Regression Models ( 6 problems )

Quantitative Analysis for Management twelfth edition

Barry Render • Ralph M. Stair, Jr. • Michael E. Hanna • Trevor S. Hale

This is a special edition of an established title widely used by colleges and universities throughout the world. Pearson published this exclusive edition for the benefit of students outside the United States and Canada. If you purchased this book within the United States or Canada you should be aware that it has been imported without the approval of the Publisher or Author.

Pearson Global Edition

Q uantitative A

nalysis for M anagem

ent R

ender • Stair • H anna • H

ale t

w elft

h

ed it

io n

GlobAl edition

GlobAl edition

For these Global Editions, the editorial team at Pearson has collaborated with educators across the world to address a wide range of subjects and requirements, equipping students with the best possible learning tools. This Global Edition preserves the cutting-edge approach and pedagogy of the original, but also features alterations, customization, and adaptation from the North American version.

G lo

b A

l ed

it io

n

RENDER_129205932X_mech.indd 1 03/02/14 5:01 PM

Quantitative Analysis for Management

Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto

Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

TwelfTh ediTion Global ediTion

Barry render

Charles Harwood Professor of Management Science Crummer Graduate School of Business, Rollins College

ralph M. Stair, Jr.

Professor of Information and Management Sciences, Florida State University

Michael e. hanna

Professor of Decision Sciences, University of Houston–Clear Lake

trevor S. hale

Associate Professor of Management Sciences, University of Houston–Downtown

A01_REND9327_12_SE_FM.indd 1 11/02/14 8:19 PM

Editor in Chief: Donna Battista Head of Learning Asset Acquisition, Global Edition:

Laura Dent Acquisitions Editor: Katie Rowland Senior Acquisitions Editor, Global Edition: Steven Jackson Editorial Project Manager: Mary Kate Murray Editorial Assistant: Elissa Senra-Sargent Director of Marketing: Maggie Moylan Senior Marketing Manager: Anne Fahlgren Marketing Assistant: Gianna Sandri Managing Editor: Jeff Holcomb

Senior Production Project Manager: Kathryn Dinovo Manufacturing Buyer/Procurement Specialist:

Carol Melville Associate Project Editor, Global Edition: Uttaran Das Gupta Senior Manufacturing Controller, Production,

Global Edition: Trudy Kimber Art Director, Cover: Blair Brown Cover Designer: PreMediaGlobal Cover Image: © Evlakhov Valeriy/Shutterstock Media Project Manager: Lisa Rinaldi Full-Service Project Management: PreMediaGlobal

Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook appear on the appropriate page within text.

Pearson Education Limited Edinburgh Gate Harlow Essex CM 20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsonglobaleditions.com © Pearson Education Limited 2015

The rights of Barry Render, Ralph M. Stair, Jr., Michael E. Hanna, and Trevor S. Hale to be identified as the authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988.

Authorized adaptation from the United States edition, entitled Quantitative Analysis for Management, 12th edition, ISBN 978-0-13-350733-1, by Barry Render, Ralph M. Stair, Jr., Michael E. Hanna, and Trevor S. Hale, published by Pearson Education © 2015.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a license permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saf- fron House, 6–10 Kirby Street, London EC1N 8TS.

All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affilia- tion with or endorsement of this book by such owners.

Microsoft and/or its respective suppliers make no representations about the suitability of the information contained in the documents and related graphics published as part of the services for any purpose. All such documents and related graphics are provided “as is” without warranty of any kind. Microsoft and/or its respective suppliers hereby disclaim all warranties and conditions with regard to this information, including all warranties and conditions of merchantability, whether express, implied or statutory, fitness for a particular purpose, title and non-infringement. In no event shall Microsoft and/or its respective sup- pliers be liable for any special, indirect or consequential damages or any damages whatsoever resulting from loss of use, data or profits, whether in an action of contract, negligence or other tortious action, arising out of or in connection with the use or performance of information available from the services.

The documents and related graphics contained herein could include technical inaccuracies or typographical errors. Changes are periodically added to the information herein. Microsoft and/or its respective suppliers may make improvements and/or changes in the product(s) and/or the program(s) described herein at any time. Partial screen shots may be viewed in full within the software version specified.

Microsoft® and Windows® are registered trademarks of the Microsoft Corporation in the U.S.A. and other countries. This book is not sponsored or endorsed by or affiliated with the Microsoft Corporation.

ISBN 10: 1-292-05932-X ISBN 13: 978-1-292-05932-7

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library

10 9 8 7 6 5 4 3 2 1 14 13 12 11 10

Typeset in 10/12 Times Roman by PreMediaGlobal. Printed and bound by Courier Kendallville in The United States of America.

To my wife and sons—BR

To Lila and Leslie—RMS

To Zoe and Gigi—MEH

To Valerie and Lauren—TSH

A01_REND9327_12_SE_FM.indd 2 11/02/14 8:19 PM

3

About the Authors

Barry Render is Professor Emeritus, the Charles Harwood Distinguished Professor of Operations Management, Crummer Graduate School of Business, Rollins College, Winter Park, Florida. He received his B.S. in Mathematics and Physics at Roosevelt University and his M.S. in Operations Research and his Ph.D. in Quantitative Analysis at the University of Cincinnati. He previously taught at George Washington University, the University of New Orleans, Boston University, and George Mason University, where he held the Mason Foundation Professorship in Decision Sciences and was Chair of the Decision Science Department. Dr. Render has also worked in the aerospace industry for General Electric, McDonnell Douglas, and NASA.

Dr. Render has coauthored 10 textbooks published by Pearson, including Managerial Decision Modeling with Spreadsheets, Operations Management, Principles of Operations Management, Service Management, Introduction to Management Science, and Cases and Readings in Management Science. More than 100 articles of Dr. Render on a variety of management topics have appeared in Decision Sciences, Production and Operations Management, Interfaces, Information and Management, Journal of Management Information Systems, Socio-Economic Planning Sciences, IIE Solutions, and Operations Management Review, among others.

Dr. Render has been honored as an AACSB Fellow and was named twice as a Senior Fulbright Scholar. He was Vice President of the Decision Science Institute Southeast Region and served as software review editor for Decision Line for six years and as Editor of the New York Times Operations Management special issues for five years. From 1984 to 1993, Dr. Render was President of Management Service Associates of Virginia, Inc., whose technology clients included the FBI, the U.S. Navy, Fairfax County, Virginia, and C&P Telephone. He is currently Consulting Editor to Financial Times Press.

Dr. Render has taught operations management courses at Rollins College for MBA and Executive MBA programs. He has received that school’s Welsh Award as leading professor and was selected by Roosevelt University as the 1996 recipient of the St. Claire Drake Award for Outstanding Scholarship. In 2005, Dr. Render received the Rollins College MBA Student Award for Best Overall Course, and in 2009 was named Professor of the Year by full-time MBA students.

Ralph Stair is Professor Emeritus at Florida State University. He earned a B.S. in chemical engi- neering from Purdue University and an M.B.A. from Tulane University. Under the guidance of Ken Ramsing and Alan Eliason, he received a Ph.D. in operations management from the University of Oregon. He has taught at the University of Oregon, the University of Washington, the University of New Orleans, and Florida State University.

He has taught twice in Florida State University’s Study Abroad Program in London. Over the years, his teaching has been concentrated in the areas of information systems, operations research, and operations management.

Dr. Stair is a member of several academic organizations, including the Decision Sciences Institute and INFORMS, and he regularly participates in national meetings. He has published numer- ous articles and books, including Managerial Decision Modeling with Spreadsheets, Introduction to Management Science, Cases and Readings in Management Science, Production and Operations Management: A Self-Correction Approach, Fundamentals of Information Systems, Principles of Information Systems, Introduction to Information Systems, Computers in Today’s World, Principles

A01_REND9327_12_SE_FM.indd 3 11/02/14 8:19 PM

4  About the Authors

of Data Processing, Learning to Live with Computers, Programming in BASIC, Essentials of BASIC Programming, Essentials of FORTRAN Programming, and Essentials of COBOL Programming. Dr. Stair divides his time between Florida and Colorado. He enjoys skiing, biking, kayaking, and other outdoor activities.

Michael E. Hanna is Professor of Decision Sciences at the University of Houston–Clear Lake (UHCL). He holds a B.A. in Economics, an M.S. in Mathematics, and a Ph.D. in Operations Research from Texas Tech University. For more than 25 years, he has been teaching courses in statistics, man- agement science, forecasting, and other quantitative methods. His dedication to teaching has been recognized with the Beta Alpha Psi teaching award in 1995 and the Outstanding Educator Award in 2006 from the Southwest Decision Sciences Institute (SWDSI).

Dr. Hanna has authored textbooks in management science and quantitative methods, has pub- lished numerous articles and professional papers, and has served on the Editorial Advisory Board of Computers and Operations Research. In 1996, the UHCL Chapter of Beta Gamma Sigma presented him with the Outstanding Scholar Award.

Dr. Hanna is very active in the Decision Sciences Institute, having served on the Innovative Education Committee, the Regional Advisory Committee, and the Nominating Committee. He has served on the board of directors of the Decision Sciences Institute (DSI) for two terms and also as regionally elected vice president of DSI. For SWDSI, he has held several positions, including president, and he received the SWDSI Distinguished Service Award in 1997. For overall service to the profession and to the university, he received the UHCL President’s Distinguished Service Award in 2001.

Trevor S. Hale is Associate Professor of Management Science at the University of Houston– Downtown (UHD). He received a B.S. in Industrial Engineering from Penn State University, an M.S. in Engineering Management from Northeastern University, and a Ph.D. in Operations Research from Texas A&M University. He was previously on the faculty of both Ohio University–Athens, and Colorado State University–Pueblo.

Dr. Hale was honored three times as an Office of Naval Research Senior Faculty Fellow. He spent the summers of 2009, 2011, and 2013 performing energy security/cyber security research for the U.S. Navy at Naval Base Ventura County in Port Hueneme, California.

Dr. Hale has published dozens of articles in the areas of operations research and quantitative analysis in journals such as the International Journal of Production Research, the European Journal of Operational Research, Annals of Operations Research, the Journal of the Operational Research Society, and the International Journal of Physical Distribution and Logistics Management among several others. He teaches quantitative analysis courses in the University of Houston–Downtown MBA program and Masters of Security Management for Executives program. He is a senior mem- ber of both the Decision Sciences Institute and INFORMS.

A01_REND9327_12_SE_FM.indd 4 11/02/14 8:19 PM

5

Chapter 1 Introduction to Quantitative Analysis 19

Chapter 2 Probability Concepts and Applications 41

Chapter 3 Decision Analysis 83

Chapter 4 Regression Models 131

Chapter 5 Forecasting 167

Chapter 6 Inventory Control Models 205

Chapter 7 Linear Programming Models: Graphical and Computer Methods 257

Chapter 8 Linear Programming Applications 309

Chapter 9 Transportation, Assignment, and Network Models 341

Chapter 10 Integer Programming, Goal Programming, and Nonlinear Programming 381

Chapter 11 Project Management 413

Chapter 12 Waiting Lines and Queuing Theory Models 453

Chapter 13 Simulation Modeling 487

Chapter 14 Markov Analysis 527

Chapter 15 Statistical Quality Control 555

Appendices 575

Online MOdules

1 Analytic Hierarchy Process M1-1

2 Dynamic Programming M2-1

3 Decision Theory and the Normal Distribution M3-1

4 Game Theory M4-1

5 Mathematical Tools: Determinants and Matrices M5-1

6 Calculus-Based Optimization M6-1

7 Linear Programming: The Simplex Method M7-1

8 Transportation, Assignment, and Network Algorithms M8-1

brief Contents

A01_REND9327_12_SE_FM.indd 5 11/02/14 8:19 PM

6

Contents

Preface 13

chaPter 1 Introduction to Quantitative Analysis 19

1.1 Introduction 20 1.2 What Is Quantitative Analysis? 20 1.3 Business Analytics 21 1.4 The Quantitative Analysis Approach 22

Defining the Problem 22 Developing a Model 22 Acquiring Input Data 23 Developing a Solution 23 Testing the Solution 24 Analyzing the Results and Sensitivity Analysis 24 Implementing the Results 24 The Quantitative Analysis Approach

and Modeling in the Real World 26 1.5 How to Develop a Quantitative Analysis

Model 26 The Advantages of Mathematical Modeling 27 Mathematical Models Categorized by Risk 27

1.6 The Role of Computers and Spreadsheet Models in the Quantitative Analysis Approach 28

1.7 Possible Problems in the Quantitative Analysis Approach 31 Defining the Problem 31 Developing a Model 32 Acquiring Input Data 33 Developing a Solution 33 Testing the Solution 34 Analyzing the Results 34

1.8 Implementation—Not Just the Final Step 35 Lack of Commitment and Resistance

to Change 35 Lack of Commitment by Quantitative Analysts 35 Summary 35 Glossary 36 Key Equations 36 Self-Test 36 Discussion Questions and Problems 37 Case Study: Food and Beverages at Southwestern University Football Games 39 Bibliography 39

chaPter 2 Probability Concepts and Applications 41 2.1 Introduction 42 2.2 Fundamental Concepts 42

Two Basic Rules of Probability 42 Types of Probability 43 Mutually Exclusive and Collectively

Exhaustive Events 44 Unions and Intersections of Events 45 Probability Rules for Unions, Intersections,

and Conditional Probabilities 46 2.3 Revising Probabilities with Bayes’ Theorem 47

General Form of Bayes’ Theorem 49 2.4 Further Probability Revisions 49 2.5 Random Variables 50 2.6 Probability Distributions 52

Probability Distribution of a Discrete Random Variable 52

Expected Value of a Discrete Probability Distribution 52

Variance of a Discrete Probability Distribution 53 Probability Distribution of a Continuous

Random Variable 54 2.7 The Binomial Distribution 55

Solving Problems with the Binomial Formula 56 Solving Problems with Binomial Tables 57

2.8 The Normal Distribution 58 Area Under the Normal Curve 60 Using the Standard Normal Table 60 Haynes Construction Company Example 61 The Empirical Rule 64

2.9 The F Distribution 64 2.10 The Exponential Distribution 66

Arnold’s Muffler Example 67 2.11 The Poisson Distribution 68

Summary 70 Glossary 70 Key Equations 71 Solved Problems 72 Self-Test 74 Discussion Questions and Problems 75 Case Study: WTVX 81 Bibliography 81

Appendix 2.1: Derivation of Bayes’ Theorem 81

A01_REND9327_12_SE_FM.indd 6 11/02/14 8:19 PM

Contents  7

chaPter 3 Decision Analysis 83 3.1 Introduction 84 3.2 The Six Steps in Decision Making 84 3.3 Types of Decision-Making Environments 85 3.4 Decision Making Under Uncertainty 86

Optimistic 86 Pessimistic 87 Criterion of Realism (Hurwicz Criterion) 87 Equally Likely (Laplace) 88 Minimax Regret 88

3.5 Decision Making Under Risk 89 Expected Monetary Value 89 Expected Value of Perfect Information 90 Expected Opportunity Loss 92 Sensitivity Analysis 92

3.6 A Minimization Example 93 3.7 Using Software for Payoff Table Problems 95

QM for Windows 95 Excel QM 96

3.8 Decision Trees 97 Efficiency of Sample Information 102 Sensitivity Analysis 102

3.9 How Probability Values Are Estimated by Bayesian Analysis 103 Calculating Revised Probabilities 103 Potential Problem in Using Survey Results 105

3.10 Utility Theory 106 Measuring Utility and Constructing

a Utility Curve 107 Utility as a Decision-Making Criterion 110 Summary 112 Glossary 112 Key Equations 113 Solved Problems 113 Self-Test 118 Discussion Questions and Problems 119 Case Study: Starting Right Corporation 127 Case Study: Blake Electronics 128 Bibliography 130

chaPter 4 Regression Models 131 4.1 Introduction 132 4.2 Scatter Diagrams 132 4.3 Simple Linear Regression 133 4.4 Measuring the Fit of the Regression Model 135

Coefficient of Determination 136 Correlation Coefficient 136

4.5 Assumptions of the Regression Model 138 Estimating the Variance 139

4.6 Testing the Model for Significance 139 Triple A Construction Example 141 The Analysis of Variance (ANOVA) Table 141 Triple A Construction ANOVA Example 142

4.7 Using Computer Software for Regression 142 Excel 2013 142 Excel QM 143 QM for Windows 145

4.8 Multiple Regression Analysis 146 Evaluating the Multiple Regression Model 147 Jenny Wilson Realty Example 148

4.9 Binary or Dummy Variables 149 4.10 Model Building 150

Stepwise Regression 151 Multicollinearity 151

4.11 Nonlinear Regression 151 4.12 Cautions and Pitfalls in Regression

Analysis 154 Summary 155 Glossary 155 Key Equations 156 Solved Problems 157 Self-Test 159 Discussion Questions and Problems 159 Case Study: North–South Airline 164 Bibliography 165

Appendix 4.1: Formulas for Regression Calculations 165

chaPter 5 Forecasting 167 5.1 Introduction 168 5.2 Types of Forecasting Models 168

Qualitative Models 168 Causal Models 169 Time-Series Models 169

5.3 Components of a Time-Series 169 5.4 Measures of Forecast Accuracy 171 5.5 Forecasting Models—Random Variations

Only 174 Moving Averages 174 Weighted Moving Averages 174 Exponential Smoothing 176 Using Software for Forecasting Time Series 178

5.6 Forecasting Models—Trend and Random Variations 181 Exponential Smoothing with Trend 181 Trend Projections 183

5.7 Adjusting for Seasonal Variations 185 Seasonal Indices 186 Calculating Seasonal Indices with No

Trend 186 Calculating Seasonal Indices with Trend 187

5.8 Forecasting Models—Trend, Seasonal, and Random Variations 188 The Decomposition Method 188 Software for Decomposition 191 Using Regression with Trend and Seasonal

Components 192 5.9 Monitoring and Controlling Forecasts 193

Adaptive Smoothing 195 Summary 195 Glossary 196 Key Equations 196 Solved Problems 197 Self-Test 198 Discussion Questions and Problems 199 Case Study: Forecasting Attendance at SWU Football Games 202 Case Study: Forecasting Monthly Sales 203 Bibliography 204

A01_REND9327_12_SE_FM.indd 7 11/02/14 8:19 PM

8  Contents

chaPter 6 Inventory Control Models 205 6.1 Introduction 206 6.2 Importance of Inventory Control 207

Decoupling Function 207 Storing Resources 207 Irregular Supply and Demand 207 Quantity Discounts 207 Avoiding Stockouts and Shortages 207

6.3 Inventory Decisions 208 6.4 Economic Order Quantity: Determining How

Much to Order 209 Inventory Costs in the EOQ Situation 210 Finding the EOQ 212 Sumco Pump Company Example 212 Purchase Cost of Inventory Items 213 Sensitivity Analysis with the EOQ Model 214

6.5 Reorder Point: Determining When to Order 215

6.6 EOQ Without the Instantaneous Receipt Assumption 216 Annual Carrying Cost for Production Run

Model 217 Annual Setup Cost or Annual Ordering

Cost 217 Determining the Optimal Production

Quantity 218 Brown Manufacturing Example 218

6.7 Quantity Discount Models 220 Brass Department Store Example 222

6.8 Use of Safety Stock 224 6.9 Single-Period Inventory Models 229

Marginal Analysis with Discrete Distributions 230

Café du Donut Example 231 Marginal Analysis with the Normal

Distribution 232 Newspaper Example 232

6.10 ABC Analysis 234 6.11 Dependent Demand: The Case for Material

Requirements Planning 234 Material Structure Tree 235 Gross and Net Material Requirements

Plan 236 Two or More End Products 237

6.12 Just-In-Time Inventory Control 239 6.13 Enterprise Resource Planning 240

Summary 241 Glossary 241 Key Equations 242 Solved Problems 243 Self-Test 245 Discussion Questions and Problems 246 Case Study: Martin-Pullin Bicycle Corporation 253 Bibliography 254

Appendix 6.1: Inventory Control with QM for Windows 255

chaPter 7 Linear Programming Models: Graphical and Computer Methods 257

7.1 Introduction 258 7.2 Requirements of a Linear Programming

Problem 258 7.3 Formulating LP Problems 259

Flair Furniture Company 259 7.4 Graphical Solution to an LP Problem 261

Graphical Representation of Constraints 261 Isoprofit Line Solution Method 265 Corner Point Solution Method 268 Slack and Surplus 270

7.5 Solving Flair Furniture’s LP Problem Using QM for Windows, Excel 2013, and Excel QM 271 Using QM for Windows 271 Using Excel’s Solver Command to Solve

LP Problems 272 Using Excel QM 275

7.6 Solving Minimization Problems 277 Holiday Meal Turkey Ranch 277

7.7 Four Special Cases in LP 281 No Feasible Solution 281 Unboundedness 281 Redundancy 282 Alternate Optimal Solutions 283

7.8 Sensitivity Analysis 284 High Note Sound Company 285 Changes in the Objective Function

Coefficient 286 QM for Windows and Changes in Objective

Function Coefficients 286 Excel Solver and Changes in Objective Function

Coefficients 287 Changes in the Technological Coefficients 288 Changes in the Resources or Right-Hand-Side

Values 289 QM for Windows and Changes in Right-Hand-

Side Values 290 Excel Solver and Changes in Right-Hand-Side

Values 290 Summary 292 Glossary 292 Solved Problems 293 Self-Test 297 Discussion Questions and Problems 298 Case Study: Mexicana Wire Works 306 Bibliography 308

chaPter 8 Linear Programming Applications 309 8.1 Introduction 310 8.2 Marketing Applications 310

Media Selection 310 Marketing Research 311

8.3 Manufacturing Applications 314 Production Mix 314 Production Scheduling 315

A01_REND9327_12_SE_FM.indd 8 11/02/14 8:19 PM

Contents  9

8.4 Employee Scheduling Applications 319 Labor Planning 319

8.5 Financial Applications 321 Portfolio Selection 321 Truck Loading Problem 324

8.6 Ingredient Blending Applications 326 Diet Problems 326 Ingredient Mix and Blending Problems 327

8.7 Other Linear Programming Applications 329 Summary 331 Self-Test 331 Problems 332 Case Study: Cable & Moore 339 Bibliography 340

chaPter 9 Transportation, Assignment, and Network Models 341

9.1 Introduction 342 9.2 The Transportation Problem 343

Linear Program for the Transportation Example 343

Solving Transportation Problems Using Computer Software 343

A General LP Model for Transportation Problems 344

Facility Location Analysis 345 9.3 The Assignment Problem 348

Linear Program for Assignment Example 348 9.4 The Transshipment Problem 350

Linear Program for Transshipment Example 350 9.5 Maximal-Flow Problem 353

Example 353 9.6 Shortest-Route Problem 355 9.7 Minimal-Spanning Tree Problem 356

Summary 360 Glossary 361 Solved Problems 361 Self-Test 363 Discussion Questions and Problems 364 Case Study: Andrew–Carter, Inc. 375 Case Study: Northeastern Airlines 376 Case Study: Southwestern University Traffic Problems 377 Bibliography 378

Appendix 9.1: Using QM for Windows 378

chaPter 10 Integer Programming, Goal Programming, and Nonlinear Programming 381

10.1 Introduction 382 10.2 Integer Programming 382

Harrison Electric Company Example of Integer Programming 382

Using Software to Solve the Harrison Integer Programming Problem 384

Mixed-Integer Programming Problem Example 386

10.3 Modeling with 0–1 (Binary) Variables 388 Capital Budgeting Example 388

Limiting the Number of Alternatives Selected 390

Dependent Selections 390 Fixed-Charge Problem Example 390 Financial Investment Example 392

10.4 Goal Programming 392 Example of Goal Programming: Harrison Electric

Company Revisited 394 Extension to Equally Important Multiple

Goals 395 Ranking Goals with Priority Levels 395 Goal Programming with Weighted Goals 396

10.5 Nonlinear Programming 397 Nonlinear Objective Function and Linear

Constraints 398 Both Nonlinear Objective Function and

Nonlinear Constraints 398 Linear Objective Function with Nonlinear

Constraints 400 Summary 400 Glossary 401 Solved Problems 401 Self-Test 404 Discussion Questions and Problems 405 Case Study: Schank Marketing Research 410 Case Study: Oakton River Bridge 411 Bibliography 412

chaPter 11 Project Management 413 11.1 Introduction 414 11.2 PERT/CPM 415

General Foundry Example of PERT/CPM 415 Drawing the PERT/CPM Network 417 Activity Times 417 How to Find the Critical Path 418 Probability of Project Completion 423 What PERT Was Able to Provide 424 Using Excel QM for the General Foundry

Example 424 Sensitivity Analysis and Project Management 425

11.3 PERT/Cost 427 Planning and Scheduling Project Costs:

Budgeting Process 427 Monitoring and Controlling Project Costs 430

11.4 Project Crashing 432 General Foundary Example 433 Project Crashing with Linear Programming 434

11.5 Other Topics in Project Management 437 Subprojects 437 Milestones 437 Resource Leveling 437 Software 437 Summary 437 Glossary 438 Key Equations 438 Solved Problems 439

A01_REND9327_12_SE_FM.indd 9 11/02/14 8:19 PM

10  Contents

Self-Test 441 Discussion Questions and Problems 442 Case Study: Southwestern University Stadium Construction 447 Case Study: Family Planning Research Center of Nigeria 448 Bibliography 450

Appendix 11.1: Project Management with QM for Windows 450

chaPter 12 Waiting Lines and Queuing Theory Models 453

12.1 Introduction 454 12.2 Waiting Line Costs 454

Three Rivers Shipping Company Example 455 12.3 Characteristics of a Queuing System 456

Arrival Characteristics 456 Waiting Line Characteristics 456 Service Facility Characteristics 457 Identifying Models Using Kendall Notation 457

12.4 Single-Channel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M/1) 460 Assumptions of the Model 460 Queuing Equations 460 Arnold’s Muffler Shop Case 461 Enhancing the Queuing Environment 465

12.5 Multichannel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M/m) 465 Equations for the Multichannel Queuing

Model 466 Arnold’s Muffler Shop Revisited 466

12.6 Constant Service Time Model (M/D/1) 468 Equations for the Constant Service Time

Model 468 Garcia-Golding Recycling, Inc. 469

12.7 Finite Population Model (M/M/1 with Finite Source) 470 Equations for the Finite Population Model 470 Department of Commerce Example 471

12.8 Some General Operating Characteristic Relationships 472

12.9 More Complex Queuing Models and the Use of Simulation 472 Summary 473 Glossary 473 Key Equations 474 Solved Problems 475 Self-Test 478 Discussion Questions and Problems 479 Case Study: New England Foundry 483 Case Study: Winter Park Hotel 485 Bibliography 485

Appendix 12.1: Using QM for Windows 486

chaPter 13 Simulation Modeling 487 13.1 Introduction 488 13.2 Advantages and Disadvantages

of Simulation 489

13.3 Monte Carlo Simulation 490 Harry’s Auto Tire Example 490 Using QM for Windows for Simulation 495 Simulation with Excel Spreadsheets 496

13.4 Simulation and Inventory Analysis 498 Simkin’s Hardware Store 498 Analyzing Simkin’s Inventory Costs 501

13.5 Simulation of a Queuing Problem 502 Port of New Orleans 502 Using Excel to Simulate the Port of New Orleans

Queuing Problem 504 13.6 Simulation Model for a Maintenance

Policy 505 Three Hills Power Company 505 Cost Analysis of the Simulation 507

13.7 Other Simulation Issues 510 Two Other Types of Simulation Models 510 Verification and Validation 511 Role of Computers in Simulation 512 Summary 512 Glossary 512 Solved Problems 513 Self-Test 516 Discussion Questions and Problems 517 Case Study: Alabama Airlines 522 Case Study: Statewide Development Corporation 523 Case Study: FB Badpoore Aerospace 524 Bibliography 526

chaPter 14 Markov Analysis 527 14.1 Introduction 528 14.2 States and State Probabilities 528

The Vector of State Probabilities for Three Grocery Stores Example 529

14.3 Matrix of Transition Probabilities 530 Transition Probabilities for the Three Grocery

Stores 531 14.4 Predicting Future Market Shares 531 14.5 Markov Analysis of Machine Operations 532 14.6 Equilibrium Conditions 533 14.7 Absorbing States and the Fundamental

Matrix: Accounts Receivable Application 536 Summary 540 Glossary 541 Key Equations 541 Solved Problems 541 Self-Test 545 Discussion Questions and Problems 545 Case Study: Rentall Trucks 550 Bibliography 551

Appendix 14.1: Markov Analysis with QM for Windows 551 Appendix 14.2: Markov Analysis With Excel 553

chaPter 15 Statistical Quality Control 555 15.1 Introduction 556 15.2 Defining Quality and TQM 556 15.3 Statiscal Process Control 557

Variability in the Process 557

A01_REND9327_12_SE_FM.indd 10 11/02/14 8:19 PM

Contents  11

15.4 Control Charts for Variables 559 The Central Limit Theorem 559 Setting x-Chart Limits 560 Setting Range Chart Limits 563

15.5 Control Charts for Attributes 564 p-Charts 564 c-Charts 566 Summary 568 Glossary 568 Key Equations 568 Solved Problems 569 Self-Test 570 Discussion Questions and Problems 570 Bibliography 573

Appendix 15.1: Using QM for Windows for SPC 573

appendices 575 appendix a Areas Under the Standard

Normal Curve 576 appendix B Binomial Probabilities 578 appendix c Values of e-l for Use in the Poisson

Distribution 583 appendix d F Distribution Values 584 appendix e Using POM-QM for Windows 586 appendix F Using Excel QM and Excel Add-Ins 589 appendix G Solutions to Selected Problems 590 appendix H Solutions to Self-Tests 594

index 597

OnLine MOduLes

MOduLe 1 Analytic Hierarchy Process M1-1 M1.1 Introduction M1-2 M1.2 Multifactor Evaluation Process M1-2 M1.3 Analytic Hierarchy Process M1-4

Judy Grim’s Computer Decision M1-4 Using Pairwise Comparisons M1-5 Evaluations for Hardware M1-7 Determining the Consistency Ratio M1-7 Evaluations for the Other Factors M1-9 Determining Factor Weights M1-10 Overall Ranking M1-10 Using the Computer to Solve Analytic Hierarchy

Process Problems M1-10 M1.4 Comparison of Multifactor Evaluation and

Analytic Hierarchy Processes M1-11 Summary M1-12 Glossary M1-12 Key Equations M1-12 Solved Problems M1-12 Self-Test M1-14 Discussion Questions and Problems M1-14 Bibliography M1-16

Appendix M1.1: Using Excel for the Analytic Hierarchy Process M1-16

MOduLe 2 Dynamic Programming M2-1 M2.1 Introduction M2-2 M2.2 Shortest-Route Problem Solved Using

Dynamic Programming M2-2

M2.3 Dynamic Programming Terminology M2-6 M2.4 Dynamic Programming Notation M2-8 M2.5 Knapsack Problem M2-9

Types of Knapsack Problems M2-9 Roller’s Air Transport Service Problem M2-9 Summary M2-16 Glossary M2-16 Key Equations M2-16 Solved Problem M2-16 Self-Test M2-18 Discussion Questions and Problems M2-19 Case Study: United Trucking M2-22 Internet Case Study M2-22 Bibliography M2-22

MOduLe 3 Decision Theory and the Normal Distribution M3-1

M3.1 Introduction M3-2 M3.2 Break-Even Analysis and the Normal

Distribution M3-2 Barclay Brothers Company’s New Product

Decision M3-2 Probability Distribution of Demand M3-3 Using Expected Monetary Value to Make a

Decision M3-5 M3.3 Expected Value of Perfect Information and the

Normal Distribution M3-6 Opportunity Loss Function M3-6 Expected Opportunity Loss M3-6 Summary M3-8 Glossary M3-8 Key Equations M3-8 Solved Problems M3-9 Self-Test M3-9 Discussion Questions and Problems M3-10 Bibliography M3-11

Appendix M3.1: Derivation of the Break-Even Point M3-11 Appendix M3.2: Unit Normal Loss Integral M3-12

MOduLe 4 Game Theory M4-1 M4.1 Introduction M4-2 M4.2 Language of Games M4-2 M4.3 The Minimax Criterion M4-3 M4.4 Pure Strategy Games M4-4 M4.5 Mixed Strategy Games M4-5 M4.6 Dominance M4-6

Summary M4-7 Glossary M4-7 Solved Problems M4-7 Self-Test M4-8 Discussion Questions and Problems M4-9 Bibliography M4-10

MOduLe 5 Mathematical Tools: Determinants and Matrices M5-1

M5.1 Introduction M5-2 M5.2 Matrices and Matrix

Operations M5-2 Matrix Addition and Subtraction M5-2 Matrix Multiplication M5-3 Matrix Notation for Systems

of Equations M5-6 Matrix Transpose M5-6

M5.3 Determinants, Cofactors, and Adjoints M5-6 Determinants M5-6 Matrix of Cofactors and Adjoint M5-8

A01_REND9327_12_SE_FM.indd 11 11/02/14 9:22 PM

12  Contents

M5.4 Finding the Inverse of a Matrix M5-10 Summary M5-11 Glossary M5-11 Key Equations M5-11 Self-Test M5-12 Discussion Questions and Problems M5-12 Bibliography M5-13

Appendix M5.1: Using Excel for Matrix Calculations M5-13

MOduLe 6 Calculus-Based Optimization M6-1 M6.1 Introduction M6-2 M6.2 Slope of a Straight Line M6-2 M6.3 Slope of a Nonlinear Function M6-3 M6.4 Some Common Derivatives M6-5

Second Derivatives M6-6 M6.5 Maximum and Minimum M6-6 M6.6 Applications M6-8

Economic Order Quantity M6-8 Total Revenue M6-9 Summary M6-10 Glossary M6-10 Key Equations M6-10 Solved Problem M6-11 Self-Test M6-11 Discussion Questions and Problems M6-12 Bibliography M6-12

MOduLe 7 Linear Programming: The Simplex Method M7-1

M7.1 Introduction M7-2 M7.2 How to Set Up the Initial Simplex

Solution M7-2 Converting the Constraints to Equations M7-3 Finding an Initial Solution Algebraically M7-3 The First Simplex Tableau M7-4

M7.3 Simplex Solution Procedures M7-8 M7.4 The Second Simplex Tableau M7-9

Interpreting the Second Tableau M7-12 M7.5 Developing the Third Tableau M7-13 M7.6 Review of Procedures for Solving LP

Maximization Problems M7-16 M7.7 Surplus and Artificial Variables M7-16

Surplus Variables M7-17 Artificial Variables M7-17 Surplus and Artificial Variables in the Objective

Function M7-18 M7.8 Solving Minimization Problems M7-18

The Muddy River Chemical Company Example M7-18

Graphical Analysis M7-19 Converting the Constraints and Objective

Function M7-20 Rules of the Simplex Method for Minimization

Problems M7-21 First Simplex Tableau for the Muddy River

Chemical Corporation Problem M7-21 Developing a Second Tableau M7-23 Developing a Third Tableau M7-24 Fourth Tableau for the Muddy River Chemical

Corporation Problem M7-26

M7.9 Review of Procedures for Solving LP Minimization Problems M7-27

M7.10 Special Cases M7-28 Infeasibility M7-28 Unbounded Solutions M7-28 Degeneracy M7-29 More Than One Optimal Solution M7-30

M7.11 Sensitivity Analysis with the Simplex Tableau M7-30 High Note Sound Company Revisited M7-30 Changes in the Objective Function

Coefficients M7-31 Changes in Resources or RHS Values M7-33

M7.12 The Dual M7-35 Dual Formulation Procedures M7-37 Solving the Dual of the High Note Sound

Company Problem M7-37 M7.13 Karmarkar’s Algorithm M7-39

Summary M7-39 Glossary M7-39 Key Equation M7-40 Solved Problems M7-41 Self-Test M7-44 Discussion Questions and Problems M7-45 Bibliography M7-54

MOduLe 8 Transportation, Assignment, and Network Algorithms M8-1

M8.1 Introduction M8-2 M8.2 The Transportation Algorithm M8-2

Developing an Initial Solution: Northwest Corner Rule M8-2

Stepping-Stone Method: Finding a Least-Cost Solution M8-4

M8.3 Special Situations with the Transportation Algorithm M8-9 Unbalanced Transportation Problems M8-9 Degeneracy in Transportation Problems M8-10 More Than One Optimal Solution M8-13 Maximization Transportation Problems M8-13 Unacceptable or Prohibited Routes M8-13 Other Transportation Methods M8-13

M8.4 The Assignment Algorithm M8-13 The Hungarian Method (Flood’s

Technique) M8-14 Making the Final Assignment M8-18

M8.5 Special Situations with the Assignment Algorithm M8-18 Unbalanced Assignment Problems M8-18 Maximization Assignment Problems M8-19

M8.6 Maximal-Flow Problem M8-20 Maximal-Flow Technique M8-20

M8.7 Shortest-Route Problem M8-23 Shortest-Route Technique M8-23 Summary M8-25 Glossary M8-25 Solved Problems M8-26 Self-Test M8-32 Discussion Questions and Problems M8-33 Cases M8-42 Bibliography M8-42

A01_REND9327_12_SE_FM.indd 12 11/02/14 9:22 PM

13

Overview

Welcome to the twelfth edition of Quantitative Analysis for Management. Our goal is to provide undergraduate and graduate students with a genuine foundation in business analytics, quantitative methods, and management science. In doing so, we owe thanks to the hundreds of users and scores of reviewers who have provided invaluable counsel and pedagogical insight for more than 30 years.

To help students connect how the techniques presented in this book apply in the real world, computer-based applications and examples are a major focus of this edition. Mathematical models, with all the necessary assumptions, are presented in a clear and “plain-English” manner. The ensuing solution procedures are then applied to example problems alongside step-by-step “how-to” instruc- tions. We have found this method of presentation to be very effective and students are very apprecia- tive of this approach. In places where the mathematical computations are intricate, the details are presented in such a manner that the instructor can omit these sections without interrupting the flow of material. The use of computer software enables the instructor to focus on the managerial problem and spend less time on the details of the algorithms. Computer output is provided for many examples throughout the book.

The only mathematical prerequisite for this textbook is algebra. One chapter on probability and another on regression analysis provide introductory coverage on these topics. We employ standard notation, terminology, and equations throughout the book. Careful explanation is provided for the mathematical notation and equations that are used.

new tO this editiOn

● An introduction to business analytics is provided.

● Excel 2013 is incorporated throughout the chapters.

● The transportation, assignment, and network models have been combined into one chapter focused on modeling with linear programming.

● Specialized algorithms for the transportation, assignment, and network methods have been combined into Online Module 8.

● New examples, over 25 problems, 8 QA in Action applications, 4 Modeling in the Real World features, and 3 new Case Studies have been added throughout the textbook. Other problems and Case Studies have been updated.

PrefACe

A01_REND9327_12_SE_FM.indd 13 11/02/14 8:19 PM

14  PrefACe

speCial Features

Many features have been popular in previous editions of this textbook, and they have been updated and expanded in this edition. They include the following:

● Modeling in the Real World boxes demonstrate the application of the quantitative analysis approach to every technique discussed in the book. Four new ones have been added.

● Procedure boxes summarize the more complex quantitative techniques, presenting them as a series of easily understandable steps.

● Margin notes highlight the important topics in the text.

● History boxes provide interesting asides related to the development of techniques and the people who originated them.

● QA in Action boxes illustrate how real organizations have used quantitative analysis to solve problems. Several new QA in Action boxes have been added.

● Solved Problems, included at the end of each chapter, serve as models for students in solving their own homework problems.

● Discussion Questions are presented at the end of each chapter to test the student’s understand- ing of the concepts covered and definitions provided in the chapter.

● Problems included in every chapter are applications oriented and test the student’s ability to solve exam-type problems. They are graded by level of difficulty: introductory (one bullet), moderate (two bullets), and challenging (three bullets). More than 40 new problems have been added.

● Internet Homework Problems provide additional problems for students to work. They are available on the Companion Website.

● Self-Tests allow students to test their knowledge of important terms and concepts in prepara- tion for quizzes and examinations.

● Case Studies, at the end of each chapter, provide additional challenging managerial applications.

● Glossaries, at the end of each chapter, define important terms.

● Key Equations, provided at the end of each chapter, list the equations presented in that chapter.

● End-of-chapter bibliographies provide a current selection of more advanced books and articles.

● The software POM-QM for Windows uses the full capabilities of Windows to solve quantita- tive analysis problems.

● Excel QM and Excel 2013 are used to solve problems throughout the book.

● Data files with Excel spreadsheets and POM-QM for Windows files containing all the examples in the textbook are available for students to download from the Companion Website. Instructors can download these plus additional files containing computer solutions to the rel- evant end-of-chapter problems from the Instructor Resource Center Web site.

● Online modules provide additional coverage of topics in quantitative analysis.

● The Companion Website, at www.pearsonglobaleditions.com/render, provides the online modules, additional problems, cases, and other material for almost every chapter.

signiFiCant Changes tO the twelFth editiOn

In the twelfth edition, we have introduced Excel 2013 in all of the chapters. Screenshots are integrated in the appropriate sections so that students can easily learn how to use Excel for the calculations. The Excel QM add-in is used with Excel 2013 allowing students with limited Excel experience to easily perform the necessary calculations. This also allows students to improve their Excel skills as they see the formulas automatically written in Excel QM.

A01_REND9327_12_SE_FM.indd 14 11/02/14 8:19 PM

PrefACe  15

From the Companion Website, students can access files for all of the examples used in the textbook in Excel 2013, QM for Windows, and Excel QM. Other files with all of the end-of-chapter problems involving these software tools are available to the instructors.

Business analytics, one of the hottest topics in the business world, makes extensive use of the models in this book. A discussion of the business analytics categories is provided, and the relevant management science techniques are placed into the appropriate category.

The transportation, transshipment, assignment, and network models have been combined into one chapter focused on modeling with linear programming. The specialized algorithms for these models have been combined into a new online module.

Examples and problems have been updated, and many new ones have been added. New screen- shots are provided for almost all of the examples in the book. A brief summary of the other changes in each chapter are presented here.

Chapter 1 Introduction to Quantitative Analysis. A section on business analytics has been added, the self-test has been modified, and two new problems were added.

Chapter 2 Probability Concepts and Applications. The presentation of the fundamental concepts of probability has been significantly modified and reorganized. Two new problems have been added.

Chapter 3 Decision Analysis. A more thorough discussion of minimization problems with payoff tables has been provided in a new section. The presentation of software usage with payoff tables was expanded. Two new problems were added.

Chapter 4 Regression Models. The use of different software packages for regression analysis has been moved to the body of the textbook instead of the appendix. Five new problems and one new QA in Action item have been added.

Chapter 5 Forecasting. The presentation of time-series forecasting models was significantly revised to bring the focus on identifying the appropriate technique to use based on which time- series components are present in the data. Five new problems were added, and the cases have been updated.

Chapter 6 Inventory Control Models. The four steps of the Kanban production process have been updated and clarified. Two new QA in Action boxes, four new problems, and one new Modeling in the Real World have been added.

Chapter 7 Linear Programming Models: Graphical and Computer Methods. More discussion of Solver is presented. A new Modeling in the Real World item was added, and the solved problems have been revised.

Chapter 8 Linear Programming Applications. The transportation model was moved to Chapter 9, and a new section describing other models has been added. The self-test questions were modified; one new problem, one new QA in Action summary, and a new case study have been added.

Chapter 9 Transportation, Assignment, and Network Models. This new chapter presents all of the distribution, assignment, and network models that were previously in two separate chapters. The modeling approach is emphasized, while the special-purpose algorithms were moved to a new online module. A new case study, Northeastern Airlines, has also been added.

Chapter 10 Integer Programming, Goal Programming, and Nonlinear Programming. The use of Excel 2013 and the new screen shots were the only changes to this chapter.

Chapter 11 Project Management. Two new end-of-chapter problems and three new QA in Action boxes have been added.

Chapter 12 Waiting Lines and Queuing Theory Models. Two new end-of-chapter problems were added.

Chapter 13 Simulation Modeling. One new Modeling in the Real World vignette, one new QA in Action box, and a new case study have been added.

A01_REND9327_12_SE_FM.indd 15 11/02/14 8:19 PM

16  PrefACe

Chapter 14 Markov Analysis. One new QA in Action box and two new end-of-chapter problems have been added.

Chapter 15 Statistical Quality Control. One new Modeling in the Real World vignette, one new QA in Action box, and two new end-of-chapter problems have been added.

Modules 1–8 The only significant change to the modules is the addition of Module 8: Transportation, Assignment, and Network Algorithms. This includes the special-purpose algorithms for the transportation, assignment, and network models.

Online MOdules

To streamline the book, eight topics are contained in modules available on the Companion Website for the book, located at www.pearsonglobaleditions.com/render.

1. Analytic Hierarchy Process

2. Dynamic Programming

3. Decision Theory and the Normal Distribution

4. Game Theory

5. Mathematical Tools: Determinants and Matrices

6. Calculus-Based Optimization

7. Linear Programming: The Simplex Method

8. Transportation, Assignment, and Network Algorithms

sOFtware

excel 2013 Instructions and screen captures are provided for, using Excel 2013, throughout the book. Instructions for activating the Solver and Analysis ToolPak add-ins in Excel 2013 are pro- vided in an appendix. The use of Excel is more prevalent in this edition of the book than in previous editions.

excel QM Using the Excel QM add-in that is available on the Companion Website makes the use of Excel even easier. Students with limited Excel experience can use this and learn from the formu- las that are automatically provided by Excel QM. This is used in many of the chapters.

pOM-QM for windows This software, developed by Professor Howard Weiss, is available to students at the Companion Website. This is very user-friendly and has proven to be a very popular software tool for users of this textbook. Modules are available for every major problem type pre- sented in the textbook.

COMpaniOn website

The Companion Website, located at www.pearsonglobaleditions.com/render, contains a variety of materials to help students master the material in this course. These include the following:

Modules There are eight modules containing additional material that the instructor may choose to include in the course. Students can download these from the Companion Website.

Files for examples in excel, excel QM, and pOM-QM for windows Students can down- load the files that were used for examples throughout the book. This helps them become familiar with the software, and it helps them understand the input and formulas necessary for working the examples.

A01_REND9327_12_SE_FM.indd 16 11/02/14 8:19 PM

PrefACe  17

internet homework problems In addition to the end-of-chapter problems in the textbook, there are additional problems that instructors may assign. These are available for download at the Companion Website, located at www.pearsonglobaleditions.com/render.

internet Case studies Additional case studies are available for most chapters.

pOM-QM for windows Developed by Howard Weiss, this very user-friendly software can be used to solve most of the homework problems in the text.

excel QM This Excel add-in will automatically create worksheets for solving problems. This is very helpful for instructors who choose to use Excel in their classes but who may have students with limited Excel experience. Students can learn by examining the formulas that have been cre- ated, and by seeing the inputs that are automatically generated for using the Solver add-in for linear programming.

instruCtOr resOurCes

● Instructor Resource Center: The Instructor Resource Center contains the electronic files for the test bank, PowerPoint slides, the Solutions Manual, and data files for both Excel and POM-QM for Windows for all relevant examples and end-of-chapter problems, at www.pearsonglobaleditions.com/render.

● Register, Redeem, Login: At www.pearsonglobaleditions.com/render, instructors can access a variety of print, media, and presentation resources that are available with this text in down- loadable, digital format.

● Need help? Our dedicated technical support team is ready to assist instructors with questions about the media supplements that accompany this text. Visit http://247pearsoned.custhelp .com/ for answers to frequently asked questions and toll-free user support phone numbers. The supplements are available to adopting instructors. Detailed descriptions are provided on the Instructor Resource Center.

instructor’s solutions Manual The Instructor’s Solutions Manual, updated by the authors, is available for download from the Instructor Resource Center. Solutions to all Internet Homework Problems and Internet Case Studies are also included in the manual.

powerpoint presentation An extensive set of PowerPoint slides is available for download from the Instructor Resource Center.

test bank The updated test bank is available for download from the Instructor Resource Center.

testgen The computerized TestGen package allows instructors to customize, save, and generate classroom tests. The test program permits instructors to edit, add, or delete questions from the test bank; edit existing graphics and create new graphics; analyze test results; and organize a database of test and student results. This software allows the instructors to benefit from the extensive flexibility and ease of use. It provides many options for organizing and displaying tests, along with search and sort features. The software and the test banks can be downloaded at www.pearsonglobaleditions.com/render.

aCknOwledgMents

We gratefully thank the users of previous editions and the reviewers who provided valuable sugges- tions and ideas for this edition. Your feedback is valuable in our efforts for continuous improvement. The continued success of Quantitative Analysis for Management is a direct result of instructor and student feedback, which is truly appreciated.

The authors are indebted to many people who have made important contributions to this pro- ject. Special thanks go to Professors Faizul Huq, F. Bruce Simmons III, Khala Chand Seal, Victor E. Sower, Michael Ballot, Curtis P. McLaughlin, and Zbigniew H. Przanyski for their contributions to the excellent cases included in this edition.

A01_REND9327_12_SE_FM.indd 17 11/02/14 8:19 PM

18  PrefACe

Stephen Achtenhagen, San Jose University M. Jill Austin, Middle Tennessee State University Raju Balakrishnan, Clemson University Hooshang Beheshti, Radford University Jason Bergner, University of Central Missouri Bruce K. Blaylock, Radford University Rodney L. Carlson, Tennessee Technological University Edward Chu, California State University, Dominguez Hills John Cozzolino, Pace University–Pleasantville Ozgun C. Demirag, Penn State–Erie Shad Dowlatshahi, University of Wisconsin, Platteville Ike Ehie, Southeast Missouri State University Richard Ehrhardt, University of North Carolina–Greensboro Sean Eom, Southeast Missouri State University Ephrem Eyob, Virginia State University Mira Ezvan, Lindenwood University Wade Ferguson, Western Kentucky University Robert Fiore, Springfield College Frank G. Forst, Loyola University of Chicago Ed Gillenwater, University of Mississippi Stephen H. Goodman, University of Central Florida Irwin Greenberg, George Mason University Nicholas G. Hall, Ohio State University Robert R. Hill, University of Houston–Clear Lake Gordon Jacox, Weber State University Bharat Jain, Towson University Vassilios Karavas, University of Massachusetts Amherst Darlene R. Lanier, Louisiana State University Kenneth D. Lawrence, New Jersey Institute of Technology Jooh Lee, Rowan College Richard D. Legault, University of Massachusetts–Dartmouth Douglas Lonnstrom, Siena College Daniel McNamara, University of St. Thomas Peter Miller, University of Windsor Ralph Miller, California State Polytechnic University

Shahriar Mostashari, Campbell University David Murphy, Boston College Robert C. Myers, University of Louisville Barin Nag, Towson State University Nizam S. Najd, Oklahoma State University Harvey Nye, Central State University Alan D. Olinsky, Bryant College Savas Ozatalay, Widener University Young Park, California University of Pennsylvania Cy Peebles, Eastern Kentucky University Yusheng Peng, Brooklyn College Dane K. Peterson, Southwest Missouri State University Sanjeev Phukan, Bemidji State University Ranga Ramasesh, Texas Christian University William Rife, West Virginia University Bonnie Robeson, Johns Hopkins University Grover Rodich, Portland State University Vijay Shah, West Virginia University–Parkersburg L. Wayne Shell, Nicholls State University Thomas Sloan, University of Massachusetts–Lowell Richard Slovacek, North Central College Alan D. Smith, Robert Morris University John Swearingen, Bryant College F. S. Tanaka, Slippery Rock State University Jack Taylor, Portland State University Madeline Thimmes, Utah State University M. Keith Thomas, Olivet College Andrew Tiger, Southeastern Oklahoma State University Chris Vertullo, Marist College James Vigen, California State University, Bakersfield William Webster, University of Texas at San Antonio Larry Weinstein, Eastern Kentucky University Fred E. Williams, University of Michigan–Flint Mela Wyeth, Charleston Southern University Oliver Yu, San Jose State University

We are very grateful to all the people at Pearson who worked so hard to make this book a suc- cess. These include Donna Battista, editor in chief; Mary Kate Murray, senior project manager; and Kathryn Dinovo, senior production project manager. We are also grateful to Tracy Duff, our project manager at PreMediaGlobal. We are extremely thankful to Annie Puciloski for her tireless work in error checking the textbook. Thank you all!

Barry Render [email protected]

Ralph Stair

Michael Hanna [email protected]

Trevor S. Hale [email protected]

Pearson wishes to thank and acknowledge the following people for their work on the Global Edition:

We thank Howard Weiss for providing Excel QM and POM-QM for Windows, two of the most outstanding packages in the field of quantitative methods. We would also like to thank the reviewers who have helped to make this textbook the most widely used one in the field of quantitative analysis:

Contributors: Krish Saha, Coventry University Stefania Paladini, Coventry University Tracey Holker, Coventry University

Reviewers: Chukri Akhras, Notre Dame University and Lebanese

International University–Lebanon Rohaizan Binti Ramlan, Universiti Tun Hussein Onn–Malaysia Yong Wooi Keong, Sunway University–Malaysia

A01_REND9327_12_SE_FM.indd 18 11/02/14 8:19 PM

 19

Summary  •  Glossary  •  Key Equations  •  Self-Test  •  Discussion Questions and Problems  •  Case Study: Food and 

Beverages at Southwestern University Football Games  •  Bibliography

1.6 The Role of Computers and Spreadsheet Models in the Quantitative Analysis Approach

1.7 Possible Problems in the Quantitative Analysis Approach

1.8 Implementation—Not Just the Final Step

1.1 Introduction

1.2 What Is Quantitative Analysis?

1.3 Business Analytics

1.4 The Quantitative Analysis Approach

1.5 How to Develop a Quantitative Analysis Model

Chapter Outline

5. Use computers and spreadsheet models to perform quantitative analysis.

6. Discuss possible problems in using quantitative analysis.

7. Perform a break-even analysis.

1. Describe the quantitative analysis approach.

2. Understand the application of quantitative analysis in a real situation.

3. Describe the three categories of business analytics.

4. Describe the use of modeling in quantitative analysis.

After completing this chapter, students will be able to:

Introduction to Quantitative Analysis

1Chapter

learning ObjeCtives

M01_REND9327_12_SE_C01.indd 19 10/02/14 1:19 PM

20  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

1.1 Introduction

People have been using mathematical tools to help solve problems for thousands of years; how- ever, the formal study and application of quantitative techniques to practical decision making  is largely a product of the twentieth century. The techniques we study in this book have been  applied successfully to an increasingly wide variety of complex problems in business, govern- ment, health care, education, and many other areas. Many such successful uses are discussed  throughout this book.

It isn’t enough, though, just to know the mathematics of how a particular quantitative tech- nique works; you must also be familiar with the limitations, assumptions, and specific applica- bility of the technique. The successful use of quantitative techniques usually results in a solution  that is timely, accurate, flexible, economical, reliable, and easy to understand and use.

In this and other chapters, there are QA (Quantitative Analysis) in Action boxes that provide  success stories on the applications of management science. They show how organizations have  used quantitative techniques to make better decisions, operate more efficiently, and generate more  profits. Taco Bell has reported saving over $150 million with better forecasting of demand and bet- ter scheduling of employees. NBC television increased advertising revenue by over $200 million  between 1996 and 2000 by using a model to help develop sales plans for advertisers. Continental  Airlines saves over $40 million per year by using mathematical models to quickly recover from  disruptions caused by weather delays and other factors. These are but a few of the many companies  discussed in QA in Action boxes throughout this book.

To see other examples of how companies use quantitative analysis or operations research  methods to operate better and more efficiently, go to the website www.scienceofbetter.org. The  success stories presented there are categorized by industry, functional area, and benefit. These  success stories illustrate how operations research is truly the “science of better.”

1.2 What Is Quantitative Analysis?

Quantitative analysis is the scientific approach to managerial decision making. This field of  study has several different names including quantitative analysis, management science, and op- erations research. These terms are used interchangeably in this book. Also, many of the quantita- tive analysis methods presented in this book are used extensively in business analytics.

Whim, emotions, and guesswork are not part of the quantitative analysis approach. The ap- proach starts with data. Like raw material for a factory, these data are manipulated or processed  into information that is valuable to people making decisions. This processing and manipulating  of raw data into meaningful information is the heart of quantitative analysis. Computers have  been instrumental in the increasing use of quantitative analysis.

In solving a problem, managers must consider both qualitative and quantitative factors. For  example, we might consider several different investment alternatives, including certificates of  deposit at a bank, investments in the stock market, and an investment in real estate. We can use  quantitative analysis to determine how much our investment will be worth in the future when de- posited at a bank at a given interest rate for a certain number of years. Quantitative analysis can  also be used in computing financial ratios from the balance sheets for several companies whose  stock we are considering. Some real estate companies have developed computer programs that  use quantitative analysis to analyze cash flows and rates of return for investment property.

In addition to quantitative analysis, qualitative factors should also be considered. The  weather, state and federal legislation, new technological breakthroughs, the outcome of an elec- tion, and so on may all be factors that are difficult to quantify.

Because of the importance of qualitative factors, the role of quantitative analysis in the  decision-making process can vary. When there  is a  lack of qualitative factors and when  the  problem, model, and input data remain the same, the results of quantitative analysis can  automate the decision-making process. For example, some companies use quantitative inventory  models to determine automatically when to order additional new materials. In most cases, how- ever, quantitative analysis will be an aid to the decision-making process. The results of quantita- tive analysis will be combined with other (qualitative) information in making decisions.

Quantitative analysis has been particularly important in many areas of management. The  field of production management, which evolved into production/operations management (POM) 

Quantitative analysis uses a scientific approach to decision making.

Both qualitative and quantitative factors must be considered.

M01_REND9327_12_SE_C01.indd 20 10/02/14 1:19 PM

1.3 BusIness analytICs  21

as society became more service oriented, uses quantitative analysis extensively. While POM  focuses on internal operations of a company, the field of supply chain management takes a more  complete view of the business and considers the entire process of obtaining materials from sup- pliers, using the materials to develop products, and distributing these products to the final con- sumers. Supply chain management makes extensive use of many management science models.  Another area of management that could not exist without the quantitative analysis methods pre- sented in this book, and perhaps the hottest discipline in business today, is business analytics.

1.3 Business Analytics

Business analytics is a data-driven approach to decision making that allows companies to make  better decisions. The study of business analytics involves the use of large amounts of data, which  means that information technology related to the management of the data is very important. Sta- tistical and quantitative analysis are used to analyze the data and provide useful information to  the decision maker.

Business analytics is often broken into three categories: descriptive, predictive, and pre- scriptive. Descriptive analytics involves the study and consolidation of historical data for a  business and an industry. It helps a company measure how it has performed in the past and how  it is performing now. Predictive analytics is aimed at forecasting future outcomes based on  patterns in the past data. Statistical and mathematical models are used extensively for this pur- pose. Prescriptive analytics involves the use of optimization methods to provide new and better  ways to operate based on specific business objectives. The optimization models presented in this   book are very important to prescriptive analytics. While there are only three business analytics  categories, many business decisions are made based on information obtained from two or three  of these categories.

Many of the quantitative analysis techniques presented in the chapters of this book are used  extensively in business analytics. Table 1.1 highlights the three categories of business analytics,  and it places many of the topics and chapters in this book in the most relevant category. Keep in  mind that some topics (and certainly some chapters with multiple concepts and models) could  possibly be placed in a different category. Some of the material in this book could overlap two or  even three of these categories. Nevertheless, all of these quantitative analysis techniques are very  important tools in business analytics.

Business AnAlytics cAtegory QuAntitAtive AnAlysis techniQue

(chAPter)

Descriptive analytics ● Statistical measures such as means and standard  deviations (Chapter 2)

● Statistical quality control (Chapter 15)

Predictive analytics ● Decision analysis and decision trees (Chapter 3) ● Regression models (Chapter 4) ● Forecasting (Chapter 5) ● Project scheduling (Chapter 11) ● Waiting line models (Chapter 12) ● Simulation (Chapter 13) ● Markov analysis (Chapter 14)

Prescriptive analytics ● Inventory models such as the economic order  quantity (Chapter 6)

● Linear programming (Chapters 7, 8) ● Transportation and assignment models (Chapter 9) ● Integer programming, goal programming, and 

nonlinear programming (Chapter 10) ● Network models (Chapter 9)

Table 1.1 Business Analytics and Quantitative Analysis Models

The three categories of business analytics are descriptive, predictive, and prescriptive.

M01_REND9327_12_SE_C01.indd 21 10/02/14 1:19 PM

22  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

1.4 The Quantitative Analysis Approach

The quantitative analysis approach consists of defining a problem, developing a model, acquir- ing input data, developing a solution, testing the solution, analyzing the results, and implement- ing the results (see Figure 1.1). One step does not have to be finished completely before the  next is started; in most cases, one or more of these steps will be modified to some extent before  the final results are implemented. This would cause all of the subsequent steps to be changed.  In some cases, testing the solution might reveal that the model or the input data are not correct.  This would mean that all steps that follow defining the problem would need to be modified.

Defining the Problem The first step in the quantitative approach is to develop a clear, concise statement of the  problem. This statement will give direction and meaning to the following steps.

In many cases, defining the problem is the most important and the most difficult step. It is  essential to go beyond the symptoms of the problem and identify the true causes. One problem  may be related to other problems; solving one problem without regard to other related problems  can make the entire situation worse. Thus, it is important to analyze how the solution to one  problem affects other problems or the situation in general.

It is likely that an organization will have several problems. However, a quantitative analysis  group usually cannot deal with all of an organization’s problems at one time. Thus, it is usually  necessary to concentrate on only a few problems. For most companies, this means selecting  those problems whose solutions will result in the greatest increase in profits or reduction in costs  to the company. The importance of selecting the right problems to solve cannot be overempha- sized. Experience has shown that bad problem definition is a major reason for failure of manage- ment science or operations research groups to serve their organizations well.

When the problem is difficult to quantify, it may be necessary to develop specific, measur- able objectives. A problem might be inadequate health care delivery in a hospital. The objectives  might be to increase the number of beds, reduce the average number of days a patient spends  in the hospital, increase the physician-to-patient ratio, and so on. When objectives are used,  however, the real problem should be kept in mind. It is important to avoid obtaining specific and  measurable objectives that may not solve the real problem.

Developing a Model Once we select the problem to be analyzed, the next step is to develop a model. Simply stated, a  model is a representation (usually mathematical) of a situation.

Even though you might not have been aware of it, you have been using models most of your  life. You may have developed models about people’s behavior. Your model might be that friend- ship is based on reciprocity, an exchange of favors. If you need a favor such as a small loan, your  model would suggest that you ask a good friend.

Of course, there are many other types of models. Architects sometimes make a physical model of a building that they will construct. Engineers develop scale models of chemical plants, called  pilot plants. A schematic model is a picture, drawing, or chart of reality. Automobiles, lawn mow- ers, gears, fans, typewriters, and numerous other devices have schematic models (drawings and 

The types of models include physical, scale, schematic, and mathematical models.

Quantitative analysis has been in existence since the beginning of recorded history, but it was Frederick W. Taylor who in the early 1900s pioneered the principles of the scientific approach to man- agement. During World War II, many new scientific and quantita- tive techniques were developed to assist the military. These new developments were so successful that after World War II many companies started using similar techniques in managerial decision

making and planning. Today, many organizations employ a staff of operations research or management science personnel or con- sultants to apply the principles of scientific management to prob- lems and opportunities.

The origin of many of the techniques discussed in this book can be traced to individuals and organizations that have applied the principles of scientific management first developed by Taylor; they are discussed in History boxes scattered throughout the book.

histOry the Origin of Quantitative analysis

Defining the problem can be the most important step.

Concentrate on only a few problems.

Figure 1.1 The Quantitative Analysis Approach

Defining the Problem

Developing a Model

Acquiring Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

M01_REND9327_12_SE_C01.indd 22 10/02/14 1:19 PM

1.4 the QuantItatIve analysIs approaCh  23

pictures) that reveal how these devices work. What sets quantitative analysis apart from other  techniques is that the models that are used are mathematical. A mathematical model is a set of  mathematical relationships. In most cases, these relationships are expressed in equations and in- equalities, as they are in a spreadsheet model that computes sums, averages, or standard deviations.

Although there is considerable flexibility in the development of models, most of the models  presented in this book contain one or more variables and parameters. A variable, as the name  implies, is a measurable quantity that may vary or is subject to change. Variables can be control- lable or uncontrollable. A controllable variable is also called a decision variable. An example  would be how many inventory items to order. A parameter is a measurable quantity that is  inherent in the problem. The cost of placing an order for more inventory items is an example of  a parameter. In most cases, variables are unknown quantities, while parameters are known quan- tities. All models should be developed carefully. They should be solvable, realistic, and easy to  understand and modify, and the required input data should be obtainable. The model developer  has to be careful to include the appropriate amount of detail to be solvable yet realistic.

Acquiring Input Data Once we have developed a model, we must obtain the data that are used in the model (input data). Obtaining accurate data for the model is essential; even if the model is a perfect represen- tation of reality, improper data will result in misleading results. This situation is called garbage in, garbage out. For a larger problem, collecting accurate data can be one of the most difficult  steps in performing quantitative analysis.

There are a number of sources that can be used in collecting data. In some cases, company  reports and documents can be used to obtain the necessary data. Another source is interviews  with employees or other persons related to the firm. These individuals can sometimes provide  excellent information, and their experience and judgment can be invaluable. A production super- visor, for example, might be able to tell you with a great degree of accuracy the amount of time  it takes to produce a particular product. Sampling and direct measurement provide other sources  of data for the model. You may need to know how many pounds of raw material are used in  producing a new photochemical product. This information can be obtained by going to the plant  and actually measuring with scales the amount of raw material that is being used. In other cases,  statistical sampling procedures can be used to obtain data.

Developing a Solution Developing a solution involves manipulating the model to arrive at the best (optimal) solution  to the problem. In some cases, this requires that an equation be solved for the best decision. In  other cases, you can use a trial and error method, trying various approaches and picking the one  that results in the best decision. For some problems, you may wish to try all possible values for 

Operations research and Oil spills

Operations researchers and decision scientists have been in- vestigating oil spill response and alleviation strategies since long before the BP oil spill disaster of 2010 in the Gulf of Mexico. A four-phase classification system has emerged for disaster re- sponse research: mitigation, preparedness, response, and recov- ery. Mitigation means reducing the probability that a disaster will occur and implementing robust, forward-thinking strategies to reduce the effects of a disaster that does occur. Preparedness is any and all organization efforts that happen a priori to a disaster. Response is the location, allocation, and overall coordination of resources and procedures during the disaster that are aimed at preserving life and property. Recovery is the set of actions taken

to minimize the long-term impacts of a particular disaster after the immediate situation has stabilized.

Many quantitative tools have helped in areas of risk analysis, insurance, logistical preparation and supply management, evacu- ation planning, and development of communication systems. Recent research has shown that while many strides and discover- ies have been made, much research is still needed. Certainly each of the four disaster response areas could benefit from additional research, but recovery seems to be of particular concern and per- haps the most promising for future research.

source: Based on N. Altay and W. Green. “OR/MS Research in Disaster   Operations Management,” European Journal of Operational Research 175, 1  (2006): 475–493.

in aCtiOn

Garbage in, garbage out means that improper data will result in misleading results.

M01_REND9327_12_SE_C01.indd 23 10/02/14 1:19 PM

24  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

the variables in the model to arrive at the best decision. This is called complete enumeration.  This book also shows you how to solve very difficult and complex problems by repeating a few  simple steps until you find the best solution. A series of steps or procedures that are repeated is  called an algorithm, named after Algorismus, an Arabic mathematician of the ninth century.

The accuracy of a solution depends on the accuracy of the input data and the model. If the  input data are accurate to only two significant digits, then the results can be accurate to only two  significant digits. For example, the results of dividing 2.6 by 1.4 should be 1.9, not 1.857142857.

Testing the Solution Before a solution can be analyzed and implemented, it needs to be tested completely. Because  the solution depends on the input data and the model, both require testing.

Testing the input data and the model includes determining the accuracy and completeness of  the data used by the model. Inaccurate data will lead to an inaccurate solution. There are several  ways to test input data. One method of testing the data is to collect additional data from a differ- ent source. If the original data were collected using interviews, perhaps some additional data can  be collected by direct measurement or sampling. These additional data can then be compared  with the original data, and statistical tests can be employed to determine whether there are dif- ferences between the original data and the additional data. If there are significant differences,  more effort is required to obtain accurate input data. If the data are accurate but the results are  inconsistent with the problem, the model may not be appropriate. The model can be checked to  make sure that it is logical and represents the real situation.

Although most of the quantitative techniques discussed in this book have been computer- ized, you will probably be required to solve a number of problems by hand. To help detect both  logical and computational mistakes, you should check the results to make sure that they are con- sistent with the structure of the problem. For example, (1.96)(301.7) is close to (2)(300), which  is equal to 600. If your computations are significantly different from 600, you know you have  made a mistake.

Analyzing the Results and Sensitivity Analysis Analyzing the results starts with determining the implications of the solution. In most cases, a  solution to a problem will result in some kind of action or change in the way an organization is  operating. The implications of these actions or changes must be determined and analyzed before  the results are implemented.

Because a model is only an approximation of reality, the sensitivity of the solution to  changes in the model and input data is a very important part of analyzing the results. This type  of analysis is called sensitivity analysis or postoptimality analysis. It determines how much the  solution will change if there were changes in the model or the input data. When the solution is  sensitive to changes in the input data and the model specification, additional testing should be  performed to make sure that the model and input data are accurate and valid. If the model or data  are wrong, the solution could be wrong, resulting in financial losses or reduced profits.

The importance of sensitivity analysis cannot be overemphasized. Because input data may  not always be accurate or model assumptions may not be completely appropriate, sensitivity  analysis can become an important part of the quantitative analysis approach. Most of the chap- ters in the book cover the use of sensitivity analysis as part of the decision-making and problem- solving process.

Implementing the Results The final step is to implement the results. This is the process of incorporating the solution into  the company. This can be much more difficult than you would imagine. Even if the solution is  optimal and will result in millions of dollars in additional profits, if managers resist the new so- lution, all of the efforts of the analysis are of no value. Experience has shown that a large number  of quantitative analysis teams have failed in their efforts because they have failed to implement a  good, workable solution properly.

After the solution has been implemented, it should be closely monitored. Over time, there  may be numerous changes that call for modifications of the original solution. A changing econ- omy, fluctuating demand, and model enhancements requested by managers and decision makers  are only a few examples of changes that might require the analysis to be modified.

The input data and model determine the accuracy of the solution.

Testing the data and model is done before the results are analyzed.

Sensitivity analysis determines how the solutions will change with a different model or input data.

M01_REND9327_12_SE_C01.indd 24 10/02/14 1:19 PM

1.4 the QuantItatIve analysIs approaCh  25

Defining the Problem Founded in 1969, the Finn-Power Group is Scandinavia’s largest machine tool manufacturer, exporting about 88% of its products to more than 50 countries. One of Finn-Power’s leading sectors is machinery automation, developed in the company’s northern Italian facility. While delivering very high-quality out- puts, the machines had to be configured in more than 60,000 different ways to accommodate customers’ needs, and the need for successive modifications based on after sale requests created substantial optimiza- tion problems and delays in product delivery.

Developing a Model In 1999, Finn-Power began to study the introduction of sophisticated planning bills to produce more ac- curate forecasts about its components’ needs. The purpose of the planning bills was to simplify the master production scheduling (MPS) and the requirements of input materials.

Acquiring input Data The input data required consisted of specific details about the components, such as whether they were common to a set of products (modular bills) or specific to a single one. Bills were based both on historical data from previous sales and on estimates about the probability of use for each component.

Developing a solution In the initial solution, planning bills were implemented, and the company was able to achieve a substantial reduction of the items that required individual estimates, therefore reducing overall time. A two-level pro- duction schedule to streamline the production was also introduced.

testing the solution The planning bills’ solution was tested in the Italian subsidiary operations. Salespeople collected orders from customers, and requests for modifications were passed to the designers and buyers to be implemented.

Analyzing the results The first test was not successful as the process of updating the planning bills was not carried out with the necessary clarity of objectives. Also, the reports produced were incomplete and hard to read, and they did not convey a real picture of the modifications actually required. As a result, the company failed to deliver the scheduled models in time and in some cases had to rework some of the components. A revised model was therefore proposed to address these shortcomings.

implementing the results The revised model, which enhanced product modularity, finally yielded the desired results. It dramati- cally improved the accuracy of forecasts, streamlined the production process as originally intended, and significantly augmented the number of on-time deliveries from 38% in 1999 to 80% in 2002. Also, it significantly reduced the value of the obsolete stock by 62.5%, resulting in huge savings and improved performance.

source: P. Danese and P. Romano. “Finn-Power Italia Develops and Implements a Method to Cope with High Product Variety  and Frequent Modifications,” Interfaces 35, 6 (November–December 2005): 449–459.

MOdeling in the real WOrld railroad uses Optimization Models to save Millions

Defining the Problem

Developing a Model

Acquiring Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

M01_REND9327_12_SE_C01.indd 25 10/02/14 1:19 PM

26  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

The Quantitative Analysis Approach and Modeling in the Real World The quantitative analysis approach is used extensively in the real world. These steps, first seen in  Figure 1.1 and described in this section, are the building blocks of any successful use of quanti- tative analysis. As seen in our first Modeling in the Real World box, the steps of the quantitative  analysis approach can be used to help a large company such as CSX plan for critical scheduling  needs now and for decades into the future. Throughout this book, you will see how the steps of  the quantitative analysis approach are used to help countries and companies of all sizes save  millions of dollars, plan for the future, increase revenues, and provide higher-quality products  and services. The Modeling in the Real World boxes in every chapter will demonstrate to you the  power and importance of quantitative analysis in solving real problems for real organizations.  Using the steps of quantitative analysis, however, does not guarantee success. These steps must  be applied carefully.

1.5 How to Develop a Quantitative Analysis Model

Developing a model is an important part of the quantitative analysis approach. Let’s see how we  can use the following mathematical model, which represents profit:

Profit = Revenue - Expenses

In many cases, we can express revenues as price per unit multiplied times the number of units  sold. Expenses can often be determined by summing fixed costs and variable cost. Variable cost  is often expressed as variable cost per unit multiplied times the number of units. Thus, we can  also express profit in the following mathematical model:

Profit = Revenue - 1Fixed cost + Variable cost2 Profit = 1Selling price per unit21Number of units sold2 - 3Fixed cost + 1Variable cost per unit21Number of units sold24 Profit = sX - 3 f + nX4 Profit = sX - f - nX (1-1)

where

s = selling price per unit f = fixed cost n = variable cost per unit X = number of units sold

The parameters in this model are f, n, and s, as these are inputs that are inherent in the model.  The number of units sold (X) is the decision variable of interest.

ExAMplE: priTchETT’s prEciOus TiME piEcEs We will use the Bill Pritchett clock repair shop  example to demonstrate the use of mathematical models. Bill’s company, Pritchett’s Precious  Time Pieces, buys, sells, and repairs old clocks and clock parts. Bill sells rebuilt springs for a  price per unit of $8. The fixed cost of the equipment to build the springs is $1,000. The variable  cost per unit is $3 for spring material. In this example,

s = 8 f = 1,000 n = 3

The number of springs sold is X, and our profit model becomes

Profit = +8X - +1,000 - +3X

If sales are 0, Bill will realize a $1,000 loss. If sales are 1,000 units, he will realize a profit  of +4,000 1+4,000 = 1+8211,0002 - +1,000 - 1+3211,00022. See if you can determine the  profit for other values of units sold.

Expenses include fixed and variable costs.

M01_REND9327_12_SE_C01.indd 26 10/02/14 1:19 PM

1.5 how to develop a QuantItatIve analysIs Model  27

In addition to the profit models shown here, decision makers are often interested in the  break-even point (BEP). The BEP is the number of units sold that will result in $0 profits. We  set profits equal to $0 and solve for X, the number of units at the BEP:

0 = sX - f - nX

This can be written as

0 = 1s - n2X - f Solving for X, we have

f = 1s - n2X

X = f

s - n

This quantity (X) that results in a profit of zero is the BEP, and we now have this model for the BEP:

BEP = Fixed cost

1Selling price per unit2 - 1Variable cost per unit2

BEP = f

s - n (1-2)

For the Pritchett’s Precious Time Pieces example, the BEP can be computed as follows:

BEP = +1,000>1+8 - +32 = 200 units, or springs, at the BEP. The Advantages of Mathematical Modeling There are a number of advantages of using mathematical models:

1. Models can accurately represent reality. If properly formulated, a model can be extremely  accurate. A valid model is one that is accurate and correctly represents the problem or sys- tem under investigation. The profit model in the example is accurate and valid for many  business problems.

2. Models can help a decision maker formulate problems. In the profit model, for example,  a decision maker can determine the important factors or contributors to revenues and   expenses, such as sales, returns, selling expenses, production costs, and transportation costs.

3. Models can give us insight and information. For example, using the profit model from the  preceding section, we can see what impact changes in revenues and expenses will have on  profits. As discussed in the previous section, studying the impact of changes in a model,  such as a profit model, is called sensitivity analysis.

4. Models can save time and money in decision making and problem solving. It usually takes  less time, effort, and expense to analyze a model. We can use a profit model to analyze the  impact of a new marketing campaign on profits, revenues, and expenses. In most cases,   using models is faster and less expensive than actually trying a new marketing campaign in  a real business setting and observing the results.

5. A model may be the only way to solve some large or complex problems in a timely fash- ion. A large company, for example, may produce literally thousands of sizes of nuts, bolts,  and fasteners. The company may want to make the highest profits possible given its manu- facturing constraints. A mathematical model may be the only way to determine the highest  profits the company can achieve under these circumstances.

6. A model can be used to communicate problems and solutions to others. A decision analyst  can share his or her work with other decision analysts. Solutions to a mathematical model  can be given to managers and executives to help them make final decisions.

Mathematical Models Categorized by Risk Some mathematical models, like the profit and break-even models previously discussed, do not  involve risk or chance. We assume that we know all values used in the model with complete cer- tainty. These are called deterministic models. A company, for example, might want to minimize 

The BEP results in $0 profits.

Deterministic means with complete certainty.

M01_REND9327_12_SE_C01.indd 27 10/02/14 1:19 PM

28  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

manufacturing costs while maintaining a certain quality level. If we know all these values with  certainty, the model is deterministic.

Other models involve risk or chance. For example, the market for a new product might be  “good” with a chance of 60% (a probability of 0.6) or “not good” with a chance of 40% (a prob- ability of 0.4). Models that involve chance or risk, often measured as a probability value, are  called probabilistic models. In this book, we will investigate both deterministic and probabilis- tic models.

1.6 The Role of Computers and Spreadsheet Models in the Quantitative Analysis Approach

Developing a solution, testing the solution, and analyzing the results are important steps in the  quantitative analysis approach. Because we will be using mathematical models, these steps   require mathematical calculations. Excel 2013 can be used to help with these calculations, and  some spreadsheets developed in Excel will be shown in some chapters. However, some of the  techniques presented in this book require sophisticated spreadsheets and are quite tedious to   develop. Fortunately, there are two software programs available from the Companion Website  for this book that makes this much easier. These are:

1. POM-QM for Windows is an easy-to-use decision support program that was developed  POM and quantitative methods (QM) courses. POM for Windows and QM for Windows  were originally separate software packages for each type of course. These are now com- bined into one program called POM-QM for Windows. As seen in Program 1.1, it is   possible to display all the modules, only the POM modules, or only the QM modules. The  images shown in this textbook will typically display only the QM modules. Hence, in this  book, reference will usually be made to QM for Windows.  Appendix E at the end of the  book provides more information about QM for Windows.

To use QM for Windows to solve the break-even problem presented earlier, from  the Module drop-down menu select Breakeven/Cost-Volume Analysis. Then select New- Breakeven Analysis to enter the problem. When the window opens, enter a name for the  problem and select OK. Upon doing this, you will see the screen shown in Program 1.2A.  The solution is shown in Program 1.2B. Notice the additional output available from the  Window drop-down menu.

Program 1.1 The QM for Windows Main Menu

Select a module from the drop-down menu

To see the modules relevant for this book, select Display QM Modules

M01_REND9327_12_SE_C01.indd 28 10/02/14 1:19 PM

1.6 the role of CoMputers and spreadsheet Models In the QuantItatIve analysIs approaCh  29

Program 1.2a Entering the Data for pritchett’s precious Time pieces Example into QM for Windows Click Solve to runthe program.

Enter the data.

Program 1.2b QM for Windows solution screen for pritchett’s precious Time pieces Example

Additional output is available from the Window menu.

Files for the QM for Windows examples throughout the book can be downloaded from  the Companion Website. Opening these files will demonstrate how data are input for the  various modules of QM for Windows.

2. Excel QM, an add-in for Excel, can also be used to perform the mathematical calculations for the techniques discussed in each chapter. When installed, in Excel 2013, Excel QM  will appear as a tab on the ribbon. From this tab, the appropriate model can be selected  from a menu as shown in Program 1.3. Appendix F has more information about this.  Excel files with the example problems shown can be downloaded from the Companion  Website.

To use Excel QM in Excel 2013 to solve the break-even problem presented earlier,  from the Alphabetical Menu (see Program 1.3) select Breakeven Analysis. When this is  done, a worksheet is prepared automatically and the user simply inputs the fixed cost,   variable cost, and revenue (selling price per unit) as shown in Program 1.4. The solution  is calculated when all the inputs have been entered.

Excel 2013 contains some functions, special features, formulas, and tools that help  with some of the questions that might be posed in analyzing a business problem. Once such  feature, Goal Seek, is shown in Program 1.5 as it is applied to the break-even example.  Excel 2013 also has some add-ins that must be activated before using them the first time.  These include the Data Analysis add-in and the Solver add-in, which will be discussed in  later chapters.

M01_REND9327_12_SE_C01.indd 29 10/02/14 1:19 PM

30  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

Program 1.4 Entering the Data for pritchett’s precious Time pieces Example into Excel QM in Excel 2013

The problem data is entered here.

The results are shown here.

Program 1.3 Excel QM in Excel 2013 ribbon and Menu of Techniques

Select the Excel QM tab.

Select the Alphabetical menu to see the techniques.

M01_REND9327_12_SE_C01.indd 30 10/02/14 1:19 PM

1.7 possIBle proBleMs In the QuantItatIve analysIs approaCh  31

Program 1.5 using Goal seek in the Break-Even problem to Achieve a specified profit

From the Data tab, select What-if Analysis. From the menu that drops down, select Goal Seek.

If the goal is $175 pro�t (B23), and this is obtained by changing the volume (B13), the Goal Seek window inputs are these.

1.7 Possible Problems in the Quantitative Analysis Approach

We have presented the quantitative analysis approach as a logical, systematic means of tackling  decision-making problems. Even when these steps are followed carefully, there are many diffi- culties that can hurt the chances of implementing solutions to real-world problems. We now take  a look at what can happen during each of the steps.

Defining the Problem One view of decision makers is that they sit at a desk all day long, waiting until a problem  arises, and then stand up and attack the problem until it is solved. Once it is solved, they sit  down,  relax, and wait for the next big problem. In the worlds of business, government, and 

Major league Operations research at the department of agriculture

in 1997, the Pittsburgh Pirates signed Ross Ohlendorf because of his 95-mph sinking fastball. Little did they know that Ross pos- sessed operations research skills also worthy of national merit. Ross Ohlendorf had graduated from Princeton University with a 3.8 GPA in operations research and financial engineering.

Indeed, after the 2009 baseball season, when Ross applied for an 8-week unpaid internship with the U.S. Department of Agriculture, he didn’t need to mention his full-time employer because the Secretary of the Department of Agriculture at the time, Tom Vilsack, was born and raised in Pittsburgh and was an

avid Pittsburgh Pirates fan. Ross spent 2 months of the ensuing off- season utilizing his educational background in operations research, helping the Department of Agriculture track disease migration in livestock, a subject Ross has a vested interest in as his family runs a cattle ranch in Texas. Moreover, when ABC News asked Ross about his off-season unpaid internship experience, he replied, “This one’s been, I’d say, the most exciting off-season I’ve had.”

source: Rick Klein, “Ross Ohlendorf: From Major League Pitcher to Unpaid  Intern,” ABCnews.go.com, December 15, 2009.

in aCtiOn

M01_REND9327_12_SE_C01.indd 31 10/02/14 1:19 PM

32  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

education, problems are, unfortunately, not easily identified. There are four potential roadblocks  that quantitative analysts face in defining a problem. We use an application, inventory analysis,  throughout this section as an example.

cOnflicTinG ViEWpOinTs The first difficulty is that quantitative analysts must often consider  conflicting viewpoints in defining the problem. For example, there are at least two views that  managers take when dealing with inventory problems. Financial managers usually feel that in- ventory is too high, as inventory represents cash not available for other investments. Sales man- agers, on the other hand, often feel that inventory is too low, as high levels of inventory may  be needed to fill an unexpected order. If analysts assume either one of these statements as the  problem definition, they have essentially accepted one manager’s perception and can expect re- sistance from the other manager when the “solution” emerges. So it’s important to consider both  points of view before stating the problem. Good mathematical models should include all perti- nent information. As we shall see in Chapter 6, both of these factors are included in inventory  models.

iMpAcT On OThEr DEpArTMEnTs The next difficulty is that problems do not exist in isolation  and are not owned by just one department of a firm. Inventory is closely tied with cash flows and  various production problems. A change in ordering policy can seriously hurt cash flows and up- set production schedules to the point that savings on inventory are more than offset by increased  costs for finance and production. The problem statement should thus be as broad as possible and  include input from all departments that have a stake in the solution. When a solution is found,  the benefits to all areas of the organization should be identified and communicated to the people  involved.

BEGinninG AssuMpTiOns The third difficulty is that people have a tendency to state problems  in terms of solutions. The statement that inventory is too low implies a solution that inventory  levels should be raised. The quantitative analyst who starts off with this assumption will prob- ably indeed find that inventory should be raised. From an implementation standpoint, a “good”  solution to the right problem is much better than an “optimal” solution to the wrong problem.  If a problem has been defined in terms of a desired solution, the quantitative analyst should ask  questions about why this solution is desired. By probing further, the true problem will surface  and can be defined properly.

sOluTiOn OuTDATED Even with the best of problem statements, however, there is a fourth  danger. The problem can change as the model is being developed. In our rapidly changing busi- ness environment, it is not unusual for problems to appear or disappear virtually overnight. The  analyst who presents a solution to a problem that no longer exists can’t expect credit for provid- ing timely help. However, one of the benefits of mathematical models is that once the original  model has been developed, it can be used over and over again whenever similar problems arise.  This allows a solution to be found very easily in a timely manner.

Developing a Model fiTTinG ThE TExTBOOk MODEls One problem in developing quantitative models is that a  manager’s perception of a problem won’t always match the textbook approach. Most inventory  models involve minimizing the total of holding and ordering costs. Some managers view these  costs as unimportant; instead, they see the problem in terms of cash flow, turnover, and levels of  customer satisfaction. Results of a model based on holding and ordering costs are probably not  acceptable to such managers. This is why the analyst must completely understand the model  and not simply use the computer as a “black box” where data are input and results are given  with no understanding of the process. The analyst who understands the process can explain to  the manager how the model does consider these other factors when estimating the different  types of inventory costs. If other factors are important as well, the analyst can consider these  and use sensitivity analysis and good judgment to modify the computer solution before it is  implemented.

unDErsTAnDinG ThE MODEl A second major concern involves the trade-off between the  complexity of the model and ease of understanding. Managers simply will not use the results 

All viewpoints should be considered before formally defining the problem.

An optimal solution to the wrong problem leaves the real problem unsolved.

M01_REND9327_12_SE_C01.indd 32 10/02/14 1:19 PM

1.7 possIBle proBleMs In the QuantItatIve analysIs approaCh  33

of a model they do not understand. Complex problems, though, require complex models. One  trade-off is to simplify assumptions in order to make the model easier to understand. The model  loses some of its reality but gains some acceptance by management.

One simplifying assumption in inventory modeling is that demand is known and constant.  This means that probability distributions are not needed and it allows us to build simple, easy- to-understand models. Demand, however, is rarely known and constant, so the model we build  lacks some reality. Introducing probability distributions provides more realism but may put  comprehension beyond all but the most mathematically sophisticated managers. One approach  is for the quantitative analyst to start with the simple model and make sure that it is completely  understood. Later, more complex models can be introduced slowly as managers gain more con- fidence in using the new approach. Explaining the impact of the more sophisticated models  (e.g., carrying extra inventory called safety stock) without going into complete mathematical  details is sometimes helpful. Managers can understand and identify with this concept, even  if the specific mathematics used to find the appropriate quantity of safety stock is not totally  understood.

Acquiring Input Data Gathering the data to be used in the quantitative approach to problem solving is often not a  simple task. One-fifth of all firms in a recent study had difficulty with data access.

usinG AccOunTinG DATA One problem is that most data generated in a firm come from basic  accounting reports. The accounting department collects its inventory data, for example, in terms  of cash flows and turnover. But quantitative analysts tackling an inventory problem need to col- lect data on holding costs and ordering costs. If they ask for such data, they may be shocked to  find that the data were simply never collected for those specified costs.

Professor Gene Woolsey tells a story of a young quantitative analyst sent down to account- ing to get “the inventory holding cost per item per day for part 23456/AZ.” The accountant asked  the young man if he wanted the first-in, first-out figure, the last-in, first-out figure, the lower of  cost or market figure, or the “how-we-do-it” figure. The young man replied that the inventory  model required only one number. The accountant at the next desk said, “Hell, Joe, give the kid a  number.” The kid was given a number and departed.

VAliDiTy Of DATA A lack of “good, clean data” means that whatever data are available must  often be distilled and manipulated (we call it “fudging”) before being used in a model. Unfortu- nately, the validity of the results of a model is no better than the validity of the data that go into  the model. You cannot blame a manager for resisting a model’s “scientific” results when he or  she knows that questionable data were used as input. This highlights the importance of the ana- lyst understanding other business functions so that good data can be found and evaluated by the  analyst. It also emphasizes the importance of sensitivity analysis, which is used to determine the  impact of minor changes in input data. Some solutions are very robust and would not change at  all for certain changes in the input data.

Developing a Solution hArD-TO-unDErsTAnD MAThEMATics The first concern in developing solutions is that al- though the mathematical models we use may be complex and powerful, they may not be com- pletely understood. Fancy solutions to problems may have faulty logic or data. The aura of  mathematics often causes managers to remain silent when they should be critical. The well- known operations researcher C. W. Churchman cautions that “because mathematics has been so  revered a discipline in recent years, it tends to lull the unsuspecting into believing that he who  thinks elaborately thinks well.”1

Only OnE AnsWEr is liMiTinG The second problem is that quantitative models usually give  just one answer to a problem. Most managers would like to have a range of options and not be 

Obtaining accurate input data can be very difficult.

1C. W. Churchman. “Relativity Models in the Social Sciences,” Interfaces 4, 1 (November 1973).

Hard-to-understand mathematics and one answer can be a problem in developing a solution.

M01_REND9327_12_SE_C01.indd 33 10/02/14 1:19 PM

34  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

put in a take-it-or-leave-it position. A more appropriate strategy is for an analyst to present a  range of options, indicating the effect that each solution has on the objective function. This gives  managers a choice as well as information on how much it will cost to deviate from the optimal  solution. It also allows problems to be viewed from a broader perspective, since nonquantitative  factors can be considered.

Testing the Solution The results of quantitative analysis often take the form of predictions of how things will work  in the future if certain changes are made now. To get a preview of how well solutions will re- ally work, managers are often asked how good the solution looks to them. The problem is that  complex models tend to give solutions that are not intuitively obvious. Such solutions tend to be  rejected by managers. The quantitative analyst now has the chance to work through the model  and the assumptions with the manager in an effort to convince the manager of the validity of the  results. In the process of convincing the manager, the analyst will have to review every assump- tion that went into the model. If there are errors, they may be revealed during this review. In  addition, the manager will be casting a critical eye on everything that went into the model, and if  he or she can be convinced that the model is valid, there is a good chance that the solution results  are also valid.

Analyzing the Results Once a solution has been tested, the results must be analyzed in terms of how they will affect  the total organization. You should be aware that even small changes in organizations are often  difficult to bring about. If the results indicate large changes in organization policy, the quantita- tive analyst can expect resistance. In analyzing the results, the analyst should ascertain who must  change and by how much, if the people who must change will be better or worse off, and who  has the power to direct the change.

Assumptions should be reviewed.

platO helps 2004 Olympic games in athens

The 2004 Olympic Games were held in Athens, Greece, over a period of 16 days. More than 2,000 athletes competed in 300 events in 28 sports. The events were held in 36 different venues (stadia, competition centers, etc.), and 3.6 million tickets were sold to people who would view these events. In addition, 2,500 members of international committees and 22,000 journalists and broadcasters attended these games. Home viewers spent more than 34 billion hours watching these sporting events. The 2004 Olympic Games was the biggest sporting event in the history of the world up to that point.

In addition to the sporting venues, other noncompetitive ven- ues, such as the airport and Olympic village, had to be considered. A successful Olympics requires tremendous planning for the transpor- tation system that will handle the millions of spectators. Three years of work and planning were needed for the 16 days of the Olympics.

The Athens Olympic Games Organizing Committee (ATHOC) had to plan, design, and coordinate systems that would be de- livered by outside contractors. ATHOC personnel would later be responsible for managing the efforts of volunteers and paid staff during the operations of the games. To make the Athens Olym- pics run efficiently and effectively, the Process Logistics Advanced Technical Optimization (PLATO) project was begun. Innovative

techniques from management science, systems engineering, and information technology were used to change the planning, de- sign, and operations of venues.

The objectives of PLATO were to (1) facilitate effective orga- nizational transformation, (2) help plan and manage resources in a cost-effective manner, and (3) document lessons learned so future Olympic committees could benefit. The PLATO project de- veloped business-process models for the various venues, devel- oped simulation models that enable the generation of what-if scenarios, developed software to aid in the creation and manage- ment of these models, and developed process steps for training ATHOC personnel in using these models. Generic solutions were developed so that this knowledge and approach could be made available to other users.

PLATO was credited with reducing the cost of the 2004 Olym- pics by over $69 million. Perhaps even more important is the fact that the Athens games were universally deemed an unqualified success. The resulting increase in tourism is expected to result in economic benefit to Greece for many years in the future.

source: Based on D. A. Beis et al. PLATO Helps Athens Win Gold: Olympic  Games Knowledge Modeling for Organizational Change and Resource Man- agement,” Interfaces 36, 1 (January–February 2006): 26–42.

in aCtiOn

M01_REND9327_12_SE_C01.indd 34 10/02/14 1:19 PM

suMMary  35

1.8 Implementation—Not Just the Final Step

We have just presented some of the many problems that can affect the ultimate acceptance of the  quantitative analysis approach and use of its models. It should be clear now that implementation  isn’t just another step that takes place after the modeling process is over. Each one of these steps  greatly affects the chances of implementing the results of a quantitative study.

Lack of Commitment and Resistance to Change Even though many business decisions can be made intuitively, based on hunches and experience,  there are more and more situations in which quantitative models can assist. Some managers,  however, fear that the use of a formal analysis process will reduce their decision-making power.  Others fear that it may expose some previous intuitive decisions as inadequate. Still others just  feel uncomfortable about having to reverse their thinking patterns with formal decision making.  These managers often argue against the use of quantitative methods.

Many action-oriented managers do not like the lengthy formal decision-making process and  prefer to get things done quickly. They prefer “quick and dirty” techniques that can yield imme- diate results. Once managers see some QA results that have a substantial payoff, the stage is set  for convincing them that quantitative analysis is a beneficial tool.

We have known for some time that management support and user involvement are critical  to the successful implementation of quantitative analysis projects. A Swedish study found that  only 40% of projects suggested by quantitative analysts were ever implemented. But 70% of the  quantitative projects initiated by users, and fully 98% of projects suggested by top managers,  were implemented.

Lack of Commitment by Quantitative Analysts Just as managers’ attitudes are to blame for some implementation problems, analysts’ attitudes  are to blame for others. When the quantitative analyst is not an integral part of the department  facing the problem, he or she sometimes tends to treat the modeling activity as an end in itself.  That is, the analyst accepts the problem as stated by the manager and builds a model to solve  only that problem. When the results are computed, he or she hands them back to the manager  and considers the job done. The analyst who does not care whether these results help make the  final decision is not concerned with implementation.

Successful implementation requires that the analyst not tell the users what to do, but work  with them and take their feelings into account. An article in Operations Research describes an  inventory control system that calculated reorder points and order quantities. But instead of in- sisting that computer-calculated quantities be ordered, a manual override feature was installed.  This allowed users to disregard the calculated figures and substitute their own. The override was  used quite often when the system was first installed. Gradually, however, as users came to real- ize that the calculated figures were right more often than not, they allowed the system’s figures  to stand. Eventually, the override feature was used only in special circumstances. This is a good  example of how good relationships can aid in model implementation.

Management support and user involvement are important.

Summary

Quantitative analysis is a scientific approach to decision mak- ing. The quantitative analysis approach includes defining a  problem, developing a model, acquiring input data, develop- ing a solution, testing the solution, analyzing the results, and  implementing the results. In using the quantitative approach,  however, there can be potential problems, including conflicting  viewpoints, the impact of quantitative analysis models on other 

departments, beginning assumptions, outdated solutions, fitting  textbook models, understanding the model, acquiring good in- put data, hard-to-understand mathematics, obtaining only one  answer, testing the solution, and analyzing the results. In using  the quantitative analysis approach, implementation is not the  final step. There can be a lack of commitment to the approach  and resistance to change.

M01_REND9327_12_SE_C01.indd 35 10/02/14 1:19 PM

36  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

Glossary

Algorithm  A set of logical and mathematical operations   performed in a specific sequence.

Break-Even Point  The quantity of sales that results in zero  profit.

Business Analytics  A data-driven approach to decision   making that allows companies to make better decisions.

Descriptive Analytics  The study and consolidation of his- torical data to describe how a company has performed in  the past is performing now.

Deterministic Model  A model in which all values used in  the model are known with complete certainty.

Input Data  Data that are used in a model in arriving at the  final solution.

Mathematical Model  A model that uses mathematical equa- tions and statements to represent the relationships within  the model.

Model  A representation of reality or of a real-life situation. Parameter  A measurable input quantity that is inherent in a 

problem.

Predictive Analytics  The use of techniques to forecast how  things will be in the future based on patterns of past data.

Prescriptive Analytics  The use of optimization methods to  provide new and better ways to operate based on specific  business objectives.

Probabilistic Model  A model in which all values used in the  model are not known with certainty but rather involve some  chance or risk, often measured as a probability value.

Problem  A statement, which should come from a manager,  that indicates a problem to be solved or an objective or a  goal to be reached.

Quantitative Analysis or Management science  A scientific  approach that uses quantitative techniques as a tool in deci- sion making.

Sensitivity Analysis  A process that involves determining  how sensitive a solution is to changes in the formulation of  a problem.

Variable  A measurable quantity that is subject to change.

Key Equations

 (1-1)  Profit = sX - f - nX

where

s = selling price per unit

f = fixed cost

n = variable cost per unit

X = number of units sold

An equation to determine profit as a function of the sell- ing price per unit, fixed costs, variable costs, and num- ber of units sold.

 (1-2)  BEP = f

s - n

An equation to determine the break-even point (BEP) in  units as a function of the selling price per unit (s), fixed  costs (f), and variable costs (n).

Self-Test

●  Before taking the self-test, refer to the learning objectives at the beginning of the chapter, the notes in the margins, and the  glossary at the end of the chapter.

● Use the key at the back of the book to correct your answers. ●  Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

  1.  In analyzing a problem, you should normally study a. the qualitative aspects. b. the quantitative aspects. c. both a and b. d. neither a nor b.   2.  Quantitative analysis is a. a logical approach to decision making. b. a rational approach to decision making. c. a scientific approach to decision making. d. all of the above.

  3.  Qualitative factors a. are less important than quantitative factors. b. are difficult or impossible to quantify. c. were first developed by Frederick W. Taylor. d. all of the above.   4.  An input (such as variable cost per unit or fixed cost) for 

a model is an example of a. a decision variable. b. a parameter. c. an algorithm. d. a stochastic variable.

M01_REND9327_12_SE_C01.indd 36 10/02/14 1:19 PM

dIsCussIon QuestIons and proBleMs  37

  5.  The point at which the total revenue equals total cost  (meaning zero profit) is called the

a. zero-profit solution. b. optimal-profit solution. c. break-even point. d. fixed-cost solution.   6.  The first step in a quantitative analysis approach will 

require a. acquiring input data. b. producing a scale model. c. defining the problem. d. determining the break-even point.   7.  Sensitivity analysis is most often associated with which 

step of the quantitative analysis approach? a. defining a problem b. acquiring input data c. implementing the results d. analyzing the results   8.  The values in a probabilistic model a. will always be known with complete certainty. b. can only be determined by developing a new 

 mathematical model. c. will include those that involve an element of 

 uncertainty. d. will rarely be subject to change.

  9.  A parameter is a. a measurable input quantity that is inherent 

in a problem. b. an algorithm. c. a type of computer program. d. a schematic model.  10.  An analysis to determine how much a solution would 

change if there were changes in the model or the input  data is called

a. sensitivity or postoptimality analysis. b. schematic or iconic analysis. c. futurama conditioning. d. both b and c.  11.  Descriptive analytics a. involves the use of optimisation methods. b. is concerned with the study and consolidation of 

historical data. c. is mainly used to predict future trends. d. none of the above.  12.  Which of the following categories of business analytics 

involves the use of optimization models? a. descriptive b. predictive c. prescriptive d. none of the above

Discussion Questions and Problems

Discussion Questions   1-1  How can quantitative analysis techniques inform 

and support decision making within organizations?

  1-2  Define quantitative analysis. What are some of the  organizations that support the use of the scientific  approach?

  1-3  What are the three categories of business analytics?

  1-4  Outline the key features of each of the seven steps in  the quantitative analysis approach.

  1-5  Briefly trace the history of quantitative analysis.  What happened to the development of quantitative  analysis during World War II?

  1-6  What  is  the difference between models  that are   deterministic  and  those  that  are  probabilistic?   Provide examples of both.

  1-7  List some sources of input data.

  1-8  What is implementation, and why is it important?

  1-9  Briefly describe and discuss your understanding of  the phrase garbage in, garbage out.

  1-10  Managers are quick to claim that quantitative ana- lysts talk to them in a jargon that does not sound  like English. List four terms that might not be un- derstood by a manager. Then explain in nontechnical  terms what each term means.

  1-11  Suggest some potential actions that a quantitative  analyst could undertake to ensure that the imple- mentation stage of a project is successful.

  1-12  Should people who will be using the results of a new  quantitative model become involved in the technical  aspects of the problem-solving procedure?

  1-13  C. W. Churchman once said that “mathematics …  tends to lull the unsuspecting into believing that he  who thinks elaborately thinks well.” Do you think  that the best QA models are the ones that are most  elaborate and complex mathematically? Why?

  1-14  What is the break-even point? What parameters are  necessary to find it?

Problems 1-15  Gina Fox has started her own company, Foxy Shirts, 

which manufactures imprinted shirts for special oc- casions. Since she has just begun this operation, she  rents the equipment from a local printing shop when  necessary. The cost of using the equipment is $350.  The materials used in one shirt cost $8, and Gina can  sell these for $15 each.

(a)   If Gina sells 20 shirts, what will her total rev- enue be? What will her total variable cost be?

(b)  How many shirts must Gina sell to break even?  What is the total revenue for this?

Note:   means the problem may be solved with QM for Windows;   means the problem may be solved with 

Excel QM; and   means the problem may be solved with QM for Windows and/or Excel QM.

M01_REND9327_12_SE_C01.indd 37 10/02/14 1:19 PM

38  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

1-16  Ray Bond sells handcrafted yard decorations at  county fairs. The variable cost to make these is $20  each, and he sells them for $50. The cost to rent a  booth at the fair is $150. How many of these must  Ray sell to break even?

1-17  Ray Bond, from Problem 1-16, is trying to find a new  supplier that will reduce his variable cost of produc- tion to $15 per unit. If he was able to succeed in re- ducing this cost, what would the break-even point be?

1-18  Katherine D’Ann is planning to finance her college  education by selling programs at the football games  for State University. There is a fixed cost of $400 for  printing these programs, and the variable cost is $3.  There is also a $1,000 fee that is paid to the univer- sity for the right to sell these programs. If Katherine  was able to sell programs for $5 each, how many  would she have to sell in order to break even?

1-19  Katherine D’Ann, from Problem 1-18, has become  concerned that sales may fall, as the team is on a ter- rible losing streak, and attendance has fallen off. In  fact, Katherine believes that she will sell only 500  programs for the next game. If it was possible to  raise the selling price of the program and still sell  500, what would the price have to be for Katherine  to break even by selling 500?

1-20  Farris Billiard Supply sells all  types of billiard  equipment, and is considering manufacturing their  own brand of pool cues. Mysti Farris, the produc- tion manager, is currently investigating the produc- tion of a standard house pool cue that should be very  popular. Upon analyzing the costs, Mysti determines  that the materials and labor cost for each cue is $25,  and the fixed cost that must be covered is $2,400 per  week. With a selling price of $40 each, how many  pool cues must be sold to break even? What would  the total revenue be at this break-even point?

1-21  Mysti Farris (see Problem 1-20) is considering rais- ing the selling price of each cue to $50 instead of  $40. If this is done while the costs remain the same,  what would  the new break-even point be? What  would the total revenue be at this break-even point?

1-22  Mysti Farris (see Problem 1-20) believes that there  is a high probability that 120 pool cues can be sold  if the selling price is appropriately set. What selling  price would cause the break-even point to be 120?

1-23  Golden Age Retirement Planners  specializes  in  providing  financial  advice  for  people planning  for a comfortable retirement. The company offers 

seminars on the important topic of retirement plan- ning. For a typical seminar, the room rental at a hotel  is $1,000, and the cost of advertising and other in- cidentals is about $10,000 per seminar. The cost of  the materials and special gifts for each attendee is  $60 per person attending the seminar. The company  charges $250 per person to attend the seminar as this  seems to be competitive with other companies in the  same business. How many people must attend each  seminar for Golden Age to break even?

1-24  A couple of entrepreneurial business students at  State University decided to put their education into  practice by developing a tutoring company for busi- ness students. While private tutoring was offered, it  was determined that group tutoring before tests in  the large statistics classes would be most beneficial.  The students rented a room close to campus for $300  for 3 hours. They developed handouts based on past  tests, and these handouts (including color graphs)  cost $5 each. The tutor was paid $25 per hour, for a  total of $75 for each tutoring session.

(a)   If students are charged $20 to attend the session,  how many students must enroll for the company  to break even?

(b)  A somewhat smaller room is available for $200  for 3 hours. The company is considering this  possibility. How would this affect the break-even  point?

  1-25  Zoe Garcia is the manager of a small office support  business that supplies copying, binding, and other  services for local companies. Zoe must replace a  worn-out copy machine that is used for black and  white copying. Two machines are being considered,  and each of these has a monthly lease cost plus a cost  for each page that is copied. Machine 1 has a monthly  lease cost of $600, and there is a cost of $0.010 per  page copied. Machine 2 has a monthly lease cost of  $400, and there is a cost of $0.015 per page copied.  Customers are charged $0.05 per page for copies.

(a) What is the break-even point for each machine? (b) If Zoe expects to make 10,000 copies per month, 

what would be the cost for each machine? (c)  If Zoe expects to make 30,000 copies per month, 

what would be the cost for each machine? (d) At what volume (the number of copies) would the 

two machines have the same monthly cost? What  would the total revenue for this number of copies?

M01_REND9327_12_SE_C01.indd 38 10/02/14 1:19 PM

BIBlIography  39

Case Study

Food and beverages at southwestern university Football games Southwestern University  (SWU),  a  large  state college  in   Stephenville, Texas, 30 miles southwest of the Dallas/Fort  Worth metroplex, enrolls close to 20,000 students. The school  is the dominant force in the small city, with more students dur- ing fall and spring than permanent residents.

A longtime football powerhouse, SWU is a member of the  Big Eleven conference and is usually in the top 20 in college  football rankings. To bolster its chances of reaching the elu- sive and long-desired number-one ranking, in 2013 SWU hired  the legendary Billy Bob Dillon as its head coach. Although the  number-one ranking remained out of reach, attendance at the  five Saturday home games each year increased. Prior to Dillon’s  arrival, attendance generally averaged 25,000–29,000. Season  ticket sales bumped up by 10,000 just with the announcement  of the new coach’s arrival. Stephenville and SWU were ready to  move to the big time!

With the growth in attendance came more fame, the need  for a bigger stadium, and more complaints about seating, park- ing, long lines, and concession stand prices. Southwestern Uni- versity’s president, Dr. Marty Starr, was concerned not only  about the cost of expanding the existing stadium versus building  a new stadium but also about the ancillary activities. He wanted  to be sure that these various support activities generated revenue  adequate to pay for themselves. Consequently, he wanted the  parking lots, game programs, and food service to all be han- dled as profit centers. At a recent meeting discussing the new  stadium, Starr told the stadium manager, Hank Maddux, to de- velop a break-even chart and related data for each of the centers.  He instructed Maddux to have the food service area break-even  report ready for the next meeting. After discussion with other  facility managers and his subordinates, Maddux developed the  following table showing the suggested selling prices, and his  estimate of variable costs, and the percent revenue by item. It  also provides an estimate of the percentage of the total revenues  that would be expected for each of the items based on historical  sales data.

Maddux’s fixed costs are interesting. He estimated that  the prorated portion of the stadium cost would be as follows:  salaries for food services at $100,000 ($20,000 for each of the  five home games); 2,400 square feet of stadium space at $2 per 

square foot per game; and six people per booth in each of the  six booths for 5 hours at $7 an hour. These fixed costs will  be proportionately allocated to each of the products based on  the percentages provided in the table. For example, the revenue  from soft drinks would be expected to cover 25% of the total  fixed costs.

Maddux wants to be sure that he has a number of things  for President Starr: (1) the total fixed cost that must be covered  at each of the games; (2) the portion of the fixed cost allocated  to each of the items; (3) what his unit sales would be at break- even for each item—that is, what sales of soft drinks, coffee,  hot dogs, and hamburgers are necessary to cover the portion of  the fixed cost allocated to each of these items; (4) what the dol- lar sales for each of these would be at these break-even points;  and (5) realistic sales estimates per attendee for attendance of  60,000 and 35,000. (In other words, he wants to know how  many dollars each attendee is spending on food at his projected  break-even sales at present and if attendance grows to 60,000.)  He felt this last piece of information would be helpful to under- stand how realistic the assumptions of his model are, and this  information could be compared with similar figures from previ- ous seasons.

discussion Question   1. Prepare a brief report with the items noted so it is ready 

for Dr. Starr at the next meeting.

Adapted from J. Heizer and B. Render. Operations Management, 6th ed. Up- per Saddle River, NJ: Prentice Hall, 2000, pp. 274–275.

Bibliography

Ackoff, R. L. Scientific Method: Optimizing Applied Research Decisions. New  York: John Wiley & Sons, Inc., 1962.

Beam, Carrie. “ASP, the Art and Science of Practice: How I Started an OR/MS  Consulting Practice with a Laptop, a Phone, and a PhD,” Interfaces 34  (July–August 2004): 265–271.

Board, John, Charles Sutcliffe, and William T. Ziemba. “Applying Operations  Research Techniques to Financial Markets,” Interfaces 33 (March–April  2003): 12–24.

Churchman, C. W. “Relativity Models in the Social Sciences,” Interfaces 4, 1  (November 1973).

Churchman, C. W. The Systems Approach. New York: Delacort Press, 1968.

Dekker, Rommert, Jacqueline Bloemhof, and Ioannis Mallidis . “Operations  Research for Green Logistics—An Overview of Aspects, Issues, Con- tributions and Challenges,” European Journal of Operational Research 219, 3 (June 2012): 671–679.

Dutta, Goutam. “Lessons for Success in OR/MS Practice Gained from Experi- ences in Indian and U.S. Steel Plants,” Interfaces 30, 5 (September– October 2000): 23–30.

Eom, Sean B., and Eyong B. Kim. “A Survey of Decision Support System  Applications (1995–2001),” Journal of the Operational Research Society 57, 11 (2006): 1264–1278.

ITEM SELLING 

PRICE/UNIT VARIABLE  COST/UNIT

PERCENT  REVENUE

Soft drink $1.50 $0.75 25%

Coffee 2.00 0.50 25

Hot dogs 2.00 0.80 20

Hamburgers 2.50 1.00 20

Misc. snacks 1.00 0.40 10

M01_REND9327_12_SE_C01.indd 39 10/02/14 1:19 PM

40  Chapter 1 • IntroduCtIon to QuantItatIve analysIs

Evans, James. R. “Business Analytics: The Next Frontier for Decision   Sciences,” Decision Line 43, 2 (March 2012).

Horowitz, Ira. “Aggregating Expert Ratings Using Preference-Neutral  Weights: The Case of the College Football Polls,” Interfaces 34  (July–August 2004): 314–320.

Keskinocak, Pinar, and Sridhar Tayur. “Quantitative Analysis for Internet- Enabled Supply Chains,” Interfaces 31, 2 (March–April 2001): 70–89.

Laval, Claude, Marc Feyhl, and Steve Kakouros. “Hewlett-Packard Combined  OR and Expert Knowledge to Design Its Supply Chains,” Interfaces 35  (May–June 2005): 238–247.

Pidd, Michael. “Just Modeling Through: A Rough Guide to Modeling,” Inter- faces 29, 2 (March–April 1999): 118–132.

Saaty, T. L. “Reflections and Projections on Creativity in Operations Research  and Management Science: A Pressing Need for a Shifting Paradigm,”  Operations Research 46, 1 (1998): 9–16.

Salveson, Melvin. “The Institute of Management Science: A Prehistory and  Commentary,” Interfaces 27, 3 (May–June 1997): 74–85.

Wright, P. Daniel, Matthew J. Liberatore, and Robert L. Nydick. “A Survey of  Operations Research Models and Applications in Homeland Security,”  Interfaces 36 (November–December 2006): 514–529.

M01_REND9327_12_SE_C01.indd 40 10/02/14 1:19 PM

 41

After completing this chapter, students will be able to:

1. Understand the basic foundations of probability analysis.

2. Describe statistically dependent and independent events.

3. Use Bayes’ theorem to establish posterior probabilities.

2.1 Introduction

2.2 Fundamental Concepts

2.3 Revising Probabilities with Bayes’ Theorem

2.4 Further Probability Revisions

2.5 Random Variables

2.6 Probability Distributions

2.7 The Binomial Distribution

2.8 The Normal Distribution

2.9 The F Distribution

2.10 The Exponential Distribution

2.11 The Poisson Distribution

4. Describe and provide examples of both discrete and continuous random variables.

5. Explain the difference between discrete and continuous probability distributions.

6. Calculate expected values and variances and use the normal table.

Summary  •  Glossary  •  Key Equations  •  Solved Problems  •  Self-Test  •  Discussion Questions and Problems  •  Internet 

Homework Problems  •  Case Study: WTVX  •  Bibliography

Appendix 2.1: Derivation of Bayes’ Theorem

ChApter Outline

leArning ObjeCtives

Probability Concepts and Applications

2ChApter

M02_REND9327_12_SE_C02.indd 41 10/02/14 1:20 PM

42  Chapter 2 • probability ConCepts and appliCations

2.1 Introduction

Life would be simpler if we knew without doubt what was going to happen in the future. The  outcome of any decision would depend only on how logical and rational the decision was. If  you lost money in the stock market, it would be because you failed to consider all the infor- mation or to make a logical decision. If you got caught in the rain, it would be because you  simply forgot your umbrella. You could always avoid building a plant that was too large, in- vesting in a company that would lose money, running out of supplies, or losing crops because  of bad weather. There would be no such thing as a risky investment. Life would be simpler,  but boring.

It wasn’t until the sixteenth century that people started to quantify risks and to apply this  concept to everyday situations. Today, the idea of risk or probability is a part of our lives. “There  is a 40% chance of rain in Omaha today.” “The Florida State University Seminoles are favored  2 to 1 over the Louisiana State University Tigers this Saturday.” “There is a 50–50 chance that  the stock market will reach an all-time high next month.”

A probability is a numerical statement about the likelihood that an event will occur. In this  chapter, we examine the basic concepts, terms, and relationships of probability and probability  distributions that are useful in solving many quantitative analysis problems. Table 2.1 lists some  of the topics covered in this book that rely on probability theory. You can see that the study of  quantitative analysis and business analytics would be quite difficult without it.

2.2 Fundamental Concepts

There are several rules, definitions, and concepts associated with probability that are very   important in understanding the use of probability in decision making. These will be briefly   presented with some examples to help clarify them.

Two Basic Rules of Probability There are two basic rules regarding the mathematics of probability:

  1. The probability, P, of any event or state of nature occurring is greater than or equal to 0 and  less than or equal to 1. That is,

0 … P1event2 … 1 (2-1)     A probability of 0 indicates that an event is never expected to occur. A probability of 

1 means that an event is always expected to occur.   2. The sum of the simple probabilities for all possible outcomes of an activity must equal 1. 

Regardless of how probabilities are determined, they must adhere to these two rules.

A probability is a numerical statement about the chance that an event will occur.

People often misuse the two basic rules of probabilities when they use statements such as, “I’m 110% sure we’re going to win the big game.”

Table 2.1 Chapters in This Book That Use Probability

Chapter title

3 Decision Analysis

4 Regression Models

5 Forecasting

6 Inventory Control Models

11 Project Management

12 Waiting Lines and Queuing Theory Models

13 Simulation Modeling

14 Markov Analysis

15 Statistical Quality Control

Module 3 Decision Theory and the Normal Distribution

Module 4 Game Theory

M02_REND9327_12_SE_C02.indd 42 10/02/14 1:20 PM

2.2 Fundamental ConCepts  43

Types of Probability There are two different ways to determine probability: the objective approach and the subjec- tive approach.

The relative frequency approach is an objective probability assessment. The probability   assigned to an event is the relative frequency of that occurrence. In general,

P1event2 = Number of occurrences of the event Total number of trials or outcomes

Here is an example. Demand for white latex paint at Diversey Paint and Supply has  always been  0, 1, 2, 3, or 4 gallons per day. (There are no other possible outcomes and when one  occurs, no  other can.) Over the past 200 working days, the owner notes the frequencies of demand as shown  in Table 2.2. If this past distribution is a good indicator of future sales, we can find the probability  of each possible outcome occurring in the future by converting the data into percentages.

Thus, the probability that sales are 2 gallons of paint on any given day is P12 gallons2 = 0.25 = 25%. The probability of any level of sales must be greater than or equal to 0 and less  than or equal to 1. Since 0, 1, 2, 3, and 4 gallons exhaust all possible events or outcomes, the  sum of their probability values must equal 1.

Objective probability can also be set using what is called the classical or logical method.  Without performing a series of trials, we can often logically determine what the probabilities of  various events should be. For example, the probability of tossing a fair coin once and getting a  head is

P1head2 = 1 2

Number of ways of getting a head

Number of possible outcomes (head or tail)

Similarly, the probability of drawing a spade out of a deck of 52 playing cards can be logically set as

P1spade2 = 13 52

Number of chances of drawing a spade

Number of possible outcomes = 1>4 = 0.25 = 25%

When logic and past history are not available or appropriate, probability values can be assessed  subjectively. The accuracy of subjective probabilities depends on the experience and judgment of  the person making the estimates. A number of probability values cannot be determined  unless the  subjective approach is used. What is the probability that the price of gasoline will be more than  $4 in the next few years? What is the probability that our economy will be in a severe depres- sion in 2020? What is the probability that you will be president of a major corporation within  20 years?

There are several methods for making subjective probability assessments. Opinion polls can  be used to help in determining subjective probabilities for possible election returns and potential  political candidates. In some cases, experience and judgment must be used in making subjec- tive assessments of probability values. A production manager, for example, might believe that  the probability of manufacturing a new product without a single defect is 0.85. In the Delphi  method, a panel of experts is assembled to make their predictions of the future. This approach is  discussed in Chapter 5.

Table 2.2 Relative Frequency Approach to Probability for Paint Sales

Quantity DemanDeD (Gallons)

number of Days

probability

0 40 0.20 1= 40>2002 1 80 0.40 1= 80>2002 2 50 0.25 1= 50>2002 3 20 0.10 1= 20>2002 4  10 0.05 1= 10>2002

Total 200 1.00 1= 200>2002

Where do probabilities come from? Sometimes they are subjective and based on personal experiences. Other times they are objectively based on logical observations such as the roll of a die. Often, probabilities are derived from historical data.

M02_REND9327_12_SE_C02.indd 43 10/02/14 1:20 PM

44  Chapter 2 • probability ConCepts and appliCations

Mutually Exclusive and Collectively Exhaustive Events Events are said to be mutually exclusive if only one of the events can occur on any one trial.  They are called collectively exhaustive if the list of outcomes includes every possible outcome.  Many common experiences involve events that have both of these properties.

In tossing a coin, the possible outcomes are a head or a tail. Since both of them cannot occur  on any one toss, the outcomes head and tail are mutually exclusive. Since obtaining a head and  obtaining a tail represent every possible outcome, they are also collectively exhaustive.

Figure 2.1 provides a Venn diagram representation of mutually exclusive events. Let A be  the event that a head is tossed, and let B be the event that a tail is tossed. The circles representing  these events do not overlap, so the events are mutually exclusive.

The following situation provides an example of events that are not mutually exclusive. You  are asked to draw one card from a standard deck of 52 playing cards. The following events are  defined:

A = event that a 7 is drawn B = event that a heart is drawn

Defining the problem The scarcity of liver organs for transplants has reached critical levels in the United States; 1,131 individu- als died in 1997 while waiting for a transplant. With only 4,000 liver donations per year, there are 10,000 patients on the waiting list, with 8,000 being added each year. There is a need to develop a model to evaluate policies for allocating livers to terminally ill patients who need them.

Developing a model Doctors, engineers, researchers, and scientists worked together with Pritsker Corp. consultants in the pro- cess of creating the liver allocation model, called ULAM. One of the model’s jobs would be to evaluate whether to list potential recipients on a national basis or regionally.

acquiring input Data Historical information was available from the United Network for Organ Sharing (UNOS), from 1990 to 1995. The data were then stored in ULAM. “Poisson” probability processes described the arrivals of donors at 63 organ procurement centers and arrival of patients at 106 liver transplant centers.

Developing a solution ULAM provides probabilities of accepting an offered liver, where the probability is a function of the pa- tient’s medical status, the transplant center, and the quality of the offered liver. ULAM also models the daily probability of a patient changing from one status of criticality to another.

testing the solution Testing involved a comparison of the model output to actual results over the 1992–1994 time period. Model results were close enough to actual results that ULAM was declared valid.

analyzing the results ULAM was used to compare more than 100 liver allocation policies and was then updated in 1998, with more recent data, for presentation to Congress.

implementing the results Based on the projected results, the UNOS committee voted 18–0 to implement an allocation policy based on regional, not national, waiting lists. This decision is expected to save 2,414 lives over an 8-year period.

source: Based on A. A. B. Pritsker. “Life and Death Decisions,” OR/MS Today (August 1998): 22–28.

MOdeling in the reAl WOrld liver transplants in the united states

Defining the Problem

Developing a Model

Acquiring Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

BA

Figure 2.1 Venn Diagram for Events That Are Mutually Exclusive

M02_REND9327_12_SE_C02.indd 44 10/02/14 1:20 PM

2.2 Fundamental ConCepts  45

The probabilities can be assigned to these using the relative frequency approach. There are four  7s in the deck and thirteen hearts in the deck. Thus, we have

P1a 7 is drawn2 = P1A2 = 4>52 P1a heart is drawn2 = P1B2 = 13>52

These events are not mutually exclusive as the 7 of hearts is common to both event A and event B.  Figure 2.2 provides a Venn diagram representing this situation. Notice that the two circles in- tersect, and this intersection is whatever is in common to both. In this example, the intersection  would be the 7 of hearts.

Unions and Intersections of Events The intersection of two events is the set of all outcomes that are common to both events. The  word and is commonly associated with the intersection, as is the symbol ¨. There are several  notations for the intersection of two events:

Intersection of event A and event B = A and B = A ¨ B = AB

The notation for the probability would be

P1Intersection of event A and event B2 = P1A and B2 = P1A ¨ B2 = P1AB2

The probability of the intersection is sometimes called a joint probability, which implies that  both events are occurring at the same time or jointly.

The union of two events is the set of all outcomes that are contained in either of these two  events. Thus, any outcome that is in event A is in the union of the two events, and any outcome  that is in event B is also in the union of the two events. The word or is commonly associated with  the union, as is the symbol  ´. Typical notation for the union of two events would be

Union of event A and event B = 1A or B2 The notation for the probability of the union of events would be

P1Union of event A and event B2 = P1A or B2 = P1A ´ B2

In the previous example, the intersection of event A and event B would be

(A and B) = the 7 of hearts is drawn

The notation for the probability would be

P1A and B2 = P17 of hearts is drawn2 = 1>52

BA

Figure 2.2 Venn Diagram for Events That Are Not Mutually Exclusive

M02_REND9327_12_SE_C02.indd 45 10/02/14 1:20 PM

46  Chapter 2 • probability ConCepts and appliCations

Also, the union of event A and event B would be

1A or B2 = 1either a 7 is drawn or a heart is drawn2 and the probability would be

P1A or B2 = P1any 7 or any heart is drawn2 = 16>52 To see why P1A or B2 = 16>52  and not 17>52  (which is P1A2 + P1B2), count all of the cards that  are in the union, and you will see there are 16. This will help you understand the general rule for  the probability of the union of two events that is presented next.

Probability Rules for Unions, Intersections, and Conditional Probabilities The general rule for the probability of the union of two events (sometimes called the additive  rule) is the following:

P1A or B2 = P1A2 + P1B2 - P1A and B2 (2-2) To illustrate this with the example we have been using, to find the probability of the union of the  two events (a 7 or a heart is drawn), we have

P1A or B2 = P1A2 + P1B2 - P1A and B2 = 4>52 + 13>52 - 1>52 = 16>52

One of the most important probability concepts in decision making is the concept of a con- ditional probability. A conditional probability is the probability of an event occurring given  that another event has already happened. The probability of event A given that event B has oc- curred is written as P1A � B2. When businesses make decisions, they often use market research  of some type to help determine the likelihood of success. Given a good result from the market  research, the probability of success would increase.

The probability that A will occur given that event B has occurred can be found by dividing  the probability of the intersection of the two events (A and B) by the probability of the event that  has occurred (B):

P1A � B2 = P1AB2 P1B2 (2-3)

From this, the formula for the intersection of two events can be easily derived and written as

P1AB2 = P1A � B2P1B2 (2-4) In the card example, what is the probability that a 7 is drawn (event A) given that we know 

that the card drawn is a heart (event B)? With what we already know, and given the formula for   conditional probability, we have

P1A � B2 = P1AB2 P1B2 =

1>52 13>52

= 1>13

With this card example, it might be possible to determine this probability without using the  formula. Given that a heart was drawn and there are 13 hearts with only one of these being a 7,  we can determine that the probability is 1>13. In business, however, we sometimes do not have  this complete information and the formula is absolutely essential.

Two events are said to be independent if the occurrence of one has no impact on the occur- rence of the other. Otherwise, the events are dependent.

For example, suppose a card is drawn from a deck of cards and it is then returned to the  deck and a second drawing occurs. The probability of drawing a seven on the second draw is  4>52 regardless of what was drawn on the first draw because the deck is exactly the same as it was on  the first draw. Now contrast to a similar situation with two draws from a deck of cards, but the  first card is not returned to the deck. Now there are only 51 cards left in the deck, and there are  either three or four 7s in the deck depending on what the first card drawn happens to be.

M02_REND9327_12_SE_C02.indd 46 10/02/14 1:20 PM

2.3 revising probabilities with bayes’ theorem  47

A more precise definition of statistical independence would be the following: Event A and  event B are independent if

P1A � B2 = P1A2 Independence is a very important condition in probability as many calculations are simplified.  One of these is the formula for the intersection of two events. If A and B are independent, then  the probability of the intersection is

P1A and B2 = P1A2P1B2 Suppose a fair coin is tossed twice. The events are defined as:

A = event that a head is the result of the first toss B = event that a head is the result of the second toss

These events are independent because the probability of a head on the second toss will be the  same regardless of the result on the first toss. Because it is a fair coin, we know there are two  equally likely outcomes on each toss (head or tail), so

P1A2 = 0.5 and

P1B2 = 0.5 Because A and B are independent,

P1AB2 = P1A2P1B2 = 0.510.52 = 0.25 Thus, there is a 0.25 probability that two tosses of a coin will result in two heads.

If events are not independent, then finding probabilities may be a bit more difficult.   However, the results may be very valuable to a decision maker. A market research study about  opening a new store in a particular location may have a positive outcome, and this would cause  a revision of our probability assessment that the new store would be successful. The next section  provides a means of revising probabilities based on new information.

2.3 Revising Probabilities with Bayes’ Theorem

Bayes’ theorem is used to incorporate additional information as it is made available and help   create revised or posterior probabilities from the original or prior probabilities. This means that  we can take new or recent data and then revise and improve upon our old probability estimates  for an event (see Figure 2.3). Let us consider the following example.

A cup contains two dice identical in appearance. One, however, is fair (unbiased) and the  other is loaded (biased). The probability of rolling a 3 on the fair die is  1�6, or 0.166. The prob- ability of tossing the same number on the loaded die is 0.60.

We have no idea which die is which, but select one by chance and toss it. The result is a 3.  Given this additional piece of information, can we find the (revised) probability that the die  rolled was fair? Can we determine the probability that it was the loaded die that was rolled?

Figure 2.3 Using Bayes’ Process

New Information

Bayes’ Process

Posterior Probabilities

Prior Probabilities

M02_REND9327_12_SE_C02.indd 47 10/02/14 1:20 PM

48  Chapter 2 • probability ConCepts and appliCations

The answer to these questions is yes, and we do so by using the formula for joint probability  under statistical dependence and Bayes’ theorem. First, we take stock of the information and  probabilities available. We know, for example, that since we randomly selected the die to roll,  the probability of it being fair or loaded is 0.50:

P1fair2 = 0.50 P1loaded2 = 0.50 We also know that

P13 � fair2 = 0.166 P13 � loaded2 = 0.60 Next, we compute joint probabilities P(3 and fair) and P(3 and loaded) using the formula  P1AB2 = P1A � B2 * P1B2:

P13 and fair2 = P13 � fair2 * P1fair2 = 10.166210.502 = 0.083

P13 and loaded2 = P13 � loaded2 * P1loaded2 = 10.60210.502 = 0.300

A 3 can occur in combination with the state “fair die” or in combination with the state “loaded  die.” The sum of their probabilities gives the unconditional or marginal probability of a 3 on the  toss, namely, P132 = 0.083 + 0.300 = 0.383.

If a 3 does occur, and if we do not know which die it came from, the probability that the die  rolled was the fair one is

P1fair � 32 = P1fair and 32 P132 =

0.083

0.383 = 0.22

The probability that the die rolled was loaded is

P1loaded � 32 = P1loaded and 32 P132 =

0.300

0.383 = 0.78

These two conditional probabilities are called the revised or posterior probabilities for the next  roll of the die.

Before the die was rolled in the preceding example, the best we could say was that there was  a 50–50 chance that it was fair (0.50 probability) and a 50–50 chance that it was loaded. After  one roll of the die, however, we are able to revise our prior probability estimates. The new  posterior estimate is that there is a 0.78 probability that the die rolled was loaded and only a 0.22  probability that it was not.

Using a table is often helpful in performing the calculations associated with Bayes’ theo- rem. Table 2.3 provides the general layout for this, and Table 2.4 provides this specific example.

Table 2.3 Tabular Form of Bayes’ Calculations Given That Event B Has Occurred

state of nature

P 1B � state of nature 2

prior probability

Joint probability

posterior probability

A P1B � A2 *P1A2 =P1B and A2 P1B and A2>P1B2 = P1A � B2 A′ P1B � A′2 *P1A′2 =P1B and A′2

P1B2 P1B and A′2>P1B2 = P1A′ � B2

Table 2.4 Bayes’ Calculations Given That a 3 Is Rolled in Example 7

state of nature

P 13 � state of nature 2

prior probability

Joint probability

posterior probability

Fair die 0.166 *0.5 = 0.083 0.083>0.383 = 0.22 Loaded die 0.600 *0.5 = 0.300

P132 = 0.383 0.300>0.383 = 0.78

M02_REND9327_12_SE_C02.indd 48 10/02/14 1:20 PM

2.4 Further probability revisions  49

General Form of Bayes’ Theorem Revised probabilities can also be computed in a more direct way using a general form for bayes’ theorem:

P1A � B2 = P1B � A2P1A2 P1B � A2P1A2 + P1B � A′2P1A′2 (2-5)

where A′ =  the complement of the event A;

for example, if A is the event “fair die,” then A′ is “loaded die”

We originally saw in Equation 2-3 the conditional probability of event A, given event B, is

P1A � B2 = P1AB2 P1B2

Thomas Bayes derived his theorem from this. Appendix 2.1 shows the mathematical steps lead- ing to Equation 2-5. Now let’s return to the example.

Although it may not be obvious to you at first glance, we used this basic equation to com- pute the revised probabilities. For example, if we want the probability that the fair die was rolled  given the first toss was a 3, namely, P(fair die  � 3 rolled), we can let

event ;fair die< replace A in Equation 2@5

event ;loaded die< replace A′ in Equation 2@5 event ;3 rolled< replace B in Equation 2@5

We can then rewrite Equation 2-5 and solve as follows:

P1fair die � 3 rolled2

= P13 � fair2P1fair2

P13 � fair2P1fair2 + P13 � loaded2P1loaded2

= 10.166210.502

10.166210.502 + 10.60210.502 =

0.083

0.383 = 0.22

This is the same answer that we computed earlier. Can you use this alternative approach to  show that P1loaded die � 3 rolled2 = 0.78? Either method is perfectly acceptable, but when we  deal with probability revisions again in Chapter 3, we may find that Equation 2-5 or the tabular   approach is easier to apply.

2.4 Further Probability Revisions

Although one revision of prior probabilities can provide useful posterior probability estimates,  additional information can be gained from performing the experiment a second time. If it is   financially worthwhile, a decision maker may even decide to make several more revisions.

Returning to the previous example, we now attempt to obtain further information about the  posterior probabilities as to whether the die just rolled is fair or loaded. To do so, let us toss the  die a second time. Again, we roll a 3. What are the further revised probabilities?

To answer this question, we proceed as before, with only one exception. The  probabilities  P1fair2 = 0.50 and P1loaded2 = 0.50 remain the same, but now we must compute P13, 3 � fair2 = 10.166210.1662 = 0.027 and P13, 3 � loaded2 = 10.6210.62 = 0.36. With these joint probabili- ties of two 3s on successive rolls, given the two types of dice, we may revise the probabilities:

P13, 3 and fair2 = P13, 3 � fair2 * P1fair2 = 10.027210.52 = 0.013

P13, 3 and loaded2 = P13, 3 � loaded2 * P1loaded2 = 10.36210.52 = 0.18

Another way to compute revised probabilities is with Bayes’ theorem.

A Presbyterian minister, Thomas Bayes (1702–1761), did the work leading to this theorem.

M02_REND9327_12_SE_C02.indd 49 10/02/14 1:20 PM

50  Chapter 2 • probability ConCepts and appliCations

Thus, the probability of rolling two 3s, a marginal probability, is 0.013 + 0.18 = 0.193, the  sum of the two joint probabilities:

P1fair � 3, 32 = P13, 3 and fair2 P13, 32

= 0.013

0.193 = 0.067

P1loaded � 3, 32 = P13, 3 and loaded2 P13, 32

= 0.18

0.193 = 0.933

What has this second roll accomplished? Before we rolled the die the first time, we knew  only that there was a 0.50 probability that it was either fair or loaded. When the first die was  rolled in the previous example, we were able to revise these probabilities:

probability the die is fair = 0.22 probability the die is loaded = 0.78

Now, after the second roll in this example, our refined revisions tell us that

probability the die is fair = 0.067 probability the die is loaded = 0.933

This type of information can be extremely valuable in business decision making.

2.5 Random Variables

We have just discussed various ways of assigning probability values to the outcomes of an ex- periment. Let us now use this probability information to compute the expected outcome, vari- ance, and standard deviation of the experiment. This can help select the best decision among a  number of alternatives.

Flight safety and probability Analysis

With the horrific events of September 11, 2001, and the use of airplanes as weapons of mass destruction, airline safety has become an even more important international issue. How can we reduce the impact of terrorism on air safety? What can be done to make air travel safer overall? One answer is to evaluate various air safety programs and to use probability theory in the analysis of the costs of these programs.

Determining airline safety is a matter of applying the con- cepts of objective probability analysis. The chance of getting killed in a scheduled domestic flight is about 1 in 5 million. This is probability of about .0000002. Another measure is the number of deaths per passenger mile flown. The number is about 1 pas- senger per billion passenger miles flown, or a probability of about .000000001. Without question, flying is safer than many other forms of transportation, including driving. For a typical weekend, more people are killed in car accidents than a typical air disaster.

Analyzing new airline safety measures involves costs and the subjective probability that lives will be saved. One airline expert proposed a number of new airline safety measures. When the

costs involved and probability of saving lives were taken into account, the result was about a $1 billion cost for every life saved on average. Using probability analysis will help determine which safety programs will result in the greatest benefit, and these pro- grams can be expanded.

In addition, some proposed safety issues are not completely certain. For example, a Thermal Neutron Analysis device to detect explosives at airports had a probability of .15 of giving a false alarm, resulting in a high cost of inspection and long flight delays. This would indicate that money should be spent on develop- ing more reliable equipment for detecting explosives. The result would be safer air travel with fewer unnecessary delays.

Without question, the use of probability analysis to deter- mine and improve flight safety is indispensable. Many transporta- tion experts hope that the same rigorous probability models used in the airline industry will some day be applied to the much more deadly system of highways and the drivers who use them.

sources: Based on Robert Machol. “Flying Scared,” OR/MS Today (October  1997): 32–37; and Arnold Barnett. “The Worst Day Ever,” OR/MS Today ( December 2001): 28–31.

in ACtiOn

M02_REND9327_12_SE_C02.indd 50 10/02/14 1:20 PM

2.5 random variables  51

A random variable assigns a real number to every possible outcome or event in an   experiment. It is normally represented by a letter such as X or Y. When the outcome itself is   numerical or quantitative, the outcome numbers can be the random variable. For example, con- sider refrigerator sales at an appliance store. The number of refrigerators sold during a given  day can be the random variable. Using X to represent this random variable, we can express this  relationship as follows:

X = number of refrigerators sold during the day

In general, whenever the experiment has quantifiable outcomes, it is beneficial to define these  quantitative outcomes as the random variable. Examples are given in Table 2.5.

When the outcome itself is not numerical or quantitative, it is necessary to define a random  variable that associates each outcome with a unique real number. Several examples are given in  Table 2.6.

There are two types of random variables: discrete random variables and continuous random variables. Developing probability distributions and making computations based on these distri- butions depends on the type of random variable.

A random variable is a discrete random variable if it can assume only a finite or limited  set of values. Which of the random variables in Table 2.5 are discrete random variables? Look- ing at Table 2.5, we can see that the variables associated with stocking 50 Christmas trees, in- specting 600 items, and sending out 5,000 letters are all examples of discrete random variables.  Each of these random variables can assume only a finite or limited set of values. The number of  Christmas trees sold, for example, can only be integer numbers from 0 to 50. There are 51 values  that the random variable X can assume in this example.

A continuous random variable is a random variable that has an infinite or an unlimited set  of values. Are there any examples of continuous random variables in Table 2.5 or 2.6? Looking 

Try to develop a few more examples of discrete random variables to be sure you understand this concept.

Table 2.5 Examples of Random Variables

experiment

outCome

ranDom Variables

ranGe of ranDom Variables

Stock 50 Christmas trees  Number of Christmas trees sold  X = number of Christmas trees sold  0, 1, 2, c, 50 Inspect 600 items  Number of acceptable items  Y = number of  acceptable items 0, 1, 2, c, 600 Send out 5,000 sales letters Number of people  responding 

to the letters Z = number of people responding 

to the letters 0, 1, 2, c, 5,000

Build an apartment building Percent of building completed  after 4 months

R = percent of building completed  after 4 months

0 … R … 100

Test the lifetime of a lightbulb  (minutes)

Length of time the bulb lasts up  to 80,000 minutes

S = time the bulb burns 0 … S … 80,000

Table 2.6 Random Variables for Outcomes That Are Not Numbers

experiment outCome ranGe of ranDom

Variables ranDom

Variables

Students respond to  a questionnaire

Strongly agree (SA)

Agree (A)

Neutral (N)

Disagree (D)

Strongly disagree (SD)

X =

5 if SA

4 if A

3 if N

2 if D

1 if SD

1, 2, 3, 4, 5

One machine is  inspected

Defective

Not defective Y =

0 if defective

1 if not defective

0, 1

Consumers respond  to how they like a  product

Good

Average

Poor

Z = 3 if good

2 if average

1 if poor

1, 2, 3

c h

e

M02_REND9327_12_SE_C02.indd 51 10/02/14 1:20 PM

52  Chapter 2 • probability ConCepts and appliCations

at Table 2.5, we can see that testing the lifetime of a lightbulb is an experiment whose results can  be described with a continuous random variable. In this case, the random variable, S, is the time  the bulb burns. It can last for 3,206 minutes, 6,500.7 minutes, 251.726 minutes, or any other  value between 0 and 80,000 minutes. In most cases, the range of a continuous random variable is  stated as: lower value … S … upper value, such as 0 … S … 80,000. The random variable R in  Table 2.5 is also continuous. Can you explain why?

2.6 Probability Distributions

Earlier we discussed the probability values of an event. We now explore the properties of  probability distributions. We see how popular distributions, such as the normal, Poisson, bi- nomial, and exponential probability distributions, can save us time and effort. Since a random  variable may be discrete or continuous, we consider discrete probability distributions and  continuous probability distributions seperately.

Probability Distribution of a Discrete Random Variable When we have a discrete random variable, there is a probability value assigned to each event.  These values must be between 0 and 1, and they must sum to 1. Let’s look at an example.

The 100 students in Pat Shannon’s statistics class have just completed a math quiz that he  gives on the first day of class. The quiz consists of five very difficult algebra problems. The  grade on the quiz is the number of correct answers, so the grades theoretically could range from  0 to 5. However, no one in this class received a score of 0, so the grades ranged from 1 to 5.  The random variable X is defined to be the grade on this quiz, and the grades are summarized  in Table 2.7. This discrete probability distribution was developed using the relative frequency   approach presented earlier.

The distribution follows the three rules required of all probability distributions: (1) the  events are mutually exclusive and collectively exhaustive, (2) the individual probability values  are between 0 and 1 inclusive, and (3) the total of the probability values sum to 1.

Although listing the probability distribution as we did in Table 2.7 is adequate, it can be  difficult to get an idea about characteristics of the distribution. To overcome this problem, the  probability values are often presented in graph form. The graph of the distribution in Table 2.7 is  shown in Figure 2.4.

The graph of this probability distribution gives us a picture of its shape. It helps us identify  the central tendency of the distribution, called the mean or expected value, and the amount of  variability or spread of the distribution, called the variance.

Expected Value of a Discrete Probability Distribution Once we have established a probability distribution, the first characteristic that is usually of  interest is the central tendency of the distribution. The expected value, a measure of central ten- dency, is computed as the weighted average of the values of the random variable:

E1X2 = gn i=1

Xi P1Xi 2 = X1P1X12 + X2P1X22 + Á + XnP1Xn2 (2-6)

Table 2.7 Probability Distribution for Quiz Scores

ranDom Variable (X)-sCore number probability P1X2

5 10 0.1 = 10>100 4 20 0.2 = 20>100 3 30 0.3 = 30>100 2 30 0.3 = 30>100 1 10 0.1 = 10>100

Total 100 1.0 = 100>100

The expected value of a discrete distribution is a weighted average of the values of the random variable.

M02_REND9327_12_SE_C02.indd 52 10/02/14 1:20 PM

2.6 probability distributions  53

where

Xi = random variable’s possible values P1Xi2 = probability of each of the random variable’s possible values gn

i=1 = summation sign indicating we are adding all n possible values

E1X2 = expected value or mean of the random variable The expected value or mean of any discrete probability distribution can be computed by 

multiplying each possible value of the random variable, Xi, times the probability, P1Xi2, that out- come will occur and summing the results, g . Here is how the expected value can be computed  for the quiz scores:

E1X2 = g5 i=1

XiP1Xi2 = X1P1X12 + X2P1X22 + X3P1X32 + X4P1X42 + X5P1X52 = 15210.12 + 14210.22 + 13210.32 + 12210.32 + 11210.12 = 2.9

The expected value of 2.9 is the mean score on the quiz.

Variance of a Discrete Probability Distribution In addition to the central tendency of a probability distribution, most people are interested in the vari- ability or the spread of the distribution. If the variability is low, it is much more likely that the outcome  of an experiment will be close to the average or expected value. On the other hand, if the variability of  the distribution is high, which means that the probability is spread out over the various random vari- able values, there is less chance that the outcome of an experiment will be close to the expected value.

The variance of a probability distribution is a number that reveals the overall spread or   dispersion of the distribution. For a discrete probability distribution, it can be computed using  the following equation:

s2 = Variance = gn i=1

3Xi - E1X242P1Xi2 (2-7)

where

Xi = random variable’s possible values E1X2 = expected value of the random variable

3Xi - E1X24 = difference between each value of the random variable and the expected value P1Xi2 = probability of each possible value of the random variable

Figure 2.4 Probability Distribution for Dr. Shannon’s Class

0.4

0.3

0.2

0.1

0 1 2 3

X

P (X

)

4 5

A probability distribution is often described by its mean and variance. Even if most of the men in class (or the United States) have heights between 5 feet 6 inches and 6 feet 2 inches, there is still some small probability of outliers.

M02_REND9327_12_SE_C02.indd 53 10/02/14 1:20 PM

54  Chapter 2 • probability ConCepts and appliCations

To compute the variance, each value of the random variable is subtracted from the expected  value, squared, and multiplied times the probability of occurrence of that value. The results are then  summed to obtain the variance. Here is how this procedure is done for Dr. Shannon’s quiz scores:

Variance = g5 i=1

3Xi - E1X242P1Xi2

Variance = 15 - 2.92210.12 + 14 - 2.92210.22 + 13 - 2.92210.32 + 12 - 2.92210.32 + 11 - 2.92210.12

= 12.12210.12 + 11.12210.22 + 10.12210.32 + 1-0.92210.32 + 1-1.92210.12 = 0.441 + 0.242 + 0.003 + 0.243 + 0.361 = 1.29

A related measure of dispersion or spread is the standard deviation. This quantity is also  used in many computations involved with probability distributions. The standard deviation is  just the square root of the variance:

s = 1Variance = 2s2 (2-8) where

1 = square root s = standard deviation

The standard deviation for the random variable X in the example is

s = 1Variance = 11.29 = 1.14

These calculations are easily performed in Excel. Program 2.1A provides the output for this ex- ample. Program 2.1B shows the inputs and formulas in Excel for calculating the mean, variance,  and standard deviation in this example.

Probability Distribution of a Continuous Random Variable There are many examples of continuous random variables. The time it takes to finish a project,  the number of ounces in a barrel of butter, the high temperature during a given day, the exact  length of a given type of lumber, and the weight of a railroad car of coal are all examples of con- tinuous random variables. Since random variables can take on an infinite number of values, the  fundamental probability rules for continuous random variables must be modified.

As with discrete probability distributions, the sum of the probability values must equal 1.  Because there are an infinite number of values of the random variables, however, the probability  of each value of the random variable must be 0. If the probability values for the random variable  values were greater than 0, the sum would be infinitely large.

Program 2.1a Excel 2013 Output for the Dr. Shannon Example

Program 2.1b Formulas in an Excel Spreadsheet for the Dr. Shannon Example

M02_REND9327_12_SE_C02.indd 54 10/02/14 1:20 PM

2.7 the binomial distribution  55

With a continuous probability distribution, there is a continuous mathematical function that  describes the probability distribution. This function is called the probability density function or simply the probability function. It is usually represented by  f 1X2. When working with con- tinuous probability distributions, the probability function can be graphed, and the area under- neath the curve represents probability. Thus, to find any probability, we simply find the area  under the curve associated with the range of interest.

We now look at the sketch of a sample density function in Figure 2.5. This curve represents  the probability density function for the weight of a particular machined part. The weight could  vary from 5.06 to 5.30 grams, with weights around 5.18 grams being the most likely. The shaded  area represents the probability the weight is between 5.22 and 5.26 grams.

If we wanted to know the probability of a part weighing exactly 5.1300000 grams, for ex- ample, we would have to compute the area of a line of width 0. Of course, this would be 0. This  result may seem strange, but if we insist on enough decimal places of accuracy, we are bound to  find that the weight differs from 5.1300000 grams exactly, be the difference ever so slight.

This is important because it means that, for any continuous distribution, the probability does  not change if a single point is added to the range of values that is being considered. In Figure 2.5,  this means the following probabilities are all exactly the same:

P15.22 6 X 6 5.262 = P15.22 6 X … 5.262 = P15.22 … X 6 5.262 = P15.22 … X … 5.262

The inclusion or exclusion of either endpoint (5.22 or 5.26) has no impact on the probability. In this section, we have investigated the fundamental characteristics and properties of proba-

bility distributions in general. In the next three sections, we introduce three important continuous  distributions—the normal distribution, the F distribution, and the exponential  distribution—and  two discrete distributions—the Poisson distribution and the binomial distribution.

2.7 The Binomial Distribution

Many business experiments can be characterized by the bernoulli process. The probability of   obtaining specific outcomes in a Bernoulli process is described by the binomial probability distri- bution. In order to be a Bernoulli process, an experiment must have the following characteristics:

  1. Each trial in a Bernoulli process has only two possible outcomes. These are typically called  a success and a failure, although examples might be yes or no, heads or tails, pass or fail,  defective or good, and so on.

  2. The probability stays the same from one trial to the next.   3. The trials are statistically independent.   4. The number of trials is a positive integer.

A common example of this process is tossing a coin.

A probability density function, f (X ), is a mathematical way of describing the probability distribution.

Figure 2.5 Graph of Sample Density Function

P ro

ba bi

lit y

5.06 5.10 5.14 5.18 5.22 5.26 5.30

Weight (grams)

M02_REND9327_12_SE_C02.indd 55 10/02/14 1:20 PM

56  Chapter 2 • probability ConCepts and appliCations

The binomial distribution is used to find the probability of a specific number of suc- cesses out of n trials of a Bernoulli process. To find this probability, it is necessary to know the  following:

n = the number of trials p = the probability of a success on any single trial

We let

r = the number of successes q = 1 - p = the probability of a failure

The binomial formula is

Probability of r successes in n trials = n!

r!1n - r2! p r qn - r (2-9)

The symbol ! means factorial, and n! = n1n - 121n - 22c (1). For example, 4! = 142132122112 = 24

Also, 1! = 1, and 0! = 1 by definition.

Solving Problems with the Binomial Formula A common example of a binomial distribution is the tossing of a coin and counting the number  of heads. For example, if we wished to find the probability of 4 heads in 5 tosses of a coin, we  would have

n = 5, r = 4, p = 0.5, and q = 1 - 0.5 = 0.5

Thus,

P14 successes in 5 trials2 = 5! 4!15 - 42! 0.5

40.55-4

= 5142132122112 413212211211!2 10.0625210.52 = 0.15625

Thus, the probability of 4 heads in 5 tosses of a coin is 0.15625 or about 16%. Using Equation 2-9, it is also possible to find the entire probability distribution (all the pos-

sible values for r and the corresponding probabilities) for a binomial experiment. The probabil- ity distribution for the number of heads in 5 tosses of a fair coin is shown in Table 2.8 and then  graphed in Figure 2.6.

Table 2.8 Binomial Probability Distribution for n = 5 and p = 0.50

number of heaDs (r) probability =

5! r! 15 − r 2! 10.5 2

r 10.5 25 - r

0 0.03125 =

5!

0!15 - 02!10.52 010.525-0

1 0.15625 =

5!

1!15 - 12!10.52 110.525-1

2 0.31250 =

5!

2!15 - 22!10.52 210.525- 2

3 0.31250 =

5!

3!15 - 32!10.52 310.525- 3

4 0.15625 =

5!

4!15 - 42!10.52 410.525-4

5 0.03125 =

5!

5!15 - 52!10.52 510.525-5

M02_REND9327_12_SE_C02.indd 56 10/02/14 1:20 PM

2.7 the binomial distribution  57

Solving Problems with Binomial Tables MSA Electronics is experimenting with the manufacture of a new type of transistor that is very  difficult to mass produce at an acceptable quality level. Every hour a supervisor takes a random  sample of 5 transistors produced on the assembly line. The probability that any one transistor  is defective is considered to be 0.15. MSA wants to know the probability of finding 3, 4, or 5  d efectives if the true percentage defective is 15%.

For this problem, n = 5, p = 0.15, and r = 3, 4, or 5. Although we could use the formula  for each of these values, it is easier to use binomial tables for this. Appendix B gives a bi- nomial table for a broad range of values for n, r, and p. A portion of this appendix is shown  in Table 2.9. To find these probabilities, we look through the n = 5 section and find the 

Figure 2.6 Binomial Probability Distribution for n = 5 and p = 0.50

0.4

0.3

0.2

0.1

0 1 2 3 4 5 6

P ro

ba bi

lit y,

P (r

)

Values of r (number of successes)

Table 2.9 A Sample Table for the Binomial Distribution

P

n r 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

1 0 0.9500 0.9000 0.8500 0.8000 0.7500 0.7000 0.6500 0.6000 0.5500 0.5000

1 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000

2 0 0.9025 0.8100 0.7225 0.6400 0.5625 0.4900 0.4225 0.3600 0.3025 0.2500

1 0.0950 0.1800 0.2500 0.3200 0.3750 0.4200 0.4550 0.4800 0.4950 0.5000

2 0.0025 0.0100 0.0225 0.0400 0.0625 0.0900 0.1225 0.1600 0.2025 0.2500

3 0 0.8574 0.7290 0.6141 0.5120 0.4219 0.3430 0.2746 0.2160 0.1664 0.1250

1 0.1354 0.2430 0.3251 0.3840 0.4219 0.4410 0.4436 0.4320 0.4084 0.3750

2 0.0071 0.0270 0.0574 0.0960 0.1406 0.1890 0.2389 0.2880 0.3341 0.3750

3 0.0001 0.0010 0.0034 0.0080 0.0156 0.0270 0.0429 0.0640 0.0911 0.1250

4 0 0.8145 0.6561 0.5220 0.4096 0.3164 0.2401 0.1785 0.1296 0.0915 0.0625

1 0.1715 0.2916 0.3685 0.4096 0.4219 0.4116 0.3845 0.3456 0.2995 0.2500

2 0.0135 0.0486 0.0975 0.1536 0.2109 0.2646 0.3105 0.3456 0.3675 0.3750

3 0.0005 0.0036 0.0115 0.0256 0.0469 0.0756 0.1115 0.1536 0.2005 0.2500

4 0.0000 0.0001 0.0005 0.0016 0.0039 0.0081 0.0150 0.0256 0.0410 0.0625

5 0 0.7738 0.5905 0.4437 0.3277 0.2373 0.1681 0.1160 0.0778 0.0503 0.0313

1 0.2036 0.3281 0.3915 0.4096 0.3955 0.3602 0.3124 0.2592 0.2059 0.1563

2 0.0214 0.0729 0.1382 0.2048 0.2637 0.3087 0.3364 0.3456 0.3369 0.3125

3 0.0011 0.0081 0.0244 0.0512 0.0879 0.1323 0.1811 0.2304 0.2757 0.3125

4 0.0000 0.0005 0.0022 0.0064 0.0146 0.0284 0.0488 0.0768 0.1128 0.1563

5 0.0000 0.0000 0.0001 0.0003 0.0010 0.0024 0.0053 0.0102 0.0185 0.0313

6 0 0.7351 0.5314 0.3771 0.2621 0.1780 0.1176 0.0754 0.0467 0.0277 0.0156

1 0.2321 0.3543 0.3993 0.3932 0.3560 0.3025 0.2437 0.1866 0.1359 0.0938

2 0.0305 0.0984 0.1762 0.2458 0.2966 0.3241 0.3280 0.3110 0.2780 0.2344

3 0.0021 0.0146 0.0415 0.0819 0.1318 0.1852 0.2355 0.2765 0.3032 0.3125

4 0.0001 0.0012 0.0055 0.0154 0.0330 0.0595 0.0951 0.1382 0.1861 0.2344

5 0.0000 0.0001 0.0004 0.0015 0.0044 0.0102 0.0205 0.0369 0.0609 0.0938

6 0.0000 0.0000 0.0000 0.0001 0.0002 0.0007 0.0018 0.0041 0.0083 0.0156

M02_REND9327_12_SE_C02.indd 57 10/02/14 1:20 PM

58  Chapter 2 • probability ConCepts and appliCations

Program 2.2a Excel Output for the Binomial Example

Using the cell references eliminates the need to retype the formula if you change a parameter such as p or r.

Program 2.2b Function in an Excel 2013 Spreadsheet for Binomial Probabilities

The function BINOM.DIST (r,n,p,TRUE) returns the cumulative probability.

The normal distribution affects a large number of processes in our lives (e.g., filling boxes of cereal with 32 ounces of corn flakes). Each normal distribution depends on the mean and standard deviation.

p = 0.15 column. In the row where r = 3, we see 0.0244. Thus, P1r = 32 = 0.0244. Simi- larly, P1r = 42 = 0.0022, and P1r = 52 = 0.0001. By adding these three probabilities, we  have the probability that the number of defects is 3 or more:

P13 or more defects2 = P132 + P142 + P152 = 0.0244 + 0.0022 + 0.0001 = 0.0267

The expected value (or mean) and the variance of a binomial random variable may be easily  found. These are

Expected value 1mean2 = np (2-10) Variance = np11 - p2 (2-11)

The expected value and variance for the MSA Electronics example are computed as follows:

Expected value = np = 510.152 = 0.75 Variance = np11 - p2 = 510.15210.852 = 0.6375

Programs 2.2A and 2.2B illustrate how Excel is used for binomial probabilities.

2.8 The Normal Distribution

One of the most popular and useful continuous probability distributions is the normal distribu- tion. The probability density function of this distribution is given by the rather complex formula

f 1X2 = 1 s12p e - 1x -m222s2 (2-12)

The normal distribution is specified completely when values for the mean, m, and the stan- dard deviation, s, are known. Figure 2.7 shows several different normal distributions with the  same standard deviation and different means. As shown, differing values of m will shift the aver- age or center of the normal distribution. The overall shape of the distribution remains the same.  On the other hand, when the standard deviation is varied, the normal curve either flattens out or  becomes steeper. This is shown in Figure 2.8.

As the standard deviation, s, becomes smaller, the normal distribution becomes steeper.  When the standard deviation becomes larger, the normal distribution has a tendency to flatten  out or become broader.

M02_REND9327_12_SE_C02.indd 58 10/02/14 1:20 PM

2.8 the normal distribution  59

Figure 2.7 Normal Distribution with Different Values for M

Smaller m, same s

Larger m, same s

m 5 50 6040

50 60m 5 40

50 m 5 6040

probability Assessments of Curling Champions

Probabilities are used every day in sporting activities. In many sporting events, there are questions involving strategies that must be answered to provide the best chance of winning the game. In baseball, should a particular batter be intentionally walked in key situations at the end of the game? In football, should a team elect to try for a two-point conversion after a touchdown? In soccer, should a penalty kick ever be aimed directly at the goal keeper? In curling, in the last round, or “end” of a game, is it better to be behind by one point and have the hammer or is it better to be ahead by one point and not have the hammer? An attempt was made to answer this last question.

In curling, a granite stone, or “rock,” is slid across a sheet of ice 14 feet wide and 146 feet long. Four players on each of two teams take alternating turns sliding the rock, trying to get it as close as possible to the center of a circle called the “house.” The team with the rock closest to this scores points. The team that is behind at the completion of a round or end has the advantage in

the next end by being the last team to slide the rock. This team is said to “have the hammer.” A survey was taken of a group of ex- perts in curling, including a number of former world champions. In this survey, about 58% of the respondents favored having the hammer and being down by one going into the last end. Only about 42% preferred being ahead and not having the hammer.

Data were also collected from 1985 to 1997 at the Canadian Men’s Curling Championships (also called the Brier). Based on the results over this time period, it is better to be ahead by one point and not have the hammer at the end of the ninth end rather than be behind by one and have the hammer, as many people prefer. This differed from the survey results. Apparently, world champions and other experts preferred to have more control of their des- tiny by having the hammer even though it put them in a worse situation.

source: Based on Keith A. Willoughby and Kent J. Kostuk. “Preferred Sce- narios in the Sport of Curling,” Interfaces 34, 2 (March–April 2004): 117–122.

in ACtiOn

Figure 2.8 Normal Distribution with Different Values for S

Same m, smaller s

Same m, larger s

m

M02_REND9327_12_SE_C02.indd 59 10/02/14 1:20 PM

60  Chapter 2 • probability ConCepts and appliCations

Area Under the Normal Curve Because the normal distribution is symmetrical, its midpoint (and highest point) is at the mean.  Values on the X axis are then measured in terms of how many standard deviations they lie from  the mean. As you may recall from our earlier discussion of probability distributions, the area un- der the curve (in a continuous distribution) describes the probability that a random variable has  a value in a specified interval. When dealing with the uniform distribution, it is easy to compute  the area between any points a and b. The normal distribution requires mathematical calculations  beyond the scope of this book, but tables that provide areas or probabilities are readily available.

Using the Standard Normal Table When finding probabilities for the normal distribution, it is best to draw the normal curve and  shade the area corresponding to the probability being sought. The normal distribution table can  then be used to find probabilities by following two steps.

Step 1.  Convert the normal distribution to what we call a standard normal distribution. A stan- dard normal distribution has a mean of 0 and a standard deviation of 1. All normal tables are set  up to handle random variables with m = 0 and s = 1. Without a standard normal distribution,  a different table would be needed for each pair of m and s values. We call the new standard  random variable Z. The value for Z for any normal distribution is computed from this equation:

Z = X - m

s (2-13)

where

X = value of the random variable we want to measure m = mean of the distribution s = standard deviation of the distribution Z = number of standard deviations from X to the mean, m

For example, if m = 100, s = 15, and we are interested in finding the probability that the  random variable X (IQ) is less than 130, we want P1X 6 1302:

Z = X - m

s =

130 - 100 15

= 30

15 = 2 standard deviations

This means that the point X is 2.0 standard deviations to the right of the mean. This is shown in  Figure 2.9.

Figure 2.9 Normal Distribution Showing the Relationship Between Z Values and X Values

m 5 100 s 5 15P(X < 130)

55 70 85 100 115 130 145

–3 –2 –1 0 1 2 3 Z 5

X 5 IQ

x – m s

m

M02_REND9327_12_SE_C02.indd 60 10/02/14 1:20 PM

2.8 the normal distribution  61

Step 2.  Look up the probability from a table of normal curve areas. Table 2.10, which also ap- pears as Appendix A, is such a table of areas for the standard normal distribution. It is set up to  provide the area under the curve to the left of any specified value of Z.

Let’s see how Table 2.10 can be used. The column on the left lists values of Z, with the sec- ond decimal place of Z appearing in the top row. For example, for a value of Z = 2.00 as just  computed, find 2.0 in the left-hand column and 0.00 in the top row. In the body of the table, we  find that the area sought is 0.97725, or 97.7%. Thus,

P1X 6 1302 = P1Z 6 2.002 = 97.7% This suggests that if the mean IQ score is 100, with a standard deviation of 15 points, the 

probability that a randomly selected person’s IQ is less than 130 is 97.7%. This is also the prob- ability that the IQ is less than or equal to 130. To find the probability that the IQ is greater than  130, we simply note that this is the complement of the previous event and the total area under the  curve (the total probability) is 1. Thus,

P1X 7 1302 = 1 - P1X … 1302 = 1 - P1Z … 22 = 1 - 0.97725 = 0.02275 While Table 2.10 does not give negative Z values,  the symmetry of  the normal distribu- tion can be used  to  find probabilities  associated with negative Z  values. For example,  P1Z 6 -22 = P1Z 7 22.

To feel comfortable with the use of the standard normal probability table, we need to work a  few more examples. We now use the Haynes Construction Company as a case in point.

Haynes Construction Company Example Haynes Construction Company builds primarily three- and four-unit apartment buildings (called  triplexes and quadraplexes) for investors, and it is believed that the total construction time in  days follows a normal distribution. The mean time to construct a triplex is 100 days, and the  standard deviation is 20 days. Recently, the president of Haynes Construction signed a contract  to complete a triplex in 125 days. Failure to complete the triplex in 125 days would result in  severe penalty fees. What is the probability that Haynes Construction will not be in violation of  their construction contract? The normal distribution for the construction of triplexes is shown in  Figure 2.10.

To compute this probability, we need to find the shaded area under the curve. We begin by  computing Z for this problem:

Z = X - m

s

= 125 - 100

20

= 25

20 = 1.25

Looking in Table 2.10 for a Z value of 1.25, we find an area under the curve of 0.89435.  (We do this by looking up 1.2 in the left-hand column of the table and then moving to the 0.05  column to find the value for Z = 1.25.) Therefore, the probability of not violating the contract is  0.89435, or about an 89% chance.

To be sure you understand the concept of symmetry in Table 2.10, try to find the probability such as P1X * 852. Note that the standard normal table shows only positive Z values.

Figure 2.10 Normal Distribution for Haynes Construction

s 5 20 days

X 5 125 days

Z 5 1.25

m 5 100 days

P (X # 125)

0.89435

M02_REND9327_12_SE_C02.indd 61 10/02/14 1:20 PM

62  Chapter 2 • probability ConCepts and appliCations

Table 2.10 Standardized Normal Distribution Function

area unDer the normal CurVe

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 .50000 .50399 .50798 .51197 .51595 .51994 .52392 .52790 .53188 .53586

0.1 .53983 .54380 .54776 .55172 .55567 .55962 .56356 .56749 .57142 .57535

0.2 .57926 .58317 .58706 .59095 .59483 .59871 .60257 .60642 .61026 .61409

0.3 .61791 .62172 .62552 .62930 .63307 .63683 .64058 .64431 .64803 .65173

0.4 .65542 .65910 .66276 .66640 .67003 .67364 .67724 .68082 .68439 .68793

0.5 .69146 .69497 .69847 .70194 .70540 .70884 .71226 .71566 .71904 .72240

0.6 .72575 .72907 .73237 .73565 .73891 .74215 .74537 .74857 .75175 .75490

0.7 .75804 .76115 .76424 .76730 .77035 .77337 .77637 .77935 .78230 .78524

0.8 .78814 .79103 .79389 .79673 .79955 .80234 .80511 .80785 .81057 .81327

0.9 .81594 .81859 .82121 .82381 .82639 .82894 .83147 .83398 .83646 .83891

1.0 .84134 .84375 .84614 .84849 .85083 .85314 .85543 .85769 .85993 .86214

1.1 .86433 .86650 .86864 .87076 .87286 .87493 .87698 .87900 .88100 .88298

1.2 .88493 .88686 .88877 .89065 .89251 .89435 .89617 .89796 .89973 .90147

1.3 .90320 .90490 .90658 .90824 .90988 .91149 .91309 .91466 .91621 .91774

1.4 .91924 .92073 .92220 .92364 .92507 .92647 .92785 .92922 .93056 .93189

1.5 .93319 .93448 .93574 .93699 .93822 .93943 .94062 .94179 .94295 .94408

1.6 .94520 .94630 .94738 .94845 .94950 .95053 .95154 .95254 .95352 .95449

1.7 .95543 .95637 .95728 .95818 .95907 .95994 .96080 .96164 .96246 .96327

1.8 .96407 .96485 .96562 .96638 .96712 .96784 .96856 .96926 .96995 .97062

1.9 .97128 .97193 .97257 .97320 .97381 .97441 .97500 .97558 .97615 .97670

2.0 .97725 .97778 .97831 .97882 .97932 .97982 .98030 .98077 .98124 .98169

2.1 .98214 .98257 .98300 .98341 .98382 .98422 .98461 .98500 .98537 .98574

2.2 .98610 .98645 .98679 .98713 .98745 .98778 .98809 .98840 .98870 .98899

2.3 .98928 .98956 .98983 .99010 .99036 .99061 .99086 .99111 .99134 .99158

2.4 .99180 .99202 .99224 .99245 .99266 .99286 .99305 .99324 .99343 .99361

2.5 .99379 .99396 .99413 .99430 .99446 .99461 .99477 .99492 .99506 .99520

2.6 .99534 .99547 .99560 .99573 .99585 .99598 .99609 .99621 .99632 .99643

2.7 .99653 .99664 .99674 .99683 .99693 .99702 .99711 .99720 .99728 .99736

2.8 .99744 .99752 .99760 .99767 .99774 .99781 .99788 .99795 .99801 .99807

2.9 .99813 .99819 .99825 .99831 .99836 .99841 .99846 .99851 .99856 .99861

3.0 .99865 .99869 .99874 .99878 .99882 .99886 .99889 .99893 .99896 .99900

3.1 .99903 .99906 .99910 .99913 .99916 .99918 .99921 .99924 .99926 .99929

3.2 .99931 .99934 .99936 .99938 .99940 .99942 .99944 .99946 .99948 .99950

3.3 .99952 .99953 .99955 .99957 .99958 .99960 .99961 .99962 .99964 .99965

3.4 .99966 .99968 .99969 .99970 .99971 .99972 .99973 .99974 .99975 .99976

3.5 .99977 .99978 .99978 .99979 .99980 .99981 .99981 .99982 .99983 .99983

3.6 .99984 .99985 .99985 .99986 .99986 .99987 .99987 .99988 .99988 .99989

3.7 .99989 .99990 .99990 .99990 .99991 .99991 .99992 .99992 .99992 .99992

3.8 .99993 .99993 .99993 .99994 .99994 .99994 .99994 .99995 .99995 .99995

3.9 .99995 .99995 .99996 .99996 .99996 .99996 .99996 .99996 .99997 .99997

M02_REND9327_12_SE_C02.indd 62 10/02/14 1:20 PM

2.8 the normal distribution  63

Now let us look at the Haynes problem from another perspective. If the firm finishes this  triplex in 75 days or less, it will be awarded a bonus payment of $5,000. What is the probability  that Haynes will receive the bonus?

Figure 2.11 illustrates the probability we are looking for in the shaded area. The first step is  again to compute the Z value:

Z = X - m

s

= 75 - 100

20

= -25 20

= -1.25

This Z value indicates that 75 days is -1.25 standard deviations to the left of the mean. But the  standard normal table is structured to handle only positive Z values. To solve this problem, we  observe that the curve is symmetric. The probability that Haynes will finish in 75 days or less is equivalent to the probability that it will finish in more than 125 days. A moment ago (in Fig- ure 2.10) we found the probability that Haynes will finish in less than 125 days. That value is  0.89435. So the probability it takes more than 125 days is

P1X 7 1252 = 1.0 - P1X … 1252 = 1.0 - 0.89435 = 0.10565

Thus, the probability of completing the triplex in 75 days or less is 0.10565, or about 11%. One final example: What is the probability that the triplex will take between 110 and 125 

days? We see in Figure 2.12 that

P1110 6 X 6 1252 = P1X … 1252 - P1X 6 1102 That is, the shaded area in the graph can be computed by finding the probability of completing  the building in 125 days or less minus the probability of completing it in 110 days or less.

Figure 2.11 Probability That Haynes Will Receive the Bonus by Finishing in 75 Days or Less

P(X # 75 days) Area of Interest

X 5 75 days m 5 100 days

Z 5 21.25

0.89435

Figure 2.12 Probability That Haynes Will Complete in 110 to 125 Days

s 5 20 days

m 5 100 days

110 days

125 days

M02_REND9327_12_SE_C02.indd 63 10/02/14 1:20 PM

64  Chapter 2 • probability ConCepts and appliCations

Program 2.3b Function in an Excel 2013 Spreadsheet for the Normal Distribution Example

Recall that P1X … 125 days2 is equal to 0.89435. To find P1X 6 110 days2, we follow the  two steps developed earlier:

  1.    Z = X - m

s =

110 - 100 20

= 10

20

= 0.5 standard deviations

  2. From Table 2.10, the area for Z = 0.50 is 0.69146. So the probability the triplex can be  completed in less than 110 days is 0.69146. Finally,

P1110 … X … 1252 = 0.89435 - 0.69146 = 0.20289 The probability that it will take between 110 and 125 days is about 20%. Programs 2.3A and  2.3B show how Excel can be used for this.

The Empirical Rule While the probability tables for the normal distribution can provide precise probabilities, many  situations require less precision. The empirical rule was derived from the normal distribution  and is an easy way to remember some basic information about normal distributions. The empiri- cal rule states that for a normal distribution

approximately 68% of the values will be within {1 standard deviation of the mean approximately 95% of the values will be within {2 standard deviations of the mean almost all (about 99.7%) of the values will be within {3 standard deviations of the mean

Figure 2.13 illustrates the empirical rule. The area from point a to point b in the first drawing  represents the probability, approximately 68%, that the random variable will be within {1 stan- dard deviation of the mean. The middle drawing illustrates the probability, approximately 95%,  that the random variable will be within {2 standard deviations of the mean. The last drawing  illustrates the probability, about 99.7% (almost all), that the random variable will be within {3 standard deviations of the mean.

2.9 The F Distribution

The F distribution is a continuous probability distribution that is helpful in testing hypotheses  about variances. The F distribution will be used in Chapter 4 when regression models are tested  for significance. Figure 2.14 provides a graph of the F distribution. As with a graph for any con- tinuous distribution, the area underneath the curve represents probability. Note that for a large  value of F, the probability is very small.

The F statistic is the ratio of two sample variances from independent normal distributions.  Every F distribution has two sets of degrees of freedom associated with it. One of the degrees  of freedom is associated with the numerator of the ratio, and the other is associated with the de- nominator of the ratio. The degrees of freedom are based on the sample sizes used in calculating  the numerator and denominator.

Program 2.3a Excel 2013 Output for the Normal Distribution Example

M02_REND9327_12_SE_C02.indd 64 10/02/14 1:20 PM

2.9 The F distribution  65

Appendix D provides values of F associated with the upper tail of the distribution for cer- tain probabilities (denoted by a) and degrees of freedom for the numerator 1df12 and degrees of  freedom for the denominator 1df22.

To find the F value that is associated with a particular probability and degrees of freedom,  refer to Appendix D. The following notation will be used:

df1 = degrees of freedom for the numerator df2 = degrees of freedom for the denominator

Consider the following example:

df1 = 5 df2 = 6

a = 0.05

Managers often speak of 95% and 99% confidence intervals, which roughly refer to ±2 and 3 standard deviation graphs.

Figure 2.13 Approximate Probabilities from the Empirical Rule

Figure 2.13 is very important, and you should comprehend the meanings of ±1, 2, and 3 standard deviation symmetrical areas.

16% 16%

2.5% 2.5%

68%

a m b 2 1s 1 1s

95%

a m b

22s 12s

99.7%

a m b

23s 13s

0.15% 0.15%

Figure 2.14 The F Distribution

Fα

M02_REND9327_12_SE_C02.indd 65 10/02/14 1:20 PM

66  Chapter 2 • probability ConCepts and appliCations

From Appendix D, we get

Fa, df1, df2 = F0.05, 5, 6 = 4.39

This means

P1F 7 4.392 = 0.05 The probability is very low (only 5%) that the F value will exceed 4.39. There is a 95% prob- ability that it will not exceed 4.39. This is illustrated in Figure 2.15. Appendix D also provides  F values associated with a = 0.01. Programs 2.4A and 2.4B illustrate Excel functions for the  F distribution.

2.10 The Exponential Distribution

The exponential distribution, also called the negative exponential distribution, is used in dealing  with queuing problems. The exponential distribution often describes the time required to service a  customer. The exponential distribution is a continuous distribution. Its probability function is given  by

f 1X2 = me-mx (2-14) where

X = random variable (service times) m = average number of units the service facility can handle in a specific period of time e =  2.718 (the base of the natural logarithm)

F = 4.39 0.05

Figure 2.15 F Value for 0.05 Probability with 5 and 6 Degrees of Freedom

Program 2.4a Excel 2013 Output for the F Distribution

Program 2.4b Functions in an Excel 2013 Spreadsheet for the F Distribution

M02_REND9327_12_SE_C02.indd 66 10/02/14 1:20 PM

2.10 the exponential distribution  67

The general shape of the exponential distribution is shown in Figure 2.16. Its expected value  and variance can be shown to be

Expected value = 1 m

= Average service time (2-15)

Variance = 1

m2 (2-16)

As with any other continuous distribution, probabilities are found by determining the area under  the curve. For the normal distribution, we found the area by using a table of probabilities. For the  exponential distribution, the probabilities can be found using the exponent key on a calculator  with the formula below. The probability that an exponentially distributed time (X) required to  serve a customer is less than or equal to time t is given by the formula

P1X … t2 = 1 - e-mt (2-17) The time period used in describing m determines the units for the time t. For example, if m is the  average number served per hour, the time t must be given in hours. If m is the average number  served per minute, the time t must be given in minutes.

Arnold’s Muffler Example Arnold’s Muffler Shop installs new mufflers on automobiles and small trucks. The mechanic can  install new mufflers at a rate of about three per hour, and this service time is exponentially dis- tributed. What is the probability that the time to install a new muffler would be 1�2 hour or less?  Using Equation 2-17, we have

X = exponentially distributed service time m = average number that can be served per time period = 3 per hour t = 1�2 hour = 0.5 hour

P1X … 0.52 = 1 - e-310.52 = 1 - e-1.5 = 1 - 0.2231 = 0.7769 Figure 2.17 shows the area under the curve from 0 to 0.5 to be 0.7769. Thus, there is about a  78% chance the time will be no more than 0.5 hour and about a 22% chance that the time will be  longer than this. Similarly, we could find the probability that the service time is no more 1>3  hour  or  2>3  hour, as follows:

P aX … 1 3 b = 1 - e-31132 = 1 - e-1 = 1 - 0.3679 = 0.6321

P aX … 2 3 b = 1 - e-31232 = 1 - e-2 = 1 - 0.1353 = 0.8647

Figure 2.16 Exponential Distribution f(X )

X

M02_REND9327_12_SE_C02.indd 67 10/02/14 1:20 PM

68  Chapter 2 • probability ConCepts and appliCations

While Equation 2-17 provides the probability that the time (X ) is less than or equal to a particu- lar value t, the probability that the time is greater than a particular value t is found by observing  that these two events are complementary. For example, to find the probability that the mechanic  at Arnold’s Muffler Shop would take longer than 0.5 hour, we have

P1X 7 0.52 = 1 - P1X … 0.52 = 1 - 0.7769 = 0.2231 Programs 2.5A and 2.5B illustrate how a function in Excel can find exponential probabilities.

2.11 The Poisson Distribution

An important discrete probability distribution is the poisson distribution.1 We examine it  because of its key role in complementing the exponential distribution in queuing theory in Chap- ter 12. The distribution describes situations in which customers arrive independently during a  certain time interval, and the number of arrivals depends on the length of the time interval. Ex- amples are patients arriving at a health clinic, customers arriving at a bank window, passengers  arriving at an airport, and telephone calls going through a central exchange.

The formula for the Poisson distribution is

P1X2 = l xe-l

X! (2-18)

Figure 2.17 Probability That the Mechanic Will Install a Muffler in 0.5 Hour

0 0

0.5

1

1.5

0.7769

P(service time # 0.5) = 0.7769

2

2.5

0.5 1 1.5 2 2.5

Program 2.5a Excel 2013 Output for the Exponential Distribution

Program 2.5b Function in an Excel 2013 Spreadsheet for the Exponential Distribution

1This distribution, derived by Siméon Denis Poisson in 1837, is pronounced “pwah-sahn.”

The Poisson probability distribution is used in many queuing models to represent arrival patterns.

M02_REND9327_12_SE_C02.indd 68 10/02/14 1:20 PM

2.11 the poisson distribution  69

where

P1X2 = probability of exactly X arrivals or occurrences l =  average number of arrivals per unit of time (the mean arrival rate), pronounced “lambda” e = 2.718,  the base of the natural logarithm X = number of occurrences 10, 1, 2, c2

The mean and variance of the Poisson distribution are equal and are computed simply as

Expected value = l (2-19)

Variance = l (2-20)

With the help of the table in Appendix C, the values of e-l are easy to find. We can use these in  the formula to find probabilities. For example, if l = 2, from Appendix C we find e-2 = 0.1353. The Poisson probabilities that X is 0, 1, and 2 when l = 2 are as follows:

P1X2 = e -llx

X!

P102 = e -220

0! =

10.135321 1

= 0.1353 ≈ 14%

P112 = e -221

1! =

e-22

1 =

0.1353122 1

= 0.2706 ≈ 27%

P122 = e -222

2! =

e-24

2112 = 0.1353142

2 = 0.2706 ≈ 27%

These probabilities, as well as others for l = 2 and l = 4, are shown in Figure 2.18. Notice  that the chances that 9 or more customers will arrive in a particular time period are virtually nil.  Programs 2.6A and 2.6B illustrate how Excel can be used to find Poisson probabilities.

It should be noted that the exponential and Poisson distributions are related. If the number of oc- currences per time period follows a Poisson distribution, then the time between occurrences follows  an exponential distribution. For example, if the number of phone calls arriving at a customer service  center followed a Poisson distribution with a mean of 10 calls per hour, the time between each phone  call would be exponentially distributed with a mean time between calls of 1>10  hour (6 minutes).

Figure 2.18 Sample Poisson Distributions with L = 2 and L = 4

0.30

0.25

0.20

0.15

0.10

0.05

0.00 0 1 2 3 4 5

X

λ 5 2 Distribution λ 5 4 Distribution

P ro

ba bi

lit y

6 7 8 9

0.25

0.20

0.15

0.10

0.05

0.00 0 1 2 3 4 5

X

P ro

ba bi

lit y

6 7 8 9

M02_REND9327_12_SE_C02.indd 69 10/02/14 1:20 PM

70  Chapter 2 • probability ConCepts and appliCations

This chapter presents the fundamental concepts of probability  and probability distributions. Probability values can be ob- tained objectively or subjectively. A single probability value  must be between 0 and 1, and the sum of all probability val- ues for all possible outcomes must be equal to 1. In addition,  probability values and events can have a number of properties.  These properties include mutually exclusive, collectively ex- haustive, statistically independent, and statistically dependent  events. Rules for computing probability values depend on these  fundamental properties. It is also possible to revise probability  values when new information becomes available. This can be  done using Bayes’ theorem.

We also covered the topics of random variables, discrete  probability distributions (such as Poisson and binomial), and  continuous probability distributions (such as normal, F, and  exponential). A probability distribution is any statement of a  probability function having a set of collectively exhaustive and  mutually exclusive events. All probability distributions follow  the basic probability rules mentioned previously.

The topics presented here will be very important in many  of the chapters to come. Basic probability concepts and distri- butions are used for decision theory, inventory control, Markov  analysis, project management, simulation, and statistical qual- ity control.

Summary

Glossary

Bayes’ Theorem  A formula that is used to revise probabili- ties based on new information.

Bernoulli Process  A process with two outcomes in each of a  series of independent trials in which the probabilities of the  outcomes do not change.

Binomial Distribution  A discrete distribution that describes  the number of successes in independent trials of a Bernoulli  process.

Classical or Logical Method  An objective way of assessing  probabilities based on logic.

Collectively Exhaustive Events  A collection of all possible  outcomes of an experiment.

Conditional Probability  The probability of one event occur- ring given that another has taken place.

Continuous Probability Distribution  A probability distribu- tion with a continuous random variable.

Continuous Random Variable  A random variable that can  assume an infinite or unlimited set of values.

Discrete Probability Distribution  A probability distribution  with a discrete random variable.

Discrete Random Variable  A random variable that can only  assume a finite or limited set of values.

Expected Value  The (weighted) average of a probability  distribution.

F Distribution  A continuous probability distribution that is  the ratio of the variances of samples from two independent  normal distributions.

Independent Events  The situation in which the occurrence  of one event has no effect on the probability of occurrence  of a second event.

Intersection  The set of all outcomes that are common to  both events.

Joint Probability  The probability of events occurring to- gether (or one after the other).

Mutually Exclusive Events  A situation in which only one  event can occur on any given trial or experiment.

Negative Exponential Distribution  A continuous probabil- ity distribution that describes the time between customer  arrivals in a queuing situation.

Normal Distribution  A continuous bell-shaped distribution  that is a function of two parameters, the mean and standard  deviation of the distribution.

Objective Approach  A method of determining probability  values based on historical data or logic.

Poisson Distribution  A discrete probability distribution  used in queuing theory.

Prior Probability  A probability value determined before  new or additional information is obtained. It is sometimes  called an a priori probability estimate.

Program 2.6a Excel 2013 Output for the Poisson Distribution

Program 2.6b Functions in an Excel 2013 Spreadsheet for the Poisson Distribution

M02_REND9327_12_SE_C02.indd 70 10/02/14 1:20 PM

Key equations  71

Key Equations

 (2-1)  0 … P1event2 … 1 A basic statement of probability.

 (2-2)  P1A or B2 = P1A2 + P1B2 - P1A and B2 Probability of the union of two events.

 (2-3)  P1A � B2 = P1AB2 P1B2

Conditional probability.

 (2-4)  P1AB2 = P1A � B2P1B2 Probability of the intersection of two events.

 (2-5)  P1A � B2 = P1B � A2P1A2 P1B � A2P1A2 + P1B � A′2P1A′2

Bayes’ theorem in general form.

 (2-6)  E1X2 = gn i=1

XiP1Xi2

An equation that computes the expected value (mean)  of a discrete probability distribution.

 (2-7)  s2 = Variance = gn i=1

3Xi - E1X242P1Xi2

An equation that computes the variance of a discrete  probability distribution.

 (2-8)  s = 1Variance = 2s2 An equation that computes the standard deviation from  the variance.

 (2-9)  Probability of r successes in n trials = n!

r!1n - r2!p r qn - r

A formula that computes probabilities for the binomial  probability distribution.

 (2-10)  Expected value 1mean2 = np The expected value of the binomial distribution.

 (2-11)  Variance = np11 - p2 The variance of the binomial distribution.

 (2-12)  f 1X2 = 1 s12p e - 1x -m222s2

The  density  function  for  the  normal  probability  distribution.

 (2-13)  Z = X - m

s

An equation that computes the number of standard  deviations, Z, the point X is from the mean m.

 (2-14)  f 1X ) = me-mx The exponential distribution.

 (2-15)  Expected value = 1 m

The expected value of an exponential distribution.

 (2-16)  Variance = 1

m2

The variance of an exponential distribution.

 (2-17)  P1X … t2 = 1 - e-mt Formula to find the probability that an exponential  random variable (X) is less than or equal to time t.

 (2-18)  P1X2 = l xe-l

X! The Poisson distribution.

 (2-19)  Expected value = l

The mean of a Poisson distribution.

 (2-20)  Variance = l

The variance of a Poisson distribution.

Probability  A statement about the likelihood of an event   occurring. It is expressed as a numerical value between 0  and 1, inclusive.

Probability Density Function  The mathematical function  that describes a continuous probability distribution. It is  represented by f 1X2.

Probability Distribution  The set of all possible values of a  random variable and their associated probabilities.

Random Variable  A variable that assigns a number to every  possible outcome of an experiment.

Relative Frequency Approach  An objective way of deter- mining probabilities based on observing frequencies over a  number of trials.

Revised or Posterior Probability  A probability value that re- sults from new or revised information and prior probabilities.

Standard Deviation  The square root of the variance. Subjective Approach  A method of determining probability 

values based on experience or judgment. Union  The set of all outcomes that are contained in either of 

these two events. Variance  A measure of dispersion or spread of the 

 probability distribution.

M02_REND9327_12_SE_C02.indd 71 10/02/14 1:20 PM

72  Chapter 2 • probability ConCepts and appliCations

Solved Problem 2-1 In the past 30 days, Roger’s Rural Roundup has sold 8, 9, 10, or 11 lottery tickets. It never sold fewer than 8  or more than 11. Assuming that the past is similar to the future, find the probabilities for the number of tick- ets sold if sales were 8 tickets on 10 days, 9 tickets on 12 days, 10 tickets on 6 days, and 11 tickets on 2 days.

solution

sales no. Days probability

8 10 0.333

9 12 0.400

10 6 0.200

11 2 0.067

Total 30 1.000

Solved Problem 2-2 A class contains 30 students. Ten are female (F) and U.S. citizens (U); 12 are male (M) and U.S. citi- zens; 6 are female and non-U.S. citizens (N); 2 are male and non-U.S. citizens.

A name is randomly selected from the class roster and it is female. What is the probability that the  student is a U.S. citizen?

solution P1FU2 = 10>30 = 0.333 P1FN2 = 6>30 = 0.200

P1MU2 = 12>30 = 0.400 P1MN2 = 2>30 = 0.067

P1F2 = P1FU2 + P1FN2 = 0.333 + 0.200 = 0.533 P1M2 = P1MU2 + P1MN2 = 0.400 + 0.067 = 0.467 P1U2 = P1FU2 + P1MU2 = 0.333 + 0.400 = 0.733 P1N2 = P1FN2 + P1MN2 = 0.200 + 0.067 = 0.267

P1U �F2 = P1FU2 P1F2 =

0.333

0.533 = 0.625

Solved Problem 2-3 Your professor tells you that if you score an 85 or better on your midterm exam, then you have a 90%  chance of getting an A for the course. You think you have only a 50% chance of scoring 85 or better.  Find the probability that both your score is 85 or better and you receive an A in the course.

solution

P1A and 852 = P1A � 852 * P1852 = 10.90210.502 = 45%

Solved Problem 2-4 A statistics class was asked if it believed that all tests on the Monday following the football game win  over their archrival should be postponed automatically. The results were as follows:

Strongly agree 40

Agree 30

Neutral 20

Disagree 10

Strongly disagree 0

100

Solved Problems

M02_REND9327_12_SE_C02.indd 72 10/02/14 1:20 PM

solved problems  73

Transform this into a numeric score, using the following random variable scale, and find a prob- ability distribution for the results:

Strongly agree 5

Agree 4

Neutral 3

Disagree 2

Strongly disagree 1

solution

outCome probability, P 1X2 Strongly agree (5) 0.4 = 40>100 Agree (4) 0.3 = 30>100 Neutral (3) 0.2 = 20>100 Disagree (2) 0.1 = 10>100 Strongly disagree (1) 0.0 = 0>100 Total 1.0 = 100>100

Solved Problem 2-5 For Solved Problem 2-4, let X be the numeric score. Compute the expected value of X.

solution

E1X2 = g5 i=1

XiP1Xi2 = X1P1X12 + X2P1X22 + X3P1X32 + X4P1X42 + X5P1X52

= 510.42 + 410.32 + 310.22 + 210.12 + 1102 = 4.0

Solved Problem 2-6 Compute the variance and standard deviation for the random variable X in Solved Problems 2-4 and 2-5.

solution

Variance = g5 i=1

1Xi - E1X222P1Xi2 = 15 - 42210.42 + 14 - 42210.32 + 13 - 42210.22 + 12 - 42210.12 + 11 - 42210.02 = 112210.42 + 102210.32 + 1-12210.22 + 1-22210.12 + 1-32210.02 = 0.4 + 0.0 + 0.2 + 0.4 + 0.0 = 1.0

The standard deviation is s = 1Variance = 11 = 1

Solved Problem 2-7 A candidate for public office has claimed that 60% of voters will vote for her. If 5 registered voters were  sampled, what is the probability that exactly 3 would say they favor this candidate?

solution We use the binomial distribution with n = 5, p = 0.6, and r = 3:

P1exactly 3 successes in 5 trials2 = n! r!1n - r2!p

r qn - r = 5!

3!15 - 32!10.62 310.425- 3 = 0.3456

M02_REND9327_12_SE_C02.indd 73 10/02/14 1:20 PM

74  Chapter 2 • probability ConCepts and appliCations

Solved Problem 2-8 The length of the rods coming out of our new cutting machine can be said to approximate a normal dis- tribution with a mean of 10 inches and a standard deviation of 0.2 inch. Find the probability that a rod  selected randomly will have a length

(a) of less than 10.0 inches (b) between 10.0 and 10.4 inches (c) between 10.0 and 10.1 inches (d) between 10.1 and 10.4 inches (e) between 9.6 and 9.9 inches (f ) between 9.9 and 10.4 inches (g) between 9.886 and 10.406 inches

solution First, compute the standard normal distribution, the Z value:

Z = X - m

s

Next, find the area under the curve for the given Z value by using a standard normal distribution table.

(a) P1X 6 10.02 = 0.50000 (b) P110.0 6 X 6 10.42 = 0.97725 - 0.50000 = 0.47725 (c) P110.0 6 X 6 10.12 = 0.69146 - 0.50000 = 0.19146 (d) P110.1 6 X 6 10.42 = 0.97725 - 0.69146 = 0.28579 (e) P19.6 6 X 6 9.92 = 0.97725 - 0.69146 = 0.28579 (f) P19.9 6 X 6 10.42 = 0.19146 + 0.47725 = 0.66871 (g) P19.886 6 X 6 10.4062 = 0.47882 + 0.21566 = 0.69448

Self-Test

●  Before taking the self-test, refer to the learning objectives at the beginning of the chapter, the notes in the margins, and the  glossary at the end of the chapter.

● Use the key at the back of the book to correct your answers. ●  Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

  1.  If the list of outcomes includes every possible outcome,  then the events are said to be

a. independent. b. collectively exhaustive. c. mutually exclusive. d. continuous.   2.  New probabilities that have been found using Bayes’ 

theorem are called a. prior probabilities. b. posterior probabilities. c. Bayesian probabilities. d. joint probabilities.   3.  A measure of central tendency is a. expected value. b. variance. c. standard deviation. d. all of the above.

  4.  To compute the variance, of a discrete random variable  you need to know the

a. variable’s possible values. b. expected value of the variable. c. probability of each possible value of the variable. d. all of the above.   5.  The square root of the variance is the a. expected value. b. standard deviation. c. area under the normal curve. d. all of the above.   6.  Which of the following is an example of a continuous 

distribution? a. the binomial distribution. b. the exponential distribution. c. the Poisson distribution. d. the Z distribution.

M02_REND9327_12_SE_C02.indd 74 10/02/14 1:20 PM

disCussion questions and problems  75

  7.  The total area under the curve for any continuous  distribution must equal

a. 1. b. 0. c. 0.5. d. none of the above.   8.  Probabilities for all the possible values of a discrete 

random variable a. may be greater than 1. b. may be negative on some occasions. c. must sum to 1. d. are represented by area underneath the curve.   9.  In a standard normal distribution, the mean is equal to a. 1. b. 0. c. the variance. d. the standard deviation.  10.  The probability of an event occurring given that another 

event has taken place is a. priory probability. b. empirical probability. c. conditional probability. d. subjective probability. e. none of the above.  11.  In the normal distribution, 99.7% of the population lies 

within a. 1 standard deviation of the mean. b. 2 standard deviations of the mean. c. 3 standard deviations of the mean. d. 4 standard deviations of the mean.

 12.  If a normal distribution has a mean of 100 and a standard  deviation of 15, 95.4% of the population falls within  what range of values?

a. 70–130 b. 75–130 c. 70–135 d. 80–140 e. 60–120  13.  If two events are mutually exclusive, then the probability 

of the intersection of these two events will equal a. 0. b. 0.5. c. 1.0. d. cannot be determined without more information.  14.  If P1A2 = 0.4 and P1B2 = 0.5 and P1A and B2 = 0.2,

then P1B � A2 = a. 0.80. b. 0.50. c. 0.10. d. 0.40. e. none of the above.  15.  If P1A2 = 0.4 and P1B2 = 0.5 and P1A and B2 = 0.2,

then P1A or B2 = a. 0.7. b. 0.9. c. 1.1. d. 0.2. e. none of the above.

Discussion Questions and Problems

Discussion Questions   2-1  What are the two basic laws of probability?

  2-2  What is the meaning of mutually exclusive events?  What is meant by collectively exhaustive? Give an  example of each.

  2-3  Describe the classical method to determine probabil- ity with example.

  2-4  Describe the addition law for events that are mutu- ally exclusive and events that are not.

  2-5  Describe what it means for two events to be statisti- cally dependent.

  2-6  Bayes’ theorem is an extension of the original prob- ability. Explain.

  2-7  Describe the characteristics of a Bernoulli process.  How is a Bernoulli process associated with the bino- mial distribution?

  2-8  What is a random variable? What are the various  types of random variables?

  2-9  What is the difference between a discrete probability  distribution and a continuous probability distribu- tion? Give your own example of each.

  2-10  What is the expected value, and what does it mea- sure? How is it computed for a discrete probability  distribution?

  2-11  What is the variance, and what does it measure?  How  is  it  computed  for  a  discrete  probability  distribution?

  2-12  Why is normal distribution converted to a standard  normal distribution?

  2-13  A card is drawn from a standard deck of playing  cards. For each of the following pairs of events, indi- cate if the events are mutually exclusive and indicate  if the events are exhaustive.

(a) Draw a spade and draw a club. (b) Draw a face card and draw a number card. (c) Draw an ace and draw a three.

M02_REND9327_12_SE_C02.indd 75 10/02/14 1:20 PM

76  Chapter 2 • probability ConCepts and appliCations

(d) Draw a red card and draw a black card. (e) Draw a five and draw a diamond. (f)  Draw a red card and draw a diamond.

Problems  2-14  A  student  taking Management  Science  301  at 

East Haven University will receive one of the five   possible grades for the course: A, B, C, D, or F. The  distribution of grades over the past 2 years is as  follows:

GraDe number of stuDents

A 80

B 75

C 90

D 30

F 25

Total 300

    If this past distribution is a good indicator of future  grades, what is the probability of a student receiving  a C in the course?

 2-15  A silver dollar is flipped twice. Calculate the prob- ability of each of the following occurring:

(a)  a head on the first flip (b) a tail on the second flip given that the first toss 

was a head (c)  two tails (d) a tail on the first and a head on the second (e)  a tail on the first and a head on the second or a 

head on the first and a tail on the second (f)  at least one head on the two flips

 2-16  An urn contains 8 red chips, 10 green chips, and  2 white chips. A chip is drawn and replaced, and  then a second chip drawn. What is the probability of

(a)  a white chip on the first draw? (b) a white chip on the first draw and a red on the 

second? (c)  two green chips being drawn? (d) a red chip on the second, given that a white chip 

was drawn on the first?

 2-17  Evertight, a leading manufacturer of quality nails,  produces 1-, 2-, 3-, 4-, and 5-inch nails for various  uses. In the production process, if there is an over- run or the nails are slightly defective, they are placed  in a common bin. Yesterday, 651 of the 1-inch nails,  243 of the 2-inch nails, 41 of the 3-inch nails, 451  of the 4-inch nails, and 333 of the 5-inch nails were  placed in the bin.

(a) What is the probability of reaching into the bin  and getting a 4-inch nail?

(b) What is the probability of getting a 5-inch nail?

(c)  If a particular application requires a nail that is  3 inches or shorter, what is the probability of  getting a nail that will satisfy the requirements of  the application?

 2-18  Last year, at Northern Manufacturing Company, 200  people had colds during the year. One hundred fifty- five people who did no exercising had colds, and the  remainder of the people with colds were involved  in a weekly exercise program. Half of the 1,000   employees were involved in some type of exercise.

(a) What is the probability that an employee will  have a cold next year?

(b) Given that an employee is involved in an exer- cise program, what is the probability that he or  she will get a cold next year?

(c) What is the probability that an employee who is  not involved in an exercise program will get a  cold next year?

(d) Are exercising and getting a cold independent  events? Explain your answer.

 2-19  The Springfield Kings, a professional basketball team,  has won 12 of its last 20 games and is expected to con- tinue winning at the same percentage rate. The team’s  ticket manager is anxious to attract a large crowd to  tomorrow’s game but believes that depends on how  well the Kings perform tonight against the Galves- ton Comets. He assesses the probability of drawing a  large crowd to be 0.90 should the team win tonight.  What is the probability that the team wins tonight and  that there will be a large crowd at tomorrow’s game?

 2-20  David Mashley teaches two undergraduate statistics  courses at Kansas College. The class for Statistics  201 consists of 7 sophomores and 3 juniors. The  more advanced course, Statistics 301, has 2 sopho- mores and 8 juniors enrolled. As an example of a  business sampling technique, Professor Mashley  randomly selects, from the stack of Statistics 201  registration cards, the class card of one student and  then places that card back in the stack. If that student  was a sophomore, Mashley draws another card from  the Statistics 201 stack; if not, he randomly draws  a card from the Statistics 301 group. Are these two  draws independent events? What is the probability of

(a)  a junior’s name on the first draw? (b) a junior’s name on the second draw, given that a 

sophomore’s name was drawn first? (c)  a junior’s name on the second draw, given that a 

junior’s name was drawn first? (d) a sophomore’s name on both draws? (e)  a junior’s name on both draws? (f)  one sophomore’s name and one junior’s name on 

the two draws, regardless of order drawn?

Note:   means the problem may be solved with QM for Windows;   means the problem may be solved with Excel QM; and   means the problem may be  solved with QM for Windows and/or Excel QM.

M02_REND9327_12_SE_C02.indd 76 10/02/14 1:20 PM

disCussion questions and problems  77

 2-21  The oasis outpost of Abu Ilan, in the heart of the   Negev desert, has a population of 20 Bedouin tribes- men and 20 Farima tribesmen. El Kamin, a nearby  oasis, has a population of 32 Bedouins and 8 Farima.  A lost Israeli soldier, accidentally separated from his  army unit, is wandering through the desert and ar- rives at the edge of one of the oases. The soldier has  no idea which oasis he has found, but the first person  he spots at a distance is a Bedouin. What is the prob- ability that he wandered into Abu Ilan? What is the  probability that he is in El Kamin?

 2-22  The lost Israeli soldier mentioned in Problem 2-21  decides to rest for a few minutes before entering the  desert oasis he has just found. Closing his eyes, he  dozes off for 15 minutes, wakes, and walks toward  the center of the oasis. The first person he spots this  time he again recognizes as a Bedouin. What is the  posterior probability that he is in El Kamin?

 2-23  Ace Machine Works estimates that the probability  its lathe tool is properly adjusted is 0.8. When the  lathe is properly adjusted, there is a 0.9 probability  that the parts produced pass inspection. If the lathe  is out of adjustment, however, the probability of a  good part being produced is only 0.2. A part ran- domly chosen is inspected and found to be accept- able. At this point, what is the posterior probability  that the lathe tool is properly adjusted?

 2-24  The Boston South Fifth Street Softball League  consists of three teams: Mama’s Boys, team 1; the   Killers, team 2; and the Machos, team 3. Each team  plays the other teams just once during the season.  The win–loss record for the past 5 years is as follows:

Winner (1) (2) (3)

Mama’s Boys (1) X 3 4

The Killers (2) 2 X 1

The Machos (3) 1 4 X

    Each row represents the number of wins over the  past 5 years. Mama’s Boys beat the Killers 3 times,  beat the Machos 4 times, and so on.

(a) What is the probability that the Killers will win  every game next year?

(b) What is the probability that the Machos will win  at least one game next year?

(c) What is the probability that Mama’s Boys will  win exactly one game next year?

(d) What is the probability that the Killers will win  fewer than two games next year?

 2-25  The schedule for the Killers next year is as follows  (refer to Problem 2-24):

    Game 1: The Machos

    Game 2: Mama’s Boys

(a) What is the probability that the Killers will win  their first game?

(b) What is the probability that the Killers will win  their last game?

(c) What is the probability that the Killers will break  even—win exactly one game?

(d) What is the probability that the Killers will win  every game?

(e) What is the probability that the Killers will lose  every game?

(f)  Would you want to be the coach of the Killers?

 2-26  The Northside Rifle team has two markspersons,  Dick and Sally. Dick hits a bull’s-eye 90% of the  time, and Sally hits a bull’s-eye 95% of the time.

(a) What is the probability that either Dick or Sally or  both will hit the bull’s-eye if each takes one shot?

(b) What is the probability that Dick and Sally will  both hit the bull’s-eye?

(c) Did you make any assumptions in answering  the preceding questions? If you answered yes,  do you think that you are justified in making the  assumption(s)?

 2-27  In a sample of 1,000 representing a survey from the  entire population, 650 people were from Laketown,  and the rest of the people were from River City. Out  of the sample, 19 people had some form of cancer.  Thirteen of these people were from Laketown.

(a) Are the events of living in Laketown and having  some sort of cancer independent?

(b) Which city would you prefer to live in, assum- ing that your main objective was to avoid having  cancer?

 2-28  Compute the probability of “loaded die, given that a  3 was rolled,” as shown in the example in Section 2.3,  this time using the general form of Bayes’ theorem  from Equation 2-5.

 2-29  Which of the following are probability distributions?  Why?

(a) 

ranDom Variable X probability

2 0.1

-1 0.2 0 0.3

1 0.25

2 0.15

(b) 

ranDom Variable Y probability

1 1.1

1.5 0.2

2 0.3

2.5 0.25

3 -1.25

M02_REND9327_12_SE_C02.indd 77 10/02/14 1:20 PM

78  Chapter 2 • probability ConCepts and appliCations

(c) 

ranDom Variable z probability

1 0.1

2 0.2

3 0.3

4 0.4

5 0.0

 2-30  Harrington Health Food stocks 5 loaves of Neutro- Bread. The probability distribution for the sales of  Neutro-Bread is listed in the following table. How  many loaves will Harrington sell on average?

number of loaVes solD probability

0 0.05

1 0.15

2 0.20

3 0.25

4 0.20

5 0.15

 2-31  What are the expected value and variance of the   following probability distribution?

ranDom Variable X probability

1 0.05

2 0.05

3 0.10

4 0.10

5 0.15

6 0.15

7 0.25

8 0.15

 2-32  There are 10 questions on a true–false test. A student  feels unprepared for this test and randomly guesses  the answer for each of these.

(a) What  is  the probability  that  the student gets   exactly 7 correct?

(b) What  is  the probability  that  the student gets   exactly 8 correct?

(c) What  is  the probability  that  the student gets   exactly 9 correct?

(d) What  is  the probability  that  the student gets   exactly 10 correct?

(e) What is the probability that the student gets more  than 6 correct?

 2-33  Gary Schwartz is the top salesman for his company.  Records indicate that he makes a sale on 70% of his  sales calls. If he calls on four potential clients, what  is the probability that he makes exactly 3 sales? What  is the probability that he makes exactly 4 sales?

 2-34  If 10% of all disk drives produced on an assembly  line are defective, what is the probability that there 

will be exactly one defect in a random sample of 5  of these? What is the probability that there will be no  defects in a random sample of 5?

 2-35  Trowbridge Manufacturing produces cases for per- sonal computers and other electronic equipment. The  quality control inspector for this company believes  that a particular process is out of control. Normally,  only 5% of all cases are deemed defective due to   discolorations. If 6 such cases are sampled, what is  the probability that there will be 0 defective cases if  the process is operating correctly? What is the prob- ability that there will be exactly 1 defective case?

 2-36  Refer to the Trowbridge Manufacturing example in  Problem 2-35. The quality control inspection pro- cedure is to select 6 items, and if there are 0 or 1  defective cases in the group of 6, the process is said  to be in control. If the number of defects is more  than 1, the process is out of control. Suppose that  the true proportion of defective items is 0.15. What  is the probability that there will be 0 or 1 defects in a  sample of 6 if the true proportion of defects is 0.15?

 2-37  An industrial oven used to cure sand cores for a fac- tory manufacturing engine blocks for small cars is  able to maintain fairly constant temperatures. The  temperature range of the oven follows a normal dis- tribution with a mean of 450°F and a standard devia- tion of 25°F. Leslie Larsen, president of the factory,  is concerned about the large number of defective  cores that have been produced in the past several  months. If the oven gets hotter than 475°F, the core  is defective. What is the probability that the oven  will cause a core to be defective? What is the prob- ability that the temperature of the oven will range  from 460° to 470°F?

 2-38  Steve Goodman, production foreman for the Florida  Gold Fruit Company, estimates that the average sale  of oranges is 4,700 and the standard deviation is 500  oranges. Sales follow a normal distribution.

(a) What is the probability that sales will be greater  than 5,500 oranges?

(b) What is the probability that sales will be greater  than 4,500 oranges?

(c) What is the probability that sales will be less  than 4,900 oranges?

(d) What is the probability that sales will be less  than 4,300 oranges?

 2-39  Susan Williams has been the production manager  of Medical Suppliers, Inc., for the past 17 years.  Medical Suppliers, Inc., is a producer of bandages  and arm slings. During the past 5 years, the demand  for No-Stick bandages has been fairly constant. On  the average, sales have been about 87,000 pack- ages of No-Stick. Susan has reason to believe that  the distribution of No-Stick follows a normal curve,  with a standard deviation of 4,000 packages. What  is the probability that sales will be less than 81,000  packages?

M02_REND9327_12_SE_C02.indd 78 10/02/14 1:20 PM

disCussion questions and problems  79

 2-40  Armstrong Faber produces a standard number-two  pencil called Ultra-Lite. Since Chuck Armstrong  started Armstrong Faber, sales have grown steadily.  With the increase in the price of wood products,  however, Chuck has been forced  to  increase  the  price of the Ultra-Lite pencils. As a result, the de- mand for Ultra-Lite has been fairly stable over the  past 6 years. On the average, Armstrong Faber has  sold 457,000 pencils each year. Furthermore, 90%  of the time sales have been between 454,000 and  460,000 pencils. It is expected that the sales follow a  normal distribution with a mean of 457,000 pencils.  Estimate the standard deviation of this distribution.  (Hint: Work backward from the normal table to find  Z. Then apply Equation 2-13.)

 2-41  The time to complete a construction project is nor- mally distributed with a mean of 60 weeks and a  standard deviation of 4 weeks.

(a) What  is  the  probability  the  project will  be   finished in 62 weeks or less?

(b) What  is  the  probability  the  project will  be   finished in 66 weeks or less?

(c) What  is  the probability  the project will  take   longer than 65 weeks?

 2-42  A new integrated computer system is to be installed  worldwide for a major corporation. Bids on this  project are being solicited, and the contract will be  awarded to one of the bidders. As a part of the pro- posal for this project, bidders must specify how long  the project will take. There will be a significant pen- alty for finishing late. One potential contractor de- termines that the average time to complete a project  of this type is 40 weeks with a standard deviation of  5 weeks. The time required to complete this project  is assumed to be normally distributed.

(a)  If the due date of this project is set at 40 weeks,  what is the probability that the contractor will  have to pay a penalty (i.e., the project will not be  finished on schedule)?

(b) If the due date of this project is set at 43 weeks,  what is the probability that the contractor will  have to pay a penalty (i.e., the project will not be  finished on schedule)?

(c)  If the bidder wishes to set the due date in the  proposal so that there is only a 5% chance of be- ing late (and consequently only a 5% chance of  having to pay a penalty), what due date should  be set?

 2-43  Patients arrive at  the emergency  room of Costa   Valley Hospital at an average of 5 per day. The de- mand for emergency room treatment at Costa Valley  follows a Poisson distribution.

(a) Using Appendix C, compute the probability of  exactly 0, 1, 2, 3, 4, and 5 arrivals per day.

(b) What is the sum of these probabilities, and why  is the number less than 1?

 2-44  Using the data in Problem 2-43, determine the prob- ability of more than 3 visits for emergency room ser- vice on any given day.

 2-45  Cars arrive at Carla’s Muffler shop for repair work  at an average of 3 per hour, following an exponential  distribution.

(a) What is the expected time between arrivals? (b) What is the variance of the time between arrivals?

 2-46  A particular test for the presence of steroids is to be  used after a professional track meet. If steroids are  present, the test will accurately indicate this 95% of  the time. However, if steroids are not present, the test  will indicate this 90% of the time (so it is wrong 10%  of the time and predicts the presence of steroids).  Based on past data, it is believed that 2% of the ath- letes do use steroids. This test is administered to one  athlete, and the test is positive for steroids. What is  the probability that this person actually used steroids?

 2-47  Market Researchers, Inc., has been hired to perform  a study to determine if the market for a new product  will be good or poor. In similar studies performed  in the past, whenever the market actually was good,  the market research study indicated that it would  be good 85% of the time. On the other hand, when- ever the market actually was poor, the market study  incorrectly predicted it would be good 20% of the  time. Before the study is performed, it is believed  there  is a 70% chance  the market will be good.  When Market Researchers, Inc., performs the study  for this product, the results predict the market will  be good. Given the results of this study, what is the  probability that the market actually will be good?

 2-48  Policy Pollsters is a market research firm specializ- ing in political polls. Records indicate in past elec- tions, when a candidate was elected, Policy Pollsters  had accurately predicted this 80% of the time and  they were wrong 20% of  the time. Records also  show, for losing candidates, Policy Pollsters accu- rately predicted they would lose 90% of the time and  they were only wrong 10% of the time. Before the  poll is taken, there is a 50% chance of winning the  election. If Policy Pollsters predicts a candidate will  win the election, what is the probability that the can- didate will actually win? If Policy Pollsters predicts  that a candidate will lose the election, what is the  probability that the candidate will actually lose?

 2-49  Burger City is a large chain of fast-food restaurants  specializing in gourmet hamburgers. A mathemati- cal model is now used to predict the success of new  restaurants based on location and demographic in- formation for that area. In the past, 70% of all res- taurants  that were opened were successful. The  mathematical model has been tested in the existing  restaurants to determine how effective it is. For the  restaurants that were successful, 90% of the time the  model predicted they would be, while 10% of the 

M02_REND9327_12_SE_C02.indd 79 10/02/14 1:20 PM

80  Chapter 2 • probability ConCepts and appliCations

time the model predicted a failure. For the restau- rants that were not successful, when the mathemati- cal model was applied, 20% of the time it incorrectly  predicted a successful restaurant while 80% of the  time it was accurate and predicted an unsuccessful  restaurant. If the model is used on a new location  and predicts the restaurant will be successful, what  is the probability that it actually is successful?

 2-50  A mortgage lender attempted to increase its business  by marketing its subprime mortgage. This mortgage  is designed for people with a less-than-perfect credit  rating, and the interest rate is higher to offset the ex- tra risk. In the past year, 20% of these mortgages re- sulted in foreclosure as customers defaulted on their  loans. A new screening system has been developed to  determine whether to approve customers for the sub- prime loans. When the system is applied to a credit  application, the system will classify the application as  “Approve for loan” or “Reject for loan.” When this  new system was applied to recent customers who  had defaulted on their loans, 90% of these customers  were classified as “Reject.” When this same system  was applied to recent loan customers who had not de- faulted on their loan payments, 70% of these custom- ers were classified as “Approve for loan.”

(a)  If a customer did not default on a loan, what is  the probability that the rating system would have  classified the applicant in the reject category?

(b) If the rating system had classified the applicant  in the reject category, what is the probability that  the customer would not default on a loan?

 2-51  Use the F table in Appendix D to find the value of F for the upper 5% of the F distribution with

(a) df1 = 5, df2 = 10 (b) df1 = 8, df2 = 7 (c) df1 = 3, df2 = 5 (d) df1 = 10, df2 = 4

 2-52  Use the F table in Appendix D to find the value of F for the upper 1% of the F distribution with

(a) df1 = 15, df2 = 6 (b) df1 = 12, df2 = 8 (c) df1 = 3, df2 = 5 (d) df1 = 9, df2 = 7

 2-53  For each of the following F values, determine whether  the probability indicated is greater than or less than 5%:

(a) P1F3,4 7 6.82 (b) P1F7,3 7 3.62 (c) P1F20,20 7 2.62 (d) P1F7,5 7 5.12 (e) P1F7,5 6 5.12

 2-54  For each of the following F values, determine whether  the probability indicated is greater than or less than 1%:

(a) P1F5,4 7 142 (b) P1F6,3 7 302 (c) P1F10,12 7 4.22 (d) P1F2,3 7 352 (e) P1F2,3 6 352

 2-55  Nite Time Inn has a toll-free telephone number so  that customers can call at any time to make a reser- vation. A typical call takes about 4 minutes to com- plete, and the time required follows an exponential  distribution. Find the probability that a call takes

(a)  3 minutes or less (b) 4 minutes or less (c)  5 minutes or less (d) longer than 5 minutes

 2-56  During normal business hours on the east coast, calls  to the toll-free reservation number of the Nite Time Inn  arrive at a rate of 5 per minute. It has been determined  that the number of calls per minute can be described  by the Poisson distribution. Find the probability that in  the next minute, the number of calls arriving will be

(a)  exactly 5 (b) exactly 4 (c)  exactly 3 (d) exactly 6 (e)  less than 2

 2-57  In the Arnold’s Muffler example for the exponential  distribution in this chapter, the average rate of service  was given as 3 per hour, and the times were expressed  in hours. Convert the average service rate to the num- ber per minute and convert the times to minutes. Find  the probabilities that the service times will be less  than 1>2  hour, 1>3 hour, and 2>3  hour. Compare these  probabilities to the probabilities found in the example.

See our Internet home page, at www.pearsonglobaleditions.com/render,  for additional   homework problems, Problems 2-58 to 2-65.

Internet Homework Problems

M02_REND9327_12_SE_C02.indd 80 10/02/14 1:21 PM

appendix 2.1: derivation oF bayes’ theorem  81

WTVX, Channel 6, is located in Eugene, Oregon, home of the  University of Oregon’s football team. The station was owned  and operated by George Wilcox, a former Duck (University  of Oregon football player). Although there were other televi- sion stations in Eugene, WTVX was the only station that had  a weatherperson who was a member of the American Meteo- rological Society (AMS). Every night, Joe Hummel would be  introduced as the only weatherperson in Eugene who was a  member of the AMS. This was George’s idea, and he believed  that this gave his station the mark of quality and helped with  market share.

In addition to being a member of AMS, Joe was also the  most popular person on any of the local news programs. Joe  was always trying to find innovative ways to make the weather  interesting, and this was especially difficult during the winter  months when the weather seemed to remain the same over long  periods of time. Joe’s forecast for next month, for example, was  that there would be a 70% chance of rain every day, and that  what happens on one day (rain or shine) was not in any way  dependent on what happened the day before.

One of Joe’s most popular features of the weather report  was to invite questions during the actual broadcast. Questions 

would be phoned in, and they were answered on the spot by Joe.  Once a 10-year-old boy asked what caused fog, and Joe did an  excellent job of describing some of the various causes.

Occasionally, Joe would make a mistake. For example, a  high school senior asked Joe what the chances were of getting 15  days of rain in the next month (30 days). Joe made a quick calcu- lation: 170%2 * 115 days>30 days2 = 170%211>22 = 35,. Joe quickly found out what it was like being wrong in a univer- sity town. He had over 50 phone calls from scientists, mathema- ticians, and other university professors, telling him that he had  made a big mistake in computing the chances of getting 15 days  of rain during the next 30 days. Although Joe didn’t understand  all of the formulas the professors mentioned, he was determined  to find the correct answer and make a correction during a future  broadcast.

discussion Questions   1. What are the chances of getting 15 days of rain during the 

next 30 days?   2. What do you think about Joe’s assumptions concerning 

the weather for the next 30 days?

Case Study

WtvX

Bibliography

Berenson, Mark, David Levine, and Timothy Krehbiel. Basic Business Statis- tics, 10th ed. Upper Saddle River, NJ: Prentice Hall, 2006.

Campbell, S. Flaws and Fallacies in Statistical Thinking. Upper Saddle River,  NJ: Prentice Hall, 1974.

Feller, W. An Introduction to Probability Theory and Its Applications, Vols. 1  and 2. New York: John Wiley & Sons, Inc., 1957 and 1968.

Groebner, David, Patrick Shannon, Phillip Fry, and Kent Smith. Business Sta- tistics, 8th ed. Upper Saddle River, NJ: Prentice Hall, 2011.

Hanke, J. E., A. G. Reitsch, and D. W. Wichern. Business Forecasting, 9th ed.  Upper Saddle River, NJ: Prentice Hall, 2008.

Huff, D. How to Lie with Statistics. New York: W. W. Norton & Company,  Inc., 1954.

Newbold, Paul, William Carlson, and Betty Thorne. Statistics for Business and Economics, 6th ed. Upper Saddle River, NJ: Prentice Hall, 2007.

Appendix 2.1: Derivation of Bayes’ Theorem

We know that the following formulas are correct:

P1A � B2 = P1AB2 P1B2   (1)

P1B � A2 = P1AB2 P1A2

3which can be rewritten as P1AB2 = P1B � A2P1A24 and  (2)

P1B � A′2 = P1A′B2 P1A′2

3which can be rewritten as P1A′B2 = P1B � A′2P1A′24. (3)

M02_REND9327_12_SE_C02.indd 81 10/02/14 1:21 PM

82  Chapter 2 • probability ConCepts and appliCations

Furthermore, by definition, we know that

P1B2 = P1AB2 + P1A′B2 = P1B � A2P1A2 + P1B � A′2P1A′2  (4)

from (2) from (3) Substituting Equations 2 and 4 into Equation 1, we have

P1A � B2 = P1AB2 P1B2 from (2)

= P1B � A2P1A2

P1B � A2P1A2 + P1B � A′2P1A′2  (5)i from (4)

This is the general form of Bayes’ theorem, shown as Equation 2-5 in this chapter.

M02_REND9327_12_SE_C02.indd 82 10/02/14 1:21 PM

 83

6. Revise probability estimates using Bayesian analysis.

7. Use computers to solve basic decision-making problems.

8. Understand the importance and use of utility theory in decision making.

1. List the steps of the decision-making process.

2. Describe the types of decision-making environments.

3. Make decisions under uncertainty.

4. Use probability values to make decisions under risk.

5. Develop accurate and useful decision trees.

After completing this chapter, students will be able to:

Summary  •  Glossary  •  Key Equations  •  Solved Problems  •  Self-Test  •  Discussion Questions and Problems  •  Internet 

Homework Problems  •  Case Study: Starting Right Corporation  •  Case Study: Blake Electronics  •  Internet Case Studies

Bibliography

3.7 Using Software for Payoff Table Problems

3.8 Decision Trees

3.9 How Probability Values Are Estimated by Bayesian Analysis

3.10 Utility Theory

3.1 Introduction

3.2 The Six Steps in Decision Making

3.3 Types of Decision-Making Environments

3.4 Decision Making Under Uncertainty

3.5 Decision Making Under Risk

3.6 A Minimization Example

Chapter Outline

Chapter 3 Decision Analysis

learning ObjeCtives

M03_REND9327_12_SE_C03.indd 83 10/02/14 1:21 PM

84  Chapter 3 • DeCision analysis

3.1 Introduction

To a great extent, the successes or failures that a person experiences in life depend on the deci- sions that he or she makes. The person who managed the ill-fated space shuttle Challenger is no  longer working for NASA. The person who designed the top-selling Mustang became president  of Ford. Why and how did these people make their respective decisions? In general, what is in- volved in making good decisions? One decision may make the difference between a successful  career and an unsuccessful one. Decision theory is an analytic and systematic approach to the  study of decision making. In this chapter, we present the mathematical models useful in helping  managers make the best possible decisions.

What makes the difference between good and bad decisions? A good decision is one that is  based on logic, considers all available data and possible alternatives, and applies the quantita- tive approach we are about to describe. Occasionally, a good decision results in an unexpected  or unfavorable outcome. But if it is made properly, it is still a good decision. A bad decision  is one that is not based on logic, does not use all available information, does not consider all  alternatives, and does not employ appropriate quantitative techniques. If you make a bad deci- sion but are lucky and a favorable outcome occurs, you have still made a bad decision. Although   occasionally good decisions yield bad results, in the long run, using decision theory will result  in successful outcomes.

3.2 The Six Steps in Decision Making

Whether you are deciding about getting a haircut today, building a multimillion-dollar plant, or  buying a new camera, the steps in making a good decision are basically the same:

Six Steps in Decision Making

  1. Clearly define the problem at hand.   2. List the possible alternatives.   3.  Identify the possible outcomes or states of nature.   4. List the payoff (typically profit) of each combination of alternatives and outcomes.   5. Select one of the mathematical decision theory models.   6. Apply the model and make your decision.

We use the Thompson Lumber Company case as an example to illustrate these decision  theory steps. John Thompson is the founder and president of Thompson Lumber Company, a  profitable firm located in Portland, Oregon.

Step 1. The problem that John Thompson identifies is whether to expand his product line by  manufacturing and marketing a new product, backyard storage sheds.

Thompson’s second step is to generate the alternatives that are available to him. In decision  theory, an alternative is defined as a course of action or a strategy that the decision maker can  choose.

Step 2. John decides that his alternatives are to construct (1) a large new plant to manufacture  the storage sheds, (2) a small plant, or (3) no plant at all (i.e., he has the option of not developing  the new product line).

One of the biggest mistakes that decision makers make is to leave out some important alter- natives. Although a particular alternative may seem to be inappropriate or of little value, it might  turn out to be the best choice.

The next step involves identifying the possible outcomes of the various alternatives. A com- mon mistake is to forget about some of the possible outcomes. Optimistic decision makers tend  to ignore bad outcomes, whereas pessimistic managers may discount a favorable outcome. If  you don’t consider all possibilities, you will not be making a logical decision, and the results  may be undesirable. If you do not think the worst can happen, you may design another Edsel  automobile. In decision theory, those outcomes over which the decision maker has little or no  control are called states of nature.

Decision theory is an analytic and systematic way to tackle problems.

A good decision is based on logic.

The first step is to define the problem.

The second step is to list alternatives.

M03_REND9327_12_SE_C03.indd 84 10/02/14 1:21 PM

3.3 types of DeCision-Making environMents  85

Step 3. Thompson determines that there are only two possible outcomes: the market for the stor- age sheds could be favorable, meaning that there is a high demand for the product, or it could be  unfavorable, meaning that there is a low demand for the sheds.

Once the alternatives and states of nature have been identified, the next step is to express  the payoff resulting from each possible combination of alternatives and outcomes. In decision  theory, we call such payoffs or profits conditional values. Not every decision, of course, can be  based on money alone—any appropriate means of measuring benefit is acceptable.

Step 4. Because Thompson wants to maximize his profits, he can use profit to evaluate each  consequence.

John Thompson has already evaluated the potential profits associated with the various   outcomes. With a favorable market, he thinks a large facility would result in a net profit of  $200,000 to his firm. This $200,000 is a conditional value because Thompson’s receiving the  money is conditional upon both his building a large factory and having a good market. The con- ditional value if the market is unfavorable would be a $180,000 net loss. A small plant would  result in a net profit of $100,000 in a favorable market, but a net loss of $20,000 would occur  if the market was unfavorable. Finally, doing nothing would result in $0 profit in either market.  The easiest way to present these values is by constructing a decision table, sometimes called  a payoff table. A decision table for Thompson’s conditional values is shown in Table 3.1. All  of the alternatives are listed down the left side of the table, and all of the possible outcomes or  states of nature are listed across the top. The body of the table contains the actual payoffs.

Steps 5 and 6. The last two steps are to select a decision theory model and apply it to the data to  help make the decision. Selecting the model depends on the environment in which you’re oper- ating and the amount of risk and uncertainty involved.

3.3 Types of Decision-Making Environments

The types of decisions people make depend on how much knowledge or information they have  about the situation. There are three decision-making environments:

● Decision making under certainty

● Decision making under uncertainty

● Decision making under risk

In the environment of decision making under certainty, decision makers know with cer- tainty the consequence of every alternative or decision choice. Naturally, they will choose the  alternative that will maximize their well-being or will result in the best outcome. For example,  let’s say that you have $1,000 to invest for a 1-year period. One alternative is to open a savings  account paying 4% interest and another is to invest in a government Treasury bond paying 6%  interest. If both investments are secure and guaranteed, there is a certainty that the Treasury  bond will pay a higher return. The return after 1 year will be $60 in interest.

In decision making under uncertainty, there are several possible outcomes for each   alternative, and the decision maker does not know the probabilities of the various outcomes. As  an example, the probability that a Democrat will be president of the United States 25 years from  now is not known. Sometimes it is impossible to assess the probability of success of a new under- taking or product. The criteria for decision making under uncertainty are explained in Section 3.4.

The third step is to identify possible outcomes.

The fourth step is to list payoffs.

During the fourth step, the decision maker can construct decision or payoff tables.

The last two steps are to select and apply the decision theory model.

Probabilities are not known.

Table 3.1 Decision Table with Conditional Values for Thompson Lumber

State of Nature

alterNative favorable Market

($) uNfavorable Market

($)

Construct a large plant 200,000 −180,000

Construct a small plant 100,000 −20,000

Do nothing 0 0

Note: It is important to include all alternatives, including “do nothing.”

M03_REND9327_12_SE_C03.indd 85 10/02/14 1:21 PM

86  Chapter 3 • DeCision analysis

In decision making under risk, there are several possible outcomes for each alternative,  and the decision maker knows the probability of occurrence of each outcome. We know, for  example, that when playing cards using a standard deck, the probability of being dealt a club is  0.25. The probability of rolling a 5 on a die is 1/6. In decision making under risk, the decision  maker usually attempts to maximize his or her expected well-being. Decision theory models for  business problems in this environment typically employ two equivalent criteria: maximization of  expected monetary value and minimization of expected opportunity loss.

In the Thompson Lumber example, John Thompson is faced with decision making under  uncertainty. If either a large plant or a small plant is constructed, the actual payoff depends on  the state of nature, and probabilities are not known. If probabilities for a favorable market and  for an unfavorable market were known, the environment would change from uncertainty to risk.  For the third alternative, do nothing, the payoff does not depend on the state of nature and is  known with certainty.

3.4 Decision Making Under Uncertainty

The presentation in this section of the criteria for decision making under uncertainty (and also  for decision making under risk) is based on the assumption that the payoff is something in which  larger values are better and high values are desirable. For payoffs such as profit, total sales, total  return on investment, and interest earned, the best decision would be one that  resulted in some  type of maximum payoff. However, there are situations in which lower payoff values (e.g., cost)  are better, and these payoffs would be minimized rather than maximized. The statement of the  decision criteria would be modified slightly for such minimization problems. These differences  will be mentioned in this section, and an example will be provided in a later section.

Several criteria exist for making decisions under conditions of uncertainty. The ones that we  cover in this section are as follows:

  1. Optimistic   2. Pessimistic   3. Criterion of realism (Hurwicz)   4. Equally likely (Laplace)   5. Minimax regret

The first four criteria can be computed directly from the decision (payoff) table, whereas the  minimax regret criterion requires use of the opportunity loss table. Let’s take a look at each of  the five models and apply them to the Thompson Lumber example.

Optimistic In using the optimistic criterion, the best (maximum) payoff for each alternative is considered  and the alternative with the best (maximum) of these is selected. Hence, the optimistic criterion  is sometimes called the maximax criterion. In Table 3.2, we see that Thompson’s optimistic  choice is the first alternative, “construct a large plant.” By using this criterion, the highest of all  possible payoffs ($200,000 in this example) may be achieved, while if any other alternative were  selected, it would be impossible to achieve a payoff this high.

In using the optimistic criterion for minimization problems in which lower payoffs (e.g.,  cost) are better, you would look at the best (minimum) payoff for each alternative and choose the  alternative with the best (minimum) of these.

Probabilities are known.

Maximax is an optimistic approach.

Table 3.2 Thompson’s Maximax Decision

State of Nature

alterNative

favorable Market

($)

uNfavorable Market

($)

MaxiMuM iN a row

($)

Construct a large plant 200,000 -180,000 200,000  Maximax

Construct a small plant 100,000 -20,000 100,000 Do nothing 0 0 0

M03_REND9327_12_SE_C03.indd 86 10/02/14 1:21 PM

3.4 DeCision Making UnDer UnCertainty  87

Pessimistic In using the pessimistic criterion, the worst (minimum) payoff for each alternative is considered  and the alternative with the best (maximum) of these is selected. Hence, the pessimistic criterion  is sometimes called the maximin criterion. This criterion guarantees the payoff will be at least  the maximin value (the best of the worst values). Choosing any other alternative may allow a  worse (lower) payoff to occur.

Thompson’s maximin choice, “do nothing,”  is shown in Table 3.3. This decision  is   associated with the maximum of the minimum number within each row or alternative.

In using the pessimistic criterion for minimization problems in which lower payoffs (e.g.,  cost) are better, you would look at the worst (maximum) payoff for each alternative and choose  the alternative with the best (minimum) of these.

Both the maximax and maximin criteria consider only one extreme payoff for each alterna- tive, while all other payoffs are ignored. The next criterion considers both of these extremes.

Criterion of Realism (Hurwicz Criterion) Often called the weighted average, the criterion of realism (the Hurwicz criterion) is a   compromise between an optimistic and a pessimistic decision. To begin, a coefficient of realism, A, is selected. This measures the degree of optimism of the decision maker and is between 0 and 1.  When a is 1, the decision maker is 100% optimistic about the future. When a is 0, the  decision  maker is 100% pessimistic about the future. The advantage of this approach is that it allows  the decision maker to build in personal feelings about relative optimism and pessimism. The  weighted average is computed as follows:

Weighted average = a1best in row2 + 11 - a21worst in row2 For a maximization problem, the best payoff for an alternative is the highest value, and the worst  payoff is the lowest value. Note that when a = 1, this is the same as the optimistic criterion, and  when a = 0 this is the same as the pessimistic criterion. This value is computed for each alter- native, and the alternative with the highest weighted average is then chosen.

If we assume that John Thompson sets his coefficient of realism, a, to be 0.80, the best de- cision would be to construct a large plant. As seen in Table 3.4, this alternative has the highest  weighted average: $124,000 = 10.8021$200,0002 = 10.2021-$180,0002.

In using the criterion of realism for minimization problems, the best payoff for an alterna- tive would be the lowest payoff in the row and the worst would be the highest payoff in the row.  The alternative with the lowest weighted average is then chosen.

Maximin is a pessimistic approach.

Criterion of realism uses the weighted average approach.

Table 3.3 Thompson’s Maximin Decision

State of Nature

alterNative

favorable Market

($)

uNfavorable Market

($)

MiNiMuM iN a row

($)

Construct a large plant 200,000 -180,000 -180,000 Construct a small plant 100,000 -20,000 -20,000 Do nothing 0 0 0

Maximin

Table 3.4 Thompson’s Criterion of Realism Decision

State of Nature

alterNative

favorable Market

($)

uNfavorable Market

($)

CriterioN of realiSM or weigHteD average

1A = 0.8 2 $ Construct a large plant 200,000 -180,000 124,000 

Realism

Construct a small plant 100,000 -20,000 76,000 Do nothing 0 0 0

M03_REND9327_12_SE_C03.indd 87 10/02/14 1:21 PM

88  Chapter 3 • DeCision analysis

Because there are only two states of nature in the Thompson Lumber example, only two  payoffs for each alternative are present and both are considered. However, if there are more than  two states of nature, this criterion will ignore all payoffs except the best and the worst. The next  criterion will consider all possible payoffs for each decision.

Equally Likely (Laplace) One criterion that uses all the payoffs for each alternative is the equally likely, also called  laplace,  decision criterion. This involves finding the average payoff for each alternative, and selecting the  alternative with the best or highest average. The equally likely approach assumes that all probabili- ties of occurrence for the states of nature are equal, and thus each state of nature is equally likely.

The equally likely choice for Thompson Lumber is the second alternative, “construct a  small plant.” This strategy, shown in Table 3.5, is the one with the maximum average payoff.

In using the equally likely criterion for minimization problems, the calculations are exactly  the same, but the best alternative is the one with the lowest average payoff.

Minimax Regret The next decision criterion that we discuss is based on opportunity loss or regret. Opportunity  loss refers to the difference between the optimal profit or payoff for a given state of nature and  the actual payoff received for a particular decision for that state of nature. In other words, it’s the  amount lost by not picking the best alternative in a given state of nature.

The first step is to create the opportunity loss table by determining the opportunity loss for  not choosing the best alternative for each state of nature. Opportunity loss for any state of nature,  or any column, is calculated by subtracting each payoff in the column from the best payoff in the  same column. For a favorable market, the best payoff is $200,000 as a result of the first alterna- tive, “construct a large plant.” The opportunity loss is 0, meaning that it is impossible to achieve  a higher payoff in this state of nature. If the second alternative is selected, a profit of $100,000  would be realized in a favorable market, and this is compared to the best payoff of $200,000.  Thus, the opportunity loss is 200,000 - 100,000 = 100,000. Similarly, if “do nothing” is se- lected, the opportunity loss would be 200,000 - 0 = 200,000.

For an unfavorable market, the best payoff is $0 as a result of the third alternative, “do noth- ing,” so this has 0 opportunity loss. The opportunity losses for the other alternatives are found  by subtracting the payoffs from this best payoff ($0) in this state of nature as shown in Table 3.6.  Thompson’s opportunity loss table is shown as Table 3.7.

Equally likely criterion uses the average outcome.

Minimax regret criterion is based on opportunity loss.

Table 3.5 Thompson’s Equally Likely Decision

State of Nature

alterNative

favorable Market

($)

uNfavorable Market

($) row average

($)

Construct a large plant 200,000 -180,000 10,000 Construct a small plant 100,000 -20,000 40,000 

Equally likely

Do nothing 0 0 0

Table 3.6 Determining Opportunity Losses for Thompson Lumber

State of Nature

favorable Market

($)

uNfavorable Market

($)

200,000 – 200,000 0 – (–180,000)

200,000 – 100,000 0 – (–20,000)

200,000 – 0 0 – 0

Table 3.7 Opportunity Loss Table for Thompson Lumber

State of Nature

alterNative

favorable Market

($)

uNfavorable Market

($)

Construct a large plant 0 180,000

Construct a small plant 100,000 20,000

Do nothing 200,000 0

M03_REND9327_12_SE_C03.indd 88 10/02/14 1:21 PM

3.5 DeCision Making UnDer risk  89

Using the opportunity loss (regret) table, the minimax regret criterion first consider the maxi- mum (worst) opportunity loss for each alternative. Next, looking at these maximum values, pick that  alternative with the minimum (or best) number. By doing this, the opportunity loss actually realized  is guaranteed to be no more than this minimax value. In Table 3.8, we can see that the minimax regret  choice is the second alternative, “construct a small plant.” When this a lternative is selected, we know  the maximum opportunity loss cannot be more than 100,000 (the minimum of the maximum regrets).

In calculating the opportunity loss for minimization problems such as those involving costs,  the best (lowest) payoff or cost in a column is subtracted from each payoff in that column. Once  the opportunity loss table has been constructed, the minimax regret criterion is applied in exactly  the same way as just described. The maximum opportunity loss for each alternative is found,  and the alternative with the minimum of these maximums is selected. As with maximization   problems, the opportunity loss can never be negative.

We have considered several decision-making criteria to be used when probabilities of the  states of nature are not known and cannot be estimated. Now we will see what to do if the   probabilities are available.

3.5 Decision Making Under Risk

Decision making under risk is a decision situation in which several possible states of nature may  occur, and the probabilities of these states of nature are known. In this section, we consider one  of the most popular methods of making decisions under risk: selecting the alternative with the  highest expected monetary value (or simply expected value). We also use the probabilities with  the opportunity loss table to minimize the expected opportunity loss.

Expected Monetary Value Given a decision table with conditional values (payoffs) that are monetary values, and prob- ability assessments for all states of nature, it is possible to determine the expected monetary value (eMv) for each alternative. The expected value, or the mean value, is the long-run aver- age value of that decision. The EMV for an alternative is just the sum of possible payoffs of the  alternative, each weighted by the probability of that payoff occurring.

This could also be expressed simply as the expected value of X, or E(X), which was   discussed in Section 2.6 of Chapter 2.

EMV1alternative2 = gXiP1Xi2 (3-1)

Ford uses Decision theory to Choose parts suppliers

Ford Motor Company manufactures about 5 million cars and trucks annually and employs more than 200,000 people at about 100 facilities around the globe. Such a large company often needs to make large supplier decisions under tight deadlines.

This was the situation when researchers at MIT teamed up with Ford management and developed a data-driven supplier se- lection tool. This computer program aids in decision making by applying some of the decision-making criteria presented in this

chapter. Decision makers at Ford are asked to input data about their suppliers (part costs, distances, lead times, supplier reliability, etc.) as well as the type of decision criterion they want to use. Once these are entered, the model outputs the best set of suppli- ers to meet the specified needs. The result is a system that is now saving Ford Motor Company over $40 million annually.

Source: Based on E. Klampfl, Y. Fradkin, C. McDaniel, and M. Wolcott. “Ford  Uses OR to Make Urgent Sourcing Decisions in a Distressed Supplier Environ- ment,” Interfaces 39, 5 (2009): 428–442.

in aCtiOn

EMV is the weighted sum of possible payoffs for each alternative.

Table 3.8 Thompson’s Minimax Decision Using Opportunity Loss

State of Nature

alterNative

favorable Market

($)

uNfavorable Market

($)

MaxiMuM iN a row

($)

Construct a large plant 0 180,000 180,000

Construct a small plant 100,000 20,000 100,000  Minimax

Do nothing 200,000 0 200,000

M03_REND9327_12_SE_C03.indd 89 10/02/14 1:21 PM

90  Chapter 3 • DeCision analysis

where

Xi = payoff for the alternative in state of nature i P1Xi2 = probability of achieving payoff Xi (i.e., probability of state of nature i)

g = summation symbol If this were expanded, it would become

EMV 1alternative2 = 1payoff in first state of nature2 * 1probability of first state of nature2

+ 1payoff in second state of nature2 * 1probability of second state of nature2 + Á + 1payoff in last state of nature2 * 1probability of last state of nature2

The alternative with the maximum EMV is then chosen. Suppose that John Thompson now believes that the probability of a favorable market is 

exactly the same as the probability of an unfavorable market; that is, each state of nature has a  0.50 probability. Which alternative would give the greatest EMV? To determine this, John has  expanded the decision table, as shown in Table 3.9. His calculations follow:

EMV 1large plant2 = 1$200,000210.502 + 1-$180,000210.502 = $10,000 EMV 1small plant2 = 1$100,000210.502 + 1-$20,000210.502 = $40,000 EMV 1do nothing2 = 1$0210.502 + 1$0210.502 = $0

The largest expected value ($40,000) results from the second alternative, “construct a small plant.”  Thus, Thompson should proceed with the project and put up a small plant to manufacture storage  sheds. The EMVs for the large plant and for doing nothing are $10,000 and $0, respectively.

When using the EMV criterion with minimization problems, the calculations are the same,  but the alternative with the smallest EMV is selected.

Expected Value of Perfect Information John Thompson has been approached by Scientific Marketing, Inc., a firm that proposes to help  John make the decision about whether to build the plant to produce storage sheds.  Scientific  Marketing claims that its technical analysis will tell John with certainty whether the market  is favorable for his proposed product. In other words, it will change his environment from  one of decision making under risk to one of decision making under certainty. This informa- tion could prevent John from making a very expensive mistake. Scientific Marketing would  charge  Thompson $65,000 for the information. What would you recommend to John? Should he  hire the firm to make the marketing study? Even if the information from the study is perfectly  accurate, is it worth $65,000? What would it be worth? Although some of these questions are dif- ficult to answer, determining the value of such perfect information can be very useful. It places  an upper bound on what you should be willing to spend on information such as that being sold  by Scientific Marketing. In this section, two related terms are investigated: the expected value of perfect information (evPi) and the expected value with perfect information (evwPi).  These techniques can help John make his decision about hiring the marketing firm.

The expected value with perfect information is the expected or average return, in the long  run, if we have perfect information before a decision has to be made. To calculate this value, we 

Table 3.9 Decision Table with Probabilities and EMVs for Thompson Lumber

State of Nature

alterNative

favorable Market

($)

uNfavorable Market

($) eMv ($)

Construct a large plant 200,000 -180,000 10,000 Construct a small plant 100,000 -20,000 40,000

Best EMV

Do nothing 0 0 0

Probabilities 0.50 0.50

EVPI places an upper bound on what to pay for information.

M03_REND9327_12_SE_C03.indd 90 10/02/14 1:21 PM

3.5 DeCision Making UnDer risk  91

choose the best alternative for each state of nature and multiply its payoff times the probability  of occurrence of that state of nature.

EVwPI = g1best payoff in state of nature i21probability of state of nature i2 (3-2) If this were expanded, it would become

EVwPI

= 1best payoff in first state of nature2 * 1probability of first state of nature2 + 1best payoff in second state of nature2 * 1probability of second state of nature2 + Á + 1best payoff in last state of nature2 * 1probability of last state of nature2

The EVPI is the expected value with perfect information minus the expected value without  perfect information (i.e., the best or maximum EMV). Thus, the EVPI is the improvement in  EMV that results from having perfect information.

EVPI = EVwPI - Best EMV (3-3)

By referring to Table 3.9, Thompson can calculate the maximum that he would pay for in- formation, that is, the EVPI. He follows a three-stage process. First, the best payoff in each state  of nature is found. If the perfect information says the market will be favorable, the large plant  will be constructed, and the profit will be $200,000. If the perfect information says the market  will be unfavorable, the “do nothing” alternative is selected, and the profit will be 0. These val- ues are shown in the “with perfect information” row in Table 3.10. Second, the expected value  with perfect information is computed. Then, using this result, EVPI is calculated.

The expected value with perfect information is

EVwPI = 1$200,000210.502 + 1$0210.502 = $100,000 Thus, if we had perfect information, the payoff would average $100,000.

The maximum EMV without additional information is $40,000 (from Table 3.9). Therefore,  the increase in EMV is

EVPI = EVwPI - maximum EMV = $100,000 - $40,000 = $60,000

Thus, the most Thompson would be willing to pay for perfect information is $60,000. This, of  course, is again based on the assumption that the probability of each state of nature is 0.50.

This EVPI also tells us that the most we would pay for any information (perfect or imperfect)  is $60,000. In a later section, we’ll see how to place a value on imperfect or sample information.

In finding the EVPI for minimization problems, the approach is similar. The best payoff in  each state of nature is found, but this is the lowest payoff for that state of nature rather than the  highest. The EVwPI is calculated from these lowest payoffs, and this is compared to the best  (lowest) EMV without perfect information. The EVPI is the improvement that results, and this is  the best EMV - EVwPI.

EVPI is the expected value with perfect information minus the maximum EMV.

Table 3.10 Decision Table with Perfect Information

State of Nature

alterNative

favorable Market

($)

uNfavorable Market

($) eMv ($)

Construct a large plant 200,000 -180,000 10,000 Construct a small plant 100,000 -20,000 40,000

Do nothing 0 0 0

With perfect information 200,000 0 100,000  EVwPI

Probabilities 0.50 0.50

M03_REND9327_12_SE_C03.indd 91 10/02/14 1:21 PM

92  Chapter 3 • DeCision analysis

Expected Opportunity Loss An alternative approach to maximizing EMV is to minimize expected opportunity loss (EOL).  First, an opportunity loss table is constructed. Then the EOL is computed for each alternative by  multiplying the opportunity loss by the probability and adding these together. In Table 3.7, we  presented the opportunity loss table for the Thompson Lumber example. Using these opportu- nity losses, we compute the EOL for each alternative by multiplying the probability of each state  of nature times the appropriate opportunity loss value and adding these together:

EOL1construct large plant2 = 10.521$02 + 10.521$180,0002 = $90,000

EOL1construct small plant2 = 10.521$100,0002 + 10.521$20,0002 = $60,000

EOL1do nothing2 = 10.521$200,0002 + 10.521$02 = $100,000

Table 3.11 gives these results. Using minimum EOL as the decision criterion, the best decision  would be the second alternative, “construct a small plant.”

It is important to note that minimum EOL will always result in the same decision as maxi- mum EMV, and that the EVPI will always equal the minimum EOL. Referring to the Thompson  case, we used the payoff table to compute the EVPI to be $60,000. Note that this is the minimum  EOL we just computed.

Sensitivity Analysis In previous sections, we determined that the best decision (with the probabilities known) for  Thompson Lumber was to construct the small plant, with an expected value of $40,000. This  conclusion depends on the values of the economic consequences and the two probability val- ues of a favorable and an unfavorable market. Sensitivity analysis investigates how our decision  might change given a change in the problem data. In this section, we investigate the impact that a  change in the probability values would have on the decision facing Thompson Lumber. We first  define the following variable:

P = probability of a favorable market

Because there are only two states of nature, the probability of an unfavorable market must be 1 - P. We can now express the EMVs in terms of P, as shown in the following equations. A graph 

of these EMV values is shown in Figure 3.1.

EMV1large plant2 = $200,000P - $180,00011 - P2 = $200,000P - $180,000 + 180,000P = $380,000P - $180,000

EMV1small plant2 = $100,000P - $20,00011 - P2 = $100,000P - $20,000 + 20,000P = $120,000P - $20,000

EMV1do nothing2 = $0P + $011 - P2 = $0

EOL is the cost of not picking the best solution.

Table 3.11 EOL Table for Thompson Lumber

State of Nature

alterNative

favorable Market

($)

uNfavorable Market

($) eol

Construct a large plant 0 180,000 90,000

Construct a small plant 100,000 20,000 60,000

Best EOL

Do nothing 200,000 0 100,000

Probabilities 0.50 0.50

EOL will always  result in the same decision as the maximum EMV.

Sensitivity analysis investigates how our decision might change with different input data.

M03_REND9327_12_SE_C03.indd 92 10/02/14 1:21 PM

3.6 a MiniMization exaMple  93

As you can see in Figure 3.1, the best decision is to do nothing as long as P is between 0 and  the probability associated with point 1, where the EMV for doing nothing is equal to the EMV  for the small plant. When P is between the probabilities for points 1 and 2, the best decision is to  build the small plant. Point 2 is where the EMV for the small plant is equal to the EMV for the  large plant. When P is greater than the probability for point 2, the best decision is to construct  the large plant. Of course, this is what you would expect as P increases. The value of P at points  1 and 2 can be computed as follows:

Point 1: EMV 1do nothing2 = EMV 1small plant2

0 = $120,000P - $20,000 P = 20,000

120,000 = 0.167

Point 2: EMV 1small plant2 = EMV 1large plant2 $120,000P - $20,000 = $380,000P - $180,000

260,000P = 160,000 P = 160,000

260,000 = 0.615

The results of this sensitivity analysis are displayed in the following table:

beSt alterNative raNge of P valueS

Do nothing Less than 0.167

Construct a small plant 0.167 - 0.615 Construct a large plant Greater than 0.615

3.6 A Minimization Example

The previous examples have illustrated how to apply the decision-making criterion when the  payoffs are to be maximized. The following example illustrates how the criteria are applied to  problems in which the payoffs are costs that should be minimized.

The Decision Sciences department at State University will be signing a 3-year lease for a  new copy machine, and three different machines are being considered. For each of these, there is  a monthly fee, which includes service on the machine, plus a charge for each copy. The number  of copies that would be made each month is uncertain, but the department has estimated that the  number of copies per month could be 10,000 or 20,000 or 30,000 per month. The monthly cost  for each machine based on each of the three levels of activity is shown in Table 3.12.

0

EMV (large plant)

EMV (small plant)

EMV (do nothing)

Values of P

Point 2

Point 1

EMV Values

$300,000

$200,000

$100,000

0

2$100,000

2$200,000

1.615.167

Figure 3.1 Sensitivity Analysis

M03_REND9327_12_SE_C03.indd 93 10/02/14 1:21 PM

94  Chapter 3 • DeCision analysis

Which machine should be selected? To determine the best alternative, a specific criterion  must be chosen. If the decision maker is optimistic, only the best (minimum) payoff for each  decision is considered. These are shown in Table 3.13, and the best (minimum) of these is  700. Thus, Machine C would be selected, allowing the possibility of achieving this best cost  of $700.

If the decision maker is pessimistic, only the worst (maximum) payoff for each decision is  considered. These are also shown in Table 3.13, and the best (minimum) of these is 1,150. Thus,  Machine A would be selected based on the pessimistic criterion. This would guarantee that the  cost would be no more than 1,150, regardless of which state of nature occurred.

Using Hurwicz criterion, if we assume that the decision maker is 70% optimistic (the coeffi- cient of realism is 0.7), the weighted average of the best and the worst payoff for each alternative  would be calculated using the formula:

Weighted average = 0.71best payoff2 + 11 - 0.721worst payoff2 For each of the three copy machines, we would get the following values.

Machine A: 0.719502 + 0.311,1502 = 1,010 Machine B: 0.718502 + 0.311,3502 = 1,000 Machine C: 0.717002 + 0.311,3002 = 880

The decision would be to select Machine C based on this criterion because it has the lowest  weighted average cost.

For the equally likely criterion, the average payoff for each machine would be calculated.

Machine A: 1950 + 1,050 + 1,1502>3 = 1,050 Machine B: 1850 + 1,100 + 1,3502>3 = 1,100 Machine C: 1700 + 1,000 + 1,3002>3 = 1,000

Based on the equally likely criterion, Machine C would be selected because it has the lowest  average cost.

To apply EMV criterion, probabilities must be known for each state of nature. Past records  indicate that 40% of the time the number of copies made in a month was 10,000, while 30% of  the time it was 20,000 and 30% of the time it was 30,000. The probabilities for the three states  of nature would be 0.4, 0.3, and 0.3. We can use these to calculate the EMVs, and the results  are shown in Table 3.14. Machine C would be selected because it has the lowest EMV. The  monthly cost would average $970 with this machine, while the other machines would average  a higher cost.

Table 3.12 Payoff Table with Monthly Copy Costs for Business Analytics Department

Table 3.13 Best and Worst Payoffs (Costs) for Business Analytics Department

10,000 CoPieS Per MoNtH

20,000 CoPieS Per MoNtH

30,000 CoPieS Per MoNtH

Machine A 950 1,050 1,150

Machine B 850 1,100 1,350

Machine C 700 1,000 1,300

10,000 CoPieS Per

MoNtH

20,000 CoPieS Per

MoNtH

30,000 CoPieS Per

MoNtH

beSt Payoff

(MiNiMuM)

worSt Payoff

(MaxiMuM)

Machine A 950 1,050 1,150 950 1,150

Machine B 850 1,100 1,350 850 1,350

Machine C 700 1,000 1,300 700 1,300

M03_REND9327_12_SE_C03.indd 94 10/02/14 1:21 PM

3.7 Using software for payoff table probleMs  95

To find EVPI, we first find the payoffs (costs) that would be experienced with perfect infor- mation. The best payoff in each state of nature is the lowest value (cost) in that state of nature,  as shown in the bottom row of Table 3.14. These values are used to calculate EVwPI. With these  costs, we find

EVwPI = $925 Best EMV without perfect information = $970

EVPI = 970 - 925 = $45

Perfect information would lower the expected value by $45. To apply criteria based on opportunity loss, we must first develop the opportunity loss table. 

In each state of nature, the opportunity loss indicates how much worse each payoff is than the  best possible payoff in that state of nature. The best payoff (cost) would be the lowest cost. Thus,  to get the opportunity loss in this case, we subtract the lowest value in each column from all the  values in that column and we obtain the opportunity loss table.

Once the opportunity loss table has been developed, the minimax regret criterion is  applied exactly as it was for the Thompson Lumber example. The maximum regret for each  alternative is found and the alternative with the minimum of these maximums is selected. As  seen in Table 3.15, the minimum of these maximums is 150 so Machine C would be selected  based on this criterion.

The probabilities are used to compute the expected opportunity losses as shown in Table 3.15.  Machine C has the lowest EOL of $45, so it would be selected based on the minimum EOL  criterion. As previously noted, the minimum EOL is equal to the expected value of perfect  information.

3.7 Using Software for Payoff Table Problems

It is easy to use QM for Windows or Excel QM or even Excel 2013 without any add-ins to per- form the calculations associated with payoff tables. The Thompson Lumber example is used in  the following examples for illustration.

QM for Windows To use QM for Windows for the Thompson Lumber example, select Modules – Decision The- ory. Then enter New-Payoff Tables, and a window appears allowing you to set up the problem.  Enter a title, the number of options (alternatives), the number of scenarios (states of nature), 

Table 3.14 Expected Monetary Values and Expected Value with Perfect Information for Business Analytics Department

Table 3.15 Opportunity Loss Table for Business Analytics Department

10,000 CoPieS Per MoNtH

20,000 CoPieS Per MoNtH

30,000 CoPieS Per MoNtH eMv

Machine A 950 1,050 1,150 1,040

Machine B 850 1,100 1,350 1,075

Machine C 700 1,000 1,300 970

With perfect information 700 1,000 1,150 925

Probability 0.4 0.3 0.3

10,000 CoPieS Per MoNtH

20,000 CoPieS Per MoNtH

30,000 CoPieS Per MoNtH MaxiMuM eol

Machine A 250 50 0 250 115

Machine B 150 100 200 200 150

Machine C 0 0 150 150 45

Probability 0.4 0.3 0.3

M03_REND9327_12_SE_C03.indd 95 10/02/14 1:21 PM

96  Chapter 3 • DeCision analysis

and specify if it is to be maximized or minimized. In this example, there are three alterna- tives or options, and two states of nature or scenarios. When these are entered, click OK and a  window appears (Program 3.1A) allowing you to enter the probabilities and payoffs. You can  also change the names of the alternatives and states of nature by simply typing over the default  names. When you click Solve, the output in Program 3.1B appears. Additional output such as  EVPI and opportunity loss results are available by clicking Window on the screen where the  output is displayed.

Excel QM To use Excel QM for the Thompson Lumber example, from the Excel QM ribbon in Excel  2013, click Alphabetical Menu and select Decision Analysis – Decision Tables. When the win- dow opens to enter the parameters of the problem, you should enter a title, the number of al- ternatives, the number of states of nature, and specify if it is to be maximized or minimized.  Click OK and an empty payoff table is created. You simply enter the probabilities, payoffs, and  names for the rows and columns. The calculations are performed automatically and the results  are shown in Program 3.2A. While the formulas are automatically developed in Excel QM,   Program 3.2B shows the important formulas in this example. To see the formulas in Excel, sim- ply hold down the control key and press the grave accent ( ̀ ) key (usually found above the Tab  key). To return Excel 2013 spreadsheet display to the numbers instead of the formulas, simply  press the keys again.

Enter data into table and type the row and column names.

Program 3.1a QM for Windows Input for Thompson Lumber Example

Click Window to see more information.

Program 3.1b QM for Windows Output Screen for Thompson Lumber Example

M03_REND9327_12_SE_C03.indd 96 10/02/14 1:21 PM

3.8 DeCision trees  97

3.8 Decision Trees

Any problem that can be presented in a decision table can also be graphically illustrated in  a decision tree. All decision trees are similar in that they contain decision points or decision nodes and state-of-nature points or state-of-nature nodes:

● A decision node from which one of several alternatives may be chosen

● A state-of-nature node out of which one state of nature will occur

In drawing the tree, we begin at the left and move to the right. Thus, the tree presents the  decisions and outcomes in sequential order. Lines or branches from the squares (decision  nodes) represent alternatives, and branches from the circles represent the states of nature.  Figure 3.2 gives the basic decision tree for the Thompson Lumber example. First, John  decides whether to construct a large plant, a small plant, or no plant. Then, once that deci- sion is made, the possible states of nature or outcomes (favorable or unfavorable market)  will occur. The next step is to put the payoffs and probabilities on the tree and begin the  analysis.

To see the formulas, hold down the control key (Ctrl) and press the ` (grave accent) key, which is usually found above the Tab key.

Program 3.2a Excel QM Results for Thompson Lumber Example

Program 3.2b Key Formulas in Excel QM for Thompson Lumber Example

M03_REND9327_12_SE_C03.indd 97 10/02/14 1:21 PM

98  Chapter 3 • DeCision analysis

Analyzing problems with decision trees involves five steps:

Five Steps of Decision Tree Analysis

  1. Define the problem.   2. Structure or draw the decision tree.   3. Assign probabilities to the states of nature.   4. Estimate payoffs for each possible combination of alternatives and states of nature.   5. Solve the problem by computing EMVs for each state-of-nature node. This is done by 

working backward, that is, starting at the right of the tree and working back to deci- sion nodes on the left. Also, at each decision node, the alternative with the best EMV is  selected.

The final decision tree with the payoffs and probabilities for John Thompson’s decision  situation is shown in Figure 3.3. Note that the payoffs are placed at the right side of each  of the tree’s branches. The probabilities are shown in parentheses next to each state of na- ture. Beginning with the payoffs on the right of the figure, the EMVs for each state-of-nature  node are then calculated and placed by their respective nodes. The EMV of the first node is  $10,000. This represents the branch from the decision node to construct a large plant. The 

Figure 3.2 Thompson’s Decision Tree A Decision Node A State-of-Nature Node

Favorable Market

Unfavorable Market

Favorable Market

Unfavorable Market

Construct Small Plant

Co nst

ruc t

Lar ge

Pla nt

Do Nothing

1

2

Figure 3.3 Completed and Solved Decision Tree for Thompson Lumber

Favorable Market (0.5)

Unfavorable Market (0.5)

Favorable Market (0.5)

Unfavorable Market (0.5)

Construct Small Plant Con

stru ct L

arg e P

lan t

Do Nothing

= (0.5)($200,000) + (0.5)( –$180,000)

Payoffs

$200,000

–$180,000

$100,000

–$20,000

$0

1

2

EMV for Node 1 = $10,000

EMV for Node 2 = $40,000

= (0.5)($100,000) + (0.5)( –$20,000)

Alternative with best EMV is selected

M03_REND9327_12_SE_C03.indd 98 10/02/14 1:21 PM

3.8 DeCision trees  99

EMV for node 2, to construct a small plant, is $40,000. Building no plant or doing nothing  has, of course, a payoff of $0. The branch leaving the decision node leading to the state-of-na- ture node with the highest EMV should be chosen. In Thompson’s case, a small plant should  be built.

A MORE COMPLEx DECISIOn FOR ThOMPSOn LUMBER—SAMPLE InFORMATIOn When  sequential decisions need to be made, decision trees are much more powerful tools than deci- sion tables. Let’s say that John Thompson has two decisions to make, with the second decision  dependent on the outcome of the first. Before deciding about building a new plant, John has the  option of conducting his own marketing research survey, at a cost of $10,000. The information  from his survey could help him decide whether to construct a large plant, a small plant, or not to  build at all. John recognizes that such a market survey will not provide him with perfect infor- mation, but it may help quite a bit nevertheless.

John’s new decision tree is represented in Figure 3.4. Let’s take a careful look at this more  complex tree. Note that all possible outcomes and alternatives are included in their logical se- quence. This is one of the strengths of using decision trees in making decisions. The user is  forced to examine all possible outcomes, including unfavorable ones. He or she is also forced to  make decisions in a logical, sequential manner.

All outcomes and alternatives must be considered.

Figure 3.4 Larger Decision Tree with Payoffs and Probabilities for Thompson Lumber

Favorable Market (0.78) $190,000

Unfavorable Market (0.22) –$190,000

Favorable Market (0.78) $90,000

Unfavorable Market (0.22) –$30,000

No Plant

Small Plant

Lar ge

Pla nt

–$10,000

Favorable Market (0.27) $190,000

Unfavorable Market (0.73) –$190,000

Favorable Market (0.27) $90,000

Unfavorable Market (0.73) –$30,000

No Plant

Small Plant

Lar ge

Pla nt

–$10,000

Favorable Market (0.50) $200,000

Unfavorable Market (0.50) –$180,000

Favorable Market (0.50) $100,000

Unfavorable Market (0.50) –$20,000

No Plant

Small Plant

Lar ge

Pla nt

$0

S

ur ve

y ( 0.

45 )

Re su

lts Fa

vo ra

ble Survey (0.55)

Results

Negative

C on

du ct

M ar

ke t S

ur ve

y

Do Not Conduct Survey

First Decision Point

Second Decision Point

Payoffs

2

3

5

6

7

4

1

M03_REND9327_12_SE_C03.indd 99 10/02/14 1:21 PM

100  Chapter 3 • DeCision analysis

Examining the tree, we see that Thompson’s first decision point is whether to conduct the  $10,000 market survey. If he chooses not to do the study (the lower part of the tree), he can  either construct a large plant, a small plant, or no plant. This is John’s second decision point.  The market will be either favorable (0.50 probability) or unfavorable (also 0.50 probability) if  he builds. The payoffs for each of the possible consequences are listed along the right side. As a  matter of fact, the lower portion of John’s tree is identical to the simpler decision tree shown in  Figure 3.3. Why is this so?

The upper part of Figure 3.4 reflects the decision to conduct the market survey. State -of-  nature node 1 has two branches. There is a 45% chance that the survey results will indicate a   favorable market for storage sheds. We also note that the probability is 0.55 that the survey  results will be negative. The derivation of this probability will be discussed in the next section.

The rest of the probabilities shown in parentheses in Figure 3.4 are all conditional proba- bilities or posterior probabilities (these probabilities will also be discussed in the next section).  For example, 0.78 is the probability of a favorable market for the sheds given a favorable result  from the market survey. Of course, you would expect to find a high probability of a favorable  market given that the research indicated that the market was good. Don’t forget, though, there is  a chance that John’s $10,000 market survey didn’t result in perfect or even reliable information.  Any market research study is subject to error. In this case, there is a 22% chance that the market  for sheds will be unfavorable given that the survey results are positive.

We note that there is a 27% chance that the market for sheds will be favorable given that  John’s survey results are negative. The probability is much higher, 0.73, that the market will   actually be unfavorable given that the survey was negative.

Finally, when we look to the payoff column in Figure 3.4, we see that $10,000, the cost of  the marketing study, had to be subtracted from each of the top 10 tree branches. Thus, a large  plant with a favorable market would normally net a $200,000 profit. But because the market  study was conducted, this figure is reduced by $10,000 to $190,000. In the unfavorable case, the  loss of $180,000 would increase to a greater loss of $190,000. Similarly, conducting the survey  and building no plant now results in a -$10,000 payoff.

With all probabilities and payoffs specified, we can start calculating the EMV at each state- of-nature node. We begin at the end, or right side of the decision tree and work back toward the  origin. When we finish, the best decision will be known.

  1. Given favorable survey results,

EMV1node 22 = EMV1large plant � positive survey2 = 10.7821$190,0002 + 10.2221-$190,0002 = $106,400

EMV1node 32 = EMV1small plant � positive survey2 = 10.7821$90,0002 + 10.2221-$30,0002 = $63,600

    The EMV of no plant in this case is -$10,000. Thus, if the survey results are  favorable,  a large plant should be built. Note that we bring the expected value of this decision  ($106,400) to the decision node to indicate that, if the survey results are positive, our   expected value will be $106,400. This is shown in Figure 3.5.

  2. Given negative survey results,

EMV1node 42 = EMV1large plant � negative survey2 = 10.2721$190,0002 + 10.7321-$190,0002 = -$87,400

EMV1node 52 = EMV1small plant � negative survey2 = 10.2721$90,0002 + 10.7321-$30,0002 = $2,400

    The EMV of no plant is again -$10,000 for this branch. Thus, given a negative survey  result, John should build a small plant with an expected value of $2,400, and this figure is  indicated at the decision node.

  3. Continuing on the upper part of the tree and moving backward, we compute the expected  value of conducting the market survey:

EMV1node 12 = EMV1conduct survey2 = 10.4521$106,4002 + 10.5521$2,4002 = $47,880 + $1,320 = $49,200

Most of the probabilities are conditional probabilities.

The cost of the survey had to be subtracted from the original payoffs.

We start by computing the EMV of each branch.

EMV calculations for favorable survey results are made first.

EMV calculations for unfavorable survey results are done next.

We continue working backward to the origin, computing EMV values.

M03_REND9327_12_SE_C03.indd 100 10/02/14 1:21 PM

3.8 DeCision trees  101

  4.  If the market survey is not conducted,

EMV1node 62 = EMV1large plant2 = 10.5021$200,0002 + 10.5021-$180,0002 = $10,000

EMV1node 72 = EMV1small plant2 = 10.5021$100,0002 + 10.5021-$20,0002 = $40,000

    The EMV of no plant is $0. Thus, building a small plant is the best choice, given that the marketing research is not 

performed, as we saw earlier.   5. We move back to the first decision node and choose the best alternative. The EMV 

of  conducting the survey is $49,200, versus an EMV of $40,000 for not conducting  the study, so the best choice is to seek marketing information. If the survey results are   favorable, John should construct a large plant, but if the research is negative, John should  construct a small plant.

Su rv

ey Re

su lts

Fa vo

ra ble

(0 .4

5)

Favorable Market (0.78) $190,000

Unfavorable Market (0.22) –$190,000

Favorable Market (0.78) $90,000

Unfavorable Market (0.22) –$30,000

No Plant

Small Plant

Lar ge

Pla nt

–$10,000

Favorable Market (0.27) $190,000

Unfavorable Market (0.73) –$190,000

Favorable Market (0.27) $90,000

Unfavorable Market (0.73) –$30,000

No Plant

Small Plant

–$10,000

Favorable Market (0.50) $200,000

Unfavorable Market (0.50) –$180,000

Favorable Market (0.50) $100,000

Unfavorable Market (0.50) –$20,000

No Plant

Small Plant

$0

Survey

Results

Negative

(0.55)

C on

du ct

M ar

ke t S

ur ve

y

Do Not Conduct Survey

First Decision Point

Second Decision Point

Payoffs

$106,400

$63,600

–$87,400

$2,400

$10,000

Lar ge

Pla nt

Lar ge

Pla nt

$40,000

49,200

2

3

5

6

7

1

4

$4 0,

00 0

$2 ,4

00 $1

06 ,4

00

$4 9,

20 0

Figure 3.5 Thompson’s Decision Tree with EMVs Shown

M03_REND9327_12_SE_C03.indd 101 10/02/14 1:21 PM

102  Chapter 3 • DeCision analysis

In Figure 3.5, these expected values are placed on the decision tree. Notice on the tree that a  pair of slash lines / / through a decision branch indicates that a particular alternative is dropped  from further consideration. This is because its EMV is lower than the EMV for the best alterna- tive. After you have solved several decision tree problems, you may find it easier to do all of  your computations on the tree diagram.

ExPECTED VALUE OF SAMPLE InFORMATIOn With the market survey he intends to conduct,  John Thompson knows that his best decision will be to build a large plant if the survey is favor- able or a small plant if the survey results are negative. But John also realizes that conducting  the market research is not free. He would like to know what the actual value of doing a survey  is. One way of measuring the value of market information is to compute the expected value of sample information (EVSI) which is the increase in expected value resulting from the sample  information.

The expected value with sample information (EV with SI) is found from the decision  tree, and the cost of the sample information is added to this since this was subtracted from  all the payoffs before the EV with SI was calculated. The expected value without sample  information (EV without SI) is then subtracted from this to find the value of the sample  information.

EVSI = 1EV with SI + cost2 - 1EV without SI2 (3-4) where

EVSI = expected value of sample information EV with SI = expected value with sample information EV without SI = expected value without sample information

In John’s case, his EMV would be $59,200 if he hadn’t already subtracted the $10,000 study  cost from each payoff. (Do you see why this is so? If not, add $10,000 back into each payoff,  as in the original Thompson problem, and recompute the EMV of conducting the market study.)  From the lower branch of Figure 3.5, we see that the EMV of not gathering the sample informa- tion is $40,000. Thus,

EVSI = 1$49,200 + $10,0002 - $40,000 = $59,200 - $40,000 = $19,200 This means that John could have paid up to $19,200 for a market study and still come out ahead.  Since it costs only $10,000, the survey is indeed worthwhile.

Efficiency of Sample Information There may be many types of sample information available to a decision maker. In developing  a new product, information could be obtained from a survey, from a focus group, from other  market research techniques, or from actually using a test market to see how sales will be. While  none of these sources of information are perfect, they can be evaluated by comparing the EVSI  with the EVPI. If the sample information was perfect, then the efficiency would be 100%. The  efficiency of sample information is

Efficiency of sample information = EVSI

EVPI 100% (3-5)

In the Thompson Lumber example,

Efficiency of sample information = 19,200

60,000 100% = 32%

Thus, the market survey is only 32% as efficient as perfect information.

Sensitivity Analysis As with payoff tables, sensitivity analysis can be applied to decision trees as well. The overall  approach is the same. Consider the decision tree for the expanded Thompson Lumber problem 

EVSI measures the value of sample information.

M03_REND9327_12_SE_C03.indd 102 10/02/14 1:21 PM

3.9 how probability valUes are estiMateD by bayesian analysis  103

shown in Figure 3.5. How sensitive is our decision (to conduct the marketing survey) to the  probability of favorable survey results?

Let p be the probability of favorable survey results. Then 11 - p2 is the probability of   negative survey results. Given this information, we can develop an expression for the EMV of  conducting the survey, which is node 1:

EMV1node 12 = 1$106,4002p + 1$2,400211 - p2 = $104,000p + $2,400

We are indifferent when the EMV of conducting the marketing survey, node 1, is the same  as the EMV of not conducting the survey, which is $40,000. We can find the indifference point  by equating EMV(node 1) to $40,000:

$104,000p + $2,400 = $40,000 $104,000p = $37,600

p = $37,600

$104,000 = 0.36

As long as the probability of favorable survey results, p, is greater than 0.36, our decision will  stay the same. When p is less than 0.36, our decision will be not to conduct the survey.

We could also perform sensitivity analysis for other problem parameters. For example, we  could find how sensitive our decision is to the probability of a favorable market given favorable  survey results. At this time, this probability is 0.78. If this value goes up, the large plant becomes  more attractive. In this case, our decision would not change. What happens when this probabil- ity goes down? The analysis becomes more complex. As the probability of a favorable market  given favorable survey results goes down, the small plant becomes more attractive. At some  point, the small plant will result in a higher EMV (given favorable survey results) than the large  plant. This, however, does not conclude our analysis. As the probability of a favorable market  given favorable survey results continues to fall, there will be a point where not conducting the  survey, with an EMV of $40,000, will be more attractive than conducting the marketing survey.  We leave the actual calculations to you. It is important to note that sensitivity analysis should  consider all possible consequences.

3.9 How Probability Values Are Estimated by Bayesian Analysis

There are many ways of getting probability data for a problem such as Thompson’s. The  numbers  (such as 0.78, 0.22, 0.27, 0.73 in Figure 3.4) can be assessed by a manager based on experience  and intuition. They can be derived from historical data, or they can be computed from other  available data using Bayes’ theorem. The advantage of Bayes’ theorem is that it incorporates  both our initial estimates of the probabilities and information about the accuracy of the informa- tion source (e.g., market research survey).

The Bayes’ theorem approach recognizes that a decision maker does not know with   certainty what state of nature will occur. It allows the manager to revise his or her initial  or prior probability assessments based on new information. The revised probabilities are  called posterior probabilities. (Before continuing, you may wish to review Bayes’ theorem in  Chapter 2.)

Calculating Revised Probabilities In the Thompson Lumber case solved in Section 3.8, we made the assumption that the following  four conditional probabilities were known:

P1favorable market1FM2 � survey results positive2 = 0.78 P1unfavorable market1UM2 � survey results positive2 = 0.22

P1favorable market1FM2 � survey results negative2 = 0.27 P1unfavorable market1UM2 � survey results negative2 = 0.73

Bayes’ theorem allows decision makers to revise probability values.

M03_REND9327_12_SE_C03.indd 103 10/02/14 1:21 PM

104  Chapter 3 • DeCision analysis

We now show how John Thompson was able to derive these values with Bayes’ theorem. From  discussions with market research specialists at the local university, John knows that  special  surveys such as his can either be positive (i.e., predict a favorable market) or be negative  (i.e., predict an unfavorable market). The experts have told John that, statistically, of all new  products with a favorable market (FM), market surveys were positive and predicted success cor- rectly 70% of the time. Thirty percent of the time the surveys falsely predicted negative results  or an unfavorable market (UM). On the other hand, when there was actually an unfavorable  market for a new product, 80% of the surveys correctly predicted negative results. The surveys  incorrectly predicted positive results the remaining 20% of the time. These conditional prob- abilities are summarized in Table 3.16. They are an indication of the accuracy of the survey that  John is thinking of undertaking.

Recall that without any market survey information, John’s best estimates of a favorable and  unfavorable market are

P1FM2 = 0.50 P1UM2 = 0.50

These are referred to as the prior probabilities. We are now ready to compute Thompson’s revised or posterior probabilities. These desired 

probabilities are the reverse of the probabilities in Table 3.16. We need the probability of a favor- able or unfavorable market given a positive or negative result from the market study. The general  form of Bayes’ theorem presented in Chapter 2 is

P1A � B2 = P1B � A2P1A2 P1B � A2P1A2 + P1B � A′2P1A′2 (3-6)

where

A, B = any two events A′ = complement of A

We can let A represent a favorable market and B represent a positive survey. Then, substi- tuting the appropriate numbers into this equation, we obtain the conditional probabilities, given  that the market survey is positive:

P1FM � survey positive2 = P1survey positive �FM2P1FM2 P1survey positive �FM2P1FM2 + P1survey positive �UM2P1UM2

= 10.70210.502

10.70210.502 + 10.20210.502 = 0.35

0.45 = 0.78

P1UM � survey positive2 = P1survey positive �UM2P1UM2 P1survey positive �UM2P1UM2 + P1survey positive �FM2P1FM2

= 10.20210.502

10.20210.502 + 10.70210.502 = 0.10

0.45 = 0.22

Note that the denominator (0.45) in these calculations is the probability of a positive survey. An  alternative method for these calculations is to use a probability table as shown in Table 3.17.

Table 3.16 Market Survey Reliability in Predicting States of nature

State of Nature

reSult of Survey favorable Market

(fM) uNfavorable Market

(uM)

Positive (predicts favorable  market for product)

P1survey positive �FM2 = 0.70 P1survey positive �UM2 = 0.20

Negative (predicts unfavorable  market for product)

P1survey negative �FM2 = 0.30 P1survey negative �UM2 = 0.80

M03_REND9327_12_SE_C03.indd 104 10/02/14 1:21 PM

3.9 how probability valUes are estiMateD by bayesian analysis  105

The conditional probabilities, given that the market survey is negative, are

P1FM � survey negative2 = P1survey negative �FM2P1FM2 P1survey negative �FM2P1FM2 + P1survey negative �UM2P1UM2

= 10.30210.502

10.30210.502 + 10.80210.502 = 0.15

0.55 = 0.27

P1UM � survey negative2 = P1survey negative �UM2P1UM2 P1survey negative �UM2P1UM2 + P1survey negative �FM2P1FM2

= 10.80210.502

10.80210.502 + 10.30210.502 = 0.40

0.55 = 0.73

Note that the denominator (0.55) in these calculations is the probability of a negative  survey.  These computations given a negative survey could also have been performed in a table instead,  as in Table 3.18.

The calculations  shown  in Tables 3.17 and 3.18 can easily be performed  in   Excel   spreadsheets. Programs 3.3A and 3.3B show the final output for this example and the formulas.

The posterior probabilities now provide John Thompson with estimates for each state of  nature if the survey results are positive or negative. As you know, John’s prior probability of  success without a market survey was only 0.50. Now he is aware that the probability of success- fully marketing storage sheds will be 0.78 if his survey shows positive results. His chances of  success drop to 27% if the survey report is negative. This is valuable management information,  as we saw in the earlier decision tree analysis.

Potential Problem in Using Survey Results In many decision-making problems, survey results or pilot studies are done before an ac- tual decision (such as building a new plant or taking a particular course of action) is made.  As discussed earlier in this section, Bayes’ analysis is used to help determine the correct 

Table 3.17 Probability Revisions Given a Positive Survey

PoSterior Probability

State of Nature

CoNDitioNal Probability

P 1Survey PoSitive � State of Nature 2

Prior Probability

JoiNt Probability

P 1State of Nature � Survey

PoSitive 2 FM 0.70 * 0.50 = 0.35 0.35>0.45 = 0.78 UM 0.20 * 0.50 = 0.10 0.10>0.45 = 0.22

P1survey results positive2 = 0.45 1.00

Table 3.18 Probability Revisions Given a negative Survey

PoSterior Probability

State of Nature

CoNDitioNal Probability

P 1 Survey Negative � State of Nature)

Prior Probability

JoiNt Probability

P 1State of Nature � Survey

Negative)

FM 0.30 * 0.50 = 0.15 0.15>0.55 = 0.27 UM 0.80 * 0.50 = 0.40 0.40>0.55 = 0.73

P1survey results negative2 = 0.55 1.00

New probabilities provide valuable information.

M03_REND9327_12_SE_C03.indd 105 10/02/14 1:21 PM

106  Chapter 3 • DeCision analysis

conditional probabilities that are needed to solve these types of decision theory problems.  In computing these conditional probabilities, we need to have data about the surveys and  their accuracies. If a decision to build a plant or to take another course of action is actu- ally made, we can determine the accuracy of our surveys. Unfortunately, we cannot always  get data about those situations in which the decision was not to build a plant or not to take  some course of action. Thus, sometimes when we use survey results, we are basing our  probabilities only on those cases in which a decision to build a plant or take some course of  action is actually made. This means that, in some situations, conditional probability infor- mation may not be not quite as accurate as we would like. Even so, calculating conditional  probabilities helps to refine the decision-making process and, in general, to make better  decisions.

3.10 Utility Theory

We have focused on the EMV criterion for making decisions under risk. However, there are  many occasions in which people make decisions that would appear to be inconsistent with  the EMV criterion. When people buy insurance, the amount of the premium is greater than  the expected payout to them from the insurance company because the premium includes  the expected payout, the overhead cost, and the profit for the insurance company. A person  involved in a lawsuit may choose to settle out of court rather than go to trial even if the  expected value of going to trial is greater than the proposed settlement. A person buys a  lottery ticket even though the expected return is negative. Casino games of all types have  negative expected returns for the player, and yet millions of people play these games. A  businessperson may rule out one potential decision because it could bankrupt the firm if  things go bad, even though the expected return for this decision is better than that of all  other alternatives.

Why do people make decisions that don’t maximize their EMV? They do this because  the monetary value is not always a true indicator of the overall value of the result of the 

Program 3.3a Results of Bayes’ Calculations in Excel 2013

Program 3.3b Formulas Used for Bayes’ Calculations in Excel 2013

M03_REND9327_12_SE_C03.indd 106 10/02/14 1:21 PM

3.10 Utility theory  107

decision. The overall worth of a particular outcome is called utility, and rational people  make decisions that maximize the expected utility. Although at times the monetary value is  a good indicator of utility, there are other times when it is not. This is particularly true when  some of the values involve an extremely large payoff or an extremely large loss. Suppose  that you are the lucky holder of a lottery ticket. Five minutes from now a fair coin could  be flipped, and if it comes up tails, you would win $5 million. If it comes up heads, you  would win nothing. Just a moment ago a wealthy person offered you $2 million for your  ticket. Let’s assume that you have no doubts about the validity of the offer. The person will  give you a certified check for the full amount, and you are absolutely sure the check would  be good.

A decision tree for this situation is shown in Figure 3.6. The EMV for rejecting the offer  indicates that you should hold on to your ticket, but what would you do? Just think, $2 million  for sure instead of a 50% chance at nothing. Suppose you were greedy enough to hold on to  the ticket, and then lost. How would you explain that to your friends? Wouldn’t $2 million be  enough to be comfortable for a while?

Most people would choose to sell the ticket for $2 million. Most of us, in fact, would proba- bly be willing to settle for a lot less. Just how low we would go is, of course, a matter of personal  preference. People have different feelings about seeking or avoiding risk. Using the EMV alone  is not always a good way to make these types of decisions.

One way to incorporate your own attitudes toward risk is through utility theory. In the next  section, we explore first how to measure utility and then how to use utility measures in decision  making.

Measuring Utility and Constructing a Utility Curve The first step in using utility theory is to assign utility values to each monetary value in a given  situation. It is convenient to begin utility assessment by assigning the worst outcome a utility  of 0 and the best outcome a utility of 1. Although any values may be used as long as the utility  for the best outcome is greater than the utility for the worst outcome, using 0 and 1 has some  benefits. Because we have chosen to use 0 and 1, all other outcomes will have a utility value be- tween 0 and 1. In determining the utilities of all outcomes, other than the best or worst outcome,  a standard gamble is considered. This gamble is shown in Figure 3.7.

In Figure 3.7, p is the probability of obtaining the best outcome, and 11 - p2 is the prob- ability of obtaining the worst outcome. Assessing the utility of any other outcome involves de- termining the probability (p), which makes you indifferent between alternative 1, which is the  gamble between the best and worst outcomes, and alternative 2, which is obtaining the other 

The overall value of the result of a decision is called utility.

EMV is not always the best approach.

Figure 3.6 Your Decision Tree for the Lottery Ticket Accept

Offer

Reject Offer

$2,000,000

$0

Heads (0.5)

Tails (0.5)

$5,000,000EMV = $2,500,000

Utility assessment assigns the worst outcome a utility of 0 and the best outcome a utility of 1.

When you are indifferent, the expected utilities are equal.

M03_REND9327_12_SE_C03.indd 107 10/02/14 1:21 PM

108  Chapter 3 • DeCision analysis

outcome for sure. When you are indifferent between alternatives 1 and 2, the expected utilities  for these two alternatives must be equal. This relationship is shown as

Expected utility of alternative 2 = Expected utility of alternative 1

Utility of other outcome = 1p21utility of best outcome, which is 12 + 11 - p21utility of the worst outcome, which is 02

Utility of other outcome = 1p2112 + 11 - p2102 = p (3-7) Now all you have to do is to determine the value of the probability (p) that makes you 

indifferent between alternatives 1 and 2. In setting the probability, you should be aware that  utility  assessment is completely subjective. It’s a value set by the decision maker that can’t be   measured on an objective scale. Let’s take a look at an example. 

Jane Dickson would like to construct a utility curve revealing her preference for money  b etween $0 and $10,000. A utility curve is a graph that plots utility value versus monetary  value. She can invest her money either in a bank savings account or in a real estate deal.

If the money is invested in the bank, in 3 years Jane would have $5,000. If she invested in the  real estate, after 3 years she could either have nothing or $10,000. Jane, however, is very conserva- tive. Unless there is an 80% chance of getting $10,000 from the real estate deal, Jane would prefer  to have her money in the bank, where it is safe. What Jane has done here is to  assess her utility for  $5,000. When there is an 80% chance (this means that p is 0.8) of getting $10,000, Jane is indiffer- ent between putting her money in real estate and putting it in the bank. Jane’s utility for $5,000 is  thus equal to 0.8, which is the same as the value for p. This utility  assessment is shown in Figure 3.8.

Other utility values can be assessed in the same way. For example, what is Jane’s utility for  $7,000? What value of p would make Jane indifferent between $7,000 and the gamble that would  result in either $10,000 or $0? For Jane, there must be a 90% chance of getting the $10,000. Oth- erwise, she would prefer the $7,000 for sure. Thus, her utility for $7,000 is 0.90. Jane’s utility for  $3,000 can be determined in the same way. If there were a 50% chance of obtaining the $10,000,  Jane would be indifferent between having $3,000 for sure and taking the gamble of either winning 

Once utility values have been determined, a utility curve can be constructed.

Figure 3.7 Standard Gamble for Utility Assessment

Other Outcome Utility = ?

(p)

Alte rna

tive 1

Alternative 2

(1 – p) Worst Outcome Utility = 0

Best Outcome Utility = 1

Figure 3.8 Utility of $5,000

$5,000 U ($5,000) = p = 0.80

p = 0.80

Inv est

in Re

al E sta

te

Invest in Bank

$0 U ($0.00) = 0.0

$10,000 U ($10,000) = 1.0

Utility for $5,000 = U ($5,000) = pU ($10,000) + (1 – p) U ($0) = (0.8)(1) + (0.2)(0) = 0.8

(1 – p) = 0.20

M03_REND9327_12_SE_C03.indd 108 10/02/14 1:21 PM

3.10 Utility theory  109

the $10,000 or getting nothing. Thus, the utility of $3,000 for Jane is 0.5. Of course, this process can  be continued until Jane has assessed her utility for as many monetary values as she wants. These  assessments, however, are enough to get an idea of Jane’s feelings toward risk. In fact, we can plot  these points in a utility curve, as is done in Figure 3.9. In the figure, the assessed utility points of  $3,000, $5,000, and $7,000 are shown by dots, and the rest of the curve is inferred from these.

Jane’s utility curve is typical of a risk avoider. A risk avoider is a decision maker who gets  less utility or pleasure from a greater risk and tends to avoid situations in which high losses might  occur. As monetary value increases on her utility curve, the utility increases at a slower rate.

Figure 3.10 illustrates that a person who is a risk seeker has an opposite-shaped utility  curve. This decision maker gets more utility from a greater risk and higher potential payoff.  As monetary value increases on his or her utility curve, the utility increases at an increasing  rate. A person who is indifferent to risk has a utility curve that is a straight line. The shape of a  person’s utility curve depends on the specific decision being considered, the monetary values  involved in the situation, the person’s psychological frame of mind, and how the person feels  about the  future. It may well be that you have one utility curve for some situations you face and   completely different curves for others.

The shape of a person’s utility curve depends on many factors.

Figure 3.9 Utility Curve for Jane Dickson

$0 $1,000 $3,000 $5,000 $7,000 $10,000

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

U til

ity

Monetary Value

U ($10,000) = 1.0

U ($7,000) = 0.90

U ($5,000) = 0.80

U ($3,000) = 0.50

U ($0) = 0

Figure 3.10 Preferences for Risk

U til

ity

Monetary Outcome

Risk Avoider

Risk Seeker

Ri sk

In dif

fe re

nc e

M03_REND9327_12_SE_C03.indd 109 10/02/14 1:21 PM

110  Chapter 3 • DeCision analysis

Utility as a Decision-Making Criterion After a utility curve has been determined, the utility values from the curve are used in making  decisions. Monetary outcomes or values are replaced with the appropriate utility values and then  decision analysis is performed as usual. The expected utility for each alternative is computed in- stead of the EMV. Let’s take a look at an example in which a decision tree is used and expected  utility values are computed in selecting the best alternative.

Mark Simkin loves to gamble. He decides to play a game that involves tossing thumbtacks  in the air. If the point on the thumbtack is facing up after it lands, Mark wins $10,000. If the  point on the thumbtack is down, Mark loses $10,000. Should Mark play the game (alternative 1)  or should he not play the game (alternative 2)?

Alternatives 1 and 2 are displayed in the tree shown in Figure 3.11. As can be seen, alterna- tive 1 is to play the game. Mark believes that there is a 45% chance of winning $10,000 and a 55%  chance of suffering the $10,000 loss. Alternative 2 is not to gamble. What should Mark do? Of  course, this depends on Mark’s utility for money. As stated previously, he likes to gamble. Using  the procedure just outlined, Mark was able to construct a utility curve showing his preference for  money. Mark has a total of $20,000 to gamble, so he has constructed the utility curve based on a  best payoff of $20,000 and a worst payoff of a $20,000 loss. This curve appears in Figure 3.12.

We see that Mark’s utility for –$10,000 is 0.05, his utility for not playing ($0) is 0.15,  and his utility for $10,000 is 0.30. These values can now be used in the decision tree. Mark’s  o bjective is to maximize his expected utility, which can be done as follows:

Step 1. U1-$10,0002 = 0.05 U1$02 = 0.15

U1$10,0002 = 0.30

Figure 3.11 Decision Facing Mark Simkin

$0

Tack Lands Point Up (0.45)

Alte rna

tive 1

Ma rk P

lay s th

e G am

e

Alternative 2

–$10,000

$10,000 Tack Lands Point Down (0.55)

Mark Does Not Play the Game

Figure 3.12 Utility Curve for Mark Simkin

–$20,000

U til

ity

Monetary Outcome

0.75

1.00

0.50

0.30 0.25

0.15

0.05 0

–$10,000 $0 $10,000 $20,000

Mark’s objective is to maximize expected utility.

Utility values replace monetary values.

M03_REND9327_12_SE_C03.indd 110 10/02/14 1:21 PM

3.10 Utility theory  111

Step 2. Replace monetary values with utility values. Refer to Figure 3.13. Here are the expected  utilities for alternatives 1 and 2:

E1alternative 1: play the game2 = 10.45210.302 + 10.55210.052 = 0.135 + 0.027 = 0.162

E1alternative 2: don>t play the game2 = 0.15 Therefore, alternative 1 is the best strategy using utility as the decision criterion. If EMV had  been used, alternative 2 would have been the best strategy. The utility curve is a risk-seeker util- ity curve, and the choice of playing the game certainly reflects this preference for risk.

Multiattribute utility Model aids in Disposal of nuclear Weapons

When the Cold War between the United States and the USSR ended, the two countries agreed to dismantle a large number of nuclear weapons. The exact number of weapons is not known, but the total number has been estimated to be over 40,000. The plutonium recovered from the dismantled weapons presented several concerns. The National Academy of Sciences character- ized the possibility that the plutonium could fall into the hands of terrorists as a very real danger. Also, plutonium is very toxic to the environment, so a safe and secure disposal process was critical. Deciding what disposal process would be used was no easy task.

Due to the long relationship between the United States and the USSR during the Cold War, it was necessary that the pluto- nium disposal process for each country occur at approximately the same time. Whichever method was selected by one country would have to be approved by the other country. The U.S. Department of Energy (DOE) formed the Office of Fissile Materials Disposition (OFMD) to oversee the process of selecting the approach to use for disposal of the plutonium. Recognizing that the decision could be controversial, the OFMD used a team of operations research analysts associated with the Amarillo National Research Center. This OR group used a multiattribute utility (MAU) model to com- bine several performance measures into one single measure.

A total of 37 performance measures were used in evaluat- ing 13 different possible alternatives. The MAU model combined these measures and helped to rank the alternatives as well as identify the deficiencies of some alternatives. The OFMD recom- mended two of the alternatives with the highest rankings, and development was begun on both of them. This parallel devel- opment permitted the United States to react quickly when the USSR’s plan was developed. The USSR used an analysis based on this same MAU approach. The United States and the USSR chose to convert the plutonium from nuclear weapons into mixed oxide fuel, which is used in nuclear reactors to make electricity. Once the plutonium is converted to this form, it cannot be used in nu- clear weapons.

The MAU model helped the United States and the USSR deal with a very sensitive and potentially hazardous issue in a way that considered economic, nonproliferation, and ecology issues. The framework is now being used by Russia to evaluate other policies related to nuclear energy.

Source: Based on John C. Butler et al. “The United States and Russia Evalu- ate Plutonium Disposition Options with Multiattribute Utility Theory,” Interfaces 35, 1 (January–February 2005): 88–101.

in aCtiOn

Figure 3.13 Using Expected Utilities in Decision Making

0.15

Tack Lands Point Up (0.45)

Alte rna

tive 1

Pla y th

e G am

e

Alternative 2

0.05

0.30 Tack Lands Point Down (0.55)

Don’t Play

Utility

M03_REND9327_12_SE_C03.indd 111 10/02/14 1:21 PM

112  Chapter 3 • DeCision analysis

Decision theory is an analytic and systematic approach to  studying decision making. Six steps are usually involved in  making decisions in three environments: decision making un- der certainty, uncertainty, and risk. In decision making under  uncertainty, decision tables are constructed to compute criteria  such as maximax, maximin, criterion of realism, equally likely,  and minimax regret. Methods such as determining expected  monetary value (EMV), expected value of perfect information  (EVPI), expected opportunity loss (EOL), and sensitivity anal- ysis are used in decision making under risk.

Decision trees are another option, particularly for larger  decision problems, when one decision must be made before  other decisions can be made. For example, a decision to take a  sample or to perform market research is made before we decide  to construct a large plant, a small one, or no plant. In this case,  we can also compute the expected value of sample information  (EVSI) to determine the value of the market research. The effi- ciency of sample information compares the EVSI to the EVPI.  Bayesian analysis can be used to revise or update probability  values using both the prior probabilities and other probabilities  related to the accuracy of the information source.

Summary

Glossary

Alternative  A course of action or a strategy that may be cho- sen by a decision maker.

Coefficient of Realism (A)  A number from 0 to 1. When the  coefficient is close to 1, the decision criterion is optimistic.  When the coefficient is close to 0, the decision criterion is  pessimistic.

Conditional Probability  A posterior probability. Conditional Value or Payoff  A consequence, normally 

expressed in a monetary value, that occurs as a result of a  particular alternative and state of nature.

Criterion of Realism  A decision-making criterion that uses  a weighted average of the best and worst possible payoffs  for each alternative.

Decision Making under Certainty  A decision-making  environment in which the future outcomes or states of   nature are known.

Decision Making under Risk  A decision-making environ- ment in which several outcomes or states of nature may  occur as a result of a decision or alternative. The probabili- ties of the outcomes or states of nature are known.

Decision Making under Uncertainty  A decision-making  environment in which several outcomes or states of nature  may occur. The probabilities of these outcomes, however,  are not known.

Decision Node (Point)  In a decision tree, this is a point  where the best of the available alternatives is chosen. The  branches represent the alternatives.

Decision Table  A payoff table. Decision Theory  An analytic and systematic approach to 

decision making. Decision Tree  A graphical representation of a decision-

making situation. Efficiency of Sample Information  A measure of how good 

the sample information is relative to perfect information. Equally Likely  A decision criterion that places an equal 

weight on all states of nature. Expected Monetary Value (EMV)  The average value of 

a decision if it can be repeated many times. This is deter- mined by multiplying the monetary values by their respec- tive probabilities. The results are then added to arrive at the  EMV.

Expected Value of Perfect Information (EVPI)  The aver- age or expected value of information if it were completely  accurate. The increase in EMV that results from having  perfect information.

Expected Value of Sample Information (EVSI)  The in- crease in EMV that results from having sample or imperfect  information.

Expected Value with Perfect Information (EVwPI)  The  average or expected value of a decision if perfect knowl- edge of the future is available.

Hurwicz Criterion  The criterion of realism. Laplace Criterion  The equally likely criterion. Maximax  An optimistic decision-making criterion. This 

selects the alternative with the highest possible return. Maximin  A pessimistic decision-making criterion. This 

alternative maximizes the minimum payoff. It selects the  alternative with the best of the worst possible payoffs.

Minimax Regret  A criterion that minimizes the maximum  opportunity loss.

Opportunity Loss  The amount you would lose by not pick- ing the best alternative. For any state of nature, this is the  difference between the consequences of any alternative and  the best possible alternative.

Optimistic Criterion  The maximax criterion. Payoff Table  A table that lists the alternatives, states of 

nature, and payoffs in a decision-making situation. Posterior Probability  A conditional probability of a state of 

nature that has been adjusted based on sample information.  This is found using Bayes’ theorem.

Prior Probability  The initial probability of a state of nature  before sample information is used with Bayes’ theorem to  obtain the posterior probability.

Regret  Opportunity loss. Risk Avoider  A person who avoids risk. On the utility curve, 

as the monetary value, the utility increases at a decreasing  rate. This decision maker gets less utility for a greater risk  and higher potential returns.

Risk Seeker  A person who seeks risk. On the utility curve,  as the monetary value increases, the utility increases at an  increasing rate. This decision maker gets more pleasure for  a greater risk and higher potential returns.

M03_REND9327_12_SE_C03.indd 112 10/02/14 1:21 PM

solveD probleMs  113

Solved Problem 3-1 Maria Rojas is considering the possibility of opening a small dress shop on Fairbanks Avenue, a few  blocks from the university. She has located a good mall that attracts students. Her options are to open a  small shop, a medium-sized shop, or no shop at all. The market for a dress shop can be good, average, or  bad. The probabilities for these three possibilities are 0.2 for a good market, 0.5 for an average market,  and 0.3 for a bad market. The net profit or loss for the medium-sized and small shops for the various  market conditions are given in the following table. Building no shop at all yields no loss and no gain.

a.  What do you recommend? b.  Calculate the EVPI. c.   Develop the opportunity loss table for this situation. What decisions would be made using the mini-

max regret criterion and the minimum EOL criterion?

alterNative

gooD Market

($)

average Market

($)

baD Market

($)

Small shop 75,000 25,000 -40,000 Medium-sized shop 100,000 35,000 -60,000 No shop 0 0 0

Sequential Decisions  Decisions in which the outcome of  one decision influences other decisions.

Standard Gamble  The process used to determine utility  values.

State of Nature  An outcome or occurrence over which the  decision maker has little or no control.

State-of-Nature Node  In a decision tree, a point where the  EMV is computed. The branches coming from this node  represent states of nature.

Utility  The overall value or worth of a particular outcome. Utility Assessment  The process of determining the utility of 

various outcomes. This is normally done using a standard 

gamble between any outcome for sure and a gamble be- tween the worst and best outcomes.

Utility Curve  A graph or curve that reveals the relationship  between utility and monetary values. When this curve has  been constructed, utility values from the curve can be used  in the decision-making process.

Utility Theory  A theory that allows decision makers to  incorporate their risk preference and other factors into the  decision-making process.

Weighted Average Criterion  Another name for the criterion  of realism.

Key Equations

 (3-1)  EMV1alternative i2 = gXiP1Xi2 An equation that computes expected monetary value.

 (3-2)  EVwPI = g1best payoff in state of nature i2 * 1probability of state of nature i2 An equation that computes the expected value with   perfect information.

 (3-3)  EVPI = EVwPI - Best EMV An equation that computes the expected value of perfect  information.

 (3-4)  EVSI = 1EV with SI + cost2 - 1EV without SI2 An equation that computes the expected value (EV) of  sample information (SI).

 (3-5)  Efficiency of sample information = EVSI

EVPI 100%

An equation  that  compares  sample  information  to  perfect information.

 (3-6)  P1A � B2 = P1B � A2P1A2 P1B � A2P1A2 + P1B � A′2P1A′2

Bayes’ theorem—the conditional probability of event A given that event B has occurred.

 (3-7)  Utility of other outcome = 1p2112 + 11 - p2102 = p An equation that determines the utility of an intermedi- ate outcome.

Solved Problems

M03_REND9327_12_SE_C03.indd 113 10/02/14 1:21 PM

114  Chapter 3 • DeCision analysis

Solution a.   Since the decision-making environment is risk (probabilities are known), it is appropriate to use the 

EMV criterion. The problem can be solved by developing a payoff table that contains all alterna- tives, states of nature, and probability values. The EMV for each alternative is also computed, as in  the following table:

State of Nature

alterNative

gooD Market

($)

average Market

($)

baD Market

($) eMv

($)

Small shop 75,000 25,000 -40,000 15,500 Medium-sized shop 100,000 35,000 -60,000 19,500

No shop 0 0 0 0

Probabilities 0.20 0.50 0.30

EMV1small shop2 = 10.221$75,0002 + 10.521$25,0002 + 10.321-$40,0002 = $15,500 EMV1medium shop2 = 10.221$100,0002 + 10.521$35,0002 + 10.321-$60,0002 = $19,500

EMV1no shop2 = 10.221$02 + 10.521$02 + 10.321$02 = $0 As can be seen, the best decision is to build the medium-sized shop. The EMV for this alternative is  $19,500.

b.   EVwPI = 10.22$100,000 + 10.52$35,000 + 10.32$0 = $37,500 EVPI = $37,500 - $19,500 = $18,000

c.  The opportunity loss table is shown here.

State of Nature

alterNative

gooD Market

($)

average Market

($)

baD Market

($) MaxiMuM

($) eol ($)

Small shop 25,000 10,000 40,000 40,000 22,000

Medium-sized shop 0 0 60,000 60,000 18,000

No shop 100,000 35,000 0 100,000 37,500

Probabilities 0.20 0.50 0.30

The best payoff in a good market is 100,000, so the opportunity losses in the first column indicate how  much worse each payoff is than 100,000. The best payoff in an average market is 35,000, so the oppor- tunity losses in the second column indicate how much worse each payoff is than 35,000. The best payoff  in a bad market is 0, so the opportunity losses in the third column indicate how much worse each payoff  is than 0.

The minimax regret criterion considers the maximum regret for each decision, and the decision cor- responding to the minimum of these is selected. The decision would be to build a small shop since the  maximum regret for this is 40,000, while the maximum regret for each of the other two alternatives is  higher as shown in the opportunity loss table.

The decision based on the EOL criterion would be to build the medium shop. Note that the mini- mum EOL ($18,000) is the same as the EVPI computed in part b. The calculations are

EOL1small2 = 10.2225,000 + 10.5210,000 + 10.3240,000 = 22,000 EOL1medium2 = 10.220 + 10.520 + 10.3260,000 = 18,000 EOL1no shop2 = 10.22100,000 + 10.5235,000 + 10.320 = 37,500

M03_REND9327_12_SE_C03.indd 114 10/02/14 1:21 PM

solveD probleMs  115

Solved Problem 3-2 Cal Bender and Becky Addison have known each other since high school. Two years ago they entered the  same university and today they are taking undergraduate courses in the business school. Both hope to gradu- ate with degrees in finance. In an attempt to make extra money and to use some of the knowledge gained from  their business courses, Cal and Becky have decided to look into the possibility of starting a small company  that would provide word processing services to students who needed term papers or other reports prepared in  a professional manner. Using a systems approach, Cal and Becky have identified three strategies. Strategy 1  is to invest in a fairly expensive microcomputer system with a high-quality laser printer. In a favorable mar- ket, they should be able to obtain a net profit of $10,000 over the next 2 years. If the market is unfavorable,  they can lose $8,000. Strategy 2 is to purchase a less expensive system. With a favorable market, they could  get a return during the next 2 years of $8,000. With an unfavorable market, they would incur a loss of $4,000.  Their final strategy, strategy 3, is to do nothing. Cal is basically a risk taker, whereas Becky tries to avoid risk.

a.  What type of decision procedure should Cal use? What would Cal’s decision be? b.  What type of decision maker is Becky? What decision would Becky make? c.   If Cal and Becky were indifferent to risk, what type of decision approach should they use? What 

would you recommend if this were the case?

Solution The problem is one of decision making under uncertainty. Before answering the specific questions, a  decision table should be developed showing the alternatives, states of nature, and related consequences.

alterNative

favorable Market

($)

uNfavorable Market

($)

Strategy 1 10,000 -8,000

Strategy 2 8,000 -4,000

Strategy 3 0 0

a.   Since Cal is a risk taker, he should use the maximax decision criteria. This approach selects the row  that has the highest or maximum value. The $10,000 value, which is the maximum value from the ta- ble, is in row 1. Thus, Cal’s decision is to select strategy 1, which is an optimistic decision approach.

b.   Becky should use the maximin decision criteria because she wants to avoid risk. The minimum or  worst outcome for each row, or strategy, is identified. These outcomes are -$8,000 for strategy 1,  -$4,000 for strategy 2, and $0 for strategy 3. The maximum of these values is selected. Thus, Becky  would select strategy 3, which reflects a pessimistic decision approach.

c.   If Cal and Becky are indifferent to risk, they could use the equally likely approach. This ap- proach selects the alternative that maximizes the row averages. The row average for strategy 1 is  $1,000 3i.e., $1,000 = 1$10,000 - $8,0002>24. The row average for strategy 2 is $2,000, and the  row average for strategy 3 is $0. Thus, using the equally likely approach, the decision is to select  strategy 2, which maximizes the row averages.

Solved Problem 3-3 Monica Britt has enjoyed sailing small boats since she was 7 years old, when her mother started sailing  with her. Today, Monica is considering the possibility of starting a company to produce small sailboats  for the recreational market. Unlike other mass-produced sailboats, however, these boats will be made  specifically for children between the ages of 10 and 15. The boats will be of the highest quality and ex- tremely stable, and the sail size will be reduced to prevent problems of capsizing.

Her basic decision is whether to build a large manufacturing facility, a small manufacturing facility,  or no facility at all. With a favorable market, Monica can expect to make $90,000 from the large facil- ity or $60,000 from the smaller facility. If the market is unfavorable, however, Monica estimates that  she would lose $30,000 with a large facility, and she would lose only $20,000 with the small facility.  Because of the expense involved in developing the initial molds and acquiring the necessary equipment 

M03_REND9327_12_SE_C03.indd 115 10/02/14 1:21 PM

116  Chapter 3 • DeCision analysis

to produce fiberglass sailboats for young children, Monica has decided to conduct a pilot study to make  sure that the market for the sailboats will be adequate. She estimates that the pilot study will cost her  $10,000. Furthermore, the pilot study can be either favorable or unfavorable. Monica estimates that the  probability of a favorable market given a favorable pilot study is 0.8. The probability of an unfavorable  market given an unfavorable pilot study result is estimated to be 0.9. Monica feels that there is a 0.65  chance that the pilot study will be favorable. Of course, Monica could bypass the pilot study and simply  make the decision as to whether to build a large plant, small plant, or no facility at all. Without doing  any testing in a pilot study, she estimates that the probability of a favorable market is 0.6. What do you  recommend? Compute the EVSI.

Solution Before Monica starts to solve this problem, she should develop a decision tree that shows all alterna- tives, states of nature, probability values, and economic consequences. This decision tree is shown in  Figure 3.14.

The EMV at each of the numbered nodes is calculated as follows:

EMV1node 22 = 60,00010.62 + 1-20,00020.4 = 28,000 EMV1node 32 = 90,00010.62 + 1-30,00020.4 = 42,000 EMV1node 42 = 50,00010.82 + 1-30,00020.2 = 34,000 EMV1node 52 = 80,00010.82 + 1-40,00020.2 = 56,000 EMV1node 62 = 50,00010.12 + 1-30,00020.9 = -22,000 EMV1node 72 = 80,00010.12 + 1-40,00020.9 = -28,000 EMV1node 12 = 56,00010.652 + 1-10,00020.35 = 32,900

Figure 3.14 Monica’s Decision Tree, Listing Alternatives, States of nature, Probability Values, and Financial Outcomes for Solved Problem 3-3

(0.6) Market Favorable

(0.6) Market Favorable

Do Not Conduct Study

Conduct Study (0.35)

U nfavorable

Study

(0 .6

5) F

av or

ab le

St

ud y

(0.4) Market Unfavorable

(0.4) Market Unfavorable

$60,000

–$20,000

$90,000

–$30,000

No Facility

(0.8) Market Favorable

(0.8) Market Favorable

(0.2) Market Unfavorable

(0.2) Market Unfavorable

No Facility

(0.1) Market Favorable

(0.1) Market Favorable

(0.9) Market Unfavorable

(0.9) Market Unfavorable

No Facility

Sma ll Fa

cility

Large Facility

Sma ll Fa

cility

Large Facility

Sma ll Fa

cility

Large Facility

$0

2

3

$50,000

–$30,000

$80,000

–$40,000

–$10,000

4

5

$50,000

–$30,000

$80,000

–$40,000

–$10,000

6

1

7

A

B

C

D

M03_REND9327_12_SE_C03.indd 116 10/02/14 1:22 PM

solveD probleMs  117

At each of the square nodes with letters, the decisions would be:

Node B: Choose Large Facility since the EMV = $42,000. Node C: Choose Large Facility since the EMV = $56,000. Node D: Choose No Facility since the EMV = -$10,000. Node A: Choose Do Not Conduct Study since the EMV 1$42,0002 for this is higher than EMV1node 12, which is $32,900.

Based on the EMV criterion, Monica would select Do Not Conduct Study and then select Large Facility.  The EMV of this decision is $42,000. Choosing to conduct the study would result in an EMV of only  $32,900. Thus, the expected value of sample information is

EVSI = $32,900 + $10,000 - $42,000 = $900

Solved Problem 3-4 Developing a small driving range for golfers of all abilities has long been a desire of John Jenkins. John,  however, believes that the chance of a successful driving range is only about 40%. A friend of John’s  has suggested that he conduct a survey in the community to get a better feeling of the demand for such  a facility. There is a 0.9 probability that the research will be favorable if the driving range facility will  be successful. Furthermore, it is estimated that there is a 0.8 probability that the marketing research will  be unfavorable if indeed the facility will be unsuccessful. John would like to determine the chances of a  successful driving range given a favorable result from the marketing survey.

Solution This problem requires the use of Bayes’ theorem. Before we start to solve the problem, we will define  the following terms:

P1SF2 = probability of successful driving range facility P1UF2 = probability of unsuccessful driving range facility

P1RF � SF2 = probability that the research will be favorable given a successful driving range facility P1RU � SF2 = probability that the research will be unfavorable given a successful driving range facility P1RU �UF2 = probability that the research will be unfavorable given an unsuccessful driving range facility P1RF �UF2 = probability that the research will be favorable given an unsuccessful driving range facility Now, we can summarize what we know:

P1SF2 = 0.4 P1RF � SF2 = 0.9

P1RU �UF2 = 0.8 From this information we can compute three additional probabilities that we need to solve the problem:

P1UF2 = 1 - P1SF2 = 1 - 0.4 = 0.6 P1RU � SF2 = 1 - P1RF � SF2 = 1 - 0.9 = 0.1 P1RF �UF2 = 1 - P1RU �UF2 = 1 - 0.8 = 0.2

Now we can put these values into Bayes’ theorem to compute the desired probability:

P1SF �RF2 = P1RF � SF2 * P1SF2 P1RF � SF2 * P1SF2 + P1RF �UF2 * P1UF2

= 10.9210.42

10.9210.42 + 10.2210.62 =

0.36

10.36 + 0.122 = 0.36

0.48 = 0.75

M03_REND9327_12_SE_C03.indd 117 10/02/14 1:22 PM

118  Chapter 3 • DeCision analysis

In addition to using formulas to solve John’s problem, it is possible to perform all calculations in a table:

Revised Probabilities Given a Favorable Research Result

State of Nature

CoNDitioNal Probability

Prior Probability

JoiNt Probability

PoSterior Probability

Favorable market 0.9 * 0.4 = 0.36 0.36>0.48 = 0.75 Unfavorable market 0.2 * 0.6 = 0.12 0.12>0.48 = 0.25 0.48

As you can see from the table, the results are the same. The probability of a successful driving range  given a favorable research result is 0.36>0.48, or 0.75.

Self-Test

●  Before taking the self-test, refer to the learning objectives at the beginning of the chapter, the notes in the margins, and the  glossary at the end of the chapter.

●  Use the key at the back of the book to correct your answers. ●  Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

  1.  In decision theory, payoffs or profits resulting from each  possible combination of alternatives and outcomes are  called

a. conditional values. b. alternatives. c. state of nature. d. none of the above.   2.  In decision theory, probabilities are associated with a. payoffs. b. alternatives. c. states of nature. d. none of the above.   3.  The decision-making environment is called uncertainty if 

the probabilities are a. impossible. b. unknown. c. certain. d. all the above.   4.  Which of the following is a decision-making criterion 

that is used in uncertainty? a. minimax regret b. criterion of realism c. optimistic (maximax) d. all of the above   5.  The expected monetary value (EMV) is a. the weighted sum of possible payoffs for each 

 alternative. b. an upper bound on what to pay for information. c. the cost of not picking the best solution. d. computed when finding the minimax regret decision.   6.  In using the criterion of realism (Hurwicz criterion), the 

coefficient of realism (a) a. is the probability of a good state of nature.

b. describes the degree of optimism of the decision  maker.

c. describes the degree of pessimism of the decision  maker.

d. is usually less than zero.   7.  The most that a person should pay for perfect 

information is a. the EVPI. b. the maximum EMV minus the minimum EMV. c. the maximum EOL. d. the maximum EMV.   8.  The minimum EOL criterion will always result in the 

same decision as a. the maximax criterion. b. the minimax regret criterion. c. the maximum EMV criterion. d. the equally likely criterion.   9.  A decision tree is preferable to a decision table when a. a number of sequential decisions are to be made. b. probabilities are available. c. the maximax criterion is used. d. the objective is to maximize regret.  10.  Bayes’ theorem is used to revise probabilities. The new 

(revised) probabilities are called a. prior probabilities. b. sample probabilities. c. survey probabilities. d. posterior probabilities.  11.  On a decision tree, at each state-of-nature node, a. the alternative with the greatest EMV is selected. b. an EMV is calculated. c. all probabilities are added together. d. the branch with the highest probability is selected.

M03_REND9327_12_SE_C03.indd 118 10/02/14 1:22 PM

DisCUssion QUestions anD probleMs  119

 12.  The EVSI a. is found by subtracting the EMV without sample infor-

mation from the EMV with sample information. b. is always equal to the expected value of perfect 

information. c. equals the EMV with sample information assuming no 

cost for the information minus the EMV without  sample information.

d. is usually negative.  13.  The efficiency of sample information a. is the EVSI/(maximum EMV without SI) expressed as 

a percentage. b. is the EVPI/EVSI expressed as a percentage. c. would be 100% if the sample information were 

perfect. d. is computed using only the EVPI and the maximum 

EMV.  14.  On a decision tree, once the tree has been drawn and the 

payoffs and probabilities have been placed on the tree, 

the analysis (computing EMVs and selecting the best  alternative)

a. working backward (starting on the right and moving to  the left).

b. working forward (starting on the left and moving to the  right).

c. starting at the top of the tree and moving down. d. starting at the bottom of the tree and moving up.  15.  In assessing utility values, a. the worst outcome is given a utility of -1. b. the best outcome is given a utility of 0. c. the worst outcome is given a utility of 0. d. the best outcome is given a value of -1.  16.  A decision maker who gets less utility from a greater risk 

and avoids situations in which high losses might occur is a. indifferent to risk. b. a risk seeker. c. a risk avoider. d. a rational decision maker.

Discussion Questions and Problems

Discussion Questions   3-1  Give an example of a good decision that you made 

that resulted in a bad outcome. Also give an example  of a bad decision that you made that had a good out- come. Why was each decision good or bad?

  3-2  What  is decision  theory? What are  the  types of   decision-making environments?

  3-3  Discuss the challenges of listing possible alternatives.

  3-4  Discuss the differences among decision making un- der certainty, decision making under risk, and deci- sion making under uncertainty.

  3-5  What techniques are used to solve decision-making  problems under uncertainty? Which technique re- sults in an optimistic decision? Which technique re- sults in a pessimistic decision?

  3-6  What is sensitivity analysis? Why should sensitivity  analysis consider all possible consequences?

  3-7  What are the strengths of using decision trees?

  3-8  Describe how you would determine the best decision  using the EMV criterion with a decision tree.

  3-9  What is the difference between prior and posterior  probabilities?

  3-10  What is the purpose of Bayesian analysis?  Describe  how  you  would  use  Bayesian  analysis  in  the  decision- making process.

  3-11  What is the problem of using survey results?

  3-12  How is the efficiency of sample information computed?

  3-13  What is the overall purpose of utility theory?

  3-14  Briefly discuss how a utility function can be as- sessed. What is a standard gamble, and how is it  used in determining utility values?

  3-15  How is a utility curve used in selecting the best deci- sion for a particular problem?

  3-16  What is a risk seeker? What is a risk avoider? How  does the utility curve for these types of decision  makers differ?

Problems 3-17  Kenneth Brown is the principal owner of Brown Oil, 

Inc. After quitting his university teaching job, Ken  has been able to increase his annual salary by a fac- tor of over 100. At the present time, Ken is forced  to consider purchasing some more equipment for  Brown Oil because of competition. His alternatives  are shown in the following table:

equiPMeNt

favorable Market

($)

uNfavorable Market

($)

Sub 100 300,000 -200,000 Oiler J 250,000 -100,000 Texan 75,000 -18,000

    For example,  if Ken purchases a Sub 100 and if  there is a favorable market, he will realize a profit  of $300,000. On the other hand, if  the market is 

Note:   means the problem may be solved with QM for Windows;   means the problem may be solved with Excel QM; and   means the problem may be  solved with QM for Windows and/or Excel QM.

M03_REND9327_12_SE_C03.indd 119 10/02/14 1:22 PM

120  Chapter 3 • DeCision analysis

unfavorable, Ken will suffer a loss of $200,000.  But Ken has always been a very optimistic decision  maker.

(a) What type of decision is Ken facing? (b) What decision criterion should he use? (c) What alternative is best?

3-18  Although Ken Brown (discussed in Problem 3-17) is  the principal owner of Brown Oil, his brother Bob is  credited with making the company a financial suc- cess. Bob is vice president of finance. Bob attributes  his success to his pessimistic attitude about business  and the oil industry. Given the information from  Problem 3-17, it is likely that Bob will arrive at a  different decision. What decision criterion should  Bob use, and what alternative will he select?

3-19  The Lubricant is an expensive oil newsletter to which  many oil giants subscribe, including Ken Brown (see  Problem 3-17 for details). In the last issue, the letter  described how the demand for oil products would be  extremely high. Apparently, the American consumer  will continue to use oil products even if the price of  these products doubles. Indeed, one of the articles in  the Lubricant states that the chances of a favorable  market for oil products was 70%, while the chance  of an unfavorable market was only 30%. Ken would  like to use these probabilities in determining the best  decision.

(a) What decision model should be used? (b) What is the optimal decision? (c)   Ken believes that the $300,000 figure for the Sub 

100 with a favorable market is too high. How  much lower would this figure have to be for Ken  to change his decision made in part b?

3-20  Mickey Lawson  is  considering  investing  some  money that he inherited. The following payoff table  gives the profits that would be realized during the  next year for each of three investment alternatives  Mickey is considering:

State of Nature

DeCiSioN alterNative

gooD eCoNoMy

Poor eCoNoMy

Stock market 80,000 -20,000 Bonds 30,000 20,000

CDs 23,000 23,000

Probability 0.5 0.5

(a) What decision would maximize expected profits? (b) What is the maximum amount that should be 

paid for a perfect forecast of the economy?

3-21  Develop an opportunity loss table for the investment  problem that Mickey Lawson faces in Problem 3-20.  What decision would minimize the expected oppor- tunity loss? What is the minimum EOL?

3-22  Allen Young has always been proud of his personal  investment strategies and has done very well over  the past several years. He invests primarily in the  stock market. Over the past several months, how- ever, Allen has become very concerned about the  stock market as a good investment. In some cases, it  would have been better for Allen to have his money  in a bank than in the market. During the next year,  Allen must decide whether to invest $10,000 in the  stock market or in a certificate of deposit (CD) at an  interest rate of 9%. If the market is good, Allen be- lieves that he could get a 14% return on his money.  With a fair market, he expects to get an 8% return. If  the market is bad, he will most likely get no  return  at all—in other words,  the return would be 0%.  Allen estimates that the probability of a good market  is 0.4, the probability of a fair market is 0.4, and the  probability of a bad market is 0.2, and he wishes to  maximize his long-run average return.

(a) Develop a decision table for this problem. (b) What is the best decision?

3-23  In Problem 3-22, you helped Allen Young determine  the best investment strategy. Now, Young is thinking  about paying for a stock market newsletter. A friend  of Young said that these types of letters could predict  very accurately whether the market would be good,  fair, or poor. Then, based on these predictions, Allen  could make better investment decisions.

(a) What is the most that Allen would be willing to  pay for a newsletter?

(b) Young now believes that a good market will give  a return of only 11% instead of 14%. Will this  information change the amount that Allen would  be willing to pay for the newsletter? If your an- swer is yes, determine the most that Allen would  be willing to pay, given this new information.

3-24  Today’s Electronics specializes in manufacturing  modern electronic components. It also builds the  equipment that produces the components.  Phyllis  Weinberger, who is responsible for advising the  president of Today’s Electronics on electronic manu- facturing equipment, has developed the following  table concerning a proposed facility:

Profit ($)

StroNg Market

fair Market

Poor Market

Large facility 550,000 110,000 -310,000 Medium-sized  facility

300,000 129,000 -100,000

Small facility 200,000 100,000 -32,000 No facility 0 0 0

(a) Develop an opportunity loss table. (b) What is the minimax regret decision?

M03_REND9327_12_SE_C03.indd 120 10/02/14 1:22 PM

DisCUssion QUestions anD probleMs  121

3-25  Brilliant Color  is a  small  supplier of chemicals  and equipment that are used by some photographic  stores to process 35mm film. One product that Bril- liant Color supplies is BC-6. John Kubick, president  of Brilliant Color, normally stocks 11, 12, or 13  cases of BC-6 each week. For each case that John  sells, he receives a profit of $35. Like many photo- graphic chemicals, BC-6 has a very short shelf life,  so if a case is not sold by the end of the week, John  must discard it. Since each case costs John $56, he  loses $56 for every case that is not sold by the end of  the week. There is a probability of 0.45 of selling 11  cases, a probability of 0.35 of selling 12 cases, and a  probability of 0.2 of selling 13 cases.

(a) Construct a decision table for this problem. In- clude all conditional values and probabilities in  the table.

(b) What is your recommended course of action? (c)  If John is able to develop BC-6 with an ingredi-

ent that stabilizes it so that it no longer has to be  discarded, how would this change your recom- mended course of action?

3-26  Megley Cheese Company is a small manufacturer of  several different cheese products. One of the prod- ucts is a cheese spread that is sold to retail outlets.  Jason Megley must decide how many cases of cheese  spread to manufacture each month. The probability  that the demand will be six cases is 0.1, seven cases  is 0.3, eight cases is 0.5, and nine cases is 0.1. The  cost of every case is $45, and the price that Jason  gets for each case is $95. Unfortunately, any cases  not sold by the end of the month are of no value, due  to spoilage. How many cases of cheese should Jason  manufacture each month?

3-27  Farm Grown, Inc., produces cases of perishable food  products. Each case contains an assortment of veg- etables and other farm products. Each case costs $5  and sells for $15. If there are any cases not sold by  the end of the day, they are sold to a large food pro- cessing company for $3 a case. The probability that  daily demand will be 100 cases is 0.3, the probabil- ity that daily demand will be 200 cases is 0.4, and  the probability that daily demand will be 300 cases  is 0.3. Farm Grown has a policy of always satisfying  customer demands. If its own supply of cases is less  than the demand, it buys the necessary vegetables  from a competitor. The estimated cost of doing this  is $16 per case.

(a) Draw a decision table for this problem. (b) What do you recommend?

3-28  In Problem 3-27, Farm Grown, Inc. has reason to  believe the probabilities may not be reliable due  to changing conditions. If these probabilities are  ignored, what decision would be made using the 

optimistic criterion? What decision would be made  using the pessimistic criterion?

3-29  Mick Karra is the manager of MCZ Drilling Prod- ucts, which produces a variety of specialty valves for  oil field equipment. Recent activity in the oil fields  has caused demand to increase drastically, and a  decision has been made to open a new manufactur- ing facility. Three locations are being considered,  and the size of the facility would not be the same  in each location. Thus, overtime might be necessary  at times. The following table gives the total monthly  cost (in $1,000s) for each possible location under  each demand possibility. The probabilities for the  demand levels have been determined to be 20% for  low demand, 30% for medium demand, and 50% for  high demand.

DeMaND iS low

DeMaND iS MeDiuM

DeMaND iS HigH

Ardmore, OK 85 110 150

Sweetwater, TX 90 100 120

Lake Charles, LA 110 120 130

(a) Which location would be selected based on the  optimistic criterion?

(b) Which  location would be selected based on  the  pessimistic criterion?

(c) Which  location would be selected based on  the  minimax regret criterion?

(d) Which  location  should be  selected  to minimize  the expected cost of operation?

(e) How much  is  a  perfect  forecast  of  the  demand  worth?

(f)  Which location would minimize the expected op- portunity loss?

(g) What is the expected value of perfect informa- tion in this situation?

3-30  Even though independent gasoline stations have  been having a difficult time, Susan Solomon has  been thinking about starting her own independent  gasoline station. Susan’s problem is to decide how  large her station should be. The annual returns will  depend on both the size of her station and a number  of marketing factors related to the oil industry and  demand for gasoline. After a careful analysis, Susan  developed the following table:

Size of firSt StatioN

gooD Market

($)

fair Market

($)

Poor Market

($)

Small 50,000 20,000 -10,000 Medium 80,000 30,000 -20,000 Large 100,000 30,000 -40,000 Very large 300,000 25,000 -160,000

M03_REND9327_12_SE_C03.indd 121 10/02/14 1:22 PM

122  Chapter 3 • DeCision analysis

    For example, if Susan constructs a small station  and the market is good, she will realize a profit of  $50,000.

(a) Develop a decision table for this decision. (b) What is the maximax decision? (c) What is the maximin decision? (d) What is the equally likely decision? (e)   What is the criterion of realism decision? Use an 

a value of 0.8. (f)  Develop an opportunity loss table. (g) What is the minimax regret decision?

3-31  Beverly Mills has decided to lease a hybrid car to  save on gasoline expenses and to do her part to help  keep the environment clean. The car she selected  is available from only one dealer in the local area,  but that dealer has several leasing options to accom- modate a variety of driving patterns. All the leases  are for 3 years and require no money at the time of  signing the lease. The first option has a monthly cost  of $330, a total mileage allowance of 36,000 miles  (an average of 12,000 miles per year), and a cost of  $0.35 per mile for any miles over 36,000. The fol- lowing table summarizes each of the three lease  options:

3-year leaSe

MoNtHly CoSt

Mileage allowaNCe

CoSt Per exCeSS Mile

Option 1 $330 36,000 $0.35

Option 2 $380 45,000 $0.25

Option 3 $430 54,000 $0.15

    Beverly has estimated that, during the 3 years of the  lease, there is a 40% chance she will drive an aver- age of 12,000 miles per year, a 30% chance she will  drive an average of 15,000 miles per year, and a 30%  chance that she will drive 18,000 miles per year. In  evaluating these lease options, Beverly would like to  keep her costs as low as possible.

(a) Develop a payoff (cost) table for this situation. (b) What decision would Beverly make if she were 

optimistic? (c) What decision would Beverly make if she were 

pessimistic? (d) What decision would Beverly make if she wanted 

to minimize her expected cost (monetary value)? (e) Calculate the expected value of perfect informa-

tion for this problem. 3-32  Refer to the leasing decision facing Beverly Mills in 

Problem 3-31. Develop the opportunity loss table for  this situation. Which option would be chosen based  on the minimax regret criterion? Which alternative  would result  in  the  lowest expected opportunity  loss?

3-33  The game of roulette is popular in many casinos  around the world. In Las Vegas, a typical roulette  wheel has the numbers 1–36 in slots on the wheel.  Half of these slots are red, and the other half are  black. In the United States, the roulette wheel typi- cally also has the numbers 0 (zero) and 00 (double  zero), and both of these are on the wheel in green  slots. Thus, there are 38 slots on the wheel. The  dealer spins the wheel and sends a small ball  in  the opposite direction of the spinning wheel. As the  wheel slows, the ball falls into one of the slots, and  that is the winning number and color. One of the bets  available is simply red or black, for which the odds  are 1 to 1. If the player bets on either red or black,  and that happens to be the winning color, the player  wins  the amount of her bet. For example,  if  the  player bets $5 on red and wins, she is paid $5 and  she still has her original bet. On the other hand, if  the winning color is black or green when the player  bets red, the player loses the entire bet.

(a) What is the probability that a player who bets  red will win the bet?

(b) If a player bets $10 on red every time the game  is  played, what  is  the  expected monetary value  (expected win)?

(c)  In Europe,  there  is usually no 00 on  the wheel,  just  the  0. With  this  type  of  game, what  is  the  probability  that  a  player who bets  red will win  the bet?  If  a player bets $10 on  red every  time  in this game (with no 00), what  is  the expected  monetary value?

(d) Since  the expected profit  (win)  in a  roulette  game is negative, why would a rational person  play the game?

3-34  Refer to the Problem 3-33 for details about the game  of roulette. Another bet in a roulette game is called a  “straight up” bet, which means that the player is bet- ting that the winning number will be the number that  she chose. In a game with 0 and 00, there are a total  of 38 possible outcomes (the numbers 1 to 36 plus  0 and 00), and each of these has the same chance of  occurring. The payout on this type of bet is 35 to 1,  which means the player is paid 35 and gets to keep  the original bet. If a player bets $10 on the number  7 (or any single number), what is the expected mon- etary value (expected win)?

3-35  The Technically Techno company has several pat- ents for a variety of different Flash memory devices  that are used in computers, cell phones, and a variety  of other things. A competitor has recently introduced  a product based on technology very similar to some- thing patented by Technically Techno last year. Con- sequently, Technically Techno has sued the other  company for copyright infringement. Based on the  facts in the case as well as the record of the lawyers  involved, Technically Techno believes there is a 40% 

M03_REND9327_12_SE_C03.indd 122 10/02/14 1:22 PM

DisCUssion QUestions anD probleMs  123

chance that it will be awarded $300,000 if the law- suit goes to court. There is a 30% chance that they  would be awarded only $50,000 if they go to court  and win, and there is a 30% chance they would lose  the case and be awarded nothing. The estimated cost  of legal fees if they go to court is $50,000. However,  the other company has offered to pay Technically  Techno $75,000 to settle the dispute without going  to court. The estimated legal cost of this would only  be $10,000. If Technically Techno wished to maxi- mize the expected gain, should they accept the set- tlement offer?

3-36  A group of medical professionals is considering the  construction of a private clinic. If the medical de- mand is high (i.e., there is a favorable market for  the clinic), the physicians could realize a net profit  of $100,000. If the market is not favorable, they  could lose $40,000. Of course, they don’t have to  proceed at all, in which case there is no cost. In the  absence of any market data, the best the physicians  can guess is that there is a 50–50 chance the clinic  will be successful. Construct a decision tree to help  analyze this problem. What should the medical pro- fessionals do?

3-37  The physicians  in Problem 3-36 have been ap- proached by a market research firm that offers to  perform a study of the market at a fee of $5,000. The  market researchers claim their experience enables  them to use Bayes’ theorem to make the following  statements of probability:

probability of a favorable market given a favorable study = 0.82

probability of an unfavorable market given a favorable study = 0.18

probability of a favorable market given an unfavorable study = 0.11

probability of an unfavorable market given an unfavorable study = 0.89

probability of a favorable research study = 0.55 probability of an unfavorable research study = 0.45 (a) Develop a new decision tree for the medical pro-

fessionals to reflect the options now open with  the market study.

(b) Use the EMV approach to recommend a strategy. (c) What  is  the  expected value of  sample  informa-

tion? How much might the physicians be willing  to pay for a market study?

(d) Calculate  the  efficiency  of  this  sample  information.

 3-38  Jerry Smith is thinking about opening a bicycle shop  in his hometown. Jerry loves to take his own bike  on 50-mile trips with his friends, but he believes that  any small business should be started only if there is 

a good chance of making a profit. Jerry can open a  small shop, a large shop, or no shop at all. The prof- its will depend on the size of the shop and whether  the market is favorable or unfavorable for his prod- ucts. Because there will be a 5-year lease on the  building that Jerry is thinking about using, he wants  to make sure that he makes the correct decision.  Jerry is also thinking about hiring his old marketing  professor to conduct a marketing research study. If  the study is conducted, the study could be favorable  (i.e., predicting a favorable market) or unfavorable  (i.e., predicting an unfavorable market). Develop a  decision tree for Jerry.

3-39  Jerry Smith  (see Problem 3-38) has done  some  analysis about the profitability of the bicycle shop.  If Jerry builds the large bicycle shop, he will earn  $60,000 if the market is favorable, but he will lose  $40,000 if the market is unfavorable. The small shop  will return a $30,000 profit in a favorable market  and a $10,000 loss in an unfavorable market. At the  present time, he believes that there is a 50–50 chance  that the market will be favorable. His old marketing  professor will charge him $5,000 for the marketing  research. It is estimated that there is a 0.6 probability  that the survey will be favorable. Furthermore, there  is a 0.9 probability that the market will be favorable  given a favorable outcome from the study. However,  the marketing professor has warned Jerry that there  is only a probability of 0.12 of a favorable market  if the marketing research results are not favorable.  Jerry is confused.

(a)  Should Jerry use the marketing research? (b) Jerry, however, is unsure the 0.6 probability of 

a favorable marketing research study is correct.  How sensitive is Jerry’s decision to this prob- ability value? How far can this probability value  deviate from 0.6 without causing Jerry to change  his decision?

3-40  Bill Holliday is not sure what she should do. He can  build a quadplex (i.e., a building with four apart- ments), build a duplex, gather additional informa- tion, or simply do nothing. If he gathers additional  information, the results could be either favorable or  unfavorable, but it would cost him $3,000 to gather  the information. Bill believes that there is a 50–50  chance that the information will be favorable. If the  rental market is favorable, Bill will earn $15,000  with the quadplex or $5,000 with the duplex. Bill  doesn’t have the financial resources to do both. With  an unfavorable rental market, however, Bill could  lose $20,000 with the quadplex or $10,000 with  the duplex. Without gathering additional informa- tion, Bill estimates that the probability of a favor- able rental market is 0.7. A favorable report from the  study would increase the probability of a favorable 

M03_REND9327_12_SE_C03.indd 123 10/02/14 1:22 PM

124  Chapter 3 • DeCision analysis

rental market to 0.9. Furthermore, an unfavorable re- port from the additional information would decrease  the probability of a favorable rental market to 0.4.  Of course, Bill could forget all of these numbers and  do nothing. What is your advice to Bill?

 3-41  Peter Martin is going to help his brother who wants  to open a food store. Peter initially believes that there  is a 50–50 chance that his brother’s food store would  be a success. Peter is considering doing a market re- search study. Based on historical data, there is a 0.8  probability that the marketing research will be favor- able given a successful food store. Moreover, there  is a 0.7 probability that the marketing research will  be unfavorable given an unsuccessful food store.

(a)  If the marketing research is favorable, what is  Peter’s revised probability of a successful food  store for his brother?

(b) If the marketing research is unfavorable, what is  Peter’s  revised  probability  of  a  successful  food  store for his brother?

(c)  If  the initial probability of a successful food  store is 0.60 (instead of 0.50), find the probabili- ties in parts a and b.

3-42  Mark Martinko has been a class A racquetball player  for the past 5 years, and one of his biggest goals is to  own and operate a racquetball facility. Unfortunately,  Mark’s thinks that the chance of a successful rac- quetball facility is only 30%. Mark’s lawyer has rec- ommended that he employ one of the local marketing  research groups to conduct a survey concerning the  success or failure of a racquetball facility. There is  a 0.8 probability that the research will be favorable  given a successful racquetball facility. In addition,  there is a 0.7 probability that the research will be un- favorable given an unsuccessful facility. Compute re- vised probabilities of a successful racquetball facility  given a favorable and given an unfavorable survey.

3-43  A financial advisor has recommended two possible  mutual funds for investment: Fund A and Fund B. The  return that will be achieved by each of these depends  on whether the economy is good, fair, or poor. A payoff  table has been constructed to illustrate this situation:

State of Nature

iNveStMeNt gooD

eCoNoMy fair

eCoNoMy Poor

eCoNoMy

Fund A $10,000 $2,000 -$5,000 Fund B $6,000 $4,000 0

Probability 0.2 0.3 0.5

(a) Draw the decision tree to represent this situation. (b) Perform the necessary calculations to determine 

which of the two mutual funds is better. Which one  should you choose to maximize the expected value?

(c)  Suppose there is question about the return of  Fund A in a good economy. It could be higher or 

lower than $10,000. What value for this would  cause a person to be indifferent between Fund A  and Fund B (i.e., the EMVs would be the same)?

3-44  Jim Sellers is thinking about producing a new type  of electric razor for men. If the market were favor- able, he would get a return of $100,000, but if the  market for this new type of razor were unfavorable,  he would lose $60,000. Since Ron Bush is a good  friend of Jim Sellers, Jim is considering the possibil- ity of using Bush Marketing Research to gather ad- ditional information about the market for the razor.  Ron has suggested that Jim use either a survey or  a pilot study to test the market. The survey would  be a sophisticated questionnaire administered to a  test market. It will cost $5,000. Another alternative  is to run a pilot study. This would involve producing  a limited number of the new razors and trying to sell  them in two cities that are typical of American cit- ies. The pilot study is more accurate but is also more  expensive. It will cost $20,000. Ron Bush has sug- gested that it would be a good idea for Jim to con- duct either the survey or the pilot before Jim makes  the decision concerning whether to produce the new  razor. But Jim is not sure if the value of the survey or  the pilot is worth the cost.

Jim estimates that the probability of a successful  market without performing a survey or pilot study is  0.5. Furthermore, the probability of a favorable sur- vey result given a favorable market for razors is 0.7,  and the probability of a favorable survey result given  an unsuccessful market for razors is 0.2. In addition,  the probability of an unfavorable pilot study given  an unfavorable market is 0.9, and the probability of  an unsuccessful pilot study result given a favorable  market for razors is 0.2.

(a) Draw the decision tree for this problem without  the probability values.

(b) Compute the revised probabilities needed to  complete the decision, and place these values in  the decision tree.

(c) What is the best decision for Jim? Use EMV as  the decision criterion.

3-45  Jim Sellers has been able to estimate his utility for  a number of different values. He would like to use  these utility values in making the decision in Prob- lem 3-44: U1-$80,0002 = 0, U1-$65,0002 = 0.5,

U1-$60,0002 = 0.55, U1-$20,0002 = 0.7, U1-$5,0002 = 0.8, U1$02 = 0.81, U1$80,0002 = 0.9, U1$95,0002 = 0.95, and U1$100,0002 = 1.     Resolve Problem 3-44 using utility values. Is Jim a 

risk avoider?

3-46.  Two states of nature exist for a particular situation:  a good economy and a poor economy. An economic  study may be performed to obtain more informa- tion about which of these will actually occur in the  coming year. The study may forecast either a good 

M03_REND9327_12_SE_C03.indd 124 10/02/14 1:22 PM

DisCUssion QUestions anD probleMs  125

economy or a poor economy. Currently there is a  60% chance that the economy will be good and a  40% chance that it will be poor. In the past, when- ever the economy was good, the economic study  predicted it would be good 80% of the time. (The  other 20% of the time the prediction was wrong.) In  the past, whenever the economy was poor, the eco- nomic study predicted it would be poor 90% of the  time. (The other 10% of the time the prediction was  wrong.)

(a) Use Bayes’ theorem and find the following:

P1good economy � prediction of good economy2 P1poor economy � prediction of good economy2 P1good economy � prediction of poor economy2 P1poor economy � prediction of poor economy2

(b) Suppose the initial (prior) probability of a good  economy is 70% (instead of 60%), and the prob- ability of a poor economy is 30% (instead of  40%). Find the posterior probabilities in part a based on these new values.

3-47  The Long Island Life Insurance Company sells a  term life insurance policy. If the policy holder dies  during the term of the policy, the company pays  $100,000. If the person does not die, the company  pays out nothing and there is no further value to  the policy. The company uses actuarial tables to de- termine the probability that a person with certain  characteristics will die during the coming year. For  a particular individual, it is determined that there is  a 0.001 chance that the person will die in the next  year and a 0.999 chance that the person will live and  the company will pay out nothing. The cost of this  policy is $200 per year. Based on the EMV criterion,  should the individual buy this  insurance policy?  How would utility theory help explain why a person  would buy this insurance policy?

3-48  In Problem 3-37, you helped the medical profession- als analyze their decision using expected monetary  value as the decision criterion. This group has also  assessed their utility for money: U1-$45,0002 = 0, U1-$40,0002 = 0.1, U1-$5,0002 = 0.7, U1$02= 0.9, U1$95,0002 = 0.99,   a nd   U1$100,0002 = 1. Use expected utility as the decision criterion, and  determine the best decision for the medical profes- sionals. Are the medical professionals risk seekers or  risk avoiders?

3-49  In this chapter, a decision tree was developed for  John Thompson (see Figure 3.5 for the complete de- cision tree analysis). After completing the analysis,  John was not completely sure that he is indifferent  to risk. After going through a number of standard  gambles,  John was able  to assess his utility  for  money. Here are some of the utility assessments: 

U1-$190,0002 = 0, U1-$180,0002 = 0.05, U1-$30,0002 = 0.10, U1-$20,0002 = 0.15, U1-$10,0002 = 0.2,U1$02 = 0.3, U1$90,0002 =

0.15, U1$100,0002 = 0.6, U1$190,0002 = 0.95, and U1$200,0002 = 1.0. If John maximizes his ex- pected utility, does his decision change?

3-50  In the past few years, the traffic problems in Lynn  McKell’s hometown have gotten worse. Now, Broad  Street is congested about half the time. The normal  travel  time to work for Lynn is only 15 minutes  when Broad Street  is used and  there  is no con- gestion. With congestion, however, it takes Lynn  40 minutes to get to work. If Lynn decides to take  the expressway, it will take 30 minutes regardless of  the traffic conditions. Lynn’s utility for travel time  is: U115 minutes2 = 0.9, U130 minutes2 = 0.7, and  U140 minutes2 = 0.2. (a) Which  route will minimize Lynn’s expected 

travel time? (b) Which route will maximize Lynn’s utility? (c) When it comes to  travel  time,  is Lynn a risk 

seeker or a risk avoider?

3-51  Coren Chemical, Inc., develops industrial chemicals  that are used by other manufacturers to produce pho- tographic chemicals, preservatives, and lubricants.  One of their products, K-1000, is used by several  photographic companies to make a chemical that  is used in the film-developing process. To produce  K-1000 efficiently, Coren Chemical uses the batch  approach, in which a certain number of gallons is  produced at one time. This reduces setup costs and  allows Coren Chemical  to produce K-1000 at a  competitive price. Unfortunately, K-1000 has a very  short shelf life of about 1 month.

Coren Chemical produces K-1000 in batches of  500 gallons, 1,000 gallons, 1,500 gallons, and 2,000  gallons. Using historical data, David Coren was able  to determine that the probability of selling 500 gal- lons of K-1000 is 0.2. The probabilities of selling  1,000, 1,500, and 2,000 gallons are 0.3, 0.4, and 0.1,  respectively. The question facing David is how many  gallons to produce of K-1000 in the next batch run.  K-1000 sells for $20 per gallon. Manufacturing cost  is $12 per gallon, and handling costs and warehous- ing costs are estimated to be $1 per gallon. In the  past, David has allocated advertising costs to K-1000  at $3 per gallon. If K-1000 is not sold after the batch  run, the chemical loses much of its important prop- erties as a developer. It can, however, be sold at a  salvage value of $13 per gallon. Furthermore, David  has guaranteed to his suppliers that there will always  be an adequate supply of K-1000. If David does run  out, he has agreed to purchase a comparable chemi- cal from a competitor at $25 per gallon. David sells  all of the chemical at $20 per gallon, so his shortage  means that David loses the $5 to buy the more ex- pensive chemical.

(a) Develop a decision tree of this problem. (b) What is the best solution? (c) Determine the expected value of perfect information.

M03_REND9327_12_SE_C03.indd 125 10/02/14 1:22 PM

126  Chapter 3 • DeCision analysis

3-52  The Jamis Corporation is involved with waste man- agement. During the past 10 years it has become  one of the largest waste disposal companies in the  Midwest, serving primarily Wisconsin, Illinois, and  Michigan. Bob Jamis, president of the company, is  considering the possibility of establishing a waste  treatment plant in Mississippi. From past experi- ence, Bob believes that a small plant in northern  Mississippi would yield a $500,000 profit regardless  of the market for the facility. The success of a me- dium-sized waste treatment plant would depend on  the market. With a low demand for waste treatment,  Bob expects a $200,000 return. A medium demand  would yield a $700,000 return in Bob’s estimation,  and a high demand would return $800,000. Although  a large facility is much riskier, the potential return is  much greater. With a high demand for waste treat- ment in Mississippi, the large facility should return  a million dollars. With a medium demand, the large  facility will return only $400,000. Bob estimates that  the large facility would be a big loser if there were  a low demand for waste treatment. He estimates that  he would lose approximately $200,000 with a large  treatment facility if demand were indeed low. Look- ing at the economic conditions for the upper part of  the state of Mississippi and using his experience in  the field, Bob estimates that the probability of a low  demand for treatment plants is 0.15. The probabil- ity for a medium-demand facility is approximately  0.40, and the probability of a high demand for a  waste treatment facility is 0.45.

Because of the large potential investment and the  possibility of a loss, Bob has decided to hire a market  research team that is based in Jackson, Mississippi.  This team will perform a survey to get a better feel- ing for the probability of a low, medium, or high de- mand for a waste treatment facility. The cost of the  survey is $50,000. To help Bob determine whether to  go ahead with the survey, the marketing research firm  has provided Bob with the following information:

PSsurvey results ∣ possible outcomesT

SURVEY RESULTS

PoSSible outCoMe

low Survey reSultS

MeDiuM Survey reSultS

HigH Survey reSultS

Low demand 0.7 0.2 0.1

Medium  demand

0.4 0.5 0.1

High demand 0.1 0.3 0.6

As you see, the survey could result in three pos- sible outcomes. Low survey results mean that a low  demand is likely. In a similar fashion, medium sur- vey results or high survey results would mean a me- dium or a high demand, respectively. What should  Bob do?

3-53  Mary is considering opening a new grocery store in  town. She is evaluating three sites: downtown, the  mall, and out at the busy traffic circle. Mary cal- culated the value of successful stores at these lo- cations as follows: downtown, $250,000; the mall,  $300,000; the circle, $400,000. Mary calculated the  losses if unsuccessful to be $100,000 at either down- town or the mall and $200,000 at the circle. Mary  figures her chance of success to be 50% downtown,  60% at the mall, and 75% at the traffic circle.

(a) Draw a decision tree for Mary and select her best  alternative.

(b) Mary  has  been  approached  by  a  marketing   research  firm  that  offers  to  study  the  area  to   determine if another grocery store is needed. The  cost of this study is $30,000. Mary believes there  is  a 60% chance  that  the  survey  results will  be  positive (show a need for another grocery store).  SRP = survey  results  positive,  SRN =  survey  results  negative,  SD = success  downtown,  SM = success at mall, SC = success at circle,  SD′ = don’t succeed downtown, and so on. For  studies of this nature:

P1SRP � success2 = 0.7, P1SRN � success2 = 0.3, P1SRP � not success2 = 0.2; and  P1SRN � not success2 = 0.8.

  Calculate  the  revised  probabilities  for  success  (and  not  success)  for  each  location,  depending  on survey results.

(c) How much is the marketing research worth to  Mary? Calculate the EVSI.

3-54.  Sue Reynolds has to decide if she should get informa- tion (at a cost of $20,000) to invest in a retail store. If she  gets the information, there is a 0.6 probability that the  information will be favorable and a 0.4 probability that  the information will not be favorable. If the informa- tion is favorable, there is a 0.9 probability that the store  will be a success. If the information is not favorable, the  probability of a successful store is only 0.2. Without any  information, Sue estimates that the probability of a suc- cessful store will be 0.6. A successful store will give a  return of $100,000. If the store is built but is not suc- cessful, Sue will see a loss of $80,000. Of course, she  could always decide not to build the retail store.

(a) What do you recommend? (b) What impact would a 0.7 probability of obtaining 

favorable  information  have  on  Sue’s  decision?  The  probability  of  obtaining  unfavorable  infor- mation would be 0.3.

(c)  Sue believes that the probabilities of a successful and  an unsuccessful retail store given favorable informa- tion might be 0.8 and 0.2,  respectively,  instead of  0.9 and 0.1, respectively. What impact, if any, would  this have on Sue’s decision and the best EMV?

(d) Sue had to pay $20,000 to get information. Would  her decision change if the cost of the information  increased to $30,000?

M03_REND9327_12_SE_C03.indd 126 10/02/14 1:22 PM

Case stUDy  127

(e) Using the data in this problem and the following  utility table, compute the expected utility. Is this  the curve of a risk seeker or a risk avoider?

MoNetary value utility

$100,000 1

$80,000 0.4

$0 0.2

-$20,000 0.1 -$80,000 0.05

-$100,000 0

(f)  Compute the expected utility given the follow- ing utility table. Does this utility table represent  a risk seeker or a risk avoider?

MoNetary value utility

$100,000 1

$80,000 0.9

$0 0.8

-$20,000 0.6 -$80,000 0.4

-$100,000 0

See our Internet home page, at www.pearsonglobaleditions.com/render,  for additional  homework problems, Problems 3-55 to 3-60.

Internet Homework Problems

After watching a movie about a young woman who quit a suc- cessful corporate career to start her own baby food company,  Julia Day decided that she wanted to do the same. In the movie,  the baby food company was very successful. Julia knew, how- ever, that it is much easier to make a movie about a successful  woman starting her own company than to actually do it. The  product had to be of the highest quality, and Julia had to get the  best people involved to launch the new company. Julia resigned  from her job and launched her new company—Starting Right.

Julia decided to target the upper end of the baby food mar- ket by producing baby food that contained no preservatives but  had a great taste. Although the price would be slightly higher  than for existing baby food, Julia believed that parents would  be willing to pay more for a high-quality baby food. Instead of  putting baby food in jars, which would require preservatives to  stabilize the food, Julia decided to try a new approach. The baby  food would be frozen. This would allow for natural ingredients,  no preservatives, and outstanding nutrition.

Getting good people to work for the new company was also  important. Julia decided to find people with experience in fi- nance, marketing, and production to get involved with Starting  Right. With her enthusiasm and charisma, Julia was able to find  such a group. Their first step was to develop prototypes of the  new frozen baby food and to perform a small pilot test of the  new product. The pilot test received rave reviews.

The  final  key  to getting  the young company off  to  a  good start was to raise funds. Three options were considered: 

corporate bonds, preferred stock, and common stock. Julia de- cided that each investment should be in blocks of $30,000. Fur- thermore, each investor should have an annual income of at least  $40,000 and a net worth of $100,000 to be eligible to invest in  Starting Right. Corporate bonds would return 13% per year for  the next 5 years. Julia furthermore guaranteed that investors in  the corporate bonds would get at least $20,000 back at the end  of 5 years. Investors in preferred stock should see their initial in- vestment increase by a factor of 4 with a good market or see the  investment worth only half of the initial investment with an un- favorable market. The common stock had the greatest potential.  The initial investment was expected to increase by a factor of 8  with a good market, but investors would lose everything if the  market was unfavorable. During the next 5 years, it was expected  that inflation would increase by a factor of 4.5% each year.

Discussion Questions   1. Sue Pansky, a retired elementary school teacher, is con-

sidering investing in Starting Right. She is very conserva- tive and is a risk avoider. What do you recommend?

  2. Ray Cahn, who is currently a commodities broker,  is  also considering an investment, although he believes that  there is only an 11% chance of success. What do you  recommend?

  3. Lila Battle  has  decided  to  invest  in Starting Right.  While she believes that Julia has a good chance of being 

Case Study

starting right Corporation

M03_REND9327_12_SE_C03.indd 127 10/02/14 1:22 PM

128  Chapter 3 • DeCision analysis

successful, Lila is a risk avoider and very conservative.  What is your advice to Lila?

  4. George Yates believes  that  there  is an equally  likely  chance for success. What is your recommendation?

  5. Peter Metarko is extremely optimistic about the market  for the new baby food. What is your advice for Pete?

  6. Julia Day has been told that developing the legal docu- ments for each fundraising alternative is expensive. Julia  would like to offer alternatives for both risk-averse and  risk-seeking investors. Can Julia delete one of the finan- cial alternatives and still offer investment choices for risk  seekers and risk avoiders?

Case Study

blake electronics

In 1979, Steve Blake founded Blake Electronics  in Long  Beach, California, to manufacture resistors, capacitors, induc- tors, and other electronic components. During the Vietnam  War, Steve was a radio operator, and it was during this time  that he became proficient at repairing radios and other commu- nications equipment. Steve viewed his 4-year experience with  the army with mixed feelings. He hated army life, but this ex- perience gave him the confidence and the initiative to start his  own electronics firm.

Over the years, Steve kept the business relatively un- changed. By 1992, total annual sales were in excess of $2  million. In 1996, Steve’s son, Jim, joined the company after  finishing high school and 2 years of courses in electronics at  Long Beach Community College. Jim was always aggressive  in high school athletics, and he became even more aggres- sive as general sales manager of Blake Electronics. This ag- gressiveness bothered Steve, who was more conservative. Jim  would make deals to supply companies with electronic compo- nents before he bothered to find out if Blake Electronics had  the ability or capacity to produce the components. On several  occasions this behavior caused the company some embarrass- ing moments when Blake Electronics was unable to produce  the electronic components for companies with which Jim had  made deals.

In 2000, Jim started to go after government contracts for  electronic components. By 2002, total annual sales had in- creased to more than $10 million, and the number of employ- ees exceeded 200. Many of these employees were electronic  specialists and graduates of electrical engineering programs  from top colleges and universities. But Jim’s tendency to  stretch Blake Electronics to contracts continued as well, and  by 2007, Blake Electronics had a reputation with govern- ment agencies as a company that could not deliver what it  promised. Almost overnight, government contracts stopped,  and Blake Electronics was left with an idle workforce and un- used manufacturing equipment. This high overhead started to  melt away profits, and in 2009, Blake Electronics was faced  with the possibility of sustaining a loss for the first time in its  history.

In 2010, Steve decided to look at the possibility of man- ufacturing electronic components for home use. Although  this was a totally new market for Blake Electronics, Steve  was convinced  that  this was  the only way  to keep Blake 

Master Control Box

Outlet Adapter

Light Switch Adapter

Lightbulb Disk

B LA

K E

Figure 3.15 Master Control Center

Electronics from dipping into the red. The research team at  Blake  Electronics was given the task of developing new elec- tronic devices for home use. The first idea from the research  team was the  Master Control Center. The basic components for  this system are shown in Figure 3.15.

The heart of the system is the master control box. This unit,  which would have a retail price of $250, has two rows of five  buttons. Each button controls one light or appliance and can be  set as either a switch or a rheostat. When set as a switch, a light  finger touch on the button turns a light or appliance on or off.  When set as a rheostat, a finger touching the button controls the  intensity of the light. Leaving your finger on the button makes  the light go through a complete cycle ranging from off to bright  and back to off again.

To allow for maximum flexibility, each master control box  is powered by two D-sized batteries that can last up to a year,  depending on usage. In addition, the research team has devel- oped three versions of the master control box—versions A, B,  and C. If a family wants to control more than 10 lights or appli- ances, another master control box can be purchased.

M03_REND9327_12_SE_C03.indd 128 10/02/14 1:22 PM

Case stUDy  129

The lightbulb disk, which would have a retail price of  $2.50, is controlled by the master control box and is used to  control the intensity of any light. A different disk is available for  each button position for all three master control boxes. By in- serting the lightbulb disk between the lightbulb and the socket,  the appropriate button on the master control box can completely  control the intensity of the light. If a standard light switch  is used, it must be on at all times for the master control box  to work.

One disadvantage of using a standard light switch is that  only the master control box can be used to control the particular  light. To avoid this problem, the research team developed a spe- cial light switch adapter that would sell for $15. When this de- vice is installed, either the master control box or the light switch  adapter can be used to control the light.

When used to control appliances other than lights, the mas- ter control box must be used in conjunction with one or more  outlet adapters. The adapters are plugged into a standard wall  outlet, and the appliance is then plugged into the adapter. Each  outlet adapter has a switch on top that allows the appliance to be  controlled from the master control box or the outlet adapter. The  price of each outlet adapter would be $25.

The research team estimated that it would cost $500,000 to  develop the equipment and procedures needed to manufacture  the master control box and accessories. If successful, this ven- ture could increase sales by approximately $2 million. But will  the master control boxes be a successful venture? With a 60%  chance of success estimated by the research team, Steve had  serious doubts about trying to market the master control boxes  even though he liked the basic idea. Because of his reservations,  Steve decided to send requests for proposals (RFPs) for addi- tional marketing research to 30 marketing research companies  in southern California.

The first RFP to come back was from a small company  called Marketing Associates, Inc. (MAI), which would charge  $100,000 for the survey. According to its proposal, MAI has  been in business for about 3 years and has conducted about 100  marketing research projects. MAI’s major strengths appeared to  be individual attention to each account, experienced staff, and 

fast work. Steve was particularly interested in one part of the  proposal, which revealed MAI’s success record with previous  accounts. This is shown in Table 3.19.

The only other proposal to be returned was by a branch  office of Iverstine and Walker, one of the largest marketing  research firms in the country. The cost for a complete survey  would be $300,000. While the proposal did not contain the same  success record as MAI, the proposal from Iverstine and Walker  did contain some interesting information. The chance of get- ting a favorable survey result, given a successful venture, was  90%. On the other hand, the chance of getting an unfavorable  survey result, given an unsuccessful venture, was 80%. Thus,  it appeared to Steve that Iverstine and Walker would be able to  predict the success or failure of the master control boxes with a  great amount of certainty.

Steve pondered the situation. Unfortunately, both market- ing research teams gave different types of information in their  proposals. Steve concluded that there would be no way that the  two proposals could be compared unless he got additional infor- mation from Iverstine and Walker. Furthermore, Steve wasn’t  sure what he would do with the information, and if it would be  worth the expense of hiring one of the marketing research firms.

Discussion Questions   1. Does Steve need additional information from Iverstine 

and Walker?   2. What would you recommend?

See our Internet home page, at www.pearsonglobaleditions.com/render, for these additional  case studies:

(1)  Drink-at-Home, inc.: This case involves the development and marketing of a new beverage. (2)   ruth Jones’ Heart bypass operation: This case deals with a medical decision regarding 

surgery.

(3)  Ski right: This case involves the development and marketing of a new ski helmet. (4)  Study time: This case is about a student who must budget time while studying for a final exam.

Internet Case Studies

Table 3.19 Success Figures for MAI

Survey reSultS

outCoMe favorable uNfavorable total

Successful  venture

35 20 55

Unsuccessful  venture

15 30 45

M03_REND9327_12_SE_C03.indd 129 10/02/14 1:22 PM

130  Chapter 3 • DeCision analysis

Abbas, Ali E. “Invariant Utility Functions and Certain Equivalent Transforma- tions,” Decision Analysis 4, 1(March 2007): 17–31.

Carassus, Laurence, and Miklós Rásonyi. “Optimal Strategies and Utility- Based Prices Converge When Agents’ Preferences Do,” Mathematics of Operations Research 32, 1 (February 2007): 102–117.

Congdon, Peter. Bayesian Statistical Modeling. New York: John Wiley &  Sons, Inc., 2001.

Ewing, Paul L., Jr. “Use of Decision Analysis in the Army Base Realignment  and Closure (BRAC) 2005 Military Value Analysis,” Decision Analysis 3 (March 2006): 33–49.

Hammond, J. S., R. L. Kenney, and H. Raiffa. “The Hidden Traps in Decision  Making,” Harvard Business Review (September–October 1998): 47–60.

Hurley, William J. “The 2002 Ryder Cup: Was Strange’s Decision to Put  Tiger Woods in the Anchor Match a Good One?” Decision Analysis 4,  1 (March 2007): 41–45.

Kirkwood, C. W. “An Overview of Methods for Applied Decision Analysis,”  Interfaces 22, 6 (November–December 1992): 28–39.

Kirkwood, C. W. “Approximating Risk Aversion in Decision Analysis Appli- cations,” Decision Analysis 1 (March 2004): 51–67.

Luce, R., and H. Raiffa. Games and Decisions. New York: John Wiley & Sons,  Inc., 1957.

Maxwell, Daniel T. “Improving Hard Decisions,” OR/MS Today 33, 6  (December 2006): 51–61.

Patchak, William M. “Software Survey: Decision Analysis.” OR/MS Today 39,  5 (October 2012).

Paté-Cornell, M. Elisabeth, and Robin L. Dillon. “The Respective Roles of  Risk and Decision Analyses in Decision Support,” Decision Analysis 3 (December 2006): 220–232.

Pennings, Joost M. E., and Ale Smidts. “The Shape of Utility Functions and  Organizational Behavior,” Management Science 49, 9 (September 2003):  1251–1263.

Raiffa, Howard, John W. Pratt, and Robert Schlaifer. Introduction to Statistical Decision Theory. Boston: MIT Press, 1995.

Raiffa, Howard, and Robert Schlaifer. Applied Statistical Decision Theory.  New York: John Wiley & Sons, Inc., 2000.

Render, B., and R. M. Stair. Cases and Readings in Management Science,  2nd ed. Boston: Allyn & Bacon, Inc., 1988.

Schlaifer, R. Analysis of Decisions Under Uncertainty. New York: McGraw- Hill Book Company, 1969.

Smith, James E., and Robert L. Winkler. “The Optimizer’s Curse: Skepticism  and Postdecision Surprise in Decision Analysis,” Management Science 52 (March 2006): 311–322.

van Binsbergen, Jules H., and Leslie M. Marx. “Exploring Relations Between  Decision Analysis and Game Theory,” Decision Analysis 4, 1 (March  2007): 32–40.

Wallace, Stein W. “Decision Making Under Uncertainty: Is Sensitivity Analy- sis of Any Use?” Operations Research 48, 1 (2000): 20–25.

Bibliography

M03_REND9327_12_SE_C03.indd 130 10/02/14 1:22 PM

 131

6. Develop a multiple regression model and use it for prediction purposes.

7. Use dummy variables to model categorical data.

8. Determine which variables should be included in a multiple regression model.

9. Transform a nonlinear function into a linear one for use in regression.

10. Understand and avoid mistakes commonly made in the use of regression analysis.

1. Identify variables and use them in a regression model.

2. Develop simple linear regression equations from sample data and interpret the slope and intercept.

3. Compute the coefficient of determination and the coefficient of correlation and interpret their meanings.

4. Interpret the F test in a linear regression model.

5. List the assumptions used in regression and use residual plots to identify problems.

After completing this chapter, students will be able to:

Summary  •  Glossary  •  Key Equations  •  Solved Problems  •  Self-Test  •  Discussion Questions and Problems 

Case Study: North–South Airline  •  Bibliography

Appendix 4.1: Formulas for Regression Calculations

4.7 Using Computer Software for Regression

4.8 Multiple Regression Analysis

4.9 Binary or Dummy Variables

4.10 Model Building

4.11 Nonlinear Regression

4.12 Cautions and Pitfalls in Regression Analysis

4.1 Introduction

4.2 Scatter Diagrams

4.3 Simple Linear Regression

4.4 Measuring the Fit of the Regression Model

4.5 Assumptions of the Regression Model

4.6 Testing the Model for Significance

ChApter Outline

ChApter 4 Regression Models

leArning ObjeCtives

M04_REND9327_12_SE_C04.indd 131 10/02/14 1:22 PM

132  Chapter 4 • regression Models

4.1 Introduction

Regression analysis is a very valuable tool for today’s manager. Regression has been used to  model things such as the relationship between level of education and income, the price of a  house and the square footage, and the sales volume for a company relative to the dollars spent  on advertising. When businesses are trying to decide which location is best for a new store or  branch office, regression models are often used. Cost estimation models are often regression  models. The applicability of regression analysis is virtually limitless.

There are generally two purposes for regression analysis. The first is to understand the rela- tionship between variables such as advertising expenditures and sales. The second purpose is to  predict the value of one variable based on the value of the other. Because of this, regression is a  very important forecasting technique and will be mentioned again in Chapter 5.

In this chapter, the simple linear regression model will first be developed, and then a  more complex multiple regression model will be used to incorporate even more variables  into our model. In any regression model, the variable to be predicted is called the dependent variable or response variable. The value of this is said to be dependent upon the value of an  independent variable, which is sometimes called an explanatory variable or a predictor variable.

4.2 Scatter Diagrams

To investigate the relationship between variables, it is helpful to look at a graph of the data. Such  a graph is often called a scatter diagram or a scatter plot. Normally the independent variable  is plotted on the horizontal axis and the dependent variable is plotted on the vertical axis. The  following example will illustrate this.

Triple A Construction Company renovates old homes in Albany. Over time, the company  has found that its dollar volume of renovation work is dependent on the Albany area payroll. The  figures for Triple A’s revenues and the amount of money earned by wage earners in Albany for  the past 6 years are presented in Table 4.1. Economists have predicted the local area payroll to  be $600 million next year, and Triple A wants to plan accordingly.

Figure 4.1 provides a scatter diagram for the Triple A Construction data given in Table 4.1.  This graph indicates that higher values for the local payroll seem to result in higher sales for the  company. There is not a perfect relationship because not all the points lie in a straight line, but  there is a relationship. A line has been drawn through the data to help show the relationship that  exists between the payroll and sales. The points do not all lie on the line, so there would be some  error involved if we tried to predict sales based on payroll using this or any other line. Many  lines could be drawn through these points, but which one best represents the true relationship?  Regression analysis provides the answer to this question.

Two purposes of regression analysis are to understand the relationship between variables and to predict the value of one based on the other.

A scatter diagram is a graph of the data.

TRiple A’s sAles ($100,000s)

locAl pAyRoll

($100,000,000s)

6 3

8 4

9 6

5 4

4.5 2

9.5 5

Table 4.1 Triple A Construction Company Sales and Local Payroll

M04_REND9327_12_SE_C04.indd 132 10/02/14 1:22 PM

4.3 siMple linear regression  133

4.3 Simple Linear Regression

In any regression model, there is an implicit assumption (which can be tested) that a relation- ship exists between the variables. There is also some random error that cannot be predicted. The  underlying simple linear regression model is

Y = b0 + b1X + P (4-1)

where

Y = dependent variable (response variable) X = independent variable (predictor variable or explanatory variable)

b0 = intercept (value of Y when X = 0) b1 = slope of regression line P = random error

The true values for the intercept and slope are not known, and therefore they are estimated  using sample data. The regression equation based on sample data is given as

Yn = b0 + b1X (4-2)

where

Yn = predicted value of Y b0 = estimate of b0, based on sample results b1 = estimate of b1, based on sample results

In the Triple A Construction example, we are trying to predict the sales, so the dependent  variable (Y) would be sales. The variable we use to help predict sales is the Albany area payroll,  so this is the independent variable (X). Although any number of lines can be drawn through these  points to show a relationship between X and Y in Figure 4.1, the line that will be chosen is the  one that in some way minimizes the errors. error is defined as

Error = 1Actual value2 - 1Predicted value2 e = Y - Yn (4-3)

Since errors may be positive or negative, the average error could be zero even though there are  extremely large errors—both positive and negative. To eliminate the difficulty of negative er- rors canceling positive errors, the errors can be squared. The best regression line will be defined 

Figure 4.1 Scatter Diagram of Triple A Construction Company Data

0 1 2 0

2

4

6

8

10

12

3 4 5 6 7 8

Payroll ($100 million)

S al

es (

$1 00

,0 00

)

The dependent variable is Y and the independent variable is X.

Estimates of the slope and intercept are found from sample data.

The regression line minimizes the sum of the squared errors.

M04_REND9327_12_SE_C04.indd 133 10/02/14 1:22 PM

134  Chapter 4 • regression Models

as the one with the minimum sum of the squared errors. For this reason, regression analysis is  sometimes called least-squares regression.

Statisticians have developed formulas that we can use to find the equation of a straight line  that would minimize the sum of the squared errors. The simple linear regression equation is

Yn = b0 + b1X

The following formulas can be used to compute the intercept and the slope:

X = gX n

= average 1mean2 of X values

Y = gY n

= average 1mean2 of Y values

b1 = g1X - X21Y - Y2

g1X - X22 (4-4) b0 = Y - b1X (4-5)

The preliminary calculations are shown in Table 4.2. There are other “shortcut” formulas that are  helpful when doing the computations on a calculator, and these are presented in Appendix 4.1.  They will not be shown here, as computer software will be used for most of the other examples  in this chapter.

Computing the slope and intercept of the regression equation for the Triple A Construction  Company example, we have

X = gX 6

= 24

6 = 4

Y = gX 6

= 42

6 = 7

b1 = g1X - X21Y - Y2

g1X - X22 = 12.5

10 = 1.25

b0 = Y - b1X = 7 - 11.252142 = 2 The estimated regression equation therefore is

Yn = 2 + 1.25X

or

sales = 2 + 1.251payroll2 If the payroll next year is $600 million 1X = 62, then the predicted value would be

Yn = 2 + 1.25162 = 9.5 or $950,000.

Y X 1X − X 22 1X − X 2 1Y − Y 2 6 3 13 - 422 = 1 13 - 4216 - 72 = 1 8 4 14 - 422 = 0 14 - 4218 - 72 = 0 9 6 16 - 422 = 4 16 - 4219 - 72 = 4 5 4 14 - 422 = 0 14 - 4215 - 72 = 0 4.5 2 12 - 422 = 4 12 - 4214.5 - 72 = 5 9.5 5 15 - 422 = 1 15 - 4219.5 - 72 = 2.5

gY = 42 Y = 42>6 = 7

gX = 24 X = 24>6 = 4

g1X - X22 = 10 g1X - X21Y - Y2 = 12.5

Table 4.2 Regression Calculations for Triple A Construction

M04_REND9327_12_SE_C04.indd 134 10/02/14 1:22 PM

4.4 Measuring the Fit oF the regression Model  135

One of the purposes of regression is to understand the relationship among variables. This  model tells us that for each $100 million (represented by X) increase in the payroll, we would  expect the sales to increase by $125,000 since b1 = 1.25 ($100,000s). This model helps Triple A  Construction see how the local economy and company sales are related.

4.4 Measuring the Fit of the Regression Model

A regression equation can be developed for any variables X and Y, even random numbers. We  certainly would not have any confidence in the ability of one random number to predict the value  of another random number. How do we know that the model is actually helpful in predicting Y based on X? Should we have confidence in this model? Does the model provide better predic- tions (smaller errors) than simply using the average of the Y values?

In the Triple A Construction example, sales figures (Y) varied from a low of 4.5 to a high  of 9.5, and the mean was 7. If each sales value is compared with the mean, we see how far they  deviate from the mean and we could compute a measure of the total variability in sales. Because  Y is sometimes higher and sometimes lower than the mean, there may be both positive and nega- tive deviations. Simply summing these values would be misleading because the negatives would  cancel out the positives, making it appear that the numbers are closer to the mean than they ac- tually are. To prevent this problem, we will use the sum of squares total (ssT) to measure the  total variability in Y:

SST = g1Y - Y22 (4-6) If we did not use X to predict Y, we would simply use the mean of Y as the prediction, and 

the SST would measure the accuracy of our predictions. However, a regression line may be used  to predict the value of Y, and while there are still errors involved, the sum of these squared errors  will be less than the total sum of squares just computed. The sum of squares error (sse) is

SSE = ge2 = g1Y - Yn22 (4-7) Table 4.3 provides the calculations for the Triple A Construction example. The mean 

1Y = 72 is compared to each value and we get SST = 22.5

The prediction 1Yn2 for each observation is computed and compared to the actual value. This  results in

SSE = 6.875

The SSE is much lower than the SST. Using the regression line has reduced the variability  in the sum of squares by 22.5 - 6.875 = 15.625. This is called the sum of squares regression (ssR) and indicates how much of the total variability in Y is explained by the regression model.  Mathematically, this can be calculated as

SSR = g1Yn - Y22 (4-8)

Deviations (errors) may be positive or negative.

The SST measures the total variability in y about the mean.

The SSE measures the variability in Y about the regression line.

Table 4.3 Sum of Squares for Triple A Construction

Y X 1Y − Y 22 Yn 1Y − Yn 22 1Yn - Y 22

6 3 16 - 722 = 1 2 + 1.25132 = 5.75 0.0625 1.563 8 4 18 - 722 = 1 2 + 1.25142 = 7.00 1 0 9 6 19 - 722 = 4 2 + 1.25162 = 9.50 0.25 6.25 5 4 15 - 722 = 4 2 + 1.25142 = 7.00 4 0 4.5 2 14.5 - 722 = 6.25 2 + 1.25122 = 4.50 0 6.25 9.5 5 19.5 - 722 = 6.25 2 + 1.25152 = 8.25 1.5625 1.563

Y = 7 g1Y - Y22 = 22.5

SST = 22.5 g1Y - Yn22 = 6.875

SSE = 6.875 g1Yn - Y22 = 15.625

SSR = 15.625

M04_REND9327_12_SE_C04.indd 135 10/02/14 1:22 PM

136  Chapter 4 • regression Models

  Table 4.3 indicates

SSR = 15.625

There is a very important relationship between the sums of squares that we have computed:

1Sum of squares total2 = 1Sum of squares due to regression2 + 1Sum of squares error2 SST = SSR + SSE (4-9)

Figure 4.2 displays the data for Triple A Construction. The regression line is shown, as is a  line representing the mean of the Y values. The errors used in computing the sums of squares are  shown on this graph. Notice how the sample points are closer to the regression line than they are  to the mean.

Coefficient of Determination The SSR is sometimes called the explained variability in Y while the SSE is the unexplained  variability in Y. The proportion of the variability in Y that is explained by the regression equation  is called the coefficient of determination and is denoted by r2. Thus,

r2 = SSR

SST = 1 -

SSE

SST (4-10)

Thus, r2 can be found using either the SSR or the SSE. For Triple A Construction, we have

r2 = 15.625

22.5 = 0.6944

This means that about 69% of the variability in sales (Y) is explained by the regression equation  based on payroll (X).

If every point in the sample were on the regression line (meaning all errors are 0), then 100%  of the variability in Y could be explained by the regression equation, so r2 = 1 and SSE = 0. The lowest possible value of r2 is 0, indicating that X explains 0% of the variability in Y. Thus,  r2 can range from a low of 0 to a high of 1. In developing regression equations, a good model  will have an r2 value close to 1.

Correlation Coefficient Another measure related to the coefficient of determination is the coefficient of correlation.  This measure also expresses the degree or strength of the linear relationship. It is usually ex- pressed as r and can be any number between and including +1 and -1. Figure 4.3 illustrates 

Figure 4.2 Deviations from the Regression Line and from the Mean

0 1 2 0

2

4

6

8

10

12

3 4 5 6 7 8

Payroll ($100 million)

S al

es (

$1 00

,0 00

)

Y – Y ^ Y

Y – Y Y – Y

^

Y = 2 1 1.25 X ^

Y

X

r2 is the proportion of variability in Y that is explained by the regression equation.

If every point lies on the regression line, r2 = 1 and SSE = 0.

The correlation coefficient ranges from -1 to +1.

M04_REND9327_12_SE_C04.indd 136 10/02/14 1:22 PM

4.4 Measuring the Fit oF the regression Model  137

possible scatter diagrams for different values of r. The value of r is the square root of r2. It is  negative if the slope is negative, and it is positive if the slope is positive. Thus,

r = ±2r2 (4-11) For the Triple A Construction example with r2 = 0.6944,

r = 10.6944 = 0.8333 We know it is positive because the slope is +1.25.

(a) Perfect Positive Correlation: r 5 11

Y

X (b) Positive Correlation: 0 , r ,1

Y

X

(c) No Correlation: r 5 0

Y

X (d ) Perfect Negative Correlation: r 5 21

Y

X

Figure 4.3 Four Values of the Correlation Coefficient

using regression as part of improvement initiative at trane/ingersoll rand

Trane U.S. Inc. is a brand of Ingersoll Rand and is a leader in both home and commercial air conditioning. In an effort to improve quality and efficiency, the company consolidated its man- ufacturing of evaporator and condenser coils into one location in South Carolina. A key part of the coil unit is a header, which controls the flow of water or refrigerant through the coil. With about 120 unique headers produced on each shift, with increased demand for their air conditioners due to federal refrigerant stan- dards, and with growing pressure to reduce costs, Trane began an improvement initiative in 2010.

Value stream mapping, a lean manufacturing technique, was used to visualize the entire process and identify waste in the system. A new layout was suggested to improve processing and reduce the overall time of production. To predict how well the new layout would perform in the manufacture of the headers, data was collected about the time to move materials into posi- tion, time to set up a job, time to set up a machine, and time to process a job. Several regression models were developed to

determine which job characteristics were most significant in the various times involved in processing, and to measure the variabil- ity in times. These were used in simulation models that were de- veloped to determine the overall impact of the new layout and to determine where bottlenecks might occur. One of the bottle- necks, at a robotic machine for welding, was eliminated by re- tooling the robot to allow the machine operator to operate both sides of the robot. This enables the robot to process two headers simultaneously.

The improvement initiatives from this project are estimated to save over $700,000 per year. This project highlighted the im- portance of data-based analytics and lean manufacturing tools in identifying the improvement tactics for complex value streams. The company plans similar improvement initiatives for other value streams in the future.

source: Based on John B. Jensen, Sanjay L. Ahire, and Manoj K. Malhotra.  “Trane/Ingersoll Rand Combines Lean and Operations Research Tools to Rede- sign Feeder Manufacturing Operations,” Interfaces 43, 4 (July–August 2013):  325–340.

in ACtiOn

M04_REND9327_12_SE_C04.indd 137 10/02/14 1:22 PM

138  Chapter 4 • regression Models

4.5 Assumptions of the Regression Model

If we can make certain assumptions about the errors in a regression model, we can perform statisti- cal tests to determine if the model is useful. The following assumptions are made about the errors:

1. The errors are independent. 2. The errors are normally distributed. 3. The errors have a mean of zero. 4. The errors have a constant variance (regardless of the value of X).

It is possible to check the data to see if these assumptions are met. Often a plot of the residuals  will highlight any glaring violations of the assumptions. When the errors (residuals) are plotted  against the independent variable, the pattern should appear random.

Figure 4.4 presents some typical error patterns, with Figure 4.4A displaying a pattern that is  expected when the assumptions are met and the model is appropriate. The errors are random and  no discernible pattern is present. Figure 4.4B demonstrates an error pattern in which the errors  increase as X increases, violating the constant variance assumption. Figure 4.4C shows errors 

A plot of the errors may highlight problems with the model.

Figure 4.4a Pattern of Errors Indicating Randomness

E rr

or

X

E rr

or

X

Figure 4.4b Nonconstant Error Variance

E rr

or

X

Figure 4.4C Errors Indicate Relationship Is Not Linear

M04_REND9327_12_SE_C04.indd 138 10/02/14 1:22 PM

4.6 testing the Model For signiFiCanCe  139

consistently increasing at first, and then consistently decreasing. A pattern such as this would  indicate that the model is not linear and some other form (perhaps quadratic) should be used. In  general, patterns in the plot of the errors indicate problems with the assumptions or the model  specification.

Estimating the Variance While the errors are assumed to have constant variance 1s22, this is usually not known. It can be  estimated from the sample results. The estimate of s2 is the mean squared error (Mse) and is  denoted by s2. The MSE is the sum of squares due to error divided by the degrees of freedom:1

s2 = MSE = SSE

n - k - 1 (4-12)

where

n = number of observations in the sample k = number of independent variables

In the Triple A Construction example, n = 6 and k = 1. So

s2 = MSE = SSE

n - k - 1 =

6.8750

6 - 1 - 1 =

6.8750

4 = 1.7188

From this we can estimate the standard deviation as

s = 1MSE (4-13) This is called the standard error of the estimate or the standard deviation of the regres-

sion. In this example,

s = 1MSE = 11.7188 = 1.31 This is used in many of the statistical tests about the model. It is also used to find interval 

estimates for both Y and regression coefficients.2

4.6 Testing the Model for Significance

Both the MSE and r2 provide a measure of accuracy in a regression model. However, when the  sample size is too small, it is possible to get good values for both of these even if there is no re- lationship between the variables in the regression model. To determine whether these values are  meaningful, it is necessary to test the model for significance.

To see if there is a linear relationship between X and Y, a statistical hypothesis test is per- formed. The underlying linear model was given in Equation 4-1 as

Y = b0 + b1X + P

If b1 = 0, then Y does not depend on X in any way. The null hypothesis says there is no linear  relationship between the two variables (i.e., b1 = 0). The alternate hypothesis is that there is a  linear relationship (i.e., b1 ≠ 0). If the null hypothesis can be rejected, then we have proven  that a linear relationship does exist, so X is helpful in predicting Y. The F distribution is used for  testing this hypothesis. Appendix D contains values for the F distribution that can be used when  calculations are performed by hand. See Chapter 2 for a review of the F distribution. The results  of the test can also be obtained from both Excel and QM for Windows.

The F statistic used in the hypothesis test is based on the MSE (seen in the previous section)  and the mean squared regression (MSR). The MSR is calculated as

MSR = SSR

k (4-14)

1See bibliography at end of this chapter for books with further details. 

The error variance is estimated by the MSE.

2The MSE is a common measure of accuracy in forecasting. When used with techniques besides regression, it is com- mon to divide the SSE by n rather than n - k - 1.

An F test is used to determine if there is a relationship between X and Y.

M04_REND9327_12_SE_C04.indd 139 10/02/14 1:22 PM

140  Chapter 4 • regression Models

where

k = number of independent variables in the model

The F statistic is

F = MSR

MSE (4-15)

Based on the assumptions regarding the errors in a regression model, this calculated F statistic is  described by the F distribution with

degrees of freedom for the numerator = df1 = k degrees of freedom for the denominator = df2 = n - k - 1

where

k = the number of independent 1X2 variables If there is very little error, the denominator (MSE) of the F statistic is very small relative 

to the numerator (MSR), and the resulting F statistic would be large. This would be an indica- tion that the model is useful. A significance level related to the value of the F statistic is then  found. Whenever the F value is large, the observed significance level (p-value) will be low,  indicating that it is extremely unlikely that this could have occurred by chance. When the F value is large (with a resulting small significance level), we can reject the null hypothesis that  there is no linear relationship. This means that there is a linear relationship and the values of  MSE and r2 are meaningful.

The hypothesis test just described is summarized here:

Steps in Hypothesis Test for a Significant Regression Model

1. Specify null and alternative hypotheses:

H0:b1 = 0 H1:b1 ≠ 0

2. Select the level of significance 1a). Common values are 0.01 and 0.05. 3. Calculate the value of the test statistic using the formula

F = MSR

MSE

4. Make a decision using one of the following methods: (a)   Reject the null hypothesis if the test statistic is greater than the F value from the table in

Appendix D. Otherwise, do not reject the null hypothesis:

Reject if Fcalculated 7 Fa,df1,df2 df1 = k df2 = n - k - 1

(b)   Reject the null hypothesis if the observed significance level, or p-value, is less than the  level of significance (a). Otherwise, do not reject the null hypothesis:

p@value = P1F 7 calculated test statistic2 Reject if p@value 6 a

If the significance level for the F test is low, there is a relationship between X and Y.

M04_REND9327_12_SE_C04.indd 140 10/02/14 1:23 PM

4.6 testing the Model For signiFiCanCe  141

Triple A Construction Example To illustrate the process of testing the hypothesis about a significant relationship, consider the  Triple A Construction example. Appendix D will be used to provide values for the F distribution.

Step 1.

H0 : b1 = 0 1no linear relationship between X and Y2 H1 : b1 ≠ 0 1linear relationship exists between X and Y2

Step 2.

Select a = 0.05

Step 3. Calculate the value of the test statistic. The MSE was already calculated to be 1.7188.  The MSR is then calculated so that F can be found:

MSR = SSR

k =

15.6250

1 = 15.6250

F = MSR

MSE =

15.6250

1.7188 = 9.09

Step 4. (a)  Reject the null hypothesis if the test statistic is greater than the F value from the table  in Appendix D:

df1 = k = 1 df2 = n - k - 1 = 6 - 1 - 1 = 4

The value of F associated with a 5% level of significance and with degrees of freedom 1 and 4 is  found in Appendix D. Figure 4.5 illustrates this:

F0.05,1,4 = 7.71 Fcalculated = 9.09 Reject H0 because 9.09 7 7.71

Thus, there is sufficient data to conclude that there is a statistically significant relationship  between X and Y, so the model is helpful. The strength of this relationship is measured by  r2 = 0.69. Thus, we can conclude that about 69% of the variability in sales (Y) is explained by  the regression model based on local payroll (X).

The Analysis of Variance (ANOVA) Table When software such as Excel or QM for Windows is used to develop regression models, the out- put provides the observed significance level, or p-value, for the calculated F value. This is then  compared to the level of significance 1a) to make the decision.

F 5 7.71 9.09

0.05

Figure 4.5 F Distribution for Triple A Construction Test for Significance

M04_REND9327_12_SE_C04.indd 141 10/02/14 1:23 PM

142  Chapter 4 • regression Models

Table 4.4 provides summary information about the ANOVA table. This shows how the num- bers in the last three columns of the table are computed. The last column of this table, labeled  Significance F, is the p-value, or observed significance level, which can be used in the hypoth- esis test about the regression model.

Triple A Construction ANOVA Example The Excel output that includes the ANOVA table for the Triple A Construction data is shown  in the next section. The observed significance level for F = 9.0909 is given to be 0.0394. This  means

P1F 7 9.09092 = 0.0394 Because this probability is less than 0.05 (a), we would reject the hypothesis of no linear rela- tionship and conclude that there is a linear relationship between X and Y. Note in Figure 4.5 that  the area under the curve to the right of 9.09 is clearly less than 0.05, which is the area to the right  of the F value associated with a 0.05, level of significance.

4.7 Using Computer Software for Regression

Software such as Excel 2013, Excel QM, and QM for Windows is usually used for the regression  calculations. All of these will be illustrated using the Triple A Construction example.

Excel 2013 To use Excel 2013 for regression, enter the data in columns. Click the Data tab to access the data  ribbon. Then select Data Analysis, as shown in Program 4.1A. If Data Analysis does not appear  on the Data ribbon, the Excel add-in for this has not yet been activated. See Appendix F at the  end of the book for instructions on how to activate this. Once this add-in is activated, it will re- main on the Data ribbon for future use.

When the Data Analysis window opens, scroll down to and highlight Regression and click  OK, as illustrated in Program 4.1A. The Regression window will open, as shown in Program 4.1B,  and you can input the X and Y ranges. Check the Labels box because the cells with the variable  name were included in the first row of the X and Y ranges. To have the output presented on this  page rather than on a new worksheet, select Output Range and give a cell address for the start of  the output. Click the OK button, and the output appears in the output range specified. Program 4.1C  shows the intercept (2), slope (1.25), and other information that was previously calculated for  the Triple A Construction example.

The sums of squares are shown in the column headed by SS. Another name for error is  residual. In Excel, the sum of squares error is shown as the sum of squares residual. The val- ues in this output are the same values shown in Table 4.3:

Sum of squares regression = SSR = 15.625 Sum of squares error (residual) = 6.8750 Sum of squares total = SST = 22.5

The coefficient of determination 1r2) is shown to be 0.6944. The coefficient of correlation  (r) is called Multiple R in the Excel output, and this is 0.8333.

DF ss Ms F siGNiFicANce F

Regression k SSR MSR = SSR>k MSR/MSE P1F 7 MSR>MSE2 Residual n - k - 1 SSE MSE = SSE>1n - k - 12 Total n - 1 SST

Table 4.4 Analysis of Variance Table for Regression

M04_REND9327_12_SE_C04.indd 142 10/02/14 1:23 PM

4.7 using CoMputer soFtware For regression  143

Program 4.1a Accessing the Regression Option in Excel 2013

Go to the Data tab.Click OK.

Select Data Analysis.

When Data Analysis window opens, scroll down to Regression.

Excel QM To use Excel QM for regression, from the Excel QM ribbon select the module by click- ing Alphabetical - Forecasting - Multiple Regression as shown in Program 4.2A. A win- dow will open  to allow you to specify  the size of  the problem, as shown in Program  4.2B.  Enter a name or title for the problem, the number of past observations, and the  number of  independent (X) variables. For the Triple A Construction example, there are  6 past  periods of data and 1 independent variable. When you click OK, a spreadsheet will   appear for you to enter the X and Y data in the shaded area. The calculations are automati- cally performed, so there is nothing to do after entering the data. The results are seen in   Program 4.2C.

Specify the X and Y ranges.

Click OK to have Excel develop the regression model.

Check the Labels box if the �rst row in the X and Y ranges includes the variable names.

Specify location for the output. To put this on the current worksheet, click Output Range and give a cell location for this to begin.

Program 4.1b Data Input for Regression in Excel 2013

The regression coef�cients are given here.

A high r2 (close to 1) is desirable.

A low (e.g., less than 0.05) Signi�cance F (p-value for overall model) indicates a signi�cant relationship between X and Y.

The SSR (regression), SSE (residual or error), and SST (total) are shown in the SS column of the ANOVA table.

Program 4.1C Excel 2013 Output for Triple A Construction Example

M04_REND9327_12_SE_C04.indd 143 10/02/14 1:23 PM

144  Chapter 4 • regression Models

Program 4.2a Using Excel QM for Regression

Go to the Excel QM tab in Excel 2013.

When options appear, click Multiple Regression.

On the Menu group, left-click Alphabetical.

Point the cursor at Forecasting.

Program 4.2C Input and Results for Regression in Excel QM

Enter the past observations of Y and X. Results appear automatically.

To forecast Y based on any value of X, simply input the value of X here.

Correlation coef�cient is given here.

The intercept and slope are shown here.

Input a title.

Click OK.

Input the number of past observations.

Input the number of independent (X) variables.

Program 4.2b Initializing the Spreadsheet in Excel QM

M04_REND9327_12_SE_C04.indd 144 10/02/14 1:23 PM

4.7 using CoMputer soFtware For regression  145

QM for Windows To develop a regression equation using QM for Windows for the Triple A Construction  Company data, select the Forecasting module and select New. Then choose Least Squares— Simple and Multiple Regression, as illustrated in Program 4.3A. This opens the window  shown in Program 4.3B.  Enter the number of observations (6 in this example) and spec- ify the number of independent variables (1 in this example). Left-click OK and a window  opens allowing you to input the data as shown in Program 4.3C. After entering the data,  click Solve, and the forecasting results shown in Program 4.3D are displayed. The equa- tion as well as other information is provided on this screen. Note the regression equation is  displayed across two lines. Additional output, including the ANOVA table, is available by  clicking the Window option on the toolbar.

After selecting the Forecasting module, click New and select Least Squares – Simple and Multiple Regression.

Program 4.3a QM for Windows Regression Option in Forecasting Module

Specify the number of observations.

Specify the number of variables.

Click OK.

Program 4.3b QM for Windows Screen to Initialize the Problem

Program 4.3C Data Input for Triple A Construction Example

M04_REND9327_12_SE_C04.indd 145 10/02/14 1:23 PM

146  Chapter 4 • regression Models

4.8 Multiple Regression Analysis

The multiple regression model is a practical extension of the model we just observed. It allows  us to build a model with several independent variables. The underlying model is

Y = b0 + b1X1 + b2X2 + Á + bkXk + P (4-16)

where

Y = dependent variable (response variable) Xi = ith independent variable (predictor variable or explanatory variable) b0 = intercept (value of Y when all Xi = 0) bi = coefficient of the ith independent variable k = number of independent variables P= random error

To estimate the values of these coefficients, a sample is taken and the following equation is  developed:

Yn = b0 + b1X1 + b2X2 + Á + bkXk (4-17)

where

Yn = predicted value of Y b0 = sample intercept (and is an estimate of b0) bi = sample coefficient of ith variable (and is an estimate of bi)

Consider  the case of Jenny Wilson Realty, a  real estate company  in Montgomery,   Alabama. Jenny Wilson, owner and broker for this company, wants to develop a model to  determine a suggested listing price for houses based on the size of the house and the age of  the house. She selects a sample of houses that have sold recently in a particular area, and she  records the selling price, the square footage of the house, the age of the house, and also the 

A multiple regression model has more than one independent variable.

Additional information is available by clicking Window.

Regression equation is across two lines.

Program 4.3D QM for Windows Output for Triple A Construction Example

M04_REND9327_12_SE_C04.indd 146 10/02/14 1:23 PM

4.8 Multiple regression analysis  147

Table 4.5 Jenny Wilson Real Estate Data

selliNG pRice ($)

squARe FooTAGe AGe coNDiTioN

95,000 1,926 30 Good

119,000 2,069 40 Excellent

124,800 1,720 30 Excellent

135,000 1,396 15 Good

142,800 1,706 32 Mint

145,000 1,847 38 Mint

159,000 1,950 27 Mint

165,000 2,323 30 Excellent

182,000 2,285 26 Mint

183,000 3,752 35 Good

200,000 2,300 18 Good

211,000 2,525 17 Good

215,000 3,800 40 Excellent

219,000 1,740 12 Mint

condition (good, excellent, or mint) of each house as shown in Table 4.5. Initially Jenny plans  to use only the square  footage and age to develop a model, although she wants to save the  information on condition of the house to use later. She wants to find the coefficients for the  following multiple regression model:

Yn = b0 + b1X1 + b2X2

where

Yn = predicted value of dependent variable (selling price) b0 = Y  intercept

X1 and X2 = value of the two independent variables (square footage and age),  respectively

b1 and  b2 = slopes for X1 and X2, respectively

The mathematics of multiple regression becomes quite complex, so we leave formulas for  b0, b1, and b2 to regression textbooks.

3 Excel can be used to develop a multiple regression model  just as it was used for a simple linear regression model. When entering the data in Excel, it  is  important that all of the independent variables are in adjoining columns to facilitate the in- put. From the Data tab in Excel, select Data Analysis and then Regression, as shown earlier, in   Program 4.1A. This opens the regression window to allow the input, as shown in Program 4.4A.  Note that the X Range includes the data in two columns (B and C) because there are two inde- pendent variables. The Excel output that Jenny Wilson obtains is shown in Program 4.4B, and it  provides the following equation:

Yn = b0 + b1X1 + b2X2 = 146,630.89 + 43.82 X1 - 2898.69 X2

Evaluating the Multiple Regression Model A multiple regression model can be evaluated in a manner similar to the way a simple  linear regression model is evaluated. Both the p-value for the F test and r2 can be inter- preted the same with multiple regression models as they are with simple linear regression 

Excel can be used to develop multiple regression models.

3See, for example, Norman R. Draper and Harry Smith. Applied Regression Analysis, 3rd ed. New York: John Wiley &  Sons, Inc., 1998.

M04_REND9327_12_SE_C04.indd 147 10/02/14 1:23 PM

148  Chapter 4 • regression Models

models. However, as there is more than one independent variable, the hypothesis that is  being tested with the F test is that all the coefficients are equal to 0. If all these are 0,  then none of the independent variables in the model is helpful in predicting the dependent  variable.

To determine which of the independent variables in a multiple regression model is  significant, a significance test on the coefficient for each variable is performed. While  statistics textbooks can provide the details of these tests, the results of these tests are   automatically displayed in the Excel output. The null hypothesis is that the coefficient is  0 1H0:bi = 02 and the alternate hypothesis is that it is not 0 1H1:bi ≠ 02. The test statistic  is calculated in Excel, and the p-value is given. If the p-value is lower than the level of sig- nificance 1a), then the null hypothesis is  rejected and it can be concluded that the variable  is significant.

Jenny Wilson Realty Example In the Jenny Wilson Realty example in Program 4.4B, the overall model is statistically  significant and useful in predicting the selling price of the house because the p-value for  the F test is 0.002. The r2 value is 0.6719, so 67% of the variability in selling price for  these houses can be explained by the regression model. However, there were two inde- pendent variables in the model—square footage and age. It is possible that one of these is  significant and the other is not. The F test simply indicates that the model as a whole is  significant.

Program 4.4a Input Screen for Jenny Wilson Realty Multiple Regression in Excel 2013

The columns containing data for the X variables should be adjoining.

The X range is from the start of the �rst column to the bottom of the second column.

Program 4.4b Excel 2013 Output Screen for Jenny Wilson Realty Multiple Regression Example The coef�cients are here.

The coef�cient of determination is 0.67.

A low Signi�cance F value indicates that the model is statistically signi�cant.

M04_REND9327_12_SE_C04.indd 148 10/02/14 1:23 PM

4.9 Binary or duMMy VariaBles  149

Two significance tests can be performed to determine if square footage or age (or both) are  significant. In Program 4.4B, the results of two hypothesis tests are provided. The first test for  variable X1 (square footage) is

H0:b1 = 0 H1:b1 ≠ 0

Using a 5% level of significance 1a = 0.052, the null hypothesis is rejected because the p-value  for this is 0.0013. Thus, square footage is helpful in predicting the price of a house.

Similarly, the variable X2 (age) is tested using the Excel output, and the p-value is 0.0039.  The null hypothesis is rejected because this is less than 0.05. Thus, age is also helpful in predict- ing the price of a house.

4.9 Binary or Dummy Variables

All of the variables we have used in regression examples have been quantitative variables  such as sales figures, payroll numbers, square footage, and age. These have all been eas- ily  measurable and have had numbers associated with them. There are many times when  we believe a qualitative variable rather than a quantitative variable would be helpful in  predicting the dependent variable Y. For example, regression may be used to find a re- lationship between annual income and certain characteristics of the employees. Years of  experience at a particular job would be a quantitative variable. However, information re- garding whether or not a person has a college degree might also be important. This would  not be a measurable value or quantity, so a special variable called a dummy variable (or a  binary variable or an indicator variable) would be used. A dummy variable is assigned a  value of 1 if a particular condition is met (e.g., a person has a college degree), and a value  of 0 otherwise.

Return to the Jenny Wilson Realty example. Jenny believes that a better model can  be  developed if the condition of the property is included. To incorporate the condition of  the house into the model, Jenny looks at the information available (see Table 4.5), and  sees that the three categories are good condition, excellent condition, and mint condi- tion. Since these are not quantitative variables, she must use dummy variables. These are   defined as

X3 = 1 if house is in excellent condition = 0 otherwise

X4 = 1 if house is in mint condition = 0 otherwise

Notice there is no separate variable for “good” condition. If X3 and X4 are both 0, then the house  cannot be in excellent or mint condition, so it must be in good condition. When using dummy  variables, the number of variables must be 1 less than the number of categories. In this problem,  there were three categories (good, excellent, and mint condition) so we must have two dummy  variables. If we had mistakenly used too many variables and the number of dummy variables  equaled the number of categories, then the mathematical computations could not be performed  or would not give reliable values.

These dummy variables will be used with the two previous variables (X1 - square footage,  and X2 - age) to try to predict the selling prices of houses for Jenny Wilson. Programs 4.5A  and 4.5B provide the Excel input and output for this new data, and this shows how the dummy   variables were coded. The significance level for the F test is 0.00017, so this model is statisti- cally significant. The coefficient of determination 1r22 is 0.898, so this is a much better model  than the previous one. The regression equation is

Yn = 121,658 + 56.43X1 - 3,962X2 + 33,162X3 + 47,369X4

The number of dummy variables must equal 1 less than the number of categories of a qualitative variable.

A dummy variable is also called an indicator variable or a binary variable.

M04_REND9327_12_SE_C04.indd 149 10/02/14 1:23 PM

150  Chapter 4 • regression Models

This indicates that a house in excellent condition 1X3 = 1, X4 = 02 would sell for about  $33,162 more than a house in good condition 1X3 = 0, X4 = 02. A house in mint condition  1X3 = 0, X4 = 12 would sell for about $47,369 more than a house in good condition.

4.10 Model Building

In developing a good regression model, possible independent variables are identified and the  best ones are selected to be used in the model. The best model is a statistically significant model  with a high r2 and few variables.

As more variables are added  to a  regression model, r2 will usually  increase, and  it  cannot decrease. It is tempting to keep adding variables to a model to try to increase  r2. However, if too many independent variables are included in the model, problems can  arise. For this reason, the adjusted r2 value is often used (rather than r2) to determine if an   additional independent variable is beneficial. The adjusted r2 takes into account the number  of independent variables in the model, and it is possible for the adjusted r2 to decrease. The  formula for r2 is

r2 = SSR

SST = 1 -

SSE

SST

The adjusted r2 is

Adjusted r2 = 1 - SSE>1n - k - 12

SST>1n - 12 (4-18)

Program 4.5a Input Screen for Jenny Wilson Realty Example with Dummy Variables in Excel 2013

The X range includes columns B, C, D, and E but not F.

Program 4.5b Output Screen for Jenny Wilson Realty Example with Dummy Variables in Excel 2013

The overall model is helpful since the Signi�cance F probability is low (much less than 5%).

Each of the variables individually is helpful since the p-values for each of these is low (much less than 5%).

The coef�cient of age is negative indicating that the price decreases as the house gets older.

The value of r2 can never decrease when more variables are added to the model.

The adjusted r2 may decrease when more variables are added to the model.

M04_REND9327_12_SE_C04.indd 150 10/02/14 1:23 PM

4.11 nonlinear regression  151

Notice that as the number of variables (k) increases, n - k - 1 will decrease. This causes  SSE>1n - k - 12 to increase, and consequently the adjusted r2 will decrease unless the extra  variable in the model causes a significant decrease in the SSE. Thus, the reduction in error (and  SSE) must be sufficient to offset the change in k.

As a general rule of thumb, if the adjusted r2 increases when a new variable is added to the  model, the variable should probably remain in the model. If the adjusted r2 decreases when a  new variable is added, the variable should not remain in the model. Other factors should also  be considered when trying to build the model, but they are beyond the introductory level of this  chapter.

Stepwise Regression While the process of model building may be tedious, there are many statistical software pack- ages that include stepwise regression procedures to do this. stepwise regression is an auto- mated process to systematically add or delete independent variables from a regression model. A  forward stepwise procedure puts the most significant variable in the model first and then adds  the next variable that will improve the model the most, given that the first variable is already in  the model. Variables continue to be added in this fashion until all the variables are in the model  or until any remaining variables do not significantly improve the model. A backward stepwise procedure begins with all independent variables in the model, and one by one the least helpful  variables are deleted. This continues until only significant variables remain. Many variations of  these stepwise models exist.

Multicollinearity In the Jenny Wilson Realty example illustrated in Program 4.5B, we saw an r2 of about  0.90 and an adjusted r2 of 0.85. While other variables such as the size of the lot,  the   number of bedrooms, and the number of bathrooms might be related to the selling price  of a house, we may not want to include these in the model. It is likely that these vari- ables would be correlated with the square footage of the house (e.g., more bedrooms  usually means a larger house), which is  already  included in the model. Thus, the informa- tion provided by these additional variables might be duplication of information already in  the model.

When an independent variable is correlated with one other independent variable, the vari- ables are said to be collinear. If an independent variable is correlated with a combination of  other independent variables, the condition of multicollinearity exists. This can create problems  in interpreting the coefficients of the variables as several variables are providing duplicate infor- mation. For example, if two independent variables were monthly salary expenses for a company  and annual salary expenses for a company, the information provided in one is also provided in  the other. Several sets of regression coefficients for these two variables would yield exactly the  same results. Thus, individual interpretation of these variables would be questionable, although  the model itself is still good for prediction purposes. When multicollinearity exists, the overall F test is still valid, but the hypothesis tests related to the individual coefficients are not. A variable  may appear to be significant when it is insignificant, or a variable may appear to be insignificant  when it is significant.

4.11 Nonlinear Regression

The regression models we have seen are linear models. However, at times there exist nonlinear  relationships between variables. Some simple variable transformations can be used to create an  apparently linear model from a nonlinear relationship. This allows us to use Excel and other  linear regression programs to perform the calculations. We will demonstrate this in the following  example.

On every new automobile sold in the United States, the fuel efficiency (as measured  by miles per gallon [MPG] of gasoline) of the automobile is prominently displayed on the 

A variable should not be added to the model if it causes the adjusted r2 to decrease.

Multicollinearity exists when a variable is correlated to other variables.

Transformations may be used to turn a nonlinear model into a linear model.

M04_REND9327_12_SE_C04.indd 151 10/02/14 1:23 PM

152  Chapter 4 • regression Models

window sticker. The MPG is related to several factors, one of which is the weight of the au- tomobile.  Engineers at Colonel Motors, in an attempt to improve fuel efficiency, have been  asked to study the impact of weight on MPG. They have decided that a regression model  should be used to do this.

A sample of 12 new automobiles was selected, and the weight and MPG rating were  recorded.  Table 4.6 provides this data. A scatter diagram of this data in Figure 4.6A shows the weight and  MPG. A linear regression line is drawn through the points. Excel was used to  develop a simple  linear regression equation to relate the MPG (Y) to the weight in 1,000 lb. 1X12 in the form

Yn = b0 + b1X1

The Excel output is shown in Program 4.6. From this we get the equation

Yn = 47.6 - 8.2X1

or

MPG = 47.6 - 8.21weight in 1,000 lb.2 The model is useful since the significance level for the F test is small and r2 = 0.7446.

However, further examination of the graph in Figure 4.6A brings into question the use of a lin- ear model. Perhaps a nonlinear relationship exists, and maybe the model should be modified 

Table 4.6 Automobile Weight Versus MPG

MpG WeiGhT (1,000 lB.) MpG

WeiGhT (1,000 lB.)

12 4.58 20 3.18

13 4.66 23 2.68

15 4.02 24 2.65

18 2.53 33 1.70

19 3.09 36 1.95

19 3.11 42 1.92

Figure 4.6a Linear Model for MPG Data

1.00 2.00 3.00 4.00 5.00 0

5

10

25

15

20

30

35

40

45

Weight (1,000 lb.)

M P

G

M04_REND9327_12_SE_C04.indd 152 10/02/14 1:23 PM

4.11 nonlinear regression  153

Figure 4.6b Nonlinear Model for MPG Data

1.00 2.00 3.00 4.00 5.00 0

5

10

25

15

20

30

35

40

45

Weight (1,000 lb.)

M P

G

to account for this. A quadratic model is illustrated in Figure 4.6B. This model would be of  the form

MPG = b0 + b11weight2 + b21weight22

The easiest way to develop this model is to define a new variable

X2 = 1weight22

This gives us the model

Yn = b0 + b1X1 + b2X2

We can create another column in Excel, and again run the regression tool. The output is  shown in Program 4.7. The new equation is

Yn = 79.8 - 30.2X1 + 3.4X2

The significance  level for F  is  low (0.0002) so  the model  is useful, and r2 = 0.8478. The  adjusted r2 increased from 0.719 to 0.814, so this new variable definitely improved  the model.

Program 4.6 Excel 2013 Output with Linear Regression Model for MPG Data

A low significance value for F and a high r2 are indications of a good model.

M04_REND9327_12_SE_C04.indd 153 10/02/14 1:23 PM

154  Chapter 4 • regression Models

This model is good for prediction purposes. However, we should not try to interpret  the coefficients of the variables due to the correlation between X1 (weight) and X2 (weight  squared). Normally we would interpret the coefficient for X1 as the change in Y that results  from a 1-unit change in X1, while holding all other variables constant. Obviously holding  one variable constant while changing the other is impossible in this example since X2 = X21. If X1 changes, then X2 must change also. This is an example of a problem that exists when  multicollinearity is present.

Other types of nonlinearities can be handled using a similar approach. A number of  transformations exist that may help to develop a linear model from variables with nonlinear  relationships.

4.12 Cautions and Pitfalls in Regression Analysis

This chapter has provided a brief introduction into regression analysis, one of the most widely  used quantitative  techniques  in business. However, some common errors are made with   regression models, so caution should be observed when using this.

If the assumptions are not met, the statistical tests may not be valid. Any interval estimates  are also invalid, although the model can still be used for prediction purposes.

Correlation does not necessarily mean causation. Two variables (such as the price of auto- mobiles and your annual salary) may be highly correlated to one another, but one is not causing  the other to change. They may both be changing due to other factors such as the economy in  general or the inflation rate.

If multicollinearity is present in a multiple regression model, the model is still good for pre- diction, but interpretation of individual coefficients is questionable. The individual tests on the  regression coefficients are not valid.

Using a regression equation beyond the range of X is very questionable. A linear relation- ship may exist within the range of values of X in the sample. What happens beyond this range  is unknown; the linear relationship may become nonlinear at some point. For example, there  is usually a linear relationship between advertising and sales within a limited range. As more  money is spent on advertising, sales tend to increase even if everything else is held constant.  However, at some point, increasing advertising expenditures will have less impact on sales   unless the company does other things to help, such as opening new markets or expanding the  product offerings. If advertising is increased and nothing else changes, the sales will probably  level off at some point.

Related to the limitation regarding the range of X is the interpretation of the intercept 1b02. Since the lowest value for X in a sample is often much greater than 0, the intercept is a point on  the regression line beyond the range of X. Therefore, we should not be concerned if the t-test for  this coefficient is not significant as we should not be using the regression equation to predict a 

Program 4.7 Excel 2013 Output with Nonlinear Regression Model for MPG Data

A high correlation does not mean one variable is causing a change in the other.

The regression equation should not be used with values of X that are below the lowest value of X or above the highest value of X found in the sample.

M04_REND9327_12_SE_C04.indd 154 10/02/14 1:23 PM

glossary  155

value of Y when X = 0. This intercept is merely used in defining the line that fits the sample  points the best.

Using the F test and concluding a linear regression model is helpful in predicting Y does not mean that this is the best relationship. While this model may explain much of  the variability in Y, it is possible that a nonlinear relationship might explain even more.  Similarly, if it is concluded that no linear relationship exists, another type of relationship  could exist.

A statistically significant relationship does not mean it has any practical value. With large  enough samples, it is possible to have a statistically significant relationship, but r2 might be  0.01. This would normally be of little use to a manager. Similarly, a high r2 could be found due  to random chance if the sample is small. The F test must also show significance to place any  value in r2.

A significant F value may occur even when the relationship is not strong.

Regression  analysis  is  an  extremely valuable quantita- tive tool. Using scatter diagrams helps to see relationships  between variables. The F  test is used to determine if the  results can be considered useful. The coefficient of determi- nation 1r22 is used to measure the proportion of  variability  in Y that is explained by the regression model. The corre- lation coefficient measures the relationship between two  variables.

Multiple regression involves the use of more than one   independent variable. Dummy variables (binary or indicator  variables) are used with qualitative or categorical data. Nonlin- ear models can be transformed into linear models.

We saw how to use Excel to develop regression models.  Interpretation of computer output was presented, and several  examples were provided.

Summary

Adjusted r2  A measure of the explanatory power of a   regression model that takes into consideration the number  of independent variables in the model.

Binary Variable  See Dummy Variable. Coefficient of Correlation 1r 2   A measure of the strength of 

the relationship between two variables. Coefficient of Determination 1r2 2   The percent of the 

 variability in the dependent variable 1Y2 that is explained  by the regression equation.

Collinearity  A condition that exists when one   independent variable is correlated with another   independent variable.

Dependent Variable  The Y variable in a regression model.  This is what is being predicted.

Dummy Variable  A variable used to represent a  qualitative  factor or condition. Dummy variables have values of 0  or 1. This is also called a binary variable or an indicator  variable.

Error  The difference between the actual value (Y) and the  predicted value 1Yn2.

Explanatory Variable  The independent variable in a   regression equation.

Independent Variable  The X variable in a regression  equation. This is used to help predict the dependent  variable.

Least Squares  A reference to the criterion used to select  the regression line, to minimize the squared distances  between the estimated straight line and the observed  values.

Mean Squared Error (MSE)  An estimate of the error  variance.

Multicollinearity  A condition that exists when one   independent variable is correlated with other independent  variables.

Multiple Regression Model  A regression model that has  more than one independent variable.

Observed Significance Level  Another name for p-value.  p-Value  A probability value that is used when testing a 

hypothesis. The hypothesis is rejected when this is low. Predictor Variable  Another name for explanatory 

variable. Regression Analysis  A forecasting procedure that uses 

the least-squares approach on one or more independent   variables to develop a forecasting model.

Residual  Another term for error.

Glossary

M04_REND9327_12_SE_C04.indd 155 10/02/14 1:23 PM

156  Chapter 4 • regression Models

Response Variable  The dependent variable in a regression  equation.

Scatter Diagrams or Scatter Plot  Diagrams of the variable  to be forecasted, plotted against another variable, such as  time. Also called scatter plots.

Standard Error of the Estimate  An estimate of the standard  deviation of the errors and is sometimes called the standard  deviation of the regression.

Stepwise Regression  An automated process to systemati- cally add or delete independent variables from a regression  model.

Sum of Squares Error (SSE)  The total sum of the squared  differences between each observation (Y) and the predicted  value 1Yn2.

Sum of Squares Regression (SSR)  The total sum of the  squared differences between each predicted value 1Yn2 and  the mean 1Y2.

Sum of Squares Total (SST)  The total sum of the  squared differences between each observation (Y) and  the mean 1Y2.

Key Equations

 (4-1)  Y = b0 + b1X + P

Underlying linear model for simple linear regression.

 (4-2)  Yn = b0 + b1X

Simple  linear  regression model  computed  from  a  sample.

 (4-3)  e = Y - Yn

Error in regression model.

(4-4)  b1 = g1X - X21Y - Y2

g1X - X22 Slope in the regression line.

 (4-5)  b0 = Y - b1X

The intercept in the regression line.

 (4-6)  SST = g1Y - Y22 Total sums of squares.

 (4-7)  SSE = ge2 = g1Y - Yn22 Sum of squares due to error.

 (4-8)  SSR = g1Yn - Y22 Sum of squares due to regression.

 (4-9)   SST = SSR + SSE

 Relationship among sums of squares in regression.

 (4-10)   r2 = SSR

SST = 1 -

SSE

SST

 Coefficient of determination.

 (4-11)   r = {2r2

   Coefficient of correlation. This has the same sign as the  slope.

 (4-12)   s2 = MSE = SSE

n - k - 1

   An estimate of the variance of the errors in regression;  n is the sample size and k is the number of independent  variables.

 (4-13)   s = 1MSE    An estimate of the standard deviation of the errors.  Also called the standard error of the estimate.

 (4-14)  MSR = SSR

k

   Mean square regression. k is the number of indepen- dent variables.

 (4-15)  F = MSR

MSE

F statistic used to test significance of overall regression  model.

 (4-16)  Y = b0 + b1X1 + b2X2 + Á + bkXk + P

  Underlying model for multiple regression model.

( 4-17)  Yn = b0 + b1X1 + b2X2 + Á + bkXk  Multiple regression model computed from a sample.

 (4-18)   Adjusted r2 = 1 - SSE>1n - k - 12

SST>1n - 12   Adjusted  r2  used  in  building multiple  regression  models.

M04_REND9327_12_SE_C04.indd 156 10/02/14 1:23 PM

solVed proBleMs  157

Solved Problem 4-1 Judith Thompson runs a florist shop on the Gulf Coast of Texas, specializing in floral arrangements for  weddings and other special events. She advertises weekly in the local newspapers and is  considering   increasing her advertising budget. Before doing so, she decides to evaluate the past effectiveness of  these ads. Five weeks are sampled, and the advertising dollars and sales volume for each of these are  shown in the following table. Develop a regression equation that would help Judith evaluate her adver- tising. Find the coefficient of determination for this model.

sAles ($1,000) ADveRTisiNG ($100)

11 5

  6 3

10 7

  6 2

12 8

solution

sAles Y ADveRTisiNG X 1X − X 22 1X − X 2 1Y − Y 2 11 5 15 - 522 = 0 15 - 52111 - 92 = 0   6 3 13 - 522 = 4 13 - 5216 - 92 = 6 10 7 17 - 522 = 4 17 - 52110 - 92 = 2   6 2 12 - 522 = 9 12 - 5216 - 92 = 9 12 8 18 - 522 = 9 18 - 52112 - 92 = 9

gY = 45 gX = 25 g1X - X22 = 26 g1X - X21Y - Y2 = 26 Y = 45>5 X = 25>5 = 9 = 5

b1 = g1X - X21Y - Y2

g1X - X22 = 26

26 = 1

b0 = Y - b1X = 9 - 112152 = 4 The regression equation is

Yn = 4 + 1X

Solved Problems

M04_REND9327_12_SE_C04.indd 157 10/02/14 1:23 PM

158  Chapter 4 • regression Models

To compute r2, we use the following table:

Y X Yn = 4 + 1X 1Yn − Y 22 1Y − Y 22

11 5 9 111 - 922 = 4 111 - 922 = 4   6 3 7 16 - 722 = 1 16 - 922 = 9 10 7 11 110 - 1122 = 1 110 - 922 = 1 6 2   6 16 - 622 = 0 16 - 922 = 9 12 8 12 112 - 1222 = 0 112 - 922 = 9

gY = 45 gX = 25 g1Y - Yn22 = 6 g1Y - Y22 = 32 Y = 9 X = 5 SSE SST

The slope 1b1 = 12 tells us that for each 1 unit increase in X (or $100 in advertising), sales increase  by 1 unit (or $1,000). Also, r2 = 0.8125 indicating that about 81% of the variability in sales can be   explained by the regression model with advertising as the independent variable.

Solved Problem 4-2 Use Excel with the data in Solved Problem 4-1 to find the regression model. What does the F test say  about this model?

solution Program 4.8 provides the Excel output for this problem. We see the equation is

Yn = 4 + 1X

The coefficient of determination 1r22 is shown to be 0.8125. The significance level for the F test is  0.0366, which is less than 0.05. This indicates the model is statistically significant. Thus, there is suf- ficient evidence in the data to conclude that the model is useful, and there is a relationship between X (advertising) and Y (sales).

Program 4.8 Excel 2013 Output for Solved Problem 4-2

M04_REND9327_12_SE_C04.indd 158 10/02/14 1:23 PM

Discussion Questions anD Problems  159

Self-Test

● Before taking the self-test, refer to the learning objectives at the beginning of the chapter, the notes in the margins, and the glossary at the end of the chapter.

● Use the key at the back of the book to correct your answers. ● Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

1. One of the assumptions in regression analysis is that a. the errors have a mean of 1. b. the errors have a mean of 0. c. the observations (Y) have a mean of 1. d. the observations (Y) have a mean of 0. 2. In a scatter diagram, the independent variable is plotted

on the a. vertical axis. b. trend line. c. horizontal axis. d. origin of the axis. 3. When using regression, an error is also called a. an intercept. b. a prediction. c. a coefficient. d. a residual. 4. Which of the following is X in regression analysis? a. control variable b. dummy variable c. independent variable d. both a and c 5. The best regression line is defined as the one with the a. minimum sum of the squared error. b. maximum sum of the squared error. c. average of the squared error. d. none of the above. 6. The percentage of the variation in the dependent variable

that is explained by a regression equation is measured by a. the coefficient of correlation. b. the MSE. c. the coefficient of determination. d. the slope. 7. Correlation coefficient measures the a. degree or strength of the linear relationship. b. variability in Y that is explained by the regression

equation.

c. both a and b. d. none of the above. 8. When using dummy variables in a regression equation to

model a qualitative or categorical variable, the number of dummy variables should equal

a. the number of categories. b. 1 more than the number of categories. c. 1 less than the number of categories. d. the number of other independent variables in the

model. 9. A multiple regression model differs from a simple linear

regression model because the multiple regression model has more than one

a. independent variable. b. dependent variable. c. intercept. d. error. 10. The overall significance of a regression model is tested

using an F test. The model is significant if a. the F value is low. b. the significance level of the F value is low. c. the r2 value is low. d. the slope is lower than the intercept. 11. Hypothesis tests related to the individual coefficients are

not valid when a. multicollinearity exist. b. collinearity exist. c. F test is still valid. d. a variable may appear as significant. 12. A good regression model should have a. a low r2 and a low significance level for the F test. b. a high r2 and a high significance level for the F test. c. a high r2 and a low significance level for the F test. d. a low r2 and a high significance level for the F test.

Discussion Questions and Problems

Discussion Questions 4-1 What are the purposes of regression analysis? How

does it help businesses?

4-2 How do we use qualitative variables in predicting the value variable Y?

4-3 What assumptions are necessary about the errors to determine the usefulness of a regression model?

4-4 Explain how a scatter diagram can be used to iden- tify the type of regression to use.

4-5 Explain how the adjusted r2 value is used in devel- oping a regression model.

4-6 Explain what information is provided by the F test.

4-7 How to build a linear model from a nonlinear relationship?

4-8 Explain how a statistical significant relationship may not be practical for a management decision.

M04_REND9327_12_SE_C04.indd 159 11/02/14 8:22 PM

160  Chapter 4 • regression Models

Problems 4-9  John Smith has developed the following forecasting 

model:

Yn = 36 + 4.3X1

    where

Yn = Demand for K10 air conditioners X1 = the outside temperature 1°F2

(a)  Forecast the demand for K10 when the tempera- ture is 70°F.

(b) What is the demand for a temperature of 80°F? (c) What is the demand for a temperature of 90°F?

  4-10  The operations manager of a musical instrument  distributor feels that demand for a particular type  of guitar may be related to the number of YouTube  views for a music video by the popular rock group  Marble Pumpkins during the preceding month. The  manager has collected the data shown in the follow- ing table:

youTube vieWs (1,000s)

GuiTAR sAles

30 8

40 11

70 12

60 10

80 15

50 13

(a) Graph these data to see whether a linear equa- tion might describe the relationship between the  views on YouTube and guitar sales.

(b) Using the equations presented in this chapter,  compute the SST, SSE, and SSR. Find the least- squares regression line for these data.

(c) Using the regression equation, predict guitar  sales if there were 40,000 views last month.

  4-11  Using the data in Problem 4-10, test to see if there is  a statistically significant relationship between sales  and YouTube views at the 0.05 level of significance.  Use the formulas in this chapter and Appendix D.

  4-12  Using computer software, find the least-squares re- gression line for the data in Problem 4-10. Based on  the F test, is there a statistically significant relation- ship between the demand for guitars and the number  of YouTube views?

  4-13  Students in a management science class have just re- ceived their grades on the first test. The instructor  has provided information about the first test grades  in some previous classes as well as the final average 

for the same students. Some of these grades have  been sampled and are as follows:

STUDENT 1 2 3 4 5 6 7 8 9

1st test grade 98 77 88 80 96 61 66 95 69

Final average 93 78 84 73 84 64 64 95 76

(a) Develop a regression model that could be used to  predict the final average in the course based on  the first test grade.

(b) Predict the final average of a student who made  an 83 on the first test.

(c) Give the values of r and r2 for this model. Inter- pret the value of r2 in the context of this problem.

  4-14  Using the data in Problem 4-13, test to see if there  is a statistically significant relationship between the  grade on the first test and the final average at the  0.05 level of significance. Use the formulas in this  chapter and Appendix D.

  4-15  Using computer software, find the least-squares regres- sion line for the data in Problem 4-13. Based on the F test, is there a statistically significant relationship between  the first test grade and the final average in the course?

  4-16  Steve Caples, a real estate appraiser in Lake Charles,  Louisiana, has developed a regression model to help  appraise residential housing in the Lake Charles  area. The model was developed using recently sold  homes in a particular neighborhood. The price (Y) of  the house is based on the square footage (X) of the  house. The model is

Yn = 33,478 + 62.4X

    The coefficient of correlation for the model is 0.63.

(a) Use the model to predict the selling price of a  house that is 1,860 square feet.

(b) A house with 1,860 square feet recently sold  for $165,000. Explain why this is not what the  model predicted.

(c)  If you were going to use multiple regression to  develop an appraisal model, what other quantita- tive variables might be included in the model?

(d) What is the coefficient of determination for this  model?

  4-17  Accountants at the firm Walker and Walker believed  that several traveling executives submit unusually  high travel vouchers when they return from busi- ness trips. The accountants took a sample of 200  vouchers submitted from the past year; they then de- veloped the following multiple regression equation  relating expected travel cost (Y) to number of days  on the road 1X12 and distance traveled 1X22 in miles:

Yn = $90.00 + $48.50X1 + $0.40X2

Note:   means the problem may be solved with QM for Windows;   means the problem may be  solved with Excel QM; and   means the problem may be solved with QM for Windows and/or Excel QM.

M04_REND9327_12_SE_C04.indd 160 10/02/14 1:23 PM

disCussion Questions and proBleMs  161

    The coefficient of correlation computed was 0.68.

(a)  If Thomas Williams returns from a 300-mile  trip that took him out of town for 5 days, what  is the expected amount that he should claim as  expenses?

(b) Williams submitted a reimbursement request for  $685; what should the accountant do?

(c)  Comment on the validity of this model. Should any  other variables be included? Which ones? Why?

  4-18  Thirteen  students  entered  the  undergraduate   business program at Rollins College 2 years ago.  The  following table indicates what their grade-point  averages (GPAs) were after being in the program for  2 years and what each student scored on the SAT  exam (maximum 2400) when he or she was in high  school. Is there a meaningful relationship between  grades and SAT scores? If a student scores a 1200 on  the SAT, what do you think his or her GPA will be?  What about a student who scores 2400?

sTuDeNT sAT scoRe GpA sTuDeNT sAT scoRe GpA

A 1263 2.90 H 1443 2.53

B 1131 2.93 I 2187 3.22

C 1755 3.00 J 1503 1.99

D 2070 3.45 K 1839 2.75

E 1824 3.66 L 2127 3.90

F 1170 2.88 M 1098 1.60

G 1245 2.15

  4-19  Bus and subway ridership in Washington, D.C., dur- ing the summer months is believed to be heavily tied  to the number of tourists visiting the city. During the  past 12 years, the following data have been obtained:

yeAR

NuMBeR oF TouRisTs (1,000,000s)

RiDeRship (100,000s)

  1 7 15

2 2 10

3   6 13

  4   4 15

  5 14 25

  6 15 27

7 16 24

8 12 20

9 14 27

10 20 44

11 15 34

12 7 17

(a)  Plot these data and determine whether a linear  model is reasonable.

(b) Develop a regression model. (c) What is expected ridership if 10 million tourists 

visit the city?

(d) If there are no tourists at all, explain the pre- dicted ridership.

  4-20  Use computer software to develop a regression model  for  the data  in Problem 4-19. Explain what  this   output indicates about the usefulness of this model.

  4-21  The following data give the starting salary for stu- dents who recently graduated from a local university  and accepted jobs soon after graduation. The starting  salary, grade-point average (GPA), and major (busi- ness or other) are provided.

SALARY $29,500 $46,000 $39,800 $36,500

GPA 3.1 3.5 3.8 2.9

Major Other Business Business Other

SALARY $42,000 $31,500 $36,200

GPA 3.4 2.1 2.5

Major Business Other Business

(a) Using a computer, develop a regression model  that could be used to predict starting salary based  on GPA and major.

(b) Use this model to predict the starting salary for a  business major with a GPA of 3.0.

(c) What does the model say about the starting sal- ary for a business major compared to a nonbusi- ness major?

(d) Do you believe this model is useful in predicting  the starting salary? Justify your answer, using in- formation provided in the computer output.

  4-22  The following data give the selling price, square  footage, number of bedrooms, and age of houses  that  have  sold  in  a  neighborhood  in  the  past  6 months. Develop three regression models to pre- dict the  selling price based upon each of the other  factors individually. Which of these is best?

selliNG pRice ($)

squARe FooTAGe

BeDRooMs

AGe (yeARs)

  84,000 1,670 2 30

  79,000 1,339 2 25

  91,500 1,712 3 30

120,000 1,840 3 40

127,500 2,300 3 18

132,500 2,234 3 30

145,000 2,311 3 19

164,000 2,377 3 7

155,000 2,736 4 10

168,000 2,500 3   1

172,500 2,500 4 3

174,000 2,479 3 3

175,000 2,400 3   1

177,500 3,124 4   0

184,000 2,500 3 2

195,500 4,062 4 10

195,000 2,854 3 3

M04_REND9327_12_SE_C04.indd 161 10/02/14 1:23 PM

162  Chapter 4 • regression Models

  4-23  Use the data in Problem 4-22 and develop a regres- sion model to predict selling price based on the  square footage and number of bedrooms. Use this  to predict the selling price of a 2,000-square-foot  house with three bedrooms. Compare this model  with the models in Problem 4-22. Should the num- ber of bedrooms be included in the model? Why or  why not?

  4-24  Use the data in Problem 4-22 and develop a regres- sion model  to predict selling price based on the  square footage, number of bedrooms, and age. Use  this to predict the selling price of a 10-year-old,  2,000-square-foot house with three bedrooms.

  4-25  The total expenses of a hospital are related to many  factors. Two of these factors are the number of beds  in the hospital and the number of admissions. Data  were collected on 14 hospitals, as shown in the fol- lowing table:

hospiTAl NuMBeR oF BeDs

ADMissioNs (100s)

ToTAl expeNses (MillioNs)

  1 215 77   57

2 336 160 127

3 520 230 157

  4 135   43   24

  5   35 9   14

  6 210 155 93

7 140   53   45

8   90     6     6

9 410 159 99

10   50   18   12

11   65   16   11

12   42 29   15

13 110 28   21

14 305 98   63

    Find the best regression model to predict the total  expenses of a hospital. Discuss the accuracy of this  model. Should both variables be included in the  model? Why or why not?

4-26  A sample of 20 automobiles was  taken, and  the  miles per gallon  (MPG), horsepower,  and  total  weight were recorded. Develop a linear regression  model to predict MPG, using horsepower as the only  independent variable. Develop another model with  weight as the independent variable. Which of these  two models is better? Explain.

MpG hoRsepoWeR WeiGhT

44   67 1,844

44   50 1,998

40   62 1,752

MpG hoRsepoWeR WeiGhT

37   69 1,980

37   66 1,797

34   63 2,199

35   90 2,404

32 99 2,611

30   63 3,236

28   91 2,606

26   94 2,580

26 88 2,507

25 124 2,922

22 97 2,434

20 114 3,248

21 102 2,812

18 114 3,382

18 142 3,197

16 153 4,380

16 139 4,036

  4-27  Use the data in Problem 4-26 to develop a multiple  linear regression model. How does this compare  with each of the models in Problem 4-26?

  4-28  Use the data in Problem 4-26 to find the best qua- dratic regression model. (There is more than one to  consider.) How does this compare to the models in  Problems 4-26 and 4-27?

  4-29  A sample of nine public universities and nine private  universities was taken. The total cost for the year (in- cluding room and board) and the median SAT score  (maximum total is 2400) at each school were recorded.  It was felt that schools with higher median SAT scores  would have a better reputation and would charge more  tuition as a result of that. The data are in the follow- ing table. Use regression to help answer the follow- ing questions based on this sample data. Do schools  with higher SAT scores charge more in tuition and  fees? Are private schools more expensive than pub- lic schools when SAT scores are taken into consider- ation? Discuss how accurate you believe these results  are using information related to the regression models.

cATeGoRy ToTAl cosT ($) MeDiAN sAT

Public 21,700 1990

Public 15,600 1620

Public 16,900 1810

Public 15,400 1540

Public 23,100 1540

Public 21,400 1600

Public 16,500 1560

Public 23,500 1890

(Continued on next page)

M04_REND9327_12_SE_C04.indd 162 10/02/14 1:23 PM

disCussion Questions and proBleMs  163

cATeGoRy ToTAl cosT ($) MeDiAN sAT

Public 20,200 1620

Private 30,400 1630

Private 41,500 1840

Private 36,100 1980

Private 42,100 1930

Private 27,100 2130

Private 34,800 2010

Private 32,100 1590

Private 31,800 1720

Private 32,100 1770

  4-30  In 2012, the total payroll for the New York Yankees  was almost $200 million, while the total payroll for  the Oakland Athletics (a team known for using base- ball analytics or sabermetrics) was about $55 million,  less than one-third of the Yankees payroll. In the fol- lowing table, you will see the payrolls (in millions)  and the total number of victories for the baseball  teams in the American League in the 2012 season.  Develop a regression model to predict the total num- ber of victories based on the payroll. Use the model to  predict the number of victories for a team with a pay- roll of $79 million. Based on the results of the com- puter output, discuss the relationship between payroll  and victories.

TeAM pAyRoll

($MillioNs) NuMBeR oF vicToRies

Baltimore Orioles 81.4 93

Boston Red Sox 173.2 69

Chicago White Sox 96.9 85

Cleveland Indians 78.4 68

Detroit Tigers 132.3 88

Kansas City Royals 60.9 72

Los Angeles Angels 154.5 89

Minnesota Twins 94.1 66

New York Yankees 198.0 95

Oakland Athletics 55.4 94

Seattle Mariners 82.0 75

Tampa Bay Rays 64.2 90

Texas Rangers 120.5 93

Toronto Blue Jays 75.5 73

  4-31  The number of victories (W), earned run average  (ERA), runs scored (R), batting average (AVG),  and on-base percentage (OBP) for each team in the  American League in the 2012 season are provided  in the following table. The ERA is one measure of  the effectiveness of the pitching staff, and a lower  number is better. The other statistics are measures of  effectiveness of the hitters, and higher numbers are  better for each of these.

TeAM W eRA R AvG oBp

Baltimore Orioles 93 3.90 712 0.247 0.311

Boston Red Sox 69 4.70 734 0.260 0.315

Chicago White Sox 85 4.02 748 0.255 0.318

Cleveland Indians 68 4.78 667 0.251 0.324

Detroit Tigers 88 3.75 726 0.268 0.335

Kansas City Royals 72 4.30 676 0.265 0.317

Los Angeles Angels 89 4.02 767 0.274 0.332

Minnesota Twins 66 4.77 701 0.260 0.325

New York Yankees 95 3.85 804 0.265 0.337

Oakland Athletics 94 3.48 713 0.238 0.310

Seattle Mariners 75 3.76 619 0.234 0.296

Tampa Bay Rays 90 3.19 697 0.240 0.317

Texas Rangers 93 3.99 808 0.273 0.334

Toronto Blue Jays 73 4.64 716 0.245 0.309

(a) Develop a regression model that could be used to  predict the number of victories based on the ERA.

(b) Develop a regression model that could be used to  predict the number of victories based on the runs  scored.

(c) Develop a regression model that could be used to  predict the number of victories based on the bat- ting average.

(d) Develop a regression model that could be used to  predict the number of victories based on the on- base percentage.

(e) Which of the four models is better for predicting  the number of victories?

(f)  Find the best multiple regression model to pre- dict the number of wins. Use any combination of  the variables to find the best model.

  4-32  The closing  stock price  for  each of  two  stocks  was recorded over a 12-month period. The clos- ing price  for  the Dow Jones  Industrial Average  (DJIA) was also recorded over this same time pe- riod. These  values are shown in the following table:

MoNTh DJiA sTock 1 sTock 2

  1 11,168 48.5 32.4

2 11,150 48.2 31.7

3 11,186 44.5 31.9

  4 11,381 44.7 36.6

  5 11,679 49.3 36.7

  6 12,081 49.3 38.7

7 12,222 46.1 39.5

8 12,463 46.2 41.2

9 12,622 47.7 43.3

10 12,269 48.3 39.4

11 12,354 47.0 40.1

12 13,063 47.9 42.1

13 13,326 47.8 45.2

M04_REND9327_12_SE_C04.indd 163 10/02/14 1:23 PM

164  Chapter 4 • regression Models

(a) Develop a regression model to predict the price  of stock 1 based on the Dow Jones Industrial  Average.

(b) Develop a regression model to predict the price  of stock 2 based on the Dow Jones Industrial  Average.

(c) Which of the two stocks is most highly corre- lated to the Dow Jones Industrial Average over  this time period?

In January 2012, Northern Airlines merged with Southeast Air- lines to create the fourth largest U.S. carrier. The new North– South Airline inherited both an aging fleet of Boeing 727-300  aircraft and Stephen Ruth. Stephen was a tough former Secre- tary of the Navy who stepped in as new president and chairman  of the board.

Stephen’s first concern in creating a financially solid com- pany was maintenance costs. It was commonly surmised in the  airline industry that maintenance costs rise with the age of the  aircraft. He quickly noticed that historically there had been a  significant difference in the reported B727-300 maintenance  costs (from ATA Form 41s) in both the airframe and engine  areas between Northern Airlines and Southeast Airlines, with  Southeast having the newer fleet.

On February 12, 2012, Peg Jones, vice president for op- erations and maintenance, was called into Stephen’s office  and asked to study the issue. Specifically, Stephen wanted to  know whether the average fleet age was correlated to direct  airframe maintenance costs, and whether there was a relation- ship between average fleet age and direct engine maintenance  costs. Peg was to report back by February 26 with the an- swer, along with quantitative and graphical descriptions of  the relationship.

Peg’s first step was to have her staff construct the aver- age age of Northern and Southeast B727-300 fleets, by quarter, 

since the introduction of that aircraft to service by each airline  in late 1993 and early 1994. The average age of each fleet was  calculated by first multiplying the total number of calendar days  each aircraft had been in service at the pertinent point in time by  the average daily utilization of the respective fleet to total fleet  hours flown. The total fleet hours flown was then divided by the  number of aircraft in service at that time, giving the age of the  “average” aircraft in the fleet.

The average utilization was found by taking the actual total  fleet hours flown on September 30, 2011, from Northern and  Southeast data, and dividing by the total days in service for all  aircraft at that time. The average utilization for Southeast was  8.3 hours per day, and the average utilization for Northern was  8.7 hours per day. Because the available cost data were calcu- lated for each yearly period ending at the end of the first quarter,  average fleet age was calculated at the same points in time. The  fleet data are shown in the following table. Airframe cost data  and engine cost data are both shown paired with fleet average  age in that table.

Discussion Question   1. Prepare Peg Jones’s response to Stephen Ruth.

Note: Dates and names of airlines and individuals have been changed in this  case to maintain confidentiality. The data and issues described here are real.

Case Study

north–south Airline

North–South Airline Data for Boeing 727-300 Jets

NoRTheRN AiRliNe DATA souTheAsT AiRliNe DATA

yeAR

AiRFRAMe cosT peR

AiRcRAFT ($)

eNGiNe cosT peR

AiRcRAFT ($) AveRAGe

AGe (houRs)

AiRFRAMe cosT peR

AiRcRAFT ($)

eNGiNe cosT peR

AiRcRAFT ($) AveRAGe

AGe (houRs)

2001 51.80 43.49 6,512 13.29 18.86 5,107

2002 54.92 38.58 8,404 25.15 31.55 8,145

2003 69.70 51.48 11,077 32.18 40.43 7,360

2004 68.90 58.72 11,717 31.78 22.10 5,773

2005 63.72 45.47 13,275 25.34 19.69 7,150

2006 84.73 50.26 15,215 32.78 32.58 9,364

2007 78.74 79.60 18,390 35.56 38.07 8,259

M04_REND9327_12_SE_C04.indd 164 10/02/14 1:23 PM

appendix 4.1: ForMulas For regression CalCulations  165

Bibliography

Berenson, Mark L., David M. Levine, and Timothy C. Kriehbiel. Basic Busi- ness Statistics, 12th ed. Upper Saddle River, NJ: Prentice Hall, 2012.

Black, Ken. Business Statistics: For Contemporary Decision Making, 7th ed.  John Wiley & Sons, Inc., 2012.

Draper, Norman R., and Harry Smith. Applied Regression Analysis, 3rd ed.  New York: John Wiley & Sons, Inc., 1998.

Kutner, Michael, John Neter, and Chris J. Nachtsheim. Applied Linear Regres- sion Models, 4th ed. Boston; New York: McGraw-Hill/Irwin, 2004.

Mendenhall, William, and Terry L. Sincich. A Second Course in Statistics: Re- gression Analysis, 7th ed. Upper Saddle River, NJ: Prentice Hall, 2012.

Appendix 4.1: Formulas for Regression Calculations

When performing regression calculations by hand, there are other formulas that can make the task  easier and are mathematically equivalent to the ones presented in the chapter. These, however,  make it more difficult to see the logic behind the formulas and to understand what the results  actually mean.

When using these formulas, it helps to set up a table with the columns shown in Table 4.7,  which has the Triple A Construction Company data that were used earlier in the chapter. The  sample size (n) is 6. The totals for all columns are shown, and the averages for X and Y are  calculated. Once this is done, we can use the following formulas for computations in a simple  linear regression model (one independent variable). The simple linear regression equation is  again given as

Yn = b0 + b1X

Slope of regression equation:

b1 = gXY - nXY gX2 - nX2

b1 = 180.5 - 6142172

106 - 61422 = 1.25

Intercept of regression equation:

b0 = Y - b1X b0 = 7 - 1.25142 = 2

Sum of squares of the error:

SSE = gY2 - b0gY - b1gXY SSE = 316.5 - 21422 - 1.251180.52 = 6.875

Table 4.7 Preliminary Calculations for Triple A Construction

y x Y2 X2 XY

6 3 62 = 36 32 = 9 3162 = 18 8 4 82 = 64 42 = 16 4182 = 32 9 6 92 = 81 62 = 36 6192 = 54 5 4 52 = 25 42 = 16 4152 = 20 4.5 2 4.52 = 20.25 22 = 4 214.52 = 9 9.5 5 9.52 = 90.25 52 = 25 519.52 = 47.5

gY = 42 gX = 24 gY2 = 316.5 gX2 = 106 gXY = 180.5 Y = 42>6 = 7 X = 24>6 = 4

M04_REND9327_12_SE_C04.indd 165 10/02/14 1:23 PM

166  Chapter 4 • regression Models

Estimate of the error variance:

s2 = MSE = SSE

n - 2

s2 = 6.875

6 - 2 = 1.71875

Estimate of the error standard deviation:

s = 1MSE s = 21.71875 = 1.311

Coefficient of determination:

r2 = 1 - SSE

gY2 - nY 2 r2 = 1 -

6.875

316.5 - 61722 = 0.6944

This formula for the correlation coefficient automatically determines the sign of r. This could  also be found by taking the square root of r2 and giving it the same sign as the slope:

r = ngXY - gXgY23ngX2 - 1gX2243ngY2 - 1gY224

r = 61180.52 - 1242142223611062 - 2424361316.52 - 4224 = 0.833

M04_REND9327_12_SE_C04.indd 166 10/02/14 1:23 PM

 167

Summary  •  Glossary  •  Key Equations  •  Solved Problems  •  Self-Test  •  Discussion Questions and Problems  •  Internet 

Homework Problems  •  Case Study: Forecasting Attendance at SWU Football Games  •  Case Study: Forecasting Monthly 

Sales  •  Internet Case Study  •  Bibliography 

Chapter Outline

5.6 Forecasting Models—Trend and Random Variations

5.7 Adjusting for Seasonal Variations

5.8 Forecasting Models—Trend, Seasonal, and Random Variations

5.9 Monitoring and Controlling Forecasts

5.1 Introduction

5.2 Types of Forecasting Models

5.3 Components of a Time Series

5.4 Measures of Forecast Accuracy

5.5 Forecasting Models—Random Variations Only

4. Understand Delphi and other qualitative decision- making approaches.

5. Compute a variety of error measures.

1. Understand and know when to use various families of forecasting models.

2. Compare moving averages, exponential smoothing, and other time-series models.

3. Seasonally adjust data.

After completing this chapter, students will be able to:

learning ObjeCtives

Forecasting

5Chapter

M05_REND9327_12_SE_C05.indd 167 11/02/14 9:25 AM

168  Chapter 5 • ForeCasting

5.1 Introduction

Every day, managers make decisions without knowing what will happen in the future. Inventory  is ordered though no one knows what sales will be, new equipment is purchased though no one  knows the demand for products, and investments are made though no one knows what profits  will be. Managers are always trying to reduce this uncertainty and to make better estimates of  what will happen in the future. Accomplishing this is the main purpose of forecasting.

There are many ways to forecast the future. In numerous firms (especially smaller ones),  the entire process is subjective, involving seat-of-the-pants methods, intuition, and years of   experience.

There are also more formal forecasting techniques, both quantitative and qualitative in   nature. The primary focus of this chapter will be understanding time-series models and deter- mining which model works best with a particular set of data.

5.2 Types of Forecasting Models

Figure 5.1 lists some of the more common forecasting models, and these are categorized by the  type of model. The first category involves qualitative models, while the others are quantitative in  nature and use mathematical models to develop better forecasts.

Qualitative Models Qualitative models are forecasting techniques based on judgmental or subjective factors. When a  totally new product such as the iPad is introduced, forecasting demand is very difficult due to the  lack of any historical sales data on that particular product or on similar products. The company  must rely on expert opinion, individual experiences and judgment, and other subjective factors.

Here is a brief overview of four different qualitative forecasting techniques:

  1. Delphi method. This iterative group process allows experts, who may be located in differ- ent places, to make forecasts. There are three different types of participants in the Delphi process: decision makers, staff personnel, and respondents. The decision-making group usually consists of 5 to 10 experts who will be making the actual forecast. The staff person- nel assist the decision makers by preparing, distributing, collecting, and summarizing a  series of questionnaires and survey results. The respondents are a group of people whose  judgments are valued and are being sought. This group provides inputs to the decision  makers before the forecast is made.

Figure 5.1 Forecasting Models

Forecasting Techniques

Qualitative Models

Causal Methods

Delphi Method

Moving Averages

Regression Analysis

Exponential Smoothing

Multiple Regression

Trend Projections

Jury of Executive Opinion

Sales Force Composite

Consumer Market Survey

Decomposition

Time-Series Methods

M05_REND9327_12_SE_C05.indd 168 11/02/14 9:25 AM

5.3 Components oF a time-series  169

In the Delphi method, when the results of the first questionnaire are obtained, the  results are summarized and the questionnaire is modified. Both the summary of the re- sults and the new questionnaire are then sent to the same respondents for a new round of  responses. The respondents, upon seeing the results from the first questionnaire, may view  things differently and may modify their original responses. This process is repeated with  the hope that a consensus is reached.

  2. Jury of executive opinion. This method takes the opinions of a small group of high-level  managers, often in combination with statistical models, and results in a group estimate of  demand.

  3. Sales force composite. In this approach, each salesperson estimates what sales will be in  his or her region; these forecasts are reviewed to ensure that they are realistic and are then  combined at the district and national levels to reach an overall forecast.

  4. Consumer market survey. This method solicits input from customers or potential customers  regarding their future purchasing plans. It can help not only in preparing a forecast but also  in improving product design and planning for new products.

Causal Models A variety of quantitative forecasting models are available when past numerical data are available.  Forecasting models are identified as causal models if the variable to be forecast is influenced by  or correlated with other variables included in the model. For example, daily sales of bottled water  might depend on the average temperature, the average humidity, and so on. A causal model would  include factors such as these in the mathematical model. Regression models (see Chapter 4) and  other more complex models would be classified as causal models.

Time-Series Models Time series are also quantitative models, and many time-series methods are available. A time- series model is a forecasting technique that attempts to predict the future values of a variable by  using only historical data on that one variable. These models are extrapolations of past values of  that series. While other factors may have influenced these past values, the impact of those other  factors is captured in the previous values of the variable being predicted. Thus, if we are fore- casting weekly sales for lawn mowers, we use the past weekly sales for lawn mowers in making  the forecast for future sales, ignoring other factors such as the economy, competition, and even  the selling price of the lawn mowers.

5.3 Components of a Time-Series

A time series is sequence of values recorded at successive intervals of time. The intervals can be  days, weeks, months, years, or other time unit. Examples include weekly sales of HP personal  computers, quarterly earnings reports of Microsoft Corporation, daily shipments of Eveready  batteries, and annual U.S. consumer price indices. A time series may consist of four possible  components—trend, seasonal, cyclical, and random.

The trend (T) component is the general upward or downward movement of the data over  a relatively long period of time. For example, total sales for a company may be increasing con- sistently over time. The consumer price index, one measure of inflation, is increasing over time.  While consistent upward (or downward) movements are indicative of a trend, there may be a  positive (or negative) trend present in a time series and yet the values do not increase (or de- crease) in every time period. There may be an occasional movement that seems inconsistent with  the trend due to random or other fluctuations.

Figure 5.2 shows scatter diagrams for several time series. The data for all the series are quar- terly, and there are four years of data. Series 3 has both trend and random components present.

The seasonal (S) component is a pattern of fluctuations above or below an average value  that repeats at regular intervals. For example, with monthly sales data on snow blowers, sales  tend to be high in December and January and lower in summer months and this pattern is ex- pected to repeat every year. Quarterly sales for a consumer electronics store may be higher in  the fourth quarter of every year and lower in other quarters. Daily sales in a retail store may  be higher on Saturdays than on other days of the week. Hourly sales in a fast-food restaurant are  usually expected to be higher around the lunch hour and the dinner hour, while other times are  not as busy. Series 2 in Figure 5.2 illustrates seasonal variations for quarterly data.

Overview of four qualitative or judgmental approaches: Delphi, jury of executive opinion, sales force composite, and consumer market survey.

M05_REND9327_12_SE_C05.indd 169 11/02/14 9:25 AM

170  Chapter 5 • ForeCasting

If a seasonal component is present in a set of data, the number of seasons depends on the  type of data. With quarterly data, there are four seasons because there are four quarters in a year.  With monthly data, there would be 12 seasons because there are 12 months in a year. With daily  sales data for a retail store that is open seven days a week, there would be seven seasons. With  hourly data, there could be 24 seasons if the business is open 24 hours a day.

A cyclical (C) component of a time series is a pattern in annual data that tends to repeat   every several years. The cyclical component of a time series is only used when making very  long-range forecasts, and it is usually associated with the business cycle. Sales or economic ac- tivity might reach a peak and then begin to recede and contract, reaching a bottom or trough. At  some point after the trough is reached, activity would pick up as recovery and expansion takes  place. A new peak would be reached and the cycle would begin again. Figure 5.3 shows 16 years  of annual data for a time series that has a cyclical as well as a random component.

While a time series with cyclical variations and a time series with seasonal variations may  look similar on a graph, cyclical variations tend to be irregular in length (from a few years to 

Figure 5.3 Scatter Diagram of a Times Series with Cyclical and Random Components

0 2 4 6 8 10

Time Period (Years)

S al

es

12 14 16 18

Figure 5.2 Scatter Diagram for Four Time Series of Quarterly Data

Series 4: Trend, Seasonal and Random Variations

Series 3: Trend and Random Variations

Series 2: Seasonal Variations Only

Series 1: Random Variations Only

1 2 3 4 5 6 7 8

Time Period (Quarters)

S al

es

9 10 11 12 13 14 15 16

M05_REND9327_12_SE_C05.indd 170 11/02/14 9:25 AM

5.4 measures oF ForeCast aCCuraCy  171

10 or more years) and magnitude, and they are very difficult to predict. Seasonal variations are  very consistent in length and magnitude, are much shorter as the seasons repeat in a year or less,  and are much easier to predict. The forecasting models presented in this chapter will be for short- range or medium-range forecasts and the cyclical component will not be included in these models.

The random (R) component consists of irregular, unpredictable variations in a time series.  Any variation in a times series that cannot be attributed to trend, seasonal, or cyclical variations  would fall into this category. If data for a time series tends to be level with no discernible trend  or seasonal pattern, random variations would be the cause for any changes from one time period  to the next. Series 1 in Figure 5.2 is an example of a time series with only a random component.

There are two general forms of time-series models in statistics. The first is a multiplicative  model, which assumes that demand is the product of the four components. It is stated as follows:

Demand = T * S * C * R

An additive model adds the components together to provide an estimate. Multiple regres- sion is often used to develop additive models. This additive relationship is stated as follows:

Demand = T + S + C + R

There are other models that may be a combination of these. For example, one of the com- ponents (such as trend) might be additive while another (such as seasonality) could be multi- plicative. Understanding the components of a time series will help in selecting an appropriate  forecasting technique to use. The methods presented in this chapter will be grouped according to  the components considered when the forecast is developed.

5.4 Measures of Forecast Accuracy

We discuss several different forecasting models in this chapter. To see how well one model  works, or to compare that model with other models, the forecast values are compared with the  actual or observed values. The forecast error (or deviation) is defined as follows:

Forecast error = Actual value - Forecast value

One measure of accuracy is the mean absolute deviation (MAD). This is computed by tak- ing the sum of the absolute values of the individual forecast errors and dividing by the numbers  of errors (n):

MAD = g � forecast error �

n (5-1)

Table 5.1 Computing the Mean Absolute Deviation (MAD)

Month

ActuAl SAleS of WireleSS

SpeAkerS

forecASt

SAleS

AbSolute VAlue of errorS (DeViAtion). ∣ActuAl–forecASt ∣

  1 110 — —

2 100 110 �100 - 110 � = 10

3 120 100 �120 - 100 � = 20

4 140 120 �140 - 120 � = 20

  5 170 140 �170 - 140 � = 30

  6 150 170 �150 - 170 � = 20

  7 160 150 �160 - 150 � = 10

8 190 160 �190 - 160 � = 30

  9 200 190 � 200 - 190 � = 10

10 190 200 �190 - 200 � = 10 11 — 190 —

Sum of  � errors � = 160

MAD = 160>9 = 17.8

M05_REND9327_12_SE_C05.indd 171 11/02/14 9:25 AM

172  Chapter 5 • ForeCasting

To illustrate how to compute the measures of accuracy, the naïve model will be used to fore- cast one period into the future. The naïve model does not attempt to address any of the com- ponents of a time series, but it is very quick and easy to use. A naïve forecast for the next time  period is the actual value that was observed in the current time period. For example, suppose  someone asked you to forecast the high temperature in your city for tomorrow. If you forecast  the temperature to be whatever the high temperature is today, your forecast would be a naïve  forecast. This technique is often used if a forecast is needed quickly and accuracy is not a   major concern.

Consider the Walker Distributors sales of wireless speakers shown in Table 5.1. Suppose that  in the past, Walker had forecast sales for each month to be the sales that were actually achieved in 

hurricane landfall location Forecasts and the Mean absolute Deviation

Scientists at the National Hurricane Center (NHC) of the Na- tional Weather Service have the very difficult job of predicting where the eye of a hurricane will hit land. Accurate forecasts are extremely important to coastal businesses and residents who need to prepare for a storm or perhaps even evacuate. They are also important to local government officials, law enforcement agencies, and other emergency responders who will provide help once a storm has passed. Over the years, the NHC has tremen- dously improved the forecast accuracy (measured by the mean absolute deviation [MAD]) in predicting the actual landfall loca- tion for hurricanes that originate in the Atlantic Ocean.

The NHC provides forecasts and periodic updates of where the hurricane eye will hit land. Such landfall location predictions

are recorded when a hurricane is 72 hours, 48 hours, 36 hours, 24 hours, and 12 hours away from actually reaching land. Once the hurricane has come ashore, these forecasts are compared to the actual landfall location, and the error (in miles) is recorded. At the end of the hurricane season, the errors for all the hurricanes in that year are used to calculate the MAD for each type of fore- cast (12 hours away, 24 hours away, etc.). The graph below shows how the landfall location forecast has improved since 1989. Dur- ing the early 1990s, the landfall forecast when the hurricane was 48 hours away had an MAD close to 200 miles; in 2009, this number was down to about 75 miles. Clearly, there has been vast improvement in forecast accuracy, and this trend is continuing.

Source: Based on National Hurricane Center, http://www.nhc.noaa.gov.

in aCtiOn

M05_REND9327_12_SE_C05.indd 172 11/02/14 9:25 AM

5.4 measures oF ForeCast aCCuraCy  173

The naïve forecast for the next period is the actual value observed in the current period.

Defining the problem To drive production at each of Tupperware’s 15 plants in the United States, Latin America, Africa, Europe, and Asia, the firm needs accurate forecasts of demand for its products.

Developing a Model A variety of statistical models are used, including moving averages, exponential smoothing, and regression analysis. Qualitative analysis is also employed in the process.

Acquiring input Data At world headquarters in Orlando, Florida, huge databases are maintained that map the sales of each product, the test market results of each new product (since 20% of the firm’s sales come from products less than 2 years old), and where each product falls in its own life cycle.

Developing a Solution Each of Tupperware’s 50 profit centers worldwide develops computerized monthly, quarterly, and 12-month sales projections. These are aggregated by region and then globally.

testing the Solution Reviews of these forecasts take place in sales, marketing, finance, and production departments.

Analyzing the results Participating managers analyze forecasts with Tupperware’s version of a “jury of executive opinion.”

implementing the results Forecasts are used to schedule materials, equipment, and personnel at each plant.

Source: Interviews by the authors with Tupperware executives.

MODeling in the real WOrlD Forecasting at tupperware international

the previous month. Table 5.1 gives these forecasts as well as the absolute value of the errors. In  forecasting for the next time period (month 11), the forecast would be 190. Notice that there is no  error computed for month 1 since there was no forecast for this month, and there is no error for  month 11 since the actual value of this is not yet known. Thus, the number of  errors (n) is 9.

From this, we see the following:

MAD = g � forecast error �

n =

160

9 = 17.8

This means that on the average, each forecast missed the actual value by 17.8 units. Other measures of the accuracy of historical errors in forecasting are sometimes used be-

sides the MAD. One of the most common is the mean squared error (MSe), which is the aver- age of the squared errors:*

MSE = g1error22

n (5-2)

*In regression analysis, the MSE formula is usually adjusted to provide an unbiased estimator of the error variance.  Throughout this chapter, we will use the formula provided here.

Defining the Problem

Developing a Model

Acquiring Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

M05_REND9327_12_SE_C05.indd 173 11/02/14 9:25 AM

174  Chapter 5 • ForeCasting

Weights can be used to put more emphasis on recent periods.

Besides the MAD and MSE, the mean absolute percent error (MApe) is sometimes used.  The MAPE is the average of the absolute values of the errors expressed as percentages of the  actual values. This is computed as follows:

MAPE = g ` error

actual `

n 100% (5-3)

There is another common term associated with error in forecasting. bias is the average error  and tells whether the forecast tends to be too high or too low and by how much. Thus, bias may  be negative or positive. It is not a good measure of the actual size of the errors because the nega- tive errors can cancel out the positive errors.

5.5 Forecasting Models—Random Variations Only

If all variations in a time series are due to random variations, with no trend, seasonal, or cycli- cal component, some type of averaging or smoothing model would be appropriate. The averag- ing techniques in this chapter are moving average, weighted moving average, and exponential  smoothing. These methods will smooth out the forecasts and not be too heavily influenced by  random variations. However, if there is a trend or seasonal pattern present in the data, then a  technique which incorporates that particular component into the forecast should be used.

Moving Averages Moving averages are useful if we can assume that market demands will stay fairly steady over  time. For example, a four-month moving average is found simply by summing the demand dur- ing the past four months and dividing by 4. With each passing month, the most recent month’s  data are added to the sum of the previous three months’ data, and the earliest month is dropped.  This tends to smooth out short-term irregularities in the data series.

An n-period moving average forecast, which serves as an estimate of the next period’s   demand, is expressed as follows:

Moving average forecast = Sum of demands in previous n periods

n (5-4)

Mathematically, this is written as

Ft+1 = Yt + Yt-1 + Á + Yt- n +1

n (5-5)

where

Ft+1 = forecast for time period t + 1 Yt = actual value in time period t n = number of periods to average

A 4-month moving average has n = 4; a 5-month moving average has n = 5.

WAllACe GARDen Supply exAMple Storage shed sales at Wallace Garden Supply are shown  in the middle column of Table 5.2. A 3-month moving average is indicated on the right. The  forecast for the next January, using this technique, is 16. Were we simply asked to find a forecast  for next January, we would only have to make this one calculation. The other forecasts are nec- essary only if we wish to compute the MAD or another measure of accuracy.

Weighted Moving Averages A simple moving average gives the same weight 11>n2 to each of the past observations being  used to develop the forecast. On the other hand, a weighted moving average allows different  weights to be assigned to the previous observations. As the weighted moving average method  typically assigns greater weight to more recent observations, this forecast is more responsive to  changes in the pattern of the data that occur. However, this is also a potential drawback to this  method because the heavier weight would also respond just as quickly to random fluctuations.

Moving averages smooth out variations when forecasting demands are fairly steady.

Three common measures of error are MAD, MSE, and MAPE. Bias gives the average error and may be positive or negative.

M05_REND9327_12_SE_C05.indd 174 11/02/14 9:25 AM

5.5 ForeCasting models—random Variations only  175

A weighted moving average may be expressed as

Ft+1 = g1Weight in period i21Actual value in period i2

g1Weights2 (5-6) Mathematically, this is

Ft+1 = w1Yt + w2Yt-1 + Á + wnYt- n +1

w1 + w2 + Á + wn (5-7)

where

wi =  weight for ith observation

Wallace Garden Supply decides to use a 3-month weighted moving average forecast with  weights of 3 for the most recent observation, 2 for the next observation, and 1 for the most dis- tant observation. This would be implemented as follows:

WeightS ApplieD perioD

3 Last month

2 2 months ago

1 3 months ago

3 * Sales last month + 2 * Sales 2 months ago + 1 * Sales 3 months ago

6

Sum of the weights

The results of the Wallace Garden Supply weighted average forecast are shown in Table 5.3. In  this particular forecasting situation, you can see that weighting the latest month more heavily  provides a much more accurate projection, and calculating the MAD for each of these would  verify this.

Choosing the weights obviously has an important impact on the forecasts. One way to  choose weights is to try various combinations of weights, calculate the MAD for each, and  select  the set of weights that results in the lowest MAD. Some forecasting software has an option to  search for the best set of weights, and forecasts using these weights are then provided. The best  set of weights can also be found by using nonlinear programming, as will be seen in a later  chapter.

Month ActuAl SheD SAleS 3-Month MoVing AVerAge

January 10

February 12

March 13

April 16 110 + 12 + 132>3 = 11.67 May 19 112 + 13 + 162>3 = 13.67 June 23 113 + 16 + 192>3 = 16.00 July 26 116 + 19 + 232>3 = 19.33 August 30 119 + 23 + 262>3 = 22.67 September 28 123 + 26 + 302>3 = 26.33 October 18 126 + 30 + 282>3 = 28.00 November 16 130 + 28 + 182>3 = 25.33 December 14 128 + 18 + 162>3 = 20.67 January — 118 + 16 + 142>3 = 16.00

Table 5.2 Wallace Garden Supply Shed Sales

M05_REND9327_12_SE_C05.indd 175 11/02/14 9:25 AM

176  Chapter 5 • ForeCasting

Some software packages require that the weights add to 1, and this would simplify Equa- tion 5-7 because the denominator would be 1. Forcing the weights to sum to 1 is easily achieved  by dividing each of the weights by the sum of the weights. In the Wallace Garden Supply ex- ample in Table 5.3, the weights are 3, 2, and 1, which add to 6. These weights could be revised to  the new weights 3/6, 2/6, and 1/6, which add to 1. Using these weights gives the same forecasts  shown in Table 5.4.

Both simple and weighted moving averages are effective in smoothing out sudden fluctua- tions in the demand pattern in order to provide stable estimates. Moving averages do, however,  have two problems. First, increasing the size of n (the number of periods averaged) does smooth  out fluctuations better, but it makes the method less sensitive to real changes in the data should  they occur. Second, moving averages cannot pick up trends very well. Because they are aver- ages, they will always stay within past levels and will not predict a change to either a higher or  a lower level.

Exponential Smoothing exponential smoothing is a forecasting method that is easy to use and is handled efficiently by  computers. Although it is a type of moving average technique, it involves little record keeping of  past data. The basic exponential smoothing formula can be shown as follows:

New forecast = Last period’s forecast (5-8) + a1Last period’s actual demand - Last period’s forecast2 where a is a weight (or smoothing constant) that has a value between 0 and 1, inclusive.

Equation 5-8 can also be written mathematically as

Ft+1 = Ft + a1Yt - Ft2 (5-9) where

Ft+1 = new forecast 1for time period t + 12 Ft = previous forecast (for time period t) a = smoothing constant 10 … a … 12 Yt = previous period’s actual demand

The concept here is not complex. The latest estimate of demand is equal to the old estimate ad- justed by a fraction of the error (last period’s actual demand minus the old estimate).

Table 5.3 Weighted Moving Average Forecast for Wallace Garden Supply

Moving averages have two problems: the larger number of periods may smooth out real changes, and they don’t pick up trend.

Month ActuAl SheD SAleS 3-Month WeighteD MoVing AVerAge

January 10

February 12

March 13

April 16 313 * 132 + 12 * 122 + 11024>6 = 12.17 May 19 313 * 162 + 12 * 132 + 11224>6 = 14.33 June 23 313 * 192 + 12 * 162 + 11324>6 = 17.00 July 26 313 * 232 + 12 * 192 + 11624>6 = 20.5 August 30 313 * 262 + 12 * 232 + 11924>6 = 23.83 September 28 313 * 302 + 12 * 262 + 12324>6 = 27.5 October 18 313 * 282 + 12 * 302 + 12624>6 = 28.33 November 16 313 * 182 + 12 * 282 + 13024>6 = 23.33 December 14 313 * 162 + 12 * 182 + 12824>6 = 18.67 January — 313 * 142 + 12 * 162 + 11824>6 = 15.33

M05_REND9327_12_SE_C05.indd 176 11/02/14 9:25 AM

5.5 ForeCasting models—random Variations only  177

The smoothing constant, a, can be changed to give more weight to recent data when the  value is high or more weight to past data when it is low. For example, when a = 0.5, it can be  shown mathematically that the new forecast is based almost entirely on demand in the past three  periods. When a = 0.1, the forecast places little weight on any single period, even the most re- cent, and it takes many periods (about 19) of historic values into account.*

For example, in January, a demand for 142 of a certain car model for February was predicted  by a dealer. Actual February demand was 153 autos, so the forecast error was 153 - 142 = 9. Using a smoothing constant a = 0.20, the exponential smoothing forecast for March demand  would be found by adding 20% of this error to the previous forecast of 142. Substituting into the  formula, we obtain

New forecast 1for March demand2 = 142 + 0.21153 - 1422 = 144.2

Thus, the forecast for the demand for cars in March would be 144 after being rounded by the  dealer.

Suppose that actual demand for the cars in March was 136. A forecast for the demand in  April, using the exponential smoothing model with a constant of a = 0.20, can be made:

New forecast 1for April demand2 = 144.2 + 0.21136 - 144.22 = 142.6, or 143 autos

GeTTinG STARTeD The exponential smoothing approach is easy to use and has been applied  successfully by banks, manufacturing companies, wholesalers, and other organizations. To   develop a forecast with this method, a smoothing constant must be selected and a previous  forecast must be known or assumed. If a company has been routinely using this method, there  is no problem. However, if a company is just beginning to use the method, these issues must be  addressed. 

The appropriate value of the smoothing constant, a, can make the difference between an  accurate forecast and an inaccurate forecast. In picking a value for the smoothing constant, the  objective is to obtain the most accurate forecast. Several values of the smoothing constant may  be tried, and the one with the lowest MAD could be selected. This is analogous to how weights  are selected for a weighted moving average forecast. Some forecasting software will automati- cally select the best smoothing constant. QM for Windows will display the MAD that would be  obtained with values of a ranging from 0 to 1 in increments of 0.01.

When the exponential smoothing model is used for the first time, there is no prior forecast  for the current time period to use in developing a forecast for the next time period. Therefore, a  previous value for a forecast must be assumed. It is common practice to assume a forecast for  time period 1, and to use this to develop a forecast for time period 2. The time period 2 forecast  is then used to forecast time period 3, and this continues until the forecast for the current time  period is found. Unless there is reason to assume another value, the forecast for time period 1  is assumed to be equal to the actual value in time period 1 (i.e., the error is zero). The value of  the forecast for time period 1 usually has very little impact on the value of forecasts many time  periods into the future.

poRT oF BAlTiMoRe exAMple Let us apply this concept with a trial-and-error testing of two  values of a in an example. The port of Baltimore has unloaded large quantities of grain from  ships during the past eight quarters. The port’s operations manager wants to test the use of   exponential smoothing to see how well the technique works in predicting tonnage unloaded. He  assumes that the forecast of grain unloaded in the first quarter was 175 tons. Two values of a are examined: a = 0.10 and a = 0.50. Table 5.4 shows the detailed calculations for a = 0.10 only.

To evaluate the accuracy of each smoothing constant, we can compute the absolute devia- tions and MADs (see Table 5.5). Based on this analysis, a smoothing constant of a = 0.10 is  preferred to a = 0.50 because its MAD is smaller.

*The term exponential smoothing is used because the weight of any one period’s demand in a forecast decreases expo- nentially over time. See an advanced forecasting book for algebraic proof.

The smoothing constant, A, allows managers to assign weight to recent data.

M05_REND9327_12_SE_C05.indd 177 11/02/14 9:25 AM

178  Chapter 5 • ForeCasting

Using Software for Forecasting Time Series The two software packages available with this book, Excel QM and QM for Windows, are very  easy to use for forecasting. The previous examples will be solved using these.

In using Excel QM with Excel 2013 for forecasting, from the Excel QM ribbon, select  Alphabetical as shown in Program 5.1A. Then simply select whichever method you wish to use,  and an initialization window opens to allow you to set the size of the problem. Program 5.1B  shows this window for the Wallace Garden Supply weighted moving average example with the  number of past periods and the number of periods in the weighted average already entered.  Clicking OK results in the screen shown in Program 5.1C to appear allowing you to enter the  data and the weights. Once all numbers are entered, the calculations are automatically performed  as you see in Program 5.1C.

To use QM for Windows to forecast, under Modules select Forecasting. Then select New and Time Series Analysis as shown in Program 5.2A. This would be for all of the models 

Table 5.5 Absolute Deviations and MADs for the port of Baltimore example

QuArter

ActuAl tonnAge

unloADeD

forecASt With

A = 0.10

AbSolute DeViAtionS for A = 0.10

forecASt With

A = 0.50

DeViAtionS AbSolute

for A = 0.50

1 180 175 5 175 5

2 168 175.5 7.5 177.5 9.5

3 159 174.75 15.75 172.75 13.75

4 175 173.18 1.82 165.88 9.12

5 190 173.36 16.64 170.44 19.56

6 205 175.02 29.98 180.22 24.78

7 180 178.02 1.98 192.61 12.61

8 182 178.22 3.78 186.30 4.3

Sum of absolute deviations 82.45 98.63

MAD = g �deviation �

n = 10.31

MAD = 12.33

Table 5.4 port of Baltimore exponential Smoothing Forecasts for A = 0.10 and A = 0.50

QuArter ActuAl tonnAge

unloADeD forecASt

uSing A = 0.10 forecASt

uSing A = 0.50

1 180 175 175

2 168 175.5 = 175.00 + 0.101180 - 1752 177.5 3 159 174.75 = 175.50 + 0.101168 - 175.502 172.75 4 175 173.18 = 174.75 + 0.101159 - 174.752 165.88 5 190 173.36 = 173.18 + 0.101175 - 173.182 170.44 6 205 175.02 = 173.36 + 0.101190 - 173.362 180.22 7 180 178.02 = 175.02 + 0.101205 - 175.022 192.61 8 182 178.22 = 178.02 + 0.101180 - 178.022 186.30 9 ? 178.60 = 178.22 + 0.101182 - 178.222 184.15

M05_REND9327_12_SE_C05.indd 178 11/02/14 9:25 AM

5.5 ForeCasting models—random Variations only  179

Program 5.1a Selecting the Forecasting Model in Wallace Garden Supply problem

Click the Excel QM Tab.

With cursor on Forecasting, a menu appears to the right. Select the model you want and an initialization window will open.

Click Alphabetical and scroll down to Forecasting.

Enter the number of periods to be averaged.

Click OK.

Enter a title. Enter the number of past periods of data. Program 5.1b initializing excel QM Spreadsheet for Wallace Garden Supply problem

presented in this chapter. In the window that opens (Program 5.2B), specify the number of  past periods of data. For the Port of Baltimore example, this is 8. Click OK and an input  screen appears allowing you to enter the data and select the forecasting technique as shown  in Program 5.2C. Based on the method you choose, other windows may open to allow you to  specify the necessary parameters for that model. In this example, the exponential smoothing  model is selected, so the smoothing constant must be specified. After entering the data, click  Solve and the output shown in  Program 5.2D appears. From this output screen, you can se- lect Window to see additional information such as the details of the calculations and a graph  of the time series and the forecasts. 

M05_REND9327_12_SE_C05.indd 179 11/02/14 9:25 AM

180  Chapter 5 • ForeCasting

Enter the demand data and the weights. The calculations will automatically be performed.

The measures of accuracy are shown here.

The forecast for the next period is here.

Program 5.1C excel QM output for Wallace Garden Supply problem

Program 5.2a Selecting Time-Series Analysis in QM for Windows in the Forecasting Module

In the Forecasting module, click New and Time-Series Analysis.

Program 5.2b entering Data for port of Baltimore example in QM for Windows

Enter a title.

Enter the number of past periods of data.

Click OK.

Program 5.2C Selecting the Model and entering Data for port of Baltimore example in QM for Windows

Enter the value for the smoothing constant.

Enter the data. Then click Solve at the top of the page.

Click here to see the models. Other input areas appear based on the model. Select Exponential Smoothing and a window opens for you to enter the smoothing constant.

M05_REND9327_12_SE_C05.indd 180 11/02/14 9:25 AM

5.6 ForeCasting models—trend and random Variations  181

Program 5.2D output for port of Baltimore example in QM for Windows

Additional output is available under Window.

The measures of accuracy are shown here.

The forecast for next period is here.

5.6 Forecasting Models—Trend and Random Variations

If there is trend present in a time series, the forecasting model must account for this and cannot  simply average past values. Two very common techniques will be presented here. The first is ex- ponential smoothing with trend, and the second is called trend projection or simply a trend line.

Exponential Smoothing with Trend An extension of the exponential smoothing model that will explicitly adjust for trend is called  the exponential smoothing with trend model. The idea is to develop an exponential smooth- ing forecast and then adjust this for trend. Two smoothing constants, a and b, are used in this  model, and both of these values must be between 0 and 1. The level of the forecast is adjusted  by multiplying the first smoothing constant, a, by the most recent forecast error and adding it to  the previous forecast. The trend is adjusted by multiplying the second smoothing constant, b, by  the most recent error or excess amount in the trend. A higher value gives more weight to recent  observations and thus responds more quickly to changes in the patterns.

As with simple exponential smoothing, the first time a forecast is developed, a previ- ous forecast 1Ft2 must be given or estimated. If none is available, often the initial forecast is  assumed to be perfect. In addition, a previous trend 1Tt2 must be given or estimated. This is  often estimated using other past data, if available, or by using subjective means, or by calcu- lating the increase (or decrease) observed during the first few time periods of the data avail- able. Without such an estimate available, the trend is sometimes assumed to be 0  initially,  although this may lead to poor forecasts if the trend is large and b is small. Once these initial  conditions have been set, the exponential smoothing forecast including trend 1FITt2 is devel- oped using three steps:

Step 1.   Compute the smoothed forecast 1Ft+12 for time period t + 1 using the equation Smoothed forecast = Previous forecast including trend + a1Last error2

Ft+1 = FITt + a1Yt - FITt2 (5-10)

Step 2.  Update the trend 1Tt+12 using the equation Smoothed trend = Previous trend + b1Error or excess in trend2

Tt+1 = Tt + b1Ft+1 - FITt2 (5-11)

Step 3.   Calculate the trend-adjusted exponential smoothing forecast 1FITt+12 using the equation Forecast including trend 1FITt+12 = Smoothed forecast 1Ft+12 + Smoothed trend 1Tt+12

FITt+1 = Ft+1 + Tt+1 (5-12)

Two smoothing constants are used.

Estimate or assume initial values for Ft and Tt

M05_REND9327_12_SE_C05.indd 181 11/02/14 9:25 AM

182  Chapter 5 • ForeCasting

where

Tt = smoothed trend for time period t Ft = smoothed forecast for time period t FITt = forecast including trend for time period t a = smoothing constant for forecasts b = smoothing constant for trend

Consider the case of Midwestern Manufacturing Company, which has a demand for electri- cal generators over the period 2007 to 2013 as shown in Table 5.6. To use the trend-adjusted ex- ponential smoothing method, first set initial conditions (previous values for F and T) and choose  a and b. Assuming that F1 is perfect and T1 is 0, and picking 0.3 and 0.4 for the smoothing  constants, we have

F1 = 74 T1 = 0 a = 0.3 b = 0.4

This results in

FIT1 = F1 + T1 = 74 + 0 = 74

Following the three steps to get the forecast for 2008 (time period 2), we have

Step 1.  Compute Ft+1 using the equation

Ft+1 = FITt + a1Yt - FITt2 F2 = FIT1 + 0.31Y1 - FIT12 = 74 + 0.3174 - 742 = 74

Step 2.  Update the trend 1Tt+12 using the equation Tt+1 = Tt + b1Ft+1 - FITt2

T2 = T1 + 0.41F2 - FIT12 = 0 + 0.4174 - 742 = 0 Step 3.  Calculate the trend-adjusted exponential smoothing forecast 1FITt+12 using the  equation

FIT2 = F2 + T2 = 74 + 0 = 74

For 2009 (time period 3) we have

Step 1.

F3 = FIT2 + 0.31Y2 - FIT22 = 74 + 0.3179 - 742 = 75.5 Step 2.

T3 = T2 + 0.41F3 - FIT22 = 0 + 0.4175.5 - 742 = 0.6 Step 3.

FIT3 = F3 + T3 = 75.5 + 0.6 = 76.1

The other results are shown in Table 5.7. The forecast for 2014 would be about 131.35.

Table 5.6 Midwestern Manufacturing’s Demand

YeAr electricAl generAtorS SolD

2007   74

2008   79

2009   80

2010   90

2011 105

2012 142

2013 122

M05_REND9327_12_SE_C05.indd 182 11/02/14 9:25 AM

5.6 ForeCasting models—trend and random Variations  183

Table 5.7 Midwestern Manufacturing exponential Smoothing with Trend Forecasts

tiMe 1 t 2

DeMAnD 1Yt 2 Ft+ 1 = FITt + 0.3 1Yt − FITt 2 Tt+ 1 = Tt + 0.4 1Ft+ 1 − FITt 2 FITt+ 1 = Ft+ 1 + Tt+ 1

1   74 74 0 74

2   79 74 = 74 + 0.3174 - 742 0 = 0 + 0.4174 - 742 74 = 74 + 0 3   80 75.5 = 74 + 0.3179 - 742 0.6 = 0 + 0.4175.5 - 742 76.1 = 75.5 + 0.6 4   90 77.270

= 76.1 + 0.3180 - 76.12 1.068

= 0.6 + 0.4177.27 - 76.12 78.338 = 77.270 + 1.068

5 105 81.837 = 78.338 + 0.3190 - 78.3382

2.468 = 1.068 + 0.4181.837 - 78.3382

84.305 = 81.837 + 2.468

6 142 90.514 = 84.305 + 0.31105 - 84.3052

4.952 = 2.468 + 0.4190.514 - 84.3052

95.466 = 90.514 + 4.952

7 122 109.426 = 95.466 + 0.31142 - 95.4662

10.536 = 4.952 + 0.41109.426 - 95.4662

119.962 = 109.426 + 10.536

8 120.573 = 119.962 + 0.31122 - 119.9622

10.780 = 10.536 + 0.41120.573 - 119.9622

131.353 = 120.573 + 10.780

To have Excel QM perform the calculations in Excel 2013, from the Excel QM ribbon, select  the alphabetical list of techniques and choose Forecasting and then Trend Adjusted Exponential Smoothing. After specifying the number of past observations, enter the data and the values for  a and b as shown in Program 5.3.

Trend Projections Another method for forecasting time series with trend is called trend projection. This tech- nique fits a trend line to a series of historical data points and then projects the line into the  future for medium- to long-range forecasts. There are several mathematical trend equations that  can be developed (e.g., exponential and quadratic), but in this section we look at linear (straight  line) trends only. A trend line is simply a linear regression equation in which the independent 

A trend line is a regression equation with time as the independent variable.

Program 5.3 output from excel QM in excel 2013 for Trend Adjusted exponential Smoothing example

The forecast for next period is here.

M05_REND9327_12_SE_C05.indd 183 11/02/14 9:25 AM

184  Chapter 5 • ForeCasting

variable (X) is the time period. The first time period will be time period 1. The second time  period will be time period 2, and so forth. The last time period will be time period n. The form  of this is

Yn = b0 + b1X

where

Yn = predicted value b0 = intercept b1 = slope of the line X = time period (i.e., X = 1, 2, 3, c, n)

The least squares regression method may be applied to find the coefficients that  minimize  the sum of the squared errors, thereby also minimizing the mean squared error (MSE).  Chapter 4  provides detailed explanation of  least  squares  regression and  formulas  to calculate  the   coefficients by hand. In this section, we use computer software to perform the calculations.

Let us consider the case of Midwestern Manufacturing’s demand for generators that was  presented in Table 5.6. A trend line can be used to predict demand (Y ) based on the time period  using a regression equation. For the first time period, which was 2007, we let X = 1. For 2008,  we let X = 2, and so forth. Using computer software to develop a regression equation as we did  in Chapter 4, we get the following equation:

Yn = 56.71 + 10.54X

To project demand in 2014, we first denote the year 2014 in our new coding system as  X = 8:

1sales in 20142 = 56.71 + 10.54182 = 141.03, or 141 generators

We can estimate demand for 2015 by inserting X = 9 in the same equation:

1sales in 20152 = 56.71 + 10.54192 = 151.57, or 152 generators

Program 5.4 provides output from Excel QM in Excel 2013. To run this model, from the Excel  QM ribbon, select Alphabetical to see the techniques. Then select Forecasting and  Regression/ Trend Analysis. When the input window opens, enter the number of past observations (7), and  the spreadsheet is initialized allowing you to input the seven Y values and the seven X values as  shown in Program 5.4.

Program 5.4 output from excel QM in excel 2013 for Trend line example

To forecast other time periods, enter the time period here.

The forecast for next period is here.

M05_REND9327_12_SE_C05.indd 184 11/02/14 9:25 AM

5.7 adjusting For seasonal Variations  185

This problem could also be solved using QM for Windows. To do this, select the Forecast- ing module, and then enter a new problem by selecting New – Time Series Analysis. When the  next window opens, enter the number of observations (7) and press OK. Enter the values (Y) for  the seven past observations when the input window opens. It is not necessary to enter the values  for X (1, 2, 3, . . . , 7) because QM for Windows will automatically use these numbers. Then  click Solve to see the results shown in Program 5.5.

Figure 5.4 provides a scatter diagram and a trend line for this data. The projected demand in  each of the next three years is also shown on the trend line.

5.7 Adjusting for Seasonal Variations

Time-series forecasting such as that in the example of Midwestern Manufacturing involves  looking at the trend of data over a series of time observations. Sometimes, however, recurring  variations at certain seasons of the year make a seasonal adjustment in the trend line forecast  necessary. Demand for coal and fuel oil, for example, usually peaks during cold winter months. 

Program 5.5 output from QM for Windows in excel 2013 for Trend line example

Forecasts for future time periods are shown here.

The trend line is shown over two lines.

Figure 5.4 Generator Demand and projections for next Three years Based on Trend line

0 0

20

40

60

80

100

120

140

160

180

1 2 3 4 5 6 7 8 9 10 11

Time Period

G en

er at

or D

em an

d

Trend Line Y = 56.71 1 10.54X ^

Projected demand for next 3 years is shown on the trend line.

M05_REND9327_12_SE_C05.indd 185 11/02/14 9:25 AM

186  Chapter 5 • ForeCasting

Demand for golf clubs or suntan lotion may be highest in summer. Analyzing data in monthly  or quarterly terms usually makes it easy to spot seasonal patterns. A seasonal index is often used  in multiplicative time-series forecasting models to make an adjustment in the forecast when a  seasonal component exists. An alternative is to use an additive model such as a regression model  that will be introduced in a later section.

Seasonal Indices A seasonal index indicates how a particular season (e.g., month or quarter) compares with an  average season. An index of 1 for a season would indicate that the season is average. If the index  is higher than 1, the values of the time series in that season tend to be higher than average. If the  index is lower than 1, the values of the time series in that season tend to be lower than average.

Seasonal indices are used with multiplicative forecasting models in two ways. First, each  observation in a time series are divided by the appropriate seasonal index to remove the impact of  seasonality. The resulting values are called deseasonalized data. Using these deseasonalized val- ues, forecasts for future values can be developed using a variety of forecasting techniques. Once  the forecasts of future deseasonalized values have been developed, they are multiplied by the  seasonal indices to develop the final forecasts, which now include variations due to seasonality.

There are two ways to compute seasonal indices. The first method, based on an overall average,  is easy and may be used when there is no trend present in the data. The second method, based on a  centered-moving-average approach, is more difficult but it must be used when there is trend present.

Calculating Seasonal Indices with No Trend When no trend is present, the index can be found by dividing the average value for a particular  season by the average of all the data. Thus, an index of 1 means the season is average. For ex- ample, if the average sales in January were 120 and the average sales in all months were 200,  the seasonal index for January would be 120>200 = 0.60, so January is below average. The  next example illustrates how to compute seasonal indices from historical data and to use these in  forecasting future values.

Monthly sales of one brand of telephone answering machine at Eichler Supplies are shown  in Table 5.8, for the two most recent years. The average demand in each month is computed, and  these values are divided by the overall average (94) to find the seasonal index for each month.  We then use the seasonal indices from Table 5.8 to adjust future forecasts. For example, suppose  we expected the third year’s annual demand for answering machines to be 1,200 units, which is 

An average season has an index of 1.

Month

SAleS DeMAnD AVerAge 2-YeAr

DeMAnD MonthlY DeMAnDa

AVerAge SeASonAl

inDexbYeAr 1 YeAr 2

January   80 100   90 94 0.957

February   85   75   80 94 0.851

March   80   90   85 94 0.904

April 110   90 100 94 1.064

May 115 131 123 94 1.309

June 120 110 115 94 1.223

July 100 110 105 94 1.117

August 110   90 100 94 1.064

September   85   95   90 94 0.957

October   75   85   80 94 0.851

November   85   75   80 94 0.851

December   80   80   80 94 0.851

 Total average demand = 1,128

a Average monthly demand = 1,128

12 months = 94 b Seasonal index =

Average 2 year demand

Average monthly demand

Table 5.8 Answering Machine Sales and Seasonal indices

M05_REND9327_12_SE_C05.indd 186 11/02/14 9:25 AM

5.7 adjusting For seasonal Variations  187

100 per month. We would not forecast each month to have a demand of 100, but we would adjust  these based on the seasonal indices as follows:

Jan. 1,200

12 * 0.957 = 96 July

1,200

12 * 1.117 = 112

Feb. 1,200

12 * 0.851 = 85 Aug.

1,200

12 * 1.064 = 106

Mar. 1,200

12 * 0.904 = 90 Sept.

1,200

12 * 0.957 = 96

Apr. 1,200

12 * 1.064 = 106 Oct.

1,200

12 * 0.851 = 85

May 1,200

12 * 1.309 = 131 Nov.

1,200

12 * 0.851 = 85

June 1,200

12 * 1.223 = 122 Dec.

1,200

12 * 0.851 = 85

Calculating Seasonal Indices with Trend When both trend and seasonal components are present in a time series, a change from one month  to the next could be due to a trend, to a seasonal variation, or simply to random fluctuations.  To help with this problem, the seasonal indices should be computed using a centered moving average (CMA) approach whenever trend is present. Using this approach prevents a variation  due to trend from being incorrectly interpreted as a variation due to the season. Consider the   following example.

Quarterly sales figures for Turner Industries are shown in Table 5.9. Notice that there is a  definite trend as the total is increasing each year, and there is an increase for each quarter from  one year to the next as well. The seasonal component is obvious as there is a definite drop from  the fourth quarter of one year to the first quarter of the next. A similar pattern is observed in  comparing the third quarters to the fourth quarters immediately following.

If a seasonal index for quarter 1 were computed using the overall average, the index would  be too low and misleading, since this quarter has less trend than any of the others in the sample.  If the first quarter of year 1 were omitted and replaced by the first quarter of year 4 (if it were  available), the average for quarter 1 (and consequently the seasonal index for quarter 1) would  be considerably higher. To derive an accurate seasonal index, we should use a CMA.

Consider quarter 3 of year 1 for the Turner Industries example. The actual sales in that   quarter were 150. To determine the magnitude of the seasonal variation, we should compare this  with an average quarter centered at that time period. Thus, we should have a total of four quar- ters (1 year of data) with an equal number of quarters before and after quarter 3 so the trend is  averaged out. Thus, we need 1.5 quarters before quarter 3 and 1.5 quarters after it. To obtain the  CMA, we take quarters 2, 3, and 4 of year 1, plus one-half of quarter 1 for year 1 and one-half of  quarter 1 for year 2. The average will be

CMA 1quarter 3 of year 12 = 0.511082 + 125 + 150 + 141 + 0.511162 4

= 132.00

We compare the actual sales in this quarter to the CMA and we have the following seasonal ratio:

Seasonal ratio = Sales in quarter 3

CMA =

150

132.00 = 1.136

Centered moving averages are used to compute seasonal indices when there is trend.

Table 5.9 Quarterly Sales ($1,000,000s) for Turner industries

QuArter YeAr 1 YeAr 2 YeAr 3 AVerAge

1 108 116 123 115.67

2 125 134 142 133.67

3 150 159 168 159.00

4 141 152 165 152.67

Average 131.00 140.25 149.50 140.25

M05_REND9327_12_SE_C05.indd 187 11/02/14 9:25 AM

188  Chapter 5 • ForeCasting

Thus, sales in quarter 3 of year 1 are about 13.6% higher than an average quarter at this time. All  of the CMAs and the seasonal ratios are shown in Table 5.10.

Since there are two seasonal ratios for each quarter, we average these to get the seasonal  index. Thus,

Index for quarter 1 = I1 = 10.851 + 0.8482>2 = 0.85 Index for quarter 2 = I2 = 10.965 + 0.9602>2 = 0.96 Index for quarter 3 = I3 = 11.136 + 1.1272>2 = 1.13 Index for quarter 4 = I4 = 11.051 + 1.0632>2 = 1.06

The sum of these indices should be the number of seasons (4) since an average season should  have an index of 1. In this example, the sum is 4. If the sum were not 4, an adjustment would be  made. We would multiply each index by 4 and divide this by the sum of the indices.

Steps used to Compute Seasonal indices Based on CMAs

  1. Compute a CMA for each observation (where possible).   2. Compute seasonal ratio = Observation/CMA for that observation.   3. Average seasonal ratios to get seasonal indices.   4.  If seasonal indices do not add to the number of seasons, multiply each index by (Number 

of seasons)/(Sum of the indices).

The seasonal indices found using the CMAs will be used to remove the impact of seasonal  so that the trend is easier to identify. Later, the seasonal indices will be used again to adjust the  trend forecast based on the seasonal variations.

5.8 Forecasting Models—Trend, Seasonal, and Random Variations

When both trend and seasonal components are present in a time series, the forecasting model  selected must address these. The decomposition method, which uses seasonal indices, is a very  common approach. Multiple regression models are also common, with dummy variables used to  adjust for seasonal variations in an additive time-series model.

The Decomposition Method The process of isolating linear trend and seasonal factors to develop more accurate forecasts is  called decomposition. The first step is to compute seasonal indices for each season as we have  done with the Turner Industries data. Then, the data are deseasonalized by dividing each  number 

Table 5.10 Centered Moving Averages and Seasonal Ratios for Turner industries

YeAr QuArter SAleS ($1,000,000s) cMA SeASonAl rAtio

1 1 108

2 125

3 150 132.000 1.136

4 141 134.125 1.051

2 1 116 136.375 0.851

2 134 138.875 0.965

3 159 141.125 1.127

4 152 143.000 1.063

3 1 123 145.125 0.848

2 142 147.875 0.960

3 168

4 165

M05_REND9327_12_SE_C05.indd 188 11/02/14 9:25 AM

5.8 ForeCasting models—trend, seasonal, and random Variations  189

by its seasonal index, as shown in Table 5.11. Figure 5.5 provides a graph of the original data as  well as a graph of the deseasonalized data. Notice how smooth the deseasonalized data is.

A trend line is then found using the deseasonalized data. Using computer software with this  data, we have*

b1 = 2.34 b0 = 124.78

The trend equation is

Yn = 124.78 + 2.34X

where

X = time

Table 5.11 Deseasonalized Data for Turner industries

SAleS ($1,000,000s) SeASonAl inDex DeSeASonAlizeD SAleS ($1,000,000s)

108 0.85 127.059

125 0.96 130.208

150 1.13 132.743

141 1.06 133.019

116 0.85 136.471

134 0.96 139.583

159 1.13 140.708

152 1.06 143.396

123 0.85 144.706

142 0.96 147.917

168 1.13 148.673

165 1.06 155.660

*If you do the calculations by hand, the numbers may differ slightly from these due to rounding.

Figure 5.5 Scatterplot of Turner industries’s original Sales Data and Deseasonalized Data

0 1 2 3 4 5 6 7 8 9 10 11 12 13 0

50

100

150

200

Time Period (Quarters)

S al

es

Sales Data

Deseasonalized Sales Data

M05_REND9327_12_SE_C05.indd 189 11/02/14 9:25 AM

190  Chapter 5 • ForeCasting

This equation is used to develop the forecast based on trend, and the result is multiplied by  the appropriate seasonal index to make a seasonal adjustment. For the Turner Industries data, the  forecast for the first quarter of year 4 (time period X = 13 and seasonal index I1 = 0.85) would  be found as follows:

Yn = 124.78 + 2.34X = 124.78 + 2.341132 = 155.2 1forecast before adjustment for seasonality2

We multiply this by the seasonal index for quarter 1 and we get

Yn * I1 = 155.2 * 0.85 = 131.92

Using the same procedure, we find the forecasts for quarters 2, 3, and 4 of the next year.  Table 5.12 shows the unadjusted forecast based on the trend line, for each of the four quarters of  year 4. The last column of this table gives the final forecasts that were found by multiplying each  trend forecast by the appropriate seasonal index. A scatter diagram showing the original data and  the deseasonalized data as well as the unadjusted forecast and the final forecast for these next  four quarters is provided in Figure 5.6.

Steps to Develop a Forecast using the Decomposition Method

  1. Compute seasonal indices using CMAs.   2. Deseasonalize the data by dividing each number by its seasonal index.   3. Find the equation of a trend line using the deseasonalized data.   4. Forecast for future periods using the trend line.   5. Multiply the trend line forecast by the appropriate seasonal index.

Table 5.12 Turner industry Forecasts for Four Quarters of year 4

YeAr QuArter tiMe perioD (X) trenD forecASt Yn = 124.78 + 2.34X SeASonAl inDex

finAl ( ADjuSteD) forecASt

4 1 13 155.20 0.85 131.92

2 14 157.54 0.96 151.24

3 15 159.88 1.13 180.66

4 16 162.22 1.06 171.95

Figure 5.6 Scatterplot of Turner industries original Sales Data and Deseasonalized Data with unadjusted and Adjusted Forecasts

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0

50

100

150

200

Time Period (Quarters)

S al

es

Sales Data

Deseasonalized Sales Data

Unadjusted Trend Forecast

Forecast Adjusted for Season

M05_REND9327_12_SE_C05.indd 190 11/02/14 9:25 AM

5.8 ForeCasting models—trend, seasonal, and random Variations  191

Most forecasting software, including Excel QM and QM for Windows, includes the decompo- sition method as one of the available techniques. This will automatically compute the CMAs,  deseasonalize the data, develop the trend line, make the forecast using the trend equation, and  adjust the final forecast for seasonality.

Software for Decomposition Program 5.6A shows the QM for Windows input screen that appears after selecting the Forecast- ing Module, selecting New-Time Series, and specifying 12 past observations. The multiplicative  decomposition method is selected, and you can see the inputs that are used for this problem.  When this is solved, the output screen in Program 5.6B appears. Additional output is available  by selecting details or graph from the Windows drop-down menu (not shown in Program 5.6B)  that appears after the problem is solved. The forecasts are slightly different from the ones shown  in Table 5.12 due to round-off error.

Excel QM can be used in Excel 2013 for this problem also. From the QM Ribbon, select the  menus By Chapter and select Chapter 5-Forecasting-Decomposition. Specify there are 12 past  observations, 4 seasons, and use a centered moving average.

Program 5.6a QM for Windows input Screen for Turner industries example

Select Rescale: set average to 1

Specify Centered Moving Average approach

Specify the number of seasons (4 for quarterly data)

Select Multiplicative Decomposition from the drop-down menu.

Input the data

Program 5.6b QM for Windows output Screen for Turner industries example

The �nal forecast is obtained by multiplying the trend (unadjusted) forecast by the seasonal indices (factors).

The seasonal indices (factors) are shown here.

The trend line is shown here over two lines.

M05_REND9327_12_SE_C05.indd 191 11/02/14 9:25 AM

192  Chapter 5 • ForeCasting

Using Regression with Trend and Seasonal Components Multiple regression may be used to forecast with both trend and seasonal components present  in a time series. One independent variable is time, and other independent variables are dummy  variables to indicate the season. If we forecast quarterly data, there are four categories (quarters)  so we would use three dummy variables. The basic model is an additive decomposition model  and is expressed as follows:

Yn = a + b1X1 + b2X2 + b3X3 + b4X4 where

X1 = time period X2 = 1 if quarter 2

= 0 otherwise X3 = 1 if quarter 3

= 0 otherwise X4 = 1 if quarter 4

= 0 otherwise

If X2 = X3 = X4 = 0, then the quarter would be quarter 1. It is an arbitrary choice as to which  of the quarters would not have a specific dummy variable associated with it. The forecasts will  be the same regardless of which quarter does not have a specific dummy variable.

Using Excel QM in Excel 2013 to develop the regression model, from the Excel QM  ribbon,  we select the menu for Chapter 5—Forecasting, and then choose the Multiple Regression method. The input screen in Program 5.7A appears, and the number of observations (12) and  the number of independent variables (4) are entered. An initialized spreadsheet appears and the  values for all the variables are entered in columns B through F as shown in Program 5.7B. Once  this has been done, the regression coefficients appear. The regression equation (with rounded  coefficients) is

Yn = 104.1 + 2.3X1 + 15.7X2 + 38.7X3 + 30.1X4 If this is used to forecast sales in the first quarter of the next year, we get

Yn = 104.1 + 2.31132 + 15.7102 + 38.7102 + 30.1102 = 134 For quarter 2 of the next year, we get

Yn = 104.1 + 2.31142 + 15.7112 + 38.7102 + 30.1102 = 152 Notice these are not the same values we obtained using the multiplicative decomposition  method. We could compare the MAD or MSE for each method and choose the one that is better.

Program 5.7a excel QM Multiple Regression initialization Screen for Turner industries

Enter the number of past period of data.

Enter the number of independent variables.

Multiple regression can be used to develop an additive decomposition model.

M05_REND9327_12_SE_C05.indd 192 11/02/14 9:25 AM

5.9 monitoring and Controlling ForeCasts  193

5.9 Monitoring and Controlling Forecasts

After a forecast has been completed, it is important that it not be forgotten. No manager wants to  be reminded when his or her forecast is horribly inaccurate, but a firm needs to determine why the  actual demand (or whatever variable is being examined) differed significantly from that projected.*

One way to monitor forecasts to ensure that they are performing well is to employ a  tracking signal. A tracking signal is a measurement of how well the forecast is predicting actual  values. As forecasts are updated every week, month, or quarter, the newly available demand data  are compared to the forecast values.

*If the forecaster is accurate, he or she usually makes sure that everyone is aware of his or her talents. Very seldom does  one read articles in Fortune, Forbes, or the Wall Street Journal, however, about money managers who are consistently  off by 25% in their stock market forecasts.

Program 5.7b excel QM Multiple Regression output Screen for Turner industries

Enter the values of Y and X1–X4 as shown.

The regression coef�cients are shown here.

Enter the values of the variables to obtain any forecast.

new route Determination at turkish airlines using Forecasting

Turkish Airlines (TA) flies to more than 170 destinations world- wide, with 152 aircraft and about 15,000 employees. It utilizes quantitative methods, including forecasting, to make complex decisions relating to new route determination, and fleet and capacity planning. It uses multiple regression to estimate coun- try traffic density and flow. In the methodology developed, top cities worldwide are selected based on six criteria: level of gross national product, trade volume, population, purchasing power, number of tourists, and airport constraints. The countries of these cities constitute the sample for regression analysis. Consequently, data about more than 100 indicators related to passenger vol- umes of countries are extracted from several databases.

After conducting correlation analyses, F tests, and t tests, TA ended up with seven meaningful variables for its regression model in 2010: Turkish population in the country, total young population, total traffic between Turkey, gross domestic product per capita, the number of landings, trade volume with Turkey, and distance to Turkey.

The model was implemented using MS Excel and SAS and tested on the countries to which TA has been flying. It was then run to forecast demand for other countries. The results

were refined using feedback from TA country managers in the region, senior vice presidents, and the executive vice president for sales and marketing. Based on these results, TA decided to launch direct or code share flights to eight new destinations in Asia, Europe, the Middle East, and Africa. TA has been the fast- est-growing airline of the Association of European Airlines since 2009. The company utilizes this model to add 10 to 20 new des- tinations to its network each year. It regularly uses regression and time-series models, a Kalman filter, and the Holt-Winters method to forecast weekly, monthly, and seasonal demand at all its in- ternational destinations. In 2010, most forecasts were within the 4% to 5% error range from the observed results. Similar results were also observed for the forecasts of one- to five-year hori- zons when compared with the market expectations of the air- craft manufacturers. In 2009, the company decided to acquire 89 aircrafts, worth billions of dollars, based on forecasts and many other inputs. In addition, TA estimates a 1% to 2% improvement in total revenue with more accurate demand forecasts.

Source: Based on interviews carried out by Dr. Arzu Tektas, Bogazici Univer- sity, with Mehmet Sukru Nenem, Strategic Planning and Investments Manager,  and Zeliha Akca, Strategic Planning Engineer, October 19, October 22, and  November 25, 2010, Turkish Airlines, www.turkishairlines.com.

in aCtiOn

A tracking signal measures how well predictions fit actual data.

M05_REND9327_12_SE_C05.indd 193 11/02/14 9:25 AM

194  Chapter 5 • ForeCasting

The tracking signal is computed as the running sum of the forecast errors (rSfe)  divided  by the mean absolute deviation:

Tracking signal = RSFE

MAD (5-13)

= g1forecast error2

MAD

where

MAD = g � forecast error �

n

as seen earlier in Equation 5-1. Positive tracking signals indicate that demand is greater than the forecast. Negative signals mean 

that demand is less than forecast. A good tracking signal—that is, one with a low RSFE—has about  as much positive error as it has negative error. In other words, small deviations are okay, but the posi- tive and negative deviations should balance so that the tracking signal centers closely around zero.

When tracking signals are calculated, they are compared with predetermined control limits.  When a tracking signal exceeds an upper or lower limit, a signal is tripped. This means that there  is a problem with the forecasting method, and management may want to reevaluate the way it  forecasts demand. Figure 5.7 shows the graph of a tracking signal that is exceeding the range of  acceptable variation. If the model being used is exponential smoothing, perhaps the smoothing  constant needs to be readjusted.

How do firms decide what the upper and lower tracking limits should be? There is no single  answer, but they try to find reasonable values—in other words, limits not so low as to be trig- gered with every small forecast error and not so high as to allow bad forecasts to be regularly  overlooked. George Plossl and Oliver Wight, two inventory control experts, suggested using  maximums of {4 MADs for high-volume stock items and {8 MADs for lower-volume items.*

Other forecasters suggest slightly lower ranges. One MAD is equivalent to approximately  0.8 standard deviation, so that {2 MADs = 1.6 standard deviations, {3 MADs = 2.4 stan- dard deviations, and {4 MADs = 3.2 standard deviations. This suggests that for a forecast to  be “in control,” 89% of the errors are expected to fall within {2 MADs, 98, within {3 MADs,  or 99.9% within {4 MADs whenever the errors are approximately normally distributed.**

*See G. W. Plossl and O. W. Wight. Production and Inventory Control. Upper Saddle River, NJ: Prentice Hall, 1967. **To prove these three percentages to yourself, just set up a normal curve for {1.6 standard deviations (Z values).   Using the normal table in Appendix A, you find that the area under the curve is 0.89. This represents {2 MADs. Simi- larly, {3 MADs = 2.4 standard deviations encompasses 98% of the area, and so on for {4 MADs.

Setting tracking limits is a matter of setting reasonable values for upper and lower limits.

Figure 5.7 plot of Tracking Signals

1

2

0 MADs

Upper Control Limit

Lower Control Limit

Signal Tripped

Tracking Signal

Acceptable Range

Time

M05_REND9327_12_SE_C05.indd 194 11/02/14 9:25 AM

summary  195

Here is an example that shows how the tracking signal and RSFE can be computed. Kim- ball’s Bakery’s quarterly sales of croissants (in thousands), as well as forecast demand and error  computations, are in the following table. The objective is to compute the tracking signal and  determine whether forecasts are performing adequately.

tiMe perioD

forecASt DeMAnD

ActuAl DeMAnD error rSfe `

forecASt error `

cuMulAtiVe error MAD

trAcking SignAl

1 100   90 -10 -10 10 10 10.0 -1 2 100   95 -5 -15   5 15   7.5 -2

3 100 115 +15     0 15 30 10.0     0

4 110 100 -10 -10 10 40 10.0 -1 5 110 125 +15 +5 15 55 11.0 +0.5 6 110 140 +30 +35 30 85 14.2 +2.5

In period 6, the calculations are

MAD = g � forecast error �

n =

85

6

= 14.2

Tracking signal = RSFE

MAD =

35

14.2

= 2.5 MADs

This tracking signal is within acceptable limits. We see that it drifted from  -2.0 MADs  to +2.5 MADs.

Adaptive Smoothing A lot of research has been published on the subject of adaptive forecasting. This refers to   computer monitoring of tracking signals and self-adjustment if a signal passes its preset limit. In  exponential smoothing, the a and b coefficients are first selected based on values that minimize  error forecasts and are then adjusted accordingly whenever the computer notes an errant tracking  signal. This is called adaptive smoothing.

Forecasts are a critical part of a manager’s function. Demand  forecasts drive the production, capacity, and scheduling sys- tems in a firm and affect the financial, marketing, and person- nel planning functions.

In this chapter we introduced three types of forecasting  models: time series, causal, and qualitative. Moving averages,  exponential smoothing, trend projection, and decomposition  time-series models were developed. Regression and multiple  regression models were recognized as causal models. Four  qualitative models were briefly discussed. In addition, we 

explained the use of scatter diagrams and measures of fore- casting accuracy. In future chapters you will see the useful- ness of these techniques in determining values for the various   decision-making models.

As we learned in this chapter, no forecasting method is  perfect under all conditions. Even when management has  found a satisfactory approach, it must still monitor and control  its forecasts to make sure that errors do not get out of hand.  Forecasting can often be a very challenging but rewarding part  of managing.

Summary

M05_REND9327_12_SE_C05.indd 195 11/02/14 9:25 AM

196  Chapter 5 • ForeCasting

Glossary

Adaptive Smoothing  The process of automatically monitor- ing and adjusting the smoothing constants in an exponential  smoothing model.

Bias  A technique for determining the accuracy of a forecast- ing model by measuring the average error and its direction.

Causal Models  Models that forecast using variables and fac- tors in addition to time.

Centered Moving Average  An average of the values cen- tered at a particular point in time. This is used to compute  seasonal indices when trend is present.

Cyclical Component  A component of a time series is a pat- tern in annual data that tends to repeat every several years.

Decision-Making Group  A group of experts in a  Delphi technique that has the responsibility of making  the forecast.

Decomposition  A forecasting model that decomposes a time  series into its seasonal and trend components.

Delphi  A judgmental forecasting technique that uses deci- sion makers, staff personnel, and respondents to determine  a forecast.

Deseasonalized Data  Time-series data in which each value  has been divided by its seasonal index to remove the effect  of the seasonal component.

Deviation  A term used in forecasting for error. Error  The difference between the actual value and the fore-

cast value. Exponential Smoothing  A forecasting method that is a 

combination of the last forecast and the last observed value. Least Squares  A procedure used in trend projection and 

regression analysis to minimize the squared distances be- tween the estimated straight line and the observed values.

Mean Absolute Deviation (MAD)  A technique for deter- mining the accuracy of a forecasting model by taking the  average of the absolute deviations.

Mean Absolute Percent Error (MAPE)  A technique for  determining the accuracy of a forecasting model by taking  the average of the absolute errors as a percentage of the  observed values.

Mean Squared Error (MSE)  A technique for determining  the accuracy of a forecasting model by taking the average  of the squared error terms for a forecasting model.

Moving Average  A forecasting technique that averages past  values in computing the forecast.

Naïve Model  A time-series forecasting model in which the  forecast for next period is the actual value for the current  period.

Qualitative Models  Models that forecast using judgments,  experience, and qualitative and subjective data.

Random Component  The irregular, unpredictable variations  in a time series.

Running Sum of Forecast Errors (RSFE)  Used to de- velop a tracking signal for time-series forecasting models,  this is a running total of the errors and may be positive or  negative.

Seasonal Component  A pattern of fluctuations in a time se- ries above or below an average value that repeats at regular  intervals.

Seasonal Index  An index number that indicates how a par- ticular season compares with an average time period (with  an index of 1 indicating an average season).

Smoothing Constant  A value between 0 and 1 that is used  in an exponential smoothing forecast.

Time-series Model  A forecasting technique that predicts the  future values of a variable by using only historical data on  that one variable.

Tracking Signal  A measure of how well the forecast is pre- dicting actual values.

Trend Component  The general upward or downward move- ment of the data in a time series over relatively long period  of time.

Trend Projection  The use of a trend line to forecast a   time-series with trend present. A linear trend line is a  r egression line with time as the independent variable.

Weighted Moving Average  A moving average  forecasting method that places different weights on  past values.

Key Equations

 (5-1)  MAD = g � forecast error �

n

A measure of overall forecast error called mean absolute  deviation.

 (5-2)  MSE = g1error22

n

A measure of forecast accuracy called mean squared  error.

 (5-3)  MAPE = g ` error

actual `

n 100%

A measure of forecast accuracy called mean absolute  percent error.

 (5-4)  Moving  =

Sum of demands in previous n periods n

  average   forecast

An equation for computing a moving average forecast.

M05_REND9327_12_SE_C05.indd 196 11/02/14 9:25 AM

solVed problems  197

 (5-5)  Ft+1 = Yt + Yt-1 + Á + Yt- n +1

n

A mathematical expression for a moving average forecast.

 (5-6)  Ft+1= g1Weight in period i21Actual value in period i2

g1Weights2 An equation for computing a weighted moving average  forecast.

 (5-7)  Ft+1 = w1Yt + w2Yt-1 + Á + wnYt- n +1

w1 + w2 + Á + wn A mathematical  expression  for a weighted moving   average forecast.

 (5-8)   New forecast = Last period>s forecast + a1Last period>s actual demand - Last period>s forecast2

An equation for computing an exponential smoothing  forecast.

 (5-9)  Ft+1 = Ft + a1Yt - Ft2 Equation 5-8 rewritten mathematically.

 (5-10) Ft+1 = FITt + a1Yt - FITt2 Equation to update the smoothed forecast 1Ft+12 used in  the trend-adjusted exponential smoothing model.

 (5-11) Tt+1 = Tt + b1Ft+1 - FITt2 Equation to update the smoothed trend value 1Tt+12 used  in the trend-adjusted exponential smoothing model.

 (5-12) FITt+1 = Ft+1 + Tt+1 Equation to develop forecast including trend (FIT) in the  trend-adjusted exponential smoothing model.

 (5-13) Tracking signal = RSFE

MAD

= g1forecast error2

MAD

An equation for monitoring forecasts with a tracking  signal.

Solved Problems

Solved Problem 5-1 Demand for patient surgery at Washington General Hospital has increased steadily in the past few years,  as seen in the following table:

YeAr outpAtient SurgerieS perforMeD

1 45

2 50

3 52

4 56

5 58

6 —

The director of medical services predicted six years ago that demand in year 1 would be 42 surger- ies. Using exponential smoothing with a weight of a = 0.20, develop forecasts for years 2 through 6.  What is the MAD?

Solution

YeAr ActuAl forecASt (SMootheD) error |error|

1 45 42 +3 3

2 50 42.6 = 42 + 0.2145 - 422 +7.4 7.4 3 52 44.1 = 42.6 + 0.2150 - 42.62 +7.9 7.9 4 56 45.7 = 44.1 + 0.2152 - 44.12 +10.3 10.3 5 58 47.7 = 45.7 + 0.2156 - 45.72 +10.3 10.3 6 — 49.8 = 47.7 + 0.2158 - 47.72 — —

MAD = g � errors �

n =

38.9

5 = 7.78

38.9

M05_REND9327_12_SE_C05.indd 197 11/02/14 9:25 AM

198  Chapter 5 • ForeCasting

Solved Problem 5-2 Quarterly demand for Jaguar XJ8 at a New York auto dealership is forecast with the equation

Yn = 10 + 3X where

X = time period (quarter): quarter 1 of last year = 0 quarter 2 of last year = 1 quarter 3 of last year = 2 quarter 4 of last year = 3 quarter 1 of this year = 4, and so on

and

Yn = predicted quarterly demand

The demand for luxury sedans is seasonal, and the indices for quarters 1, 2, 3, and 4 are 0.80, 1.00, 1.30,  and 0.90, respectively. Using the trend equation, forecast the demand for each quarter of next year. Then  adjust each forecast to adjust for seasonal (quarterly) variations.

Solution Quarter 2 of this year is coded X = 5; quarter 3 of this year, X = 6; and quarter 4 of this year, X = 7. Hence, quarter 1 of next year is coded X = 8; quarter 2, X = 9; and so on.

Yn 1next year quarter 12 = 10 + 132182 = 34 Adjusted forecast = 10.8021342 = 27.2 Yn 1next year quarter 22 = 10 + 132192 = 37 Adjusted forecast = 11.0021372 = 37 Yn 1next year quarter 32 = 10 + 1321102 = 40 Adjusted forecast = 11.3021402 = 52 Yn 1next year quarter 42 = 10 + 1321112 = 43 Adjusted forecast = 10.9021432 = 38.7

Self-Test

●  Before taking the self-test, refer to the learning objectives at the beginning of the chapter, the notes in the margins, and the  glossary at the end of the chapter.

●  Use the key at the back of the book to correct your answers. ●  Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

  1.  Which ones are forecasting techniques? a. regression analysis b. Delphi c. time-series models d. all of the above   2.  One of the following methods is a causal forecasting 

model. Which one? a. exponential smoothing b. regression analysis c. Delphi model d. sales force composite   3.  Who are the typical participants in a Delphi forecasting 

process? a. decision makers b. staff personnel c. respondents d. all of the above   4.  Trend projection is an example of a. Delphi process. b. time series.

c. causal methods. d. regression analysis.   5.  Least squares is a technique used a. in regression analysis. b. to determine the accuracy of a forecasting model. c. to determine an exponential smoothing model. d. none of the above.   6.  Which of the following may be negative? a. MAD b. bias c. MAPE d. MSE   7.  When comparing several forecasting models to determine 

which one best fits a particular set of data, the model that  should be selected is the one

a. with the highest MSE. b. with the MAD closest to 1. c. with a bias of 0. d. with the lowest MAD.

M05_REND9327_12_SE_C05.indd 198 11/02/14 9:25 AM

disCussion Questions and problems  199

  8.  In exponential smoothing, if you wish to give a  significant weight to the most recent observations, then  the smoothing constant should be

a. close to 0. b. close to 1. c. close to 0.5. d. less than the error.   9.  A trend equation is a regression equation in which a. there are multiple independent variables. b. the intercept and the slope are the same. c. the dependent variable is time. d. the independent variable is time.  10.  Sales for a company are typically higher in the summer 

months than in the winter months. This variation would  be called a

a. trend. b. seasonal factor. c. random factor. d. cyclical factor.  11.  An exponential smoothing forecast with a smoothing 

constant of 1 would result in the same forecast as a. a trend line forecast. b. an exponential smoothing forecast with a smoothing 

constant of 0. c. a naïve forecast. d. a 2-period moving average forecast.

 12.  If the seasonal index for January is 0.80, then a. January sales tend to be 80% higher than an average month. b. January sales tend to be 20% higher than an average 

month. c. January sales tend to be 80% lower than an average month. d. January sales tend to be 20% lower than an average month.  13.  If both trend and seasonal components are present in a 

time-series, then the seasonal indices a. should be computed based on an overall average. b. should be computed based on CMAs. c. will all be greater than 1. d. should be ignored in developing the forecast.  14.  Which of the following is used to alert the user of a 

forecasting model that a significant error occurred in one  of the periods?

a. a seasonal index b. a smoothing constant c. a tracking signal d. a regression coefficient  15.  If the multiplicative decomposition model is used to 

forecast daily sales for a retail store, how many seasons  will there be?

a. 4 b. 7 c. 12 d. 365

Discussion Questions and Problems

Discussion Questions   5-1  Describe briefly the steps used to develop a forecast-

ing system.

  5-2  What is a time-series forecasting model?

  5-3  What  are  the  two general  forms of  time-series   models in statistics?

  5-4  What is a qualitative forecasting model, and when is  it appropriate?

  5-5  What are some of the problems and drawbacks of  the moving average forecasting model?

  5-6  What effect does the value of the smoothing con- stant have on the weight given to the past forecast  and the past observed value?

  5-7  Is Delphi a quantitative forecasting method?

  5-8  What variations does the random component consist  of in a time series?

  5-9  What are moving averages used for? Provide an  example.

  5-10  A seasonal index may be less than one, equal to one,  or greater than one. Explain what each of these val- ues would mean.

  5-11  What are  the main problems with using moving  averages in forecasting?

  5-12  In using the decomposition method, the forecast based  on trend is found using the trend line. How is the sea- sonal index used to adjust this forecast based on trend?

  5-13  Explain what would happen if the smoothing con- stant in an exponential smoothing model was equal  to zero. Explain what would happen if the smooth- ing constant was equal to one.

  5-14  Explain when a CMA (rather than an overall aver- age) should be used in computing a seasonal index.  Explain why this is necessary.

Problems 5-15  Develop a four-month moving average forecast for 

Wallace Garden Supply and compute the MAD. A  three-month moving average forecast was developed  in the section on moving averages in Table 5.2.

5-16  Using MAD, determine whether  the  forecast  in  Problem 5-15 or the forecast in the section concern- ing Wallace Garden Supply is more accurate.

Note:   means the problem may be solved with QM for Windows;   means the problem may be solved with Excel QM; and   means the problem may be  solved with QM for Windows and/or Excel QM.

M05_REND9327_12_SE_C05.indd 199 11/02/14 9:25 AM

200  Chapter 5 • ForeCasting

5-17  Data collected on the yearly demand for 50-pound  bags of  fertilizer at Wallace Garden Supply are  shown in the following table. Develop a 3-year mov- ing average to forecast sales. Then estimate demand  again with a weighted moving average in which  sales in the most recent year are given a weight of 2  and sales in the other 2 years are each given a weight  of 1. Which method do you think is best?

YeAr DeMAnD for fertilizer

(1,000s of bAgS)

  1 4

2   6

3 4

4   5

  5 10

  6 8

  7   7

8   9

  9 12

10 14

11 15

5-18  Develop a trend line for the demand for fertilizer in  Problem 5-17, using any computer software.

5-19  In Problems 5-17 and 5-18, three different fore- casts were developed for the demand for fertilizer.  These three forecasts are a 3-year moving average,  a weighted moving average, and a trend line. Which  one would you use? Explain your answer.

5-20  Use exponential smoothing with a smoothing con- stant of 0.3 to forecast  the demand for fertilizer  given in Problem 5-17. Assume that last period’s  forecast for year 1 is 5,000 bags to begin the pro- cedure. Would you prefer to use, the exponential  smoothing model or the weighted average model   developed in Problem 5-17? Explain your answer.

5-21  Sales of Cool-Man air conditioners have grown  steadily during the past 5 years:

YeAr SAleS

1 450

2 495

3 518

4 563

5 584

6 ?

    The sales manager had predicted, before the business  started, that year 1’s sales would be 410 air condi- tioners. Using exponential smoothing with a weight  of a = 0.30, develop forecasts for years 2 through 6.

5-22  Using smoothing constants of 0.6 and 0.9, develop  forecasts for the sales of Cool-Man air conditioners  (see Problem 5-21).

5-23  What effect did the smoothing constant have on the  forecast for Cool-Man air conditioners? (See Prob- lems 5-21 and 5-22.) Which smoothing constant  gives the most accurate forecast?

5-24  Use a three-year moving average forecasting model  to forecast the sales of Cool-Man air conditioners  (see Problem 5-21).

5-25  Using the trend projection method, develop a fore- casting model for the sales of Cool-Man air condi- tioners (see Problem 5-21).

5-26  Would  you  use  exponential  smoothing with  a  smoothing constant of 0.3, a 3-year moving average,  or a trend to predict the sales of Cool-Man air condi- tioners? Refer to Problems 5-21, 5-24, and 5-25.

5-27  Sales of industrial vacuum cleaners at R. Lowenthal  Supply Co. over the past 13 months are as follows:

SAleS ($1,000s) Month SAleS ($1,000s) Month

11 January 14 August

14 February 17 September

16 March 12 October

10 April 14 November

15 May 16 December

17 June 11 January

11 July

(a) Using a moving average with  three periods,   determine the demand for vacuum cleaners for  next February.

(b) Using a weighted moving average with three   periods,  determine  the  demand  for  vacuum  cleaners for February. Use 3, 2, and 1 for the  weights of the most recent, second most recent,  and third most recent periods, respectively. For  example, if you were forecasting the demand  for February, November would have a weight  of 1, December would have a weight of 2, and   January would have a weight of 3.

(c)  Evaluate the accuracy of each of these methods. (d) What other factors might R. Lowenthal consider 

in forecasting sales?

5-28  Passenger miles flown on Northeast Airlines, a com- muter firm serving the Boston hub, are as follows  for the past 12 weeks:

ActuAl pASSenger ActuAl pASSenger

Week MileS (1,000s) Week MileS (1,000s)

1 17 7 20

2 21 8 18

3 19 9 22

4 23 10 20

5 18 11 15

6 16 12 22

(a) Assuming an initial forecast for week 1 of 17,000  miles, use exponential smoothing to compute  miles for weeks 2 through 12. Use a = 0.2.

M05_REND9327_12_SE_C05.indd 200 11/02/14 9:25 AM

disCussion Questions and problems  201

(b) What is the MAD for this model? (c) Compute the RSFE and tracking signals. Are 

they within acceptable limits? 5-29  Emergency  calls  to Winter Park, Florida’s  911 

 system, for the past 24 weeks are as follows:

Week cAllS Week cAllS Week cAllS

1 50   9 35 17 55

2 35 10 20 18 40

3 25 11 15 19 35

4 40 12 40 20 60

5 45 13 55 21 75

6 35 14 35 22 50

7 20 15 25 23 40

8 30 16 55 24 65

(a) Compute the exponentially smoothed forecast of  calls for each week. Assume an initial forecast of  50 calls in the first week and use a = 0.1. What  is the forecast for the 25th week?

(b) Reforecast each period using a = 0.6. (c) Actual calls during the 25th week were 85. Which 

smoothing constant provides a superior forecast?

5-30  How would the forecast for week 25 of the previous  problem change if the initial forecast was 40 instead  of 50? How would the forecast for week 25 change  if the forecast for week 1 were assumed to be 60?

5-31  Consulting income at Kate Walsh Associates for the  period February–July has been as follows:

Month incoMe ($1,000s)

February 70.0

March 68.5

April 64.8

May 71.7

June 71.3

July 72.8

    Use exponential smoothing to forecast August’s in- come. Assume that the initial forecast for February is  $65,000. The smoothing constant selected is a = 0.1.

5-32  Resolve Problem 5-31 with a = 0.3. Using MAD,  which smoothing constant provides a better forecast?

5-33  A major source of revenue in Texas is a state sales  tax on certain types of goods and services. Data  are compiled and the state comptroller uses them  to project future revenues for the state budget. One  particular category of goods is classified as Retail  Trade. Four years of quarterly data (in $millions) for  one particular area of southeast Texas follow:

QuArter YeAr 1 YeAr 2 YeAr 3 YeAr 4

1 218 225 234 250

2 247 254 265 283

3 243 255 264 289

4 292 299 327 356

(a) Compute seasonal indices for each quarter based  on a CMA.

(b) Deseasonalize the data and develop a trend line  on the deseasonalized data.

(c) Use the trend line to forecast the sales for each  quarter of year 5.

(d) Use the seasonal indices to adjust the forecasts  found in part (c) to obtain the final forecasts.

5-34  Using the data in Problem 5-33, develop a multiple  regression model to predict sales (both trend and  seasonal components), using dummy variables to  incorporate the seasonal factor into the model. Use  this model to predict sales for each quarter of the  next year. Comment on the accuracy of this model.

5-35  Trevor Harty, an avid mountain biker, always wanted  to start a business selling top-of-the-line moun- tain bikes and other outdoor supplies. A little over  6 years ago, he and a silent partner opened a store  called Hale and Harty Trail Bikes and Supplies.  Growth was rapid in the first 2 years, but since that  time, growth in sales has slowed a bit, as expected.  The quarterly sales (in $1,000s) for the past 4 years  are shown in the table below:

YeAr 1 YeAr 2 YeAr 3 YeAr 4

QUARTER 1 274 282 282 296

QUARTER 2 172 178 182 210

QUARTER 3 130 136 134 158

QUARTER 4 162 168 170 182

(a) Develop a trend line using the data in the table.  Use this to forecast sales for each quarter of year  5. What does the slope of this line indicate?

(b) Use the multiplicative decomposition model to  incorporate both trend and seasonal components  into the forecast. What does the slope of this line  indicate?

(c) Compare the slope of the trend line in part (a)  to the slope in the trend line for the decomposi- tion model that was based on the deseasonalized  sales figures. Discuss why these are so different  and explain which one is best to use.

5-36  The unemployment rates in the United States  during  a 10-year period are given in the following table.  Use exponential smoothing to find the best forecast  for next year. Use smoothing constants of 0.2, 0.4,  0.6, and 0.8. Which one had the lowest MAD?

YeAr 1 2 3 4 5 6 7 8 9 10

Unemployment  rate (%)

7.2 7.0 6.2 5.5 5.3 5.5 6.7 7.4 6.8 6.1

  5-37  Management of Davis’s Department Store has used  time-series extrapolation  to  forecast  retail sales  for the next four quarters. The sales estimates are  $100,000, $120,000, $140,000, and $160,000 for  the respective quarters before adjusting for seasonal- ity. Seasonal indices for the four quarters have been 

M05_REND9327_12_SE_C05.indd 201 11/02/14 9:25 AM

202  Chapter 5 • ForeCasting

found to be 1.30, 0.90, 0.70, and 1.10, respectively.  Compute a seasonalized or adjusted sales forecast.

  5-38  In the past, Judy Holmes’s tire dealership sold an  average of 1,000 radials each year. In the past two  years, 200 and 250, respectively, were sold in fall,  350 and 300 in winter, 150 and 165 in spring, and  300 and 285 in summer. With a major expansion  planned, Judy projects sales next year to increase to  1,200 radials. What will the demand be each season?

5-39  The following table provides the Dow Jones Indus- trial Average (DJIA) opening index value on the first  working day of 1994–2013:

    Develop a trend line and use it to predict the opening  DJIA index value for years 2014, 2015, and 2016.  Find the MSE for this model.

YeAr DjiA YeAr DjiA

2013 13,104 2003 8,342

2012 12,392 2002 10,022

2011 11,577 2001 10,791

2010 10,431 2000 11,502

2009 8,772 1999 9,213

2008 13,262 1998 7,908

2007 12,460 1997 6,448

2006 10,718 1996 5,117

2005 10,784 1995 3,834

2004 10,453 1994 3,754

5-40  Using the DJIA data in Problem 5-39, use expo- nential smooth with trend adjustment to forecast  the opening DJIA value for 2014. Use a = 0.8 and  b = 0.2. Compare the MSE for this technique with  the MSE for the trend line.

5-41  Refer to the DJIA data in Problem 5-39.

(a)  Use an exponential  smoothing model with a  smoothing constant of 0.4 to predict the opening  DJIA index value for 2014. Find the MSE for this.

(b) Use QM for Windows or Excel and find the smooth- ing constant that would provide the lowest MSE.

  5-42  The following table gives the average monthly ex- change rate between the U.S. dollar and the euro for  2009. It shows that 1 euro was equivalent to 1.289  U.S. dollars in January 2009. Develop a trend line  that could be used to predict the exchange rate for  2010. Use this model to predict the exchange rate for  January 2010 and February 2010.

Month exchAnge rAte

January 1.289

February 1.324

March 1.321

April 1.317

May 1.280

June 1.254

July 1.230

August 1.240

September 1.287

October 1.298

November 1.283

December 1.311

5-43  For the data in Problem 5-42, develop an exponen- tial smoothing model with a smoothing constant of  0.3. Using the MSE, compare this with the model in  Problem 5-42.

See our Internet home page, at www.pearsonglobaleditions.com/render,  for additional  homework problems, Problems 5-44 to 5-52.

Internet Homework Problems

Southwestern University  (SWU),  a  large  state college  in   Stephenville, Texas, 30 miles southwest of the Dallas/Fort  Worth metroplex, enrolls close to 20,000 students. In a typical  town–gown relationship, the school is a dominant force in the  small city, with more students during fall and spring than per- manent residents.

A longtime football powerhouse, SWU is a member of the  Big Eleven conference and is usually in the top 20 in college  football rankings. To bolster its chances of reaching the elu- sive and long-desired number-one ranking, in 2008 SWU hired  the legendary Billy Bob Taylor as its head coach. Although 

the number-one ranking remained out of reach, attendance  at the five Saturday home games each year increased. Prior  to  Taylor’s arrival, attendance generally averaged 25,000 to  29,000 per game. Season ticket sales bumped up by 10,000 just  with the announcement of the new coach’s arrival. Stephenville  and SWU were ready to move to the big time!

The  immediate  issue  facing SWU, however, was not  NCAA ranking. It was capacity. The existing SWU stadium,  built in 1953, has seating for 54,000 fans. The following table  indicates attendance at each game for the past six years.

Case Study

Forecasting attendance at sWu Football games

M05_REND9327_12_SE_C05.indd 202 11/02/14 9:25 AM

Case study  203

One of Taylor’s demands upon joining SWU had been  a  stadium expansion, or possibly even a new stadium. With   attendance  increasing, SWU administrators began  to face  the issue head-on. Taylor had wanted dormitories solely for  his   athletes  in the stadium as an additional feature of any  expansion.

SWU’s president, Dr. Marty Starr, decided it was time for  his vice president of development to forecast when the existing  stadium would “max out.” He also sought a revenue projection, 

assuming an average ticket price of $20 in 2014 and a 5% in- crease each year in future prices.

Discussion Questions   1. Develop a forecasting model, justify its selection over 

other techniques, and project attendance through 2015.   2. What revenues are to be expected in 2014 and 2015?   3. Discuss the school’s options.

Southwestern university Football Game Attendance, 2008–2013

2008 2009 2010

gAMe AttenDeeS opponent AttenDeeS opponent AttenDeeS opponent

1 34,200 Baylor 36,100 Oklahoma 35,900 TCU

2* 39,800 Texas 40,200 Nebraska 46,500 Texas Tech

3 38,200 LSU 39,100 UCLA 43,100 Alaska

4** 26,900 Arkansas 25,300 Nevada 27,900 Arizona

5 35,100 USC 36,200 Ohio State 39,200 Rice

2011 2012 2013

gAMe AttenDeeS opponent AttenDeeS opponent AttenDeeS opponent

1 41,900 Arkansas 42,500 Indiana 46,900 LSU

2* 46,100 Missouri 48,200 North Texas 50,100 Texas

3 43,900 Florida 44,200 Texas A&M 45,900 Prairie View  A&M

4** 30,100 Miami 33,900 Southern 36,300 Montana

5 40,500 Duke 47,800 Oklahoma 49,900 Arizona State

*Homecoming games **During the fourth week of each season, Stephenville hosted a hugely popular southwestern crafts festival. This event brought tens  of thousands of tourists to the town, especially on weekends, and had an obvious negative impact on game attendance.

Source: J. Heizer and B. Render. Operations Management, 11th ed. Upper Saddle River, NJ: Pearson, 2014, p. 148.

For years The Glass Slipper restaurant has operated in a re- sort community near a popular ski area of New Mexico. The  restaurant is busiest during the first 3 months of the year,  when the ski slopes are crowded and tourists flock to the area.

When James and Deena Weltee built The Glass Slipper,  they had a vision of the ultimate dining experience. As the  view of surrounding mountains was breathtaking, a high pri- ority was placed on having large windows and providing a  spectacular view from anywhere inside the restaurant. Special  attention was also given to the lighting, colors, and overall  ambiance, resulting in a truly magnificent experience for all 

who came to enjoy gourmet dining. Since its opening, The  Glass Slipper has developed and maintained a reputation as  one of the “must visit” places in that region of New Mexico.

While James loves to ski and truly appreciates the moun- tains and all that they have to offer, he also shares Deena’s  dream of retiring to a tropical paradise and enjoying a more  relaxed lifestyle on the beach. After some careful analysis of  their financial condition, they knew that retirement was many  years away. Nevertheless, they were hatching a plan to bring  them closer to their dream. They decided to sell The Glass  Slipper and open a bed and breakfast on a beautiful beach in 

Case Study

Forecasting Monthly sales

M05_REND9327_12_SE_C05.indd 203 11/02/14 9:25 AM

204  Chapter 5 • ForeCasting

Mexico. While this would mean that work was still in their   future, they could wake up in the morning to the sight of the  palm trees blowing in the wind and the waves lapping at the 

shore. They also knew that hiring the right manager would   allow James and Deena the time to begin a semi-retirement in  a corner of paradise.

To make this happen, James and Deena would have to sell  The Glass Slipper for the right price. The price of the business  would be based on the value of the property and equipment, as  well as projections of future income. A forecast of sales for the  next year is needed to help in the determination of the value of  the restaurant. Monthly sales for each of the past 3 years are  provided in Table 5.13.

Discussion Questions   1. Prepare a graph of the data. On this same graph, plot a 

12-month moving average forecast. Discuss any apparent  trend and seasonal patterns.

  2. Use regression to develop a trend line that could be used  to forecast monthly sales for the next year. Is the slope of  this line consistent with what you observed in question 1?  If not, discuss a possible explanation.

  3. Use the multiplicative decomposition model on these  data. Use this model to forecast sales for each month of  the next year. Discuss why the slope of the trend equation  with this model is so different from that of the trend equa- tion in question 2.

Table 5.13 Monthly Revenue (in $1,000s)

Month 2011 2012 2013

january 438 444 450

february 420 425 438

March 414 423 434

April 318 331 338

May 306 318 331

june 240 245 254

july 240 255 264

August 216 223 231

September 198 210 224

october 225 233 243

november 270 278 289

December 315 322 335

See our Internet home page, at www.pearsonglobaleditions.com/render, for the additional case  study on Akron zoological park. This case involves forecasting attendance at Akron’s zoo.

Internet Case Study

Bibliography

Billah, Baki, Maxwell L. King, Ralph D. Snyder, and Anne B. Koehler.  “ Exponential Smoothing Model Selection for Forecasting,” International Journal of Forecasting 22, 2, (April–June 2006): 239–247.

Black, Ken. Business Statistics: For Contemporary Decision Making, 7th ed.  John Wiley & Sons, Inc., 2012.

Diebold, F. X. Elements of Forecasting, 4th ed. Cincinnati: Cengage College  Publishing, 2007.

Gardner, Everette Jr. “Exponential Smoothing: The State of the Art—Part II,”  International Journal of Forecasting 22, 4 (October 2006): 637–666.

Granger, Clive W., and J. M. Hashem Pesaran. “Economic and  Statistical   Measures of Forecast Accuracy,” Journal of Forecasting 19, 7  ( December 2000): 537–560.

Hanke, J. E., and D. W. Wichern. Business Forecasting, 9th ed. Upper Saddle  River, NJ: Prentice Hall, 2009.

Heizer, J., and B. Render. Operations Management, 9th ed. Upper Saddle  River, NJ: Prentice Hall, 2008.

Hyndman, Rob J. “The Interaction Between Trend and Seasonality,”  International Journal of Forecasting 20, 4 (October–December 2004):  561–563.

Hyndman, Rob J., and Anne B. Koehler. “Another Look at Measures of   Forecast Accuracy,” International Journal of Forecasting 22, 4 (October  2006): 679–688.

Levine, David M., David F. Stephan, and Kathryn A. Szabat. Statistics for Managers Using Microsoft Excel, 7th ed. Upper Saddle River, NJ:   Pearson, 2014.

Meade, Nigel. “Evidence for the Selection of Forecasting Methods,” Journal of Forecasting 19, 6 (November 2000): 515–535.

Snyder, Ralph D., and Roland G. Shami. “Exponential Smoothing of  Seasonal  Data: A Comparison,” Journal of Forecasting 20, 3 (April 2001):  197–202.

Yurkiewicz, J. “Forecasting Software Survey,” OR/MS Today 39, 3 (June  2012): 52–61.

M05_REND9327_12_SE_C05.indd 204 11/02/14 9:26 AM

 205

Learning Objectives

After completing this chapter, students will be able to:

1. Understand the importance of inventory control and ABC analysis.

2. Use the economic order quantity (EOQ) to determine how much to order.

3. Compute the reorder point (ROP) in determining when to order more inventory.

4. Handle inventory problems that allow quantity discounts or noninstantaneous receipt.

6.1 Introduction

6.2 Importance of Inventory Control

6.3 Inventory Decisions

6.4 Economic Order Quantity: Determining How Much to Order

6.5 Reorder Point: Determining When to Order

6.6 EOQ Without the Instantaneous Receipt Assumption

6.7 Quantity Discount Models

6.8 Use of Safety Stock

6.9 Single-Period Inventory Models

6.10 ABC Analysis

6.11 Dependent Demand: The Case for Material Requirements Planning

6.12 Just-In-Time Inventory Control

6.13 Enterprise Resource Planning

5. Understand the use of safety stock.

6. Describe the use of material requirements planning in solving dependent-demand inventory problems.

7. Discuss just-in-time inventory concepts to reduce inventory levels and costs.

8. Discuss enterprise resource planning systems.

Summary  •  Glossary  •  Key Equations  •  Solved Problems  •  Self-Test  •  Discussion Questions and Problems  •  Internet 

Homework Problems  •  Case Study: Martin-Pullin Bicycle Corporation  •  Internet Case Studies  •  Bibliography

appendix 6.1: Inventory Control with QM for Windows

chapter OutLine

6chapter Inventory

Control Models

M06_REND9327_12_SE_C06.indd 205 10/02/14 1:23 PM

206  Chapter 6 • Inventory Control Models

6.1 Introduction

Inventory is one of the most expensive and important assets to many companies,  representing  as much as 50% of total invested capital. Managers have long recognized that good  inventory   control is crucial. On one hand, a firm can try to reduce costs by reducing on-hand  inventory   levels. On the other hand, customers become dissatisfied when frequent inventory  outages,  called stockouts, occur. Thus, companies must make the balance between low and high   inventory levels. As you would expect, cost minimization is the major factor in obtaining this  delicate balance.

Inventory is any stored resource that is used to satisfy a current or a future need. Raw ma- terials, work-in-process, and finished goods are examples of inventory. Inventory levels for fin- ished goods are a direct function of demand. When we determine the demand for completed  clothes dryers, for example, it is possible to use this information to determine how much sheet  metal, paint, electric motors, switches, and other raw materials and work-in-process are needed  to produce the finished product.

All organizations have some type of inventory planning and control system. A bank has  methods to control its inventory of cash. A hospital has methods to control blood supplies and  other important items. State and federal governments, schools, and virtually every manufactur- ing and production organization are concerned with inventory planning and control. Studying  how organizations control their inventory is equivalent to studying how they achieve their objec- tives by supplying goods and services to their customers. Inventory is the common thread that  ties all the functions and departments of the organization together.

Figure 6.1 illustrates the basic components of an inventory planning and control system.  The planning phase is concerned primarily with what inventory is to be stocked and how it  is to be acquired (whether it is to be manufactured or purchased). This information is then  used in forecasting demand for the inventory and in controlling inventory levels. The feedback  loop in  Figure 6.1 provides a way of revising the plan and forecast based on experiences and  observation.

Through inventory planning, an organization determines what goods and/or services are  to be produced. In cases of physical products, the organization must also determine whether  to produce these goods or to purchase them from another manufacturer. When this has been  determined, the next step is to forecast the demand. As discussed in Chapter 5, there are many  mathematical techniques that can be used in forecasting demand for a particular product. The  emphasis in this chapter is on inventory control, that is, how to maintain adequate inventory  levels within an organization.

Inventory is any stored resource that is used to satisfy a current or future need.

Figure 6.1 Inventory Planning and Control

Planning on what to stock and how to

get it

Forecasting parts/product

demand

Controlling inventory

levels

Feedback metrics to

revise plans and forecasts

M06_REND9327_12_SE_C06.indd 206 10/02/14 1:23 PM

6.2 IMportanCe of Inventory Control  207

6.2 Importance of Inventory Control

Inventory control serves several important functions and adds a great deal of flexibility to the  operation of the firm. Consider the following five uses of inventory:

  1. The decoupling function   2. Storing resources   3.  Irregular supply and demand   4. Quantity discounts   5. Avoiding stockouts and shortages

Decoupling Function One of the major functions of inventory is to decouple manufacturing processes within the or- ganization. If you did not store inventory, there could be many delays and inefficiencies. For  example, when one manufacturing activity has to be completed before a second activity can be  started, it could stop the entire process. If, however, you have some stored inventory between  processes, it could act as a buffer.

Storing Resources Agricultural and seafood products often have definite seasons over which they can be harvested  or caught, but the demand for these products is somewhat constant during the year. In these and  similar cases, inventory can be used to store these resources.

In a manufacturing process, raw materials can be stored by themselves, in work-in-process,  or in the finished product. Thus, if your company makes lawn mowers, you might obtain lawn  mower tires from another manufacturer. If you have 400 finished lawn mowers and 300 tires in  inventory, you actually have 1,900 tires stored in inventory. Three hundred tires are stored by  themselves, and 1,600 (1,600 = 4 tires per lawn mower * 400 lawn mowers) tires are stored in  the finished lawn mowers. In the same sense, labor can be stored in inventory. If you have 500  subassemblies and it takes 50 hours of labor to produce each assembly, you actually have 25,000  labor hours stored in inventory in the subassemblies. In general, any resource, physical or other- wise, can be stored in inventory.

Irregular Supply and Demand When the supply or demand for an inventory item is irregular, storing certain amounts in   inventory can be important. If the greatest demand for Diet-Delight beverage is during the   summer, you will have to make sure that there is enough supply to meet this irregular demand.  This might require that you produce more of the soft drink in the winter than is actually needed  to meet the winter demand. The inventory levels of Diet-Delight will gradually build up over  the winter, but this inventory will be needed in the summer. The same is true for irregular  supplies.

Quantity Discounts Another use of inventory is to take advantage of quantity discounts. Many suppliers offer dis- counts for large orders. For example, an electric jigsaw might normally cost $20 per unit. If  you order 300 or more saws in one order, your supplier may lower the cost to $18.75. Purchas- ing in larger quantities can substantially reduce the cost of products. There are, however, some  disadvantages of buying in larger quantities. You will have higher storage costs and higher costs  due to spoilage, damaged stock, theft, insurance, and so on. Furthermore, by investing in more  inventory, you will have less cash to invest elsewhere.

Avoiding Stockouts and Shortages Another important function of inventory is to avoid shortages or stockouts. If you are repeatedly  out of stock, customers are likely to go elsewhere to satisfy their needs. Lost goodwill can be an  expensive price to pay for not having the right item at the right time.

Inventory can act as a buffer.

Resources can be stored in work-in-process.

Inventory can alleviate irregular demand and supply patterns.

Buying in bulk can be an economic advantage.

Carrying safety stock can increase customer satisfaction.

M06_REND9327_12_SE_C06.indd 207 10/02/14 1:23 PM

208  Chapter 6 • Inventory Control Models

6.3 Inventory Decisions

Even though there are literally millions of different types of products produced in our society,  there are only two fundamental decisions that you have to make when controlling inventory:

  1. How much to order   2. When to order

The purpose of all inventory models and techniques is to determine rationally how much  to order and when to order. As you know, inventory fulfills many important functions within an  organization. But as the inventory levels go up to provide these functions, the cost of storing and  holding inventory also increases. Thus, you must reach a fine balance in establishing inventory  levels. A major objective in controlling inventory is to minimize total inventory costs. Some of  the most significant inventory costs follow:

  1. Cost of the items (purchase cost or material cost)   2. Cost of ordering   3. Cost of carrying, or holding, inventory   4. Cost of stockouts

A major objective of all inventory models is to minimize inventory costs.

Defining the Problem Managing today’s energy supply chain is an ever increasing complex problem. Yet, to survive in today’s marketplace, oil and gas companies must optimize their supply chains.

Developing a Model Researchers from Portugal developed a uni-directional model of oil and gas products from refineries to distribution center. The model incorporated many different parameters including number of products, maximum storage capacities for each product, pipeline capacities, as well as the demand for each product among them.

Acquiring the Input Data The input data consisted of the current inventory levels of specific oil and gas products at different points along the supply chain.

Developing a Solution A large mixed integer-linear program (see Chapter 10 in this book) was developed as part of the solution procedure to optimize the supply chain.

Testing the Solution All the problems that were solved by the Portuguese researchers were based on real data from Companhia Logistica Combustiveis, a large Portuguese oil and gas distribution company. Testing was done and the results were compared to real-world scenarios.

Analyzing the Results Results obtained from the model were compared to the real-world operation. The results identified opportunities for marked improvements in flow rates and reductions in pumping costs.

Implementing the Results As a result, oil and gas companies can better plan and manage their supply chains with less disruptions and lower overall costs.

Source: S. Relvas, S.N.B. Magatão, A.P.F.D. Barbosa-Póvoa, and F. Neves. “Integrated Scheduling and Inventory Management  of an Oil Products Distribution System,” Omega 41 (2013): 955–968.

MOdeLing in the reaL WOrLd Managing today’s energy supply chain

Defining the Problem

Developing a Model

Acquiring the Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

M06_REND9327_12_SE_C06.indd 208 10/02/14 1:23 PM

6.4 eConoMIC order QuantIty: deterMInIng how MuCh to order  209

The most common factors associated with ordering costs and holding costs are shown in  Table 6.1. Notice that the ordering costs are generally independent of the size of the order, and  many of these involve personnel time. An ordering cost is incurred each time an order is placed,  whether the order is for 1 unit or 1,000 units. The time to process the paperwork, pay the bill,  and so forth does not depend on the number of units ordered.

On the other hand, the holding cost varies as the size of the inventory varies. If 1,000 units  are placed into inventory, the taxes, insurance, cost of capital, and other holding costs will be  higher than if only 1 unit was put into inventory. Similarly, if the inventory level is low, there is  little chance of spoilage and obsolescence.

The cost of the items, or the purchase cost, is what is paid to acquire the inventory. The  stockout cost indicates the lost sales and goodwill (future sales) that result from not having the  items available for the customers. This is discussed later in the chapter.

6.4 Economic Order Quantity: Determining How Much to Order

The economic order quantity (EOQ) is one of the oldest and most commonly known inventory  control techniques. Research on its use dates back to a 1915 publication by Ford W. Harris. This  technique is still used by a large number of organizations today. It is relatively easy to use, but it  does make a number of assumptions. Some of the most important assumptions follow:

  1. Demand is known and constant over time.   2. The lead time—that is, the time between the placement of the order and the receipt of the 

order—is known and constant.   3. The receipt of inventory is instantaneous. In other words, the inventory from an order 

 arrives in one batch, at one point in time.   4. The purchase cost per unit is constant throughout the year. Quantity discounts are not 

possible.   5. The only variable costs are the cost of placing an order, ordering cost, and the cost of 

 holding or storing inventory over time, holding or carrying cost. The holding cost per unit  per year and the ordering cost per order are constant throughout the year.

  6. Orders are placed so that stockouts or shortages are avoided completely.

When these assumptions are not met, adjustments must be made to the EOQ model. These  are discussed later in this chapter.

With  these assumptions,  inventory usage has a sawtooth shape, as  in Figure 6.2.  In   Figure 6.2, Q represents the amount that is ordered. If this amount is 500 dresses, all 500 dresses  arrive at one time when an order is received. Thus, the inventory level jumps from 0 to 500  dresses. In general, an inventory level increases from 0 to Q units when an order arrives.

Because demand is constant over time, inventory drops at a uniform rate over time. (Refer  to the sloped lines in Figure 6.2.) Another order is placed such that when the inventory level  reaches 0, the new order is received and the inventory level again jumps to Q units, represented  by the vertical lines. This process continues indefinitely over time.

Table 6.1 Inventory Cost Factors

The inventory usage curve has a sawtooth shape.

ORDERIng COST FACTORS CARRyIng COST FACTORS

Developing and sending purchase orders Cost of capital

Processing and inspecting incoming inventory Taxes

Bill paying Insurance

Inventory inquiries Spoilage

Utilities, phone bills, and so on for the  purchasing department Theft

Salaries and wages for purchasing department employees Obsolescence

Supplies such as forms and paper for the  purchasing department Salaries and wages for warehouse employees

Utilities and building costs for the warehouse

Supplies such as forms and paper for the warehouse

M06_REND9327_12_SE_C06.indd 209 10/02/14 1:23 PM

210  Chapter 6 • Inventory Control Models

Inventory Costs in the EOQ Situation The objective of most inventory models is to minimize the total costs. With the assumptions just  given, the relevant costs are the ordering cost and the carrying, or holding cost. All other costs,  such as the cost of the inventory itself (the purchase cost), are constant. Thus, if we minimize the  sum of the ordering and carrying costs, we are also minimizing the total costs.

The annual ordering cost is simply the number of orders per year times the cost of placing  each order. Since the inventory level changes daily, it is appropriate to use the average inventory  level to determine annual holding or carrying cost. The annual carrying cost will equal the aver- age inventory times the inventory carrying cost per unit per year. Again looking at Figure 6.2, we  see that the maximum inventory is the order quantity (Q), and the average inventory will be one- half of that. Table 6.2 provides a numerical example to illustrate this. Notice that for this situation,  if the order quantity is 10, the average inventory will be 5, or one-half of Q. Thus:

Average inventory level = Q

2 (6-1)

Using the following variables, we can develop mathematical expressions for the annual  ordering  and carrying costs:

Q = number of pieces of order EOQ = Q* = optimal number of pieces to order

D = annual demand in units for the inventory item Co = ordering cost of each order Ch = holding or carrying cost per unit per year

The objective of the simple EOQ model is to minimize total inventory cost. The relevant costs are the ordering and holding costs.

The average inventory level is one-half the maximum level.

Figure 6.2 Inventory Usage over Time

0

Inventory Level

Minimum Inventory

Time

Order Quantity = Q = Maximum Inventory Level

Table 6.2 Computing Average Inventory

InvEnTORy LEvEL

DAy BEgInnIng EnDIng AvERAgE

April 1 (order received) 10 8 9

April 2 8 6 7

April 3 6 4 5

April 4 4 2 3

April 5 2 0 1

Maximum level April 1 = 10 units Total of daily averages = 9 + 7 + 5 + 3 + 1 = 25 Number of days = 5 Average inventory level = 25>5 = 5 units

M06_REND9327_12_SE_C06.indd 210 10/02/14 1:23 PM

6.4 eConoMIC order QuantIty: deterMInIng how MuCh to order  211

Annual ordering cost = 1Number of orders placed per year2 * 1Ordering cost per order2

= Annual demand

Number of units in each order * 1Ordering cost per order2

= D

Q Co

Annual holding or carrying cost = 1Average inventory2 * 1Carrying cost per unit per year2

= Order quantity

2 * 1Carrying cost per unit per year2

= Q

2 Ch

A graph of  the holding cost,  the ordering cost,  and  the  total of  these  two  is  shown  in  Figure 6.3. The lowest point on the total cost curves occurs where the ordering cost is equal to  the carrying cost. Thus, to minimize total costs given this situation, the order quantity should   occur where these two costs are equal.

global Fashion Firm Fashions inventory Management system

Founded in 1975, the Spanish retailer Zara currently has more than 1,600 stores worldwide, launches more than 10,000 new designs each year, and is recognized as one of the world’s principal fashion retailers. Goods were shipped from two central warehouses to each of the stores, based on requests from individual store managers. These local decisions inevitably led to inefficient warehouse, shipping, and logistics operations when assessed on a global scale. Recent production overruns, inefficient supply chains, and an ever-changing marketplace (to say the least) caused Zara to tackle this problem.

A variety of operations research models were used in redesigning and implementing an entirely new inventory

management system. The new centralized decision-making system replaced all store-level inventory decisions, thus providing results that were more globally optimal. Having the right products in the right places at the right time for customers has increased sales from 3% to 4% since implementation. This translated into an increase in revenue of over $230 million in 2007 and over $350 million in 2008. Talk about fashionistas!

Source: Based on F. Caro, J. Gallien, M. Díaz, J. García, J. M. Corredoira,  M. Montes, J.A. Ramos, and J. Correa. “Zara Uses Operations Research to Re- engineer Its Global Distribution Process,” Interfaces 40, 1 (January–February  2010): 71–84.

in actiOn

Figure 6.3 Total Cost as a Function of Order Quantity

Cost

Minimum Total Cost

Curve for Total Cost of Carrying and Ordering

Carrying Cost Curve

Ordering Cost Curve

Order QuantityOptimal Order Quantity

M06_REND9327_12_SE_C06.indd 211 10/02/14 1:23 PM

212  Chapter 6 • Inventory Control Models

Finding the EOQ When the EOQ assumptions are met, total cost is minimized when:

Annual holding cost = Annual ordering cost

Q

2 Ch =

D

Q Co

Solving this for Q gives the optimal order quantity:

Q2 Ch = 2DCo

Q2 = 2DCo

Ch

Q = A2DCoCh This optimal order quantity is often denoted by Q*. Thus, the economic order quantity is given  by the following formula:

EOQ = Q* = A2DCoCh This EOQ formula is the basis for many more advanced models, and some of these are discussed  later in this chapter.

Economic Order Quantity (EOQ) Model

Annual ordering cost = D

Q Co (6-2)

Annual holding cost = Q

2 Ch (6-3)

EOQ = Q* = A2DCoCh (6-4) Sumco Pump Company Example Sumco, a company that sells pump housings to other manufacturers, would like to reduce its  inventory cost by determining the optimal number of pump housings to obtain per order. The   annual demand is 1,000 units, the ordering cost is $10 per order, and the average carrying cost  per unit per year is $0.50. Using these figures, if the EOQ assumptions are met, we can calculate  the optimal number of units per order:

Q* = A2DCoCh = B 211,000211020.50 = 140,000 = 200 units

The relevant total annual inventory cost is the sum of the ordering costs and the carrying costs:

Total annual cost = Order cost + Holding cost

We derive the EOQ equation by setting ordering cost equal to carrying cost.

M06_REND9327_12_SE_C06.indd 212 10/02/14 1:23 PM

6.4 eConoMIC order QuantIty: deterMInIng how MuCh to order  213

In terms of the variables in the model, the total cost (TC) can now be expressed as

TC = D

Q Co +

Q

2 Ch (6-5)

The total annual inventory cost for Sumco is computed as follows:

TC = D

Q Co +

Q

2 Ch

= 1,000

200 1102 + 200

2 10.52

= $50 + $50 = $100

The number of orders per year 1D>Q2 is 5, and the average inventory 1Q>22 is 100. As you might expect, the ordering cost is equal to the carrying cost. You may wish to try 

 different values for Q, such as 100 or 300 pumps. You will find that the minimum total cost   occurs when Q is 200 units. The EOQ, Q*, is 200 pumps.

UsIng ExCEl QM FOr BAsIC EOQ InvEnTOry PrOBlEMs The Sumco Pump Company exam- ple, and a variety of other inventory problems we address in this chapter, can be easily solved  using Excel QM. Program 6.1A shows the input data for Sumco and the Excel formulas needed  for the EOQ model. Program 6.1B contains the solution for this example, including the optimal  order quantity, maximum inventory level, average inventory level, and the number of setups or  orders.

Purchase Cost of Inventory Items Sometimes the total inventory cost expression is written to include the actual cost of the  material  purchased. With the EOQ assumptions, the purchase cost does not depend on the particular  order  policy found to be optimal, because regardless of how many orders are placed each year, we still  incur the same annual purchase cost of D * C, where C is the purchase cost per unit and D is  the annual demand in units.*

The total annual inventory cost is equal to ordering plus holding costs for the simple EOQ model.

*Later in this chapter, we discuss the case in which price can affect order policy, that is, when quantity discounts  are offered.

Program 6.1a Input Data and Excel QM Formulas for the sumco Pump Company Example

Enter the demand rate, the ordering cost, the holding cost, and if it is available, the unit price here.

Total unit cost is given here.

M06_REND9327_12_SE_C06.indd 213 10/02/14 1:23 PM

214  Chapter 6 • Inventory Control Models

It is useful to know how to calculate the average inventory level in dollar terms when the  price per unit is given. This can be done as follows. With the variable Q representing the quantity  of units ordered, and assuming a unit cost of C, we can determine the average dollar value of  inventory:

Average dollar level = 1CQ2

2 (6-6)

This formula is analogous to Equation 6-1. Inventory carrying costs for many businesses and industries are also often expressed as an 

annual percentage of the unit cost or price. When this is the case, a new variable is introduced.  Let I be the annual inventory holding charge as a percent of unit price or cost. Then the cost of  storing one unit of inventory for the year, Ch, is given by Ch = IC, where C is the unit price or  cost of an inventory item. Q* can be expressed, in this case, as

Q* = B 2DCoIC (6-7) Sensitivity Analysis with the EOQ Model The EOQ model assumes that all input values are fixed and known with certainty. However,  since these values are often estimated or may change over time, it is important to understand  how the order quantity might change if different input values are used. Determining the effects  of these changes is called sensitivity analysis.

The EOQ formula is given as follows:

EOQ = B 2DCoCh Because of the square root in the formula, any changes in the inputs 1D, Co, Ch2 will result in  relatively minor changes in the optimal order quantity. For example, if Co were to increase by  a factor of 4, the EOQ would only increase by a factor of 2. Consider the Sumco example just  presented. The EOQ for this company is as follows:

EOQ = B 211,000211020.50 = 200

I is the annual carrying cost as a percentage of the cost per unit.

Total cost including holding, ordering, and, if it is given, unit price is computed here.

Program 6.1b Excel QM solution for the sumco Pump Company Example

M06_REND9327_12_SE_C06.indd 214 10/02/14 1:23 PM

6.5 reorder poInt: deterMInIng when to order  215

If we increased Co from $10 to $40,

EOQ = B 211,000214020.50 = 400 In general, the EOQ changes by the square root of a change in any of the inputs.

6.5 Reorder Point: Determining When to Order

Now that we have decided how much to order, we look at the second inventory question: when  to order. The time between the placing and receipt of an order, called the lead time or delivery  time, is often a few days or even a few weeks. Inventory must be available to meet the demand  during this time, and this inventory can either be on hand now or on order but not yet received.  The total of these is called the inventory position. Thus, the when to order decision is usually  expressed in terms of a reorder point (ROP), the inventory position at which an order should be  placed. The ROP is given as

ROP = 1Demand per day2 * 1Lead time for a new order in days2 = d * L (6-8)

Figure 6.4 has two graphs showing the ROP. One of these has a relatively small lead time, while  the other has a relatively large lead time. When the inventory position reaches the ROP, a new  order should be placed. While waiting for that order to arrive, the demand will be met with either 

The reorder point (ROP) determines when to order inventory. It is found by multiplying the daily demand times the lead time in days.

Figure 6.4 reorder Point graphs

0

Inventory Level

ROP

Lead time = L ROP < Q

Q

Time

0

Inventory Level

On hand

Lead time = L ROP > Q

Q

Time

On Order

M06_REND9327_12_SE_C06.indd 215 10/02/14 1:23 PM

216  Chapter 6 • Inventory Control Models

inventory currently on hand or with inventory that already has been ordered but will arrive when  the on-hand inventory falls to zero. Let’s look at an example.

PrOCOMP’s COMPUTEr ChIP ExAMPlE Procomp’s demand for computer chips is 8,000 per  year. The firm has a daily demand of 40 units, and the order quantity is 400 units. Delivery of an  order takes three working days. The reorder point for chips is calculated as follows:

ROP = d * L = 40 units per day * 3 days = 120 units

Hence, when the inventory stock of chips drops to 120, an order should be placed. The order  will arrive three days later, just as the firm’s stock is depleted to 0. Since the order quantity is  400 units, the ROP is simply the on-hand inventory. This is the situation in the first graph in  Figure 6.4.

Suppose the lead time for Procomp Computer Chips was 12 days instead of 3 days. The  reorder point would be:

ROP = 40 units per day * 12 days = 480 units

Since the maximum on-hand inventory level is the order quantity of 400, an inventory position  of 480 would be:

Inventory position = 1Inventory on hand2 + 1Inventory on order2 480 = 80 + 400

Thus, a new order would have to be placed when the on-hand inventory fell to 80 while  there was one other order in-transit. The second graph in Figure 6.4 illustrates this type of  situation.

6.6 EOQ Without the Instantaneous Receipt Assumption

When a firm receives its inventory over a period of time, a new model is needed that does not   require the instantaneous inventory receipt assumption. This new model is applicable when   inventory continuously flows or builds up over a period of time after an order has been placed  or when units are produced and sold simultaneously. Under these circumstances, the daily   demand rate must be taken into account. Figure 6.5 shows inventory levels as a function of time.   Because this model is especially suited to the production environment, it is commonly called the  production run model.

In this model, instead of having an ordering cost, there will be a setup cost. This is the cost  of setting up the production facility to manufacture the desired product. It normally includes the  salaries and wages of employees who are responsible for setting up the equipment, engineering  and design costs of making the setup, paperwork, supplies, utilities, and so on. The carrying  cost per unit is composed of the same factors as the traditional EOQ model, although the annual  c arrying cost equation changes due to a change in average inventory.

The production run model eliminates the instantaneous receipt assumption.

Figure 6.5 Inventory Control and the Production Process

Inventory Level

Time

Maximum Inventory

t

Part of Inventory Cycle During Which Production Is Taking Place

There Is No Production During This Part of the Inventory Cycle

M06_REND9327_12_SE_C06.indd 216 10/02/14 1:23 PM

6.6 eoQ wIthout the Instantaneous reCeIpt assuMptIon  217

The optimal production quantity can be derived by setting setup costs equal to holding  or carrying costs and solving for the order quantity. Let’s start by developing the expression  for carrying cost. You should note, however, that making setup cost equal to carrying cost  does not always guarantee optimal solutions for models more complex than the production  run model.

Annual Carrying Cost for Production Run Model As with the EOQ model, the carrying costs of the production run model are based on the average  inventory, and the average inventory is one-half the maximum inventory level. However, since  the replenishment of inventory occurs over a period of time and demand continues during this  time, the maximum inventory will be less than the order quantity Q. We can develop the annual  carrying, or holding, cost expression using the following variables:

Q = number of pieces per order, or production run Cs = setup cost Ch = holding or carrying cost per unit per year p = daily production rate d = daily demand rate t = length of production run in days

The maximum inventory level is as follows:

1Total produced during the production run2 - 1Total used during production run2 = 1Daily production rate21Number of days of production2

- 1Daily demand21Number of days of production2 = 1pt2 - 1dt2

Since

total produced = Q = pt,

we know that

t = Q p

Maximum inventory level = pt - dt = p Q p

- d Q p

= Q a1 - d p b

Since the average inventory is one-half of the maximum, we have

Average inventory = Q

2 a1 - d

p b (6-9)

and

Annual holding cost = Q

2 a1 - d

p bCh (6-10)

Annual Setup Cost or Annual Ordering Cost When a product is produced over time, setup cost replaces ordering cost. Both of these are inde- pendent of the size of the order and the size of the production run. This cost is simply the num- ber of orders (or production runs) times the ordering cost (setup cost). Thus,

Annual setup cost = D

Q Cs (6-11)

and

Annual ordering cost = D

Q Co (6-12)

Solving the production run model involves setting setup costs equal to holding costs and solving for Q.

The maximum inventory level in the production model is less than Q.

M06_REND9327_12_SE_C06.indd 217 10/02/14 1:24 PM

218  Chapter 6 • Inventory Control Models

Determining the Optimal Production Quantity When the assumptions of the production run model are met, costs are minimized when the setup  cost equals the holding cost. We can find the optimal quantity by setting these costs equal and  solving for Q. Thus,

Annual holding cost = Annual setup cost

Q

2 a1 - d

p bCh = DQ Cs

Solving this for Q, we get the optimal production quantity 1Q*2:

Q* = H 2DCsCh a1 - dp b (6-13) It should be noted that if the situation does not involve production but rather involves the receipt  of inventory gradually over a period of time, this same model is appropriate, but Co replaces Cs in the formula.

Production run Model

Annual holding cost = Q

2 a1 - d

p bCh

Annual setup cost = D

Q Cs

Optimal production quantity Q* = H 2DCsCh a1 - dp b Brown Manufacturing Example Brown Manufacturing produces commercial refrigeration units in batches. The firm’s estimated  demand for the year is 10,000 units. It costs about $100 to set up the manufacturing process, and  the carrying cost is about 50 cents per unit per year. When the production process has been set  up, 80 refrigeration units can be manufactured daily. The demand during the production period  has traditionally been 60 units each day. Brown operates its refrigeration unit production area 167  days per year. How many refrigeration units should Brown Manufacturing produce in each batch?  How long should the production part of the cycle shown in Figure 6.5 last? Here is the solution:

Annual demand = D = 10,000 units Setup cost = Cs = $100

Carrying cost = Ch = $0.50 per unit per day Daily production rate = p = 80 units daily

Daily demand rate = d = 60 units daily

  1.  Q* = R 2DCsCh a1 - dp b   2.  Q* = R2 * 10,000 * 1000.5 a1 - 6080 b

= B 2,000,0000.511>42 = 116,000,000 = 4,000 units

Here is the formula for the optimal production quantity. Notice the similarity to the basic EOQ model.

M06_REND9327_12_SE_C06.indd 218 10/02/14 1:24 PM

6.6 eoQ wIthout the Instantaneous reCeIpt assuMptIon  219

If Q* = 4,000 units and we know that 80 units can be produced daily, the length of each pro- duction cycle will be Q>p = 4,000>80 = 50 days. Thus, when Brown decides to produce re- frigeration units, the equipment will be set up to manufacture the units for a 50-day time span.  The number of production runs per year will be D>Q = 10,000>4,000 = 2.5. This means that  the average number of production runs per year is 2.5. There will be three production runs in  one year with some inventory carried to the next year, so only two production runs are needed  in the second year.

UsIng ExCEl QM FOr PrODUCTIOn rUn MODEls The Brown Manufacturing production run  model can also be solved using Excel QM. Program 6.2A contains the input data and the  Excel  formulas for this problem. Program 6.2B provides the solution results, including the optimal   production quantity, maximum inventory level, average inventory level, and the number of  setups.

Program 6.2a Excel QM Formulas and Input Data for the Brown Manufacturing Problem

Enter the demand rate, setup cost, and holding cost. Notice that the holding cost is a xed dollar amount rather than a percentage of the unit price.

Enter the daily production rate and the daily demand rate.

Calculates the optimal production quantity.

Calculates the maximum inventory level. Calculates the average number of setups.

Calculates the annual holding costs and setup costs.

Program 6.2b Excel QM solutions for the Brown Manufacturing Problem

M06_REND9327_12_SE_C06.indd 219 10/02/14 1:24 PM

220  Chapter 6 • Inventory Control Models

6.7 Quantity Discount Models

In developing the EOQ model, we assumed that quantity discounts were not available. However,  many companies do offer quantity discounts. If such a discount is possible, and all of the other  EOQ assumptions are met, it is possible to find the quantity that minimizes the total inventory  cost by using the EOQ model and making some adjustments.

When quantity discounts are available, the purchase cost or material cost becomes a  relevant  cost, as it changes based on the order quantity. The total relevant costs are as follows:

Total cost = Material cost + Ordering cost + Carrying cost

Total cost = DC + D

Q Co +

Q

2 Ch (6-14)

where

D = annual demand in units Co = ordering cost of each order C = cost per unit

Ch = holding or carrying cost per unit per year

Since holding cost per unit per year is based on the cost of the items, it is convenient to express  this as

Ch = IC

where

I = holding cost as a percentage of the unit cost 1C2 For a specific purchase cost (C), given the assumptions we have made, ordering the EOQ will  minimize total inventory costs. However, in the discount situation, this quantity may not be large  enough to qualify for the discount, so we must also consider ordering this minimum quantity for  the discount. A typical quantity discount schedule is shown in Table 6.3.

As can be seen in the table, the normal cost for the item is $5. When 1,000 to 1,999 units are  ordered at one time, the cost per unit drops to $4.80, and when the quantity ordered at one time  is 2,000 units or more, the cost is $4.75 per unit. As always, management must decide when and  how much to order. But with quantity discounts, how does the manager make these decisions?

Fortune 100 Firm improves inventory policy for service vehicles

Most manufacturers of home appliances provide in-home repair of the appliances that are under warranty. One such For- tune 100 firm had about 70,000 different parts that were used in the repair of its appliances. The annual value of the parts in- ventory was over $7 million. The company had more than 1,300 service vehicles that were dispatched when service requests were received. Due to limited space on these vehicles, only about 400 parts were typically carried on each one. If a service person ar- rived to repair an appliance and did not have necessary part (i.e., a stockout occured), a special order was made to have the order delivered by air so that the repair person could return and fix the appliance as soon as possible.

Deciding which parts to carry was a particularly difficult prob- lem. A project was begun to find a better way to forecast the demand for parts and identify which parts should be stocked on each vehicle. Initially, the intent was to reduce the parts inventory

on each truck as this represented a significant cost of holding the inventory. However, upon further analysis, it was decided that the goal should be to minimize the overall cost—including the costs of special deliveries of parts, revisiting the customer for the repair if the part was not initially available, and overall cus- tomer satisfaction.

The project team improved the forecasting system used to project the number of parts needed on each vehicle. As a result, the actual number of parts carried on each vehicle increased. However, the number of first-visit repairs increased from 86% to 90%. This resulted in a savings of $3 million per year in the cost of these repairs. It also improved customer satisfaction because the problem was fixed without the service person having to return a second time.

Source: Based on Michael F. Gorman and Sanjay Ahire. “A Major Appliance  Manufacturer Rethinks Its Inventory Policies for Service Vehicles,” Interfaces 36, 5 (September–October 2006): 407–419.

in actiOn

The overall objective of the quantity discount model is to minimize total inventory costs, which now include actual material costs.

M06_REND9327_12_SE_C06.indd 220 10/02/14 1:24 PM

6.7 QuantIty dIsCount Models  221

As with other inventory models discussed so far, the overall objective will be to minimize  the total cost. Because the unit cost for the third discount in Table 6.3 is lowest, you might be  tempted to order 2,000 units or more to take advantage of the lower material cost. Placing an  order for that quantity with the greatest discount cost, however, might not minimize the total  inventory cost. As the discount quantity goes up, the material cost goes down, but the carrying  cost increases because the orders are large. Thus, the major trade-off when considering quantity  discounts is between the reduced material cost and the increased carrying cost.

Figure 6.6 provides a graphical representation of the total cost for this situation. Notice  the cost curve drops considerably when the order quantity reaches the minimum for each dis- count. With the specific costs in this example, we see that the EOQ for the second price category  11,000 … Q … 1,9992 is less than 1,000 units. Although the total cost for this EOQ is less than  the total cost for the EOQ with the cost in category 1, the EOQ is not large enough to obtain this  discount. Therefore, the lowest possible total cost for this discount price occurs at the minimum  quantity required to obtain the discount 1Q = 1,0002. The process for determining the mini- mum cost quantity in this situation is summarized in the following box.

Quantity Discount Model

  1. For each discount price (C), compute EOQ = B 2DCoIC .   2.  If EOQ 6 Minimum for discount, adjust the quantity to Q = Minimum for discount.

  3. For each EOQ or adjusted Q, compute Total cost = DC + D

Q Co +

Q

2 Ch.

  4. Choose the lowest-cost quantity.

Table 6.3 Quantity Discount schedule

DISCOunT nuMBER

DISCOunT QuAnTITy

DISCOunT (%)

DISCOunT COST ($)

1 0 to 999 0 5.00

2 1,000 to 1,999 4 4.80

3 2,000 and over 5 4.75

Figure 6.6 Total Cost Curve for the Quantity Discount Model

Total Cost

$ TC Curve for Discount 1

TC Curve for Discount 2

EOQ for Discount 2

TC Curve for Discount 3

0 1,000 2,000

Order Quantity

M06_REND9327_12_SE_C06.indd 221 10/02/14 1:24 PM

222  Chapter 6 • Inventory Control Models

Brass Department Store Example Let’s see how this procedure can be applied by showing an example. Brass Department Store  stocks toy race cars. Recently, the store was given a quantity discount schedule for the cars; this  quantity discount schedule is shown in Table 6.3. Thus, the normal cost for the toy race cars is $5.  For orders between 1,000 and 1,999 units, the unit cost is $4.80, and for orders of 2,000 or more  units, the unit cost is $4.75. Furthermore, the ordering cost is $49 per order, the annual demand  is 5,000 race cars, and the inventory carrying charge as a percentage of cost, I, is 20% or 0.2.  What order quantity will minimize the total inventory cost?

The first step is to compute EOQ for every discount in Table 6.3. This is done as follows:

EOQ1 = B 12215,0002149210.2215.002 = 700 cars per order EOQ2 = B 12215,0002149210.2214.802 = 714 cars per order EOQ3 = B 12215,0002149210.2214.752 = 718 cars per order

The second step is to adjust those quantities that are below the allowable discount range. Since  EOQ1 is between 0 and 999, it does not have to be adjusted. EOQ2 is below the allowable range  of 1,000 to 1,999, and therefore, it must be adjusted to 1,000 units. The same is true for EOQ3; it  must be adjusted to 2,000 units. After this step, the following order quantities must be tested in  the total cost equation:

Q1 = 700 Q2 = 1,000 Q3 = 2,000

The third step is to use Equation 6-14 and compute a total cost for each of the order quantities.  This is accomplished with the aid of Table 6.4.

Table 6.4 Total Cost Computations for Brass Department store

EOQ values are computed.

EOQ values are adjusted.

The total cost is computed.

intel improves inventory Operations

Faced with skyrocketing distribution costs, Intel turned to quantitative analysis to design, construct, and implement a multi- echelon inventory optimization (MEIO) system. The system spanned all of Intel’s global supply chain and included suppliers, third party logistics (3PL) providers, dealers, and retailers among others. Because of the inherent complexities, designing and con- structing the system took several years by itself.

However, as a result, Intel was able to optimize inventory levels worldwide. After one year of operation, the MEIO system

reduced global inventory levels by approximately 11 percent without sacrificing delivery schedules, service levels improved by eight percentage points, and an order of magnitude reduction in the number of costly expedites was realized.

Source: Based on B. Wieland, P. Mastrantonio, S. Willems, and K.G. Kempf.  “ Optimizing  Inventory Levels within  Intel’s Channel  Supply Demand   Operations,” Interfaces 42, 6 (November–December 2012): 517–527.

in actiOn

DISCOunT nuMBER

unIT PRICE (C)

ORDER QuAnTITy (Q)

AnnuAL MATERIAL

COST 1$ 2 = DC

AnnuAL ORDERIng

COST 1$ 2 = D Q

Co

AnnuAL CARRyIng

COST 1$ 2 = Q 2

Ch TOTAL ($)

1 $5.00 700 25,000 350.00 350.00 25,700.00

2 4.80 1,000 24,000 245.00 480.00 24,725.00

3 4.75 2,000 23,750 122.50 950.00 24,822.50

M06_REND9327_12_SE_C06.indd 222 10/02/14 1:24 PM

6.7 QuantIty dIsCount Models  223

The fourth step is to select that order quantity with the lowest total cost. Looking at  Table 6.4, you can see that an order quantity of 1,000 toy race cars minimizes the total  cost. It should be recognized, however, that the total cost for ordering 2,000 cars is only  slightly greater than the total cost for ordering 1,000 cars. Thus, if the third discount cost is  lowered to $4.65, for  example, this order quantity might be the one that minimizes the total   inventory cost.

UsIng ExCEl QM FOr QUAnTITy DIsCOUnT PrOBlEMs As seen in the previous analysis, the  quantity discount model is more complex than the inventory models discussed so far in this  chapter. Fortunately, we can use the computer to simplify the calculations. Program 6.3A shows  the Excel formulas and input data needed for Excel QM for the Brass Department Store  problem.  Program 6.3B provides the solution to this problem, including adjusted order quantities and total  costs for each price break.

Q* is selected.

Program 6.3a Excel QM Formulas and Input Data for the Brass Department store Quantity Discount Problem

Enter the quantity discount schedule of quantities and unit prices for each price break.

Computes the order quantity for each price break and adjusts them upward if necessary.

Enter the demand rate, setup cost, and holding cost.

Computes the holding and setup costs.

Computes the total cost of each price break.

Determines the optimal order quantity, Q*.

Program 6.3b Excel QM solutions for the Brass Department store Quantity Discount Problem

M06_REND9327_12_SE_C06.indd 223 10/02/14 1:24 PM

224  Chapter 6 • Inventory Control Models

6.8 Use of Safety Stock

When the EOQ assumptions are met, it is possible to schedule orders to arrive so that stockouts  are completely avoided. However, if the demand or the lead time is uncertain, the exact demand  during the lead time (which is the ROP in the EOQ situation) will not be known with certainty.  Therefore, to prevent stockouts, it is necessary to carry additional inventory called safety stock.

When demand is unusually high during the lead time, you dip into the safety stock instead  of encountering a stockout. Thus, the main purpose of safety stock is to avoid stockouts when the  demand is higher than expected. Its use is shown in Figure 6.7. Note that although stockouts can  often be avoided by using safety stock, there is still a chance that they may occur. The demand  may be so high that all the safety stock is used up, and thus there is still a stockout.

One of the best ways to implement a safety stock policy is to adjust the reorder point. In  the EOQ situation where the demand and lead time are constant, the reorder point is simply the  amount of inventory that would be used during the lead time (i.e., the daily demand times the  lead time in days). This is assumed to be known with certainty, so there is no need to place an  order when the inventory position is more than this. However, when the daily demand or the  lead time fluctuate and are uncertain, the exact amount of inventory that will be used during the  lead time is uncertain. The average inventory usage during the lead time should be computed  and some safety stock should be added to this to avoid stockouts. The reorder point becomes

ROP = 1Average demand during lead time2 + 1Safety stock2 ROP = 1Average demand during lead time2 + SS (6-15)

where

SS = safety stock

Safety stock helps in avoiding stockouts. It is extra stock kept on hand.

Figure 6.7 Use of safety stock Inventory

on Hand

Time

Inventory on Hand

Time

Safety Stock, SS

0 Units

0 Units

Stockout Is Avoided

Stockout

M06_REND9327_12_SE_C06.indd 224 10/02/14 1:24 PM

6.8 use of safety stoCk  225

How to determine the correct amount of safety stock is the only remaining question. Two  important factors in this decision are the stockout cost and the holding cost. The stockout cost  usually involves lost sales and lost goodwill, which results in loss of future sales. If holding  cost is low while stockout cost is high, a large amount of safety stock should be carried to avoid  stockouts as it costs little to carry this, while stockouts are expensive. On the other hand, if  stockout cost is low but holding cost is high, a lower amount of safety stock would be preferred,  as having a stockout would cost very little, but too much safety stock will result in much higher  annual holding costs.

How is the optimum stock level determined? If demand fluctuates while lead time is con- stant, and if both the stockout cost per unit and the holding cost per unit are known, the use of a  payoff/cost table might be considered. With only a small number of possible demand values dur- ing the lead time, a cost table could be constructed in which the different possible demand levels  would be the states of nature, and the different amounts of safety stock as the alternatives. Us- ing the techniques discussed in Chapter 3, the expected cost could be calculated for each safety  stock level, and the minimum cost solution could be found.

However, a more general approach is to determine what service level is desired and then to  find the safety stock level that would accomplish this. A prudent manager will look at the hold- ing cost and the stockout cost to help determine an appropriate service level. A service level indicates what percentage of the time customer demand is met. In other words, the service level  is the percentage of time that stockouts are avoided. Thus,

Service level = 1 - Probability of a stockout or

Probability of a stockout = 1 - Service level

Once the desired service level is established, the amount of safety stock to carry can be found  using the probability distribution of demand during the lead time.

sAFETy sTOCk wITh ThE nOrMAl DIsTrIBUTIOn Equation 6-15 provides the general formula  for determining the reorder point. When demand during the lead time is normally distributed, the  reorder point becomes

ROP = 1Average demand during lead time2 + Zsd LT (6-16) where

Z = number of standard deviations for a given service level sd LT = standard deviation of demand during the lead time

Thus, the amount of safety stock is simply Zsd LT. The following example looks at how to deter- mine the appropriate safety stock level when demand during the lead time is normally distrib- uted and the mean and standard deviation are known.

hInsDAlE COMPAny ExAMPlE The Hinsdale Company carries a variety of electronic inven- tory items, and these are typically identified by SKU. One particular item, SKU A3378, has a  demand that is normally distributed during the lead time, with a mean of 350 units and a stan- dard deviation of 10. Hinsdale wants to follow a policy that results in stockouts occurring only  5% of the time on any order. How much safety stock should be maintained and what is the reor- der point? Figure 6.8 helps visualize this example.

From the normal distribution table (Appendix A) we have Z = 1.65:

ROP = 1Average demand during lead time2 + ZsdLT = 350 + 1.651102 = 350 + 16.5 = 366.5 units 1or about 367 units2

So the reorder point is 366.5, and the safety stock is 16.5 units.

CAlCUlATIng lEAD TIME DEMAnD AnD sTAnDArD DEvIATIOn If the mean and standard de- viation of demand during the lead time are not known, they must be calculated from historical  demand and lead time data. Once these are found, Equation 6-16 can be used to find the safety 

Safety stock is included in the ROP.

M06_REND9327_12_SE_C06.indd 225 10/02/14 1:24 PM

226  Chapter 6 • Inventory Control Models

stock and reorder point. Throughout this section, we assume that lead time is in days, although  the same procedure can be applied to weeks, months, or any other time period. We will also  assume that if demand fluctuates, the distribution of demand each day is identical to and inde- pendent of demand on other days. If both daily demand and lead time fluctuate, they are also  assumed to be independent.

There are three situations to consider. In each of the following ROP formulas, the average  demand during the lead time is the first term and the safety stock 1ZsdLT2 is the second term.   1. Demand is variable but lead time is constant:

ROP = dL + Z1sd2L2 (6-17) where

d = average daily demand sd = standard deviation of daily demand L = lead time in days

  2. Demand is constant but lead time is variable:

ROP = dL + Z1dsL2 (6-18) where

L = average lead time sL = standard deviation of lead time d = daily demand

  3. Both demand and lead time are variable:

ROP = d L + Z12Ls2d + d2s2L2 (6-19) Notice that the third situation is the most general case and the others can be derived from that.  If either demand or lead time is constant, the standard deviation and variance for that would be  0, and the average would just equal the constant amount. Thus, the formula for ROP in situation  3 can be simplified to the ROP formula given for that situation.

hInsDAlE COMPAny ExAMPlE, COnTInUED Hinsdale has decided to determine the safety  stock and ROP for three other items: SKU F5402, SKU B7319, and SKU F9004.

For SKU F5402, the daily demand is normally distributed, with a mean of 15 units and a  standard deviation of 3. Lead time is exactly 4 days. Hinsdale wants to maintain a 97% service  level. What is the reorder point, and how much safety stock should be carried?

SS

5% Area of Normal Curve

m = 350 X = ?

m = Mean demand = 350

s = Standard deviation = 10

= Mean demand + Safety stockX

SS = Safety stock = X – m 5 Zs

X – m Z =

s

Figure 6.8 safety stock and the normal Distribution

M06_REND9327_12_SE_C06.indd 226 10/02/14 1:24 PM

6.8 use of safety stoCk  227

From Appendix A, for a 97% service level Z = 1.88. Since demand is variable but lead  time is constant, we find

ROP = dL + Z1sd1L2 = 15142 + 1.8813142 = 15142 + 1.88162 = 60 + 11.28 = 71.28

So the average demand during the lead time is 60, and the safety stock is 11.28 units. For SKU B7319, the daily demand is constant at 25 units per day, and the lead time is nor-

mally distributed, with a mean of 6 days and a standard deviation of 3. Hinsdale wants to main- tain a 98% service level on this particular product. What is the reorder point?

From Appendix A, for a 98% service level Z = 2.05. Since demand is constant but lead  time is variable, we find

ROP = dL + Z1dsL2 = 25162 + 2.051252132 = 150 + 2.051752 = 150 + 153.75 = 303.75

So the average demand during the lead time is 150, and the safety stock is 154.03 units. For SKU F9004, the daily demand is normally distributed, with a mean of 20 units and a 

standard deviation of 4, and the lead time is normally distributed, with a mean of 5 days and a  standard deviation of 2. Hinsdale wants to maintain a 94% service level on this particular prod- uct. What is the reorder point?

From Appendix A, for a 94% service level, Z = 1.55. Since both demand and lead time are  variable, we find

ROP = d L + Z12Ls2d + d2s2L2 = 1202152 + 1.551251422 + 1202212222 = 100 + 1.5511680 = 100 + 1.55140.992 = 100 + 63.53 = 163.53

So the average demand during the lead time is 100, and the safety stock is 63.53 units. As the service level increases, the safety stock increases at an increasing rate. Table 6.5 

illustrates how the safety stock level would change in the Hinsdale Company (SKU A3378)  example for changes in the service level. As the amount of safety stock increases, the annual  holding cost increases as well.

CAlCUlATIng AnnUAl hOlDIng COsT wITh sAFETy sTOCk When the EOQ assumptions  of constant demand and constant lead time are met, the average inventory is Q>2, and the  annual holding cost is 1Q>22Ch. When safety stock is carried because demand fluctuates, the 

A safety stock level is determined for each service level.

Table 6.5 safety stock for skU A3378 at Different service levels

SERvICE LEvEL (%)

Z vALuE FROM nORMAL CuRvE TABLE

SAFETy STOCk (unITS)

90 1.28 12.8

91 1.34 13.4

92 1.41 14.1

93 1.48 14.8

94 1.55 15.5

95 1.65 16.5

96 1.75 17.5

97 1.88 18.8

98 2.05 20.5

99 2.33 23.3

99.99 3.72 37.2

M06_REND9327_12_SE_C06.indd 227 10/02/14 1:24 PM

228  Chapter 6 • Inventory Control Models

holding cost for this safety stock is added to the holding cost of the regular inventory to get  the total annual holding cost. If demand during the lead time is normally distributed and safety  stock is used, the average inventory on the order quantity (Q) is still Q>2, but the average  amount of safety stock carried is simply the amount of the safety stock (SS) and not one-half  this amount. Since demand during the lead time is normally distributed, there would be times  when inventory usage during the lead time exceeded the expected amount and some safety  stock would be used. But it is just as likely that the inventory usage during the lead time would  be less than the expected amount and the order would arrive while some regular inventory  remained in addition to all the safety stock. Thus, on the average, the company would always  have this full amount of safety stock in inventory, and a holding cost would apply to all of this.  From this we have

Total annual holding cost = Holding cost of regular inventory + Holding cost of safety stock

THC = Q

2 Ch + 1SS2Ch (6-20)

where

THC = total annual holding cost Q = order quantity

Ch = holding cost per unit per year SS = safety stock

In the Hinsdale example for SKU A3378, let’s assume that the holding cost is $2 per unit per  year. The amount of safety stock needed to achieve various service levels is shown in Table 6.5.  The holding cost for the safety stock would be these amounts times $2 per unit. As illus- trated in  Figure 6.9, this holding cost would increase extremely rapidly once the service level  reached 98%.

UsIng ExCEl QM FOr sAFETy sTOCk PrOBlEMs To use Excel QM to determine the safety  stock and reorder point, select Excel QM from the Add-Ins tab and select Inventory—Reorder Point/Safety Stock (normal distribution). Enter a title when the input window appears and click 

Carrying cost increases at an increasing rate as the service level increases.

($) 80

70

60

50

40

30

20

90 91 92 93 94 95 96

Service Level (%)

In ve

nt or

y C

ar ry

in g

C os

ts (

$)

97 98 99 99.99 (%)

Figure 6.9 service level versus Annual Carrying Costs

M06_REND9327_12_SE_C06.indd 228 10/02/14 1:24 PM

6.9 sIngle-perIod Inventory Models  229

OK. Program 6.4A shows the input screen and formulas for the first three Hinsdale Company  examples. Program 6.4B shows the solution.

6.9 Single-Period Inventory Models

So far, we have considered inventory decisions in which demand continues in the future, and  future orders will be placed for the same product. There are some products for which a decision  to meet the demand for a single time period is made, and items that do not sell during this time  period are of no value or have a greatly reduced value in the future. For example, a daily news- paper is worthless after the next paper is available. Other examples include weekly magazines,  programs printed for athletic events, certain prepared foods that have a short life, and some 

The standard deviation of demand during the lead time is calculated here.

The standard deviation of demand during the lead time is calculated here.

The average demand and standard deviation during the lead time are entered here, if available.

If daily demand is normally distributed but lead time is constant, the data are entered here.

If daily demand and/or lead time is normally distributed, the data are entered here. If one or the other is constant, enter zero for the standard deviation, respectively.

Program 6.4a Excel QM Formulas and Input Data for the hinsdale safety stock Problem

Solution to the �rst Hinsdale example when standard deviation of demand during lead time was given.

Solution to second Hinsdale example where daily demand was normally distributed.

Solution to third Hinsdale example where daily demand was constant but lead time was normally distributed.

Program 6.4b Excel QM solutions for the hinsdale safety stock Problem

M06_REND9327_12_SE_C06.indd 229 10/02/14 1:24 PM

230  Chapter 6 • Inventory Control Models

seasonal clothes that have greatly reduced value at the end of the season. This type of problem is  often called the news vendor problem or a single-period inventory model.

For example, a large restaurant might be able to stock from 20 to 100 cartons of doughnuts  to meet a demand that ranges from 20 to 100 cartons per day. While this could be modeled us- ing a payoff table (see Chapter 3), we would have to analyze 101 possible alternatives and states  of nature, which would be quite tedious. A simpler approach for this type of decision is to use  marginal, or incremental, analysis.

A decision-making approach using marginal profit and marginal loss is called marginal analysis. Marginal profit (MP) is the additional profit achieved if one additional unit is stocked  and sold. Marginal loss (ML) is the loss that occurs when an additional unit is stocked but can- not be sold.

When there are a manageable number of alternatives and states of nature and we know the  probabilities for each state of nature, marginal analysis with discrete distributions can be used.  When there are a very large number of possible alternatives and states of nature and the prob- ability distribution of the states of nature can be described with a normal distribution, marginal  analysis with the normal distribution are appropriate.

Marginal Analysis with Discrete Distributions Finding inventory level with the lowest cost is not difficult when we follow the marginal analy- sis procedure. This approach says that we would stock an additional unit only if the expected  marginal profit for that unit equals or exceeds the expected marginal loss. This relationship is  expressed symbolically as follows:

P = probability that demand will be greater than or equal to a given supply (or the prob- ability of selling at least one additional unit)

1 - P = probability that demand will be less than supply (or the probability that one ad- ditional unit will not sell)

The expected marginal profit is found by multiplying the probability that a given unit will  be sold by the marginal profit, P1MP2. Similarly, the expected marginal loss is the probability of  not selling the unit multiplied by the marginal loss, or 11 - P21ML2.

The optimal decision rule is to stock the additional unit if

P1MP2 Ú 11 - P2ML With some basic mathematical manipulations, we can determine the level of P for which 

this relationship holds:

P1MP2 Ú ML - P1ML2 P1MP2 + P1ML2 Ú ML P1MP + ML2 Ú ML

or

P Ú ML

ML + MP (6-21)

In other words, as long as the probability of selling one more unit (P) is greater than or  equal to ML>1MP + ML2, we would stock the additional unit.

steps of Marginal Analysis with Discrete Distributions

  1. Determine the value of  ML

ML + MP  for the problem.

  2. Construct a probability table and add a cumulative probability column.   3. Keep ordering inventory as long as the probability (P) of selling at least one additional unit 

    is greater than  ML

ML + MP .

M06_REND9327_12_SE_C06.indd 230 10/02/14 1:24 PM

6.9 sIngle-perIod Inventory Models  231

Café du Donut Example Café du Donut is a popular New Orleans dining spot on the edge of the French Quarter. Its spe- cialty is coffee and doughnuts; it buys the doughnuts fresh daily from a large industrial bakery.  The café pays $4 for each carton (containing two dozen doughnuts) delivered each morning.  Any cartons not sold at the end of the day are thrown away, for they would not be fresh enough  to meet the café’s standards. If a carton of doughnuts is sold, the total revenue is $6. Hence, the  marginal profit per carton of doughnuts is

MP = Marginal profit = $6 - $4 = $2

The marginal loss is ML = $4 since the doughnuts cannot be returned or salvaged at day’s end. From past sales, the café’s manager estimates that the daily demand will follow the prob-

ability distribution shown in Table 6.6. The manager then follows the three steps to find the opti- mal number of cartons of doughnuts to order each day.

Step 1. Determine the value of  ML

ML + MP  for the decision rule

P Ú ML

ML + MP =

$4

$4 + $2 =

4

6 = 0.67

P Ú 0.67

So the inventory stocking decision rule is to stock an additional unit if P Ú 0.67.

Step 2. Add a new column to the table to reflect the probability that doughnut sales will be  at each level or greater. This is shown in the right-hand column of Table 6.7. For example,  the probability that demand will be 4 cartons or greater is 1.00 1= 0.05 + 0.15 + 0.15 + 0.20 + 0.25 + 0.10 + 0.102. Similarly, the probability that sales will be 8 cartons or greater is  0.45 1= 0.25 + 0.10 + 0.102: namely, the sum of probabilities for sales of 8, 9, or 10 cartons.

DAILy SALES (CARTOnS OF DOughnuTS)

PROBABILITy (P) ThAT DEMAnD WILL BE AT ThIS LEvEL

4 0.05

5 0.15

6 0.15

7 0.20

8 0.25

9 0.10

10 0.10

Total 1.00

Table 6.6 Café du Donut’s Probability Distribution

Table 6.7 Marginal Analysis for Café du Donut

DAILy SALES (CARTOnS OF DOughnuTS)

PROBABILITy (P) ThAT DEMAnD WILL BE AT

ThIS LEvEL

PROBABILITy (P) ThAT DEMAnD WILL BE AT ThIS

LEvEL OR gREATER

4 0.05 1.00 Ú 0.66 5 0.15 0.95 Ú 0.66 6 0.15 0.80 Ú 0.66 7 0.20 0.65

8 0.25 0.45

9 0.10 0.20

10 0.10 0.10

Total 1.00

M06_REND9327_12_SE_C06.indd 231 10/02/14 1:24 PM

232  Chapter 6 • Inventory Control Models

Step 3. Keep ordering additional cartons as long as the probability of selling at least one   additional carton is greater than P, which is the indifference probability. If Café du Donut orders  6 cartons, marginal profits will still be greater than marginal loss since

P at 6 cartons = 0.80 7 0.67

Marginal Analysis with the Normal Distribution When product demand or sales follow a normal distribution, which is a common business situ- ation, marginal analysis with the normal distribution can be applied. First, we need to find four  values:

  1. The average or mean sales for the product, m   2. The standard deviation of sales, s   3. The marginal profit for the product, MP   4. The marginal loss for the product, ML

Once these quantities are known, the process of finding the best stocking policy is somewhat  similar to marginal analysis with discrete distributions. We let X* = optimal stocking level.

steps of Marginal Analysis with the normal Distribution

  1. Determine the value of  ML

ML + MP  for the problem.

  2. Locate P on the normal distribution (Appendix A) and find the associated Z-value.   3. Find X* using the relationship

Z = X* - m

s

    to solve for the resulting stocking policy:

X* = m + Zs

Newspaper Example Demand for copies of the Chicago Tribune newspaper at Joe’s Newsstand is normally distrib- uted and has averaged 60 papers per day, with a standard deviation of 10 papers. With a marginal  loss of 20 cents and a marginal profit of 30 cents, what daily stocking policy should Joe follow?

Step 1. Joe should stock the Chicago Tribune as long as the probability of selling the last unit  is at least ML>1ML + MP2:

ML

ML + MP =

20 cents

20 cents + 30 cents =

20

50 = 0.40

Let P = 0.40.

Step 2. Figure 6.10 shows the normal distribution. Since the normal table has cumulative areas  under the curve between the left side and any point, we look for 0.60 1= 1.0 - 0.402 in order  to get the corresponding Z value:

Z = 0.25 standard deviations from the mean

Step 3. In this problem, m = 60 and s = 10, so

0.25 = X* - 60

10

or

X* = 60 + 0.251102 = 62.5, or 62 newspapers. Thus, Joe should order 62 copies of the Chicago Tribune daily since the probability of selling 63  is slightly less than 0.40.

M06_REND9327_12_SE_C06.indd 232 10/02/14 1:24 PM

6.9 sIngle-perIod Inventory Models  233

When P is greater than 0.50, the same basic procedure is used, although caution should  be used when looking up the Z value. Let’s say that Joe’s Newsstand also stocks the Chicago Sun-Times, and its marginal loss is 40 cents and marginal profit is 10 cents. The daily sales have  averaged 100 copies of the Chicago Sun-Times, with a standard deviation of 10 papers. The   optimal stocking policy is as follows:

ML

ML + MP =

40 cents

40 cents + 10 cents =

40

50 = 0.80

The normal curve is shown in Figure 6.11. Since the normal curve is symmetrical, we find  Z for an area under the curve of 0.80 and multiply this number by -1 since all values below the  mean are associated with a negative Z value:

Z = -0.84 standard deviations from the mean for an area of 0.80

With m = 100 and s = 10,

-0.84 = X* - 100

10

or

X* = 100 - 0.841102 = 91.6, or 91 newspapers. So Joe should order 91 copies of the Chicago Sun-Times every day.

The optimal stocking policies in these two examples are intuitively consistent. When mar- ginal profit is greater than marginal loss, we would expect X* to be greater than the average  demand, m and when marginal profit is less than marginal loss, we would expect the optimal  stocking policy, X*, to be less than m.

Mean Daily Sales

Area under the Curve Is 0.40

Area under the Curve is 1 – 0.40 = 0.60 (Z = 0.25)

X*

Optimal Stocking Policy (62 Newspapers)

m 5 60 X = Demand

Figure 6.10 Joe’s stocking Decision for the Chicago Tribune

Area under the Curve is 0.80 (Z = –0.84)

X*

Optimal Stocking Policy (91 Newspapers)

m 5 100 X = Demand

Figure 6.11 Joe’s stocking Decision for the Chicago Sun-Times

M06_REND9327_12_SE_C06.indd 233 10/02/14 1:24 PM

234  Chapter 6 • Inventory Control Models

6.10 ABC Analysis

Earlier, we showed how to develop inventory policies using quantitative techniques. There are  also some very practical considerations that should be incorporated into the implementation of  inventory decisions, such as ABC analysis.

The purpose of ABC analysis is to divide all of a company’s inventory items into three  groups (group A, group B, and group C) based on the overall inventory value of the items. A  prudent manager should spend more time managing those items representing the greatest dol- lar inventory cost because this is where the greatest potential savings are. A brief description of  each group follows, with general guidelines as to how to categorize items.

The inventory items in the A group account for a major portion of the inventory costs of  the organization. As a result, their inventory levels must be monitored carefully. These items  typically make up more than 70% of the company’s business in dollars, but may consist of only  10% of all inventory items. In other words, a few inventory items are very costly to the company.  Thus, great care should be taken in forecasting the demand and developing good inventory man- agement policies for this group of items (refer to Table 6.8). Since there are relatively few of  these, the time involved would not be excessive.

The items in the B group are typically moderately priced items and represent much less  investment than the A items. Thus, it may not be appropriate to spend as much time develop- ing optimal inventory policies for this group as with the A group since these inventory costs are  much lower. Typically, the group B items represent about 20% of the company’s business in dol- lars, and about 20% of the items in inventory.

The items in the C group are the very low-cost items that represent very little in terms of the  total dollars invested in inventory. These items may constitute only 10% of the company’s busi- ness in dollars, but they may consist of 70% of the items in inventory. From a cost–benefit per- spective, it would not be good to spend as much time managing these items as the A and B items.

For the group C items, the company should develop a very simple inventory policy, and this  may include a relatively large safety stock. Since the items cost very little, the holding cost asso- ciated with a large safety stock will also be very low. More care should be taken in determining  the safety stock with the higher priced group B items. For the very expensive group A items, the  cost of carrying the inventory is so high, it is beneficial to carefully analyze the demand for these  so that safety stock is at an appropriate level. Otherwise, the company may have exceedingly  high holding costs for the group A items.

6.11 Dependent Demand: The Case for Material Requirements Planning

In all the inventory models discussed earlier, we assume that the demand for one item is inde- pendent of the demand for other items. For example, the demand for refrigerators is usually in- dependent of the demand for toaster ovens. Many inventory problems, however, are interrelated;  the demand for one item is dependent on the demand for another item. Consider a manufacturer  of small power lawn mowers. The demand for lawn mower wheels and spark plugs is dependent  on the demand for lawn mowers. Four wheels and one spark plug are needed for each finished  lawn mower. Usually when the demand for different items is dependent, the relationship be- tween the items is known and constant. Thus, you should forecast the demand for the final prod- ucts and compute the requirements for component parts.

As with the inventory models discussed previously, the major questions that must be  answered are how much to order and when to order. But with dependent demand, inventory 

InvEnTORy gROuP

DOLLAR uSAgE (%)

InvEnTORy ITEMS (%)

ARE QuAnTITATIvE COnTROL TEChnIQuES uSED?

A 70 10 Yes

B 20 20 In some cases

C 10 70 No

Table 6.8 summary of ABC Analysis

M06_REND9327_12_SE_C06.indd 234 10/02/14 1:24 PM

6.11 dependent deMand: the Case for MaterIal reQuIreMents plannIng  235

scheduling and planning can be very complex indeed. In these situations, materials require- ments palnning (MRP) can be employed effectively. Some of the benefits of MRP follow:

  1.  Increased customer service and satisfaction   2. Reduced inventory costs   3. Better inventory planning and scheduling   4. Higher total sales   5. Faster response to market changes and shifts   6. Reduced inventory levels without reduced customer service

Although most MRP systems are computerized, the analysis is straightforward and similar  from one computerized system to the next. Here is the typical procedure.

Material Structure Tree We begin by developing a bill of materials (BOM). The BOM identifies the components, their  descriptions, and the number required in the production of one unit of the final product. From  the BOM, we develop a material structure tree. Let’s say that demand for product A is 50 units.  Each unit of A requires 2 units of B and 3 units of C. Now, each unit of B requires 2 units of  D and 3 units of E. Furthermore, each unit of C requires 1 unit of E and 2 units of F. Thus, the  demand for B, C, D, E, and F is completely dependent on the demand for A. Given this informa- tion, a material structure tree can be developed for the related inventory items (see Figure 6.12).

The structure tree has three levels: 0, 1, and 2. Items above any level are called parents, and  items below any level are called components. There are three parents: A, B, and C. Each parent  item has at least one level below it. Items B, C, D, E, and F are components because each item  has at least one level above it. In this structure tree, B and C are both parents and components.

Note that the number in the parentheses in Figure 6.12 indicates how many units of that  particular item are needed to make the item immediately above it. Thus, B(2) means that it takes  2 units of B for every unit of A, and F(2) means that it takes 2 units of F for every unit of C.

After the material structure tree has been developed, the number of units of each item re- quired to satisfy demand can be determined. This information can be displayed as follows:

Part B:  2 * number of A>s = 2 * 50 = 100 Part C:  3 * number of A>s = 3 * 50 = 150 Part D:  2 * number of B>s = 2 * 100 = 200 Part E:  3 * number of B>s + 1 * number of C>s = 3 * 100 + 1 * 150 = 450 Part F:  2 * number of C>s = 2 * 150 = 300

Thus, for 50 units of A we need 100 units of B, 150 units of C, 200 units of D, 450 units  of E, and 300 units of F. Of course, the numbers in this table could have been determined di- rectly from the material structure tree by multiplying the numbers along the branches times the 

B(2) C(3)

Material Structure Tree for Item A

D(2) E(3) E(1) F(2)

Level 0

1

2

A

Figure 6.12 Material structure Tree for Item A

Parents and components are identified in the material structure tree.

The material structure tree shows how many units are needed at every level of production.

M06_REND9327_12_SE_C06.indd 235 10/02/14 1:24 PM

236  Chapter 6 • Inventory Control Models

demand for A, which is 50 units for this problem. For example, the number of units of D needed  is simply 2 * 2 * 50 = 200 units.

Gross and Net Material Requirements Plan Once the materials structure tree has been developed, we construct a gross material requirements  plan. This is a time schedule that shows when an item must be ordered from suppliers when there  is no inventory on hand, or when the production of an item must be started in order to satisfy the  demand for the finished product at a particular date. Let’s assume that all of the items are pro- duced or manufactured by the same company. It takes one week to make A, two weeks to make B,  one week to make C, one week to make D, two weeks to make E, and three weeks to make F. With  this information, the gross material requirements plan can be constructed to reveal the production  schedule needed to satisfy the demand of 50 units of A at a future date. (Refer to Figure 6.13.)

The interpretation of the material in Figure 6.13 is as follows: If you want 50 units of A at  week 6, you must start the manufacturing process in week 5. Thus, in week 5 you need 100 units  of B and 150 units of C. These two items take 2 weeks and 1 week to produce. (See the lead  times.) Production of B should be started in week 3, and C should be started in week 4. (See the  order release for these items.) Working backward, the same computations can be made for all  the other items. The material requirements plan graphically reveals when each item should be  started and completed in order to have 50 units of A at week 6. Now, a net requirements plan can  be developed given the on-hand inventory in Table 6.9; here is how it is done.

Lead Time � 1 Week

Lead Time � 2 Weeks

Lead Time � 1 Week

Lead Time � 1 Week

Lead Time � 2 Weeks

Lead Time � 3 Weeks

50

50

100

100

150

150

200

200

300 150

300 150

300

300

A

B

C

D

E

F

Week

1 2 4 5 6

Required Date

Order Release

Every part A requires 2 part B’s in week 5

Required Date

Order Release

Required Date

Order Release

Required Date

Order Release

Required Date

Order Release

Required Date

Order Release

3

Figure 6.13 gross Material requirements Plan for 50 Units of A

ITEM On-hAnD InvEnTORy

A 10

B 15

C 20

D 10

E 10

F 5

Table 6.9 On-hand Inventory

Using on-hand inventory to compute net requirements.

M06_REND9327_12_SE_C06.indd 236 10/02/14 1:24 PM

6.11 dependent deMand: the Case for MaterIal reQuIreMents plannIng  237

Using these data, we can develop a net material requirements plan that includes gross re- quirements, on-hand inventory, net requirements, planned-order receipts, and planned-order re- leases for each item. It is developed by beginning with A and working backward through the  other items. Figure 6.14 shows a net material requirements plan for product A.

The net requirements plan is constructed like the gross requirements plan. Starting with  item A, we work backward determining net requirements for all items. These computations are  done by referring constantly to the structure tree and lead times. The gross requirements for A  are 50 units in week 6. Ten items are on hand, and thus the net requirements and planned-order  receipt are both 40 items in week 6. Because of the one-week lead time, the planned-order re- lease is 40 items in week 5. (See the arrow connecting the order receipt and order release.) Look  down column 5 and refer to the structure tree in Figure 6.13. Eighty 12 * 402 items of B and  120 = 3 * 40 items of C are required in week 5 in order to have a total of 50 items of A in  week 6. The letter A in the upper-right corner for items B and C means that this demand for B  and C was generated as a result of the demand for the parent, A. Now the same type of analysis  is done for B and C to determine the net requirements for D, E, and F.

Two or More End Products So far, we have considered only one end product. For most manufacturing companies, there are  normally two or more end products that use some of the same parts or components. All of the  end products must be incorporated into a single net material requirements plan.

In the MRP example just discussed, we developed a net material requirements plan for  product A. Now, we show how to modify the net material requirements plan when a second 

Item 1 2 3 4 5 6 Lead Time

Week

A

B

C

D

E

F

Gross On-Hand 10 Net Order Receipt Order Release

Gross On-Hand 15 Net Order Receipt Order Release

Gross On-Hand 20 Net Order Receipt Order Release

Gross On-Hand 10 Net Order Receipt Order Release

Gross On-Hand 10 Net Order Receipt Order Release

Gross On-Hand 5 Net Order Receipt Order Release

1

2

1

1

2

3

50 10 40 40

40

80 15 65 65

65

100

120

100185

195

120 20

100 100

195 10

185 185

100 0

100 100

200 5

195 195

A

A

B C

C

130 10

120 120

B

Figure 6.14 net Material requirements Plan for 50 Units of A

M06_REND9327_12_SE_C06.indd 237 10/02/14 1:24 PM

238  Chapter 6 • Inventory Control Models

end product is introduced. Let’s call the second end product AA. The material structure tree for  product AA is as follows:

AA

D(3) F(2)

Let’s assume that we need 10 units of AA. With this information we can compute the gross   requirements for AA:

Part D:  3 * number of AA>s = 3 * 10 = 30 Part F:  2 * number of AA>s = 2 * 10 = 20

To develop a net material requirements plan, we need to know the lead time for AA. Let’s  assume that it is one week. We also assume that we need 10 units of AA in week 6 and that we  have no units of AA on hand.

Now, we are in a position to modify the net material requirements plan for product A to in- clude AA. This is done in Figure 6.15.

Item 1 2 3 4 5 6 Lead Time

Week

AA

A

C

D

E

F

Gross On-Hand: 0 Net Order Receipt Order Release

Gross On-Hand: 10 Net Order Receipt Order Release

Gross On-Hand: 20 Net Order Receipt Order Release

Gross On-Hand: 10 Net Order Receipt Order Release

Gross On-Hand: 10 Net Order Receipt Order Release

Gross On-Hand: 5 Net Order Receipt Order Release

B Gross On-Hand: 15 Net Order Receipt Order Release

Inventory

10 0

10 10

1 Week

1 Week

1 Week

1 Week

2 Weeks

3 Weeks

10

50 10 40 40

40

100

120

100185

195

120 20

100 100

130 10

120 120

195 10

185 185

100 0

100 100

200 5

195 195

2 Weeks

65

80 15 65 65

30

30 0

30 30

20

20 0

20 20

C AA

AA

A

A

B

B C

Figure 6.15 net Material requirements Plan, Including AA

M06_REND9327_12_SE_C06.indd 238 10/02/14 1:24 PM

6.12 Just-In-tIMe Inventory Control  239

Look at the top row of the figure. As you can see, we have a gross requirement of 10 units  of AA in week 6. We don’t have any units of AA on hand, so the net requirement is also 10 units  of AA. Because it takes one week to make AA, the order release of 10 units of AA is in week 5.  This means that we start making AA in week 5 and have the finished units in week 6.

Because we start making AA in week 5, we must have 30 units of D and 20 units of F in  week 5. See the rows for D and F in Figure 6.15. The lead time for D is one week. Thus, we must  give the order release in week 4 to have the finished units of D in week 5. Note that there was no  inventory on hand for D in week 5. The original 10 units of inventory of D were used in week 5  to make B, which was subsequently used to make A. We also need to have 20 units of F in week  5 to produce 10 units of AA by week 6. Again, we have no on-hand inventory of F in week 5.  The original 5 units were used in week 4 to make C, which was subsequently used to make A.  The lead time for F is three weeks. Thus, the order release for 20 units of F must be in week 2.  (See the F row in Figure 6.15.)

This example shows how the inventory requirements of two products can be reflected  in the same net material requirements plan. Some manufacturing companies can have more  than 100 end products that must be coordinated in the same net material requirements plan.  Although such a situation can be very complicated, the same principles we used in this exam- ple are  employed. Remember that computer programs have been developed to handle large and  complex  manufacturing operations.

In addition to using MRP to handle end products and finished goods, MRP can also be used  to handle spare parts and components. This is important because most manufacturing companies  sell spare parts and components for maintenance. A net material requirements plan should also  reflect these spare parts and components.

6.12 Just-In-Time Inventory Control

During the past several decades, there has been a trend to make the manufacturing process more  efficient. One objective is to have less in-process inventory on hand. This is known as JIT in- ventory. With this approach, inventory arrives just in time to be used during the manufacturing  process to produce subparts, assemblies, or finished goods. One technique of implementing JIT  is a manual procedure called kanban. Kanban in Japanese means “card.” With a dual-card kan- ban system, there is a conveyance kanban, or C-kanban, and a production kanban, or P-kanban.  The kanban system is very simple. Here is how it works:

Four steps of kanban

Arrow 1:   As shown in Figure 6.16, full containers along with their C-kanban card are  taken from the storage area to a user area, typically on a manufacturing line  (see arrow 1).

Arrow 2:   During the manufacturing process, parts in the container are used up by the user.  When the container is empty, the empty container along with the same C-kanban  card is taken back to the storage area (depicted by arrow number 2). Here the user  picks up a new full container, detaches the P-kanban card from it, attaches his or  her C-kanban card to it, and returns with it to the user area (depicted by arrow num- ber 1 again).

Arrow 3:   This detached P-kanban card is then attached to an empty container in the storage  area and then—and only then—is the empty container taken back to the upstream  producer area (depicted by arrow number 3).

Arrow 4:   This empty container is then refilled with parts and taken with its P-kanban card  back to the storage area (depicted by arrow number 4). This kanban process con- tinuously cycles throughout the day. Kanban is sometimes known as a “pull”   production system.

With JIT, inventory arrives just before it is needed.

M06_REND9327_12_SE_C06.indd 239 10/02/14 1:24 PM

240  Chapter 6 • Inventory Control Models

At a minimum, two containers are required using the kanban system. One container is used  at the user area, and another container is being refilled for future use. In reality, there are usually  more than two containers. This is how inventory control is accomplished. Inventory managers  can introduce additional containers and their associated P-kanbans into the system. In a similar  fashion, the inventory manager can remove containers and the P-kanbans to have tighter control  over inventory buildups.

In addition to being a simple, easy-to-implement system, the kanban system can also be  very effective in controlling inventory costs and in uncovering production bottlenecks. Inventory  arrives at the user area or on the manufacturing line just before it is needed. Inventory does not  build up unnecessarily, cluttering the production line or adding to unnecessary inventory ex- pense. The kanban system reduces inventory levels and makes for a more effective operation. It  is like putting the production line on an inventory diet. Like any diet, the inventory diet imposed  by the kanban system makes the production operation more streamlined. Furthermore, produc- tion bottlenecks and problems can be uncovered. Many production managers remove containers  and their associated P-kanban from the kanban system in order to “starve” the production line to  uncover bottlenecks and potential problems.

In implementing a kanban system, a number of work rules or kanban rules are normally  implemented. One typical kanban rule is that no containers are filled without the appropriate  P-kanban. Another rule is that each container must hold exactly the specified number of parts or  inventory items. These and similar rules make the production process more efficient. Only those  parts that are actually needed are produced. The production department does not produce inven- tory just to keep busy. It produces inventory or parts only when they are needed in the user area  or on an actual manufacturing line.

6.13 Enterprise Resource Planning

Over the years, MRP has evolved to include not only the materials required in production but  also the labor hours, material cost, and other resources related to production. When approached  in this fashion, the term MRP II is often used, and the word resource replaces the word require- ments. As this concept evolved and sophisticated computer software programs were developed,  these systems were called enterprise resource planning (ERP) systems.

The objective of an ERP system is to reduce costs by integrating all of the operations of a  firm. This starts with the supplier of the materials needed and flows through the organization to  include invoicing the customer of the final product. Data are entered once into a database, and  then these data can be quickly and easily accessed by anyone in the organization. This benefits  not only the functions related to planning and managing inventory but also other business pro- cesses such as accounting, finance, and human resources.

The benefits of a well-developed ERP system are reduced transaction costs and increased  speed and accuracy of information. However, there are drawbacks as well. The software is ex- pensive to buy and costly to customize. The implementation of an ERP system may require a  company to change its normal operations, and employees are often resistant to change. Also,  training employees on the use of the new software can be expensive.

There are many ERP systems available. The most common ones include SAP, Oracle, and  PeopleSoft. Even small systems can cost hundreds of thousands of dollars. The larger systems  can cost hundreds of millions of dollars.

P-kanban and

Container

C-kanban and

Container

Producer Area

Storage Area

User Area

4 1

23

Figure 6.16 The kanban system

M06_REND9327_12_SE_C06.indd 240 10/02/14 1:24 PM

glossary  241

This chapter introduces the fundamentals of inventory control  theory. We show that the two most important problems are  (1) how much to order and (2) when to order.

We investigate the economic order quantity, which de- termines how much to order, and the reorder point, which  determines when to order. In addition, we explore the use of  sensitivity analysis to determine what happens to computations  when one or more of the values used in one of the equations  changes.

The basic EOQ inventory model presented in this chap- ter makes a number of assumptions: (1) known and constant  demand and lead times, (2) instantaneous receipt of inventory,  (3) no quantity discounts, (4) no stockouts or shortages, and  (5) the only variable costs are ordering costs and carrying  costs. If these assumptions are valid, the EOQ inventory model  provides optimal solutions. On the other hand, if these assump- tions do not hold, the basic EOQ model does not apply. In these 

cases, more complex models are needed, including the produc- tion run, quantity discount, and safety stock models. When the  inventory item is for use in a single time period, the marginal  analysis approach is used. ABC analysis is used to determine  which items represent the greatest potential inventory cost so  these items can be more carefully managed.

When the demand for inventory is not independent of  the demand for another product, a technique such as MRP is  needed. MRP can be used to determine the gross and net mate- rial requirements for products. Computer software is necessary  to implement major inventory systems including MRP systems  successfully. Today, many companies are using ERP software  to integrate all of the operations within a firm, including inven- tory, accounting, finance, and human resources.

JIT can lower inventory levels, reduce costs, and make a  manufacturing process more efficient. Kanban, a Japanese word  meaning “card,” is one way to implement the JIT approach.

Summary

Glossary

ABC Analysis  An analysis that divides inventory into three  groups. Group A is more important than group B, which  is more important than group C.

Annual Setup Cost  The cost to set up the manufacturing or  production process for the production run model.

Average Inventory  The average inventory on hand. In this  chapter the average inventory is Q>2 for the EOQ model.

Bill of Materials (BOM)  A material structure tree of the  components in a product, with a description and the quan- tity required to make one unit of that product.

Economic Order Quantity (EOQ)  The amount of inventory  ordered that will minimize the total inventory cost. It is also  called the optimal order quantity, or Q*.

Enterprise Resource Planning (ERP)  A computerized   information system that integrates and coordinates the   operations of a firm.

Instantaneous Inventory Receipt  A system in which inven- tory is received or obtained at one point in time and not  over a period of time.

Inventory Position  The amount of inventory on hand plus  the amount in any orders that have been placed but not yet  received.

Just-In-Time (JIT) Inventory  An approach whereby inven- tory arrives just in time to be used in the manufacturing  process.

Kanban  A manual JIT system developed by the Japanese.  Kanban means “card” in Japanese.

Lead Time  The time it takes to receive an order after it is  placed (called L in the chapter).

Marginal Analysis  A decision-making technique that uses  marginal profit and marginal loss in determining optimal  decision policies. Marginal analysis is used when the num- ber of alternatives and states of nature is large.

Marginal Loss  The loss that would be incurred by stocking  and not selling an additional unit.

Marginal Profit  The additional profit that would be realized  by stocking and selling one more unit.

Material Requirements Planning (MRP)  An inventory  model that can handle dependent demand.

Production Run Model  An inventory model in which inven- tory is produced or manufactured over time instead of being  ordered or purchased. This model eliminates the instanta- neous  receipt assumption.

Quantity Discount  The cost per unit when large orders of  an inventory item are placed.

Reorder Point (ROP)  The number of units on hand when an  order for more inventory is placed.

Safety Stock  Extra inventory that is used to help avoid stockouts. Sensitivity Analysis  The process of determining how sensi-

tive the optimal solution is to changes in the values used  in the equations.

Service Level  The chance, expressed as a percent, that there  will not be a stockout. Service level = 1 - Probability of  a stockout.

Stockout  A situation that occurs when there is no inventory  on hand.

M06_REND9327_12_SE_C06.indd 241 10/02/14 1:24 PM

242  Chapter 6 • Inventory Control Models

Equations 6-1 through 6-8 are associated with the economic order quantity (EOQ).

  (6-1)  Average inventory level = Q

2

  (6-2)  Annual ordering cost = D

Q Co

  (6-3)  Annual holding cost = Q

2 Ch

  (6-4)  EOQ = Q* = A2DCoCh   (6-5)  TC =

D

Q Co +

Q

2 Ch

    Total relevant inventory cost.

  (6-6)  Average dollar level = 1CQ2

2

  (6-7)  Q = B 2DCoIC     EOQ with Ch expressed as percentage of unit cost.

  (6-8)  ROP = d * L     Reorder point: d is the daily demand and L is the lead 

time in days.

Equations 6-9 through 6-13 are associated with the production run model.

  (6-9)  Average inventory = Q

2 a1 - d

p b

 (6-10)  Annual holding cost = Q

2 a1 - d

p bCh

 (6-11)  Annual setup cost = D

Q Cs

 (6-12)  Annual ordering cost = D

Q Co

 (6-13)  Q* = H 2DCsCh a1 - dp b     Optimal production quantity.

Equation 6-14 is used for the quantity discount model.

 (6-14)  Total cost = DC + D

Q Co +

Q

2 Ch

    Total inventory cost (including purchase cost).

Equations 6-15 to 6-20 are used when safety stock is required.

 (6-15)  ROP = 1Average demand during lead time2 + SS     General reorder point formula for determining when 

safety stock (SS) is carried.

 (6-16)  ROP = 1Average demand during lead time2 + Zsd LT     Reorder point formula when demand during lead time 

is normally distributed with a standard deviation of  sdLT.

 (6-17)  ROP = dL + Z1sd2L2     Formula for determining the reorder point when daily 

demand is normally distributed but lead time is con- stant, where d is the average daily demand, L is the  constant lead time in days, and sd is the standard de- viation of daily demand.

 (6-18)  ROP = dL + Z1dsL2     Formula for determining the reorder point when daily 

demand is constant but lead time is normally distrib- uted, where L is the average lead time in days, d is the  constant daily demand, and sL is the standard deviation  of lead time.

 (6-19)  ROP = d L + Z12Ls2d + d2s2L2     Formula for determining reorder point when both daily 

demand and lead time are normally distributed; where  d is the average daily demand, L is the average lead  time in days, sL is the standard deviation of lead time,  and sd is the standard deviation of daily demand.

 (6-20)  THC = Q

2 Ch + 1SS2Ch

    Total annual holding cost formula when safety stock is  carried.

Equation 6-21 is used for marginal analysis.

 (6-21)  P Ú ML

ML + MP Decision rule in marginal analysis for stocking addi- tional units.

Key Equations

M06_REND9327_12_SE_C06.indd 242 10/02/14 1:24 PM

solved probleMs  243

Solved Problems

Solved Problem 6-1 Patterson Electronics supplies microcomputer circuitry to a company that incorporates microproces- sors into refrigerators and other home appliances. One of the components has an annual demand of 250  units, and this is constant throughout the year. Carrying cost is estimated to be $1 per unit per year, and  the ordering cost is $20 per order.

a.  To minimize cost, how many units should be ordered each time an order is placed? b.  How many orders per year are needed with the optimal policy? c.  What is the average inventory if costs are minimized? d.   Suppose the ordering cost is not $20, and Patterson has been ordering 150 units each time an order is 

placed. For this order policy to be optimal, what would the ordering cost have to be?

Solution a.  The EOQ assumptions are met, so the optimal order quantity is

EOQ = Q* = B 2DCoCh = B 212502201 = 100 units b.  Number of orders per year  =

D

Q =

250

100 = 2.5 orders per year

Note that this would mean in one year the company places three orders and in the next year it would  only need two orders, since some inventory would be carried over from the previous year. It aver- ages 2.5 orders per year.

c.  Average inventory  = Q

2 =

100

2 = 500 units

d.   Given an annual demand of 250, a carrying cost of $1, and an order quantity of 150, Patterson Elec- tronics must determine what the ordering cost would have to be for the order policy of 150 units to  be optimal. To find the answer to this problem, we must solve the traditional EOQ equation for the  ordering cost. As you can see in the calculations that follow, an ordering cost of $45 is needed for  the order quantity of 150 units to be optimal.

Q = A2DCoCh Co = Q2

Ch 2D

= 115022112

212502

= 22,500

500 = $45

Solved Problem 6-2 Flemming Accessories produces paper slicers used in offices and in art stores. The minislicer has  been one of its most popular items: Annual demand is 6,750 units and is constant throughout the  year. Kristen Flemming, owner of the firm, produces the minislicers in batches. On average, Kristen  can manufacture 125 minislicers per day. Demand for these slicers during the production process is  30 per day. The setup cost for the equipment necessary to produce the minislicers is $150. Carry- ing costs are $1 per minislicer per year. How many minislicers should Kristen manufacture in each  batch?

M06_REND9327_12_SE_C06.indd 243 10/02/14 1:24 PM

244  Chapter 6 • Inventory Control Models

Solution The data for Flemming Accessories are summarized as follows:

D = 6,750 units Cs = $150 Ch = $1 d = 30 units p = 125 units

This is a production run problem that involves a daily production rate and a daily demand rate. The ap- propriate calculations are shown here:

Q* = B 2DCsCh11 - d>p2 = B 216,750211502111 - 30>1252 = 1,632

Solved Problem 6-3 Dorsey Distributors has an annual demand for a metal detector of 1,400. The cost of a typical detector  to Dorsey is $400. Carrying cost is estimated to be 20% of the unit cost, and the ordering cost is $25 per  order. If Dorsey orders in quantities of 300 or more, it can get a 5% discount on the cost of the detectors.  Should Dorsey take the quantity discount? Assume the demand is constant.

Solution The solution to any quantity discount model involves determining the total cost of each alternative after  quantities have been computed and adjusted for the original problem and every discount. We start the  analysis with no discount:

EOQ 1no discount2 = B 211,400212520.214002 = 29.6 units

Total cost 1no discount2 = Material cost + Ordering cost + Carrying cost = $40011,4002 + 1,4001$252

29.6 +

29.61$400210.22 2

= $560,000 + $1,183 + $1,183 = $562,366

The next step is to compute the total cost for the discount:

EOQ 1with discount2 = B 211,400212520.21$3802 = 30.3 units

Q1adjusted2 = 300 units Because this last economic order quantity is below the discounted price, we must adjust the order quan- tity to 300 units. The next step is to compute total cost:

Total cost 1with discount2 = Material cost + Ordering cost + Carrying cost = $38011,4002 + 1,4001252

300 +

3001$380210.22 2

= $532,000 + $117 + $11,400 = $543,517

The optimal strategy is to order 300 units at a total cost of $543,517.

M06_REND9327_12_SE_C06.indd 244 10/02/14 1:24 PM

self-test  245

Solved Problem 6-4 The F. W. Harris Company sells an industrial cleaner to a large number of manufacturing plants in the   Houston area. An analysis of the demand and costs has resulted in a policy of ordering 300 units of this  product every time an order is placed. The demand is constant, at 25 units per day. In an agreement with the  supplier, F. W. Harris is willing to accept a lead time of 20 days since the supplier has provided an excellent  price. What is the reorder point? How many units are actually in inventory when an order should be placed?

Solution The reorder point is

ROP = dxL = 251202 = 500 units This means that an order should be placed when the inventory position is 500. Since the ROP is greater  than the order quantity, Q = 300, an order must have been placed already but not yet delivered. So the  inventory position must be

Inventory position = 1Inventory on hand2 + 1Inventory on order2 500 = 200 + 300

There would be 200 units on hand and an order of 300 units in transit.

Solved Problem 6-5 The B. N. Thayer and D. N. Thaht Computer Company sells a desktop computer that is popular among  gaming enthusiasts. In the past few months, demand has been relatively consistent, although it does  fluctuate from day to day. The company orders the computer cases from a supplier. It places an order for  5,000 cases at the appropriate time to avoid stockouts. The demand during the lead time is normally dis- tributed, with a mean of 1,000 units and a standard deviation of 200 units. The holding cost per unit per  year is estimated to be $4. How much safety stock should the company carry to maintain a 96% service  level? What is the reorder point? What would the total annual holding cost be if this policy is followed?

Solution Using the table for the normal distribution, the Z value for a 96% service level is about 1.75. The stan- dard deviation is 200. The safety stock is calculated as

SS = ZsdLT = 1.7512002 = 375 units For a normal distribution with a mean of 1,000, the reorder point is

ROP = 1Average demand during lead time2 + SS = 1000 + 350 = 1,350 units

The total annual holding cost is

THC = Q

2 Ch + 1SS2Ch = 5,0002 4 + 137524 = $11,500

Self-Test

●  Before taking the self-test, refer to the learning objectives at the beginning of the chapter, the notes in the margins, and the  glossary at the end of the chapter.

●  Use the key at the back of the book to correct your answers. ●  Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

  1.  Which of the following phases is part of an inventory  system?

a. planning b. forecasting c. control d. feedback and revisions e. all of the above

  2.  What is not an inventory use? a. decoupling function b. storing resources c. irregular supply and demand d. production monitoring e. avoiding stockouts and shortages

M06_REND9327_12_SE_C06.indd 245 10/02/14 1:24 PM

246  Chapter 6 • Inventory Control Models

  3.  How are items like raw materials stored in inventories? a. by themselves b. in work-in-process c. in the finished product d. all of the above   4.  Some of the costs are minimised by the use of inventory. 

Which one is not affected by it? a. cost of purchasing items b. cost of advertisement c. cost of carrying or holding inventory d. cost of stockouts   5.  If the EOQ assumptions are met and a company orders 

more than the economic order quantity, then a. total annual holding cost will be greater than the total 

annual ordering cost. b. total annual holding cost will be less than the total 

 annual ordering cost. c. total annual holding cost will be equal to the total 

 annual ordering cost. d. total annual holding cost will be equal to the total 

 annual purchase cost.   6.  The reorder point is a. the quantity that is reordered each time an order is 

placed. b. the amount of inventory that would be needed to meet 

demand during the lead time. c. equal to the average inventory when the EOQ assump-

tions are met. d. assumed to be zero if there is instantaneous 

 replenishment.   7.  If the EOQ assumptions are met, then a. annual stockout cost will be zero. b. total annual holding cost will equal total annual order-

ing cost. c. average inventory will be one-half the order quantity. d. all of the above are true.   8.  Which variables are a part of the maximum inventory-

level expression? a. number of pieces per order b. setup cost c. holding or carrying cost per unit per year d. all of the above

  9.  Why is the annual purchase (material) cost not  considered to be a relevant inventory cost if the EOQ  assumptions are met?

a. This cost will be zero. b. This cost is constant and not affected by the order 

quantity. c. This cost is insignificant compared with the other 

inventory costs. d. This cost is never considered to be an inventory cost.  10.  A JIT system will usually result in a. a low annual holding cost. b. very few orders per year. c. frequent shutdowns in an assembly line. d. high levels of safety stock.  11.  Manufacturers use MRP when a. the demand for one product is dependent on the 

demand for other products. b. the demand for each product is independent of the 

demand for other products. c. demand is totally unpredictable. d. purchase cost is extremely high.  12.  In using marginal analysis, an additional unit should be 

stocked if a. MP = ML. b. the probability of selling that unit is greater than or 

equal to MP>1MP + ML2. c. the probability of selling that unit is less than or equal 

to ML>1MP + ML2. d. the probability of selling that unit is greater than or 

equal to ML>1MP + ML2.  13.  In using marginal analysis with the normal distribution, if 

marginal profit is less than marginal loss, we expect the  optimal stocking quantity to be

a. greater than the standard deviation. b. less than the standard deviation. c. greater than the mean. d. less than the mean.  14.  The inventory position is defined as a. the amount of inventory needed to meet the demand 

during the lead time. b. the amount of inventory on hand. c. the amount of inventory on order. d. the total of the on-hand inventory plus the on-order 

inventory.

Discussion Questions and Problems

Discussion Questions   6-1  Give  a working definition of  inventory  and  its 

purposes.

  6-2  What is the purpose of decoupling?

  6-3  Under what circumstances can inventory be used as  a hedge against inflation?

  6-4  Why wouldn’t a company always store large quan- tities  of  inventory  to  eliminate  shortages  and  stockouts?

  6-5  What is EOQ (economic order quantity)?

  6-6  List  and  explain  the  main  assumption  of  the   economic order quantity.

  6-7  What is the annual ordering cost?

  6-8  If the ROP is greater than the order quantity, explain  how the ROP is implemented. Can the ROP be more  than twice the order quantity and, if so, how is such  a situation handled?

M06_REND9327_12_SE_C06.indd 246 10/02/14 1:24 PM

dIsCussIon QuestIons and probleMs  247

  6-9  Let the annual demand for an arbitrary commod- ity be 1,000 units per year and the associated EOQ  be 400 units per order. Under this circumstance the  number of orders per year would be 1D>Q2 = 2.5 orders per year. How can this be so?

  6-10  What is the purpose of sensitivity analysis?

  6-11  What assumptions are made in the production run  model?

  6-12  What happens to the production run model when the  daily production rate becomes very large?

  6-13  Briefly describe what is involved in solving a quan- tity discount model.

  6-14  When using safety stock, how is the standard devia- tion of demand during the lead time calculated if  daily demand is normally distributed but lead time  is constant? How is it calculated if daily demand is  constant but lead time is normally distributed? How  is it calculated if both daily demand and lead time  are normally distributed?

  6-15  Briefly explain the marginal analysis approach to the  single period inventory problem.

  6-16  Briefly describe what is meant by ABC analysis.  What is the purpose of this inventory technique?

  6-17  What is the overall purpose of MRP?

  6-18  What is the difference between the gross and net ma- terial requirements plan?

  6-19  What is the objective of JIT?

Problems 6-20  Lila Battle has determined that the annual demand 

for number 6 screws is 100,000 screws. Lila, who  works in her brother’s hardware store, is in charge  of purchasing. She estimates that it costs $10 ev- ery time an order is placed. This cost includes her  wages, the cost of the forms used in placing the or- der, and so on. Furthermore, she estimates that the  cost of carrying one screw in inventory for a year is  one-half of 1 cent. Assume that the demand is con- stant throughout the year.

(a) How many number 6 screws should Lila order at  a time if she wishes to minimize total inventory  cost?

(b) How many orders per year would be placed?  What would the annual ordering cost be?

(c) What would the average inventory be? What  would the annual holding cost be?

6-21  It takes approximately 8 working days for an order  of number 6 screws to arrive once the order has been  placed. (Refer to Problem 6-20.) The demand for  number 6 screws is fairly constant, and on the aver- age, Lila has observed that her brother’s hardware 

Note:  means the problem may be solved with QM for Windows;   means the problem may be solved with Excel QM; and  means the problem may be  solved with QM for Windows and/or Excel QM.

store sells 500 of these screws each day. Because  the demand is fairly constant, Lila believes that she  can avoid stockouts completely if she only orders  the number 6 screws at the correct time. What is the  ROP?

  6-22  Lila’s brother believes that she places too many or- ders for screws per year. He believes that an order  should be placed only twice per year. If Lila follows  her brother’s policy, how much more would this  cost every year over the ordering policy that she de- veloped in Problem 6-20? If only two orders were  placed each year, what effect would this have on the  ROP?

6-23  Barbara Bright  is  the purchasing agent for West  Valve Company. West Valve sells industrial valves  and fluid control devices. One of the most popular  valves is the Western, which has an annual demand  of 4,000 units. The cost of each valve is $90, and  the inventory carrying cost is estimated to be 10% of  the cost of each valve. Barbara has made a study of  the costs involved in placing an order for any of the  valves that West Valve stocks, and she has concluded  that the average ordering cost is $25 per order. Fur- thermore, it takes about two weeks for an order to  arrive from the supplier, and during this time the de- mand per week for West valves is approximately 80.

(a) What is the EOQ? (b) What is the ROP? (c)  Is the ROP greater than the EOQ? If so, how is 

this situation handled? (d) What  is  the  average  inventory? What  is  the 

 annual holding cost? (e) How many orders per year would be placed? 

What is the annual ordering cost?

  6-24  Ken Ramsing has been in the lumber business for  most of his life. Ken’s biggest competitor is Pacific  Woods. Through many years of experience, Ken  knows that the ordering cost for an order of plywood  is $25 and that the carrying cost is 25% of the unit  cost. Both Ken and Pacific Woods receive plywood  in loads that cost $100 per load. Furthermore, Ken  and Pacific Woods use the same supplier of ply- wood, and Ken was able  to find out  that Pacific  Woods orders in quantities of 4,000 loads at a time.  Ken also knows that 4,000 loads is  the EOQ for  Pacific Woods. What is the annual demand in loads  of plywood for Pacific Woods?

  6-25  Shoe Shine is a local retail shoe store located on the  north side of Centerville. Annual demand for a pop- ular sandal is 500 pairs, and John Dirk, the owner  of Shoe Shine, has been in the habit of ordering 100  pairs at a time. John estimates that the ordering cost 

M06_REND9327_12_SE_C06.indd 247 10/02/14 1:24 PM

248  Chapter 6 • Inventory Control Models

is $10 per order. The cost of the sandal is $5 per  pair. For John’s ordering policy to be correct, what  would the carrying cost as a percentage of the unit  cost have to be? If the carrying cost were 10% of the  cost, what would the optimal order quantity be?

6-26  In Problem 6-20 you helped Lila Battle determine  the optimal order quantity for number 6 screws. She  had estimated that the ordering cost was $10 per or- der. At this time, though, she believes that this esti- mate was too low. Although she does not know the  exact ordering cost, she believes that it could be as  high as $40 per order. How would the optimal order  quantity change if the ordering cost were $20, $30,  and $40?

6-27  Ross White’s machine shop uses 2,500 brackets  during the course of a year, and this usage is rela- tively constant throughout the year. These brackets  are purchased from a supplier 100 miles away for  $15 each, and the lead time is 2 days. The holding  cost per bracket per year is $1.50 (or 10% of the  unit cost) and the ordering cost per order is $18.75.  There are 250 working days per year.

(a) What is the EOQ? (b) Given the EOQ, what is the average inventory? 

What is the annual inventory holding cost? (c)  In minimizing cost, how many orders would be 

made each year? What would be the annual or- dering cost?

(d) Given the EOQ, what is the total annual inven- tory cost (including purchase cost)?

(e) What is the time between orders? (f)  What is the ROP?

6-28  Ross White (see Problem 6-27) wants to reconsider  his decision of buying the brackets and is consid- ering making the brackets in-house. He has deter- mined that setup costs would be $25 in machinist  time and lost production time, and 50 brackets could  be produced in a day once the machine has been set  up. Ross estimates that the cost (including labor time  and materials) of producing one bracket would be  $14.80. The holding cost would be 10% of this cost.

(a) What is the daily demand rate? (b) What is the optimal production quantity? (c) How long will  it  take to produce the optimal 

quantity? How much inventory is sold during  this time?

(d) If Ross uses the optimal production quantity,  what would be the maximum inventory level?  What would be  the average  inventory  level?  What is the annual holding cost?

(e) How many production runs would there be each  year? What would be the annual setup cost?

(f)  Given the optimal production run size, what is  the total annual inventory cost?

(g) If the lead time is one-half day, what is the ROP?

6-29  Upon hearing that Ross White (see Problems 6-27  and 6-28) is considering producing the brackets in- house, the vendor has notified Ross that the purchase  price would drop from $15 per bracket to $14.50 per  bracket if Ross will purchase the brackets in lots  of 1,000. Lead times, however, would increase to  3 days for this larger quantity.

(a) What is the total annual inventory cost plus pur- chase cost if Ross buys the brackets in lots of  1,000 at $14.50 each?

(b) If Ross does buy in lots of 1,000 brackets, what  is the new ROP?

(c) Given the options of purchasing the brackets at  $15 each, producing them in-house at $14.80,  and taking advantage of the discount, what is  your recommendation to Ross White?

  6-30  After analyzing the costs of various options for ob- taining brackets, Ross White (see Problems 6.27  through 6.29) recognizes that although he knows  that lead time is 2 days and demand per day aver- ages 10 units, the demand during the lead time often  varies. Ross has kept very careful records and has  determined lead time demand is normally distributed  with a standard deviation of 1.5 units.

(a) What Z value would be appropriate for a 98%  service level?

(b) What safety stock should Ross maintain if he  wants a 98% service level?

(c) What is the adjusted ROP for the brackets? (d) What is the annual holding cost for the safety 

stock if the annual holding cost per unit is $1.50?

6-31  Douglas Boats is a supplier of boating equipment for  the states of Oregon and Washington. It sells 5,000  White Marine WM-4 diesel engines every year.  These engines are shipped to Douglas in a ship- ping container of 100 cubic feet, and Douglas Boats  keeps the warehouse full of these WM-4 motors.  The warehouse can hold 5,000 cubic feet of boating  supplies. Douglas estimates that the ordering cost is  $10 per order, and the carrying cost is estimated to  be $10 per motor per year. Douglas Boats is consid- ering the possibility of expanding the warehouse for  the WM-4 motors. How much should Douglas Boats  expand, and how much would it be worth for the  company to make the expansion? Assume demand is  constant throughout the year.

6-32  Northern Distributors is a wholesale organization  that supplies retail stores with lawn care and house- hold products. One building is used to store Never- fail lawn mowers. The building is 25 feet wide by  40 feet deep by 8 feet high. Anna Oldham, manager  of the warehouse, estimates that about 60% of the  warehouse can be used to store the Neverfail lawn  mowers. The remaining 40% is used for walkways  and a small office. Each Neverfail  lawn mower 

M06_REND9327_12_SE_C06.indd 248 10/02/14 1:24 PM

comes in a box that  is 5 feet by 4 feet by 2 feet  high. The annual demand for these lawn mowers  is 12,000, and the ordering cost for Northern Dis- tributors is $30 per order. It is estimated that it costs  Northern $2 per lawn mower per year for storage.  Northern Distributors is thinking about increasing  the size of the warehouse. The company can only do  this by making the warehouse deeper. At the pres- ent time, the warehouse is 40 feet deep. How many  feet of depth should be added on to the warehouse  to minimize the annual inventory costs? How much  should the company be willing to pay for this addi- tion? Remember that only 60% of the total area can  be used to store Neverfail lawn mowers. Assume all  EOQ conditions are met.

6-33  Lisa Surowsky was asked to help in determining the  best ordering policy for a new product. Currently,  the demand for the new product has been projected  to be about 1,000 units annually. To get a handle  on the carrying and ordering costs, Lisa prepared a  series of average inventory costs. Lisa thought that  these costs would be appropriate for the new prod- uct. The results are summarized in the following ta- ble. These data were compiled for 10,000 inventory  items that were carried or held during the year and  were ordered 100 times during the past year. Help  Lisa determine the EOQ.

COST FACTOR COST ($)

Taxes 2,000

Processing and inspection 1,500

New product development 2,500

Bill paying 500

Ordering supplies 50

Inventory insurance 600

Product advertising 800

Spoilage 750

Sending purchasing orders 800

Inventory inquiries 450

Warehouse supplies 280

Research and development 2,750

Purchasing salaries 3,000

Warehouse salaries 2,800

Inventory theft 800

Purchase order supplies 500

Inventory obsolescence 300

6-34  Jan Gentry is the owner of a small company that pro- duces electric scissors used to cut fabric. The annual  demand is for 8,000 scissors, and Jan produces the  scissors in batches. On the average, Jan can produce 

150 scissors per day, and during the production pro- cess, demand for scissors has been about 40 scissors  per day. The cost to set up the production process is  $100, and it costs Jan 30 cents to carry one pair of  scissors for one year. How many scissors should Jan  produce in each batch?

6-35  Jim Overstreet, inventory control manager for Itex,  receives wheel bearings from Wheel-Rite, a small  producer of metal parts. Unfortunately, Wheel-Rite  can only produce 500 wheel bearings per day. Itex  receives 10,000 wheel bearings from Wheel-Rite  each year. Since Itex operates 200 working days  each year, its average daily demand for wheel bear- ings is 50. The ordering cost for Itex is $40 per  order, and the carrying cost is 60 cents per wheel  bearing per year. How many wheel bearings should  Itex order from Wheel-Rite at one time? Wheel-Rite  has agreed to ship the maximum number of wheel  bearings that it produces each day to Itex when an  order has been received.

6-36  North Manufacturing has a demand for 1,000 pumps  each year. The cost of a pump is $50. It costs North  Manufacturing $40 to place an order, and the carry- ing cost is 25% of the unit cost. If pumps are ordered  in quantities of 200, North Manufacturing can get a  3% discount on the cost of the pumps. Should North  Manufacturing order 200 pumps at a time and take  the 3% discount?

  6-37  Linda Lechner is in charge of maintaining hospital  supplies at General Hospital. During the past year,  the mean lead time demand for bandage BX-5 was  60. Furthermore, the standard deviation for BX-5  was 7. Linda would like to maintain a 90% service  level. What safety stock level do you recommend for  BX-5?

  6-38  Linda Lechner has  just been severely chastised  for her inventory policy. (See Problem 6-37.) Sue   Surrowski, her boss, believes that the service level  should be either 95% or 98%. Compute the safety  stock  levels for a 95% and a 98% service  level.  Linda knows that the carrying cost of BX-5 is 50  cents per unit per year. Compute the carrying cost  that is associated with a 90%, a 95%, and a 98% ser- vice level.

6-39  Ralph Janaro simply does not have time to ana- lyze all of the items in his company’s inventory. As  a young manager, he has more important things to  do. The following is a table of six items in inventory  along with the unit cost and the demand in units.

(a)  Find the total amount spent on each item during  the year. What is the total investment for all of  these?

(b) Find the percentage of the total investment in   inventory that is spent on each item.

dIsCussIon QuestIons and probleMs  249

M06_REND9327_12_SE_C06.indd 249 10/02/14 1:24 PM

250  Chapter 6 • Inventory Control Models

IDEnTIFICATIOn CODE

unIT COST ($)

DEMAnD In unITS

XX1 5.84 1,200

B66 5.40 1,110

3CPO 1.12 896

33CP 74.54 1,104

R2D2 2.00 1,110

RMS 2.08 961

(c) Based on  the percentages  in part  (b), which  item(s) would be classified in categories A, B,  and C using ABC analysis?

(d) Which item(s) should Ralph most carefully con- trol using quantitative techniques?

6-40  Thaarugo, Inc., produces a GPS device that is be- coming popular  in parts  of Scandinavia. When  Thaarugo produces one of these, a printed circuit  board (PCB) is used, and it is populated with several  electronic components. Thaarugo determines that it  needs about 16,000 of this type of PCB each year.  Demand is relatively constant throughout the year,  and the ordering cost is about $25 per order; the  holding cost is 20% of the price of each PCB. Two  companies are competing to become the dominant  supplier of the PCBs, and both have now offered  discounts, as shown in the following table. Which  of the two suppliers should be selected if Thaarugo  wishes to minimize total annual  inventory cost?  What would be the annual inventory cost?

SuPPLIER A SuPPLIER B

QuAnTITy PRICE QuAnTITy PRICE

1–199 38.40 1–299 39.50

200–499 35.80 300–999 35.40

500 or more 34.70 1000 or more 34.60

6-41  Dillard Travey  receives 5,000  tripods  annually  from Quality Suppliers to meet his annual demand.   Dillard runs a large photographic outlet, and the tri- pods are used primarily with 35-mm cameras. The  ordering cost is $15 per order, and the carrying cost  is 50 cents per unit per year. Quality is starting a new  option for its customers. When an order is placed,  Quality will ship one-third of the order every week  for three weeks instead of shipping the entire order  at one time. Weekly demand over the lead time is  100 tripods.

(a) What is the order quantity if Dillard has the en- tire order shipped at one time?

(b) What  is  the order quantity  if Dillard has  the  order shipped over three weeks using the new 

option from Quality Suppliers, Inc.? To simplify  your calculations, assume that the average inven- tory is equal to one-half of the maximum inven- tory level for Quality’s new option.

(c) Calculate the total cost for each option. What do  you recommend?

  6-42  Quality Suppliers, Inc., has decided to extend its  shipping option. (Refer  to Problem 6-41 for de- tails.) Now, Quality Suppliers is offering to ship the  amount ordered in five equal shipments, one each  week. It will take five weeks for this entire order to  be received. What are the order quantity and total  cost for this new shipping option?

6-43  The Hardware Warehouse is evaluating the safety  stock policy for all its items, as identified by the  SKU code. For SKU M4389, the company always  orders 80 units each time an order is placed. The  daily demand is constant, at 5 units per day; the  lead time is normally distributed, with a mean of  3 days and a standard deviation of 2. Holding cost  is $3 per unit per year. A 95% service level is to be  maintained.

(a) What is the standard deviation of demand during  the lead time?

(b) How much safety stock should be carried, and  what should be the reorder point?

(c) What is the total annual holding cost?

6-44  For SKU A3510 at the Hardware Warehouse, the  order quantity has been set at 150 units each time  an order is placed. The daily demand is normally  distributed, with a mean of 12 units and a standard  deviation of 4. It always takes exactly 5 days for an  order of this item to arrive. Holding cost has been  determined to be $10 per unit per year. Due to the  large volume of this item sold, management wants to  maintain a 99% service level.

(a) What is the standard deviation of demand during  the lead time?

(b) How much safety stock should be carried, and  what should be the reorder point?

(c) What is the total annual holding cost?

6-45  H & K Electronic Warehouse sells a 12-pack of AAA  batteries, and this is a very popular item. Demand  for this is normally distributed, with an average of  50 packs per day and a standard deviation of 16.  The average delivery time is 5 days, with a standard  deviation of 2 days. Delivery time has been found  to be normally distributed. A 96% service level is  desired.

(a) What is the standard deviation of demand during  the lead time?

(b) How much safety stock should be carried, and  what should be the reorder point?

M06_REND9327_12_SE_C06.indd 250 10/02/14 1:24 PM

6-46  Xemex has collected the following inventory data  for the six items that it stocks:

ITEM CODE

unIT COST

($)

AnnuAL DEMAnD (unITS)

ORDERIng COST ($)

CARRyIng COST AS A PERCEnTAgE

OF unIT COST

1 10.60 600 40 20

2 11.00 450 30 25

3 2.25 500 50 15

4 150.00 560 40 15

5 4.00 540 35 16

6 4.10 490 40 17

    Lynn Robinson, Xemex’s inventory manager, does  not feel that all of the items can be controlled. What  ordered quantities do you recommend for which in- ventory product(s)?

6-47  Georgia Products offers  the  following discount  schedule for its 4- by 8-foot sheets of good-quality  plywood:

ORDER unIT COST ($)

9 sheets or less 18.00

10 to 50 sheets 17.50

More than 50 sheets 17.25

Home Sweet Home Company orders plywood from  Georgia Products. Home Sweet Home has an order- ing cost of $45. The carrying cost is 20%, and the an- nual demand is 100 sheets. What do you recommend?

6-48  Sunbright Citrus Products produces orange juice,  grapefruit juice, and other citrus-related items. Sun- bright obtains fruit concentrate from a cooperative in  Orlando consisting of approximately 50 citrus grow- ers. The cooperative will sell a minimum of 100 cans  of fruit concentrate to citrus processors such as Sun- bright. The cost per can is $9.90.

Last year, the cooperative developed the Incen- tive Bonus Program to give an incentive to their  large customers to buy in quantity. It works like this:  If 200 cans of concentrate are purchased, 10 cans of  free concentrate are included in the deal. In addition,  the names of the companies purchasing the concen- trate are added to a drawing for a new personal com- puter. The personal computer has a value of about  $3,000, and currently about 1,000 companies are  eligible for this drawing. At 300 cans of concentrate,  the cooperative will give away 30 free cans and will  also place the company name in the drawing for  the personal computer. When the quantity goes up  to 400 cans of concentrate, 40 cans of concentrate  will be given away free with the order. In addition,  the company is also placed in a drawing for the per- sonal computer and a free trip for two. The value of 

the trip for two is approximately $5,000. About 800  companies are expected to qualify and to be in the  running for this trip.

Sunbright estimates that its annual demand for  fruit concentrate is 1,000 cans. In addition, the order- ing cost is estimated to be $10, while the carrying cost  is estimated to be 10%, or about $1 per unit. The firm  is intrigued with the incentive bonus plan. If the com- pany decides that it will keep the free cans, the trip, or  the computer if they are won, what should it do?

6-49  John Lindsay sells CDs that contain 25 software  packages that perform a variety of financial func- tions, including net present value, internal rate of re- turn, and other financial programs typically used by  business students majoring in finance. Depending on  the quantity ordered, John offers the following price  discounts. The annual demand is 2,000 units on av- erage. His setup cost to produce the CDs is $250. He  estimates holding costs to be 10% of the price, or  about $1 per unit per year.

QuAnTITy ORDERED

PRICE RAngES FROM TO PRICE

1 500 $10.00

501 1,000 9.95

1,001 1,500 9.90

1,500 2,000 9.85

(a) What is the optimal number of CDs to produce  at a time?

(b) What is the impact of the following quantity– price schedule on the optimal order quantity?

QuAnTITy ORDERED

PRICE RAngES FROM TO PRICE

1    500 $10.00

501 1,000 9.99

1,001 1,500 9.98

1,500 2,000 9.97

  6-50  Teresa Granger is the manager of Chicago Cheese,  which produces cheese spreads and other cheese-  related products. E-Z Spread Cheese is a product  that has always been popular. The probability of  sales, in cases, is as follows:

DEMAnD (CASES) PROBABILITy

10 0.2

11 0.3

12 0.2

13 0.2

14 0.1

dIsCussIon QuestIons and probleMs  251

M06_REND9327_12_SE_C06.indd 251 10/02/14 1:24 PM

252  Chapter 6 • Inventory Control Models

A case of E-Z Spread Cheese sells for $100 and has  a cost of $75. Any cheese that is not sold by the end  of the week is sold to a local food processor for $50.  Teresa never sells cheese that is more than a week  old. Use marginal analysis to determine how many  cases of E-Z Spread Cheese to produce each week to  maximize average profit.

6-51  Harry’s Hardware does a brisk business during the  year. During Christmas, Harry’s Hardware sells  Christmas trees for a substantial profit. Unfortu- nately, any trees not sold at the end of the season are  totally worthless. Thus, the number of trees that are  stocked for a given season is a very important de- cision. The following table reveals the demand for  Christmas trees:

DEMAnD FOR ChRISTMAS TREES PROBABILITy

50 0.05

75 0.1

100 0.2

125 0.3

150 0.2

175 0.1

200 0.05

Harry sells trees for $80 each, but his cost is only  $20.

(a) Use marginal analysis to determine how many  trees Harry should stock at his hardware store.

(b) If the cost increased to $35 per tree and Harry  continues to sell trees for $80 each, how many  trees should Harry stock?

(c) Harry is thinking about increasing the price to  $100 per tree. Assume that the cost per tree is  $20. With the new price, it is expected that the  probability of selling 50, 75, 100, or 125 trees  will be 0.25 each. Harry does not expect to sell  more than 125 trees with this price increase.  What do you recommend?

  6-52  In addition to selling Christmas trees during the  Christmas holidays, Harry’s Hardware sells all the  ordinary hardware items (see Problem 6-51). One of  the most popular items is Great Glue HH, a glue that  is made just for Harry’s Hardware. The selling price  is $4 per bottle, but unfortunately, the glue gets hard  and unusable after one month. The cost of the glue  is $1.20. During the past several months, the mean  sales of glue have been 60 units, and the standard  deviation is 7. How many bottles of glue should  Harry’s Hardware stock? Assume that sales follow a  normal distribution.

  6-53  The marginal loss on Washington Reds, a brand of  apples from the state of Washington, is $35 per case. 

The marginal profit is $15 per case. During the past  year, the mean sales of Washington Reds in cases  was 45,000 cases, and the standard deviation was  4,450. How many cases of Washington Reds should  be brought to market? Assume that sales follow a  normal distribution.

6-54  Linda Stanyon has been the production manager for  Plano Produce for over eight years. Plano Produce  is a small company located near Plano, Illinois. One  produce item, tomatoes, is sold in cases, with daily  sales averaging 400 cases. Daily sales are assumed  to be normally distributed. In addition, 85% of the  time the sales are between 350 and 450 cases. Each  case costs $10 and sells for $15. All cases that are  not sold must be discarded.

(a) Using the information provided, estimate the  standard deviation of sales.

(b) Using the standard deviation in part (a), deter- mine how many cases of tomatoes Linda should  stock.

  6-55  Paula Shoemaker produces a weekly stock market  report for an exclusive readership. She normally  sells 3,000 reports per week, and 70% of the time  her sales range from 2,900 to 3,100. The report costs  Paula $15 to produce, but Paula is able to sell reports  for $350 each. Of course, any reports not sold by the  end of the week have no value. How many reports  should Paula produce each week?

  6-56  Emarpy Appliance produces all kinds of major ap- pliances. Richard Feehan, the president of Emarpy,  is concerned about the production policy for the  company’s best selling refrigerator. The demand for  this has been relatively constant at about 8,000 units  each year. The production capacity for this product  is 200 units per day. Each time production starts,  it costs the company $120 to move materials into  place, reset the assembly line, and clean the equip- ment. The holding cost of a refrigerator is $50 per  year. The current production plan calls for 400 re- frigerators to be produced in each production run.  Assume there are 250 working days per year.

(a) What is the daily demand of this product? (b) If the company were to continue to produce 400 

units each  time production starts, how many  days would production continue?

(c) Under the current policy, how many production  runs per year would be required? What would  the annual setup cost be?

(d) If the current policy continues, how many refrig- erators would be in inventory when production  stops? What would the average inventory level  be?

(e)  If the company produces 400 refrigerators at a  time, what would the total annual setup cost and  holding cost be?

M06_REND9327_12_SE_C06.indd 252 10/02/14 1:24 PM

  6-57  Consider the Emarpy Appliance situation in Prob- lem 6-56. If Richard Feehan wants to minimize the  total annual inventory cost, how many refrigerators  should be produced in each production run? How  much would this save the company in  inventory  costs compared with the current policy of producing  400 in each production run?

  6-58  This chapter presents a material structure tree for  item A in Figure 6.12. Assume that it now takes 1  unit of item B to make every unit of item A. What  impact does  this have on  the material structure  tree and the number of items of D and E that are  needed?

  6-59  Given the information in Problem 6-58, develop  a gross material requirements plan for 50 units of  item A.

  6-60  Using the data from Figures 6.12–6.14, develop a  net material requirements plan for 50 units of item A  assuming that it only takes 1 unit of item B for each  unit of item A.

  6-61  The demand for product S is 100 units. Each unit of  S requires 1 unit of T and 0.5 ounce of U. Each unit  of T requires 1 unit of V, 2 units of W, and 1 unit of  X. Finally, each unit of U requires 0.5 ounce of Y  and 3 units of Z. All items are manufactured by the  same firm. It takes two weeks to make S, one week to  make T, two weeks to make U, two weeks to make V,  three weeks to make W, one week to make X, two  weeks to make Y, and one week to make Z.

(a) Construct a material structure tree and a gross  material requirements plan for the dependent in- ventory items.

(b) Identify all levels, parents, and components.

(c) Construct a net material requirements plan using  the following on-hand inventory data:

ITEM S T u v W X y Z

On-Hand  Inventory 20 20 10 30 30 25 15 10

  6-62  The Webster Manufacturing Company produces a  popular type of serving cart. This product, the SL72,  is made from the following parts: 1 unit of Part A,  1 unit of Part B, and 1 unit of Subassembly C. Each  subassembly C is made up of 2 units of Part D,  4 units of Part E, and 2 units of Part F. Develop a  material structure tree for this.

  6-63  Blair H. Dodds, III, runs a medium-to-large-sized  home eBay business dealing in vintage photographs.  The annual demand for his photos is approximately  50,000. The annual overhead cost (excluding the  purchase price) to buy the photographs is $4,000 per  year. Given that this cost represents the optimal an- nual ordering cost and the optimal ordering quantity  is 400 photographs at a time determine the ordering  cost and the average inventory level.

  6-64  The lead time for each of the parts in the SL72 (Prob- lem 6-62) is one week, except for Part B, which has  a lead time of two weeks. Develop a net materials re- quirements plan for an order of 800 SL72s. Assume  that currently there are no parts in inventory.

  6-65  Refer to Problem 6-64. Develop a net material require- ments plan assuming that there are currently 150 units  of Part A, 40 units of Part B, 50 units of Subassembly  C, and 100 units of Part F currently in inventory.

See our Internet home page, at www.pearsonglobaleditions.com/render, for additional  homework problems, Problems 6-66 to 6-73.

Internet Homework Problems

Case Study

Martin-pullin bicycle corporation

Martin-Pullin Bicycle Corp. (MPBC), located in Dallas, is a  wholesale distributor of bicycles and bicycle parts. Formed  in 1981 by cousins Ray Martin and Jim Pullin, the firm’s pri- mary retail outlets are located within a 400-mile radius of the  distribution center. These retail outlets receive the order from 

Martin-Pullin within two days after notifying the distribution  center, provided that the stock is available. However, if an or- der is not fulfilled by the company, no backorder is placed; the  retailers arrange to get their shipment from other distributors,  and MPBC loses that amount of business.

Case study  253

M06_REND9327_12_SE_C06.indd 253 10/02/14 1:24 PM

254  Chapter 6 • Inventory Control Models

Demands for AirWing Model

MONTH 2009 2010 FORECAST FOR 2011

January 6 7 8

February 12 14 15

March 24 27 31

April 46 53 59

May 75 86 97

June 47 54 60

July 30 34 39

August 18 21 24

September 13 15 16

October 12 13 15

November 22 25 28

December 38 42 47

Total 343 391 439

The company distributes a wide variety of bicycles. The  most popular model, and the major source of revenue to the  company, is the AirWing. MPBC receives all the models from 

a single manufacturer overseas, and shipment takes as long as  four weeks from the time an order is placed. With the cost of  communication, paperwork, and customs clearance included,  MPBC estimates that each time an order is placed, it incurs a  cost of $65. The purchase price paid by MPBC, per bicycle,  is roughly 60% of the suggested retail price for all the styles  available, and the inventory carrying cost is 1% per month (12%  per year) of the purchase price paid by MPBC. The retail price  (paid by the customers) for the AirWing is $170 per bicycle.

MPBC is interested in making an inventory plan for 2011.  The firm wants to maintain a 95% service level with its custom- ers to minimize the losses on the lost orders. The data collected  for the past two years are summarized in the following table. A  forecast for AirWing model sales in the upcoming year 2011  has been developed and will be used to make an inventory plan  for MPBC.

discussion Questions   1. Develop an inventory plan to help MPBC.   2. Discuss ROPs and total costs.   3. How can you address demand that is not at the level of the 

planning horizon?

Source: Professor Kala Chand Seal, Loyola Marymount University.

See our Internet home page, at www.pearsonglobaleditions.com/render, for these additional  case studies:

(1) LaPlace Power and Light: This case involves a public utility in Louisiana and its use of elec- tric cables for connecting houses to power lines.

(2) Western Ranchman Outfitters: This case involves managing the inventory of a popular style  of jeans when the delivery date is sometimes unpredictable.

(3) Professional video Management: This case involves a videotape system in which discounts  from suppliers are possible.

(4) Drake Radio: This case involves ordering FM tuners.

Internet Case Studies

Bibliography

Anderson, Eric T., Gavan J. Fitzsimons, and Duncan Simester. “Measuring  and Mitigating the Costs of Stockouts,” Management Science 52, 11  (November 2006): 1751–1763.

Bradley, James R., and Richard W. Conway. “Managing Cyclic Inventories,”  Production and Operations Management 12, 4 (Winter 2003):  464–479.

Chan, Lap Mui Ann, David Simchi-Levi, and Julie Swann. “Pricing, Produc- tion, and Inventory Policies for Manufacturing with Stochastic Demand  and Discretionary Sales,” Manufacturing & Service Operations Manage- ment 8, 2 (Spring 2006): 149–168.

Chickering, David Maxwell, and David Heckerman. “Targeted Advertising on  the Web with Inventory Management,” Interfaces 33, 5 (September–  October 2003): 71–77.

Chopra, Sunil, Gilles Reinhardt, and Maqbool Dada. “The Effect of Lead  Time Uncertainty on Safety Stocks,” Decision Sciences 35, 1  (Winter 2004): 1–24.

Desai, Preyas S., and Oded Koenigsberg. “The Role of Production Lead Time  and Demand Uncertainty in Marketing Durable Goods,” Management Science 53, 1 (January 2007): 150–158.

Jayaraman, Vaidyanathan, Cindy Burnett, and Derek Frank. “Separating  Inventory Flows in the Materials Management Department of Hancock  Medical Center,” Interfaces 30, 4 (July–August 2000): 56–64.

Kapuscinski, Roman, Rachel Q. Zhang, Paul Carbonneau, Robert Moore, and  Bill Reeves. “Inventory Decisions in Dell’s Supply Chain,” Interfaces 34, 3 (May–June 2004): 191–205.

M06_REND9327_12_SE_C06.indd 254 10/02/14 1:24 PM

Appendix 6.1: Inventory Control with QM for Windows

A variety of  inventory control models are covered in this chapter. Each model makes different  assumptions and uses slightly different approaches. The use of QM for Windows is similar for these  different types of inventory problems. As you can see in the inventory menu for QM for Windows,  most of the inventory problems discussed in this chapter can be solved using your computer.

To demonstrate QM for Windows, we start with the basic EOQ model. Sumco, a manufacturing  company discussed in the chapter, has an annual demand of 1,000 units, an ordering cost of $10 per  unit, and a carrying cost of $0.50 per unit per year. With these data, we can use QM for Windows to  determine the economic order quantity. The results are shown in Program 6.5.

The production run inventory problem, which requires the daily production and demand rate in  addition to the annual demand, the ordering cost per order, and the carrying cost per unit per year, is  also covered in this chapter. The Brown Manufacturing example is used in this chapter to show how  the calculations can be made manually. We can use QM for Windows on these data. Program 6.6  shows the results.

Karabakal, Nejat, Ali Gunal, and Warren Witchie. “Supply-Chain Analysis at Volkswagen of America,” Interfaces 30, 4 (July–August, 2000):  46–55.

Kök, A. Gürhan, and Kevin H. Shang. “Inspection and Replenishment Policies  for Systems with Inventory Record Inaccuracy,” Manufacturing & Service Operations Management 9, 2 (Spring 2007): 185–205.

Ramasesh, Ranga V. and Ram Rachmadugu. “Lot-Sizing Decision Under  Limited-Time Price Reduction,” Decision Sciences 32, 1 (Winter 2001):  125–143.

Ramdas, Kamalini and Robert E. Spekman. “Chain or Shackles: Unders- tanding What Drives Supply-Chain Performance,” Interfaces 30, 4  (July– August 2000): 3–21.

Rubin, Paul A., and W. C. Benton. “A Generalized Framework for Quantity  Discount Pricing Schedules,” Decision Sciences 34, 1 (Winter 2003):  173–188.

Shin, Hojung, and W. C. Benton. “Quantity Discount-Based Inventory Coor- dination: Effectiveness and Critical Environmental Factors,” Production and Operations Management 13, 1 (Spring 2004): 63–76.

Program 6.5 QM for windows results for EOQ Model

Program 6.6 QM for windows results for the Production run Model

appendIx 6.1: Inventory Control wIth QM for wIndows  255

M06_REND9327_12_SE_C06.indd 255 10/02/14 1:24 PM

256  Chapter 6 • Inventory Control Models

When an organization has a large number of inventory items, ABC analysis is often used. As   discussed in this chapter, total dollar volume for an inventory item is one way to determine if quantita- tive control techniques should be used. Performing the necessary calculations is done in Program 6.8,  which shows how QM for Windows can be used to compute dollar volume and determine if quantita- tive control techniques are justified for each inventory item with this new example.

The quantity discount model allows the material cost to vary with the quantity ordered. In this  case the model must consider and minimize material, ordering, and carrying costs by examining  each price discount. Program 6.7 shows how QM for Windows can be used to solve the quantity  discount model discussed in the chapter. Note that the program output shows the input data in addi- tion to the results.

Program 6.7 QM for windows results for the Quantity Discount Model

Program 6.8 QM for windows results for ABC Analysis

M06_REND9327_12_SE_C06.indd 256 10/02/14 1:24 PM

 257

Learning Objectives

Summary  •  Glossary  •  Solved Problems  •  Self-Test  •  Discussion Questions and Problems  •  Internet Homework 

 Problems  •  Case Study: Mexicana Wire Works  •  Internet Case Study  •  Bibliography

1. Understand the basic assumptions and properties of linear programming (LP).

2. Graphically solve any LP problem that has only two variables by both the corner point and isoprofit line methods.

3. Understand special issues in LP such as infeasibility, unboundedness, redundancy, and alternative optimal solutions.

4. Understand the role of sensitivity analysis.

5. Use Excel spreadsheets to solve LP problems.

After completing this chapter, students will be able to:

7.5 Solving Flair Furniture’s LP Problem Using QM for Windows, Excel 2013, and Excel QM

7.6 Solving Minimization Problems

7.7 Four Special Cases in LP

7.8 Sensitivity Analysis

7.1 Introduction

7.2 Requirements of a Linear Programming Problem

7.3 Formulating LP Problems

7.4 Graphical Solution to an LP Problem

chapter OutLine

Linear Programming Models: Graphical and Computer

Methods

7chapter

M07_REND9327_12_SE_C07.indd 257 10/02/14 1:25 PM

258  Chapter 7 • Linear programming modeLs: graphiCaL and Computer methods

7.1 Introduction

Many management decisions involve trying to make the most effective use of an organization’s  resources. Resources typically include machinery, labor, money, time, warehouse space, and raw  materials. These resources may be used to make products (such as machinery, furniture, food, or  clothing) or services (such as schedules for airlines or production, advertising policies, or invest- ment decisions). Linear programming (LP) is a widely used mathematical modeling technique  designed to help managers in planning and decision making relative to resource allocation. We  devote this and the next chapter to illustrating how and why linear programming works.

Despite its name, LP and the more general category of techniques called “mathematical” programming have very little to do with computer programming. In the world of management  science, programming refers to modeling and solving a problem mathematically.  Computer   programming has, of course, played an important role in the advancement and use of LP.   Real-life LP problems are too cumbersome to solve by hand or with a calculator. So throughout  the chapters on LP we give examples of how valuable a computer program can be in solving an  LP problem.

7.2 Requirements of a Linear Programming Problem

In the past 60 years, LP has been applied extensively to military, industrial, financial, marketing,  accounting, and agricultural problems. Even though these applications are diverse, all LP prob- lems have several properties and assumptions in common.

All problems seek to maximize or minimize some quantity, usually profit or cost. We refer to  this property as the objective function of an LP problem. The major objective of a typical man- ufacturer is to maximize dollar profits. In the case of a trucking or railroad distribution system,  the objective might be to minimize shipping costs. In any event, this objective must be stated  clearly and defined mathematically. It does not matter, by the way, whether profits and costs are  measured in cents, dollars, or millions of dollars.

The second property that LP problems have in common is the presence of restrictions, or  constraints, that limit the degree to which we can pursue our objective. For example, deciding  how many units of each product in a firm’s product line to manufacture is restricted by available  personnel and machinery. Selection of an advertising policy or a financial portfolio is limited by  the amount of money available to be spent or invested. We want, therefore, to maximize or mini- mize a quantity (the objective function) subject to limited resources (the constraints).

There must be alternative courses of action to choose from. For example, if a company  produces three different products, management may use LP to decide how to allocate among  them its limited production resources (of personnel, machinery, and so on). Should it devote all  manufacturing capacity to make only the first product, should it produce equal amounts of each  product, or should it allocate the resources in some other ratio? If there were no alternatives to  select from, we would not need LP.

The objective and constraints  in LP problems must be expressed in  terms of  linear  equations or inequalities. Linear mathematical relationships just mean that all terms used in  the  objective function and constraints are of the first degree (i.e., not squared, or to the third  or higher power, or appearing more than once). Hence, the equation 2A + 5B = 10 is an   acceptable linear  function, while the equation 2A2 + 5B3 + 3AB = 10 is not linear because  the variable A is squared, the variable B is cubed, and the two variables appear again as a   product of each other.

The term linear implies both proportionality and additivity. Proportionality means that if  production of 1 unit of a product uses 3 hours, production of 10 units would use 30 hours.   Additivity means that the total of all activities equals the sum of the individual activities. If the  production of one product generated $3 profit and the production of another product generated  $8 profit, the total profit would be the sum of these two, which would be $11.

We assume that conditions of certainty exist: that is, number in the objective and constraints  are known with certainty and do not change during the period being studied.

We make the divisibility assumption that solutions need not be in whole numbers (integers).  Instead, they are divisible and may take any fractional value. In production problems, we often  define variables as the number of units produced per week or per month, and a fractional value 

Linear programming is a technique that helps in resource allocation decisions.

Problems seek to maximize or minimize an objective.

Constraints limit the degree to which the objective can be obtained.

There must be alternatives available.

Mathematical relationships are linear.

M07_REND9327_12_SE_C07.indd 258 10/02/14 1:25 PM

7.3 FormuLating Lp probLems  259

Product mix problems use LP to decide how much of each product to make, given a series of resource restrictions.

Table 7.1 LP Properties and Assumptions

ProPerties of Linear Programs

1.  One objective function

2.  One or more constraints

3.  Alternative courses of action

4.   Objective function and constraints are  linear—proportionality and divisibility

5.  Certainty

6.  Divisibility

7.  Nonnegative variables

*Technically, we maximize total contribution margin, which is the difference between unit selling price and costs  that vary in proportion to the quantity of the item produced. Depreciation, fixed general expense, and advertising are   excluded from calculations.

(e.g., 0.3 chair) would simply mean that there is work in process. Something that was started  in one week can be finished in the next. However, in other types of problems, fractional values  do not make sense. If a fraction of a product cannot be purchased (for example, one-third of a  submarine), an integer programming problem exists. Integer programming is discussed in more  detail in Chapter 10.

Finally, we assume that all answers or variables are nonnegative. Negative values of physi- cal quantities are impossible; you simply cannot produce a negative number of chairs, shirts,  lamps, or computers. However, there are some variables that can have negative values, such as  profit, where a negative value indicates a loss. A simple mathematical operation can transform  such a variable into two nonnegative variables, and that process can be found in books on lin- ear programming. However, when working with linear programming in this book, we will only  work with nonnegative variables. Table 7.1 summarizes these properties and assumptions.

7.3 Formulating LP Problems

Formulating a linear program involves developing a mathematical model to represent the mana- gerial problem. Thus, in order to formulate a linear program, it is necessary to completely under- stand the managerial problem being faced. Once this is understood, we can begin to develop the  mathematical statement of the problem. The steps in formulating a linear program follow:

  1. Completely understand the managerial problem being faced.   2.  Identify the objective and the constraints.   3. Define the decision variables.   4. Use the decision variables to write mathematical expressions for the objective function and 

the constraints.

One of the most common LP applications is the product mix problem. Two or more prod- ucts are usually produced using limited resources such as personnel, machines, raw materials,  and so on. The profit that the firm seeks to maximize is based on the profit contribution per unit  of each product. (Profit contribution, you may recall, is just the selling price per unit minus the  variable cost per unit.*) The company would like to determine how many units of each product  it should produce so as to maximize overall profit given its limited resources. A problem of this  type is formulated in the following example.

Flair Furniture Company The Flair Furniture Company produces inexpensive tables and chairs. The production process  for each is similar in that both require a certain number of hours of carpentry work and a certain  number of labor hours in the painting and varnishing department. Each table takes 4 hours of  carpentry and 2 hours in the painting and varnishing shop. Each chair requires 3 hours in car- pentry and 1 hour in painting and varnishing. During the current production period, 240 hours  of carpentry time are available and 100 hours in painting and varnishing time are available. Each  table sold yields a profit of $70; each chair produced is sold for a $50 profit.

M07_REND9327_12_SE_C07.indd 259 10/02/14 1:25 PM

260  Chapter 7 • Linear programming modeLs: graphiCaL and Computer methods

Flair Furniture’s problem is to determine the best possible combination of tables and chairs  to manufacture in order to reach the maximum profit. The firm would like this production mix  situation formulated as an LP problem.

We begin by summarizing the information needed to formulate and solve this problem (see  Table 7.2). This helps us understand the problem being faced. Next we identify the objective and  the constraints. The objective is

Maximize profit

The constraints are

  1. The hours of carpentry time used cannot exceed 240 hours per week.   2. The hours of painting and varnishing time used cannot exceed 100 hours per week.

The decision variables that represent the actual decisions we will make are defined as

T = number of tables to be produced per week C = number of chairs to be produced per week

Now we can create the LP objective function in terms of T and C. The objective function is  Maximize profit = $70T + $50C.

Our next step is to develop mathematical relationships to describe the two constraints in  this problem. One general relationship is that the amount of a resource used is to be less than or  equal to 1…2 the amount of resource available.

In the case of the carpentry department, the total time used is

14 hours per table21Number of tables produced2 + 13 hours per chair21Number of chairs produced2

Prior to 1945, some conceptual problems regarding the allo- cation of scarce resources had been suggested by economists and others. Two of these people, Leonid Kantorovich and Tjalling Koopmans, shared the 1975 Nobel Prize in Economics for advanc- ing the concepts of optimal planning in their work that began during the 1940s. In 1945, George Stigler proposed the “diet problem” which is now the name given to a major category of linear programming applications. However, Stigler relied on heu- ristic techniques to find a good solution to this problem as there was no method available to find the optimal solution.

Major progress in the field, however, took place in 1947 when George D. Dantzig, often called the “Father of Linear Programming,” published his work on the solution procedure known as the simplex algorithm. Dantzig had been an Air Force

mathematician during World War II and was assigned to work on logistics problems. He saw that many problems involving limited resources and more than one demand could be set up in terms of a series of equations and inequalities, and he subsequently devel- oped the simplex algorithm for solving these problems.

Although early applications of linear programming were military in nature, industrial applications rapidly became appar- ent with the widespread use of business computers. As problems became larger, research continued to find even better ways to solve linear programs. The work of Leonid Kachiyan in 1979 and Narendra Karmarkar in 1984 spurred others to study the use of interior point methods for solving linear programs, some of which are used today. However, the simplex algorithm developed by Dantzig is still the basis for much of the software used for solving linear programs today.

histOry the beginning of Linear programming

Table 7.2 Flair Furniture Company Data

Hours required to Produce 1 unit

dePartment (T)

tabLes (C)

cHairs avaiLabLe Hours

tHis Week

Carpentry 4 3 240

Painting and varnishing 2 1 100

Profit per unit $70 $50

M07_REND9327_12_SE_C07.indd 260 10/02/14 1:25 PM

7.4 graphiCaL soLution to an Lp probLem  261

So the first constraint may be stated as follows:

Carpentry time used …  Carpentry time available 4T + 3C … 240 1hours of carpentry time2

Similarly, the second constraint is as follows: Painting and varnishing time used …  Painting and varnishing time available

○2 T + 1C … 100 1hours of painting and varnishing time2 (This means that each table produced takes two hours of the painting and varnishing 

resource.) Both of these constraints represent production capacity restrictions and, of course, affect the 

total profit. For example, Flair Furniture cannot produce 80 tables during the production period  because if T = 80, both constraints will be violated. It also cannot make T = 50 tables and  C = 10 chairs. Why? Because this would violate the second constraint that no more than 100  hours of painting and varnishing time be allocated.

To obtain meaningful solutions, the values for T and C must be nonnegative numbers. That is,  all potential solutions must represent real tables and real chairs. Mathematically, this means that

T Ú 0 1number of tables produced is greater than or equal to 02 C Ú 0 1number of chairs produced is greater than or equal to 02

The complete problem may now be restated mathematically as

Maximize profit = $70T + $50C

subject to the constraints

4T + 3C … 240 1carpentry constraint2 2T + 1C … 100 1painting and varnishing constraint2

T Ú 0 1first nonnegativity constraint2 C Ú 0 1second nonnegativity constraint2

While the nonnegativity constraints are technically separate constraints, they are often writ- ten on a single line with the variables separated by commas. In this example, this would be  written as

T, C Ú 0

7.4 Graphical Solution to an LP Problem

The easiest way to solve a small LP problem such as that of the Flair Furniture Company is  with the graphical solution approach. The graphical procedure is useful only when there are two  decision variables (such as number of tables to produce, T, and number of chairs to produce, C)  in the problem. When there are more than two variables, it is not possible to plot the solution  on a two-dimensional graph and we must turn to more complex approaches. But the graphical  method is invaluable in providing us with insights into how other approaches work. For that  reason alone, it is worthwhile to spend the rest of this chapter exploring graphical solutions as an  intuitive basis for the chapters on mathematical programming that follow.

Graphical Representation of Constraints To find the optimal solution to an LP problem, we must first identify a set, or region, of feasible  solutions. The first step in doing so is to plot each of the problem’s constraints on a graph. The  variable T (tables) is plotted as the horizontal axis of the graph and the variable C (chairs) is  plotted as the vertical axis. The notation 1T, C2 is used to identify the points on the graph. The  nonnegativity constraints mean that we are always working in the first (or northeast) quadrant  of a graph (see Figure 7.1).

To represent the first constraint graphically, 4T + 3C … 240, we must first graph the equal- ity portion of this, which is

4T + 3C = 240

The resource constraints put limits on the carpentry labor resource and the painting labor resource mathematically.

Here is a complete mathematical statement of the LP problem.

The graphical method works only when there are two decision variables, but it provides valuable insight into how larger problems are structured.

Nonnegativity constraints mean T # 0 and C # 0.

M07_REND9327_12_SE_C07.indd 261 10/02/14 1:25 PM

262  Chapter 7 • Linear programming modeLs: graphiCaL and Computer methods

As you may recall from elementary algebra, a linear equation in two variables is a straight line.  The easiest way to plot the line is to find any two points that satisfy the equation, then draw a  straight line through them.

The two easiest points to find are generally the points at which the line intersects the T and  C axes.

When Flair Furniture produces no tables, namely T = 0, it implies that

4102 + 3C = 240 or

3C = 240

or

C = 80

In other words, if all of the carpentry time available is used to produce chairs, 80 chairs could be  made. Thus, this constraint equation crosses the vertical axis at 80.

To find the point at which the line crosses the horizontal axis, we assume that the firm  makes no chairs, that is, C = 0. Then

4T + 3102 = 240 or

4T = 240

or

T = 60

Hence, when C = 0, we see that 4T = 240 and that T = 60. The carpentry constraint is illustrated in Figure 7.2. It is bounded by the line running from 

point 1T = 0, C = 802 to point 1T = 60, C = 02. Recall, however, that the actual carpentry constraint was the inequality 4T + 3C … 240.

How can we identify all of the solution points that satisfy this constraint? It turns out that  there are three possibilities. First, we know that any point that lies on the line 4T + 3C = 240  satisfies the constraint. Any combination of tables and chairs on the line will use up all 240  hours of carpentry time.* Now we must find the set of solution points that would use less than 

Plotting the first constraint involves finding points at which the line intersects the T and C axes.

Figure 7.1 Quadrant Containing All Positive Values

200 40 60 80 100 T

20

40

60

N um

be r

of C

ha irs 80

100

C

Number of Tables

This Axis Represents the Constraint T ≥ 0

This Axis Represents the Constraint C ≥ 0

*Thus, what we have done is to plot the constraint equation in its most binding position, that is, using all of the   carpentry resource.

M07_REND9327_12_SE_C07.indd 262 10/02/14 1:25 PM

7.4 graphiCaL soLution to an Lp probLem  263

the 240 hours. The points that satisfy the 6  portion of the constraint (i.e., 4T + 3C 6 240) will  be all the points on one side of the line, while all the points on the other side of the line will not  satisfy this condition. To determine which side of the line this is, simply choose any point on  either side of the constraint line shown in Figure 7.2 and check to see if it satisfies this condition.  For example, choose the point (30, 20), as illustrated in Figure 7.3:

41302 + 31202 = 180 Since 180 6 240, this point satisfies the constraint, and all points on this side of the line will  also satisfy the constraint. This set of points is indicated by the shaded region in Figure 7.3.

To see what would happen if the point did not satisfy the constraint, select a point on the  other side of the line, such as (70, 40). This constraint would not be met at this point as

41702 + 31402 = 400 Since 400 7 240, this point and every other point on that side of the line would not satisfy this  constraint. Thus, the solution represented by the point 170, 402 would require more than the  240 hours that are available. There are not enough carpentry hours to produce 70 tables and  40 chairs.

Figure 7.2 Graph of Carpentry Constraint Equation 4T + 3C = 240

200 40 60 80 100 T

20

40

60

N um

be r

of C

ha irs 80

100

C

Number of Tables

( T C

( =60, =0)T C

=0, =80)

Figure 7.3 Region That Satisfies the Carpentry Constraint

200 40 60 80 100 T

20

40

60

N um

be r

of C

ha irs 80

100

C

Number of Tables

(70, 40)

(30, 20)

M07_REND9327_12_SE_C07.indd 263 10/02/14 1:25 PM

264  Chapter 7 • Linear programming modeLs: graphiCaL and Computer methods

Next, let us identify the solution corresponding to the second constraint, which  limits  the time available in the painting and varnishing department. That constraint was given as  2T + 1C … 100. As before, we start by graphing the equality portion of this constraint, which is

2T + 1C = 100

To find two points on the line, select T = 0 and solve for C:

2102 + 1C = 100 C = 100

So, one point on the line is 10, 1002. To find the second point, select C = 0 and solve for T: 2T + 1102 = 100

T = 50

The second point used to graph the line is (50, 0). Plotting this point, (50, 0), and the other point,  (0, 100), results in the line representing all the solutions in which exactly 100 hours of painting  and varnishing time are used, as shown in Figure 7.4.

To find the points that require less than 100 hours, select a point on either side of this line to  see if the inequality portion of the constraint is satisfied. Selecting (0, 0) gives us

2102 + 1102 = 0 6 100 This indicates that this and all the points below the line satisfy the constraint, and this region is  shaded in Figure 7.4.

Now that each individual constraint has been plotted on a graph, it is time to move on to  the next step. We recognize that to produce a chair or a table, both the carpentry and paint- ing and varnishing departments must be used. In an LP problem we need to find that set  of  solution points that satisfies all of the constraints simultaneously. Hence, the constraints  should be  redrawn on one graph (or superimposed one upon the other). This is shown in  Figure 7.5.

The shaded region now represents the area of solutions that does not exceed either of the  two Flair Furniture constraints. It is known by the term area of feasible solutions or, more sim- ply, the feasible region. The feasible region in an LP problem must satisfy all conditions speci- fied by the problem’s constraints, and is thus the region where all constraints overlap. Any point  in the region would be a feasible solution to the Flair Furniture problem; any point outside the  shaded area would represent an infeasible solution. Hence, it would be feasible to manufacture 

In LP problems we are interested in satisfying all constraints at the same time.

The feasible region is the set of points that satisfy all the constraints.

20 40 60 80 100 T

20

40

60

N um

be r

of C

ha irs 80

100

C

Number of Tables

( =0, =100)CT

( =50, =0)CT

0

Figure 7.4 Region That Satisfies the Painting and Varnishing Constraint

M07_REND9327_12_SE_C07.indd 264 10/02/14 1:25 PM

7.4 graphiCaL soLution to an Lp probLem  265

30 tables and 20 chairs 1T = 30, C = 202 during a production period because both constraints  are observed:

Carpentry constraint 4T + 3C … 240 hours available 1421302 + 1321202 = 180 hours used ✓◯

Painting constraint 2T + 1C … 100 hours available 1221302 + 1121202 = 80 hours used ✓◯ But it would violate both of the constraints to produce 70 tables and 40 chairs, as we see here  mathematically:

Carpentry constraint 4T + 3C … 240 hours available 1421702 + 1321402 = 400 hours used ⊗

Painting constraint 2T + 1C … 100 hours available 1221702 + 1121402 = 180 hours used ⊗ Furthermore, it would also be infeasible to manufacture 50 tables and 5 chairs 1T = 50, C = 52. Can you see why?

Carpentry constraint 4T + 3C … 240 hours available 1421502 + 132152 = 215 hours used ✓◯ Painting constraint 2T + 1C … 100 hours available

1221502 + 112152 = 105 hours used ⊗ This possible solution falls within the time available in carpentry but exceeds the time available  in painting and varnishing and thus falls outside the feasible region.

Isoprofit Line Solution Method Now that the feasible region has been graphed, we may proceed to find the optimal solution to  the problem. The optimal solution is the point lying in the feasible region that produces the high- est profit. Yet there are many, many possible solution points in the region. How do we go about  selecting the best one, the one yielding the highest profit?

There are a few different approaches that can be taken in solving for the optimal solution  when the feasible region has been established graphically. The speediest one to apply is called  the isoprofit line method.

We start the technique by letting profits equal some arbitrary but small dollar amount. For  the Flair Furniture problem we may choose a profit of $2,100. This is a profit level that can be 

The isoprofit method is the first method we introduce for finding the optimal solution.

Figure 7.5 Feasible Solution Region for the Flair Furniture Company Problem

20 40 60 80 100 T

20

40

60

N um

be r

of C