarter
Year 2Year 1 Year 3 Year 4
S ea
so n
al ly
A d
ju st
ed A
ve ra
ge D
ai ly
C al
l V ol
u m
e
Seasonally
adjusted
value
Seasonally
adjusted
forecast
6,589
7,180
8,315
6,544
6,471
7,227
8,708
6,865
6,657
7,341
8,816
7,003
13 : : : :
FIGURE 10.6 The Excel template in your MS Courseware for the averaging method with seasonal adjustments is applied here to the CCW problem.
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which is considerably larger than the 295 obtained for the last-value forecasting method. Similarly, the average of the square of the forecasting errors given in cell K26 is
MSE 5 242,876
which also is considerably larger than the corresponding value of 145,909 for the last-value forecasting method.
Lydia is quite surprised, since she expected an average to do much better than a sample size of one. The consultant agrees that averaging should perform considerably better if con- ditions remain the same throughout the time series. However, it appears that the conditions affecting the CCW call volume have been changing significantly over the past three years. The call volume was quite a bit higher in year 2 than in year 1, and then jumped up again late in year 3, apparently as popular new products became available. Therefore, the Year 1 values were not very relevant for forecasting under the changed conditions of years 2 and 3. Includ- ing the year 1 call volumes in the overall average caused every forecast for years 2 and 3 to be too low, sometimes by large amounts.
The Moving-Average Forecasting Method Rather than using old data that may no longer be relevant, the moving-average forecasting method averages the data for only the most recent time periods. Let
n 5 Number of most recent periods considered particularly relevant for forecasting the next period
Then the forecast for the next period is
Forecast 5 Average of last n values
Lydia and the consultant decide to use n 5 4, since conditions appear to be relatively stable for only about four quarters (one year) at a time.
With n 5 4, the first forecast becomes available after four quarters of call volumes have been observed. Thus, the initial seasonally adjusted forecasts in cells F10:F12 of Figure 10.7 are
Y2, Q1: Seas. adj. forecast 5 7,322 1 7,186 1 6,635 1 7,005
4 5 7,036
The averaging forecasting method is a good one to use when conditions are very stable, which is not the case for CCW.
Taco Bell Corporation has over 6,500 quick-service restau- rants in the United States and a growing international market. It serves approximately 2 billion meals per year, generating about $5.4 billion in annual sales income.
At each Taco Bell restaurant, the amount of business is highly variable throughout the day (and from day to day), with a heavy concentration during the normal meal times. Therefore, determining how many employees should be scheduled to perform what functions in the restaurant at any given time is a complex and vexing problem.
To attack this problem, Taco Bell management instructed a team of management scientists (including several con- sultants) to develop a new labor–management system. The team concluded that the system needed three major components: (1) a forecasting model for predicting cus- tomer transactions at any time, (2) a computer simulation model (such as those described in Chapters 12 and 13) to translate customer transactions to labor requirements, and (3) an integer programming model to schedule employees to satisfy labor requirements and minimize payroll.
To enable application of a forecasting model in each restaurant, a procedure is needed to continually gather
data on the number of customer transactions during each 15-minute interval throughout the day, each day of the week. Therefore, the management science team devel- oped and implemented a rolling database containing six weeks of in-store and drive-through transaction data to be stored on each restaurant’s computer. After some testing of alternative forecasting methods, the team concluded that a six-week moving average is best. In other words, the forecast of the number of transactions in a particular 15-minute period on a particular day of the week should be the average of the number of transactions during the corresponding period in the preceding six weeks. However, the restaurant manager has the authority to modify the forecast if unusual events distorted the data being used.
The implementation of this forecasting procedure along with the other components of the labor–management sys- tem has provided Taco Bell with documented savings of $13 million per year in labor costs.
Source: J. Hueter and W. Swart, “An Integrated Labor–Manage- ment System for Taco Bell,” Interfaces 28, no. 1 (January–February 1998), pp. 75–91. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
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10.3 Applying Time-Series Forecasting Methods to the Case Study 399
Moving-Average Forecasting Method with Seasonality for CCW1
A B C
Year Quarter
D E F G H I J K
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
1
1
1
1
2
2
2
2
3
3
3
3
4
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
Quarter
Type of Seasonality
1
2
3
4
True
Value
Seasonally
Adjusted
Value
Seasonally
Adjusted
Forecast
Actual
Forecast
Forecasting
Error
7,036
7,157
7,323
7,630
7,727
7,656
7,589
7,630
7,826
713
623
534
279
194
68
436
646
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
7,322
7,183
6,635
7,005
7,803
7,849
7,863
7,393
7,518
7,580
8,029
8,178
Quarterly
4
Seasonal Factor
0.93
0.90
0.99
1.18
Mean Absolute Deviation
periods to consider Number of previous
MAD =
n =
437
Mean Square Error
MSE = 238,816
ActualForecast
ForecastingError
MAD
MSE
NumberOfPeriods
SeasonalFactor
SeasonallyAdjustedForecast
SeasonallyAdjustedValue
TrueValue
TypeOfSeasonality
G6:G30
H6:H30
K26
K29
K6
K12:K23
F6:F30
E6:E30
D6:D30
K9
Range Name Cells Seasonally
Adjusted
Value
Seasonally
Adjusted
Forecast
Actual
Forecast
Forecasting
Error
3
4
5
6
7
8
9
10
11
12
=D6/K12
=D7/K13
=D8/K14
=D9/K15
=D10/K12
=D11/K13
=D12/K14
=D13/K15
=D14/K12
=D15/K13
=K12*F10
=K13*F11
=K14*F12
=K15*F13
=K12*F14
=K13*F15
=ABS(D10-G10)
=ABS(D11-G11)
=ABS(D12-G12)
=ABS(D13-G13)
=ABS(D14-G14)
=ABS(D15-G15)
:
=AVERAGE(E6:E9)
=AVERAGE(E7:E10)
=AVERAGE(E8:E11)
=AVERAGE(E9:E12)
=AVERAGE(E10:E13)
=AVERAGE(E11:E14)
: : :
13
14
15
16
E F G H
=AVERAGE(ForecastingError)MAD = MSE =26
J K
=SUMSQ(ForecastingError)/COUNT(ForecastingError)29
J K
6,000
1 2 3 4 1 2 3 4 1 2 3 4 1
6,500
7,000
7,500
8,000
8,500
9,000
Quarter
Year 2Year 1 Year 3 Year 4
S ea
so n
al ly
A d
ju st
ed A
ve ra
ge D
ai ly
C al
l V ol
u m
e
Seasonally
adjusted
value
Seasonally
adjusted
forecast
6,544
6,441
7,250
9,003
7,186
6,890
7,513
9,004
7,279
17 : : : :
FIGURE 10.7 The Excel template in your MS Courseware for the moving-average method with seasonal adjustments is applied here to the CCW problem.
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Y2, Q2: Seas. adj. forecast 5 7,183 1 6,635 1 7,005 1 7,803
4 5 7,157
Y2, Q3: Seas. adj. forecast 5 6,635 1 7,005 1 7,803 1 7,849
4 5 7,323
Note how each forecast is updated from the preceding one by lopping off one observation (the oldest one) and adding one new one (the most recent observation).
Column F of Figure 10.7 shows all the seasonally adjusted forecasts obtained in this way with the equations at the bottom. For each of these forecasts, note in the plot how it lies at the average of the four preceding (seasonally adjusted) call volumes. Consequently, each time there is a large shift in the call volume, it takes four quarters for the forecasts to fully catch up with this shift (by which time another shift may already have occurred). Consequently, the average of the eight forecasting errors in column H is
MAD 5 437
the highest of any of the methods so far, including even the 25 percent rule. The average of the square of the forecasting errors is somewhat better at
MSE 5 238,816
since this is slightly lower than for the averaging method and considerably below the value for the 25 percent rule, but it is still substantially higher than for the last-value method.
Lydia is very puzzled about this surprisingly high MAD value. The moving-average method seemed like a very sensible approach to forecasting, with more rationale behind it than any of the previous methods. (It uses only recent history and it uses multiple observa- tions.) So why should it do so poorly?
The consultant explains that this is indeed a very good forecasting method when condi- tions remain pretty much the same over n time periods (or four quarters in this case). For example, the seasonally adjusted call volumes remained reasonably stable throughout year 2 and the first half of year 3. Consequently, the forecasting error dropped all the way down to 68 (cell H15) for the last of these six quarters. However, when conditions shift sharply, as with the big jump up in call volumes at the beginning of year 2, and then again in the middle of year 3, the next few forecasting errors tend to be very large.
Thus, the moving-average method is somewhat slow to respond to changing conditions. One reason is that it places the same weight on each of the last n values in the time series even though the older values may be less representative of current conditions than the last value observed.
The next method corrects this weighting defect.
The Exponential Smoothing Forecasting Method The exponential smoothing forecasting method modifies the moving-average method by placing the greatest weight on the last value in the time series and then progressively smaller weights on the older values. However, rather than needing to calculate a weighted average each time, it uses a simpler formula to obtain the same result.
This formula for forecasting the next value in the time series combines the last value and the last forecast (the one used one time period ago to forecast this last value) as follows:
Forecast 5 a(Last value) 1 (1 2 a)(Last forecast)
where a (the Greek letter alpha) is a constant between 0 and 1 called the smoothing constant . For example, if the last value in a time series (not the CCW time series) is 24, the last forecast is 20, and a 5 0.25, then
Forecast 5 0.25(24) 1 0.75(20)
5 21
Two Excel templates (one without seasonal adjustments and one with) are available in your MS Courseware for applying this formula to generate a series of forecasts (period by period) for a time series when you specify the value of a .
The moving-averaging forecasting method is a good one to use when conditions don’t change much over the number of time periods included in the average.
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10.3 Applying Time-Series Forecasting Methods to the Case Study 401
The choice of the value for the smoothing constant a has a substantial effect on the fore- cast, so the choice should be made with care. A small value (say, a 5 0.1) is appropriate if conditions are remaining relatively stable. However, a larger value (say, a 5 0.3) is needed if significant changes in the conditions are occurring relatively frequently. Because of the frequent shifts in the CCW seasonally adjusted time series, Lydia and the consultant conclude that a 5 0.5 would be an appropriate value. (The values selected for most applications are between 0.1 and 0.3, but a larger value can be used in this kind of situation.)
When making the first forecast, there is no last forecast available to plug into the right-hand side of the above formula. Therefore, to get started, a reasonable approach is to make an initial estimate of the average value anticipated for the time series. This initial estimate is used as the forecast for the first value, and then the formula is used to forecast the second value onward.
CCW call volumes have averaged just over 7,500 for the past three years, and the level of busi- ness just prior to Year 1 was comparable. Consequently, Lydia and the consultant decide to use
Initial estimate 5 7,500
to begin retrospectively generating the forecasts over the past three years. Recall that the first few seasonally adjusted call volumes are 7,322, 7,183, and 6,635. Thus, using the above for- mula with a 5 0.5 for the second quarter onward, the first few seasonally adjusted forecasts are
Y1, Q1: Seas. adj. forecast 5 7,500 Y1, Q2: Seas. adj. forecast 5 0.5(7,322) 1 0.5(7,500) 5 7,411 Y1, Q3: Seas. adj. forecast 5 0.5(7,183) 1 0.5(7,411) 5 7,297 Y1, Q4: Seas. adj. forecast 5 0.5(6,635) 1 0.5(7,297) 5 6,966
To see why these forecasts are weighted averages of the time series values to date, look at the calculations for Quarters 2 and 3. Since
0.5(7,322) 1 0.5(7,500) 5 7,411
the forecast for Quarter 3 can be written as
Seas. adj. forecast 5 0.5(7,183) 1 0.5(7,411) 5 0.5(7,183) 1 0.530.5(7,322) 1 0.5(7,500) 4 5 0.5(7,183) 1 0.25(7,322) 1 0.25(7,500) 5 7,297
Similarly, the forecast for Quarter 4 is
Seas. adj. forecast 5 0.5(6,635) 1 0.5(7,297) 5 0.5(6,635) 1 0.530.5(7,183) 1 0.25(7,322) 1 0.25(7,500) 4 5 0.5(6,635) 1 0.25(7,183) 1 0.125(7,322) 1 0.125(7,500) 5 6,966
Thus, this latter forecast places a weight of 0.5 on the last value, 0.25 on the next-to-last value, and 0.125 on the next prior value (the first one), with the remaining weight on the initial estimate. In subsequent quarters as more and more values are considered by the forecast, the same pattern continues of placing less and less weight on the values as they get older. Thus, with any choice of a , the weights prior to the initial estimate would be a , a (1 2 a ), a (1 2 a ) 2 , and so forth.
Therefore, choosing the value of a amounts to using this pattern to choose the desired progression of weights on the time series values. With frequent shifts in the time series, a large weight needs to be placed on the most recent value, with rapidly decreasing weights on older values. However, with a relatively stable time series, it is desirable to place a significant weight on many values in order to have a large sample size.
Further insight into the choice of a is provided by an alternative form of the forecasting formula.
Forecast 5 a(Last value) 1 (1 2 a)(Last forecast) 5 a(Last value) 1 Last forecast 2 a(Last forecast) 5 Last forecast 1 a(Last value 2 Last forecast)
The more unstable the conditions, the larger the smoothing constant a needs to be (but never bigger than 1).
The exponential smooth- ing forecasting method places the greatest weight on the last value and then decreases the weights as the values get older.
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402 Chapter Ten Forecasting
where the absolute value of (Last value 2 Last forecast) is just the last forecasting error. Therefore, the bottom form of this formula indicates that each new forecast is adjusting the last forecast by adding or subtracting the quantity a times the last forecasting error. If the forecasting error usually is mainly due to random fluctuations in the time-series values, then only a small value of a should be used for this adjustment. However, if the forecasting error often is largely due to a shift in the time series, then a large value of a is needed to make a substantial adjustment quickly.
Using a 5 0.5, the Excel template in Figure 10.8 provides all the results for CCW with this forecasting method. Rows 22–32 show a plot of all the seasonally adjusted forecasts next to the true values of the seasonally adjusted call volumes. Note how each forecast lies midway between the preceding call volume and the preceding forecast. Therefore, each time there is a large shift in the call volume, the forecasts largely catch up with the shift rather quickly. The resulting average of the forecasting errors in column H is given in cell K28 as
MAD 5 324
This is significantly smaller than for the previous forecasting methods, except for the value of MAD 5 295 for the last-value forecasting method. The same comparison also holds for the average of the square of the forecasting errors, which is calculated in cell K31 as
MSE 5 157,836
Lydia is somewhat frustrated at this point. She feels that she needs a method with average forecasting errors well below 295. Realizing that the last-value forecasting method is con- sidered the naive method, she had expected that such a popular and sophisticated method as exponential smoothing would beat it easily.
The consultant is somewhat surprised also. However, he points out that the difference between MAD 5 324 for exponential smoothing and MAD 5 295 for last-value forecasting is really too small to be statistically significant. If the same two methods were to be applied the next three years, exponential smoothing might come out ahead. Lydia is not impressed.
Although he isn’t ready to mention it to Lydia yet, the consultant is beginning to develop an idea for a whole new approach that might give her the forecasting precision she needs. But first he has one more time-series forecasting method to present.
To lay the groundwork for this method, the consultant explains a major reason why expo- nential smoothing did not fare well in this case. Look at the plot of seasonally adjusted call volumes in Figure 10.8 . Note the distinct trend downward in the first three quarters, then a sharp trend upward for the next two, and finally a major trend upward for the last five quar- ters. Also note the large gap between the two plots (meaning large forecasting errors) by the end of each of these trends. The reason for these large errors is that exponential smoothing forecasts lag well behind such a trend because they place significant weight on values near the beginning of the trend. Although a large value of a 5 0.5 helps, exponential smoothing forecasts tend to lag further behind such a trend than last-value forecasts.
The next method adjusts exponential smoothing by also estimating the current trend and then projects this trend forward to help forecast the next value in the time series.
Exponential Smoothing with Trend Exponential smoothing with trend uses the recent values in the time series to estimate any current upward or downward trend in these values. It is especially designed for the kind of time series depicted in Figure 10.9 where an upward (or downward) trend tends to continue for a considerable number of periods (but not necessarily indefinitely). This particular figure shows the estimated population of a certain state of the United States at midyear over a series of years. The line in the figure (commonly referred to as a trend line ) shows the basic trend that the time series is following, but with fluctuations on both sides of the line. Because the basic trend is upward in this case, forecasts based on any of the preceding forecasting meth- ods would tend to be considerably too low. However, by developing an estimate of the current slope of this trend line, and then adjusting the forecast to consider this slope, considerably more accurate forecasts should be obtained. This is the basic idea behind exponential smooth- ing with trend.
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10.3 Applying Time-Series Forecasting Methods to the Case Study 403
Exponential Smoothing Forecasting Method with Seasonality for CCW1
A B C
Year Quarter
D E F G H I J K
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
1
1
1
1
2
2
2
2
3
3
3
3
4
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
Quarter
Type of Seasonality
1
2
3
4
True
Value
Seasonally
Adjusted
Value
Seasonally
Adjusted
Forecast
Actual
Forecast
Forecasting
Error
7,500
7,411
7,297
6,966
6,986
7,394
7,622
7,742
7,568
7,543
7,561
7,795
7,987
166
205
655
46
760
409
239
412
46
33
463
451
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
7,322
7,183
6,635
7,005
7,803
7,849
7,863
7,393
7,518
7,580
8,029
8,178
Quarterly
Seasonal Factor
0.93
0.90
0.99
1.18
Mean Absolute Deviation
Initial Estimate
MAD =
Average =
Smoothing Constant
α =
324
Mean Square Error
MSE = 157,836
ActualForecast
Alpha
ForecastingError
InitialEstimate
MAD
MSE
SeasonalFactor
SeasonallyAdjustedForecast
SeasonallyAdjustedValue
TrueValue
TypeOfSeasonality
G6:G30
K5
H6:H30
K8
K28
K31
K14:K25
F6:F30
E6:E30
D6:D30
K11
Range Name Cells Seasonally
Adjusted
Value
Seasonally
Adjusted
Forecast
Actual
Forecast
Forecasting
Error
3
4
5
6
7
8
9
10
11
12
=D6/K14
=D7/K15
=D8/K16
=D9/K17
=D10/K14
=D11/K15
=K14*F6
=K15*F7
=K16*F8
=K17*F9
=K14*F10
=K15*F11
=ABS(D6-G6)
=ABS(D7-G7)
=ABS(D8-G8)
=ABS(D9-G9)
=ABS(D10-G10)
=ABS(D11-G11)
:
=InitialEstimate
=Alpha*E6+(1-Alpha)*F6
=Alpha*E7+(1-Alpha)*F7
=Alpha*E8+(1-Alpha)*F8
=Alpha*E9+(1-Alpha)*F9
=Alpha*E10+(1-Alpha)*F10
: : :
: : : :13
E F G H
=AVERAGE(ForecastingError)MAD = MSE =28
J K
=SUMSQ(ForecastingError)/COUNT(ForecastingError)31
J K
0.5
7,500
6,000
1 2 3 4 1 2 3 4 1 2 3 4 1
6,500
7,000
7,500
8,000
8,500
9,000
Quarter
Year 2Year 1 Year 3 Year 4
S ea
so n
al ly
A d
ju st
ed A
ve ra
ge D
ai ly
C al
l V ol
u m
e
Seasonally
adjusted
value
Seasonally
adjusted
forecast
6,975
6,670
7,224
8,220
6,497
6,655
7,545
9,136
7,038
6,789
7,486
9,199
7,428
FIGURE 10.8 The Excel template in your MS Courseware for the exponential smoothing method with seasonal adjustments is applied here to the CCW problem.
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404 Chapter Ten Forecasting
Trend is defined as
Trend 5 Average change from one time-series value to the next if the current pattern continues
The formula for forecasting the next value in the time series, then, is modified from the pre- ceding method by adding the estimated trend. Thus, the new formula is
Forecast 5 a(Last value) 1 (1 2 a)(Last forecast) 1 Estimated trend
(A separate box describes how this formula can be easily modified to forecast beyond the next value in the time series as well.)
Adding the estimated trend enables the forecast to keep up with the current trend in the data.
Population (Millions)
Year
5.4
5.2
5.0
4.8
2005 2010 2015
Trend line
FIGURE 10.9 A time series that gives the estimated population of a certain state of the United States over a series of years. The trend line shows the basic upward trend of the population.
FORECASTING MORE THAN ONE TIME PERIOD AHEAD We have focused thus far on forecasting what will happen in the next time period (the next quarter in the case of CCW). However, managers sometimes need to forecast further into the future. How can the various time-series forecasting methods be adapted to do this?
In the case of the last-value, averaging, moving-average, and exponential smoothing methods, the forecast for the next period also is the best available forecast for subsequent periods as well. However, when there is a trend in the data, it is important to take this trend into account for long-range forecasts. Exponential smoothing with trend provides a way of doing this. In particular, after determining the estimated trend, this method’s fore- cast for n time periods into the future is
Forecast for n periods from now 5 a(Last value) 1 (1 2 a)(Last forecast) 1 n 3 Estimated trend
Exponential smoothing also is used to obtain and update the estimated trend each time. The formula is
Estimate trend 5 b(Latest trend) 1 (1 2 b)(Last estimate of trend)
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10.3 Applying Time-Series Forecasting Methods to the Case Study 405
where b (the Greek letter beta) is the trend smoothing constant , which, like a , must be between 0 and 1. Latest trend refers to the trend based on just the last two values in the time series and the last two forecasts. Its formula is
Latest trend 5 a(Last value 2 Next-to-last value) 1 (1 2 a)(Last forecast 2 Next-to-last forecast)
Getting started with this forecasting method requires making two initial estimates about the status of the time series just prior to beginning forecasting. These initial estimates are
1. Initial estimate of the average value of the time series if the conditions just prior to begin- ning forecasting were to remain unchanged without any trend.
2. Initial estimate of the trend of the time series just prior to beginning forecasting.
The forecast for the first period being forecasted then is
First forecast 5 Initial estimate of average value 1 Initial estimate of trend
The second forecast is obtained from the above formulas, where the initial estimate of trend is used as the last estimate of trend in the formula for estimated trend and the initial estimate of average value is used as both the next-to-last value and the next-to-last forecast in the formula for latest trend. The above formulas then are used directly to obtain subsequent forecasts.
Since the calculations involved with this method are relatively involved, a computer com- monly is used to implement the method. Your MS Courseware includes two Excel templates (one without seasonal adjustments and one with) for this method.
The considerations involved in choosing the trend smoothing constant b are similar to those for a . A large value of b (say, b 5 0.3) is more responsive to recent changes in the trend, whereas a relatively small value (say, b 5 0.1) uses more data in a significant way to estimate trend.
After trying various combinations of a and b on the CCW problem, the consultant con- cludes that a 5 0.3 and b 5 0.3 perform about as well as any. Both values are on the high end of the typically used range (0.1 to 0.3), but the frequent changes in the CCW time series call for large values. However, lowering a from the 0.5 value used with the preceding method seems justified since incorporating trend into the analysis would help respond more quickly to changes.
When applying exponential smoothing without trend earlier, Lydia and the consultant chose 7,500 as the initial estimate of the average value of the seasonally adjusted call vol- umes. They now note that there was no noticeable trend in these call volumes just prior to the retrospective generation of forecasts three years ago. Therefore, to apply exponential smooth- ing with trend, they decide to use
Initial estimate of average value 5 7,500 Initial estimate of trend 5 0
Working with the seasonally adjusted call volumes given in several recent figures, these initial estimates lead to the following seasonally adjusted forecasts.
Y1, Q1: Seas. adj. forecast 5 7,500 1 0 5 7,500 Y1, Q2: Latest trend 5 0.3(7,322 2 7,500) 1 0.7(7,500 2 7,500) 5 2 53.4 Estimated trend 5 0.3( 2 53.4) 1 0.7(0) 5 2 16 Seas. adj. forecast 5 0.3(7,322) 1 0.7(7,500) 2 16 5 7,431 Y1, Q3: Latest trend 5 0.3(7,183 2 7,322) 1 0.7(7,431 2 7,500) 5 2 90 Estimated trend 5 0.3( 2 90) 1 0.7( 2 16) 5 2 38.2 Seas. adj. forecast 5 0.3(7,183) 1 0.7(7,431) 2 38.2 5 7,318
The Excel template in Figure 10.10 shows the results from these calculations for all 12 quar- ters over the past three years, as well as for the upcoming quarter. The middle of the figure shows the plots of all the seasonally adjusted call volumes and seasonally adjusted forecasts. Note how each trend up or down in the call volumes causes the forecasts to gradually trend in
The trend smoothing con- stant b is used to apply exponential smoothing to estimating the trend.
The Excel templates for this method perform the calcu- lations for you.
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406 Chapter Ten Forecasting
Exponential Smoothing with Trend Forecasting Method with Seasonality for CCW1
A B C
Year Quarter
D E F G H I J K L M
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
1
1
1
1
2
2
2
2
3
3
3
3
4
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
Quarter
Type of Seasonality
1
2
3
4
True
Value
Seasonally
Adjusted
Value
Latest
Trend
Estimated
Trend
Forecasting
Error
-54
-90
-243
-102
167
187
179
13
32
34
155
176
0
-16
-38
-100
-100
-20
42
83
62
53
47
80
108
Actual
Forecast
166
222
676
10
830
622
451
276
93
55
355
378
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
7,322
7,183
6,635
7,005
7,803
7,849
7,863
7,393
7,518
7,580
8,029
8,178
Seasonally
Adjusted
Value
7,500
7,430
7,318
7,013
6,910
7,158
7,407
7,627
7,619
7,642
7,670
7,858
8,062
Quarterly
Seasonal Factor
0.93
0.90
0.99
1.18
Mean Absolute Deviation
Initial Estimate
MAD =
Average =
Smoothing Constant
α =
345
Mean Square Error
MSE = 180,796
ActualForecast Alpha Beta ForecastingError InitialEstimateAverage InitialEstimateTrend MAD MSE SeasonalFactor SeasonallyAdjustedForecast SeasonallyAdjustedValue TrueValue TypeOfSeasonality
I6:I30 M5 M6 J6:J30 M9 M10 M30 M33 M16:M27 H6:H30 E6:E30 D6:D30 M13
Range Name Cells
Seasonally
Adjusted
Value
Latest
Trend
Estimated
Trend
3
4
5
6
7
8
9
10
11
12
=D6/M16
=D7/M17
=D8/M18
=D9/M19
=D10/M16
=D11/M17
=InitialEstimateTrend
=Beta*F7+(1-Beta)*G6
=Beta*F8+(1-Beta)*G7
=Beta*F9+(1-Beta)*G8
=Beta*F10+(1-Beta)*G9
=Beta*F11+(1-Beta)*G10
Actual
Forecast
Forecasting
Error
=M16*H6
=M17*H7
=M18*H8
=M19*H9
=M16*H10
=M17*H11
=ABS(D6-I6)
=ABS(D7-I7)
=ABS(D8-I8)
=ABS(D9-I9)
=ABS(D10-I10)
=ABS(D11-I11)
=Alpha*(E6-InitialEstimateAverage)+(1-Alpha)*(H6-InitialEstimateAverage)
=Alpha*(E7-E6)+(1-Alpha)*(H7-H6)
=Alpha*(E8-E7)+(1-Alpha)*(H8-H7)
=Alpha*(E9-E8)+(1-Alpha)*(H9-H8)
=Alpha*(E10-E9)+(1-Alpha)*(H10-H9 )
Seasonally
Adjusted
Forecast
=InitialEstimateAverage+InitialEstimateTrend
=Alpha*E6+(1-Alpha)*H6+G7
=Alpha*E7+(1-Alpha)*H7+G8
=Alpha*E8+(1-Alpha)*H8+G9
=Alpha*E9+(1-Alpha)*H9+G10
=Alpha*E10+(1-Alpha)*H10+G11
E F JIG H
: : ::: :
=AVERAGE(ForecastingError)MAD =
MSE =
30
L M
=SUMSQ(ForecastingError)/COUNT(ForecastingError)33
L M
0.3
7,500
Trend =
β = 0.3
0
6,975
6,687
7,245
8,276
6,427
6,442
7,333
9,000
7,085
6,877
7,594
9,272
7,498
6,000
1 2 3 4 1 2 3 4 1 2 3 4 1
6,500
7,000
7,500
8,000
8,500
9,000
Quarter
Year 2Year 1 Year 3 Year 4
S ea
so n
al ly
A d
ju st
ed A
ve ra
ge D
ai ly
C al
l V ol
u m
e
Seasonally
adjusted
value
Seasonally
adjusted
forecast
FIGURE 10.10 The Excel template in your MS Courseware for the exponential smoothing with trend method with seasonal adjustments is applied here to the CCW problem.
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10.3 Applying Time-Series Forecasting Methods to the Case Study 407
the same direction, but then the trend in the forecasts takes a couple quarters to turn around when the trend in call volumes suddenly reverses direction. The resulting forecasting errors in column J then give an average forecasting error (cell M30) of
MAD 5 345
a little above the 324 value for regular exponential smoothing and 295 for last-value forecast- ing. A similar result is obtained when using the square of the forecasting errors since the mean square error given in cell M33,
MSE 5 180,796
also is a little above the MSE values for these other two forecasting methods. Table 10.2 summarizes the values of MAD and MSE for all the forecasting methods so far.
Here is Lydia’s reaction to the large MAD value for exponential smoothing with trend.
Lydia: I’m very discouraged. These time-series forecasting methods just aren’t doing the job I need. I thought this one would. It sounded like an excellent method that also would deal with the trends we keep encountering. Consultant: Yes, it is a very good method under the right circumstances. When you have trends that may occasionally shift some over time, it should do a great job. Lydia: So what went wrong here? Consultant: Well, look at the trends you have here in the seasonally adjusted time series. You have a fairly sharp downward trend the first three quarters and then suddenly a very sharp upward trend for a couple quarters. Then it flattens out before a big drop in the eighth quarter. Then suddenly it is going up again. It is really tough to keep up with such abrupt big shifts in the trends. This method is better suited for much more gradual shifts in the trends. Lydia: OK. But aren’t there any other methods? None of these will do. Consultant: There is one other main time-series forecasting method. It is called the ARIMA method, which is an acronym for AutoRegressive Integrated Moving Average. It also is sometimes called the Box-Jenkins method, in honor of its founders. It is a very sophisticated method, but some excellent software is available for implementing it. Another nice feature is that it is well-suited for dealing with strong seasonal patterns. Lydia: Sounds good. So shouldn’t we be trying this ARIMA method? Consultant: Not at this point. It is such a sophisticated method that it requires a great amount of past data, say, a minimum of 50 time periods. We don’t have nearly enough data. Lydia: A pity. So what are we going to do? I haven’t seen anything that will do the job. Consultant: Cheer up. I have an idea for how we can use one of these time-series fore- casting methods in a different way that may do the job you want. Lydia: Really? Tell me more. Consultant: Well, let me hold off on the details until we can check out whether this is going to work. What I would like you to do is contact CCW’s marketing manager and set up a meeting between the three of us. Also, send him your data on call volumes for the past three years. Ask him to dig out his sales data for the same period and compare it to your data. Lydia: OK. What should I tell him is the purpose of the meeting? Consultant: Explain what we’re trying to accomplish here about forecasting call volumes. Then tell him that we’re trying to understand better what has been causing these sudden
When the trend in the data suddenly reverses direction, it takes a little while for the estimated trend to turn around.
The ARIMA method is another good forecasting method, but it requires much more data than CCW currently has available.
Forecasting Method MAD MSE
CCW’s 25 percent rule 424 317,815 Last-value method 295 145,909 Averaging method 400 242,876 Moving-average method 437 238,816 Exponential smoothing 324 157,836 Exponential smoothing with trend 345 180,796
TABLE 10.2 The Average Forecasting Error (MAD) and Mean Square Error (MSE) for the Various Time-Series Forecasting Methods When Forecasting CCW Call Volumes
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shifts in call volumes. He knows more about what has been driving sales up or down than anybody. We just want to pick his brain about this. Lydia: OK. Will do.
The Meeting with the Marketing Manager This meeting takes place a few days later. As you eavesdrop (after the preliminaries), you will find it helpful to refer to the call volume data in one of the recent spreadsheets, such as Figure 10.10 . Lydia: Did you receive the call volume data I mailed to you? Marketing manager: Yes, I did. Consultant: How does it compare with your own sales data for these three years? Marketing manager: Your data track mine pretty closely. I see the same ups and downs in both sets of data. Lydia: That makes sense, since it’s the calls to my call center that generate those sales. Marketing manager: Right. Consultant: Now, let me check on what caused the ups and downs. Three years ago, what we labeled as Year 1 in our data, there was a definite trend down for most of the year. What caused that? Marketing manager: Yes, I remember that year all too well. It wasn’t a very good year. The new Klugman operating system had been scheduled to come out early that year. Then they kept pushing the release date back. People kept waiting. They didn’t manage to get it out until the beginning of the next year, so we even missed the Christmas sales. Lydia: But our call volume did jump up a little more than usual during that holiday season. Marketing manager: Yes. So did sales. I remember that we came out with a new net- working tool, one with faster data transfer, in time for the holiday season. It turned out to be very popular for a few months. It really bailed us out during that slow period. Consultant: Then the Klugman operating system was released and sales jumped the next year. Marketing manager: Right. Lydia: What happened late in the year? We weren’t as busy as we expected to be. Marketing manager: I assume that most people already had updated to the new operating system by then. There wasn’t any major change in our product mix during that period. Consultant: Then sales moved back up the next year. Last year. Marketing manager: Yes, last year was a rather good year. We had a couple new prod- ucts that did very well. One was a new data storage device that came out early in the year. Very inexpensive. The other was a color plain-paper printer that was released in July. We were able to offer it at a very competitive price and our customers gobbled it up. Consultant: Thanks. That really clarifies what lies behind those call volume numbers we’ve been working with. Now I have another key question for you. When you look at your sales data and do your own forecasting, what do you see as the key factors that drive total sales up or down? Marketing manager: There really is just one big factor: Do we have any hot new prod- ucts out there? We have well over a hundred products. But most of them just fill a small niche in the market. Many of them are old standbys that, with updates, just keep going indef- initely. All these small-niche products together provide most of our total sales. A nice stable market base. Then, in addition, we should have three or four major new products out there. Maybe a couple that have been out for a few months but still have some life left in them. Then one or two just coming out that we hope will do very well. Consultant: I see. A large market base and then three or four major new products. Marketing manager: That’s what we shoot for. Consultant: Are you able to predict how well a major new product will do?
The one big factor that drives total sales up or down is whether the com- pany has just released any hot new products.
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Marketing manager: I try. I’ve gotten better at it. I’m usually fairly close on what the ini- tial response will be, but it is difficult to predict how long the product will hold up. I would like to have a better handle on it. Consultant: Thanks very much for all your information. It has verified what I’ve been suspecting for awhile now. Lydia: What’s that? Consultant: That we really need to coordinate directly with what is driving sales in order to do a better job of forecasting call volumes. Lydia: Good thought! Consultant: How would the two of you feel about coordinating in developing better pro- cedures for forecasting both sales and call volumes? Lydia: You bet. Marketing manager: Sounds good.
We will return to this story in the next two sections after introducing some useful forecast- ing software below.
Helpful Educational Software A helpful forecasting module is included in your Interactive Management Science Modules at www.mhhe.com/hillier5e . An offline version also is included in your MS Courseware on the CD-ROM.
This module includes all the time-series forecasting methods presented in this section (as well as in Section 10.5). All you need to do is select a forecasting method, enter the time- series data from which you want to obtain a forecast, and then press the Forecast button. (It does not explicitly make seasonal adjustments, so you will need to enter seasonally adjusted data if such adjustments are necessary.) In addition to listing the forecasts and forecasting errors period by period, the module also plots a graph that shows both the time-series data (in blue dots) and the resulting forecasts (in red dots).
What is unique about this software is its interactive graphing feature that immediately shows you graphically how the forecasts will change as you change any piece of data. You move your mouse onto the blue dot that corresponds to the piece of data and then drag the dot vertically to change its value. As you drag the blue dot, the red dots corresponding to the fore- casts instantaneously change accordingly. The purpose is to allow you to play with the data and gain a better feeling for how the forecasts perform with various configurations of data for each of the forecasting methods. Thus, this module is designed primarily to be an educational tool rather than professional forecasting software. It is limited to dealing with small, textbook- sized problems.
Forecasting call volumes better requires coordinating directly with what is driv- ing sales.
1. What does a seasonal factor measure? 2. What is the formula for calculating the seasonally adjusted call volume from the actual call
volume and the seasonal factor? 3. What is the formula for calculating the forecast of the actual call volume from the seasonal
factor and the seasonally adjusted forecast? 4. Why is the last-value forecasting method sometimes called the naive method? 5. Why did the averaging forecasting method not perform very well on the case study? 6. What is the rationale for replacing the averaging forecasting method by the moving-average
forecasting method? 7. How does the exponential smoothing forecasting method modify the moving-average fore-
casting method? 8. With exponential smoothing, when is a small value of the smoothing constant appropriate?
A larger value? 9. What is the formula for obtaining the next forecast with exponential smoothing? What is
added to this formula when using exponential smoothing with trend? 10. What does the marketing manager say is the one big factor that drives CCW’s total sales up
or down?
Review Questions
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10.4 THE TIME-SERIES FORECASTING METHODS IN PERSPECTIVE
Section 10.3 presented several methods for forecasting the next value of a time series in the context of the CCW case study. We now will take a step back to place into perspective just what these methods are trying to accomplish. After providing this perspective, CCW’s con- sultant then will give his recommendation for setting up a forecasting system.
The Goal of the Forecasting Methods It actually is something of a misnomer to talk about forecasting the value of the next observa- tion in a time series (such as CCW’s call volume in the next quarter). It is impossible to pre- dict the value precisely, because this next value can turn out to be anything over some range. What it will be depends upon future circumstances that are beyond our control.
In other words, the next value that will occur in a time series is a random variable. It has some probability distribution. For example, Figure 10.11 shows a typical probability distri- bution for the CCW call volume in a future quarter in which the mean of this distribution happens to be 7,500. This distribution indicates the relative likelihood of the various possible values of the call volume. Nobody can say in advance which value actually will occur.
So what is the meaning of the single number that is selected as the “forecast” of the next value in the time series? If possible, we would like this number to be the mean of the distribution. The reason is that random observations from the distribution tend to cluster around the mean of the distribution. Therefore, using the mean as the forecast would tend to minimize the average forecasting error.
Unfortunately, we don’t actually know what this probability distribution is, let alone its mean. The best we can do is use all the available data (past values from the time series) to estimate the mean as closely as possible.
The goal of time-series forecasting methods is to estimate the mean of the underlying probabil- ity distribution of the next value of the time series as closely as possible.
Given some random observations from a single probability distribution, the best estimate of its mean is the sample average (the average of all these observations). Therefore, if a time series has exactly the same distribution for each and every time period, then the averaging forecasting method provides the best estimate of the mean.
However, other forecasting methods commonly are used instead because the distribution may be changing over time.
Problems Caused by Shifting Distributions Section 10.3 began by considering seasonal effects. This then led to estimating CCW’s sea- sonal factors as 0.93, 0.90, 0.99, and 1.18 for Quarters 1, 2, 3, and 4, respectively.
If the overall average daily call volume for a year is 7,500, these seasonal factors imply that the probability distributions for the four quarters of that year fall roughly as shown in
The next value in a time series cannot be forecasted with certainty because it has a probability distribution rather than a fixed value that definitely will occur.
7,250 7,500
Mean
7,750
FIGURE 10.11 A typical probability dis- tribution of what the aver- age daily call volume will be for CCW in a quarter when the mean is 7,500.
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10.4 The Time-Series Forecasting Methods in Perspective 411
Figure 10.12 . Since these distributions have different means, we should no longer simply average the random observations (observed call volumes) from all four quarters to estimate the mean for any one of these distributions.
This complication is why the preceding section seasonally adjusted the time series. Divid- ing each quarter’s call volume by its seasonal factor shifts the distribution of this seasonally adjusted call volume over to basically the distribution shown in Figure 10.11 with a mean of 7,500. This allows averaging the seasonally adjusted values to estimate this mean.
Unfortunately, even after seasonally adjusting the time series, the probability distribution may not remain the same from one year to the next (or even from one quarter to the next). For example, as CCW’s marketing manager explained, total sales jumped substantially at the beginning of Year 2 when the new Klugman operating system became available. This also caused the average daily call volume to increase by about 10 percent, from just over 7,000 in Year 1 to over 7,700 in Year 2. Figure 10.13 compares the resulting distributions for typical quarters (seasonally adjusted) in the two years.
Random observations from the Year 1 distribution in this figure provide a poor basis for estimating the mean of the Year 2 distribution. Yet, except for the last-value method, each of the forecasting methods presented in the preceding section placed at least some weight on these observations from Year 1 to estimate the mean for each quarter in Year 2. This was a major part of the reason why both the average forecasting errors (MAD) and the mean square errors (MSE) were higher for these methods than for the last-value method.
Judging from the marketing manager’s information, it appears that some shift in the distri- bution also occurred several times from just one quarter to the next. This further added to the forecasting errors.
Comparison of the Forecasting Methods Section 10.3 presented five methods for forecasting the next value in a time series. Which of these methods is particularly suitable for a given application depends greatly on how stable the time series is.
A time series is said to be stable if its underlying probability distribution usually remains the same from one time period to the next. (Any shifts that do occur in the distribution are both infrequent and small.) A time series is unstable if both frequent and sizable shifts in the distri- bution tend to occur.
When the probability dis- tribution for a time series shifts frequently, recent data quickly become outdated for forecasting purposes.
6,500 7,000 7,500 8,000 8,500 9,000
Quarter 4Quarter 3Quarter 1Quarter 2FIGURE 10.12 Typical probability distri- butions of CCW’s average daily call volumes in the four quarters of a year in which the overall average is 7,500.
6,500 7,000 7,500 8,000
Year 1 Year 2FIGURE 10.13 Comparison of typical probability distributions of CCW’s average daily call volumes (seasonally adjusted) in years 1 and 2.
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CCW’s seasonally adjusted time series shown in Figure 10.4 (and many subsequent fig- ures) appears to have had several shifts in the distribution, including the sizable one depicted in Figure 10.13 . Therefore, this time series is an example of a relatively unstable one.
Here is a summary of which type of time series fits each of the forecasting methods.
Last-value method: Suitable for a time series that is so unstable that even the next-to-last value is not considered relevant for forecasting the next value. Averaging method: Suitable for a very stable time series where even its first few values are considered relevant for forecasting the next value. Moving-average method: Suitable for a moderately stable time series where the last few values are considered relevant for forecasting the next value. The number of values included in the moving average reflects the anticipated degree of stability in the time series. Exponential smoothing method: Suitable for a time series in the range from somewhat unstable to rather stable, where the value of the smoothing constant needs to be adjusted to fit the anticipated degree of stability. Refines the moving-average method by placing the greatest weight on the most recent values, but is not as readily understood by manag- ers as the moving-average method. Exponential smoothing with trend: Suitable for a time series where the mean of the distri- bution tends to follow a trend either up or down, provided that changes in the trend occur only occasionally and gradually.
Unfortunately for CCW, its seasonally adjusted time series proved to be a little too unsta- ble for any of these methods except the last-value method, which is considered to be the least powerful of these forecasting methods. Even when using exponential smoothing with trend, the changes in the trend occurred too frequently and sharply.
In light of these considerations, the consultant now is ready to present his recommenda- tions to Lydia for a new forecasting procedure.
The Consultant’s Recommendations 1. Forecasting should be done monthly rather than quarterly in order to respond more
quickly to changing conditions. 2. Hiring and training of new agents also should be done monthly instead of quarterly in
order to fine-tune staffing levels to meet changing needs. 3. Recently retired agents should be offered the opportunity to work part-time on an on-call
basis to help meet current staffing needs more closely. 4. Since sales drive call volume, the forecasting process should begin by forecasting sales. 5. For forecasting purposes, total sales should be broken down into the major components
described by the marketing manager, namely, (1) the relatively stable market base of numerous small-niche products and (2) each of the few (perhaps three or four) major new products whose success or failure can significantly drive total sales up or down. These major new products would be identified by the marketing manager on an ongoing basis.
6. Exponential smoothing with a relatively small smoothing constant is suggested for fore- casting sales of the marketing base of numerous small-niche products. However, before making a final decision on the forecasting method, retrospective testing should be con- ducted to check how well this particular method would have performed over the past three years. This testing also should guide selection of the value of the smoothing constant.
7. Exponential smoothing with trend, with relatively large smoothing constants, is sug- gested for forecasting sales of each of the major new products. Once again, retrospec- tive testing should be conducted to check this decision and to guide choosing values for the smoothing constants. The marketing manager should be asked to provide the initial estimate of anticipated sales in the first month for a new product. He also should be asked to check the subsequent exponential smoothing forecasts and make any adjust- ments he feels are appropriate based on his knowledge of what is happening in the marketplace.
The key factor in choosing a forecasting method is how stable the time series is.
Forecasting call volume should begin by separately forecasting the major com- ponents of total sales.
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8. Because of the strong seasonal sales pattern, seasonally adjusted time series should be used for each application of these forecasting methods.
9. After separately obtaining forecasts of actual sales for each of the major components of total sales identified in recommendation 5, these forecasts should be summed to obtain a forecast of total sales.
10. Causal forecasting with linear regression (as described in the next section) should be used to obtain a forecast of call volume from this forecast of total sales.
Lydia accepts these recommendations with considerable enthusiasm. She also agrees to work with the marketing manager to gain his cooperation.
Read on to see how the last recommendation is implemented.
The next section describes how to obtain a forecast of call volume from a forecast of total sales.
1. What kind of variable is the next value that will occur in a time series? 2. What is the goal of time-series forecasting methods? 3. Is the probability distribution of CCW’s average daily call volume the same for every quarter? 4. What is the explanation for why the average forecasting errors were higher for the other time-
series forecasting methods than for the supposedly less powerful last-value method? 5. What is the distinction between a stable time series and an unstable time series? 6. What is the consultant’s recommendation regarding what should be forecasted instead of call
volumes to begin the forecasting process? 7. What are the major components of CCW’s total sales?
Review Questions
10.5 CAUSAL FORECASTING WITH LINEAR REGRESSION
We have focused so far on time-series forecasting methods, that is, methods that forecast the next value in a time series based on its previous values. These methods have been used retrospectively in Section 10.3 to forecast CCW’s call volume in the next quarter based on its previous call volumes.
Causal Forecasting However, the consultant’s last recommendation suggests another approach to forecasting. It is really sales that drive call volume, and sales can be forecasted considerably more precisely than call volume. Therefore, it should be possible to obtain a better forecast of call volume by relating it directly to forecasted sales. This kind of approach is called causal forecasting.
Causal forecasting obtains a forecast of the quantity of interest (the dependent variable ) by relating it directly to one or more other quantities (the independent variables ) that drive the quantity of interest.
Table 10.3 shows some examples of the kinds of situations where causal forecasting some- times is used. In each of the first four cases, the indicated dependent variable can be expected to go up or down rather directly with the independent variable(s) listed in the rightmost col- umn. The last case also applies when some quantity of interest (e.g., sales of a product) tends to follow a steady trend upward (or downward) with the passage of time (the independent variable that drives the quantity of interest).
For the CCW problem, call volume is the dependent variable and total sales is the independent variable.
Type of Forecasting Possible Dependent Variable
Possible Independent Variables
Sales Sales of a product Amount of advertising Spare parts Demand for spare parts Usage of equipment Economic trends Gross domestic product Various economic factors CCW call volume Call volume Total sales Any quantity This same quantity Time
TABLE 10.3 Possible Examples of Causal Forecasting
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Linear Regression At Lydia’s request, the marketing manager brought sales data for the past three years to their recent meeting. These data are summarized in Figure 10.14 . In particular, column D gives the average daily sales (in units of thousands of dollars) for each of the 12 past quarters. Column E repeats the data given previously on average daily call volumes. None of the data have been seasonally adjusted.
The right side of the figure was generated by selecting the data in D5:E16 and then either selecting an XY Scatter Chart from the Insert tab (for Excel 2010) or Chart tab (for Excel 2011). This graph shows a plot of the data in columns D and E on a two-dimensional graph. Thus, each of the 12 points in the graph shows the combination of sales and call volume for one of the 12 quarters (without identifying which quarter).
This graph shows a close relationship between call volume and sales. Each increase or decrease in sales is accompanied by a roughly proportional increase or decrease in call volume. This is not surprising since the sales are being made through the calls to the call center.
It appears from this graph that the relationship between call volume and sales can be approximated by a straight line. Figure 10.15 shows such a line. (This line was generated by right-clicking on one of the data points in the graph in Figure 10.14 , choosing Add Trendline and then specifying Linear Trend. The equation above the line is added by choosing Display Equation on Chart (this option is available in the main dialog box for Excel 2010, or under the Options tab for Excel 2011). This line is referred to as a linear regression line.
When doing causal forecasting with a single independent variable, linear regression involves approximating the relationship between the dependent variable (call volume for CCW) and the independent variable (sales for CCW) by a straight line. This linear regression line is drawn on a graph with the independent variable on the horizontal axis and the dependent vari- able on the vertical axis. The line is constructed after plotting a number of points showing each observed value of the independent variable and the corresponding value of the dependent variable.
Thus, the linear regression line in Figure 10.15 can be used to estimate what the call vol- ume should be for a particular value of sales. In general, the equation for the linear regression line has the form
y 5 a 1 bx
A linear regression line estimates what the value of the dependent variable should be for any particular value of the independent variable.
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FIGURE 10.14 The data needed to do causal forecasting for the CCW problem by relating call volume to sales.
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y = 1.6324x – 1,223.9
FIGURE 10.15 Figure 10.14 has been modified here by adding a trend line to the graph.
where
y 5 Estimated value of the dependent variable, as given by the linear regression line
a 5 Intercept of the linear regression line with the y -axis
b 5 Slope of the linear regression line
x 5 Value of the independent variable
(If there is more than one independent variable, then this regression equation has a term, a constant times the variable, added on the right-hand side for each of these variables.) For the linear regression line in this figure, the exact values of a and b happen to be
a 5 21,223.9 b 5 1.6324
Figure 10.16 shows the Excel template in your MS Courseware that can also be used to find these values of a and b. You need to input all the observed values of the independent variable (sales) and of the dependent variable (call volume) in columns C and D, and then the template performs all the calculations. On the right, note that you have the option of inserting a value for x (sales) in cell J10 and then the template calculates the corresponding value of y (call volume) that lies on the linear regression line. This calculation can be repeated for as many values of x as desired. In addition, column E already shows these calculations for each value of x in column C, so each cell in column E gives the estimate of call volume provided by the linear regression line for the corresponding sales level in column C. The difference between this estimate and the actual call volume in column D gives the estimation error in column F. The square of this error is shown in column G.
The procedure used to obtain a and b is called the method of least squares . This method chooses the values of a and b that minimize the sum of the square of the estima- tion errors given in column G of Figure 10.16 . Thus, the sum of the numbers in column G (22,051) is the minimum possible. Any significantly different values of a and b would give different estimation errors that would cause this sum to be larger.
The average of the square of the estimation errors in column G (1,838) has an interesting interpretation. Suppose that the sales for a quarter could be known in advance (either because of advance orders or an exact prediction). In this case, using the linear regression line to forecast call volume would give a mean square error (MSE) of 1,838. What the method of least squares has done is place the linear regression line exactly where it would minimize MSE in this situ- ation. Note that this minimum value of MSE 5 1,838 is roughly 1 percent of the MSE values given earlier in Table 10.2 for the time-series forecasting methods presented in Section 10.3.
The equation for a linear regression line is called the regression equation.
The method of least squares chooses the values of a and b that make the sum of the resulting numbers in col- umn G as small as possible.
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The numbers in column F also are interesting. Averaging these numbers reveals that the average estimation error for the 12 quarters is only 35. This indicates that if the sales for a quarter were known in advance (or could be predicted exactly), then using the linear regres- sion line to forecast call volume would give an average forecasting error (MAD) of only 35. This is only about 10 percent of the values of MAD obtained for the various time-series fore- casting methods in Section 10.3.
For some applications of causal forecasting, the value of the independent variable will be known in advance. This is not the case here, where the independent variable is the total sales for the upcoming time period. However, the consultant is confident that a very good forecast of total sales can be obtained by following his recommendations. This forecast can then be used as the value of the independent variable for obtaining a good forecast of call volume from the linear regression line.
CCW’s New Forecasting Procedure 1. Obtain a forecast of total (average daily) sales for the upcoming month by implementing
the consultant’s recommendations presented at the end of Section 10.4. 2. Use this forecast as the value of total sales in forecasting the average daily call volume for
the upcoming month from the linear regression line identified in Figures 10.15 and 10.16 .
Seasonal adjustments play a role in step 1 of this procedure, but not in step 2. Based on the consultant’s recommendations (see recommendation no. 8), seasonal adjustments should be incorporated into whatever forecasting method is used in step 1. However, a forecast of
a b DependentVariable
Estimate
EstimationError
IndependentVariable
SquareOfError
x y
J5
J6
D5:D34
E5:E34
F5:F34
C5:C34
G5:G34
J10
J12
Range Name Cells
Estimate Error
Estimation3
4
5
6
7
8
9
10
11
=a+b*C5
=a+b*C6
=a+b*C7
=a+b*C8
=a+b*C9
:
:
=ABS(D5-E5)
=ABS(D6-E6)
=ABS(D7-E7)
=ABS(D8-E8)
=ABS(D9-E9)
:
:
:
:
E F
of Error
Square
=F5^2
=F6^2
=F7^2
=F8^2
=F9^2
G
=INTERCEPT(DependentVariable,IndependentVariable)
=SLOPE(DependentVariable,IndependentVariable)
a =
b =
then y =
5
6
I J
=a+b*x12
I J
2
1
3
4
5
6
7
8
9
10
11
12
13
14
15
16
A B C D
1
2
3
4
5
6
7
8
9
10
11
12
Dependent
Variable
Independent
Variable
Linear Regression of Call Volume vs. Sales Volume for CCW E
Estimate
F G H I J
6,809
6,465
6,569
8,266
7,257
7,064
7,784
8,724
6,992
6,822
7,949
9,650
6,765
6,453
6,527
8,316
7,252
7,079
7,772
8,745
7,023
6,914
7,878
9,627
Estimation
Error
43.85
11.64
42.18
49.93
5.40
14.57
11.66
21.26
31.07
91.70
70.55
23.24
Square
of Error
Linear Regression Line
y = a + bx
a =
b =
1,923
136
1,780
2,493
29
212
136
452
965
8,408
4,977
540
-1,223.86
1.63
Estimator
if x =
then y = 6,938.18
4,894
4,703
4,748
5,844
5,192
5,086
5,511
6,107
5,052
4,985
5,576
6,647
Time
Period
5,000
FIGURE 10.16 The Excel template in your MS Courseware for doing causal forecasting with linear regression, as illustrated here for the CCW problem.
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10.5 Causal Forecasting with Linear Regression 417
seasonally adjusted sales would then be converted back into a forecast of actual sales for use in step 2. By using a forecast of actual sales, the forecast of call volume obtained from the linear regression line in step 2 will be the desired forecast of actual call volume rather than seasonally adjusted call volume.
This application of linear regression to the CCW problem involves only one indepen- dent variable (total sales) that drives the dependent variable (call volume). As mentioned at the beginning of this section, some applications of causal forecasting with linear regres- sion involve multiple independent variables that together drive the dependent variable. For example, when using linear regression to forecast the nation’s gross domestic product (the dependent variable) for the next quarter, the independent variables might include such leading indicators of future economic performance as the current level of the stock market, the cur- rent index of consumer confidence, the current index of business activity (measuring orders placed), and so forth. If there are, say, two independent variables, the regression equation would have the form
y 5 a 1 b1x1 1 b2x2
where x 1 and x 2 are the independent variables with b 1 and b 2 as their respective coefficients. The corresponding linear regression line now would lie in a three-dimensional graph with y (the dependent variable) as the vertical axis and both x 1 and x 2 as horizontal axes in the other two dimensions. Regardless of the number of independent variables, the method of least squares still can be used to choose the value of a, b 1 , b 2 , etc., that minimizes the sum of the square of the estimation errors when comparing the values of the dependent variable at the various data points with the corresponding estimates given by the regression equation. How- ever, we will not delve further into this more advanced topic.
The CCW Case Study a Year Later A year after implementing the consultant’s recommendations, Lydia gives him a call.
Lydia: I just wanted to let you know how things are going. And to congratulate you on the great job you did for us. Remember that the 25 percent rule was giving us MAD values over 400? And then the various time-series forecasting methods were doing almost as badly? Consultant: Yes, I do remember. You were pretty discouraged there for a while. Lydia: I sure was! But I’m feeling a lot better now. I just calculated MAD for the first year under your new procedure. 120. Only 120! Consultant: Great. That’s the kind of improvement we like to see. What do you think has made the biggest difference? Lydia: I think the biggest factor was tying our forecasting into forecasting sales. We never had much feeling for where call volumes were headed. But we have a much better handle on what sales will be because, with the marketing manager’s help, we can see what is causing the shifts. Consultant: Yes. I think that is a real key to successful forecasting. You saw that we can get a lot of garbage by simply applying a time-series forecasting method to historical data without understanding what is causing the shifts. You have to get behind the numbers and see what is really going on, then design the forecasting procedure to catch the shifts as they occur, like we did by having the marketing manager identify the major new products that impact total sales and then separately forecasting sales for each of them. Lydia: Right. Bringing the marketing manager in on this was a great move. He’s a real supporter of the new procedure now, by the way. He says it is giving him valuable informa- tion as well. Consultant: Good. Is he making adjustments in the statistical forecasts, based on his knowledge of what is going on in the marketplace, like I recommended? Lydia: Yes, he is. You have a couple fans here. We really appreciate the great job you did for us.
Linear regression with mul- tiple independent variables is sometimes referred to as multiple linear regression.
A key to successful fore- casting is to understand what is causing shifts so that they can be caught as soon as they occur.
A good forecasting procedure combines a well-constructed statisti- cal forecasting procedure and a savvy manager who understands what is driving the numbers and so is able to make appropriate adjust- ments in the forecasts.
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418 Chapter Ten Forecasting
1. What is causal forecasting? 2. When applying causal forecasting to the CCW problem, what is the dependent variable and
what is the independent variable? 3. When doing causal forecasting with a single independent variable, what does linear regression
involve? 4. What is the form of the equation for a linear regression line with a single independent vari-
able? With more than one independent variable? 5. What is the name of the method for obtaining the value of the constants, a and b, for a linear
regression line? 6. How does the MAD value for CCW’s new forecasting procedure compare with that for the old
procedure that used the 25 percent rule?
Review Questions
10.6 JUDGMENTAL FORECASTING METHODS
We have focused so far on statistical forecasting methods that base the forecast on histori- cal data. However, such methods cannot be used if no data are available, or if the data are not representative of current conditions. In such cases, judgmental forecasting methods can be used instead.
Even when good data are available, some managers prefer a judgmental method instead of a formal statistical method. In many other cases, a combination of the two may be used. For example, in the CCW case study, the marketing manager uses his judgment, based on his long experience and his knowledge of what is happening in the marketplace, to adjust the sales forecasts obtained from time-series forecasting methods.
Here is a brief overview of the main judgmental forecasting methods.
1. Manager’s opinion : This is the most informal of the methods, because it simply involves a single manager using his or her best judgment to make the forecast. In some cases, some data may be available to help make this judgment. In others, the manager may be drawing solely on experience and an intimate knowledge of the current conditions that drive the forecasted quantity.
2. Jury of executive opinion : This method is similar to the first one, except now it involves a small group of high-level managers who pool their best judgments to collectively make the forecast. This method may be used for more critical forecasts for which several executives share responsibility and can provide different types of expertise.
3. Salesforce composite : This method is often used for sales forecasting when a company employs a salesforce to help generate sales. It is a bottom-up approach whereby each salesperson provides an estimate of what sales will be in his or her region. These estimates then are sent up through the corporate chain of command, with managerial review at each level, to be aggregated into a corporate sales forecast.
4. Consumer market survey : This method goes even further than the preceding one in adopting a grass-roots approach to sales forecasting. It involves surveying customers and potential customers regarding their future purchasing plans and how they would respond to various new features in products. This input is particularly helpful for designing new prod- ucts and then in developing the initial forecasts of their sales. It also is helpful for planning a marketing campaign.
5. Delphi method : This method employs a panel of experts in different locations who inde- pendently fill out a series of questionnaires. However, the results from each questionnaire are provided with the next one, so each expert then can evaluate this group information in adjusting his or her responses next time. The goal is to reach a relatively narrow spread of conclusions from most of the experts. The decision makers then assess this input from the panel of experts to develop the forecast. This involved process normally is used only at the highest levels of a corporation or government to develop long-range forecasts of broad trends.
Judgmental forecasting methods use expert opinion to make forecasts.
The salesforce composite method uses a bottom-up approach.
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419
Compañia Sud Americana de Vapores (CSAV) is one of the world’s largest shipping companies. It is the most global firm in Chile, with regional offices on four continents. It has opera- tions in over 100 countries while using more than 2,000 termi- nals and depots worldwide. The company’s main business is shipping cargo using containers on large container ships. The container fleet transports about 700,000 20-foot containers.
Once loaded containers reach their destination and are unloaded, it is important to quickly send the empty con- tainers to where they will be most needed for subsequent shipments. In 2006, CSAV and researchers from the Uni- versity of Chile began a project to apply management sci- ence to more effectively make these decisions quickly on a global basis. The result was the development of an Empty Container Logistics Optimization System (ECO).
ECO uses both an advanced network optimization model and a sophisticated inventory model. However, both of these models need to be driven by relatively pre- cise forecasts of the demand for empty containers. The
forecasting methods used to generate these forecasts are a combination of certain time-series forecasting methods (moving average and seasonally adjusted moving average with trend) and two types of judgmental forecasting meth- ods. One of the latter types is sales force composite—a con- sensus forecast model in which the sales agents worldwide register their demand expectations. The other is manager’s opinion—whereby logistics planners provide information that must be considered in the forecasts.
This extensive application of management science, including a combination of forecasting methods, resulted in savings of $81 million in its first full year. These cost reductions were mainly the result of reducing empty con- tainer inventories by 50 percent and increasing container turnover by 60 percent.
Source: R. Epstein and 14 co-authors, “A Strategic Empty Con- tainer Logistics Optimization in a Major Shipping Company,” Inter- faces 42, no. 1 (January–February 2012), pp. 5–16. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
1. Statistical forecasting methods be used under what circumstances? 2. Are judgmental forecasting only used when statistical forecasting methods cannot be used? 3. How does the jury of executive opinion method differ from the manager’s opinion method? 4. How does the salesforce composite method begin? 5. When is a market survey particularly helpful? 6. When might the Delphi method be used?
Review Questions
The future success of any business depends heavily on the ability of its management to forecast well. Forecasting may be needed in several areas, including sales, the need for spare parts, production yields, economic trends, and staffing needs.
The Computer Club Warehouse (CCW) case study illustrates a variety of approaches to forecasting, some of which prove unsatisfactory for this particular application. Ultimately, it becomes necessary to get behind the CCW data to understand just what is driving the call volumes at its call center in order to develop a good forecasting system.
A time series is a series of observations over time of some quantity of interest. Several statistical forecasting methods use these observations in some way to forecast what the next value will be. These methods include the last-value method, the averaging method, the moving-average method, the expo- nential smoothing method, and exponential smoothing with trend.
The goal of all these methods is to estimate the mean of the underlying probability distribution of the next value of the time series as closely as possible. This may require using seasonal factors to season- ally adjust the time series, as well as identifying other factors that may cause this underlying probability distribution to shift from one time period to the next.
Another statistical forecasting approach is called causal forecasting. This approach obtains a forecast of the quantity of interest (the dependent variable) by relating it directly to one or more other quanti- ties (the independent variables) that drive the quantity of interest. Frequently, this involves using linear regression to approximate the relationship between the dependent variable and each independent vari- able by a straight line.
The software accompanying this book includes Excel templates for the various statistical forecasting methods, as well as a forecasting module in the Interactive Management Science Modules.
Still another key category of forecasting methods is judgmental methods. This category involves basing the forecast on a manager’s opinion, a jury of executive opinion, a salesforce composite, a con- sumer market survey, or the Delphi method.
10.7 Summary
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420 Chapter Ten Forecasting
Glossary ARIMA An acronym for the AutoRegressive Integrated Moving Average method, a sophisti- cated time-series forecasting method commonly referred to as the Box-Jenkins method. (Section 10.3), 407 averaging forecasting method A method that uses the average of the past observations from a time series as a forecast of the next value. (Sec- tions 10.1 and 10.3), 385, 396 causal forecasting Obtaining a forecast of the dependent variable by relating it directly to one or more independent variables. (Section 10.5), 413 consumer market survey A judgmental fore- casting method that uses surveys of customers and potential customers. (Section 10.6), 418 Delphi method A judgmental forecasting method that uses input from a panel of experts in different locations. (Section 10.6), 418 dependent variable The quantity of interest when doing causal forecasting. (Section 10.5), 413 exponential smoothing forecasting method A method that uses a weighted average of the last value from a time series and the last forecast to obtain the forecast of the next value. (Sections 10.1 and 10.3), 385, 400 exponential smoothing with trend An adjust- ment of the exponential smoothing forecasting method that projects the current trend forward to help forecast the next value of a time series (and perhaps subsequent values as well). (Sections 10.1 and 10.3), 385, 402 forecasting error The deviation of the forecast from the realized quantity. (Sections 10.1 and 10.2), 386, 389 independent variable A quantity that drives the value of the dependent variable in causal fore- casting. (Section 10.5), 413 judgmental forecasting methods Methods that use expert judgment to make forecasts. (Introduc- tion and Section 10.6), 384, 418 jury of executive opinion A judgmental fore- casting method that involves a small group of high-level managers pooling their best judgment to collectively make the forecast. (Section 10.6), 418 last-value forecasting method A method that uses the last value of a time series as the forecast of the next value. (Sections 10.1 and 10.3), 385, 394 linear regression Approximating the relation- ship between the dependent variable and each independent variable by a straight line. (Sections 10.1 and 10.5), 385, 414 linear regression line The line that approxi- mates the relationship between the dependent variable and each independent variable when using causal forecasting. (Section 10.5), 414 MAD An acronym for mean absolute deviation, the average forecasting error. (Sections 10.1 and 10.2), 386, 389
manager’s opinion A judgmental forecasting method that involves using a single manager’s best judgment to make the forecast. (Section 10.6), 418 mean absolute deviation (MAD) The average forecasting error. (Sections 10.1 and 10.2), 386, 389 mean square error (MSE) The average of the square of the forecasting errors. (Sections 10.1 and 10.2), 386, 390 method of least squares The procedure used to obtain the constants in the equation for a linear regression line. (Section 10.5), 415 moving-average forecasting method A method that uses the average of the last n obser- vations from a time series as a forecast of the next value. (Sections 10.1 and 10.3), 385, 398 MSE An acronym for mean square error, the average of the square of the forecasting errors. (Sections 10.1 and 10.2), 386, 390 naive method Another name for the last-value forecasting method. (Section 10.3), 396 regression equation The equation for a linear regression line. (Section 10.5), 415 salesforce composite A judgmental forecasting method that aggregates the sales forecasts of the salesforce from their various regions. (Section 10.6), 418 seasonal factor A factor for any period of a year that measures how that period compares to the over- all average for an entire year. (Section 10.3), 391 seasonally adjusted time series An adjustment of the original time series that removes seasonal effects. (Section 10.3), 393 smoothing constant A parameter of the expo- nential smoothing forecasting method that gives the weight to be placed on the last value in the time series. (Section 10.3), 400 stable time series A time series whose underlying probability distribution usually remains the same from one time period to the next. (Section 10.4), 411 statistical forecasting methods Methods that use historical data to forecast future quantities. (Introduction and Sections 10.1–10.5), 384, 418 time series A series of observations over time of some quantity of interest. (Section 10.2), 390 time-series forecasting methods Methods that use the past observations in a time series to fore- cast what the next value will be. (Sections 10.2 and 10.3), 390, 393 trend The average change from one time series value to the next if the current pattern continues. (Section 10.3), 402 trend smoothing constant A smoothing con- stant for estimating the trend when using expo- nential smoothing with trend. (Section 10.3), 405 unstable time series A time series that has frequent and sizable shifts in its underlying prob- ability distribution. (Section 10.4), 411
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Chapter 10 Solved Problem 421
Summary of Key Formulas Forecasting error 5 Difference between a forecasted value and the true value then obtained (Section 10.2)
MAD 5 Sum of forecasting errors
Number of forecasts (Section 10.2)
MSE 5 Sum of square of forecasting errors
Number of forecasts (Section 10.2)
Seasonal factor 5 Average for the period
Overall average (Section 10.3)
Seasonally adjusted value 5 Actual value
Seasonal factor (Section 10.3)
Last-Value Method: (Section 10.3)
Forecast 5 Last value
Averaging Method: (Section 10.3)
Forecast 5 Average of all data to date
Moving-Average Method: (Section 10.3)
Forecast 5 Average of last n values
Exponential Smoothing Method: (Section 10.3)
Forecast 5 a(Last value) 1 (1 2 a)(Last forecast)
Exponential Smoothing with Trend: (Section 10.3)
Forecast 5 a(Last value) 1 (1 2 a)(Last forecast) 1 Estimated trend
Estimated trend 5 b(Latest trend) (12b)(Last estimate of trend)
Latest trend 5 a(Last value 2 Next-to-last value) 1 (1 2 a)(Last forecast 2 Next-to-last forecast)
Linear Regression Line: (Section 10.5)
y 5 a 1 bx
Chapter 10 Excel Files:
Template for Seasonal Factors
Templates for Last-Value Method (with and without Seasonality)
Templates for Averaging Method (with and without Seasonality)
Templates for Moving-Average Method (with and without Seasonality)
Templates for Exponential Smoothing Method (with and with- out Seasonality)
Templates for Exponential Smoothing with Trend (with and without Seasonality)
Template for Linear Regression
Interactive Management Science Module:
Module for Forecasting
Learning Aids for This Chapter in Your MS Courseware
exponential smoothing method with trend (with initial esti- mates of 275 for the average value, 2 for the trend, along with smoothing constants of a 5 0.2 and b 5 0.2) when they are applied retrospectively to the years 2011–2013.
10.S1. Forecasting Charitable Donations at the Union Mission Cash donations (in thousands of dollars) at the Union Mission for 2011–2013 were as shown below.
Solved Problem (See the CD-ROM or Website for the Solution)
Quarter Donations Quarter Donations Quarter Donations
Q1 2011 242 Q1 2012 253 Q1 2013 270 Q2 2011 282 Q2 2012 290 Q2 2013 286 Q3 2011 254 Q3 2012 262 Q3 2013 271 Q4 2011 345 Q4 2012 352 Q4 2013 378
a. Ignoring seasonal effects, compare both the MAD and MSE values for the last-value method, the averaging method, the moving-average method (based on the most recent 4 quar- ters), the exponential smoothing method (with an initial es- timate of 275 and a smoothing constant of a 5 0.2), and the
b. Determine the seasonal factors for the four quarters. c. Repeat part a, but now consider the seasonal effects. d. Using the forecasting method from part a or c with the lowest
MAD value, make long-range forecasts for charitable dona- tions in each of the quarters of 2014.
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422 Chapter Ten Forecasting
a. Calculate and compare MAD for these two fore- casting methods. Then do the same with MSE.
b. Sharon is uncomfortable with choosing between these two methods based on such limited data, but she also does not want to delay further before mak- ing her choice. She does have similar sales data for the three years prior to using these forecasting methods the past five months. How can these older data be used to further help her evaluate the two methods and choose one?
10.5. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 10.2. Briefly describe how forecasting was applied in this study. Then list the various financial and nonfi- nancial benefits that resulted from this study. 10.6. Figure 10.1 shows CCW’s average daily call volume for each quarter of the past three years and Figure 10.4 gives the seasonally adjusted call volumes. Lydia Weigelt now wonders what these seasonally adjusted call volumes would have been had she started using seasonal factors two years ago rather than applying them retrospectively now. a. Use only the call volumes in year 1 to determine
the seasonal factors for year 2 (so that the “average” call volume for each quarter is just the actual call volume for that quarter in year 1).
b. Use these seasonal factors to determine the season- ally adjusted call volumes for year 2.
c. Use the call volumes in years 1 and 2 to determine the seasonal factors for year 3.
d. Use the seasonal factors obtained in part c to determine the seasonally adjusted call volumes for year 3.
10.7. Even when the economy is holding steady, the unem- ployment rate tends to fluctuate because of seasonal effects. For example, unemployment generally goes up in Quarter 3 (sum- mer) as students (including new graduates) enter the labor mar- ket. The unemployment rate then tends to go down in Quarter 4 (fall) as students return to school and temporary help is hired for the Christmas season. Therefore, using seasonal factors to obtain a seasonally adjusted unemployment rate is helpful for painting a truer picture of economic trends.
Over the past 10 years, one state’s average unemployment rates (not seasonally adjusted) in Quarters 1, 2, 3, and 4 have been 6.2 percent, 6.0 percent, 7.5 percent, and 5.5 percent, respectively. The overall average has been 6.3 percent. a. Determine the seasonal factors for the four quarters. b. Over the next year, the unemployment rates (not
seasonally adjusted) for the four quarters turn out
The first 18 problems should be done by hand without using the templates in this chapter’s Excel files. To the left of the subse- quent problems (or their parts), we have inserted the symbol E (for Excel) to indicate that one of these templates can be helpful. (The forecasting module in your Interactive Management Sci- ence Modules should be used for certain problems, but this will be specified in the statement of the problem whenever needed.) An asterisk on the problem number indicates that at least a par- tial answer is given in the back of the book.
10.1.* The Hammaker Company’s newest product has had the following sales during its first five months: 5, 17, 29, 41, 39. The sales manager now wants a forecast of sales in the next month a. Use the last-value method. b. Use the averaging method. c. Use the moving-average method with the three
most recent months. d. Given the sales pattern so far, do any of these meth-
ods seem inappropriate for obtaining the forecast? Why?
10.2. Sales of stoves have been going well for the Good- Value Department Store. These sales for the past five months have been 15, 18, 12, 17, 13. Use the following methods to obtain a forecast of sales for the next month. a. The last-value method. b. The averaging method. c. The moving-average method with three months. d. If you feel that the conditions affecting sales next
month will be the same as in the last five months, which of these methods do you prefer for obtaining the forecast? Why?
10.3.* You have been forecasting sales the last four quar- ters. These forecasts and the true values that subsequently were obtained are shown below.
Quarter Forecast True Value
1 327 345 2 332 317 3 328 336 4 330 311
Calculate the forecasting error for each quarter. Then calculate MAD and MSE. 10.4. Sharon Johnson, sales manager for the Alvarez-Baines Company, is trying to choose between two methods for forecast- ing sales that she has been using during the past five months. During these months, the two methods obtained the forecasts shown next for the company’s most important product, where the subsequent actual sales are shown on the right.
Problems
Forecast
Month Method 1 Method 2 Actual Sales
1 5,324 5,208 5,582 2 5,405 5,377 4,906 3 5,195 5,462 5,755 4 5,511 5,414 6,320 5 5,762 5,549 5,153
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Chapter 10 Problems 423
to be 7.8 percent, 7.4 percent, 8.7 percent, and 6.1 percent. Determine the seasonally adjusted unem- ployment rates for the four quarters. What does this progression of rates suggest about whether the state’s economy is improving?
10.8. Ralph Billett is the manager of a real estate agency. He now wishes to develop a forecast of the number of houses that will be sold by the agency over the next year.
The agency’s quarter-by-quarter sales figures over the last three years are shown below.
Quarter Year 1 Year 2 Year 3
1 23 19 21 2 22 21 26 3 31 27 32 4 26 24 28
a. Determine the seasonal factors for the four quarters. b. After considering seasonal effects, use the last-
value method to forecast sales in Quarter 1 of next year.
c. Assuming that each of the quarterly forecasts is cor- rect, what would the last-value method forecast as the sales in each of the four quarters next year?
d. Based on his assessment of the current state of the housing market, Ralph’s best judgment is that the agency will sell 100 houses next year. Given this forecast for the year, what is the quarter-by-quarter forecast according to the seasonal factors?
10.9.* You are using the moving-average forecasting method based on the last four observations. When making the forecast for the last period, the oldest of the four observations was 1,945 and the forecast was 2,083. The true value for the last period then turned out to be 1,977. What is your new forecast for the next period? 10.10. You are using the moving-average forecasting method based on sales in the last three months to forecast sales for the next month. When making the forecast for last month, sales for the third month before were 805. The forecast for last month was 782 and then the actual sales turned out to be 793. What is your new forecast for next month? 10.11. After graduating from college with a degree in math- ematical statistics, Ann Preston has been hired by the Monty Ward Company to apply statistical methods for forecasting the company’s sales. For one of the company’s products, the mov- ing-average method based on sales in the 10 most recent months already is being used. Ann’s first task is to update last month’s forecast to obtain the forecast for next month. She learns that the forecast for last month was 1,551 and that the actual sales then turned out to be 1,532. She also learns that the sales for the 10th month before last month were 1,632. What is Ann’s forecast for next month? 10.12. The J. J. Bone Company uses exponential smooth- ing to forecast the average daily call volume at its call cen- ter. The forecast for last month was 782, and then the actual value turned out to be 792. Obtain the forecast for next month for each of the following values of the smoothing constant: a 5 0.1, 0.3, 0.5.
10.13.* You are using exponential smoothing to obtain monthly forecasts of the.sales of a certain product. The forecast for last month was 2,083, and then the actual sales turned out to be 1,973. Obtain the forecast for next month for each of the follow- ing values of the smoothing constant: a 5 0.1, 0.3, 0.5. 10.14. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 10.3. Briefly describe how forecasting was applied in this study. Then list the various financial and nonfi- nancial benefits that resulted from this study. 10.15. Three years ago, the Admissions Office for Ivy College began using exponential smoothing with a smoothing constant of 0.25 to forecast the number of applications for admission each year. Based on previous experience, this process was begun with an initial estimate of 5,000 applications. The actual number of applications then turned out to be 4,600 in the first year. Thanks to new favorable ratings in national surveys, this number grew to 5,300 in the second year and 6,000 last year.
a. Determine the forecasts that were made for each of the past three years.
b. Calculate MAD and MSE for these three years.
c. Determine the forecast for next year. 10.16. Reconsider Problem 10.15. Notice the steady trend upward in the number of applications over the past three years— from 4,600 to 5,300 to 6,000. Suppose now that the Admis- sions Office of Ivy College had been able to foresee this kind of trend and so had decided to use exponential smoothing with trend to do the forecasting. Suppose also that the initial estimates just over three years ago had been average value 5 3,900 and trend 5 700. Then, with any values of the smoothing constants, the forecasts obtained by this forecasting method would have been exactly correct for all three years.
Illustrate this fact by doing the calculations to obtain these forecasts when the smoothing constant is a 5 0.25 and the trend smoothing constant is b 5 0.25.
10.17.* Exponential smoothing with trend, with a smoothing constant of a 5 0.2 and a trend smoothing constant of b 5 0.3, is being used to forecast values in a time series. At this point, the last two values have been 535 and then 550. The last two forecasts have been 530 and then 540. The last estimate of trend has been 10. Use this information to forecast the next value in the time series. 10.18. The Healthwise Company produces a variety of exer- cise equipment. Healthwise management is very pleased with the increasing sales of its newest model of exercise bicycle. The sales during the last two months have been 4,655 and then 4,935.
Management has been using exponential smoothing with trend, with a smoothing constant of a 5 0.1 and a trend smooth- ing constant of b 5 0.2, to forecast sales for the next month each time. The forecasts for the last two months were 4,720 and then 4,975. The last estimate of trend was 240.
Calculate the forecast of sales for next month. 10.19.* Ben Swanson, owner and manager of Swanson’s Department Store, has decided to use statistical forecasting to get a better handle on the demand for his major products. How- ever, Ben now needs to decide which forecasting method is most appropriate for each category of product. One category is major household appliances, such as washing machines, which have a
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b. Ben Swanson now has found that an error was made in determining the sales for April, but he has not
relatively stable sales level. Monthly sales of washing machines last year are shown below.
Month Sales Month Sales Month Sales
January 23 May 22 September 21 February 24 June 27 October 29 March 22 July 20 November 23 April 28 August 26 December 28
a. Considering that the sales level is relatively stable, which of the most basic forecasting methods—the last-value method, the averaging method, or the moving-average method—do you feel would be most appropriate for forecasting future sales? Why?
E b. Use the last-value method retrospectively to deter- mine what the forecasts would have been for the last 11 months of last year. What are MAD and MSE?
E c. Use the averaging method retrospectively to deter- mine what the forecasts would have been for the last 11 months of last year. What are MAD and MSE?
E d. Use the moving-average method with n 5 3 retro- spectively to determine what the forecasts would have been for the last nine months of last year. What are MAD and MSE?
e. Use their MAD values to compare the three methods.
yet obtained the corrected sales figure. For each of the three forecasting methods, Ben wants to know which of the original monthly forecasts would change now because of changing the sales figure for April. Answer this question by dragging vertically the blue dot that corresponds to April sales and observing which of the red dots (corresponding to monthly forecasts) move.
c. Repeat part a if the sales for April change from 28 to 16.
d. Repeat part a if the sales for April change from 28 to 40.
10.22. Management of the Jackson Manufacturing Corpora- tion wishes to choose a statistical forecasting method for fore- casting total sales for the corporation. Total sales (in millions of dollars) for each month of last year are shown below.
Month Sales Month Sales Month Sales
January 126 May 153 September 147 February 137 June 154 October 151 March 142 July 148 November 159 April 150 August 145 December 166
f. Use their MSE values to compare the three methods. g. Do you feel comfortable in drawing a definitive
conclusion about which of the three forecasting methods should be the most accurate in the future based on these 12 months of data?
E10.20. Reconsider Problem 10.19. Ben Swanson now has decided to use the exponential smoothing method to forecast future sales of washing machines, but he needs to decide on which smoothing constant to use. Using an initial estimate of 24, apply this method retrospectively to the 12 months of last year with a 5 0.1, 0.2, 0.3, 0.4, and 0.5. Compare MAD for these five values of the smoothing constant a . Then do the same with MSE. 10.21. Reconsider Problem 10.19. For each of the forecasting methods specified in parts b, c, and d, use the forecasting mod- ule in your Interactive Management Science Modules to obtain the requested forecasts. Then use the accompanying graph that plots both the sales data and forecasts to answer the following questions for these forecasting methods. a. Based on your examination of the graphs for the
three forecasting methods, which method do you feel is doing the best job of forecasting with the given data? Why?
a. Note how the sales level is shifting significantly from month to month—first trending upward and then dipping down before resuming an upward trend. Assuming that similar patterns would con- tinue in the future, evaluate how well you feel each of the five forecasting methods introduced in Sec- tion 10.3 would perform in forecasting future sales.
E b. Apply the last-value method, the averaging method, and the moving-average method (with n 5 3) retro- spectively to last year’s sales and compare their MAD values. Then compare their MSE values.
E c. Using an initial estimate of 120, apply the exponen- tial smoothing method retrospectively to last year’s sales with a 5 0.1, 0.2, 0.3, 0.4, and 0.5. Compare both MAD and MSE for these five values of the smoothing constant a .
E d. Using initial estimates of 120 for the average value and 10 for the trend, apply exponential smooth- ing with trend retrospectively to last year’s sales. Use all combinations of the smoothing constants where a 5 0.1, 0.3, or 0.5 and b 5 0.1, 0.3, or 0.5. Compare both MAD and MSE for these nine combinations.
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a. 52, 55, 55, 58, 59, 63, 64, 66, 67, 72, 73, 74 b. 52, 55, 59, 61, 66, 69, 71, 72, 73, 74, 73, 74 c. 52, 53, 51, 50, 48, 47, 49, 52, 57, 62, 69, 74 10.27. The Andes Mining Company mines and ships copper ore. The company’s sales manager, Juanita Valdes, has been using the moving-average method based on the last three years of sales to forecast the demand for the next year. However, she has become dissatisfied with the inaccurate forecasts being pro- vided by this method.
The annual demands (in tons of copper ore) over the past 10 years are 382, 405, 398, 421, 426, 415, 443, 451, 446, 464.
a. Explain why this pattern of demands inevitably led to significant inaccuracies in the moving-average forecasts.
E b. Determine the moving-average forecasts for the past seven years. What are MAD and MSE? What is the forecast for next year?
E c. Determine what the forecasts would have been for the past 10 years if the exponential smoothing method had been used instead with an initial esti- mate of 380 and a smoothing constant of a 5 0.5. What are MAD and MSE? What is the forecast for next year?
E d. Determine what the forecasts would have been for the past 10 years if exponential smoothing with trend had been used instead. Use initial estimates of 370 for the average value and 10 for the trend, with smoothing constants a 5 0.25 and b 5 0.25.
e. Based on the MAD and MSE values, which of these three methods do you recommend using hereafter?
10.28. Reconsider Problem 10.27. For each of the forecasting methods specified in parts b, c, and d, use the forecasting mod- ule in your Interactive Management Science Modules to obtain the requested forecasts. After examining the accompanying graph that plots both the demand data and forecasts, write a one-sentence description for each method regarding whether its plot of forecasts tends to lie below or above or at about the same level as the demands being forecasted. Then use these conclusions to select one of the methods to recommend using hereafter. E10.29.* The Pentel Microchip Company has started produc- tion of its new microchip. The first phase in this production is the wafer fabrication process. Because of the great difficulty in fabricating acceptable wafers, many of these tiny wafers must be rejected because they are defective. Therefore, management places great emphasis on continually improving the wafer fabri- cation process to increase its production yield (the percentage of wafers fabricated in the current lot that are of acceptable quality for producing microchips).
So far, the production yields of the respective lots have been 15 percent, 21 percent, 24 percent, 32 percent, 37 percent, 41 percent, 40 percent, 47 percent, 51 percent, and 53 percent. Use exponential smoothing with trend to forecast the production yield of the next lot. Begin with initial estimates of 10 percent for the average value and 5 percent for the trend. Use smoothing constants of a 5 0.2 and b 5 0.2.
10.30. The Centerville Water Department provides water for the entire town and outlying areas. The number of acre-feet of
e. Which one of the above forecasting methods would you recommend that management use? Using this method, what is the forecast of total sales for Janu- ary of the new year?
10.23. Reconsider Problem 10.22. Use the lessons learned from the CCW case study to address the following questions. a. What might be causing the significant shifts in total
sales from month to month that were observed last year?
b. Given your answer to part a, how might the basic statistical approach to forecasting total sales be improved?
c. Describe the role of managerial judgment in apply- ing the statistical approach developed in part b.
10.24. Reconsider Problem 10.22. For each of the forecasting methods specified in parts b, c, and d (with smoothing constants a 5 0.5 and b 5 0.5 as needed), use the forecasting module in your Interactive Management Science Modules to obtain the requested forecasts. Then use the accompanying graph that plots both the sales data and forecasts to answer the following ques- tions for these forecasting methods. a. Based on your examination of the graphs for the
five forecasting methods, which method do you feel is doing the best job of forecasting with the given data? Why?
b. Management now has been informed that an error was made in calculating the sales for April, but a corrected sales figure has not yet been obtained. Therefore, for each of the five forecasting methods, management wants to know which of the original monthly forecasts would change now because of changing the sales figure for April. Answer this question by dragging vertically the blue dot that corresponds to April sales and observing which of the red dots (corresponding to monthly forecasts) move.
c. Repeat part a if the sales for April change from 150 to 125.
d. Repeat part a if the sales for April change from 150 to 175.
E10.25. Choosing an appropriate value of the smoothing con- stant a is a key decision when applying the exponential smooth- ing method. When relevant historical data exist, one approach to making this decision is to apply the method retrospectively to these data with different values of a and then choose the value of a that gives the smallest MAD. Use this approach for choosing a with each of the following time series representing monthly sales. In each case, use an initial estimate of 50 and compare a 5 0.1, 0.2, 0.3, 0.4, and 0.5. a. 51, 48, 52, 49, 53, 49, 48, 51, 50, 49 b. 52, 50, 53, 51, 52, 48, 52, 53, 49, 52 c. 50, 52, 51, 55, 53, 56, 52, 55, 54, 53 E10.26. The choice of the smoothing constants, a and b , has a considerable effect on the accuracy of the forecasts obtained by using exponential smoothing with trend. For each of the follow- ing time series, set a 5 0.2 and then compare MAD obtained with b 5 0.1, 0.2, 0.3, 0.4, and 0.5. Begin with initial estimates of 50 for the average value and 2 for the trend.
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E e. Apply the exponential smoothing method with an initial estimate of 25 and a smoothing constant of a 5 0.25.
E f. Apply exponential smoothing with trend with smooth- ing constants of a 5 0.25 and b 5 0.25. Use initial esti- mates of 25 for the average value and 0 for the trend.
E g. Compare both the MAD and MSE values for these methods. Use the one with the smallest MAD to forecast sales in Quarter 1 of next year.
h. Use the forecast in part g and the seasonal factors to make long-range forecasts now of the sales in the remaining quarters of next year.
E10.32. Transcontinental Airlines maintains a computerized forecasting system to forecast the number of customers in each fare class who will fly on each flight in order to allocate the avail- able reservations to fare classes properly. For example, consider economy-class customers flying in midweek on the noon flight from New York to Los Angeles. The following table shows the average number of such passengers during each month of the year just completed. The table also shows the seasonal factor that has been assigned to each month based on historical data.
water consumed in each of the four seasons of the three preced- ing years is shown below.
Season Year 1 Year 2 Year 3
Winter 25 27 24 Spring 47 46 49 Summer 68 72 70 Fall 42 39 44
E a. Determine the seasonal factors for the four seasons. E b. After considering seasonal effects, use the last-value
method to forecast water consumption next winter. c. Assuming that each of the forecasts for the next
three seasons is correct, what would the last-value method forecast as the water consumption in each of the four seasons next year?
E d. After considering seasonal effects, use the averaging method to forecast water consumption next winter.
E e. After considering seasonal effects, use the moving- average method based on four seasons to forecast water consumption next winter.
Month Average Number
Seasonal Factor Month
Average Number
Seasonal Factor
January 68 0.90 July 94 1.17 February 71 0.88 August 96 1.15 March 66 0.91 September 80 0.97 April 72 0.93 October 73 0.91 May 77 0.96 November 84 1.05 June 85 1.09 December 89 1.08
Month Average Number Month
Average Number Month
Average Number
January 75 May 85 September 94 February 76 June 99 October 90 March 81 July 107 November 106 April 84 August 108 December 110
E f. After considering seasonal effects, use the exponen- tial smoothing method with an initial estimate of 46 and a smoothing constant of a 5 0.1 to forecast water consumption next winter.
E g. Compare both the MAD and MSE values of these four forecasting methods when they are applied ret- rospectively to the last three years.
10.31. Reconsider Problem 10.8. Ralph Billett realizes that the last-value method is considered to be the naive forecast- ing method, so he wonders whether he should be using another method. Therefore, he has decided to use the available Excel templates that consider seasonal effects to apply various statisti- cal forecasting methods retrospectively to the past three years of data and compare both their MAD and MSE values. E a. Determine the seasonal factors for the four quarters. E b. Apply the last-value method. E c. Apply the averaging method. E d. Apply the moving-average method based on the
four most recent quarters of data.
a. After considering seasonal effects, compare both the MAD and MSE values for the last-value method, the averaging method, the moving-average method (based on the most recent three months), and the exponential smoothing method (with an initial esti- mate of 80 and a smoothing constant of a 5 0.2) when they are applied retrospectively to the past year.
b. Use the forecasting method with the smallest MAD value to forecast the average number of these pas- sengers flying in January of the new year.
10.33. Reconsider Problem 10.32. The economy is beginning to boom so the management of Transcontinental Airlines is pre- dicting that the number of people flying will steadily increase this year over the relatively flat (seasonally adjusted) level of last year. Since the forecasting methods considered in Problem 10.32 are relatively slow in adjusting to such a trend, consideration is being given to switching to exponential smoothing with trend.
Subsequently, as the year goes on, management’s prediction proves to be true. The following table shows the corresponding average number of passengers in each month of the new year.
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E c. After considering seasonal effects, apply the expo- nential smoothing method to forecast monthly sales this year. Use an initial estimate of 420 and a smoothing constant of a 5 0.2.
E d. After considering seasonal effects, apply exponen- tial smoothing with trend to forecast monthly sales this year. Use initial estimates of 420 for the aver- age value and 0 for the trend, along with smoothing constants of a 5 0.2 and b 5 0.2.
e. Compare both the MAD and MSE values obtained in parts b, c, and d.
f. Calculate the combined forecast for each month by averaging the forecasts for that month obtained in parts b, c, and d. Then calculate MAD for these combined forecasts.
g. Based on these results, what is your recommenda- tion for how to do the forecasts next year?
10.35.* Long a market leader in the production of heavy machinery, the Spellman Corporation recently has been enjoy- ing a steady increase in the sales of its new lathe. The sales over the past 10 months are shown below.
Month Sales Month Sales
1 430 6 514 2 446 7 532 3 464 8 548 4 480 9 570 5 498 10 591
Because of this steady increase, management has decided to use causal forecasting, with the month as the independent variable and sales as the dependent variable, to forecast sales in the com- ing months. a. Plot these data on a two-dimensional graph with the
month on the horizontal axis and sales on the verti- cal axis.
E a. Repeat part a of Problem 10.32 for the two years of data.
E b. After considering seasonal effects, apply exponen- tial smoothing with trend to just the new year. Use initial estimates of 80 for the average value and 2 for the trend, along with smoothing constants of a 5 0.2 and b 5 0.2. Compare MAD for this method to the MAD values obtained in part a. Then do the same with MSE.
E c. Repeat part b when exponential smoothing with trend is begun at the beginning of the first year and then applied to both years, just like the other fore- casting methods in part a. Use the same initial esti- mates and smoothing constants except change the initial estimate of trend to 0.
d. Based on these results, which forecasting method would you recommend that Transcontinental Air- lines use hereafter?
10.34. Quality Bikes is a wholesale firm that specializes in the distribution of bicycles. In the past, the company has maintained ample inventories of bicycles to enable filling orders immedi- ately, so informal rough forecasts of demand were sufficient to make the decisions on when to replenish inventory. However, the company’s new president, Marcia Salgo, intends to run a tighter ship. Scientific inventory management is to be used to reduce inventory levels and minimize total variable inventory costs. At the same time, Marcia has ordered the development of a computerized forecasting system based on statistical fore- casting that considers seasonal effects. The system is to gener- ate three sets of forecasts—one based on the moving-average method, a second based on the exponential smoothing method, and a third based on exponential smoothing with trend. The average of these three forecasts for each month is to be used for inventory management purposes.
The following table gives the available data on monthly sales of 10-speed bicycles over the past three years. The last column also shows monthly sales this year, which is the first year of operation of the new forecasting system.
Past Sales
Month Year 1 Year 2 Year 3 Current Sales
This Year
January 352 317 338 364 February 329 331 346 343 March 365 344 383 391 April 358 386 404 437 May 412 423 431 458 June 446 472 459 494 July 420 415 433 468 August 471 492 518 555 September 355 340 309 387 October 312 301 335 364 November 567 629 594 662 December 533 505 527 581
E a. Determine the seasonal factors for the 12 months based on past sales.
E b. After considering seasonal effects, apply the moving- average method based on the most recent three months to forecast monthly sales for each month of this year.
E b. Find the formula for the linear regression line that fits these data.
c. Plot this line on the graph constructed in part a. d. Use this line to forecast sales in month 11. e. Use this line to forecast sales in month 20.
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a. Plot the data for the past 10 years (years 1 through 10) on a two-dimensional graph with the year on the horizontal axis and the demand on the vertical axis.
E b. Find the formula for the linear regression line that fits these data.
c. Plot this line on the graph constructed in part a. d. Use this line to forecast demand next year (year 11). e. Use this line to forecast demand in year 15. f. What does the formula for the linear regression line
indicate is roughly the average growth in demand per year?
g. Use the forecasting module in your Interactive Man- agement Science Modules to generate a graph of the data and the linear regression line. Then experiment with the data to see how the linear regression line shifts as you drag any of the data points up or down.
10.38. Luxury Cruise Lines has a fleet of ships that travel to Alaska repeatedly every summer (and elsewhere during other times of the year). A considerable amount of advertising is done each winter to help generate enough passenger business for that summer. With the coming of a new winter, a decision needs to be made about how much advertising to do this year.
The following table shows the amount of advertising (in thousands of dollars) and the resulting sales (in thousands of passengers booked for a cruise) for each of the past five years.
f. What does the formula for the linear regression line indicate is roughly the average growth in sales per month?
10.36. Reconsider Problems 10.15 and 10.16. Since the num- ber of applications for admission submitted to Ivy College has been increasing at a steady rate, causal forecasting can be used to forecast the number of applications in future years by letting the year be the independent variable and the number of applica- tions be the dependent variable. a. Plot the data for years 1, 2, and 3 on a two- dimensional
graph with the year on the horizontal axis and the number of applications on the vertical axis.
b. Since the three points in this graph line up in a straight line, this straight line is the linear regres- sion line. Draw this line.
E c. Find the formula for this linear regression line. d. Use this line to forecast the number of applications
for each of the next five years (years 4 through 8). e. As these next years go on, conditions change for the
worse at Ivy College. The favorable ratings in the national surveys that had propelled the growth in applications turn unfavorable. Consequently, the num- ber of applications turn out to be 6,300 in year 4 and 6,200 in year 5, followed by sizable drops to 5,600 in year 6 and 5,200 in year 7. Does it still make sense to use the forecast for year 8 obtained in part d? Explain.
Amount of advertising ($1,000s) 225 400 350 275 450
Sales (thousands of passengers) 16 21 20 17 23
E f. Plot the data for all seven years. Find the formula for the linear regression line based on all these data and plot this line. Use this formula to forecast the number of applications for year 8. Does the linear regression line provide a close fit to the data? Given this answer, do you have much confidence in the forecast it provides for year 8? Does it make sense to continue to use a linear regression line when changing conditions cause a large shift in the under- lying trend in the data?
E g. Apply exponential smoothing with trend to all seven years of data to forecast the number of applications in year 8. Use initial estimates of 3,900 for the aver- age and 700 for the trend, along with smoothing constants of a 5 0.5 and b 5 0.5. When the under- lying trend in the data stays the same, causal fore- casting provides the best possible linear regression line (according to the method of least squares) for making forecasts. However, when changing con- ditions cause a shift in the underlying trend, what advantage does exponential smoothing with trend have over causal forecasting?
10.37. Reconsider Problem 10.27. Despite some fluctuations from year to year, note that there has been a basic trend upward in the annual demand for copper ore over the past 10 years. There- fore, by projecting this trend forward, causal forecasting can be used to forecast demands in future years by letting the year be the independent variable and the demand be the dependent variable.
a. To use causal forecasting to forecast sales for a given amount of advertising, which need to be the dependent variable and the independent variable?
b. Plot the data on a graph.
E c. Find the formula for the linear regression line that fits these data. Then plot this line on the graph con- structed in part b.
d. Forecast the sales that would be attained by expend- ing $300,000 on advertising.
e. Estimate the amount of advertising that would need to be done to attain a booking of 22,000 passengers.
f. According to the linear regression line, about how much increase in sales can be attained on the average per $1,000 increase in the amount of advertising?
10.39. Reconsider Problem 10.38. Use the forecasting module in your Interactive Management Science Modules to generate the lin- ear regression line. On the resulting graph that shows this line and the five data points (as blue dots), note that the leftmost data point, the middle data point, and the rightmost data point all lie very close to the line. You can see how the linear regression line shifts as any one of these data points moves up or down by moving your mouse onto the blue dot at this point and dragging it vertically.
For each of these three data points, determine whether the linear regression line shifts above or below this point or whether it still passes essentially through it when the following change is made in one of these data points (but none of the others).
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Case 10-1 Finagling the Forecasts 429
of the data and the linear regression line. Then experiment with the data to see how the linear regression line shifts as you drag any of the data points up or down.
E10.41. Joe Barnes is the owner of Standing Tall, one of the major roofing companies in town. Much of the company’s busi- ness comes from building roofs on new houses. Joe has learned that general contractors constructing new houses typically will subcontract the roofing work about two months after construc- tion begins. Therefore, to help him develop long-range schedules for his work crews, Joe has decided to use county records on the number of housing construction permits issued each month to forecast the number of roofing jobs on new houses he will have two months later.
a. Change the sales from 16 to 19 when the amount of advertising is 225.
b. Change the sales from 23 to 26 when the amount of advertising is 450.
c. Change the sales from 20 to 23 when the amount of advertising is 350.
10.40. To support its large fleet, North American Airlines maintains an extensive inventory of spare parts, including wing flaps. The number of wing flaps needed in inventory to replace damaged wing flaps each month depends partially on the num- ber of flying hours for the fleet that month, since increased usage increases the chances of damage.
The following table shows both the number of replacement wing flaps needed and the number of thousands of flying hours for the entire fleet for each of several recent months.
Thousands of flying hours 162 149 185 171 138 154
Number of wing flaps needed 12 9 13 14 10 11
Month Permits Jobs Month Permits Jobs
January 323 19 July 446 34 February 359 17 August 407 37 March 396 24 September 374 33 April 421 23 October 343 30 May 457 28 November 311 27 June 472 32 December 277 22
a. Identify the dependent variable and the independent variable for doing causal forecasting of the number of wing flaps needed for a given number of flying hours.
b. Plot the data on a graph. E c. Find the formula for the linear regression line. d. Plot this line on the graph constructed in part b.
Joe has now gathered the following data for each month over the past year, where the second column gives the number of housing construction permits issued in that month and the third column shows the number of roofing jobs on new houses that were subcontracted out to Standing Tall in that month. Use a causal forecasting approach to develop a forecasting procedure for Joe to use hereafter.
e. Forecast the average number of wing flaps needed in a month in which 150,000 flying hours are planned.
f. Repeat part e for 200,000 flying hours. g. Use the forecasting module in your Interactive
Management Science Modules to generate a graph
10.42. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 10.6. Briefly describe how forecasting methods, including judgmental forecasting methods, were applied as part of this study. Then list the various financial and nonfinancial benefits that resulted from this study.
Case 10-1
Finagling the Forecasts
Mark Lawrence has been pursuing a vision for more than two years. This pursuit began when he became frustrated in his role as director of Human Resources at Cutting Edge, a large com- pany manufacturing computers and computer peripherals. At that time, the Human Resources Department under his direc- tion provided records and benefits administration to the 60,000 Cutting Edge employees throughout the United States, and 35 separate records and benefits administration centers existed
across the country. Employees contacted these records and ben- efits centers to obtain information about dental plans and stock options, change tax forms and personal information, and process leaves of absence and retirements. The decentralization of these administration centers caused numerous headaches for Mark. He had to deal with employee complaints often since each cen- ter interpreted company policies differently—communicating inconsistent and sometimes inaccurate answers to employees.
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serviced 15,000 customers and had received 2,000 calls per month. He concluded that since the new centralized center would service four times the number of customers—60,000 customers—it would receive four times the number of calls—8,000 calls per month.
Mark slowly checked off the items on his “to do” list, and the centralized records and benefits administration center opened one year after Mark had received the go-ahead from corporate headquarters.
Now, after operating the new center for 13 weeks, Mark’s call center forecasts are proving to be terribly inaccurate. The number of calls the center receives is roughly three times as large as the 8,000 calls per month that Mark had forecasted. Because of demand overload, the call center is slowly going to hell in a handbasket. Customers calling the center must wait an average of five minutes before speaking to a representative, and Mark is receiving numerous complaints. At the same time, the customer service representatives are unhappy and on the verge of quitting because of the stress created by the demand overload. Even corporate headquarters has become aware of the staff and service inadequacies, and executives have been breathing down Mark’s neck demanding improvements.
Mark needs help, and he approaches you to forecast demand for the call center more accurately.
Luckily, when Mark first established the call center, he real- ized the importance of keeping operational data, and he provides you with the number of calls received on each day of the week over the last 13 weeks. The data (shown next) begins in week 44 of the last year and continues to week 5 of the current year.
His department also suffered high operating costs since operat- ing 35 separate centers created inefficiency.
His vision? To centralize records and benefits administra- tion by establishing one administration center. This centralized records and benefits administration center would perform two distinct functions: data management and customer service. The data management function would include updating employee records after performance reviews and maintaining the human resource management system. The customer service function would include establishing a call center to answer employee questions concerning records and benefits and to process records and benefits changes over the phone.
One year after proposing his vision to management, Mark received the go-ahead from Cutting Edge corporate headquar- ters. He prepared his “to do” list—specifying computer and phone systems requirements, installing hardware and software, integrating data from the 35 separate administration centers, standardizing record-keeping and response procedures, and staffing the administration center. Mark delegated the systems requirements, installation, and integration jobs to a competent group of technology specialists. He took on the responsibility of standardizing procedures and staffing the administration center.
Mark had spent many years in human resources and there- fore had little problem with standardizing record-keeping and response procedures. He encountered trouble in determining the number of representatives needed to staff the center, how- ever. He was particularly worried about staffing the call center since the representatives answering phones interact directly with
Monday Tuesday Wednesday Thursday Friday
Week 44 1,130 851 859 828 726 Week 45 1,085 1,042 892 840 799 Week 46 1,303 1,121 1,003 1,113 1,005 Week 47 2,652 2,825 1,841 0 0 Week 48 1,949 1,507 989 990 1,084 Week 49 1,260 1,134 941 847 714 Week 50 1,002 847 922 842 784 Week 51 823 0 0 401 429 Week 52/1 1,209 830 0 1,082 841 Week 2 1,362 1,174 967 930 853 Week 3 924 954 1,346 904 758 Week 4 886 878 802 945 610 Week 5 910 754 705 729 772
customers—the 60,000 Cutting Edge employees. The customer service representatives would receive extensive training so that they would know the records and benefits policies backwards and forwards—enabling them to answer questions accurately and process changes efficiently. Overstaffing would cause Mark to suffer the high costs of training unneeded representatives and paying the surplus representatives the high salaries that go along with such an intense job. Understaffing would cause Mark to continue to suffer the headaches from customer complaints— something he definitely wanted to avoid.
The number of customer service representatives Mark needed to hire depended on the number of calls that the records and ben- efits call center would receive. Mark therefore needed to forecast the number of calls that the new centralized center would receive. He approached the forecasting problem by using judgmental fore- casting. He studied data from one of the 35 decentralized admin- istration centers and learned that the decentralized center had
Mark indicates that the days where no calls were received were holidays.
a. Mark first asks you to forecast daily demand for the next week using the data from the past 13 weeks. You should make the forecasts for all the days of the next week now (at the end of week 5), but you should provide a different fore- cast for each day of the week by treating the forecast for a single day as being the actual call volume on that day.
1. From working at the records and benefits administration cen- ter, you know that demand follows “seasonal” patterns within the week. For example, more employees call at the beginning of the week when they are fresh and productive than at the end of the week when they are planning for the weekend. You therefore realize that you must account for the seasonal pat- terns and adjust the data that Mark gave you accordingly. What is the seasonally adjusted call volume for the past 13 weeks?
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c. At the end of the hour, Mark arrives at your desk with two data sets: weekly case volumes for the decentralized center and weekly case volumes for the centralized center. You ask Mark if he has data for daily case volumes, and he tells you that he does not. You therefore first have to forecast the weekly demand for the next week and then break this weekly demand into daily demand. The decentralized center was shut down last year when the
new centralized center opened, so you have the decentralized
For each of the forecasting methods, calculate the mean absolute deviation for the method and evaluate the performance of the method. When calculating the mean absolute deviation, you should use the actual forecasts you found in part a above. You should not recalculate the forecasts based on the actual val- ues. In your evaluation, provide an explanation for the effective- ness or ineffectiveness of the method.
You realize that the forecasting methods that you have inves- tigated do not provide a great degree of accuracy, and you decide
You simply need to understand how the decentralized volumes relate to the new centralized volumes. Once you understand this relationship, you can use the call volumes from the decentralized center to forecast the call volumes for the centralized center.
You approach Mark and ask him whether call center data exist for the decentralized center. He tells you that data exist, but data do not exist in the format that you need. Case volume data—not call volume data—exist. You do not understand the distinction, so Mark continues his explanation. There are two types of demand data—case volume data and call volume data. Case volume data count the actions taken by the representatives at the call center. Call volume data count the number of calls answered by the representatives at the call center. A case may require one call or multiple calls to resolve it. Thus, the number of cases is always less than or equal to the number of calls.
You know you only have case volume data for the decen- tralized center, and you certainly do not want to compare apples and oranges. You therefore ask if case volume data exist for the new centralized center. Mark gives you a wicked grin and nods his head. He sees where you are going with your forecasts, and he tells you that he will have the data for you within the hour.
2. Using the seasonally adjusted call volume, forecast the daily demand for the next week using the last-value fore- casting method.
3. Using the seasonally adjusted call volume, forecast the daily demand for the next week using the averaging fore- casting method.
4. Using the seasonally adjusted call volume, forecast the daily demand for the next week using the moving-average forecasting method. You decide to use the five most recent days in this analysis.
5. Using the seasonally adjusted call volume, forecast the daily demand for the next week using the exponen- tial smoothing forecasting method. You decide to use a smoothing constant of 0.1 because you believe that demand without seasonal effects remains relatively stable. Use the daily call volume average over the past 13 weeks for the initial estimate.
b. After one week, the period you have forecasted passes. You realize that you are able to determine the accuracy of your forecasts because you now have the actual call volumes from the week you had forecasted. The actual call volumes are shown below.
Decentralized Case Volume Centralized Case Volume
Week 44 612 2,052 Week 45 721 2,170 Week 46 693 2,779 Week 47 540 2,334 Week 48 1,386 2,514 Week 49 577 1,713 Week 50 405 1,927 Week 51 441 1,167 Week 52/1 655 1,549 Week 2 572 2,126 Week 3 475 2,337 Week 4 530 1,916 Week 5 595 2,098
Monday Tuesday Wednesday Thursday Friday
Week 6 723 677 521 571 498
to use a creative approach to forecasting that combines the statis- tical and judgmental approaches. You know that Mark had used data from one of the 35 decentralized records and benefits admin- istration centers to perform his original forecasting. You there- fore suspect that call volume data exists for this decentralized center. Because the decentralized centers performed the same functions as the new centralized center currently performs, you decide that the call volumes from the decentralized center will help you forecast the call volumes for the new centralized center.
case data spanning from week 44 of two years ago to week 5 of last year. You compare this decentralized data to the centralized data spanning from week 44 of last year to week 5 of this year. The weekly case volumes are shown in the above table.
1. Find a mathematical relationship between the decentralized case volume data and the centralized case volume data.
2. Now that you have a relationship between the weekly decen- tralized case volume and the weekly centralized case volume,
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3. Using the actual call volumes given in part b, calculate the mean absolute deviation and evaluate the effectiveness of this forecasting method.
4. Which forecasting method would you recommend Mark use and why? As the call center continues its operation, how would you recommend improving the forecasting procedure?
Additional Cases Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
you are able to forecast the weekly case volume for the new center. Unfortunately, you do not need the weekly case vol- ume; you need the daily call volume. To calculate call volume from case volume, you perform further analysis and deter- mine that each case generates an average of 1.5 calls. To cal- culate daily call volume from weekly call volume, you decide to use the seasonal factors as conversion factors. Given the following case volume data from the decentralized center for week 6 of last year, forecast the daily call volume for the new center for week 6 of this year.
Week 6
Decentralized case volume 613
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Chapter Eleven
Queueing Models Learning Objectives
After completing this chapter, you should be able to
1. Describe the elements of a queueing model.
2. Identify the characteristics of the probability distributions that are commonly used in queueing models.
3. Give many examples of various types of queueing systems that are commonly encountered.
4. Identify the key measures of performance for queueing systems and the relationships between these measures.
5. Describe the main types of basic queueing models.
6. Determine which queueing model is most appropriate from a description of the queueing system being considered.
7. Apply a queueing model to determine the key measures of performance for a queueing system.
8. Describe how differences in the importance of customers can be incorporated into priority queueing models.
9. Describe some key insights that queueing models provide about how queueing systems should be designed.
10. Apply economic analysis to determine how many servers should be provided in a queueing system.
Queues (waiting lines) are a part of everyday life. We all wait in queues to buy a movie ticket, make a bank deposit, pay for groceries, mail a package, obtain food in a cafeteria, start a ride in an amusement park, and so on. We have become accustomed to considerable amounts of waiting, but still we get annoyed by unusually long waits.
However, having to wait is not just a petty personal annoyance. The amount of time that a nation’s populace wastes by waiting in queues is a major factor in both the quality of life there and the efficiency of the nation’s economy.
Great inefficiencies also occur because of waiting that doesn’t involve people standing in line. For example, making machines wait to be repaired may result in lost production. Vehicles (including ships and trucks) that need to wait to be unloaded may delay subsequent shipments. Airplanes waiting to take off or land may disrupt later travel schedules. Delays in telecommunication transmissions due to saturated lines may cause data glitches. Causing manufacturing jobs to wait to be performed may disrupt subsequent production. Delaying service jobs beyond their due dates may result in lost future business.
Queueing theory is the study of waiting in all these various guises. It uses queueing models to represent the various types of queueing systems (systems that involve queues of some kind) that arise in practice. Formulas for each model indicate how the corresponding queueing system should perform, including the average amount of waiting that will occur, under a variety of circumstances.
Therefore, these queueing models are very helpful for determining how to operate a queueing system in the most effective way. Providing too much service capacity to operate the system involves excessive costs. But not providing enough service results in excessive
In Great Britain, waiting lines are referred to as queues, so this term has been adopted by manage- ment scientists.
Making customers, employ- ees, or jobs wait very long in a queue can have seri- ous consequences for any business.
Queueing models often are used to determine how much service capacity should be provided to a queue to avoid excessive waiting.
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waiting and all its unfortunate consequences. The models enable finding an appropriate bal- ance between the cost of service and the amount of waiting.
The first three sections of this chapter describe the elements of queueing models, give various examples of important queueing systems to which these models can be applied, and present measures of performance for these queueing systems. Section 11.4 then introduces a case study that will be carried through most of the chapter. Three subsequent sections present the most important queueing models in the context of analyzing the case study. Section 11.8 summarizes some key insights from the case study for designing queueing systems and Section 11.9 describes how economic analysis can be used to determine the number of servers to provide in a queueing system. Additional queueing models are described in the supplement to this chapter on the CD-ROM.
11.1 ELEMENTS OF A QUEUEING MODEL
We begin by describing the basic type of queueing system assumed by the queueing models in this chapter.
A Basic Queueing System Figure 11.1 depicts a typical queueing system. Customers arrive individually to receive some kind of service. If an arrival cannot be served immediately, that customer joins a queue (waiting line) to await service. (The queue does not include the customers who are currently being served.) One or more servers at the service facility provide the service. Each customer is individually served by one of the servers and then departs. You can see a demonstration of a queueing system in action by viewing the Waiting Line module in your Interactive Manage- ment Science Modules at www.mhhe.com/hillier5e or on the CD-ROM.
For some queueing systems, the customers are people. However, in other cases, the cus- tomers might be vehicles (e.g., airplanes waiting to take off on a runway), machines (e.g., machines waiting to be repaired), or other items (e.g., jobs waiting for a manufacturing operation).
A server commonly is an individual person. However, it might instead be a crew of people working together to serve each customer. The server can also be a machine, a vehicle, an electronic device, and so forth.
In most cases, the queue is just an ordinary waiting line. However, it is not necessary for the customers to be standing in line in front of a physical structure that constitutes the service facility. They might be sitting in a waiting room. They might even be scattered throughout an area waiting for a server to come to them (e.g., stationary machines needing repair).
The next section presents many more examples of important queueing systems that fit Figure 11.1 and the above description. All the queueing models in this chapter also are based on this figure.
The customers coming to some queueing systems are vehicles or machines or jobs instead of people.
Customers C C C C C C C
C C C C
S S S S
Queue
Queueing system
Served customers
Served customers
Service facility
FIGURE 11.1 A basic queueing system, where each customer is indicated by C and each server by S. Although this figure shows four servers, some queueing systems (including the example in this section) have only a single server.
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However, we also should mention that more complicated kinds of queueing systems some- times do arise in practice. For example, a server might serve a group of customers simultane- ously. Customers also might arrive in a group rather than individually. Impatient customers might leave before receiving service. The queueing system might include multiple queues, one for each server, with customers occasionally switching queues. It might include multiple service facilities, where some customers need to go to more than one of the facilities to obtain all the required service. (This last type of queueing system is referred to as a queueing net- work. ) Such queueing systems also are quite important, but we will not delve into the more complicated queueing models that have been developed to deal with them. The next two chapters will describe another technique (computer simulation) that often is used to analyze complex queueing systems.
An Example Herr Cutter is a German barber who runs a one-person barber shop. Thus, his shop is a basic queueing system for which he is the only server.
Herr Cutter opens his shop at 8:00 am each weekday morning. Table 11.1 shows his queueing system in action over the beginning of a typical morning. For each of his first five customers, the table indicates when the customer arrived, when his haircut began, how long the haircut took, and when the haircut was finished.
Figure 11.2 plots the number of customers in this queueing system over the first 100 min- utes. This number includes both the customers waiting to begin a haircut and the one already under way. Thus, the number of customers in the queue (only those waiting to begin) is one less (except that this number is zero when the number of customers in the queueing system is zero).
TABLE 11.1 The Data for Herr Cutter’s First Five Customers
Customer Time of Arrival Haircut Begins Duration of Haircut Haircut Ends
1 8:03 8:03 17 minutes 8:20 2 8:15 8:20 21 minutes 8:41 3 8:25 8:41 19 minutes 9:00 4 8:30 9:00 15 minutes 9:15 5 9:05 9:15 20 minutes 9:35 6 9:43 — — —
Number of customers in the system
4
3
2
1
0 20 40 60 80 100
Time (in minutes)
FIGURE 11.2 The evolution of the number of customers in Herr Cutter’s barber shop over the first 100 minutes (from 8:00 to 9:40), given the data in Table 11.1 .
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Referring to this example, let us now look at the kinds of assumptions that the queueing models make about the different parts of a basic queueing system.
Arrivals The times between consecutive arrivals to a queueing system are called the interarrival times . For Herr Cutter’s barber shop, the second column of Table 11.1 indicates that the interarrival times on this particular morning are 12 minutes, 10 minutes, 5 minutes, 35 min- utes, and 38 minutes.
This high variability in the interarrival times is common for queueing systems. As with Herr Cutter, it usually is impossible to predict just how long until the next customer will arrive.
However, after gathering a lot more data such as in the second column of Table 11.1 , it does become possible to do two things:
1. Estimate the expected number of arrivals per unit time. This quantity is normally referred to as the mean arrival rate . (The symbol for this quantity is l , which is the Greek letter lambda.)
2. Estimate the form of the probability distribution of interarrival times.
The mean of this distribution actually comes directly from item 1. Since
l 5 Mean arrival rate for customers coming to the queueing system
the mean of the probability distribution of interarrival times is
1 l
5 Expected interarrival time
For example, after gathering more data, Herr Cutter finds that 300 customers have arrived over a period of 100 hours. 1 Therefore, the estimate of l is
l 5 300 customers
100 hours 5 3 customers per hour on the average
The corresponding estimate of the expected interarrival time is
1 l
5 1 3
hour between customers on the average
Most queueing models assume that the form of the probability distribution of interarrival times is an exponential distribution, as explained below.
The Exponential Distribution for Interarrival Times Figure 11.3 shows the shape of an exponential distribution, where the height of the curve at various times represents the relative likelihood of those times occurring. Note in the figure how the highest points on the curve are at very small times and then the curve drops down “exponentially” as time increases. This indicates a high likelihood of small interarrival times, well under the mean. However, the long tail of the distribution also indicates a small chance of a very large interarrival time, much larger than the mean. All this is characteristic of inter- arrival times observed in practice. Several customers may arrive in quick succession. Then there may be a long pause until the next arrival.
This variability in interarrival times makes it impossible to predict just when future arriv- als will occur. When the variability is as large as for the exponential distribution, this is referred to as having random arrivals.
The times between consecutive arrivals to a queueing system (called interarrival times ) commonly are highly variable.
1 The count of 300 arrivals includes those customers who enter the barber shop but decide not to stay because the wait would be too long. The effect of these immediate departures is analyzed in the supplement to this chapter on the CD-ROM.
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The Nature of Random Arrivals: For most queueing systems, the servers have no control over when customers will arrive. In this case, the customers generally arrive randomly. Having ran- dom arrivals means that arrival times are completely unpredictable in the sense that the chance of an arrival in the next minute always is just the same (no more and no less) as for any other minute. It does not matter how long it has been since the last arrival occurred. The only distribu- tion of interarrival times that fits having random arrivals is the exponential distribution.
The fact that the probability of an arrival in the next minute is completely uninfluenced by when the last arrival occurred is called the lack-of-memory property (or the Markovian property ). This is a strange property, because it implies that the probability distribution of the remaining time from now until the next arrival occurs always is the same, regardless of whether the last arrival occurred just now or a long time ago. Therefore, this distribution of the remaining time from now is the same as the distribution of the total interarrival time given in Figure 11.3 . (This is what causes the probability of an arrival in the next minute to always be the same.) Although this concept of a lack-of-memory property takes some getting used to, it is an integral part of what is meant by having random arrivals.
The Queue The queue is where customers wait before being served. For Herr Cutter’s barber shop, the customers in the queue sit in chairs (other than the barber’s chair) while waiting to begin a haircut.
Because there are two ways of counting the customers, queueing models distinguish between them with the following terminology.
The number of customers in the queue (or queue size for short) is the number of custom- ers waiting for service to begin. The number of customers in the system is the number in the queue plus the number currently being served.
For example, Figure 11.1 shows 7 customers in the queue plus 4 more being served by the 4 servers, so a total of 11 customers are in the system. Since Herr Cutter is the only server for his queueing system, the number of customers in his queue is one less than the number of cus- tomers in the system shown in Figure 11.2 (except that the number in the queue is zero when the number in the system is zero).
The queue capacity is the maximum number of customers that can be held in the queue. An infinite queue is one in which, for all practical purposes, an unlimited number of cus- tomers can be held there. When the capacity is small enough that it needs to be taken into account, then the queue is called a finite queue . During those times when a finite queue is full, any arriving customers will immediately leave.
Herr Cutter’s queue actually is a finite queue. The queue capacity is three since he pro- vides only three chairs (other than the barber’s chair) for waiting. (He has found that his customers typically are unwilling to wait for a haircut when there already are three customers waiting in front of them.)
Interarrival times have an exponential distribution when the customers arrive randomly. In this case, the time of the next arrival always is completely unin- fluenced by when the last arrival occurred (called the lack-of-memory property).
The queue does not include the customers who are already being served.
All the queueing models in this chapter assume an infinite queue, so no limit is placed on the number of customers that can be held in the queue.
0 Mean Time
FIGURE 11.3 The shape of an exponential distribution, commonly used in queueing models as the distribution of interarrival times (and sometimes as the distribution of service times as well).
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Unless specified otherwise, queueing models conventionally assume that the queue is an infinite queue. (All the models in this chapter make this assumption, but the chapter supple- ment on the CD-ROM introduces a model that assumes a finite queue and then applies this model to analyze Herr Cutter’s barber shop.)
The queue discipline refers to the order in which members of the queue are selected to begin service. The most common is first-come, first-served (FCFS). However, other possibili- ties include random selection, some priority procedure, or even last-come, first-served. (This last possibility occurs, for example, when jobs brought to a machine are piled on top of the preceding jobs and then the machine operator takes the next job to be performed off the top of the pile.)
Section 11.7 will focus on priority queueing models. Otherwise, the queueing models throughout the chapter make the conventional assumption that the queue discipline is first- come, first-served.
Service For a basic queueing system, each customer is served individually by one of the servers. A system with more than one server is called a multiple-server system, whereas a single-server system has just one server (as for Herr Cutter’s barber shop).
When a customer enters service, the elapsed time from the beginning to the end of the service is referred to as the service time . Service times generally vary from one customer to the next. However, basic queueing models assume that the service time has a particular prob- ability distribution, independent of which server is providing the service.
The symbol used for the mean of the service-time distribution is
1 m
5 Expected service time
where m is the Greek letter mu. The interpretation of m itself is
m 5 Expected number of service completions per unit time for a single continuously busy server
where this quantity is called the mean service rate . For example, Herr Cutter’s expected time to give a haircut is
1 m
5 20 minutes 5 1 3
hour per customer
so his mean service rate is
m 5 3 customers per hour
Different queueing models provide a choice of service-time distributions, as described next.
Some Service-Time Distributions The most popular choice for the probability distribution of service times is the exponential distribution , which has the shape already shown in Figure 11.3 . The main reason for this choice is that this distribution is much easier to analyze than any other. Although this distribu- tion provides an excellent fit for interarrival times for most situations, this is much less true for service times. Depending on the nature of the queueing system, the exponential distribu- tion can provide either a reasonable approximation or a gross distortion of the true service- time distribution. Caution is needed.
As suggested by Figure 11.3 , the exponential distribution implies that many of the ser- vice times are quite short (considerably less than the mean) but occasional service times are very long (far more than the mean). This accurately describes the kind of queueing system
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where many customers have just a small amount of business to transact with the server but occasional customers have a lot of business. For example, if the server is a bank teller, many customers have just a single check to deposit or cash, but occasional customers have many transactions.
However, the exponential distribution is a poor fit for the kind of queueing system where service consists basically of a fixed sequence of operations that require approximately the same time for every customer. For example, this describes the situation where the server is an ATM machine. Although there may be small variations in service times from one customer to the next, these times generally are just about the same.
For the latter kind of queueing system, a much better approximation would be to assume constant service times , that is, the same service time for every customer. (This also is referred to as having a degenerate distribution for service times.)
Other probability distributions also can be used to represent service times. For example, the Erlang distribution allows the amount of variability in the service times to fall somewhere between those for the exponential and degenerate distributions. The Erlang distribution is described further in the chapter supplement on the CD-ROM. Unfortunately, these other pos- sibilities for service-time distributions such as the Erlang distribution are not nearly as conve- nient to work with as the exponential and degenerate distributions.
Labels for Queueing Models To identify which probability distribution is being assumed for service times (and for inter- arrival times), a queueing model for a basic queueing system conventionally is labeled as follows:
The symbols used for the possible distributions (for either service times or interarrival times) are
M 5 Exponential distribution (Markovian) D 5 Degenerate distribution (constant times)
For example, the M/M/ 1 model is the single-server model that assumes that both interar- rival times and service times have an exponential distribution. The M/M/ 2 model is the cor- responding model with two servers. Letting s be the symbol that represents the number of servers, the M/M/s model is the corresponding model that permits any number of servers. Similarly, the M/D/s model has exponential interarrival times, constant service times, and any desired number of servers.
Interarrival times also can have a degenerate distribution instead of an exponential distri- bution. The D/M/s model has constant interarrival times, exponential service times, and any desired number of servers.
All the queueing models mentioned above will be considered at least briefly later in the chapter, along with results on how well such queueing systems perform.
There even are queueing models (with limited results) that permit choosing any probabil- ity distribution for the interarrival times or for the service times. The symbols used in these cases are
GI 5 General independent interarrival-time distribution (any arbitrary distribution allowed)
G 5 General service-time distribution (any arbitrary distribution allowed)
Thus, the GI/M/s model allows any interarrival-time distribution (with independent interar- rival times), exponential service times, and any desired number of servers. The M/G/ 1 model has exponential interarrival times and one server but allows any service-time distribution. (We will touch later on just the latter model.)
For some queueing sys- tems, the service times have much less variability than implied by the exponential distribution, so queueing models that use other distributions should be considered.
The first symbol identifies the distribution of interar- rival times and the second symbol identifies the distri- bution of service times.
Distribution of service times
— / — / — ← Number of servers Distribution of interarrival times
←←
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Summary of Model Assumptions To summarize, we list below the assumptions generally made by queueing models of a basic queueing system. Each of these assumptions should be taken for granted unless a model explicitly states otherwise.
1. Interarrival times are independent and identically distributed according to a specified prob- ability distribution.
2. All arriving customers enter the queueing system and remain there until service has been completed.
3. The queueing system has a single infinite queue, so that the queue will hold an unlimited number of customers (for all practical purposes).
4. The queue discipline is first-come, first-served. 5. The queueing system has a specified number of servers, where each server is capable of
serving any of the customers. 6. Each customer is served individually by any one of the servers. 7. A server becomes available to begin serving another customer immediately after complet-
ing the service of the current customer. 8. Service times are independent and identically distributed according to a specified prob-
ability distribution.
1. What might the customers of a queueing system be other than people? 2. What might the server of a queueing system be other than an individual person? 3. What is the relationship between the mean arrival rate and the mean of the probability distri-
bution of interarrival times? 4. What is the shape of the exponential distribution? 5. How would you characterize the amount of variability in the times given by the exponential
distribution? 6. What is meant by customers arriving randomly? Which distribution of interarrival times cor-
responds to random arrivals? 7. What is the distinction between the number of customers in the queue and the number in
the system? 8. What is the conventional assumption made by most queueing models about the queue
capacity? About the queue discipline? 9. What is the relationship between the mean of the service-time distribution and the mean
service rate for a single continuously busy server? 10. What are the two most important service-time distributions? 11. What information is provided by the three parts of the label for queueing models?
Review Questions
11.2 SOME EXAMPLES OF QUEUEING SYSTEMS
Our description of queueing systems in the preceding section may appear relatively abstract and applicable to only rather special practical situations. On the contrary, queueing systems are surprisingly prevalent in a wide variety of contexts. To broaden your horizons on the applicability of queueing models, let us take a brief look at a variety of examples of real queueing systems.
One important class of queueing systems that we all encounter in our daily lives is commercial service systems , where outside customers receive service from commercial organizations. The first column of Table 11.2 lists a sampling of typical commercial service systems. Each of these is a queueing system whose customers and servers are identified in the second and third columns.
Most of these examples involve the customers coming to the server at a fixed location, where a physical queue forms if customers need to wait to begin service. However, for the plumbing services and roofing services examples, the server comes to the customers, so the
A commercial service sys- tem is a queueing system where a commercial organi- zation provides a service to customers from outside the organization.
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11.2 Some Examples of Queueing Systems 441
customers in the queue are geographically dispersed. (In this kind of situation, the time a server spends in traveling to the customer is considered to be part of the service time.) In several other cases, the service is performed over the telephone, perhaps after some customers have been placed on hold (the queue).
Organizations also have their own internal service systems , where the customers receiving service are internal to the organization. As the examples in Table 11.3 indicate, these too are queueing systems. In some cases, the customers are employees of the organiza- tions. In other examples, the customers are loads to be moved, machines to be repaired, items to be inspected, jobs to be performed, and so forth.
Transportation service systems provide another important category of queueing sys- tems. Table 11.4 gives some examples. For several of the cases, the vehicles involved are the customers. For others, each vehicle is a server. A few of the examples go beyond the basic kind of queueing system described in the preceding section. In particular, the airline flight
An internal service system is a queueing system where the customers receiving service are internal to the organization providing the service.
TABLE 11.2 Examples of Commercial Service Systems That Are Queueing Systems
Type of System Customers Server(s)
Barber shop People Barber Bank teller service People Teller ATM service People ATM Checkout at a store People Checkout clerk Plumbing services Clogged pipes Plumber Ticket window at a movie theater People Cashier Check-in counter at an airport People Airline agent Brokerage service People Stockbroker Gas station Cars Pump Call center for ordering goods People Telephone agent Call center for technical assistance People Technical representative Travel agency People Travel agent Automobile repair shop Car owners Mechanic Vending services People Vending machine Dental services People Dentist Roofing services Roofs Roofer
Type of System Customers Server(s)
Secretarial services Employees Secretary Copying services Employees Copy machine Computer programming services Employees Programmer Mainframe computer Employees Computer First-aid center Employees Nurse Faxing services Employees Fax machine Materials-handling system Loads Materials-handling unit Maintenance system Machines Repair crew Inspection station Items Inspector Production system Jobs Machine Semiautomatic machines Machines Operator Tool crib Machine operators Clerk
TABLE 11.3 Examples of Internal Service Systems That Are Queueing Systems
TABLE 11.4 Examples of Transportation Service Stations That Are Queueing Systems
Type of System Customers Server(s)
Highway tollbooth Cars Cashier Truck loading dock Trucks Loading crew Port unloading area Ships Unloading crew Airplanes waiting to take off Airplanes Runway Airplanes waiting to land Airplanes Runway Airline flight People Airplane Taxicab service People Taxicab Elevator service People Elevator Fire department Fires Fire truck Parking lot Cars Parking space Ambulance service People Ambulance
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442 Chapter Eleven Queueing Models
and elevator service examples involve a server that commonly serves a group of customers simultaneously rather than just one at a time. The queue in the parking lot example has zero capacity because arriving cars (customers) go elsewhere to park if all the parking spaces are occupied (all the servers are busy).
There are many additional examples of important queueing systems that may not fit nicely into any of the above categories. For example, a judicial system is a queueing network, where the courts are service facilities, the judges (or panels of judges) are the servers, and the cases waiting to be tried are the customers. Various health care systems, such as hospital emergency rooms, also are queueing systems. For example, x-ray machines and hospital beds can be viewed as servers in their own queueing systems. The initial applications of queueing theory were to telephone engineering, and the general area of telecommunications continues to be a very important area of application. Furthermore, we all have our own personal queues— homework assignments, books to be read, and so forth. Queueing systems do indeed pervade many areas of society.
A transportation service system is a queueing system involving transportation so that either the customers or the server(s) are vehicles.
1. What are commercial service systems? Also give a new example (not included in Table 11.2 ) of such a system, including identifying the customers and server.
2. What are internal service systems? Also give a new example (not included in Table 11.3 ) of such a system, including identifying the customers and server.
3. What are transportation service systems? Also give a new example (not included in Table 11.4 ) of such a system, including identifying the customers and server.
Review Questions
11.3 MEASURES OF PERFORMANCE FOR QUEUEING SYSTEMS
Managers who oversee queueing systems are mainly concerned with two types of measures of performance:
1. How many customers typically are waiting in the queueing system? 2. How long do these customers typically have to wait?
These measures are somewhat related, since how long a customer has to wait is partially determined by how many customers are already there when this customer arrives. Which measure is of greater concern depends on the situation.
Choosing a Measure of Performance When the customers are internal to the organization providing the service (internal service systems), the first measure tends to be more important. In this situation, forcing customers to wait causes them to be unproductive members of the organization during the wait. For exam- ple, this is the case for machine operators waiting at a tool crib or for machines that are down waiting to be repaired. Having such customers wait causes lost productivity, where the amount of lost productivity is directly related to the number of waiting customers. The active mem- bers of the organization may be able to fill in for one or two idle members, but not for more.
Commercial service systems (where outside customers receive service from commercial organizations) tend to place greater importance on the second measure. For such queueing systems, an important goal is to keep customers happy so they will return again. Customers are more concerned with how long they have to wait than with how many other customers are there. The consequence of making customers wait too long may be lost profit from lost future business.
Defining the Measures of Performance The two measures of performance commonly are expressed in terms of their expected values (in the statistical sense). To do this, it is necessary to clarify whether we are counting custom- ers only while they are in the queue (i.e., before service begins) or while they are anywhere in the queueing system (i.e., either in the queue or being served). These two ways of defining the
Having customers wait in an internal service system causes lost productivity.
Making customers wait too long in a commercial ser- vice system may result in lost profit from lost future business.
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11.3 Measures of Performance for Queueing Systems 443
two types of measures thereby give us four measures of performance. These four measures and their symbols are shown below.
L 5 Expected number of customers in the system , including those being served (the symbol L comes from L ine L ength)
L q 5 Expected number of customers in the queue , which excludes customers being served
W 5 Expected waiting time in the system (includes service time) for an individual cus- tomer (the symbol W comes from W aiting time)
W q 5 Expected waiting time in the queue (excludes service time) for an individual customer
These definitions assume that the queueing system is in a steady-state condition , that is, the system is in its normal condition after operating for some time. During the initial startup period after a queueing system opens up with no customers there, it takes awhile for the expected number of customers to reach its normal level. After essentially reaching this level, the system is said to be in a steady-state condition. (This condition also rules out such abnormal operating conditions as a temporary “rush hour” jump in the mean arrival rate.)
The choice of whether to focus on the entire queueing system ( L or W ) or just on the queue ( L q or W q ) depends on the nature of the queueing system. For a hospital emergency room or a fire department, the queue (the time until service can begin) probably is more important. For an internal service system, the entire queueing system (the total number of members of the organization that are idle there) may be more important.
Relationships betwee n L, W, L q , and W q The only difference between W and W q is that W includes the expected service time and W q does not. Therefore, since 1/ m is the symbol for the expected service time (where m is called the mean service rate ),
W 5 Wq 1 1 m
For example, if
W q 5 ¾ hour waiting in the queue on the average
1 m
5 ¼ hour service time on the average
then
W 5 ¾ hour 1 ¼ hour
5 1 hour waiting in the queueing system on the average
Perhaps the most important formula in queueing theory provides a direct relationship between L and W. This formula is
L 5 lW
where
l 5 Mean arrival rate for customers coming to the queueing system
This is called Little’s formula , in honor of the eminent management scientist John D. C. Little (a long-time faculty member at MIT), who provided the first rigorous proof of the for- mula in 1961.
To illustrate the formula, suppose that
W 5 1 hour waiting in the queueing system on the average l 5 3 customers per hour arrive on the average
These are four key mea- sures of performance for any queueing system.
This is a very handy for- mula for immediately obtaining either L or W from the other one.
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444 Chapter Eleven Queueing Models
It then follows that
L 5 (3 customers/hour)(1 hour)
5 3 customers in the queueing system on the average
Here is an intuitive way to view Little’s formula. Since L is the expected number of cus- tomers in the queueing system at any time, a customer looking back at the system after com- pleting service should see L customers there on the average. With a first-come, first-served queue discipline, all L customers there normally would have arrived during this customer’s waiting time in the queueing system. This waiting time is W on the average. Since l is the expected number of arrivals per unit time, l W is the expected number of arrivals during this customer’s waiting time in the system. Therefore, L 5 l W.
Professor Little’s proof that L 5 l W also applies to the relationship between L q and W q . Therefore, another version of Little’s formula is
Lq 5 lWq
For example, if
W q 5 ¾ hour waiting in the queue on the average l 5 ¼ customers per hour arrive on the average
then
Lq 5 (3 customers/hour)(¾ hour)
5 2¼ customers in the queue on the average
Combining the above relationships also gives the following direct relationship between L and L q .
L 5 lW 5 laWq 1 1 m b
5 Lq 1 l
m
For example, if L q 5 2¼, l 5 3, and m 5 4, then
L 5 2¼ 1 ¾ 5 3 customers in the system on the average
These relationships are extremely important because they enable all four of the funda- mental quantities— L, W, L q , and W q —to be immediately determined as soon as one is found analytically. This situation is fortunate because some of these quantities often are much easier to find than others when a queueing model is solved from basic principles.
Using Probabilities as Measures of Performance Managers frequently are interested in more than what happens on the average in a queueing system. In addition to wanting L, L q , W, and W q not to exceed target values, they also may be concerned with worst-case scenarios. What will be the maximum number of customers in the system (or in the queue) that will only be exceeded a small fraction of the time (that is, with a small probability)? What will be the maximum waiting time of customers in the system (or in the queue) that will only be exceeded a small fraction of the time? A manager might specify that the queueing system should be designed in such a way that these maximum numbers do not exceed certain values.
Meeting such a goal requires using the steady-state probability distribution of these quan- tities (the number of customers and the waiting time). For example, suppose that the goal is to have no more than three customers in the system at least 95 percent of the time. Using the notation
Pn 5 Steady-state probability of having exactly n customers in the system (for n 5 0, 1, 2, c)
These expressions show the relationships between all four measures of performance.
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11.4 A Case Study: The Dupit Corp. Problem 445
meeting this goal requires that
P0 1 P1 1 P2 1 P3 $ 0.95
Similarly, suppose that another goal is that the waiting time in the system should not exceed two hours for at least 95 percent of the customers. Let the random variable w be the waiting time in the system for an individual customer while the system is in a steady-state condition. (Thus, W is the expected value of this random variable.) Using the probability dis- tribution for this random variable, meeting the goal requires that
P(w # 2 hours) $ 0.95
If the goal is stated in terms of the waiting time in the queue instead, then a different random variable wq representing this waiting time would be used in the same way.
Formulas are available for calculating at least some of these probabilities for several of the queueing models considered later in the chapter. Excel templates in your MS Courseware will perform these calculations for you.
Since waiting times vary from customer to customer, w has a probability distri- bution, whereas W is the mean of this distribution.
1. Which type of measure of performance of queueing systems tends to be more important when the customers are internal to the organization?
2. Which type of measure of performance tends to be more important for commercial service systems?
3. What are the four basic measures of performance based on expected values? What are their symbols?
4. What is meant by a queueing system being in a steady-state condition? 5. What is the formula that relates W and W q ? 6. What is Little’s formula that relates L and W? That relates L q and W q ? 7. What is the formula that relates L and L q ? 8. What kinds of probabilities can also be used as measures of performance of queueing systems?
Review Questions
11.4 A CASE STUDY: THE DUPIT CORP. PROBLEM
The Dupit Corporation is a long-time leader in the office photocopier marketplace. One rea- son for this leadership position is the service the company provides its customers. Dupit has enjoyed a reputation of excellent service and intends to maintain that reputation.
Some Background Dupit has a service division that is responsible for providing high-quality support to the com- pany’s customers by promptly repairing the Dupit machines when needed. This work is done on the customer’s site by the company’s service technical representatives, more commonly known as tech reps .
Each tech rep is given responsibility for a specified territory. This enables providing per- sonalized service, since a customer sees the same tech rep on each service call. The tech rep generally feels like a one-person territory manager and takes pride in this role.
John Phixitt is the Dupit senior vice president in charge of the service division. He has spent his entire career with the company and actually began as a tech rep. While in this initial position, John took classes in the evening for several years to earn his business degree. Since then, he has moved steadily up the corporate ladder. He is well respected for his sound judg- ment and his thorough understanding of the company’s business from the ground up.
John’s years as a tech rep impressed upon him the importance of the tech rep’s role as an ambassador of the company to its customers. He continues to preach this message regularly. He has established high personnel standards for becoming and remaining a tech rep and has built up the salaries accordingly. The morale in the division is quite high, largely through his efforts.
Each tech rep currently is assigned his or her own territory for servicing the machines.
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John also emphasizes obtaining regular feedback from a random sample of the company’s customers on the quality of the service being provided. He likes to refer to this as keeping his ear to the ground. The customer feedback is channeled to both the tech reps and management for their information.
Another of John’s themes is the importance of not overloading the tech reps. When he was a tech rep himself, the company policy had been to assign each tech rep enough machines in his or her territory that the tech rep would be active repairing machines 90 percent of the time (during an eight-hour working day). The intent was to maintain a high utilization of expensive personnel while providing some slack so that customers would not have to wait very long for repairs. John’s own experience was that this did not work very well. He did have his idle periods about 10 percent of the time, which was helpful for catching up on his paperwork and maintaining his equipment. However, he also had frequent busy periods with many repair requests, including some long ones, and a large backlog of unhappy customers waiting for repairs would build up.
Therefore, when he was appointed to his current position, one of his first moves was to make the case to Dupit top management that tech reps needed to have more slack time to ensure providing prompt service to customers. A major part of his argument was that cus- tomer feedback indicated that the company was failing to deliver on the second and third parts of the company slogan given below.
1. High-quality products. 2. High-quality service. 3. All delivered efficiently.
The company president had been promoting this slogan for years and so found this argument persuasive. Despite continuing pressure to hold costs down, John won approval for changing company policy regarding tech reps as summarized below.
Current Policy: Each tech rep’s territory should be assigned enough machines so that the tech rep will be active repairing machines (or traveling to the repair site) approximately 75 percent of the time. When working continuously, each tech rep should be able to repair an average of four machines per day (an average of two hours per machine, including travel time). Therefore, to minimize customer waiting times, the goal is to have an average of three repair calls per working day. Since the company’s machines now are averaging 50 work days between needing repairs, the target is to assign approximately 150 machines to each tech rep’s territory.
Under this policy, the company now has nearly 10,000 tech reps, with a total payroll (includ- ing benefits) of approximately $600 million per year.
The Issue Facing Top Management A long succession of very successful products has helped Dupit maintain its position as a market leader for many years. Furthermore, its latest product has been a particularly big win- ner. It is a color printer-copier that collates, staples, and so on, as well as having advanced networking capabilities. Thus, it is a state-of-the-art, all-in-one copier for the modern office. Sales have even exceeded the optimistic predictions made by the vice president for marketing.
However, the success of this product also has brought problems. The fact that the machine performs so many key functions makes it a vital part of the purchaser’s office. The owner has great difficulty in getting along without it for even a few hours when it is down requiring repair. Unfortunately, the average waiting time until the tech rep becomes available to come do the repair now runs about 6 hours and some waiting times are even considerably longer. Consequently, even though the tech reps are giving the same level of service as they have in the past, complaints about intolerable waits for repairs have skyrocketed.
This crisis has led to an emergency meeting of top management, with John Phixitt the man on the spot. He assures his colleagues that service has not deteriorated in the least. There is agreement that the company is a victim of its own success. The new machine is so valuable that a much higher level of service is required.
Tech reps need consider- able slack time to ensure providing prompt service to customers.
Each tech rep’s territory currently has approxi- mately 150 machines, which requires a tech rep to be busy on service calls approximately 75 percent of the time.
The company’s new color printer-copier is such a vital part of each purchaser’s office that a much higher level of service is required to reduce downtime.
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11.4 A Case Study: The Dupit Corp. Problem 447
After considerable discussion about how to achieve the needed service, Dupit’s president suggests the following four-step approach to dealing with the problem:
1. Agree on a tentative new standard for the level of service that needs to be provided. 2. Develop some proposals for alternative approaches that might achieve this standard. 3. Have a management science team work with John Phixitt to analyze these alternative
approaches in detail to evaluate the effectiveness and cost of each one. 4. Reconvene this group of top management to make a final decision on what to do.
The group agrees. Discussion then turns to what the new standard should be for the level of service. John
proposes that this standard should specify that a customer’s average waiting time before the tech rep can respond to the request for a repair should not exceed some maximum quantity. The customer relations manager agrees and argues that this average waiting time should not exceed two hours (versus about six hours now). The group agrees to adopt two hours as the tentative standard, pending further analysis by the management science team.
Proposed New Service Standard: The average waiting time of customers before the tech rep begins the trip to the customer site to repair the machine should not exceed two hours.
Alternative Approaches to the Problem After further discussion of various ideas about how to meet this service standard, the meeting concludes. The president asks the participants who had proposed some approach to think fur- ther about their idea. If they conclude that their idea should be a particularly sound approach to the problem, they are to send him a memorandum supporting that approach.
The president subsequently receives four memoranda supporting the approaches summa- rized below:
Approach Suggested by John Phixitt: Modify the current policy by decreasing the per- centage of time that tech reps are expected to be active repairing machines. This involves simply decreasing the number of machines assigned to each tech rep and adding more tech reps. Thus, each tech rep would continue to have sole responsibility for servicing a specified territory, but each such one-person territory now would be a smaller one. This approach would enable continuing the mode of operation for the service division that has served the company so well in the past while increasing the level of service to meet the new demands of the marketplace. Approach Suggested by the Vice President for Engineering: Provide new state-of- the-art equipment to the tech reps that would substantially reduce the time required for the longer repairs. Although expensive, this would significantly reduce the average repair time. Perhaps more importantly, it would greatly reduce the variability of repair times, which might decrease average waiting times for repairs. Approach Suggested by the Chief Financial Officer: Replace the current one-person tech rep territories by larger territories that would be served by multiple tech reps. Having teams of tech reps to back each other up during busy periods might decrease average wait- ing times for repairs enough that the company would not need to hire additional tech reps. Approach Suggested by the Vice President for Marketing: Give owners of the new printer-copier priority for receiving repairs over the company’s other customers. Since the complaints about slow service are coming mainly from these owners, this approach might give them the service they require while still giving adequate service to other customers.
The president is pleased to have four promising approaches to consider. As previously agreed, his next step is to set up a team of management scientists (three from the company plus an outside consultant) to work with John Phixitt in analyzing these approaches in detail. They are to report back to top management with their results and recommendations in six weeks.
Before reading further, we suggest that you think about these four alternative approaches and decide which one seems most promising. You then will be able to compare your conclu- sions with the results from the management science study.
The proposal is to reduce average waiting times before the repair process begins from six hours to two hours.
Queueing models will be used to analyze each of the four proposed approaches.
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The Management Science Team’s View of the Problem The management science team quickly recognizes that queueing theory will be a key tech- nique for analyzing this problem. In particular, each tech rep’s territory can be viewed as including the basic queueing system described below.
The Queueing System for Each Tech Rep 1. The customers: The machines needing repair. 2. Customer arrivals: The calls to the tech rep on his or her cellular telephone requesting
repairs.
3. The queue: The machines waiting for repair to begin at their sites. 4. The server: The tech rep. 5. Service time: The total time the tech rep is tied up with a machine, either traveling to the
machine site or repairing the machine. (Thus, a machine is viewed as leaving the queue and entering service when the tech rep begins the trip to the machine site.)
With the approach suggested by the chief financial officer (enlarge the territories with multiple tech reps for each territory), this single-server queueing system would be changed to a multiple-server queueing system.
The management science team now needs to decide which specific queueing model is most appropriate for analyzing each of the four approaches. You will see this story unfold in the next few sections while we are presenting various important queueing models.
1. What is the company’s current policy regarding the workload for tech reps? 2. What is the issue currently facing top management? 3. What is the proposed new service standard? 4. How many alternative approaches have been suggested for dealing with the issue facing top
management? 5. In the queueing system interpretation of this problem, what are the customers? The server?
Review Questions
11.5 SOME SINGLE-SERVER QUEUEING MODELS
Using the background on the elements of queueing models presented in Section 11.1, this section focuses on models of basic queueing systems having just one server. Key symbols introduced in Section 11.1 that will continue to be used here (and throughout the remainder of the chapter) are
l = Mean arrival rate for customers coming to the queueing system
= Expected number of arrivals per unit time
m = Mean service rate(for a continuously busy server
= Expected number of service completions per unit time
Also recall that 1/ l is the expected interarrival time (the average time between the arrival of consecutive customers) and 1/ m is the expected service time for each customer.
A new symbol for this section is
r 5 l
m
where r is the Greek letter rho. This quantity r is referred to as the utilization factor , because it represents the average fraction of time that the server is being utilized serving customers.
The utilization factor plays a key role in the efficiency of a queueing system.
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449
In the Dupit Corp. case study, under the company’s current policy, a typical tech rep experiences
l 5 3 customers (machines needing repair) arriving per day on the average
m 5 4 service completions (repair completions) per day on the average when the tech rep is continuously busy
Since
r 5 3 4
5 0.75
the tech rep is active repairing machines 75 percent of the time. For each of the queueing models, we will consider the measures of performance intro-
duced in Section 11.3. Because of the relationships between the four basic measures— L, L q , W, and W q —including Little’s formula given in that section, recall that all four quantities can be calculated easily as soon as one of their values has been determined. Therefore, we some- times will be focusing on just one of these measures of performance for the following models.
The M/M/ 1 Model Using the labels for queueing models given near the end of Section 11.1, recall that the first symbol ( M ) in the M/M/ 1 label identifies the probability distribution of interarrival times, the second symbol ( M ) indicates the distribution of service times, and the third symbol (1) gives the number of servers. Since M is the symbol used for the exponential distribution, the M/M/ 1 model makes the following assumptions.
For many decades, General Motors Corporation (GM) enjoyed its position as the world’s largest automotive man- ufacturer. However, since the late 1980s, when the produc- tivity of GM’s plants ranked near the bottom in the industry, the company’s market position was steadily eroding due to ever-increasing foreign competition.
To counter this foreign competition, GM management initiated an ongoing management science project to predict and improve the throughput performance of the company’s several hundred production lines throughout the world. The goal was to greatly increase the company’s productivity throughout its manufacturing operations and thereby pro- vide GM with a strategic competitive advantage.
The most important analytical tool used in this project has been a complicated queueing model that uses a simple single-server model as a building block. The overall model begins by considering a two-station production line where each station is modeled as a single-server queueing sys- tem with constant interarrival times and constant service times with the following exceptions. The server (commonly a machine) at each station occasionally breaks down and does not resume serving until a repair is completed. The server at the first station also shuts down when it com- pletes a service and the buffer between the stations is full. The server at the second station shuts down when it com- pletes a service and has not yet received a job from the first station.
The next step in the analysis is to extend this queue- ing model for a two-station production line to one for a
production line with any number of stations. This larger queueing model then is used to analyze how production lines should be designed to maximize their throughput. (The technique of computer simulation described in the next two chapters also is used for this purpose for relatively complex production lines.)
This application of queueing theory (and computer sim- ulation), along with supporting data-collection systems, has reaped remarkable benefits for GM. According to impar- tial industry sources, its plants that once were among the least productive in the industry now rank among the very best. The resulting improvements in production through- put in over 30 vehicle plants and 10 countries yielded over $2.1 billion in documented savings and increased revenue by 2005. Then, following the near-collapse of the company in 2009 during the Great Recession, the exceptional pro- ductivity of its plants is one of the factors that enabled GM to regain its position as the world’s largest automotive manufacturer in 2011.
This innovative application of management science, including a single-server queueing model, led to GM win- ning the prestigious first prize in the 2005 international competition for the Franz Edelman Award for Achievement in Operations Research and the Management Sciences.
Source: J. M. Alden, L. D. Burns, T. Costy, R. D. Hutton, C. A. Jackson, D. S. Kim, K. A. Kohls, J. H. Owen, M. A. Turnquist, and D. J. Vander Veen, “General Motors Increases Its Production Throughput,” Inter- faces 36, no. 1 (January–February 2006), pp. 6–25. (A link to this article is provided on our website, www.mhhe.com/hillier5e . )
An Application Vignette
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Assumptions 1. Interarrival times have an exponential distribution with a mean of 1/ l . (See Figure 11.3
and the description of this distribution in Section 11.1.) 2. Service times have an exponential distribution with a mean of 1/ m . 3. The queueing system has 1 server.
As discussed in Section 11.1, the first assumption corresponds to having customers arrive randomly. Consequently, this assumption commonly is a valid one for real queueing systems.
The second assumption also is a reasonable one for those queueing systems where many service times are quite short (well under the mean) but occasional service times are very long. Some queueing systems fit this description, but some others do not even come close.
Along with its multiple-server counterpart (considered in Section 11.6), the M/M/ 1 model is the most widely used queueing model. (It is even sometimes used for queueing systems that don’t fit the second assumption very well.) A key reason is that this model has the most results readily available. Because the formulas are relatively simple, we give them for all the measures of performance below. (All these measures assume that the queueing system is in a steady-state condition. )
Formulas for the M/M/ 1 Model Using r 5 l / m , two equivalent formulas for the expected number of customers in the system are
L 5 r
1 2 r 5
l
m 2 l
Because of Little’s formula ( L 5 l W ), the expected waiting time in the system is
W 5 1 l
L 5 1
m 2 l
Therefore, the expected waiting time in the queue (excludes service time) is
Wq 5 W 2 1 m
5 1
m 2 l 2
1 m
5 m 2 (m 2 l)
m(m 2 l)
5 l
m(m 2 l)
Applying the other version of Little’s formula again ( L q 5 l W q ), the expected number of cus- tomers in the queue (excludes customers being served) is
Lq 5 lWq 5 l
2
m(m 2 l) 5
r 2
1 2 r
Even the formulas for the various probabilities are relatively simple. The probability of having exactly n customers in the system is
Pn 5 (1 2 r)p n for n 5 0, 1, 2, c
Thus,
P0 5 1 2 r
P1 5 (1 2 r)r
P2 5 (1 2 r)r 2
#
#
# The probability that the waiting time in the system exceeds some amount of time t is
P(w . t) 5 e2m(12r)t for t $ 0
where e (approximately equal to 2.718) is a special number widely used in mathematics.
Although the second assumption is sometimes questionable, this model is widely used because it provides so many useful results.
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The corresponding probability that the waiting time in the queue exceeds t is
P(wq . t) 5 re 2m(12r)t for t $ 0
Since this waiting time in the queue is 0 if there are no customers in the system when an arrival occurs,
P(wq 5 0) 5 P0 = 1 2 r All these formulas assume that the server has a manageable utilization factor ( r 5 l / m ), that is, that
r , 1
( All single-server queueing models of the basic queueing system described in Section 11.1 make this same assumption.) When r > 1, so that the mean arrival rate l exceeds the mean service rate m , the server is not able to keep up with the arrivals so the queueing system never reaches a steady-state condition. (This is even technically true when r 5 1.)
Some Helpful Software Although these formulas for the M/M /1 model are relatively simple ones, there are a lot of them. Fortunately, there is an Excel template for the M/M/s model in your MS Course- ware that will calculate all these measures of performance for you, including even the waiting time probabilities, if you wish. As illustrated in Figure 11.4 for the case study, all you have to do is set s 5 1 and then specify the values of l and m (plus the value of t if you want the waiting time probabilities). Since l and m are the estimated values of the mean arrival rate and mean service rate, respectively, you then can conduct sensitivity analysis on l and m by rerunning the template for various other possible values. All this can be done in a matter of seconds.
In addition, an animated demonstration of an M/M/s queueing system in action— showing customers arriving, waiting in the system, and departing when their service is completed—is provided by the Waiting Line module in your Interactive Management Science Modules at www.mhhe.com/hillier5e or on the CD-ROM. After setting the values of s, l , and m and then viewing the queueing system in action (repeatedly if desired), the module will calculate estimates of L, L q , W, and W q by using the averages for the customers that have arrived so far. If the queueing system were to run an extremely long time, these estimates would be very close to the exact values for these measures of performance that are provided by the Excel template for this model.
Applying the M/M/ 1 Model to the Case Study under the Current Policy The Dupit management science team begins its study by gathering some data on the experi- ences of some representative tech reps. They determine that the company’s current policy regarding tech rep workloads (they are supposed to be busy repairing machines 75 percent of the time) is operating basically as intended. Although there is some variation from one tech rep to the next, they typically are averaging about three calls requesting repairs per day. They also are averaging about two hours per repair (including a little travel time), and so can average four repairs for each eight-hour working day that they are continuously repairing machines. This verifies that the best estimates of the daily rates for a typical tech rep’s queue- ing system (where the tech rep is the server and the machines needing repairs are the custom- ers) are a mean arrival rate of l 5 3 customers per day and a mean service rate of m 5 4 customers per day (so r 5 l / m 5 0.75), just as assumed under the current policy. (Other time units, such as hourly rates rather than daily rates, could be used for l and m , but it is essential that the same time units be used for both.)
The team also concludes that the customer arrivals (calls requesting repairs) are occurring randomly, so the first assumption of the M/M/ 1 model (an exponential distribution for inter- arrival times) is a good one for this situation. The team is less comfortable with the second assumption (an exponential distribution for service times), since the total service time (travel time plus repair time) never is extremely short as allowed by the exponential distribution.
The Excel template for the M/M/s model can be applied to this M/M/ 1 model by setting s 5 1.
Under the current policy, each tech rep is the server for his or her own queueing system where the machines needing repair are the customers.
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A B C D E F G
Data
M/M/1 Queueing Model for the Dupit Corp. Problem
Results
L =
Lq=
W =
Wq=
ρ =
L =
Lq=
W =
Wq=
ρ =
n Pn
=Lambda/(Mu-Lambda)
=Lambda)^2/(Mu*(Mu-Lambda)
=1/(Mu-Lambda)
=Lambda/(Mu*(Mu-Lambda))
=Lambda/Mu
=1-Rho
=(1-Rho)*Rho^n
=(1-Rho)*Rho^n
:
:
Pnn
0
1
2
3
4
5
6
7
8
9
10
(mean arrival rate)
(mean service rate)
(# servers)
λ = μ = s =
Pr(0> t) =
when t =
Prob(0q> t) =
when t =
L Lambda
Lq Mu
n P0 Pn Rho
s Time1
Time2
W Wq
G4
C4
G5
C5
F13:F38
G13
G13:G38
G10
C6
C9
C12
G7
G8
Range Name Cells
4
5
6
7
8
9
10
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13 0
1
2
3
4
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17
F G
8 Pr(0>t) = =EXP(-Mu*(1-Rho)*C9)
B C
11 Prob(0q>t) = =Rho*EXP(-Mu*(1-Rho)*C12)
B C
0.0000 0
0.0500
0.1000
0.1500
0.2000
0.2500
0.3000
1 2 3 4 5 6 7 8 9 10 11 12
Number of Customers in System
13 14 15 16 17 18 19 20 21 22 23 24 25
P ro
b a b il
it y
3
4
1
0.368
1
0.276
1
3
2.25
1
0.75
0.75
0.2500
0.1875
0.1406
0.1055
0.0791
0.0593
0.0445
0.0334
0.0250
0.0188
0.0141
FIGURE 11.4 This spreadsheet shows the results from applying the M/M/1 model with l 5 3 and m 5 4 to the Dupit case study under the current policy. The equations for the M/M/1 model have been entered into the corresponding output cells, as shown at the bottom of the figure.
However, many service times are at least fairly short (well under the mean) and occasional service times are very long, which does fit the exponential distribution reasonably well. Therefore, the team decides that it is reasonable to use the M/M/ 1 model to represent a typical tech rep’s queueing system under the current policy.
The Excel template in Figure 11.4 shows the results from applying the various formu- las for this model to this queueing system. Look first at the results at the top of column G. The expected number of machines needing repairs is L 5 3. When excluding any machine
The M/M/ 1 model with l 5 3 and m 5 4 provides a reasonable representation of each tech rep’s queueing system.
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11.5 Some Single-Server Queueing Models 453
that is currently being repaired, the expected number of machines waiting to begin service is L q 5 2.25. The expected waiting time of a machine, measured from when the service request is submitted to the tech rep until the repair is completed, is W 5 1 day. When excluding the repair time, the expected waiting time to begin service is W q 5 0.75 day 5 6 hours during an eight-hour working day. (These results are treating a machine as moving from the queue into service when the tech rep begins the trip to the site of the machine.)
While gathering data, the management science team found that the average waiting time of customers until service begins on their failed machines is approximately six hours of an eight-hour workday (i.e., ¾ of a workday). The fact that this time agrees with the value of W q yielded by the model gives further credence to the validity of the model for this application.
Now look at the results in column G for P n (the probability of having exactly n customers in the system). With P 0 5 0.25, the tech rep will only be busy repairing machines 75 percent of the time (as indicated by the utilization factor of r 5 0.75). Since P 0 1 P 1 1 P 2 5 0.58, the tech rep will have no more than two machines needing repair (including the one being worked on) well over half the time. However, he or she also will have much larger backlogs with some frequency. For example, P 0 1 P 1 1 P 2 1 . . . 1 P 7 5 0.9, which indicates that the tech rep will have at least eight machines needing repair (about two days’ work or more) 10 percent of the time. With all the randomness inherent in such a queueing system (the great variability in both interarrival times and service times), these very big backlogs (and many unhappy customers) will occur occasionally despite the tech rep only having a utilization fac- tor of 0.75.
Finally, look at the results in cells C8:C12. By setting t 5 1, the probability that a customer has to wait more than one day (eight work hours) before a failed machine is operational again is given as P(w . 1 day) 5 0.368. The probability of waiting more than one day before the repair begins is P(wq . 1 day) 5 0.276.
Upon being shown all these results, John Phixitt comments that he understands better now why the complaints have been pouring in. No owner of such a vital machine as the new printer-copier should be expected to go more than a day (or even most of a day) before it is repaired.
Applying the M/M/ 1 Model to John Phixitt’s Suggested Approach The management science team now is ready to begin analyzing each of the suggested approaches for lowering to two hours (¼ workday) the average waiting time before service begins. Thus, the new constraint is that
Wq # ¼ day
The first approach, suggested by John Phixitt, is to modify the current policy by lowering a tech rep’s utilization factor sufficiently to meet this new service requirement. This involves decreasing the number of machines assigned to each tech rep from about 150 to some smaller number. Since each machine needs repair about once every 50 work days on the average, decreasing the number of machines in a tech rep’s territory results in decreasing the mean arrival rate l from 3 to
l 5 Number of machines assigned to tech rep
50
With m fixed at four, this decrease in l will decrease the utilization factor, r 5 l / m . Since decreasing l decreases W q , the largest value of l that has W q ≤ ¼ day is the one that
makes W q equal to ¼ day. The easiest way to find this l is by trial and error with the Excel template, trying various values of l until one is found where W q 5 0.25. Figure 11.5 shows the template that gives this value of W q by setting l 5 2. (By using the formula for W q , it also is possible to solve algebraically to find l 5 2.)
Decreasing l from three to two would require decreasing the target for the number of machines assigned to each tech rep from 150 to 100. This 100 is the maximum number that
All the results from the M/M/ 1 model indicate that unacceptably long delays in repairing failed machines will occur too frequently under the current policy.
To meet the proposed new service standard that Wq # ¼ day, John Phixitt suggests reducing the num- ber of machines assigned to each tech rep from 150 to some smaller number.
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would satisfy the requirement that W q ≤ ¼ day. With l 5 2 and m 5 4, the utilization factor for each tech rep would be only
r l
m 5
2 4
5 0.5
Recall that the company’s payroll (including benefits) for its nearly 10,000 tech reps cur- rently is about $600 million annually. Decreasing the number of machines assigned to each tech rep from 150 to 100 would require hiring nearly 5,000 more tech reps to cover all the machines. The additional payroll cost would be about $270 million annually. (It is a little less than half the current payroll cost because the new tech reps would have less seniority than the current ones.) However, the management science team estimates that the additional costs of hiring and training the new tech reps, covering their work expenses, providing them with equipment, and adding more field service managers to administer them would be equivalent to about $30 million annually.
Total Additional Cost of the Approach Suggested by John Phixitt: Approximately $300 million annually.
The M/G/ 1 Model This queueing model differs from the M/M/ 1 model only in the second of its assumptions summarized below.
Assumptions 1. Interarrival times have an exponential distribution with a mean of 1/ l . 2. Service times can have any probability distribution. It is not even necessary to determine
the form of this distribution. You just need to estimate the mean (1/ m ) and standard devia- tion ( s ) of the distribution.
3. The queueing system has one server.
Thus, this is an extremely flexible model that only requires the common situation of random arrivals (equivalent to the first assumption) and a single server, plus estimates of l , m , and s .
The M/M/ 1 model indi- cates that the number of machines assigned to each tech rep would need to be reduced to 100 with John Phixitt’s approach.
You now need to also esti- mate s , the standard devia- tion of the service-time distribution.
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A B C D E F G
Data
M/M/1 Queueing Model for John Phixitt's Approach (Reduce Machines/Rep)
Results
L =
Lq=
W =
Wq=
ρ =
n
0
1
2
3
4
5
6
7
8
9
10
(mean arrival rate)
(mean service rate)
(# servers)
λ = μ = s =
Pr(0> t) =
when t =
Prob(0q> t) =
when t =
0 0
0.1
0.2
0.3
0.4
0.5
0.6
1 2 3 4 5 6 7 8 9 10 11 12
Number of Customers in System
13 14 15 16 17 18 19 20 21 22 23 24 25
P ro
b a b il
it y
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4
1
0.135
1
0.0677
1
1
0.5
0.5
0.25
0.5
Pn 0.5
0.25
0.125
0.0625
0.0313
0.0156
0.00781
0.00391
0.00195
0.00098
0.00049
FIGURE 11.5 This application of the spreadsheet in Figure 11.4 shows that, when m 5 4, the M/M/1 model gives an expected waiting time to begin service of Wq 5 0.25 day (the larg- est value that satisfies Dupit’s proposed new service standard) when l is changed from l 5 3 to l 5 2.
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11.5 Some Single-Server Queueing Models 455
Formulas for the M/G/ 1 Model Using r 5 l / m , here are the available formulas for this model.
P0 5 1 2 r
Lq 5 l
2 s
2 1 r
2
2(1 2 r)
L 5 Lq 1 r
Wq 5 Lq l
W 5 Wq 1 1 m
These steady-state measures of performance require only that r < 1, which allows the queue- ing system to reach a steady-state condition.
To illustrate the formulas, suppose that the service-time distribution is the exponential distribution with mean 1/ m . Then, since the standard deviation s is
s 5 mean 5 1 m
for the exponential distribution
the formula for L q indicates that
Lq 5 l
2a 1 m
2b 1 r2 2(1 2 r)
5 r
2 1 r
2
2(1 2 p)
5 r
2
(1 2 r)
just as for the M/M/ 1 model. Having s 5 1/ m also causes the formulas for L, W q , and W to reduce algebraically to those given earlier for the M/M/ 1 model. In fact, the M/M/ 1 model is just the special case of the M/G/ 1 model where s 5 1/ m . (However, the M/M/ 1 model yields some results that are not available from the M/G/ 1 model.)
Insights Provided by the M / G /1 Model Another important special case of the M/G/ 1 model is the M/D/ 1 model, which assumes that the service-time distribution is the degenerate distribution (constant service times). Since
s 5 0 for the degenerate distribution
the formula for L q yields
Lq 5 l
2(0) 1 r2
2(1 2 r) 5
1 2
r
2
1 2 r
which is just half that for the M/M/ 1 model. Thus, going from a service-time distribution that has high variability (the exponential distribution) to one that has no variability (the degener- ate distribution) has a dramatic effect on reducing L q .
As this illustrates, the L q formula for the M/G/ 1 model is an enlightening one, because it reveals what effect the variability of the service-time distribution has on this measure of performance. With fixed values of l , m , and r , decreasing this variability (i.e., decreasing s ) definitely decreases L q . The same thing happens with L, W, and W q . Thus, the consistency of the server has a major bearing on the performance of the queueing system. Given the choice between two servers with the same average speed (the same value of 1/ m ), the one with less variability (smaller s ) definitely should be preferred over the other one. (We will discuss this further in Section 11.8.)
Decreasing the variability of the service-time distribution has the beneficial effect of decreasing L q , L, W, and W q .
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Considering the complexity involved in analyzing a model that permits any service-time distribution, it is remarkable that such a simple formula can be obtained for L q . This formula is one of the most important results in queueing theory because of its ease of use and the prev- alence of M/G/ 1 queueing systems in practice. This equation for L q (or its counterpart for W q ) commonly is referred to as the Pollaczek-Khintchine formula, named after two pioneers in the development of queueing theory who derived the formula independently in the early 1930s.
Applying the M/G/ 1 Model to the Approach Suggested by the Vice President for Engineering Dupit’s vice president for engineering has suggested providing the tech reps with new state- of-the-art equipment that would substantially reduce the time required for the longer repairs. This would decrease the average repair time somewhat and also would substantially decrease the variability of the repair times.
After gathering more information from this vice president and analyzing it further, the management science team makes the following estimates about the effect of this approach on the service-time distribution.
The mean would decrease from ¼ day to 1⁄5 day. The standard deviation would decrease from ¼ day to 1⁄10 day.
Thus, the standard deviation would decrease from equaling the previous mean (as for the exponential distribution) to being just half the new mean. Since m 5 1/mean, we now have m 5 5 instead of m 5 4.
With s 5 0.1, the Excel template for the M/G/ 1 model in your MS Courseware yields the results shown in Figure 11.6 . Note that W q 5 0.188 day. This big reduction from W q 5 0.75 day under the current policy (as given in Figure 11.4 ) is largely due to the big decrease in s . If the service-time distribution continued to be an exponential distribution, then increasing m from 4 to 5 would decrease W q from 0.75 day to 0.3 day. The additional reduction from 0.3 day to 0.188 day is because of the large reduction in the variability of service times.
The new state-of-the-art equipment suggested by the vice president for engineer- ing would substantially reduce both the mean and the standard devia- tion of the service-time distribution.
The M/G/ 1 model indicates that the proposed new ser- vice standard would be met easily by this approach.
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Data
M/G/1 Model for VP of Engineering's Approach (New Equipment)
3
0.2
0.1
1
Results
L =
Lq=
W =
Wq=
ρ =
P0 =
L =
Lq=
W =
Wq=
ρ =
P0 =
=Lq+Rho
=((Lambda)^2)*(Sigma^2)+(Rho^2))/(2*(1-Rho))
=Wq+OneOverMu
=Lq/Lambda
=Lambda*OneOverMu
=1-Rho
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
λ = 1/μ =
σ = s =
L
Lambda
Lq OneOverMu
Rho
s
Sigma
W
Wq
G4
C4
G5
C5
G10
C7
C6
G7
G8
Range Name Cell
4
5
6
7
8
9
10
11
12
F G
1.163
0.563
0.388
0.188
0.6
0.4
FIGURE 11.6 This Excel template for the M/G/1 model shows the results from applying this model to the approach suggested by Dupit’s vice president for engineering to use new state-of-the-art equipment.
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11.6 Some Multiple-Server Queueing Models 457
Recall that the proposed new service standard is W q ≤ 0.25 day. Therefore, the approach suggested by the vice president for engineering would satisfy this standard.
Unfortunately, the management science team also determines that this approach would be expensive, as summarized below.
Total Additional Cost of the Approach Suggested by the Vice President for Engineering: A one-time cost of approximately $500 million (about $50,000 for new equipment per tech rep).
1. What are represented by the symbols l and m ? By 1/ l and 1/ m ? By r ? 2. What are the assumptions of the M/M/ 1 model? 3. For which measures of performance (both expected values and probabilities) are formulas
available for the M/M/ 1 model? 4. Which values of r correspond to the server in a single-server queueing system having a man-
ageable utilization factor that allows the system to reach a steady-state condition? 5. Under Dupit’s current policy, what is the average waiting time of customers until service
begins on their failed machines? 6. How much more would it cost Dupit to reduce this average waiting time to 1/4 workday by
decreasing the number of machines assigned to each tech rep? 7. How does the M/G/ 1 model differ from the M/M/ 1 model? 8. Which service-time distribution is assumed by the M/D/ 1 model? 9. For the M/G/ 1 model, what is the effect on L q , L, W, and W q of decreasing the standard devia-
tion of the service-time distribution? 10. What is the total additional cost of the approach suggested by Dupit’s vice president for
engineering?
Review Questions
11.6 SOME MULTIPLE-SERVER QUEUEING MODELS
Many queueing systems have more than one server, so we now turn our attention to multiple- server queueing models. In particular, we will discuss what results are available for the multiple- server counterparts of the single-server models introduced in the preceding section.
Recall that the third symbol in the label for a queueing model indicates the number of serv- ers. For example, the M/M/ 2 model has two servers. The M/M/s model allows the choice of any number of servers, where s is the symbol for this number.
Also recall that r ( 5 l / m ) was the symbol used for the utilization factor for the server in a single-server queueing system. With multiple servers, the formula for this symbol changes to
r 5 l
sm (utilization factor)
where l continues to be the mean arrival rate (so 1/ l still is the expected interarrival time) and m continues to be the mean service rate for a single continuously busy server (so 1/ m still is the expected service time). The models assume that all the servers have the same service-time distribution, so m is the same for every server. Since
l 5 Expected number of arrivals per unit time
sm 5 Expected number of service completions per unit time when all s servers are con- tinuously busy
it follows that r 5 l / s m is indeed the average fraction of time that individual servers are being utilized serving customers.
In order for the servers to have a manageable utilization factor, it is again necessary that
r , 1
All multiple-server models of the basic queueing system described in Section 11.1 make this assumption to enable the queueing system to reach a steady-state condition.
As with single-server queueing models, the uti- lization factor r still is the average fraction of time that the individual servers are being utilized serving customers.
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Of the three previously considered single-server models ( M/M/ 1, M/G/ 1, and M/D/ 1), the M/G/ 1 model is the only one whose multiple-server counterpart yields no useful analytical results. Combining the complication of multiple servers with the complication of allowing the choice of any service-time distribution is too much to handle.
We begin with the M/M/s model, including its application to the Dupit case study. We then mention the limited results available for the M/D/s model.
The M/M/s Model Except for the last one, the assumptions are the same as for the M/M/ 1 model.
Assumptions 1. Interarrival times have an exponential distribution with a mean of 1/ l .
2. Service times have an exponential distribution with a mean of 1/ m .
3. Any number of servers (denoted by s ) can be chosen for the queueing system.
Explicit formulas are available for all the measures of performance (including the prob- abilities) considered for the M/M/ 1 model. However, when s > 1, the formulas are too tedious to want to do by hand. Therefore, you should use the Excel template for the M/M/s model (as demonstrated earlier in Figures 11.4 and 11.5 ) to generate all these results. (You also can see a demonstration of the M/M/s queueing system in action by viewing the Waiting Line mod- ule in your Interactive Management Science Modules at www.mhhe.com/hillier5e or on the CD-ROM.)
Another alternative is to use Figure 11.7 , which shows the values of L versus the utiliza- tion factor for various values of s. (Be aware that the vertical axis uses a logarithmic scale, so you need to refer to the notches to determine the value along this axis.) By estimating L from this graph, you then can use Little’s formula ( L 5 l W and L q 5 l W q ), plus W 5 Wq 1
1⁄m to calculate W, W q , and L q .
The Excel template for the M/M/s model provides all the measures of perfor- mance that were described in Section 11.5 for the M/M/ 1 model.
S te
ad y-
st at
e ex
p ec
te d
n u
m b
er o
f cu
st om
er s
in t
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q u
eu ei
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sy st
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1.0
0.5
0.2
0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Utilization factor
L
r = l—
sm
s = 25
s = 1
s = 2
s = 3
s = 4
s = 5
s = 7
s = 10
s = 15
s = 20
FIGURE 11.7 Values of L for the M/M/s model for various val- ues of s, the number of servers.
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459
Applying These Models to the Approach Suggested by the Chief Financial Officer Dupit’s chief financial officer has suggested combining the current one-person tech rep ter- ritories into larger territories that would be served jointly by multiple tech reps. The hope is that, without changing the total number of tech reps, this reorganization might decrease W q sufficiently from its current value ( W q 5 0.75 day) to satisfy the proposed new service stan- dard ( W q ≤ 0.25 day).
Let us first try it with two tech reps assigned to each territory.
A Territory with Two Tech Reps
Number of machines: 300 (versus 150 before) Mean arrival rate: l 5 6 (versus l 5 3 before) Mean service rate: m 5 4 (same as before) Number of servers: s 5 2 (versus s 5 1 before)
Utilization factor: r 5 l
sm 5 0.75 (same as before)
Applying the Excel template for the M/M/s model with these data yields the results shown in Figure 11.8 , including W q 5 0.321 day. (The equations entered into the output cells are not given in this figure because they are very complicated, but they can be viewed in this chap- ter’s Excel file that contains this template.)
This is a very big improvement over the current value of W q 5 0.75 day, but it does not quite satisfy the service standard of W q ≤ 0.25 day. So let us next see what would happen if three tech reps were assigned to each territory.
A Territory with Three Tech Reps
Number of machines: 450 (versus 150 before) Mean arrival rate: l 5 9 (versus l 5 3 before) Mean service rate: m 5 4 (same as before)
KeyCorp is a Fortune 500 company headquartered in Cleveland, Ohio. It is the 13th largest bank holding com- pany in the United States with 19,000 employees, assets of $93 billion, and annual revenues of $6.7 billion. The com- pany emphasizes consumer banking and has 2.4 million customers across more than 1,300 branch banks and many additional affiliate offices.
To help grow its business, KeyCorp management initi- ated an extensive management science study to deter- mine how to improve customer service (defined primarily as reducing customer waiting time before beginning ser- vice) while also providing cost-effective staffing. A service- quality goal was set that at least 90 percent of the customers should have waiting times of less than five minutes.
The key tool in analyzing this problem was the M/M/s queueing model, which proved to fit this application very well. To apply this model, data were gathered that revealed that the average service time required to process a customer was a distressingly high 246 seconds. With this average service time and typical mean arrival rates, the model indicated that a 30 percent increase in the num- ber of tellers would be needed to meet the service-quality
goal. This prohibitively expensive option led management to conclude that an extensive campaign needed to be undertaken to drastically reduce the average service time by both reengineering the customer session and providing better management of staff. Over a period of three years, this campaign led to a reduction in the average service time all the way down to 115 seconds. Frequent reapplication of the M/M/s model then revealed how the service-quality goal can be substantially surpassed while actually reducing personnel levels through improved scheduling of the per- sonnel in the various branch banks.
The net result has been savings of nearly $20 million per year with vastly improved service that enables 96 percent of the customers to wait less than five minutes. This improve- ment extended throughout the company as the percentage of branch banks that met the service-quality goal increased from 42 percent to 94 percent. Surveys also confirm a great increase in customer satisfaction.
Source: S. K. Kotha, M. P. Barnum, and D. A. Bowen, “KeyCorp Service Excellence Management System,” Interfaces 26, no. 1 (January–February 1996), pp. 54–74. (A link to this article is pro- vided on our website, www.mhhe.com/hillier5e . )
An Application Vignette
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Number of servers: s 5 3 (versus s 5 1 before)
Utilization factor: r 5 l
sm 5 0.75 (same as before)
With this utilization factor, Figure 11.7 indicates that L is very close to 4. Using 4 as the approximate value, and applying the relationships given in Section 11.3 (Little’s formula, etc.),
W 5 L l
5 4 9
5 0.44 day
Wq 5 W 2 1 m
5 0.44 2 0.19 day
More precisely, the Excel template in Figure 11.9 gives L 5 3.953 and W q 5 0.189 day. Since a workday is eight hours, this expected waiting time converts to just over one hour and 30 minutes.
Consequently, three-person territories would easily satisfy the proposed new service stan- dard of W q ≤ 0.25 workday (two hours). Even considering that these larger territories would modestly increase the travel times for the tech reps, they still would comfortably satisfy the service standard.
Table 11.5 summarizes the data and values of W q for territories with one, two, and three tech reps. Note how sharply W q decreases as the number of tech reps (servers) increases with- out changing the utilization factor. In fact, W q for s 5 2 is well under half that for s 5 1, and W q for s 5 3 is about a fourth of that for s 5 1.
These results suggest that further enlarging the territories by assigning four or more tech reps to each one would decrease W q even further. However, there also are disadvantages to enlarging the territories. One is the possibility of a significant increase in the average time required for a tech rep to travel to the site of a failed machine. When combining only two or three one-person tech rep territories into a single joint territory, the average travel times should not increase much since the tech reps can effectively coordinate in dividing up the repair jobs based on the proximity of the jobs to the current locations of the tech reps. How- ever, this becomes more difficult with even more tech reps in an even larger territory, so
The M/M/s model indicates that the proposed new service standard would be comfortably satisfied by combining three one-person tech rep territories into a single larger territory that would be served jointly by all three tech reps.
Combining too many tech reps into a very large ter- ritory can cause excessive travel times, among other problems.
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A B C D E F G
Data
M/M/s Model for CFO's Approach (Combine into Teams of Two)
Results
L =
Lq =
W =
Wq =
ρ =
n
0
1
2
3
4
5
6
7
8
9
10
(mean arrival rate)
(mean service rate)
(# servers)
λ = μ = s =
Pr(0 > t) =
when t =
Prob(0q > t) =
when t =
0.0000 0
0.0500
0.1000
0.1500
0.2000
0.2500
1 2 3 4 5 6 7 8 9 10 11 12
Number of Customers in System
13 14 15 16 17 18 19 20 21 22 23 24 25
P ro
b a b il
it y
6
4
2
0.169
1
0.087
1
3.429
1.929
0.571
0.321
0.75
Pn 0.1429
0.2143
0.1607
0.1205
0.0904
0.0678
0.0509
0.0381
0.0286
0.0215
0.0161
FIGURE 11.8 This Excel template for the M/M/s model shows the results from applying this model to the approach suggested by Dupit’s chief financial officer with two tech reps assigned to each territory.
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11.6 Some Multiple-Server Queueing Models 461
occasional travel times might become excessive. Since time traveling to a repair site is part of the total time a tech rep must devote to a repair, the mean service rate m may decrease slightly from the four repairs per day assumed in Table 11.5 when the number of tech reps is more than three. For any given number of tech reps, decreasing m increases W q . Therefore, it is unclear how much further W q can be decreased, if at all, by increasing the number of tech reps per territory beyond three.
Assigning a large number of tech reps to each territory has a number of practical draw- backs as well. Coordination between tech reps becomes more difficult. Customers lose the feeling of receiving personalized service when they are visited by so many different tech reps. Furthermore, tech reps lose the pride of ownership in managing their own territory and dealing with “their” customers. Personal or professional conflicts between tech reps also can arise when they share the same territory, and the opportunities for such conflicts increase with larger teams.
For all these reasons, John Phixitt concludes that normally assigning three tech reps to each territory would provide the best trade-off between minimizing these disadvantages of large territories and reducing W q to a satisfactory level.
Conclusion: The approach suggested by the chief financial officer would indeed satisfy the proposed new service standard ( W q ≤ 0.25 day) if each three contiguous one-person tech rep territories are combined into a larger territory served jointly by the same three tech reps. Since the total number of tech reps does not change, there would be no significant additional cost from implementing this approach other than the disadvantages of larger territories just cited. To mini- mize these disadvantages, the territories should not be enlarged any further than having three tech reps per territory.
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A B C D E F G
Data
M/M/s Model for CFO's Approach (Combine into Teams of Three)
Results
L =
Lq =
W =
Wq =
ρ =
n
0
1
2
3
4
5
6
7
8
9
10
(mean arrival rate)
(mean service rate)
(# servers)
λ = μ = s =
Pr(0 > t) =
when t =
Prob(0q > t) =
when t =
0.0000
0
0.0500
0.1000
0.1500
0.2000
1 2 3 4 5 6 7 8 9 10 11 12
Number of Customers in System
13 14 15 16 17 18 19 20 21 22 23 24 25
P ro
b a b il
it y
9
4
3
0.0898
1
0.0283
1
3.9533
1.7033
0.4393
0.1893
0.75
Pn 0.0748
0.1682
0.1893
0.1419
0.1065
0.0798
0.0599
0.0449
0.0337
0.0253
0.0189
FIGURE 11.9 This Excel template modifies the results in Figure 11.8 by assigning three tech reps to each territory.
TABLE 11.5 Comparison of W q Values with Territories of Different Sizes for the Dupit Problem
Number of Tech Reps
Number of Machines l m s r Wq
1 150 3 4 1 0.75 0.75 workday (6 hours) 2 300 6 4 2 0.75 0.321 workday (2.57 hours) 3 450 9 4 3 0.75 0.189 workday (1.51 hours)
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462 Chapter Eleven Queueing Models
The M/D/s Model The service times in many queueing systems have much less variability than is assumed by the M/M/s model. In some cases, there may be no variability (or almost no variability) at all in the service times. The M/D/s model is designed for these cases.
Assumptions: Same as for the M/M/s model, except now all the service times are the same. This constant service time is denoted by 1/ m . (This is referred to as having a degenerate service- time distribution, which provides the symbol D for the model label.)
Constant service times arise when exactly the same work is being performed to serve each customer. When the servers are machines, there may literally be no variability at all in the ser- vice times. The assumption of constant service times also can be a reasonable approximation with human servers if they are performing the same routine task for all customers.
We found in the preceding section that, when s 5 1, the value of L q for the M/D/ 1 model is only half that for the M/M/ 1 model. Similar differences in L q between the two models also occur when s > 1 (especially with larger values of the utilization factor r ). Substantial differ- ences between the models also occur for W q , W, and L.
These large differences emphasize the importance of using the model that best fits the queueing system under study. Because the M/M/s model is the most convenient one, it is common practice to routinely use this model for most applications. However, doing so when there is little or no variability in the service times causes a large error in some measures of performance.
The procedures for calculating the various measures of performance for the M/D/s model are far more complicated than for the M/M/s model, so no Excel template is available in the case when s > 1. However, special projects have been conducted to calculate the measures. Figure 11.10 shows the values of L versus r for many values of s. The other main mea- sures ( W, W q , and L q ) then can be obtained from L by using Little’s formula, and so forth (as described in Section 11.3).
Just as for single-server queueing systems, elimi- nating the variability of service times substantially increases the efficiency of multiple-server queueing systems.
FIGURE 11.10 Values of L for the M/D/s model for various val- ues of s, the number of servers.
S te
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2
1.0
0.5
0.2
0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Utilization factor
L
s = 1
s = 2
s = 7
s = 10
s = 20
s = 25
s = 15
s = 3
s = 4 s =
5
r = l—
sm
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11.7 Priority Queueing Models 463
1. For multiple-server queueing models, what is the formula for the utilization factor r ? What is the interpretation of r in terms of how servers use their time?
2. Which values of r correspond to the servers having a manageable utilization factor that allows the system to reach a steady-state condition?
3. Are there any measures of performance that can be calculated for the M/M/ 1 model but not the M/M/s model?
4. How many one-person tech rep territories need to be combined into a larger territory in order to satisfy Dupit’s proposed new service standard?
5. Compare the M/M/s and M/D/s models in terms of the amount of variability in the service times.
Review Questions
11.7 PRIORITY QUEUEING MODELS
All the queueing models presented so far assume that the customers are served on a first- come, first-served basis. Not all queueing systems operate that way. In some systems, the more important customers are served ahead of others who have waited longer. Management may want certain special customers to be given priority over others. In some cases, the cus- tomers in the queueing system are jobs to be performed, and the different deadlines for the jobs dictate the order in which these customers are served. Rush jobs need to be done before routine jobs.
A hospital emergency room is an example of a queueing system where priorities automati- cally are used. An arriving patient who is in critical condition naturally will be treated ahead of a routine patient who was already there waiting.
The models for such queueing systems generally make the following general assumptions.
General Assumptions 1. There are two or more categories of customers. Each category is assigned to a priority
class . Customers in priority class 1 are given priority for receiving service over customers in priority class 2. If there are more than two priority classes, customers in priority class 2 then are given priority over customers in priority class 3, and so on.
2. After deferring to higher priority customers, the customers within each priority class are served on a first-come, first-served basis. Thus, within a priority class, priority for receiv- ing service is based on the time already spent waiting in the queueing system.
There are two types of priorities, as described below.
Nonpreemptive priorities : Once a server has begun serving a customer, the service must be completed without interruption even if a higher priority customer arrives while this service is in process. However, once service is completed, if there are customers in the queue, priorities are applied to select the one to begin service. In particular, the one selected is that member of the highest priority class represented in the queue who has waited longest. Preemptive priorities : The lowest priority customer being served is preempted (ejected back into the queue) whenever a higher priority customer enters the queueing system. A server is thereby freed to begin serving the new arrival immediately. Whenever a server does succeed in finishing a service, the next customer to begin receiving service is selected just as described above for nonpreemptive priorities. (The preempted customer becomes the member of its priority class in the queue who has waited longest, so it hopefully will get back into service soon and, perhaps after additional preemptions, will eventually finish.)
This section includes a basic queueing model for each of these two types of priorities.
A Preemptive Priorities Queueing Model Along with the general assumptions about priorities given above, this model makes the fol- lowing assumptions.
Priority queueing models are used when high-priority customers are served ahead of others who have waited longer.
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464 Chapter Eleven Queueing Models
Additional Assumptions 1. Preemptive priorities are used as just described. (Let n denote the number of priority
classes.) 2. For priority class i ( i 5 1, 2, . . . , n ), the interarrival times of customers in that class have
an exponential distribution with a mean of 1/ l i . 3. All service times have an exponential distribution with a mean of 1/ m , regardless of the
priority class involved. 4. The queueing system has a single server.
Thus, except for the complication of using preemptive priorities, the assumptions are the same as for the M/M/ 1 model.
Since l i is the mean arrival rate for customers in priority class i ( i 5 1, 2, . . . , n ), l 5 ( l 1 1 l 2 1 . . . 1 l n ) is the overall mean arrival rate for all customers. Therefore, the utilization factor for the server is
r 5 l1 1 l2 1
c1 ln
m
As with the previous models, r > 1 is required to enable the queueing system to reach a steady-state condition for all priority classes.
The reason for using priorities is to decrease the waiting times for high-priority custom- ers. This is accomplished at the expense of increasing the waiting times for low-priority customers.
Assuming r > 1, formulas are available for calculating the main measures of performance ( L, W, L q , and W q ) for each of the priority classes. An Excel template in your MS Courseware quickly performs all these calculations for you.
A Nonpreemptive Priorities Queueing Model Along with the general assumptions given at the beginning of this section, this model makes the following assumptions.
Additional Assumptions 1. Nonpreemptive priorities are used as described earlier in this section. (Again, let n be the
number of priority classes.) 2. and 3. Same as for the preemptive priorities queueing model. 4. The queueing system can have any number of servers.
Except for using nonpreemptive priorities, these assumptions are the same as for the M/M/s model.
The utilization factor for the servers is
r 5 l1 1 l2 1
c1 ln
sm
Again, r > 1 is needed to enable the queueing system to reach a steady-state condition for all the priority classes.
As before, an Excel template is available in your MS Courseware to calculate all the main measures of performance for each of the priority classes.
Applying the Nonpreemptive Priorities Queueing Model to the Approach Suggested by the Vice President for Marketing Now we come to the last of the four approaches being investigated by Dupit’s management science team. The vice president for marketing has proposed giving the printer-copiers prior- ity over other machines for receiving service. In other words, whenever a tech rep finishes a repair, if there are both printer-copiers and other machines still waiting to be repaired, the tech rep always should choose a printer-copier (the one that has waited longest) to be repaired next, even if other machines have waited longer.
This model fits the M/M/ 1 model except for also hav- ing preemptive priorities.
This model fits the M/M/s model except for also having nonpreemptive priorities.
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The rationale for this proposal is that the printer-copier performs so many vital functions that its owners cannot tolerate being without it as long as other machines. Indeed, nearly all the complaints about excessive waiting for repairs have come from these owners even though other machines wait just as long. Therefore, the vice president for marketing feels that the proposed new service standard ( W q ≤ 2 hours) only needs to be applied to the printer-copiers. Giving them priority for service hopefully will result in meeting this standard while still pro- viding satisfactory service to other machines.
To investigate this, the management science team is applying the nonpreemptive priorities queueing model. There are two priority classes.
Priority class 1: Printer-copiers. Priority class 2: Other machines.
Therefore, a distinction is made between these two types of arriving customers (machines needing repairs) for the queueing system in each tech rep territory. To determine the mean arrival rate for each of these two priority classes (denoted by l 1 and l 2 , respectively), the team has ascertained that about a third of the machines assigned to tech reps currently are printer-copiers. Each printer-copier requires service with about the same frequency (approxi- mately once every 50 workdays) as other machines. Consequently, since the total mean arrival rate for all the machines in a one-person tech rep territory typically is three machines per day,
l1 5 1 customer (printer-copier) per workday (now)
l2 5 2 customers (other machines) per workday (now)
However, the proportion of the machines that are printer-copiers is expected to gradually increase until it peaks at about half in a couple of years. At that point, the mean arrival rates will have changed to
l1 5 1.5 customers (printer-copiers) per workday (later)
l2 5 1.5 customers (other machines) per workday (later)
The mean service rate for each tech rep is unchanged by applying priorities, so its best esti- mate continues to be m 5 4 customers per workday. Under the company’s current policy of one-person tech rep territories, the queueing system for each territory has a single server ( s 5 1). Since ( l 1 1 l 2 ) 5 3 both now and later, the value of the utilization factor will con- tinue to be
r 5 l1 1 l2
sm 5
3 4
Figure 11.11 shows the results obtained by applying the Excel template for the nonpreemp- tive priorities model to this queueing system now ( l 1 5 1 and l 2 5 2). Figure 11.12 does the same under the conditions expected later ( l 1 5 1.5 and l 2 5 1.5).
The management science team is particularly interested in the values of W q , the expected waiting time in the queue, given in the last column of these two figures. These values are sum- marized in Table 11.6 , where the first row comes from Figure 11.11 and the second comes from Figure 11.12 .
For the printer-copiers, note that W q 5 0.25 workday now, which barely meets the pro- posed new service standard of W q ≤ 0.25 workday, but this expected waiting time would dete- riorate later to 0.3 workday. Thus, this approach falls a little short. Furthermore, the expected waiting time before service begins for the other machines would go from W q 5 1 workday now to W q 5 1.2 workdays later. This large increase from the average waiting times being experienced under the current policy of W q 5 0.75 workday (as given in Figure 11.4 ) is likely to alienate a considerable number of customers.
Table 11.5 in the preceding section demonstrated what a great impact combining one- person tech rep territories into larger territories has on decreasing expected waiting times. Therefore, the management science team decides to investigate combining this approach with applying nonpreemptive priorities.
The suggestion of the vice president for marketing is to apply the proposed new service standard to the printer-copiers only and then to give them non- preemptive priority over the other machines.
Since assigning nonpre- emptive priorities doesn’t help enough (especially later), let’s also try combin- ing one-person tech rep territories into larger joint territories.
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0.825
2.175
0.45
1.8
0.55
1.45
0.3
1.2
Nonpreemptive Priorities Model for VP of Marketing's Approach (Future Arrival Rates)
2
4
1
λi 1.5
1.5
3
0.75
(# of priority classes)
(mean service rate)
(# servers)
n =
μ = s =
Priority Class 1
Priority Class 2
λ = ρ =
FIGURE 11.12 The modification of Figure 11.11 that applies the same model to the later version of the Dupit problem.
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(# of priority classes)
(mean service rate)
(# servers)
n =
μ = s =
Priority Class 1
Priority Class 2
λ = ρ =
0.5
2.5
0.25
2
0.5
1.25
0.25
1
FIGURE 11.11 This Excel template applies the nonpreemptive priorities queueing model to the Dupit problem now under the approach suggested by the vice president for marketing to give priority to the printer-copiers.
TABLE 11.6 Expected Waiting Times* when Nonpreemptive Priorities Are Applied to the Dupit Problem
s When l1 l2 m r Wq for Printer-Copiers Wq for Other Machines
1 Now 1 2 4 0.75 0.25 workday (2 hrs.) 1 workday (8 hrs.) 1 Later 1.5 1.5 4 0.75 0.3 workday (2.4 hrs.) 1.2 workdays (9.6 hrs.) 2 Now 2 4 4 0.75 0.107 workday (0.86 hr.) 0.429 workday (3.43 hrs.) 2 Later 3 3 4 0.75 0.129 workday (1.03 hrs.) 0.514 workday (4.11 hrs.) 3 Now 3 6 4 0.75 0.063 workday (0.50 hr.) 0.252 workday (2.02 hrs.) 3 Later 4.5 4.5 4 0.75 0.076 workday (0.61 hr.) 0.303 workday (2.42 hrs.)
*These times are obtained in units of workdays, consisting of eight hours each, and then converted to hours.
Combining pairs of one-person tech rep territories into single two-person tech rep ter- ritories doubles the mean arrival rates for both priority classes ( l 1 and l 2 ) for each new territory. Since the number of servers also doubles (from s 5 1 to s 5 2) without any change in m (the mean service rate for each server), the utilization factor r remains the same. These
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11.7 Priority Queueing Models 467
values now and later are shown in the third and fourth rows of Table 11.6 . Applying the nonpreemptive priorities queueing model then yields the expected waiting times given in the last two columns.
These large reductions in the W q values due to increasing s from s 5 1 to s 5 2 result in rather reasonable waiting times. Both now and later, W q for printer-copiers is only about half of the maximum under the proposed new service standard ( W q # 2 hours). Although W q for the other machines is somewhat over this maximum both now and later, these waiting times also are somewhat under the average waiting times currently being experienced (6 hours) without many complaints from members of this priority class. John Phixitt’s reaction is favor- able. He feels that the service standard of W q # 2 hours really was proposed with the printer- copiers in mind and that the other members of top management probably will also be satisfied with the values of W q shown in the third and fourth rows of Table 11.6 .
Since the analytical results reported in Table 11.5 were so favorable for three-person tech rep territories without priorities, the management science team decides to investigate this option with priorities as well. The last two rows of Table 11.6 show the results for this case. Note that these W q values for s 5 3 are even smaller than for s 5 2. In fact, even the W q values for other machines nearly satisfy the proposed new service standard at this point. However, John Phixitt points out that three-person territories have substantial disadvantages compared to two-person territories. One is longer travel times to machine sites. Another is that cus- tomers would feel that service is considerably less personalized when they are seeing three different tech reps coming for repairs instead of just two. Another perhaps more important disadvantage is that three tech reps would have considerably more difficulty coordinating their work than two. John does not feel that the decreases in W q values for s 5 3 are worth these (and related) disadvantages.
Conclusion: Since the high-priority need is to improve service for the printer-copiers, strong consideration should be given to giving these machines priority over others for receiving repairs. However, the waiting times for both printer-copiers and other machines will remain somewhat unsatisfactory (especially later) if the current one-person tech rep territories continue to be used. Enlarging to two-person territories would reduce these waiting times to levels that appear to be satisfactory, without any significant additional (monetary) costs. Enlarging the territories even further probably would not be worthwhile in light of the disadvantages of large territories.
Management’s Conclusions Having been assigned by Dupit’s president to study the four suggested approaches to the company’s problem, the management science team and John Phixitt were asked to report back to the top management group dealing with the problem in six weeks. They now do so by sending their report to each member of the group. The report presents their conclusions (as stated above and in the preceding sections) on each of the four approaches they were asked to investigate. Also included are the projected measures of performance (such as in Tables 11.5 and 11.6 ) for these approaches.
Table 11.7 summarizes the four approaches as they now have been refined by the manage- ment science team.
Two-person tech rep terri- tories with priorities reduce waiting times to satisfactory levels.
Three-person tech rep territories with priorities reduce waiting times even further but have substantial disadvantages compared to two-person territories.
TABLE 11.7 The Four Approaches Being Considered by Dupit Management
Proposer Proposal Additional Cost
John Phixitt Maintain one-person territories, but reduce number of machines assigned to each from 150 to 100
$300 million per year
Vice president for engineering
Keep current one-person territo- ries, but provide new state-of-the- art equipment to the tech reps
One-time cost of $500 million
Chief financial officer Change to three-person territories None, except disadvantages of larger territories
Vice president for marketing
Change to two-person territories, with priority given to the printer- copiers for repairs
None, except disadvantages of larger territories
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At this point, the president reconvenes his top management group (including John Phixitt). The meeting begins with a brief (and well-rehearsed) presentation by the head of the manage- ment science team summarizing the analysis and conclusions of the team. The presentation is interrupted frequently by comments and questions from the group. The president next asks John Phixitt to present his recommendations.
John begins by emphasizing the many advantages of the current system of one-person ter- ritories. The first two proposals in Table 11.7 would enable continuing this system, but at a very high cost. He then concedes that he has concluded that the cost would be too high and that the time has come to modify the system in order to efficiently deliver the service that the marketplace now is demanding. (A brief discussion reveals strong agreement from the group on this point.)
This leaves the third and fourth proposals in Table 11.7 under consideration. John repeats the arguments he had given earlier to the management science team about the important advantages of two-person territories over three-person territories. He then points out that the fourth proposal not only would provide two-person territories but also would result in the printer-copiers having smaller average waiting times for repairs than under the third pro- posal. When the customer relations manager objects that the average waiting times of other machines would not meet the proposed new service standard (a maximum of two hours), John emphasizes that these waiting times still would decrease substantially from current levels and that the owners of these machines aren’t even complaining now. In conclusion, John recom- mends adoption of the fourth proposal.
Some minor concerns are raised in the subsequent discussion, including the possibility that owners of other machines might feel that they are being treated as second-class custom- ers. However, John indicates that the new policy would not be publicized, but, if discovered, could be easily justified to a customer. The group soon concurs with John’s recommendation.
Decision: Adopt the fourth proposal in Table 11.7 .
Finally, John points out that there currently are a relatively few one-person territories that are so sparsely populated that combining them into two-person territories would cause exces- sive travel times for the tech reps. Since this would defeat the purpose of the new policy, he suggests adopting the second proposal for these territories and then using the experience with the new equipment to make future decisions on which equipment to provide to all tech reps as service demands further increase. The group agrees.
Decision: As an exception to the new policy, the second proposal in Table 11.7 is adopted just for current one-person territories that are particularly sparsely populated. The experience with the new equipment will be closely monitored to help guide future equipment purchase decisions for all tech reps.
The president thanks John Phixitt and the management science team for their outstanding work in pointing the way toward what appears to be an excellent resolution of a critical prob- lem for the company. John graciously states that the real key was the insights obtained by the management science team by making effective use of the appropriate queueing models. The president smiles and makes a mental note to seek John’s advice more often.
Although the first two proposals retain the many advantages of one-person territories, both proposals are too costly, so the choice is between the third and fourth proposals.
1. How does using priorities differ from serving customers on a first-come, first-served basis? 2. What is the difference between nonpreemptive priorities and preemptive priorities? 3. Except for using preemptive priorities, the assumptions of the preemptive priorities model are
the same as for which basic queueing model? 4. Except for using nonpreemptive priorities, the assumptions of the nonpreemptive priorities
model are the same as for which basic queueing model? 5. For these models, which values of the utilization factor r enable the queueing system to reach
a steady-state condition for all priority classes? 6. When applying the nonpreemptive priorities queueing model to the Dupit case study, what are
the two priority classes?
Review Questions
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11.8 Some Insights about Designing Queueing Systems 469
7. For this application, what is the conclusion about the minimum number of tech reps per terri- tory needed to reduce waiting times for repairs to levels that appear to be satisfactory?
8. What is the decision of Dupit’s top management regarding which of the four proposed approaches will be adopted (except for particularly sparsely populated territories)?
11.8 SOME INSIGHTS ABOUT DESIGNING QUEUEING SYSTEMS
The Dupit case study illustrates some key insights that queueing models provide about how queueing systems should be designed. This section highlights these insights in a broader context.
There are four insights presented here. Each one was first seen when analyzing one of the four approaches proposed for the Dupit problem. After summarizing each insight, we will briefly review its application to the case study and then describe the insight in general terms.
Insight 1: When designing a single-server queueing system, beware that giving a relatively high utilization factor (workload) to the server provides surprisingly poor measures of performance for the system. 2
This insight arose in Section 11.5 when analyzing John Phixitt’s suggested approach of decreasing the utilization factor r for each tech rep sufficiently to meet the proposed new service standard (a maximum average waiting time for repairs of two hours). The current r 5 0.75 gave average waiting times of six hours, which falls far short of this standard. It was necessary to decrease r all the way down to r 5 0.5 to meet this standard.
To further demonstrate this insight, we have used the Excel template for the M/M/s model (previously shown in Figures 11.8 and 11.9 ), with s 5 1 and m 5 1 (so the utilization factor r equals l ), to generate the data table in Figure 11.13 . Here are the steps to follow to do this. First make a table with the column headings shown in columns I, J, and K in Figure 11.13 . In the first column of the table (I5:I16), list the trial values for the data cell (the mean arrival rate, or equivalently, the utilization factor), except leave the first row blank. The headings of the next columns specify which output will be evaluated. For each of these columns, use the first row of the table (cells J4:K4) to write an equation that refers to the relevant output cell. In this case, the cells of interest are the expected number of customers in the system ( L ) and in the queue ( L q ), so the equations for J4:K4 are those shown below the spreadsheet in Figure 11.13 .
Next, select the entire table (I4:K16) and then choose Data Table from the What-If Analy- sis menu of the Data tab. In the Data Table dialog box (as shown at the bottom left-hand side of Figure 11.13 ), indicate the column input cell (Lambda or C4), which refers to the data cell that is being changed in the first column of the table. Nothing is entered for the row input cell because no row is being used to list the trial values of a data cell in this case.
Clicking OK then generates the data table shown in Figure 11.13 . For each trial value for the data cell listed in the first column of the table, the corresponding output cell values are calculated and displayed in the other columns of the table. (The numbers in the first row of the table come from the original solution in the spreadsheet.)
Note in this data table how rapidly L q and L increase with even small increases in r . For example, L triples when r is increased from 0.5 to 0.75, and then triples again when increasing r from 0.75 to 0.9. As r is increased above 0.9, L q and L grow astronomically. (Although this data table has been generated with m 5 1, the same values of L q and L would be obtained with any other value of m as well when the numbers in the first column are the utilization factor r 5 l / m .)
Managers normally strive for a high utilization factor for their employees, machines, equipment, and so on. This is an important part of running an efficient business. A utiliza- tion factor of 0.9 or higher would be considered desirable. However, all this should change when the employee or machine or piece of equipment is the server in a single-server queueing
The average number of cus- tomers waiting in a queue- ing system ( L ) increases rapidly with even small increases in the utilization factor r , so r should be kept well under 1.
2 The one exception is a queueing system that has constant (or nearly constant) interarrival times and service times. Such a system will perform very well with a high utilization factor.
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H I J K M NL O
Data Table Demonstrating the Effect of Increasing ρ on Lq and L for M/M/1
L
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0.010
0.333
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1.5
2.333
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5.667
9
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99
999
Lq 0.5
0.000
0.083
0.5
0.9
1.633
2.25
3.2
4.817
8.1
18.05
98.01
998.001
λ = ρ
0.01
0.25
0.5
0.6
0.7
0.75
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0.85
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0.99
0.999
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A B C D E F G
Data
Template for M/M/s Queueing Model
0.5
1
Results
L =
Lq= (mean arrival rate)
(mean service rate)
λ = μ =
0 0 0.2 0.4 0.6
System Utilization (p) 0.8
A v e ra
g e L
in e L
e n g th
( L
)
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20
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60
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L Lambda
Lq
G4
C4
G5
Range Name Cell
3
4
L
=L
Lq =Lq
J K
Select entire table (I4:K16), before
choosing Table from the Data menu.
1.0000
0.5000
FIGURE 11.13 This data table demon- strates Insight 1 in Section 11.8.
system that has considerable variability in its interarrival times and service times (such as for an M/M/ 1 system). For most such systems, the cognizant manager would consider it unac- ceptable to average having nine customers wait in the system ( L 5 9 with r 5 0.9). If so, a utilization factor somewhat less (perhaps much less) than 0.9 would be needed. For example, we just mentioned that meeting Dupit’s proposed service standard with John Phixitt’s original suggested approach required reducing the utilization factor all the way down to r 5 0.5.
Insight 2: Decreasing the variability of service times (without any change in the mean) improves the performance of a single-server queueing system substantially. (This also tends to be true for multiple-server queueing systems, especially with higher utilization factors.)
This insight was found in the Dupit study while analyzing the proposal by the vice president for engineering to provide new state-of-the-art equipment to all the tech reps. As described at the end of Section 11.5, this approach would decrease both the mean and standard deviation of the service-time distribution. Decreasing the mean also decreased the utilization factor, which decreased the expected waiting time W q . Decreasing the standard deviation s (which measures the amount of variability) then provided an additional 37.5 percent reduction in W q . Insight 2 refers to this latter substantial improvement in W q (and in the other measures of performance).
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11.8 Some Insights about Designing Queueing Systems 471
The two-way data table in cells C19:F23 of Figure 11.14 demonstrates the effect on L q of decreasing the standard deviation s of the service-time distribution for any M/G/ 1 queueing system. (This table was generated from the Excel template introduced in Figure 11.6 for the M/G/ 1 model.)
To create this two-way data table, make a table with column and row headings as shown in rows 19–23 of the spreadsheet in Figure 11.14 . In the upper left-hand corner of the table (C19), write an equation that refers to the output cell for which you are interested in seeing the results ( 5 L q or G5). In the first column of the table (column C, below the equation in cell C19), insert all the different values for the first changing data cell ( l ). In the first row of the table (row 19, to the right of the equation in cell C19), insert all the different values for the second changing data cell ( s ).
Next, select the entire table (C19:F23) and then choose Data Table from the What-If Analy- sis menu of the Data tab. In the Data Table dialog box (shown at the bottom left-hand side of Figure 11.14 ), indicate which data cells are being changed simultaneously. The column input cell refers to the data cell whose various values are indicated in the first column of the table (Lambda, or cell C4), while the row input cell refers to the data cell whose various values are indicated in the first row of the table (Sigma, or cell C6).
The data table shown in Figure 11.14 is then generated automatically by clicking OK. For each pair of values of the input cell indicated in the first row and column of the table, Excel determines the corresponding value of the output cell referred to in the upper left-hand corner of the table. These values are then filled into the body of the table.
Lambda
Lq Sigma
C4
G5
C6
Range Name Cell
19 =Lq
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ρ (=λ)
0.3125
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σ
A B C D E F G H I
Data
Template for the M/G/1 Queueing Model
Data Table Demonstrating the Effect of Decreasing σ on Lq for M/G/1
0.5
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0.5
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Results
L =
Lq=
W =
Wq=
ρ =
P0 =
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
λ = 1/μ =
σ = s =
Select entire table
(C19:F23), before
choosing Table from
the Data menu.
1
0.500
2.250
8.100
98.010
0
0.250
1.125
4.050
49.005
0.5
0.313
1.406
5.063
61.256
0.8125
0.3125
1.625
0.625
0.5
0.5
FIGURE 11.14 This data table demon- strates Insight 2 in Section 11.8.
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Before generating this data table, the mean of the service-time distribution has been set in cell C5 at 1/ m 5 1 (which makes r 5 l ), so the column headings of s 5 1, s 5 0.5, and s 5 0 correspond to s 5 mean, s 5 0.5 mean, and s 5 0, respectively. Therefore, as you read the values of L q in the table from left to right, s decreases from equaling the mean of the distri- bution (as for the M/M/ 1 model) to being half the mean and then to s 5 0 (as for the M/D/ 1 model). If the mean were to be changed to some value different from 1, the same values of L q still would be obtained for each value of the utilization factor r 5 l / m listed in cells C20:C23 as long as the values of s for the respective columns are s 5 mean, s 5 0.5 mean, and s 5 0.
In each row of this table, the value in the s 5 0 column is only half that in the s 5 1 column, so completely eliminating the variability of the service times gives a large improve- ment. However, the value in the s 5 0.5 column is only 62.5 percent of that in the s 5 1 column, so even cutting the variability in half provides most of the improvement from com- pletely eliminating the variability. Therefore, whatever can be done to reduce the variability even modestly is going to improve the performance of the system significantly.
Insight 3: Multiple-server queueing systems can perform satisfactorily with somewhat higher utilization factors than can single-server queueing systems. For example, pooling servers by combining separate single-server queueing systems into one multiple-server queueing system (without changing the utilization factor) greatly improves the measures of performance.
This insight was gained during the Dupit study while investigating the proposal by the chief financial officer to combine one-person territories into larger territories served jointly by multiple tech reps. Table 11.5 in Section 11.6 summarizes the great impact that this approach would have on improving average waiting times to begin repairs ( W q ). In particu- lar, W q for two-person territories is well under half that for one-person territories, and W q for three- person territories is about a fourth of that for one-person territories, even though the utilization factor is the same for all these cases.
These dramatic improvements are not unusual. In fact, it has been found that pooling serv- ers as described below always provides similar improvements.
The Impact of Pooling Servers: Suppose you have a number (denoted by n ) of identical sin- gle-server queueing systems that fit the M/M/ 1 model. Suppose you then combine these n sys- tems (without changing the utilization factor) into a single queueing system that fits the M/M/s model, where the number of servers is s 5 n. This change always improves the value of W q by more than dividing by n, that is,
Wq(for combined system) , Wq(for each single-server system)
n
Although this inequality is not guaranteed to hold if these queueing systems do not fit the M/M/ 1 and M/M/s models, the improvement in W q by combining systems still will be very substantial for other models as well.
Insight 4: Applying priorities when selecting customers to begin service can greatly improve the measures of performance for high-priority customers.
This insight became evident during the Dupit study while investigating the proposal by the vice president for marketing to give higher (nonpreemptive) priority to repairing the printer- copiers than to repairing other machines. Table 11.6 in the preceding section gives the val- ues of W q for the printer-copiers and for the other machines under this proposal. Comparing these values to those in Table 11.5 without priorities shows that giving priority to the printer- copiers would reduce their waiting times now dramatically (but would also increase the wait- ing times for the other machines). Later, as printer-copiers become a larger proportion of the machines being serviced (half instead of a third), the reduction in their waiting times would not be quite as large.
For other queueing systems as well, the impact of applying priorities depends somewhat on the proportion of the customers in the respective priority classes. If the proportion in the top priority class is small, the measures of performance for these customers will improve tre- mendously. If the proportion is large, the improvement will be more modest.
If service times are highly variable, eliminating this variability can reduce L q by about half.
Here is a way to reduce waiting times dramatically.
Applying priorities can reduce waiting times dra- matically for high-priority customers but will increase waiting times for low- priority customers.
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Preemptive priorities give an even stronger preference to high-priority customers than do the nonpreemptive priorities used for the Dupit problem. Therefore, applying preemptive pri- orities improves the measures of performance for customers in the top priority class even more than applying nonpreemptive priorities.
1. What is the effect of giving a relatively large utilization factor (workload) to the server in a single-server queueing system?
2. What happens to the values of L q and L for the M/M/ 1 model when r is increased well above 0.9?
3. What is the effect of decreasing the variability of service times (without any change in the mean) on the performance of a single-server queueing system?
4. For an M/G/ 1 queueing system, does cutting the variability (standard deviation) of service times in half provide most of the improvement that would be achieved by completely eliminating the variability?
5. What is the effect of combining separate single-server queueing systems into one multiple- server queueing system (without changing the utilization factor)?
6. What is the effect of applying priorities when selecting customers to begin service?
7. Do preemptive priorities or nonpreemptive priorities give the greater improvement in the mea- sures of performance for customers in the top priority class?
Review Questions
11.9 ECONOMIC ANALYSIS OF THE NUMBER OF SERVERS TO PROVIDE
When designing a queueing system, a key question often is how many servers to provide. Providing too many causes excessive costs. Providing too few causes excessive waiting by the customers. Therefore, choosing the number of servers involves finding an appropriate trade-off between the cost of the servers and the amount of waiting.
In many cases, the consequences to an organization of making its customers wait can be expressed as a waiting cost . This is especially true when the customers are internal to the organization, such as the employees of a company. Making one’s own employees wait (both while waiting for service to begin and during service) causes lost productivity, which results in lost profit. This lost profit is the waiting cost.
A manager is interested in minimizing the total cost. Let
TC 5 Expected total cost per unit time
SC 5 Expected service cost per unit time
WC 5 Expected waiting cost per unit time
Then the objective is to choose the number of servers so as to
Minimize TC 5 SC 1 WC
When each server costs the same, the service cost is
SC 5 Css
where
C s 5 Cost of a server per unit time s 5 Number of servers
When the waiting cost is proportional to the amount of waiting (including during service), this cost can be expressed as
WC 5 CwL
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where
C w 5 Waiting cost per unit time for each customer in the queueing system L 5 Expected number of customers in the queueing system
Therefore, after estimating the constants C s and C w , the goal is to choose the value of s so as to
Minimize TC = Cs s + Cw L
By choosing the queueing model that fits the queueing system, the value of L can be obtained for various values of s. Increasing s decreases L, at first rapidly and then gradually more slowly.
Figure 11.15 shows the general shape of the SC, WC, and TC curves versus the number of servers s. (For better conceptualization, we have drawn these as smooth curves even though the only feasible values of s are s 5 1, 2, . . .). By calculating TC for consecutive values of s until TC stops decreasing and starts increasing instead, it is straightforward to find the num- ber of servers that minimizes total cost. The following example illustrates this process.
An Example The Acme Machine Shop has a tool crib for storing tools required by the shop mechanics. Two clerks run the tool crib. The tools are handed out by the clerks as the mechanics arrive and request them and are returned to the clerks when they are no longer needed. There have been complaints from supervisors that their mechanics have had to waste too much time wait- ing to be served at the tool crib, so it appears that there should be more clerks. On the other hand, management is exerting pressure to reduce overhead in the plant, and this reduction would lead to fewer clerks. To resolve these conflicting pressures, a management science study is being conducted to determine just how many clerks the tool crib should have.
The tool crib constitutes a queueing system, with the clerks as its servers and the mechan- ics as its customers. After gathering some data on interarrival times and service times, the management science team has concluded that the queueing model that fits this queueing
Calculate TC for consecu- tive values of s until TC stops decreasing to find the optimal number of servers.
E xp
ec te
d c
os t
p er
u n
it t
im e
Number of servers (s)
Total cost
Service cost
Waiting cost
FIGURE 11.15 The shape of the cost curves for determining the number of servers to provide.
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system best is the M/M/s model. The estimates of the mean arrival rate l and the mean service rate (per server) m are
l 5 120 customers per hour m 5 80 customers per hour
so the utilization factor for the two clerks is
r 5 l
sm 5
120 2(80)
5 0.75
The total cost to the company of each tool crib clerk is about $20 per hour, so C s 5 $20. While a mechanic is busy, the value to the company of his or her output averages about $48 per hour, so C w 5 $48. Therefore, the management science team now needs to find the number of servers (tool crib clerks) s that will
Minimize TC = $20 s + $48 L
An Excel template has been provided in your MS Courseware for calculating these costs with the M/M/s model. All you need to do is enter the data for the model along with the unit service cost C s , the unit waiting cost C w , and the number of servers s you want to try. The tem- plate then calculates SC, WC, and TC. This is illustrated in Figure 11.16 with s 5 3 for this example. By repeatedly entering alternative values of s, the template then can reveal which value minimizes TC in a matter of seconds.
Figure 11.17 shows a data table that has been generated from this template by repeating these calculations for s 5 1, 2, 3, 4, and 5. (See Section 11.8 for more information about
Your MS Courseware includes an Excel template that will calculate TC for you.
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Economic Analysis of Acme Machine Shop Example
Results
L =
Lq=
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Wq=
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Pnn
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(mean arrival rate)
(mean service rate)
(# servers)
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Pr(0> t) =
when t =
Prob(0q> t) =
when t =
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Cw =
Cost of Service
Cost of Waiting
Total Cost
= Cs *s
= Cw *L
= CostOfService+CostOfWaiting
Cost of Service
Cost of Waiting
Total Cost
CostOfService
CostOfWaiting
Cs Cw L s TotalCost
C18
C19
C15
C16
G4
C6
C20
Range Name Cell
18
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B C
$20.00
$48.00
$60.00
$83.37
$143.37
120
80
3
0.02581732
0.05
0.00058707
0.05
1.736842105
0.236842105
0.014473684
0.001973684
0.5
0.210526316
0.315789474
0.236842105
0.118421053
0.059210526
0.029605263
0.014802632
0.007401316
FIGURE 11.16 This Excel template for using economic analysis to choose the number of servers with the M/M/s model is applied here to the Acme Machine Shop example with s 5 3.
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H I J K M NL O P Q R S
Data Table for Expected Total Cost for Alternatives in the Acme Machine Shop Example
ρ 0.50
1.50
0.75
0.50
0.38
0.30
L
1.74
#N/A
3.43
1.74
1.54
1.51
Cost of
Service
$60.00
$20.00
$40.00
$60.00
$80.00
$100.00
Cost of
Waiting
$83.37
#N/A
$164.57
$83.37
$74.15
$72.41
Total
Cost
$143.37
#N/A
$204.57
$143.37
$154.15
$172.41
s
1
2
3
4
5
CostOfService
CostOfWaiting
L Rho
s TotalCost
C18
C19
G4
G10
C6
C20
Range Name Cell
3
4 ρ =Rho
L
=L5
Cost of
Service
=CostOfService
Cost of
Waiting
=CostOfWaiting
Total
Cost
=TotalCost
J K L M N
Select entire table (I5:N10), before
choosing Table from the Data menu.
$0 0 1 2 3
Number of Servers (s)
4 5
$50
$100
$150
C o st
( $ /h
o u r) $200
$250
Cost of Service
Cost of Waiting
Total Cost
FIGURE 11.17 This data table compares the expected hourly costs with various alterna- tive numbers of clerks assigned to the Acme Machine Shop tool crib.
generating data tables.) Since the utilization factor for s 5 1 is r 5 1.5, a single clerk would be unable to keep up with the customers (as indicated by #N/A in cells K6 and M6:N6), so this option is ruled out. All larger values of s are feasible, but s 5 3 has the smallest total cost. Furthermore, s 5 3 would decrease the current total cost for s 5 2 by $62 per hour. There- fore, despite management’s current drive to reduce overhead (which includes the cost of tool crib clerks), the management science team recommends that a third clerk be added to the tool crib. Note that this recommendation would decrease the utilization factor for the clerks from an already modest 0.75 all the way down to 0.5. However, because of the large improvement in the productivity of the mechanics (who are much more expensive than the clerks) through decreasing their time wasted waiting at the tool crib, management adopts the recommendation.
A low utilization factor of 0.5 is best for the tool crib clerks because this greatly reduces the time wasted by expensive mechanics wait- ing at the tool crib.
1. What is the trade-off involved in choosing the number of servers for a queueing system? 2. What is the nature of the waiting cost when the customers for the queueing system are the
company’s own employees? 3. When the waiting cost is proportional to the amount of waiting, what is an expression for the
waiting cost? 4. What does the Acme Machine Shop example demonstrate about the advisability of always
maintaining a relatively high utilization factor for the servers in a queueing system?
Review Questions
Queueing systems are prevalent throughout society. The adequacy of these systems can have an impor- tant effect on the quality of life and the productivity of the economy.
Key components of a queueing system are the arriving customers, the queue in which they wait for service, and the servers that provide the service. A queueing model representing a queueing system needs to specify the number of servers, the distribution of interarrival times, and the distribution of service times. An exponential distribution usually is chosen for the distribution of interarrival times because this corresponds to the common phenomenon of arrivals occurring randomly. An exponential distribution sometimes provides a reasonable fit to the service-time distribution as well, and is a particu- larly convenient choice in terms of ease of analysis. Other probability distributions sometimes used for the service-time distribution include the degenerate distribution (constant service times).
11.10 Summary
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Key measures of the performance of queueing systems are the expected values of the number of customers in the queue or in the system (the latter adds on customers currently being served) and of the waiting time of a customer in the queue or in the system. General relationships between these expected values, including Little’s formula, enable all four values to be determined immediately as soon as one has been found. In addition to the expected values, the probability distributions of these quantities are sometimes used as measures of performance as well.
This chapter’s case study has the top management of the Dupit Corporation grappling with a difficult issue. The company’s customers now are demanding a much higher level of service in promptly repair- ing the photocopiers (and particularly a new printer-copier) purchased from the company. Dupit already is spending $600 million per year servicing these machines. Each tech rep territory includes a queueing system with the tech rep as the server and the machines needing repairs as the customers. A manage- ment science team finds that the M/M/ 1 model, the M/G/ 1 model, the M/M/s model, and a nonpreemp- tive priorities model enable analyzing the alternative approaches to redesigning this queueing system. This analysis leads to top management adopting a policy of combining pairs of one-person tech rep ter- ritories into two-person territories that give priority to repairing the new printer-copiers. This provides the needed level of service without a significant increase in cost.
Other queueing models discussed in the chapter include the M/D/ 1 and M/D/s models, as well as a preemptive priorities model. A supplement to this chapter on the CD-ROM also introduces the finite queue variation and the finite calling population variation of the M/M/s model, as well as models that use another service-time distribution (the Erlang distribution) that allow the amount of variability in the service times to fall somewhere between that for the exponential and degenerate distributions.
Section 11.8 presents four key insights that queueing models provide about how queueing systems should be designed. Each of these insights also is illustrated by the Dupit case study.
A key question when designing queueing systems frequently is how many servers to provide. Sec- tion 11.9 describes how to determine the number of servers that will minimize the expected total cost of the queueing system, including the cost of providing the servers and the cost associated with making customers wait.
Glossary commercial service system A queueing system where a commercial organization provides a service to customers from outside the organization. (Section 11.2), 440 constant service times Every customer has the same service time. (Section 11.1), 439 customers A generic term that refers to whichever kind of entity (people, vehicles, machines, items, etc.) is coming to the queueing system to receive service. (Section 11.1), 434 exponential distribution The most popular choice for the probability distribution of both interarrival times and service times. Its shape is shown in Figure 11.3 . (Section 11.1), 438 finite queue A queue that can hold only a limited number of customers. (Section 11.1), 437 infinite queue A queue that can hold an essentially unlimited number of customers. (Section 11.1), 437 interarrival time The elapsed time between consecutive arrivals to a queueing system. (Section 11.1), 436 internal service system A queueing system where the customers receiving service are internal to the organization providing the service. (Section 11.2), 441 lack-of-memory property When referring to arrivals, this property is that the time of the next arrival is completely uninfluenced by when the last arrival occurred. Also called the Markovian property. (Section 11.1), 437
Little’s formula The formula L 5 l W, or L q 5 l W q . (Section 11.3), 443 mean arrival rate The expected number of arrivals to a queueing system per unit time. (Section 11.1), 436 mean service rate The expected number of service completions per unit time for a single continuously busy server. (Section 11.1), 438 nonpreemptive priorities Priorities for selecting the next customer to begin service when a server becomes free. However, these priorities do not affect customers who already have begun service. (Section 11.7), 463 number of customers in the queue The number of customers who are waiting for service to begin. (Sections 11.1, 11.3), 437, 443 number of customers in the system The total number of customers in the queueing system, either waiting for service to begin or currently being served. (Sections 11.1, 11.3), 437, 443 preemptive priorities Priorities for serving customers that include ejecting the lowest priority customer being served back into the queue in order to serve a higher priority customer that has just entered the queueing system. (Section 11.7), 463 priority classes Categories of customers that are given different priorities for receiving service. (Section 11.7), 463 queue The waiting line in a queueing system. The queue does not include customers who are already being served. (Section 11.1), 434
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tech rep An abbreviated name for the service technical representatives in the Dupit case study. (Section 11.4), 445 transportation service system A queueing system involving transportation, so that either the customers or the server(s) are vehicles. (Section 11.2), 441 utilization factor The average fraction of time that the servers are being utilized serving cus- tomers. (Sections 11.5, 11.6), 448, 457 waiting cost The cost associated with making customers wait in a queueing system. (Section 11.9), 473 waiting time in the queue The elapsed time that an individual customer spends in the queue waiting for service to begin. (Section 11.3), 443 waiting time in the system The elapsed time that an individual customer spends in the queue- ing system both before service begins and during service. (Section 11.3), 443
queue capacity The maximum number of customers that can be held in the queue. (Section 11.1), 437 queue discipline The rule for determining the order in which members of the queue are selected to begin service. (Section 11.1), 438 queueing system A place where customers receive some kind of service from a server, per- haps after waiting in a queue. (Section 11.1), 434 server An entity that is serving the customers coming to a queueing system. (Section 11.1), 434 service cost The cost associated with providing the servers in a queueing system. (Section 11.9), 473 service time The elapsed time from the beginning to the end of a customer’s service. (Section 11.1), 438 steady-state condition The normal condition that a queueing system is in after operating for some time with a fixed utilization factor less than one. (Section 11.3), 443
L q 5 Expected number of customers in the queue (Section 11.3)
W 5 Expected waiting time in the system (Section 11.3)
W q 5 Expected waiting time in the queue (Section 11.3)
r 5 Utilization factor for the servers (Sections 11.5 and 11.6)
l 5 Mean arrival rate (Section 11.1)
m 5 Mean service rate (Section 11.1)
s 5 Number of servers (Section 11.1)
L 5 Expected number of customers in the system (Section 11.3)
Key Symbols
Chapter 11 Excel Files:
Template for M/M/s Model
Template for M/G/1 Model
Template for M/D/1 Model
Template for Nonpreemptive Priorities Model
Template for Preemptive Priorities Model
Template for M/M/s Economic Analysis of Number of Servers
Interactive Management Science Modules:
Waiting Line Module
Supplement to This Chapter on the CD-ROM:
Additional Queueing Models
Supplement to Chapter 11 Excel Files:
Template for Finite Queue Variation of M/M/s Model
Template for Finite Calling Population Variation of M/M/s Model
Template for M / E k /1 Model
Learning Aids for This Chapter in Your MS Courseware
Solved Problem (See the CD-ROM or Website for the Solution) 11.S1. Managing Waiting Lines at First Bank of Seattle Sally Gordon has just completed her MBA degree and is proud to have earned a promotion to Vice President for Customer Services at the First Bank of Seattle. One of her responsibilities is to manage how tellers provide services to customers, so she is taking a hard look at this area of the
bank’s operations. Customers needing teller service arrive randomly at a mean rate of 30 per hour. Customers wait in a single line and are served by the next available teller when they reach the front of the line. Each service takes a variable amount of time (assume an exponential distribution), but on average can be completed in three minutes. The tellers earn an average wage of $18 per hour.
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customer is 15 per hour. Both types of customers currently wait in the same line and are served by the same tellers with the same average service time. However, Sally is considering changing this. The new system she is considering would have two lines—one for merchant customers and one for regular customers. There would be a single teller serving each line. What would be the average waiting time for each type of cus- tomer before reaching a teller? On average, how many total customers would be in the bank, including those currently be- ing served? How do these results compare to those from part a?
e. Sally feels that if the tellers are specialized into merchant tellers and regular tellers, they would be more efficient and could serve customers in an average of 2.5 minutes instead of 3 minutes. Answer the questions for part d again with this new average service time.
a. If two tellers are used, what will be the average waiting time for a customer before reaching a teller? On average, how many customers will be in the bank, including those currently being served?
b. Company policy is to have no more than a 10 percent chance that a customer will need to wait more than five minutes be- fore reaching a teller. How many tellers need to be used in order to meet this standard?
c. Sally feels that a significant cost is incurred by making a customer wait because of potential lost future business. Sally estimates the cost to be $0.50 for each minute a customer spends in the bank, counting both waiting time and service time. Given this cost, how many tellers should Sally employ?
d. First Bank has two types of customers: merchant customers and regular customers. The mean arrival rate for each type of
times, label the statement as true or false and then justify your answer by referring to a specific statement in the chapter. a. It generally provides an excellent approximation of
the true service-time distribution. b. Its mean and variance are always equal. c. It represents a rather extreme case regarding the
amount of variability in the service times.
11.5. For each of the following statements about the queue in a queueing system, label the statement as true or false and then justify your answer by referring to a specific statement in the chapter. a. The queue is where customers wait in the queueing
system until their service is completed. b. Queueing models conventionally assume that the
queue can hold only a limited number of customers. c. The most common queue discipline is first-come,
first-served.
11.6. Midtown Bank always has two tellers on duty. Cus- tomers arrive to receive service from a teller at a mean rate of 40 per hour. A teller requires an average of two minutes to serve a customer. When both tellers are busy, an arriving customer joins a single line to wait for service. Experience has shown that customers wait in line an average of one minute before service begins. a. Describe why this is a queueing system. b. Determine the basic measures of performance—
W q , W, L q , and L —for this queueing system. ( Hint: We don’t know the probability distributions of interarrival times and service times for this queue- ing system, so you will need to use the relationships between these measures of performance to help answer the question.)
11.7. Mom-and-Pop’s Grocery Store has a small adjacent parking lot with three parking spaces reserved for the store’s customers. During store hours, when the lot is not full, cars enter the lot and use one of the spaces at a mean rate of two per hour. When the lot is full, arriving cars leave and do not return. For n 5 0, 1, 2, 3, the probability P n that exactly n spaces currently are being used is P 0 5 0.2, P 1 5 0.3, P 2 5 0.3, P 3 5 0.2.
To the left of the following problems (or their parts), we have inserted the symbol E (for Excel) whenever one of the above templates can be helpful. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
11.1. Consider a typical hospital emergency room. a. Describe why it is a queueing system. b. What is the queue in this case? Describe how you
would expect the queue discipline to operate. c. Would you expect random arrivals? d. What are service times in this context? Would you
expect much variability in the service times? 11.2. Identify the customers and the servers in the queueing system in each of the following situations. a. The checkout stand in a grocery store. b. A fire station. c. The toll booth for a bridge. d. A bicycle repair shop. e. A shipping dock. f. A group of semiautomatic machines assigned to one
operator. g. The materials-handling equipment in a factory area. h. A plumbing shop. i. A job shop producing custom orders. j. A secretarial word processing pool. 11.3.* For each of the following statements about using the exponential distribution as the probability distribution of interar- rival times, label the statement as true or false and then justify your answer by referring to a specific statement in the chapter. a. It is the only distribution of interarrival times that
fits having random arrivals. b. It has the lack-of-memory property because it can-
not remember when the next arrival will occur. c. It provides an excellent fit for interarrival times for
most situations. 11.4. For each of the following statements about using the exponential distribution as the probability distribution of service
Problems
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480 Chapter Eleven Queueing Models
a. Use the formulas for the M/M/ 1 model to calculate L, W, W q , L q , P 0 , P 1 , and P 2 for the current mode of operation. What is the probability of having more than two customers at the checkout stand?
E b. Use the Excel template for this model to check your answers in part a. Also, find the probability that the waiting time before beginning service exceeds five minutes, and the probability that the waiting time before finishing service exceeds seven minutes.
c. Repeat part a for the alternative being considered by the manager.
E d. Repeat part b for this alternative. e. Which approach should the manager use to satisfy
her criteria as closely as possible?
11.12.* The 4M Company has a single turret lathe as a key work center on its factory floor. Jobs arrive randomly at this work center at a mean rate of two per day. The processing time to perform each job has an exponential distribution with a mean of ¼ day. Because the jobs are bulky, those not being worked on are currently being stored in a room some distance from the machine. However, to save time in fetching the jobs, the pro- duction manager is proposing to add enough in-process storage space next to the turret lathe to accommodate three jobs in addi- tion to the one being processed. (Excess jobs will continue to be stored temporarily in the distant room.) Under this proposal, what proportion of the time will this storage space next to the turret lathe be adequate to accommodate all waiting jobs? a. Use available formulas to calculate your answer. E b. Use an Excel template to obtain the information
needed to answer the question.
11.13. Jerry Jansen, materials handling manager at the Casper-Edison Corporation’s new factory, needs to decide whether to purchase a small tractor-trailer train or a heavy-duty forklift truck for transporting heavy goods between certain pro- ducing centers in the factory. Calls for the materials-handling unit to move a load would come essentially at random at a mean rate of four per hour. The total time required to move a load has an exponential distribution, where the expected time would be 12 minutes for the tractor-trailer train and 9 min- utes for the forklift truck. The total equivalent uniform hourly cost (capital recovery cost plus operating cost) would be $50 for the tractor-trailer train and $150 for the forklift truck. The estimated cost of idle goods (waiting to be moved or in tran- sit) because of increased in-process inventory is $20 per load per hour.
Jerry also has established certain criteria that he would like the materials-handling unit to satisfy in order to keep production flowing on schedule as much as possible. He would like to aver- age no more than half an hour for completing the move of a load after receiving the call requesting the move. He also would like the time for completing the move to be no more than one hour 80 percent of the time. Finally, he would like to have no more than three loads waiting to start their move at least 80 percent of the time. E a. Obtain the various measures of performance if the
tractor-trailer train were to be chosen. Evaluate how well these measures meet the above criteria.
E b. Repeat part a if the forklift truck were to be chosen.
a. Describe how this parking lot can be interpreted as being a queueing system. In particular, identify the customers and the servers. What is the service being provided? What constitutes a service time? What is the queue capacity? ( Hint: See Table 11.4 .)
b. Determine the basic measures of performance— L, L q , W, and W q —for this queueing system. ( Hint: You can use the given probabilities to determine the average number of parking spaces that are being used.)
c. Use the results from part b to determine the average length of time that a car remains in a parking space.
11.8.* Newell and Jeff are the two barbers in a barber shop they own and operate. They provide two chairs for customers who are waiting to begin a haircut, so the number of customers in the shop varies between 0 and 4. For n 5 0, 1, 2, 3, 4, the probability P n that exactly n customers are in the shop is P0 5 P1 5 1⁄16, P2 5
6⁄16, P3 5 4⁄16, P4 5
1⁄16. a. Use the formula L 5 0 P 0 1 1 P 1 1 2 P 2 1 3 P 3 1 4 P 4 to
calculate L. How would you describe the meaning of L to Newell and Jeff?
b. For each of the possible values of the number of customers in the queueing system, specify how many customers are in the queue. For each of the possible numbers in the queue, multiply by its prob- ability, and then add these products to calculate L q . How would you describe the meaning of L q to New- ell and Jeff?
c. Given that an average of four customers per hour arrive and stay to receive a haircut, determine W and W q . Describe these two quantities in terms meaningful to Newell and Jeff.
d. Given that Newell and Jeff are equally fast in giving haircuts, what is the average duration of a haircut?
11.9. Explain why the utilization factor r for the server in a single-server queueing system must equal 1 2 P 0 , where P 0 is the probability of having 0 customers in the system.
11.10 Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 11.5. Briefly describe how queueing mod- els were applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study.
11.11. The Friendly Neighbor Grocery Store has a single check- out stand with a full-time cashier. Customers arrive randomly at the stand at a mean rate of 30 per hour. The service-time distribu- tion is exponential, with a mean of 1.5 minutes. This situation has resulted in occasional long lines and complaints from customers. Therefore, because there is no room for a second checkout stand, the manager is considering the alternative of hiring another per- son to help the cashier by bagging the groceries. This help would reduce the expected time required to process a customer to 1 min- ute, but the distribution still would be exponential.
The manager would like to have the percentage of time that there are more than two customers at the checkout stand down below 25 percent. She also would like to have no more than 5 percent of the customers needing to wait at least five minutes before beginning service, or at least seven minutes before finish- ing service.
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b. Use the cost figures to determine which grinder speed minimizes the expected total cost per minute.
E11.17. The Centerville International Airport has two run- ways, one used exclusively for takeoffs and the other exclu- sively for landings. Airplanes arrive randomly in the Centerville air space to request landing instructions at a mean rate of 10 per hour. The time required for an airplane to land after receiving clearance to land has an exponential distribution with a mean of three minutes, and this process must be completed before giving clearance to land to another airplane. Airplanes awaiting clear- ance must circle the airport.
The Federal Aviation Administration has a number of crite- ria regarding the safe level of congestion of airplanes waiting to land. These criteria depend on a number of factors regard- ing the airport involved, such as the number of runways avail- able for landing. For Centerville, the criteria are (1) the average number of airplanes waiting to receive clearance to land should not exceed one, (2) 95 percent of the time, the actual number of airplanes waiting to receive clearance to land should not exceed four, (3) for 99 percent of the airplanes, the amount of time spent circling the airport before receiving clearance to land should not exceed 30 minutes (since exceeding this amount of time often would require rerouting the plane to another airport for an emer- gency landing before its fuel runs out). a. Evaluate how well these criteria are currently being
satisfied. b. A major airline is considering adding this airport
as one of its hubs. This would increase the mean arrival rate to 15 airplanes per hour. Evaluate how well the above criteria would be satisfied if this happens.
c. To attract additional business (including the major airline mentioned in part b ), airport management is considering adding a second runway for landings. It is estimated that this eventually would increase the mean arrival rate to 25 airplanes per hour. Evaluate how well the above criteria would be satisfied if this happens.
11.18.* Consider the M/G/ 1 model. What is the effect on L q and W q if 1/ l , 1/ m , and s are all reduced by half? Explain.
11.19. Consider the M/G/ 1 model with l 5 0.2 and m 5 0.25. E a. Use the Excel template for this model to gener-
ate a data table that gives the main measures of performance— L, L q , W, W q —for each of the fol- lowing values of s : 4, 3, 2, 1, 0.
b. What is the ratio of L q with s 5 4 to L q with s 5 0? What does this say about the importance of reduc- ing the variability of the service times?
c. Calculate the reduction in L q when s is reduced from 4 to 3, from 3 to 2, from 2 to 1, and from 1 to 0. Which is the largest reduction? Which is the smallest?
E d. Use trial and error with the template to see approx- imately how much m would need to be increased with s 5 4 to achieve the same L q as with m 5 0.25 and s 5 0.
E e. Use the template to generate a data table that gives the value of L q with s 5 4 when m increases in increments of 0.01 from 0.25 to 0.35.
c. Compare the two alternatives in terms of their expected total cost per hour (including the cost of idle goods).
d. Which alternative do you think Jerry should choose? E11.14. Suppose a queueing system fitting the M/M/ 1 model has W 5 120 minutes and L 5 8 customers. Use these facts (and the formula for W ) to find l and m . Then find the various other measures of performance for this queueing system. 11.15.* The Seabuck and Roper Company has a large ware- house in southern California to store its inventory of goods until they are needed by the company’s many furniture stores in that area. A single crew with four members is used to unload and/or load each truck that arrives at the loading dock of the warehouse. Management currently is downsizing to cut costs, so a decision needs to be made about the future size of this crew.
Trucks arrive randomly at the loading dock at a mean rate of one per hour. The time required by a crew to unload and/or load a truck has an exponential distribution (regardless of crew size). The mean of this distribution with the four-member crew is 15 minutes. If the size of the crew were to be changed, it is estimated that the mean service rate of the crew (now m 5 4 cus- tomers per hour) would be proportional to its size.
The cost of providing each member of the crew is $20 per hour. The cost that is attributable to having a truck not in use (i.e., a truck standing at the loading dock) is estimated to be $30 per hour. a. Identify the customers and servers for this queueing
system. How many servers does it currently have? E b. Find the various measures of performance of this
queueing system with four members on the crew. (Set t 5 1 hour in the Excel template for the waiting-time probabilities.)
E c. Repeat b with three members. E d. Repeat part b with two members. e. Should a one-member crew also be considered?
Explain. f. Given the previous results, which crew size do you
think management should choose? g. Use the cost figures to determine which crew size
would minimize the expected total cost per hour. 11.16. Jake’s Machine Shop contains a grinder for sharpen- ing the machine cutting tools. A decision must now be made on the speed at which to set the grinder.
The grinding time required by a machine operator to sharpen the cutting tool has an exponential distribution, where the mean 1/ m can be set at 1 minute, 1.5 minutes, or 2 minutes, depending upon the speed of the grinder. The running and maintenance costs go up rapidly with the speed of the grinder, so the esti- mated cost per minute is $1.60 for providing a mean of 1 min- ute, $0.90 for a mean of 1.5 minutes, and $0.40 for a mean of 2 minutes.
The machine operators arrive randomly to sharpen their tools at a mean rate of one every two minutes. The estimated cost of an operator being away from his or her machine to the grinder is $0.80 per minute. E a. Obtain the various measures of performance for
this queueing system for each of the three alterna- tive speeds for the grinder. (Set t 5 5 minutes in the Excel template for the waiting-time probabilities.)
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The subassemblies arrive randomly at the inspection sta- tion at a mean rate of three per hour. The cost of having the subassemblies wait at the inspection station (thereby increasing in-process inventory and possibly disrupting subsequent produc- tion) is estimated to be $20 per hour for each subassembly.
Management now needs to make a decision about whether to continue the status quo or adopt the proposal. E a. Find the main measures of performance— L, L q , W,
W q —for the current queueing system. E b. Repeat part a for the proposed queueing system. c. What conclusions can you draw about what man-
agement should do from the results in parts a and b? d. Determine and compare the expected total cost per
hour for the status quo and the proposal. E 11.24. The Security & Trust Bank employs four tellers to serve its customers. Customers arrive randomly at a mean rate of two per minute. However, business is growing and management projects that the mean arrival rate will be three per minute a year from now. The transaction time between the teller and customer has an exponential distribution with a mean of one minute.
Management has established the following guidelines for a satisfactory level of service to customers. The average number of customers waiting in line to begin service should not exceed one. At least 95 percent of the time, the number of customers waiting in line should not exceed five. For at least 95 percent of the customers, the time spent in line waiting to begin service should not exceed five minutes. a. Use the M/M/s model to determine how well these
guidelines are currently being satisfied. b. Evaluate how well the guidelines will be satisfied
year from now if no change is made in the number of tellers.
c. Determine how many tellers will be needed a year from now to completely satisfy these guidelines.
E11.25. Consider the M/M/s model. For each of the following two cases, generate a data table that gives the values of L, L q , W, W q , and P{w . 5} for the following mean arrival rates: 0.5, 0.9, and 0.99 customers per minute. a. Suppose there is one server and the expected ser-
vice time is one minute. Compare L for the cases where the mean arrival rate is 0.5, 0.9, and 0.99 customers per minute, respectively. Do the same for L q , W, W q , and P{w . 5}. What conclusions do you draw about the impact of increasing the utiliza- tion factor r from small values (e.g., r 5 0.5) to fairly large values (e.g., r 5 0.9) and then to even larger values very close to 1 (e.g., r 5 0.99)?
b. Now suppose there are two servers and the expected service time is two minutes. Follow the instructions for part a.
E11.26. Consider the M/M/s model with a mean arrival rate of 10 customers per hour and an expected service time of five minutes. Use the Excel template for this model to print out the various measures of performance (with t 5 10 and t 5 0, respec- tively, for the two waiting-time probabilities) when the number of servers is one, two, three, four, and five. Then, for each of the following possible criteria for a satisfactory level of service
E f. Use the template to generate a two-way data table that gives the value of L q for the various combina- tions of values of m and s where m 5 0.22, 0.24, 0.26, 0.28, 0.3 and s 5 4, 3, 2, 1, 0.
11.20. Consider the following statements about the M/G/ 1 queueing model, where s 2 is the variance of service times. Label each statement as true or false, and then justify your answer. a. Increasing s 2 (with fixed l and m ) will increase L q
and L, but will not change W q and W. b. When the choice is between a tortoise (small m and
s 2 ) and a hare (large m and s 2 ) to be the server, the tortoise always wins by providing a smaller L q .
c. With l and m fixed, the value of L q with an expo- nential service-time distribution is twice as large as with constant service times.
11.21. Marsha operates an espresso stand. Customers arrive randomly at a mean rate of 30 per hour. The time needed by Marsha to serve a customer has an exponential distribution with a mean of 75 seconds. E a. Use the Excel template for the M/G/ 1 model to find
L, L q , W, and W q . E b. Suppose Marsha is replaced by an espresso vending
machine that requires exactly 75 seconds to operate for each customer. Find L, L q , W, and W q .
c. What is the ratio of L q in part b to L q in part a? E d. Use trial and error with the template to see approxi-
mately how much Marsha would need to reduce her expected service time to achieve the same L q as with the espresso vending machine.
E e. Use the template to generate a data table that gives the value of L q when Marsha is serving with the fol- lowing values (in seconds) for her expected service time: 75, 70, 65, 64, 63, 62, 61, 60.
11.22. Read the referenced article that fully describes the management science study summarized in the applica- tion vignette presented in Section 11.6. Briefly describe how a queueing model was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 11.23.* The production of tractors at the Jim Buck Com- pany involves producing several subassemblies and then using an assembly line to assemble the subassemblies and other parts into finished tractors. Approximately three tractors per day are produced in this way. An in-process inspection station is used to inspect the subassemblies before they enter the assembly line. At present, there are two inspectors at the station, and they work together to inspect each subassembly. The inspection time has an exponential distribution, with a mean of 15 minutes. The cost of providing this inspection system is $40 per hour.
A proposal has been made to streamline the inspection pro- cedure so that it can be handled by only one inspector. This inspector would begin by visually inspecting the exterior of the subassembly, and she would then use new efficient equipment to complete the inspection. Although this process with just one inspector would slightly increase the mean of the distribution of inspection times from 15 minutes to 16 minutes, it also would reduce the variance of this distribution to only 40 percent of its current value. The cost would be $30 per hour.
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add these two products) to calculate the expected waiting time in the system for a random arrival of either type of slip.
b. If the recommendations are adopted, determine the expected waiting time in the system for arriving slips.
c. Now suppose that adopting the recommendations would result in a slight increase in the expected pro- cessing time. Use the Excel template for this model to determine by trial and error the expected process- ing time (within 0.01 minute) that would cause the expected waiting time in the system for a random arrival to be essentially the same under current pro- cedures and under the recommendations.
E 11.29. People’s Software Company has just set up a call cen- ter to provide technical assistance on its new software package. Two technical representatives are taking the calls, where the time required by either representative to answer a customer’s ques- tions has an exponential distribution with a mean of eight min- utes. Calls are arriving randomly at a mean rate of 10 per hour.
By next year, the mean arrival rate of calls is expected to decline to five per hour, so the plan is to reduce the number of technical representatives to one then. Determine L, L q , W, and W q for both the current queueing system and next year’s system. For each of these four measures of performance, which system yields the smaller value?
11.30. The Southern Railroad Company has been subcon- tracting for the painting of its railroad cars as needed. However, management has decided that the company can save money by doing this work itself. A decision now needs to be made to choose between two alternative ways of doing this.
Alternative 1 is to provide two paint shops, where painting is done by hand (one car at a time in each shop), for a total hourly cost of $70. The painting time for a car would be six hours. Alternative 2 is to provide one spray shop involving an hourly cost of $100. In this case, the painting time for a car (again done one at a time) would be three hours. For both alternatives, the cars arrive randomly with a mean rate of 1 every 5 hours. The cost of idle time per car is $100 per hour.
a. Use Figure 11.10 to estimate L, L q , W, and W q for Alternative 1.
E b. Find these same measures of performance for Alter- native 2.
c. Determine and compare the expected total cost per hour for these alternatives.
11.31.* Southeast Airlines is a small commuter airline. Its ticket counter at one of its airports is staffed by a single ticket agent. There are two separate lines—one for first-class passen- gers and one for coach-class passengers. When the ticket agent is ready for another customer, the next first-class passenger is served if there are any in line. If not, the next coach-class pas- senger is served. Service times have an exponential distribution with a mean of three minutes for both types of customers. Dur- ing the 12 hours per day that the ticket counter is open, passen- gers arrive randomly at a mean rate of 2 per hour for first-class passengers and 10 per hour for coach-class passengers.
a. What kind of queueing model fits this queueing system?
(where the unit of time is one minute), use the printed results to determine how many servers are needed to satisfy this criterion.
a. L q # 0.25
b. L # 0.9
c. W q # 0.1
d. W # 6
e. P{wq . 0} # 0.01
f. P{w . 10} # 0.2
g. a s
n 5 0 Pn $ 0.95
11.27. Greg is making plans to open a new fast-food restau- rant soon. He is estimating that customers will arrive randomly at a mean rate of 150 per hour during the busiest times of the day. He is planning to have three employees directly serving the customers. He now needs to make a decision about how to orga- nize these employees.
Option 1 is to have three cash registers with one employee at each to take the orders and get the food and drinks. In this case, it is estimated that the average time to serve each customer would be one minute, and the distribution of service times is assumed to be exponential.
Option 2 is to have one cash register with the three employ- ees working together to serve each customer. One would take the order, a second would get the food, and the third would get the drinks. Greg estimates that this would reduce the average time to serve each customer down to 20 seconds, with the same assumption of exponential service times.
Greg wants to choose the option that would provide the best service to his customers. However, since Option 1 has three cash registers, both options would serve the customers at a mean rate of three per minute when everybody is busy serving customers, so it is not clear which option is better.
E a. Use the main measures of performance— L, L q , W, W q —to compare the two options.
b. Explain why these comparisons make sense intuitively.
c. Which measure do you think would be most impor- tant to Greg’s customers? Why? Which option is better with respect to this measure?
E*11.28. In the Blue Chip Life Insurance Company, the deposit and withdrawal functions associated with a certain investment product are separated between two clerks. Deposit slips arrive randomly at Clerk Clara’s desk at a mean rate of 16 per hour. Withdrawal slips arrive randomly at Clerk Clarence’s desk at a mean rate of 14 per hour. The time required to process either transaction has an exponential distribution with a mean of three minutes. In order to reduce the expected waiting time in the sys- tem for both deposit slips and withdrawal slips, the Actuarial Department has made the following recommendations: (1) Train each clerk to handle both deposits and withdrawals; (2) put both deposit and withdrawal slips into a single queue that is accessed by both clerks.
a. Determine the expected waiting time in the system under current procedures for each type of slip. Then combine these results (multiply W for deposit slips by 16⁄30 multiply W for withdrawal slips by 14⁄30 and
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contends that five machines are required, as opposed to the three machines that he now has. However, because of pressure from management to hold down capital expenditures, only one addi- tional machine will be authorized unless there is solid evidence that a second one is necessary.
This shop does three kinds of jobs, namely, government jobs, commercial jobs, and standard products. Whenever a turret lathe operator finishes a job, he starts a government job if one is wait- ing; if not, he starts a commercial job if any are waiting; if not, he starts on a standard product if any are waiting. Jobs of the same type are taken on a first-come, first-served basis.
Although much overtime work is required currently, man- agement wants the turret lathe department to operate on an eight- hour, five-day-per-week basis. The probability distribution of the time required by a turret lathe operator for a job appears to be approximately exponential, with a mean of 10 hours. Jobs come into the shop randomly at a mean rate of six per week for government jobs, four per week for commercial jobs, and two per week for standard products. (These figures are expected to remain the same for the indefinite future.)
Management feels that the average waiting time before work begins in the turret lathe department should not exceed 0.25 (working) days for government jobs, 0.5 days for commercial jobs, and 2 days for standard products. a. Determine how many additional turret lathes need to
be obtained to satisfy these management guidelines. b. It is worth about $750, $450, and $150 to avoid a
delay of one additional (working) day in finishing a government, commercial, and standard job, respec- tively. The incremental capitalized cost of providing each turret lathe (including the operator and so on) is estimated to be $250 per working day. Determine the number of additional turret lathes that should be obtained to minimize the expected total cost.
E11.34. When describing economic analysis of the num- ber of servers to provide in a queueing system, Section 11.9 introduces a cost model where the objective is to minimize TC 5 C s s 1 C w L. The purpose of this problem is to enable you to explore the effect that the relative sizes of C s and C w have on the optimal number of servers.
Suppose that the queueing system under consideration fits the M/M/s model with l 5 8 customers per hour and m 5 10 customers per hour. Use the Excel template for economic analy- sis with the M/M/s model to find the optimal number of servers for each of the following cases. a. C s 5 $100 and C w 5 $10. b. C s 5 $100 and C w 5 $100. c. C s 5 $10 and C w 5 $100. d. For each of these three cases, generate a data table
that compares the expected hourly costs with vari- ous alternative numbers of servers.
E11.35.* Jim McDonald, manager of the fast-food hamburger restaurant McBurger, realizes that providing fast service is a key to the success of the restaurant. Customers who have to wait very long are likely to go to one of the other fast-food restau- rants in town next time. He estimates that each minute a cus- tomer has to wait in line before completing service costs him an average of 30¢ in lost future business. Therefore, he wants to be sure that enough cash registers always are open to keep
E b. Find the main measures of performance— L, L q , W, and W q —for both first-class passengers and coach- class passengers.
c. What is the expected waiting time before service begins for first-class customers as a fraction of this waiting time for coach-class customers?
d. Determine the average number of hours per day that the ticket agent is busy.
11.32. The County Hospital emergency room always has one doctor on duty. In the past, having just a single doctor there has been sufficient. However, because of a growing tendency for emergency cases to use these facilities rather than go to a private doctor, the number of emergency room visits has been steadily increasing. By next year, it is estimated that patients will arrive randomly at a mean rate of two per hour during peak usage hours (the early evening). Therefore, a proposal has been made to assign a second doctor to the emergency room next year during those hours. Hospital management (an HMO) is resisting this proposal, but has asked a management scientist (you) to analyze whether a single doctor will continue to be sufficient next year.
The patients are not treated on a first-come, first-served basis. Rather, the admitting nurse divides the patients into three categories: (1) critical cases, where prompt treatment is vital for survival; (2) serious cases, where early treatment is important to prevent further deterioration; and (3) stable cases, where treat- ment can be delayed without adverse medical consequences. Patients are then treated in this order of priority, where those in the same category are normally taken on a first-come, first- served basis. A doctor will interrupt treatment of a patient if a new case in a higher priority category arrives. Approximately 10 percent of the patients fall into the first category, 30 percent into the second, and 60 percent into the third. Because the more serious cases will be sent to the hospital for further care after receiving emergency treatment, the average treatment time by a doctor in the emergency room actually does not differ greatly among these categories. For all of them, the treatment time can be approximated by an exponential distribution with a mean of 20 minutes.
Hospital management has established the following guide- lines. The average waiting time in the emergency room before treatment begins should not exceed 2 minutes for critical cases, 15 minutes for serious cases, and 2 hours for stable cases. a. What kind of queueing model fits this queueing
system? E b. Use this model to determine if the management
guidelines would be satisfied next year by continu- ing to have just a single doctor on duty.
c. Use the formula for W q for the M/M/ 1 model to determine if these guidelines would be satisfied if treatment were given on a first-come, first-served basis instead.
E d. The mean arrival rate of two patients per hour dur- ing peak usage hours next year is only an estimate. Perform sensitivity analysis by repeating part b if this mean arrival rate were to turn out to be 2.25 patients per hour instead.
E11.33. The Becker Company factory has been experiencing long delays in jobs going through the turret lathe department because of inadequate capacity. The head of this department
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However, due to recent complaints about considerable time being wasted waiting for a copier to become free, management is considering adding one or more additional copy machines.
During the 2,000 working hours per year, employees arrive randomly at the copying room at a mean rate of 30 per hour. The time each employee needs with a copy machine is believed to have an exponential distribution with a mean of five minutes. The lost productivity due to an employee spending time in the copying room is estimated to cost the company an average of $25 per hour. Each copy machine is leased for $3,000 per year.
Determine how many copy machines the company should have to minimize its expected total cost per hour.
waiting to a minimum. Each cash register is operated by a part- time employee who obtains the food ordered by each customer and collects the payment. The total cost for each such employee is $9 per hour.
During lunchtime, customers arrive randomly at a mean rate of 66 per hour. The time needed to serve a customer is esti- mated to have an exponential distribution with a mean of two minutes.
Determine how many cash registers Jim should have open during lunchtime to minimize his expected total cost per hour. E11.36. The Garrett-Tompkins Company provides three copy machines in its copying room for the use of its employees.
Case 11-1
Queueing Quandary
A SEQUEL TO CASE 10.1
Never dull. That is how you would describe your job at the cen- tralized records and benefits administration center for Cutting Edge, a large company manufacturing computers and computer peripherals. Since opening the facility six months ago, you and Mark Lawrence, the director of Human Resources, have endured one long roller-coaster ride. Receiving the go-ahead from corpo- rate headquarters to establish the centralized records and bene- fits administration center was definitely an up. Getting caught in the crossfire of angry customers (all employees of Cutting Edge) because of demand overload for the records and benefits call center was definitely a down. Accurately forecasting the demand for the call center provided another up.
And today you are faced with another down. Mark approaches your desk with a not altogether attractive frown on his face.
He begins complaining immediately, “I just don’t under- stand. The forecasting job you did for us two months ago really allowed us to understand the weekly demand for the center, but we still have not been able to get a grasp on the staffing problem. We used both historical data and your forecasts to calculate the average weekly demand for the call center. We transformed this average weekly demand into average hourly demand by dividing the weekly demand by the number of hours in the workweek. We then staffed the center to meet this average hourly demand by taking into account the average number of calls a representa- tive is able to handle per hour.
But something is horribly wrong. Operational data records show that over 35 percent of the customers wait over four min- utes for a representative to answer the call! Customers are still sending me numerous complaints, and executives from corpo- rate headquarters are still breathing down my neck! I need help!”
You calm Mark down and explain to him that you think you know the problem: The number of calls received in a cer- tain hour can be much greater (or much less) than the average because of the stochastic nature of the demand. In addition, the number of calls a representative is able to handle per hour can be much less (or much greater) than the average depending upon the types of calls received.
You then tell him to have no fear; you have the problem under control. You have been reading about the successful appli- cation of queueing theory to the operation of call centers, and you decide that the queueing models you learned about in school will help you determine the appropriate staffing level.
a. You ask Mark to describe the demand and service rate. He tells you that calls are randomly received by the call center and that the center receives an average of 70 calls per hour. The computer system installed to answer and hold the calls is so advanced that its capacity far exceeds the demand. Because the nature of a call is random, the time required to process a call is random, where the time frequently is small but occa- sionally can be much longer. On average, however, represen- tatives can handle six calls per hour. Which queueing model seems appropriate for this situation? Given that slightly more than 35 percent of customers wait over four minutes before a representative answers the call, use this model to estimate how many representatives Mark currently employs.
b. Mark tells you that he will not be satisfied unless 95 percent of the customers wait only one minute or less for a represen- tative to answer the call. Given this customer service level and the average arrival rates and service rates from part a, how many representatives should Mark employ?
c. Each representative receives an annual salary (including ben- efits) of $30,000, and Mark tells you that he simply does not have the resources available to hire the number of representa- tives required to achieve the customer service level desired in part b. He asks you to perform sensitivity analysis. How many representatives would he need to employ to ensure that 80 percent of customers wait one minute or less? How many would he need to employ to ensure that 95 percent of cus- tomers wait 90 seconds or less? How would you recommend Mark choose a customer service level? Would the decision criteria be different if Mark’s call center were to serve exter- nal customers (not connected to the company) instead of internal customers (employees)?
d. Mark tells you that he is not happy with the number of repre- sentatives required to achieve a high customer service level.
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Mark have to employ and train to achieve the customer ser- vice level desired in part b? Do you prefer this alternative to simply hiring additional representatives? Why or why not?
e. Mark realizes that queueing theory helps him only so much in determining the number of representatives needed. He realizes that the queueing models will not provide accurate answers if the inputs used in the models are inaccurate. What inputs do you think need reevaluation? How would you go about estimating these inputs?
He therefore wants to explore alternatives to simply hiring additional representatives. The alternative he considers is instituting a training program that will teach representatives to more efficiently use computer tools to answer calls. He believes that this alternative will increase the average num- ber of calls a representative is able to handle per hour from six calls to eight calls. The training program will cost $2,500 per employee per year since employees’ knowledge will have to be updated yearly. How many representatives will
Case 11-2
Reducing In-Process Inventory
Jim Wells, vice president for manufacturing of the Northern Airplane Company, is exasperated. His walk through the com- pany’s most important plant this morning has left him in a foul mood. However, he now can vent his temper at Jerry Carstairs, the plant’s production manager, who has just been summoned to Jim’s office.
“Jerry, I just got back from walking through the plant, and I am very upset.”
“What is the problem, Jim?” “Well, you know how much I have been emphasizing the
need to cut down on our in-process inventory.” “Yes, we’ve been working hard on that,” responds Jerry. “Well, not hard enough!” Jim raises his voice even higher.
“Do you know what I found by the presses?” “No.” “Five metal sheets still waiting to be formed into wing sec-
tions. And then, right next door at the inspection station, 13 wing sections! The inspector was inspecting one of them, but the other 12 were just sitting there. You know we have a couple hun- dred thousand dollars tied up in each of those wing sections. So between the presses and the inspection station, we have a few million bucks’ worth of terribly expensive metal just sitting there. We can’t have that!”
The chagrined Jerry Carstairs tries to respond. “Yes, Jim, I am well aware that that inspection station is a bottleneck. It usu- ally isn’t nearly as bad as you found it this morning, but it is a bottleneck. Much less so for the presses. You really caught us on a bad morning.”
“I sure hope so,” retorts Jim, “but you need to prevent any- thing nearly this bad happening even occasionally. What do you propose to do about it?”
Jerry now brightens noticeably in his response. “Well, actu- ally, I’ve already been working on this problem. I have a couple proposals on the table and I have asked a management scien- tist on my staff to analyze these proposals and report back with recommendations.”
“Great,” responds Jim, “glad to see you are on top of the problem. Give this your highest priority and report back to me as soon as possible.”
“Will do,” promises Jerry. Here is the problem that Jerry and his management scien-
tist are addressing. Each of 10 identical presses is being used to form wing sections out of large sheets of specially processed metal. The sheets arrive randomly at a mean rate of seven per
hour. The time required by a press to form a wing section out of a sheet has an exponential distribution with a mean of one hour. When finished, the wing sections arrive randomly at an inspec- tion station at the same mean rate as the metal sheets arrived at the presses (seven per hour). A single inspector has the full-time job of inspecting these wing sections to make sure they meet specifications. Each inspection takes her 71/2 minutes, so she can inspect eight wing sections per hour. This inspection rate has resulted in a substantial average amount of in-process inventory at the inspection station (i.e., the average number of wing sheets waiting to complete inspection is fairly large), in addition to that already found at the group of machines.
The cost of this in-process inventory is estimated to be $8 per hour for each metal sheet at the presses or each wing section at the inspection station. Therefore, Jerry Carstairs has made two alternative proposals to reduce the average level of in-process inventory.
Proposal 1 is to use slightly less power for the presses (which would increase their average time to form a wing section to 1.2 hours), so that the inspector can keep up with their output better. This also would reduce the cost for each machine (operating cost plus capital recovery cost) from $7.00 to $6.50 per hour. (By contrast, increasing to maximum power would increase this cost to $7.50 per hour while decreasing the average time to form a wing section to 0.8 hours.)
Proposal 2 is to substitute a certain younger inspector for this task. He is somewhat faster (albeit with some variability in his inspection times because of less experience), so he should keep up better. (His inspection time would have a probability distribution with a mean of 7.2 minutes and a standard deviation of 5 minutes.) This inspector is in a job classification that calls for a total compensation (including benefits) of $19 per hour, whereas the current inspector is in a lower job classification where the compensation is $17 per hour. (The inspection times for each of these inspectors are typical of those in the same job classifications.)
You are the management scientist on Jerry Carstair’s staff who has been asked to analyze this problem. He wants you to “use the latest management science techniques to see how much each proposal would cut down on in-process inventory and then make your recommendations.”
a. To provide a basis of comparison, begin by evaluating the status quo. Determine the expected amount of in-process
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of machines. Be specific in your recommendations, and sup- port them with quantitative analysis like that done in part a. Make specific comparisons to the results from part a, and cite the improvements that your recommendations would yield.
Additional Cases Additional cases for this chapter also are available at the Uni- versity of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
inventory at the presses and at the inspection station. Then calculate the expected total cost per hour of the in-process inventory, the presses, and the inspector.
b. What would be the effect of Proposal 1? Why? Make specific comparisons to the results from part a. Explain this outcome to Jerry Carstairs.
c. Determine the effect of Proposal 2. Make specific compari- sons to the results from part a. Explain this outcome to Jerry Carstairs.
d. Make your recommendations for reducing the average level of in-process inventory at the inspection station and at the group
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Chapter Twelve
Computer Simulation: Basic Concepts Learning Objectives
After completing this chapter, you should be able to
1. Describe the basic concept of computer simulation.
2. Describe the role computer simulation plays in many management science studies.
3. Use random numbers to generate random events that have a simple discrete distribution.
4. Use Excel to perform basic computer simulations on a spreadsheet.
5. Use the Queueing Simulator to perform computer simulations of basic queueing sys- tems and interpret the results.
6. Describe and use the building blocks of a simulation model for a stochastic system.
7. Outline the steps of a major computer simulation study.
In this chapter, we now are ready to focus on the last of the key techniques of management science. Computer simulation ranks very high among the most widely used of these tech- niques. Furthermore, because it is such a flexible, powerful, and intuitive tool, it is continuing to rapidly grow in popularity. Many managers consider it one of their most valuable decision- making aids.
This technique involves using a computer to imitate (simulate) the operation of an entire process or system. For example, computer simulation is frequently used to perform risk anal- ysis on financial processes by repeatedly imitating the evolution of the transactions involved to generate a profile of the possible outcomes. Computer simulation also is widely used to analyze systems that will continue operating indefinitely. For such systems, the computer ran- domly generates and records the occurrences of the various events that drive the system just as if it were physically operating. Because of its speed, the computer can simulate even years of operation in a matter of seconds. Recording the performance of the simulated operation of the system for a number of alternative designs or operating procedures then enables evaluat- ing and comparing these alternatives before choosing one. For many processes and systems, all this now can be done with spreadsheet software.
The range of applications of computer simulation has been quite remarkable. The case study in this chapter will illustrate how it is used for the design and operation of queueing systems. A variety of applications have involved the design and operation of manufacturing systems, as well as the design and operation of distribution systems. Some more specific areas of application include managing inventory systems and estimating the probability of complet- ing a project by the deadline. Financial risk analysis is a particularly active area of applica- tion. Health care applications also abound. The list goes on and on.
The first section of this chapter describes and illustrates the essence of computer simula- tion. The case study for this chapter (a revisit of Herr Cutter’s barber shop from the pre- ceding chapter) is discussed and analyzed in Sections 12.2 and 12.3. The following section then outlines the overall procedure for applying computer simulation. Chapter 13 will expand further on the application of computer simulation by describing how to apply Risk Solver
By imitating the operation of a proposed system, a computer can simulate years of operation in a matter of seconds and then record the performance.
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Platform for Education (RSPE) to efficiently perform fairly complicated computer simula- tions on spreadsheets. (One of the supplementary chapters on the CD-ROM, Chapter 20, does the same thing with another popular Excel add-in called Crystal Ball.)
12.1 THE ESSENCE OF COMPUTER SIMULATION
The technique of simulation has long been an important tool of the designer. For example, simulating airplane flight in a wind tunnel is standard practice when a new airplane is designed. Theoretically, the laws of physics could be used to obtain the same information about how the performance of the airplane changes as design parameters are altered, but, as a practical matter, the analysis would be too complicated to do it all. Another alternative would be to build real airplanes with alternative designs and test them in actual flight to choose the final design, but this would be far too expensive (as well as unsafe). Therefore, after some preliminary theoretical analysis is performed to develop a rough design, simu- lating flight in a wind tunnel is a vital tool for experimenting with specific designs. This simulation amounts to imitating the performance of a real airplane in a controlled environ- ment in order to estimate what its actual performance will be. After a detailed design is developed in this way, a prototype model can be built and tested in actual flight to fine-tune the final design.
The Role of Computer Simulation Computer simulation plays essentially this same role in many management science studies. However, rather than designing an airplane, the management science team is concerned with developing a design or operating procedure for some system. In many cases, the system is a stochastic system, as defined below.
A stochastic system is a system that evolves over time according to one or more probability distributions. For example, the queueing systems described in the preceding chapter are stochas- tic systems because both the interarrival times and service times occur according to probability distributions.
Computer simulation imitates the operation of such a system by using the corresponding prob- ability distributions to randomly generate the various events that occur in the system (e.g., the arrivals and service completions in a queueing system). However, rather than literally operat- ing a physical system, the computer is just recording the occurrences of the simulated events and the resulting performance of this simulated system.
When computer simulation is used as part of a management science study, commonly it is preceded and followed by the same steps described earlier for the design of an airplane. In particular, some preliminary analysis is done first (perhaps with approximate mathematical models) to develop a rough design of the system (including its operating procedures). Then computer simulation is used to experiment with specific designs to estimate how well each will perform. After a detailed design is developed and selected in this way, the system prob- ably is tested in actual use to fine-tune the final design.
When dealing with relatively complex systems, computer simulation tends to be a rela- tively expensive procedure. To get started, a detailed model must be formulated to describe the operation of the system of interest and how it is to be simulated. Then considerable time often is required to develop and debug the computer programs needed to run the simulation. Next, many long computer runs may be needed to obtain good estimates of how well all the alternative designs of the system would perform. Finally, all these data should be carefully analyzed before drawing any final conclusions. This entire process typically takes a lot of time and effort. Therefore, computer simulation should not be used when a less-expensive procedure is available that can provide the same information.
Computer simulation typically is used when the stochastic system involved is too complex to be analyzed satisfactorily by the kinds of mathematical models (e.g., queueing models) described in the preceding chapters. One of the main strengths of a mathematical model is that it abstracts the essence of the problem and reveals its underlying structure, thereby providing
Computer simulation uses probability distributions to randomly generate the vari- ous events that occur in a system.
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490 Chapter Twelve Computer Simulation: Basic Concepts
insight into the cause-and-effect relationships within the system. Therefore, if the modeler is able to construct a mathematical model that is both a reasonable approximation of the problem and amenable to solution, this approach usually is superior to computer simulation. However, many problems are too complex to permit this approach. Fortunately, computer simulation is an exceptionally versatile tool so it can be used to study even highly complex stochastic systems. Thus, computer simulation often provides the only practical approach to a problem.
Now let us look at a few examples to illustrate the basic ideas of computer simulation. These examples have been kept considerably simpler than the usual application of this tech- nique in order to highlight the main ideas more readily. This also will enable us to obtain ana- lytical solutions for the performance of the systems involved to compare with the estimates of the performance provided by computer simulation.
Example 1: A Coin-Flipping Game You are the lucky winner of a sweepstakes contest. Your prize is an all-expense-paid vacation at a major hotel in Las Vegas, including some chips for gambling in the hotel casino.
Upon entering the casino, you find that, in addition to the usual games (blackjack, roulette, etc.), they are offering an interesting new game with the following rules.
Rules of the Game 1. Each play of the game involves repeatedly flipping an unbiased coin until the difference
between the number of heads tossed and the number of tails is three. 2. If you decide to play the game, you are required to pay $1 for each flip of the coin. You are
not allowed to quit during a play of the game. 3. You receive $8 at the end of each play of the game.
Thus, you win money if the number of flips required is fewer than eight, but you lose money if more than eight flips are required. Here are some examples (where H denotes a head and T a tail).
HHH 3 flips You win $5 THTTT 5 flips You win $3 THHTHTHTTTT 11 flips You lose $3
How would you decide whether to play this game? Many people would base this decision on simulation, although they probably would not
call it by that name. In this case, simulation amounts to nothing more than playing the game alone many times until it becomes clear whether it is worthwhile to play for money. Half an hour spent in repeatedly flipping a coin and recording the earnings or losses that would have resulted might be sufficient. This is a true simulation because you are imitating the actual play of the game without actually winning or losing any money.
Since the topic of this chapter is computer simulation, let us see now how a computer can be used to perform this same simulated experiment. Although a computer cannot flip coins, it can simulate doing so. It accomplishes this by generating a sequence of random numbers, as defined below.
A number is a random number between 0 and 1 if it has been generated in such a way that every possible number within this interval has an equal chance of occurring. For example, if numbers with four decimal places are being used, every one of the 10,000 numbers between 0.0000 and 0.9999 has an equal chance of occurring. Thus, a random number between 0 and 1 is a random observation from a uniform distribution between 0 and 1. (Hereafter, we will delete the phrase between 0 and 1 when referring to these random numbers.)
An easy way to generate random numbers is to use the RAND() function in Excel. For example, the lower left-hand corner of Figure 12.1 indicates that 5 RAND() has been entered into cell C13 and then copied into the range C14:C62. (The parentheses need to be included with this function, but nothing is inserted between them.) This causes Excel to generate the random numbers shown in cells C13:C62 of the spreadsheet. (Rows 27–56 have been hidden to save space in the figure.)
Computer simulation can predict the performance of proposed systems that are too complex to be analyzed by other mathematical models.
Random numbers play a key role in performing computer simulations, so Excel uses the RAND() function to generate them.
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1
2
3
4
5
6
7
8
Required Difference
Cash at End of Game
Number of Flips
Winnings
Summary of Game
9
10
11
Flip
1
2
3
4
5
6
7
8
9
10
11
12
13
14
45
46
47
48
49
50
Random
Number
0.3039
0.7914
0.8543
0.6902
0.3004
0.0383
0.3883
0.6052
0.2231
0.4250
0.3729
0.7983
0.2340
0.0082
0.7539
0.2989
0.6427
0.2824
0.2124
0.6420
Result
Heads
Tails
Tails
Tails
Heads
Heads
Heads
Tails
Heads
Heads
Heads
Tails
Heads
Heads
Tails
Heads
Tails
Heads
Heads
Tails
Total
Heads
1
1
1
1
2
3
4
4
5
6
7
7
8
9
26
27
27
28
29
29
Total
Tails
0
1
2
3
3
3
3
4
4
4
4
5
5
5
19
19
20
20
20
21
Stop?
Stop
NA
NA
NA
NA
NA
NA
NA
NA
NA
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
57
58
59
60
61
62
A B C D E F G
Coin-Flipping Game
CashAtEndOfGame
Flip
NumberOfFlips
RandomNumber
RequiredDifference
Result
Stop?
TotalHeads
TotalTails
Winnings
D4
B13:B62
D7
C13:C62
D3
D13:D62
G13:G62
E13:E62
F13:F62
D8
Range Name Cells
6
Number of Flips
Winnings
Summary of Game
7 =COUNTBLANK(Stop?)+1
=CashAtEndOfGame–NumberOfFlips8
C D
11 Random
Number Result
Total
Heads
Total
Tails
=RAND()
=RAND()
=RAND()
:
:
=IF(RandomNumber<0.5,"Heads","Tails")
=IF(RandomNumber<0.5,"Heads","Tails")
=IF(RandomNumber<0.5,"Heads","Tails")
:
:
=IF(Results="Heads",1,0)
=E13+IF(Results="Heads",1,0)
=E14+IF(Results="Heads",1,0)
:
:
=Flip-TotalHeads
=Flip-TotalHeads
=Flip-TotalHeads
:
:
12
13
14
15
16
17
C D E F
12 Stop?
=IF(ABS(TotalHeads-TotalTails)>=RequiredDifference,"Stop","")
=IF(G15="",IF(ABS(TotalHeads-TotalTails)>=RequiredDifference,"Stop",""),"NA")
=IF(G16="",IF(ABS(TotalHeads-TotalTails)>=RequiredDifference,"Stop",""),"NA")
13
14
15
16
17
18
19
:
:
G
3
$8
11
–$3
FIGURE 12.1 A spreadsheet model for a computer simulation of the coin-flipping game (Example 1).
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Although these numbers in cells C13:C62 have all the important properties of random num- bers, Excel actually uses a fixed formula to calculate each random number from the preceding one, starting with a seed value to initialize the process. Since the sequence of random numbers is predictable in the sense that it can be reproduced by using the same seed value again (which some- times is advantageous), these numbers sometimes are referred to as pseudo-random numbers.
The probabilities for the outcome of flipping a coin are
P(heads) 5 ½ P(tails) 5 ½
Therefore, to simulate the flipping of a coin, the computer can just let any half of the possible random numbers correspond to heads and the other half correspond to tails. To be specific, we will use the following correspondence.
0.0000 to 0.4999 correspond to heads 0.5000 to 0.9999 correspond to tails
By using the formula
5 IF(RandomNumber , 0.5, “Heads”, “Tails”)
in each of the column D cells in Figure 12.1 , Excel inserts Heads if the random number is less than 0.5 and inserts Tails otherwise. Consequently, the first 11 random numbers generated in column C yield the following sequence of heads (H) and tails (T):
HTTTHHHTHHH
at which point the game stops because the number of heads (seven) exceeds the number of tails (four) by three. Cells D7 and D8 record the total number of flips (11) and resulting win- nings ($8 2 $11 5 2 $3).
Thus, Figure 12.1 records the computer simulation of one complete play of the game. To virtually ensure that the game will be completed, 50 flips of the coin have been simulated. Columns E and F record the cumulative number of heads and tails after each flip. The equa- tions entered into the column G cells leave each cell blank until the difference in the numbers of heads and tails reaches 3, at which point Stop is inserted into the cell. Thereafter, NA (for Not Applicable) is inserted instead.
Such simulations of plays of the game can be repeated as often as desired with this spread- sheet. Pressing the F9 key causes Excel to recalculate the entire spreadsheet, including calcu- lating new random numbers in cells C13:C62. In fact, making any change in the spreadsheet also causes Excel to recalculate, including generating new random numbers. (If you ever want to fix a given set of random numbers, e.g., to save one particular outcome of the game, select the range of random numbers, right-click and choose Copy, right-click again and choose Paste Special, and then select the Values option.)
Computer simulations normally are repeated many times to obtain a more reliable estimate of an average outcome. Figure 12.2 shows how a data table can be used to trick Excel into repeating the simulation 14 times. You first make a table with the column headings shown in columns J, K, and L. The first column of the table (J7:J20) is used to label the 14 plays of the game, leaving the first row blank. The headings of the next two columns specify which output will be evaluated. For each of these two columns, use the first row of the table (cells K6:L6) to write an equation that refers to the relevant output cell. In this case, the cells of interest are the number of flips and the winnings, so the equations for K6:L6 are those shown to the right of the spreadsheet in Figure 12.2 .
The next step is to select the entire table (J6:L20) and then choose Data Table from the What-If Analysis menu of the Data tab. In the Data Table dialog box (as shown on the right- hand side of Figure 12.2 ), choose any blank cell for the column input cell (for example, E4) but do not enter anything for the row input cell. Clicking OK then generates the data table shown in Figure 12.2 .
The first thing Excel does while generating the data table is to enter the numbers in the first column of the table (J7:J20), one at a time, into the column input cell (E4), which has no direct impact on the simulation. However, each time a new number is entered into the column input cell, Excel recalculates the entire original spreadsheet ( Figure 12.1 ) in cells C13:G62
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and then enters the resulting numbers in the output cells, NumberOfFlips (D7) and Winnings (D8), into the corresponding row of the data table. In essence, we have tricked Excel into repeating the simulation 14 times, each time generating new random numbers in column C to perform a completely new simulation.
Cell K22 shows that this sample of 14 plays of the game gives a sample average of 7.14 flips. The sample average provides an estimate of the true mean of the underlying probability distribution of the number of flips required for a play of the game. Hence, this sample average of 7.14 would seem to indicate that, on the average, you should win about $0.86 (cell L22) each time you play the game. Therefore, if you do not have a relatively high aversion to risk, it appears that you should choose to play this game, preferably a large number of times.
However, beware! One common error in the use of computer simulation is that conclu- sions are based on overly small samples, because statistical analysis was inadequate or totally lacking. A qualified statistician can be very helpful with designing the experiments to be per- formed with computer simulation. In this case, careful statistical analysis (using confidence intervals, etc.) would indicate that hundreds of simulated plays of the game would be needed before any conclusions should be drawn about whether you are likely to win or lose by play- ing this game numerous times.
It so happens that the true mean of the number of flips required for a play of this game is 9. (This mean can be found analytically, but not easily.) Thus, in the long run, you actually would average losing about $1 each time you played the game. Part of the reason that the above simulated experiment failed to draw this conclusion is that you have a small chance of a very large loss on any play of the game, but you can never win more than $5 each time. However, 14 simulated plays of the game were not enough to obtain any observations far out in the tail of the probability distribution of the amount won or lost on one play of the game. Only one simulated play gave a loss of more than $3, and that was only $7.
Figure 12.3 gives the results of running the simulation for 1,000 plays of the game (with rows 17–1,000 not shown). Cell K1008 records the average number of flips as 8.97, very
At least hundreds of simu- lated plays of this game are needed to obtain a reason- ably reliable estimate of an average outcome.
NumberOfFlips
Winnings
D7
D8
Range Name Cell
4 Number
of Flips
=NumberOfFlips
Winnings
=Winnings
5
6
K L
22 Average
J
=AVERAGE(K7:K20) =AVERAGE(L7:L20)
K L
1
2
3
4
Play
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Average
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
I J K L M
Data Table for Coin-Flipping Game (14 Replications)
Select the
whole table
(J6:L20),
before
choosing
Table from
the Data
menu.
Number
of Flips
3
9
5
7
11
5
3
3
11
7
15
3
7
9
5
7.14
Winnings
$5
–$1
$3
$1
–$3
$3
$5
$5
–$3
$1
–$7
$5
$1
–$1
3
$0.86
FIGURE 12.2 A data table that records the results of performing 14 replications of a computer simulation with the spreadsheet in Figure 12.1.
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494 Chapter Twelve Computer Simulation: Basic Concepts
close to the true mean of 9. With this number of replications, the average winnings of –$0.97 in cell L1008 now provide a reliable basis for concluding that this game will not win you money in the long run. (You can bet that the casino already has used computer simulation to verify this fact in advance.)
Example 2: Corrective Maintenance versus Preventive Maintenance The Heavy Duty Company has just purchased a large machine for a new production process. The machine is powered by a motor that occasionally breaks down and requires a major over- haul. Therefore, the manufacturer of the machine also provides a second standby motor. The two motors are rotated in use, with each one remaining in the machine until it is removed for an overhaul and replaced by the other one.
Given the planned usage of the machine, its manufacturer has provided the company with information about the durability of the motors (the number of days of usage until a break- down occurs). This information is shown in the first two columns of Table 12.1 . The first column lists the number of days the current machine has been in use. For each of these days, the second column then gives the probability that the breakdown will occur on that day. Since these probabilities are 0 except for days 4, 5, and 6, the breakdown always occurs on the fourth, fifth, or sixth day.
1
2
3
4
Play
1
2
3
4
5
6
7
8
9
10
995
996
997
998
999
1,000
Average
5
6
7
8
9
10
11
12
13
14
15
16
1001
1002
1003
1004
1005
1006
1007
1008
I J K L M
Data Table for Coin-Flipping Game (1,000 Replications)
Number
of Flips
5
3
3
7
11
13
7
3
7
3
9
5
27
7
3
9
17
8.97
Winnings
$3
$5
$5
$1
–$3
–$5
$1
$5
$1
$5
–$1
$3
–$19
$1
$5
–$1
–$9
–$0.97
FIGURE 12.3 This data table improves the reliability of the com- puter simulation recorded in Figure 12.2 by per- forming 1,000 replications instead of only 14.
Day Probability of a Breakdown
Corresponding Random Numbers
1, 2, 3 0 4 0.25 0.0000 to 0.2499 5 0.5 0.2500 to 0.7499 6 0.25 0.7500 to 0.9999 7 or more 0
TABLE 12.1 The Probability Distri- bution of Breakdowns for Heavy Duty’s Motors, and the Cor- responding Random Numbers
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Fortunately, the time required to overhaul a motor never exceeds three days, so a replace- ment motor always is ready when a breakdown occurs. When this happens, the remainder of the day (plus overtime if needed) is used to remove the failed motor and install the replace- ment motor, so the machine then is ready to begin operation again at the beginning of the next day. The average costs incurred during each replacement cycle (the time from when a replacement of a motor begins until just before another replacement is needed) are summa- rized below.
Cost of a Replacement Cycle that Begins with a Breakdown
Replace a motor $ 2,000 Lost production during replacement 5,000 Overhaul a motor 4,000 Total $11,000
Using Computer Simulation Computer simulation can be used to estimate what the average daily cost will be for replacing the motors as needed. This requires using random numbers to determine when breakdowns occur in the simulated process. Using the probabilities in the second column of Table 12.1 , 25 percent of the possible random numbers need to correspond to a breakdown on day 4, 50 percent to a breakdown on day 5, and the remaining 25 percent to a breakdown on day 6. The rightmost column of Table 12.1 shows the natural way of doing this.
Excel provides a convenient VLOOKUP function for implementing this correspondence between a random number and the associated event. Figure 12.4 illustrates how it works. One step is to create the table shown in columns I, J, and K, where columns K and I come directly from the first two columns of Table 12.1 . Column J gives the cumulative probability prior to the number of days in column K, so J8 5 I7 and J9 5 I7 1 I8. Cells J7:K9 then constitute the lookup table for the VLOOKUP function. The bottom of the figure displays how the VLOOKUP command has been entered into the column D cells. The first argument of this function indicates that the cell in the same row of RandomNumber (C5:C34) provides the random number being used. The second argument gives the range for the lookup table. The third argument (2) indicates that column 2 of the lookup table is providing the number being entered into this cell in column D. The choice of the number in column 2 of the lookup table is based on where the random number falls within the ranges between rows in column 1 of this table. In particular, the three possible choices are
if 0 # RAND() , 0.25 choose 4 days if 0.25 # RAND() , 0.75 choose 5 days if 0.75 # RAND() , 1 choose 6 days
which is precisely the correspondence indicated in Table 12.1 . By generating 30 simulated breakdowns in this way in column D of Figure 12.4 , columns
E, F, and G then show the resulting cumulative number of days, the estimated cost for each replacement cycle, and the cumulative cost for the corresponding replacement cycles. (In a more detailed computer simulation, random numbers also could be used to generate the exact costs with each simulated breakdown.) Since the total number of days in this simulation (cell E34) is 153 and the cumulative cost (cell G34) is $330,000, the average daily cost is calcu- lated in cell J34 as
Average cost per day 5 $330,000
153 5 $2,157
Comparisons with Example 1 Comparing this computer simulation with the ones run for the coin-flipping game reveal a couple of interesting differences. One is that the IF function was used to generate each simu- lated coin flip from a random number (see the equations entered into the column D cells in Figure 12.1 ), whereas the VLOOKUP function has just been used here to generate the
VLOOKUP(a, B, c) looks in the first column of B (where B is a range of cells) for the row containing the largest value that is less than or equal to a. It then returns the value in column c from that row of B.
The lookup table in cells J7:K9 provides this cor- respondence between a ran- dom number and the time until a breakdown occurs.
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1
2
3
4
5
6
7
8
9
10
11
12
13
14
30
31
32
33
34
A B C
Random
Number
0.7142
0.4546
0.3142
0.1722
0.0932
0.3645
0.1636
0.7572
0.3067
0.9520
0.8548
0.7464
0.9781
0.6584
0.8829
D E F G H I KJ
Breakdown
1
2
3
4
5
6
7
8
9
10
26
27
28
29
30
Time since Last
Breakdown
5
5
5
4
4
5
4
6
5
6
6
5
6
5
6
Cumulative
Day
5
10
15
19
23
28
32
38
43
49
131
136
142
147
153
Cumulative
Cost
$11,000
$22,000
$33,000
$44,000
$55,000
$66,000
$77,000
$88,000
$99,000
$110,000
$286,000
$297,000
$308,000
$319,000
$330,000
Distribution of
Time between BreakdownsCost
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
$11,000
Heavy Duty Company Corrective Maintenance Simulation
Average Cost per Day
$2,157
Cumulative
0
0.25
0.75
$11,000
Probability
0.25
0.5
0.25
Breakdown Cost
Number
of Days
4
5
6
3
4
5 =RAND()
=RAND()
=RAND()
:
:
=VLOOKUP(RandomNumber,$J$7:$K$9,2)
=VLOOKUP(RandomNumber,$J$7:$K$9,2)
=VLOOKUP(RandomNumber,$J$7:$K$9,2)
:
:
=TimeSinceLastBreakdown
=E5+TimeSinceLastBreakdown
=E6+TimeSinceLastBreakdown
:
:
=BreakdownCost
=BreakdownCost
=BreakdownCost
:
:
=Cost
=G5+Cost
=G6+Cost
:
:
6
7
8
9
C
Random
Number
D E F G
Time Since Last
Breakdown
Cumulative
Day
Cumulative
CostCost
AverageCostPerDay
Breakdown
BreakdownCost
Cost
CumulativeCost
CumulativeDay
RandomNumber
TimeSinceLastBreakdown
J34
B5:B34
J11
F5:F34
G5:G34
E5:E34
C5:C34
D5:D34
Range Name Cells
33 Average Cost per Day
34 =CumulativeCost/CumulativeDay
J
FIGURE 12.4 A spreadsheet model for a computer simulation of performing corrective maintenance on the Heavy Duty Co. motors.
simulated results. Actually, the VLOOKUP function could have been used instead for the coin flips, but the IF function was more convenient. Conversely, a nested IF function could have been used instead for the current example, but the VLOOKUP function was more con- venient. In general, we prefer using the IF function to generate a random observation from a probability distribution that has only two possible values, whereas we prefer the VLOOKUP function when the distribution has more than two possible values.
A second difference arose in the way the replications of the two computer simulations were recorded. For the coin-flipping game, simulating a single play of the game involved using the spreadsheet with 62 rows shown in Figure 12.1 . Therefore, to record many replica- tions, this same spreadsheet was used to generate the data table in Figure 12.2 , which summa- rized the results of each replication in a single row. For the current example, no separate data table was needed because each replication could be executed and displayed in a single row of the original spreadsheet in Figure 12.4 .
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However, one similarity between the two examples is that we purposely kept each one suf- ficiently simple that an analytical solution is available to compare with the simulation results. In fact, it is quite straightforward to obtain the analytical solution for the current version of the Heavy Duty Co. problem. Using the probabilities in Table 12.1 , the expected number of days until a breakdown occurs is
E(time until a breakdown) 5 0.25(4 days) 1 0.5(5 days) 1 0.25(6 days) 5 5 days
Therefore, the expected value (in the statistical sense) of the cost per day is
E(cost per day) 5 $11,000 5 days
5 $2,200 per day
The average cost of $2,157 per day obtained by computer simulation (cell J34 of Figure 12.4 ) is an estimate of this true expected value.
The fact that computer simulation actually was not needed to analyze this version of the Heavy Duty Co. problem illustrates a possible pitfall with this technique. Computer simula- tion is easy enough to use on small problems that there occasionally is a tendency to rush into using this technique when a bit of careful thought and analysis first could provide all the needed information more precisely (and perhaps more quickly) than computer simulation. In other cases, starting with a simple analytical model sometimes can provide important insights as a prelude to using computer simulation to refine the analysis with a more precise formula- tion of the problem.
Some Preventive Maintenance Options So far, we have assumed that the company will use a corrective maintenance policy. This means that the motor in the machine will be removed and overhauled only after it has broken down. However, many companies use a preventive maintenance policy instead. Such a policy in this case would involve scheduling the motor to be removed (and replaced) for an overhaul at a certain time even if a breakdown has not occurred. The goal is to provide maintenance early enough to prevent a breakdown. Scheduling the overhaul also enables removing and replacing the motor at a convenient time when the machine would not be in use otherwise, so that no production is lost. For example, by paying overtime wages for the removal and replacement, this work can be done after the normal workday ends so that the machine will be ready by the beginning of the next day. One possibility is to do this at the end of day 3, which would definitely be in time to prevent a breakdown. Other options are to do it at the end of day 4 or day 5 (if a breakdown has not yet occurred) in order to prevent disrupting production with a breakdown in the very near future. Computer simulation can be used to evaluate and compare each of these options (along with a corrective maintenance policy) when analytical solutions are not available.
Consider the option of removing (and replacing) the motor for an overhaul at the end of day 3. The average cost each time this is done happens to be the following.
Cost of a Replacement Cycle that Begins without a Breakdown
Replace a motor on overtime $3,000 Lost production during replacement 0 Overhaul a motor before a breakdown 3,000 Total $6,000
Since this total cost of $6,000 occurs every three days, the expected cost per day of this option would be
E(cost per day) 5 $6,000 3 days
5 $2,000 per day
Since this cost has been obtained analytically, computer simulation is not needed in this case.
The goal of preventive maintenance is to provide maintenance early enough to prevent a breakdown.
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498 Chapter Twelve Computer Simulation: Basic Concepts
Now consider the remaining two options of removing (and replacing) the motor after day 4 or after day 5 if a breakdown has not yet occurred. Since it is somewhat more difficult to find the expected cost per day analytically for these options, we now will use computer simulation. For either case, the average cost during a replacement cycle depends on whether the replace- ment began before or after a breakdown occurred. As outlined earlier, these average costs are
Cost of a replacement cycle that begins with a breakdown 5 $11,000
Cost of a replacement cycle that begins without a breakdown 5 $ 6,000
Figure 12.5 shows the use of computer simulation for the option of scheduling the replace- ment of each motor after four days. The times until 30 consecutive motors would have broken down without the replacements are obtained from column D (except rows 15–29 are hidden). The cases where this time is four days (indicating a breakdown during day 4) correspond to a motor breaking down before it is replaced. (This occurs in rows 6, 9, 13–14 and in five of the hidden rows.) The first cycle concludes with the replacement of the first motor after four days, as shown in row 5. Column G gives the cumulative number of days at the end of each cycle. Column F indicates whether each cycle ends with a breakdown or with a replacement that is soon enough to avoid a breakdown, and column H gives the resulting cost. Column I then cumulates these costs. Since the 30 cycles last 120 days (cell G34) and have a total cost of $225,000 (cell I34), this simulation yields
Average cost per day 5 $225,000
120 5 $1,875
as the estimate of the expected cost per day (which actually is $1,812 per day) for this option. Figure 12.6 shows the corresponding simulation for the option of scheduling the replace-
ment of each motor after five days. Thus, if the time until a breakdown would be on the sixth day (as indicated in column D), the replacement is made in time to avoid the breakdown (as indicated in column F). Since most of the times in column D are 4 or 5 instead, most of the cycles conclude with a breakdown. This leads to a much higher total cost for the 30 cycles of $300,000, along with a somewhat longer total time of 141 days. Therefore, the estimate of the expected cost per day for this option is
Average cost per day 5 $300,000
141 5 $2,128
(The true expected cost per day is $2,053.) Based on all the above results, the clear choice for the least expensive option is the one
that schedules the replacement of each motor after four days, since its estimated expected cost per day is only $1,875. Although this estimate based on the simulation in Figure 12.5 overestimates the true expected cost per day by $63, this option still is the least expensive one by a wide margin.
In practice, the simulation runs usually would be considerably longer than those shown in Figures 12.4 , 12.5 , and 12.6 in order to obtain more precise estimates of the true costs for the alternative options. The simulations typically also would include more details, such as when during a day a breakdown occurs and the resulting cost of lost production that day.
Both Examples 1 and 2 used random numbers to generate random observations from dis- crete probability distributions. Many computer simulations require generating random obser- vations from continuous distributions instead. We next describe a general method for doing this with either continuous or discrete distributions.
Generating Random Observations from a Probability Distribution Both Examples 1 and 2 demonstrated how to generate random observations from a discrete probability distribution. As illustrated in Example 2, Excel’s VLOOKUP function can be use- ful for doing this for any discrete distribution.
However, many computer simulations require generating random observations from continu- ous distributions instead. A general mathematical procedure called the inverse transformation method is available for generating random observations from either discrete or continuous dis- tributions. This procedure is described in the supplement to this chapter on the CD-ROM.
Longer and more detailed simulation runs are com- monly conducted.
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C onfirm
ing P ages
1 2 .1
T he E
ssence of C om
puter Sim ulation 4
9 9
1
2
3
4
5
6
7
8
9
10
11
12
13
14
30
31
32
33
34
A B C D E
Random
Number
0.7861
0.0679
0.9296
0.4430
0.1223
0.4530
0.3972
0.9289
0.2195
0.0706
0.8720
0.8902
0.3839
0.7404
0.7264
Time until
Breakdown
6
4
6
5
4
5
5
6
4
4
6
6
5
5
5
Scheduled Time
until Replacement
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
F G H I J K ML
Cycle
1
2
3
4
5
6
7
8
9
10
26
27
28
29
30
Event That
Concludes Cycle
Replacement
Breakdown
Replacement
Replacement
Breakdown
Replacement
Replacement
Replacement
Breakdown
Breakdown
Replacement
Replacement
Replacement
Replacement
Replacement
Cumulative
Day
4
8
12
16
20
24
28
32
36
40
104
108
112
116
120
Cumulative
Cost
$6,000
$17,000
$23,000
$29,000
$40,000
$46,000
$52,000
$58,000
$69,000
$80,000
$201,000
$207,000
$213,000
$219,000
$225,000
Distribution of Time between BreakdownsCost
$6,000
$11,000
$6,000
$6,000
$11,000
$6,000
$6,000
$6,000
$11,000
$11,000
$6,000
$6,000
$6,000
$6,000
$6,000
Heavy Duty Company Preventive Maintenance Simulation (Replace After 4 Days)
Average Cost per Day
$1,875
Cumulative
0
0.25
0.75
$11,000
$6,000
4
Probability
0.25
0.5
0.25
Breakdown Cost
Replacement Cost
Replace After
Number
of Days
4
5
6
days
3
4
5 =RAND()
=RAND()
=RAND()
:
:
=VLOOKUP(RandomNumber,$L$7:$M$9,2)
=VLOOKUP(RandomNumber,$L$7:$M$9,2)
=VLOOKUP(RandomNumber,$L$7:$M$9,2)
:
:
=ReplaceAfter
=ReplaceAfter
=ReplaceAfter
=IF(TimeUntilBreakdown<=ScheduledTimeUntilReplacement,"Breakdown","Replacement")
=IF(TimeUntilBreakdown<=ScheduledTimeUntilReplacement,"Breakdown","Replacement")
=IF(TimeUntilBreakdown<=ScheduledTimeUntilReplacement,"Breakdown","Replacement")
6
7
8 :
:
:
:9
C
Random
Number
D E F
Time until
Breakdown
Scheduled Time
until Replacement
Event That
Concludes Cycle
3
4
5 =MIN(D5,E5)
=G5+MIN(D6,E6)
=G6+MIN(D7,E7) :
:
=IF(Event="Breakdown",BreakdownCost,ReplacementCost)
=IF(Event="Breakdown",BreakdownCost,ReplacementCost)
=IF(Event="Breakdown",BreakdownCost,ReplacementCost)
=Cost
=I5+Cost
=I6+Cost
6
7
8 :
:
:
:9
G
Cumulative
Day
H I
Cost
Cumulative
Cost
AverageCostPerDay
BreakdownCost
Cost
CumulativeCost
CumulativeDay
Cycle
Event
RandomNumber
ReplaceAfter
ReplacementCost
ScheduledTimeUntilReplacement
TimeUntilBreakdown
L34
L11
H5:H34
I5:I34
G5:G34
B5:B34
F5:F34
C5:C34
L14
L12
E5:E34
D5:D34
Range Name Cells
33 Average Cost per Day
34 =CumulativeCost/CumulativeDay
L
FIGURE 12.5 A spreadsheet model for a computer simulation of performing preventive maintenance (replace after four days) on the Heavy Duty Co. motors.
h il2
4 0 6 4 _ ch
1 2 _ 4 8 8 -5
2 4 .in
d d 4
9 9
h il2
4 0 6 4 _ ch
1 2 _ 4 8 8 -5
2 4 .in
d d 4
9 9
1 0 /1
1 /1
2 5
:2 1 P
M 1 0 /1
1 /1
2 5
:2 1 P
M
C onfirm
ing P ages
5 0 0 C
h a p
te r T
w e lve
C om
puter Sim ulation: B
asic C oncepts
1
2
3
4
5
6
7
8
9
10
11
12
13
14
30
31
32
33
34
A B C D E
Random
Number
0.0558
0.0690
0.1889
0.9471
0.9173
0.3541
0.7035
0.0350
0.5755
0.8910
0.7386
0.2648
0.6239
0.9988
0.0061
Time until
Breakdown
4
4
4
6
6
5
5
4
5
6
5
5
5
6
4
Scheduled Time
until Replacement
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
F G H I J K ML
Cycle
1
2
3
4
5
6
7
8
9
10
26
27
28
29
30
Event That
Concludes Cycle
Breakdown
Breakdown
Breakdown
Replacement
Replacement
Breakdown
Breakdown
Breakdown
Breakdown
Replacement
Breakdown
Breakdown
Breakdown
Replacement
Breakdown
Cumulative
Day
4
8
12
17
22
27
32
36
41
46
122
127
132
137
141
Cumulative
Cost
$11,000
$22,000
$33,000
$39,000
$45,000
$56,000
$67,000
$78,000
$89,000
$95,000
$261,000
$272,000
$283,000
$289,000
$300,000
Distribution of Time between BreakdownsCost
$11,000
$11,000
$11,000
$6,000
$6,000
$11,000
$11,000
$11,000
$11,000
$6,000
$11,000
$11,000
$11,000
$6,000
$11,000
Heavy Duty Company Preventive Maintenance Simulation (Replace After 5 Days)
Average Cost per Day
$2,128
Cumulative
0
0.25
0.75
$11,000
$6,000
5
Probability
0.25
0.5
0.25
Breakdown Cost
Replacement Cost
Replace after
Number
of Days
4
5
6
days
FIGURE 12.6 A revision of Figure 12.5 to schedule the replacement of the motors after five days instead of four.
h il2
4 0 6 4 _ ch
1 2 _ 4 8 8 -5
2 4 .in
d d 5
0 0
h il2
4 0 6 4 _ ch
1 2 _ 4 8 8 -5
2 4 .in
d d 5
0 0
1 0 /1
1 /1
2 5
:2 1 P
M 1 0 /1
1 /1
2 5
:2 1 P
M
Confirming Pages
12.2 A Case Study: Herr Cutter’s Barber Shop (Revisited) 501
However, for relatively complicated continuous distributions, even the inverse transforma- tion method becomes difficult to apply. One example is the normal distribution. This distri- bution is such an important one that more convenient special methods have been developed to generate random observations from this distribution. In particular, Excel uses the function
NORMINV(RAND(), m, s)
to do this after you substitute the numerical values for the mean m and standard deviation s of the distribution. The next section describes how to generate random observations from two other important continuous distributions—the uniform distribution and the exponential distribution.
Several Excel add-ins have been developed to extend the simulation capabilities of the stan- dard Excel package, including providing special functions to immediately generate random observations from a wide variety of probability distributions. For example, the next chapter will demonstrate how Risk Solver Platform for Education (RSPE) fully includes these capabilities in addition to all the other functionalities described in the current chapter. (Another popular add-in with these capabilities, Crystal Ball, will be used in Chapter 20 on the CD-ROM.)
This is a very handy Excel function to generate a ran- dom observation from a normal distribution.
1. How does computer simulation imitate the operation of a stochastic system? 2. Why does computer simulation tend to be a relatively expensive procedure? 3. When is computer simulation typically used despite being relatively expensive? 4. What is a random number? For what purpose is it used?
Review Questions
12.2 A CASE STUDY: HERR CUTTER’S BARBER SHOP (REVISITED)
In case you have already studied the preceding chapter, you hopefully recall the brief descrip- tion in Section 11.1 of Herr Cutter’s barber shop as an example of a basic kind of queueing system. (A queueing system is a place where customers receive some kind of service from a server, perhaps after waiting in a line called a queue.) If you have not studied the preceding chapter, rest easy. The current chapter does not assume any prior knowledge of Chapter 11 and so will point out the few bits of needed information about queueing systems. Otherwise, you do not need to know anything about the contents of Chapter 11 except for the following facts: Herr Cutter is a German barber who runs a one-person barber shop. He opens his shop at 8:00 am each weekday morning. His customers arrive randomly at an average rate of two customers per hour. He requires an average of 20 minutes for each haircut. (You do not need to refer to Section 11.1 for any other details.)
The case study concerns the problem described below.
The Decision Facing Herr Cutter Herr Cutter has run his barber shop in the same location for nearly 25 years. Although his parents had wanted him to follow in his father’s footsteps as a medical doctor, he has never regretted his decision to follow this more modest career path. He enjoys the relaxed working environment, the regular hours, and the opportunity to visit with his customers.
Over the years, he has built up a loyal clientele. He is a fine barber who takes pride in his work. As his business has increased, his customers now often need to wait a while (sometimes over half an hour) to begin a haircut. However, his long-time customers are willing to do so.
The shop is in a growing city. As the pace of life has increased, Herr Cutter has noticed that new customers are much less likely to return than in the early years, especially if they have had to wait very long. He attributes this to a decreasing tolerance for waiting. However, since he is not gaining many new regular customers, his volume of business has leveled off at a steady average of two customers per hour.
As he has grown older, Herr Cutter has wondered increasingly about whether he should add an associate to share the workload. He also would enjoy the company, as well as the addi- tional flexibility. A second barber should reduce the waiting times of the customers consider- ably, so an additional benefit would be that the total volume of business for the shop should increase somewhat.
However, what has always held him back from adding an associate is the fear of decreas- ing his personal income from the business. He needs to be putting away considerable money
The decision facing Herr Cutter is whether to add an associate to share the work- load in the barber shop.
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toward retirement and really can’t afford a significant decrease in his already modest income. Given the salary and commission he would need to pay an associate, business would need to almost double just to maintain his current level of income. (We will spell out the financial details in the next section when the analysis takes place.) He is doubtful that business would increase nearly this much.
But now opportunity has come knocking on the door. A fellow barber (and friend) in the city has decided to retire and close his shop. This friend has had the same associate for sev- eral years, and he now has invited Herr Cutter to hire this fine young man. The friend highly recommends him, and also points out that the associate would bring considerable business with him.
So now Herr Cutter is in a quandary as to whether he should take the plunge in hiring this associate.
Fortunately, help is at hand for making this decision. This friend has shown Herr Cutter an interesting recent article in The Barber’s Journal. The article describes a study that has been done of barber shops and how long customers now are willing to wait for haircuts to begin. The article concludes with two rules of thumb.
First Rule of Thumb: In a well-run barber shop with a long-established clientele, these loyal cus- tomers are willing to tolerate an average waiting time of about 20 minutes until the haircut begins.
Herr Cutter feels that this description fits his situation. He has never tried to estimate the waiting times of his customers, but guesses that an average of 20 minutes sounds about right.
Second Rule of Thumb: In a well-run barber shop, new customers are willing to tolerate an average waiting time of about 10 minutes before the haircut begins. (With longer waits, they tend to take their business elsewhere in the future.)
Again, Herr Cutter feels that this rule of thumb agrees with his own experience. This second rule of thumb has given Herr Cutter a good idea about how to view his deci-
sion. With his current clientele, adding an associate probably would reduce their average waiting time to less than 10 minutes. This prompt service then should help to gradually attract and retain new customers (including some of the associate’s customers from the barber shop that is closing). According to the rule of thumb, the level of business should increase until it reaches the point where the average waiting time before the haircut begins has increased to about 10 minutes. Estimating the level of business at that point would indicate the new level of income to the shop and his share of that income. The dilemma is that he does not see how to estimate this level of business in advance.
Herr Cutter asks for advice from his nephew Fritz (a university student majoring in busi- ness) about how to resolve this dilemma. Fritz excitedly responds that he thinks he knows just the right approach to use. Computer simulation.
Fritz recently took a course in management science. In fact, he has a copy of MS Course- ware, including its Queueing Simulator for simulating queueing systems like his uncle’s barber shop. Although not as sophisticated as expensive commercial software packages for performing computer simulations, Fritz explains to his uncle how this routine can indeed provide a good estimate in advance of what the level of business would be with an associate.
Fritz proposes spending a little time with his uncle to gather some data and develop a simu- lation model in preparation for performing the computer simulations. His first simulation will be of the barber shop under its current mode of operation (without an associate) to estimate the current average waiting time. Comparing the results from this simulation with what is actu- ally happening in the barber shop also will help to test the validity of the simulation model. If necessary, the model will be adjusted to better represent the real system. The subsequent simulations will be run of the barber shop with an associate. These simulations will assume that the associate’s speed in giving a haircut is the same as Herr Cutter’s. Different means of the interarrival-time distribution will be tried to determine which mean (i.e., which level of business) would lead to an average waiting time of 10 minutes before the haircut begins.
Fritz asks his uncle if he should proceed with this plan. Herr Cutter urges him to do so. The remainder of this section describes the execution of this plan, including the mechan-
ics of how these simulations are performed. The next section then presents the results of the actual computer simulations and the analysis of what Herr Cutter should do.
With an associate, the level of business should reach the point where the aver- age waiting time to begin a haircut is about 10 minutes.
Here is the plan for using computer simulation to esti- mate what the new level of business would be with an associate.
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12.2 A Case Study: Herr Cutter’s Barber Shop (Revisited) 503
Gathering Data The key events for this barber shop are service (haircut) completions and customer arrivals. For example, Table 11.1 (in Section 11.1) records the times at which these events occurred over a typical early morning. Figure 11.2 displays these data in a different form by plotting the num- ber of customers in the shop (a basic measure of performance) over this same early morning.
By observing the barber shop over an extended period of time, extensive data of the same kind could be gathered to estimate various measures of performance for the barber shop under its current mode of operation. However, it is not necessary to spend months or years gather- ing such data. Once it has been set up on a computer, computer simulation can accomplish the same thing in a matter of seconds by simulating the operation of the barber shop over a lengthy period (even years if desired). However, performing this simulation does require gathering a bit of other data first.
In particular, it is necessary to estimate the probability distributions involving the random events (service completions and customer arrivals) in the system. These probability distribu- tions are the distribution of service times (the times required to give a haircut) and the distri- bution of interarrival times (the times between consecutive arrivals).
Herr Cutter has found that the time required to give a haircut varies between 15 and 25 minutes, depending on the customer’s amount of hair, the desired hair style, and so forth. Furthermore, his best estimate is that the times between 15 and 25 minutes are equally likely, which indicates the following distribution.
Estimated distribution of service times: The uniform distribution over the interval from 15 minutes to 25 minutes.
Since the barber shop has random arrivals of customers, Section 11.1 points out that the distribution of interarrival times must be an exponential distribution.
Estimated distribution of interarrival times: An exponential distribution (described in Sec- tion 11.1) with a mean of 30 minutes.
Generating Random Observations from These Probability Distributions A computer simulation of the operation of the barber shop requires generating a series of random observations from the distributions identified above. As usual, random numbers will be used to do this. However, since these probability distributions are continuous distribu- tions, it is not very convenient to use random numbers in the ways described for the discrete distributions in Examples 1 and 2 (which employed Excel’s IF and VLOOKUP functions, respectively).
Fortunately, it is relatively straightforward to generate a random observation from any uniform distribution, including this one over the interval from 15 minutes to 25 minutes. The key is that a random number between 0 and 1 is, in fact, a random observation from a uniform distribution between 0 and 1. Thus, the Excel function RAND() generates such a random observation. Similarly, 10 RAND() generates a random observation from a uniform distribu- tion between 0 and 10. Adding 15 to such a random observation thereby provides a random observation from a uniform distribution between 15 and 25.
Therefore, when developing a simulation model for the operation of Herr Cutter’s barber shop, the equation to be entered into each cell receiving a random observation from this uni- form distribution is
5 15 1 10 * RAND()
For a uniform distribution with lower and upper bounds different from 15 and 25, the lower bound would be substituted for 15 in this equation and the difference between the two bounds would be substituted for 10. (This is a simple example of how the inverse transformation method is used to generate a random observation from a probability distribution.)
Although the exponential distribution is more complicated than the uniform distribution, an Excel equation also is available for generating a random observation from an exponen- tial distribution with a mean of 30 minutes (the distribution of interarrival times for Herr Cutter’s barber shop). The supplement to this chapter on the CD-ROM describes how the
Estimates are needed of the probability distributions of both the time required to give a haircut (the “ser- vice time”) and the time between consecutive arriv- als of customers (the “inter- arrival time”).
The Excel equation for a random observation from a uniform distribu- tion over the interval from a to b is 5 a 1 (b 2 a) RAND().
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inverse transformation method is used to derive this equation by using the Excel function LN(), which calculates the natural logarithm of whatever quantity is inside the parentheses.
In particular, for each cell in a spreadsheet model that receives a random observation from this exponential distribution, the Excel equation to be entered into this cell is
5 230 * LN(RAND())
For an exponential distribution with a different mean, this mean would be substituted for 30 in this equation.
RSPE also includes functions to generate random values from any of 46 different probabil- ity distributions. These formulas can be entered into a cell by choosing the distribution from the Distributions menu on the RSPE ribbon, which then brings up a dialog box in which the parameters of the distribution can be specified. This results in a function being entered into the cell that calculates a random value from the specified distribution. (The formulas for the two distributions discussed here would be 5 PsiUniform(15, 25) and 5 PsiExponential(30)). RSPE will be discussed in much more detail in the next chapter. The examples in the current chapter are simple enough that they can be done without RSPE.
The Building Blocks of a Simulation Model for a Stochastic System Recall that the beginning of Section 12.1 defines “stochastic system” as a system that evolves over time according to one or more probability distributions. Herr Cutter’s barber shop is a stochastic system because it evolves over time according to the two probability distributions discussed above.
With its multiple probability distributions, the Herr Cutter case study has some of the com- plications that are typical of the stochastic systems for which many computer simulations are performed. When preparing for a relatively complex simulation of this type, it is sometimes helpful to develop a formal simulation model.
A simulation model is a representation of the system to be simulated that also describes how the simulation will be performed.
Here are the basic building blocks of a typical simulation model for a stochastic system.
1. A description of the components of the system, including how they are assumed to operate and interrelate.
2. A simulation clock. 3. A definition of the state of the system.
After replacing 30 by the mean, this is the Excel equation for generating a random observation from any exponential distribution.
The U.S. Federal Aviation Administration (FAA) is charged with managing air traffic in the national airspace. Air traffic controllers are used to guide individual flights to keep them safely separated from every other flight. In addition, the FAA controls aggregate flows of flights to keep arrivals at each airport within manageable levels and to adjust to adverse weather conditions by rerouting traffic as needed. When bad weather or congestions occurs, traffic managers are used to decide which flights should be held on the ground and which flights already airborne should be rerouted.
A particularly difficult problem for traffic managers arises when extended lines of thunderstorms block major flight routes. Such severe weather across a wide area can result in enormous, systemwide disruptions, leading to bil- lions of dollars annually in increased operating costs and revenue loss to airlines as well as great inconvenience for the flying public. Therefore, in 2005, the FAA commis- sioned a year-long simulation study by a management science team to develop better operating procedures for traffic managers in this situation.
The resulting simulation model was a very complex one that incorporated the actions and interactions of hundreds or thousands of flights that were being controlled by the FAA infrastructure. For many months, this model was used to test various proposed operating procedures under typi- cal severe weather conditions to determine the best of these procedures. These conclusions then were incorpo- rated into a computerized decision-support system that traffic managers would use thereafter to guide their deci- sions under such weather conditions.
This innovation has been estimated to save aircraft oper- ators $1 billion to $3 billion in operating costs by reducing the delays and cancellations over the first decade of use. It also is estimated to reduce passenger delays by more than a million hours per year.
Source: V. P. Sud, M. Tanino, J. Wetherly, M. Brennan, M. Lehky, K. Howard, and R. Oiesen, “Reducing Flight Delays through Bet- ter Traffic Management,” Interfaces 39, no. 1 (January–February 2009), pp. 35–45. (A link to this article is provided on our website, www.mhhe.com/hillier5e .)
An Application Vignette
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12.2 A Case Study: Herr Cutter’s Barber Shop (Revisited) 505
4. A method for randomly generating the (simulated) events that occur over time. 5. A method for changing the state of the system when an event occurs. 6. A procedure for advancing the time on the simulation clock.
We will use the case study to illustrate each of these building blocks. As first described in Section 11.1, Herr Cutter’s barber shop is a basic kind of stochas-
tic system, namely, a single-server queueing system. The components of this system are the arriving customers, the customers in the shop awaiting service (a haircut), and Herr Cutter as the server. The assumed distributions of service times and interarrival times were described earlier in this section.
Once a computer simulation is under way, it is necessary to keep track of the passage of time in the system being simulated. Starting at time 0, let
t 5 Amount of simulated time that has elapsed so far
The variable t in the computer program is referred to as the simulation clock . The program continually updates the current value of this variable as the simulation proceeds. With today’s powerful computers, simulated time typically proceeds millions of times faster than running time on the computer.
For Herr Cutter’s barber shop, the simulation clock records the amount of simulated time (in minutes) that has elapsed so far since the shop opened at 8:00 am A precise simulation then would start anew for each successive day of simulated operation of the shop. (The next section describes a simplifying assumption that Fritz makes at this point.) A small amount of running time can simulate years of operation.
The key information that defines the current status of the system is called the state of the system . For Herr Cutter’s barber shop (the system in this case), the state of the system is
N(t) 5 Number of customers in the barber shop at time t
The computer program for the simulation typically records the cumulative amount of time that the system spends in each state, as well as other measures of performance (e.g., the wait- ing times of the customers).
For Herr Cutter’s barber shop, the key events are the arrivals of customers and the comple- tions of service (haircuts). The preceding subsection describes how these events are randomly generated in a computer simulation by generating random observations from the distributions of interarrival times and service times.
Both of these types of events change the state of the system. The method used to adjust the state accordingly is to
Reset N(t) 5 eN(t) 1 1 if an arrival occurs at time t N(t) 2 1 If a service completion occur at time t
The main procedure for advancing the time on the simulation clock is called next-event time advance . Here is how it works.
The Next-Event Time-Advance Procedure 1. Observe the current time t on the simulation clock and the randomly generated times of the
next occurrence of each event type that can occur next. Determine which event will occur first. 2. Advance the time on the simulation clock to the time of this next event. 3. Update the system by determining its new state as a result of this event and by randomly
generating the time until the next occurrence of any event type that can occur from this state (if not previously generated). Also record desired information about the performance of the system. Then return to step 1.
This process continues until the computer simulation has run as long as desired.
Illustrating the Computer Simulation Process The Excel spreadsheet in Figure 12.7 shows a computer simulation of the operation of Herr Cutter’s barber shop (without an associate) over a period when 100 customers arrive. The
Whenever a customer arrives or a haircut is completed, the computer program adjusts the number of customers in the barber shop in this way.
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506 Chapter Twelve Computer Simulation: Basic Concepts
1
2
3
4 Mean Interarrival Time
Min Service Time
Max Service Time
Average Time in Line (Wq) Average Time in System (W)
minutes
minutes
minutes
minutes
minutes
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
111
112
113
114
115
A B C
Interarrival
Time
0.5
45.6
2.6
1.5
27.6
29.2
18.6
21.8
7.4
12.1
48.8
12.3
82.7
4.0
85.2
D E F G H I
Customer
Arrival
1
2
3
4
5
6
7
8
9
10
96
97
98
99
100
Time
of
Arrival
0.5
46.0
48.6
50.1
77.7
106.9
125.5
147.4
154.8
166.9
3,132.8
3,145.1
3,227.7
3,231.8
3,317.0
Time
Service
Begins
0.5
46.0
67.9
88.8
107.7
124.9
141.2
165.8
188.6
212.9
3,132.8
3,155.0
3,227.7
3,246.7
3,317.0
Time
Service
Ends
22.3
67.9
88.8
107.7
124.9
141.2
165.8
188.6
212.9
231.3
3,155.0
3,180.0
3,246.7
3,266.8
3,336.5
Time
in
Line
0.0
0.0
19.4
38.7
30.0
18.0
15.7
18.5
33.8
46.0
0.0
9.9
0.0
14.9
0.0
Time
in
System
21.9
21.9
40.2
57.6
47.2
34.3
40.3
41.2
58.1
64.4
22.2
34.9
19.0
35.0
19.5
Service
Time
21.9
21.9
20.8
18.9
17.2
16.3
24.6
22.7
24.3
18.4
22.2
25.0
19.0
20.1
19.5
Herr Cutter's Barber Shop
13
14
15
=-MeanInterarrivalTime*LN(RAND())
=-MeanInterarrivalTime*LN(RAND())
=-MeanInterarrivalTime*LN(RAND())
:
:
=MinServiceTime+(MaxServiceTime-MinServiceTime)*RAND()
=MinServiceTime+(MaxServiceTime-MinServiceTime)*RAND()
=MinServiceTime+(MaxServiceTime-MinServiceTime)*RAND()
:
:
=InterarrivalTime
=D16+InterarrivalTime
=D17+InterarrivalTime
:
:
=TimeOfArrival
=MAX(G16,D17)
=MAX(G17,D18)
:
:
16
17
18
19
20
C
Interarrival
Time
ED F
Time
of
Arrival
Time
Service
Begins
Service
Time
13
14
15
=TimeServiceBegins+ServiceTime
=TimeServiceBegins+ServiceTime
=TimeServiceBegins+ServiceTime
:
:
=TimeServiceBegins-TimeOfArrival
=TimeServiceBegins-TimeOfArrival
=TimeServiceBegins-TimeOfArrival
:
:
=TimeServiceEnds-TimeOfArrival
=TimeServiceEnds-TimeOfArrival
=TimeServiceEnds-TimeOfArrival
:
:
16
17
18
19
20
G
Time
Service
Ends
IH
Time
in
Line
Time
in
System
10
11
C
Average Time in Line (Wq) Average Time in System (W)
D
=AVERAGE(TimeInLine)
=AVERAGE(TimeInSystem)
AverageTimeInLine
AverageTimeInSystem
InterarrivalTime
MaxServiceTime
MeanInterarrivalTime
MinServiceTime
ServiceTime
TimeInLine
TimeInSystem
TimeOfArrival
TimeServiceBegins
TimeServiceEnds
D10
D11
C16:C115
D8
D4
D7
F16:F115
H16:H115
I16:I115
D16:D115
E16:E115
G16:G115
Range Name Cells
(exponential)
30
(uniform)
15
25
12.8
33.0
FIGURE 12.7 A computer simulation of Herr Cutter’s barber shop (as currently operated) over a period of 100 customer arrivals.
pertinent data regarding each customer is recorded on a single row of the spreadsheet (where the rows for customers 11–95 are hidden). All the times are in minutes. As indicated by the equations at the bottom of the figure, the formulas given earlier are being used to generate ran- dom observations for the interarrival times and service times in columns C and F. These two
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12.2 A Case Study: Herr Cutter’s Barber Shop (Revisited) 507
times then enable calculating the other pertinent times for each customer in order. Column H records the waiting time before the haircut begins for each customer and column I gives the total waiting time in the barber shop (including the haircut) for the customer.
The next-event time-advance procedure is used to carry out this simulation. The procedure focuses on the two key types of events—arrivals and service completions—being recorded in columns D and G, and then moves chronologically through these events. To start, t 5 0 and N ( t ) 5 0 (no customers are in the shop at the instant it opens). Since no service comple- tions can occur without any customers there, the only type of event that can occur next is a customer arrival, so the time on the simulation clock is advanced next to t 5 0.5 minute (cell D16), the time when the first customer arrives. Subsequently, the clock is moved ahead to t 5 22.3 minutes (cell G16 indicates a service completion for this customer), then to t 5 46.0 (arrival of customer 2 according to cell D17), then to t 5 48.6 and then to t 5 50.1 (arrival of customers 3 and 4 according to cells D18 and D19, respectively), then to t 5 67.9 (service completion for customer 2 according to cell G17), and so forth.
Figure 12.8 shows the evolution of the state of this system (the number of customers in the barber shop) throughout the first 100 minutes of simulated operation. Thus, the number of customers in the barber shop fluctuates between 0 and 3 during this period.
Estimating Measures of Performance The purpose of performing a computer simulation of a system is to estimate the measures of performance of the system. The most important measure of performance for Herr Cutter’s barber shop (the system of interest here) is the expected waiting time of his customers before beginning a haircut. By averaging the waiting times in column H, cell D10 in Figure 12.7 provides an estimate of 12.8 minutes for this quantity. Similarly, cell D11 averages the times in column I to give an estimate of 33.0 minutes for the expected total waiting time in the shop, including the haircut.
Various other measures of performance also could be estimated from this simulation. For example, the probability that a customer has to wait more than 20 minutes to begin a haircut is estimated by the fraction of the customers with a time greater than 20 in column H. An estimate of the expected number of customers in the system (including receiving a haircut) is obtained by summing the numbers in column I and dividing by the total simulated time. The expected number of customers waiting to begin a haircut is estimated from column H in the same way.
The probability distribution of the number of customers in the system also might be of inter- est. As suggested by Figure 12.8 , the probability of any particular number of customers in the system can be estimated by the fraction of time that the simulated system spends in that state.
The computer simulation displayed in Figure 12.7 is a rather short one, so it only provides fairly rough estimates of the desired measures of performance. To obtain relatively precise estimates, a computer simulation might run for some years of simulated operation (as we will demonstrate in the next section).
The next-event time- advance procedure moves chronologically through the arrivals and service comple- tions as they occur.
The numbers in columns H and I provide estimates of various other measures of performance as well.
0
1
2
3
Number of
customers
in the system
20 40 60
Elapsed time (in minutes)
80 100
FIGURE 12.8 This graph shows the evolution of the number of customers in Herr Cut- ter’s barber shop over the first 100 minutes of the computer simulation in Figure 12.7.
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508 Chapter Twelve Computer Simulation: Basic Concepts
Simulating the Barber Shop with an Associate Figures 12.7 and 12.8 have illustrated the simulation of the barber shop under its current mode of operation ( without an associate). In most respects, the procedure for the shop with an associate is the same. Each time an arrival occurs (or the shop opens), the next interarrival time needs to be randomly generated. Similarly, each time a customer enters service (begins a haircut), this service time needs to be randomly generated.
The only difference comes when the next-event time-advance procedure is determining which event occurs next. Instead of just two possibilities for this next event, there now are the following three:
1. A departure because Herr Cutter completes a haircut. 2. A departure because the associate completes a haircut. 3. An arrival.
However, other than needing to separately keep track of the time until the next departure of each of these two kinds, the simulation proceeds in basically the same way.
The next section presents the results of several lengthy computer simulations of the barber shop, both with and without an associate.
When adding a second server, the simulation procedure remains the same except for needing to also keep track of service completions by the second server.
1. What is the decision facing Herr Cutter? 2. What are the two rules of thumb that will help guide this decision? 3. Which probability distributions need to be estimated in order to apply computer simulation to
this case study? 4. What is a simulation clock? 5. What is the name of the main procedure used to advance the time on the simulation clock? 6. What is the state of the system for Herr Cutter’s barber shop? 7. What is the basic difference in the procedure between simulating Herr Cutter’s barber shop
without an associate and with an associate?
Review Questions
12.3 ANALYSIS OF THE CASE STUDY
Recall that the decision facing Herr Cutter is whether to add an associate to work with him as a second barber in his shop. The basic issue is whether he would still be able to at least main- tain his current income level if he were to add the associate.
The Financial Factors Here are the main financial factors (converted from German currency to American dollars) for addressing this decision.
Revenue 5 $15 per haircut
Average tip 5 $2 per haircut
Cost of maintaining the shop 5 $50 per working day (with or without an associate)
Salary of an associate 5 $120 per working day
Commission for an associate 5 $5 per haircut given by the associate
In addition to his salary and commission, the associate would keep his own tips. Otherwise, the revenue would go to Herr Cutter.
The shop opens at 8:00 am and closes its door to new customers at 5:00 pm, so it admits customers for nine hours. Herr Cutter and any associate eat their sack lunches and take other breaks only during times when no customers are waiting. Thus, any customer who wants to enter the shop at any time during the nine hours is welcomed by a barber on duty.
Analysis of Continuing without an Associate As indicated in the preceding section, the current distribution of interarrival times has a mean of 30 minutes. Thus, Herr Cutter is averaging two customers per hour, or an average of
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12.3 Analysis of the Case Study 509
18 customers per working day. Therefore, after subtracting the cost of maintaining the shop, his average net income per working day is
Net daily income 5 ($15 1 $2)(18 customers) 2 $50
5 $306 2 $50
5 $256
Herr Cutter’s nephew Fritz is helping his uncle analyze his decision by using the Queue- ing Simulator in your MS Courseware to run computer simulations of the barber shop. This routine is specifically designed to efficiently run long simulations for a variety of queueing systems. It operates basically as illustrated in Figure 12.7 , but with more flexibility as to the type of system and with far more output, as outlined below.
Features of the Queueing Simulator 1. Can run computer simulations of various kinds of basic queueing systems described in
Section 11.1. 2. Can have any number of servers up to a maximum of 25. 3. Can use any of the following probability distributions for either interarrival times or ser-
vice times: a. Constant time (also called the degenerate distribution). b. Exponential distribution (described in Section 11.1). c. Translated exponential distribution (the sum of a constant time and a time from an
exponential distribution). d. Uniform distribution. e. Erlang distribution (described in the supplement to Chapter 11).
4. Provides estimates of various key measures of performance described in Section 11.3 for queueing systems, namely,
L 5 Expected number of customers in the system, including those being served
L q 5 Expected number of customers in the queue, which exclude customers being served
W 5 Expected waiting time in the system (includes service time) for an individual customer
W q 5 Expected waiting time in the queue (excludes service time) for an individual customer
P n 5 Probability of exactly n customers in the system (for n 5 0, 1, 2, . . . , 10)
(If you have not previously studied Chapter 11 to learn about queueing systems, you might find it helpful to see the live demonstration of a queueing system in action that is provided by the Waiting Line module in your Interactive Management Science Modules at www. mhhe.com/hillier5e . An offline version also is included in your MS Courseware on the CD-ROM.)
Largely to help test the validity of his simulation model (described in the preceding sec- tion), Fritz is beginning by simulating the current operation of the shop. Although Figure 12.7 already did this for roughly a week of simulated operation (100 customer arrivals), he now wishes to simulate several years of operation (100,000 arrivals).
Figure 12.9 shows the output that Fritz obtains from this computer simulation. If you wish, you can duplicate this simulation run by using the Queueing Simulator yourself. You should obtain very similar results, although they will be slightly different because different random numbers are used each time.
The measures of performance in column E are the ones defined above. Column F gives the point estimate , the single number that is the best estimate of the measure from this simula- tion run. Using statistical theory, columns G and H then provide a 95 percent confidence interval for each measure. Thus, there is a 95 percent chance that the true value of the mea- sure lies within this interval. Because the simulation run was so long (100,000 arrivals), each of these confidence intervals is quite narrow.
The Queueing Simulator is available in one of this chapter’s Excel files.
A confidence interval is an interval within which the true value of a measure of performance is likely to lie.
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510 Chapter Twelve Computer Simulation: Basic Concepts
Testing the Validity of the Simulation Model When starting a management science study that will use computer simulation, it is a good idea to first run the simulation model on a simple version of the system for which analytical results are available (if such a version exists). Comparing the results from this simulation run with the analytical results then provides a good test of the validity of the simulation model.
Fritz recalls that the M / G /1 queueing model presented in Section 11.5 provides some exact analytical results for the same queueing system that has been assumed for the simulation run in Figure 12.9 . This queueing model uses four parameters:
l 5 Mean arrival rate 5 ⅓0 customer per minute
m 5 Mean service rate 5 ½0 customer per minute
r 5 l
m 5
1/30 1/20
5 1 3
s 5 Standard deviation of the distribution of service times
Because the standard deviation of the uniform distribution from 0 to 1 is 1/"12, the standard deviation of the service-time distribution (the uniform distribution between 15 and 25) is
s 5 10
"12 5 2.887 After entering these values of l , 1/ m , and s , the Excel template for the M / G /1 model in the
Chapter 11 portion of your MS Courseware yields the results shown in Figure 12.10 . Note how each of these exact results for the measures of performance fall well within the corre- sponding 95 percent confidence interval in Figure 12.9 . This provides some reassurance that the simulation model and the computer simulation are operating as intended.
To further test the validity of the simulation model, Fritz shows the results in column F of Figure 12.9 to Herr Cutter and asks whether these numbers seem consistent with what he has been experiencing in the barber shop. Although Herr Cutter has not been keeping such data, his impression is that the numbers seem about right. He also points out that the average waiting time of about 20 minutes before beginning a haircut is consistent with the first rule of thumb in the article in The Barber’s Journal (described at the beginning of the preceding section).
Since a queueing model is available for the single- server version of the system, Fritz will use its results to test the validity of his simulation model.
To make sure that no big mistake has been made when constructing a simulation model, its results should be checked for rea- sonableness by someone familiar with the system being simulated.
1
A B C
Number of Servers
Distribution
Mean
Distribution
Minimum
Maximum
Number of Arrivals
Interarrival Times
Service Times
Length of Simulation Run
L =
Lq =
W =
Wq =
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
Point
Estimate
1.358
0.689
40.582
20.577
0.330
0.310
0.183
0.0942
0.0451
0.0206
0.00950
0.00432
0.00219
0.000876
0.000372
95% Confidence Interval
D E F G H
2
3
4 Low High
1.332
0.666
39.983
19.980
0.326
0.307
0.180
0.0920
0.0433
0.0192
0.00849
0.00360
0.00163
0.000540
0.000165
1.385
0.712
41.180
21.174
0.335
0.313
0.185
0.0963
0.0469
0.0220
0.0105
0.00503
0.00274
0.00121
0.000579
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Queueing Simulator for Herr Cutter's Barber Shop
Run Simulation
1
Exponential
30
Uniform
15
25
100,000
FIGURE 12.9 The output obtained by using the Queueing Simulator in one of this chapter’s Excel files to perform a computer simu- lation of Herr Cutter’s barber shop (without an associate) over a period of 100,000 customer arrivals.
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12.3 Analysis of the Case Study 511
Unfortunately, no queueing model yielding useful analytical results is available for the two-server queueing system that corresponds to Herr Cutter’s barber shop with an associate. (None of the multiple-server queueing models presented in Chapter 11 allows a service-time distribution even close to the one in this barber shop.) Therefore, even though it was not actu- ally needed for the option of Herr Cutter continuing without an associate, it will be necessary to use computer simulation to obtain good estimates of how the barber shop would perform with an associate. However, after the above testing of the validity of his simulation model, Fritz now is confident that this model will indeed provide good estimates.
Fritz does recognize that his simulation model (just like the M / G /1 queueing model) makes two simplifying assumptions that are only approximations of how the barber shop actually operates. (These assumptions are incorporated into the Queueing Simulator.)
Simplifying Assumptions 1. The system (barber shop) has an infinite queue, so arriving customers always enter the sys-
tem regardless of how many customers already are there. (In reality, Herr Cutter has found that arriving customers normally do not stay if three customers already are there waiting to begin a haircut, so he now only provides three chairs for waiting customers.)
2. Once started, the system operates continually without ever closing and reopening. (In real- ity, the barber shop closes its door at 5:00 pm each working day and reopens at 8:00 am the next day.)
To evaluate the effect of the first assumption, Fritz notes that the results in Figure 12.9 estimate that
P0 1 P1 1 P2 1 P3 1 P4 5 0.330 1 0.310 1 0.183 1 0.094 1 0.045
5 0.962
Thus, the simulation run exceeds the actual maximum of four customers in the barber shop (one receiving a haircut and three waiting to begin) less than 4 percent of the time. The effect of exceeding the actual maximum so infrequently is to slightly inflate the estimates of
1
2
3
4
5
6
7
8
9
10
11
12
A B C D E F G
Data
Analytical M/G/1 Queueing Results for Herr Cutter
0.0333
20
2.887
1
Results
L =
Lq=
W =
Wq=
ρ =
P0 =
L =
Lq=
W =
Wq=
ρ =
P0 =
=Lq+Rho
=((Lambda^2)*(Sigma^2)+(Rho^2))/(2*(1-Rho))
=Wq +OneOverMu
=Lq /Lambda
=Lambda*OneOverMu
=1-Rho
(Mean arrival rate)
(Expected service time)
(Standard deviation)
(# servers)
λ = 1/μ =
σ = s =
L
Lambda
Lq OneOverMu
Rho
s
Sigma
W
Wq
G4
C4
G5
C5
G10
C7
C6
G7
G8
Range Name Cell
4
5
6
7
8
9
10
11
12
F G
1.344
0.678
40.356
20.356
0.666
0.334
FIGURE 12.10 This Excel template for the M/G/1 model shows the basic measures of performance for Herr Cut- ter’s barber shop without an associate.
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512 Chapter Twelve Computer Simulation: Basic Concepts
L, L q , W, and W q above their true values for the barber shop. Thus, the numbers in Figure 12.9 provide conservative estimates (which are preferable to overly optimistic estimates). If Herr Cutter does add an associate, he would provide three additional chairs for waiting customers. There also would be less waiting to begin a haircut, so having arriving customers not stay would become very unusual. Therefore, the first simplifying assumption seems very reason- able for simulating the barber shop with an associate.
The effect of the second simplifying assumption also is to slightly inflate the estimates of L, L q , W, and W q above their true values. The reason is that the barber shop begins empty each morning and then gradually builds up to a steady-state condition, whereas the simulation model has the shop operating in a steady-state condition for all but the very beginning of the simulation run. Fortunately, adding an associate would tend to keep the number of customers in the shop down to minimal levels, even in a steady-state condition (which would be nearly reached early in the day). Therefore, the estimation errors from using this assumption to simu- late the shop with an associate should be reasonably small.
By obtaining a more expensive computer simulation package and devoting additional preparation time, Fritz would be able to closely simulate the actual operation of the barber shop without making these two approximations. A key advantage of computer simulation is the ability to incorporate as many realistic features into the model as desired.
However, just as with the mathematical model for any other management science tech- nique, there always is a trade-off between the amount of realism incorporated into a model and the ease with which the model can be used. A simulation model does not need to be a completely realistic representation of the real system. Many simulation models err on the side of being overly realistic rather than overly idealistic. An overly realistic model includes unimportant details that do not significantly affect the estimates obtained from the simulation runs. Such a model often is very difficult to debug, and may never be completely debugged. It also is likely to require a great deal of programming and computer time to obtain a small amount of information. The goal should be to incorporate only the important features of the system into the model in order to generate reasonably accurate information that will enable management to make well-informed decisions in a timely fashion.
Fritz feels that his current simulation model meets this goal.
Analysis of the Option of Adding an Associate As described near the beginning of Section 12.2, Herr Cutter and his nephew Fritz have agreed on a plan for analyzing the option of adding an associate. They assume that the probability dis- tribution of service times (the times required to give a haircut) for the associate would be the same as for Herr Cutter. Based on the second rule of thumb given in Section 12.2, they also are assuming that adding the associate would (1) reduce the average waiting time before a haircut begins to less than 10 minutes and (2) then gradually attract new business until this average waiting time reaches about 10 minutes. The level of business (say, the average number of customers per day) determines the mean of the probability distribution of interarrival times. Therefore, a number of computer simulations will be run with different means of this distribu- tion to determine which mean would result in an average waiting time of about 10 minutes. Given the corresponding level of business, a financial analysis can then be conducted.
The current mean (without an associate) of the distribution of interarrival times is 30 min- utes. Therefore, proceeding by trial and error, Fritz tries the series of means shown in the first column of Table 12.2 . To quickly hone in on the neighborhood for the right mean, he uses the Queueing Simulator to run computer simulations of only moderate length, namely, 10,000 arrivals each (roughly half a year of simulated operation). The point estimates of W q (the aver- age waiting time until a haircut begins) in the second column indicate that the mean that gives a true value of W q of 10 minutes should be somewhere close to 14.3 minutes. The 95 percent confidence intervals for W q in the rightmost column further suggest that this mean should be within about half a minute of 14.3 minutes.
To check this further, Fritz next does a long simulation run (100,000 arrivals) with a mean of 14.3 minutes for the interarrival-time distribution. The complete results for all the measures of per- formance are shown in Figure 12.11 . The point estimate of W q (and most of the 95 percent confi- dence interval for W q ) now is slightly over 10. However, Fritz also recalls that the two simplifying
Simplifying assumptions that provide conservative estimates are preferable to those that lead to overly optimistic estimates.
Although some software packages enable adding many realistic features into a simulation model, unim- portant details that make the model overly complex should be avoided.
The level of business is expected to increase to the point where the aver- age waiting time to begin a haircut is 10 minutes, so computer simulations will be run to estimate this level of business.
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assumptions discussed in the preceding subsection cause this estimate to slightly overstate the true value of W q for the barber shop. Therefore, he concludes that 14.3 minutes is the best available estimate of the mean that would result in an average waiting time of about 10 minutes.
Fritz realizes that he could spend more time running long computer simulations with means slightly different from 14.3 minutes in order to pin down this estimate even better. However, he already knows from the confidence intervals in Table 12.2 that 14.3 minutes is at least very close. Furthermore, given the slight inaccuracies known to be in the simulation model due to the two simplifying assumptions, there is no point in trying to obtain an estimate of the mean that is more precise than the model is. This would only give a false sense of accuracy. He is content that 14.3 minutes provide a very adequate and conservative estimate of the mean for purposes of analysis.
Based on this estimate, Fritz concludes that having his uncle add an associate should grad- ually increase the level of business to around the point where
Average interarrival time 5 14.3 minutes
which would yield
Mean arrival rate 5 60
14.3 customers per hour
5 4.2 customers per hour 5 4.2(9) customers per day 5 37.8 customers per day
A conservative estimate is that the level of business will increase to the point where customers are arriv- ing at an average of one every 14.3 minutes.
1
A B C
Number of Servers
Distribution
Mean
Distribution
Minimum
Maximum
Number of Arrivals
Interarrival Times
Service Times
Length of Simulation Run
L =
Lq =
W =
Wq =
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
Point
Estimate
2.126
0.719
30.212
10.211
0.163
0.266
0.233
0.1541
0.0877
0.0467
0.02417
0.01282
0.00634
0.003208
0.001546
95% Confidence Interval
D E F G H
2
3
4 Low High
2.090
0.689
29.833
9.834
0.160
0.262
0.230
0.1518
0.0855
0.0448
0.02264
0.01162
0.00546
0.002530
0.001076
2.163
0.748
30.591
10.588
0.166
0.270
0.235
0.1564
0.0898
0.0487
0.0257
0.01401
0.00722
0.00389
0.002017
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Queueing Simulator for Herr Cutter's Barber Shop with an Associate
Run Simulation
2
Exponential
14.3
Uniform
15
25
100,000
FIGURE 12.11 The results obtained by using the Queueing Simu- lator to perform a com- puter simulation of Herr Cutter’s barber shop with an associate over a period of 100,000 customer arrivals.
Mean of Interarrival Times
Point Estimate of Wq
95 Percent Confidence Interval for Wq
20 minutes 3.33 minutes 3.05 to 3.61 minutes 15 minutes 8.10 minutes 6.98 to 9.22 minutes 14 minutes 10.80 minutes 9.51 to 12.08 minutes 14.2 minutes 9.83 minutes 8.83 to 10.84 minutes 14.3 minutes 9.91 minutes 8.76 to 11.05 minutes
TABLE 12.2 The Estimates of Wq Obtained by Using the Queueing Simulator to Simulate Herr Cut- ter’s Barber Shop with an Associate for 10,000 Arrivals for Different Means of the Distribution of Interarrival Times
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This level of business would be more than double the current average of 18 customers per day for the shop. Herr Cutter would plan to divide the customers equally with the associate, so each would average 18.9 customers per day.
Therefore, using the cost factors given at the beginning of this section, Herr Cutter’s aver- age net income per working day would become
Net daily income 5 37.8($15) (shop revenue)
118.9($2) (his tips)
2$50 (shop maintenance)
2$120 (associate’s salary)
218.9 ($5)(associate’s commission)
5 $567 1 $37.80 2 $50 2 $120 2 $94.50
5 $340.30
This compares with Herr Cutter’s current net daily income of $256. Thus, it is estimated that the change in his net daily income from adding an associate would eventually become
Change in net daily income 5 $340.30 2 $256
5 $84.30
Thus, he actually would increase his income significantly. When presenting this analysis to his uncle, Fritz emphasizes that this $84.30 figure is just
an estimate of what will happen after the level of business gradually increases to its new level. It may take awhile, even a year or two, to reach this new level. Meanwhile, Herr Cut- ter’s income may start off less than it has been before gradually increasing. Furthermore, the optimistic conclusion of a substantial increase in income eventually is based largely on the rather shaky premise that the second rule of thumb in the article in The Barber’s Journal will prove to be valid and applicable to his shop. This premise leads to an estimate that his level of business would more than double eventually. Achieving this big increase in business would seem realistic only if the associate is able to bring a considerable number of customers with him from his current shop and then the two of them are able to attract many additional new customers.
Herr Cutter feels confident that they can accomplish this. This associate was highly rec- ommended by his friend. Furthermore, he feels that his own skill as a barber already would have attracted many new customers if he didn’t already have as much business as he could handle alone. In this growing city, the opportunity is there. He also likes the fact that adding an associate would enable him to improve the level of service for his current loyal clientele by substantially decreasing their average waiting time before beginning a haircut. Finally, he also sees many personal advantages to having a good associate that cannot be measured in monetary terms. Therefore, he wouldn’t mind a temporary decrease in income as long as he probably would at least equal his current income level in a year or two. Actually increasing his income would be a pleasant bonus.
On these grounds, Herr Cutter decides to hire the associate. He also thanks his nephew for the invaluable help that Fritz’s computer simulations provided him in making his decision.
This estimate of increased income needs to be inter- preted carefully.
Herr Cutter decides to hire the associate.
1. What did Fritz simulate in his first simulation run? For what purpose? 2. What are the two types of estimates of a measure of performance obtained by the Queueing
Simulator? 3. What were the two ways with which Fritz tested the validity of his simulation model? 4. Does Fritz’s simulation model make any simplifying assumptions? Is it necessary for a simula-
tion model to be a completely realistic representation of the real system? 5. Does Fritz’s analysis estimate that Herr Cutter’s income will eventually increase or decrease
(compared to its current level) if he adds an associate?
Review Questions
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515
12.4 OUTLINE OF A MAJOR COMPUTER SIMULATION STUDY
Thus far, this chapter has focused mainly on the process of performing a computer simulation and its illustration by a case study. We now place this material into broader perspective by briefly outlining all the typical steps involved when a management science team performs a major study that is based on applying computer simulation. (Nearly the same steps also apply when the study is applying other management science techniques instead.)
Step 1: Formulate the Problem and Plan the Study The management science team needs to begin by meeting with management to address the following kinds of questions.
1. What is the problem that management wants studied? 2. What are the overall objectives for the study? 3. What specific issues should be addressed? 4. What kinds of alternative system configurations should be considered? 5. What measures of performance of the system are of interest to management? 6. What are the time constraints for performing the study?
In addition, the team needs to meet with engineers and operational personnel to learn the details of just how the system would operate. (The team generally will also include one or more members with a firsthand knowledge of the system.)
Step 2: Collect the Data and Formulate the Simulation Model The types of data needed depend on the nature of the system to be simulated. For Herr Cut- ter’s barber shop, the key pieces of data were the distribution of interarrival times and the distribution of service times (times needed to give a haircut). For most other cases as well, it is the probability distributions of the relevant quantities that are needed. Generally, it will only be possible to estimate these distributions, but it is important to do so. In order to gen- erate representative scenarios of how a system will perform, it is essential for a computer simulation to generate random observations from these distributions rather than simply using averages.
These are key questions that management should answer to initiate any man- agement science study.
Computer simulations should use probability distributions of the rel- evant quantities rather than averages.
Sasol is an integrated energy and chemicals company that is based in South Africa and operates in 38 countries. It had a market capitalization of over $23 billion in 2009.
Historically, the petrochemical industry based business decisions on the average results throughout its produc- tion processes. However, Sasol’s management science team recognized that these production processes actually are stochastic systems that involve substantial variability and dynamic interactions. Therefore, for the first time in the industry, this team introduced the use of computer simula- tions to much more adequately consider the effect of all this variability and dynamic interaction.
Three large computer simulation models were devel- oped to meet Sasol’s needs. The gas factory model cov- ers the process from raw materials to the production of synthetic crude oil. The liquid factory model simulates the refining of the synthetic crude oil and the associated chemical production processes. The fuels blending model
blends the different fuel components into multiple grades of gasoline and diesel.
This industry is one where frequent changes need to be made in its facilities and production processes because of changes in government regulations, fuel specifica- tions, availability of raw materials, prices of these materi- als, etc. Sasol uses one or more of its computer simulation models to evaluate the viable options for changes in its facilities and production processes whenever the need arises.
This industry-leading use of computer simulation has enabled Sasol to radically improve its decision making. This use during its first decade (2000 to 2009) has resulted in an estimated value addition to Sasol in excess of $230 million.
Source: M. Meyer and 11 other co-authors, “Innovative Decision Support in a Petrochemical Production Environment,” Interfaces 41, no. 1 (January–February 2011), pp. 79–92. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
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Step 3: Check the Accuracy of the Simulation Model Before constructing a computer program, the management science team should engage the people most intimately familiar with how the system will operate in checking the accuracy of the simulation model. This often is done by performing a structured walk-through of the conceptual model, using a projection onto a large screen before an audience of all the key people. Typically at such meetings, several erroneous model assumptions will be discovered and corrected, a few new assumptions will be added, and some issues will be resolved about how much detail is needed in the various parts of the model.
Step 4: Select the Software and Construct a Computer Program There are several major classes of software used for computer simulations. One is spread- sheet software. Section 12.1 described how Excel is able to perform some basic computer simulations on a spreadsheet. In addition, some excellent Excel add-ins now are available to enhance this kind of spreadsheet modeling. Chapter 13 focuses on the use of one of these add- ins, Risk Solver Platform for Education (RSPE), in your MS Courseware. (Chapter 20 on the CD-ROM does the same for another add-in called Crystal Ball.)
Other classes of software for computer simulations are intended for more extensive appli- cations where it is no longer convenient to use spreadsheet software. One such class is a general-purpose programming language, such as C 1 1 , Visual Basic, Java, and so on. Such languages (and their predecessors) often were used in the early history of the field because of their great flexibility for programming any sort of simulation. However, because of the con- siderable programming time required, they are not used nearly as much now.
Many commercial software packages that don’t use spreadsheets also have been devel- oped specifically to perform computer simulations. Historically, these simulation software packages have been classified into two categories, general-purpose simulation languages and application-oriented simulators. General-purpose simulation languages provide many of the features needed to program any simulation model efficiently. Application-oriented simula- tors (or just simulators for short) are designed for simulating fairly specific types of sys- tems. However, as time has gone on, the distinction between these two categories has become increasingly blurred. General-purpose simulation languages now may include some special features that make them almost as well suited as simulators for certain specific kinds of appli- cations. Conversely, today’s simulators tend to include more flexibility than they previously had for dealing with a broader class of systems.
Another way of categorizing simulation software packages is by whether they use an event-scheduling approach or a process approach to discrete-event simulation modeling. The event-scheduling approach closely follows the next-event time-advance procedure described in Section 12.2. The process approach still uses the next-event time-advance procedure in the background but focuses the modeling instead on describing the processes that generate the events. Most contemporary simulation software packages now use the process approach.
It has become increasingly common for simulation software packages to include animation capabilities for displaying computer simulations in action. In an animation, key ele- ments of a system are represented in a computer display by icons that change shape, color, or position when there is a change in the state of the simulation system. (One example of an ani- mation of a computer simulation of a queueing system is provided by the Waiting Line module in your Interactive Management Science Modules at www.mhhe.com/hillier5e or on the CD- ROM.) The major reason for the popularity of animation is its ability to communicate the essence of a simulation model (or of a computer simulation run) to managers and other key personnel.
Step 5: Test the Validity of the Simulation Model After the computer program has been constructed and debugged, the next key step is to test whether the simulation model incorporated into the program is providing valid results for the system it is representing. Specifically, will the measures of performance for the real system be closely approximated by the values of these measures generated by the simulation model?
In some cases, a mathematical model may be available to provide results for a simple ver- sion of the system. If so, these results also should be compared with the simulation results.
For example, in the case study, the barber shop currently is in operation with Herr Cutter as the only barber. Therefore, as described in Section 12.3 (see the subsection entitled “Testing
Animation capabilities for displaying computer simu- lations in action are very useful for communicating the essence of a simulation model to managers and other key personnel.
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12.4 Outline of a Major Computer Simulation Study 517
the Validity of the Simulation Model”), Fritz compared the results from an applicable queue- ing model with a simulation of this current version of the barber shop.
When no real data are available to compare with simulation results, one possibility is to conduct a field test to collect such data. This would involve constructing a small prototype of some version of the proposed system and placing it into operation.
Another useful validation test is to have knowledgeable operational personnel check the credibility of how the simulation results change as the configuration of the simulated system is changed. Watching animations of simulation runs also is a useful way of checking the validity of the simulation model.
Step 6: Plan the Simulations to Be Performed At this point, you need to begin making decisions as to which system configurations to simu- late. This often is an evolutionary process, where the initial results for a range of configura- tions help you to hone in on which specific configurations warrant detailed investigation.
Decisions also need to be made now on such issues as the lengths of simulation runs. Keep in mind that computer simulation does not produce exact values for the measures of performance of a system. Instead, each simulation run can be viewed as a statistical experi- ment that is generating statistical observations of the performance of the simulated system. These observations are used to produce statistical estimates of the measures of performance. Increasing the length of a run increases the precision of these estimates.
The statistical theory for designing statistical experiments conducted through computer sim- ulation is little different than for experiments conducted by directly observing the performance of a physical system. Therefore, the services of a professional statistician (or at least an expe- rienced simulation analyst with a strong statistical background) can be invaluable at this step.
Step 7: Conduct the Simulation Runs and Analyze the Results The output from the simulation runs now provides statistical estimates of the desired mea- sures of performance for each system configuration of interest. In addition to a point estimate of each measure, a confidence interval normally should be obtained to indicate the range of likely values of the measure (just as was done for the case study).
These results might immediately indicate that one system configuration is clearly superior to the others. More often, they will identify the few strong candidates to be the best one. In the latter case, some longer simulation runs would be conducted to better compare these candidates. Addi- tional runs also might be used to fine-tune the details of what appears to be the best configuration.
Step 8: Present Recommendations to Management After completing its analysis, the management science team needs to present its recommendations to management. This usually would be done through both a written report and a formal oral pre- sentation to the managers responsible for making the decisions regarding the system under study.
The report and presentation should summarize how the study was conducted, including documentation of the validation of the simulation model. A demonstration of the animation of a simulation run might be included to better convey the simulation process and add credibil- ity. Numerical results that provide the rationale for the recommendations need to be included.
Management usually involves the management science team further in the initial imple- mentation of the new system, including the indoctrination of the affected personnel.
A field test of a small pro- totype of the proposed sys- tem is sometimes used to collect real data to compare with the simulation results and to fine-tune the design.
Each simulation run gener- ates statistical observations of the performance of the simulated system, so sta- tistical theory should guide the planning of the runs.
After identifying the few best system configurations, longer simulation runs should be used to select the single best one and to fine- tune its design.
Both the written report and the oral presentation should highlight the recommenda- tions and the rationale for those recommendations.
1. When beginning a computer simulation study, with whom should a management science team meet to address some key questions and then to learn the details of how the system would operate?
2. Who should the team engage to help check the accuracy of the simulation model? 3. What is the difference between a general-purpose simulation language and an applications-
oriented simulator? 4. When using animation to display a computer simulation in action, how are the key elements of
the system represented? 5. What is the specific question being addressed when testing the validity of a simulation model? 6. A simulation run can be viewed as what kind of statistical experiment? 7. What kinds of estimates are obtained from simulation runs? 8. What are the two ways in which a management science team usually presents its recommen-
dations to management?
Review Questions
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Computer simulation is one of the most popular management science techniques because it is such a flexible, powerful, and intuitive tool. It involves using a computer to imitate (simulate) the operation of an entire process or system. For a stochastic system (a system that evolves over time according to one or more probability distributions), random observations are generated from these distributions to generate the various events that occur in the simulated system over time. This provides a relatively quick way of investigating how well a proposed system configuration would perform without incurring the great expense of actually constructing and operating the system. Therefore, many alternative system configu- rations can be investigated and compared in advance before choosing the one to use.
Herr Cutter’s barber shop provides a case study of how computer simulation was able to provide the needed information to decide whether to change this stochastic system by adding a second barber. Like so many others, this stochastic system is a queueing system, but one that is too complicated to be analyzed solely by using queueing models.
This case study also illustrates the building blocks of a simulation model that represents the system to be simulated and describes how the simulation will be performed. One key building block is a simula- tion clock, which is the variable in the computer program that records the amount of simulated time that has elapsed so far. The next-event time-advance procedure advances the time on the simulation clock by repeatedly moving from the current event to the next event that will occur in the simulated system.
In a matter of seconds or minutes, a computer simulation can simulate even years of operation of a typical system. Each simulation run generates a series of statistical observations about the performance of the system over the period of time simulated. These observations then are used to estimate the inter- esting measures of performance of the system. Both a point estimate and a confidence interval can be obtained for each measure.
Some computer simulation studies can be done relatively quickly by a single individual, who might be the manager concerned with the problem. For a more extensive study, however, the manager might want to assign a staff member, or even a full-fledged management science team, to the project. A major management science study based on computer simulation requires a series of important steps before the team is ready to obtain results from simulation runs. A series of questions must be addressed to manage- ment to properly define the problem from their viewpoint. Collecting good data generally is a difficult and time-consuming process. Another big task is formulating the simulation model, checking its accu- racy, and then testing the validity of the model for closely approximating the system being simulated. One of the team’s most important decisions is the choice of the software to be used. Most simulation software vendors now offer a version of their software with animation capabilities. Animation is very useful for illustrating the results of computer simulation for managers and other key personnel, which can add much credibility to the study.
Even after the computer program is ready to go, the management science team needs to design the statistical experiments to be conducted through computer simulation. Then the simulation runs can be conducted and the results analyzed. Finally, the team usually needs to both prepare a written report and make a formal oral presentation to present its recommendations to management.
12.5 Summary
Glossary animation A computer display with icons that shows what is happening in a computer simula- tion. (Section 12.4), 516 confidence interval An interval within which the true value of a measure of performance is likely to lie. (Section 12.3), 509 inverse transformation method A method for generating random observations from a probabil- ity distribution. (Section 12.1), 498 next-event time advance A procedure for advancing the time on the simulation clock by repeatedly moving from the current event to the next event that will occur in the simulated sys- tem. (Section 12.2), 505 point estimate The single number that provides the best estimate of a measure of performance. (Section 12.3), 509
random number A random observation from the uniform distribution over the interval from 0 to 1. (Section 12.1), 490 simulation clock A variable in the computer program that records how much simulated time has elapsed so far. (Section 12.2), 505 simulation model A representation of the sys- tem to be simulated that also describes how the simulation will be performed. (Section 12.2), 504 state of the system The key information that defines the current status of the system. (Section 12.2), 505 stochastic system A system that evolves over time according to one or more probability distri- butions. (Section 12.1), 489
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Chapter 12 Problems 519
Chapter 12 Excel Files:
Coin-Flipping Game Example
Heavy Duty Co. Examples (3)
Herr Cutter’s Barber Shop Case Study
Queueing Simulator
Template for M/G/1 Queueing Model
Excel Add-in:
RSPE (to be featured in the next chapter)
Routine:
Queueing Simulator (in an Excel file)
Interactive Management Science Module:
Waiting Line Module
Supplement to This Chapter on the CD-ROM:
The Inverse Transformation Method for Generating Random Observations
Learning Aids for This Chapter in Your MS Courseware
Solved Problem (See the CD-ROM or Website for the Solution) 12.S1. Estimating the Cost of Insurance Claims The employees of General Manufacturing Corp. receive health insurance through a group plan issued by Wellnet. During the past year, 40 percent of the employees did not file any health insurance claims, 40 percent filed only a small claim, and 20 percent filed a large claim. The small claims were spread uniformly between 0 and $2,000, whereas the large claims were spread uniformly between $2,000 and $20,000.
On the basis of this experience, Wellnet now is negotiat- ing the corporation’s premium payment per employee for the upcoming year. You are a management science analyst for the insurance carrier, and you have been assigned the task of esti- mating the average cost of insurance coverage for the corpora- tion’s employees.
a. Use the random numbers 0.4071, 0.5228, 0.8185, 0.5802, and 0.0193 to simulate whether each of five employees files no claim, a small claim, or a large claim. Then use the random num- bers 0.9823, 0.0188, 0.8771, 0.9872, and 0.4129 to simulate the size of the claim (including zero if no claim was filed). Calculate the average of these claims to estimate the mean of the overall distribution of the size of employees’ health insurance claims.
b. Formulate and apply a spreadsheet model to simulate the cost for 300 employees’ health insurance claims. Calculate the av- erage of these random observations.
c. The true mean of the overall probability distribution of the size of an employee’s health insurance claim is $2,600. Com- pare the estimates of this mean obtained in parts a and b with the true mean of the distribution.
a. Simulate one play of this game by repeatedly flip- ping your own coin until the game ends. Record your results in the format shown in columns B, D, E, F, and G of Figure 12.1 . How much would you have won or lost if this had been a real play of the game?
E* b. Revise the spreadsheet model in Figure 12.1 by using Excel’s VLOOKUP function instead of the IF function to generate each simulated flip of the coin. Then perform a computer simulation of one play of the game.
E* c. Use this revised spreadsheet model to generate a data table with 14 replications like Figure 12.2 .
E* d. Repeat part c with 1,000 replications (like Figure 12.3 ).
12.3. Each time an unbiased coin is flipped three times, the probability of getting 0, 1, 2, and 3 heads is ⅛, ⅜, ⅜, and ⅛, respectively. Therefore, with eight groups of three flips each, on the average, one group will yield no heads, three groups will yield one head, three groups will yield two heads, and one group will yield three heads.
The symbols to the left of some of the problems (or their parts) have the following meaning:
E*: Use Excel. Q*: Use the Queueing Simulator.
An asterisk on the problem number indicates that at least a par- tial answer is given in the back of the book.
12.1.* Use the random numbers in cells C13:C18 of Fig- ure 12.1 to generate six random observations for each of the fol- lowing situations. a. Throwing an unbiased coin. b. A baseball pitcher who throws a strike 60 percent of
the time and a ball 40 percent of the time. c. The color of a traffic light found by a randomly
arriving car when it is green 40 percent of the time, yellow 10 percent of the time, and red 50 percent of the time.
12.2. Reconsider the coin-flipping game introduced in Sec- tion 12.1 and analyzed with computer simulation in Figures 12.1 , 12.2 , and 12.3 .
Problems
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a. Use these data to estimate the probability distribu- tion of daily sales.
b. Calculate the mean of the distribution obtained in part a.
c. Describe how random numbers can be used to sim- ulate daily sales.
d. Use the random numbers 0.4476, 0.9713, and 0.0629 to simulate daily sales over three days. Com- pare the average with the mean obtained in part b.
E* e. Formulate a spreadsheet model for performing a computer simulation of the daily sales. Perform 300 replications and obtain the average of the sales over the 300 simulated days.
12.7. Generate three random observations from the uniform distribution between 2 10 and 40 by using the following random numbers: 0.0965, 0.5692, 0.6658. 12.8. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 12.2. Briefly describe how computer simu- lation was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 12.9. Eddie’s Bicycle Shop has a thriving business repair- ing bicycles. Trisha runs the reception area where customers check in their bicycles to be repaired and then later pick up their bicycles and pay their bills. She estimates that the time required to serve a customer on each visit has a uniform distribution between three minutes and eight minutes. a. Simulate the service times for five customers by
using the following five random numbers: 0.6505, 0.0740, 0.8443, 0.4975, 0.8178.
b. Calculate the average of the five service times and compare it to the mean of the service-time distribution.
E* c. Use Excel to generate 500 random observations and calculate the average. Compare this average to the mean of the service-time distribution.
12.10.* Reconsider Eddie’s Bicycle Shop described in the pre- ceding problem. Forty percent of the bicycles require only a minor repair. The repair time for these bicycles has a uniform distribution between zero and one hour. Sixty percent of the bicycles require a major repair. The repair time for these bicy- cles has a uniform distribution between one hour and two hours. You now need to estimate the mean of the overall probability distribution of the repair times for all bicycles by using the fol- lowing alternative methods. a. Use the random numbers 0.7256, 0.0817, and
0.4392 to simulate whether each of three bicycles requires minor repair or major repair. Then use the random numbers 0.2243, 0.9503, and 0.6104 to simulate the repair times of these bicycles. Calcu- late the average of these repair times to estimate the mean of the overall distribution of repair times.
b. Repeat part a with the complements of the ran- dom numbers used there, so the new random num- bers are 0.2744, 0.9183, 0.5608, and then 0.7757, 0.0497, and 0.3896.
c. Combine the random observations from parts a and b and calculate the average of these six observations
a. Using your own coin, flip it 24 times divided into eight groups of three flips each, and record the num- ber of groups with no head, with one head, with two heads, and with three heads.
b. Use random numbers in the order in which they are given in column C of Figure 12.4 and then in cells C5:C13 of Figure 12.5 to simulate the flips speci- fied in part a and record the information indicated in part a.
E* c. Formulate a spreadsheet model for performing a computer simulation of three flips of the coin and recording the number of heads. Perform one repli- cation of this simulation.
E* d. Use this spreadsheet to generate a data table with eight replications of the simulation. Compare this frequency distribution of the number of heads with the probability distribution of the number of heads with three flips.
E* e. Repeat part d with 800 replications.
12.4. The weather can be considered a stochastic system, because it evolves in a probabilistic manner from one day to the next. Suppose for a certain location that this probabilistic evolu- tion satisfies the following description:
The probability of rain tomorrow is 0.6 if it is raining today. The probability of its being clear (no rain) tomorrow is 0.8 if it is clear today.
a. Use the random numbers in cells C17:C26 of Fig- ure 12.1 to simulate the evolution of the weather for 10 days, beginning the day after a clear day.
E* b. Now use a computer with the random numbers gen- erated by Excel to perform the simulation requested in part a on a spreadsheet.
12.5.* The game of craps requires the player to throw two dice one or more times until a decision has been reached as to whether he (or she) wins or loses. He wins if the first throw results in a sum of seven or 11 or, alternatively, if the first sum is 4, 5, 6, 8, 9, or 10 and the same sum reappears before a sum of seven has appeared. Conversely, he loses if the first throw results in a sum of 2, 3, or 12 or, alternatively, if the first sum is 4, 5, 6, 8, 9, or 10 and a sum of 7 appears before the first sum reappears.
E* a. Formulate a spreadsheet model for performing a computer simulation of the throw of two dice. Per- form one replication.
E* b. Perform 25 replications of this simulation.
c. Trace through these 25 replications to determine the number of times the simulated player would have won the game of craps when each play starts with the next throw after the previous play ends.
12.6. Jessica Williams, manager of kitchen appliances for the Midtown Department Store, feels that her inventory levels of stoves have been running higher than necessary. Before revising the inventory policy for stoves, she records the number sold each day over a period of 25 days, as summarized below.
Number sold 2 3 4 5 6 Number of days 4 7 8 5 1
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over a period of 100 breakdowns for each of the three crew sizes under consideration. What do these results suggest the crew size should be?
Q* c. Use the Queueing Simulator to perform this com- puter simulation over 10,000 breakdowns for each of the three crew sizes.
E* d. Use the Excel template for the M / G /1 queueing model in this chapter’s Excel file to obtain the expected waiting time analytically for each of the three crew sizes. Which crew size should be used?
12.13. Refer to the first 100 minutes of the computer simu- lation of the current operation of Herr Cutter’s barber shop presented in Figure 12.7 and summarized in Figure 12.8 . Now consider the alternative of adding an associate. Perform a simu- lation of this alternative by hand by using exactly the same inter- arrival times (in the same order) and exactly the same service times (in the same order) as in Figure 12.7 .
a. Determine the new waiting time before beginning a haircut for each of the 5 customers who arrive in the first 100 minutes. Use these results to estimate W q , the expected waiting time before the haircut.
b. Plot the new version of Figure 12.8 to show the evolution of the number of customers in the barber shop over these 100 minutes.
12.14. While performing a computer simulation of a single- server queueing system, the number of customers in the system is zero for the first 10 minutes, one for the next 17 minutes, two for the next 24 minutes, one for the next 15 minutes, two for the next 16 minutes, and one for the next 18 minutes. After this total of 100 minutes, the number becomes 0 again. Based on these results for the first 100 minutes, perform the following analysis (using the notation for queueing models defined in Sections 11.3 and 12.3). a. Draw a figure like Figure 12.8 showing the evolu-
tion of the number of customers in the system. b. Develop estimates of P 0 , P 1 , P 2 , P 3 . c. Develop estimates of L and L q . d. Develop estimates of W and W q . 12.15. Read the referenced article that fully describes the man- agement science study summarized in the application vignette presented in Section 12.4. Briefly describe how computer simu- lation was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 12.16. A major banking institution, Best Bank, plans to open a new branch office in Littletown. Preliminary estimates suggest that two tellers (and teller windows) should be provided, but this decision now awaits further analysis.
Marketing surveys indicate that the new Littletown bank will attract enough business that customers requiring teller service will enter the bank at the rate of about one per minute on the average. Thus, the average time between consecutive customer arrivals is estimated to be one minute.
No parking is available near the bank, so a special parking lot for bank customers only will be provided. A parking lot attendant will be on duty to validate each customer’s parking before he or she leaves the car to enter the bank. This validation process takes at least 0.5 minutes, so the minimum time between consecutive arrivals of customers into the bank is 0.5 minutes. The amount by which the interarrival time exceeds 0.5 minutes is estimated
to estimate the mean of the overall distribution of repair times. (This is referred to as the method of complementary random numbers. )
d. The true mean of the overall probability distribution of repair times is 1.1. Compare the estimates of this mean obtained in parts a, b, and c. For the method that provides the closest estimate, give an intuitive explanation for why it performed so well.
E* e. Formulate a spreadsheet model to apply the method of complementary random numbers described in part c. Use 600 random numbers and their comple- ments to generate 600 random observations of the repair times and calculate the average of these ran- dom observations. Compare this average with the true mean of the distribution.
12.11. The William Graham Entertainment Company will be opening a new box office where customers can come to make ticket purchases in advance for the many entertainment events being held in the area. Computer simulation is being used to ana- lyze whether to have one or two clerks on duty at the box office.
While simulating the beginning of a day at the box office, the first customer arrives five minutes after it opens and then the interarrival times for the next four customers (in order) are three minutes, nine minutes, one minute, and four minutes, after which there is a long delay until the next customer arrives. The service times for these first five customers (in order) are eight minutes, six minutes, two minutes, four minutes, and seven minutes.
a. For the alternative of a single clerk, draw a figure like Figure 12.8 that shows the evolution of the number of customers at the box office over this period.
b. Use this figure to estimate the usual measures of performance— L, L q , W, W q , and the P n (as defined in Sections 11.3 and 12.3)—for this queueing system.
c. Repeat part a for the alternative of two clerks. d. Repeat part b for the alternative of two clerks.
12.12. The Rustbelt Manufacturing Company employs a main- tenance crew to repair its machines as needed. Management now wants a computer simulation study done to analyze what the size of the crew should be, where the crew sizes under consideration are two, three, and four. The time required by the crew to repair a machine has a uniform distribution over the interval from zero to twice the mean, where the mean depends on the crew size. The mean is four hours with two crew members, three hours with three crew members, and two hours with four crew members. The time between breakdowns of some machine has an expo- nential distribution with a mean of five hours. When a machine breaks down and so requires repair, management wants its aver- age waiting time before repair begins to be no more than three hours. Management also wants the crew size to be no larger than necessary to achieve this.
a. Develop a simulation model for this problem by describing its six basic building blocks listed in Sec- tion 12.2 as they would be applied to this situation.
E* b. Formulate a spreadsheet model to perform a com- puter simulation to estimate the average waiting time before repair begins. Perform this simulation
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Q* e. Hugh is considering hiring a second mechanic who specializes in German cars so that two such cars can be repaired simultaneously. (Only one mechanic works on any one car.) Use the Queueing Simulator with 10,000 arrivals of German cars to evaluate this option.
Q* f. Another option is to train the two current mechanics to work on either kind of car. This would increase the mean repair time by 10 percent, from 0.2 days to 0.22 days. Use the Queueing Simulator with 20,000 arrivals of cars of either kind to evaluate this option.
E* g. Because both the interarrival-time and service- time distributions are exponential, the M/M/ 1 and M/M/s queueing models introduced in Sections 11.5 and 11.6 can be used to evaluate all the above options analytically. Use the template for the M/M/s queueing model (with s 5 1 or 2) in an Excel file for Chapter 11 to determine W, the expected wait- ing time until repair is completed for each of the cases considered in parts b through f. For each case, compare the estimate of W obtained by computer simulation with the analytical value. What does this say about the number of car arrivals that should be included in the computer simulation?
h. Based on the above results, which option would you select if you were Hugh? Why?
12.18. Vistaprint produces monitors and printers for comput- ers. In the past, only some of them were inspected on a sampling basis. However, the new plan is that they all will be inspected before they are released. Under this plan, the monitors and print- ers will be brought to the inspection station one at a time as they are completed. For monitors, the interarrival time will have a uniform distribution between 10 and 20 minutes. For printers, the interarrival time will be a constant 15 minutes.
The inspection station has two inspectors. One inspector works on only monitors and the other one inspects only comput- ers. In either case, the inspection time has an exponential distri- bution with a mean of 10 minutes.
Before beginning the new plan, management wants an evalu- ation made of how long the monitors and printers will be held up waiting at the inspection station. E* a. Formulate a spreadsheet model to perform a com-
puter simulation to estimate the average waiting times (both before beginning inspection and after completing inspection) for either the monitors or the printers.
E* b. Perform this simulation for the monitors over a period of 100 arrivals.
E* c. Repeat part b for the printers. Q* d. Use the Queueing Simulator to repeat parts b and c
with 10,000 arrivals in each case. Q* e. Management is considering the option of providing
new inspection equipment to the inspectors. This equipment would not change the mean time to per- form an inspection but it would decrease the vari- ability of the times. In particular, for either product, the inspection time would have an Erlang distribu- tion with a mean of 10 minutes and shape parameter k 5 4. Use the Queueing Simulator to repeat part d
to have an exponential distribution with a mean of 0.5 minutes. Therefore, the total interarrival time has a translated exponential distribution with a mean of (0.5 1 0.5) 5 1.0 minute. (A trans- lated exponential distribution is just an exponential distribution with a constant added.)
Based on past experience in other branch offices, it is known that the time required by a teller to serve a customer will vary widely from customer to customer, but the average time is about 1.5 minutes. This experience also indicates that service time has approximately an Erlang distribution with a mean of 1.5 min- utes and a shape parameter of k 5 4, which provides a standard deviation of 0.75 minute (half that for an exponential distribu- tion with the same mean).
These data suggest that two tellers should be able to keep up with the customers quite well. However, management wants to be sure that customers will not frequently encounter a long wait- ing line and an excessive wait before receiving service. There- fore, computer simulation will be used to study these measures of performance. Q* a. Use the Queueing Simulator with 5,000 customer
arrivals to estimate the usual measures of perfor- mance for this queueing system if two tellers are provided.
Q* b. Repeat part a if three tellers are provided. Q* c. Now perform some sensitivity analysis by checking
the effect if the level of business turns out to be even higher than projected. In particular, assume that the average time between customer arrivals turns out to be only 0.9 minutes (0.5 minutes plus a mean of only 0.4 minutes). Evaluate the alternatives of two tellers and three tellers under this assumption.
d. Suppose you were the manager of this bank. Use your computer simulation results as the basis for a managerial decision on how many tellers to pro- vide. Justify your answer.
12.17.* Hugh’s Repair Shop specializes in repairing German and Japanese cars. The shop has two mechanics. One mechanic works on only German cars and the other mechanic works on only Japanese cars. In either case, the time required to repair a car has an exponential distribution with a mean of 0.2 days. The shop’s business has been steadily increasing, especially for German cars. Hugh projects that, by next year, German cars will arrive randomly to be repaired at a mean rate of four per day, so the time between arrivals will have an exponential distribution with a mean of 0.25 days. The mean arrival rate for Japanese cars is projected to be two per day, so the distribution of interar- rival times will be exponential with a mean of 0.5 days.
For either kind of car, Hugh would like the average waiting time in the shop before the repair is completed to be no more than 0.5 days. E* a. Formulate a spreadsheet model to perform a com-
puter simulation to estimate what the average wait- ing time until repair is completed will be next year for either kind of car.
E* b. Perform this simulation for German cars over a period of 100 car arrivals.
E* c. Repeat part b for Japanese cars. Q* d. Use the Queueing Simulator to do parts b and c with
10,000 car arrivals in each case.
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Case 12-1 Planning Planers 523
There always is some travel time and then some setup time to start the actual repair, so the total time generally is at least 40 minutes ( 1/12 workday).
A key advantage of computer simulation over mathematical models is that it is not necessary to make simplifying approxi- mations like this one. For example, one of the options available in the Queueing Simulator is to use a translated exponential distribution, which has a certain minimum time and then the additional time has an exponential distribution with some mean. (Commercial packages for computer simulation have an even greater variety of options.)
Use computer simulation to refine the results obtained by queueing models as given by the Excel templates in the figures indicated below. Use a translated exponential distribution for the repair times where the minimum time is 1/12 workday and the additional time has an exponential distribution with a mean of 1/6 workday (80 minutes). In each case, use a run size of 25,000 arrivals and compare the point estimate obtained for W q (the key measure of performance for this case study) with the value of W q obtained by the queueing model. Q* a. Figure 11.4. Q* b. Figure 11.5. Q* c. Figure 11.8. Q* d. Figure 11.9. e. What conclusion do you draw about how sensitive
the results from a computer simulation of a queue- ing system can be to the assumption made about the probability distribution of service times?
under this option. Compare the results with those obtained in part d.
12.19. Consider the case study introduced in Section 12.2. After observing the operation of the barber shop, Herr Cutter’s nephew Fritz is concerned that his uncle’s estimate that the time required to give a haircut has a uniform distribution between 15 and 25 minutes appears to be a poor approximation of the actual probability distribution of haircut times. Based on the data he has gathered, Fritz’s best estimate is that the actual distribution is an Erlang distribution with a mean of 20 minutes and a shape parameter of k 5 8. a. Repeat the simulation run that Fritz previously used
to obtain Figure 12.9 (with a mean of 30 minutes for the interarrival-time distribution) except substitute this new distribution of haircut times.
b. Repeat the simulation run that Fritz previously used to obtain Figure 12.11 (with a mean of 14.3 minutes for the interarrival-time distribution) except substi- tute this new distribution of haircut times.
12.20. For the Dupit Corp. case study introduced in Section 11.4, the management science team was able to apply a vari- ety of queueing models by making the following simplifying approximation. Except for the approach suggested by the vice president for engineering, the team assumed that the total time required to repair a machine (including travel time to the machine site) has an exponential distribution with a mean of two hours (¼ workday). However, the team was somewhat uncomfort- able in making this assumption because the total repair times are never extremely short, as allowed by the exponential distribution.
Case 12-1
Planning Planers
This was the first time that Carl Schilling had been summoned to meet with the bigwigs in the fancy executive offices upstairs. And he hopes it will be the last time. Carl doesn’t like the pres- sure. He has had enough pressure just dealing with all the prob- lems he has been encountering as the foreman of the planer department on the factory floor. What a nightmare this last month has been!
Fortunately, the meeting had gone better than Carl had feared. The bigwigs actually had been quite nice. They explained that they needed to get Carl’s advice on how to deal with a problem that was affecting the entire factory. The origin of the problem is that the planer department has had a difficult time keeping up with its workload. Frequently there are a number of workpieces waiting for a free planer. This waiting has seriously disrupted the production schedule for subsequent operations, thereby greatly increasing the cost of in-process inventory as well as the cost of idle equipment and resulting lost production. They understood that this problem was not Carl’s fault. However, they needed to get his ideas on what changes were needed in the planer depart- ment to relieve this bottleneck. Imagine that! All these bigwigs with graduate degrees from the fanciest business schools in the country asking advice from a poor working slob like him who
had barely made it through high school. He could hardly wait to tell his wife that night.
The meeting had given Carl an opportunity to get two pet peeves off his chest. One peeve is that he has been telling his boss for months that he really needs another planer, but nothing ever gets done about this. His boss just keeps telling him that the planers he already has aren’t being used 100 percent of the time, so how can adding even more capacity be justified? Doesn’t his boss understand about the big backlogs that build up during busy times?
Then there is the other peeve—all those peaks and valleys of work coming to his department. At times, the work just pours in and a big backlog builds up. Then there might be a long pause when not much comes in so the planers stand idle part of the time. If only those departments that are feeding castings to his department could get their act together and even out the work flow, many of his backlog problems would disappear.
Carl was pleased that the bigwigs were nodding their heads in seeming agreement as he described these problems. They really appeared to understand. And they seemed very sincere in thanking him for his good advice. Maybe something is actually going to get done this time.
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be $30 per hour. (This estimate takes into account the fact that, even with an additional planer, the total running time for all the planers will remain the same.) Proposal 2: Eliminate the variability in the interarrival times of the castings, so that the castings would arrive regularly, one every 15 minutes, alternating between platen castings and housing castings. This would require making some changes in the preceding production processes, with an incremental cost of $60 per hour.
These proposals are not mutually exclusive, so any combination can be adopted.
It is estimated that the total cost associated with castings hav- ing to wait to be processed (including processing time) is $200 per hour for each platen casting and $100 per hour for each housing casting, provided the waits are not excessive. To avoid excessive waits for either kind of casting, all the castings are processed as soon as possible on a first-come, first-served basis.
Management’s objective is to minimize the expected total cost per hour.
Use computer simulation to evaluate and compare all the alternatives, including the status quo and the various com- binations of proposals. Then make your recommendation to management.
Are there any other alternatives you would recommend considering?
Here are the details of the situation that Carl and his “big- wigs” are addressing. The company has two planers for cutting flat smooth surfaces in large castings. The planers currently are being used for two purposes. One is to form the top surface of the platen for large hydraulic lifts. The other is to form the mat- ing surface of the final drive housing for a large piece of earth- moving equipment. The time required to perform each type of job varies somewhat, depending largely upon the number of passes that must be made. In particular, for each platen or each housing, the time required by a planer has a translated exponen- tial distribution, where the minimum time is 10 minutes and the additional time beyond 10 minutes has an exponential distribu- tion with a mean of 10 minutes. (A distribution of this type is one of the options in the Queueing Simulator in this chapter’s Excel file.)
Castings of both types arrive one at a time to the planer department. For the castings for forming platens, the arrivals occur randomly with a mean rate of two per hour. For the cast- ings for forming housings, the arrivals again occur randomly with a mean rate of two per hour.
Based on Carl Schilling’s advice, management has asked a management scientist (you) to analyze the following two pro- posals for relieving the bottleneck in the planer department:
Proposal 1: Obtain one additional planer. The total incre- mental cost (including capital recovery cost) is estimated to
Case 12-2
Reducing In-Process Inventory (Revisited)
Reconsider Case 11-2. The current and proposed queueing sys- tems in this case were to be analyzed with the help of queueing models to determine how to reduce in-process inventory as much as possible. However, these same queueing systems also can be effectively analyzed by applying computer simulation with the help of the Queueing Simulator in your MS Courseware.
Use computer simulation to perform all the analysis requested in this case.
Additional Cases Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
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Chapter Thirteen
Computer Simulation with Risk Solver Platform Learning Objectives
After completing this chapter, you should be able to
1. Describe the role of Risk Solver Platform for Education (RSPE) in performing computer simulations.
2. Use RSPE to perform various basic computer simulations that cannot be readily performed with the standard Excel package.
3. Interpret the results generated by RSPE when performing a computer simulation.
4. Use an RSPE feature that enables stopping a simulation run after achieving the desired level of precision.
5. Describe the characteristics of many of the probability distributions that can be incorporated into a computer simulation when using RSPE.
6. Use an RSPE procedure that identifies the continuous distribution that best fits historical data.
7. Use RSPE to generate a parameter analysis report and a trend chart as an aid to decision making.
8. Use RSPE’s Solver to search for an optimal solution for a simulation model.
The preceding chapter presented the basic concepts of computer simulation. Its emphasis throughout was on the use of spreadsheet modeling to perform basic computer simulations. Except for the use of the Queueing Simulator to deal with queueing systems, all of the computer simulations in Chapter 12 were executed using nothing more than the standard Excel package.
Although the standard Excel package has some basic simulation capabilities, an exciting development in recent years has been the development of powerful Excel add-ins that greatly extend these capabilities. One such add-in is the very versatile Frontline Systems product Risk Solver Platform. You already have seen a student-friendly version of this product, Risk Solver Platform for Education (RSPE), in action for various applications in several preceding chap- ters. You will see throughout this chapter that RSPE also has powerful capabilities for per- forming computer simulations. (Chapter 20 on the CD-ROM also does the same for another popular Excel add-in called Crystal Ball.) Instructions for installing RSPE are on a supplemen- tary insert included with the book and also on the book’s website, www.mhhe.com/hillier5e .
This chapter focuses on describing and illustrating the advances in spreadsheet simulation modeling that are made possible by RSPE. Other Excel add-ins for spreadsheet simulation modeling, including Crystal Ball featured in Chapter 20, provide some of the same func- tionality. Section 13.1 begins with a case study that will be revisited in Sections 13.7–13.8. Sections 13.2–13.6 present several other examples of important business problems that can be effectively addressed by using computer simulations with RSPE. Section 13.7 focuses on how to choose the right probability distributions as inputs for a computer simulation. Section 13.8 then describes how parameter analysis reports can be constructed and applied to make a deci- sion about the problem being simulated. Finally, Section 13.9 discusses and illustrates how to use the Solver in RSPE in combination with computer simulation to seek optimal solutions.
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13.1 A CASE STUDY: FREDDIE THE NEWSBOY’S PROBLEM
This case study concerns a newsstand in a prominent downtown location of a major city. The newsstand has been there longer than most people can remember. It has always been run by a well-known character named Freddie. (Nobody seems to know his last name.) His many customers refer to him affectionately as Freddie the newsboy, even though he is considerably older than most of them.
Freddie sells a wide variety of newspapers and magazines. The most expensive of the newspapers is a large national daily called the Financial Journal. Our case study involves this newspaper.
Freddie’s Problem The day’s copies of the Financial Journal are brought to the newsstand early each morning by a distributor. Any copies unsold at the end of the day are returned to the distributor the next morning. However, to encourage ordering a large number of copies, the distributor does give a small refund for unsold copies.
Here are Freddie’s cost figures:
Freddie pays $1.50 per copy delivered. Freddie charges $2.50 per copy. Freddie’s refund is $0.50 per unsold copy.
Partially because of the refund, Freddie always has taken a plentiful supply. However, he has become concerned about paying so much for copies that then have to be returned unsold, particularly since this has been occurring nearly every day. He now thinks he might be better off ordering only a minimal number of copies and saving this extra cost.
To investigate this further, Freddie has been keeping a record of his daily sales. This is what he has found:
• Freddie sells anywhere between 40 and 70 copies on any given day. • The frequency of the numbers between 40 and 70 are roughly equal.
Freddie needs to determine how many copies to order per day from the distributor. His objec- tive is to maximize his average daily profit.
If you have previously studied inventory management in an operations management course, you might recognize this problem as being an example of what is called the news- vendor problem. In fact, we use a basic inventory model to analyze a simplified version of this same case study in Chapter 19 (one of the supplementary chapters on the CD-ROM). However, we will use computer simulation to analyze this problem in this chapter.
A Spreadsheet Model for This Problem Figure 13.1 shows a spreadsheet model for this problem. Given the data cells C4:C6, the deci- sion variable is the order quantity to be entered in cell C9. (The number 60 has been entered arbitrarily in this figure as a first guess of a reasonable value.) The bottom of the figure shows the equations used to calculate the output cells C14:C16. These output cells are then used to calculate the output cell Profit (C18).
The only uncertain input quantity in this spreadsheet is the day’s demand in cell C12. This quantity can be anywhere between 40 and 70. Since the frequency of the numbers between 40 and 70 are about the same, the probability distribution of the day’s demand can be reason- ably assumed to be an integer uniform distribution between 40 and 70 (so all integer values between 40 and 70 are assumed to be equally likely), as indicated in cells D12:F12. Rather than enter a single number permanently into Demand (C12), what RSPE does is enter this probability distribution into this cell. By using RSPE to generate a random observation from this probability distribution, the spreadsheet can calculate the output cells in the usual way. Each time this is done is referred to as a trial by RSPE. By running the number of trials speci- fied by the user (typically hundreds or thousands), the computer simulation thereby generates the same number of random observations of the values in the output cells. RSPE records this
Freddie’s problem involves determining the order quan- tity that will maximize his average daily profit.
A day’s demand for the Financial Journal appears to have a discrete uniform distribution between 40 and 70.
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information for the output cell(s) of particular interest (Freddie’s daily profit) and then, at the end, displays it in a variety of convenient forms that reveal an estimate of the underlying probability distribution of Freddie’s daily profit. (More about this later.)
The Application of RSPE Five steps are taken to use the spreadsheet in Figure 13.1 to perform the computer simulation with RSPE. They are:
1. Define the uncertain variable cells. 2. Define the results cells. 3. Define any statistic cells as desired (e.g., the mean profit). 4. Set the simulation options. 5. Run the simulation.
We now describe each of these five steps in turn.
Define the Uncertain Variable Cells An uncertain variable cell is a cell that has a random value (such as the daily demand for the Financial Journal ). Therefore, an assumed probability distribution must be entered into
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A B C D E F
Freddie the Newsboy
Data Unit Sale Price $2.50
Unit Purchase Cost $1.50
Unit Salvage Value $0.50
Decision Variable Order Quantity 60
Simulation Limit Lower
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Demand 44 Integer Uniform 40 70
Sales Revenue $110.00
Purchasing Cost $90.00
Salvage Value $8.00
Profit $28.00
Mean Profit $46.45
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B C
Sales Revenue
Demand
=UnitSalePrice*MIN(OrderQuantity,Demand)
=PsilntUniform(E12,F12)
Purchasing Cost =UnitPurchaseCost*OrderQuantity
Salvage Value =UnitSalvageValue*MAX(OrderQuantity-Demand,0)
Profit
Mean Profit
=SalesRevenue-PurchasingCost+SalvageValue + PsiOutput()
=PsiMean(C18)
Range Name Cell
Demand C12
MeanProfit
OrderQuantity
C20
C9
Profit C18
PurchasingCost C15
SalesRevenue C14
SalvageValue C16
UnitPurchaseCost C5
UnitSalePrice C4
UnitSalvageValue C6
FIGURE 13.1 A spreadsheet model for applying computer simulation to the case study that involves Freddie the newsboy. The uncertain variable cell is Demand (C12), the results cell is Profit (C18), the statistic cell is MeanProfit (C20), and the decision variable is Order Quantity (C9).
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528 Chapter Thirteen Computer Simulation with Risk Solver Platform
the cell instead of permanently entering a single number. The only uncertain variable cell in Figure 13.1 is Demand (C12).
The following procedure is used to define the uncertain variable cells.
Procedure for Defining an Uncertain Variable Cell 1. Select the cell by clicking on it. 2. Select a probability distribution to enter into the cell by choosing from the Distribution
menu on the RSPE ribbon as shown in Figure 13.2 . 3. Use this distribution dialog to enter the parameters for the distribution, preferably by refer-
ring to the cells in the spreadsheet that contain the values of these parameters. 4. Click on Save.
The Distributions menu mentioned in step 2 provides a wide variety of 46 probability dis- tributions, divided into several submenus, from which to choose. Figure 13.2 displays the eight distributions in the Discrete submenu, but many more distributions are available under the other submenus. (Section 13.7 will focus on the question of how to choose the right distribution.)
Which distribution is appropriate in Freddie’s case? Since newspaper sales are always inte- ger, the Discrete submenu is the right place to look to choose an appropriate distribution. (The various continuous distributions under the other submenus assume fractional values are pos- sible.) Since the frequency of sales between 40 and 70 are all roughly equal, the IntUniform (short for integer uniform) distribution is the best choice. The integer uniform distribution assumes all integer values between some minimum and maximum value are equally likely.
The Distributions menu includes 46 probability distributions.
FIGURE 13.2 The Distributions menu on the RSPE ribbon showing the distributions available under the Discrete submenu. In addition to the 8 distributions displayed here, 38 more distributions are available in the other submenus.
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Selecting the integer uniform distribution in the Distributions menu brings up the dialog box shown in Figure 13.3 , which is used to enter the parameters of the distribution. For each of the parameters (lower and upper), we refer to the data cells in E12 and F12 on the spread- sheet. After clicking Save, RSPE puts a formula in the cell that is used to calculate the random values from the distribution. For the uniform distribution in Demand (C12), that formula is 5 PsiUniform(E12, F12). This formula calculates a random value from the uniform distribution with parameters lower 5 E12 and upper 5 F12. The formula can be copied and pasted just like any other Excel function. (We will see how this is very handy in the example in Section 13.2.)
Define the Results Cells Each output cell that is being used by a computer simulation to forecast a measure of perfor- mance is referred to as a results cell . The spreadsheet model for a computer simulation often does not include an objective cell, but a results cell plays roughly the same role.
The measure of performance of interest to Freddie the newsboy is his daily profit from selling the Financial Journal, so the only results cell in Figure 13.1 is Profit (C18). The fol- lowing procedure is used to define such a results cell.
Procedure for Defining a Results Cell 1. Select the cell by clicking on it. 2. Choose Output/In Cell from the Results menu on the RSPE ribbon.
In Figure 13.1 , the results cell (C18) shows a profit value of $50. It is important to note that this is only the result for the particular random value of the uncertain variable that is currently showing in the spreadsheet (a Demand of 55). This is not the result for the entire simulation run. It is not even the mean profit from the entire run. It is just one single random outcome (a trial). To obtain the results for the entire simulation run, hovering on this cell will reveal a chart that shows all of the results (more on this later).
Define a Statistic Cell (or Cells) Since the number in the results cell only gives the result for a single trial of the simula- tion (before hovering over the cell to show more results), it can be useful to show statistics
RSPE Tip: Rather than entering raw numbers, use cell references for the dis- tribution parameters (e.g., E12 and F12). This allows changes to be made directly on the spreadsheet rather than having to dig into RSPE dialog boxes.
The value in the results cell shows just one single possible outcome from a single trial of the simulation for the particular random value(s) that are currently in the uncertain variable cell(s). This is not the result of the entire simulation run. Hover over the results cell to reveal a chart showing a summary of the complete simulation results.
FIGURE 13.3 The dialog box used to specify the parameters for the integer uniform distribution in the uncertain variable cell, Demand (C12), for the spreadsheet model in Figure 13.1. The two parameters for the integer uniform distribution are lower and upper and are entered here as cell references to E12 (40) and F12 (70), respectively.
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(measures of performance) directly on the spreadsheet that summarize the results of the entire simulation run. RSPE refers to such cells as statistic cells. In Figure 13.1 , cell C20 is defined as a statistic cell to show the mean value of profit ($46.45). RSPE uses the following procedure to define a statistic cell:
Procedure for Defining a Statistic Cell 1. Click on the results cell for which you want to show a statistic. 2. Choose the statistic you want to show (e.g., Mean) under the Statistic submenu of the
Results menu on the RSPE ribbon, as shown in Figure 13.4 . 3. Click on the statistic cell in which you want the value of the statistic to be shown.
Set the Simulation Options The fourth step—setting simulation options—refers to such things as choosing the number of trials to run and deciding on other options regarding how to perform the computer simulation. This step begins by clicking on the options button on the RSPE ribbon and selecting the Sim- ulation tab. This brings up the Simulation Options dialog box shown in Figure 13.5 . Perhaps the most important option is how many trials to run in the simulation. The figure indicates that 1,000 trials will be run. Other options allow you to change the sampling method or the random number generator that is used by RSPE. We will keep these at their default values.
Run the Simulation At this point, the stage is set to run the computer simulation. In fact, the simulation may already have been run behind the scenes. As seen in either Figure 13.2 or 13.4 , the Simulate button on the RSPE ribbon contains a lightbulb. If the lightbulb is lit (appears yellow), this means RSPE is in interactive simulation mode. In this mode, every time a change in the model is made, the simulation runs automatically in the background and the results are imme- diately updated. So if the lightbulb is lit, the simulation has already been run and the results
FIGURE 13.4 The Results menu on the RSPE ribbon that shows the statistics available under the Statistic submenu. Choosing a statistic from this submenu will cause that statistic to be calculated for the current simulation run. The value of this statistic then will appear within a specified statistic cell.
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are ready for viewing. For small and medium-sized models, the simulation runs so quickly that you will not even notice the work going on in the background.
If the lightbulb is not lit (appears gray), then RSPE will only run the simulation when it is instructed to do so. To run the simulation, you can turn on interactive simulation by clicking on the Simulate button. Alternatively, you can run the simulation model just once by clicking and holding on the Simulate button to reveal its menu, and then choosing Run Once.
With interactive simulation mode on, the statistic cells will always show the results of the latest simulation run. For example, in Figure 13.1 , the statistic cell MeanProfit (C20) shows that the mean value of Freddie’s daily profit is $46.45. To view more extensive results, hover the mouse over the results cell Profit (C18). This will cause a chart to appear that shows a quick summary of all of the results along with a button labeled Click here to open full chart. Clicking on this button reveals the results shown in Figure 13.6 .
The default view is a frequency chart shown on the left side and a statistics table shown on the right side. The height of the vertical lines in the frequency chart indicates the relative fre- quency of the various profit values that were obtained during the simulation run. For example, consider the tall vertical line at $60. The right-hand side of the chart indicates a frequency of about 350 there, which means that about 350 of the 1,000 trials led to a profit of $60. Thus, the left-hand side of the chart indicates that the estimated probability of a profit of $60 is 350/1000 5 0.35. This is the profit that results whenever the demand equals or exceeds the order quantity of 60. The remainder of the time, the profit was scattered fairly evenly between $20 and $60. These profit values correspond to trials where the demand was between 40 and 60 units, with lower profit values corresponding to demands closer to 40 and higher profit values corresponding to demands closer to 60.
The statistics table on the right side of Figure 13.6 summarizes the outcome of the 1,000 trials of the computer simulation. These 1,000 trials provide a sample of 1,000 random
RSPE Tip: If either a simu- lation model is very large or a huge number of simu- lation trials are being run, interactive simulation can cause delays as you build your model. In this case, you may want to turn off interactive simulation until the model is completely built.
FIGURE 13.5 The RSPE Options dialog box after showing the Simulation tab.
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observations from the underlying probability distribution of Freddie’s daily profit. The most interesting statistics about this sample provided by the table include the mean of $46.45, the standard deviation of $13.67 (a measure of the variability of the profit values from the trials) and the mode of $60 (meaning that this was the profit value that occurred most frequently). The information further down the table regarding the minimum and maximum profit values also is particularly useful.
In addition to the frequency chart and statistics table presented in Figure 13.6 , there are other useful ways of displaying the results of a simulation run. By clicking on the appropriate tab at the top of the frequency chart, you can display a cumulative frequency, reverse cumula- tive frequency, sensitivity, or scatter plot chart. Also, the menu above the statistics table lets you choose whether to show statistics or a percentiles table (as well as giving choices for changing various options in the charts). Figure 13.7 shows the cumulative frequency chart on the left and the percentiles table on the right that resulted from the current simulation run. The percentiles table is based on listing the profit values generated by the 1,000 trials from small- est to largest, dividing this list into 100 equal parts (10 values in each), and then recording the value at the end of each part. Thus, the value 5 percent through the list is $22, the value 10 percent through the list is $26, and so forth. (For example, the intuitive interpretation of the 10 percent percentile of $26 is that 10 percent of the trials have profit values less than or equal to $26 and the other 90 percent of the trials have profit values greater than or equal to $26, so $26 is the dividing line between the smallest 10 percent of the values and the largest 90 percent.) The cumulative frequency chart on the left of Figure 13.7 provides similar (but more detailed) information about this same list of the smallest-to-largest profit values. The
In general, the x percent percentile is the dividing line between the smallest x percent of the values and the rest of the values.
FIGURE 13.6 The frequency chart and statistics table provided by RSPE to summarize the results of running the simulation model in Figure 13.1 for the case study that involves Freddie the newsboy.
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13.1 A Case Study: Freddie the Newsboy’s Problem 533
horizontal axis shows the entire range of values from the smallest possible profit value ($20) to the largest possible profit value ($60). For each value in this range, the chart cumulates the number of actual profits generated by the 1,000 trials that are less than or equal to that value. This number equals the frequency shown on the right or, when divided by the number of tri- als, the probability shown on the left.
Figure 13.8 illustrates another of the many helpful ways provided by RSPE for extracting helpful information from the results of a simulation run. Freddie the newsboy feels that he has
FIGURE 13.7 Two more ways (a cumulative frequency chart and a percentiles table) RSPE can display the results of running the simulation model in Figure 13.1 for the case study that involves Freddy the newsboy.
FIGURE 13.8 After setting a lower cutoff of $40 for desirable profit values, the Likelihood box reveals that 64.5 percent of the trials in Feddie’s simulation run provided a profit at least this high.
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had a reasonably successful day if he obtains a profit of at least $40 from selling the Finan- cial Journal. Therefore, he would like to know the percentage of days that he could expect to achieve this much profit if he were to adopt the order quantity currently being analyzed (60). To obtain an estimate of this percentage with RSPE, enter $40 as the Lower Cutoff in the Chart Statistics on the right side of Figure 13.8 . The estimate of this percentage (64.5 percent) then appears in the Likelihood box just below (and is also displayed above the chart on the left side). If desired, the probability of obtaining a profit between any two values also could be estimated by entering both a Lower Cutoff and an Upper Cutoff.
How Accurate Are the Simulation Results? An important number provided by Figure 13.6 is the mean of $46.45. This number was calcu- lated as the average of the 1,000 random observations from the underlying probability distri- bution of Freddie’s daily profit that were generated by the 1,000 trials. This sample average of $46.45 thereby provides an estimate of the true mean of this distribution. The true mean might deviate somewhat from $46.45. How accurate can we expect this estimate to be?
The answer to this key question is provided by the standard error of $0.43 given at the bottom of the statistics table in Figure 13.6 . In particular, the true mean can readily deviate from the sample mean by any amount up to the mean standard error, but most of the time (approximately 68 percent of the time), it will not deviate by more than that. Thus, the inter- val from $46.45 2 $0.43 5 $46.02 to $46.45 1 $0.43 5 $46.88 is a 68 percent confidence interval for the true mean. Similarly, a larger confidence interval can be obtained by using an appropriate multiple of the mean standard error to subtract from the sample mean and then to add to the sample mean. For example, the appropriate multiple for a 95 percent confidence interval is 1.965, so such a confidence interval ranges from $46.45 2 1.965($0.43) 5 $45.60 to $46.45 1 1.965($0.43) 5 $47.30. (This multiple of 1.965 will change slightly if the number of trials is different from 1,000.) Therefore, it is very likely that the true mean is somewhere between $45.60 and $47.30. RSPE provides a shortcut for calculating the 95 percent confidence interval. The Mean Confidence 95% value of $0.85 in the statistics table shows that the 95 percent confidence interval ranges from $46.45 2 $0.85 5 $45.60 to $46.45 1 $0.85 5 $47.30.
If greater precision is required, the mean standard error normally can be reduced by increasing the number of trials in the simulation run. However, the reduction tends to be small unless the number of trials is increased substantially. For example, cutting the mean standard error in half requires approximately quadrupling the number of trials. Thus, a surprisingly large number of trials may be required to obtain the desired degree of precision.
Freddie’s Conclusions The results presented in Figure 13.6 were from a simulation run that fixed Freddie’s daily order quantity at 60 copies of the Financial Journal (as indicated in cell C9 of the spread- sheet in Figure 13.1 ). Freddie wanted this order quantity tried first because it seems to pro- vide a reasonable compromise between being able to fully meet the demand on many days (about two-thirds of them) while often not having many unsold copies on those days. How- ever, the results obtained do not reveal whether 60 is the optimal order quantity that would maximize his average daily profit. Many more simulation runs with other order quantities will be needed to determine (or at least estimate) the optimal order quantity. Utilizing the interactive simulation mode of RSPE, it is fairly straightforward to do this by trial-and-error. Enter a new value for OrderQuantity (C9) in the model, as shown in Figure 13.9 , and (when the interactive simulation mode is turned on) a new simulation is immediately run and a new value for the mean profit appears nearly instantaneously in the statistic cell, Mean- Profit (C20). With just a single decision variable to vary, trial-and-error can be used to fairly quickly zero in on the optimal order quantity. (Try it!) Decreasing the order quantity from 60 by one unit at a time reveals that the mean profit increases. This most likely would continue until the order quantity is decreased below 55, at which point the mean profit instead starts to decrease. (Remember, however, that the calculated mean profit from a simulation run is only an estimate, so the change in direction might not occur exactly at 55 because of random errors.) Thus, as shown in Figure 13.9 , trial-and-error would suggest that when the order
The Likelihood box gives the percentage of the trials that generated values between the lower cutoff and the upper cutoff.
The standard error speci- fies how close the mean obtained in a simulation run is likely to be to the true mean.
With the interactive simula- tion mode turned on, it is fairly straightforward to use trial-and-error to search for an optimal solution. Simply try different values in the decision variable cell(s) and observe the simulation results that are automati- cally calculated in the sta- tistic cell(s).
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quantity is 55, the mean profit reaches its maximum value of $47.26 according to the current series of simulation runs.
Section 13.8 will describe a more systematic way to compare different order quantities with a parameter analysis report (as used in Chapter 5 for optimization models) and a trend chart. Section 13.9 then describes how the Solver included with RSPE (as discussed in Chap- ters 2 through 8) can be used in combination with simulation to search for the optimal order quantity. That presentation will bring the case study to a close.
Although there is still much more to come, Freddie already has learned from Figure 13.6 that an order quantity of 60 would provide a nice average daily profit of approximately $46.45. This is only an estimate, but Freddie also has learned from the 95 percent confidence interval that the average daily profit probably would turn out to be somewhere between $45.60 and $47.30. Freddie has also learned from Figure 13.9 that a smaller order quantity of 55 would provide an even higher average daily profit of approximately $47.26.
However, these profit figures provided by computer simulation are based on the assump- tion in the spreadsheet model (see cells D12:F12 in Figure 13.1 ) that demand has an integer uniform distribution between 40 and 70. Therefore, these profit figures will be correct only if this assumption is a valid one. More work is needed to either verify that this assumed distribu- tion is the appropriate one or to identify another probability distribution that provides a better fit to the data of daily demands that Freddie actually has been experiencing. This issue will be explored further in Section 13.7.
Computer simulation with RSPE is an extremely versatile tool that can address a myriad of managerial issues. Therefore, before continuing the case study in Sections 13.7–13.8, we turn in the intervening sections to presenting five additional examples of the application of computer simulation with RSPE.
Freddie now has 95 percent confidence that an order quantity of 60 would pro- vide an average daily profit between $45.60 and $47.30 in the long run.
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A B C D E F
Freddie the Newsboy
Data Unit Sale Price $2.50
Unit Purchase Cost $1.50
Unit Salvage Value $0.50
Decision Variable Order Quantity 55
Simulation Limit Lower
Limit
Upper
Demand 40 Integer Uniform 40 70
Sales Revenue $100.00
Purchasing Cost $82.50
Salvage Value $7.50
Profit $25.00
Mean Profit $47.26
FIGURE 13.9 This figure illustrates what can happen when the decision variable, OrderQuantity (C20), is changed by trial-and-error and the statis- tic cell, MeanProfit (C20), is observed. When the OrderQuantity is 55, the MeanProfit (C20) reaches its maximum value of $47.26.
1. What is the decision that Freddie the newsboy needs to make? 2. What is an uncertain variable cell when using RSPE? 3. What is a results cell when using RSPE? 4. What is entered into the results cell in Freddie’s spreadsheet model?
Review Questions
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5. What is a statistic cell when using RSPE? 6. What kind of information is provided by a frequency chart when using RSPE? 7. What are the key statistics provided by the statistics table when using RSPE? 8. What is the significance of the standard error of the results from a computer simulation? 9. What has Freddie the newsboy learned so far about what his order quantity should be?
13.2 BIDDING FOR A CONSTRUCTION PROJECT: A PRELUDE TO THE RELIABLE CONSTRUCTION CO. CASE STUDY
Managers frequently must make decisions whose outcomes will be greatly affected by the corresponding decisions being made by the management of competitor firms. For example, marketing decisions often fall into this category. To illustrate, consider the case in which a manager must determine the price for a new product being brought to market. How well this decision works out will depend greatly on the pricing decisions being made nearly simultane- ously by other firms marketing competitive new products. Similarly, the success of a decision on how soon to market a product under development will be determined largely by whether this product reaches the market before competitive products are released by other firms.
When a decision must be made before learning the corresponding decisions being made by competitors, the analysis needs to take into account the uncertainty surrounding what com- petitors’ decisions will be. Computer simulation provides a natural way of doing this by using uncertain variable cells to represent competitors’ decisions.
The following example illustrates this process by considering a situation where the deci- sion being made is the bid to submit on a construction project while three other companies are simultaneously preparing their own bids.
The Reliable Construction Co. Bidding Problem The case study carried throughout Chapter 16 on the CD-ROM involves the Reliable Con- struction Co. and its project to construct a new plant for a major manufacturer. That chapter describes how the project manager (David Perty) made extensive use of PERT/CPM models to help guide his management of the project.
As the opening sentence of Section 16.1 indicates, this case study begins as the company has just made the winning bid of $5.4 million to do this project. We now will back up in time to describe how the company’s management used computer simulation with RSPE to guide its choice of $5.4 million as its bid for the project. You will not need to review the case study in Chapter 16 to follow this example.
Reliable’s first step in this process was to estimate what the company’s total cost would be if it were to undertake the project. This was determined to be $4.55 million. (This amount excludes the penalty for missing the deadline for completion of the project, as well as the bonus for completion well before the deadline, since management considers either event to be rela- tively unlikely.) There also is an additional cost of approximately $50,000 for preparing the bid, including estimating the project cost and analyzing the bidding strategies of the competition.
Three other construction companies also were invited to submit bids for this project. All three have been long-standing competitors of the Reliable Construction Co., so the company has had a great deal of experience in observing their bidding strategies. A veteran analyst in the bid preparation office has taken on the task of estimating what bid each of these competi- tors will submit. Since there is so much uncertainty in this process, the analyst has determined that each of these estimates needs to be in the form of a probability distribution. Competitor 1 is known to use a 30 percent profit margin above the total (direct) cost of a project in set- ting its bid. However, competitor 1 also is a particularly unpredictable bidder because of an inability to estimate the true costs of a project with much accuracy. Its actual profit margin on past bids has ranged from as low as minus 5 percent to as high as 60 percent. Competitor 2 uses a 25 percent profit margin and is somewhat more accurate than competitor 1 in estimat- ing project costs, but it still has set bids in the past that have missed this profit margin by as much as 15 percent in either direction. On the other hand, competitor 3 is unusually accurate
This section reveals how the company chose the bid of $5.4 million, which won the contract.
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13.2 Bidding for a Construction Project: A Prelude to the Reliable Construction Co. Case Study 537
in estimating project costs (as is the Reliable Construction Co.). Competitor 3 also is adept at adjusting its bidding strategy, so it is equally likely to set its profit margin anywhere between 20 and 30 percent, depending on its assessment of the competition, its current backlog of work, and various other factors.
This information about the competitors is invaluable, but the analyst who developed it knew that her work wasn’t quite done yet. Based on these numbers, she still needed to develop an estimate of the probability distribution of what the bid will be for each of the competitors.
This task is relatively straightforward in the case of competitor 3. Because the analyst esti- mates that this competitor is equally likely to set its profit margin anywhere between 20 and 30 percent, its bid then is equally likely to be anywhere between 120 and 130 percent of the total project cost. The probability distribution that fits this is the uniform distribution between 120 and 130 percent.
However, this task is not as easy when considering competitors 1 and 2. Fortunately, the analyst has been able to estimate three key numbers for each competitor—a minimum value, a most likely value, and a maximum value—for the profit margin and so (by adding 100 percent) for the bid as a percentage of the total project cost. For example, the analyst has estimated that the bid of competitor 1 (expressed as a percentage of total project cost) has a minimum value of 95 percent, a most likely value of 130 percent, and a maximum value of 160 percent. (The corresponding numbers for competitor 2 are 110 percent, 125 percent, and 140 percent, respectively.) There is a particularly convenient type of probability distribution called the tri- angular distribution that is based on these same three kinds of numbers. Figure 13.10 shows the shape of a triangular distribution. Its three parameters are min (the minimum value), likely (the most likely value), and max (the maximum value). ( Figure 13.10 shows likely as being much closer to min than to max, but it actually can be anywhere between min and max.) These three parameters are a perfect fit for the distributions of the bids from competitors 1 and 2, so the analyst has chosen a triangular distribution as her best estimate of these distribu- tions. (This is not surprising since triangular distributions are a particularly popular choice for performing computer simulations.)
In summary, the estimated probability distributions of the bids that the three competitors will submit, expressed as a percentage of Reliable’s assessment of the total project cost ($4.55 million), are as follows.
Competitor 1: A triangular distribution with a minimum value of 95 percent, a most likely value of 130 percent, and a maximum value of 160 percent. Competitor 2: A triangular distribution with a minimum value of 110 percent, a most likely value of 125 percent, and a maximum value of 140 percent. Competitor 3: A uniform distribution between 120 percent and 130 percent.
A Spreadsheet Model for Applying Computer Simulation Figure 13.11 shows the spreadsheet model that has been formulated to evaluate any pos- sible bid that Reliable might submit. Since there is uncertainty about what the competitors’
min likely max
Triangular distribution
FIGURE 13.10 The shape of a trian- gular distribution and the location of its three parameters: (1) min (the minimum possible value), (2) likely (the most likely value), and (3) max (the maximum possible value).
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Our Project Cost ($million)
Our Bid Cost ($million)
Bid ($million)
Distribution
Minimum
Most Likely
Maximum
Minimum
Most Likely
Maximum
Minimum Competitor Bid ($million)
Our Bid ($million)
Win Bid?
Profit ($million)
Data
Competitor Bids
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TriangularTriangular Uniform
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Competitor Distribution Parameters (Proportion of Our Project Cost)
Competitor Distribution Parameters ($millions)
18 5.005
5.688
6.370
(1=yes, 0=no)
5.460
5.915
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A B C D E
Reliable Construction Co. Contract Bidding
CompetitorBids
MeanProfit
MinimumCompetitorBid OurBid
OurBidCost
OurProjectCost
Profit
WinBid?
C8:E8
C31
C23 C25
C5
C4
C29
C27
Range Name Cells
18 Minimum
Most Likely
Maximum
19
=OurProjectCost*C13
=OurProjectCost*C14
=OurProjectCost*C15
=OurProjectCost*D13
=OurProjectCost*D14
=OurProjectCost*D15
=OurProjectCost*E13
=OurProjectCost*E1520
B C D
Bid ($million) =PsiTriangular(C18,C19,C20)
Competitor 1 Competitor 2 Competitor 3
=PsiTriangular(D18,D19,D20) =PsiUniform(E18,E20)
E
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B C
0.4872
=PsiMean(C29)
D E
22 Minimum Competitor Bid ($million)
Our Bid ($million)
Win Bid?
Profit ($million)
23 =MIN(C8:E8)
5.4
=IF(OurBid<MinimumCompetitorBid,1,0)
=WinBid?*(OurBid-OurProjectCost)-OurBidCost + PsiOutput()
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B C
4.550
0.050
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6.810
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95%
130%
160%
110%
125%
140%
120%
130%
4.323
5.915
7.280
5.771
5.4
1
0.800
Competitor Bids
Mean Profit ($million)
Mean Profit ($million)
FIGURE 13.11 A spreadsheet model for applying computer simulation to the Reliable Construction Co.’s contract bidding problem. The uncertain variable cells are CompetitorBids (C8:E8), the results cell is Profit (C29), the statistic cell is MeanProfit (C31), and the decision variable is OurBid (C25).
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13.2 Bidding for a Construction Project: A Prelude to the Reliable Construction Co. Case Study 539
bids will be, this model needs CompetitorBids (C8:E8) to be uncertain variable cells, so the above probability distributions are entered into these cells. As described in the preced- ing section, this is done by selecting each cell in turn, choosing the appropriate distribution from the Distributions menu on the RSPE ribbon (in this case under the Common sub- menu), which brings up the dialog box for that distribution. Figure 13.12 shows the Trian- gular Distribution dialog box that has been used to set the parameter values (min, likely, and max) for competitor 1, and competitor 2 would be handled similarly. These parameter values for competitor 1 come from cells C18:C20, where the parameters in percentage terms (cells C13:C15) have been converted to dollars by multiplying them by OurProject- Cost (C4). The Uniform Distribution dialog box is used instead to set the parameter values for competitor 3 in cell E8.
MinimumCompetitorBid (C23) records the smallest of the competitors’ bids for each trial of the computer simulation. The company wins the bid on a given trial only if the quantity entered into OurBid (C25) is less than the smallest of the competitors’ bids. The IF function entered into WinBid? (C27) then returns a 1 if this occurs and a 0 otherwise.
Since management wants to maximize the expected profit from the entire process of deter- mining a bid (if the bid wins) and then doing the project, the results cell in this model is Profit (C29). The profit achieved on a given trial depends on whether the company wins the bid. If not, the profit actually is a loss of $50,000 (the bid cost). However, if the bid wins, the profit is the amount by which the bid exceeds the sum of the project cost and the bid cost. The equation entered into Profit (C29) performs this calculation for whichever case applies. Profit (C29) is defined as a results cell by clicking on the cell and then choosing Output/In Cell from the Results menu on the RSPE ribbon. Finally, MeanProfit (C31) is defined as a statistic cell by selecting the Profit cell (C29), choosing Mean from the Statistic submenu of the Results menu, and then clicking in cell C31. This will show the mean value of the profit after the simulation is run.
Here is a summary of the key cells in this model.
Uncertain variable cells: CompetitorBids (C8:E8) Decision variable: OurBid (C25) Results cell: Profit (C29) Statistic cell: MeanProfit (C31) (See Section 13.1 for the details regarding
how to define these kinds of cells.)
The bids that other compa- nies will submit are uncer- tain, so they need to be uncertain variable cells.
The objective is to deter- mine the bid that would maximize the resulting mean profit.
FIGURE 13.12 The Triangular Distribution dialog box. It is being used here to enter a triangular distribution with the parameters min 5 C18 (4.323), likely 5 C19 (5.915), and max 5 C20 (7.280) into the uncertain variable cell C8 in the spreadsheet model in Figure 13.11.
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540 Chapter Thirteen Computer Simulation with Risk Solver Platform
The Simulation Results To evaluate a possible bid of $5.4 million entered into OurBid (C25), a computer simulation of this model ran for 1,000 trials. Figure 13.13 shows the results in the form of a frequency chart and a statistics table. Using units of millions of dollars, the profit on each trial has only two possible values, namely, a loss shown as –0.050 in these figures (if the bid loses) or a profit of 0.800 (if the bid wins). The frequency chart indicates that this loss of $50,000 occurred on about 380 of the 1,000 trials whereas the profit of $800,000 occurred on the other 620 trials. This resulted in a mean profit of 0.487 ($487,000) from all 1,000 trials, as well as the other statistics recorded in the statistics table.
By themselves, these results do not show that $5.4 million is the best bid to submit. We still need to estimate with additional simulation runs whether a larger expected profit could be obtained with another bid value. As seen in the preceding section, this can be done with trial- and-error. Alternatively, Section 13.8 will describe how doing this with a parameter analysis report leads to choosing $5.4 million as the bid. This turned out to be the winning bid for the Reliable Construction Co., which then led into the case study for Chapter 16.
This story will continue in Section 13.8.
FIGURE 13.13 The frequency chart and statistics table that summarize the results of running the simulation model in Figure 13.11 for the Reliable Construction Co. contract bidding problem.
1. What is the project for which the Reliable Construction Co. is submitting a bid? 2. The bids of the competitors are being estimated in what form? 3. What are the quantities in the uncertain variable cells in this example’s spreadsheet model for
applying computer simulation? 4. What quantity appears in the results cell for this spreadsheet model? 5. What are the possible outcomes on each trial of this computer simulation?
Review Questions
13.3 PROJECT MANAGEMENT: REVISITING THE RELIABLE CONSTRUCTION CO. CASE STUDY
One of the most important responsibilities of a project manager is to meet the deadline that has been set for the project. Therefore, a skillful project manager will revise the plan for con- ducting the project as needed to ensure a strong likelihood of meeting the deadline. But how does the project manager estimate the probability of meeting the deadline with any particular
Computer simulation improves upon a PERT/ CPM method for estimating the probability of complet- ing a project by the deadline.
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13.3 Project Management: Revisiting the Reliable Construction Co. Case Study 541
plan? Section 16.4 describes one method provided by PERT/CPM. We now will illustrate how computer simulation provides a better method.
This example illustrates a common role for computer simulation—refining the results from a preliminary analysis conducted with approximate mathematical models. You also will get a first look at uncertain variable cells where the values shown are times. Another interesting feature of this example is its use of a special kind of RSPE chart called the sensitivity chart. This chart will provide a key insight into how the project plan should be revised.
The Problem Being Addressed Like the example in the preceding section, this one also revolves around the Reliable Con- struction Co. case study introduced in Section 16.1 and continued throughout Chapter 16. However, rather than preceding the part of the story described in Chapter 16, this example arises in the middle of the case study. In particular, Section 16.4 discusses how a PERT/CPM procedure was used to obtain a rough approximation of the probability of meeting the dead- line for the Reliable Construction Co. project. It then was pointed out that computer simula- tion could be used to obtain a better approximation. We now are in a position to describe how this is done.
Here are the essential facts about the case study that are needed for the current example. (There is no need for you to refer to Chapter 16 for further details.) The Reliable Construc- tion Company has just made the winning bid to construct a new plant for a major manufac- turer. However, the contract includes a large penalty if construction is not completed by the deadline 47 weeks from now. Therefore, a key element in evaluating alternative construc- tion plans is the probability of meeting this deadline under each plan. There are 14 major activities involved in carrying out this construction project, as listed on the right-hand side of Figure 13.14 (which repeats Figure 16.1 for your convenience). The project network in this figure depicts the precedence relationships between the activities. Thus, there are six
The deadline for complet- ing the project is 47 weeks from now.
Start
Finish
A
B
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D I
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K L
E
FG
M
H
0
2
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6 4
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54
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Activity Code
A. Excavate B. Foundation C. Rough wall D. Roof E. Exterior plumbing F. Interior plumbing G. Exterior siding H. Exterior painting I. Electrical work J. Wallboard K. Flooring L. Interior painting M. Exterior fixtures N. Interior fixtures
FIGURE 13.14 The project network for the Reliable Construction Co. project.
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542 Chapter Thirteen Computer Simulation with Risk Solver Platform
sequences of activities (paths through the network), all of which must be completed to finish the project. These six sequences are listed below.
Path 1: Start → A → B → C → D → G → H → M → Finish Path 2: Start → A → B → C → E → H → M → Finish Path 3: Start → A → B → C → E → F → J → K → N → Finish Path 4: Start → A → B → C → E → F → J → L → N → Finish Path 5: Start → A → B → C → I → J → K → N → Finish Path 6: Start → A → B → C → I → J → L → N → Finish
The numbers next to the activities in the project network represent the estimates of the num- ber of weeks the activities will take if they are carried out in the normal manner with the usual crew sizes, and so forth. Adding these times over each of the paths (as done in Table 16.2) reveals that path 4 is the longest path, requiring a total of 44 weeks. Since the project is fin- ished as soon as its longest path is completed, this indicates that the project can be completed in 44 weeks, 3 weeks before the deadline.
Now we come to the crux of the problem. The times for the activities in Figure 13.14 are only estimates, and there actually is considerable uncertainty about what the duration of each activity will be. Therefore, the duration of the entire project could well differ substantially from the estimate of 44 weeks, so there is a distinct possibility of missing the deadline of 47 weeks. What is the probability of missing this deadline? To estimate this probability, we need to learn more about the probability distribution of the duration of the project.
This is the reason for the PERT three-estimate approach described in Section 16.4. This approach involves obtaining three estimates—a most likely estimate, an optimistic estimate, and a pessimistic estimate —of the duration of each activity. (Table 16.4 lists these estimates for all 14 activities for the project under consideration.) These three quantities are intended to estimate the most likely duration, the minimum duration, and the maximum duration, respec- tively. Using these three quantities, PERT assumes (somewhat arbitrarily) that the form of the probability distribution of the duration of an activity is a beta distribution. By also mak- ing three simplifying approximations (described in Section 16.4), this leads to an analytical method for roughly approximating the probability of meeting the project deadline.
One key advantage of computer simulation is that it does not need to make most of the simplifying approximations that may be required by analytical methods. Another is that there is great flexibility about which probability distributions to use. It is not necessary to choose an analytically convenient one.
When dealing with the duration of an activity, computer simulations commonly use a tri- angular distribution as the distribution of this duration. The triangular distribution fits the PERT three-estimate approach very well because it has three parameters that correspond to the three estimates in a very natural way. Figure 13.10 shows the shape of this distribution and its three parameters— min (the minimum possible value), likely (the most likely value), and max (the maximun possible value). Thus, the duration of an activity is assumed to have a triangular distribution where min 5 optimistic estimate, likely 5 most likely estimate, and max 5 pessimistic estimate. For each uncertain variable cell containing this distribution, a Triangular Distribution dialog box (such as the one shown in Figure 13.12 ) is used to enter the values of the three estimates by entering their respective cell references into the min, likely, and max boxes.
A Spreadsheet Model for Applying Computer Simulation Figure 13.15 shows a spreadsheet model for simulating the duration of the Reliable Construc- tion Co. project. The values of o, m, and p in columns D, E, and F are obtained directly from Table 16.4. The equations entered into the cells in columns G and I give the start times and finish times for the respective activities. For each trial of the simulation, the maximum of the finish times for the last two activities (M and N) gives the duration of the project (in weeks), which goes into the results cell ProjectCompletion (I21).
Since the activity times generally are variable, the cells H6:H19 all need to be uncer- tain variable cells. Figure 13.16 shows the Triangular Distribution dialog box after it has
Computer simulation has two key advantages over analytical methods like PERT/CPM.
Variable activity times need to be uncertain variable cells.
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13.3 Project Management: Revisiting the Reliable Construction Co. Case Study 543
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Project Completion
Mean Project Completion
A B C D E F G H I
Start
Time
0
2.28
5.80
21.17
21.17
25.35
28.04
34.83
21.17
30.80
37.44
37.44
42.16
42.89
Finish
Time
2.28
5.80
21.17
28.04
25.35
30.80
34.83
42.16
24.72
37.44
41.44
42.89
44.63
48.67
Simulation of Reliable Construction Co. Project
Activity
A
B
C
D
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G
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I
J
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L
M
N
Immediate
Predecessor
—
A
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o
1
2
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1
1
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Time Estimates
m
2
3.5
9
5.5
4.5
4
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8
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2
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p
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Activity
Time
(triangular)
2.28
3.52
15.37
6.86
4.18
5.44
6.80
7.33
3.55
6.64
48.67
4.00
5.45
2.47
5.77
46.26
3 Activity
Time
(triangular)
4 Start
Time
Finish
Time5
6 0
=AFinish
=BFinish
=CFinish
=CFinish
=EFinish
=DFinish
=MAX(EFinish,GFinish)
=CFinish
=MAX(FFinish,IFinish)
=JFinish
=JFinish
=HFinish
=MAX(KFinish,LFinish)
=PsiTriangular(D6,E6,F6)
=PsiTriangular(D7,E7,F7)
=PsiTriangular(D8,E8,F8)
=AStart+ATime
=BStart+BTime
=CStart+CTime
=DStart+DTime
=EStart+ETime
=FStart+FTime
=GStart+GTime
=HStart+HTime
=IStart+ITime
=JStart+JTime
=KStart+KTime
=LStart+LTime
=MStart+MTime
=NStart+NTime
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Project Completion
Mean Project Completion =PsiMean(I21)
G H I
=MAX(MFinish,NFinish) + PsiOutput()
=PsiTriangular(D9,E9,F9)
=PsiTriangular(D10,E10,F10)
=PsiTriangular(D11,E11,F11)
=PsiTriangular(D12,E12,F12)
=PsiTriangular(D13,E13,F13)
=PsiTriangular(D14,E14,F14)
=PsiTriangular(D15,E15,F15)
=PsiTriangular(D16,E16,F16)
=PsiTriangular(D17,E17,F17)
=PsiTriangular(D18,E18,F18)
=PsiTriangular(D19,E19,F19)
AFinish AStart ATime BFinish BStart BTime CFinish CStart CTime DFinish DStart DTime EFinish EStart ETime FFinish FStart FTime GFinish GStart GTime HFinish HStart HTime IFinish IStart ITime JFinish JStart JTime KFinish KStart KTime LFinish LStart LTime MFinish MeanProjectCompletion MStart MTime NFinish NStart NTime ProjectCompletion
I6 G6 H6 I7 G7 H7 I8 G8 H8 I9 G9 H9 I10 G10 H10 I11 G11 H11 I12 G12 H12 I13 G13 H13 I14 G14 H14 I15 G15 H15 I16 G16 H16 I17 G17 H17 I18 I23 G18 H18 I19 G19 H19 I21
Range Name Cell
FIGURE 13.15 A spreadsheet model for applying computer simulation to the Reliable Construction Co. project scheduling problem. The uncertain variable cells are cells H6:H15. The results cell is ProjectCompletion (I21). The statistic cell is MeanProjectCompletion (I23).
been used to specify the parameters for the first uncertain variable cell, which records the time of activity A with a range name of ATime (H6). The right side of Figure 13.16 notes that RSPE has automatically entered a formula ( 5 PsiTriangular(D6,E6,F6)) into ATime (H6) to calculate a random value from this distribution. Rather than repeating this process
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544 Chapter Thirteen Computer Simulation with Risk Solver Platform
for all the other uncertain variable cells, it is quicker to simply copy and paste. To copy the formula in H6 down to H7 through H19, select cell H6 and drag the fill handle (the small box on the lower right corner of the cell cursor) down to cell H19. This copies the formula in H6 ( 5 PsiTriangular(D6,E6,F6), used by RSPE to calculate a random value from the triangular distribution, with parameters min 5 D6, likely 5 E6, and max 5 F6) into cells H7 through H19. Since the parameters in cell H6 (D6, E6, and F6) are relative references, the row numbers of the parameters will update appropriately to refer to the data in the cor- rect rows during the copy-and-paste process. For example, the formula in H7 will update to 5 PsiTriangular(D7,E7,F7).
Here is a summary of the key cells in this model.
Uncertain variable cells: Cells H6:H15 Results cell: ProjectCompletion (I21) Statistic cell: MeanProjectCompletion (I23)
(See Section 13.1 for the details regarding how to define uncertain variable cells, results cells, and statistic cells.)
The Simulation Results We now are ready to evaluate the computer simulation of the spreadsheet model in Figure 13.15 . After running a simulation of 1,000 trials, Figure 13.17 shows the results in the form of a frequency chart and a statistics table. These results show a very wide range of possible project durations. Out of the 1,000 trials, the statistics table indicates that one trial had a dura- tion as short as 36.74 weeks while another was as long as 60.66 weeks. The frequency chart indicates that the duration that occurred most frequently during the 1,000 trials is close to 47 weeks (the project deadline), but that many other durations up to a few weeks either shorter or longer than this also occurred with considerable frequency. The mean is 46.26 weeks, which is much too close to the deadline of 47 weeks to leave much margin for slippage in the project schedule.
A statistic of special interest to Reliable’s management is the probability of meeting the deadline of 47 weeks under the current project plan. Figure 13.17 shows that all you need to do to identify the exact percentage is to type the deadline of 47 in the Upper Cutoff box. The Likelihood box then reveals that about 57.7 percent of the trials met the deadline.
FIGURE 13.16 A triangular distribution with parameters D6 (51), E6 (52), and F6 (53) is being entered into the first uncertain variable cell ATime (H6) in the spreadsheet model in Figure 13.15.
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13.3 Project Management: Revisiting the Reliable Construction Co. Case Study 545
If the simulation run were to be repeated with another 1,000 trials, this percentage prob- ably would change a little. However, with such a large number of trials, the difference in the percentages should be slight. Therefore, the probability of 0.577 provided by the Likelihood box in Figure 13.17 is a close estimate of the true probability of meeting the deadline under the assumptions of the spreadsheet model in Figure 13.15 . Note how much smaller this rela- tively precise estimate is than the rough estimate of 0.84 obtained by the PERT three-estimate approach in Section 16.4. Thus, the simulation estimate provides much better guidance to management in deciding whether the project plan should be changed to improve the chances of meeting the deadline. This illustrates how useful computer simulation can be in refining the results obtained by approximate analytical results.
A Key Insight Provided by the Sensitivity Chart Given such a low probability (0.577) of meeting the project deadline, Reliable’s project man- ager (David Perty) will want to revise the project plan to improve the probability substan- tially. RSPE has another tool, called the sensitivity chart, that provides strong guidance in identifying which revisions in the project plan would be most beneficial.
To view a sensitivity chart after running a simulation, click on the Sensitivity tab above the chart for the results cell. This reveals a sensitivity chart, as shown in Figure 13.18 . Using range names, the left side of the chart identifies various uncertain variable cells (activity times) in column H of the spreadsheet model in Figure 13.15 .
The bars in the chart give the correlation coefficient (based on moment values) between each uncertain variable cell and the results cell. A correlation coefficient between two vari- ables measures the strength of the relationship between those variables. Thus, each correla- tion coefficient in Figure 13.18 measures how strongly that activity time is influencing the project completion time. The higher the correlation coefficient, the stronger is this influence. Therefore, the activities with the highest correlation coefficients are those where the greatest effort should be made to reduce their activity times.
Figure 13.18 indicates that CTime has a far higher correlation coefficient than the times for any of the other activities. An examination of Figures 13.14 and 13.15 suggests why. Figure 13.14 shows that activity C precedes all the other activities except activities A and B, so any delay in completing activity C would delay the start time for all these other activities.
Computer simulation pro- vides a close estimate of the probability of meeting the project deadline (0.577) that is much smaller than the rough PERT/CPM estimate (0.84).
FIGURE 13.17 The frequency chart and statistics table that summarize the results of running the simulation model in Figure 13.15 for the Reliable Construction Co. project scheduling problem.
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546 Chapter Thirteen Computer Simulation with Risk Solver Platform
FIGURE 13.18 This sensitivity chart shows how strongly various activity times in the Reliable Construction Co. project are influencing the project completion time.
Furthermore, cells D8:F8 in Figure 13.15 indicate that CTime is highly variable, with an unusually large spread of 9 weeks between its most likely estimate and its pessimistic esti- mate, so long delays beyond the most likely estimate may well occur.
This very high correlation coefficient for CTime suggest that the best way to reduce the project completion time (and its variability) is to focus on reducing this activity time (and its variability). This can be accomplished by revising the project plan to assign activity C more personnel, better equipment, stronger supervision, and so forth. RSPE’s sensitivity chart clearly highlights this insight into where the project plan needs to be revised.
The sensitivity chart reveals that reducing the time for activity C would most improve the chances of completing the project before the deadline.
1. What is the project that is being considered in this section’s example? 2. A PERT/CPM procedure can obtain only a rough approximation of a certain key quantity, so com-
puter simulation is used to obtain a much closer estimate of the quantity. What is this quantity? 3. Computer simulations commonly use which kind of probability distribution as the distribution
of the duration of an activity? 4. What are the three estimates provided by the PERT three-estimate approach that give the
parameters of this distribution of the duration of an activity? 5. What can be done to quickly enter these parameters into all the uncertain variable cells after
using the dialog box for the distribution of the duration of only the first activity? 6. What quantity appears in the results cell of the spreadsheet model for this example? 7. What needs to be done to identify the exact percentage of trials of a computer simulation of
this spreadsheet model that will result in meeting the project completion deadline? 8. What kind of chart is used to highlight where a project plan needs to be revised to best
improve the chances of meeting the project deadline?
Review Questions
13.4 CASH FLOW MANAGEMENT: REVISITING THE EVERGLADE GOLDEN YEARS COMPANY CASE STUDY
Many applications of computer simulation involve scenarios that evolve far into the future. Since nobody can predict the future with certainty, computer simulation is needed to take future uncertainties into account. For example, businesses typically have great uncertainty
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13.4 Cash Flow Management: Revisiting the Everglade Golden Years Company Case Study 547
about what their future cash flows will be. An attempt often is made to predict these future cash flows as a first step toward making decisions about what should be done (e.g., arrang- ing for loans) to meet cash flow needs. However, effective cash management requires going a step further to consider the effect of the uncertainty in the future cash flows. This is where computer simulation comes in, with uncertain variable cells being used for the cash flows in various future periods. This process is illustrated by the following example.
The Everglade Cash Flow Management Problem The case study analyzed in Chapter 4 involves the Everglade Golden Years Company (which operates upscale retirement communities) and its efforts to manage its cash flow problems. In particular, because of both a temporary decline in business and some current or future con- struction costs, the company is facing some negative cash flows in the next few years as well as in some more distant years. As first provided in Table 4.1, Table 13.1 shows the projected net cash flows over the next 10 years (2014 to 2023). The company has some new retirement communities opening within the 10 years, so it is anticipated (or at least hoped) that a large positive cash flow will occur in 2023. Therefore, the problem confronting Everglade manage- ment is how to best arrange Everglade’s financing to tide the company over until its invest- ments in new retirement communities can start to pay off.
Chapter 4 describes how a decision was made to combine taking a long-term (10-year) loan now (the beginning of 2014) and a series of short-term (1-year) loans as needed to maintain a positive cash balance of at least $500,000 (as dictated by company policy) throughout the 10 years. Assuming no deviation from the projected cash flows shown in Table 13.1 , linear pro- gramming was used to optimize the size of both the long-term loan and the short-term loans so as to maximize the company’s cash balance at the beginning of 2024 when all of the loans have been paid off. Figure 4.5 in Chapter 4 shows the complete spreadsheet model after using Solver to obtain the optimal solution. For your convenience, Figure 4.5 is repeated here as Figure 13.19 . The changing cells, LTLoan (D11) and STLoan (E11:E20), give the sizes of the long-term loan and the short-term loans at the beginning of the various years. The objective cell EndBalance (J21) indicates that the resulting cash balance at the end of the 10 years (the begin- ning of 2024) would be $5.39 million. Since this is the cell that is being maximized, any other plan for the sizes of the loans would result in a smaller cash balance at the end of the 10 years.
Obtaining the “optimal” financing plan presented in Figure 13.19 is an excellent first step in developing a final plan. However, the drawback of the spreadsheet model in Figure 13.19 is that it makes no allowance for the inevitable deviations from the projected cash flows shown in Table 13.1 . The actual cash flow for the first year (2014) probably will turn out to be quite close to the projection. However, it is difficult to predict the cash flows in even the second and third years with much accuracy, let alone up to 10 years into the future. Computer simulation is needed to assess the effect of these uncertainties.
A Spreadsheet Model for Applying Computer Simulation Figure 13.20 shows the modification of the spreadsheet model in Figure 13.19 that is needed to apply computer simulation. One key difference is that the constants in CashFlow (C11:C20)
The drawback of the linear programming solution in Figure 13.19 is that it does not assess the effect of the great uncertainty regarding future cash flows, so com- puter simulation is needed to do this.
Year Projected Net Cash Flow
(millions of dollars)
2014 28 2015 22 2016 24 2017 3 2018 6 2019 3 2020 24 2021 7 2022 22 2023 10
TABLE 13.1 Projected Net Cash Flows for the Everglade Golden Years Company over the Next 10 Years
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Interest
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-0.23
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-0.23
-0.23
-0.23
ST
Interest
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-0.37
-0.69
-0.55
-0.18
0
-0.30
0
0
0
LT
Payback
-6.65
ST
Payback
-2.85
-5.28
-9.88
-7.81
-2.59
0
-4.23
0
0
0
Minimum
Balance
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
13
14
15
16
17
18
19
20
21
Everglade Cash Flow Management Problem When Applying Linear Programming
5%
7%
1
0.5
Cash
Flow
-8
-2
-4
3
6
3
-4
7
-2
10
LT
Loan
4.65
ST
Loan
2.85
5.28
9.88
7.81
2.59
0
4.23
0
0
0
Ending
Balance
0.50
0.50
0.50
0.50
0.50
0.50
0.50
2.74
0.51
10.27
5.39
(all cash figures in millions of dollars)
≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥
≥ ≥ ≥
F G H I J K L
9
10
11
12
LT
Interest
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=-LTRate*LTLoan
ST
Interest
=–STRate*E11
=–STRate*E12
=–STRate*E13
=–STRate*E14
=–STRate*E15
=–STRate*E16
=–STRate*E17
=–STRate*E18
=–STRate*E19
=–STRate*E20
LT
Payback
=–LTLoan
ST
Payback
=–E11
=–E12
=–E13
=–E14
=–E15
=–E16
=–E17
=–E18
=–E19
=–E20
=StartBalance+SUM(C11:I11)
=J11+SUM(C12:I12)
=J12+SUM(C13:I13)
=J13+SUM(C14:I14)
=J14+SUM(C15:I15)
=J15+SUM(C16:I16)
=J16+SUM(C17:I17)
=J17+SUM(C18:I18)
=J18+SUM(C19:I19)
=J19+SUM(C20:I20)
=J20+SUM(C21:I21)
Ending
Balance
Minimum
Balance
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
13
14
15
16
17
18
19
20
21
≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥
Solver Parameters Set Objective Cell: EndBalance To: Max By Changing Variable Cells:
LTLoan, STLoan
Subject to the Constraints: EndingBalance >= MinimumBalance
Solver Options: Make Variables Nonnegative
Solving Method: Simplex LP
FIGURE 13.19 The spreadsheet model that used linear programming in Chapter 4 (Figure 4.5) to analyze the Everglade Golden Years Company cash flow management problem without taking the uncertainty in future cash flows into account.
h il2
4 0 6 4 _ ch
1 3 _ 5 2 5 -5
9 8 .in
d d 5
4 8
h il2
4 0 6 4 _ ch
1 3 _ 5 2 5 -5
9 8 .in
d d 5
4 8
2 3 /1
1 /1
2 4
:1 7 P
M 2 3 /1
1 /1
2 4
:1 7 P
M
C onfirm
ing P ages
1 3 .4
C ash F
low M
anagem ent: R
evisiting the E verglade G
olden Y ears C
om pany C
ase Study 5 4 9
BalanceBeforeSTLoan
CashFlow
EndBalance
EndingBalance
LTLoan
LTRate
MeanEndBalance
MinimumBalance
MinimumCash
StartBalance
STLoan
STRate
L12:L22
F12:F21
N22
N12:N22
G12
C3
N24
P12:P22
C7
C6
M12:M22
C4
Range Name Cells
1
A B C D E F G H I N O P
2
3
4
5
6
7
8
9
10
11
12
LT Rate
ST Rate
Start Balance
Minimum Cash
Year
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
LT
Interest
-0.23
-0.23
-0.23
-0.23
-0.23
-0.23
-0.23
-0.23
-0.23
-0.23
ST
Interest
-0.21
-0.51
-0.68
-0.54
-0.21
0
-0.16
0
0
0
J
LT
Payback
-4.65
K
ST
Payback
-3.05
-7.25
-9.65
-7.78
-3.03
0
-2.35
0
0
0
L
Balance
Before
ST Loan
-2.55
-6.75
-9.15
-7.28
-2.53
1.67
-1.85
4.52
5.60
12.45
7.57
M
ST
Loan
3.05
7.25
9.65
7.78
3.03
0.00
2.35
0.00
0.00
0.00
Minimum
Balance
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
13
14
15
16
17
18
19
20
21
22
23
24
Everglade Cash Flow Management Problem When Applying Simulation
Mean 2024 Ending Balance
5%
7%
1
0.5
Min
-9
-4
-7
0
3
1
-6
4
-5
5
Cash Flow (Triangular Distribution)
(all cash figures in millions of dollars)
Max
-7
1
0
7
9
5
-2
12
4
18
Likely
-8
-2
-4
3
6
3
-4
7
-2
10
Ending
Balance
0.50
0.50
0.50
0.50
0.50
1.67
0.50
4.52
5.60
12.45
7.57
Simulated
Cash
Flow
-8.20
-3.76
-1.65
2.78
5.53
4.64
-3.29
6.77
1.31
7.08
LT
Loan
4.65 >= >= >= >= >= >= >= >= >= >= >=
HGF I J K L
Balance
Mean 2024 Ending Balance =PsiMean(N22)
M O P
22
23
24
10
9
11 Loan
LT
4.6512 =MAX(MinimumBalance–L12,0)
=MAX(MinimumBalance–L13,0)
=MAX(MinimumBalance–L14,0)
=MAX(MinimumBalance–L15,0)
=MAX(MinimumBalance–L16,0)
=MAX(MinimumBalance–L17,0)
=MAX(MinimumBalance–L18,0)
=MAX(MinimumBalance–L19,0)
=MAX(MinimumBalance–L20,0)
=MAX(MinimumBalance–L21,0)
LT
Interest
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
Simulated
Cash
Flow
=PsiTriangular(C12,D12,E12)
=PsiTriangular(C13,D13,E13)
=PsiTriangular(C14,D14,E14)
=PsiTriangular(C15,D15,E15)
=PsiTriangular(C16,D16,E16)
=PsiTriangular(C17,D17,E17)
=PsiTriangular(C18,D18,E18)
=PsiTriangular(C19,D19,E19)
=PsiTriangular(C20,D20,E20)
=PsiTriangular(C21,D21,E21)
ST
Interest
=–STRate*M12
=–STRate*M13
=–STRate*M14
=–STRate*M15
=–STRate*M16
=–STRate*M17
=–STRate*M18
=–STRate*M19
=–STRate*M20
=–STRate*M21
LT
Payback
=–LTLoan
ST
Payback
=–M12
=–M13
=–M14
=–M15
=–M16
=–M17
=–M18
=–M19
=–M20
=–M21
=StartBalance+SUM(F12:K12)
=N12+SUM(F13:K13)
=N13+SUM(F14:K14)
=N14+SUM(F15:K15)
=N15+SUM(F16:K16)
=N16+SUM(F17:K17)
=N17+SUM(F18:K18)
=N18+SUM(F19:K19)
=N19+SUM(F20:K20)
=N20+SUM(F21:K21)
=N21+SUM(F22:K22)
Before
ST Loan
ST
Loan
N
=L12+M12
=L13+M13
=L14+M14
=L15+M15
=L16+M16
=L17+M17
=L18+M18
=L19+M19
=L20+M20
=L21+M21
=L22+M22
Ending
Balance
Minimum
Balance
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
13
14
15
16
17
18
19
20
21
>= >= >= >= >= >= >= >= >= >= >=
9.18
FIGURE 13.20 A spreadsheet model for applying computer simulation to the Everglade Golden Years Company cash flow management problem. The uncertain variable cells are CashFlow (F12:F21), the results cell is EndBalance (N22), and the statistic cell is MeanEndBalance (N24).
h il2
4 0 6 4 _ ch
1 3 _ 5 2 5 -5
9 8 .in
d d 5
4 9
h il2
4 0 6 4 _ ch
1 3 _ 5 2 5 -5
9 8 .in
d d 5
4 9
2 3 /1
1 /1
2 4
:1 7 P
M 2 3 /1
1 /1
2 4
:1 7 P
M
Confirming Pages
550 Chapter Thirteen Computer Simulation with Risk Solver Platform
in Figure 13.19 have turned into random inputs in CashFlow (F12:F21) in Figure 13.20 . Thus, the latter cells, CashFlow (F12:F21), are uncertain variable cells. (The numbers appearing in these cells are just one possible random outcome—the last trial of the simulation run.) As indicated in cells D9:E9, the assumption has been made that each of the cash flows has a triangular distribution. Estimates have been made of the three parameters of this distribution (min, likely, and max) for each of the years, as presented in cells C12:E21.
The number 4.65 entered into LTLoan (G12) is the size of the long-term loan (in millions of dollars) that was obtained in Figure 13.19 . However, because of the variability in the cash flows, it no longer makes sense to lock in the sizes of the short-term loans that were obtained in STLoan (E11:E20) in Figure 13.19 . It is better to be flexible and adjust these sizes based on the actual cash flows that occur in the preceding years. If the balance at the beginning of a year (as calculated in BalanceBeforeSTLoan [L12:L22]) already exceeds the required mini- mum balance of $0.50 million, then there is no need to take any short-term loan at that point. However, if the balance is not this large, then a sufficiently large short-term loan should be taken to bring the balance up to $0.50 million. This is what is done by the equations entered into STLoan (M12:M22) that are shown at the bottom of Figure 13.20 .
The objective cell EndBalance (J21) in Figure 13.19 becomes the results cell End- Balance (N22) in Figure 13.20 . A statistic cell, MeanEndBalance (N24), is defined to determine the mean value of EndBalance for the simulation run. On any trial of the com- puter simulation, if the simulated cash flows in CashFlow (F12:F21) in Figure 13.20 are more favorable than the projected cash flows given in Table 13.1 (as is the case for the current numbers in Figure 13.20 ), then EndBalance (N22) in Figure 13.20 would be larger than EndBalance (J21) in Figure 13.19 . However, if the simulated cash flows are less favorable than the projections, then EndBalance (N22) in Figure 13.20 might even be a negative number. For example, if all the simulated cash flows are close to the cor- responding minimum values given in cells C12:C21, then the required short-term loans will become so large that paying off the last one at the beginning of 2024 (along with paying off the long-term loan then) will result in a very large negative number in End- Balance (N22). This would spell serious trouble for the company. Computer simulation will reveal the relative likelihood of this occurring versus a favorable outcome.
Here is a summary of the key cells in this model.
Uncertain variable cells: CashFlow (F12:F21) Results cell: EndBalance (N22) Statistic cell: MeanEndBalance (N24)
(See Section 13.1 for the details regarding how to define uncertain variable cells, results cells, and statistic cells.)
The Simulation Results Figure 13.21 shows the results from applying computer simulation with 1,000 trials. Because Everglade management is particularly interested in learning how likely it is that the current financing plan would result in a positive cash balance at the end of the 10 years, the number 0 has been entered into the Lower Cutoff box in the statistics table. The Likelihood box then indicates that over 95 percent of the trials resulted in a positive cash balance at the end. Fur- thermore, the frequency chart shows that many of these positive cash balances are reasonably large, with many exceeding $10 million. The overall mean is $9.18 million.
On the other hand, it is worrisome that nearly 5 percent of the trials resulted in a negative cash balance at the end. Although huge losses were rare, some of these negative cash balances were quite significant, ranging up to $5 million.
Conclusions Everglade management is pleased that the simulation results indicate that the proposed financ- ing plan is likely to lead to a favorable outcome at the end of the 10 years. At the same time, management feels that it would be prudent to take steps to reduce the 5 percent chance of an unfavorable outcome.
The uncertain future cash flows need to be uncertain variable cells.
A key question is the like- lihood of achieving a positive cash balance at the end of the 10 years.
The Certainty box in the frequency chart reveals that over 95 percent of the trials resulted in a positive cash balance at the end of the 10 years.
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13.4 Cash Flow Management: Revisiting the Everglade Golden Years Company Case Study 551
One possibility would be to increase the size of the long-term loan, since this would reduce the sizes of the higher interest short-term loans that would be needed in the later years if the cash flows are not as good as currently projected. This possibility is investigated in Problem 13.19.
The scenarios that would lead to a negative cash balance at the end of the 10 years are those where the company’s retirement communities fail to achieve full occupancy because of overestimating the demand for this service. Therefore, Everglade management concludes that it should take a more cautious approach in moving forward with its current plans to build more retirement communities over the next 10 years. In each case, the final decisions regard- ing the start date for construction and the size of the retirement community should be made only after obtaining and carefully assessing a detailed forecast of the trends in the demand for this service.
After adopting this policy, Everglade management approves the financing plan that is incor- porated into the spreadsheet model in Figure 13.20 . In particular, a 10-year loan of $4.65 million will be taken now (the beginning of 2014). In addition, a one-year loan will be taken at the begin- ning of each year from 2014 to 2023 if it is needed to bring the cash balance for that year up to the level of $500,000 required by company policy.
A more cautious expansion plan is needed to improve the chances of ending with a positive cash balance.
FIGURE 13.21 The frequency chart and statistics table that summarize the results of running the simulation model in Figure 13.20 for the Everglade Golden Years cash flow management problem.
1. What is the cash flow management problem that is currently confronting the management of the Everglade Golden Years Company?
2. Which management science technique was previously used to address this cash flow manage- ment problem before applying computer simulation?
3. What aspect of the problem does computer simulation take into account that this prior man- agement science technique could not?
4. What are the quantities in the uncertain variable cells in this example’s spreadsheet model for applying computer simulation?
5. How are the sizes of the short-term loans determined in this spreadsheet model?
Review Questions
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552 Chapter Thirteen Computer Simulation with Risk Solver Platform
6. What can happen in a trial of the computer simulation that would result in a negative cash balance at the end of 10 years?
7. What percentage of the trials actually resulted in a negative cash balance at the end of 10 years? 8. What policy did Everglade management adopt to reduce the possibility of having a negative
cash balance at the end of 10 years?
13.5 FINANCIAL RISK ANALYSIS: REVISITING THE THINK-BIG DEVELOPMENT CO. PROBLEM
One of the earliest areas of application of computer simulation, dating back to the 1960s, was financial risk analysis. This continues today to be one of the most important areas of application.
When assessing any financial investment (or a portfolio of investments), the key trade-off is between the return from the investment and the risk associated with the investment. Of these two quantities, the less difficult one to determine is the return that would be obtained if everything evolves as currently projected. However, assessing the risk is relatively difficult. Fortunately, computer simulation is ideally suited to perform this risk analysis by obtaining a risk profile , namely, a frequency distribution of the return from the investment. The por- tion of the frequency distribution that reflects an unfavorable return clearly describes the risk associated with the investment.
The following example illustrates this approach in the context of real estate investments. Like the Everglade example in the preceding section, you will see computer simulation being used to refine prior analysis done by linear programming because this prior analysis was unable to take the uncertainty in future cash flows into account.
The Think-Big Financial Risk Analysis Problem As introduced in Section 3.2, the Think-Big Development Co. is a major investor in com- mercial real estate development projects. It has been considering taking a share in three large construction projects—a high-rise office building, a hotel, and a shopping center. In each case, the partners in the project would spend three years with the construction, then retain ownership for three years while establishing the property, and then sell the property in the seventh year. By using estimates of expected cash flows, Section 3.2 describes how linear programming has been applied to obtain the following proposal for how big a share Think- Big should take in each of these projects.
Proposal
Do not take any share of the high-rise building project. Take a 16.50 percent share of the hotel project. Take a 13.11 percent share of the shopping center project.
This proposal is estimated to return a net present value (NPV) of $18.11 million to Think-Big. However, Think-Big management understands very well that such decisions should not be
made without taking risk into account. These are very risky projects since it is unclear how well these properties will compete in the marketplace when they go into operation in a few years. Although the construction costs during the first three years can be estimated fairly roughly, the net incomes during the following three years of operation are very uncertain. Consequently, there is an extremely wide range of possible values for each sale price in year 7. Therefore, management wants risk analysis to be performed in the usual way (with computer simulation) to obtain a risk profile of what the total NPV might actually turn out to be with this proposal.
To perform this risk analysis, Think-Big staff now has devoted considerable time to esti- mating the amount of uncertainty in the cash flows for each project over the next seven years. These data are summarized in Table 13.2 (in units of millions of dollars) for a 100 percent share of each project. Thus, when taking a smaller percentage share of a project, the numbers in the table should be reduced proportionally to obtain the relevant numbers for Think-Big.
Management needs a risk profile of the proposal to assess whether the likeli- hood of a sizable profit justifies the risk of possible large losses.
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553
For nearly a century after its founding in 1914, Merrill Lynch was a leading full-service financial service firm that strove to bring Wall Street to Main Street by making finan- cial markets accessible to everyone. It then was purchased in 2008 by the Bank of America Corporation and given the new name of Merrill Lynch Wealth Management as part of the merged corporate and investment bank now called Bank of America Merrill Lynch.
Prior to this merger, Merrill Lynch employed a highly trained salesforce of over 15,000 financial advisors throughout the United States as well as operating in 36 countries. A Fortune 100 company with net revenues of $26 billion in 2005, it managed client assets that totaled over $1.7 trillion.
Faced with increasing competition from discount bro- kerage firms and electronic brokerage firms, a task force was formed in late 1998 to recommend a product or ser- vice response to the marketplace challenge. Merrill Lynch’s strong management science group was charged with doing the detailed analysis of two potential new pricing options for clients. One option would replace charging for trades individually by charging a fixed percentage of a client’s assets at Merrill Lynch and then allowing an unlimited number of free trades and complete access to a financial advisor. The other option would allow self-directed inves- tors to invest online directly for a fixed low fee per trade without consulting a financial advisor.
The great challenge facing the management science group was to determine a “sweet spot” for the prices for these options that would be likely to grow the firm’s busi- ness and increase its revenues while minimizing the risk of losing revenue instead. A key tool in attacking this problem proved to be computer simulation. To undertake a major computer simulation study, the group assembled and evalu- ated an extensive volume of data on the assets and trading activity of the firm’s 5 million clients. For each segment of the client base, a careful analysis was done of its offer-adoption behavior by using managerial judgment, market research, and experience with clients. With this input, the group then formulated and ran a computer simulation model with vari- ous pricing scenarios to identify the pricing sweet spot.
The implementation of these results had a profound impact on Merrill Lynch’s competitive position, restoring it to a leadership role in the industry. Instead of continu- ing to lose ground to the fierce new competition, client assets managed by the company had increased by $22 bil- lion and its incremental revenue reached $80 million within 18 months. The CEO of Merrill Lynch called the new strat- egy “the most important decision we as a firm have made (in the last 20 years).”
Source: S. Altschuler, D. Batavia, J. Bennett, R. Labe, B. Liao, R. Nigam, and J. Oh, “Pricing Analysis for Merrill Lynch Integrated Choice,” Interfaces 32, no. 1 (January–February 2002), pp. 5–19. (A link to this article is provided on our website, www.mhhe.com/hillier5e.)
An Application Vignette
Hotel Project Shopping Center Project
Year Cash Flow ($1,000,000s) Year Cash Flow ($1,000,000s)
0 280 0 290 1 Normal (280, 5) 1 Normal (250, 5) 2 Normal (280, 10) 2 Normal (220, 5) 3 Normal (270, 15) 3 Normal (260, 10) 4 Normal (130, 20) 4 Normal (115, 15) 5 Normal (140, 20) 5 Normal (125, 15) 6 Normal (150, 20) 6 Normal (140, 15) 7 Uniform (1200, 844) 7 Uniform (160, 600)
TABLE 13.2 The Estimated Cash Flows for 100 Percent of the Hotel and Shopping Center Projects
In years 1 through 6 for each project, the probability distribution of cash flow is assumed to be a normal distribution, where the first number shown is the estimated mean and the second number is the estimated standard deviation of the distribution. In year 7, the income from the sale of the property is assumed to have a uniform distribution over the range from the first number shown to the second number shown.
To compute NPV, a cost of capital of 10 percent per annum is being used. Thus, the cash flow in year n is divided by (1.1) n before adding these discounted cash flows to obtain NPV.
A Spreadsheet Model for Applying Computer Simulation A spreadsheet model has been formulated for this problem in Figure 13.22 . There is no uncertainty about the immediate (year 0) cash flows appearing in cells D6 and D16, so these are data cells. However, because of the uncertainty for years 1–7, cells D7:D13 and D17:D23 containing the simulated cash flows for these years need to be uncertain vari- able cells. (The numbers in these cells in Figure 13.22 represent one possible random
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554 Chapter Thirteen Computer Simulation with Risk Solver Platform
Hotel Project:
Shopping Center Project
Construction Costs:
Revenue per Share
Selling Price per Share
Construction Costs:
Revenue per Share
Selling Price per Share
1
2
3
4
5
6 Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
Year 7
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
Year 7
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
Year 7
7 (mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(lower,upper)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(lower,upper)
Normal
Normal
Normal
Normal
Normal
Normal
Uniform
Normal
Normal
Normal
Normal
Normal
Normal
Uniform
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28 Hotel
Shopping Center
Cost of Capital
Share
29
30
31
32
33
34
35
36
37
38
39
Net Present Value ($millions)
MeanNPV ($millions)
A B C D E F G H
Simulation of Think-Big Development Co. Problem
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
C D
Think Big’s
Simulated Cash Flow
($millions)
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
Year 7
Net Present Value ($millions)
=HotelShare*D6+ShoppingCenterShare*D16
=HotelShare*D7+ShoppingCenterShare*D17
=HotelShare*D8+ShoppingCenterShare*D18
=HotelShare*D9+ShoppingCenterShare*D19
=HotelShare*D10+ShoppingCenterShare*D20
=HotelShare*D11+ShoppingCenterShare*D21
=HotelShare*D12+ShoppingCenterShare*D22
=HotelShare*D13+ShoppingCenterShare*D23
=CashFlowYear0+NPV(CostOfCapital,CashFlowYear1To7) +PsiOutput()
CashFlowYear0 CashFlowYear1To7 CostOfCapital HotelShare MeanNPV NetPresentValue ShoppingCenterShare
D28 D29:D35 H31 H28 D39 D37 H29
Range Name Cells
-80
-80
-70
30
40
50
200
-50
-20
-60
15
25
40
160
5
10
15
20
20
20
844
5
5
10
15
15
15
615
Project Simulated
Cash Flow
($millions)
-80
-79.057
-80.343
-73.063
14.059
29.746
81.373
395.247
-90
-42.329
-15.124
-54.653
21.923
10.122
14.780
494.378
Think-Big's
Simulated Cash Flow
($millions)
-24.999
-18.594
-15.239
-19.221
5.194
6.235
15.364
130.029
13.879
16.50%
13.11%
10%
3
C D
Year 0 -80
-90
=PsiNormal(F7,G7)
=PsiNormal(F8,G8)
=PsiNormal(F9,G9)
=PsiNormal(F10,G10)
=PsiNormal(F11,G11)
=PsiNormal(F12,G12)
=PsiNormal(F17,G17)
=PsiNormal(F18,G18)
=PsiNormal(F19,G19)
=PsiNormal(F20,G20)
=PsiNormal(F21,G21)
=PsiNormal(F22,G22)
=PsiUniform(F13,G13)
=PsiUniform(F23,G23)
Project Simulated
Cash Flow
($millions)
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
Year 7
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
Year 7
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
18.120
Mean NPV ($millions) =PsiMean(D37)
FIGURE 13.22 A spreadsheet model for applying computer simulation to the Think-Big Development Co. financial risk analysis problem. The uncertain variable cells are cells D7:D13 and D17:D23, the results cell is NetPresentValue (D37), the statistic cell is MeanNPV (D39), and the decision variables are HotelShare (H28) and ShoppingCenterShare (H29).
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13.5 Financial Risk Analysis: Revisiting the Think-Big Development Co. Problem 555
FIGURE 13.23 A normal distribution with parameters F7 (5280) and G7 (55) is being entered into the first uncertain variable cell D7 in the spreadsheet model in Figure 13.22.
outcome—the last trial of the simulation run.) Table 13.2 specifies the probability distribu- tions and their parameters that have been estimated for these cash flows, so the form of the distributions has been recorded in cells E7:E13 and E17:E23 while entering the correspond- ing parameters in cells F7:G13 and F17:G23. Figure 13.23 shows the Normal Distribution dialog box that is used to enter the parameters (mean and standard deviation) for the normal distribution into the first uncertain variable cell D7 by referencing cells F7 and G7. The for- mula in D7 is then copied and pasted into cells D8:D12 and D17:D22 to define these uncer- tain variable cells. The Uniform Distribution dialog box (like the similar one displayed earlier in Figure 13.3 for an integer uniform distribution) is used in a similar way to enter the parameters (minimum and maximum) for this kind of distribution into the uncertain variable cells D13 and D23.
The simulated cash flows in cells D6:D13 and D16:D23 are for 100 percent of the hotel project and the shopping center project, respectively, so Think-Big’s share of these cash flows needs to be reduced proportionally based on its shares in these projects. The proposal being analyzed is to take the shares shown in cells H28:H29. The equations entered into cells D28:D35 (see the bottom of Figure 13.22 ) then gives Think-Big’s total cash flow in the respective years for its share of the two projects.
Think-Big’s management wants to obtain a risk profile of what the total net present value (NPV) might be with this proposal. Therefore, the results cell is NetPresent Value (D37). To show the mean NPV over the simulation run, MeanNPV (D39) is defined as a statistic cell.
Here is a summary of the key cells in this model.
Uncertain variable cells: Cells D7:D13 and D17:D23 Decision variables: HotelShare (H28) and ShoppingCenterShare (H29) Results cell: NetPresentValue (D37) Statistic cell: MeanNPV (D39)
(See Section 13.1 for the details regarding how to define uncertain variable cells, results cells, and statistic cells.)
The Simulation Results Using the Simulation Options dialog box to specify 1,000 trials, Figure 13.24 shows the results of applying computer simulation to the spreadsheet model in Figure 13.22 . The fre- quency chart in Figure 13.24 provides the risk profile for the proposal since it shows the
Excel Tip: The NPV (discount rate, cash flows) function calculates the net present value of a stream of future cash flows at regular intervals (e.g., annually) in a range of cells (cash flows) using the specified discount rate per interval.
The frequency chart provides the risk profile for the proposal.
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relative likelihood of the various values of NPV, including those where NPV is negative. The mean is $18.120 million, which is very attractive. However, the 1,000 trials gener- ated an extremely wide range of NPV values, all the way from about 2 $28 million to over $62 million. Thus, there is a significant chance of incurring a huge loss. By entering 0 into the box in the Upper Cutoff box of the statistics table, the Likelihood box indicates that 81 percent of the trials resulted in a profit (a positive value of NPV). This also gives the bad news that there is roughly a 19 percent chance of incurring a loss of some size. The lightly shaded portion of the chart to the left of 0 shows that most of the trials with losses involved losses up to about $10 million, but that quite a few trials had losses that ranged from $10 million to nearly $30 million.
Armed with all this information, a managerial decision now can be made about whether the likelihood of a sizable profit justifies the significant risk of incurring a loss and perhaps even a very substantial loss. Thus, the role of computer simulation is to provide the informa- tion needed for making a sound decision, but it is management that uses its best judgment to make the decision.
FIGURE 13.24 The frequency chart and statistics table that summarize the results of running the simulation model in Figure 13.22 for the Think- Big Development Co. financial risk analysis problem. The Likelihood box in the statistics table reveals that 81 percent of the trials resulted in a positive net present value.
1. What is a risk profile for an investment (or a portfolio of investments)? 2. What is the investment proposal that the management of the Think-Big Development Co.
needs to evaluate? 3. What are the estimates that need to be made to prepare for applying computer simulation to
this example? 4. What quantities appear in the uncertain variable cells of the spreadsheet model for this
example? 5. What quantity appears in the results cell for this example? 6. Does computer simulation indicate a significant chance of incurring a loss if Think-Big manage-
ment approves the investment proposal?
Review Questions
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13.6 Revenue Management in the Travel Industry 557
13.6 REVENUE MANAGEMENT IN THE TRAVEL INDUSTRY
One of the most prominent areas for the application of management science in recent years has been in improving revenue management in the travel industry. Revenue management refers to the various ways of increasing the flow of revenues through such devices as setting up different fare classes for different categories of customers. The objective is to maximize total income by setting fares that are at the upper edge of what the different market segments are willing to pay and then allocating seats appropriately to the various fare classes.
As the example in this section will illustrate, one key area of revenue management is over- booking, that is, accepting a slightly larger number of reservations than the number of seats available. There usually are a small number of no-shows, so overbooking will increase rev- enue by essentially filling the available seating. However, there also are costs incurred if the number of arriving customers exceeds the number of available seats. Therefore, the amount of overbooking needs to be set carefully so as to achieve an appropriate trade-off between filling seats and avoiding the need to turn away customers who have a reservation.
American Airlines was the pioneer in making extensive use of management science for improving its revenue management. The guiding motto was “selling the right seats to the right customers at the right time.” This work won the 1991 Franz Edelman Award as that year’s best application of management science anywhere throughout the world. This application was credited with increasing annual revenues for American Airlines by over $500 million. Nearly half of these increased revenues came from the use of a new overbooking model.
Following this breakthrough at American Airlines, other airlines quickly stepped up their use of management science in similar ways. These applications to revenue management then spread to other segments of the travel industry (train travel, cruise lines, rental cars, hotels, etc.) around the world. Our example below involves overbooking by an airline company.
The Transcontinental Airlines Overbooking Problem Transcontinental Airlines has a daily flight (excluding weekends) from San Francisco to Chi- cago that is mainly used by business travelers. There are 150 seats available in the single cabin. The average fare per seat is $300. This is a nonrefundable fare, so no-shows forfeit the entire fare. The fixed cost for operating the flight is $30,000, so more than 100 reservations are needed to make a profit on any particular day.
For most of these flights, the number of requests for reservations considerably exceeds the number of seats available. The company’s management science group has been compiling data on the number of reservation requests per flight for the past several months. The aver- age number has been 195, but with considerable variation from flight to flight on both sides of this average. Plotting a frequency chart for these data suggests that they roughly follow a bell-shaped curve. Therefore, the group estimates that the number of reservation requests per flight has a normal distribution with a mean of 195. A calculation from the data estimates that the standard deviation is 30.
The company’s policy is to accept 10 percent more reservations than the number of seats available on nearly all its flights, since roughly 10 percent of all its customers making reser- vations end up being no-shows. However, if its experience with a particular flight is much dif- ferent from this, then an exception is made and the management science group is called in to analyze what the overbooking policy should be for that particular flight. This is what has just happened regarding the daily flight from San Francisco to Chicago. Even when the full quota of 165 reservations has been reached (which happens for most of the flights), there usually are a significant number of empty seats. While gathering its data, the management science group has discovered the reason why. On the average, only 80 percent of the customers who make reservations for this flight actually show up to take the flight. The other 20 percent forfeit the fare (or, in most cases, allow their company to do so) because their plans have changed.
Now that the data have been gathered, the management science group decides to begin its analysis by investigating the option of increasing the number of reservations to accept for this flight to 190. If the number of reservation requests for a particular day actually reaches this level, then this number should be large enough to avoid many, if any, empty seats.
A new overbooking model increased annual revenues for American Airlines by about $225 million.
An unusually high 20 per- cent no-show rate requires developing a special over- booking policy for this particular flight.
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Furthermore, this number should be small enough that there will not be many occasions when a significant number of customers need to be bumped from the flight because the number of arrivals exceeds the number of seats available (150). Thus, 190 appears to be a good first guess for an appropriate trade-off between avoiding many empty seats and avoiding bumping many customers.
When a customer is bumped from this flight, Transcontinental Airlines arranges to put the customer on the next available flight to Chicago on another airline. The company’s average cost for doing this is $150. In addition, the company gives the customer a voucher worth $200 for use on a future flight. The company also feels that an additional $100 should be assessed for the intangible cost of a loss of goodwill on the part of the bumped customer. Therefore, the total cost of bumping a customer is estimated to be $450.
The management science group now wants to investigate the option of accepting 190 res- ervations by using computer simulation to generate frequency charts for the following three measures of performance for each day’s flight:
1. The profit. 2. The number of filled seats. 3. The number of customers denied boarding.
A Spreadsheet Model for Applying Computer Simulation Figure 13.25 shows a spreadsheet model for this problem. Because there are three measures of interest here, the spreadsheet model needs three results cells. These results cells are Profit (F23), NumberOfFilledSeats (C20), and NumberDeniedBoarding (C21). In addition, three statistic cells are defined in cells C23:C25 to measure the mean value of each of the results cells for the simulation run. The decision variable ReservationsToAccept (C13) has been set at 190 for investigating this current option. Some basic data have been entered near the top of the spreadsheet in cells C4:C7.
Each trial of the computer simulation will correspond to one day’s flight. There are two random inputs associated with each flight, namely, the number of customers requesting res- ervations (abbreviated as Ticket Demand in cell B10) and the number of customers who actually arrive to take the flight (abbreviated as Number That Show in cell B17). Thus, the two uncertain variable cells in this model are SimulatedTicketDemand (C10) and Number- ThatShow (C17).
Since the management science group has estimated that the number of customers request- ing reservations has a normal distribution with a mean of 195 and a standard deviation of 30, this information has been entered into cells D10:F10. The Normal Distribution dialog box (shown earlier in Figure 13.23 ) then has been used to enter this distribution with these param- eters into SimulatedTicketDemand (C10). Because the normal distribution is a continuous distribution, whereas the number of reservations must have an integer value, Demand (C11) uses Excel’s ROUND function to round the number in SimulatedTicketDemand (C10) to the nearest integer.
The random input for the second uncertain variable cell NumberThatShow (C17) depends on two key quantities. One is TicketsPurchased (E17), which is the minimum of Demand (C11) and ReservationsToAccept (C13). The other key quantity is the probability that an indi- vidual making a reservation actually will show up to take the flight. This probability has been set at 80 percent in ProbabilityToShowUp (F17) since this is the average percentage of those who have shown up for the flight in recent months.
However, the actual percentage of those who show up on any particular day may vary somewhat on either side of this average percentage. Therefore, even though NumberThatShow (C17) would be expected to be fairly close to the product of cells E17 and F17, there will be some variation according to some probability distribution. What is the appropriate distribu- tion for this uncertain variable cell? Section 13.7 will describe the characteristics of various distributions. The one that has the characteristics to fit this uncertain variable cell turns out to be the binomial distribution.
As indicated in Section 13.7, the binomial distribution gives the distribution of the number of times a particular event occurs out of a certain number of opportunities. In this case, the
Having three measures of performance means that the simulation model needs three results cells.
Excel Tip: The Excel func- tion ROUND(x, 0) rounds the value x to the nearest integer. The zero speci- fies that 0 digits after the decimal point should be included when rounding.
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13.6 Revenue Management in the Travel Industry 559
1
2 3 4 5 6 7 8
Available Seats Fixed Cost
Avg. Fare / Seat Cost of Bumping
Ticket Demand Demand (rounded)
Reservations to Accept
Number That Show
Number of Filled Seats Number Denied Boarding
Data
9 10 11
Normal
Binomial
12 13 14 15 16 17 18 19 20 Ticket Revenue
Bumping Cost Fixed Cost
Profit
21 22 23 24 25
A B C D E F
Transcontinental Airlines overbooking
AvailableSeats AverageFare BumpingCost CostOfBumping Demand FixedCost MeanDeniedBoarding MeanFilled Seats MeanProfit NumberDeniedBoarding NumberOfFilledSeats NumberThatShow ProbabilityToShowUp Profit ReservationsToAccept SimulatedTicketDemand TicketRevenue TicketsPurchased
C4 C6 F21 C7 C11 C5 C24 C23 C25 C21 C20 C17 F17 F23 C13 C10 F20 E17
Range Name Cell 11 Demand (rounded) =ROUND(SimulatedTicketDemand,0)
B C
20 Ticket Revenue Bumping Cost
Fixed Cost Profit
21 =AverageFare*NumberOfFilledSeats =CostOfBumping*NumberDeniedBoarding =FixedCost =TicketRevenue - BumpingCost - FixedCost + PsiOutput()
22 23
E F
20 Number of Filled Seats Number Denied Boarding21
22
23
24
25
=MIN(AvailableSeats,NumberThatShow) + PsiOutput() =MAX(0,NumberThatShow - AvailableSeats) + PsiOutput()
B C
15 Tickets Purchased16
=MIN(Demand,ReservationsToAccept)17
E
Mean 195
Tickets Purchased
180
Standard Dev. 30
Probability to Show Up
80%
$45,000 $1,350 $30,000 $13,650
150 $30,000
$300 $450
179.74 180
190
153
150 3
Mean Filled Seats Mean Denied Boarding
Mean Profit
142.27 2.02
$11,775
Mean Filled Seats
Mean Denied Boarding
Mean Profit
=PsiMean(C20)
=PsiMean(C21)
=PsiMean(F23)
FIGURE 13.25 A spreadsheet model for applying computer simulation to the Trans- continental Airlines overbooking problem. The uncertain var iable cells are Simulated- TicketDemand (C10) and NumberThatShow (C17). The results cells are Profit (F23), Number- OfFilledSeats (C20), and NumberDeniedBoarding (C21). The statistic cells are MeanFilledSeats (C23), MeanDenied- Boarding (C24), and MeanProfit (C25). The decision variable is ReservationsTo-Accept (C13).
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event of interest is a passenger showing up to take the flight. The opportunity for this event to occur arises when a customer makes a reservation for the flight. These opportunities are conventionally referred to as trials (not to be confused with a trial of a computer simulation). The binomial distribution assumes that the trials are statistically independent and that, on each trial, there is a fixed probability (80 percent in this case) that the event will occur. The parameters of the distribution are this fixed probability and the number of trials.
Figure 13.26 displays the Binomial Distribution dialog box that enters this distribution into NumberThatShow (C17) by referencing the parameters TicketsPurchased (E17) and ProbabilityToShowUp (F17). The actual value in Trials for the binomial distribution will vary from simulation trial to simulation trial because it depends on the number of tickets purchased which in turn depends on the ticket demand which is random. RSPE therefore must determine the value for TicketsPurchased (E17) before it can randomly generate Num- berThatShow (C17). Fortunately, RSPE automatically takes care of the order in which to generate the various uncertain variable cells so that this is not a problem.
The equations entered into all the output cells, results cells, and statistic cells are given at the bottom of Figure 13.25 .
Here is a summary of the key cells in this model.
Uncertain variable cells: SimulatedTicketDemand (C10) and NumberThatShow (C17) Decision variable: ReservationsToAccept (C13) Results cells: Profit (F23), NumberOfFilledSeats (C20), and
NumberDeniedBoarding (C21) Statistic cells: MeanFilledSeats (C23), MeanDeniedBoarding (C24), MeanProfit (C25)
(See Section 13.1 for the details regarding how to define uncertain variable cells, results cells, and statistic cells.)
The Simulation Results Figure 13.27 shows the frequency chart obtained for each of the three results cells after applying computer simulation for 1,000 trials to the spreadsheet model in Figure 13.25 , with ReservationsToAccept (C13) set at 190.
The profit results estimate that the mean profit per flight would be $11,775. However, this mean is a little less than the profits that had the highest frequencies. The reason is that a small number of trials had profits far below the mean, including even a few that incurred losses, which dragged the mean down somewhat. By entering 0 into the Lower Cutoff box, the Likelihood box reports that 98.6 percent of the trials resulted in a profit for that day’s flight.
These characteristics of the binomial distribution are just what is needed for the uncertain variable cell NumberThatShow (C17).
FIGURE 13.26 A binomial distribution with parameters TicketsPurchased (E17) and ProbabilityToShowUp (F17) is being entered into the uncertain variable cell NumberThatShow (C17).
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13.6 Revenue Management in the Travel Industry 561
FIGURE 13.27 The frequency charts and statistics tables that summarize the results for the respective results cells—Profit (F23), NumberOfFilled- Seats (C20), and NumberDeniedBoarding (C21)—from running the simulation model in Figure 13.25 for the Transcontinental Airlines overbooking problem. The Likelihood box in the first statistics table reveals that 98.6 percent of the trials resulted in a positive profit.
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The frequency chart for NumberOfFilledSeats (C20) indicates that almost half of the 1,000 trials resulted in all 150 seats being filled. Furthermore, most of the remaining trials had at least 130 seats filled. The fact that the mean of 142.273 is so close to 150 shows that a policy of accepting 190 reservations would do an excellent job of filling seats.
The price that would be paid for filling seats so well is that a few customers would need to be bumped from some of the flights. The frequency chart for NumberDeniedBoarding (C21) indicates that this occurred about 40 percent of the time. On nearly all of these trials, the number ranged between 1 and 10. Considering that no customers were denied boarding for 60 percent of the trials, the mean number is only 2.015.
Although these results suggest that a policy of accepting 190 reservations would be an attractive option for the most part, they do not demonstrate that this is necessarily the best option. Additional simulation runs are needed with other numbers entered in ReservationsTo- Accept (C13) to pin down the optimal value of this decision variable. This can be done fairly easily with trial-and-error. We also will demonstrate how to do this efficiently with the help of a parameter analysis table in Section 13.8.
We will return to this example in Section 13.8 to further evaluate how many reservations to accept.
1. What is meant by revenue management in the travel industry? 2. In the pioneering application of management science to revenue management at American
Airlines, what increase in annual revenues was achieved? 3. What problem is being addressed by the management science group at Transcontinental Air-
lines in this section’s example? 4. What trade-off must be considered in addressing this problem? 5. What is the decision variable for this problem? 6. What quantities appear in the results cells of the spreadsheet model for this problem? 7. What quantities appear in the uncertain variable cells of the spreadsheet model? 8. What are the parameters of a binomial distribution? 9. Did the simulation results obtained in this section determine how many reservations should be
accepted for the flight under consideration?
Review Questions
13.7 CHOOSING THE RIGHT DISTRIBUTION
As mentioned in Section 13.1, RSPE’s Distributions menu provides a wealth of choices. Any of 46 probability distributions can be selected as the one to be entered into any uncertain vari- able cell. In the preceding sections, we have illustrated the use of five of these distributions (the integer uniform, uniform, triangular, normal, and binomial distributions). However, not much was said about why any particular distribution was chosen.
In this section, we focus on the issue of how to choose the right distribution. We begin by surveying the characteristics of many of the 46 distributions and how these characteristics help to identify the best choice. We next describe a special feature of RSPE for creating one of the 7 available custom distributions when none of the other 39 choices in the Distribu- tions menu will do. We then return to the case study presented in Section 13.1 to illustrate another special feature of RSPE. When historical data are available, this feature will identify which of the available distributions provides the best fit to these data while also estimating the parameters of this distribution. If you do not like this choice, it will even identify which of the distributions provides the second best fit, the third best fit, and so on.
Characteristics of the Available Distributions The probability distribution of any random variable describes the relative likelihood of the possible values of the random variable. A continuous distribution is used if any values are possible, including both integer and fractional numbers, over the entire range of pos- sible values. A discrete distribution is used if only certain specific values (e.g., only the integer numbers over some range) are possible. However, if the only possible values are
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13.7 Choosing the Right Distribution 563
integer numbers over a relatively broad range, a continuous distribution may be used as an approximation by rounding any fractional value to the nearest integer. (This approximation was used in cells C10:C11 of the spreadsheet model in Figure 13.25 .) RSPE’s Distributions menu includes both continuous and discrete distributions. We will begin by looking at the continuous distributions.
The right-hand side of Figure 13.28 shows the dialog box for three popular continuous dis- tributions from the Common submenu of the Distributions menu. The dark figure in each dia- log box displays a typical probability density function for that distribution. The height of the probability density function at the various points shows the relative likelihood of the corre- sponding values along the horizontal axis. Each of these distributions has a most likely value where the probability density function reaches a peak. Furthermore, all the other relatively high points are near the peak. This indicates that there is a tendency for one of the central val- ues located near the most likely value to be the one that occurs. Therefore, these distributions are referred as central-tendency distributions. The characteristics of each of these distribu- tions are listed on the left-hand side of Figure 13.28 .
The Normal Distribution The normal distribution is widely used by both management scientists and others because it describes so many natural phenomena. One reason that it arises so frequently is that the sum of many random variables tends to have a normal distribution (approximately) even when the individual random variables do not. Using this distribution requires estimating the mean and the standard deviation. The mean coincides with the most likely value because this is a symmetric distribution. Thus, the mean is a very intuitive quantity that can be readily esti- mated, but the standard deviation is not. About two-thirds of the distribution lies within one standard deviation of the mean. Therefore, if historical data are not available for calculating an estimate of the standard deviation, a rough estimate can be elicited from a knowledgeable individual by asking for an amount such that the random value will be within that amount of the mean about two-thirds of the time.
One danger with using the normal distribution for some applications is that it can give negative values even when such values actually are impossible. Fortunately, it can give negative values with significant frequency only if the mean is less than three standard devia- tions. For example, consider the situation where a normal distribution was entered into an uncertain variable cell in Figure 13.25 to represent the number of customers requesting a reservation. A negative number would make no sense in this case, but this was no problem since the mean (195) was much larger than three standard deviations (3 3 30 5 90) so a negative value essentially could never occur. (When normal distributions were entered into uncertain variable cells in Figure 13.22 to represent cash flows, the means were small or even negative, but this also was no problem since cash flows can be either negative or positive.)
The Triangular Distribution A comparison of the shapes of the triangular and normal distributions in Figure 13.28 reveals some key differences. One is that the triangular distribution has a fixed minimum value and a fixed maximum value, whereas the normal distribution allows rare extreme values far into the tails. Another is that the triangular distribution can be asymmetric (as shown in the figure), because the most likely value does not need to be midway between the bounds, whereas the normal distribution always is symmetric. This provides additional flexibility to the triangular distribution. Another key difference is that all its parameters — min (the minimum value), likely (the most likely value), and max (the maximum value) — are intuitive ones, so they are relatively easy to estimate.
These advantages have made the triangular distribution a popular choice for computer sim- ulations. They are the reason why this distribution was used in previous examples to represent competitors’ bids for a construction contract (in Figure 13.11 ), activity times (in Figure 13.15 ), and cash flows (in Figure 13.20 ).
The normal distribution allows negative values, which is not appropriate for some applications.
The parameters of a triangu- lar distribution are relatively easy to estimate because they are so intuitive.
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Popular Central-Tendency Distributions
Normal Distribution:
• Some value most likely (the mean)
• Values close to mean more likely
• Symmetric (as likely above as below mean)
• Extreme values possible, but rare
Triangular Distribution:
• Some value most likely
• Values close to most likely value more common
• Can be asymmetric
• Fixed upper and lower bound
Lognormal Distribution:
• Some value most likely
• Positively skewed (below mean more likely)
• Values cannot fall below zero
• Extreme values (high end only) possible, but rare
FIGURE 13.28 The characteristics and dialog boxes for three popular central-tendency distributions in RSPE’s Common submenu of the Distributions menu: (1) the normal distribution, (2) the triangular distribution, and (3) the lognormal distribution.
However, the triangular distribution also has certain disadvantages. One is that, in many situations, rare extreme values far into the tails are possible, so it is quite artificial to have fixed minimum and maximum values. This also makes it difficult to develop meaningful estimates of the bounds. Still another disadvantage is that a curve with a gradually changing slope, such as the bell-shaped curve for the normal distribution, should describe the true dis- tribution more accurately than the straight line segments in the triangular distribution.
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13.7 Choosing the Right Distribution 565
The Lognormal Distribution The lognormal distribution shown at the bottom of Figure 13.28 combines some of the advan- tages of the normal and triangular distributions. It has a curve with a gradually changing slope. It also allows rare extreme values on the high side. At the same time, it does not allow negative values, so it automatically fits situations where this is needed. This is particularly advantageous when the mean is less than three standard deviations and the normal distribu- tion should not be used.
This distribution always is “positively skewed,” meaning that the long tail always is to the right. This forces the most likely value to be toward the left side (so the mean is on its right), so this distribution is less flexible than the triangular distribution. Another disadvantage is that it has the same parameters as the normal distribution (the mean and the standard devia- tion), so the less intuitive one (the standard deviation) is difficult to estimate unless historical data are available.
When a positively skewed distribution that does not allow negative values is needed, the lognormal distribution provides an attractive option. That is why this distribution frequently is used to represent stock prices or real estate prices.
The Uniform and Integer Uniform Distributions Although the preceding three distributions are all central-tendency distributions, the uniform distributions shown in Figure 13.29 definitely are not. They have a fixed minimum value and a fixed maximum value. Otherwise, they say that no value between these bounds is any more likely than any other possible value. Therefore, these distributions have more variability than the central-tendency distributions with the same range of possible values (excluding rare extreme values).
The choice between these two distributions depends on which values between the mini- mum and maximum values are possible. If any values in this range are possible, including even fractional values, then the uniform distribution would be preferred over the integer uni- form distribution. If only integer values are possible, then the integer uniform distribution would be the preferable one.
Either of these distributions is a particularly convenient one because it has only two param- eters (lower and upper limit) and both are very intuitive. These distributions receive consider- able use for this reason. Earlier in this chapter, the integer uniform distribution was used to represent the demand for a newspaper (in Figure 13.1 ), whereas the uniform distribution was used to generate the bid for a construction project by one competitor (in Figure 13.11 ), and the future sale price for real estate property (in Figure 13.22 ).
The disadvantage of this distribution is that it usually is only a rough approximation of the true distribution. It is uncommon for either the minimum value or the maximum value to be just as likely as any other value between these bounds while any value barely outside these bounds is impossible.
The Exponential Distribution If you have studied Chapter 11 on queueing models, you hopefully will recall that the most commonly used queueing models assume that the time between consecutive arrivals of customers to receive a particular service has an exponential distribution. The reason for this assumption is that, in most such situations, the arrivals of customers are random events and the exponential distribution is the probability distribution of the time between random events. Section 11.1 describes this property of the exponential distribution in some detail.
As first depicted in Figure 11.3, this distribution has the unusual shape shown in Figure 13.30 . In particular, the peak is at 0 but there is a long tail to the right. This indicates that the most likely times are short ones well below the mean but that very long times also are possible. This is the nature of the time between random events.
Since the only parameter is the mean time until the next random event occurs, this distribu- tion is a relatively easy one to use.
The lognormal distribution has a long tail to the right but does not allow negative values to the left.
The uniform distribution is easy to use but usually is only a rough approximation of the true distribution.
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Uniform and Integer Uniform Distribution
Uniform Distribution:
• Fixed lower and upper limit
• All values equally likely
Integer Uniform Distribution:
• Fixed lower and upper limit
• All integer values equally likely
FIGURE 13.29 The characteristics and dialog boxes for the uniform distributions in RSPE’s Distributions menu.
The Poisson Distribution Although the exponential distribution (like most of the preceding ones) is a continuous dis- tribution, the Poisson distribution is a discrete distribution. The only possible values are nonnegative integers: 0, 1, 2. . . . However, it is natural to pair this distribution with the expo- nential distribution for the following reason. If the time between consecutive events has an exponential distribution (i.e., the events are occurring at random), then the number of events that occur within a certain period of time has a Poisson distribution. This distribution has some other applications as well.
When considering the number of events that occur within a certain period of time, the mean to be entered into the one parameter field in the dialog box should be the average num- ber of events that occur within that period of time.
The Bernoulli and Binomial Distribution The Bernoulli distribution is a very simple discrete distribution with only two possible values (1 or 0) as shown in Figure 13.31 . It is used to simulate whether a particular event occurs or not. The only parameter of the distribution is the probability that the event occurs. The Bernoulli distribution gives a value of 1 (representing yes) with this probability; otherwise, it gives a value of 0 (representing no).
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Distributions for Random Events
Exponential Distribution:
• Widely used to describe time between random
events (e.g., time between arrivals)
• Events are independent
• Mean = average time until the next event occurs
Poisson Distribution:
• Describes the number of times an event occurs
during a given period of time or space
• Occurrences are independent
• Any number of events is possible
• Mean = average number of events during period
of time (e.g., arrivals per hour), assumed
constant over time
FIGURE 13.30 The characteristics and dialog boxes for two distributions that involve random events. These distributions in RSPE’s Distributions menu are (1) the exponential distribution and (2) the Poisson distribution.
The binomial distribution is an extension of the Bernoulli distribution for when an event might occur a number of times. The binomial distribution, as shown in Figure 13.31 , gives the probability distribution of the number of times a particular event occurs, given the number of independent opportunities (called trials) for the event to occur, where the probability of the event occurring remains the same from trial to trial. For example, if the event of interest is getting heads on the flip of a coin, the binomial distribution (with Prob. 5 0.5) gives the dis- tribution of the number of heads in a given number of flips of the coin. Each flip constitutes a trial where there is an opportunity for the event (heads) to occur with a fixed probability (0.5). The binomial distribution is equivalent to the Bernoulli distribution when the number of trials is equal to 1.
You have seen another example in the preceding section when the binomial distribution was entered into the uncertain variable cell NumberThatShow (C17) in Figure 13.25 . In this airline overbooking example, the events are customers showing up for the flight and the trials are customers making reservations, where there is a fixed probability that a customer making a reservation actually will arrive to take the flight.
The only parameters for this distribution are the number of trials and the probability of the event occurring on a trial.
The Geometric and Negative Binomial Distributions These two distributions displayed in Figure 13.32 are related to the binomial distribution because they again involve trials where there is a fixed probability on each trial that the event will occur. The geometric distribution gives the distribution of the number of trials until the event occurs for the first time. After entering a positive integer into the suc field in its dialog
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Distributions for Number of Times an Event Occurs
Binomial Distribution:
• Describes number of times an event
occurs in a fixed number of trials
(e.g., number of heads in 10 flips
of a coin)
• For each trial, only two outcomes
possible
• Trials independent
• Probability remains same for
each trial
Bernoulli Distribution:
Describes whether an event
occurs or not
Two possible outcomes: 1 (Yes)
or 0 (No)
•
•
FIGURE 13.31 The characteristics and dialog boxes for the Bernoulli and binomial distributions in RSPE’s Distributions menu.
box, the negative binomial distribution gives the distribution of the number of trials until the event occurs the number of times specified in the suc field (suc is the number of successful events that must occur). Thus, suc is a parameter for this distribution and the fixed probability of the event occurring on a trial is a parameter for both distributions.
To illustrate these distributions, suppose you are again interested in the event of getting heads on a flip of a coin (a trial). The geometric distribution (with Prob. 5 0.5) gives the dis- tribution of the number of flips until the first head occurs. If you want five heads, the negative binomial distribution (with Prob. 5 0.5 and suc 5 5) gives the distribution of the number of flips until heads have occurred five times.
Similarly, consider a production process with a 50 percent yield, so each unit produced has an 0.5 probability of being acceptable. The geometric distribution (with Prob. 5 0.5) gives the distribution of the number of units that need to be produced to obtain one accept- able unit. If a customer has ordered five units, the negative binomial distribution (with Prob. 5 0.5 and suc 5 5) gives the distribution of the production run size that is needed to fulfill this order.
Other Distributions The Distributions menu includes many other distributions as well, such as beta, gamma, Weibull, Pert, Pareto, Erlang, and many more. These distributions are not as widely used in computer simulations, so they will not be discussed further.
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Distributions for Number of Trials Until Event Occurs
Geometric Distribution:
• Describes number of trials until an event
occurs (e.g., number of times to spin
roulette wheel until you win)
• Probability same for each trial
• Continue until succeed
• Number of trials unlimited
Negative Binomial Distribution:
• Describes number of trials until an event
occurs n times
• Same as geometric when suc = n = 1
• Probability same for each trial
• Continue until nth success
• Number of trials unlimited
FIGURE 13.32 The characteristics and dialog boxes for two distributions that involve the number of trials until events occur. These distributions in RSPE’s Distributions menu are (1) the geometric distribution and (2) the negative binomial distribution.
There is also a Custom submenu that enables you to design your own distribution when none of the other distributions will do. The next subsection will focus on how this is done.
The Custom Distribution Of the 46 probability distributions included in the Distributions menu, 39 of them are stan- dard types that might be discussed in a course on probability and statistics. In most cases, one of these standard distributions will be just what is needed for an uncertain variable cell. However, unique circumstances occasionally arise where none of the standard distributions fit the situation. This is where the distributions in the Custom submenu of the Distributions menu enter the picture.
The custom distributions actually are not probability distributions until you make them one. Rather, choosing a member of the Custom submenu triggers a process that enables you to custom-design your own probability distribution to fit almost any unique situation you might encounter.
There are seven choices in the Custom submenu: Cumul (short for cumulative), Discrete, DisUniform (short for discrete uniform), General, Histogram, Sip, and Slurp. The custom cumulative, custom general, and custom histogram distributions are all similar in that they are all used to create a continuous distribution with a fixed minimum and maximum value. With the custom cumulative distribution, you enter several values in between the minimum
Choosing a custom distri- bution from the Custom submenu enables you to custom-design your own distribution to fit a special situation.
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and maximum, along with the corresponding cumulative probability at those values. With the custom general distribution, you also enter several values in between the minimum and maxi- mum, but instead of cumulative probabilities, you enter relative weights that represent how likely it should be for outcomes near the listed values to occur (relative to outcomes near the other values in the list). Finally, with the custom histogram distribution, the range between the minimum and maximum is divided into a number of equal-sized segments and weights are provided for each segment to indicate how likely it should be (relative to the other segments) for a random outcome to fall within that segment.
The custom discrete and the custom discrete uniform distributions are also similar. With both, you enter a set of discrete values and these values are assumed to be the only possible outcomes. With the custom discrete distribution, each value (or outcome) is assigned its own probability, whereas the custom discrete uniform distribution assumes that all the discrete values have the same probability.
Finally, the custom sip and custom slurp distribution are used when you have a set of his- torical data and you want the uncertain variable to sample directly from the historical data. This might be appropriate if you expect the future to behave similarly to the past.
We will show two examples that use distributions from the Custom submenu. The first utilizes the custom discrete distribution whereas the second utilizes the custom general distribution.
In the first example, a company is developing a new product but it is unclear which of three production processes will be needed to produce the product. The unit production cost will be $10, $12, or $14, depending on which process is needed. The probabilities for these individual discrete values of the cost are the following.
20 percent chance of $10 50 percent chance of $12 30 percent chance of $14
To enter this distribution, first choose Discrete from the Custom submenu under the Distri- butions menu on the RSPE ribbon. Each discrete value and weight (expressed as a decimal number representing the probability) is then entered in the values and weights boxes as a list within curly brackets, as shown in Figure 13.33 .
The second example also involves a company that is developing a new product. However, the complication in this case is that our company’s management has learned that another firm
FIGURE 13.33 This dialog box illustrates how RSPE’s custom discrete distribution can enable you to custom-design your own distribution by enter- ing a set of discrete values and their weights.
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13.7 Choosing the Right Distribution 571
is developing a competitive product. It is unclear which company will be able to bring its product to market first and thereby capture most of the sales. In this light, here are the pre- dicted thousands of sales for our company’s new product.
0–20 (with 10 the most likely) if the competitive product comes to market first. 20–30 (all equally likely) if both products reach the market at the same time. 30–50 (with 40 the most likely) if our company’s product comes to market first.
The two products are believed to have an equal chance of reaching the market first. Each case is considered about three times as likely as both products reaching the market at the same time.
To enter this distribution, first choose General from the Custom submenu of the Distri- butions menu to bring up the dialog box shown in Figure 13.34 . The first two parameters, min 5 0 and max 5 50, are used to specify the smallest and largest possible value for sales (in thousands). In the values box, any number of values between the minimum and maximum can be entered as a list inside of curly brackets. For each value in this list, a corresponding weight needs to be entered into the weights box. The weight is a relative value used to specify the likelihood of each value relative to the other values in the list. Since sales around 10 thousand or 40 thousand are each about three times as likely as sales between 20 thousand and 30 thou- sand, the weights for the values {10, 20, 30, 40} are entered as {3, 1, 1, 3}. The net result is what is known as a bimodal distribution, with two distinct peaks, as shown in the chart on the left side of the dialog box.
Identifying the Continuous Distribution That Best Fits Historical Data We now have at least mentioned most of the probability distributions in the Distributions menu and have described the characteristics of many of them. This brings us to the question of how to identify which distribution is best for a particular uncertain variable cell. When historical data are available, RSPE provides a powerful feature for doing this by using the Fit button on the RSPE ribbon. We will illustrate this feature next by returning to the case study presented in Section 13.1.
Recall that one of the most popular newspapers that Freddie the newsboy sells from his newsstand is the daily Financial Journal. Freddie purchases copies from his distributor early
If you don’t know which continuous distribution should be chosen for an uncertain variable cell, RSPE can do it for you if historical data are available.
FIGURE 13.34 This dialog box illustrates how RSPE’s custom general distribution can enable you to custom-design your own continuous distribu- tion. The minimum and maximum value are specified as 0 and 50. Values near 10 and 40 are roughly 3 times as likely to occur as values near 20 or 30.
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each morning. Since excess copies left over at the end of the day represent a loss for Freddie, he is trying to decide what his order quantity should be in the future. This led to the spread- sheet model in Figure 13.1 that was shown earlier in Section 13.1. This model includes the uncertain variable cell Demand (C12). To get started, a discrete uniform distribution between 40 and 70 has been entered into this uncertain variable cell.
To better guide his decision on what the order quantity should be, Freddie has been keep- ing a record of the demand (the number of customers requesting a copy) for this newspaper each day. Figure 13.35 shows a portion of the data he has gathered over the last 60 days in cells F4:F63, along with part of the original spreadsheet model from Figure 13.1 . These data indicate a lot of variation in sales from day to day—ranging from about 40 copies to 70 cop- ies. However, it is difficult to tell from these numbers which distribution in the Distributions menu best fits these data.
RSPE provides the following procedure for addressing this issue.
Procedure for Fitting the Best Distribution to Data 1. Gather the data needed to identify the best distribution to enter into an uncertain vari-
able cell.
2. Enter the data into the spreadsheet containing your simulation model.
1
2 3 4 5 6 7 8
Unit Sale Price Unit Purchase Cost Unit Salvage Value
Order Quantity
Demand
Sales Revenue Purchasing Cost
Salvage Value
Profit
Mean Profit
Decision Variable
Data
9 10 11 12 13 14 15 16 17 18 19 20 21
60 61 62 63 64 65
A B C D E F
Freddie the Newsboy
$2.50 $1.50 $0.50
60
Simulation 44
$110.00 $90.00 $8.00
$28.00
$46.45
Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
55 56 57 58 59 60
Historical Demand
Data 62 45 59 65 50 64 56 51 55 61 40 47 63 68 67 67
41 42 64 45 59 70
FIGURE 13.35 Cells F4:F63 contain the historical demand data that have been collected for the case study involv- ing Freddie the newsboy that was introduced in Section 13.1. Columns B and C come from the sim- ulation model for this case study in Figure 13.1.
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13.7 Choosing the Right Distribution 573
3. Select the cells containing the data.
4. Click the Fit button on the RSPE ribbon, which brings up the Fit Options dialog box.
5. Make sure the Range box in this dialog box is correct for the range of the historical data in your worksheet.
6. Specify which type of distributions are being considered for fitting (continuous or discrete).
7. Indicate whether to allow shifted distributions and whether to run a sample indepen- dence test.
8. Also use this dialog box to select which ranking method should be used to evaluate how well a distribution fits the data.
9. Click Fit, which brings up the Fit Results chart that identifies the distribution that best fits the data.
10. If desired, check the box to select distributions to view that are lower on the list on the left side of the dialog box. This identifies the other types of distributions (including their parameter values) that are next in line for fitting the data well.
11. After choosing the distribution (from steps 9 and 10) that you want to use, close the dia- log box by using the close box in the upper-right-hand corner and then click Yes to accept the distribution.
12. Click the cell where you want the uncertain variable cell to be. This then enters the cho- sen distribution into the uncertain variable cell.
Since Figure 13.35 already includes the needed data in cells F4:F63, applying this procedure to Freddie’s problem begins by selecting the data. Then clicking the Fit but- ton brings up the Fit Options dialog box displayed in Figure 13.36 . The range F4:F63 of the data in Figure 13.35 is already entered into the Range box of this dialog box. When deciding which type of distributions should be considered for fitting, the default option of continuous distributions has been selected here. Sales will always be integer, so dis- crete might seem the more logical choice. However, when all the integer values over a
FIGURE 13.36 This Fit Options dialog box specifies (1) the range of the data in Figure 13.35 for the case study, (2) only continuous distributions will be considered, (3) shifted distributions will be allowed, (4) a sample independence test will be run, and (5) which ranking method will be used (the chi-square test) to evalu- ate how well each of the distributions fit the data.
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FIGURE 13.37 This Fit Results dialog box identifies the continuous distributions that provide the best fit, ranked top-to-bottom from best to worst on the left side. For the distribution that provides the best fit (Uniform), the distribution is plotted (the horizontal line at the top of the chart) so that it can be compared with the frequency distribution of the historical demand data. The value of the Fit Statistic (chi-square) is 4.4.
wide range are possible (all 31 integer values between 40 and 70 in this case), the form of the distribution begins to resemble a continuous distribution. Furthermore, there are many more continuous distributions (31) available in RSPE than discrete distributions (8). Thus, there may a better chance of finding a continuous distribution that is a good fit. This continuous distribution can then be made to give only integer values by rounding each number in the uncertain variable cell to the nearest integer (as was done in the airline overbooking example of Section 13.6 with the ticket demand in cell C11 of Figure 13.25 ). The chi-square test also has been selected for the ranking method. Clicking Fit then brings up the Fit Results chart displayed in Figure 13.37 .
The left side of the Fit Results chart in Figure 13.37 identifies the best-fitting distributions, ranked according to the Chi-Square test. This is a widely used test in statistics where smaller values indicate a better fit. It appears that the uniform distribution would be a good fit. In combination with the fact that demand actually must be integer, this confirms that the choice made in Freddie’s original spreadsheet model in Figure 13.1 to enter the integer uniform distribution into the uncertain variable cell Demand (C12) was reasonable. In fact, if we had chosen Discrete instead of Continuous as the type of distribution to fit in Figure 13.36 , RSPE would have found the integer uniform distribution to be the best fit. Choosing either Continu- ous or Discrete (or both) would have been reasonable in this case and would have led to the same type of distribution (uniform).
There are many more contin- uous distributions available in RSPE than discrete dis- tributions. Therefore, even when a discrete (or integer) distribution is needed, RSPE may have a better chance of finding a continuous distri- bution that is a good fit. A continuous distribution can always be rounded to create a distribution that gives only integer values.
1. How many probability distributions are available in RSPE’s Distributions menu? 2. What is the difference between a continuous distribution and a discrete distribution? 3. What is a possible danger with choosing the normal distribution to enter into an uncertain vari-
able cell? 4. What are some advantages of choosing the triangular distribution instead?
Review Questions
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13.8 Decision Making with Parameter Analysis Reports and Trend Charts 575
13.8 DECISION MAKING WITH PARAMETER ANALYSIS REPORTS AND TREND CHARTS
Many simulation models include at least one decision variable. For example, here are some of the decision variables encountered in this chapter.
The case study: OrderQuantity (C9) in Figure 13.1 Bidding example: OurBid (C25) in Figure 13.11 Overbooking example: ReservationsToAccept (C13) in Figure 13.25
In each of these cases, you have seen how well computer simulation with RSPE can evalu- ate a particular value of the decision variable by providing a wealth of output for the results cell(s). However, in contrast to many of the management science techniques presented in previous chapters (including linear programming and decision analysis), this approach has not identified an optimal solution for the decision variable(s). Fortunately, RSPE provides a way to systematically perform multiple simulations by using parameter cells. This makes it easy to identify at least an approximation of an optimal solution for problems with only one or two decision variables. In this section, we describe this approach and illustrate it by apply- ing it in turn to the three decision variables listed above. (The next section will present still another approach, using the Solver in RSPE to search for an optimal solution for simulation models.)
An intuitive approach for searching for an optimal solution is to use trial and error. Try different values of the decision variable(s), run a simulation for each, and see which one provides the best estimate of the chosen measure of performance. The interactive simulation mode in RSPE makes this especially easy, since the results in the statistic cells are available immediately after changing the value of a decision variable. Using parameter cells allows you to do the same thing in a more systematic way. After defining a parameter cell, all the desired simulations are run and the results soon are displayed nicely in the parameter analysis report . If desired, you also can view an enlightening trend chart , which can provide addi- tional details about the results.
If you have previously used parameter cells with the Solver in RSPE to generate parameter analysis reports for performing sensitivity analysis systematically (as was done in Chapter 5), the parameter analysis reports in simulation models work in much the same way. Two is the maximum number of decision variables that can be varied simultaneously in a parameter analy- sis report.
Let us begin by returning to this chapter’s case study and use a parameter cell to run mul- tiple simulations.
A Parameter Analysis Report and Trend Chart for the Case Study Recall that Freddie the newsboy wants to determine what his daily order quantity should be for copies of the Financial Journal. Figures 13.1 – 13.9 in Section 13.1 show the appli- cation of computer simulation for evaluating the option of using an order quantity of 60. The final estimate of the average daily profit that would be obtained with this order quantity is $46.45. As indicated in Figure 13.35 , the number of copies that Freddie’s
Defining a parameter cell and then generating a parameter analysis report applies computer simula- tion over a range of values of one or two decision vari- ables and then displays the results in a table.
5. How does the lognormal distribution compare with the normal and triangular distributions? 6. Why is the uniform distribution sometimes a convenient choice for an uncertain variable cell? 7. The binomial distribution gives the probability distribution of what? 8. What does choosing one of the custom distributions from the Distributions menu enable you
to do? 9. What procedure does RSPE provide for helping to identify the best distribution to enter into an
uncertain variable cell?
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customers want to purchase varies widely from day to day. The Fit Results chart in Figure 13.37 suggests that the probability distribution that best describes this variabil- ity is the integer uniform distribution between 40 and 70. Given such a high degree of variability, it is unclear where the order quantity should be set within the range between 40 and 70. Could an average daily profit larger than $46.45 be obtained by choosing an order quantity different from 60? Which order quantity between 40 and 70 would maxi- mize the average daily profit?
To address these questions, it would seem sensible to begin by trying a sampling of possi- ble order quantities, say, 40, 45, 50, 55, 60, 65, and 70. To do this, the first step is to define the decision variable being investigated — OrderQuantity (C9) in Figure 13.1 — as a parameter cell by using the following procedure.
Procedure for Defining a Decision Variable as a Parameter Cell 1. Select the cell containing the decision variable by clicking on it. 2. Choose Simulation from the Parameters menu on the RSPE ribbon. 3. Enter the lower limit and the upper limit of the range of values to be simulated for the deci-
sion variable. 4. Click on OK.
Figure 13.38 shows the application of this procedure to Freddie’s problem. Since simula- tions will be run for order quantities ranging from 40 to 70, these limits for the range have been entered.
Now we are ready to generate a parameter analysis report by running simulations for different values of the parameter cell. First choose Parameter Analysis from the Reports. Simulation menu on the RSPE ribbon. This brings up the dialog box in Figure 13.39 that allows you to specify which parameter cells to vary and which results to show after the simulations are run. The choice of which parameter cells to vary is made under Parameters in the bottom half of the dialog box. Clicking on (..) will select all of the parameter cells defined so far (moving them to the box on the right). In this case, only one parameter has been defined, so this causes the single parameter cell (OrderQuantity) to appear on the right. If more parameter cells had been defined, particular parameter cells can be chosen for immediate analysis by clicking on them and using (.) to move these individual parameter cells to the list on the right.
The choice of which simulation results to show as the parameter cell is varied is made in the upper half of the dialog box. By selecting the box next to Mean, the mean profit observed during the simulation run will be displayed for each different value of the parameter cell.
FIGURE 13.38 This parameter cell dialog box specifies the charac- teristics of the decision variable OrderQuantity (C9) in the simulation model in Figure 13.1 for the case study that involves Freddie the newsboy.
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Finally, enter the number of Major Axis Points to specify how many different values of the parameter cell will be included in the parameter analysis report. The values will be spread evenly between the lower and upper values specified in the parameter cell dialog box in Figure 13.38 . With seven major axis points, a lower value of 40, and an upper value of 70, a simulation will be run with order quantities of 40, 45, 50, 55, 60, 65, and 70. Clicking OK causes RSPE to run each of these simulations.
After RSPE runs the simulations, the parameter analysis report is created in a new spread- sheet as shown in Figure 13.40 . For each of the order quantities shown in column A, column B gives the mean of the values of the results cell, Profit (C18), obtained in all the trials of that simulation run. Cells B2:B8 reveal that an order quantity of 55 achieved the largest mean profit of $47.26, while order quantities of 50 and 60 essentially tied for the second largest mean profit.
After specifying the number of values of a parameter cell to consider, RSPE dis- tributes the values evenly over the range of values specified in the parameter cell dialog box.
FIGURE 13.39 This Parameter Analysis dialog box allows you to specify which parameter cells to vary and which results to show after the simulation run. Here the OrderQuantity (C9) parameter cell will be varied over seven differ- ent values and the value of the mean will be displayed for each of the seven simulation runs.
A
40
45
50
55
60
65
70 $40.00
$44.03
$46.45
$47.26
$46.45
$44.03
$40.00
B
1
2
3
4
5
6
7
8
MeanOrderQuantity
FIGURE 13.40 The parameter analysis report for the case study introduced in Section 13.1.
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The sharp drop-off in mean profits on both sides of these order quantities virtually guar- antees that the optimal order quantity lies between 50 and 60 (and probably close to 55). To pin this down better, the logical next step would be to generate another parameter analysis report that considers all integer order quantities between 50 and 60. You are asked to do this in Problem 13.14.
RSPE can also generate a variety of charts that show the results over simulation runs for different values of a parameter cell. After defining a parameter cell, the number of its values to receive a simulation run needs to be specified. To do this, click on the Options button on the RSPE ribbon and choose the simulation tab to bring up the Simulation Options dialog box shown in Figure 13.41 . The desired number of values of the parameter cell to simulate then is entered in the Simulations to Run box. This number plays the same role as the number of Major Axis points in Figure 13.39 when generating a parameter analysis report. The resulting values of the parameter are spread evenly between the lower and upper values specified in the param- eter cell dialog box in Figure 13.38 . For example, with seven simulations to run (as specified in Figure 13.41 ), the order quantities once again will be 40, 45, 50, 55, 60, 65, and 70.
Once the number of simulations to run has been specified, a variety of charts can be gener- ated by choosing a chart from the Charts.Multiple Simulations menu on the RSPE ribbon. For example, choosing Parameter Analysis from this menu gives the same information as the parameter analysis report in Figure 13.40 in graphical form.
A particularly interesting type of chart is the trend chart. Choosing Trend Chart from the Charts.Multiple Simulations menu brings up the dialog box shown in Figure 13.42 . This dialog box is used to choose which of the simulations should appear in the trend chart. Click- ing on (..) specifies that all seven simulations should be shown in the trend chart. Clicking OK then generates the trend chart shown in Figure 13.43 .
The report in Figure 13.40 is a one-dimensional parameter analysis report because the problem has only one decision variable. For problems where two decision variables have been defined as parameter cells, the resulting param- eter analysis report will be a two-dimensional table, with one parameter changing in the rows and the other in the columns.
FIGURE 13.41 This Simulation Options dialog box allows you to specify how many simulations to run before choosing a chart to show the results of running simulations for that num- ber of different values of a parameter cell.
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FIGURE 13.42 This trend chart dialog box is used to specify which simulations should be used to show results. Clicking (..) causes the results from all of the simulations to appear in the trend chart.
FIGURE 13.43 This trend chart shows the trend in the mean and in the range of the frequency distribution as the order quantity increases for Freddie’s problem.
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FIGURE 13.44 This parameter cell dialog box specifies the charac- teristics of the decision variable OurBid (C25) in Figure 13.11 for the Reliable Construction Co. contract bidding problem.
The horizontal axis of the trend chart shows the seven values of the parameter cell (order quantities of 40, 45, . . . , 70) for which simulations were run. The vertical axis gives the profit values obtained during the simulation runs. The curved line through the middle shows the mean profit for the simulations run at each of the different order quantities. Surrounding the mean line are two bands summarizing information about the frequency distribution of the profit values from each simulation run. (On a color monitor, the bands appear light gray and dark green.) The middle gray band contains the middle 75 percent of the profit values while the outer dark green band (in combination with the gray band within it) contains the middle 90 percent of the profit values. (These percentages are listed above the trend chart.) Thus, 5 percent of the profit values generated in the trials of each simulation run lie above the top band and 5 percent lie below the bottom band.
The trend chart received its name because it shows the trends graphically as the value of the decision variable (the order quantity in this case) increases. In Figure 13.43 , for example, consider the mean line. In going from an order quantity of 40 to 55, the mean line is trending upward, but then it is trending downward thereafter. Thus, the mean profit reaches its peak near an order quantity of 55. The fact that the gray and green bands are spreading out as it moves to the right provides the further insight that the variability of the profit values increases as the order quantity is increased. Although the largest order quantities provide some chance of particularly high profits on occasional days, they also give some likelihood to obtaining an unusually low profit on any given day. This risk profile may be relevant to Freddie if he is concerned about the variability of his daily profits.
We return to this case study again in the next section when the Solver in RSPE is used to search for the optimal order quantity.
A Parameter Analysis Report for the Reliable Construction Co. Bidding Problem We turn now to generating a parameter analysis report for the Reliable Construction Co. bid- ding problem presented in Section 13.2. Since the procedure for how to generate a parameter analysis report already has been presented in the preceding subsection, our focus here is on summarizing the results.
Recall that the management of the company is concerned with determining what bid it should submit for a project that involves constructing a new plant for a major manufacturer. Therefore, the decision variable in the spreadsheet model in Figure 13.11 is OurBid (C25). The parameter cell dialog box in Figure 13.44 is used to further describe this decision vari- able. Management feels that the bid should be in the range between $4.8 million and $5.8 million, so these are the numbers (in units of millions of dollars) that are entered into the entry boxes for Bounds in this dialog box.
A trend chart shows the trends graphically as the value of a decision variable increases.
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FIGURE 13.45 This Parameter Analysis dialog box allows you to specify which parameter cells to vary and which results to show after each simulation run. Here the OurBid (C25) parameter cell will be varied over six different values and the value of the mean will be displayed for each of the six simulation runs.
1
A
4.8 0.188
0.356
0.472
0.482
0.257
0.024
5.0
5.2
5.4
5.6
5.8
OurBid Mean
B
2 3
4
5
6
7
FIGURE 13.46 The parameter analysis report for the Reliable Construction Co. con- tract bidding problem described in Section 13.2.
Management wants to choose the bid that would maximize its expected profit. Conse- quently, the results cell in the spreadsheet model is Profit (C29). After choosing Parameter Analysis from the Reports.Simulation menu on the RSPE ribbon, the corresponding dialog box in Figure 13.45 is used to specify that the mean of the Profit should be shown as the parameter cell OurBid is varied over six major axis points. The six values automatically are distributed evenly over the range specified in Figure 13.44 , so simulations will be run for bids of 4.8, 5.0, 5.2, 5.4, 5.6, and 5.8 (in millions of dollars).
Figure 13.46 shows the resulting parameter analysis report. A bid of $5.4 million gives the largest mean value of the profits obtained on the 1,000 trials of the simulation run. This mean value of $482,000 in cell B5 should be a close estimate of the expected profit from using this bid. The case study in Chapter 16 begins with the company having just won the contract by submitting this bid.
Problem 13.17 asks you to refine this analysis by generating a parameter analysis report that considers all bids between $5.2 million and $5.6 million in multiples of $0.05 million.
A bid of $5.4 million was the winning bid for the Reliable Construction Co. in the case study for Chapter 16.
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A Parameter Analysis Report and Trend Chart for the Transcontinental Airlines Overbooking Problem As described in Section 13.6, Transcontinental Airlines has a popular daily flight from San Francisco to Chicago with 150 seats available. The number of requests for reservations usu- ally exceeds the number of seats by a considerable amount. However, even though the fare is non-refundable, an average of only 80 percent of the customers who make reservations actually show up to take the flight, so it seems appropriate to accept more reservations than can be flown. At the same time, significant costs are incurred if customers with reservations are not allowed to take the flight. Therefore, the company’s management science group is analyzing what number of reservations should be accepted to maximize the expected profit from the flight.
In the spreadsheet model in Figure 13.25 , the decision variable is ReservationsToAccept (C13) and the results cell is Profit (F23). The management science group wants to consider integer values of the decision variable over the range between 150 and 200, so the parameter cell dialog box is used in the usual way to specify these bounds on the variable. The decision is made to test 11 values of ReservationsToAccept (C13), so simulations will be run for val- ues in intervals of five between 150 and 200.
The results are shown in Figure 13.47 . The parameter analysis report on the left side of the figure reveals that the mean of the profit values obtained in the respective simulation runs climbs rapidly as ReservationsToAccept (C13) increases until the mean reaches a peak of $11,912 at 185 reservations, after which it starts to drop. Only the means at 180 and 190 res- ervations are close to this peak, so it seems clear that the most profitable number of reserva- tions lies somewhere between 180 and 190. (Now that the range of numbers that need to be
The decision variable in Figure 13.25 is ReservationsToAccept (C13).
1
150 $5,789
$6,896
$7,968
$9,001
$9,982
$10,880
$11,592
$11,712
$11,124
$10,395
$11,912
155
160
165
170
175
180
185
190
195
200
ReservationsToAccept Mean
A B
2
3
4
5
6
7
8
9
10
11
12
FIGURE 13.47 The parameter analysis report and trend chart for the Transcontinental Airlines overbooking problem described in Section 13.6.
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considered has been narrowed down this far, Problem 13.21 asks you to continue that analysis by generating a parameter analysis report that considers all integer values over this range.)
The trend chart on the right side of Figure 13.47 provides additional insight. The bands in this chart trend upward until the number of reservations to accept reaches approximately 185; then they start trending slowly downward. This indicates that the entire frequency distribution from the respective simulation runs keeps shifting upward until the run for 185 reservations and then starts shifting downward. Also note that the width of the entire set of seven bands increases until about the simulation run for 180 reservations and then remains about the same thereafter. This indicates that the amount of variability in the profit values also increases until the simulation run for 180 reservations and then remains about the same thereafter.
1. What does a parameter analysis report enable you to do that a single simulation run with one value of the decision variable(s) does not?
2. What are the advantages of using a parameter analysis report instead of simply using trial and error to try different values of the decision variable(s) and running a simulation for each?
3. What is the maximum number of decision variables that can be varied simultaneously in a parameter analysis report?
4. What procedure needs to be used before choosing Parameter Analysis from the Reports. Simulation menu?
5. What kind of information is summarized by the bands in a trend chart? 6. After a parameter analysis report has been used to narrow down the range of values of a deci-
sion variable that need to be considered, how can another parameter analysis report be used to better approximate the optimal value of the decision variable?
Review Questions
13.9 OPTIMIZING WITH COMPUTER SIMULATION USING RSPE’S SOLVER
In the preceding section, you have seen how parameter analysis reports and trend charts sometimes can be used to find at least a close approximation of an optimal solution. The three examples presented there illustrate the kind of problem where these tools work well. All three examples had only a single decision variable. (Remember that a parameter analysis report can consider a maximum of only two decision variables and a trend chart can only consider a single decision variable.) Furthermore, in two of these examples (the case study involving Freddie the newsboy and the airline overbooking example), the single decision variable was a discrete variable that had only a moderate number of possible values that needed consideration (namely, integers over a reasonably small range). This enabled using a parameter analysis report to identify a small range of values that provide the best solutions. If desired, a second parameter analysis report can then be generated to evaluate every possible value of the decision variable within this small range.
However, this approach does not work as well when the single decision variable is either a continuous variable or a discrete variable with a large range of possible values. It also is more difficult with two decision variables. It is not feasible at all for larger problems with more than two decision variables and numerous possible solutions. Many problems in practice fall into these categories.
Fortunately, RSPE includes a tool called Solver that automatically searches for an optimal solution for simulation models with any number of decision variables. This Solver was first introduced in Section 2.6. It includes some solving methods in common with the standard Excel Solver (first introduced in Section 2.5) and these solving methods were used in sev- eral chapters to find optimal solutions for linear programming models and (in Chapter 8) nonlinear programming models. However, the RSPE Solver also includes some additional functionalities, including its capabilities in the simulation area, that are not available with the Excel Solver. In particular, by using RSPE’s simulation tools, the RSPE Solver can be used to search for an optimal solution for a simulation model in much the same way that it deals with linear and nonlinear programming models. Hereafter in this chapter, we will simply use the term Solver to mean RSPE’s Solver.
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Solver conducts its search by executing a series of simulation runs to try a series of leading candidates to be an optimal solution, where the results so far are used to determine the most promising remaining candidate to try next. Solver cannot guarantee that the best solution it finds will literally be an optimal solution. However, given enough time, it often will find an optimal solution and, if not, usually will find a solution that is close to optimal. For prob- lems with only a few discrete decision variables, it frequently will find an optimal solution fairly early in the process and then spend the rest of the time ruling out other candidate solu- tions. Thus, although Solver cannot tell when it has found an optimal solution, it can estimate (within the range of precision provided by simulation runs) that the other leading candidates are not better than the best solution found so far.
To illustrate how to use Solver, we begin with a problem that it can handle extremely eas- ily, namely, the case study with Freddie the newsboy. After summarizing the overall proce- dure, we then turn to a more challenging example involving the selection of projects.
Application of Computer Simulation and Solver to the Case Study In the preceding section, the parameter analysis report generated in Figure 13.40 indicated that Freddie the newsboy should order between 50 and 60 copies of the Financial Journal each day. Now let us see how Solver can estimate which specific order quantity would maxi- mize his average daily profit.
Before using Solver, the initial steps are the same as described in Section 13.1 for prepar- ing to begin a single simulation run. Thus, after formulating the simulation model in a spread- sheet, as shown in Figure 13.1 , RSPE is used to define the uncertain variable cell Demand (C12), the results cell Profit (C18), and the statistic cell MeanProfit (C20). The Simulation Options dialog box is used in the usual way. These definitions and options set in RSPE are the ones that will be used when the model is solved.
The goal in Freddie’s problem is to choose the value of the order quantity that would maximize the mean profit that Freddie will earn each day. MeanProfit (C20) records the mean profit during the simulation run for a given value of the order quantity. Selecting this cell and then choosing Max.Normal from the Objective menu specifies that the objective is to maxi- mize the quantity in this cell.
Next, the decision variables need to be defined. In Freddie’s problem, the only decision to be made is the value for OrderQuantity (C9), so there is only one decision variable. Selecting this cell and choosing Normal from the Decisions menu defines this cell as a (normal) decision variable. Solver uses a search engine to search for the best value of the decision variable(s). Therefore, the smaller the search space (as measured by the number of possible values Solver must search), the faster Solver will be able to solve the problem. Thus, we should take into account any constraints on the possible values of the decision variable. Since OrderQuantity must be integer, select the cell again and choose Integer from the Constraints.Variable Type/ Bound menu. This greatly reduces the number of possible values to search since only the inte- gers will need to be considered. In addition, since demand is random between a lower limit of 40 and an upper limit of 70, it is clear that the order quantity also should be somewhere in this range. To specify that OrderQuantity should be between 40 and 70, we add a pair of bound constraints. First select the cell and choose . 5 from the Constraints.Variable Type/Bound menu. This brings up the first Add Constraint dialog box shown in Figure 13.48 . Click in the Constraint box and then on cell E12 to specify that OrderQuantity . 5 E12 ( 5 40) and then
An alternative way to maximize the mean profit is to select Profit (C18), the results cell that measures the profit of each individual trial of the simulation, and then choose Max.Expected from the Objective menu. Choosing Max.Expected for a results cell is equiva- lent to defining a statistic cell that calculates the mean (an estimate of the expected value) of the results cell and then maximizing that by choosing Max.Normal for the statistic cell.
FIGURE 13.48 These two Add Constraint dialog boxes allow you to specify bounds on the decision variable, Order- Quantity (C9). The top dialog box specifies that OrderQuantity .5 E12 (540). The bottom dialog box specifies that Order- Quantity <5 F12 (570).
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click OK. Similarly, choose , 5 from the Constraints.Variable Type/Bound menu and use the Constraint box to specify that OrderQuantity , 5 F12 ( 5 70). The net result of these three constraints are that OrderQuantity must be an integer between 40 and 70. This has reduced the search space to just 31 possible values.
The Model tab of the Model pane should now appear as seen on the left side of Figure 13.49 . (If the Model pane is not showing on the right side of the spreadsheet, it can be toggled on and off by clicking on the Model button on the RSPE ribbon.) The Model pane shows that (1) the objective is to maximize MeanProfit (C20), (2) the decision variable is OrderQuantity (C9), and (3) OrderQuantity should be integer and between 40 and 70. It also shows the simulation settings, which indicate that Demand (C12) is the uncertain variable, the results cell is Profit (C18), and MeanProfit (C20) is defined as a statistic cell.
Before running Solver to optimize Freddie’s problem, we need to consider the settings on the Engine tab of the Model pane, as seen on the right side of Figure 13.49 . In particular, the checkbox Automatically Select Engine should be checked to have Solver automatically choose which search engine is most appropriate for the problem. Second, the Max Time and/ or Max Time without Improvement should be specified. Max Time sets a limit (in seconds) for how long you would like the search to proceed. Leaving this quantity blank in Figure 13.49 means that no limit has been placed on the length of the search. This is OK because we instead have set Max Time without Improvement to 10 seconds. This will keep the search engine searching until Solver has not improved the solution within the last 10 seconds.
At this point, clicking on Optimize on the RSPE ribbon begins the search for an optimal solution. Solver searches over different order quantities in the search space. For each trial solution, it runs a simulation to determine the mean profit. Solver then evaluates the results so far to determine the most promising candidates for the order quantity to try next. This con- tinues until it has either considered all values for the order quantity or it reaches one of the stopping rules ( Max Time or Max Time without Improvement ). RSPE will then put the best value for the order quantity (the one with the largest mean profit) directly in the spreadsheet. In Freddie’s case, it usually will find the exact optimal solution, namely, an order quantity of 55 leading to a mean profit of approximately $47.26. In this case, the spreadsheet looks just like Figure 13.9 , where this optimal solution was found by trial and error.
When possible, reducing the search space by add- ing integer constraints and/or putting bounds on the decision variable(s) will improve how fast Solver can find an optimal solution.
FIGURE 13.49 The Model tab and Engine tab of the Model pane for Freddie’s problem. The Model tab on the left shows the Solver optimization settings and the simulation set- tings. The objective is to maximize MeanProfit (C20) by changing the decision variable Order- Quantity (C9) subject to OrderQuantity being both integer and between 40 and 70. The Engine tab on the right specifies that RSPE will automatically select the search engine to solve the model and that it will keep searching until it hasn’t found an improved solution for at least 10 seconds.
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Here is a summary of the entire procedure for applying Solver that has just been illustrated for Freddie’s problem.
Procedure for Applying Solver 1. Formulate your simulation model on a spreadsheet. 2. Use RSPE to define your uncertain variable cells, results cells, and statistic cells, as well as
to set Simulation Options. 3. Use RSPE to define your decision variables and the objective. 4. If possible, define constraints on the decision variables to reduce the search space. 5. Use the Engine tab of the Model pane to have RSPE automatically select the search engine
and to set the stopping rule ( Max Time and/or Max Time without Improvement ). 6. Click on Optimize to run the optimization.
Application of Computer Simulation and Solver to a Project Selection Example We now turn to a more challenging example for applying Solver. This example is based on Case 3-7 at the end of Chapter 3. Here are the essential facts.
Tazer Corp., a pharmaceutical manufacturing company, is beginning the search for a new breakthrough drug. The following five potential research and development projects have been identified for attempting to develop such a drug.
Project Up: Develop a more effective antidepressant that does not cause serious mood swings.
Project Stable: Develop a drug that addresses manic depression. Project Choice: Develop a less intrusive birth control method for women. Project Hope: Develop a vaccine to prevent HIV infection. Project Release: Develop a more effective drug to lower blood pressure.
In contrast to Case 3-7, Tazer management now has concluded that the company cannot devote enough money to research and development to undertake all of these projects. Only $1.2 billion is available, which will be enough for only two or three of the projects. The second column of Table 13.3 shows the amount needed (in millions of dollars) for each of these projects. The third column estimates the probability that each project would be able to develop a successful drug. If a project is successful, it is quite uncertain how much revenue would be generated by the drug. The estimate of the amount of revenue (in millions of dol- lars) is that it has a normal distribution with the mean and standard deviation given in the last two columns of the table.
Tazer management now wants to determine which of these projects should be undertaken to maximize the expected total profit from the resulting revenues (if any). Because of the great uncertainty in what the total profit will turn out to be, management also would like to have a reasonably high probability of achieving a satisfactory total profit (at least $100 million).
Figure 13.50 shows a simulation model on a spreadsheet for this problem. The data in Table 13.3 have been transferred directly into the data cells C7:F11. The cells in the next col- umn, Success? (G7:G11) are uncertain variable cells that will have a value of 0 or 1 for each trial of a simulation run. (The values shown in column G in Figure 13.50 are one possible random outcome—the result of the last simulation trial.) This value indicates whether the cor- responding project would fail (a value of 0) or succeed (a value of 1) on that trial if it were to
R&D Investment Revenue ($millions) if Successful
Project ($millions) Success Rate Mean Standard Deviation
Up 400 50% 1,400 400 Stable 300 35% 1,200 400 Choice 600 35% 2,200 600 Hope 500 20% 3,000 900 Release 200 45% 600 200
TABLE 13.3 Data for the Tazer Proj- ect Selection Problem
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be undertaken. Thus, the probability distribution entered into each of these uncertain variable cells needs to be a Bernoulli distribution (as described in Section 13.7), where the parameter is the probability of a success on this trial given in column D. The cells in column H, Revenue (H7:H11), also are uncertain variable cells. The probability distribution for each is a normal distribution with the parameters given in columns E and F.
The cells in column J, Decisions (J7:J11), are the decision variables for the model. Each of these decision variables is a binary variable, that is, a variable whose only possible values are 0 and 1. To define these decision variables, select the cells and choose Normal from the Deci- sions menu on the RSPE ribbon. Then, with the cells still selected, constrain them to be binary by choosing binary from the Constraints.Variable Type/Bound menu. For each project listed in column B, the corresponding decision variable in column J has the following interpretation.
Decision variable 5 b1, if approve the project 0, if reject the project
Budget (C15) gives the maximum amount that can be invested in these research and devel- opment projects. The output cell Invested (C13) records the total amount invested in the proj- ects, given the decisions regarding which ones are approved. The equation entered into this cell is shown under the spreadsheet on the left side of Figure 13.50 . The limited budget means that the decision variables must satisfy the constraint that
Invested (C 13) # Budget (C 15)
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
A B C D E F G H I J
Budget-Constrained Project Selection
R&D Investment Success Revenue ($millions)
Project ($millions) Rate Mean St. Dev. Success? (if Successful) Profit Decisions Up 400 50% 1,400 400 1 1,677 0.00 0
Stable 300 35% 1,200 400 1 1,547 0.00 0 Choice 600 35% 2,200 600 0 1,975 0.00 0 Hope 500 20% 3,000 900 0 3,852 0.00 0
Release 200 45% 600 200 1 135 0.00 0
Invested 0 Total Profit ($millions)
Mean Profit ($millions)
Min Acceptable Profit ($million) Prob(Profit > Min Acceptable)
0.00
0.00
100 0.000
<= Budget 1,200
$millions if Successful (Normal Distribution)
Estimated Revenue
6 5
G H
Success? Revenue ($millions)
(if Successful)
=PsiBernoulli(D7) =PsiNormal(E7,F7)
=PsiNormal(E8,F8)
=PsiNormal(E9,F9)
=PsiNormal(E10,F10)
Total Profit ($millions)
Mean Profit ($millions)
Min Acceptable Profit($million)
Prob(Profit > Min Acceptable)
=PsiBernoulli(D8)
=PsiBernoulli(D9) =PsiBernoulli(D10)
=PsiBernoulli(D11)
7 8 9
10 11 12 13 14 15 16 17 18
I
Profit
=Decisions*(Success?*Revenue-RandDInvestment)
=Decisions*(Success?*Revenue-RandDInvestment)
=Decisions*(Success?*Revenue-RandDInvestment)
=Decisions*(Success?*Revenue-RandDInvestment) =Decisions*(Success?*Revenue-RandDInvestment)
=SUM(Profit)
100
=PsiMean(Profit)
=1-PsiTarget(TotalProfit, MinAcceptable Profit)
13
B C
Invested =SUMPRODUCT(RandDInvestment,Decisions)
Range Name Cells
Budget C15 Decisions J7:J11 Invested C13
Profit
MeanProfit MinAccetableProfit
ProbProfitAcceptable I7:I11
I17 I15 I18
RandDInvestment C7:C11 Revenue H7:H11 Success? G7:G11 TotalProfit I13
FIGURE 13.50 A spreadsheet model for applying computer simulation to the Tazer Corp. project selection problem. The uncertain variable cells are Success? (G7:G11) and Revenue (H7:H11), the decision variables are Decisions (J7:J11), the results cell is TotalProfit (I13), and the two statistic cells are MeanProfit (I15) and ProbProfitAcceptable (I18).
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To enter this constraint in Solver, select the Invested (C13) cell and choose , 5 under the Constraints.Normal Constraint menu. Then, in the Constraint box of the Add Constraint dialog box, select Budget (C15) for the right-hand side of this constraint.
The output cells Profit (I7:I11) give the profit (revenue minus investment) from each proj- ect on each trial of a simulation run. The profit from a project is 0 if the project is rejected. If it is approved, the revenue is 0 if the project is not successful (as would be indicated by a 0 in the corresponding row of column G). If the project is successful (as would be indicated by a 1 in its row of column G), the revenue on that trial will be the random value that appears in the corresponding row of column H. Therefore, the equations entered into Profit (I7:I11) are those shown in the lower-right-hand corner of Figure 13.50 . Also note that SUM (Profit) gives the value in the results cell TotalProfit (I13).
Tazer management is seeking a solution that maximizes the mean of TotalProfit (I13) in Figure 13.50 . Therefore, MeanProfit (I15) is defined as a statistic cell to measure the mean value of the TotalProfit (I13) over the simulation run. MeanProfit is defined as the objective by choosing Max.Normal from the Objective menu.
After using RSPE to define all the uncertain variable cells, the results cell, and the sta- tistic cells in the usual way (along with the decision variables, constraints, and objective as described above), the complete model is shown in the Model tab of the Model pane, as dis- played in Figure 13.51 . The Engine tab is used to specify that RSPE should automatically select the search engine and that the search should be allowed to proceed until it has not found an improved solution within the last 10 seconds.
Clicking on the Optimize button then causes Solver to search for the value of the Decisions (J7:J11) that maximizes MeanProfit (I15). The resulting solution is
Choose Project Choice, Project Release, and Project Up Mean of total profit 5 $540.61 million
Figure 13.52 shows the frequency chart and statistics table that was obtained with the simulation run that used the best solution. This chart shows a high degree of variability in the profit values obtained during the various trials of the simulation run. There is a substantial probability of actually incurring a loss from the selected research and development projects (which is fairly common in this industry). In fact, nearly 200 of the 1,000 trials resulted in a loss of $1.2 billion because all three projects failed. Fortunately, there also is a good chance
FIGURE 13.51 The Model tab and Engine tab of the Model pane for the Tazer project selec- tion problem. The Model tab on the left shows the Solver optimization set- tings and the simulation settings. The objective is to maximize MeanProfit (I15) by selecting the best values of the deci- sion variables Decisions (J7:J11), subject to the constraints that Invested (C13) <5 Budget (C15) and Decisions are binary. The engine tab on the right specifies that RSPE will automatically select the search engine and that it will keep searching until it has not found an improved solution within the last 10 seconds.
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of reaping extremely big profits. Because Tazer management would like to have a high prob- ability of obtaining a total profit of at least $100 million, this quantity has been entered in the Lower Cutoff box above the statistics table. The Likelihood box indicates that 58.5 percent of the trials did at least this well.
Tazer management had hoped to have a higher probability of obtaining a total profit of at least $100 million. Therefore, the question is raised whether there might be another combina- tion of research and development projects that would increase this probability.
Addressing this question will require defining a new kind of statistic cell. In addition to knowing the mean profit (using the MeanProfit statistic cell), we need a statistic to measure the probability that the profit will be at least $100 million. RSPE provides such a statistic. To begin defining this statistic cell, select the results cell Profit (I13) and choose Target from the Results. Range menu, and then click in cell I18. This inserts the formula 5 PsiTarget(TotalProfit, 0) into cell I18. The PsiTarget function measures the cumulative probability of the results cell, so here it calculates the probability that TotalProfit will be 0 or less (since the second parameter in the formula is entered as 0 by default). This is not exactly what we want, since we are interested in the probability that TotalProfit will be 100 or more. Thus, this formula needs to be changed in two ways. First, change the 0 to 100 (by referring to the data cell MinAcceptableProfit in Figure 13.50 ). The formula 5 PsiTarget(TotalProfit, MinAcceptableProfit) then would calcu- late the probability of earning no more than the MinAcceptableProfit of $100 million. To calculate the probability of earning no less than the MinAcceptableProfit, we subtract this probability from 1, resulting in the final formula used in Figure 13.50 , ProbProfitAcceptable (I18) 5 1 2 PsiTarget(TotalProfit, MinAcceptableProfit).
Next, change the objective in Solver to instead maximize the probability that the profit will be at least $100 million. Select the cell ProbProfitAcceptable (I18) that measures this prob- ability and then choose Max.Normal from the Objective menu. Rerunning Solver with the new objective leads to the following solution:
Choose Project Up and Project Stable
62.8% certainty of total profit $ $100 million
By changing from the solution analyzed in Figure 13.52 , this more conservative solution has managed to boost the estimated probability of achieving a satisfactory total profit from 58.5 percent to 62.8 percent.
The formula PsiTarget(Result, x) mea- sures the probability that Result (a results cell) will be no more than x.
FIGURE 13.52 The frequency chart and statistics table for the best solution found (choose Project Choice, Project Release, and Project Up) when maximizing MeanProfit in the Tazer project selection problem. The Likelihood box in the statistics table reveals that 58.5 percent of the trials provided a profit of at least $100 million.
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However, the frequency chart and statistics table shown in Figure 13.53 reveals a disad- vantage of this conservative solution. The mean of the total profit from this solution was only $419.19 million versus $540.61 million for the best solution found in Figure 13.52 when the objective was to maximize this quantity. At the same time, the conservative solution has reduced the maximum possible loss from $1.2 billion to $700 million.
In conclusion, Solver has provided Tazer management with two different kinds of solu- tions from which to choose, along with considerable information about each. One appears to be the best high-risk, high-reward solution that is available because it maximizes the total profit that would be obtained on the average. The other appears to be the best available con- servative solution because it maximizes the chances of obtaining a satisfactory profit. By evaluating the trade-off between risk and reward, management now can make a judgment decision about which solution to use.
FIGURE 13.53 The frequency chart and statistics table for the best solution found (choose Project Up and Project Stable) when maximizing the probability of achieving at least $100 million in profit. The Likelihood box in the statistics table reveals that 62.8 percent of the trials provided a profit of at least $100 million.
1. What does Solver search for in a simulation model with decision variables? 2. What kinds of problems can Solver deal with that a parameter analysis report cannot? 3. Does Solver always find an optimal solution to a simulation model? 4. The decision variables for the project selection example needed to be of what kind? 5. What type of statistic cell can measure the probability that the results cell will be less than (or
more than) a specified value?
Review Questions
Increasingly, spreadsheet software is being used to perform computer simulations. As described in the preceding chapter, the standard Excel package sometimes is sufficient to do this. In addition, some Excel add-ins now are available that greatly extend these capabilities. RSPE is an especially powerful add-in of this kind.
When using RSPE, each input cell that has a random value is referred to as an uncertain variable cell. The procedure for defining an uncertain variable cell includes selecting one of 46 types of probabil- ity distributions from the Distributions menu to enter into the cell. When historical data are available, RSPE also has a procedure for identifying which continuous distribution fits the data best.
An output cell that is used to forecast a measure of performance is called a results cell. Each trial of a simulation run generates a value in each results cell. When the simulation run is completed, RSPE
13.10 Summary
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provides the results in a variety of useful forms, including a frequency distribution, a statistics table, a percentiles table, and a cumulative chart.
When a simulation model has one or two decision variables, RSPE provides a parameter analysis report that systematically applies computer simulation to identify at least an approximation of an opti- mal solution. A trend chart also provides additional insights to aid in decision making.
In addition, RSPE includes a powerful optimization module called Solver. This module efficiently uses a series of simulation runs to search for an optimal solution for a simulation model with any num- ber of decision variables.
The availability of such powerful software now enables managers to add computer simulation to their personal tool kit of management science techniques for analyzing some key managerial problems. A variety of examples in this chapter illustrate some of the many possibilities for important applications of computer simulation.
Glossary Distributions menu A menu on the RSPE rib- bon that includes 46 probability distributions from which one is chosen to enter into any uncertain variable cell. (Sections 13.1 and 13.7), 528 parameter analysis report An RSPE module that systematically applies computer simulation over a range of values of one or two decision variables and then displays the results in a table. (Section 13.8), 575 results cell An output cell that is being used by a computer simulation to calculate a measure of performance. (Section 13.1), 529 risk profile A frequency distribution of the return from an investment. (Section 13.5), 552 Solver A component of RSPE that auto- matically searches for an optimal solution for a simulation model with any number of decision variables. (Section 13.9), 583
statistic cell A cell that shows a measure of performance that summarizes the results of an entire simulation run. trend chart A chart that shows the trend of the values in a results cell as a decision variable increases. (Section 13.8), 575 trial A single application of the process of generating a random observation from each prob- ability distribution entered into the spreadsheet and then calculating the output cells in the usual way and recording the results of interest. (Section 13.1), 526 uncertain variable cell An input cell that has a random value so that a probability distribution must be entered into the cell instead of perma- nently entering a single number. (Section 13.1), 527
Chapter 13 Excel Files:
Freddie the Newsboy Case Study
Reliable Co. Bidding Example
Reliable Co. Project Scheduling Example
Everglade Co. Cash Flow Linear Programming
Everglade Co. Cash Flow Management Example
Think-Big Co. Financial Risk Analysis Example
Transcontinental Airlines Overbooking Example
Tazer Corp. Project Selection Example
Sales Data 1
Sales Data 2
An Excel Add-in:
Risk Solver Platform for Education (RSPE)
Learning Aids for This Chapter in Your MS Courseware
Solved Problem (See the CD-ROM or Website for the Solution) 13.S1. Saving for Retirement Patrick Gordon is 10 years away from retirement. He has accumu- lated a $100,000 nest egg that he would like to invest for his golden years. Furthermore, he is confident that he can invest $10,000 more each year until retirement. He is curious about what kind of nest egg he can expect to have accumulated at retirement 10 years from now.
Patrick plans to split his investments evenly among four investments: a money market fund, a domestic stock fund, a global stock fund, and an aggressive growth fund. On the basis of past performance, Patrick expects each of these funds to earn a return in each of the upcoming 10 years according to the distri- butions shown in the following table.
Fund Distribution
Money market Uniform (minimum 5 2%, maximum 5 5%) Domestic stock Normal (mean 5 6%, standard deviation 5 5%) Global stock Normal (mean 5 8%, standard deviation 5 10%) Aggressive growth Normal (mean 5 11%, standard deviation 5 16%)
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his other sources of retirement income. Use a 1,000-trial RSPE simulation to estimate each of the following.
a. What will be the expected value (mean) of Patrick’s nest egg at year 10?
b. What will be the standard deviation of Patrick’s nest egg at year 10?
c. What is the probability that the total nest egg at year 10 will be at least $300,000?
Assume that the initial nest egg ($100,000) and the first year’s investment ($10,000) are made right now (year 0) and are split evenly among the four funds (i.e., $27,500 in each fund). The returns of each fund are allowed to accumulate (i.e., are reinvested) in the same fund and no redistribution will be done before retirement. Furthermore, nine additional investments of $10,000 will be made and split evenly among the four funds ($2,500 each) at year 1, year 2, . . . , year 9.
A financial advisor told Patrick that he can retire comfort- ably if he can accumulate $300,000 by year 10 to supplement
a. Which distribution provides the closest fit to the data? What are the parameters of the distribution?
b. Which distribution provides the second-closest fit to the data? What are the parameters of the distribution?
13.4. Consider the historical data contained in the Excel File “Sales Data 2” on your MS Courseware CD-ROM. Use RSPE to fit continuous distributions to these data. a. Which distribution provides the closest fit to the
data? What are the parameters of the distribution? b. Which distribution provides the second-closest fit to
the data? What are the parameters of the distribution? 13.5. The Aberdeen Development Corporation (ADC) is reconsidering the Aberdeen Resort Hotel project. It would be located on the picturesque banks of Grays Harbor and have its own championship-level golf course.
The cost to purchase the land would be $1 million, payable now. Construction costs would be approximately $2 million, payable at the end of year 1. However, the construction costs are uncertain. These costs could be up to 20 percent higher or lower than the estimate of $2 million. Assume that the construction costs would follow a triangular distribution.
ADC is very uncertain about the annual operating profits (or losses) that would be generated once the hotel is constructed. Its best estimate for the annual operating profit that would be gener- ated in years 2, 3, 4, and 5 is $700,000. Due to the great uncertainty, the estimate of the standard deviation of the annual operating profit in each year also is $700,000. Assume that the yearly profits are statistically independent and follow the normal distribution.
After year 5, ADC plans to sell the hotel. The selling price is likely to be somewhere between $4 and $8 million (assume a uniform distribution). ADC uses a 10 percent discount rate for calculating net present value. (For purposes of this calculation, assume that each year’s profits are received at year-end.) Use RSPE to perform 1,000 trials of a computer simulation of this project on a spreadsheet. a. What is the mean net present value (NPV) of the
project? ( Hint: The NPV (rate, cash stream) func- tion in Excel returns the NPV of a stream of cash flows assumed to start one year from now. For example, NPV(10%, C5:F5) returns the NPV at a 10% discount rate when C5 is a cash flow at the end of year 1, D5 at the end of year 2, E5 at the end of year 3, and F5 at the end of year 4.)
b. What is the estimated probability that the project will yield an NPV greater than $2 million?
RSPE should be used for all of the following problems. An asterisk on the problem number indicates that a partial answer is given in the back of the book.
13.1. The results from a simulation run are inherently ran- dom. This problem will demonstrate this fact and investigate the impact of the number of trials per simulation on this random- ness. Consider the case study involving Freddie the newsboy that was introduced in Section 13.1. The spreadsheet model is available on your MS Courseware CD-ROM. Make sure that the Monte Carlo sampling method is chosen in Simulation Options. Use an order quantity of 60. a. Set the trials per simulation to 100 in Simulation
Options and run the simulation of Freddie’s prob- lem five times. Note the mean profit for each simu- lation run.
b. Repeat part a except set the number of trials per simulation to 1,000 in Simulation Options.
c. Compare the results from part a and part b and com- ment on any differences.
13.2. Consider the Reliable Construction Co. project sched- uling example presented in Section 13.3. Recall that computer simulation was used to estimate the probability of meeting the deadline and that Figure 13.17 revealed that the deadline was met on 57.7 percent of the trials from one simulation run. As discussed while interpreting this result, the percentage of tri- als on which the project is completed by the deadline will vary from simulation run to simulation run. This problem will dem- onstrate this fact and investigate the impact of the number of trials per simulation on this randomness. The spreadsheet model is available on your MS Courseware CD-ROM. Make sure that the Monte Carlo sampling method is chosen in Simulation Options.
a. Set the trials per simulation to 100 in Simulation Options and run the simulation of the project five times. Note the mean completion time and the per- centage of trials on which the project is completed within the deadline of 47 weeks for each simula- tion run.
b. Repeat part a except set the trials per simulation to 1,000 in Simulation Options.
c. Compare the results from part a and part b and com- ment on any differences.
13.3.* Consider the historical data contained in the Excel File “Sales Data 1” on your MS Courseware CD-ROM. Use RSPE to fit continuous distributions to these data.
Problems
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c. Generate a sensitivity chart. Based on this chart, which activities have the largest impact on the proj- ect completion time?
13.7.* Consider Problem 16.12 (see Chapter 16 on the CD- ROM), which involves estimating both the duration of a proj- ect and the probability that it will be completed by the deadline. Assume now that the duration of each activity has a triangular distribution that is based on the three estimates shown in Prob- lem 16.12. Use RSPE to perform 1,000 trials of a computer sim- ulation of the project on a spreadsheet. a. What is the mean project completion time? b. What is the probability that the project can be com-
pleted within 22 months? c. Generate a sensitivity chart. Based on this chart,
which two activities have the largest impact on the project completion time?
13.8. The employees of General Manufacturing Corp. receive health insurance through a group plan issued by Wellnet. During the past year, 40 percent of the employees did not file any health insurance claims, 30 percent filed only a small claim, and 20 percent filed a large claim. The small claims were spread uniformly between 0 and $2,000, whereas the large claims were spread uniformly between $2,000 and $20,000.
Based on this experience, Wellnet now is negotiating the corporation’s premium payment per employee for the upcoming year. To obtain a close estimate of the average cost of insur- ance coverage for the corporation’s employees, use RSPE with a spreadsheet to perform 1,000 trials of a computer simulation of an employee’s health insurance experience. Generate a fre- quency chart and a statistics table. 13.9. Reconsider the Heavy Duty Co. problem that was pre- sented as Example 2 in Section 12.1. For each of the follow- ing three options in parts a through c, obtain an estimate of the expected cost per day by using RSPE to perform 1,000 trials of a computer simulation of the problem on a spreadsheet. Estimate the mean and generate a frequency chart. a. The option of not replacing a motor until a break-
down occurs. b. The option of scheduling the replacement of a
motor after four days (but replacing it sooner if a breakdown occurs).
c. ADC also is concerned about cash flow in years 2, 3, 4, and 5. Use RSPE to estimate the distribu- tion of the minimum annual operating profit (undis- counted) earned in any of the four years. What is the mean value of the minimum annual operating profit over the four years?
d. What is the probability that the annual operat- ing profit will be at least $0 in all four years of operation?
13.6. Ivy University is planning to construct a new building for its business school. This project will require completing all of the activities in the table below. For most of these activities, a set of predecessor activities must be completed before the activ- ity begins. For example, the foundation cannot be laid until the building is designed and the site prepared.
Activity Predecessors
A. Secure funding — B. Design building A C. Site preparation A D. Foundation B, C E. Framing D F. Electrical E G. Plumbing E H. Walls and roof F, G I. Finish construction H J. Landscaping H
Obtaining funding likely will take approximately six months (with a standard deviation of one month). Assume that this time has a normal distribution. The architect has estimated that the time required to design the building could be anywhere between 6 and 10 months. Assume that this time has a uniform distri- bution. The general contractor has provided three estimates for each of the construction tasks—an optimistic scenario (mini- mum time required if the weather is good and all goes well), a most likely scenario, and a pessimistic scenario (maximum time required if there are weather and other problems). These esti- mates are provided in the table that follows. Assume that each of these construction times has a triangular distribution. Finally, the landscaper has guaranteed that his work will be completed in five months.
Construction Time Estimates (months)
Activity Optimistic Scenario
Most Likely Scenario
Pessimistic Scenario
C. Site preparation 1.5 2 2.5 D. Foundation 1.5 2 3 E. Framing 3 4 6 F. Electrical 2 3 5 G. Plumbing 3 4 5 H. Walls and roof 4 5 7 I. Do the finish work 5 6 7
Use RSPE to perform 1,000 trials of a computer simulation for this project. Use the results to answer the following questions. a. What is the mean project completion time? b. What is the probability that the project will be com-
pleted in 36 months or less?
c. The option of scheduling the replacement of a motor after five days (but replacing it sooner if a breakdown occurs).
d. An analytical result of $2,000 per day is available for the expected cost per day if a motor is replaced
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Thus, in addition to option 1 (the proposal), option 2 is to take a 16.50 percent share of the hotel project only (so no participation in the shopping center project), and option 3 is to take a 13.11 percent share of the shopping center only (so no participation in the hotel project). Management wants to choose one of the three options. Risk profiles now are needed to evaluate the latter two. a. Estimate the mean NPV and the probability that the
NPV will be greater than 0 for option 2 after per- forming a computer simulation with 1,000 trials for this option.
b. Repeat part a for option 3. c. Suppose you were the CEO of the Think-Big
Development Co. Use the results in Figure 13.24 for option 1 along with the corresponding results obtained for the other two options as the basis for a managerial decision on which of the three options to choose. Justify your answer.
13.13. Reconsider Problem 12.5 involving the game of craps. Now the objective is to estimate the probability of winning a play of this game. If the probability is greater than 0.5, you will want to go to Las Vegas to play the game numerous times until you eventually win a considerable amount of money. However, if the probability is less than 0.5, you will stay home.
You have decided to perform computer simulation on a spreadsheet to estimate this probability. Use RSPE to perform the number of trials (plays of the game) indicated below. a. 100 trials. b. 1,000 trials. c. 10,000 trials. d. The true probability is 0.493. Based upon the above
simulation runs, what number of trials appears to be needed to give reasonable assurance of obtaining an estimate that is within 0.007 of the true probability?
13.14. Consider the case study involving Freddie the newsboy that was introduced in Section 13.1. The spreadsheet model is avail- able on your MS Courseware CD-ROM. The parameter analysis report generated in Section 13.8 (see Figure 13.40 ) for Freddie’s problem suggests that 55 is the best order quantity, but this table only considered order quantities that were a multiple of 5. Refine the search by generating a parameter analysis report for Freddie’s prob- lem that considers all integer order quantities between 50 and 60. 13.15.* Michael Wise operates a newsstand at a busy intersec- tion downtown. Demand for the Sunday Times averages 300 copies with a standard deviation of 50 copies (assume a normal distribution). Michael purchases the papers for $0.75 and sells them for $1.25. Any papers left over at the end of the day are recycled with no monetary return. a. Suppose that Michael buys 350 copies for his news-
stand each Sunday morning. Use RSPE to perform 1,000 trials of a computer simulation on a spread- sheet. What will be Michael’s mean profit from selling the Sunday Times? What is the probability that Michael will make at least $0 profit?
b. Generate a parameter analysis report to consider five possible order quantities between 250 and 350. Which order quantity maximizes Michael’s mean profit?
c. Generate a trend chart for the five order quantities considered in part b.
every three days. Comparing this option and the above three, which one appears to minimize the expected cost per day?
13.10. The Avery Co. factory has been having a maintenance problem with the control panel for one of its production pro- cesses. This control panel contains four identical electromechan- ical relays that have been the cause of the trouble. The problem is that the relays fail fairly frequently, thereby forcing the control panel (and the production process it controls) to be shut down while a replacement is made. The current practice is to replace the relays only when they fail. The average total cost of doing this has been $3.19 per hour. To attempt to reduce this cost, a proposal has been made to replace all four relays whenever any one of them fails to reduce the frequency with which the control panel must be shut down. Would this actually reduce the cost?
The pertinent data are the following. For each relay, the oper- ating time until failure has approximately a uniform distribution from 1,000 to 2,000 hours. The control panel must be shut down for one hour to replace one relay or for two hours to replace all four relays. The total cost associated with shutting down the control panel and replacing relays is $1,000 per hour plus $200 for each new relay.
Use computer simulation on a spreadsheet to evaluate the cost of the proposal and compare it to the current practice. Use RSPE to perform 1,000 trials (where the end of each trial coin- cides with the end of a shutdown of the control panel) and deter- mine the average cost per hour. 13.11. For one new product produced by the Aplus Company, bushings need to be drilled into a metal block and cylindrical shafts need to be inserted into the bushings. The shafts are required to have a radius of at least 1.0000 inch, but the radius should be as little larger than this as possible. With the proposed production process for producing the shafts, the probability distribution of the radius of a shaft has a triangular distribution with a minimum of 1.0000 inch, a most likely value of 1.0010 inches, and a maximum value of 1.0020 inches. With the proposed method of drilling the bushings, the probability distribution of the radius of a bushing has a normal distribution with a mean of 1.0020 inches and a stan- dard deviation of 0.0010 inches. The clearance between a bush- ing and a shaft is the difference in their radii. Because they are selected at random, there occasionally is interference (i.e., nega- tive clearance) between a bushing and a shaft to be mated.
Management is concerned about the disruption in the production of the new product that would be caused by this occasional interfer- ence. Perhaps the production processes for the shafts and bushings should be improved (at considerable cost) to lessen the chance of interference. To evaluate the need for such improvements, man- agement has asked you to determine how frequently interference is likely to occur with the currently proposed production processes.
Estimate the probability of interference by using RSPE to perform 1,000 trials of a computer simulation on a spreadsheet. 13.12. Refer to the financial risk analysis example presented in Section 13.5, including its results shown in Figure 13.24 . Think- Big management is quite concerned about the risk profile for the proposal. Two statistics are causing particular concern. One is that there is nearly a 20 percent chance of losing money (a nega- tive NPV). Second, there is more than a 6 percent chance of los- ing more than half ($10 million) as much as the mean gain ($18 million). Therefore, management is wondering whether it would be more prudent to go ahead with just one of the two projects.
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a. Suppose that RPI bids $5.7 million on the project. Use RSPE to perform 1,000 trials of a computer sim- ulation on a spreadsheet. What is the probability that RPI will win the bid? What is RPI’s mean profit?
b. Generate a parameter analysis report to consider eight possible bids between $5.3 million and $6 million and forecast RPI’s mean profit. Which bid maximizes RPI’s mean profit?
c. Generate a trend chart for the eight bids considered in part b.
d. Use RSPE’s Solver to search for the bid that maxi- mizes RPI’s mean profit.
13.19. Consider the Everglade cash flow problem analyzed in Section 13.4. The spreadsheet model is available on your MS Courseware CD-ROM. a. Generate a parameter analysis report to consider
five possible long-term loan amounts between $0 million and $20 million and forecast Everglade’s mean ending balance. Which long-term loan amount maximizes Everglade’s mean ending balance?
b. Generate a trend chart for the five long-term loan amounts considered in part a.
c. Use RSPE’s Solver to search for the long-term loan amount that maximizes Evergreen’s mean ending balance.
13.20. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 13.5. Briefly describe how com- puter simulation was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 13.21. Consider the airline overbooking problem discussed in Section 13.6. The spreadsheet model is available on the MS Courseware CD-ROM packaged with the textbook. The param- eter analysis report generated in Section 13.8 (see Figure 13.47 ) for this problem suggests that 185 is the best number of reserva- tions to accept in order to maximize profit, but the only numbers considered were a multiple of five. a. Refine the search by generating a parameter analy-
sis report for this overbooking problem that consid- ers all integer values for the number of reservations to accept between 180 and 190.
b. Generate a trend chart for the 11 forecasts consid- ered in part a.
c. Use RSPE’s Solver to search for the number of reservations to accept that maximizes the airline’s mean profit. Assume that the number of reserva- tions to accept may be any integer value between 150 and 200.
13.22. Flight 120 between Seattle and San Francisco is a popular flight among both leisure and business travelers. The airplane holds 112 passengers in a single cabin. Both a discount 7-day advance fare and a full-price fare are offered. The airline’s management is trying to decide (1) how many seats to allocate to its discount 7-day advance fare and (2) how many tickets to issue in total.
The discount ticket sells for $150 and is nonrefundable. Demand for the 7-day advance fares is typically between 50 and 150, but is most likely to be near 90. (Assume a triangular dis- tribution.) The full-price fare (no advance purchase requirement
d. Use RSPE’s Solver to search for the order quantity that maximizes Michael’s mean profit.
13.16. Susan is a ticket scalper. She buys tickets for Los Angeles Lakers games before the beginning of the season for $100 each. Since the games all sell out, Susan is able to sell the tickets for $150 on game day. Tickets that Susan is unable to sell on game day have no value. Based on past experience, Susan has predicted the probability distribution for how many tickets she will be able to sell, as shown in the following table.
Tickets Probability
10 0.05 11 0.10 12 0.10 13 0.15 14 0.20 15 0.15 16 0.10 17 0.10 18 0.05
a. Suppose that Susan buys 14 tickets for each game. Use RSPE to perform 1,000 trials of a computer simulation on a spreadsheet. What will be Susan’s mean profit from selling the tickets? What is the probability that Susan will make at least $0 profit? ( Hint: Use the Custom Discrete distribution to sim- ulate the demand for tickets.)
b. Generate a parameter analysis report to consider all nine possible quantities of tickets to purchase between 10 and 18. Which purchase quantity maxi- mizes Susan’s mean profit?
c. Generate a trend chart for the nine purchase quanti- ties considered in part b.
d. Use RSPE’s Solver to search for the purchase quan- tity that maximizes Susan’s mean profit.
13.17. Consider the Reliable Construction Co. bidding prob- lem discussed in Section 13.2. The spreadsheet model is avail- able on your MS Courseware CD-ROM. The parameter analysis report generated in Section 13.8 (see Figure 13.46 ) for this prob- lem suggests that $5.4 million is the best bid, but this table only considered bids that were a multiple of $0.2 million. a. Refine the search by generating a parameter analy-
sis report for this bidding problem that considers all bids between $5.2 million and $5.6 million in mul- tiples of $0.05 million.
b. Use RSPE’s Solver to search for the bid that maxi- mizes Reliable Construction Co.’s mean profit. Assume that the bid may be any value between $4.8 million and $5.8 million.
13.18. Road Pavers, Inc. (RPI) is considering bidding on a county road construction project. RPI has estimated that the cost of this particular job would be $5 million. The cost of putting together a bid is estimated to be $50,000. The county also will receive four other bids on the project from competitors of RPI. Past experience with these competitors suggests that each com- petitor’s bid is most likely to be 20 percent over cost, but could be as low as 5 percent over or as much as 40 percent over cost. Assume a triangular distribution for each of these bids.
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c. Use RSPE’s Solver to try to determine the maxi- mum number of discount fare tickets and the maxi- mum total number of tickets to make available so as to maximize the airline’s expected profit.
13.23. Now that Jennifer is in middle school, her parents have decided that they really must start saving for her college education. They have $6,000 to invest right now. Furthermore, they plan to save another $4,000 each year until Jennifer starts college in five years. They plan to split their investment evenly between a stock fund and a bond fund. Historically, the stock fund has had an average annual return of 8 percent with a stan- dard deviation of 6 percent. The bond fund has had an average annual return of 4 percent with a standard deviation of 3 percent. (Assume a normal distribution for both.)
Assume that the initial investment ($6,000) and the first year’s investment ($4,000) are made right now (year 0) and are split evenly between the two funds (i.e., $5,000 in each fund). The returns of each fund are allowed to accumulate (i.e., are reinvested) in the same fund and no redistribution will be done before Jennifer starts college. Furthermore, four additional investments of $4,000 will be made and split evenly between both funds ($2,000 each) at year 1, year 2, year 3, and year 4. Use a 1,000-trial RSPE simulation to estimate each of the following. a. What will be the expected value (mean) of the col-
lege fund at year 5? b. What will be the standard deviation of the college
fund at year 5? c. What is the probability that the college fund at year
5 will be at least $30,000? d. What is the probability that the college fund at year
5 will be at least $35,000?
and fully refundable prior to check-in time) is $400. Exclud- ing customers who purchase this ticket and then cancel prior to check-in time, demand is equally likely to be anywhere between 30 and 70 for these tickets (with essentially all of the demand occurring within one week of the flight). The average no-show rate is 5 percent for the nonrefundable discount tickets and 15 percent for the refundable full-price tickets. If more ticketed passengers show up than there are seats available, the extra pas- sengers must be bumped. A bumped passenger is rebooked on another flight and given a voucher for a free ticket on a future flight. The total cost to the airline for bumping a passenger is $600. There is a fixed cost of $10,000 to operate the flight.
There are two decisions to be made. First, prior to one week before flight time, how many tickets should be made available at the discount fare? Too many and the airline risks losing out on potential full-fare passengers. Too few and the airline may have a less-than-full flight. Second, how many tickets should be issued in total? Too many and the airline risks needing to bump passen- gers. Too few and the airline risks having a less-than-full flight. a. Suppose that the airline makes available a maxi-
mum of 75 tickets for the discount fare and a maxi- mum of 120 tickets in total. Use RSPE to generate a 1,000 trial forecast of the distribution of the profit, the number of seats filled, and the number of pas- sengers bumped.
b. Generate a two-dimensional parameter analysis report that gives the mean profit for all combinations of the following values of the two decision variables: (1) the maximum number of tickets made available at the discount fare is a multiple of 10 between 50 and 90 and (2) the maximum number of tickets made available for either fare is 112, 117, 122, 127, or 132.
Case 13-1
Action Adventures The Adventure Toys Company manufactures a popular line of action figures and distributes them to toy stores at the wholesale price of $10 per unit. Demand for the action figures is seasonal, with the highest sales occurring before Christmas and during the spring. The lowest sales occur during the summer and winter months.
Each month the monthly “base” sales follow a normal distri- bution with a mean equal to the previous month’s actual “base” sales and with a standard deviation of 500 units. The actual sales in any month are the monthly base sales multiplied by the sea- sonality factor for the month, as shown in the subsequent table. Base sales in December 2013 were 6,000, with actual sales equal to (1.18)(6,000) 5 7,080. It is now January 1, 2014.
Cash sales typically account for about 40 percent of monthly sales, but this figure has been as low as 28 percent and as high as 48 percent in some months. The remainder of the sales are made on a 30-day interest-free credit basis, with full payment received one month after delivery. In December 2013, 42 percent of sales were cash sales and 58 percent were on credit.
The production costs depend upon the labor and material costs. The plastics required to manufacture the action figures fluctuate in price from month to month, depending on market conditions. Because of these fluctuations, production costs can be anywhere from $6 to $8 per unit. In addition to these variable production costs, the company incurs a fixed cost of $15,000
Month Seasonality
Factor Month Seasonality
Factor
January 0.79 July 0.74 February 0.88 August 0.98 March 0.95 September 1.06 April 1.05 October 1.10 May 1.09 November 1.16 June 0.84 December 1.18
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Case 13-2 Pricing under Pressure 597
exceed) 9 percent and the savings interest rate will never drop below 2 percent.
The prime rate in December 2013 was 5 percent per annum. This rate depends upon the whims of the Federal Reserve Board. In particular, for each month there is a 70 percent chance it will stay unchanged, a 10 percent chance it will increase by 25 basis points (0.25 percent), a 10 percent chance it will decrease by 25 basis points, a 5 percent chance it will increase by 50 basis points, and a 5 percent chance it will decrease by 50 basis points.
a. Formulate a simulation model on a spreadsheet to track the company’s cash flows from month to month. Use RSPE to simulate 1,000 trials for the year 2014.
b. Adventure Toys management wants information about what the company’s net worth might be at the end of 2014, includ- ing the likelihood that the net worth will exceed $0. (The net worth is defined here as the ending cash balance plus savings interest and account receivables minus any loans and interest due.) Display the results of your simulation run from part a in the various forms that you think would be helpful to manage- ment in analyzing this issue.
c. Arrangements need to be made to obtain a specific credit limit from the bank for the short-term loans that might be needed during 2014. Therefore, Adventure Toys manage- ment also would like information regarding the size of the maximum short-term loan that might be needed during 2014. Display the results of your simulation run from part a in the various forms that you think would be helpful to manage- ment in analyzing this issue.
per month for manufacturing the action figures. The company assembles the products to order. When a batch of a particular action figure is ordered, it is immediately manufactured and shipped within a couple of days.
The company utilizes eight molding machines to mold the action figures. These machines occasionally break down and require a $5,000 replacement part. Each machine requires a replacement part with a 10 percent probability each month.
The company has a policy of maintaining a minimum cash balance of at least $20,000 at the end of each month. The bal- ance at the end of December 2013 (or equivalently, at the begin- ning of January 2014) is $25,000. If required, the company will take out a short-term (one-month) loan to cover expenses and maintain the minimum balance. The loans must be paid back the following month with interest (using the current month’s loan interest rate). For example, if March’s annual interest rate is 6 percent (so 0.5 percent per month) and a $1,000 loan is taken out in March, then $1,005 is due in April. However, a new loan can be taken out each month.
Any balance remaining at the end of a month (including the minimum balance) is carried forward to the following month and also earns savings interest. For example, if the ending bal- ance in March is $20,000 and March’s savings interest is 3 per- cent per annum (so 0.25 percent per month), then $50 of savings interest is earned in April.
Both the loan interest rate and the savings interest rate are set monthly based upon the prime rate. The loan interest rate is set at prime 1 2%, while the savings interest rate is set at prime 2 2%. However, the loan interest rate is capped at (can’t
Case 13-2
Pricing under Pressure
Elise Sullivan moved to New York City in September to begin her first job as an analyst working in the Client Services Divi- sion of FirstBank, a large investment bank providing broker- age services to clients across the United States. The moment she arrived in the Big Apple after receiving her undergraduate busi- ness degree, she hit the ground running—or, more appropriately, working. She spent her first six weeks in training, where she met new FirstBank analysts like herself and learned the basics of FirstBank’s approach to accounting, cash flow analysis, cus- tomer service, and federal regulations.
After completing training, Elise moved into her bullpen on the 40th floor of the Manhattan FirstBank building to begin work. Her first few assignments have allowed her to learn the ropes by placing her under the direction of senior staff members who delegate specific tasks to her.
Today, she has an opportunity to distinguish herself in her career, however. Her boss, Michael Steadman, has given her an assignment that is under her complete direction and control. A very eccentric, wealthy client and avid investor by the name of Emery Bowlander is interested in purchasing a European call option that provides him with the right to purchase shares of Fellare stock for $44.00 on the first of February—12 weeks from today. Fel- lare is an aerospace manufacturing company operating in France, and Mr. Bowlander has a strong feeling that the European Space
Agency will award Fellare with a contract to build a portion of the International Space Station some time in January. In the event that the European Space Agency awards the contract to Fellare, Mr. Bowlander believes the stock will skyrocket, reflecting investor confidence in the capabilities and growth of the company. If Fel- lare does not win the contract, however, Mr. Bowlander believes the stock will continue its current slow downward trend. To guard against this latter outcome, Mr. Bowlander does not want to make an outright purchase of Fellare stock now.
Michael has asked Elise to price the option. He expects a figure before the stock market closes so that if Mr. Bowlander decides to purchase the option, the transaction can take place today.
Unfortunately, the investment science course Elise took to complete her undergraduate business degree did not cover options theory; it only covered valuation, risk, capital budgeting, and market efficiency. She remembers from her valuation stud- ies that she should discount the value of the option on February 1 by the appropriate interest rate to obtain the value of the option today. Because she is discounting over a 12-week period, the formula she should use to discount the option is (Value of the Option/[1 1 Weekly Interest Rate] 12 ). As a starting point for her calculations, she decides to use an annual interest rate of 8 per- cent. But she now needs to decide how to calculate the value of the option on February 1.
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mean of the normally distributed random variable by subtracting one-half of the square of the weekly stock volatility from the weekly interest rate w. In other words:
Mean 5 w 2 0.5(Weekly Stock Volatility)2
The standard deviation of the normally distributed random vari- able is simply equal to the weekly stock volatility.
Elise is now ready to build her simulation model.
a. Build a simulation model in a spreadsheet to calculate the value of the option in today’s dollars. Use RSPE to run three separate simulations to estimate the value of the call option and hence the price of the option in today’s dollars. For the first simulation, run 100 trials of the simulation. For the sec- ond simulation, run 1,000 trials of the simulation. For the third simulation, run 10,000 trials of the simulation. For each simulation, record the price of the option in today’s dollars.
b. Elise takes her calculations and recommended price to Michael. He is very impressed, but he chuckles and indicates that a simple, closed-form approach exists for calculating the value of an option: the Black-Scholes formula. Michael grabs an investment science book from the shelf above his desk and reveals the very powerful and very complicated Black-Scholes formula:
V 5 N 3d1 4P 2 N 3d2 4PV 3K 4 where
d1 5 ln3P/PV 3K 4 4
s"t 1 s"t
2
d2 5 d1 2 s"t N [ x ] 5 The Excel function NORMSDIST ( x ) where x 5 d 1
or x 5 d 2 P 5 Current price of the stock K 5 Exercise price
PV[ K ] 5 Present value of exercise price 5 K
(1 1 w)t
t 5 Number of weeks to exercise date s 5 Weekly volatility of stock
Use the Black-Scholes formula to calculate the value of the call option and hence the price of the option. Compare this value to the value obtained in part a.
c. In the specific case of Fellare stock, do you think that a ran- dom walk as described above completely describes the price movement of the stock? Why or why not?
Additional Cases Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases , in the segment of the CaseMate area designated for this book.
Elise knows that on February 1, Mr. Bowlander will take one of two actions: either he will exercise the option and purchase shares of Fellare stock or he will not exercise the option. Mr. Bowlander will exercise the option if the price of Fellare stock on February 1 is above his exercise price of $44.00. In this case, he purchases Fel- lare stock for $44.00 and then immediately sells it for the market price on February 1. Under this scenario, the value of the option would be the difference between the stock price and the exercise price. Mr. Bowlander will not exercise the option if the price of Fellare stock is below his exercise price of $44.00. In this case, he does nothing, and the value of the option would be $0.
The value of the option is therefore determined by the value of Fellare stock on February 1. Elise knows that the value of the stock on February 1 is uncertain and is therefore represented by a probability distribution of values. Elise recalls from a manage- ment science course in college that she can use computer simu- lation to estimate the mean of this distribution of stock values. Before she builds the simulation model, however, she needs to know the price movement of the stock. Elise recalls from a prob- ability and statistics course that the price of a stock can be mod- eled as following a random walk and either growing or decaying according to a lognormal distribution. Therefore, according to this model, the stock price at the end of the next week is the stock price at the end of the current week multiplied by a growth factor. This growth factor is expressed as the number e raised to a power that is equal to a normally distributed random variable. In other words:
sn 5 e Nsc
where
s n 5 The stock price at the end of next week s c 5 The stock price at the end of the current week N 5 A random variable that has a normal distribution
To begin her analysis, Elise looks in the newspaper to find that the Fellare stock price at the end of the current week is $42.00. She decides to use this price to begin her 12-week analy- sis. Thus, the price of the stock at the end of the first week is this current price multiplied by the growth factor. She next estimates the mean and standard deviation of the normally distributed ran- dom variable used in the calculation of the growth factor. This random variable determines the degree of change (volatility) of the stock, so Elise decides to use the current annual interest rate and the historical annual volatility of the stock as a basis for estimating the mean and standard deviation.
The current annual interest rate is r 5 8 percent, and the historical annual volatility of the aerospace stock is 30 percent. But Elise remembers that she is calculating the weekly change in stock— not the annual change. She therefore needs to calculate the weekly interest rate and weekly historical stock volatility to obtain estimates for the mean and standard deviation of the weekly growth factor. To obtain the weekly interest rate w, Elise must make the following calculation:
w 5 (1 1 r)(1/52) 2 1
The historical weekly stock volatility equals the historical annual volatility divided by the square root of 52. She calculates the
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Tips for Using Microsoft Excel for Modeling Microsoft Excel is a powerful and flexible tool with a myriad of features. It is certainly not necessary to master all the features of Excel in order to successfully build models in spread- sheets. However, there are some features of Excel that are particularly useful for modeling that we will highlight here. This appendix is not designed to be a basic tutorial for Excel. It is designed instead for someone with a working knowledge of Excel (at least at a basic level) who wants to take advantage of some of the more advanced features of Excel that are useful for building models more efficiently.
ANATOMY OF THE MICROSOFT EXCEL WINDOW
When Microsoft Excel is first opened (e.g., by choosing Microsoft Excel from the Start menu), a blank spreadsheet appears in an Excel window. The various components of the Excel window are labeled in Figure A.1 .
The Excel file is called a workbook. A workbook consists of a number of worksheets or spreadsheets, identified in the sheet tabs at the bottom of the screen (Sheet1, Sheet2, and Sheet3 in Figure A.1 ). Only one spreadsheet at a time is shown in the window, with the cur- rently displayed spreadsheet highlighted in the sheet tab (Sheet1 in Figure A.1 ). To show a different spreadsheet (e.g., Sheet2 or Sheet3), click on the appropriate sheet tab.
Each spreadsheet consists of a huge grid, with many rows and columns. The rows are labeled on the left of the grid by numbers (1, 2, 3, . . .). The columns are labeled on the top of the grid by letters (A, B, C, . . .). Each element of the grid is referred to as a cell, and is referred to by its row and column label (e.g., cell C7). The currently selected cell is high- lighted by the cell cursor (a dark or colored border). A different cell can be selected either by clicking on it or by moving the cell cursor with the arrow keys.
Only a portion of the spreadsheet is shown at any one time. For example, in Figure A.1 only the first 9 columns and first 17 rows are shown. The scroll bars can be used to show a different portion of the spreadsheet.
WORKING WITH WORKBOOKS
When Microsoft Excel is first opened (e.g., by choosing Microsoft Excel from the Start menu), a new workbook is created and given a default name that is visible in the Title Bar (e.g., Book1 in Figure A.1 ). To give the workbook a different name, save it under whatever name you desire by choosing Save As under the Office Button.
To open an existing workbook that has been saved previously, choose Open from the Office Button. It is possible to have more than one workbook open at a time within Excel. This may be desirable if you want to copy worksheets from one workbook to another or if you want to see the contents of another workbook while you are working on an existing
Appendix A
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workbook. When multiple workbooks are open, some of the workbooks can become hidden behind another workbook that is being displayed. To bring any workbook to the front, select it under the Switch Windows menu on the View tab. The workbooks also can be arranged on the screen (e.g., one over the other, or one next to the other) by choosing Arrange All on the View tab.
WORKING WITH WORKSHEETS
By default, a new Excel workbook consists of a few worksheets titled Sheet1, Sheet2, Sheet3, and so on. The currently displayed sheet is highlighted in the sheet tabs. To display a different sheet, click on the appropriate tab. If the desired tab is not visible because there are more tabs than can be displayed, the list of tabs can be scrolled using the sheet tab scroll buttons.
The sheets can be given descriptive names by double-clicking on the sheet tab and typing a new name. A new sheet can be added to the workbook by choosing Insert Sheet from the Insert menu of the Cells group on the Home tab. The sheet tabs can be reordered by clicking and dragging a tab to a new location. To make a copy of a sheet, control-click (option-click on a Mac) and drag the tab. If multiple workbooks are open, you can also click (or control- click) and drag a sheet tab to a different workbook to move (or copy) a sheet to a different workbook.
Using Worksheets with Solver A model must be confined to a single sheet. When using the Excel Solver, all cell refer- ences (e.g., the objective cell, changing cells, etc.) must be on the currently displayed sheet. Thus, the different components of the Solver model cannot be spread among different sheets. RSPE's Solver does allow the model on a given sheet to have its components spread across different sheets.
The Solver information is saved with the sheet. When information is entered in Solver (including the objective cell, changing cells, and constraints), all of that information is saved with the sheet when the workbook is saved.
Separate sheets can contain separate models. Separate Solver information (e.g., the objective cell, changing cells, etc.) is kept for each sheet in the workbook. Thus, each sheet
FIGURE A.1 The Microsoft Excel window for Excel 2010.
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in a workbook can contain a separate and independent model. When Solver is run, only the model on the currently displayed sheet is solved.
Copy the whole sheet rather than just the relevant cells to copy models. To copy a model to another workbook or within the current workbook, it is important to control-click and drag the worksheet tab rather than simply selecting the cells containing the model and using copy and paste. Copying the sheet (by control-clicking and dragging the sheet tab) will copy all the contents of the sheet (formulas, data, and the Solver information). Using copy and paste copies only the formulas and data, but does not include the Solver information.
Using Worksheets with RSPE Decision Trees Separate sheets can contain separate RSPE decision trees. If the currently displayed sheet does not contain an existing RSPE decision tree, then choosing Add Node under the Decision Tree . Node menu on the RSPE ribbon will present the option of adding a new tree to the existing sheet. However, if a decision tree already exists on the sheet, then new nodes and branches can only be added to the existing decision tree. To create a new decision tree, first switch to (or add) a new sheet. A workbook can contain separate decision trees so long as they are on separate sheets.
Using Worksheets with RSPE Simulation Models The entire workbook is treated as a single simulation model for RSPE. Uncertain variable cells, results cells, and statistic cells can be defined on any or all of the different sheets of the workbook. When a simulation is run, all uncertain variable cells are randomly generated, all results cells are evaluated, and all statistic cells are calculated regardless of whether they are on the currently displayed sheet.
WORKING WITH CELLS
Selecting Cells To make any changes to a cell or range of cells, such as entering or editing data or changing the formatting, the cell or cells involved first need to be selected. The cell cursor shows the currently selected cell (or range of cells). To select a different single cell, either click on it or use the arrow keys to move the cell cursor to that location. To select an entire row or an entire column, click the row or column heading (i.e., the A, B, C along the top of the spreadsheet, or the 1, 2, 3 along the left of the spreadsheet). To select the entire spreadsheet, click in the blank box at the upper left corner of the worksheet.
There are three ways to select a range of cells within a spreadsheet, which we will illustrate by considering the 3-by-3 range of cells from A1 to C3:
1. Click on one corner of the range (A1) and, without releasing the mouse button, drag to the other corner of the range (C3).
2. Click on one corner of the range (A1) and then hold down the SHIFT key and click on the other corner of the range (C3).
3. Click on one corner of the range (A1), hold down SHIFT or press F8 to turn on the extend mode, use the arrow keys to extend the range to the other corner (C3), and then either release the SHIFT key or press F8 again to turn the extend mode off.
Entering or Editing Data, Text, and Formulas into Cells There are a number of ways to enter and edit the contents of a cell:
1. Use the Formula Bar: The contents of the currently selected cell appear in the formula bar (see Figure A.1 ). To enter data, text, or a formula into a cell, click on the cell and type or edit the contents in the formula bar. Press Enter when you are done.
2. Double-click: Double-clicking on a cell (or pressing F2) will display the contents of the cell and allow typing or editing directly within the cell on the spreadsheet. If the cell
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contains a formula, the cells referred to in the formula will be highlighted in different col- ors on the spreadsheet. The formula can be modified either by clicking and typing within the cell or by dragging the highlighted cell markers to new locations.
3. Insert Function: In an empty cell, pressing the f x button next to the formula bar will bring up a dialog box showing all of the functions available in Excel sorted by type. After choos- ing a function from the list, the function is inserted into the cell and all the parameters of the function are shown in a small window.
Moving or Copying Cells To move a cell or range of cells on the spreadsheet, first select the cell(s). To move the cell(s) a short distance on the spreadsheet (e.g., down a few rows), it is usually most convenient to use the dragging method. Click on an edge of the cell cursor and, without releasing the mouse button, drag the cell(s) to the new location. To move the cell(s) a large distance (e.g., down 100 rows, or to a different worksheet), it is usually more convenient to use Cut and Paste from the Home tab.
Similar methods can be used to make a copy of a cell or range of cells. To copy a cell (or range of cells), press ctrl (option on a Mac) while clicking on the edge of the cell cursor and dragging, or use Copy and Paste from the Home tab.
Filling Cells When building a spreadsheet, it is common to need a series of numbers or dates in a row or column. For example, Figure A.2 shows a spreadsheet that calculates the projected annual cash flow and taxes due for 2014 through 2018, based upon monthly cash flows. Rather than typing all 12 column labels for the months in cells B2:M2, the fill handle (the small box on the lower-right corner of the cell cursor) can be used to fill in the series. After entering the first couple of elements of the series, for example, Jan in cell B2 and Feb in cell C2, select cells B2:C2 and then click and drag the fill handle to cell M2. The remainder of the series (Mar, Apr, May, etc.) will be filled in automatically. The year labels in cells A3:A7 can be filled in a similar fashion. After entering the first couple of years, 2014 in A3 and 2015 in A4, select cells A3:A4 and then click and drag the fill handle down to A7. On the basis of the data in the cells selected, the fill handle will try to guess the remainder of the series.
The fill handle is also useful for copying similar formulas into adjacent cells in a row or a column. For example, the formula to calculate the annual cash flows in N3:N7 is basically the same formula for every year. After entering the formula for 2014 in cell N3, select cell N3 and then click and drag the fill handle to copy the formula down through cell N7. Similarly, the taxes due formula in cell O3 can be copied down to cells O4:O7. In fact, both the annual
1 2 3 4 5 6 7 8 9
A B C D E F G H I J K L M N O Cash Flow ($000) Annual Tax
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Cash Flow Due 2014 10 −2 4 5 4 6 8 10 12 3 −4 8 64 16.0 2015 15 3 −4 3 10 4 6 10 3 6 −2 12 66 16.5 2016 8 4 2 −3 −5 7 4 8 8 11 −3 11 52 13.0 2017 7 5 5 3 2 6 10 12 14 8 2 8 82 20.5 2018 5 2 2 −4 9 7 12 14 3 −4 6 10 62 15.5
Tax Rate 25%
1 2 3 4 5 6 7 8 9
N O Annual Tax
Cash Flow Due =SUM(B3:M3) =N3*$O$9 =SUM(B4:M4) =N4*$O$9 =SUM(B5:M5) =N5*$O$9 =SUM(B6:M6) =N6*$O$9 =SUM(B7:M7) =N7*$O$9
Tax Rate 0.25
FIGURE A.2 A simple spreadsheet to calculate projected annual cash flow and tax due.
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cash flow and tax due formulas can be copied at once by selecting both cells N3 and O3 (the range N3:O3) and then dragging the fill handle down to cell O7. This will fill both formulas down into the cells N4:O7.
Relative and Absolute References When using the fill handle, it is important to understand the difference between relative and absolute references. Consider the formula in cell N3 ( 5 SUM(B3:M3)). The references to cells in the formula (B3:M3) are based upon their relative position to the cell containing the formula. Thus, B3:M3 are treated as the 12 cells immediately to the left. This is known as a relative reference. When this formula is copied to new cells using the fill handle, the refer- ences are automatically adjusted to refer to the new cell(s) at the same relative location (the 12 cells immediately to the left). For example, the formula in N4 becomes 5 SUM(B4:M4), the formula in N5 becomes 5 SUM(B5:M5), and so on.
In contrast, the reference to the tax rate ($O$9) in the formula in cell O3 is called an abso- lute reference. These references do not change when they are filled into other cells. Thus, when the formula in cell O3 is copied into cells O4:O7, the reference still refers to cell O9.
To make an absolute reference, put $ signs in front of the letter and number of the cell reference (e.g., $O$9). Similarly, you can make the column absolute and the row relative (or vice versa) by putting a $ sign in front of only the letter (or number) of the cell reference. After entering a cell reference, repeatedly pressing the F4 key (or command-T on a Mac) will rotate among the four possibilities of relative and absolute references (e.g., O9, $O$9, O$9, $O9).
Using Range Names A block of related cells can be given a range name. Then, rather than referring to the cells by their cell addresses (e.g., L11:L21 or C3), a more descriptive name can be used (e.g., Total- Profit). To give a cell or range of cells a range name, first select the cell(s). Then click in the Name Box (see Figure A.1 ) and type a name. For example, for the spreadsheet in Figure A.2 we could define a range name for the tax rate by selecting cell O9 and typing TaxRate into the name box. Spaces are not allowed in range names, so use capital letters or underscore charac- ters to separate words in a name.
Once a range name is defined, rather than typing the cell reference (e.g., O9) when it is used in a formula, the range name can be used instead (e.g., TaxRate). If you click on a cell (or cells) to use it in a formula, the range name is automatically used rather than the cell reference. This can make the formula easier to interpret (e.g., 5 SUM(B3:M3)*TaxRate, rather than 5 SUM(B3:M3)*$ O $9). When using a range name in a formula, it is treated as an absolute reference. To make a relative reference to a cell that has a range name, type the cell address (e.g., O9) rather than either typing the range name or clicking on the cell (which then automatically uses the range name).
Formatting Cells To make formatting changes to a cell or range of cells, first select the cell(s). If a range of cells is selected, any formatting changes will apply to every cell in the range. Most common types of formatting of cells, for example, changing the font, making text bold or italic, or changing the borders or shading of a cell, can be done by using the Home tab.
Clicking on the .0 → .00 or .00 → .0 buttons changes the number of decimal places shown in a cell. Note that this only changes how the number is displayed, since Excel always uses the full precision when this cell is used in other formulas.
For more advanced types of formatting, choose Format Cells under the Format menu of the Cells group on the Home tab. A shortcut is to press ctrl-1 on a PC or command-1 on a Mac. This brings up the Format cells dialog box, as shown in Figure A.3 . Under the Numbers tab you can choose to display the contents in a cell as a number with any number of decimal places (e.g., 123.4 or 123.486), as currency (e.g., $1,234.10), as a date (e.g., 12/10/2016 or Dec 2016), and so on. The other tabs are used to change the alignment of the text (e.g., left or
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604 Appendix A Tips for Using Microsoft Excel for Modeling
right justified, printed vertically or horizontally, etc.), the font, the borders, the patterns, and the protection.
If a cell displays ####, this means that the column width is not wide enough to show the contents of the cell. To change column widths or row heights, click and drag the vertical or horizontal lines between the column or row labels. Double-clicking on the vertical line between column labels will make the column just wide enough to show the entire contents of every cell in the column.
FIGURE A.3 The Format Cells dialog box.
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Shipment Quantities Customer 1 Customer 2 Customer 3
Factory 1 300 0 100 Factory 2 0 200 300
Appendix B
Partial Answers to Selected Problems
CHAPTER 2
2.6. d. Fraction of 1 st 5 0.667, fraction of 2 nd 5 0.667. Profit 5 $6,000. 2.13. b. x 1 5 13, x 2 5 5. Profit 5 $31.
CHAPTER 3
3.3. c. 3.333 of Activity 1, 3.333 of Activity 2. Profit 5 $166.67. 3.6. d. 26 of Product 1, 54.76 of Product 2, 20 of Product 3. Profit 5 $2,904.76. 3.12. d. 1.14 kg of corn, 2.43 kg of alfalfa. Cost 5 $2.42. 3.17. b. Cost 5 $410,000.
3.19. c. $60,000 in Investment A (year 1), $84,000 in Investment A (year 3), $117,600 in Investment D (year 5). Total accumulation in year 6 5 $152,880.
3.22. a. Profit 5 $13,330.
Cargo Placement Front Center Back
Cargo 1 0 5 10 Cargo 2 7.333 4.167 0 Cargo 3 0 0 0 Cargo 4 4.667 8.333 0
CHAPTER 4
4.2. d. 0 end tables, 40 coffee tables, 30 dining room tables. Profit 5 $10,600. 4.4. e. 19% participation in Project A, 0% participation in Project B, and 100% participa-
tion in Project C. Ending Balance 5 $59.5 million.
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606 Appendix B Partial Answers to Selected Problems
4.9. d. 4 FT (8 am –4 pm ), 4 FT (12 pm –8 pm ), 4 FT (4 pm –midnight), 2 PT (8 am –12 pm ), 0 PT (12 pm –4 pm ), 4 PT (4 pm –8 pm ), 2 PT (8 pm –midnight). Total cost per day 5 $1,728.
CHAPTER 5
5.1. e. Allowable range for unit profit from producing toys: $2.50 to $5.00. Allowable range for unit profit from producing subassemblies: 2 $3.00 to 2 $1.50. 5.4. f. ( Part a ) Optimal solution does not change (within allowable increase of $10).
( Part b ) Optimal solution does change (outside of allowable decrease of $5).
( Part c ) By the 100% rule for simultaneous changes in the objective function, the optimal
solution may or may not change.
C8AM: $160 S $165 % of allowable increase 5 100 a165 2 160 10
b 5 50%
C4PM: $180 S $170 % of allowable decrease 5 100 a180 2 170 5
b 5 200% Sum 5 250%
5.11. a. Produce 2,000 toys and 1,000 sets of subassemblies. Profit 5 $3,500. b. The shadow price for subassembly A is $0.50, which is the maximum premium that
the company should be willing to pay. 5.15. a. The total expected number of exposures could be increased by 3,000 for each addi-
tional $1,000 added to the advertising budget. b. This remains valid for increases of up to $250,000. e. By the 100% rule for simultaneous changes in right-hand sides, the shadow prices
are still valid. Using units of thousands of dollars,
CA: $4,000 S $4,100 % of allowable increase 5 100 a4,100 2 4,000250 b 5 40%
CP: $1,000 S $1,100 % of allowable increase 5 100 a1,100 2 1,000450 b 5 22% Sum 5 62%
CHAPTER 6
6.2. b. 0 S1-D1, 10 S1-D2, 30 S1-D3, 30 S2-D1, 30 S2-D2, 0 S2-D3. Total cost 5 $580. 6.5. c. $2,187,000. 6.9. Maximum flow 5 15. 6.15. b. Replace after year 1. Total cost 5 $29,000.
CHAPTER 7
7.3. b. Marketing and dishwashing by Eve, cooking and laundry by Steven. Total time 5 18.4 hours.
7.8. Optimal path 5 OADT. Total distance 5 10 miles.
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Appendix B Partial Answers to Selected Problems 607
CHAPTER 8
8.7. c. Invest $46,667 in Stock 1 and $3,333 in Stock 2 for $13,000 expected profit. Invest $33,333 in Stock 1 and $16,667 in Stock 2 for $15,000 expected profit. 8.11. d. Dorwyn should produce 1 window and 1 door.
CHAPTER 9
9.4. a. Speculative investment. d. Counter-cyclical investment. 9.7. b. A 3 c. A 2 9.12. a. A 1 b. $18. 9.16. c. EVPI 5 $3,000. The credit-rating organization should not be used. 9.21. c. Choose to build computers (expected payoff is $27 million). f. They should build when p ≤ 0.722 and sell when p > 0.722. 9.22. a. EVPI 5 $7.5 million. c. P(Sell 10,000 | Predict Sell 10,000) 5 0.667. P(Sell 100,000 | Predict Sell 100,000) 5 0.667. 9.23. a. The optimal policy is to do no market research and build the computers. 9.26. c. $800,000.
f, g. Leland University should hire William. If he predicts a winning season, then they should hold the campaign. If he predicts a losing season, then they should not hold the campaign.
9.30. a. Choose to introduce the new product (expected payoff is $12.5 million). b. $7.5 million. c. The optimal policy is not to test but to introduce the new product. g. Both charts indicate that the expected profit is sensitive to both parameters, but is
somewhat more sensitive to changes in the profit if successful than to changes in the loss if unsuccessful.
9.35. a. Choose not to buy insurance (expected payoff is $249,840). b. Choose to buy insurance (expected utility is 499.82).
CHAPTER 10
10.1. a. 39. b. 26. c. 36. 10.3. MAD 5 15. 10.9. 2,091. 10.13. When a 5 0.1, forecast 5 2,072. 10.17. 552. 10.19. b. MAD 5 5.18. c. MAD 5 3. d. MAD 5 3.93. 10.29. 62 percent. 10.35. b. y 5 410 1 17.6 x. d. 604.
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608 Appendix B Partial Answers to Selected Problems
Summary of Results:
Win? (1 5 Yes, 0 5 No) 0 Number of Tosses 5 3
Simulated Tosses
Toss Die 1 Die 2 Sum
1 4 2 6 2 3 2 5 3 6 1 7 4 5 2 7 5 4 4 8 6 1 4 5 7 2 6 8
Results
Win? Lose? Continue?
0 0 Yes 0 0 Yes 0 1 No NA NA No NA NA No NA NA No NA NA No
CHAPTER 11 11.3. a. True. b. False. c. True. 11.8. a. L 5 2 b. L q 5 0.375 c. W 5 30 minutes, W q 5 5.625 minutes. 11.12. a. 96.9% of the time. 11.15. b. L 5 0.333 g. Two members. 11.18. L q is unchanged and W q is reduced by half. 11.23. a. L 5 3 d. TC (status quo) 5 $85/hour. TC (proposal) 5 $73/hour. 11.28. a. 0.211 hours. c. Approximately 3.43 minutes. 11.31. c. 0.4. d. 7.2 hours. 11.35. Jim should operate 4 cash registers. Expected cost per hour 5 $80.59.
CHAPTER 12
12.1. b. Let the numbers 0.0000 to 0.5999 correspond to strikes and the numbers 0.6000 to 0.9999 correspond to balls. The random observations for pitches are 0.3039 5 strike, 0.7914 5 ball, 0.8543 5 ball, 0.6902 5 ball, 0.3004 5 strike, 0.0383 5 strike.
12.5. a. Here is a sample replication.
12.10. a. Let the numbers 0.0000 to 0.3999 correspond to a minor repair and 0.4000 to 0.9999 correspond to a major repair. The average repair time is then (1.224 1 0.9 50 1 1.610)/3 5 1.26 hours.
12.17. b. The average waiting time should be approximately 1 day. c. The average waiting time should be approximately 0.33 days.
CHAPTER 13 13.3. a. Min Extreme distribution (Mode 5 170.3, Scale 5 50.9, Shift 5 320.0). 13.7. a. The mean project completion time should be approximately 33 months. c. Activities B and J have the greatest impact on the variability in the project com-
pletion time. 13.15. a. Mean profit should be around $107, with about a 96.5% chance of making at least $0.
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Index A
Abbink, E., 247 Absolute references, 28, 139, 603 Action Adventures (case study), 596–597 Advertising-mix problems
campaign planning in, 47 cost–benefit–trade-off problems
and, 82–83 management considerations in, 88–90 mathematical model in spreadsheet
in, 50–51 as mixed problem, 88–90 problem analysis for, 66–71 problem identification in, 65–66 Profit & Gambit Co., 46–51 resource allocation and, 71–72 Solver applied to, 49–50 spreadsheet formulation in, 91–93 spreadsheet model formulation
for, 47–49 Super Grain Corp., 65–71
Aiding Allies (case study), 224–227 Airline scheduling (case study), 229–230 Alden, H., 162 Alden, J. M., 449 Algebraic formulation, in product-mix
problem, 31–32 Algebraic models, 32–33 Algorithms
genetic, 299 for network optimization problems, 194 for Nonlinear Solver, 276 for quadratic programming, 280 special-purpose, 202
Allen, S. J., 136 Allowable range
for objective function coefficient, 158–159, 174
for the right-hand side, 172 sensitivity report to find, 9–161
Altschuler, S., 553 Analytics, 3 Andrews, B. H., 388 Angelis, D. P., 162 Animation, of computer
simulations, 516, 517 Application-oriented simulators, 516 Application vignettes
Bank Hapoalim Group, 284
Canadian Pacific Railway, 210 Compañia Sud Americana de Vapores
(CSAV), 419 ConocoPhillips, 363 Continental Airlines, 252 Deutsche Post DHL, 298 Federal Aviation Administration, 504 Federal Express, 12 function of, 16 Gassco, 206 General Motors Corporation, 449 Hewlett-Packard, 197 KeyCorp, 459 list of, 13, 14 L.L. Bean, 388 Merrill Lynch, 553 Midwest Independent Transmission
Operator, Inc., 241 Netherlands Railways, 247 Pacific Lumber Company, 162 Proctor & Gamble, 97 Samsung Electronics, 50 Sasol, 515 Swift & Company, 24 Taco Bell Corporation, 398 United Airlines, 83 Waste Management, Inc., 234 Welch’s, Inc., 136 Westinghouse Science and
Technology Center, 346 Workers’ Compensation Board of
British Columbia, 331 Arcs, 197, 210 Argüello, M., 252 Arrivals, in queueing system, 436 Assigning Art (case study), 261–263 Assigning Students to Schools (case
study), 119–120, 193, 266 Assignment problems
characteristics of, 100, 102 example of, 99–100 explanation of, 99 fixed requirements and, 88 as minimum-cost flow
problems, 194, 201 spreadsheet model for, 100, 101
Assignments, 99 Auditing tools (Excel), 143 Automobile assembly (case study), 60 Averaging forecasting method
explanation of, 385, 386, 412 formula for, 421 use of, 396–398, 410
Avriel, M., 284
B
Bank Hapoalim Group, 284 Barnum, M. P., 459 Bayes’ decision rule
explanation of, 328–330 views of, 326
Bayes’ theorem, 343 Benefit constraints
advertising-mix problems and, 82 cost–benefit–trade-off problems
and, 82, 86 explanation of, 88
Bennett, J., 553 Berkey, B. G., 346 Bernoulli distribution, 567–568, 587 Bidding for construction project.
See Reliable Construction Co. (case study)
Big M Company example, 95–99 Binary decision variables, 232 Binary integer programming (BIP)
case study of, 233–239 for crew scheduling, 246–250 explanation of, 232–233 mixed, 232, 241, 250–254 for project selection, 239–342 pure, 232, 250 for setup costs for initiating
production, 250–254 for site selection, 241–245
Binary integer programming (BIP) model, 236–237
Binary variables explanation of, 232, 587 formulation with, 240–241, 243–244,
247–248, 251–252 Binomial distribution, 558, 560, 568 BIP. See Binary integer programming (BIP) BIP model, 236–237 BMZ Company (case study), 202–204,
206–207 Bottom-up forecasting approach, 418 Bowen, D. A., 459 Brainy Business (case study), 380–381
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610 Index
Branches, 330 Break-even analysis
complete mathematical model of problem in, 9–10
expressing problem mathematically in, 8 incorporating break-even point into
spreadsheet model in, 10–11 problem analysis and, 8–9 spreadsheet model in, 6–8 what-if analysis of mathematical
model in, 10 Break-even point
explanation of, 9 incorporated into spreadsheet
model, 10–11 Brennan, M., 504 Broadcasting the Olympic Games
(case study), 230–231, 266 Burns, L. D., 449 Business analytics, 3 Byrne, J. E., 83
C
Caliente City problem, 243–245 California Manufacturing Co.
problem, 233–239 Call center case study. See Computer
Club Warehouse (case study) Canadian Pacific Railway (CPR), 210 Capacity, 197 Capacity Concerns (case study), 115–116 Capital budgeting, resource-allocation
problem, 75–80 Carlson, B., 241 Carlson, W., 12 Case, R., 210 Case studies
lists, 14–15 value of, 16
Cash flow management problem. See Everglades Golden Years Company Cash Flow Problem (case study)
Causal forecasting in case study, 417–418 examples of, 413 explanation of, 413 linear regression and, 414–416 in spreadsheets, 414 use of, 413, 416–417
Cells (Excel) entering data, text and formulas into,
601–602 explanation of, 599 filling, 602–603 formulation of, 603–604
moving or copying, 602 range names for, 603 relative and absolute references to, 603 results, 529, 539, 544, 588 selection of, 601 statistic, 529–531, 539, 544 uncertain variable cells, 528, 529,
538, 539, 542–543, 550 Central-tendency distributions, 563, 564 Changing cells, 26 Coefficient in the objective function. See
Objective function coefficient Coin-flipping game, 490–494. See also
Computer simulation Commercial service systems, 440–441 Compañia Sud Americana de Vapores
(CSAV), 419 Computer Club Warehouse (case study),
386–409, 417. See also Forecasting methods
Computer simulation accuracy of, 534 animation for, 516 coin-flipping game as example
of, 490–494 corrective maintenance vs. preventive
maintenance decision as example of, 494–500
explanation of, 489 for financial risk analysis, 552–556 generating random observations
from probability distribution and, 498, 501, 503–504
optimizing with RSPE’s Solver and, 583–590
overview of, 488 performance measures estimation
and, 507 process of, 505–507 role of, 489–490 software selection for, 516 use of, 489–490
Computer simulation (case study) analysis of, 508–514 assumptions in, 511–512 background information for, 501 building blocks of simulation model
for stochastic system in, 504–505 data collection for, 503 decision factors in, 501–502 financial factors in, 508 generating random observations
from probability distributions in, 503–504
performance measurement estimation in, 507
process used in, 505–508 simulation model validity
in, 510–512 Computer simulation models
discrete-event, 516 explanation of, 504 for stochastic system, 504–505 testing accuracy of, 516 testing validity of, 510–512, 516–517
Computer simulation process data collection simulation model
formulation as step in, 515 plan simulations to be performed
as step in, 517 problem formulation and study
plan as step in, 515 recommendations presentation
following, 517 simulation model accuracy check as
step in, 516 simulation model validity test as step
in, 516–517 simulation runs and results analysis
as step in, 517 software selection and computer pro-
gram construction as step in, 516 Computer simulation with RSPE
applications for, 535 bidding for construction project
using, 536–540 case study of, 526–536 cash flow management using,
546–551 choosing correct distribution for,
562–575 decision making with parameter
analysis reports and trend charts using, 575–583
financial risk analysis using, 552–556
optimizing with, 583–590 overview of, 525 project management using, 540–546 revenue management using, 557–562
Conditional probability, 341 Confidence intervals, 509, 517, 534 ConocoPhillips, 363 Conservation of flow, 197 Constant service times, 439 Constraint boundary equation, 35 Constraint boundary line, 35 Constraints
in advertising-mix problems, 68 benefit, 82, 86, 88 cost–benefit–trade-off problems and,
82, 86
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Index 611
Discrete distribution, 562–563 Discrete-event simulation modeling, 516 Distributions. See also Probability
distributions Bernoulli, 567–568, 587 binomial, 558, 560, 568 central tendency, 563, 564 characteristics of available, 562–563 continuous, 498, 501, 562, 572–575 custom, 570–572 discrete, 562–563 exponential, 436–437, 449, 450,
454, 565 frequency, 552 geometric, 568–570 integer uniform, 528–529, 565 lognormal, 564, 565 negative binomial, 568–570 normal, 501, 553, 557, 558, 563, 564,
586, 587 Poisson, 566, 567 time-series forecasting methods
and, 410–411 triangular, 537, 544, 563–565 uniform, 553, 565, 566
Distributions menu, 528 Distribution Unlimited Co. problem,
195–196. See also Minimum-cost flow problems
D/M/s model, 439 Downs, B., 24 Dummy destination, 216 Dupit Corp. Problem (case study),
445–448, 467–468 Dyer, J. S., 363 Dynamic problems, 126
E
Economic analysis, of number of servers to provide, 473–476
Eidesen, B. H., 206 Epstein, R., 419 Equivalent lottery method, 356, 359 Estimated trend, 404 Estimated trend formula, 421 Etzenhouser, M. J., 162 Event nodes, 330, 345 Event-scheduling approach, 516 Everglades Golden Years Company Cash
Flow Problem (case study) background of, 125–126, 547 computer simulation and, 546–551 debugging and, 141–144 modeling with spreadsheets and,
126–135, 140–141
function of, 323 information needs and, 338–340 overview of, 322–323 probability updates and, 340–344 for problems with sequence of
decisions, 344–354 sensitivity analysis and, 333–338,
351–354 utilities to reflect value of payoffs
and, 354–365 Decision conferencing, 365–366 Decision criteria
background of, 325–326 Bayes’ decision rule and, 328–330 maximax, 326 maximin, 326–327 maximum likelihood, 327
Decision making with parameter analysis reports and
trend charts, 575–583 with probabilities, 328–330 risk tolerance and, 362 role of management science in, 3 without probabilities, 326–327
Decision nodes, 330, 345 Decisions
analyzing problems with sequence of, 344–354
contingent, 236 interrelationships between, 235–236 role of management science in, 3
Decision-support system, 5 Decision trees
to analyze problem with sequence of decisions, 344–350
to analyze utility problems, 360–362 construction of, 345–346 explanation of, 330, 345 RSPE, 350, 601 sensitivity analysis with, 333–338 spreadsheet software for, 330–333
Decision variables, 8, 32 Decreasing marginal returns
explanation of, 270–272 nonlinear programming with, 277–286 separable programming for, 287
Decreasing marginal utility for money, 355 Delphi method, 418 Demand nodes, 197, 200 Dependent variables, 413 Descriptive analytics, 3 Destination
dummy, 216 explanation of, 212, 213
Deutsche Post DHL, 298 Discontinuities, 271
effect of simultaneous changes in, 175–178
effect of single changes in, 169–174 explanation of, 9–10, 30 fixed-requirement, 88, 90, 95, 99 functional, 32, 88, 169 nonnegativity, 32, 99 resource, 80, 88
Consumer market survey, 418 Continental Airlines, 252 Contingent decisions, 236. See also
Yes-or-no decisions Continuation of the Super Grain
(case study), 317–318 Continuous distribution, 498, 501, 562,
572–575 Controlling Air Pollution (case study),
189–191 Copeland, D., 12 Corrective maintenance vs. preventive
maintenance example, 494–498 Cost–benefit–trade-off problems
advertising mix and, 82–83 explanation of, 81–82 formulation procedure for, 87 nonlinear version of, 283 personnel scheduling and, 83–87
Cost graphs, use of, 271–272 Costy, T., 449 Crew scheduling problem, binary integer
programming for, 246–250 Cunningham, S. M., 388 Curve fitting method (Excel), 273 Custom distributions, 570–572 Customers. See also Queueing models
in queueing system, 434, 450 in system, 450
Cutting Cafeteria Costs (case study), 61
D
Data for computer simulation
studies, 503, 515 consolidation of, 333–334 for cost–benefit–trade-off problems, 82 organizing and identifying, 136–137 separated from formulas, 138 for spreadsheet models, 135–138
Data cells, 26, 137–138 Data tables, 335–338, 353–354 Decision analysis
applications for, 365–366 case study of, 323–325 decision criteria and, 325–330 decision trees for, 330–338, 344–350
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612 Index
nonlinear, 267, 268, 273–294 separating data from, 138 in spreadsheets, 267, 268 use of relative and absolute
references for related, 139 Freddie the Newsboy’s Problem
(case study) computer simulation and Solver
applied to, 584–586 conclusions regarding, 534–535 description of, 526 parameter analysis report
for, 576–580 RSPE applied to, 527–534 simulation results accuracy in, 534 spreadsheet model for, 526–527
Frequency distribution, 552 Freundt, T., 298 Frontline Systems, 16, 42 Functional constraints, 32, 88 Fundamental property of utility
functions, 356
G
Gassco, 206 General Motors Corporation (GM), 449 General-purpose simulation language, 516 Generation, 299 Genetic algorithms, 299 Geometric distribution, 568–570 Giehl, W., 298 GI/M/s model, 439 Global maximum, 275–277 Goferbroke Company (case study),
323–325 Graphical Linear Programming and
Sensitivity Analysis, 160 Graphical method
explanation of, 33 to solve two-variable problems, 33–37 summary of, 37 uses for, 37, 38
Grass-roots forecasting approach, 418 Group decision support
systems, 365–366
H
Hellemo, L., 206 Herr Cutter’s Barber Shop (case study). See
Computer simulation (case study) Hewlett-Packard (HP), 197 Holloran, T. J., 83 Holman, S. P., 162 Howard, K., 504
Fallis, J., 210 Farm Management (case study), 191–193 Fattedad, S. O., 331 Feasible solution
explanation of, 32 for linear programming
problems, 33, 35 Feasible solutions property, 198 Federal Aviation Administration
(FAA), 504 Federal Express, 12 Finagling the Forecasts
(case study), 429–432 Finance case studies, 15 Financial risk analysis
explanation of, 552 spreadsheet model to apply computer
simulation to, 553–556 Think-Big Development Co.
case, 552–553 Finite queue, 437 Fioole, P.-J., 247 Fischer, M., 298 Fischetti, M., 247 Fixed-requirement constraints
advertising-mix problems and, 90 explanation of, 88 transportation problems and, 95, 99
Fixed-requirements problems, 88 Fletcher, L. R., 162 Fodstad, B. H., 206 Forecasting error, 389 Forecasting errors, 386 Forecasting methods
applications for, 384 averaging, 385, 412 case study of, 386–409 causal, 413–417 comparison of, 411–412 exponential smoothing, 385, 412 exponential smoothing with trend,
385, 412 goal of, 410 judgmental, 384, 418 last-value, 385, 412 linear regression, 385–386 moving-average, 385, 412 overview of, 385–386 recommendations for, 412–413 statistical, 384, 418 time-series, 391–412
Formulas absolute or relative reference, 28 forecasting, 421 for M/G/1 model, 455 for M/M/1 model, 450–451
Evolutionary Solver advantages and disadvantages
of, 305–306 applied to traveling salesman
problem, 302–305 explanation of, 268, 299 for portfolio selection, 300–302
Excel (Microsoft) auditing tools on, 143 cells, 601–604 curve fitting method on, 273 Evolutionary Solver, 298 generating random numbers with,
491–493 guidelines for use of, 17 LN() function on, 504 for minimum-cost flow problems,
198–199 Nonlinear Solver, 268, 274–276,
297, 298 RAND( ) function on, 490, 495 ROUND function, 558 SUMPRODUCT function, 30, 31,
70, 100, 138, 267, 273 toggle function on, 141, 142 VLOOKUP function on, 495–496 window, 599 workbooks, 599–600 worksheets, 599–601
Expected monetary value (EMV) criterion, 329
Expected number of customers, 450 Expected payoff (EP)
Bayes’ decision rule and, 354, 355 explanation of, 328, 347–348
Expected utilities, 360 Expected value
of perfect information, 337–339 performance measures and, 442–443 of sample information, 348–350
Exponential distribution explanation of, 436, 565 for interarrival times, 436–437, 449,
450, 454 Exponential smoothing forecasting method
explanation of, 385, 400, 412 formula for, 421 with trend, 385, 402–408, 412, 421 use of, 400–402, 404
Exponential utility function explanation of, 362 used with RSPE, 362–365
F
Fabrics and Fall Fashions (case study), 116–118
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Index 613
How-much decisions, 232 Hueter, J., 398 Huisman, D., 247 Hutton, R. D., 449
I
Identifying features in cost–benefit–trade-off problems, 87 explanation of, 64
Increasing marginal utility for money, 355
Independent variables, 413 Infeasible solution, 32 Infinite queue, 437, 438 Influence diagram, 365 Information technology (IT), 4 Institute for Operational Research and
the Management Sciences (INFORMS), 12, 13
Integer programming problems, 75 Integer programming model, 30 Integer solutions property, 198 Integer uniform distribution, 528–529, 565 Interactive Management Science
Modules, 17, 37, 160, 409 Interarrival times
in computer simulation model, 503, 513 exponential distribution for, 436–437,
449, 450, 454 in preemptive priorities queueing
model, 464 symbols for, 439
Interfaces, 3, 12, 13 Internal service systems, 441 International Federation of Operational
Research Societies (IFORS), 12 International Investments
(case study), 319–321 Inverse transformation method, 498 Ireland, P., 210
J
Jackson, C. A., 449 Joint probabilities, 341 Judgmental forecasting methods, 384, 418 Jury of executive opinion, 418
K
Kang, J., 50 Keeping Time (case study), 20–21 KeyCorp, 459 Kim, B.-I., 234 Kim, D. S., 449 Kim, S., 234
King, P. V., 346 Kohls, K. A., 449 Kotha, S. K., 459 Krass, B., 234 Kroon, L., 247 Kuehn, J., 210
L
L.L. Bean, Inc., 388 Labe, R., 553 Lack-of-memory property, 437 Last-value forecasting method
explanation of, 385, 394–396, 412 formula for, 421
Latest trend formula, 421 Leachman, R. C., 50 Lehky, M., 504 Length, 212 Liao, B., 553 Lin, Y., 50 Linear formulas, 267, 268 Linear models, 267 Linear programming
applications for, 46–47, 51–52, 64, 84
assumptions of, 70 function of, 150 nonlinear vs., 269 overview of, 22 proportionality assumption
of, 30, 269–270 Linear programming models
analysis of, 102–103 characteristics on spreadsheets, 29–30 development of, 25–31 enrichment of, 103 formulation of, 26–31, 102 for product-mix problem, 25–31 separable programing model
as, 289 terminology for, 32 for transportation
problems, 96–97 validation of, 102
Linear regression application of, 417 explanation of, 414–416 forecasting with, 385–386
Linear regression line, 414, 421 Linear Solver, 306–307 Links, 210 Little, John D. C., 443 Little’s formula, 443–444, 450 Littletown Fire Department
Problem, 209–212
Local maximum, 275 Lognormal distribution, 564, 565
M
MAD. See Mean absolute deviation (MAD) Maintenance problem. See Computer
simulation Management information systems, 5 Management science
applications for, 12–14 break-even analysis to
illustration, 6–11 as decision making aid, 3 as discipline, 2–3 explanation of, 2 impact of, 12–17 overview of, 1–2 professional journals for, 3, 12 quantitative factors considered in, 5 scientific approach of, 3–5
Management Science, 3 Management science teams, 5 Managerial decision making. See
Decision making Manager’s opinion method, 418 Marginal returns, decreasing, 270–287 Marketing case studies, 15 Marketing costs, nonlinear, 292–296 Maróti, G., 247 Mason, R. O., 12 Mathematical models
for advertising-mix problem, 50–51 application of, 5 for break-even analysis, 9–10 function of, 4–5 in spreadsheet, 31–33 tests for, 5 what-if analysis of, 10
Maxima global, 275–277 local, 275
Maximax criterion, 326 Maximin criterion, 326–327 Maximum flow problems
applications for, 207 assumptions of, 205–206 case study of, 202–204, 206–207 as minimum-cost flow problems, 201 multiple supply and demand points
and, 206–207 solving very large, 208–209
Maximum likelihood criterion, 327 McAllister, W. J., 346 McGowan, S. M., 252 McKenney, J. L., 12
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Confi rming Pages
614 Index
difficult problems related to, 297–298 evolutionary solver and genetic
algorithms and, 299–306 linear vs., 269 portfolio selection with, 283–286 quadratic programming as, 280, 285 separable programming and, 287–296 spreadsheet models for, 280–283 use of, 268 use of RSPE for, 306–310
Nonlinear programming models constructing nonlinear formulas
for, 273–274 explanation of, 268 methods to solve, 274–277 nonproportional relationships in,
269–272 Nonlinear Solver
explanation of, 268, 274–275 for nonlinear programming
problems, 297, 298 Nonnegativity constraints, 32, 99 Nonpreemptive priorities, 463 Nonpreemptive priorities
queueing model application of, 464–478 explanation of, 464
Nonproportional relationships explanation of, 270 nonlinear programming and, 269–272
Normal distribution, 501, 553, 557, 558, 564, 586, 587
Number of customers in queue, 443 Number of customers in system, 443
O
Objective cell, 28 Objective function, 10, 32 Objective function coefficients
allowable range for, 158–159 effect of changes in one, 155–161 effect of simultaneous changes
in, 161–168 Objective function line, 36–37 Oh, J., 553 Oiesen, R., 504 100 percent rule
for changes in right-hand side, 178 for simultaneous changes in objective
function coefficient, 168 Operations Research, 3 Operations research (OR), 3. See also
Management science Optimal solutions
explanation of, 32, 35, 150 with graphical method, 33, 37
M/M/s model, 439, 458, 459 Model enrichment, 103 Modeling, use of Excel for, 599–604 Models, 4, 15. See also Mathematical
models Model validation, 102 Money. See Utility function for money Money in Motion (case study), 227–229 Morahan, G. T., 363 Moving-average forecasting method
explanation of, 385, 398, 412 formula for, 421 use of, 398–400
MS Courseware, 16 MSE. See Mean square error (MSE) Multiple-server queueing models
application of, 459–461 assumptions of, 458 explanation of, 438, 457–458, 472
Multiple-server system, 438 Mutation, 299 Mutually exclusive alternatives, 235–236
N
Naive method, 396 Natural logarithms, 504 Negative binomial distribution, 568–570 Netherlands Railways, 247 Network models
for maximum flow problem, 204 for minimum-cost flow problems, 196
Network optimization problems maximum flow problems
as, 202–209 minimum-cost flow problems
as, 195–202 overview of, 194 shortest path problems as, 209–218
Networks, 194, 197 Network simplex method, 200, 202 New Frontiers (case study), 118–119 Newsvendor problem, 526. See also Freddie
the Newsboy’s Problem (case study) Next-event time-advance procedure,
505, 516 Nigam, R., 553 Nodes, 197, 330 Nonlinear formulas
construction of, 273–274 in spreadsheets, 267, 268
Nonlinear marketing costs, 292–296 Nonlinear programming
application of, 269 challenges of, 269–277 with decreasing marginal returns,
277–286
M/D/s model, 439, 462 Mean absolute deviation (MAD)
explanation of, 386, 389–390 exponential smoothing forecasting
and, 402, 407 formula for, 421
Mean arrival rate, 436 Mean service rate, 438, 461, 465 Mean square error (MSE)
explanation of, 386, 390 formula for, 421 minimum value of, 415 time-series forecasting and, 407
Meiri, R., 284 Meketon, M., 210 Merrill Lynch, 553 Method of least squares, 415 Meyer, M., 515 M/G/1 model
applications of, 456–457, 510, 511 assumptions of, 454 explanation of, 439, 510 formulas for, 455 insights provided by, 455–456
Microsoft Excel. See Excel (Microsoft) Midwest Independent Transmission
Operator, Inc., 241 Minimizing total cost, 212–214 Minimizing total time, 214–217 Minimum acceptable level of benefits, 82 Minimum-cost flow problems
applications for, 200–201 assumptions of, 197–198 example of, 195–196 Excel to formulate and solve, 198–199 feasible solutions property and, 198 general characteristics of, 197–198 integer solutions property and, 198 method solve large, 199–200 network simplex method for, 200 types of, 194, 201–202
Mixed BIP application of, 241 explanation of, 232 for setup costs for initiating
production, 250–254 Mixed problems
advertising-mix problem, 88–90 explanation of, 88 spreadsheet formulation for, 91–95
M/M/1 model applications of, 451–454 assumptions of, 450, 464 explanation of, 439, 449 formulas for, 450–451 software for, 451
M/M/2 model, 439
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Index 615
separable programming and, 287–296 use of Solver for, 38–42 what-if analysis and, 151–155
Professional journals, 3, 12 Profit & Gambit Co. advertising-mix
problem, 46–51, 82–83. See also Advertising-mix problems
Profit graphs, use of, 271–272 Programming, 22 Project Pickings (case study), 121–123 Project selection problem, binary integer
programming for, 239–241 Proportionality assumption, of linear
programming, 30 Proportional relationships, 269 Prudent Provisions for Pensions (case
study), 148–149 Pure BIP, 232, 250 Puterman, M. L., 331
Q
Quadratic programming explanation of, 280 for portfolio selection, 285
Queue capacity, 437 Queue discipline, 438 Queueing models
arrivals and, 436 assumptions of, 440 basic queueing system and, 434–435 example of, 435–436 exponential distribution of inter-
arrival times and, 436–437 labels for, 439 multiple-server, 457–461, 472 nonpreemptive priorities, 464–478 overview of, 433–434 preemptive priorities, 463–464 priority, 463–468 single-server, 448–457
Queueing Quandary (case study), 485–486 Queueing simulator, 509, 510, 513 Queueing systems
case study of, 445–448 commercial services, 440–441 design of, 469–473 examples of, 440–442 explanation of, 434–435 internal service, 441 measures of performance
for, 442–445 server decisions for, 473–476 service and, 438 service-time distributions
and, 438–439 transportation service, 441–442
Predictive analytics, 3 Preemptive priorities, 463 Preemptive priorities queueing
model, 463–464 Prescriptive analytics, 3 Preventive maintenance vs. corrective
maintenance example, 494–498 Pricing under Pressure
(case study), 597–598 Priority classes, 463 Priority queueing models
application of, 464–468 explanation of, 463 nonpreemptive, 463–464 preemptive, 463–464
Prior probabilities Bayes’ decision rule and, 328 explanation of, 325, 326, 345
Pri-Zan, H., 284 Probabilities
conditional, 341 decision making with, 327–330 decision making without, 326–327 joint, 341 as measures of performance, 444–445 posterior, 340–345 prior, 325, 326, 328, 345 unconditional, 341
Probability density function, 563 Probability distributions
in coin-flipping game, 492–493 in computer simulation, 498, 501,
503–504, 507 in corrective vs. preventive
maintenance study, 494 problems caused by
shifting, 410–411 random observations from, 498, 501 steady-state, 444–445 triangular, 537, 544
Probability tree diagram, 341, 342 Procter & Gamble (P & G), 97 Production rates, 26, 28–29 Product-mix problems. See also Wyndor
Glass Co. Product Mix Problem (case study)
background of, 23 formulated on spreadsheet, 25–32 graphical method and, 33–37 management discussion and
issues for, 23–24 management science team work
on, 24–25 overtime costs and, 287–295 resource allocation and, 72–73 on Risk Solver Platform for
Education, 42–46
parameter analysis and trend charts and, 575, 583–584
what-if analysis and, 152 Origin, 212, 213 Output cells, 27–28 Overtime costs, 287–295 Owen, J. H., 449
P
Pacfic Lumber Company (PALCO), 162 Parameter analysis reports
decision making with, 575–583 to determine effect of making
changes to parameter in constraint, 171–172, 175–178
sensitivity analysis using, 156–159 two-way, 162–165
Parameter cells defining decision variable as, 576–578 explanation of, 157, 158
Parameters explanation of, 10, 32 of model, 151 sensitivity, 151
Payoff decision criteria and, 328, 347–348 explanation of, 325
Payoff table, 325 Pedersen, B., 206 Perdue, R. K., 346 Peretz, A., 284 Perfect information, expected value
of, 337–339 Performance measures
for advertising-mix problems, 68–69 for computer simulation problem, 507 for cost–benefit–trade-off problem, 86 of probabilities as, 444–445 for queueing systems, 442–445
Personnel scheduling cost–benefit–trade-off analysis
and, 83–87 United Airlines and, 83
PERT three-estimates approach, 541, 542 Piecewise linear, 271 Planning Planers (case study), 523–524 Point estimates, 509, 517 Point of indifference, 358, 359 Poisson distribution, 566, 567 Popov, A., Jr., 234 Population, 299 Portfolio selection
to beat market, 299–301 Evolutionary Solver for, 300–302 nonlinear program for, 283–286
Posterior probabilities, 340–345
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616 Index
with decision trees, 333–338 explanation of, 152 to find allowable range, 159–161 parameter analysis report
for, 156–159 for problems with sequence of
decisions, 351–354 spreadsheets for, 155–156
Sensitivity chart, 545–546 Sensitivity reports
to find allowable range, 159–161 gleaning additional information from,
165–167, 176–178 Separable programming
explanation of, 287 nonlinear marketing costs and, 292–295 overtime costs and, 287–295 smooth profit graphs and, 291–292
Servers, queueing systems and, 434 Service cost, 473 Service times
constant, 439 distributions for, 438–439, 455 exponential distribution for, 450, 464 M/G/1 model and, 454 in queueing systems, 438 symbols for, 439
Set covering constraint, 244 Set covering problem, 244 Setup costs, mixed BIP and, 250–254 Shadow price, 172 Shipping Wood to Market
(case study), 114–115 Shortest path problems
applications for, 212 assumptions of, 212 example of, 209–212 minimizing total cost in, 212–214 minimizing total time in, 214–217
Simplex method, 199 Simulation, 489. See also Computer
simulation Simulation clock, 505 Simulation models, 504. See also
Computer simulation models Single-server queueing models
explanation of, 448–449 M/M/1, 449–454
Single-server system, 438 Sink, 205 Site selection, binary integer
programming for, 241–245 Smart Steering Support
(case study), 382–383 Smoothing constant, 400 Smooth profit graphs, 291–292
Risk profile, 552 Risk seekers, 355 Risk Solver Platform for Education
(RSPE). See also Computer simulation with RSPE; Solver
to analyze decision trees, 330–333, 360–362
to analyze models and choose solving method, 306–310
decision tree tool, 350, 601 Excel worksheets with, 601 explanation of, 16, 42 to generate random values from
probability distributions, 504 optimizing with computer simulation
and, 583–590 parameter analysis report generated
by, 157 product-mix problem on, 42–46 Solver tool and, 583–590 use of, 162–165, 171–172
Rømo, F., 206 RSPE. See Computer simulation with
RSPE; Risk Solver Platform for Education (RSPE)
RSPE Solver, 583–590
S
Sahoo, S., 234 Salesforce composite method, 418 Sample average, 410, 534 Samsung Electronics, 50 Sasol, 515 Savvy Stock Selection
(case study), 318–319 Schrijver, A., 247 Schuster, E. W., 136 Scientific method
explanation of, 3–5 steps in, 4–5
Seasonal factor explanation of, 391–392 formula for, 421
Seasonally adjusted time series explanation of, 392–394 formula for, 421 use of, 413
Self, M., 24 Selling Soap (case study), 188–189 Sellmore Company problem, 99–102 Sensitive parameters, 151 Sensitivity analysis. See also What-if
analysis binary integer programming and, 238 Data Table use for, 353–354
Queueing theory, 433, 448 Queues
explanation of, 433, 434 finite, 437 infinite, 437, 438 nature of, 437–438
R
Random arrivals, 436, 454 Random numbers
explanation of, 490 generation of, 491–492
Random observations, from probability distribution, 498, 501
Random variables, 410, 445 Range names
for Excel cells, 603 explanation of, 6, 26, 138–139
Reclaiming Solid Wastes (case study), 120–121
Reducing In-Process Inventory (case study), 486–487, 524
References, relative and absolute, 28, 139, 603
Regression equation, 415 Relative references, 28, 139, 603 Reliable Construction Co. (case study)
background of, 536–537 computer simulation for, 540, 542–545 parameter analysis report
for, 580–581 project management issues in, 540–542 sensitivity charts and, 545–546 spreadsheet model for applying
computer simulation to, 537–539
Resource-allocation problem advertising mix and, 71–72 capital budgeting as, 75–80 characteristics of, 72 explanation of, 71 formulation procedure for, 80–81 product mix and, 72–73 TBA Airlines problem and, 73–75
Resource constraints, 80, 88 Results cell, 529, 539, 544, 588 Revenue management
explanation of, 557 overbooking problem and, 557–562
Risk, determining attitude toward, 357–359
Risk analysis, 552. See also Financial risk analysis
Risk averse, 355 Risk neutral, 355, 360
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Index 617
Super Grain Corp. advertising-mix (case study), 65–71, 88–93
Supply nodes, 197 Swart, W., 398 Swift & Company, 24
T
Taco Bell Corporation, 398 Tanino, M., 504 Tazer Corp. problem, 239–241, 586–590 Tech reps, 445, 448 Think-Big Development Co., 552–556 Time series
explanation of, 390 stable, 411 unstable, 411–412
Time-series forecasting methods averaging, 396–398 comparison of, 411–413 explanation of, 390, 393–394 exponential smoothing, 400–402 exponential smoothing with trend,
402–408 goal of, 410 last-value, 394–396 moving-average, 398–400 seasonal effects and, 391–392 seasonally adjusted, 392–394 shifting distributions and, 410–411 software for, 409
Toggle function (Excel), 141, 142 Tomasgard, A., 206 Total cost minimization, 212–214 Total profit, 26, 28, 42 Total time minimization, 214–217 Transcontinental Airlines problem
overbooking and, 557–562 parameter analysis report and trend
chart for, 582–583 Transportation problems
explanation of, 95 fixed requirements and, 88 formulated in linear programming
terms, 96–97 as minimum-cost flow
problems, 194, 201 Procter & Gamble and, 97 spreadsheet model of, 97–99
Transportation service systems, 441–442 Transshipment nodes, 197 Transshipment problems, as minimum-
cost flow problems, 201 Traveling salesman problem
applying Evolutionary Solver to, 302–305
explanation of, 302
example of poor formulation of, 140–141
explanation of, 4 guidelines for producing good,
135–141 incorporating break-even point into,
10–11 linear programming, 29–30 for maximum flow problems, 208 M/G/1 model, 456, 471 for minimum-cost flow problems, 199 for mixed problems, 91–95 M/M/1 model, 452 M/M/s model, 460, 461, 470 for nonlinear programming, 280–283 for nonpreemptive priorities, 466 overview of, 124–125 for portfolio selection, 300, 301,
303–305 procedure to debug, 141–144 process for, 126–135 for product-mix problems, 25–31 for sensitivity analysis, 351–352 sensitivity analysis using, 155–156 for shortest path problems,
211, 215, 217 for transportation problems, 97–99 for what-if analysis, 169–171
Spreadsheet software. See also Excel; Solver
applications for, 1–2 for computer simulation studies, 516 for decision trees, 330–333
Stable time series, 411 Staffing a Call Center (case study), 62 State of the system, 505 Statistical forecasting methods. See also
Forecasting methods explanation of, 384 judgmental methods vs., 384, 418
Statistic cell explanation of, 529–530 procedure to define, 530–531 in simulation model, 539, 544
Statistics table, 531–532, 540, 545 Steady-state condition, 443 Steady-state probability
distribution, 444–445 Steenbeck, A., 247 Stochastic system
explanation of, 489 simulation model for, 504–505
Stocking Sets (case study), 263–265 Sud, V. P., 504 SUMPRODUCT function, 30, 31, 70,
100, 138, 267, 273, 290, 328
Solutions, for linear programming models, 32
Solver. See also Excel (Microsoft); Risk Solver Platform for Education (RSPE)
for advertising-mix problem, 49–50 applications for, 38, 42, 306 for assignment problems, 100, 102 for BIP programming, 237 computer simulation and, 583–590 Evolutionary, 268, 299–306 explanation of, 16 gaining access to, 38 guidelines for use of, 139–140 linear, 306–307 nonlinear, 268, 274–275, 297, 298 nonlinear programming models
and, 295, 296 product-mix problem on, 38–42 for transportation problems, 98 worksheets used with, 600–601
Song, C., 252 Source, 205 Southwestern Airways
problem, 246–250 Special-purpose algorithms, 202 Spreadsheet analysis, illustration of, 6–11 Spreadsheet modeling procedure
analysis as element of, 132–135 expanding and testing full-scale
model of, 131–132 explanation of, 126–127 hand calculations as step in, 128–129 sketching out spreadsheet as
element of, 129–130 starting with small version of spread-
sheet as element of, 130, 131 testing small version as element
of, 131 visualizing where you want to
finish as, 127–128 Spreadsheet models
for advertising-mix problems, 47–49, 67, 69
for assignment problems, 100, 101 binary integer programming, 237,
242, 245, 249, 253 for break-even analysis, 6–8 for cash flow problems, 125–135 for computer simulations, 491, 496,
499, 500, 506, 510, 511, 513 for computer simulations with RSPE,
526–527, 535, 538, 543, 548, 549, 554, 559, 573, 587
for cost–benefit–trade-off problems, 85
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Confi rming Pages
618 Index
explanation of, 10, 150 of mathematical models, 10 for product-mix problem, 153–155 simultaneous change in constraints
and, 175–178 simultaneous change in objective func-
tion coefficients and, 161–169 single change in constraints and,
169–174 What-if questions, 150 White, A., 252 Who Wants to Be a Millionaire? (case
study), 379 Workbook (Excel), 599–600 Workers’ Compensation Board (WCB)
of British Columbia, 331 Worksheets (Excel)
explanation of, 599 use of, 600–601
Wyndor Glass Co. Product Mix Problem (case study), 23–46, 72–73, 136, 137, 151–156, 162–165, 250–254, 278–283, 287–296. See also Product-mix problems
Y
Ybema, R., 247 Yes-or-no decisions
binary decision variables for, 235 as contingent decisions, 236 explanation of, 232 for site selection, 242
Young, E. E., 331 Yu, G., 252
Utility functions decision trees and, 360–362 explanation of, 357 exponential, 362–365 fundamental property of, 356
Utilization factor multiple-server queueing models
and, 457 for preemptive priorities queueing
model, 464 single-server queueing models
and, 448–449
V
Validity, of simulation models, 516–517 Vander Veen, D. J., 449 Van Dyke, C., 210
W
Waiting cost, 473, 474 Waiting time in queue, 443, 447, 451 Waiting time in system, 443, 450 Walls, M. R., 363 Ward, J., 197 Waste Management, Inc., 234 Welch’s, Inc., 136 Westinghouse Science and Technology
Center, 346 Wetherly, J., 504 What-if analysis. See also Sensitivity
analysis benefits of, 151–152 change in one objective function
coefficient and, 155–161
Trend estimated, 404 explanation of, 402 exponential smoothing
with, 402–408 formulas for, 421
Trend charts application of, 579–583 explanation of, 575, 578
Trend line, 402, 404 Trend smoothing constant, 405 Trials, 526, 560 Triangular distribution, 537, 544,
563–565 Turnquist, M. A., 449 25 percent rule, 388 Two-way parameter analysis
report, 162–165
U
Uncertain variable cells, 528, 529, 538, 539, 542–543, 550
Unconditional probabilities, 341 Undirected arc, 210n Uniform distribution, 553, 565, 566 United Airlines, 83 University Toys and the Business
Professor Action Figures (case study), 379–380
Unstable time series, 411–412 Urbanovich, E., 331 Utility function for money
construction of, 359–360 explanation of, 355–357
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- Cover
- Title
- Copyright
- Contents
- Chapter 1 Introduction
- 1.1 The Nature of Management Science
- 1.2 An Illustration of the Management Science Approach: Break-Even Analysis
- 1.3 The Impact of Management Science
- 1.4 Some Special Features of This Book
- 1.5 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problem
- Problems
- Case 1-1 Keeping Time
- Chapter 2 Linear Programming: Basic Concepts
- 2.1 A Case Study: The Wyndor Glass Co. Product- Mix Problem
- 2.2 Formulating the Wyndor Problem on a Spreadsheet
- 2.3 The Mathematical Model in the Spreadsheet
- 2.4 The Graphical Method for Solving Two-Variable Problems
- 2.5 Using Excel's Solver to Solve Linear Programming Problems
- 2.6 Risk Solver Platform for Education (RSPE)
- 2.7 A Minimization Example—The Profit & Gambit Co. Advertising-Mix Problem
- 2.8 Linear Programming from a Broader Perspective
- 2.9 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problems
- Problems
- Case 2-1 Auto Assembly
- Case 2-2 Cutting Cafeteria Costs
- Case 2-3 Staffing a Call Center
- Chapter 3 Linear Programming: Formulation and Applications
- 3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem
- 3.2 Resource-Allocation Problems
- 3.3 Cost–Benefit–Trade-Off Problems
- 3.4 Mixed Problems
- 3.5 Transportation Problems
- 3.6 Assignment Problems
- 3.7 Model Formulation from a Broader Perspective
- 3.8 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problems
- Problems
- Case 3-1 Shipping Wood to Market
- Case 3-2 Capacity Concerns
- Case 3-3 Fabrics and Fall Fashions
- Case 3-4 New Frontiers
- Case 3-5 Assigning Students to Schools
- Case 3-6 Reclaiming Solid Wastes
- Case 3-7 Project Pickings
- Chapter 4 The Art of Modeling with Spreadsheets
- 4.1 A Case Study: The Everglade Golden Years Company Cash Flow Problem
- 4.2 Overview of the Process of Modeling with Spreadsheets
- 4.3 Some Guidelines for Building "Good" Spreadsheet Models
- 4.4 Debugging a Spreadsheet Model
- 4.5 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problems
- Problems
- Case 4-1 Prudent Provisions for Pensions
- Chapter 5 What-If Analysis for Linear Programming
- 5.1 The Importance of What-If Analysis to Managers
- 5.2 Continuing the Wyndor Case Study
- 5.3 The Effect of Changes in One Objective Function Coefficient
- 5.4 The Effect of Simultaneous Changes in Objective Function Coefficients
- 5.5 The Effect of Single Changes in a Constraint
- 5.6 The Effect of Simultaneous Changes in the Constraints
- 5.7 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problem
- Problems
- Case 5-1 Selling Soap
- Case 5-2 Controlling Air Pollution
- Case 5-3 Farm Management
- Case 5-4 Assigning Students to Schools (Revisited)
- Chapter 6 Network Optimization Problems
- 6.1 Minimum-Cost Flow Problems
- 6.2 A Case Study: The BMZ Co. Maximum Flow Problem
- 6.3 Maximum Flow Problems
- 6.4 Shortest Path Problems
- 6.5 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problems
- Problems
- Case 6-1 Aiding Allies
- Case 6-2 Money in Motion
- Case 6-3 Airline Scheduling
- Case 6-4 Broadcasting the Olympic Games
- Chapter 7 Using Binary Integer Programming to Deal with Yes-or-No Decisions
- 7.1 A Case Study: The California Manufacturing Co. Problem
- 7.2 Using BIP for Project Selection: The Tazer Corp. Problem
- 7.3 Using BIP for the Selection of Sites for EmergencyServices Facilities: The Caliente City Problem
- 7.4 Using BIP for Crew Scheduling: The Southwestern Airways Problem
- 7.5 Using Mixed BIP to Deal with Setup Costs forInitiating Production: The Revised Wyndor Problem
- 7.6 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problems
- Problems
- Case 7-1 Assigning Art
- Case 7-2 Stocking Sets
- Case 7-3 Assigning Students to Schools (Revisited)
- Case 7-4 Broadcasting the Olympic Games (Revisited)
- Chapter 8 Nonlinear Programming
- 8.1 The Challenges of Nonlinear Programming
- 8.2 Nonlinear Programming with Decreasing Marginal Returns
- 8.3 Separable Programming
- 8.4 Difficult Nonlinear Programming Problems
- 8.5 Evolutionary Solver and Genetic Algorithms
- 8.6 Using RSPE to Analyze a Model and Choose a Solving Method
- 8.7 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problem
- Problems
- Case 8-1 Continuation of the Super Grain Case Study
- Case 8-2 Savvy Stock Selection
- Case 8-3 International Investments
- Chapter 9 Decision Analysis
- 9.1 A Case Study: The Goferbroke Company Problem
- 9.2 Decision Criteria
- 9.3 Decision Trees
- 9.4 Sensitivity Analysis with Decision Trees
- 9.5 Checking Whether to Obtain More Information
- 9.6 Using New Information to Update the Probabilities
- 9.7 Using a Decision Tree to Analyze the Problem with a Sequence of Decisions
- 9.8 Performing Sensitivity Analysis on the Problem with a Sequence of Decisions
- 9.9 Using Utilities to Better Reflect the Values of Payoffs
- 9.10 The Practical Application of Decision Analysis
- 9.11 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problems
- Problems
- Case 9-1 Who Wants to Be a Millionaire?
- Case 9-2 University Toys and the Business Professor Action Figures
- Case 9-3 Brainy Business
- Case 9-4 Smart Steering Support
- Chapter 10 Forecasting
- 10.1 An Overview of Forecasting Techniques
- 10.2 A Case Study: The Computer Club Warehouse (CCW) Problem
- 10.3 Applying Time-Series Forecasting Methods to the Case Study
- 10.4 The Time-Series Forecasting Methods in Perspective
- 10.5 Causal Forecasting with Linear Regression
- 10.6 Judgmental Forecasting Methods
- 10.7 Summary
- Glossary
- Summary of Key Formulas
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problem
- Problems
- Case 10-1 Finagling the Forecasts
- Chapter 11 Queueing Models
- 11.1 Elements of a Queueing Model
- 11.2 Some Examples of Queueing Systems
- 11.3 Measures of Performance for Queueing Systems
- 11.4 A Case Study: The Dupit Corp. Problem
- 11.5 Some Single-Server Queueing Models
- 11.6 Some Multiple-Server Queueing Models
- 11.7 Priority Queueing Models
- 11.8 Some Insights about Designing Queueing Systems
- 11.9 Economic Analysis of the Number of Servers to Provide
- 11.10 Summary
- Glossary
- Key Symbols
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problem
- Problems
- Case 11-1 Queueing Quandary
- Case 11-2 Reducing In-Process Inventory
- Chapter 12 Computer Simulation: Basic Concepts
- 12.1 The Essence of Computer Simulation
- 12.2 A Case Study: Herr Cutter's Barber Shop (Revisited)
- 12.3 Analysis of the Case Study
- 12.4 Outline of a Major Computer Simulation Study
- 12.5 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problem
- Problems
- Case 12-1 Planning Planers
- Case 12-2 Reducing In-Process Inventory (Revisited)
- Chapter 13 Computer Simulation with Risk Solver Platform
- 13.1 A Case Study: Freddie the Newsboy's Problem
- 13.2 Bidding for a Construction Project: A Prelude to the Reliable Construction Co. Case Study
- 13.3 Project Management: Revisiting the Reliable Construction Co. Case Study
- 13.4 Cash Flow Management: Revisiting the Everglade Golden Years Company Case Study
- 13.5 Financial Risk Analysis: Revisiting the Think- Big Development Co. Problem
- 13.6 Revenue Management in the Travel Industry
- 13.7 Choosing the Right Distribution
- 13.8 Decision Making with Parameter Analysis Reports and Trend Charts
- 13.9 Optimizing with Computer Simulation Using RSPE's Solver
- 13.10 Summary
- Glossary
- Learning Aids for This Chapter in Your MS Courseware
- Solved Problem
- Problems
- Case 13-1 Action Adventures
- Case 13-2 Pricing under Pressure
- Appendix A: Tips for Using Microsoft Excel for Modeling
- Appendix B: Partial Answers to Selected Problems
- Index
- A
- B
- C
- D
- E
- F
- G
- H
- I
- J
- K
- L
- M
- N
- O
- P
- Q
- R
- S
- T
- U
- V
- W
- Y
