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9780357432907, Quantitative Methods for Business, Thirteenth Edition, David R. Anderson, - © Cengage Learning.

David R. Anderson University of Cincinnati

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13e Quantitative Methods for Business

9780357432907, Quantitative Methods for Business, Thirteenth Edition, David R. Anderson, - © Cengage Learning.

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9780357432907, Quantitative Methods for Business, Thirteenth Edition, David R. Anderson, - © Cengage Learning.

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Quantitative Methods for Business, Thirteenth edition

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9780357432907, Quantitative Methods for Business, Thirteenth Edition, David R. Anderson, - © Cengage Learning.

To My Children Krista, Justin, Mark, and Colleen

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To My Children Mark, Linda, Brad, Tim, Scott, and Lisa

DJS

To My Children Cathy, David, and Kristin

TAW

To My Family Karen, Jennifer, Stephanie, and Allison

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To My Wife Teresa

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To My Family Nicole and Ian

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To My Family Amie and Willa

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Brief Contents

Preface xvii

Chapter 1 Introduction 1

Chapter 2 Introduction to Probability 24

Chapter 3 Probability Distributions 61

Chapter 4 Decision Analysis 103

Chapter 5 Utility and Game Theory 164

Chapter 6 Time Series Analysis and Forecasting 195

Chapter 7 Introduction to Linear Programming 249

Chapter 8 Linear Programming: Sensitivity Analysis and Interpretation of Solution 308

Chapter 9 Linear Programming Applications in Marketing, Finance, and Operations Management 362

Chapter 10 Distribution and Network Models 424

Chapter 11 Integer Linear Programming 487

Chapter 12 Advanced Optimization Applications 537

Chapter 13 Project Scheduling: PERT/CPM 592

Chapter 14 Inventory Models 631

Chapter 15 Waiting Line Models 680

Chapter 16 Simulation 719

Chapter 17 Markov Processes 770

Appendix A Building Spreadsheet Models 796

Appendix B Binomial Probabilities 828

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Appendix C Poisson Probabilities 835

Appendix D Areas for the Standard Normal Distribution 841

Appendix E Values of e2λ 843

Appendix F References and Bibliography 844

Appendix G Self-Test Solutions and Answers to Even-Numbered Problems 846

Index 909

vi Brief Contents

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Contents

Preface xvii

Chapter 1 Introduction 1 1.1 Problem Solving and Decision Making 3 1.2 Quantitative Analysis and Decision Making 5 1.3 Quantitative Analysis 7

Model Development 7 Data Preparation 10 Model Solution 11 Report Generation 13 A Note Regarding Implementation 13

1.4 Models of Cost, Revenue, and Profit 14 Cost and Volume Models 14 Revenue and Volume Models 15 Profit and Volume Models 15 Breakeven Analysis 16

1.5 Quantitative Methods in Practice 17 Summary 19 Glossary 19 Problems 20 Case Problem Scheduling a Golf League 23

Chapter 2 Introduction to Probability 24 2.1 Experiments and the Sample Space 26 2.2 Assigning Probabilities to Experimental

Outcomes 28 Classical Method 28 Relative Frequency Method 29 Subjective Method 29

2.3 Events and Their Probabilities 30 2.4 Some Basic Relationships of Probability 31

Complement of an Event 31 Addition Law 32 Conditional Probability 35 Multiplication Law 39

2.5 Bayes’ Theorem 40 The Tabular Approach 43

2.6 Simpson’s Paradox 45 Summary 47 Glossary 48 Problems 49 Case Problem Hamilton County Judges 57 Case Problem College Softball Recruiting 59

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Chapter 3 Probability Distributions 61 3.1 Random Variables 63 3.2 Discrete Random Variables 64

Probability Distribution of a Discrete Random Variable 64 Expected Value 66 Variance 67

3.3 Binomial Probability Distribution 69 Nastke Clothing Store Problem 70 Expected Value and Variance for the Binomial Distribution 73

3.4 Poisson Probability Distribution 73 An Example Involving Time Intervals 74 An Example Involving Length or Distance Intervals 74

3.5 Continuous Random Variables 76 Applying the Uniform Distribution 76 Area as a Measure of Probability 77

3.6 Normal Probability Distribution 80 Standard Normal Distribution 81 Computing Probabilities for Any Normal Distribution 85 Grear Tire Company Problem 87

3.7 Exponential Probability Distribution 88 Computing Probabilities for the Exponential Distribution 89 Relationship Between the Poisson and Exponential Distributions 90

Summary 91 Glossary 91 Problems 92 Case Problem Specialty Toys 99 Appendix 3.1 Computing Discrete Probabilities with Excel 100 Appendix 3.2 Computing Probabilities for Continuous Distributions

with Excel 101

Chapter 4 Decision Analysis 103 4.1 Problem Formulation 105

Influence Diagrams 106 Payoff Tables 106 Decision Trees 107

4.2 Decision Making Without Probabilities 109 Optimistic Approach 109 Conservative Approach 109 Minimax Regret Approach 110

4.3 Decision Making with Probabilities 112 Expected Value of Perfect Information 114

4.4 Risk Analysis and Sensitivity Analysis 116 Risk Analysis 116 Sensitivity Analysis 117

4.5 Decision Analysis with Sample Information 122 Influence Diagram 122 Decision Tree 123 Decision Strategy 125 Risk Profile 130

viii Contents

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Expected Value of Sample Information 131 Efficiency of Sample Information 132

4.6 Computing Branch Probabilities with Bayes’ Theorem 133 Summary 137 Glossary 138 Problems 139 Case Problem 1 Property Purchase Strategy 153 Case Problem 2 Lawsuit Defense Strategy 154 Appendix 4.1 Using Analytic Solver Platform to Create Decision Trees 155

Chapter 5 Utility and Game Theory 164 5.1 The Meaning of Utility 165 5.2 Utility and Decision Making 167

The Expected Utility Approach 169 Summary of Steps for Determining the Utility of Money 171

5.3 Utility: Other Considerations 172 Risk Avoiders Versus Risk Takers 172

5.4 Introduction to Game Theory 177 Competing for Market Share 178 Identifying a Pure Strategy 180

5.5 Mixed Strategy Games 181 A Larger Mixed Strategy Game 183 Summary of Steps for Solving Two-Person, Zero-Sum Games 185 Extensions 185

Summary 185 Glossary 186 Problems 186 Case Problem Utility, Game Theory, and Product Line Extension Decisions 193

Chapter 6 Time Series Analysis and Forecasting 195 6.1 Time Series Patterns 197

Horizontal Pattern 197 Trend Pattern 199 Seasonal Pattern 201 Trend and Seasonal Pattern 203 Cyclical Pattern 204 Selecting a Forecasting Method 204

6.2 Forecast Accuracy 206 6.3 Moving Averages and Exponential Smoothing 211

Moving Averages 211 Weighted Moving Averages 214 Exponential Smoothing 215

6.4 Linear Trend Projection 218 6.5 Seasonality 223

Seasonality Without Trend 223 Seasonality with Trend 226 Models Based on Monthly Data 228

Contents ix

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Summary 229 Glossary 230 Problems 230 Case Problem 1 Forecasting Food and Beverage Sales 238 Case Problem 2 Forecasting Lost Sales 239 Appendix 6.1 Forecasting with Excel Data Analysis Tools 240

Chapter 7 Introduction to Linear Programming 249 7.1 A Simple Maximization Problem 251

Problem Formulation 252 Mathematical Model for the RMC Problem 254

7.2 Graphical Solution Procedure 255 A Note on Graphing Lines 264 Summary of the Graphical Solution Procedure for Maximization Problems 265 Slack Variables 266

7.3 Extreme Points and the Optimal Solution 268 7.4 Computer Solution of the RMC Problem 269

Interpretation of Answer Report 270 7.5 A Simple Minimization Problem 271

Summary of the Graphical Solution Procedure for Minimization Problems 273 Surplus Variables 274 Computer Solution of the M&D Chemicals Problem 275

7.6 Special Cases 276 Alternative Optimal Solutions 276 Infeasibility 276 Unbounded 278

7.7 General Linear Programming Notation 280 Summary 282 Glossary 283 Problems 284 Case Problem 1 Workload Balancing 299 Case Problem 2 Production Strategy 300 Case Problem 3 Hart Venture Capital 301 Appendix 7.1 Solving Linear Programs with Excel Solver 302 Appendix 7.2 Solving Linear Programs with LINGO 306

Chapter 8 Linear Programming: Sensitivity Analysis and Interpretation of Solution 308

8.1 Introduction to Sensitivity Analysis 310 8.2 Objective Function Coefficients 311 8.3 Right-Hand Sides 314

Cautionary Note on the Interpretation of Shadow Prices 318 8.4 Limitations of Classical Sensitivity Analysis 319

Simultaneous Changes 319 Changes in Constraint Coefficients 320 Nonintuitive Shadow Prices 321

8.5 More Than Two Decision Variables 323 Modified RMC Problem 324 Bluegrass Farms Problem 326

x Contents

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8.6 Electronic Communications Problem 329 Problem Formulation 330 Solution and Interpretation 331

Summary 334 Glossary 335 Problems 336 Case Problem 1 Product Mix 355 Case Problem 2 Investment Strategy 356 Case Problem 3 Truck Leasing Strategy 356 Appendix 8.1 Sensitivity Analysis with Excel Solver 357 Appendix 8.2 Sensitivity Analysis with LINGO 358

Chapter 9 Linear Programming Applications in Marketing, Finance, and Operations Management 362

9.1 Marketing Applications 364 Media Selection 364 Marketing Research 367

9.2 Financial Applications 370 Portfolio Selection 370 Financial Planning 373

9.3 Operations Management Applications 377 A Make-or-Buy Decision 377 Production Scheduling 381 Workforce Assignment 388 Blending Problems 393

Summary 397 Problems 398 Case Problem 1 Planning an Advertising Campaign 411 Case Problem 2 Schneider’s Sweet Shop 413 Case Problem 3 Textile Mill Scheduling 414 Case Problem 4 Workforce Scheduling 415 Case Problem 5 Duke Energy Coal Allocation2 417 Appendix 9.1 Excel Solution of Hewlitt Corporation Financial

Planning Problem 419

Chapter 10 Distribution and Network Models 424 10.1 Supply Chain Models 425

Transportation Problem 425 Problem Variations 428 A General Linear Programming Model 431 Transshipment Problem 432 Problem Variations 438 A General Linear Programming Model 438

10.2 Assignment Problem 440 Problem Variations 443 A General Linear Programming Model 443

10.3 Shortest-Route Problem 445 A General Linear Programming Model 448

Contents xi

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10.4 Maximal Flow Problem 449 10.5 A Production and Inventory Application 453 Summary 456 Glossary 457 Problems 458 Case Problem 1 Solutions Plus 475 Case Problem 2 Supply Chain Design for the Darby Company 477 Case Problem 3 DK Dental Care 478 Appendix 10.1 Excel Solver Solution of Transportation, Transshipment,

and Assignment Problems 480

Chapter 11 Integer Linear Programming 487 11.1 Types of Integer Linear Programming Models 489 11.2 Graphical and Computer Solutions for an All-Integer Linear Program 491

Graphical Solution of the LP Relaxation 492 Rounding to Obtain an Integer Solution 493 Graphical Solution of the All-Integer Problem 493 Using the LP Relaxation to Establish Bounds 494 Computer Solution 495

11.3 Applications Involving 0-1 Variables 496 Capital Budgeting 496 Fixed Cost 497 Supply Chain Design 499 Bank Location 504 Product Design and Market Share Optimization 506

11.4 Modeling Flexibility Provided by 0-1 Integer Variables 511 Multiple-Choice and Mutually Exclusive Constraints 511 k Out of n Alternatives Constraint 512 Conditional and Corequisite Constraints 512 A Cautionary Note About Sensitivity Analysis 514

Summary 515 Glossary 515 Problems 516 Case Problem 1 Textbook Publishing 527 Case Problem 2 Yeager National Bank 528 Case Problem 3 Production Scheduling with Changeover Costs 529 Case Problem 4 Applecore Children’s Clothing 530 Appendix 11.1 Excel Solver Solution of Integer Linear Programs 532 Appendix 11.2 LINGO Solution of Integer Linear Programs 535

Chapter 12 Advanced Optimization Applications 537 12.1 Data Envelopment Analysis 538

Evaluating the Performance of Hospitals 539 Overview of the DEA Approach 540 DEA Linear Programming Model 541 Summary of the DEA Approach 545

12.2 Revenue Management 546

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12.3 Portfolio Models and Asset Allocation 552 A Portfolio of Mutual Funds 552 Conservative Portfolio 553 Moderate Risk Portfolio 556

12.4 Nonlinear Optimization—The RMC Problem Revisited 559 An Unconstrained Problem 560 A Constrained Problem 561 Local and Global Optima 564 Shadow Prices 566

12.5 Constructing an Index Fund 568 Summary 572 Glossary 573 Problems 573 Case Problem CAFE Compliance in the Auto Industry 585 Appendix 12.1 Solving Nonlinear Problems with LINGO 587 Appendix 12.2 Solving Nonlinear Problems with Excel Solver 590

Chapter 13 Project Scheduling: PERT/CPM 592 13.1 Project Scheduling Based on Expected

Activity Times 593 The Concept of a Critical Path 595 Determining the Critical Path 596 Contributions of PERT/CPM 601 Summary of the PERT/CPM Critical Path Procedure 601

13.2 Project Scheduling Considering Uncertain Activity Times 602

The Daugherty Porta-Vac Project 602 Uncertain Activity Times 603 The Critical Path 606 Variability in Project Completion Time 608

13.3 Considering Time–Cost Trade-Offs 611 Crashing Activity Times 612 Linear Programming Model for Crashing 614

Summary 617 Glossary 617 Problems 618 Case Problem R. C. Coleman 628 Appendix 13.1 Finding Cumulative Probabilities for

Normally Distributed Random Variables 629

Chapter 14 Inventory Models 631 14.1 Economic Order Quantity (EOQ) Model 633

The How-Much-to-Order Decision 637 The When-to-Order Decision 638 Sensitivity Analysis for the EOQ Model 639 Excel Solution of the EOQ Model 640 Summary of the EOQ Model Assumptions 641

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14.2 Economic Production Lot Size Model 642 Total Cost Model 643 Economic Production Lot Size 645

14.3 Inventory Model with Planned Shortages 645 14.4 Quantity Discounts for the EOQ Model 650 14.5 Single-Period Inventory Model with Probabilistic Demand 652

Neiman Marcus 653 Nationwide Car Rental 656

14.6 Order-Quantity, Reorder Point Model with Probabilistic Demand 658 The How-Much-to-Order Decision 659 The When-to-Order Decision 660

14.7 Periodic Review Model with Probabilistic Demand 662 More Complex Periodic Review Models 665

Summary 666 Glossary 666 Problems 667 Case Problem 1 Wagner Fabricating Company 675 Case Problem 2 River City Fire Department 676 Appendix 14.1 Development of the Optimal Order Quantity (Q*)

Formula for the EOQ Model 677 Appendix 14.2 Development of the Optimal Lot Size (Q*) Formula for the

Production Lot Size Model 678

Chapter 15 Waiting Line Models 680 15.1 Structure of a Waiting Line System 682

Single-Server Waiting Line 682 Distribution of Arrivals 682 Distribution of Service Times 684 Queue Discipline 685 Steady-State Operation 685

15.2 Single-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times 685

Operating Characteristics 685 Operating Characteristics for the Burger Dome Problem 687 Managers’ Use of Waiting Line Models 687 Improving the Waiting Line Operation 688 Excel Solution of Waiting Line Model 689

15.3 Multiple-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times 690

Operating Characteristics 690 Operating Characteristics for the Burger Dome Problem 692

15.4 Some General Relationships for Waiting Line Models 694 15.5 Economic Analysis of Waiting Lines 696 15.6 Other Waiting Line Models 698 15.7 Single-Server Waiting Line Model with Poisson Arrivals and

Arbitrary Service Times 699 Operating Characteristics for the M/G/1 Model 699 Constant Service Times 700

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15.8 Multiple-Server Model with Poisson Arrivals, Arbitrary Service Times, and No Waiting Line 701

Operating Characteristics for the M/G/k Model with Blocked Customers Cleared 701

15.9 Waiting Line Models with Finite Calling Populations 703 Operating Characteristics for the M/M/1 Model with a Finite Calling Population 704

Summary 706 Glossary 708 Problems 708 Case Problem 1 Regional Airlines 716 Case Problem 2 Office Equipment, Inc. 717

Chapter 16 Simulation 719 16.1 What-If Analysis 721

Sanotronics 721 Base-Case Scenario 721 Worst-Case Scenario 722 Best-Case Scenario 722

16.2 Simulation of Sanotronics Problem 723 Use of Probability Distributions to Represent Random Variables 723 Generating Values for Random Variables with Excel 725 Executing Simulation Trials with Excel 729 Measuring and Analyzing Simulation Output 730

16.3 Inventory Simulation 732 Simulation of the Butler Inventory Problem 735

16.4 Waiting Line Simulation 737 Black Sheep Scarves 738 Customer (Scarf) Arrival Times 738 Customer (Scarf) Service (Inspection) Times 739 Simulation Model 739 Simulation of Black Sheep Scarves 742 Simulation with Two Quality Inspectors 744 Simulation Results with Two Quality Inspectors 745

16.5 Simulation Considerations 747 Verification and Validation 747 Advantages and Disadvantages of Using Simulation 747

Summary 748 Glossary 749 Problems 750 Case Problem 1 Four Corners 757 Case Problem 2 Harbor Dunes Golf Course 759 Case Problem 3 County Beverage Drive-Thru 761 Appendix 16.1 Probability Distributions for Random Variables 762 Appendix 16.2 Simulation with Analytic Solver Platform 765

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Chapter 17 Markov Processes 770 17.1 Market Share Analysis 772 17.2 Accounts Receivable Analysis 779

Fundamental Matrix and Associated Calculations 781 Establishing the Allowance for Doubtful Accounts 782

Summary 784 Glossary 785 Problems 785 Case Problem Dealer’s Absorbing State Probabilities in Blackjack 790 Appendix 17.1 Matrix Notation and Operations 791 Appendix 17.2 Matrix Inversion with Excel 794

Appendix A Building Spreadsheet Models 796

Appendix B Binomial Probabilities 828

Appendix C Poisson Probabilities 835

Appendix D Areas for the Standard Normal Distribution 841

Appendix E Values of e2λ 843

Appendix F References and Bibliography 844

Appendix G Self-Test Solutions and Answers to Even-Numbered Problems 846

Index 909

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Preface

The purpose of this thirteenth edition, as with previous editions, is to provide undergraduate and graduate students with a conceptual understanding of the role that quantitative methods play in the decision-making process. The text describes the many quantitative methods developed over the years, explains how they work, and shows how the decision maker can apply and interpret them.

This book is applications-oriented and uses our problem scenario approach to gently introduce quantitative material. In each chapter, a problem is described in conjunction with the quantitative procedure being introduced. Development of the quantitative technique or model includes applying it to the problem to generate a solution or recommendation. This approach can help to motivate the student by demonstrating not only how the procedure works, but also how it contributes to the decision-making process.

The mathematical prerequisite for this text is an algebra course. The two chapters on probability and probability distributions will provide the necessary background for the use of probability in subsequent chapters. Throughout the text we use generally accepted nota- tion for the topic being covered. As a result, students who pursue study beyond the level of this text will generally experience little difficulty reading more advanced material. To also assist in further study, a bibliography is included at the end of this book.

CHANGES IN THE THIRTEENTH EDITION

We are very excited about the changes in the thirteenth edition of Quantitative Methods for Business, and want to tell you about some of the changes we have made and why.

Updated Chapter 16: Simulation The most substantial content change in this latest edition involves the coverage of simu- lation. We maintain an intuitive introduction by continuing to use the concepts of best-, worst-, and base-case scenarios, but we have added a more elaborate treatment of uncer- tainty by using Microsoft Excel to develop spreadsheet simulation models. Within the chapter, we explain how to construct a spreadsheet simulation model using only native Excel functionality. In the chapter appendix, we describe how an Excel add-in, Analytic Solver Platform, facilitates more sophisticated simulation analyses. This new appendix on Analytic Solver Platform replaces the previous edition’s coverage of Crystal Ball, which we no longer pair with our textbook. Nine new problems are introduced, and several others have been updated to reflect the new simulation coverage.

Other Content Changes A variety of other changes have been made throughout the text in response to user sugges- tions. The most prominent of these include a new section on variability in project manage- ment in Chapter 13, new Appendix A coverage of data tables and Goal Seek functionality in Excel 2013, and adjustment of forecasting notation in Chapter 6. The software previously used to create decision trees in the Chapter 4 appendix, TreePlan, has now been incorporated into the Excel add-in Analytic Solver Platform, and we have updated Chapter 4 accordingly.

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New Q.M. in Action, Cases, and Problems Q.M. in Action is the name of the short summaries that describe how the quantitative methods being covered in the chapter have been used in practice. In this edition, you will find numerous Q.M. in Action vignettes, cases, and homework problems. We have updated many of these Q.M. in Actions to provide more recent examples. In all, we have added 15 new Q.M. in Actions.

The end of each chapter of this book contains cases for students. The cases are more in-depth and often more open-ended than the end-of-chapter homework problems. We have added three new cases to this edition: one on linear programming applications in Chapter 9, one on distribution and network models in Chapter 10, and one on integer programming in Chapter 11. Solutions to all cases are available to instructors.

We have added more than 35 new homework problems to this edition. Many other homework problems have been updated to provide more timely references.

FEATURES AND PEDAGOGY

We continued many of the features that appeared in previous editions. Some of the impor- tant ones are noted here.

● Annotations: Annotations that highlight key points and provide additional insights for the student are a continuing feature of this edition. These annotations, which appear in the margins, are designed to provide emphasis and enhance understanding of the terms and concepts presented in the text.

● Notes and Comments: We provide Notes and Comments at the end of many sections to give the student additional insights about the methodology being discussed and its application. These insights include warnings about or limitations of the method- ology, recommendations for application, brief descriptions of additional technical considerations, and other matters.

● Self-Test Exercises: Certain exercises are identified as self-test exercises. Completely worked-out solutions for these exercises are provided in Appendix G, entitled Self- Test Solutions and Answers to Even-Numbered Problems, located at the end of the book. Students can attempt the self-test problems and immediately check the solutions to evaluate their understanding of the concepts presented in the chapter. At the request of professors using our textbooks, we now provide the answers to even-numbered problems in this same appendix.

● Q.M. in Action: These articles are presented throughout the text and provide a sum- mary of an application of quantitative methods found in business today. Adaptations of materials from the popular press, academic journals such as Interfaces, and write- ups provided by practitioners provide the basis for the applications in this feature.

ANCILLARY LEARNING AND TEACHING MATERIALS

For Students Print and online resources are available to help the student work more efficiently as well as learn how to use Excel.

● LINGO: The student version of LINGO 14.0 software is available for download at no additional cost to students who purchase a new text, through a link on the student companion site.

xviii Preface

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● Analytic Solver Platform: An educational version of the latest version of the Analytic Solver Platform software is available at no cost with a new text.

For Instructors Instructor ancillaries are now provided on the website. Included in this convenient format are the following:

● Solutions Manual: The Solutions Manual, prepared by the authors, includes solu- tions for all problems in the text.

● Solutions to Case Problems: Also prepared by the authors, it contains solutions to all case problems presented in the text.

● PowerPoint Presentation Slides: Prepared by John Loucks of St. Edwards University, the presentation slides contain a teaching outline that incorporates graphics to help instructors create even more stimulating lectures. The slides may be adapted using PowerPoint software to facilitate classroom use.

● Test Bank: Also prepared by John Loucks, the Test Bank in Microsoft Word files includes multiple choice, true/false, short-answer questions, and problems for each chapter.

Cengage Learning Testing Powered by Cognero is a flexible, online system that allows you to:

● author, edit, and manage test bank content from multiple Cengage Learning solutions ● create multiple test versions in an instant ● deliver tests from your LMS, your classroom, or wherever you want

COURSE OUTLINE FLEXIBILITY

The text provides instructors with substantial flexibility in selecting topics to meet specific course needs. Although many variations are possible, the single-semester and single- quarter outlines that follow are illustrative of the options available.

Suggested One-Semester Course Outline

Introduction (Chapter 1)

Probability Concepts (Chapters 2 and 3)

Decision Analysis (Chapters 4 and 5)

Forecasting (Chapter 6)

Linear Programming (Chapters 7, 8, and 9)

Distribution and Network Models (Chapter 10)

Integer Linear Programming (Chapter 11)

Advanced Optimization Applications (Chapter 12)

Project Scheduling: PERT/CPM (Chapter 13)

Simulation (Chapter 15)

Suggested One-Quarter Course Outline

Introduction (Chapter 1)

Decision Analysis (Chapters 4 and 5)

Linear Programming (Chapters 7, 8, and 9)

Distribution and Network Models (Chapter 10)

Integer Linear Programming (Chapter 11)

Preface xix

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Advanced Optimization Applications (Chapter 12)

Project Scheduling: PERT/CPM (Chapter 13)

Simulation (Chapter 15)

Many other possibilities exist for one-term courses, depending on the time available, course objectives, and backgrounds of the students.

ACKNOWLEDGMENTS

We were fortunate in having the thoughts and comments of a number of colleagues as we began work on this thirteenth edition of Quantitative Methods for Business. Our apprecia- tion and thanks go to:

xx Preface

Larry Barchett Ohio Valley University

Stacie A. Bosley Hamline University

Gregory Chase West Liberty University

Ali A. Choudry Embry-Riddle University

Mary Fletcher Mount Mary University

Edward Gordhammer Embry-Riddle Aeronautical University

Wendy Keyes California Baptist University

Kenneth Lawrence New Jersey Institute of Technology

Holly S. Lutze Texas Lutheran University

Penina Orenstein Seton Hall University

Edward R. Sim Hardin-Simmons University

Rahmat Tavallali Walsh University

John S. Watters McKendree University

Susan M. L. Zee Southeastern Louisiana University

Ellen Parker Allen Southern Methodist University

Gopesh Anand The Ohio State University

Robert L. Armacost University of Central Florida

Daniel Asera University of Nevada, Las Vegas

Uttarayan Bagchi University of Texas at Austin

Stephen Baglione Saint Leo University

Ardith Baker Oral Roberts University

Edward Baker University of Miami

Norman Baker University of Cincinnati

David Bakuli Westfield State College

Robert T. Barrett Francis Marion University

Oded Berman University of Toronto

Gary Blau Purdue University

William Bleuel Pepperdine University

Richard G. Bradford Avila University

Thomas Bundt Hillsdale College

Heidi Burgiel Bridgewater State College

Rodger D. Carlson Morehead State University

Ying Chien University of Scranton

Renato Clavijo Robert Morris University

Ron Craig Wilfrid Laurier University

Mary M. Danaher Florida Atlantic University

Stanley Dick Babson College

Swarna D. Dutt State University of West Georgia

Writing and revising a textbook is a continuing process. We owe a debt to many of our colleagues and friends for their helpful comments and suggestions during the development of earlier editions. Among these are the following:

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Preface xxi

John Eatman University of North Carolina–Greensboro

Ronald Ebert University of Missouri– Columbia

Charlie Edmonson University of Dayton

Don Edwards University of South Carolina

Ronald Ehresman Baldwin-Wallace College

Peter Ellis Utah State University

Lawrence Ettkin University of Tennessee at Chattanooga

James Evans University of Cincinnati

Paul Ewell, Bridgewater College

Ephrem Eyob Virginia State University

Michael Ford Rochester Institute of Technology

Terri Friel Eastern Kentucky University

Phil Fry Boise State University

Christian V. Fugar Dillard University

Robert Garfinkel University of Connecticut

Alfredo Gomez, Florida Atlantic University

Bob Gregory Bellevue University

Leland Gustafson State University of West Georgia

Joseph Haimowitz Avila University

Nicholas G. Hall The Ohio State University

Michael E. Hanna University of Houston– Clear Lake

John Hanson University of San Diego

William V. Harper Otterbein College

Melanie Hatch Miami University

Harry G. Henderson Davis & Elkins College

Carl H. Hess Marymount University

Daniel G. Hotard Southeastern Louisiana University

David Hott Florida Institute of Technology

Woodrow W. Hughes Jr., Converse College

Christine Irujo Westfield State College

Barry Kadets Bryant College

Birsen Karpak Youngstown State University

William C. Keller Webb Institute of the University of Phoenix

Christos Koulamas Florida International University

M. S. Krishnamoorthy Alliant International University

Melvin H. Kuhbander Sullivan University

Anil Kukreja Xavier University of Louisiana

Alireza Lari Fayetteville State University

John Lawrence, Jr. California State University–Fullerton

Jodey Lingg City University

John S. Loucks St. Edwards University

Constantine Loucopoulos Emporia State University

Donald R. MacRitchie Framingham State College

Larry Maes Davenport University

Ka-sing Man Georgetown University

William G. Marchal University of Toledo

Barbara J. Mardis University of Northern Iowa

Kamlesh Mathur Case Western Reserve University

Joseph Mazzola Duke University

Timothy McDaniel Buena Vista University

Patrick McKeown University of Georgia

Constance McLaren Indiana State University

Mohammad Meybodi Indiana University– Kokomo

John R. Miller Mercer University

Mario Miranda The Ohio State University

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xxii Preface

Joe Moffitt University of Massachusetts

Saeed Mohaghegh Assumption College

Herbert Moskowitz Purdue University

Shahriar Mostashari Campbell University– School of Business

Alan Neebe University of North Carolina

V. R. Nemani Trinity College

William C. O’Connor University of Montana– Western

Donald A. Ostasiewski Thomas More College

David Pentico Duquesne University

John E. Powell University of South Dakota

B. Madhusudan Rao Bowling Green State University

Handanhal V. Ravinder University of New Mexico

Avuthu Rami Reddy University of Wisconsin

Donna Retzlaff-Roberts University of Memphis

Don R. Robinson Illinois State University

Richard Rosenthal Naval Postgraduate School

Kazim Ruhi University of Maryland

Susan D. Sandblom Scottsdale Community College

Tom Schmidt Simpson College

Antoinette Somers Wayne State University

Rajesh Srivastava Florida Gulf Coast University

Donald E. Stout, Jr. Saint Martin’s College

Minghe Sun University of Texas at San Antonio

Christopher S. Tang University of California– Los Angeles

Giri Kumar Tayi State University of New York–Albany

Willban Terpening Gonzaga University

Dothang Truong Fayetteville State University

Vicente A. Vargas University of San Diego

William Vasbinder Becker College

Emre Veral City University of New York–Baruch

Elizabeth J. Wark Springfield College

John F. Wellington Indiana University–Purdue University, Fort Wayne

Robert P. Wells Becker College

Laura J. White University of West Florida

Edward P. Winkofsky University of Cincinnati

Cynthia Woodburn Pittsburg State University

Neba L J Wu Eastern Michigan University

Kefeng Xu University of Texas at San Antonio

Mari Yetimyan San Jose State University

Our associates from organizations who provided application write-ups made a major contribution to the text. These individuals are cited in a credit line on the associated Q.M. in Action.

We are also indebted to our product manager, Aaron Arnsparger; our marketing man- ager, Heather Mooney; our senior content developer, Maggie Kubale; our content project manager, Jana Lewis; our media editor, Chris Valentine; and others at Cengage Learning for their counsel and support during the preparation of this text.

David R. Anderson Dennis J. Sweeney

Thomas A. Williams Jeffrey D. Camm

James J. Cochran Michael J. Fry

Jeffrey W. Ohlmann

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Introduction

CONTENTS

1.1 PROBLEM SOLVING AND DECISION MAKING

1.2 QUANTITATIVE ANALYSIS AND DECISION MAKING

1.3 QUANTITATIVE ANALYSIS Model Development Data Preparation Model Solution Report Generation A Note Regarding

Implementation

1.4 MODELS OF COST, REVENUE, AND PROFIT Cost and Volume Models Revenue and Volume Models Profit and Volume Models Breakeven Analysis

1.5 QUANTITATIVE METHODS IN PRACTICE Methods Used Most Frequently

CHAPTER 1

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2 Chapter 1 Introduction

This book is concerned with the use of quantitative methods to assist in decision making. It emphasizes not the methods themselves, but rather how they can contribute to better decisions. A variety of names exists for the body of knowledge involving quantitative approaches to deci- sion making. Today, the terms most commonly used—management science (MS), operations research (OR), decision science, and business analytics—are often used interchangeably.

The scientific management revolution of the early 1900s, initiated by Frederic W. Taylor, provided the foundation for the use of quantitative methods in management. How- ever, modern research in the use of quantitative methods in decision making, for the most part, originated during the World War II period. At that time, teams of people with diverse specialties (e.g., mathematicians, engineers, and behavioral scientists) were formed to deal with strategic and tactical problems faced by the military. After the war, many of these team members continued their research into quantitative approaches to decision making.

Two developments that occurred during the post–World War II period led to the growth and use of quantitative methods in nonmilitary applications. First, continued research re- sulted in numerous methodological developments. Arguably the most notable of these devel- opments was the discovery by George Dantzig, in 1947, of the simplex method for solving linear programming problems. At the same time these methodological developments were taking place, digital computers prompted a virtual explosion in computing power. Comput- ers enabled practitioners to use the methodological advances to solve a large variety of problems. The computer technology explosion continues, and personal computers can now be used to solve problems larger than those solved on mainframe computers in the 1990s.

To reinforce the applied nature of the text and to provide a better understanding of the variety of applications in which quantitative methods (Q.M.) have been used successfully, Q.M. in Action articles are presented throughout the text. Each Q.M. in Action article sum- marizes an application of quantitative methods in practice. The first Q.M. in Action, Revenue Management at AT&T Park, describes one of the most important applications of quantita- tive methods in the sports and entertainment industry.

Imagine the difficult position Russ Stanley, Vice Presi- dent of Ticket Services for the San Francisco Giants, found himself facing late in the 2010 baseball season. Prior to the season, his organization had adopted a dy- namic approach to pricing its tickets similar to the model successfully pioneered by Thomas M. Cook and his op- erations research group at American Airlines. Stanley desparately wanted the Giants to clinch a playoff birth, but he didn’t want the team to do so too quickly.

When dynamically pricing a good or service, an or- ganization regularly reviews supply and demand of the

product and uses operations research to determine if the price should be changed to reflect these conditions. As the scheduled takeoff date for a flight nears, the cost of a ticket increases if seats for the flight are relatively scarce. On the other hand, the airline discounts tickets for an approaching flight with relatively few ticketed passengers. Through the use of optimization to dynami- cally set ticket prices, American Airlines generates nearly $1 billion annually in incremental revenue.

The management team of the San Francisco Giants recognized similarities between their primary product (tickets to home games) and the primary product sold by airlines (tickets for flights) and adopted a similar rev- enue management system. If a particular Giants’ game is appealing to fans, tickets sell quickly and demand far

REVENUE MANAGEMENT AT AT&T PARK*

Q.M. in ACtIon

(continued)

*Based on Peter Horner, “the Sabre Story,” OR/MS Today (June 2000); Ken Belson, “Baseball tickets too Much? Check Back tomor- row,” New York Times.com (May 18, 2009); and Rob Gloster, “Giants Quadruple Price of Cheap Seats as Playoffs Drive Demand,” Bloom- berg Businessweek (September 30, 2010).

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1.1 Problem Solving and Decision Making 3

1.1 Problem Solving and Decision Making Problem solving can be defined as the process of identifying a difference between the actual and the desired state of affairs and then taking action to resolve this difference. For problems important enough to justify the time and effort of careful analysis, the problem- solving process involves the following seven steps:

1. Identify and define the problem. 2. Determine the set of alternative solutions. 3. Determine the criterion or criteria that will be used to evaluate the alternatives. 4. Evaluate the alternatives. 5. Choose an alternative. 6. Implement the selected alternative. 7. Evaluate the results to determine whether a satisfactory solution has been obtained.

Decision making is the term generally associated with the first five steps of the problem-solving process. Thus, the first step of decision making is to identify and define the problem. Decision making ends with the choosing of an alternative, which is the act of making the decision.

Let us consider the following example of the decision-making process. For the moment, assume you are currently unemployed and that you would like a position that will lead to a satisfying career. Suppose your job search results in offers from companies in Rochester, New York; Dallas, Texas; Greensboro, North Carolina; and Pittsburgh, Pennsylvania. Fur- ther suppose that it is unrealistic for you to decline all of these offers. Thus, the alternatives for your decision problem can be stated as follows:

1. Accept the position in Rochester. 2. Accept the position in Dallas. 3. Accept the position in Greensboro. 4. Accept the position in Pittsburgh.

exceeds supply as the date of the game approaches; under these conditions fans will be willing to pay more and the Giants charge a premium for the ticket. Similarly, tickets for less attractive games are discounted to reflect relatively low demand by fans. This is why Stanley found himself in a quandary at the end of the 2010 baseball season. The Giants were in the middle of a tight pennant race with the San Diego Padres that effectively increased demand for tickets to Giants’ games, and the team was actually sched- uled to play the Padres in San Fransisco for the last three games of the season. While Stanley certainly wanted his club to win its division and reach the Major League Base- ball playoffs, he also recognized that his team’s revenues would be greatly enhanced if it didn’t qualify for the play- offs until the last day of the season. “I guess financially it is

better to go all the way down to the last game,” Stanley said in a late season interview. “Our hearts are in our stomachs; we’re pacing watching these games.”

Does revenue management and operations research work? Today, virtually every airline uses some sort of revenue-management system, and the cruise, hotel, and car rental industries also now apply revenue-management methods. As for the Giants, Stanley said dynamic pricing provided a 7 to 8% increase in revenue per seat for Giants’ home games during the 2010 season. Coincidentally, the Giants did win the National League West division on the last day of the season and ultimately won the World Series. Several professional sports franchises are now looking to the Giants’ example and considering implementation of similar dynamic ticket-pricing systems.

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4 Chapter 1 Introduction

The next step of the problem-solving process involves determining the criteria that will be used to evaluate the four alternatives. Obviously, the starting salary is a factor of some importance. If salary were the only criterion important to you, the alternative selected as “best” would be the one with the highest starting salary. Problems in which the objective is to find the best solution with respect to one criterion are referred to as single-criterion decision problems.

Suppose that you also conclude that the potential for advancement and the location of the job are two other criteria of major importance. Thus, the three criteria in your decision problem are starting salary, potential for advancement, and location. Problems that involve more than one criterion are referred to as multicriteria decision problems.

The next step of the decision-making process is to evaluate each of the alternatives with respect to each criterion. For example, evaluating each alternative relative to the starting salary criterion is done simply by recording the starting salary for each job alternative. However, evaluating each alternative with respect to the potential for advancement and the location of the job is more difficult because these evaluations are based primarily on subjective factors that are often difficult to quantify. Suppose for now that you decide to measure potential for advancement and job location by rating each of these criteria as poor, fair, average, good, or excellent. The data you compile are shown in Table 1.1.

You are now ready to make a choice from the available alternatives. What makes this choice phase so difficult is that the criteria are probably not all equally important, and no one alternative is “best” with regard to all criteria. When faced with a multicriteria deci- sion problem, the third step in the decision-making process often includes an assessment of the relative importance of the criteria. Although we will present a method for dealing with situations like this one later in the text, for now let us suppose that after a careful evaluation of the data in Table 1.1, you decide to select alternative 3. Alternative 3 is thus referred to as the decision.

At this point in time, the decision-making process is complete. In summary, we see that this process involves five steps:

1. Define the problem. 2. Identify the alternatives. 3. Determine the criteria. 4. Evaluate the alternatives. 5. Choose an alternative.

Note that missing from this list are the last two steps in the problem-solving process: imple- menting the selected alternative and evaluating the results to determine whether a satisfac- tory solution has been obtained. This omission is not meant to diminish the importance

Starting Potential for Job Alternative Salary Advancement Location

1. Rochester $48,500 Average Average 2. Dallas $46,000 Excellent Good 3. Greensboro $46,000 Good Excellent 4. Pittsburgh $47,000 Average Good

TABLE 1.1 DATA FOR THE JOB EVALUATION DECISION-MAKING PROBLEM

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1.2 Quantitative Analysis and Decision Making 5

of each of these activities, but to emphasize the more limited scope of the term decision making as compared to the term problem solving. Figure 1.1 summarizes the relationship between these two concepts.

1.2 Quantitative Analysis and Decision Making Consider the flowchart presented in Figure 1.2. Note that we combined the first three steps of the decision-making process under the heading of “Structuring the Problem” and the latter two steps under the heading “Analyzing the Problem.” Let us now consider in greater detail how to carry out the activities that make up the decision-making process.

Figure 1.3 shows that the analysis phase of the decision-making process may take two basic forms: qualitative and quantitative. Qualitative analysis is based primarily on the man- ager’s judgment and experience; it includes the manager’s intuitive “feel” for the problem and is more an art than a science. If the manager has had experience with similar problems, or if the problem is relatively simple, heavy emphasis may be placed upon a qualitative analysis. However, if the manager has had little experience with similar problems, or if the problem

De�ne the

Problem

Identify the

Alternatives

Determine the

Criteria

Evaluate the

Alternatives

Choose an

Alternative

Implement the

Decision

Evaluate the

Results

Problem Solving

Decision Making

Decision

FIGURE 1.1 THE RELATIONSHIP BETWEEN PROBLEM SOLVING AND DECISION MAKING

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6 Chapter 1 Introduction

is sufficiently complex, then a quantitative analysis of the problem can be an especially important consideration in the manager’s final decision.

When using a quantitative approach, an analyst will concentrate on the quantitative facts or data associated with the problem and develop mathematical expressions that de- scribe the objectives, constraints, and other relationships that exist in the problem. Then, by using one or more mathematical methods, the analyst will make a recommendation based on the quantitative aspects of the problem.

Although skills in the qualitative approach are inherent in the manager and usually increase with experience, the skills of the quantitative approach can be learned only by studying the assumptions and methods of management science. A manager can increase decision-making effectiveness by learning more about quantitative methodology and by better understanding its contribution to the decision-making process. A manager who is knowledgeable in quantitative decision-making procedures is in a much better position to compare and evaluate the qualitative and quantitative sources of recommendations and ultimately to combine the two sources to make the best possible decision.

The box in Figure 1.3 entitled “Quantitative Analysis” encompasses most of the sub- ject matter of this text. We will consider a managerial problem, introduce the appropriate quantitative methodology, and then develop the recommended decision.

Quantitative methods are especially helpful with large, complex problems. For example, in the coordination of the thousands of tasks associated with landing the Apollo 11 safely on the moon, quantitative techniques helped to ensure that more than 300,000 pieces of work performed by more than 400,000 people were integrated smoothly.

Structuring the Problem Analyzing the Problem

Choose an

Alternative

Evaluate the

Alternatives

Determine the

Criteria

Identify the

Alternatives

De�ne the

Problem

FIGURE 1.2 A SUBCLASSIFICATION OF THE DECISION-MAKING PROCESS

Structuring the Problem

Analyzing the Problem

Make the

Decision

Summary and

Evaluation

De�ne the

Problem

Identify the

Alternatives

Determine the

Criteria

Qualitative Analysis

Quantitative Analysis

FIGURE 1.3 THE ROLE OF QUALITATIVE AND QUANTITATIVE ANALYSIS

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1.3 Quantitative Analysis 7

Some of the reasons why a quantitative approach might be used in the decision- making process include the following:

1. The problem is complex, and the manager cannot develop a good solution without the aid of quantitative analysis.

2. The problem is critical (e.g., a great deal of money is involved), and the manager desires a thorough analysis before making a decision.

3. The problem is new, and the manager has no previous experience from which to draw.

4. The problem is repetitive, and the manager saves time and effort by relying on quantitative procedures to automate routine decision recommendations.

1.3 Quantitative Analysis From Figure 1.3 we see that quantitative analysis begins once the problem has been struc- tured. It usually takes imagination, teamwork, and considerable effort to transform a rather general problem description into a well-defined problem that can be approached via quan- titative analysis. It is important to involve the stakeholders (the decision maker, users of results, etc.) in the process of structuring the problem to improve the likelihood that the ensuing quantitative analysis will make an important contribution to the decision-making process. When those familiar with the problem agree that it has been adequately structured, work can begin on developing a model to represent the problem mathematically. Solution procedures can then be employed to find the best solution for the model. This best solu- tion for the model then becomes a recommendation to the decision maker. The process of developing and solving models is the essence of the quantitative analysis process.

Model Development Models are representations of real objects or situations and can be presented in various forms. For example, a scale model of an airplane is a representation of a real airplane. Similarly, a child’s toy truck is a model of a real truck. The model airplane and toy truck are examples of models that are physical replicas of real objects. In modeling terminology, physical replicas are referred to as iconic models.

A second classification includes models that are physical in form but do not have the same physical appearance as the object being modeled. Such models are referred to as analog models. The speedometer of an automobile is an analog model; the position of the needle on the dial represents the speed of the automobile. A thermometer is another analog model representing temperature.

A third classification of models—the type we will primarily be studying—includes representations of a problem by a system of symbols and mathematical relationships or expressions. Such models are referred to as mathematical models and are a critical part of any quantitative approach to decision making. For example, the total profit from the sale of a product can be determined by multiplying the profit per unit by the quantity sold. Let x represent the number of units produced and sold, and let P represent the total profit. With a profit of $10 per unit, the following mathematical model defines the total profit earned by producing and selling x units:

P 5 10x (1.1)

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8 Chapter 1 Introduction

The purpose, or value, of any model is that it enables us to make inferences about the real situation by studying and analyzing the model. For example, an airplane designer might test an iconic model of a new airplane in a wind tunnel to learn about the potential flying characteristics of the full-size airplane. Similarly, a mathematical model may be used to make inferences about how much profit will be earned if a specified quantity of a particular product is sold. According to the mathematical model of equation (1.1), we would expect that selling three units of the product (x 5 3) would provide a profit of P 5 10(3) 5 $30.

In general, experimenting with models requires less time and is less expensive than experimenting with the real object or situation. One can certainly build and study a model airplane in less time and for less money than it would take to build and study the full-size airplane. Similarly, the mathematical model in equation (1.1) allows a quick identifica- tion of profit expectations without requiring the manager to actually produce and sell x units. Models also reduce the risks associated with experimenting with the real situation. In particular, bad designs or bad decisions that cause the model airplane to crash or the mathematical model to project a $10,000 loss can be avoided in the real situation.

The value of model-based conclusions and decisions depends on how well the model represents the real situation. The more closely the model airplane represents the real air- plane, the more accurate will be the conclusions and predictions. Similarly, the more closely the mathematical model represents the company’s true profit–volume relationship, the more accurate will be the profit projections.

Because this text deals with quantitative analysis based on mathematical models, let us look more closely at the mathematical modeling process. When initially considering a managerial problem, we usually find that the problem definition phase leads to a spe- cific objective, such as maximization of profit or minimization of cost, and possibly a set of restrictions or constraints, which express limitations on resources. The success of the mathematical model and quantitative approach will depend heavily on how accurately the objective and constraints can be expressed in mathematical equations or relationships.

The mathematical expression that defines the quantity to be maximized or minimized is referred to as the objective function. For example, suppose x denotes the number of units produced and sold each week, and the firm’s objective is to maximize total weekly profit. With a profit of $10 per unit, the objective function is 10x. A production capacity constraint would be necessary if, for instance, 5 hours are required to produce each unit and only 40 hours are available per week. The production capacity constraint is given by

5x # 40 (1.2)

The value of 5x is the total time required to produce x units; the symbol # indicates that the production time required must be less than or equal to the 40 hours available.

The decision problem or question is the following: How many units of the product should be produced each week to maximize profit? A complete mathematical model for this simple production problem is

Maximize 10x objective function

subject to ss.t.d

5x # 40

x $ 0 f constraints

The x $ 0 constraint requires the production quantity x to be greater than or equal to zero, which simply recognizes the fact that it is not possible to manufacture a negative number

Herbert A. Simon, a Nobel Prize winner in economics and an expert in decision making, said that a mathematical model does not have to be exact; it just has to be close enough to provide better results than can be obtained by common sense.

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1.3 Quantitative Analysis 9

of units. The optimal solution to this simple model can be easily calculated and is given by x 5 8, with an associated profit of $80. This model is an example of a linear programming model. In subsequent chapters we will discuss more complicated mathematical models and learn how to solve them in situations for which the answers are not nearly so obvious.

In the preceding mathematical model, the profit per unit ($10), the production time per unit (5 hours), and the production capacity (40 hours) are factors not under the control of the manager or decision maker. Such factors, which can affect both the objective function and the constraints, are referred to as uncontrollable inputs to the model. Inputs that are controlled or determined by the decision maker are referred to as controllable inputs to the model. In the example given, the production quantity x is the controllable input to the model. Controllable inputs are the decision alternatives specified by the manager and thus are also referred to as the decision variables of the model.

Once all controllable and uncontrollable inputs are specified, the objective function and constraints can be evaluated and the output of the model determined. In this sense, the output of the model is simply the projection of what would happen if those particular factors and decisions occurred in the real situation. A flowchart of how controllable and uncontrollable inputs are transformed by the mathematical model into output is shown in Figure 1.4. A similar flowchart showing the specific details for the production model is shown in Figure 1.5. Note that we have used “Max” as an abbreviation for maximize.

Uncontrollable Inputs (Environmental Factors)

Output (Projected Results)

Controllable Inputs

(Decision Variables)

Mathematical Model

FIGURE 1.4 FLOWCHART OF THE PROCESS OF TRANSFORMING MODEL INPUTS INTO OUTPUT

Value for the Production

Quantity (x = 8)

Uncontrollable Inputs

Mathematical Model

$10 pro�t per unit 5 labor-hours per unit

40 labor-hours capacity

Controllable Input

Pro�t = 80

Time Used = 40

Output

Max s.t.

(8)

(8) ≤ 40 8 ≥ 0

FIGURE 1.5 FLOWCHART FOR THE PRODUCTION MODEL

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10 Chapter 1 Introduction

As stated earlier, the uncontrollable inputs are those the decision maker cannot influ- ence. The specific controllable and uncontrollable inputs of a model depend on the par- ticular problem or decision-making situation. In the production problem, the production time available (40) is an uncontrollable input. However, if it were possible to hire more employees or use overtime, the number of hours of production time would become a con- trollable input and therefore a decision variable in the model.

Uncontrollable inputs can either be known exactly or be uncertain and subject to varia- tion. If all uncontrollable inputs to a model are known and cannot vary, the model is referred to as a deterministic model. Corporate income tax rates are not under the influence of the manager and thus constitute an uncontrollable input in many decision models. Because these rates are known and fixed (at least in the short run), a mathematical model with cor- porate income tax rates as the only uncontrollable input would be a deterministic model. The distinguishing feature of a deterministic model is that the uncontrollable input values are known in advance.

If any of the uncontrollable inputs are uncertain and subject to variation, the model is referred to as a stochastic or probabilistic model. An uncontrollable input in many produc- tion planning models is demand for the product. Because future demand may be any of a range of values, a mathematical model that treats demand with uncertainty would be con- sidered a stochastic model. In the production model, the number of hours of production time required per unit, the total hours available, and the unit profit were all uncontrollable inputs. Because the uncontrollable inputs were all known to take on fixed values, the model was deterministic. If, however, the number of hours of production time per unit could vary from 3 to 6 hours depending on the quality of the raw material, the model would be stochastic. The distinguishing feature of a stochastic model is that the value of the output cannot be determined even if the value of the controllable input is known because the specific values of the uncontrollable inputs are unknown. In this respect, stochastic models are often more difficult to analyze.

Data Preparation Another step in the quantitative analysis of a problem is the preparation of the data required by the model. Data in this sense refer to the values of the uncontrollable inputs to the model. All uncontrollable inputs or data must be specified before we can analyze the model and recommend a decision or solution for the problem.

In the production model, the values of the uncontrollable inputs or data were $10 per unit for profit, 5 hours per unit for production time, and 40 hours for produc- tion capacity. In the development of the model, these data values were known and incorporated into the model as it was being developed. If the model is relatively small with respect to the number of the uncontrollable input values, the quantitative analyst will probably combine model development and data preparation into one step. In these situations the data values are inserted as the equations of the mathematical model are developed.

However, in many mathematical modeling situations, the data or uncontrollable input values are not readily available. In these situations, the analyst may know that the model will require profit per unit, production time, and production capacity data, but the values will not be known until the accounting, production, and engineering departments can be consulted. Rather than attempting to collect the required data as the model is being devel- oped, the analyst will usually adopt a general notation for the model development step, and a separate data preparation step will then be performed to obtain the uncontrollable input values required by the model.

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1.3 Quantitative Analysis 11

Using the general notation

c 5 profit per unit

a 5 production time in hours per unit

b 5 production capacity in hours

the model development step for the production problem would result in the following gen- eral model (recall x 5 the number of units to produce and sell):

Max cx

s.t.

ax # b

x $ 0

A separate data preparation step to identify the values for c, a, and b would then be neces- sary to complete the model.

Many inexperienced quantitative analysts assume that once the problem is defined and a general model developed, the problem is essentially solved. These individuals tend to believe that data preparation is a trivial step in the process and can be easily handled by clerical staff. Actually, this is a potentially dangerous assumption that could not be further from the truth, especially with large-scale models that have numerous data input values. For example, a moderate-sized linear programming model with 50 decision vari- ables and 25 constraints could have more than 1300 data elements that must be identified in the data preparation step. The time required to collect and prepare these data and the possibility of data collection errors will make the data preparation step a critical part of the quantitative analysis process. Often, a fairly large database is needed to support a mathematical model, and information systems specialists also become involved in the data preparation step.

Model Solution Once the model development and data preparation steps are completed, we proceed to the model solution step. In this step, the analyst attempts to identify the values of the decision variables that provide the “best” output for the model. The specific decision-variable value or values providing the “best” output are referred to as the optimal solution for the model. For the production problem, the model solution step involves finding the value of the pro- duction quantity decision variable x that maximizes profit while not causing a violation of the production capacity constraint.

One procedure that might be used in the model solution step involves a trial-and-error approach in which the model is used to test and evaluate various decision alternatives. In the production model, this procedure would mean testing and evaluating the model using vari- ous production quantities or values of x. As noted in Figure 1.5, we could input trial values for x and check the corresponding output for projected profit and satisfaction of the pro- duction capacity constraint. If a particular decision alternative does not satisfy one or more of the model constraints, the decision alternative is rejected as being infeasible, regardless of the corresponding objective function value. If all constraints are satisfied, the decision alternative is feasible and is a candidate for the “best” solution or recommended decision. Through this trial-and-error process of evaluating selected decision alternatives, a decision maker can identify a good—and possibly the best—feasible solution to the problem. This solution would then be the recommended decision for the problem.

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12 Chapter 1 Introduction

Table 1.2 shows the results of a trial-and-error approach to solving the production model of Figure 1.5. The recommended decision is a production quantity of 8 because the feasible solution with the highest projected profit occurs at x 5 8.

Although the trial-and-error solution process is often acceptable and can provide valuable information for the manager, it has the drawbacks of not necessarily providing the best solution and of being inefficient in terms of requiring numerous calculations if many decision alternatives are considered. Thus, quantitative analysts have developed special solution procedures for many models that are much more efficient than the trial- and-error approach. Throughout this text, you will be introduced to solution procedures that are applicable to the specific mathematical models. Some relatively small models or problems can be solved by hand computations, but most practical applications require the use of a computer.

The model development and model solution steps are not completely separable. An analyst will want both to develop an accurate model or representation of the actual problem situation and to be able to find a solution to the model. If we approach the model develop- ment step by attempting to find the most accurate and realistic mathematical model, we may find the model so large and complex that it is impossible to obtain a solution. In this case, a simpler and perhaps more easily understood model with a readily available solu- tion procedure is preferred even though the recommended solution may be only a rough approximation of the best decision. As you learn more about quantitative solution proce- dures, you will form a better understanding of the types of mathematical models that can be developed and solved.

After obtaining a model solution, the quantitative analyst will be interested in deter- mining the quality of the solution. Even though the analyst has undoubtedly taken many precautions to develop a realistic model, often the usefulness or accuracy of the model cannot be assessed until model solutions are generated. Model testing and validation are frequently conducted with relatively small “test” problems with known or at least expected solutions. If the model generates the expected solutions, and if other output information appears correct or reasonable, the go-ahead may be given to use the model on the full-scale problem. However, if the model test and validation identify potential problems or inaccuracies inherent in the model, corrective action, such as model modifi- cation or collection of more accurate input data, may be taken. Whatever the corrective action, the model solution will not be used in practice until the model satisfactorily passes testing and validation.

Try Problem 8 to test your understanding of the concept of a mathematical model and what is referred to as the optimal solution to the model.

Decision Alternative Total Feasible (Production Quantity) Projected Hours of Solution? x Profit Production (Hours Used # 40) 0 0 0 Yes 2 20 10 Yes 4 40 20 Yes 6 60 30 Yes 8 80 40 Yes 10 100 50 No 12 120 60 No

TABLE 1.2 TRIAL-AND-ERROR SOLUTION FOR THE PRODUCTION MODEL OF FIGURE 1.5

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1.3 Quantitative Analysis 13

Report Generation An important part of the quantitative analysis process is the preparation of managerial reports based on the model’s solution. As indicated in Figure 1.3, the solution based on the quantita- tive analysis of a problem is one of the inputs the manager considers before making a final decision. Thus, the results of the model must appear in a managerial report that can be easily understood by the decision maker. The report includes the recommended decision and other pertinent information about the results that may be useful to the decision maker.

A Note Regarding Implementation As discussed in Section 1.2, the manager is responsible for integrating the quantitative solution with qualitative considerations to determine the best possible decision. After com- pleting the decision-making process, the manager must oversee the implementation and follow-up evaluation of the decision. During the implementation and follow-up, the man- ager should continue to monitor the performance of the model. At times, this process may lead to requests for model expansion or refinement that will require the quantitative analyst to return to an earlier step of the process.

Successful implementation of results is critical to any application of quantitative analy- sis. If the results of the quantitative analysis are not correctly implemented, the entire effort may be of no value. Because implementation often requires people to change the way they do things, it often meets with resistance. People may want to know, “What’s wrong with the way we’ve been doing it?” One of the most effective ways to ensure successful imple- mentation is to include users throughout the modeling process. A user who feels a part of identifying the problem and developing the solution is much more likely to enthusiastically implement the results, and the input the quantitative analyst receives from these users can substantially enhance the models being developed. The success rate for implementing the results of a quantitative analysis project is much greater for those projects characterized by extensive user involvement. The Q.M. in Action, Quantitative Analysis at Merrill Lynch, discusses some of the reasons for the success of quantitative analysis at Merrill Lynch.

For over 25 years, the Management Science Group at Merrill Lynch has successfully implemented quantita- tive models for a wide variety of decision problems. The group has applied quantitative methods for port- folio optimization, asset allocation, financial planning, marketing analysis, credit and liquidity assessment, as well as developing pricing and compensation structures. Although technical expertise and objectivity are clearly important factors in any analytical group, the manage- ment science group attributes much of its success to

communications skills, teamwork, professional devel- opment for its members, and consulting skills.

From the earliest discussion of a potential project, the group focuses on fully understanding the problem and its business impact. Each client is asked, “Whose life will this change?” and “By how much?” The answers to these questions help the group understand who really has responsibility for the project, the processes involved, and how recommendations will be implemented. Analysts assigned to a project are fully engaged from start to fin- ish. They are involved in project scope definition, data collection, analysis, development of recommendations,

QUANTITATIVE ANALYSIS AT MERRILL LYNCH*

Q.M. in ACtIon

*Based on R. nigam, “Structuring and Sustaining Excellence in Man- agement Science at Merrill Lynch,” Interfaces 38, no. 3 (May/June 2008): 202–209. (continued)

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14 Chapter 1 Introduction

1.4 Models of Cost, Revenue, and Profit Some of the most basic quantitative models arising in business and economic applications involve the relationships among a volume variable—such as production volume or sales volume—and cost, revenue, and profit. Through the use of these models, a manager can determine the projected cost, revenue, or profit associated with a planned production quan- tity or a forecasted sales volume. Financial planning, production planning, sales quotas, and other areas of decision making can benefit from such cost, revenue, and profit models.

Cost and Volume Models The cost of manufacturing or producing a product is a function of the volume produced. This cost can usually be defined as a sum of two costs: fixed cost and variable cost. Fixed cost is the portion of the total cost that does not depend on the production volume; this cost remains the same no matter how much is produced. Variable cost, on the other hand, is the portion of the total cost that depends on and varies with the production volume. To illustrate how cost and volume models can be developed, we will consider a manufacturing problem faced by Nowlin Plastics.

Nowlin Plastics produces a line of cell phone covers. Nowlin’s best-selling cover is its Viper model, a slim but very durable black and gray plastic cover. Several products are produced on the same manufacturing line, and a setup cost is incurred each time a change- over is made for a new product. Suppose the setup cost for the Viper is $3000; this setup cost is a fixed cost and is incurred regardless of the number of units eventually produced. In addition, suppose that variable labor and material costs are $2 for each unit produced. The cost–volume model for producing x units of the Viper can be written as

C(x) 5 3000 1 2x (1.3)

and marketing those recommendations to the client. The group prides itself on technology transfer; that is, it gives any models it develops to the clients with assistance and training on the use of the models. This leads to longer- term impact through ongoing use of the model. Finally, like any good organization focused on improvement, the Management Science Group seeks feedback from clients after every project it completes.

This approach to problem solving and the imple- mentation of quantitative analysis has been a hallmark

of the Management Science Group. The impact and success of the group translates into hard dollars, repeat business, and recognition through a number of presti- gious professional awards. The group received the annual Edelman Award given by the Institute for Operations Research and the Management Sciences (INFORMS) for effective use of management science for organizational success as well as the INFORMS Prize, given for long- term and high-impact use of quantitative methods within an organization.

NOTES AND COMMENTS

1. Developments in computer technology have increased the availability of quantitative meth- ods to decision makers. A variety of software packages is now available for personal comput- ers. Versions of Microsoft Excel and LINGO

are widely used to apply quantitative methods to business problems. Various chapter appen- dixes provide step-by-step instructions for us- ing Excel and LINGO to solve problems in the text.

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1.4 Models of Cost, Revenue, and Profit 15

where

x 5 production volume in units

C(x) 5 total cost of producing x units

Once a production volume is established, the model in equation (1.3) can be used to compute the total production cost. For example, the decision to produce x 5 1200 units would result in a total cost of C(1200) 5 3000 1 2(1200) 5 $5400.

Marginal cost is defined as the rate of change of the total cost with respect to production volume; that is, the cost increase associated with a one-unit increase in the production volume. In the cost model of equation (1.3), we see that the total cost C(x) will increase by $2 for each unit increase in the production volume. Thus, the marginal cost is $2. With more complex total cost models, marginal cost may depend on the production volume. In such cases, we could have marginal cost increasing or decreasing with the production volume x.

Revenue and Volume Models Management of Nowlin Plastics will also want information about projected revenue asso- ciated with selling a specified number of units. Thus, a model of the relationship between revenue and volume is also needed. Suppose that each Viper cover sells for $5. The model for total revenue can be written as

R(x) 5 5x (1.4)

where

x 5 sales volume in units

R(x) 5 total revenue associated with selling x units

Marginal revenue is defined as the rate of change of total revenue with respect to sales vol- ume, that is, the increase in total revenue resulting from a one-unit increase in sales volume. In the model of equation (1.4), we see that the marginal revenue is $5. In this case, marginal revenue is constant and does not vary with the sales volume. With more complex models, we may find that marginal revenue increases or decreases as the sales volume x increases.

Profit and Volume Models One of the most important criteria for management decision making is profit. Managers need to know the profit implications of their decisions. If we assume that we will only pro- duce what can be sold, the production volume and sales volume will be equal. We can then combine equations (1.3) and (1.4) to develop a profit–volume model that determines profit associated with a specified production-sales volume. Total profit is total revenue minus total cost; therefore, the following model provides the profit associated with producing and selling x units:

P(x) 5 R(x) 2 C(x)

5 5x 2 (3000 1 2x) 5 23000 1 3x (1.5)

Thus, the model for profit P(x) can be derived from the models of the revenue–volume and cost–volume relationships.

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16 Chapter 1 Introduction

Breakeven Analysis Using equation (1.5), we can now determine the profit associated with any production vol- ume x. For example, suppose that a demand forecast indicates that 500 units of the product can be sold. The decision to produce and sell the 500 units results in a projected profit of

P(500) 5 23000 1 3(500) 5 21500

In other words, a loss of $1500 is predicted. If sales are expected to be 500 units, the man- ager may decide against producing the product. However, a demand forecast of 1800 units would show a projected profit of

P(1800) 5 23000 1 3(1800) 5 2400

This profit may be sufficient to justify proceeding with the production and sale of the product.

We see that a volume of 500 units will yield a loss, whereas a volume of 1800 provides a profit. The volume that results in total revenue equaling total cost (providing $0 profit) is called the breakeven point. If the breakeven point is known, a manager can quickly infer that a vol- ume above the breakeven point will generate a profit, whereas a volume below the breakeven point will result in a loss. Thus, the breakeven point for a product provides valuable information for a manager who must make a yes/no decision concerning production of the product.

Let us now return to the Nowlin Plastics example and show how the profit model in equation (1.5) can be used to compute the breakeven point. The breakeven point can be found by setting the profit expression equal to zero and solving for the production volume. Using equation (1.5), we have

P(x) 5 23000 1 3x 5 0

3x 5 3000

x 5 1000

With this information, we know that production and sales of the product must exceed 1000 units before a profit can be expected. The graphs of the total cost model, the to- tal revenue model, and the location of the breakeven point are shown in Figure 1.6. In Appendix A we discuss how Excel can be used to perform a breakeven analysis for the Nowlin Plastics production example.

Try Problem 12 to test your ability to determine the breakeven point for a quantitative model.

Fixed Cost

10,000

8000

6000

4000

2000

R ev

en u

e an

d C

os t

($ )

0 400 800 1200 1600 2000

Production Volume

Loss

Total Revenue R (x) = 5 x

Breakeven Point = 1000 Units

Total Cost C(x) = 3000 + 2 x

Pro�t

FIGURE 1.6 GRAPH OF THE BREAKEVEN ANALYSIS FOR NOWLIN PLASTICS

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1.5 Quantitative Methods in Practice 17

1.5 Quantitative Methods in Practice In this section we present a brief overview of the quantitative methods covered in this text. There are numerous applications for each of the following methods.

Linear Programming Linear programming is a problem-solving approach developed for situations involving maximizing or minimizing a linear function subject to linear constraints that limit the degree to which the objective can be pursued. The production model developed in Section 1.3 (see Figure 1.5) is an example of a simple linear pro- gramming model.

Integer Linear Programming Integer linear programming is an approach used for problems that can be set up as linear programs with the additional requirement that some or all of the decision recommendations be integer values.

Project Scheduling: PERT/CPM In many situations managers are responsible for plan- ning, scheduling, and controlling projects that consist of numerous separate jobs or tasks performed by a variety of departments, individuals, and so forth. PERT (Program Evalu- ation and Review Technique) and CPM (Critical Path Method) help managers carry out their project scheduling and tracking responsibilities.

Inventory Models Inventory models are used by managers faced with the problem of maintaining sufficient inventories to meet demand for goods while incurring the lowest possible inventory holding costs.

Waiting Line or Queueing Models Waiting line or queueing models help managers understand and make better decisions concerning the operation of systems involving wait- ing lines.

Simulation Simulation is a technique used to model the operation of a complex system. This technique employs a computer program to model the operation and perform simula- tion computations.

Decision Analysis Decision analysis can be used to determine optimal strategies in situations involving several decision alternatives and an uncertain or risk-filled pattern of future events.

Forecasting Forecasting methods are techniques that can be used to predict future as- pects of a business operation.

Markov-Process Models Markov-process models are useful in studying the evolution of certain systems over repeated trials. For example, Markov processes have been used to describe the probability that a machine, functioning in one period, will function or break down in some future period.

We believe barriers to the use of quantitative methods can best be removed by increas- ing the manager’s understanding of how quantitative analysis can be applied. The text will help you develop an understanding of which quantitative methods are most useful, how they are used, and, most importantly, how they can assist managers in making better decisions.

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18 Chapter 1 Introduction

Mark Eisner, Communications Associate of the School of Operations Research and Information Engineering at Cornell University, once said that operations research “is probably the most important field nobody has ever heard of.” The impact of operations research on every- day living over the past 20 years is substantial.

Suppose you schedule a vacation to Florida and use the Orbitz website to book your flights. An algorithm developed by operations researchers will search among millions of options to find the cheapest fare. Another algorithm will schedule the flight crews and aircraft used by the airline. If you rent a car in Florida, the price you pay for the car is determined by a mathematical model that seeks to maximize revenue for the car rental firm. If you do some shopping on your trip and decide to ship your purchases home using UPS, another algorithm

determines the truck on which your packages are loaded, which route the truck should follow, and where your packages should be placed on the truck to minimize loading and unloading time.

If you enjoy watching college basketball, operations research plays a role in what games you see. Michael Trick, a professor at the Tepper School of Business at Carnegie Mellon, designed a system for scheduling each year’s Atlantic Coast Conference men’s and women’s basketball games. Even though it might initially appear that scheduling 16 games among the nine men’s teams would be easy, it requires sorting through hundreds of millions of possible combinations of possible schedules. Each of those possibilities entails some desirable and some undesirable characteristics. For example, you do not want to schedule too many consecutive home games for any team, and you want to ensure that each team plays the same number of weekend games.

IMPACT OF OPERATIONS RESEARCH ON EVERYDAY LIVING*

Q.M. in ACtIon

*Based on Virginia Postrel, “operations Everything,” The Boston Globe, June 27, 2004.

The Q.M. in Action, Impact of Operations Research on Everyday Living, describes some of the many ways quantitative analysis affects our everyday lives.

NOTES AND COMMENTS

1. In the United States, the Institute for Opera- tions Research and the Management Sciences (INFORMS) and the Decision Sciences Institute (DSI) are two flagship professional societies that publish journals and newsletters dealing with current research and applications of operations research and management science techniques. In Canada, the Canadian Operational Research Society (CORS) provides similar services.

2. Several European countries, including (but not limited to) Great Britain, France, Italy, Germany, Austria, and the Czech Republic, have their own professional operations research and management science societies, and these societies belong to the Association of European Operational Research Societies (EURO). Pro- fessional operations research and management science societies from Latin American and Iberian peninsula countries, including (but not

limited to) Chile, Brazil, Argentina, Colombia, Spain, Uruguay, Portugual, and Mexico, all be- long to the Asociación Latino-Iberoamericana de Investigación Operativa (ALIO). Profes- sional operations research and management science societies from Australia, Japan, China, India, Malaysia, Thailand, New Zealand, and other countries from Asia and the Pacific Rim belong to the Association of Asian Pacific Operational Research Societies (APORS). Af- rican operations research societies include the Operations Research Society of South Africa (ORSSA) and the Operations Research Society of Eastern Africa (ORSEA). The International Federation of Operational Research Societies (IFORS) is the global organization to which most of these (and other) professional opera- tions research and management science socie- ties belong.

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Glossary 19

Summary

This text focuses on the use of quantitative methods to help managers make better decisions. The discussion in this chapter centered on the problem orientation of the decision-making process and an overview of how mathematical models can be used in this type of analysis.

The difference between the model and the situation or managerial problem it represents is an important consideration. Mathematical models are abstractions of real-world situa- tions and, as such, cannot capture all the aspects of the real situation. However, if a model can capture the major relevant aspects of the problem and can then provide a meaningful solution recommendation, it can be a valuable aid to decision making.

One of the characteristics of quantitative analysis that will become increasingly ap- parent as we proceed through the text is the search for a best solution to the problem. In carrying out the quantitative analysis, we attempt to develop procedures for finding the “best” or optimal solution.

Glossary

Analog model Although physical in form, an analog model does not have a physical appearance similar to the real object or situation it represents. Breakeven point The volume at which total revenue equals total cost. Constraint A restriction or limitation imposed on a problem. Controllable input The decision alternatives that can be specified by the decision maker. Decision The alternative selected. Decision making The process of defining the problem, identifying the alternatives, deter- mining the criteria, evaluating the alternatives, and choosing an alternative. Decision variable Another term for controllable input. Deterministic model A model in which all uncontrollable inputs are known and cannot vary. Feasible solution A decision alternative or solution that satisfies all constraints. Fixed cost The portion of the total cost that does not depend on the volume; this cost remains the same no matter how much is produced. Iconic model A physical replica, or representation, of a real object. Infeasible solution A decision alternative or solution that violates one or more constraints. Marginal cost The rate of change of the total cost with respect to volume. Marginal revenue The rate of change of total revenue with respect to volume. Mathematical model Mathematical symbols and expressions used to represent a real situation. Model A representation of a real object or situation. Multicriteria decision problem A problem that involves more than one criterion; the objective is to find the “best” solution, taking into account all the criteria. Objective function The mathematical expression that defines the quantity to be maxi- mized or minimized. Optimal solution The specific decision variable value or values that provide the “best” output for the model. Problem solving The process of identifying a difference between the actual and the de- sired state of affairs and then taking action to resolve the difference. Single-criterion decision problem A problem in which the objective is to find the “best” solution with respect to just one criterion. Stochastic model A model in which at least one uncontrollable input is uncertain and subject to variation; stochastic models are also referred to as probabilistic models.

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20 Chapter 1 Introduction

Uncontrollable input The factors that cannot be controlled by the decision maker. Variable cost The portion of the total cost that is dependent on and varies with the volume.

Problems

1. Define the terms management science and operations research.

2. List and discuss the steps of the decision-making process.

3. Discuss the different roles played by the qualitative and quantitative approaches to mana- gerial decision making. Why is it important for a manager or decision maker to have a good understanding of both of these approaches to decision making?

4. A firm recently built a new plant that will use more than 50 production lines and machines to produce over 500 different products. The production scheduling decisions are critical because sales will be lost if customer demand is not met on time. If no individual in the firm has had experience with this production operation, and if new production schedules must be generated each week, why should the firm consider a quantitative approach to the production scheduling problem?

5. What are the advantages of analyzing and experimenting with a model as opposed to a real object or situation?

6. Suppose a manager must choose between the following two mathematical models of a given situation: (a) a relatively simple model that is a reasonable approximation of the real situation and (b) a thorough and complex model that is the most accurate mathematical representation of the real situation possible. Why might the model described in part (a) be preferred by the manager?

7. Suppose you are going on a weekend trip to a city that is d miles away. Develop a model that determines your round-trip gasoline costs. What assumptions or approximations are necessary to treat this model as a deterministic model? Are these assumptions or approxi- mations acceptable to you?

8. Recall the production model from Section 1.3:

Max 10x s.t.

5x # 40

x $ 0

Suppose the firm in this example considers a second product that has a unit profit of $5 and requires 2 hours for each unit produced. Assume total production capacity remains 40 units. Use y as the number of units of product 2 produced. a. Show the mathematical model when both products are considered simultaneously. b. Identify the controllable and uncontrollable inputs for this model. c. Draw the flowchart of the input–output process for this model (see Figure 1.5). d. What are the optimal solution values of x and y? e. Is this model a deterministic or a stochastic model? Explain.

9. Suppose we modify the production model from Section 1.3 to obtain the following math- ematical model:

Max 10x s.t.

ax # 40

x $ 0

testSELF

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Problems 21

where a is the number of hours required for each unit produced. With a 5 5, the optimal solution is x 5 8. If we have a stochastic model in which the value of a varies between 3 and 6 (i.e., a 5 3, a 5 4, a 5 5, or a 5 6) as the possible values for the number of hours required per unit, what is the optimal value for x? What problems does this stochastic model cause?

10. A retail store in Des Moines, Iowa, receives shipments of a particular product from Kansas City and Minneapolis. Let

x 5 units of product received from Kansas City

y 5 units of product received from Minneapolis

a. Write an expression for the total units of product received by the retail store in Des Moines.

b. Shipments from Kansas City cost $0.20 per unit, and shipments from Minneapolis cost $0.25 per unit. Develop an objective function representing the total cost of shipments to Des Moines.

c. Assuming the monthly demand at the retail store is 5000 units, develop a constraint that requires 5000 units to be shipped to Des Moines.

d. No more than 4000 units can be shipped from Kansas City and no more than 3000 units can be shipped from Minneapolis in a month. Develop constraints to model this situation.

e. Of course, negative amounts cannot be shipped. Combine the objective function and constraints developed to state a mathematical model for satisfying the demand at the Des Moines retail store at minimum cost.

11. For most products, higher prices result in a decreased demand, whereas lower prices result in an increased demand (economists refer to such products as normal goods). Let

d 5 annual demand for a product in units

p 5 price per unit

Assume that a firm accepts the following price–demand relationship as being a real- istic representation of its market:

d 5 800 2 10p

where p must be between $20 and $70. a. How many units can the firm sell at the $20 per-unit price? At the $70 per-unit price? b. What happens to annual units demanded for the product if the firm increases the per-

unit price from $26 to $27? From $42 to $43? From $68 to $69? What is the suggested relationship between per-unit price and annual demand for the product in units?

c. Show the mathematical model for the total revenue (TR), which is the annual demand multiplied by the unit price.

d. Based on other considerations, the firm’s management will only consider price al- ternatives of $30, $40, and $50. Use your model from part (b) to determine the price alternative that will maximize the total revenue.

e. What are the expected annual demand and the total revenue according to your recom- mended price?

12. The O’Neill Shoe Manufacturing Company will produce a special-style shoe if the order size is large enough to provide a reasonable profit. For each special-style order, the com- pany incurs a fixed cost of $2000 for the production setup. The variable cost is $60 per pair, and each pair sells for $80. a. Let x indicate the number of pairs of shoes produced. Develop a mathematical model

for the total cost of producing x pairs of shoes.

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22 Chapter 1 Introduction

b. Let P indicate the total profit. Develop a mathematical model for the total profit real- ized from an order for x pairs of shoes.

c. What is the breakeven point?

13. Micromedia offers computer training seminars on a variety of topics. In the seminars each student works at a personal computer, practicing the particular activity that the instructor is presenting. Micromedia is currently planning a two-day seminar on the use of Microsoft Excel in statistical analysis. The projected fee for the seminar is $600 per student. The cost for the conference room, instructor compensation, lab assistants, and promotion is $9600. Micromedia rents computers for its seminars at a cost of $60 per computer per day. a. Develop a model for the total cost to put on the seminar. Let x represent the number

of students who enroll in the seminar. b. Develop a model for the total profit if x students enroll in the seminar. c. Micromedia has forecasted an enrollment of 30 students for the seminar. How much

profit will be earned if its forecast is accurate? d. Compute the breakeven point.

14. Eastman Publishing Company is considering publishing a paperback textbook on spread- sheet applications for business. The fixed cost of manuscript preparation, textbook design, and production setup is estimated to be $160,000. Variable production and material costs are estimated to be $6 per book. Demand over the life of the book is estimated to be 4000 copies. The publisher plans to sell the text to college and university bookstores for $46 each. a. What is the breakeven point? b. What profit or loss can be anticipated with a demand of 3800 copies? c. With a demand of 3800 copies, what is the minimum price per copy that the publisher

must charge to break even? d. If the publisher believes that the price per copy could be increased to $50.95 and not

affect the anticipated demand of 4000 copies, what action would you recommend? What profit or loss can be anticipated?

15. Preliminary plans are underway for construction of a new stadium for a major league baseball team. City officials question the number and profitability of the luxury corporate boxes planned for the upper deck of the stadium. Corporations and selected individuals may purchase a box for $300,000. The fixed construction cost for the upper-deck area is estimated to be $4,500,000, with a variable cost of $150,000 for each box constructed. a. What is the breakeven point for the number of luxury boxes in the new stadium? b. Preliminary drawings for the stadium show that space is available for the construction

of up to 50 luxury boxes. Promoters indicate that buyers are available and that all 50 could be sold if constructed. What is your recommendation concerning the construc- tion of luxury boxes? What profit is anticipated?

16. Financial Analysts, Inc., is an investment firm that manages stock portfolios for a number of clients. A new client has requested that the firm handle an $800,000 portfolio. As an initial investment strategy, the client would like to restrict the portfolio to a mix of the following two stocks:

Price/ Estimated Annual Stock Share Return/Share Oil Alaska $50 $6 Southwest Petroleum $30 $4

Let x 5 number of shares of Oil Alaska y 5 number of shares of Southwest Petroleum

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Case Problem Scheduling a Golf League 23

a. Develop the objective function, assuming that the client desires to maximize the total annual return.

b. Show the mathematical expression for each of the following three constraints: (1) Total investment funds available are $800,000. (2) Maximum Oil Alaska investment is $500,000. (3) Maximum Southwest Petroleum investment is $450,000.

Note: Adding the x $ 0 and y $ 0 constraints provides a linear programming model for the investment problem. A solution procedure for this model will be discussed in Chapter 7.

17. Models of inventory systems frequently consider the relationships among a beginning inventory, a production quantity, a demand or sales, and an ending inventory. For a given production period j, let

sj–1 5 beginning inventory for period j (ending inventory from period j – 1, the previous period) xj 5 production quantity in period j dj 5 demand in period j sj 5 ending inventory for period j

a. Write the mathematical relationship or model that shows ending inventory as a func- tion of beginning inventory, production, and demand.

b. What constraint should be added if production capacity for period j is given by Cj? c. What constraint should be added if inventory requirements for period j mandate an

ending inventory of at least Ij?

Case Problem Scheduling a Golf League Chris Lane, the head professional at Royal Oak Country Club, must develop a schedule of matches for the couples’ golf league that begins its season at 4:00 p.m. tomorrow. Eighteen couples signed up for the league, and each couple must play every other couple over the course of the 17-week season. Chris thought it would be fairly easy to develop a schedule, but after working on it for a couple of hours, he has been unable to come up with a schedule. Because Chris must have a schedule ready by tomorrow afternoon, he has asked you to help him. A possible complication is that one of the couples told Chris that they may have to cancel for the season. They told Chris they would let him know by 1:00 p.m. tomorrow whether they will be able to play this season.

Managerial Report Prepare a report for Chris Lane. Your report should include, at a minimum, the following items:

1. A schedule that will enable each of the 18 couples to play every other couple over the 17-week season.

2. A contingency schedule that can be used if the couple that contacted Chris decides to cancel for the season.

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Introduction to Probability

CONTENTS

2.1 EXPERIMENTS AND THE SAMPLE SPACE

2.2 ASSIGNING PROBABILITIES TO EXPERIMENTAL OUTCOMES Classical Method Relative Frequency Method Subjective Method

2.3 EVENTS AND THEIR PROBABILITIES

2.4 SOME BASIC RELATIONSHIPS OF PROBABILITY Complement of an Event Addition Law Conditional Probability Multiplication Law

2.5 BAYES’ THEOREM The Tabular Approach

2.6 SIMPSON’S PARADOX

CHAPTER 2

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Chapter 2 Introduction to Probability 25

Business decisions are often based on an analysis of uncertainties such as the following:

1. What are the “chances” that sales will decrease if we increase prices? 2. What is the “likelihood” that a new assembly method will increase productivity? 3. How “likely” is it that the project will be completed on time? 4. What are the “odds” that a new investment will be profitable?

Probability is a numeric measure of the likelihood that an event will occur. Thus, prob- abilities could be used as measures of the degree of uncertainty associated with the four events previously listed. If probabilities were available, we could determine the likelihood of each event occurring.

Probability values are always assigned on a scale from 0 to 1. A probability of 0 indi- cates that an event will not occur (is an impossible event); a probability of 1 indicates that an event is certain to occur (is a certain event). Other probabilities between 0 and 1 represent varying degrees of likelihood that an event will occur. The closer a probability is to 0, the less likely the associated event is to occur; the closer a probability is to 1, the more likely the associated event is to occur. Figure 2.1 depicts this view of probability.

Probability is important in decision making because it provides a way to measure, express, and analyze the uncertainties associated with future events. The Q.M. in Action, Probability to the Rescue, describes the role that probability played in the efforts to rescue 33 Chilean miners.

Probability provides a more precise description of uncertainty than expressions such as chances are “pretty good,” chances are “fair,” and so on.

On August 5, 2010, the San José copper and gold mine suffered a cave-in. Thirty-three men were trapped over 2000 feet underground in the Atacama Desert near Copi- apó, Chile. While most feared that these men’s prospects

were grim, several attempts were made to locate the min- ers and determine if they were still alive. Seventeen days later, rescuers reached the men with a 51⁄2-inch borehole and ascertained that they were still alive.

PROBABILITY TO THE RESCUE*

Q.M. in ACtIon

*the authors are indebted to Dr. Michael Duncan and Clinton Cragg of nASA for providing input for this Q.M. in Action. (continued)

The Occurrence of the Event Is Just as Likely as It Is Unlikely

Probability:

0 .5 1.0

Increasing Likelihood

of Occurrence

FIGURE 2.1 PROBABILITY AS A NUMERIC MEASURE OF THE LIKELIHOOD OF AN EVENT OCCURRING

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26 Chapter 2 Introduction to Probability

2.1 Experiments and the Sample Space In discussing probability, we define an experiment to be any process that generates well- defined outcomes. On any single repetition of an experiment, one and only one of the possible experimental outcomes will occur. Several examples of experiments and their associated outcomes follow.

After locating the men and establishing the means to communicate and deliver food, water, and medical supplies to them, the rescue effort could proceed in ear- nest. While it was important to bring these men safely to the surface as quickly as possible, it was imperative that the rescue effort also proceed cautiously. “The mine is old and there is concern of further collapses,” Murray & Roberts Cementation Managing Director Henry Laas said in an interview with the Santiago Times. “The rescue methodology therefore has to be carefully designed and implemented.”

The Chilean government asked NASA to consult on the rescue operation. In response, NASA sent a four- person team consisting of an engineer (Clinton Cragg), two physicians (Michael Duncan and J. D. Polk), and a psychologist (Al Holland). When asked why a space agency was brought in to consult on the rescue of trapped miners, Duncan stated, “We brought our experience in vehicle design and long duration confinement to our Chilean counterparts.”

The probability of failure was prominent in the thoughts of everyone involved. “We were thinking that the rescue vehicle would have to make over forty round trips, so in consideration of the probability of part fail- ures we suggested the rescue team have three rigs and several sets of replacement parts available.” said Cragg. “We also tried to increase the probability of success by placing spring loaded rollers on the sides of the cage so the cage itself would not be damaged through direct contact with the rock wall as it moved through the res- cue portal.”

Duncan added, “While we and the Chileans would have preferred to have precise estimates of various probabilities based on historical data, the uniqueness of the situation made this infeasible. For example, a miner had to stand virtually straight up in the cage on an ascent

that was originally estimated to last two to four hours per miner, so we had to be concerned about fainting. All we could do was consider what we thought to be the facts and apply what we had learned from astronauts’ experiences in their returns from short and long dura- tion space missions.” Duncan continued, “We recom- mended the miners wear compression hosiery on lower extremities to prevent blood pooling, and that they load up on salty solutions such as chicken consommé prior to their ascents. We used all of this information to develop a subjective estimate of the probability a miner would faint on ascent,” Duncan then concluded. “It actually took fifteen minutes to bring the cage up from the bot- tom of the mine, so our estimates in this case were very conservative. Considering the risk involved, that is ex- actly what we wanted.”

Ultimately the rescue approach designed by the Chileans in consultation with the NASA team was suc- cessful. On October 13, 2010, the last of the 33 miners emerged; the 13-foot-long, 924-pound steel Fénix 2 res- cue capsule withstood over 40 trips into and out of the mine, and no miner suffered from syncope (i.e., fainting) on his ascent.

The use of subjective probabilities in unique situ- ations is common for NASA. How else could NASA estimate the probability that micrometeoroids or space debris will damage a space vehicle? With the limited space available on a space vehicle, assessing probabil- ity of failure for various components and the risks as- sociated with these potential failures becomes critical to how NASA decides which spare components will be included on a space flight. NASA also employs prob- ability to estimate the likelihood of crew health and per- formance issues arising on space exploration missions. Risk assessment teams then use these estimates in their mission design.

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2.1 Experiments and the Sample Space 27

The first step in analyzing a particular experiment is to carefully define the experimen- tal outcomes. When we define all possible experimental outcomes, we identify the sample space for the experiment; that is, the set of all possible experimental outcomes. Any one particular experimental outcome is also referred to as a sample point and is an element of the sample space.

Consider the experiment of tossing a coin. The experimental outcomes are defined by the upward face of the coin—a head or a tail. If we let S denote the sample space, we can use the following notation to describe the sample space and sample points for the coin- tossing experiment:

S 5 {Head, Tail}

Using this notation for the second experiment in the preceding table, selecting a part for inspection, provides a sample space with sample points as follows:

S 5 {Defective, Nondefective}

Finally, suppose that we consider the fourth experiment in the table, rolling a standard die. The experimental outcomes are defined as the number of dots appearing on the upward face of the die. In this experiment, the numeric values 1, 2, 3, 4, 5, and 6 rep- resent the possible experimental outcomes or sample points. Thus the sample space is denoted

S 5 {1, 2, 3, 4, 5, 6}

Try Problem 1, parts (a) and (b), for practice in listing the experimental outcomes (sample points) for an experiment.

NOTES AND COMMENTS

1. In probability, the notion of an experiment is somewhat different from the laboratory sci- ences. In the laboratory sciences, the researcher assumes that each time an experiment is re- peated in exactly the same way, the same out- come will occur. For the type of experiment we

study in probability, the outcome is determined by chance. Even though the experiment might be repeated in exactly the same way, a differ- ent outcome may occur. Because of this differ- ence, the experiments we study in probability are sometimes called random experiments.

Experiment Experimental Outcomes

Toss a coin Head, tail Select a part for inspection Defective, nondefective Conduct a sales call Purchase, no purchase Roll a standard die 1, 2, 3, 4, 5, 6 Play a football game Win, lose, tie

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28 Chapter 2 Introduction to Probability

2.2 Assigning Probabilities to Experimental Outcomes With an understanding of an experiment and the sample space, let us now see how prob- abilities for the experimental outcomes can be determined. The probability of an experi- mental outcome is a numeric measure of the likelihood that the experimental outcome will occur on a single repetition of the experiment. In assigning probabilities to experimental outcomes, two basic requirements of probability must be satisfied:

1. The probability values assigned to each experimental outcome (sample point) must be between 0 and 1. If we let Ei indicate the ith experimental outcome and P(Ei) indicate the probability of this experimental outcome, we must have

0 # PsEid # 1 sfor all id (2.1)

2. The sum of all of the experimental outcome probabilities must be 1. For example, if a sample space has k experimental outcomes, we must have

PsE1d 1 PsE2d 1 Á 1 PsEkd 5 1 (2.2)

Any method of assigning probability values to the experimental outcomes that satisfies these two requirements and results in reasonable numeric measures of the likelihood of the outcomes is acceptable. In practice, the classical method, the relative frequency method, or the subjective method is often used.

Classical Method To illustrate the classical method of assigning probabilities, let us again consider the ex- periment of flipping a coin. On any one flip, we will observe one of two experimental outcomes: head or tail. It is reasonable to assume the two possible outcomes are equally likely. Therefore, as one of the two equally likely outcomes is a head, we logically should conclude that the probability of observing a head is 1/2, or 0.50. Similarly, the probability of observing a tail is 0.50. When the assumption of equally likely outcomes is used as a basis for assigning probabilities, the approach is referred to as the classical method. If an experiment has n possible outcomes, application of the classical method would lead us to assign a probability of 1/n to each experimental outcome.

As another illustration of the classical method, consider again the experiment of roll- ing a standard die. In Section 2.1 we described the sample space and sample points for this experiment with the notation

S 5 {1, 2, 3, 4, 5, 6}

A standard die is designed so that the six experimental outcomes are equally likely, and hence each outcome is assigned a probability of 1/6. Thus, if P(1) denotes the probability that one dot appears on the upward face of the die, then P(1) 5 1/6. Similarly, P(2) 5 1/6, P(3) 5 1/6, P(4) 5 1/6, P(5) 5 1/6, and P(6) 5 1/6. Note that this probability assignment satisfies the two basic requirements for assigning probabilities. In fact, requirements (2.1)

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2.2 Assigning Probabilities to Experimental outcomes 29

and (2.2) are automatically satisfied when the classical method is used, because each of the n sample points is assigned a probability of 1/n.

The classical method was developed originally to analyze gambling probabilities for which the assumption of equally likely outcomes often is reasonable. In many business problems, however, this assumption is not valid. Hence, alternative methods of assigning probabilities are required.

Relative Frequency Method Consider a firm that is preparing to market a new product. In order to estimate the prob- ability that a customer will purchase the product, a test market evaluation has been set up wherein salespeople call on potential customers. Each sales call conducted has two pos- sible outcomes: The customer purchases the product, or the customer does not purchase the product. With no reason to assume that the two experimental outcomes are equally likely, the classical method of assigning probabilities is inappropriate.

Suppose that in the test market evaluation of the product, 400 potential customers were contacted; 100 purchased the product, but 300 did not. In effect, we have repeated the experiment of contacting a customer 400 times and have found that the product was purchased 100 times. Thus, we might decide to use the relative frequency of the number of customers that purchased the product as an estimate of the probability of a customer making a purchase. We could assign a probability of 100/400 5 0.25 to the experimental outcome of purchasing the product. Similarly, 300/400 5 0.75 could be assigned to the experimental outcome of not purchasing the product. This approach to assigning probabilities is referred to as the relative frequency method. It is important to note that if we repeated this experi- ment 400 additional times, we may not obtain exactly the same probability estimate that we obtained from the first 400 trials of the experiment. The relative frequency approach only generates an empirical estimate of the actual probability of an experimental outcome, and different executions of the relative frequency approach can yield differing estimates of the probability of the same experimental outcome.

Subjective Method The subjective method of assigning probabilities is most appropriate when we cannot realistically assume that the experimental outcomes are equally likely and when little rel- evant data are available. When the subjective method is used to assign probabilities to the experimental outcomes, we may use any information available, such as our experience or intuition. After considering all available information, a probability value that expresses our degree of belief (on a scale from 0 to 1) that the experimental outcome will occur is speci- fied. Because subjective probability expresses a person’s degree of belief, it is personal. Using the subjective method, different people may assign different probabilities to the same experimental outcome.

The subjective method requires extra care to ensure that the two basic requirements of equations (2.1) and (2.2) are satisfied. Regardless of a person’s degree of belief, the prob- ability value assigned to each experimental outcome must be between 0 and 1, inclusive, and the sum of all the probabilities for the experimental outcomes must equal 1.

Consider the case in which Tom and Judy Elsbernd just made an offer to purchase a house. Two outcomes are possible:

E1 5 their offer is accepted

E2 5 their offer is rejected

Try Problem 2, part (b), for practice with the classical method.

Try Problem 2, part (c), for practice in assigning probabilities to experimental outcomes using the relative frequency approach.

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30 Chapter 2 Introduction to Probability

Judy believes that the probability that their offer will be accepted is 0.8; thus, Judy would set P(E1) 5 0.8 and P(E2) 5 0.2. Tom, however, believes that the probability that their offer will be accepted is 0.6; hence, Tom would set P(E1) 5 0.6 and P(E2) 5 0.4. Note that Tom’s probability estimate for E1 reflects a greater pessimism that their offer will be accepted.

Both Judy and Tom assigned probabilities that satisfy the two basic requirements. The fact that their probability estimates differ reflects the personal nature of the subjective method.

These three approaches for assigning probabilities to experimental outcomes are often used in various combinations. For example, in some business situations managers may combine estimates from the classical or relative frequency approach with subjective prob- ability estimates to obtain improved probability estimates.

2.3 Events and Their Probabilities An event is a collection of sample points (experimental outcomes). For example, in the experiment of rolling a standard die, the sample space has six sample points and is denoted S 5 {1, 2, 3, 4, 5, 6}. Now consider the event that the number of dots shown on the upward face of the die is an even number. The three sample points in this event are 2, 4, and 6. Using the letter A to denote this event, we write A as a collection of sample points:

A 5 {2, 4, 6}

Thus, if the experimental outcome or sample point were 2, 4, or 6, we would say that the event A has occurred.

Much of the focus of probability analysis is involved with computing probabilities for various events that are of interest to a decision maker. If the probabilities of the sample points are defined, the probability of an event is equal to the sum of the probabilities of the sample points in the event.

Returning to the experiment of rolling a standard die, we used the classical method to conclude that the probability associated with each sample point is 1/6. Thus, the probability of rolling a 2 is 1/6, the probability of rolling a 4 is 1/6, and the probability of rolling a 6 is 1/6. The probability of event A—an even number of dots on the upward face of the die—is

PsAd 5 Ps2d 1 Ps4d 1 Ps6d

5 1

6 1

6 5

Any time that we can identify all the sample points of an experiment and assign the corresponding sample point probabilities, we can use the preceding approach to compute the probability of an event. However, in many experiments the number of sample points is large, and the identification of the sample points, as well as determining their associated probabilities, becomes extremely cumbersome if not impossible. In the remainder of this chapter we present some basic probability relationships that can be used to compute the probability of an event without knowing all the individual sample point probabilities. These probability relationships require a knowledge of the probabilities for some events in the experiment. Probabilities of other events are then computed from these known probabilities using one or more of the probability relationships.

Bayes’ theorem (see Section 2.5) provides a means for combining subjectively determined prior probabilities with probabilities obtained by other means to obtain revised, or posterior, probabilities.

Try Problem 6 for practice in assigning probabilities to events.

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2.4 Some Basic Relationships of Probability In this section we present several relationships that will be helpful in computing prob- abilities. The relationships are the complement of an event, the addition law, conditional probability, and the multiplication law.

Complement of an Event For an event A, the complement of event A is the event consisting of all sample points in sample space S that are not in A. The complement of A is denoted by Ac. Figure 2.2 provides a diagram, known as a Venn diagram, that illustrates the concept of a complement. The rectangular area represents the sample space for the experiment and as such contains all possible sample points. The circle represents event A and contains only the sample points that belong to A. The remainder of the rectangle contains all sample points not in event A, which by definition is the complement of A.

In any probability application, event A and its complement Ac must satisfy the condition

P(A) 1 P(Ac) 5 1

Solving for P(A), we have

P(A) 5 1 2 P(Ac) (2.3)

NOTES AND COMMENTS

1. The sample space, S, is itself an event. It con- tains all the experimental outcomes, so it has a probability of 1; that is, P(S) 5 1.

2. When the classical method is used to assign probabilities, the assumption is that the ex-

perimental outcomes are equally likely. In such cases, the probability of an event can be com- puted by counting the number of experimental outcomes in the event and dividing the result by the total number of experimental outcomes.

Note: The shaded region depicts the complement of A, denoted Ac

AcEvent A

Sample Space S

FIGURE 2.2 COMPLEMENT OF EVENT A

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Equation (2.3) shows that the probability of an event A can be computed by subtraction if the probability of its complement, P(Ac), is known. Similarly, subtraction can be used to compute the probability of the compliment Ac of an event A if the probability of the event P(A), is known.

Consider the case of a sales manager who, after reviewing sales reports, states that 80% of new customer contacts result in no sale. By letting A denote the event of a sale and Ac denote the event of no sale, the manager is stating that P(Ac) 5 0.80. Using equation (2.3), we see that

P(A) 5 1 2 P(Ac) 5 1 2 0.80 5 0.20

which shows that there is a 0.20 probability that a sale will be made on a new customer contact.

In another case, a purchasing agent states a 0.90 probability that a supplier will send a ship- ment that is free of defective parts. Using the complement, we can conclude a 1 2 0.90 5 0.10 probability that the shipment will contain some defective parts.

Addition Law The addition law is a useful relationship when we have two events and are interested in knowing the probability that at least one of the events occurs. That is, with events A and B, we are interested in knowing the probability that event A or event B or both will occur. Before we present the addition law, we need to discuss two concepts concerning combina- tions of events: the union of events and the intersection of events.

For two events A and B, the union of events A and B is the event containing all sample points belonging to A or B or both. The union is denoted A ´ B. The Venn diagram shown in Figure 2.3 depicts the union of events A and B; the shaded region contains all the sample points in event A, as well as all the sample points in event B. The fact that the circles overlap (or intersect) indicates that some sample points are contained in both A and B.

For two events A and B, the intersection of events A and B is the event containing the sample points belonging to both A and B. The intersection is denoted by A ¨ B. The Venn diagram depicting the intersection of the two events is shown in Figure 2.4. The area where the two circles overlap is the intersection; it contains the sample points that are in both A and B.

Event A

Sample Space S

Event B

FIGURE 2.3 UNION OF EVENTS A AND B (SHADED REGION)

Key words for the union of events (A ´ B) are “either A or B occurs” or “at least one of the two events occurs.” Note that the conjunction “or” commonly indicates a union of events.

Key words for the intersection of events (A ¨ B) are “both A and B occur.” Note that the conjunction “and” commonly indicates an intersection of events.

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2.4 Some Basic Relationships of Probability 33

The addition law provides a way to compute the probability of event A or B or both occurring. In other words, the addition law is used to compute the probability of the union of two events, A ´ B. The addition law is formally stated as follows:

PsA ø Bd 5 PsAd 1 PsBd 2 PsA ù Bd (2.4)

To obtain an intuitive understanding of the addition law, note that the first two terms in the addition law, P(A) 1 P(B), account for all the sample points in A ´ B. However, as the sample points in the intersection A ¨ B are in both A and B, when we compute P(A) 1 P(B), we in effect are counting each of the sample points in A ¨ B twice. We correct for this double counting by subtracting P(A ¨ B).

To apply the addition law, let us consider the following situations in a college course in quantitative methods for decision making. Of 200 students taking the course, 160 passed the midterm examination and 140 passed the final examination; 124 students passed both exams. Let

A 5 event of passing the midterm exam

B 5 event of passing the final exam

This relative frequency information leads to the following probabilities:

PsAd 5 160

200 5 0.80

PsBd 5 140

200 5 0.70

PsA ù Bd 5 124

200 5 0.62

After reviewing the grades, the instructor decided to give a passing grade to any student who passed at least one of the two exams; note that this implies the instructor will give a passing grade to any student who passed the midterm exam or passed the final exam. That is, any student who passed the midterm, any student who passed the final, and any student who passed both exams would receive a passing grade. What is the probability of a student receiving a passing grade in this course?

Note that P(A ´ B) 5 P(B ´ A); that is, the order of events in a union does not affect the probability of the union.

Note that P(A ¨ B) 5 P(B ¨ A); that is, the order of events in an intersection does not affect the probability of the intersection.

Event

Event A > B

Sample Space S

Event B

FIGURE 2.4 INTERSECTION OF EVENTS A AND B (SHADED REGION)

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34 Chapter 2 Introduction to Probability

Your first reaction may be to try to count how many of the 200 students passed at least one exam, but note that the probability question is about the union of the events A and B. That is, we want to know the probability that a student passed the midterm (A), passed the final (B), or passed both. Thus we want to know P(A ´ B). Using the addition law (2.4) for the events A and B, we have

PsA ø Bd 5 PsAd 1 PsBd 2 PsA ù Bd

Knowing the three probabilities on the right-hand side of this equation, we obtain

PsA ø Bd 5 0.80 1 0.70 2 0.62 5 0.88

This result indicates an 88% chance of a student passing the course because of the 0.88 probability of passing at least one of the exams.

Now consider a study involving the television-viewing habits of married couples. It was reported that 30% of the husbands and 20% of the wives were regular viewers of a particular Friday evening program. For 12% of the couples in the study, both husband and wife were regular viewers of the program. What is the probability that at least one member of a married couple is a regular viewer of the program?

Let

H 5 husband is a regular viewer

W 5 wife is a regular viewer

We have P(H) 5 0.30, P(W) 5 0.20, and P(H ¨ W) 5 0.12; thus, the addition law yields

PsH ø Wd 5 PsHd 1 PsWd 2 PsH ù Wd 5 0.30 1 0.20 2 0.12 5 0.38

This result shows a 0.38 probability that at least one member of a married couple is a regular viewer of the program.

Before proceeding, let us consider how the addition law is applied to mutually exclu- sive events. Two or more events are said to be mutually exclusive if the events do not have any sample points in common—that is, there are no sample points in the intersection of the events. For two events A and B to be mutually exclusive, P(A ̈ B) 5 0. Figure 2.5 provides

An event and its complement are mutually exclusive and their union is the entire sample space.

Event A Event B

FIGURE 2.5 MUTUALLY EXCLUSIVE EVENTS

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2.4 Some Basic Relationships of Probability 35

a Venn diagram depicting two mutually exclusive events. Because P(A ¨ B) 5 0 for the special case of mutually exclusive events, the addition law becomes

PsA ø Bd 5 PsAd 1 PsBd (2.5)

To compute the probability of the union of two mutually exclusive events, we simply add the corresponding probabilities.

Conditional Probability In many probability situations, being able to determine the probability of one event when another related event is known to have occurred is important. Suppose that we have an event A with probability P(A) and that we obtain new information or learn that another event, denoted B, has occurred. If A is related to B, we will want to take advantage of this information in computing a new or revised probability for event A.

This new probability of event A is written P(A Z B). The “Z” denotes the fact that we are considering the probability of event A given the condition that event B has occurred. Thus, the notation P(A Z B) is read “the probability of A given B.”

With two events A and B, the general definitions of conditional probability for A given B and for B given A are as follows:

PsA u Bd 5 PsA ù Bd

PsBd (2.6)

PsB u Ad 5 PsA ù Bd

PsAd (2.7)

Note that for these expressions to have meaning, P(B) cannot equal 0 in equation (2.6) and P(A) cannot equal 0 in equation (2.7). Also note that P(A Z B) Þ P(B Z A), unless P(A) 5 P(B).

To obtain an intuitive understanding of the use of equation (2.6), consider the Venn diagram in Figure 2.6. The shaded region (both light gray and dark gray) denotes that

For a conditional prob- ability such as P(A | B) 5 0.25, the probability value of 0.25 refers only to the probability of event A. No information is provided about the probability of event B.

Event A

Event A > B

Event B

FIGURE 2.6 CONDITIONAL PROBABILITY P(A Z B) 5 P(A ¨ B)/P(B)

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36 Chapter 2 Introduction to Probability

event B has occurred; the dark gray shaded region denotes the event (A ¨ B). We know that once B has occurred, the only way that we can also observe event A is for event (A ¨ B) to occur. Thus, the ratio P(A ¨ B)/P(B) provides the probability that we will observe event A when event B has already occurred.

We can apply conditional probability to the promotional status of male and female of- ficers of a major metropolitan police force. The force consists of 1200 officers: 960 men and 240 women. Over the past two years, 324 officers have been promoted. Table 2.1 shows the specific breakdown of promotions for male and female officers. Such a table is often called a contingency table or a crosstabulation.

After reviewing the promotional record, a committee of female officers filed a discrimi- nation case on the basis that only 36 female officers had received promotions during the past two years. The police administration argued that the relatively low number of promotions for female officers is due not to discrimination but to the fact that few female officers are on the force. We use conditional probability to evaluate the discrimination charge.

Let

M 5 event an officer is a man

W 5 event an officer is a woman

B 5 event an officer is promoted

Dividing the data values in Table 2.1 by the total of 1200 officers permits us to summarize the available information as follows:

PsM ù Bd 5 288

1200 5 0.24

probability that an officer is a man and is promoted

PsM ù Bcd 5 672

1200 5 0.56

probability that an officer is a man and is not promoted

PsW ù Bd 5 36

1200 5 0.03

probability that an officer is a woman and is promoted

PsW ù Bcd 5 204

1200 5 0.17

probability that an officer is a woman and is not promoted

Because each of these values gives the probability of the intersection of two events, these probabilities are called joint probabilities. Table 2.2, which provides a summary of the probability information for the police officer promotion situation, is referred to as a joint probability table.

Try Problem 12 for practice computing conditional probabilities.

Promoted Not Promoted Total Men 288 672 960 Women 36 204 240

Total 324 876 1200

TABLE 2.1 CONTINGENCY TABLE FOR PROMOTIONAL STATUS OF POLICE OFFICERS DURING THE PAST TWO YEARS

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2.4 Some Basic Relationships of Probability 37

The values in the margins of the joint probability table provide the probabilities of each single event separately: P(M) 5 0.80, P(W) 5 0.20, P(B) 5 0.27, and P(Bc) 5 0.73, which indicate that 80% of the force is male, 20% of the force is female, 27% of all offi- cers received promotions, and 73% were not promoted. These probabilities are referred to as marginal probabilities because of their location in the margins of the joint probability table. Returning to the issue of discrimination against the female officers, we see that the probability of promotion of an officer is P(B) 5 0.27 (regardless of whether that officer is male or female). However, the critical issue in the discrimination case involves the two conditional probabilities P(B Z M) and P(B Z W); that is, what is the probability of a promo- tion given that the officer is a man and what is the probability of a promotion given that the officer is a woman? If these two probabilities are equal, the discrimination case has no basis because the chances of a promotion are the same for male and female officers. However, different conditional probabilities will support the position that male and female officers are treated differently in terms of promotion.

Using equation (2.7), the conditional probability relationship, we obtain

PsB u M d 5 PsB ù Md

PsM d 5

0.24

0.80 5 0.30 15 288@1200960@1200 5

288

9602 PsB u W d 5

PsB ù Wd PsWd

5 0.03

0.20 5 0.15 15 36@1200240@1200 5

2402 What conclusions do you draw? The probability of a promotion for a male officer is 0.30, which is twice the 0.15 probability of a promotion for a female officer. Although the use of conditional probability does not in itself prove that discrimination exists in this case, the conditional probability values strongly support the argument presented by the female officers.

In this illustration, P(B) 5 0.27, P(B Z M) 5 0.30, and P(B Z W) 5 0.15. Clearly, the probability of promotion (event B) differs by gender. In particular, as P(B Z M) Z P(B), events B and M are dependent events. The probability of event B (promotion) is higher when M (the officer is male) occurs. Similarly, with P(B Z W) Z P(B), events B and W are dependent events. But, if the probability of event B was not changed by the existence of

Promoted Not Promoted Total

Men 0.24 0.56 0.80 Women 0.03 0.17 0.20

Total 0.27 0.73 1.00

TABLE 2.2 JOINT PROBABILITY TABLE FOR POLICE OFFICER PROMOTIONS

Marginal probabilities appear in the margins of the table.

Joint probabilities appear in the body of the table.

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38 Chapter 2 Introduction to Probability

event M—that is, P(B Z M) 5 P(B)—events B and M would be independent events. Two events A and B are independent if

P(B Z A) 5 P(B)

P(A Z B) 5 P(A)

Otherwise, the events are dependent. The Q.M. in Action, Product Testing for Quality Control at Morton International,

describes how a subsidiary of Morton International used conditional probability to help decide to implement a quality control test.

Morton International is a company with businesses in salt, household products, rocket motors, and specialty chemicals. Carstab Corporation, a subsidiary of Morton, produces a variety of specialty chemical products de- signed to meet the unique specifications of its custom- ers. For one particular customer, Carstab produced an expensive catalyst used in chemical processing. Some, but not all, of the product produced by Carstab met the customer’s specifications.

Carstab’s customer agreed to test each lot after receiv- ing it to determine whether the catalyst would perform the desired function. Lots that did not pass the customer’s test would be returned to Carstab. Over time, Carstab found that the customer was accepting 60% of the lots and re- turning 40%. In probability terms, each Carstab shipment to the customer had a 0.60 probability of being accepted and a 0.40 probability of being returned.

Neither Carstab nor its customer was pleased with these results. In an effort to improve service, Carstab ex- plored the possibility of duplicating the customer’s test prior to shipment. However, the high cost of the spe- cial testing equipment made that alternative infeasible. Carstab’s chemists then proposed a new, relatively low- cost test designed to indicate whether a lot would pass the customer’s test. The probability question of interest was: What is the probability that a lot will pass the customer’s test given that it passed the new Carstab test?

A sample of lots was tested under both the custom- er’s procedure and Carstab’s proposed procedure. Results were that 55% of the lots passed Carstab’s test, and 50% of the lots passed both the customer’s and Carstab’s test. In probability notation, we have

A 5 the event the lot passes the customer’s test

B 5 the event the lot passes Carstab’s test

where

P(B) 5 0.55 and P(A ¨ B) 5 0.50

The probability information sought was the conditional probability P(A | B) given by

PsA u Bd 5 PsA ù Bd

PsBd 5

0.50

0.55 5 0.909

Prior to Carstab’s new test, the probability that a lot would pass the customer’s test was 0.60. However, the new results showed that given that a lot passed Carstab’s new test, it had a 0.909 probability of passing the cus- tomer’s test. This result was good supporting evidence for the use of the test prior to shipment. Based on this probability analysis, the preshipment testing procedure was implemented at the company. Immediate results showed an improved level of customer service. A few lots were still being returned; however, the percentage was greatly reduced. The customer was more satisfied and return shipping costs were reduced.

PRODUCT TESTING FOR QUALITY CONTROL AT MORTON INTERNATIONAL*

Q.M. in ACtIon

*Based on information provided by Michael Haskell of Morton International.

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2.4 Some Basic Relationships of Probability 39

Multiplication Law The multiplication law can be used to find the probability of an intersection of two events. The multiplication law is derived from the definition of conditional probability. Using equations (2.6) and (2.7) and solving for P(A ¨ B), we obtain the multiplication law:

PsA ù Bd 5 PsA u BdPsBd (2.8)

PsA ù Bd 5 PsB u AdPsAd (2.9)

The multiplication law is useful in situations for which probabilities such as P(A), P(B), P(A Z B), and/or P(B Z A) are known but P(A ¨ B) is not. For example, suppose that a news- paper circulation department knows that 84% of its customers subscribe to the daily edition of the paper. Let D denote the event that a customer subscribes to the daily edition; hence, P(D) 5 0.84. In addition, the department knows that the conditional probability that a cus- tomer who already holds a daily subscription also subscribes to the Sunday edition (event S) is 0.75; that is, P(S Z D) 5 0.75. What is the probability that a customer subscribes to both the daily and Sunday editions of the newspaper? Using equation (2.9), we compute P(D ¨ S):

PsD ù Sd 5 PsS u DdPsDd 5 0.75s0.84d 5 0.63

This result tells us that 63% of the newspaper’s customers subscribe to both the daily and Sunday editions.

Before concluding this section, let us consider the special case of the multiplication law when the events involved are independent. Recall that independent events exist whenever P(B Z A) 5 P(B) or P(A Z B) 5 P(A). Returning to the multiplication law, equations (2.8) and (2.9), we can substitute P(A) for P(A Z B) and P(B) for P(B Z A). Hence, for the special case of independent events, the multiplication law becomes

PsA ù Bd 5 PsAdPsBd (2.10)

Thus, to compute the probability of the intersection of two independent events, we simply multiply the corresponding probabilities. For example, a service station manager knows from past experience that 40% of her customers use a credit card when purchasing gasoline. What is the probability that the next two customers purchasing gasoline will both use a credit card? If we let

A 5 the event that the first customer uses a credit card

B 5 the event that the second customer uses a credit card

the event of interest is A ¨ B. With no other information, it is reasonable to assume A and B are independent events. Thus,

PsA ù Bd 5 PsAdPsBd 5 s.40ds.40d 5 0.16

NOTES AND COMMENTS

1. Do not confuse mutually exclusive events with independent events. Two events with nonzero probabilities cannot be both mutually exclusive and independent. If one mutually exclusive event

is known to occur, the probability of the other occurring is reduced to zero. Thus, they cannot be independent.

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40 Chapter 2 Introduction to Probability

2.5 Bayes’ Theorem In the discussion of conditional probability, we indicated that revising probabilities when new information is obtained is an important phase of probability analysis. Often, we begin an analysis with initial or prior probability estimates for specific events of interest; these initial estimates are generally developed using either the relative frequency approach (applied to historical data) or the subjective approach. Then, from sources such as a sample, a special report, or a product test, we obtain some additional information about the events. With this new information, we update the prior probability values by calculating revised probabilities, referred to as posterior probabilities. Bayes’ theorem provides a means for making these probability revisions. The steps in this probability revision process are shown in Figure 2.7.

We can apply Bayes’ theorem to a manufacturing firm that receives shipments of parts from two different suppliers. Let A1 denote the event that a part is from supplier 1 and A2 denote the event that a part is from supplier 2. Currently, 65% of the parts purchased by the company are from supplier 1, and the remaining 35% are from supplier 2. Thus, if a part is selected at random, we would assign the prior probabilities P(A1) 5 0.65 and P(A2) 5 0.35.

The quality of the purchased parts varies with the source of supply. Based on historical data, the conditional probabilities of receiving good and bad parts from the two suppliers are shown in Table 2.3. Thus, if we let G denote the event that a part is good and B denote the event that a part is bad, the information in Table 2.3 provides the following conditional probability values:

P(G Z Al) 5 0.98 P(B Z A1) 5 0.02 P(G Z A2) 5 0.95 P(B Z A2) 5 0.05

The tree diagram shown in Figure 2.8 depicts the process of the firm receiving a part from one of the two suppliers and then discovering that the part is good or bad as a two-step experiment. Of the four possible experimental outcomes, two correspond to the part being good, and two correspond to the part being bad.

Each of the experimental outcomes is the intersection of two events, so we can use the multiplication rule to compute the probabilities. For instance,

P(Al ¨ G) 5 P(A1)P(G Z A1)

Prior Probabilities

Posterior Probabilities

New Information

Application of Bayes’ Theorem

FIGURE 2.7 PROBABILITY REVISION USING BAYES’ THEOREM

Good Parts Bad Parts Supplier 1 0.98 0.02 Supplier 2 0.95 0.05

TABLE 2.3 CONDITIONAL PROBABILITIES OF RECEIVING GOOD AND BAD PARTS FROM TWO SUPPLIERS

P(B | A1)

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2.5 Bayes’ theorem 41

The process of computing these joint probabilities can be depicted in what is sometimes called a probability tree, as shown in Figure 2.9. From left to right in the tree, the prob- abilities for each of the branches at step 1 are the prior probabilities, and the probabilities for each branch at step 2 are conditional probabilities. To find the probabilities of each experimental outcome, we simply multiply the probabilities on the branches leading to the outcome. Each of these joint probabilities is shown in Figure 2.9, along with the known probabilities for each branch. Note that the probabilities of the four experimental outcomes sum to 1.

Note: Step 1 shows that the part comes from one of two suppliers, and step 2 shows whether the part is good or bad.

Step 1 Step 2 (Condition)

Experimental Outcome

(A1, G )

(A1, B )

(A2, G )

(A2, B )

FIGURE 2.8 TWO-STEP TREE DIAGRAM

P(A1)

0.65

Step 1 (Supplier)

Step 2 (Condition)

Probability of Outcome

P(A2)

0.35

P(G | A1)

0.98

P(B | A1)

0.02

P(G | A2)

0.95

P(B | A2)

0.05

P(A1 G ) = P(A1)P(G | A1) = 0.6370

P(A1 B ) = P(A1)P(B | A1) = 0.0130

P(A2 G ) = P(A2)P(G | A2) = 0.3325

P(A2 B ) = P(A2)P(B | A2) = 0.0175

FIGURE 2.9 PROBABILITY TREE FOR TWO-SUPPLIER EXAMPLE

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42 Chapter 2 Introduction to Probability

Now suppose that the parts from the two suppliers are used in the firm’s manufacturing process and that a bad part causes a machine to break down. What is the probability that the bad part came from supplier 1, and what is the probability that it came from supplier 2? With the information in the probability tree (Figure 2.9), we can use Bayes’ theorem to answer these questions.

Letting B denote the event the part is bad, we are looking for the posterior probabilities P(A1 Z B) and P(A2 Z B). From the definition of conditional probability, we know that

PsA1 u Bd 5 PsA1 ù Bd

PsBd (2.11)

Referring to the probability tree, we see that

PsA1 ù Bd 5 PsA1dPsB u A1d (2.12)

To find P(B), we note that event B can occur in only two ways: (A1 ¨ B) and (A2 ¨ B). Therefore, we have

PsBd

5 PsA1 ù Bd 1 PsA2 ù Bd 5 PsA1dPsB u A1d 1 PsA2dPsB u A2d

(2.13)

Substituting from equations (2.12) and (2.13) into equation (2.11) and writing a similar result for P(A2 Z B), we obtain Bayes’ theorem for the case of two events.

PsA1 u Bd 5 PsA1dPsB u A1d

PsA1dPsB u A1d 1 PsA2dPsB u A2d (2.14)

PsA2 u Bd 5 PsA2dPsB u A2d

PsA1dPsB u A1d 1 PsA2dPsB u A2d (2.15)

Using equation (2.14) and the probability values provided in our example, we have

PsA1 u Bd 5 PsA1dPsB u A1d

PsA1dPsB u A1d 1 PsA2dPsB u A2d

5 s0.65ds0.02d

s0.65ds0.02d 1 s0.35ds0.05d 5

0.0130

0.0130 1 0.0175

5 0.0130

0.0305 5 0.4262

In addition, using equation (2.15), we obtain P(A2 Z B):

PsA2 u Bd 5 s0.35ds0.05d

s0.65ds0.02d 1 s0.35ds0.05d

5 0.0175

0.0130 1 0.0175 5

0.0175

0.0305 5 0.5738

The Reverend Thomas Bayes (1702–1761), a Presbyterian minister, is credited with the original work leading to the version of Bayes’ theorem in use today.

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2.5 Bayes’ theorem 43

Note that in this application we initially started with a probability of 0.65 that a part selected at random was from supplier 1. However, given information that the part is bad, we de- termine the probability that the part is from supplier 1 drops to 0.4262. If the part is bad, the probability that the part was from supplier 2 increases from 0.35 to 0.5738; that is, P(A2 Z B) 5 0.5738.

Bayes’ theorem is applicable when the events for which we want to compute posterior probabilities are mutually exclusive and their union is the entire sample space.1 Bayes’ theorem can be extended to the case of n mutually exclusive events A1, A2, . . . , An, whose union is the entire sample space. In such a case Bayes’ theorem for the computation of any posterior probability P(Ai Z B) becomes

PsAi u Bd 5 PsAi dPsB u Ai d

PsA1dPsB u A1d 1 PsA2 dPsB u A2 d 1 . . . 1 PsAn dPsB u An d (2.16)

With prior probabilities P(A1), P(A2), . . . , P(An) and the appropriate conditional probabili- ties P(B Z A1), P(B Z A2), . . . , P(B Z An), equation (2.16) can be used to compute the posterior probability of the events A1, A2, . . . , An.

The Tabular Approach The tabular approach is helpful in conducting the Bayes’ theorem calculations simultane- ously for all events Ai. Such an approach is shown in Table 2.4. The computations shown there involve the following steps.

Step 1. Prepare three columns: Column 1— The mutually exclusive events for which posterior probabilities

are desired Column 2— The prior probabilities for the events Column 3— The conditional probabilities of the new information given each

event Step 2. In column 4, compute the joint probabilities for each event and the new in-

formation B by using the multiplication law. To get these joint probabilities, multiply the prior probabilities in column 2 by the corresponding conditional probabilities in column 3—that is, P(Ai ¨ B) 5 P(Ai)P(B Z Ai).

Step 3. Sum the joint probabilities in column 4 to obtain the probability of the new information, P(B). In the example there is a 0.0130 probability that a part is

Try Problem 20 for practice using Bayes’ theorem to compute posterior probabilities.

(1) (2) (3) (4) (5) Prior Conditional Joint Posterior Events Probabilities Probabilities Probabilities Probabilities Ai P(Ai) P(B Z Ai) P(Ai > B) P(Ai Z B)

A1 0.65 0.02 0.0130 0.0130y0.0305 5 0.4262 A2 0.35 0.05 0.0175 0.0175y0.0305 5 0.5738 1.00 P(B) 5 0.0305 1.0000

TABLE 2.4 SUMMARY OF BAYES’ THEOREM CALCULATIONS FOR THE TWO-SUPPLIER PROBLEM

1If the union of events is the entire sample space, the events are often called collectively exhausted.

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44 Chapter 2 Introduction to Probability

from supplier 1 and is bad and a 0.0175 probability that a part is from supplier 2 and is bad. These are the only two ways by which a bad part can be obtained, so the sum 0.0130 1 0.0175 shows an overall probability of 0.0305 of finding a bad part from the combined shipments of both suppliers.

Step 4. In column 5, compute the posterior probabilities by using the basic relation- ship of conditional probability:

PsAi u Bd 5 PsAi ù Bd

PsBd

Note that the joint probabilities P(Ai ¨ B) appear in column 4, whereas P(B) is the sum of the column 4 values.

Try Problem 25 for an application of Bayes’ theorem involving the tabular approach.

NOTES AND COMMENTS

1. Bayes’ theorem is used in decision analysis (see Chapter 4). The prior probabilities often are sub- jective estimates provided by a decision maker. Sample information is obtained and posterior probabilities are computed for use in develop- ing a decision strategy.

2. An event and its complement are mutually exclusive, and their union is the entire sample space. Thus, Bayes’ theorem is always appli- cable for computing posterior probabilities of an event and its complement.

In an article published in the International Journal of Network Security & Its Applications, an algorithm for detecting spam in blogs (which is often called comment spam or blog spam) is outlined. The content available in blogs, forums, and email archives is a rich source for search engines, making these envi- ronments attractive targets for comment spammers. Some comment spammers use the information culled from these environments to identify and reach poten- tial customers, others place trackbacks or ping pongs that are designed to attract traffic to the spammer’s website or blog, while still others launch man-in-the- middle attacks to cripple network servers or to phish for information on potential victims. Comment spam also wastes a user’s time with unwanted comments and consumes bandwidth, and it ultimately can have a deleterious impact on the traffic and success of

the spammed forum. Therefore, the identification of comment spammers is critical.

The comment spam detection algorithm is based on Bayes’ theorem. First, the probability that a word is found in a blog comment spam post is determined. For example, if a post is comment spam, the word “free” has a relatively high probability of occurring in the post. Then the overall prob- ability that any post is comment spam is established. Bayes’ theorem is applied to these two probabilities to derive the probability that a post containing the word in question is comment spam. If this probability exceeds some threshold, the potential post is not allowed. The individual responsible for maintaining the blog can set the threshold prob ability to reflect her or his tolerance for comment spam.

A spam attacker using automated tools such as xRumer can send millions of spam mails and cause many problems with little effort, so blogs need a sound defense against comment spammers. Bayes’ theorem can help organizations to quickly and accurately distinguish com- ment spam from legitimate posts and avoid the deleteri- ous effects of comment spam.

USING BAYES’ THEOREM TO DEFEAT COMMENT SPAM*

Q.M. in ACtIon

*Dhinaharan nagamalai, Beatrice Cynthia Dhinakaran, and Jae Kwang Lee (2010). “Bayesian Based Comment Spam Defending tool.” Interna- tional Journal of Network Security & Its Applications, 2(4): 267–280.

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2.6 Simpson’s Paradox 45

2.6 Simpson’s Paradox As we have discussed in Sections 2.4 and 2.5, the concept of conditional probability quanti- fies the effect of an event B on the likelihood of another event A. It is possible for the apparent association between two events to be reversed upon considering a third event; this phenom- enon is referred to as Simpson’s paradox. Simpson’s paradox often occurs when data from two or more subsets of a population are combined or aggregated to produce a summary crosstabulation in an attempt to show how two events are related. In such cases, we must be careful in drawing conclusions from the aggregated data about the relationship between the two events because the conclusions based upon the aggregated data can be completely re- versed if we look at the disaggregated data. To provide an illustration of Simpson’s paradox, we consider an example involving the analysis of verdicts for two judges.

Judges Ron Luckett and Dennis Kendall presided over cases in Common Pleas Court and Municipal Court during the past three years. Some of the verdicts they rendered were appealed. In most of these cases the appeals court upheld the original verdicts, but in some cases those verdicts were reversed. Table 2.5 illustrates the crosstabulation of cases broken down by verdict (upheld or reversed) and judge (Luckett or Kendall). This crosstabulation shows the number of appeals for which the verdict was upheld and the number of appeals for which the verdict was reversed for each judge (the column percentages in parentheses next to each value).

A review of the column percentages shows that 14% of the verdicts were reversed for Judge Luckett, but only 12% of the verdicts were reversed for Judge Kendall. That is, P (reversed Z Luckett) 5 0.14 and P(reversed Z Kendall) 5 0.12. Based on this data, we might conclude that Judge Kendall is doing a better job because a lower percentage of his verdicts are being reversed upon appeal.

A closer look at the data, however, suggests there may be a problem with the conclusion that Judge Kendall’s verdicts are reversed less often. If we further break down the verdict data by the type of court (Common Pleas or Municipal) in which the cases were heard, we obtain the crosstabulations in Table 2.6; column percentages are also shown in parentheses next to each value.

From the crosstabulation and column percentages for Luckett, we see that his verdicts were reversed in 9% of the Common Pleas Court cases and in 15% of the Municipal Court cases. That is, P(reversed Z Luckett and Common Pleas Court) 5 0.09 and P(reversed | Luckett and Municipal Court) 5 0.15. From the crosstabulation and column percentages for Kendall, we see that his verdicts were reversed in 10% of the Common Pleas Court cases and in 20% of the Municipal Court cases. That is, P(reversed Z Kendall and Common Pleas Court) 5 0.10 and P(reversed Z Kendall and Municipal Court) 5 0.20. Since P(reversed | Luckett and Common Pleas Court) , P(reversed Z Kendall and Common Pleas Court) and

Judge

Verdict Luckett Kendall Total

Upheld 129 (86%) 110 (88%) 239

Reversed 21 (14%) 15 (12%) 36

Total (%) 150 (100%) 125 (100%) 275

TABLE 2.5 SUMMARY OF VERDICTS UPHELD AND REVERSED FOR JUDGES LUCKETT AND KENDALL

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46 Chapter 2 Introduction to Probability

P(reversed Z Luckett and Municipal Court) , P(reversed Z Kendall and Municipal Court), we see that Judge Luckett demonstrates a better record than Judge Kendall in each court. This result contradicts the conclusion we reached when we aggregated the data across both courts for the original crosstabulation. It appeared then that Judge Kendall had the better record. This example illustrates Simpson’s paradox.

The original crosstabulation was obtained by aggregating the data in the separate cross- tabulations for the two courts. Note that for both judges the percentage of appeals that resulted in reversals was much higher in Municipal Court than in Common Pleas Court. Because Judge Luckett tried a much higher percentage of his cases in Municipal Court, the aggregated data favored Judge Kendall. When we look at the crosstabulations for the two courts separately, however, Judge Luckett clearly shows the better record. Thus, for the original crosstabulation, we see that the type of court is a hidden variable that cannot be ignored when evaluating the records of the two judges.

Because of Simpson’s paradox, we need to be particularly careful when drawing conclusions using aggregated data. Before drawing any conclusions about the relationship between two events shown for a crosstabulation involving aggregated data, you should investigate whether any hidden variables could affect the results. The QM in Action, Accurate Comparison of Unemployment Rates, provides an interesting and real example of Simpson’s paradox.

NOTES AND COMMENTS

1. While Simpson’s paradox is named for Edward Simpson (who described the paradox in 1951), the phenomenon was previously documented in papers by Pearson, Lee, and Bramley-Moore in

1899 and again by Yule in 1903. For this reason some refer to Simpson’s paradox as the rever- sal paradox, the amalgamation paradox, or the Yule–Simpson effect.

Judge Luckett

Common Municipal Verdict Pleas Court Total

Upheld 29 (91%) 100 (85%) 129

Reversed 3 (9%) 18 (15%) 21

Total (%) 32 (100%) 118 (100%) 150

Judge Kendall

Common Municipal Verdict Pleas Court Total

Upheld 90 (90%) 20 (80%) 110

Reversed 10 (10%) 5 (20%) 15

Total (%) 100 (100%) 25 (100%) 125

TABLE 2.6 SUMMARY OF VERDICTS UPHELD AND REVERSED IN COMMON PLEAS AND MUNICIPAL COURT FOR JUDGES LUCKETT AND KENDALL

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Summary 47

In a December 2009 article in The Wall Street Journal, Cari Tuna asked, “Is the current economic slump worse than the recession of the early 1980s?” Tuna observed that if one uses the unemployment rate as the standard, on the surface the answer to this question appears to be no; the jobless rate was 10.2% in October of 2009, which is substantially below the peak of 10.8% in November and December of 1982.

Tuna then considers the question seperately for each of several education groups. Surprisingly, the unemploy- ment rate among workers in each education group (high- school dropouts, high-school graduates, some college, and college graduates) was higher in October of 2009 than it was during the 1980s recession.

How can this be? How could the overall unemploy- ment rate be lower in 2009 than it was during the reces- sion of the early 1980s, while the unemployment rate for each of these groups was higher in 2009 than it was during the early 1980s recession? The anomaly is an example of Simpson’s paradox—the phenomenon by which the

apparent association between two events is reversed upon consideration of a third event.

As Tuna explains, the 2009 overall unemployment rate appears to be lower because college graduates (who have the lowest unemployment rate among education groups) were a larger proportion of the workforce in 2009 than they were in the early 1980s (they were ap- proximately one-third of the 2009 workforce and 25% in the early 1980s). At the same time, the proportion of high-school dropouts fell from almost 20% in the early 1980s to approximately 10% in 2009. Even though the 2009 unemployment rate was higher among both groups than it was during the recession of the early 1980s, the discrepancies in contributions to the composition of the labor force by these two groups reverses the result when data from the two groups are aggregated.

In this case the data aggregated across education groups suggests the opposite of what the disaggregated data imply, and consideration of the aggregated data leads to an erroneous conclusion. Examples of Simp- son’s paradox are common across industries; Tuna cites examples from medicine, sports, education, and air travel. If we are not careful in our analyses, we can draw incorrect conclusions from aggregate data and as a result make poor decisions.

ACCURATE COMPARISON OF UNEMPLOYMENT RATES—WHICH RECESSION WAS MORE SEVERE?*

Q.M. in ACtIon

*Based on C. tuna, “When Combined Data Reveal the Flaw of Aver- ages in a Statistical Anomaly Dubbed Simpson’s Paradox: Aggregated numbers obscure trends in Job Market, Medicine and Baseball,” The Wall Street Journal, December 2, 2009 (http://online.wsj.com/article /SB125970744553071829.html).

Summary

In this chapter we introduced basic probability concepts and illustrated how probability analysis can provide helpful decision-making information. We described how probabil- ity can be interpreted as a numeric measure of the likelihood that an event will occur, and discussed various ways that probabilities can be assigned to events. In addition, we showed that the probability of an event can be computed either by summing the prob- abilities of the experimental outcomes (sample points) comprising the event or by using the basic relationships of probability. When additional information becomes available, we showed how conditional probability and Bayes’ theorem can be used to obtain revised or posterior probabilities. We also demonstrated how an apparent association between two events can be reversed upon consideration of a third event through the phenomenon of Simpson’s paradox.

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48 Chapter 2 Introduction to Probability

The probability concepts covered will be helpful in future chapters when we describe quantitative methods based on the use of probability information. Specific chapters and quantitative methods that make use of probability are as follows:

Chapter 3 Probability distributions Chapter 4 Decision analysis Chapter 5 Utility and game theory Chapter 13 Project scheduling: PERT/CPM Chapter 14 Inventory models Chapter 15 Waiting line models Chapter 16 Simulation Chapter 17 Markov processes

Glossary

Addition law A probability law used to compute the probability of a union: P(A ´ B) 5 P(A) 1 P(B) – P(A ¨ B). For mutually exclusive events, P(A ¨ B) 5 0, and the addition law simplifies to P(A ´ B) 5 P(A) 1 P(B). Basic requirements of probability Two requirements that restrict the manner in which probability assignments can be made:

1. For each experimental outcome Ei, 0 # P(Ei) # 1. 2. P(E1) 1 P(E2) 1 ? ? ? 1 P(Ek) 5 1.

Bayes’ theorem A method used to compute posterior probabilities. Certain event An event that is certain to occur. The probability of a certain event is 1. Classical method A method of assigning probabilities that is based on the assumption that the experimental outcomes are equally likely. Complement of event A The event containing all sample points that are not in A. Conditional probability The probability of an event given another event has occurred. The conditional probability of A given B is P(A Z B) 5 P(A ¨ B)/P(B). Dependent events Two events A and B for which P(A Z B) Z P(A) or P(B Z A) Z P(B); that is, the probability of one event is altered or affected by knowing whether the other event occurs. Event A collection of sample points or experimental outcomes. Experiment Any process that generates well-defined outcomes. Impossible event An event that is certain not to occur. The probability of an uncertain event is 0. Independent events Two events A and B for which P(A Z B) 5 P(A) and P(B Z A) 5 P(B); that is, the events have no influence on each other. Intersection of events A and B The event containing all sample points that are in both A and B. Joint probability The probability of the intersection of two events. Joint probability table A table used to display joint and marginal probabilities. Marginal probabilities The values in the margins of the joint probability table, which provide the probability of each event separately. Multiplication law A probability law used to compute the probability of an intersection: P(A ¨ B) 5 P(A Z B)P(B) or P(A ¨ B) 5 P(B Z A)P(A). For independent events, this simpli- fies to P(A ¨ B) 5 P(A)P(B). Mutually exclusive events Events that have no sample points in common; that is, A ¨ B is empty and P(A ¨ B) 5 0. Posterior probabilities Revised probabilities of events based on additional information.

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Problems 49

Number of Days Number Waiting Outcome Occurred

0 2 1 5 2 6 3 4 4 3

Total 20

Prior probabilities Initial probabilities of events. Probability A numeric measure of the likelihood that an event will occur. Relative frequency method A method of assigning probabilities based on experimenta- tion or historical data. Sample point An experimental outcome and an element of the sample space. Sample space The set of all sample points (experimental outcomes). Simpson’s paradox The phenomenon by which the apparent association between two events is reversed upon consideration of a third event. Subjective method A method of assigning probabilities based on judgment. Union of events A and B The event containing all sample points that are in A, in B, or in both. Venn diagram A graphical device for representing the sample space and operations in- volving events.

Problems

1. A study examined waiting times in the X-ray department for a hospital in Jacksonville, Florida. A clerk recorded the number of patients waiting for service at 9:00 a.m. on 20 consecutive days and obtained the following results.

testSELF

a. Define the experiment the clerk conducted. b. List the experimental outcomes. c. Assign probabilities to the experimental outcomes. d. What method did you use?

2. A company that franchises coffee houses conducted taste tests for a new coffee product. The company prepared four blends and randomly chose individuals to do a taste test and state which of the four blends they liked best. Results of the taste test for 100 individuals are given.

testSELF

Blend Taste-Testers’ Preference

1 20 2 30 3 35 4 15

a. Define the experiment being conducted. How many times was it repeated? b. Prior to conducting the experiment, it is reasonable to assume preferences for the four

blends are equal. What probabilities would you assign to the experimental outcomes prior to conducting the taste test? What method did you use?

c. After conducting the taste test, what probabilities would you assign to the experimen- tal outcomes? What method did you use?

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50 Chapter 2 Introduction to Probability

4. In a recent year the U.S. Internal Revenue Service (IRS) received 132,275,830 individual tax returns. The actual number of each type of individual return received by the IRS dur- ing the year is given below:

Number of Design Times Preferred

1 5 2 15 3 30 4 40 5 10

Type of Return Total Returns Filed

1040A, Income Under $25,000 31,675,935 Non 1040A, Income Under $25,000 20,295,694 Income $25,000–$50,000 30,828,932 Income $50,000–$100,000 26,463,973 Income $100,000 & Over 12,893,802 Schedule C, Receipts Under $25,000 3,376,943 Schedule C, Receipts $25,000–$100,000 3,867,743 Schedule C, Receipts $100,000 & Over 2,288,550 Schedule F, Receipts Under $100,000 318,646 Schedule F, Receipts $100,000 & Over 265,612

3. A company that manufactures toothpaste is studying five different package designs. As- suming that one design is just as likely to be selected by a consumer as any other design, what selection probability would you assign to each of the package designs? In an actual experiment, 100 consumers were asked to pick the design they preferred. The following data were obtained. Do the data confirm the belief that one design is just as likely to be selected as another? Explain.

Suppose an IRS auditor must randomly select and examine an individual return. a. What is the probability that the auditor will select an individual return from the 1040A,

Income Under $25,000 category? b. What is the probability that the selected return did not use Schedule C? c. What is the probability that the selected return reported income or reciepts of $100,000

& Over? d. In 2006 the IRS examined 1% of all individual returns. Assuming the examined re-

turns were evenly distributed across the ten categories in the above table, how many returns from the Non 1040A, Income $50,000–$100,000 category were examined?

e. When examining 2006 individual income tax returns, IRS auditors found that indi- vidual taxpayers still owed $13,045,221,000 in income taxes due to errors the in- dividual taxpayers had made on this year’s individual income tax returns (this is referred to by the IRS as recommended individual taxes). Use this information to estimate the recommended additional taxes for the Schedule C, Receipts $100,000 & Over category.

5. Strom Construction made a bid on two contracts. The owner identified the possible out- comes and subjectively assigned the following probabilities.

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Problems 51

Experimental Obtain Obtain Outcome Contract 1 Contract 2 Probability

1 Yes Yes 0.15 2 Yes No 0.15 3 No Yes 0.30 4 No No 0.25

Amount ($) Number

0–49 13 50–99 22 100–149 34 150–199 26 200–249 5

a. Are these valid probability assignments? Why or why not? b. If not, what would have to be done to make the probability assignments valid?

6. A sample of 100 customers of Montana Gas and Electric resulted in the following frequency distribution of monthly charges.testSELF

a. Let A be the event that monthly charges are $150 or more. Find P(A). b. Let B be the event that monthly charges are less than $150. Find P(B).

7. Suppose that a sample space has five equally likely experimental outcomes: E1, E2, E3, E4, E5. Let

A 5 {E1, E2}

B 5 {E3, E4}

C 5 {E2, E3, E5}

a. Find P(A), P(B), and P(C). b. Find P(A ´ B). Are A and B mutually exclusive? c. Find Ac, Cc, P(Ac), and P(Cc). d. Find A ´ Bc and P(A ´ Bc). e. Find P(B ´ C).

8. In a recent article U.S. News and World Report rated pediatric hospitals and provided data on several characteristics, including daily inpatient volume and nurse-to-patient ratio, for the top 30 hospitals on its list. Suppose we consider a daily inpatient volume of at least 200 to be high and a nurse-to-patient ratio of at least 3.0 to be high. Sixteen hospitals had a daily inpatient volume of at least 200, one-third of the hospitals had a nurse-to-patient ratio of at least 3.0, and seven of the hospitals had both a daily inpatient volume of at least 200 and a nurse-to-patient ratio of at least 3.0. a. Find the probability of a hospital having a daily inpatient volume of at least 200, the

probability of a hospital having a nurse-to-patient ratio of at least 3.0, and the prob- ability of a hospital having both a daily inpatient volume of at least 200 and a nurse- to-patient ratio of at least 3.0.

b. What is the probability that a hospital had a daily inpatient volume of at least 200 or a nurse-to-patient ratio of at least 3.0 or both?

c. What is the probability that a hospital had neither a daily inpatient volume of at least 200 nor a nurse-to-patient ratio of at least 3.0?

testSELF

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52 Chapter 2 Introduction to Probability

9. A pharmaceutical company conducted a study to evaluate the effect of an allergy relief medicine; 250 patients with symptoms that included itchy eyes and a skin rash received the new drug. The results of the study are as follows: 90 of the patients treated experienced eye relief, 135 had their skin rash clear up, and 45 experienced relief of both itchy eyes and the skin rash. What is the probability that a patient who takes the drug will experience relief of at least one of the two symptoms?

10. A quality control specialist has sampled 25 widgets from the production line. A widget can have minor or major defects. Of the 25 sampled widgets, 4 have minor defects and 2 have major defects. What is the probability that a widget has a major defect, given that it has a defect?

11. Let A be an event that a person’s primary method of transportation to and from work is an automobile and B be an event that a person’s primary method of transportation to and from work is a bus. Suppose that in a large city P(A) 5 0.45 and P(B) 5 0.35. a. Are events A and B mutually exclusive? What is the probability that a person uses an

automobile or a bus in going to and from work? b. Find the probability that a person’s primary method of transportation is some means

other than a bus.

12. For two events A and B, P(A) 5 0.5, P(B) 5 0.60, and P(A ¨ B) 5 0.40. a. Find P(A Z B). b. Find P(B Z A). c. Are A and B independent? Why or why not?

13. A survey of MBA students obtained the following data on “Students’ first reason for application to the school in which they matriculated.”

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Reason for Application

School Cost School or Quality Convenience Other Totals

Enrollment Full Time 421 393 76 890 Status Part Time 400 593 46 1039

Totals 821 986 122 1929

a. Develop a joint probability table using these data. b. Use the marginal probabilities of school quality, school cost or convenience, and other

to comment on the most important reason for choosing a school. c. If a student goes full time, what is the probability that school quality will be the first

reason for choosing a school? d. If a student goes part time, what is the probability that school quality will be the first

reason for choosing a school? e. Let A be the event that a student is full time and let B be the event that the student lists

school quality as the first reason for applying. Are events A and B independent? Justify your answer.

14. The checking accounts of Sun Bank are categorized by the age of account and the account balance. Auditors will select accounts at random from the following 1000 accounts (num- bers in the table are the number of accounts in each category):

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Problems 53

Account Balance

Age of the Account 0–$499 $500–$999 $1000 or More

Less than 2 years 120 240 90 2 years or more 75 275 200

Automobile Insurance

Yes No

Age 18 to 34 1500 340

35 and over 1900 260

a. What is the probability that an account is less than 2 years old? b. What is the probability that an account has a balance of $1000 or more? c. What is the probability that two accounts will both have a balance of $1000 or more? d. What is the probability that an account has a balance of $500–$999 given that its age

is 2 years or more? e. What is the probability that an account is less than 2 years old and has a balance of

$1000 or more? f. What is the probability that an account is at least 2 years old given that the balance is

$500–$999?

15. During a recent period of high unemployment, hundreds of thousands of drivers dropped their automobile insurance. Sample data representative of the national automobile insur- ance coverage for individuals 18 years of age and older are shown here.

a. Develop a joint probability table for these data and use the table to answer the remaining questions.

b. What do the marginal probabilities tell you about the age of the U.S. population? c. What is the probability that a randomly selected individual does not have automobile

insurance coverage? d. If the individual is between the ages of 18 and 34, what is the probability that the

individual does not have automobile insurance coverage? e. If the individual is age 35 or over, what is the probability that the individual does not

have automobile insurance coverage? f. If the individual does not have automobile insurance, what is the probability that the

individual is in the 18–34 age group? g. What does the probability information tell you about automobile insurance coverage

in the United States?

16. A purchasing agent placed a rush order for a particular raw material with two different suppliers, A and B. If neither order arrives in four days, the production process must be shut down until at least one of the orders arrives. The probability that supplier A can deliver the material in four days is 0.55. The probability that supplier B can deliver the material in four days is 0.35. a. What is the probability that both suppliers deliver the material in four days? Because

two separate suppliers are involved, assume independence. b. What is the probability that at least one supplier delivers the material in four days? c. What is the probability that the production process is shut down in four days because

of a shortage of raw material (that is, both orders are late)?

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54 Chapter 2 Introduction to Probability

17. Interested in learning more about its fans, the marketing office of the Arena Football League (AFL) conducted a survey at one of its games. The survey had 989 respondents: 759 males and 230 females. Out of the 989 total respondents, 196 stated that they had at- tended multiple AFL games. Of these 196 fans that had attended multiple games, 177 were male. Using this survey information, answer the following questions. a. What is the probability that a randomly selected fan has attended multiple games? b. Given that a randomly selected fan has attended multiple games, what is the prob-

ability of this person being male? c. What is the probability of a randomly selected fan being male and having attended

multiple games? d. Given that a randomly selected fan is male, what is the probability that this person has

attended multiple games? e. What is the probability that a randomly selected fan is male or has attended multiple

games?

18. In the evaluation of a sales training program, a firm discovered that of 50 salespeople receiving a bonus last year, 20 had attended a special sales training program. The firm employs 200 salespeople. Let B 5 the event that a salesperson makes a bonus and S 5 the event that a salesperson attends the sales training program. a. Find P(B), P(S Z B), and P(S ¨ B). b. Assume that 40% of the salespeople attended the training program. What is the prob-

ability that a salesperson makes a bonus given that the salesperson attended the sales training program, P(B Z S)?

c. If the firm evaluates the training program in terms of its effect on the probability of a salesperson’s receiving a bonus, what is your evaluation of the training program? Comment on whether B and S are dependent or independent events.

19. A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 6% of the employees had lost-time accidents last year. Management believes that a special safety program will reduce the accidents to 5% during the current year. In addition, it estimates that 15% of those employees having had lost-time accidents last year will have a lost-time accident during the cur- rent year. a. What percentage of the employees will have lost-time accidents in both years? b. What percentage of the employees will have at least one lost-time accident over the

two-year period?

20. The prior probabilities for events A1, A2, and A3 are P(A1) 5 0.20, P(A2) 5 0.50, and P(A3) 5 0.30. The conditional probabilities of event B given A1, A2, and A3 are P(B Z A1) 5 0.50, P(B | A2) 5 0.40, and P(B Z A3) 5 0.30. a. Compute P(B ¨ A1), P(B ¨ A2), and P(B ¨ A3). b. Apply Bayes’ theorem, equation (2.16), to compute the posterior probability P(A2 Z B). c. Use the tabular approach to applying Bayes’ theorem to compute P(A1 Z B), P(A2 | B),

and P(A3 Z B). 21. A consulting firm submitted a bid for a large research project. The firm’s management

initially felt there was a 50/50 chance of getting the bid. However, the agency to which the bid was submitted subsequently requested additional information on the bid. Experience indicates that on 75% of the successful bids and 40% of the unsuccessful bids the agency requested additional information. a. What is the prior probability that the bid will be successful (i.e., prior to receiving the

request for additional information)? b. What is the conditional probability of a request for additional information given that

the bid will ultimately be successful?

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Problems 55

c. Compute a posterior probability that the bid will be successful given that a request for additional information has been received.

22. Companies that do business over the Internet can often obtain probability informa- tion about website visitors from previous websites visited. For instance, Par Fore created a website to market golf equipment and apparel, and the organization has collected data from its website visitors. Management would like a certain offer to appear for female visitors and a different offer to appear for male visitors. A sample of past website visits indicates that 60% of the visitors to ParFore.com are male and 40% are female. a. What is your prior probability that the next visitor to the Par Fore website will be female? b. Suppose you know that the current visitor at the Par Fore website previously visited the

Dillard’s website and that women are three times as likely to visit this website as men. What is your revised probability that the visitor is female? Should you display the offer that has more appeal to female visitors or the one that has more appeal to male visitors?

23. An oil company purchased an option on land in Alaska. Preliminary geologic studies as- signed the following prior probabilities.

P(high quality oil) 5 0.50

P(medium quality oil) 5 0.20

P(no oil) 5 0.30

a. What is the probability of finding oil? b. After 200 feet of drilling on the first well, a soil test is made. The probabilities of find-

ing the particular type of soil identified by the test are

P(soil Z high quality oil) 5 0.20 P(soil Z medium quality oil) 5 0.20 P(soil Z no oil) 5 0.30

How should the firm interpret the soil test? What are the revised probabilities, and what is the new probability of finding oil?

24. During a recent year, speeding was reported in 12.9% of all automobile accidents in the United States. Assume the probability that speeding is reported in an accident is 0.129, the probability of an accident in which speeding is reported leading to a fatality is 0.196, and the probability of an accident in which speeding is not reported leading to a fatality is 0.05. Suppose you learn of an accident involving a fatality. What is the probability that speeding was reported?

25. The Wayne Manufacturing Company purchases a certain part from suppliers A, B, and C. Supplier A supplies 60% of the parts, B 30%, and C 10%. The quality of parts varies among the suppliers, with A, B, and C parts having 0.25%, 1%, and 2% defective rates, respectively. The parts are used in one of the company’s major products. a. What is the probability that the company’s major product is assembled with a defec-

tive part? Use the tabular approach to Bayes’ theorem to solve. b. When a defective part is found, which supplier is the likely source?

26. Bayes’ theorem and conditional probability can be used in medical diagnosis. Prior prob- abilities of diseases are based on the physician’s assessment of factors such as geographic location, seasonal influence, and occurrence of epidemics. Assume that a patient is believed to have one of two diseases, denoted D1 and D2, with P(D1) 5 0.60 and P(D2) 5 0.40,

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56 Chapter 2 Introduction to Probability

After finding that a certain symptom is present, the medical diagnosis may be aided by finding the revised probabilities that the patient has each particular disease. Compute the posterior probabilities of each disease for the following medical findings. a. The patient has symptom S1. b. The patient has symptom S2. c. The patient has symptom S3. d. For the patient with symptom S1 in part (a), suppose that symptom S2 also is present.

What are the revised probabilities of D1 and D2?

27. In an article about investment alternatives, Money magazine reported that drug stocks pro- vide a potential for long-term growth, with over 50% of the adult population of the United States taking prescription drugs on a regular basis. For adults age 65 and older, 82% take prescription drugs regularly. For adults age 18 to 64, 49% take prescription drugs regular- ly. The 18–64 age group accounts for 83.5% of the adult population (Statistical Abstract of the United States, 2008). a. What is the probability that a randomly selected adult is 65 or older? b. Given that an adult takes prescription drugs regularly, what is the probability that the

adult is age 65 or older?

28. According to the Open Doors 2011 Report, 9.5% of all full-time U.S. undergraduate stu- dents studied abroad during the 2009–2010 academic year (Institute of International Edu- cation, November 14, 2011). Assume that participation records show women make up 60% of the students who studied abroad during the 2009–2010 academic year, but women make up only 49% of the students who didn’t participate. a. Let A1 5 the student studied abroad during the 2009–2010 academic year A2 5 the student did not study abroad during the 2009–2010 academic year W 5 the student is a female student Using the given information, what are the values for P(A1), P(A2), P(W Z A1), and

P(W Z A2)? b. What is the probability that a female student studied abroad during the 2009–2010

academic year? c. What is the probability that a male student studied abroad during the 2009–2010 aca-

demic year? d. Given the preceding results, what were the percentage of women and the percentage

of men studying full-time during the 2009–2010 academic year?

29. Recently, management at Oak Tree Golf Course received a few complaints about the con- dition of the greens. Several players complained that the greens are too fast. Rather than react to the comments of just a few, the Golf Association conducted a survey of 100 male and 100 female golfers. The survey results are summarized here.

and that medical research shows a probability associated with each symptom that may accompany the diseases. Suppose that, given diseases D1 and D2, the probabilities that a patient will have symptoms S1, S2, or S3 are as follows:

Symptoms S1 S2 S3

Disease D1 0.15 0.10 0.15

D2 0.80 0.15 0.03

P(S3 | D1)

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Case Problem Hamilton County Judges 57

College of Engineering College of Business

Male Applicants Female Applicants Male Applicants Female Applicants

Accept 60 20 10 20 Deny 60 20 30 60

Male Golfers Female Golfers

Greens Condition Greens Condition

Handicap Too Fast Acceptable Handicap Too Fast Acceptable Under 15 10 40 Under 15 1 9 15 or more 25 25 15 or more 39 51

a. Combine these two crosstabulations into a single crosstabulation, with Male and Female as the row labels and Too Fast and Acceptable as the column labels. Which group shows the highest proportion saying that the greens are too fast?

b. Refer to the initial crosstabulations. For those players with low handicaps (better play- ers), which group (male or female) shows the highest proportion saying the greens are too fast?

c. Refer to the initial crosstabulations. For those players with higher handicaps, which group (male or female) shows the highest proportion saying the greens are too fast?

d. What conclusions can you draw about the preferences of men and women concerning the speed of the greens? Are the conclusions you draw from part (a) as compared with parts (b) and (c) consistent? Explain any apparent inconsistencies.

30. A small private midwestern university has been accused of favoring male applicants in its admissions process. Prior to the current academic year, 44% of male applicants were admitted by the university while only 33% of female applicants were admitted. The num- bers of applicants accepted and denied for each gender are provided in the following table for the College of Engineering and the College of Business.

a. Combine these two crosstabulations into a single crosstabulation, with Accept and Deny as the row labels and Male and Female as the column labels. Use these data to confirm the reported rates of acceptance across the university by gender (44% of male applicants were admitted and 33% of female applicants were admitted).

b. Refer to the initial crosstabulations. What are your conclusions about the possible gender bias in the admission process? Is this consistent with the conclusions you reach when assessing the aggregated data in part (a) of this question?

Case Problem Hamilton County Judges Hamilton County judges try thousands of cases per year. In an overwhelming majority of the cases disposed, the verdict stands as rendered. However, some cases are appealed, and of those appealed, some of the cases are reversed. Kristen DelGuzzi of the Cincinnati Enquirer conducted a study of cases handled by Hamilton County judges over a three-year period. Shown in Table 2.7 are the results for 182,908 cases handled (disposed) by 38 judges in Common Pleas Court, Domestic Relations Court, and Municipal Court. Two of the judges (Dinkelacker and Hogan) did not serve in the same court for the entire three-year period.

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58 Chapter 2 Introduction to Probability

Common Pleas Court

Total Cases Appealed Reversed Judge Disposed Cases Cases Fred Cartolano 3037 137 12 Thomas Crush 3372 119 10 Patrick Dinkelacker 1258 44 8 Timothy Hogan 1954 60 7 Robert Kraft 3138 127 7 William Mathews 2264 91 18 William Morrissey 3032 121 22 Norbert Nadel 2959 131 20 Arthur Ney, Jr. 3219 125 14 Richard Niehaus 3353 137 16 Thomas Nurre 3000 121 6 John O’Connor 2969 129 12 Robert Ruehlman 3205 145 18 J. Howard Sundermann 955 60 10 Ann Marie Tracey 3141 127 13 Ralph Winkler 3089 88 6

Total 43,945 1762 199

Domestic Relations Court

Total Cases Appealed Reversed Judge Disposed Cases Cases Penelope Cunningham 2729 7 1 Patrick Dinkelacker 6001 19 4 Deborah Gaines 8799 48 9 Ronald Panioto 12,970 32 3

Total 30,499 106 17

Municipal Court

Total Cases Appealed Reversed Judge Disposed Cases Cases Mike Allen 6149 43 4 Nadine Allen 7812 34 6 Timothy Black 7954 41 6 David Davis 7736 43 5 Leslie Isaiah Gaines 5282 35 13 Karla Grady 5253 6 0 Deidra Hair 2532 5 0 Dennis Helmick 7900 29 5 Timothy Hogan 2308 13 2 James Patrick Kenney 2798 6 1 Joseph Luebbers 4698 25 8 William Mallory 8277 38 9 Melba Marsh 8219 34 7 Beth Mattingly 2971 13 1 Albert Mestemaker 4975 28 9 Mark Painter 2239 7 3 Jack Rosen 7790 41 13 Mark Schweikert 5403 33 6 David Stockdale 5371 22 4 John A. West 2797 4 2

Total 108,464 500 104

TABLE 2.7 CASES DISPOSED, APPEALED, AND REVERSED IN HAMILTON COUNTY COURTS

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Case Problem College Softball Recruiting 59

The purpose of the newspaper’s study was to evaluate the performance of the judges. Appeals are often the result of mistakes made by judges, and the newspaper wanted to know which judges were doing a good job and which were making too many mistakes. You have been called in to assist in the data analysis. Use your knowledge of probabil- ity and conditional probability to help with the ranking of the judges. You also may be able to analyze the likelihood of cases handled by the different courts being appealed and reversed.

Managerial Report Prepare a report with your rankings of the judges. Also include an analysis of the likelihood of appeal and case reversal in the three courts. At a minimum, your report should include the following:

1. The probability of cases being appealed and reversed in the three different courts. 2. The probability of a case being appealed for each judge. 3. The probability of a case being reversed for each judge. 4. The probability of reversal given an appeal for each judge. 5. Rank the judges within each court. State the criteria you used and provide a rationale

for your choice.

Case Problem College Softball Recruiting College softball programs have a limited number of scholarships to offer promising high school seniors, so the programs invest a great deal of effort in evaluating these players. One measure of performance the programs commonly use to evaluate recruits is the batting average—the proportion of at-bats (excluding times when the player is walked or hit by a pitch) in which the player gets a hit. For example, a player who gets 50 hits in 150 at-bats has a batting average of

150 5 0.333

A college softball program is considering two players, Fran Hayes and Millie Marshall, who have recently completed their senior years of high school. Their respective statistics for their junior and senior years are as shown in Table 2.8.

Junior Year Senior Year

At-Bats Hits At-Bats Hits

Fran Hayes 200 70 40 15

Millie Marshall 196 67 205 76

TABLE 2.8 SUMMARY OF BATTING PERFORMANCES IN JUNIOR AND SENIOR YEARS BY HAYES AND MARSHALL

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60 Chapter 2 Introduction to Probability

Managerial Report The Athletic Director and Coach of the women’s softball team at a large public university are trying to decide to which of these two players they will offer an athletic scholarship (i.e., an opportunity to attend the university for free in exchange for playing on the university’s softball team). Take the following steps to determine which player had the better batting average over the two-year period provided in the table, and use your results to advise the Athletic Director and Coach on their decision.

1. Calculate the batting average of each player for her junior year; then also calculate the batting average of each player for her senior year. Which player would this analysis lead you to choose?

2. Calculate the batting average of each player for her combined junior and senior years. Which player would this analysis lead you to choose?

3. After considering both of your analyses, which player would you choose? Why? 4. Prepare a report on your findings for the atheletic director and coach of the college

program. Focus on clearly explaining the discrepancy in your two analyses.

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Probability Distributions

CONTENTS

3.1 RANDOM VARIABLES

3.2 DISCRETE RANDOM VARIABLES Probability Distribution of a

Discrete Random Variable Expected Value Variance

3.3 BINOMIAL PROBABILITY DISTRIBUTION Nastke Clothing Store Problem Expected Value and Variance for

the Binomial Distribution

3.4 POISSON PROBABILITY DISTRIBUTION An Example Involving Time

Intervals An Example Involving Length or

Distance Intervals

3.5 CONTINUOUS RANDOM VARIABLES Applying the Uniform

Distribution Area as a Measure of Probability

3.6 NORMAL PROBABILITY DISTRIBUTION Standard Normal Distribution Computing Probabilities for Any

Normal Distribution Grear Tire Company Problem

3.7 EXPONENTIAL PROBABILITY DISTRIBUTION Computing Probabilities for the

Exponential Distribution Relationship Between the

Poisson and Exponential Distributions

AppENdix 3.1 COMPUTING DISCRETE PROBABILITIES wITh EXCEL

AppENdix 3.2 COMPUTING PROBABILITIES fOR CONTINUOUS DISTRIBUTIONS wITh EXCEL

CHAPTER 3

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62 Chapter 3 Probability Distributions

In this chapter we continue the study of probability by introducing the concepts of random variables and probability distributions. we consider the probability distributions of both discrete and continuous random variables. Of particular interest are the binomial, Poisson, uniform, normal, and exponential probability distributions. These probability distributions are of great interest because they are used extensively in practice. The Q.M. in Action, ESPN and Probability, describes how the development and use of a probability distribution helped the organization enhance its audience’s enjoyment and understanding of sports.

The Entertainment and Sports Programming Network, known as ESPN since 1985, was originally established as a nationwide cable-television network dedicated to broadcasting and producing sports-related program- ming. The Bristol, Connecticut–based network, which provides programming 24 hours a day throughout the year, has grown rapidly since its debut on September 7, 1979. At various times its programming has included Major League Baseball, the National football League (NfL), the National Basketball Association (NBA), the National hockey League (NhL), NASCAR, NCAA football and Basketball, Major League Soccer (MLS), the Men’s and women’s professional golf associations (PGA and LPGA), and Men’s and women’s profes- sional tennis (ATP and wTA). ESPN is now in over 100 million homes in the United States, and ESPN In- ternational spans over 200 countries and territories on all seven continents and includes 46 television networks reaching over 350 million subscribers in 16 languages, plus wireless, interactive, print, radio, broadband, event management, and consumer products.

ESPN’s rapid growth has coincided with a dramatic increase in sports fans’ desire for more sophisticated analyses. The organization responded to this trend by es- tablishing its Production Analytics department, a group of analysts who provide all of ESPN’s media platforms with statistical analyses for a wide variety of sports prob- lems. Senior Director of the Production Analytics de- partment Jeff Bennett explains that “ESPN appreciates the sports fan’s passion for meaningful analytics, and we are dedicated to creating and providing this content.

Basic probability and statistics are critical tools in our analytic arsenal.”

Alok Pattani, who is an Analytics Specialist in the Production Analytics department, further describes the department’s uses of some specific probability concepts. “for example, we use very basic probability to determine the likelihood an NBA team will win one of the first three picks in the NBA draft. The league holds a Draft Lottery to determine which of its fourteen teams that didn’t qualify for the playoffs during the most recent season will receive each of the first three picks, which are considered to be extremely valuable. The lottery is weighted so that teams with worse records have better chances of obtaining early picks, and we use information on how many chances each team is allocated to calculate the probability that any of these fourteen teams will win one of those top picks.

“Conditional probability is also very important; when we look at the probability a team will win a home game or a player will get a hit when playing at night, we are using conditional probability.” Alok continues, “we use proba- bility distributions extensively in our work, especially the binomial and normal distributions. we apply the binomial distribution to all kinds of success and failure situations such as wins and losses, field goals and missed shots in basketball, complete and incomplete passes in football, and hits and outs in baseball. The binomial distribution is also useful when estimating the probability of a hitting streak in baseball or a winning streak in any sport.”

Bennett adds, “The results of these types of analyses are of great interest to ESPN’s base. They enhance the fan’s enjoyment and understanding of his or her favorite sports, and that is good business for ESPN.”

ESPN AND PROBABILITY*

Q.M. in ACtion

*the authors are indebted to Jeff Bennett and Alok Pattani of ESPn inc. for providing input for this Q.M. in Action.

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3.1 Random Variables 63

3.1 Random Variables Recall that in Chapter 2 we defined an experiment as any process that generates well- defined outcomes. we now want to concentrate on the process of assigning numeric values to experimental outcomes. To do so we introduce the notion of a random variable.

for any particular experiment a random variable can be defined so that each possible experimental outcome generates exactly one numeric value for the random variable. for example, if we consider the experiment of selling automobiles for one day at a particular dealership, we could describe the experimental outcomes in terms of the number of cars sold. In this case, if x 5 number of cars sold, x is called a random variable. The particular numeric value that the random variable assumes depends on the outcome of the experiment; that is, we will not know the specific value of the random variable until we have observed the experi- mental outcome. for example, if on a given day three cars are sold, the value of the random variable is 3; if on another day (a repeat of the experiment) four cars are sold, the value is 4. we define a random variable as follows:

A random variable is a numeric description of the outcome of an experiment.

Some additional examples of experiments and their associated random variables are given in Table 3.1. Although many experiments have experimental outcomes that lend themselves quite naturally to numeric values, others do not. for example, for the experi- ment of tossing a coin one time, the experimental outcome will be either a head or a tail, neither of which has a natural numeric value. however, we still may want to express the outcomes in terms of a random variable. Thus, we need a rule that can be used to assign a numeric value to each of the experimental outcomes. One possibility is to let the random variable x 5 1 if the experimental outcome is a head and x 5 0 if the experimental outcome is a tail. Although the numeric values for x are arbitrary, x is a random variable because it describes the experimental outcomes numerically.

A random variable may be classified as either discrete or continuous, depending on the numeric values it may assume. A random variable that may assume only a finite or an infinite sequence (e.g., 1, 2, 3, . . .) of values is a discrete random variable. The number of units sold, the number of defects observed, the number of customers that enter a bank during one day of operation, and so on are examples of discrete random variables. The first two and the last random variables in Table 3.1 are discrete. Random variables such as weight, time, and temperature that may assume any value in a certain interval or collection of intervals are continuous random variables. for instance, the third random variable in Table 3.1 is a continuous random variable because it may assume any value in the interval from 0 to 100 (for example, 56.33 or 64.223).

Random variables must assume numeric values.

possible Values for Experiment Random Variable (x) the Random Variable Make 100 sales calls Total number of sales 0, 1, 2, . . . , 100 Inspect a shipment of 70 radios Number of defective radios 0, 1, 2, . . . , 70 Build a new library Percentage of project completed 0 # x # 100 after 6 months Operate a restaurant Number of customers entering 0, 1, 2, . . . in one day

TABLE 3.1 EXAMPLES Of RANDOM VARIABLES

Try Problem 1 for practice in identifying discrete and continuous random variables.

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64 Chapter 3 Probability Distributions

3.2 Discrete Random Variables we can demonstrate the use of a discrete random variable by considering the sales of automo- biles at DiCarlo Motors, Inc., in Saratoga, New York. The owner of DiCarlo Motors is inter- ested in the daily sales volume for automobiles. Suppose that we let x be a random variable denoting the number of cars sold on a given day. Sales records show that 5 is the maximum number of cars that DiCarlo has ever sold during one day. The owner believes that the previous history of sales adequately represents what will occur in the future, so we would expect the random variable x to assume one of the numeric values 0, 1, 2, 3, 4, or 5. The possible values of the random variable are finite; thus we would classify x as a discrete random variable.

Probability Distribution of a Discrete Random Variable Suppose that in checking DiCarlo’s sales records we find that over the past year the firm was open for business 300 days. The sales volumes generated and the frequency of their occurrence are summarized in Table 3.2. with these historical data available, the owner of

NOTES AND COMMENTS

1. One way to determine whether a random variable is discrete or continuous is to think of the values of the random variable as points on a line. Choose two points representing values the random vari- able might assume. If the entire line segment between the two points also represents possible values for the random variable, the random vari- able is continuous. An alternative (but equiva- lent) way of determining whether a random variable is discrete or continuous is to choose two points that represent values the random variable might assume. If, no matter what two points you initially chose, you can always find a third point between your initial two points that also repre- sents a value of the random variable, the random variable is continuous. for example, if your random variable is the exact weight of a bag of

potato chips and you choose 16.0005 ounces and 16.0006 ounces as your original two points, the point 16.00051 (or 16.00052 or 16.00053, etc.) ounces represents a possible value of the exact weight of the bag of potato chips and lies between your intial two points. No matter what two values you initially select, you can find another value that represents a value of the random variable and lies between your initial two points. On the other hand, if your random variable is the number of customers who enter a restaurant in a day and you choose 109 customers and 110 customers as your initial two points, there is no point that represents a value of the random variable and lies between your initial two points (i.e., you cannot have 109.7 customers enter a restaurant on a particular day). This indicates that the random variable is discrete.

Sales Volume Number of days No sales 54 One car 117 Two cars 72 Three cars 42 four cars 12 five cars 3

Total 300

TABLE 3.2 CARS SOLD PER DAY AT DICARLO MOTORS

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3.2 Discrete Random Variables 65

DiCarlo Motors believes that the relative frequency method will provide a reasonable means of assessing the probabilities for the random variable x. The probability function, denoted f(x), provides the probability that the random variable x takes on a specific value. Because on 54 of the 300 days of historical data DiCarlo Motors did not sell any cars and because no sales corresponds to x 5 0, we assign to f(0) the value 54/300 5 0.18. Similarly, f(1) denotes the probability that x takes on the value 1, so we assign to f(1) the value 117/300 5 0.39. After computing the relative frequencies for the other possible values of x, we can develop a table of x and f (x) values. Table 3.3 shows a tabular presentation of the probability distribution of the random variable x.

we can also represent the probability distribution of x graphically. In figure 3.1 the val- ues of the random variable x are shown on the horizontal axis. The probability that x takes on each of these values is shown on the vertical axis. for many discrete random variables the probability distribution also can be represented as a formula that provides f (x) for every possible value of x. we illustrate this approach in the next section.

x f (x) 0 0.18 1 0.39 2 0.24 3 0.14 4 0.04 5 0.01

Total 1.00

TABLE 3.3 PROBABILITY DISTRIBUTION fOR ThE NUMBER Of CARS SOLD PER DAY

.40

0 1 2 3 4 5 x

P ro

b ab

il it

Number of Cars Sold per Day

.30

.20

.10

FIGURE 3.1 PROBABILITY DISTRIBUTION fOR ThE NUMBER Of CARS SOLD PER DAY

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66 Chapter 3 Probability Distributions

In the development of a discrete probability distribution, two requirements must always be satisfied:

f (x) $ 0 (3.1)

of (x) 5 1 (3.2)

Equation (3.1) specifies that the probabilities associated with each value of x must be greater than or equal to zero, whereas equation (3.2) indicates that the sum of the probabili- ties for all values of the random variable x must be equal to 1. Table 3.3 shows that equa- tions (3.1) and (3.2) are satisfied. Thus, the probability distribution developed for DiCarlo Motors is a valid discrete probability distribution.

After establishing a random variable and its probability distribution, we can develop a variety of additional probability information, depending on the needs and interests of the decision maker. for example, in the DiCarlo Motors problem the probability distribution shown in Table 3.3 can be used to provide the following information:

1. There is a 0.18 probability that no cars will be sold during a day. 2. The most probable sales volume is 1, with f(1) 5 0.39. 3. There is a 0.05 probability of an outstanding sales day with four or five cars being

sold.

Using probability information such as that just given, DiCarlo’s management can under- stand better the uncertainties associated with the car sales operation. Perhaps this improved understanding can serve as the basis for a new policy or decision that will increase the ef- fectiveness of the firm.

Expected Value After constructing the probability distribution for a random variable, we often want to com- pute the mean or expected value of the random variable. The expected value of a discrete random variable is a weighted average of all possible values of the random variable, where the weights are the probabilities associated with the values. The mathematical formula for computing the expected value of a discrete random variable x is

E(x) 5 m 5 ox f (x) (3.3)

As equation (3.3) shows, both the notations E(x) and µ are used to refer to the expected value of a random variable.

To compute the expected value of a discrete random variable, we must multiply each value of the random variable by its corresponding probability and then add the resulting terms. Calculation of the expected value of the random variable (number of daily sales) for DiCarlo Motors is shown in Table 3.4. The first column contains the values of the random variable x, and the second column contains their associated probabilities f (x). Multiplying each value of x by its probability f(x) provides the xf (x) values in the third column. fol- lowing equation (3.3), we sum this column, oxf (x), to find the expected value of 1.50 cars sold per day.

The expected value of a random variable is the mean, or average, value. for experiments that can be repeated numerous times, the expected value can be interpreted as the “long-run” average value for the random variable. however, the expected value is not necessarily the

In Section 2.2 we defined the two basic requirements of all probability assignments as 0 # P(Ei) # 1 and oP(Ei) 5 1. Equations (3.1) and (3.2) are the analogs of these basic requirements.

Try Problem 3 for practice in constructing a discrete probability distribution.

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3.2 Discrete Random Variables 67

number that we think the random variable will assume the next time the experiment is con- ducted. In fact, it is impossible for DiCarlo to sell exactly 1.50 cars on any day. however, if we envision selling cars at DiCarlo Motors for many days into the future, the expected value of 1.50 cars provides the mean, or average, daily sales volume.

The expected value can be important to a manager from both the planning and decision- making points of view. for example, suppose that DiCarlo Motors will be open 60 days during the next three months. how many cars will be sold during this time? Although we can’t specify the exact sales for any given day, the expected value of 1.50 cars per day provides an expected or average sales estimate of 60(1.50) 5 90 cars for the next three- month period. In terms of setting sales quotas and/or planning orders, the expected value may provide helpful decision-making information.

Variance The expected value gives us an idea of the average or central value for the random variable, but we often want a measure of the dispersion, or variability, of the possible values of the random variable. for example, if the values of the random variable range from quite large to quite small, we would expect a large value for the measure of variability. If the values of the random variable show only modest variation, we would expect a relatively small value for the measure of variability. The variance is a measure commonly used to summa- rize the variability in the values of a random variable. The mathematical expression for the variance of a discrete random variable is

Var(x) 5 s2 5 o(x 2 m)2f (x) (3.4)

As equation (3.4) shows, an essential part of the variance formula is a deviation, x 2 m, which measures how far a particular value of the random variable is from the expected value or mean, m. In computing the variance of a discrete random variable, we square the devia- tions and then weight them by the corresponding probability. The sum of these weighted squared deviations for all values of the random variable is the variance. In other words, the variance is a weighted average of the squared deviations.

The calculation of the variance for the number of daily sales in the DiCarlo Motors problem is summarized in Table 3.5. we see that the variance for the number of cars sold per day is 1.25. A related measure of variability is the standard deviation, s, which is

An alternative formula for the variance of a discrete random variable is Var(x) 5 ox2f (x) 2 µ2.

x f (x) xf (x) 0 0.18 0(0.18) 5 0.00 1 0.39 1(0.39) 5 0.39 2 0.24 2(0.24) 5 0.48 3 0.14 3(0.14) 5 0.42 4 0.04 4(0.04) 5 0.16 5 0.01 5(0.01) 5 0.05

E(x) 5 1.50

TABLE 3.4 EXPECTED VALUE CALCULATION

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68 Chapter 3 Probability Distributions

defined as the positive square root of the variance. for DiCarlo Motors, the standard devia- tion of the number of cars sold per day is

s 5 Ï 1.25 5 1.118

for the purpose of easier managerial interpretation, the standard deviation may be preferred over the variance because it is measured in the same units as the random variable (s 5 1.118 cars sold per day). The variance (s2) is measured in squared units and is thus more difficult for a manager to interpret.

At this point our interpretation of the variance and the standard deviation is limited to com- parisons of the variability of different random variables. for example, if the daily sales data from a second DiCarlo dealership in Albany, New York, provided s2 5 2.56 and s 5 1.6, we can conclude that the number of cars sold per day at this dealership exhibits more variability than at the first DiCarlo dealership, where s2 5 1.25 and s 5 1.118. Later in this chapter we discuss the normal distribution. for that probability distribution, we show that the variance and the standard deviation of the random variable are essential for making probability calculations.

The Q.M. in Action, Long Island City, New York, describes how Citbank models the length of waiting lines at its ATMs as discrete random variables in order to provide better service to its customers.

Try Problem 4 to be sure you can compute the expected value, variance, and standard deviation.

x x 2 m (x 2 m)2 f (x) (x 2 m)2f (x) 0 0 2 1.50 5 21.50 2.25 0.18 2.25(0.18) 5 0.4050 1 1 2 1.50 5 20.50 0.25 0.39 0.25(0.39) 5 0.0975 2 2 2 1.50 5 0.50 0.25 0.24 0.25(0.24) 5 0.0600 3 3 2 1.50 5 1.50 2.25 0.14 2.25(0.14) 5 0.3150 4 4 2 1.50 5 2.50 6.25 0.04 6.25(0.04) 5 0.2500 5 5 2 1.50 5 3.50 12.25 0.01 12.25(0.01) 5 0.1225

s2 5 1.2500

TABLE 3.5 VARIANCE CALCULATION

Citibank, the retail banking division of Citigroup, offers a wide range of financial services, including check- ing and saving accounts, loans and mortgages, insur- ance, and investment services. It delivers these services through a unique system referred to as Citibanking.

Citibank was one of the first banks in the United States to introduce automatic teller machines (ATMs). Citibank’s ATMs, located in Citicard Banking Centers

(CBCs), let customers do all of their banking in one place with the touch of a finger, 24 hours a day, 7 days a week. More than 150 different banking functions—from deposits to managing investments—can be performed with ease. Citibank customers use ATMs for 80% of their transactions.

Each Citibank CBC operates as a waiting line sys- tem with randomly arriving customers seeking service

CITIBANK* LONg ISLAND CITY, NEw YORK

Q.M. in ACtion

*the authors are indebted to Ms. Stacey Karter, Citibank, for providing this Q.M. in Action.

(continued)

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3.3 Binomial Probability Distribution 69

3.3 Binomial Probability Distribution In this section we consider a class of experiments that meet the following conditions:

1. The experiment consists of a sequence of n identical trials. 2. Two outcomes are possible on each trial. we refer to one outcome as a success and

the other as a failure. 3. The probabilities of the two outcomes do not change from one trial to the next. 4. The trials are independent (i.e., the outcome of one trial does not affect the outcome

of any other trial).

Experiments that satisfy conditions 2, 3, and 4 are said to be generated by a Bernoulli process. In addition, if condition 1 is satisfied (there are n identical trials), we have a bi- nomial experiment. An important discrete random variable associated with the binomial experiment is the number of outcomes labeled success in the n trials. If we let x denote the value of this random variable, then x can have a value of 0, 1, 2, 3, . . . , n, depending on the number of successes observed in the n trials. The probability distribution associated with this random variable is called the binomial probability distribution.

In cases where the binomial distribution is applicable, the mathematical formula for computing the probability of any value for the random variable is the binomial probability function

f sxd 5 n!

x!sn 2 xd! p xs1 2 pdn2x x 5 0, 1, . . . , n (3.5)

where

n 5 number of trials

p 5 probability of success on one trial

x 5 number of successes in n trials

f(x) 5 probability of x successes in n trials

Try Problem 9, parts (a–d), for practice computing binomial probabilities.

at one of the ATMs. If all ATMs are busy, the arriving customers wait in line. Periodic CBC capacity studies are used to analyze customer waiting times and to determine whether additional ATMs are needed.

Data collected by Citibank showed that the random customer arrivals followed a probability distribution known as the Poisson distribution. Using the Poisson distribution, Citibank can compute probabilities for the number of customers arriving at a CBC during any time period and make decisions concerning the number of ATMs needed. for example, let x 5 the number of cus- tomers arriving during a one-minute period. Assuming that a particular CBC has a mean arrival rate of two cus- tomers per minute, the following table shows the prob-

abilities for the number of customers arriving during a one-minute period.

x probability

0 .1353 1 .2707 2 .2707 3 .1804 4 .0902 5 or more .0527

Discrete probability distributions, such as the one used by Citibank, are the topic of this chapter. In addi- tion to the Poisson distribution, you will learn about the binomial and hypergeometric distributions and how they can be used to provide helpful probability information.

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70 Chapter 3 Probability Distributions

The term n! in the preceding expression is referred to as n factorial and is defined as

n! 5 nsn 2 1dsn 2 2d . . . s2ds1d

for example, 4! 5 (4)(3)(2)(1) 5 24. Also, by definition, the special case of zero factorial is 0! 5 1.

Nastke Clothing Store Problem To illustrate the binomial probability distribution, let us consider the experiment of custom- ers entering the Nastke Clothing Store. To keep the problem relatively small, we restrict the experiment to the next three customers. If, based on experience, the store manager estimates that the probability of a customer making a purchase is 0.30, what is the probability that exactly two of the next three customers make a purchase?

we first want to demonstrate that three customers entering the clothing store and decid- ing whether to make a purchase can be viewed as a binomial experiment. Checking the four requirements for a binomial experiment, we note the following:

1. The experiment can be described as a sequence of three identical trials, one trial for each of the three customers who will enter the store.

2. Two outcomes—the customer makes a purchase (success) or the customer does not make a purchase (failure)—are possible for each trial.

3. The probabilities of the purchase (0.30) and no purchase (0.70) outcomes are as- sumed to be the same for all customers.

4. The purchase decision of each customer is independent of the purchase decision of the other customers.

Thus, if we define the random variable x as the number of customers making a purchase (i.e., the number of successes in the three trials), we satisfy the requirements of the binomial probability distribution.

with n 5 3 trials and the probability of a purchase p 5 0.30 for each customer, we use equation (3.5) to compute the probability of two customers making a purchase. This prob- ability, denoted f (2), is

f s2d 5 3!

2!1! s0.30d2s0.70d1

5 3 3 2 3 1

2 3 1 3 1 s0.30d2s0.70d1 5 0.189

Similarly, the probability of no customers making a purchase, denoted f (0), is

f s0d 5 3!

0!3! s0.30d0s0.70d3

5 3 3 2 3 1

1 3 3 3 2 3 1 s0.30d0s0.70d3 5 0.343

Try Problem 12 for an application of the binomial distribution.

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3.3 Binomial Probability Distribution 71

Equation (3.5) can be used in a similar manner to show that the probabilities of one and three purchases are f (1) 5 0.441 and f (3) 5 0.027. Table 3.6 and figure 3.2 summarize the binomial probability distribution for the Nastke Clothing Store problem.

If we consider any variation of the Nastke problem, such as 10 customers rather than 3 customers entering the store, the binomial probability function given by equation (3.5) still applies. for example, the probability that 4 of the 10 customers make a purchase is

f s4d 5 10!

4!6! s0.30d4s0.70d6 5 0.2001

In this binomial experiment, n 5 10, x 5 4, and p 5 0.30.

x f (x) 0 0.343 1 0.441 2 0.189 3 0.027

Total 1.000

TABLE 3.6 PROBABILITY DISTRIBUTION fOR ThE NUMBER Of CUSTOMERS MAKING A PURChASE

0 1 2 3 x

P ro

b ab

il it

Number of Customers Making a Purchase

f(x)

.40

.30

.20

.10

.00

FIGURE 3.2 PROBABILITY DISTRIBUTION fOR ThE NASTKE CLOThING STORE PROBLEM

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72 Chapter 3 Probability Distributions

with the use of equation (3.5), tables have been developed that provide the probability of x successes in n trials for a binomial experiment. Such a table of binomial probability values is provided in Appendix B. we include a partial binomial table in Table 3.7. In order to use this table, specify the values of n, p, and x for the binomial experiment of interest. Check the use of this table by employing it to verify the probability of four successes in 10 trials for the Nastke Clothing Store problem. Note that the value of f (4) 5 0.2001 can be read directly from the table of binomial probabilities, making it unnecessary to perform the calculations required by equation (3.5).

with modern computers and calculators, these tables are almost unnecessary. It is easy to evaluate equation (3.5) directly.

n x p

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 9 0 0.6302 0.3874 0.2316 0.1342 0.0751 0.0404 0.0207 0.0101 0.0046 0.0020 1 0.2985 0.3874 0.3679 0.3020 0.2253 0.1556 0.1004 0.0605 0.0339 0.0176 2 0.0629 0.1722 0.2597 0.3020 0.3003 0.2668 0.2162 0.1612 0.1110 0.0703 3 0.0077 0.0446 0.1069 0.1762 0.2336 0.2668 0.2716 0.2508 0.2119 0.1641 4 0.0006 0.0074 0.0283 0.0661 0.1168 0.1715 0.2194 0.2508 0.2600 0.2461 5 0.0000 0.0008 0.0050 0.0165 0.0389 0.0735 0.1181 0.1672 0.2128 0.2461 6 0.0000 0.0001 0.0006 0.0028 0.0087 0.0210 0.0424 0.0743 0.1160 0.1641 7 0.0000 0.0000 0.0000 0.0003 0.0012 0.0039 0.0098 0.0212 0.0407 0.0703 8 0.0000 0.0000 0.0000 0.0000 0.0001 0.0004 0.0013 0.0035 0.0083 0.0176 9 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0003 0.0008 0.0020

10 0 0.5987 0.3487 0.1969 0.1074 0.0563 0.0282 0.0135 0.0060 0.0025 0.0010 1 0.3151 0.3874 0.3474 0.2684 0.1877 0.1211 0.0725 0.0403 0.0207 0.0098 2 0.0746 0.1937 0.2759 0.3020 0.2816 0.2335 0.1757 0.1209 0.0763 0.0439 3 0.0105 0.0574 0.1298 0.2013 0.2503 0.2668 0.2522 0.2150 0.1665 0.1172 4 0.0010 0.0112 0.0401 0.0881 0.1460 0.2001 0.2377 0.2508 0.2384 0.2051 5 0.0001 0.0015 0.0085 0.0264 0.0584 0.1029 0.1536 0.2007 0.2340 0.2461 6 0.0000 0.0001 0.0012 0.0055 0.0162 0.0368 0.0689 0.1115 0.1596 0.2051 7 0.0000 0.0000 0.0001 0.0008 0.0031 0.0090 0.0212 0.0425 0.0746 0.1172 8 0.0000 0.0000 0.0000 0.0001 0.0004 0.0014 0.0043 0.0106 0.0229 0.0439 9 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0005 0.0016 0.0042 0.0098 10 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0003 0.0010

11 0 0.5688 0.3138 0.1673 0.0859 0.0422 0.0198 0.0088 0.0036 0.0014 0.0005 1 0.3293 0.3835 0.3248 0.2362 0.1549 0.0932 0.0518 0.0266 0.0125 0.0054 2 0.0867 0.2131 0.2866 0.2953 0.2581 0.1998 0.1395 0.0887 0.0531 0.0269 3 0.0137 0.0710 0.1517 0.2215 0.2581 0.2568 0.2254 0.1774 0.1259 0.0806 4 0.0014 0.0158 0.0536 0.1107 0.1721 0.2201 0.2428 0.2365 0.2060 0.1611 5 0.0001 0.0025 0.0132 0.0388 0.0803 0.1321 0.1830 0.2207 0.2360 0.2256 6 0.0000 0.0003 0.0023 0.0097 0.0268 0.0566 0.0985 0.1471 0.1931 0.2256 7 0.0000 0.0000 0.0003 0.0017 0.0064 0.0173 0.0379 0.0701 0.1128 0.1611 8 0.0000 0.0000 0.0000 0.0002 0.0011 0.0037 0.0102 0.0234 0.0462 0.0806 9 0.0000 0.0000 0.0000 0.0000 0.0001 0.0005 0.0018 0.0052 0.0126 0.0269 10 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0007 0.0021 0.0054 11 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0005

TABLE 3.7 SELECTED VALUES fROM ThE BINOMIAL PROBABILITY TABLE. EXAMPLE: n 5 10, x 5 4, p 5 0.30; f (4) 5 0.2001

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3.4 Poisson Probability Distribution 73

Expected Value and Variance for the Binomial Distribution from the probability distribution in Table 3.6, we can use equation (3.3) to compute the expected value or expected number of customers making a purchase:

m 5 o xf (x) 5 0(0.343) 1 1(0.441) 1 2(0.189) 1 3(0.027) 5 0.9

Note that we could have obtained this same expected value simply by multiplying n (the number of trials) by p (the probability of success on any one trial):

np 5 3(0.30) 5 0.9

for the special case of a binomial probability distribution, the expected value of the random variable is given by

m 5 np (3.6)

Thus, if you know that the probability distribution is binomial, you do not have to make the detailed calculations required by equation (3.3) to compute the expected value.

Suppose that during the next month Nastke’s Clothing Store expects 1000 customers to enter the store. what is the expected number of customers who will make a purchase? Using equation (3.6), the answer is m 5 np 5 (1000)(0.3) 5 300. To increase the expected number of sales, Nastke’s must induce more customers to enter the store and/or somehow increase the probability that any individual customer will make a purchase after entering.

for the special case of a binomial distribution, the variance of the random variable is

s 2 5 np(1 2 p) (3.7)

for the Nastke Clothing Store problem with three customers, the variance and standard deviation for the number of customers making a purchase are

s2 5

s 5

nps1 2 pd 5 3s0.3ds0.7d 5 0.63

Ï0.63 5 0.79

3.4 Poisson Probability Distribution In this section we will consider a discrete random variable that often is useful when we are dealing with the number of occurrences of an event over a specified interval of time or space. for example, the random variable of interest might be the number of arrivals at a car wash in 1 hour, the number of repairs needed in 10 miles of highway, or the number of leaks in 100 miles of pipeline. If the following two assumptions are satisfied, the poisson probability distribution is applicable:

1. The probability of an occurrence of the event is the same for any two intervals of equal length.

2. The occurrence or nonoccurrence of the event in any interval is independent of the occurrence or nonoccurrence in any other interval.

The probability function of the Poisson random variable is given by equation (3.8):

f sxd 5 lxe2l

x! for x 5 0, 1, 2, . . . (3.8)

Try Problem 9, part (e), for practice computing the expected value, variance, and standard deviation of a binomial random variable.

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74 Chapter 3 Probability Distributions

1Values of e2l are available in Appendix E and can be easily computed with most modern calculators.

where

l 5 mean or average number of occurrences in an interval e 5 2.71828 x 5 number of occurrences in the interval f(x) 5 probability of x occurrences in the interval

Note that equation (3.8) shows no upper limit to the number of possible values that a Poisson random variable can realize. That is, x is a discrete random variable with an infinite sequence of values (x 5 0, 1, 2, . . .); the Poisson random variable has no set upper limit.

An Example Involving Time Intervals Suppose that we are interested in the number of arrivals at the drive-in teller window of a bank during a 15-minute period on weekday mornings. If we assume that the probability of a car arriving is the same for any two time periods of equal length and that the arrival or nonar- rival of a car in any time period is independent of the arrival or nonarrival in any other time period, the Poisson probability function is applicable. Then if we assume that an analysis of historical data shows that the average number of cars arriving during a 15-minute interval of time is 10, the Poisson probability function with l 5 10 applies:

f sxd 5 lxe2l

x! 5

10xe210

x! for x 5 0, 1, 2, . . .

If we wanted to know the probability of five arrivals in 15 minutes, we would set x 5 5 and obtain1

f s5d 5 105e210

5! 5 0.0378

Although we determined this probability by evaluating the probability function with l 5 10 and x 5 5, the use of Poisson probability distribution tables often is easier. These tables provide probabilities for specific values of x and l. we included such a table as Appendix C. for convenience we reproduce a portion of it as Table 3.8. To use the table of Poisson probabilities, you need know only the values of x and l. Thus, from Table 3.8, the probability of five arrivals in a 15-minute period is the value in the row corresponding to x 5 5 and the column corresponding to l 5 10. hence, f(5) 5 0.0378.

An Example Involving Length or Distance Intervals Suppose that we are concerned with the occurrence of major defects in a section of high- way one month after resurfacing. we assume that the probability of a defect is the same for any two intervals of equal length and that the occurrence or nonoccurrence of a defect in any one interval is independent of the occurrence or nonoccurrence in any other interval. Thus, the Poisson probability distribution applies.

Suppose that major defects occur at the average rate of two per mile. we want to find the probability that no major defects will occur in a particular 3-mile section of the

Bell Labs used the Poisson distribution in modeling the arrival of phone calls.

Try Problem 14 for practice computing Poisson probabilities.

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3.4 Poisson Probability Distribution 75

x 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 0 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0000 1 0.0010 0.0009 0.0009 0.0008 0.0007 0.0007 0.0006 0.0005 0.0005 0.0005 2 0.0046 0.0043 0.0040 0.0037 0.0034 0.0031 0.0029 0.0027 0.0025 0.0023 3 0.0140 0.0131 0.0123 0.0115 0.0107 0.0100 0.0093 0.0087 0.0081 0.0076 4 0.0319 0.0302 0.0285 0.0269 0.0254 0.0240 0.0226 0.0213 0.0201 0.0189

5 0.0581 0.0555 0.0530 0.0506 0.0483 0.0460 0.0439 0.0418 0.0398 0.0378 6 0.0881 0.0851 0.0822 0.0793 0.0764 0.0736 0.0709 0.0682 0.0656 0.0631 7 0.1145 0.1118 0.1091 0.1064 0.1037 0.1010 0.0982 0.0955 0.0928 0.0901 8 0.1302 0.1286 0.1269 0.1251 0.1232 0.1212 0.1191 0.1170 0.1148 0.1126 9 0.1317 0.1315 0.1311 0.1306 0.1300 0.1293 0.1284 0.1274 0.1263 0.1251

10 0.1198 0.1210 0.1219 0.1228 0.1235 0.1241 0.1245 0.1249 0.1250 0.1251 11 0.0991 0.1012 0.1031 0.1049 0.1067 0.1083 0.1098 0.1112 0.1125 0.1137 12 0.0752 0.0776 0.0799 0.0822 0.0844 0.0866 0.0888 0.0908 0.0928 0.0948 13 0.0526 0.0549 0.0572 0.0594 0.0617 0.0640 0.0662 0.0685 0.0707 0.0729 14 0.0342 0.0361 0.0380 0.0399 0.0419 0.0439 0.0459 0.0479 0.0500 0.0521

15 0.0208 0.0221 0.0235 0.0250 0.0265 0.0281 0.0297 0.0313 0.0330 0.0347 16 0.0118 0.0127 0.0137 0.0147 0.0157 0.0168 0.0180 0.0192 0.0204 0.0217 17 0.0063 0.0069 0.0075 0.0081 0.0088 0.0095 0.0103 0.0111 0.0119 0.0128 18 0.0032 0.0035 0.0039 0.0042 0.0046 0.0051 0.0055 0.0060 0.0065 0.0071 19 0.0015 0.0017 0.0019 0.0021 0.0023 0.0026 0.0028 0.0031 0.0034 0.0037

20 0.0007 0.0008 0.0009 0.0010 0.0011 0.0012 0.0014 0.0015 0.0017 0.0019 21 0.0003 0.0003 0.0004 0.0004 0.0005 0.0006 0.0006 0.0007 0.0008 0.0009 22 0.0001 0.0001 0.0002 0.0002 0.0002 0.0002 0.0003 0.0003 0.0004 0.0004 23 0.0000 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0002 0.0002 24 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001

TABLE 3.8 SELECTED VALUES fROM ThE POISSON PROBABILITY TABLE. EXAMPLE: l 5 10, x 5 5; f (5) 5 0.0378

NOTES AND COMMENTS

1. when working with the Poisson probability dis- tribution, you need to be sure that l is the mean number of occurrences for the desired interval. for instance, suppose that you know that 30 calls come into a switchboard every 15 minutes. To

compute Poisson probabilities for the number of calls coming in over a 5-minute period, you would use l 5 10; to compute probabilities for the number of calls coming in over a 1-minute period, you would use l 5 2.

highway. The interval length is 3 miles, so l 5 (2 defects/mile)(3 miles) 5 6 represents the expected number of major defects over the 3-mile section of highway. Thus, by us- ing equation (3.8) or Appendix C with l 5 6 and x 5 0, we obtain the probability of no major defects of 0.0025. Thus, finding no major defects in the 3-mile section is very unlikely. In fact, there is a 1 2 0.0025 5 0.9975 probability of at least one major defect in that section of highway.

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76 Chapter 3 Probability Distributions

3.5 Continuous Random Variables In this section we introduce probability distributions for continuous random variables. Recall from Section 3.1 that random variables that may assume any value in a certain in- terval or collection of intervals are said to be continuous. Examples of continuous random variables include the following:

1. The number of ounces of soup placed in a can labeled “8 ounces” 2. The flight time of an airplane traveling from Chicago to New York 3. The lifetime of the monitor of a new laptop computer 4. The drilling depth required to reach oil in an offshore drilling operation

To understand the nature of continuous random variables more fully, suppose that, in the first example, one can of soup has 8.2 ounces and another 8.3 ounces. Other cans could weigh 8.25 ounces, 8.225 ounces, and so on. In fact, the actual weight can be any numeric value from 0 ounces for an empty can to, say, 8.5 ounces for a can filled to capacity. Be- cause this interval contains infinitely many values, we can no longer list each value of the random variable and then identify its associated probability. In fact, for continuous random variables we need a new method for computing the probabilities associated with the values of the random variable.

Applying the Uniform Distribution Let x denote the flight time of an airplane traveling from Chicago to New York. Assume that the minimum time is 2 hours and that the maximum time is 2 hours 20 minutes. Thus, in terms of minutes, the flight time can be any value in the interval from 120 minutes to 140 minutes (e.g., 124 minutes, 125.48 minutes, etc.). As the random variable x can take on any value from 120 to 140 minutes, x is a continuous rather than a discrete random vari- able. Assume that sufficient actual flight data are available to conclude that the probability of a flight time between 120 and 121 minutes is the same as the probability of a flight time within any other 1-minute interval up to and including 140 minutes. with every 1-minute interval being equally likely, the random variable x has a uniform probability distribution. The following probability density function describes the uniform probability distribution for the flight time random variable:

f sxd 5 5 1

for 120 # x # 140

elsewhere (3.9)

figure 3.3 shows a graph of this probability density function. In general, the uniform prob- ability density function for a random variable x is

f sxd 5 5 1

b 2 a

for a # x # b

elsewhere

(3.10)

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3.5 Continuous Random Variables 77

In the flight time example, a 5 120 and b 5 140. In the graph of a probability density function, f (x) shows the height or value of

the function at any particular value of x. Because the probability density function for flight time is uniform, the height or value of the function is the same for each value of x between 120 and 140. That is, f (x) 5 1/20 for all values of x between 120 and 140. The probability density function f (x), unlike the probability function for a discrete random variable, represents the height of the function at any particular value of x and not probability. Recall that, for each value of a discrete random variable (say, x 5 2), the probability function yielded the probability of x having exactly that value [that is, f (2)]. however, a continuous random variable has infinitely many values, so we can no longer identify the probability for each specific value of x. Rather, we must consider probability in terms of the likelihood that a random variable takes on a value within a specified interval. for instance, in the flight time example an acceptable probability question is: what is the probability that the flight time is between 120 and 130 minutes? That is, what is P(120 # x # 130)? As the flight time must be between 120 and 140 minutes and as the probability is uniformly distributed over this interval, we feel comfortable saying that P(120 # x # 130) 5 0.50. Indeed, as we will show, this is correct.

Area as a Measure of Probability Refer to figure 3.4 and consider the area under the graph of f (x) over the interval from 120 to 130. Note that the region is rectangular in shape and that the area of a rectangle is simply the width times the height. with the width of the interval equal to 130 2 120 5 10 and the height of the graph f (x) 5 1/20, the area 5 width 3 height 5 10(1/20) 5

10/20 5 0.50.

0 120 125 130 135 140 x

Flight Time in Minutes

1 20

f (x)

FIGURE 3.3 UNIfORM PROBABILITY DENSITY fUNCTION fOR fLIGhT TIME

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78 Chapter 3 Probability Distributions

what observation can you make about the area under the graph of f (x) and probabil- ity? They are identical! Indeed, that is true for all continuous random variables. In other words, once you have identified a probability density function f (x) for a continuous ran- dom variable, you can obtain the probability that x takes on a value between some lower value a and some higher value b by computing the area under the graph of f (x) over the interval a to b.

with the appropriate probability distribution and the interpretation of area as probability, we can answer any number of probability questions. for example, what is the probability of a flight time between 128 and 136 minutes? The width of the interval is 136 2 128 5 8. with the uniform height of 1/20, P(128 # x # 136) 5

8/20 5 0.40. Note that P(120 # x # 140) 5 20(1/20) 5 1; the total area under the f (x) graph is

equal to 1. This property holds for all continuous probability distributions and is the analog of the requirement that the sum of the probabilities must equal 1 for a discrete probability distribution. A continuous probability distribution also requires that f (x) $ 0 for all values of x. It is the analog of the requirement that f (x) $ 0 for discrete prob- ability distributions.

Two principal differences between continuous random variables and probability distri- butions and their discrete counterparts stand out.

1. we no longer talk about the probability of the random variable taking on a particu- lar value. Instead we talk about the probability of the random variable taking on a value within some given interval.

2. The probability of the random variable taking on a value within some given interval is defined to be the area under the graph of the probability density function over the interval. This definition implies that the probability that a continuous random vari- able takes on any particular value is zero because the area under the graph of f (x) at a single point is zero.

Try Problem 18 to practice computing probabilities us- ing the uniform probability distribution.

0 120 125 130 135 140 x

Flight Time in Minutes

1 20

f (x)

P(120 x 130) = Area = 1/20(10) = 10/20 = 0.50

FIGURE 3.4 AREA PROVIDES PROBABILITY Of fLIGhT TIME

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3.5 Continuous Random Variables 79

Procter & Gamble (P&G) produces and markets such prod- ucts as detergents, disposable diapers, over-the-counter pharmaceuticals, dentifrices, bar soaps, mouthwashes, and paper towels. worldwide, it has the leading brand in more categories than any other consumer products company. Since its merger with Gillette, P&G also produces and mar- kets razors, blades, and many other personal care products.

As a leader in the application of statistical methods in decision making, P&G employs people with diverse academic backgrounds: engineering, statistics, opera- tions research, and business. The major quantitative technologies for which these people provide support are probabilistic decision and risk analysis, advanced simu- lation, quality improvement, and quantitative methods (e.g., linear programming, regression analysis, probabil- ity analysis).

The Industrial Chemicals Division of P&G is a ma- jor supplier of fatty alcohols derived from natural sub- stances such as coconut oil and from petroleum-based derivatives. The division wanted to know the economic risks and opportunities of expanding its fatty-alcohol production facilities, so it called in P&G’s experts in probabilistic decision and risk analysis to help. After structuring and modeling the problem, they determined that the key to profitability was the cost difference between the petroleum- and coconut-based raw materi- als. future costs were unknown, but the analysts were able to approximate them with the following continuous random variables.

x 5 the coconut oil price per pound of fatty alcohol

and

y 5 the petroleum raw material price per pound of fatty alcohol

Because the key to profitability was the difference between these two random variables, a third random variable, d 5 x 2 y, was used in the analysis. Experts were interviewed to determine the probability distri- butions for x and y. In turn, this information was used to develop a probability distribution for the difference in prices d. This continuous probability distribution showed a .90 probability that the price difference would be $.0655 or less and a .50 probability that the price dif- ference would be $.035 or less. In addition, there was only a .10 probability that the price difference would be $.0045 or less.†

The Industrial Chemicals Division thought that being able to quantify the impact of raw material price differences was key to reaching a consensus. The prob- abilities obtained were used in a sensitivity analysis of the raw material price difference. The analysis yielded sufficient insight to form the basis for a recommendation to management.

The use of continuous random variables and their probability distributions was helpful to P&G in analyz- ing the economic risks associated with its fatty-alcohol production. In this chapter, you will gain an under- standing of continuous random variables and their probability distributions, including one of the most important probability distributions in statistics, the normal distribution.

PROCTER & gAMBLE* CINCINNATI, OHIO

Q.M. in ACtion

*the authors are indebted to Joel Kahn of Procter & Gamble for provid- ing this Q.M. in Action.

†the price differences stated here have been modified to protect pro- prietary data.

The Q.M. in Action, Procter & Gamble, describes how a large manufacturer of consumer goods used a model of random variables to assist in deciding whether to expand one of its production facilities.

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80 Chapter 3 Probability Distributions

3.6 Normal Probability Distribution Perhaps the most important probability distribution used to describe a continuous random variable is the normal probability distribution. It is applicable in a great many practical problem situations, and its probability density function has the form of the bell-shaped curve shown in figure 3.5. The mathematical function that provides the bell-shaped curve of the normal probability density function follows:

f sxd 5 1

s Ï2p e2sx 2 md

2/2s2 for 2` , x , ` (3.11) where

m 5 mean or expected value of the random variable x

s2 5 variance of the random variable x

s 5 standard deviation of the random variable x

p 5 3.14159

e 5 2.71828

The normal distribution was first observed by Abra- ham de Moivre, a French mathematician, in the early 1700s. De Moivre’s work was motivated by the study of probability associated with gambling and games of chance.

NOTES AND COMMENTS

1. for any continuous random variable the prob- ability of any particular value is zero, so P(a # x # b) 5 P(a , x , b). Thus, the probability of a random variable assuming a value in any interval is the same whether the endpoints are included or not.

2. To see more clearly why the height of a proba- bility density function is not a probability, think

about a random variable with a uniform prob- ability distribution of

f sxd 5 520 for 0 # x # 0.5

elsewhere

The height of the probability density function is 2 for values of x between 0 and 0.5. But we know that probabilities can never be greater than 1.

f ( x )

FIGURE 3.5 NORMAL PROBABILITY DISTRIBUTION

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3.6 normal Probability Distribution 81

Recall from the previous discussion of continuous random variables that f (x) is the height of the curve at a particular value of x. Thus, once we specify the mean (m) and either the standard deviation (s) or variance (s2), we can use equation (3.11) to determine the graph for the corresponding normal distribution. figure 3.6 shows two normal distributions, one with m 5 50 and s 5 15 and another with m 5 50 and s 5 7.5. Note in particular the effect that the standard deviation s has on the general shape of the normal curve. A larger standard deviation tends to flatten and broaden the curve because larger values of s indicate greater variability in the values of the random variable.

fortunately, whenever we use the normal distribution to answer probability questions, we do not have to use the probability density function of equation (3.11). In fact, when we use the normal distribution, we will have tables of probability values [areas under the f (x) curve] that can provide the desired probability information. To show how to use the tables of areas or probabilities for the normal distribution, we must first introduce the standard normal distribution.

Standard Normal Distribution A random variable that has a normal distribution with a mean of 0 and a standard devia- tion of 1 is said to have a standard normal distribution. we use the letter z to designate this particular normal random variable. figure 3.7 shows the graph of the standard normal distribution. Note that it has the same general appearance as other normal distributions, but with the special properties of m 5 0 and s 5 1. The units on the horizontal axis (z) measure the number of standard deviations from the mean.

Recall the procedure for finding probabilities associated with a continuous random variable. we want to determine the probability of the random variable having a value in a specified interval from a to b. Thus we have to find the area under the curve in the interval from a to b. In the preceding section we showed that finding probabilities, or areas under the curve, for a uniform distribution was relatively easy. All we had to do was multiply the width of the interval by the height of the graph. however, finding areas under the normal distribution curve appears at first glance to be much more difficult because the height of the curve varies. The mathematical technique for obtaining these areas is beyond the scope of the text, but fortunately tables are available that provide the areas or probability values

.06

.05

.04

.03

.02

.01

0 10 20 30 40 50 60 70 80 90 100 x

= 7.5

= 15

FIGURE 3.6 TwO NORMAL DISTRIBUTIONS wITh m 5 50

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82 Chapter 3 Probability Distributions

for the standard normal distribution. Table 3.9 is such a table of areas. This table is also available as Appendix D.

The graphs at the top of Table 3.9 show that the area in the table is the probability of a standard normal random variable being less than or equal to a specific value of z. Such probabilities are referred to as cumulative probabilities. In using Table 3.9 to determine a cumulative probability, note that the value of z with one decimal appears in the left-hand column of the table, with the second decimal appearing in the top row. Negative values of z are provided on the first page of the table, whereas positive values of z are provided on the second page. for example, for z 5 20.85, we find 20.8 in the left-hand column and the second decimal 0.05 in the top row of the first page of the table. Then, by looking in the body of the table, we find an area or probability of 0.1977. This is the cumulative probability that the standard normal random variable is less than or equal to z 5 20.85. This area is shown graphically at the top of Table 3.9. As another example, we can use the second page of the table to determine that the cumulative probability that the standard nor- mal random variable is less than or equal to z 5 1.25. we find 1.2 in the left-hand column and the second decimal 0.05 in the top row of the second page of the table. In the body of the table we find an area or probability of 0.8944. This is the cumulative probability that the standard normal random variable is less than or equal to z 5 1.25. This area is also shown graphically at the top of Table 3.9.

Suppose that we wanted to use the cumulative standard normal distribution table to determine the probability that the standard normal random variable z will be between 21.00 and 11.00. Table 3.9 shows the cumulative probability that z is less than or equal to 11.00 is 0.8413 and the cumulative probability that z is less than or equal to 21.00 is 0.1587. Thus, the probability that z will be between 21.00 and 11.00 must be the difference between these two cumulative probabilities: 0.8413 2 0.1587 5 0.6826. This is shown graphically in figure 3.8.

Similarly, we can find the probability that the standard normal random variable z will be between 22.00 and 12.00. Using the cumulative probabilities at z 5 12.00 and z 5 22.00, the probability that z will be between 22.00 and 12.00 is 0.9772 2 0.0228 5 0.9544. In addition, we can use the cumulative probabilities at z 5 13.00 and z 5 23.00 to conclude that the probability z will be between 23.00 and 13.00 is 0.9986 2 0.0013 5 0.9973. Since the total probability or total area under the curve is equal to 1.0000, the probability of 0.9973 tells us that z will almost always fall between 23.00 and 13.00.

The probability that a standard normal random variable z is between a and b is always the difference between two cumulative probabilities: one for z 5 b and one for z 5 a.

.40

.30

.20

.10

-3 -2 -1 0 +1 +2 +3

= 1

FIGURE 3.7 STANDARD NORMAL DISTRIBUTION

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3.6 normal Probability Distribution 83

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

23.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010

22.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 22.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 22.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 22.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 22.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048

22.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 22.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084 22.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 22.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 22.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183

21.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 21.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 21.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 21.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 21.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559

21.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 21.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 21.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 21.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 21.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379

20.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611 20.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 20.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 20.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 20.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776

20.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 20.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 20.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 20.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 20.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641

TABLE 3.9 CUMULATIVE PROBABILITIES fOR ThE STANDARD NORMAL DISTRIBUTION

Cumulative probability

Entries in the table give the area under the curve to the left of the z value. For example, for z = 20.85, the cumulative probability is 0.1977.

(continued)

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84 Chapter 3 Probability Distributions

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767

2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9913 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936

2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986

3.0 0.9986 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990

TABLE 3.9 CUMULATIVE PROBABILITIES fOR ThE STANDARD NORMAL DISTRIBUTION (Continued )

Cumulative probability

Entries in the table give the area under the curve to the left of the z value. For example, for z = 1.25, the cumulative probability is 0.8944.

0 z

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3.6 normal Probability Distribution 85

As a final example, what is the probability that the normal random variable z is greater than 2.00? from Table 3.9, we find that the cumulative probability that z is less than or equal to 2.00 is 0.9772. Since the total area under the curve is equal to 1.0000, the probability that z will be greater than 2.00 must be 1.0000 2 0.9772 5 0.0228. This is shown graphically in figure 3.9. As the examples in this section have shown, you should be able to use the cumulative probabilities in Table 3.9 to answer a variety of probability questions about the standard normal random variable z.

Computing Probabilities for Any Normal Distribution The reason we discuss the standard normal distribution so extensively is that we can com- pute probabilities for any normal distribution by first converting to the standard normal distribution. Thus, when we have a normal distribution with any mean m and any standard deviation s, we can answer probability questions about this distribution by converting to

Area less than or equal to z = +1.00 is 0.8413

Area = 0.8413 2 0.1587 = 0.6826

Area less than or equal to z = 21.00

is 0.1587

23 22 21 0 +1 +2 +3 z

FIGURE 3.8 PROBABILITY Of z BETwEEN 21.00 AND 11.00

Area less than or equal to z = 2.00 is 0.9772 Area greater than z = 2.00

Is 1.0000 2 0.9772 = 0.0228

23 22 21 0 +1 +2 +3 z

FIGURE 3.9 PROBABILITY Of z GREATER ThAN 2.00

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86 Chapter 3 Probability Distributions

the standard normal distribution. we then use Table 3.9 and the appropriate z values to find the probability. The formula used to convert any normal random variable x with mean m and standard deviation s to the standard normal distribution is

z 5 x 2 m

s (3.12)

when used in this way, z is a measure of the number of standard deviations that x is from m. we can use an example to show most easily how the conversion to the z value allows

us to use the standard normal distribution to compute probabilities for any normal dis tribution. Suppose that we have a normal distribution with m 5 10 and s 5 2, as shown graphically in figure 3.10. Note that, in addition to the values of the random variable shown on the x axis, we have included a second axis (the z axis) to show that for each value of x there is a corresponding value of z. for example, when x 5 10, the corresponding z value (the number of standard deviations away from the mean) is z 5 (x 2 m)ys 5 (10 2 10)y2 5 0. Similarly, for x 5 14 we have z 5 (x 2 m)ys 5 (14 2 10)y2 5 2.

Now suppose that we want to know the probability that the random variable x is between 10 and 14; that is, P(10 # x # 14). we do not have tables that provide this prob- ability directly. however, note that in figure 3.10 the area under the curve (probability) for x between 10 and 14 is the same as the area under the curve for z between 0 and 2. Using z 5 2.00 and Table 3.9, we find that the cumulative probability of z being less than or equal to 2.00 is 0.9772. Similarly, Table 3.9 shows that the cumulative probability of z being less than or equal to 0.00 is 0.5000. Thus, the probability that the standard normal random vari- able z is between 0.00 and 2.00 is 0.9772 2 0.5000 5 0.4772. Thus, we conclude that the probability of x being between 10 and 14 is also 0.4772.

This procedure applies to any normal distribution problem. That is, for any x value a corresponding z value is given by equation (3.12). To find the probability that x is in a specified interval, simply convert the x interval to its corresponding z interval. Then use the table for the standard normal distribution to answer the probability question.

23 22 21 0 1 2 3

Corresponding z Value

4 6 8 10 12 14 16

FIGURE 3.10 NORMAL DISTRIBUTION wITh m 5 10 AND s 5 2

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3.6 normal Probability Distribution 87

Grear Tire Company Problem Suppose that Grear Tire Company just developed a new steel-belted radial tire that will be sold through a national chain of discount stores. Because the tire is a new product, Grear’s management believes that the mileage guarantee offered with the tire will be an important factor in consumer acceptance of the product. Before finalizing the tire mileage guarantee policy, Grear’s management wants some probability information concerning the number of miles the tires will last.

from actual road tests with the tires, Grear’s engineering group estimates the mean tire mileage at m 5 36,500 miles and the standard deviation at s 5 5000 miles. In addition, the data collected indicate that a normal distribution is a reasonable assumption.

what percentage of the tires, then, can be expected to last more than 40,000 miles? In other words, what is the probability that the tire mileage will exceed 40,000? To compute this probability, we need to find the area of the shaded region in figure 3.11.

At x 5 40,000 we have

z 5 x 2 m

s 5

40,000 2 36,500

5000 5

3500

5000 5 0.70

Thus the probability that the normal distribution for tire mileage will have an x value greater than 40,000 is the same as the probability that the standard normal distribution will have a z value greater than 0.70. Using Table 3.9, we find that the cumulative probability that z is less than or equal to 0.70 is 0.7580. Thus, the probability that z will be greater than 0.70 must be 1.0000 2 0.7580 5 0.2420. In terms of the tire mileage x, we can conclude that there is a 0.2420 probability that x will be greater than 40,000 miles. Thus, we can anticipate about 24.2% of the tires manufactured by Grear will last more than 40,000 miles.

Let us now assume that Grear is considering a guarantee that will provide a discount on a new set of tires if the mileage on the original tires doesn’t exceed the mileage stated on the guarantee. what should the guarantee mileage be if Grear wants no more than 10% of the tires to be eligible for the discount? This question is interpreted graphically in figure 3.12. Note that 10% of the area is below the unknown guarantee mileage. Because this 10% is the lower tail of the normal probability distribution, 0.1000 is the cumula- tive probability that the tire mileage will be less than or equal to the unknown guarantee

Probability That x Exceeds 40,000 = ?

40,000

FIGURE 3.11 GREAR TIRE COMPANY TIRE MILEAGE

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mileage. The question is now, how many standard deviations (z value) do we have to be below the mean to a achieve a 0.1000 cumulative probability? This time we will look into the body of Table 3.9 and try to find the cumulative probability 0.1000. we can- not find 0.1000 exactly, but a cumulative probability 0.1003 is close. here we find the corresponding z 5 21.28. This tells us that we must be 1.28 below the mean to find the desired tire guarantee mileage. This mileage is

Guarantee mileage 5 m 2 1.28s

5 36,500 2 1.28(5000) 5 30,100

Therefore, a guarantee of 30,100 miles will meet the requirement that approximately 10% of the tires will be eligible for the discount. with this information the firm might confidently set its tire mileage guarantee at 30,000 miles.

Again we see the important role that probability distributions play in providing decision-making information. Once a probability distribution is established for a particular problem, it can be used rather quickly and easily to provide information about the likelihood of various scenarios. Although this information does not make a decision recommendation directly, it does provide information that helps the decision maker understand the problem better. Ultimately, this information may assist the decision maker in reaching a good decision.

3.7 Exponential Probability Distribution A continuous probability distribution that is often useful in describing the time needed to complete a task is the exponential probability distribution. The exponential random vari- able can be used to describe the time between arrivals at a car wash, the time required to load a truck, the distance between major defects in a highway, and so on. The exponential probability density function is

f sxd 5 1 m

e2x/m for x $ 0, m . 0 (3.13)

Try Problem 23 for practice finding a z value that cuts off a particular probability.

10% of Tires Eligible for Discount Guarantee

Guarantee Mileage = ?

FIGURE 3.12 GREAR’S DISCOUNT GUARANTEE

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3.7 Exponential Probability Distribution 89

To provide an example of the exponential probability distribution, suppose that the loading time for a truck at a factory dock follows an exponential probability distribution. If the mean, or average, loading time is 15 minutes (m 5 15), the appropriate probability density function is

f sxd 5 1

15 e2x/15

figure 3.13 shows the graph of this density function.

Computing Probabilities for the Exponential Distribution As with any continuous probability distribution, the area under the curve corresponding to some interval provides the probability that the random variable takes on a value in that interval. for instance, for the factory loading dock example, the probability that 6 minutes or less (x # 6) are needed to load a truck is defined to be the area under the curve from x 5 0 to x 5 6. Similarly, the probability that a loading time is 18 minutes or less (x # 18) is the area under the curve from x 5 0 to x 5 18. Note also that the probability of loading a truck in between 6 and 18 minutes (6 # x # 18) is the area under the curve from x 5 6 to x 5 18.

To compute exponential probabilities such as those previously described, the following formula provides the probability of obtaining a value for the exponential random variable of less than or equal to some specific value of x, denoted by x0:

Psx # x0d 5 1 2 e 2x0/m (3.14)

f (x)

.07

.06

.05

.04

.03

.02

.01

.0 5 10 15 20 25 30 35 40 45 x

FIGURE 3.13 EXPONENTIAL DISTRIBUTION fOR LOADING TIME AT ThE LOADING DOCK wITh m 5 15

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90 Chapter 3 Probability Distributions

Thus, for the factory loading dock example, equation (3.14) becomes

Psloading time # x0d 5 1 2 e 2x0/15

hence, the probability that 6 minutes or less (x # 6) are needed to load a truck is

P (loading time # 6) 5 1 2 e26y15 5 0.3297

Note also that the probability that 18 minutes or less (x # 18) are needed to load a truck is

P (loading time # 18) 5 1 2 e218y15 5 0.6988

Thus, the probability that 6 to 18 minutes are required to load a truck is 0.6988 2 0.3297 5 0.3691. Probabilities for any other interval can be computed in a similar manner.

Relationship Between the Poisson and Exponential Distributions In Section 3.4 we introduced the Poisson distribution as a discrete probability distribution that often is useful when we are dealing with the number of occurrences over a specified interval of time or space. Recall that the Poisson probability function is

f sxd 5 lxe2l

where

l 5 expected value or mean number of occurrences in an interval

The continuous exponential probability distribution is related to the discrete Poisson dis- tribution. The Poisson distribution provides an appropriate description of the number of occurrences per interval, and the exponential distribution provides a description of the length of the interval between occurrences.

To illustrate this relationship, let us suppose that the number of cars that arrive at a car wash during 1 hour is described by a Poisson probability distribution with a mean of 10 cars per hour. Thus the Poisson probability function that provides the probability of x arrivals per hour is

f sxd 5 10xe210

The average number of arrivals is 10 cars per hour, so the average time between cars arriving is

1 hour

10 cars 5 0.1 hour/car

Thus, the corresponding exponential distribution that describes the time between the ar- rival of cars has a mean of m 5 0.1 hours per car. The appropriate exponential probability density function is

f sxd 5 1

0.1 e2x/0.1 5 10e210 x

Try Problem 29 for practice finding probabilities with the exponential probability distribution.

The Poisson and exponential probability distributions are used in the chapter on waiting line models (Chapter 15). In these models the Poisson distribution is used as the probability distribution for the number of arrivals, while the exponential probability distribution is used as the probability distribution for the service time.

Try Problems 30 and 31 for applications of the exponential probability distribution.

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Glossary 91

Summary

In this chapter we continued the discussion of probability by introducing the important concepts of random variables and probability distributions. Random variables provide numeric descriptions of the outcomes of experiments. when random variables are used, computations of the expected value, variance, and standard deviation can help the decision maker understand characteristics of the problem under study. we discussed the probability distributions for both discrete and continuous random variables.

Of particular interest are special probability distributions such as the binomial, Pois- son, uniform, normal, and exponential distributions. These distributions provide wide applicability, and special formulas and/or tables make the probability information easily available.

Through a variety of problems and applications, we illustrated the role that probabil- ity distributions play in providing decision-making information. Although the probability values generated by the techniques and methods of this chapter do not by themselves make decision recommendations, they do provide assistance to the decision maker in terms of understanding the uncertainties inherent in the problem. Ultimately, this better understand- ing may lead to new and better decisions.

Glossary

Binomial probability distribution The probability distribution for a discrete random variable, used to compute the probability of x successes in n trials. Continuous random variable A random variable that may assume any value in an inter- val or collection of intervals. Cumulative probability The probability that a random variable takes on a value less than or equal to stated value. discrete probability distribution A table, graph, or equation describing the values of the random variable and the associated probabilities. discrete random variable A random variable that may assume only a finite or infinite sequence of values. Expected value A weighted average of the values of the random variable, for which the probability function provides the weights. If an experiment can be repeated a large number of times, the expected value can be interpreted as the “long-run average.” Exponential probability distribution A continuous probability distribution that is useful in describing the time to complete a task or the time between occurrences of an event. Normal probability distribution A continuous probability distribution whose prob- ability density function is bell shaped and determined by the mean, m, and standard deviation, s. poisson probability distribution The probability distribution for a discrete random vari- able, used to compute the probability of x occurrences over a specified interval. probability density function The function that describes the probability distribution of a continuous random variable. probability function A function, denoted f (x), that provides the probability that a discrete random variable x takes on some specific value. Random variable A numeric description of the outcome of an experiment. Standard deviation The positive square root of the variance.

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92 Chapter 3 Probability Distributions

Experiment Random Variable (x)

a. Take a 20-question examination Number of questions answered correctly b. Observe cars arriving at a tollbooth Number of cars arriving at the tollbooth for 1 hour c. Audit 50 tax returns Number of returns containing errors d. Observe an employee’s work Number of nonproductive hours for 8 hours e. weigh a shipment of goods Number of pounds

x f (x)

2100 0.10 0 0.20 50 0.30 100 0.25 150 0.10 200

2. The following table shows a partial probability distribution for the MRA Company’s pro- jected profits (in thousands of dollars) for the first year of operation (the negative value denotes a loss):

a. find the missing value of f (200). what is your interpretation of this value? b. what is the probability that MRA will be profitable? c. what is the probability that MRA will make at least $100,000?

3. Data were collected on the number of operating rooms in use at Tampa General hospital over a 20-day period. On 3 of the days only one operating room was used; on 5 days, two were used; on 8 days, three were used; and on 4 days all four rooms were used. a. Use the relative frequency approach to construct a probability distribution for the

number of operating rooms in use on any given day. b. Draw a graph of the probability distribution. c. Show that your probability distribution satisfies the requirements for a valid discrete

probability distribution.

testSELF

Standard normal distribution A normal distribution with a mean of 0 and a standard deviation of 1. Uniform probability distribution A continuous probability distribution in which the probability that the random variable will assume a value in any interval of equal length is the same for each interval. Variance A measure of the dispersion or variability in the random variable. It is a weighted average of the squared deviations from the mean, m.

Problems

1. The following examples are experiments and their associated random variables. In each case identify the values the random variable can assume and state whether the random variable is discrete or continuous:

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Problems 93

4. Shown is a probability distribution for the random variable x.

a. what is the expected amount of money in a single bar of Jackpot soap? b. what is the standard deviation of the money in a single bar of Jackpot soap? c. how many bars of soap would a customer have to buy so that, on average, he or she

has purchased three bars containing a $50 or $100 bill? d. If a customer buys 8 bars of soap, what is the probability that at least one of these bars

contains a bill of $20 or larger?

6 The National Center for health Statistics reported the following data on the number of children born in individual pregnancies in 1996 and 2006 (The world Almanac, 2010):

distribution of paper Currency prizes

Bill denomination Number of Bills $1 520 $5 260 $10 120 $20 70 $50 29 $100 1

Total 1000

1996 2006 Number of Children Frequency Frequency

one child 3,671,455 3,971,276 twins 100,750 137,085 triplets 5,298 6,118 quadruplets 560 355 quintuplets or more 81 67

a. Define a random variable x 5 number of children born in a single pregnancy in 1996 and develop a probability distribution for the random variable. Let x 5 5 represent quintuplets or more.

x f (x)

3 0.25 6 0.50 9 0.25

Total 1.00

testSELF

a. Compute E(x), the expected value of x. b. Compute s2, the variance of x. c. Compute s, the standard deviation of x.

5. Brandon Lang is a creative entrepreneur who has developed a novelty soap item called Jack- pot to target consumers with a gambling habit. Inside each bar of Jackpot shower soap is a single rolled-up bill of U.S. currency. The currency (rolled up and sealed in shrink-wrap) is appropriately inserted into the soap mixture prior to the cutting and stamping procedure. The distribution of paper currency (per 1000 bars of soap) is given in the following table.

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94 Chapter 3 Probability Distributions

b. Compute the expected value and variance for the number of children born in a single pregnancy in 1996.

c. Define a random variable y 5 number of children born in a single pregnancy in 2006 and develop a probability distribution for the random variable. Let y 5 5 represent quintuplets or more.

d. Compute the expected value and variance for the number of children born in a single pregnancy in 2006.

e. Do these data support the conclusion that the increased use of fertility drugs by older women has generated an upward trend in multiple births?

7. The demand for Carolina Industries’ product varies greatly from month to month. Based on the past two years of data, the following probability distribution shows the company’s monthly demand:

Unit demand probability

300 0.20 400 0.30 500 0.35 600 0.15

a. If the company places monthly orders equal to the expected value of the monthly demand, what should Carolina’s monthly order quantity be for this product?

b. Assume that each unit demanded generates $70 in revenue and that each unit ordered costs $50. how much will the company gain or lose in a month if it places an order based on your answer to part (a) and the actual demand for the item is 300 units?

c. what are the variance and standard deviation for the number of units demanded?

8. The J. R. Ryland Computer Company is considering a plant expansion that will enable the company to begin production of a new computer product. The company’s president must determine whether to make the expansion a medium- or large-scale project. The demand for the new product involves an uncertainty, which for planning purposes may be low demand, medium demand, or high demand. The probability estimates for the demands are 0.20, 0.50, and 0.30, respectively. Letting x indicate the annual profit in $1000s, the firm’s planners developed profit forecasts for the medium- and large-scale expansion projects.

Medium-Scale Large-Scale Expansion profits Expansion profits

x f(x) y f(y)

Low 50 0.20 0 0.20 demand Medium 150 0.50 100 0.50 High 200 0.30 300 0.30

a. Compute the expected value for the profit associated with the two expansion alter- natives. which decision is preferred for the objective of maximizing the expected profit?

b. Compute the variance for the profit associated with the two expansion alternatives. which decision is preferred for the objective of minimizing the risk or uncertainty?

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Problems 95

9. Consider a binomial experiment with 2 trials and p 5 0.4. a. Compute the probability of 1 success, f (1). b. Compute f (0). c. Compute f (2). d. find the probability of at least one success. e. find the expected value, variance, and standard deviation.

10. Consider a binomial experiment with n 5 10 and p 5 0.10. Use the binomial tables (Appendix B) to answer parts (a) through (d). a. find f (0). b. find f (2). c. find P(x # 2). d. find P(x $ 1). e. find E(x). f. find Var(x) and s.

11. A recent survey by the New Statesman on British social attitudes asked respondents if they believe that inequality is too large. The survey found that 74% of the respondents do believe inequality is too large. a. In a sample of six British citizens, what is the probability that two believe inequality

is too large? b. In a sample of six British citizens, what is the probability that at least two respondents

believe that inequality is too large? c. In a sample of four British citizens, what is the probability that none believe inequality

is too large?

12. when a new machine is functioning properly, only 3% of the items produced are defec- tive. Assume that we will randomly select two parts produced on the machine and that we are interested in the number of defective parts found. a. Describe the conditions under which this situation would be a binomial experiment. b. how many experimental outcomes yield one defect? c. Compute the probabilities associated with finding no defects, one defect, and two

defects.

13. Military radar and missile detection systems are designed to warn a country of enemy attacks. A reliability question deals with the ability of the detection system to identify an attack and issue the warning. Assume that a particular detection system has a 0.90 prob- ability of detecting a missile attack. Answer the following questions using the binomial probability distribution: a. what is the probability that one detection system will detect an attack? b. If two detection systems are installed in the same area and operate independently, what

is the probability that at least one of the systems will detect the attack? c. If three systems are installed, what is the probability that at least one of the systems

will detect the attack? d. would you recommend that multiple detection systems be operated? Explain.

14. Consider a Poisson probability distribution with 2 as the average number of occurrences per time period. a. write the appropriate Poisson probability function. b. what is the average number of occurrences in three time periods? c. write the appropriate Poisson probability function to determine the probability of

x occurrences in three time periods. d. find the probability of two occurrences in one time period. e. find the probability of six occurrences in three time periods. f. find the probability of five occurrences in two time periods.

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96 Chapter 3 Probability Distributions

15. Telephone calls arrive at the rate of 48 per hour at the reservation desk for Regional Airways. a. find the probability of receiving 3 calls in a 5-minute interval. b. find the probability of receiving 10 calls in 15 minutes. c. Suppose that no calls are currently on hold. If the agent takes 5 minutes to complete

processing the current call, how many callers do you expect to be waiting by that time? what is the probability that no one will be waiting?

d. If no calls are currently being processed, what is the probability that the agent can take 3 minutes for personal time without being interrupted?

16. More than 50 million guests stayed at bed and breakfasts (B&Bs) last year. The website for the Bed and Breakfast Inns of North America, which averages approximately seven visitors per minute, enables many B&Bs to attract guests without waiting years to be mentioned in guidebooks. a. what is the probability of no website visitors in a 1-minute period? b. what is the probability of two or more website visitors in a 1-minute period? c. what is the probability of one or more website visitors in a 30-second period? d. what is the probability of five or more website visitors in a 1-minute period?

17. Airline passengers arrive randomly and independently at the passenger screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. a. what is the probability of no arrivals in a 1-minute period? b. what is the probability of 3 or fewer arrivals in a 1-minute period? c. what is the probability of no arrivals in a 15-second period? d. what is the probability of at least 1 arrival in a 15-second period?

18. A random variable x is uniformly distributed between 1.0 and 1.5. a. Show the graph of the probability density function. b. find P(x 5 1.25). c. find P(1.00 # x # 1.25). d. find P(1.20 , x , 1.50).

19. Delta Airlines quotes a flight time of 2 hours, 5 minutes for its flights from Cincinnati to Tampa. Suppose we believe that actual flight times are uniformly distributed between 2 hours and 2 hours, 20 minutes. a. Show the graph of the probability density function for flight times. b. what is the probability that the flight will be no more than 5 minutes late? c. what is the probability that the flight will be more than 10 minutes late? d. what is the expected flight time?

20. Most computer languages have a function that can be used to generate random numbers. In Microsoft’s Excel, the RAND function can be used to generate random numbers be- tween 0 and 1. If we let x denote the random number generated, then x is a continuous random variable with the probability density function:

f sxd 5 510 for 0 # x # 1elsewhere a. Graph the probability density function. b. what is the probability of generating a random number between 0.25 and 0.75? c. what is the probability of generating a random number with a value less than or equal

to 0.30? d. what is the probability of generating a random number with a value greater than 0.60?

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Problems 97

21. for the standard normal random variable z, compute the following probabilities: a. P(0 # z # 0.83) b. P(21.57 # z # 0) c. P(z . 0.44) d. P(z $ 20.23) e. P(z , 1.20) f. P(z # 20.71)

22. for the standard normal random variable z, find z for each situation. a. The area to the left of z is 0.9750. b. The area between 0 and z is 0.4750. c. The area to the left of z is 0.7291. d. The area to the right of z is 0.1314. e. The area to the left of z is 0.6700. f. The area to the right of z is 0.3300.

23. for the standard normal random variable z, find z for each situation. a. The area to the left of z is 0.2119. b. The area between 2z and z is 0.9030. c. The area between 2z and z is 0.2052. d. The area to the left of z is 0.9948. e. The area to the right of z is 0.6915.

24. The demand for a new product is estimated to be normally distributed with m 5 200 and s 5 40. Let x be the number of units demanded, and find the following probabilities: a. P(180 # x # 220) b. P(x $ 250) c. P(x # 100) d. P(225 # x # 250)

25. The College Board National Office recently reported that in 2011–2012, the 547,038 high school juniors who took the ACT achieved a mean score of 530 with a standard deviation of 123 on the mathematics portion of the test (http://media.collegeboard.com/digitalServices/pdf /research/2013/TotalGroup-2013.pdf). Assume these test scores are normally distributed. a. what is the probability that a high school junior who takes the test will score at least 610

on the mathematics portion of the test? b. what is the probability that a high school junior who takes the test will score no higher

than 460 on the mathematics portion of the test? c. what is the probability that a high school junior who takes the test will score between

460 and 550 on the mathematics portion of the test? d. how high does a student have to score to be in the top 10% of high school juniors on

the mathematics portion of the test?

26. General hospital’s patient account division has compiled data on the age of accounts receivable. The data collected indicate that the age of the accounts follows a normal dis- tribution with m 5 28 days and s 5 8 days. a. what portion of the accounts is between 20 and 40 days old—that is, P(20 # x # 40)? b. The hospital administrator is interested in sending reminder letters to the oldest 15%

of accounts. how many days old should an account be before a reminder letter is sent? c. The hospital administrator wants to give a discount to those accounts that pay their balance

by the twenty-first day. what percentage of the accounts will receive the discount?

27. To boost holiday sales, a jewelry store in Bismarck, North Dakota, is advertising the following promotion: “If more than seven inches of cumulative snow fall on December 24, 25, 26, 27, and 28, you get your money back on all purchase made on December 17.” To analyze this promotion, the store manager has collected data and determined that

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snowfall over this 5-day period in December is normally distributed with an average of 6 inches and standard deviation of 0.559 inches. what is the probability that the store will have to refund the money to its December 17 customers?

28. A machine fills containers with a particular product. The standard deviation of filling weights computed from past data is 0.6 ounces. If only 2% of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is, what must m equal? Assume that the filling weights have a normal distribution.

29. Consider the exponential probability density function:

f sxd 5 1

3 e2x/3 for x $ 0

a. write the formula for P(x # x0). b. find P(x # 2). c. find P(x $ 3). d. find P(x # 5). e. find P(2 # x # 5).

30. CopRadar.com reports that the average time for an average driver to react to a red traffic light (which includes recognizing that the light has changed, deciding to stop or to con- tinue, and engaging the brake if stopping) is 2.3 seconds. Assume the time to react to a red traffic light follows an exponential distribution. a. what is the probability that it will take less than 2 seconds for a driver to react to a red

traffic light? b. what is the probability that it will take more than 3 seconds for a driver to react to a

red traffic light? c. what is the probability that it will take between 2 and 3 seconds for a driver to react

to a red traffic light?

31. The time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 12 seconds. a. Sketch this exponential probability distribution. b. what is the probability that the time between vehicle arrivals is 12 seconds or less? c. what is the probability that the time between vehicle arrivals is 6 seconds or less? d. what is the probability that there will be 30 or more seconds between arriving vehicles?

32. The lifetime (hours) of an electronic device is a random variable with the exponential probability density function:

f sxd 5 1

50 e2x/50 for x $ 0

a. what is the mean lifetime of the device? b. what is the probability that the device fails in the first 25 hours of operation? c. what is the probability that the device operates 100 or more hours before failure?

33. The time (in minutes) between telephone calls at an insurance claims office has the expo- nential probability distribution:

f sxd 5 0.50e20.50x for x $ 0 a. what is the mean time between telephone calls? b. what is the probability of 30 seconds or less between telephone calls? c. what is the probability of 1 minute or less between telephone calls? d. what is the probability of 5 or more minutes without a telephone call?

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Case Problem Specialty toys 99

34. Sparagowski & Associates conducted a study of service times at the drive-up window of fast-food restaurants. The average time between placing an order and receiving the order at McDonald’s restaurants was 2.78 minutes. waiting times, such as these, frequently follow an exponential distribution. a. what is the probability that a customer’s service time is less than 2 minutes? b. what is the probability that a customer’s service time is more than 5 minutes? c. what is the probability that a customer’s service time is more than 2.78 minutes?

Case problem Specialty Toys Specialty Toys, Inc., sells a variety of new and innovative children’s toys and believes that the preholiday season is the best time to introduce a new toy. Many families use this time to look for new ideas for December holiday gifts. when Specialty has a new toy with good market potential, it chooses an October market entry date.

In order to get toys in its stores by October, Specialty places one-time orders with its manufacturers in June or July of each year. Demand for children’s toys can be highly volatile. If a new toy catches on, a sense of shortage in the marketplace often increases the demand to very high levels and large profits can be realized. On the other hand, new toys can also flop, leaving Specialty stuck with high levels of inventory that must be sold at reduced prices. The most important question the company faces is deciding how many units of a new toy should be purchased to meet expected sales demand. If too few are purchased, sales will be lost; if too many are purchased, profits will be reduced because of low prices realized in clearance sales.

for the coming season, Specialty plans to introduce a new product called weather Teddy. This variation of a talking teddy bear is made by a company in Taiwan. when a child presses Teddy’s hand, the bear begins to talk. with the aid of a built-in barometer, Teddy says one of five responses that predict the weather conditions. The responses range from “It looks to be a very nice day! have fun” to “I think it may rain today. Don’t forget your umbrella.” Tests with the product show that even though it is not a perfect weather predic- tor, its predictions are surprisingly good. Several of Specialty’s managers claimed Teddy gave predictions of the weather that were as good as the local television weather forecasters.

Specialty faces the decision of how many weather Teddy units to order for the com- ing holiday season. Members of the management team recommended order quantities of 15,000, 18,000, 24,000, and 28,000. Considerable disagreement concerning the market potential is evidenced by the different order quantities suggested. The product management team has asked you for an analysis of the stock-out probabilities for various order quantities, an estimate of the profit potential, and help in making an order quantity recommendation. Specialty expects to sell weather Teddy for $24, and the cost is $16 per unit. If inventory remains after the holiday season, Specialty will sell all surplus inventory for $5 per unit. After reviewing the sales history of similar products, Specialty’s senior sales forecaster predicted an expected demand of 20,000 units with a 0.95 probability that demand would be between 10,000 units and 30,000 units.

Managerial Report Prepare a managerial report that addresses the following issues and recommends an order quantity for the weather Teddy product:

1. Use the sales forecaster’s prediction to describe a normal probability distribution that can be used to approximate the demand distribution. Sketch the distribution and show its mean and standard deviation.

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2. Compute the probability of a stock-out for the order quantities suggested by mem- bers of the management team.

3. Compute the projected profit for the order quantities suggested by the management team under three scenarios. worst case: sales 5 10,000 units; most likely case: sales 5 20,000 units; and best case: sales 5 30,000 units.

4. One of Specialty’s managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs. what quantity would be ordered under this policy, and what is the projected profit under the three scenarios in part 3?

5. Provide your own recommendation for an order quantity and note the associated profit projections. Provide a rationale for your recommendation.

Appendix 3.1 Computing Discrete Probabilities with Excel Excel has the capability of computing probabilities for several discrete probability distri- butions including the binomial and Poisson. In this appendix, we describe how Excel can be used to compute the probabilities for any binomial probability distribution. The procedures for the Poisson probability distributions are similar to the one we describe for the binomial probability distribution.

Let us return to the Nastke Clothing Store problem, where the binomial probabilities of interest are based on a binomial experiment with n 5 10 and p 5 0.30. we assume that the user is interested in the probability of x 5 4 successes in the 10 trials. The following steps describe how to use Excel to produce the desired binomial probability:

Step 1. Select a cell in the worksheet where you want the binomial probability to appear

Step 2. Select the Formulas tab (see Appendix A) Step 3. Choose the insert Function option Step 4. when the insert Function dialog box appears:

Choose Statistical from the Or select a category box Choose BiNOM.diST from the Select a function box Click OK

Step 5. when the Function Arguments dialog box appears: Enter 4 in the Number_s box (the value of x) Enter 10 in the Trials box (the value of n) Enter 0.30 in the probability_s box (the value of p) Enter false in the Cumulative box2

Note: At this point the desired binomial probability of 0.2001 is automatically computed and appears near the bottom of the dialog box.

Click OK and the binomial probability will appear in the worksheet cell requested in Step 1.

A user who wants other binomial probabilities may obtain the information without repeating the steps for each probability desired. Perhaps the easiest alternative is to stay in step 5. After the four entries have been made and the first probability appears, simply return to the Number_s box and insert a new value of x. The new probability will appear. Repeated changes can be made in the dialog box, including changes to the trials, probability,

2Placing false in the cumulative box provides the probability of exactly four successes. Placing true in this box provides the cumulative probability of four or fewer successes.

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Appendix 3.2 Computing Probabilities for Continuous Distributions with Excel 101

and/or cumulative boxes. for each change, the desired probability will appear. when OK is selected, only the last binomial probability will be placed in the worksheet.

If the user wants to insert multiple binomial probabilities into the worksheet, the desired values of x are entered into the worksheet first. Then, in step 5, the user enters the cell loca- tion of one of the values of x in the numbers box. After completing the steps for one binomial probability, individuals experienced with Excel can use Excel’s Copy command to copy the binomial function into the cells where the other binomial probabilities are to appear.

The Excel procedure for generating Poisson probabilities is similar to the procedure just described. Step 4 can be used to select the pOiSSON.diST function name. The dialog box in step 5 will guide the user through the input values required to compute the desired probabilities.

Appendix 3.2 Computing Probabilities for Continuous Distributions with Excel Excel has the capability of computing probabilities for several continuous probability dis- tributions, including the normal and exponential probability distributions. In this appendix, we describe how Excel can be used to compute probabilities for any normal probability distribution. The procedures for the exponential and other continuous probability distribu- tions are similar to the one we describe for the normal probability distribution.

Let us return to the Grear Tire Company problem, where the tire mileage was described by a normal probability distribution with m 5 36,500 and s 5 5000. Assume we are in- terested in the probability that tire mileage will exceed 40,000 miles. The following steps describe how to use Excel to compute the desired normal probability:

Step 1. Select a cell in the worksheet where you want the normal probability to appear Step 2. Select the Formulas tab (see Appendix A) Step 3. Choose the insert Function option Step 4. when the insert Function dialog box appears:

Choose Statistical from the Or select a category box Choose NORM.diST from the Select a function box Click OK

Step 5. when the Function Arguments dialog box appears: Enter 40000 in the x box Enter 36500 in the Mean box Enter 5000 in the Standard_dev box Enter true in the Cumulative box Click OK

At this point, 0.7580 will appear in the cell selected in step 1, indicating that the cumulative probability that the tire mileage is less than or equal to 40,000 miles is 0.7580. Therefore, the probability that tire mileage will exceed 40,000 miles is 1 2 0.7580 5 0.2420.

Excel uses an inverse computation to convert a given cumulative normal probability into a value for the random variable. for example, what mileage guarantee should Grear offer if the company wants no more than 10% of the tires to be eligible for the guarantee? To compute the mileage guarantee by using Excel, follow the procedure just described. however, two changes are necessary: In step 4, choose NORM.iNV from the Select a function box; in step 5, enter the cumulative probability of 0.10 in the probability box and then enter the

The Excel functions NORM.DIST, NORM.INV, and EXPON.DIST are only recognized by Excel 2010. Earlier versions of Excel will use the function names NORMDIST, NORM- INV, and EXPONDIST, respectively, to compute the same values using the same steps.

The Excel functions BINOM.DIST and POISSON.DIST are only recognized by Excel 2010. Earlier versions of Excel will use the function names BINOMDIST and POISSON, respectively, to compute the same values using the same steps.

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mean and the standard deviation. when OK is selected in step 5, the tire mileage guarantee of 30,092, or approximately 30,100, miles appears in the worksheet.

The Excel procedure for generating exponential probabilities is similar to the procedure just described. Step 4 can be used to choose the ExpON.diST function name. The dialog box in step 5 will guide the user through the input values required to compute the desired probability. Note that the value entered in the Lambda box is 1/m. when OK is selected in step 5, the cumulative exponential probability appears in the worksheet.

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Decision Analysis

CONTENTS

4.1 PROBLEM FORMULATION Influence Diagrams Payoff Tables Decision Trees

4.2 DECISION MAKING WITHOUT PROBABILITIES Optimistic Approach Conservative Approach Minimax Regret Approach

4.3 DECISION MAKING WITH PROBABILITIES Expected Value of Perfect

Information

4.4 RISK ANALYSIS AND SENSITIVITY ANALYSIS Risk Analysis Sensitivity Analysis

4.5 DECISION ANALYSIS WITH SAMPLE INFORMATION Influence Diagram Decision Tree Decision Strategy Risk Profile Expected Value of Sample

Information Efficiency of Sample Information

4.6 COMPUTING BRANCH PROBABILITIES WITH BAYES’ THEOREM

AppENdix 4.1 USING ANALYTIC SOLVER PLATFORM TO CREATE DECISION TREES

CHAPTER 4

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Decision analysis can be used to develop an optimal strategy when a decision maker is faced with several decision alternatives and an uncertain or risk-filled pattern of future events. For example, Ohio Edison used decision analysis to choose the best type of particulate control equipment for coal-fired generating units when it faced future uncertainties concerning sul- fur content restrictions, construction costs, and so on. The State of North Carolina used deci- sion analysis in evaluating whether to implement a medical screening test to detect metabolic disorders in newborns. The Q.M. in Action, Natural Resource Management, discusses the use of decision analysis to evaluate alternative actions to protect endangered species.

Even when a careful decision analysis has been conducted, the uncertain future events make the final consequence uncertain. In some cases, the selected decision alternative may provide good or excellent results. In other cases, a relatively unlikely future event may occur, causing the selected decision alternative to provide only fair or even poor results. The risk associated with any decision alternative is a direct result of the uncertainty associated with the final consequence. A good decision analysis includes careful consideration of risk. Through risk analysis the decision maker is provided with probability information about the favorable as well as the unfavorable consequences that may occur.

We begin the study of decision analysis by considering problems that involve reason- ably few decision alternatives and reasonably few possible future events. Influence dia- grams and payoff tables are introduced to provide a structure for the decision problem and

Caution must be exercised when making decisions on what protective measures are taken to protect an endan- gered or threatened species. A conservative action may not be sufficient to save the species, while an aggres- sive action may have serious economic consequences— decision analysis has long been used to strike a balance between these two concerns. However, in recent years policy analysts have been giving increasing consider- ation to another issue: the potential deleterious long-run effects the decision ultimately may have on the endan- gered or threatened species’ ecosystem. Conservation- ists and policy analysts are now recognizing that the resilience of an ecological system, or the degree of dis- turbance that an ecological system can absorb without changing substantially, must be an important consider- ation when making these decisions.

In research funded by the U.S. Geological Survey and the U.S. Fish and Wildlife Service, B. Ken Williams of the Wildlife Society and Fred A. Johnson and James D. Nichols of the U.S. Geological Survey have devel- oped a means for using decision analysis that considers resilience of an ecological system when assessing alter- native strategies for protecting an endangered or threat- ened species. Another strategy, transplanting members of an endangered or threatened species to areas outside the species’ established habitat, is assessed with decision analysis by Tracy Rout of the Quantitative and Applied Ecology Group at the University of Melbourne.

Although the resilience of the ecological system and the intended ecological and social benefits of various strategies for protecting a species are difficult to mea- sure, this approach strives to consider them when select- ing from various alternative strategies. Incorporating the resilience of the ecological system into decision analy- sis of alternative strategies for protecting endangered and threatened species promises to lead to actions that simultaneously enhance the probability of the species’ survival and reduce the risk to the ecological system.

NATURAL RESOURCE MANAGEMENT*

Q.M. in ACtion

*Based on Fred A. Johnson, B. Ken Williams, and James D. nichols, “Resil- ience thinking and a Decision-Analytic Approach to Conservation: Strange Bedfellows or Essential Partners?” Ecology and Society 17(4): 28 and tracy M. Rout, Eve McDonald-Madden, tara G. Martin, nicola J. Mitchell, Hugh P. Possingham, and Doug P. Armstrong, “How to Decide Whether to Move Species threatened by Climate Change.” PLoS ONE 8(10).

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4.1 Problem Formulation 105

to illustrate the fundamentals of decision analysis. We then introduce decision trees to show the sequential nature of decision problems. Decision trees are used to analyze more complex problems and to identify an optimal sequence of decisions, referred to as an optimal decision strategy. Sensitivity analysis shows how changes in various aspects of the problem affect the recommended decision alternative.

4.1 Problem Formulation The first step in the decision analysis process is problem formulation. We begin with a verbal statement of the problem. We then identify the decision alternatives; the uncertain future events, referred to as chance events; and the consequences associated with each combination of decision alternative and chance event outcome. Let us begin by considering a construction project of the Pittsburgh Development Corporation.

Pittsburgh Development Corporation (PDC) purchased land that will be the site of a new luxury condominium complex. The location provides a spectacular view of downtown Pittsburgh and the Golden Triangle, where the Allegheny and Monongahela Rivers meet to form the Ohio River. PDC plans to price the individual condominium units between $300,000 and $1,400,000.

PDC commissioned preliminary architectural drawings for three different projects: one with 30 condominiums, one with 60 condominiums, and one with 90 condominiums. The financial success of the project depends upon the size of the condominium complex and the chance event concerning the demand for the condominiums. The statement of the PDC decision problem is to select the size of the new luxury condominium project that will lead to the largest profit given the uncertainty concerning the demand for the condominiums.

Given the statement of the problem, it is clear that the decision is to select the best size for the condominium complex. PDC has the following three decision alternatives:

d1 5 a small complex with 30 condominiums

d2 5 a medium complex with 60 condominiums

d3 5 a large complex with 90 condominiums

A factor in selecting the best decision alternative is the uncertainty associated with the chance event concerning the demand for the condominiums. When asked about the possible demand for the condominiums, PDC’s president acknowledged a wide range of possibili- ties but decided that it would be adequate to consider two possible chance event outcomes: a strong demand and a weak demand.

In decision analysis, the possible outcomes for a chance event are referred to as the states of nature. The states of nature are defined so they are mutually exclusive (no more than one can occur) and collectively exhaustive (at least one must occur); thus one and only one of the possible states of nature will occur. For the PDC problem, the chance event concerning the demand for the condominiums has two states of nature:

s1 5 strong demand for the condominiums

s2 5 weak demand for the condominiums

Management must first select a decision alternative (complex size); then a state of na- ture follows (demand for the condominiums) and finally a consequence will occur. In this case, the consequence is PDC’s profit.

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Influence Diagrams An influence diagram is a graphical device that shows the relationships among the deci- sions, the chance events, and the consequences for a decision problem. The nodes in an influence diagram represent the decisions, chance events, and consequences. Rectangles or squares depict decision nodes, circles or ovals depict chance nodes, and diamonds depict consequence nodes. The lines connecting the nodes, referred to as arcs, show the direction of influence that the nodes have on one another. Figure 4.1 shows the influence diagram for the PDC problem. The complex size is the decision node, demand is the chance node, and profit is the consequence node. The arcs connecting the nodes show that both the complex size and the demand influence PDC’s profit.

Payoff Tables Given the three decision alternatives and the two states of nature, which complex size should PDC choose? To answer this question, PDC will need to know the consequence associated with each decision alternative and each state of nature. In decision analysis, we refer to the consequence resulting from a specific combination of a decision alternative and a state of nature as a payoff. A table showing payoffs for all combinations of decision alternatives and states of nature is a payoff table.

Because PDC wants to select the complex size that provides the largest profit, profit is used as the consequence. The payoff table with profits expressed in millions of dollars is shown in Table 4.1. Note, for example, that if a medium complex is built and demand turns out to be strong, a profit of $14 million will be realized. We will use the notation Vij to de- note the payoff associated with decision alternative i and state of nature j. Using Table 4.1, V31 5 20 indicates a payoff of $20 million occurs if the decision is to build a large complex (d3) and the strong demand state of nature (s1) occurs. Similarly, V32 5 29 indicates a loss of $9 million if the decision is to build a large complex (d3) and the weak demand state of nature (s2) occurs.

Payoffs can be expressed in terms of profit, cost, time, distance, or any other measure appropriate for the decision problem being analyzed.

Complex Size

Demand

Pro�t

Decision Alternatives Small complex (d1) Medium complex (d2) Large complex (d3)

States of Nature Strong (s1) Weak (s2)

Consequence Pro�t

FIGURE 4.1 INFLUENCE DIAGRAM FOR THE PDC PROJECT

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Decision Trees A decision tree provides a graphical representation of the decision-making process. Figure 4.2 presents a decision tree for the PDC problem. Note that the decision tree shows the natural or logical progression that will occur over time. First, PDC must make a deci- sion regarding the size of the condominium complex (d1, d2, or d3). Then, after the decision is implemented, either state of nature s1 or s2 will occur. The number at each endpoint of the tree indicates the payoff associated with a particular sequence. For example, the top- most payoff of 8 indicates that an $8 million profit is anticipated if PDC constructs a small condominium complex (d1) and demand turns out to be strong (s1). The next payoff of 7 indicates an anticipated profit of $7 million if PDC constructs a small condominium com- plex (d1) and demand turns out to be weak (s2). Thus, the decision tree provides a graphical depiction of the sequences of decision alternatives and states of nature that provide the six possible payoffs for PDC.

The decision tree in Figure 4.2 shows four nodes, numbered 124. Squares are used to depict decision nodes and circles are used to depict chance nodes. Thus, node 1 is a deci- sion node, and nodes 2, 3, and 4 are chance nodes. The branches connect the nodes; those

If you have a payoff table, you can develop a decision tree. Try Problem 1, part (a).

State of Nature

decision Alternative Strong demand s1 Weak demand s2 Small complex, d1 8 7 Medium complex, d2 14 5 Large complex, d3 20 29

TABLE 4.1 PAYOFF TABLE FOR THE PDC CONDOMINIUM PROJECT (PAYOFFS IN $ MILLIONS)

Weak (s2)

Strong (s1)

Weak (s2)

Strong (s1)

Weak (s2)

Strong (s1)

–9

Small (d1)

Medium (d2)

Large (d3)

FIGURE 4.2 DECISION TREE FOR THE PDC CONDOMINIUM PROJECT (PAYOFFS IN $ MILLIONS)

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108 Chapter 4 Decision Analysis

leaving the decision node correspond to the decision alternatives. The branches leaving each chance node correspond to the states of nature. The payoffs are shown at the end of the states-of-nature branches. We now turn to the question: How can the decision maker use the information in the payoff table or the decision tree to select the best decision alternative? Several approaches may be used.

The Q.M. in Action, Using Decision Analysis to Decide Whether to Transplant Endangered or Threatened Species, describes how decision analysis is used to assess whether it is advisable to transplant members of an endangered or threatened species into a new habitat.

NOTES AND COMMENTS

1. The first step in solving a complex problem is to decompose the problem into a series of smaller subproblems. Decision trees provide a useful way to decompose a problem and illustrate the sequential nature of the decision process.

2. People often view the same problem from different perspectives. Thus, the discussion regarding the development of a decision tree may provide additional insight about the problem.

One strategy for attempting to prevent the extinction of an endangered or threatened species is to transplant members of the species to areas outside the species’ established habitat. This strategy, which is often re- ferred to as assisted migration, is controversial not only because it is difficult to assess the likelihood that the transplanted species will survive in new habitats, but also because it is difficult to anticipate the impact transplanted species will have on other species in its new habitat if it survives. Ecosystems are extremely complex, and a species that is not native to a particular habitat can potentially devastate the species that are na- tive to the habitat.

Tracy Rout, a post-doctoral research fellow in the Quantitative and Applied Ecology Group at the Univer- sity of Melbourne, uses decision analysis to address this issue. In order to make an informed decision on whether to transplant a particular endangered or threatened species to a new habitat, she considers the likelihood the trans- planted species will survive in the new habitat, the impact the transplanted species will have on its new habitat if

it survives, the probability the endangered or threatened species will become extinct if not transplanted, and the costs associated with the extinction of this species. Her decision tree features a single decision node for whether to transplant the endangered or threatened species. If the species is not transplanted, two states of nature can occur: the species becomes extinct or the species survives. If the species is transplanted, the first ensuing state of nature is whether the species survives in its new habitat. If the spe- cies does survive its introduction into the new habitat, the next state of nature is the impact the introduction of the new species has on its new habitat.

It is difficult to determine the values of inputs into this decision tree, but once reliable estimates of these val- ues are developed, the decision tree provides important information that can be used to make better decisions about whether to transplant an endangered or threatened species into a new habitat. If there are several new habitats under consideration, the decision tree can be applied to each to help decision makers understand which of these new habitats is most promising.

USING DECISION ANALYSIS TO DECIDE WHETHER TO TRANSPLANT ENDANGERED OR THREATENED SPECIES*

Q.M. in ACtion

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4.2 Decision Making Without Probabilities 109

4.2 Decision Making Without Probabilities In this section we consider approaches to decision making that do not require knowledge of the probabilities of the states of nature. These approaches are appropriate in situations in which the decision maker has little confidence in his or her ability to assess the probabili- ties, or in which a simple best-case and worst-case analysis is desirable. Because different approaches sometimes lead to different decision recommendations, the decision maker must understand the approaches available and then select the specific approach that, ac- cording to the judgment of the decision maker, is the most appropriate.

Optimistic Approach The optimistic approach evaluates each decision alternative in terms of the best payoff that can occur. The decision alternative that is recommended is the one that provides the best possible payoff. For a problem in which maximum profit is desired, as in the PDC problem, the optimistic approach would lead the decision maker to choose the alternative corresponding to the largest profit. For problems involving minimization, this approach leads to choosing the alternative with the smallest payoff.

To illustrate the optimistic approach, we use it to develop a recommendation for the PDC problem. First, we determine the maximum payoff for each decision alternative; then we select the decision alternative that provides the overall maximum payoff. These steps systematically identify the decision alternative that provides the largest possible profit. Table 4.2 illustrates these steps.

Because 20, corresponding to d3, is the largest payoff, the decision to construct the large condominium complex is the recommended decision alternative using the optimistic approach.

Conservative Approach The conservative approach evaluates each decision alternative in terms of the worst pay- off that can occur. The decision alternative recommended is the one that provides the best of the worst possible payoffs. For a problem in which the output measure is profit, as in the PDC problem, the conservative approach would lead the decision maker to choose the alternative that maximizes the minimum possible profit that could be obtained. For prob- lems involving minimization, this approach identifies the alternative that will minimize the maximum payoff.

To illustrate the conservative approach, we use it to develop a recommendation for the PDC problem. First, we identify the minimum payoff for each of the decision alternatives; then we select the decision alternative that maximizes the minimum payoff. Table 4.3 illus- trates these steps for the PDC problem.

For a maximization problem, the optimistic approach often is referred to as the maximax approach; for a minimization problem, the corresponding terminology is minimin.

For a maximization problem, the conservative approach is often referred to as the maximin approach; for a minimization problem, the corresponding terminol- ogy is minimax.

Many people think of a good decision as one in which the consequence is good. However, in some instances, a good, well-thought-out decision may still lead to a bad or undesirable consequence while a poor, ill-conceived decision may still lead to a good or desirable consequence.

decision Alternative Maximum payoff Small complex, d1 8 Medium complex, d2 14 Large complex, d3 20

TABLE 4.2 MAXIMUM PAYOFF FOR EACH PDC DECISION ALTERNATIVE

Maximum of the maxi- mum payoff values

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Because 7, corresponding to d1, yields the maximum of the minimum payoffs, the decision alternative of a small condominium complex is recommended. This decision ap- proach is considered conservative because it identifies the worst possible payoffs and then recommends the decision alternative that avoids the possibility of extremely “bad” payoffs. In the conservative approach, PDC is guaranteed a profit of at least $7 million. Although PDC may make more, it cannot make less than $7 million.

Minimax Regret Approach In decision analysis, regret is the difference between the payoff associated with a particular decision alternative and the payoff associated with the decision that would yield the most desirable payoff for a given state of nature. Thus, regret represents how much potential payoff one would forgo by selecting a particular decision alternative given that a specific state of nature will occur. This is why regret is often referred to as opportunity loss.

As its name implies, under the minimax regret approach to decision making one would choose the decision alternative that minimizes the maximum state of regret that could occur over all possible states of nature. This approach is neither purely optimistic nor purely conservative. Let us illustrate the minimax regret approach by showing how it can be used to select a decision alternative for the PDC problem.

Suppose that PDC constructs a small condominium complex (d1) and demand turns out to be strong (s1). Table 4.1 showed that the resulting profit for PDC would be $8 million. However, given that the strong demand state of nature (s1) has occurred, we realize that the decision to construct a large condominium complex (d3), yielding a profit of $20 million, would have been the best decision. The difference between the payoff for the best decision alterna- tive ($20 million) and the payoff for the decision to construct a small condominium complex ($8 million) is the regret or opportunity loss associated with decision alternative d1 when state of nature s1 occurs; thus, for this case, the opportunity loss or regret is $20 million 2 $8 million 5 $12 million. Similarly, if PDC makes the decision to construct a medium condominium com- plex (d2) and the strong demand state of nature (s1) occurs, the opportunity loss, or regret, associated with d2 would be $20 million 2 $14 million 5 $6 million.

In general, the following expression represents the opportunity loss, or regret:

R ij 5 uVj* 2 Viju (4.1)

where

Rij 5 the regret associated with decision alternative di and state of nature sj Vj* 5 the payoff value

1 corresponding to the best decision for the state of nature sj Vij 5 the payoff corresponding to decision alternative di and state of nature sj

decision Alternative Minimum payoff Small complex, d1 7 Medium complex, d2 5 Large complex, d3 29

TABLE 4.3 MINIMUM PAYOFF FOR EACH PDC DECISION ALTERNATIVE

Maximum of the mini- mum payoff values

1in maximization problems, V*j will be the largest entry in column j of the payoff table. in minimization problems, V*j will be the smallest entry in column j of the payoff table.

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4.2 Decision Making Without Probabilities 111

Note the role of the absolute value in equation (4.1). For minimization problems, the best payoff, Vj*, is the smallest entry in column j. Because this value always is less than or equal to Vij, the absolute value of the difference between Vj* and Vij ensures that the regret is always the magnitude of the difference.

Using equation (4.1) and the payoffs in Table 4.1, we can compute the regret associ- ated with each combination of decision alternative di and state of nature sj. Because the PDC problem is a maximization problem, Vj* will be the largest entry in column j of the payoff table. Thus, to compute the regret, we simply subtract each entry in a column from the largest entry in the column. Table 4.4 shows the opportunity loss, or regret, table for the PDC problem.

The next step in applying the minimax regret approach is to list the maximum regret for each decision alternative; Table 4.5 shows the results for the PDC problem. Selecting the decision alternative with the minimum of the maximum regret values—hence, the name minimax regret—yields the minimax regret decision. For the PDC problem, the alternative to construct the medium condominium complex, with a corresponding maximum regret of $6 million, is the recommended minimax regret decision.

Note that the three approaches discussed in this section provide different recom- mendations, which in itself isn’t bad. It simply reflects the difference in decision-making philosophies that underlie the various approaches. Ultimately, the decision maker will have to choose the most appropriate approach and then make the final decision accord- ingly. The main criticism of the approaches discussed in this section is that they do not consider any information about the probabilities of the various states of nature. In the next section we discuss an approach that utilizes probability information in selecting a decision alternative.

For practice in developing a decision recommenda- tion using the optimistic, conservative, and minimax regret approaches, try Problem 1, part (b).

State of Nature

decision Alternative Strong demand s1 Weak demand s2 Small complex, d1 12 0 Medium complex, d2 6 2 Large complex, d3 0 16

TABLE 4.4 OPPORTUNITY LOSS, OR REGRET, TABLE FOR THE PDC CONDOMINIUM PROJECT ($ MILLIONS)

decision Alternative Maximum Regret Small complex, d1 12 Medium complex, d2 6 Large complex, d3 16

TABLE 4.5 MAXIMUM REGRET FOR EACH PDC DECISION ALTERNATIVE

Minimum of the maximum regret

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4.3 Decision Making with Probabilities In many decision-making situations, we can obtain probability assessments for the states of nature. When such probabilities are available, we can use the expected value approach to identify the best decision alternative. Let us first define the expected value of a decision alternative and then apply it to the PDC problem.

Let

N 5 the number of states of nature

P(sj) 5 the probability of state of nature sj

Because one and only one of the N states of nature can occur, the probabilities must satisfy two conditions:

Pssjd $ 0 for all states of nature (4.2) o

j5 1

Pssjd 5 Pss1d 1 Pss2d 1 . . . 1 PssNd 5 1 (4.3)

The expected value (EV) of decision alternative di is defined as follows:

EVsdid 5 o N

j51 PssjdVij (4.4)

In words, the expected value of a decision alternative is the sum of weighted payoffs for the decision alternative. The weight for a payoff is the probability of the associated state of nature and therefore the probability that the payoff will occur. Let us return to the PDC problem to see how the expected value approach can be applied.

PDC is optimistic about the potential for the luxury high-rise condominium complex. Suppose that this optimism leads to an initial subjective probability assessment of 0.8 that de- mand will be strong (s1) and a corresponding probability of 0.2 that demand will be weak (s2). Thus, P(s1) 5 0.8 and P(s2) 5 0.2. Using the payoff values in Table 4.1 and equation (4.4), we compute the expected value for each of the three decision alternatives as follows:

EV(d1) 5 0.8(8) 1 0.2(7) 5 7.8 EV(d2) 5 0.8(14) 1 0.2(5) 5 12.2 EV(d3) 5 0.8(20) 1 0.2(29) 5 14.2

Thus, using the expected value approach, we find that the large condominium complex, with an expected value of $14.2 million, is the recommended decision.

The calculations required to identify the decision alternative with the best expected value can be conveniently carried out on a decision tree. Figure 4.3 shows the decision tree for the PDC problem with state-of-nature branch probabilities. Working backward through the decision tree, we first compute the expected value at each chance node. That is, at each chance node, we weight each possible payoff by its probability of occurrence. By doing so, we obtain the expected values for nodes 2, 3, and 4, as shown in Figure 4.4.

Because the decision maker controls the branch leaving decision node 1 and because we are trying to maximize the expected profit, the best decision alternative at node 1 is d3. Thus, the decision tree analysis leads to a recommendation of d3, with an expected value

Computer packages are available to help in constructing more complex decision trees. See Appendix 4.1.

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4.3 Decision Making with Probabilities 113

of $14.2 million. Note that this recommendation is also obtained with the expected value approach in conjunction with the payoff table.

Other decision problems may be substantially more complex than the PDC problem, but if a reasonable number of decision alternatives and states of nature are present, you can use the decision tree approach outlined here. First, draw a decision tree consisting of deci- sion nodes, chance nodes, and branches that describe the sequential nature of the problem. If you use the expected value approach, the next step is to determine the probabilities for

– 9 Weak (s2)

Strong (s1)

Weak (s2)

Strong (s1)

Weak (s2)

Strong (s1)

Small (d1)

Medium (d2 )

Large (d3)

P(s1) = 0.8

P(s2) = 0.2

P(s1) = 0.8

P(s2) = 0.2

P(s1) = 0.8

P(s2) = 0.2

FIGURE 4.3 PDC DECISION TREE WITH STATE-OF-NATURE BRANCH PROBABILITIES

Small (d1)

Medium (d 2)

Large (d3)

EV(d1) = 0.8(8) + 0.2(7) = $7.8

EV(d2) = 0.8(14) + 0.2(5) = $12.2

EV(d3) = 0.8(20) + 0.2(–9) = $14.2

FIGURE 4.4 APPLYING THE EXPECTED VALUE APPROACH USING A DECISION TREE

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Oil and natural gas are big businesses: nine of the top ten organizations in Fortune’s Global 500 are oil and gas companies. The rewards in these industries can be high, but the associated risks are also great. Oil prices at or above $100 per barrel make an oil reser- voir with potentially 1 million barrels of supply appear to be an enticing development venture. But as Adam Farris, senior vice president of business development for Drillinginfo, explains, realizing this potential $100 million stream of revenue is not simple. He points out that the acquisition, processing, and interpretation of the seismic data necessary to evaluate the potential well before drilling could cost $30 million. Typical deals involve the procurement of access to thousands of acres of land (a single well may require 120 acres), and land can cost $30,000 per acre. Drilling can cost from $5 million to $10 million for U.S. onshore wells and up to $100 million for offshore drilling. The costs

of producing the oil and getting it to market are also substantial. If the well is drilled and does not produce oil, the company has incurred the costs associated with the acquisition, processing, and interpretation of the seismic data, obtaining access to the land, and drilling with no resulting revenue.

“If you are a major integrated oil and gas com- pany, your profit on $100 million will be $1 million to $12 million (a bit higher for independent operators),” states Farris. “Many will lose money overall. Analyti- cal approaches that impact the success rate of finding or reducing the cost to develop and produce oil and gas can make energy more affordable, safer and environmentally conscious.”

Decision analysis provides a means for oil explo- ration companies to assess the complex data in a sys- tematic manner and extract information from the data that ultimately are used to decide whether to drill in a potential well site. Identifying well sites for which the potential gains exceed and justify the risk of drilling is critical to the economic success of these firms.

GUSHERS, DRY WELLS, AND DECISION ANALYSIS*

Q.M. in ACtion

*Based on Adam Farris, “How Big Data is Changing the oil & Gas industry.” Analytics (november/December 2012).

each of the states of nature and compute the expected value at each chance node. Then select the decision branch leading to the chance node with the best expected value. The decision alternative associated with this branch is the recommended decision.

The Q.M. in Action, Gushers, Dry Wells, and Decision Analysis, describes the importance of using data in the oil and natural gas industry to accurately estimate the likelihoods and the profit associated with possible outcomes in order to make wise de- velopment decisions.

Expected Value of Perfect Information Suppose that PDC has the opportunity to conduct a market research study that would help evaluate buyer interest in the condominium project and provide information that manage- ment could use to improve the probability assessments for the states of nature. To determine the potential value of this information, we begin by supposing that the study could provide perfect information regarding the states of nature; that is, we assume for the moment that PDC could determine with certainty, prior to making a decision, which state of nature is going to occur. To make use of this perfect information, we will develop a decision strategy that PDC should follow once it knows which state of nature will occur. A decision strategy is simply a decision rule that specifies the decision alternative to be selected after new information becomes available.

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4.3 Decision Making with Probabilities 115

To help determine the decision strategy for PDC, we reproduced PDC’s payoff table as Table 4.6. Note that, if PDC knew for sure that state of nature s1 would occur, the best decision alternative would be d3, with a payoff of $20 million. Similarly, if PDC knew for sure that state of nature s2 would occur, the best decision alternative would be d1, with a payoff of $7 million. Thus, we can state PDC’s optimal decision strategy when the perfect information becomes available as follows:

If s1, select d3 and receive a payoff of $20 million.

If s2, select d1 and receive a payoff of $7 million.

What is the expected value for this decision strategy? To compute the expected value with perfect information, we return to the original probabilities for the states of nature: P(s1) 5 0.8 and P(s2) 5 0.2. Thus, there is a 0.8 probability that the perfect information will indicate state of nature s1, and the resulting decision alternative d3 will provide a $20 million profit. Similarly, with a 0.2 probability for state of nature s2, the optimal decision alternative d1 will provide a $7 million profit. Thus, from equation (4.4) the expected value of the decision strategy that uses perfect information is 0.8(20) 1 0.2(7) 5 17.4.

We refer to the expected value of $17.4 million as the expected value with perfect information (EVwPI).

Earlier in this section we showed that the recommended decision using the expected value approach is decision alternative d3, with an expected value of $14.2 million. Because this decision recommendation and expected value computation were made without the benefit of perfect information, $14.2 million is referred to as the expected value without perfect information (EVwoPI).

The expected value with perfect information is $17.4 million, and the expected value without perfect information is $14.2; therefore, the expected value of the perfect informa- tion (EVPI) is $17.4 2 $14.2 5 $3.2 million. In other words, $3.2 million represents the additional expected value that can be obtained if perfect information were available about the states of nature.

Generally speaking, a market research study will not provide “perfect” information; however, if the market research study is a good one, the information gathered might be worth a sizable portion of the $3.2 million. Given the EVPI of $3.2 million, PDC might seriously consider a market survey as a way to obtain more information about the states of nature.

In general, the expected value of perfect information (EVpi) is computed as follows:

EVPI 5 uEVwPI 2 EVwoPIu (4.5)

It would be worth $3.2 million for PDC to learn the level of market acceptance before selecting a decision alternative.

State of Nature

decision Alternative Strong demand s1 Weak demand s2 Small complex, d1 8 7 Medium complex, d2 14 5 Large complex, d3 20 29

TABLE 4.6 PAYOFF TABLE FOR THE PDC CONDOMINIUM PROJECT ($ MILLIONS)

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where

EVPI 5 expected value of perfect information

EVwPI 5 expected value with perfect information about the states of nature

EVwoPI 5 expected value without perfect information about the states of nature

Note the role of the absolute value in equation (4.5). For minimization problems, the ex- pected value with perfect information is always less than or equal to the expected value without perfect information. In this case, EVPI is the magnitude of the difference between EVwPI and EVwoPI, or the absolute value of the difference as shown in equation (4.5).

For practice in determining the expected value of perfect information, try Problem 14.

NOTES AND COMMENTS

1. We restate the opportunity loss, or regret, table for the PDC problem (see Table 4.4) as follows:

State of Nature

Strong Weak demand demand decision s1 s2 Small complex, d1 12 0 Medium complex, d2 6 2 Large complex, d3 0 16

Using P(s1), P(s2), and the opportunity loss values, we can compute the expected opportu- nity loss (EOL) for each decision alternative. With P(s1) 5 0.8 and P(s2) 5 0.2, the expected

opportunity loss for each of the three decision alternatives is

EOL(d1) 5 0.8(12) 1 0.2(0) 5 9.6 EOL(d2) 5 0.8(6) 1 0.2(2) 5 5.2 EOL(d3) 5 0.8(0) 1 0.2(16) 5 3.2

Regardless of whether the decision analysis involves maximization or minimization, the minimum expected opportunity loss always pro- vides the best decision alternative. Thus, with EOL(d3) 5 3.2, d3 is the recommended deci- sion. In addition, the minimum expected oppor- tunity loss always is equal to the expected value of perfect information. That is, EOL(best deci- sion) 5 EVPI; for the PDC problem, this value is $3.2 million.

4.4 Risk Analysis and Sensitivity Analysis Risk analysis helps the decision maker recognize the difference between the expected value of a decision alternative and the payoff that may actually occur. Sensitivity analysis also helps the decision maker by describing how changes in the state-of-nature probabilities and/ or changes in the payoffs affect the recommended decision alternative.

Risk Analysis A decision alternative and a state of nature combine to generate the payoff associated with a decision. The risk profile for a decision alternative shows the possible payoffs along with their associated probabilities.

Let us demonstrate risk analysis and the construction of a risk profile by returning to the PDC condominium construction project. Using the expected value approach, we iden- tified the large condominium complex (d3) as the best decision alternative. The expected value of $14.2 million for d3 is based on a 0.8 probability of obtaining a $20 million

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4.4 Risk Analysis and Sensitivity Analysis 117

profit and a 0.2 probability of obtaining a $9 million loss. The 0.8 probability for the $20 million payoff and the 0.2 probability for the 2$9 million payoff provide the risk profile for the large complex decision alternative. This risk profile is shown graphically in Figure 4.5.

Sometimes a review of the risk profile associated with an optimal decision alterna- tive may cause the decision maker to choose another decision alternative even though the expected value of the other decision alternative is not as good. For example, the risk profile for the medium complex decision alternative (d2) shows a 0.8 probability for a $14 million payoff and a 0.2 probability for a $5 million payoff. Because no prob- ability of a loss is associated with decision alternative d2, the medium complex decision alternative would be judged less risky than the large complex decision alternative. As a result, a decision maker might prefer the less risky medium complex decision alternative even though it has an expected value of $2 million less than the large complex decision alternative.

Sensitivity Analysis Sensitivity analysis can be used to determine how changes in the probabilities for the states of nature or changes in the payoffs affect the recommended decision alternative. In many cases, the probabilities for the states of nature and the payoffs are based on subjective as- sessments. Sensitivity analysis helps the decision maker understand which of these inputs are critical to the choice of the best decision alternative. If a small change in the value of one of the inputs causes a change in the recommended decision alternative, the solution to the decision analysis problem is sensitive to that particular input. Extra effort and care should be taken to make sure the input value is as accurate as possible. On the other hand, if a modest-to-large change in the value of one of the inputs does not cause a change in the recommended decision alternative, the solution to the decision analysis problem is not sensitive to that particular input. No extra time or effort would be needed to refine the estimated input value.

1.0

–10 0 10 20 Pro�t ($ millions)

P ro

b ab

il it

FIGURE 4.5 RISK PROFILE FOR THE LARGE COMPLEX DECISION ALTERNATIVE FOR THE PDC CONDOMINIUM PROJECT

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One approach to sensitivity analysis is to select different values for the probabilities of the states of nature and the payoffs and then resolve the decision analysis problem. If the recommended decision alternative changes, we know that the solution is sensitive to the changes made. For example, suppose that in the PDC problem the probability for a strong demand is revised to 0.2 and the probability for a weak demand is revised to 0.8. Would the recommended decision alternative change? Using P(s1) 5 0.2, P(s2) 5 0.8, and equation (4.4), the revised expected values for the three decision alternatives are

EV(d1) 5 0.2(8) 1 0.8(7) 5 7.2

EV(d2) 5 0.2(14) 1 0.8(5) 5 6.8

EV(d3) 5 0.2(20) 1 0.8(29) 5 23.2

With these probability assessments, the recommended decision alternative is to construct a small condominium complex (d1), with an expected value of $7.2 million. The probability of strong demand is only 0.2, so constructing the large condominium complex (d3) is the least preferred alternative, with an expected value of 2$3.2 million (a loss).

Thus, when the probability of strong demand is large, PDC should build the large com- plex; when the probability of strong demand is small, PDC should build the small complex. Obviously, we could continue to modify the probabilities of the states of nature and learn even more about how changes in the probabilities affect the recommended decision alterna- tive. The drawback to this approach is the numerous calculations required to evaluate the effect of several possible changes in the state-of-nature probabilities.

For the special case of two states of nature, a graphical procedure can be used to de- termine how changes for the probabilities of the states of nature affect the recommended decision alternative. To demonstrate this procedure, we let p denote the probability of state of nature s1; that is, P(s1) 5 p. With only two states of nature in the PDC problem, the prob- ability of state of nature s2 is

P(s2) 5 1 2 P(s1) 5 1 2 p

Using equation (4.4) and the payoff values in Table 4.1, we determine the expected value for decision alternative d1 as follows:

EVsd1d 5 5

Pss1ds8d 1 Pss2ds7d ps8d 1 s1 2 pds7d 8p 1 7 2 7p 5 p 1 7

(4.6)

Repeating the expected value computations for decision alternatives d2 and d3, we obtain expressions for the expected value of each decision alternative as a function of p:

EV(d2) 5 9p 1 5 (4.7)

EV(d3) 5 29p 2 9 (4.8)

Thus, we have developed three equations that show the expected value of the three decision alternatives as a function of the probability of state of nature s1.

We continue by developing a graph with values of p on the horizontal axis and the associated EVs on the vertical axis. Because equations (4.6), (4.7), and (4.8) are linear equa- tions, the graph of each equation is a straight line. For each equation, we can obtain the line

Computer software packages for decision analysis make it easy to calculate these revised scenarios.

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4.4 Risk Analysis and Sensitivity Analysis 119

by identifying two points that satisfy the equation and drawing a line through the points. For instance, if we let p 5 0 in equation (4.6), EV(d1) 5 7. Then, letting p 5 1, EV(d1) 5 8. Connecting these two points, (0,7) and (1,8), provides the line labeled EV(d1) in Figure 4.6. Similarly, we obtain the lines labeled EV(d2) and EV(d3); these lines are the graphs of equa- tions (4.7) and (4.8), respectively.

Figure 4.6 shows how the recommended decision changes as p, the probability of the strong demand state of nature (s1), changes. Note that for small values of p, decision alterna- tive d1 (small complex) provides the largest expected value and is thus the recommended decision. When the value of p increases to a certain point, decision alternative d2 (medium complex) provides the largest expected value and is the recommended decision. Finally, for large values of p, decision alternative d3 (large complex) becomes the recommended decision.

The value of p for which the expected values of d1 and d2 are equal is the value of p cor- responding to the intersection of the EV(d1) and the EV(d2) lines. To determine this value, we set EV(d1) 5 EV(d2) and solve for the value of p:

p 1 7 5

8p 5

p 5

9p 1 5

8 5 0.25

-5

-10

0.2 0.4 0.80.6 1.0 p

d1 provides the highest EV

d2 provides the highest EV

d3 provides the highest EV

EV (d 3

)

EV(d2 )

EV(d1)

E xp

ec te

d V

al u

e (E

V )

FIGURE 4.6 EXPECTED VALUE FOR THE PDC DECISION ALTERNATIVES AS A FUNCTION OF p

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120 Chapter 4 Decision Analysis

Hence, when p 5 0.25, decision alternatives d1 and d2 provide the same expected value. Repeating this calculation for the value of p corresponding to the intersection of the EV(d2) and EV(d3) lines, we obtain p 5 0.70.

Using Figure 4.6, we can conclude that decision alternative d1 provides the largest expected value for p # 0.25, decision alternative d2 provides the largest expected value for 0.25 # p # 0.70, and decision alternative d3 provides the largest expected value for p $ 0.70. Because p is the probability of state of nature s1 and (1 2 p) is the probability of state of nature s2, we now have the sensitivity analysis information that tells us how changes in the state-of-nature probabilities affect the recommended decision alternative.

Sensitivity analysis calculations can also be made for the values of the payoffs. In the original PDC problem, the expected values for the three decision alternatives were as follows: EV(d1) 5 7.8, EV(d2) 5 12.2, and EV(d3) 5 14.2. Decision alternative d3 (large complex) was recommended. Note that decision alternative d2 with EV(d2) 5 12.2 was the second best decision alternative. Decision alternative d3 will remain the optimal decision alternative as long as EV(d3) is greater than or equal to the expected value of the second best decision alternative. Thus, decision alternative d3 will remain the optimal decision alternative as long as

EV(d3) $ 12.2 (4.9)

Let

S 5 the payoff of decision alternative d3 when demand is strong

W 5 the payoff of decision alternative d3 when demand is weak

Using P(s1) 5 0.8 and P(s2) 5 0.2, the general expression for EV(d3) is

EV(d3) 5 0.8S 1 0.2W (4.10)

Assuming that the payoff for d3 stays at its original value of 2$9 million when demand is weak, the large complex decision alternative will remain optimal as long as

EV(d3) 5 0.8S 1 0.2(29) $ 12.2 (4.11)

Solving for S, we have

0.8S 2 1.8 $ 12.2

0.8S $ 14

S $ 17.5

Recall that when demand is strong, decision alternative d3 has an estimated payoff of $20 million. The preceding calculation shows that decision alternative d3 will remain opti- mal as long as the payoff for d3 when demand is strong is at least $17.5 million.

Assuming that the payoff for d3 when demand is strong stays at its original value of $20 million, we can make a similar calculation to learn how sensitive the optimal solution is with regard to the payoff for d3 when demand is weak. Returning to the expected value calculation of equation (4.10), we know that the large complex decision alternative will remain optimal as long as

EV(d3) 5 0.8(20) 1 0.2W $ 12.2 (4.12)

Graphical sensitivity analysis shows how changes in the probabilities for the states of nature affect the recommended decision alternative. Try Problem 8.

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4.4 Risk Analysis and Sensitivity Analysis 121

Solving for W, we have

16 1 0.2 $ 12.2

0.2W $ 23.8

W $ 219

Recall that when demand is weak, decision alternative d3 has an estimated payoff of 2$9 million. The preceding calculation shows that decision alternative d3 will remain optimal as long as the payoff for d3 when demand is weak is at least 2$19 million.

Based on this sensitivity analysis, we conclude that the payoffs for the large complex decision alternative (d3) could vary considerably, and d3 would remain the recommended decision alternative. Thus, we conclude that the optimal solution for the PDC decision problem is not particularly sensitive to the payoffs for the large complex decision alterna- tive. We note, however, that this sensitivity analysis has been conducted based on only one change at a time. That is, only one payoff was changed and the probabilities for the states of nature remained P(s1) 5 0.8 and P(s2) 5 0.2. Note that similar sensitivity analysis calcula- tions can be made for the payoffs associated with the small complex decision alternative d1 and the medium complex decision alternative d2. However, in these cases, decision alterna- tive d3 remains optimal only if the changes in the payoffs for decision alternatives d1 and d2 meet the requirements that EV(d1) # 14.2 and EV(d2) # 14.2.

The Q.M. in Action, Reducing the Threat of Terrorism Through Decision Analysis, describes how U.S. Customs and Border Protection uses decision analysis to reduce the risk of terrorism in commercial truck crossings at land ports of entry in the southwestern United States.

Sensitivity analysis can assist management in deciding whether more time and effort should be spent obtaining better estimates of payoffs and probabilities.

Border security is a critical component of a nation’s security. For a country the size of the United States, border crossings are particularly difficult to patrol, monitor, and control. The U.S. Customs and Border Protection (CBP), the federal government agency responsible for border management and control, is one of the world’s largest law enforcement organizations. CBP has over 60,000 employees who routinely receive approximately 1 million visitors, screen almost 70,000 cargo containers, and seize approximately six tons of illicit drugs daily. Making this many decisions every day necessitates that CBT agents quickly and efficiently evaluate the information available on each individual, vehicle, boat, plane, or helicopter that is attempting to cross a border and enter the United States.

Niyazi Onur Bakır of the National Center for Risk and Economic Analysis of Terrorism Events (CREATE) at the University of Southern California has developed a decision tree approach to reduce vulnerabilities to terror- ism in commercial truck crossings at land ports of entry in the southwestern United States. The model takes into consideration potential costs and benefits associated with improving transportation security, inspections at Mexican land ports, and inspections at U.S. land ports. Costs that are considered include not only the actual costs of mak- ing these improvements but also the costs associated with security breaches and false alarms.

Ultimately, the model provides guidelines for condi- tions under which the various improvements under con- sideration should be undertaken. Results demonstrate that through the use of decision analysis, CBP can provide increased security at a decreased cost.

REDUCING THE THREAT OF TERRORISM THROUGH DECISION ANALYSIS*

Q.M. in ACtion

*Based on niyazi onur Bakır, “A Decision tree Model for Evaluating Countermeasures to Secure Cargo at United States Southwestern Ports of Entry.” Decision Analysis 5(4): 230–248.

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NOTES AND COMMENTS

1. Some decision analysis software automatically provides the risk profiles for the optimal deci- sion alternative. These packages also allow the user to obtain the risk profiles for other decision alternatives. After comparing the risk profiles, a decision maker may decide to select a decision alternative with a good risk profile even though the expected value of the decision alternative is not as good as the optimal decision alternative.

2. A tornado diagram, a graphical display, is par- ticularly helpful when several inputs combine to

determine the value of the optimal solution. By varying each input over its range of values, we obtain information about how each input affects the value of the optimal solution. To display this information, a bar is constructed for the input, with the width of the bar showing how the input affects the value of the optimal solution. The widest bar corresponds to the input that is most sensitive. The bars are arranged in a graph with the widest bar at the top, resulting in a graph that has the appearance of a tornado.

4.5 Decision Analysis with Sample Information In applying the expected value approach, we showed how probability information about the states of nature affects the expected value calculations and thus the decision recommendation. Frequently, decision makers have preliminary or prior probability assessments for the states of nature that are the best probability values available at that time. However, to make the best possible decision, the decision maker may want to seek additional information about the states of nature. This new information can be used to revise or update the prior probabilities so that the final decision is based on more accurate probabilities for the states of nature. Most often, ad- ditional information is obtained through experiments designed to provide sample information about the states of nature. Raw material sampling, product testing, and market research studies are examples of experiments (or studies) that may enable management to revise or update the state-of-nature probabilities. These revised probabilities are called posterior probabilities.

Let us return to the PDC problem and assume that management is considering a 6-month market research study designed to learn more about potential market acceptance of the PDC condominium project. Management anticipates that the market research study will provide one of the following two results:

1. Favorable report: A substantial number of the individuals contacted express interest in purchasing a PDC condominium.

2. Unfavorable report: Very few of the individuals contacted express interest in pur- chasing a PDC condominium.

Influence Diagram By introducing the possibility of conducting a market research study, the PDC problem be- comes more complex. The influence diagram for the expanded PDC problem is shown in Figure 4.7. Note that the two decision nodes correspond to the research study and the complex-size decisions. The two chance nodes correspond to the research study results and demand for the condominiums. Finally, the consequence node is the profit. From the arcs of the influence diagram, we see that demand influences both the research study results and profit. Although demand is currently unknown to PDC, some level of demand for the condominiums already exists in the Pittsburgh area. If existing demand is strong, the research study is likely to find a substantial number of individuals who express an interest in purchasing a condominium.

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4.5 Decision Analysis with Sample information 123

However, if the existing demand is weak, the research study is more likely to find a substantial number of individuals who express little interest in purchasing a condominium. In this sense, existing demand for the condominiums will influence the research study results, and clearly, demand will have an influence upon PDC’s profit.

The arc from the research study decision node to the complex-size decision node in- dicates that the research study decision precedes the complex-size decision. No arc spans from the research study decision node to the research study results node because the deci- sion to conduct the research study does not actually influence the research study results. The decision to conduct the research study makes the research study results available, but it does not influence the results of the research study. Finally, the complex-size node and the demand node both influence profit. Note that if a stated cost to conduct the research study were given, the decision to conduct the research study would also influence profit. In such a case, we would need to add an arc from the research study decision node to the profit node to show the influence that the research study cost would have on profit.

Decision Tree The decision tree for the PDC problem with sample information shows the logical sequence for the decisions and the chance events in Figure 4.8.

First, PDC’s management must decide whether the market research should be conducted. If it is conducted, PDC’s management must be prepared to make a decision about the size of the condominium project if the market research report is favorable and, possibly, a different decision about the size of the condominium project if the market research report is unfavorable. In Figure 4.8, the squares are decision nodes and the circles are chance nodes. At each decision node, the branch of the tree that is taken is based on the decision made. At each chance node, the branch of the tree that is taken is based on probability or chance. For example, decision node 1 shows that PDC must first make the decision of whether to conduct the market research study. If the market research study is undertaken, chance node 2 indicates that both the favor- able report branch and the unfavorable report branch are not under PDC’s control and will be determined by chance. Node 3 is a decision node, indicating that PDC must make the decision to construct the small, medium, or large complex if the market research report is favorable. Node 4 is a decision node showing that PDC must make the decision to construct the small, medium, or large complex if the market research report is unfavorable. Node 5 is a decision

Research Study

Complex Size

Research Study

Results

Pro�t

Demand

FIGURE 4.7 INFLUENCE DIAGRAM FOR THE PDC PROBLEM WITH SAMPLE INFORMATION

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Strong (s1)

Weak (s2)

Unfavorable Report

Market Research Study

No Market Research Study

Favorable Report

Small (d1)

Large (d3)

Medium (d2)

FIGURE 4.8 THE PDC DECISION TREE INCLUDING THE MARKET RESEARCH STUDY

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4.5 Decision Analysis with Sample information 125

node indicating that PDC must make the decision to construct the small, medium, or large complex if the market research is not undertaken. Nodes 6 to 14 are chance nodes indicating that the strong demand or weak demand state-of-nature branches will be determined by chance.

Analysis of the decision tree and the choice of an optimal strategy require that we know the branch probabilities corresponding to all chance nodes. PDC has developed the follow- ing branch probabilities:

If the market research study is undertaken

P(Favorable report) 5 0.77

P(Unfavorable report) 5 0.23

If the market research report is favorable

P(Strong demand given a favorable report) 5 0.94

P(Weak demand given a favorable report) 5 0.06

If the market research report is unfavorable

P(Strong demand given an unfavorable report) 5 0.35

P(Weak demand given an unfavorable report) 5 0.65

If the market research report is not undertaken, the prior probabilities are applicable.

P(Strong demand) 5 0.80

P(Weak demand) 5 0.20

The branch probabilities are shown on the decision tree in Figure 4.9.

Decision Strategy A decision strategy is a sequence of decisions and chance outcomes where the decisions chosen depend on the yet-to-be-determined outcomes of chance events.

The approach used to determine the optimal decision strategy is based on a backward pass through the decision tree using the following steps:

1. At chance nodes, compute the expected value by multiplying the payoff at the end of each branch by the corresponding branch probabilities.

2. At decision nodes, select the decision branch that leads to the best expected value. This expected value becomes the expected value at the decision node.

Starting the backward pass calculations by computing the expected values at chance nodes 6 to 14 provides the following results:

EV(Node 6) 5 0.94(8) 1 0.06(7) 5 7.94

EV(Node 7) 5 0.94(14) 1 0.06(5) 5 13.46

EV(Node 8) 5 0.94(20) 1 0.06(29) 5 18.26

EV(Node 9) 5 0.35(8) 1 0.65(7) 5 7.35

EV(Node 10) 5 0.35(14) 1 0.65(5) 5 8.15

EV(Node 11) 5 0.35(20) 1 0.65(29) 5 1.15

EV(Node 12) 5 0.80(8) 1 0.20(7) 5 7.80

EV(Node 13) 5 0.80(14) 1 0.20(5) 5 12.20

EV(Node 14) 5 0.80(20) 1 0.20(29) 5 14.20

We explain in Section 4.6 how the branch probabilities for P(Favorable report) and P(Unfavorable report) can be developed.

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Strong (s1)

Weak (s2)

Unfavorable Report 0.23

Market Research Study

No Market Research Study

Small (d1)

Large (d3)

Medium (d2)

0.94

0.06

0.94

0.35

0.06

0.65

0.35

0.65

0.35

0.65

0.80

0.20

0.80

0.20

0.80

Favorable Report 0.77

FIGURE 4.9 THE PDC DECISION TREE WITH BRANCH PROBABILITIES

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4.5 Decision Analysis with Sample information 127

Figure 4.10 shows the reduced decision tree after computing expected values at these chance nodes.

Next, move to decision nodes 3, 4, and 5. For each of these nodes, we select the decision alternative branch that leads to the best expected value. For example, at node 3 we have the choice of the small complex branch with EV(Node 6) 5 7.94, the medium complex branch with EV(Node 7) 5 13.46, and the large complex branch with EV(Node 8) 5 18.26. Thus,

Unfavorable Report 0.23

Market Research Study

No Market Research Study

Small (d1)

Large (d3)

Medium (d2)

EV = 7.94

EV = 13.46

EV = 18.26

EV = 7.35

EV = 8.15

EV = 1.15

EV = 7.80

EV = 12.20

EV = 14.20

Favorable Report 0.77

FIGURE 4.10 PDC DECISION TREE AFTER COMPUTING EXPECTED VALUES AT CHANCE NODES 6 TO 14

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128 Chapter 4 Decision Analysis

we select the large complex decision alternative branch and the expected value at node 3 becomes EV(Node 3) 5 18.26.

For node 4, we select the best expected value from nodes 9, 10, and 11. The best decision alternative is the medium complex branch that provides EV(Node 4) 5 8.15. For node 5, we select the best expected value from nodes 12, 13, and 14. The best decision alternative is the large complex branch that provides EV(Node 5) 5 14.20. Figure 4.11 shows the reduced decision tree after choosing the best decisions at nodes 3, 4, and 5.

The expected value at chance node 2 can now be computed as follows:

EV(Node 2) 5 0.77EV(Node 3) 1 0.23EV(Node 4)

5 0.77(18.26) 1 0.23(8.15) 5 15.93

This calculation reduces the decision tree to one involving only the two decision branches from node 1 (see Figure 4.12).

Unfavorable Report 0.23

Market Research Study

No Market Research Study

EV(d3) = 18.26

EV(d2 ) = 8.15

EV(d3) = 14.20

Favorable Report 0.77

FIGURE 4.11 PDC DECISION TREE AFTER CHOOSING BEST DECISIONS AT NODES 3, 4, AND 5

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4.5 Decision Analysis with Sample information 129

Market Research Study

No Market Research Study

EV = 15.93

EV = 14.20

FIGURE 4.12 PDC DECISION TREE REDUCED TO TWO DECISION BRANCHES

Finally, the decision can be made at decision node 1 by selecting the best expected val- ues from nodes 2 and 5. This action leads to the decision alternative to conduct the market research study, which provides an overall expected value of 15.93.

The optimal decision for PDC is to conduct the market research study and then carry out the following decision strategy:

If the market research is favorable, construct the large condominium complex.

If the market research is unfavorable, construct the medium condominium complex.

The analysis of the PDC decision tree describes the methods that can be used to ana- lyze more complex sequential decision problems. First, draw a decision tree consisting of decision and chance nodes and branches that describe the sequential nature of the problem. Determine the probabilities for all chance outcomes. Then, by working backward through the tree, compute expected values at all chance nodes and select the best decision branch at all decision nodes. The sequence of optimal decision branches determines the optimal decision strategy for the problem.

The Q.M. in Action, Decision Analysis At Bat, describes the application of decision analysis to guide a batter’s strategy during an at-bat in a baseball game.

Problem 16 will test your ability to develop an optimal decision strategy.

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130 Chapter 4 Decision Analysis

Risk Profile Figure 4.13 provides a reduced decision tree showing only the sequence of decision al- ternatives and chance events for the PDC optimal decision strategy. By implementing the optimal decision strategy, PDC will obtain one of the four payoffs shown at the terminal branches of the decision tree. Recall that a risk profile shows the possible payoffs with their associated probabilities. Thus, in order to construct a risk profile for the optimal decision strategy, we will need to compute the probability for each of the four payoffs.

Note that each payoff results from a sequence of branches leading from node 1 to the payoff. For instance, the payoff of $20 million is obtained by following the upper branch from node 1, the upper branch from node 2, the lower branch from node 3, and the upper branch from node 8. The probability of following that sequence of branches can be found by multiplying the probabilities for the branches from the chance nodes in the sequence. Thus, the probability of the $20 million payoff is (0.77)(0.94) 5 0.72. Similarly, the probabilities

Evan Gattis took a circuitous route to major league base- ball. His path to becoming a 26-year-old rookie on the 2013 Atlanta Braves’ roster has earned comparisons to Roy Hobbs, the mythical baseball player central to the novel and movie The Natural. As in the story of Roy Hobbs, Gattis dropped completely out of baseball for several years before returning in a big way. In his first month in the major leagues, he hit six home runs and was named the National League’s Rookie of the Month. He duplicated both feats in his second month—he hit six more home runs and was again named the National League’s Rookie of the Month. But what is most un- usual about Gattis is that he may be the first major league baseball player to give credit to management science and decision analysis for his success.

While in the minor leagues, Gattis began read- ing the work of University of Texas Professor J. Eric Bickel, who has used decision analysis to determine the optimal decisions for a hitter to make in each count. “One paper I wrote was how to act on different pitch counts,” Bickel said. “Sometimes the batter will just let a pitch go by on purpose. If it’s three balls, no strikes,

a lot of times the coach will say, ‘Don’t swing at the pitch, no matter what.’”

What Bickel said is that most people don’t under- stand why a batter would take a pitch on a 3–0 count. Because one more ball will result in a walk and put the batter on first base, under these circumstances the oppos- ing pitcher will usually put a very hittable fastball through the heart of the strike zone on the next pitch. However, Bickel’s research demonstrates why taking a pitch when the count is three balls and no strikes rather than swinging at what will likely be a very hittable pitch increases the probability the batter will ultimately get on base.

“About 38 percent of all batters eventually get on base,” Bickel said. “At 3–0, 77 percent of batters eventu- ally get on base. Suppose you’re sitting there with a 3–0 count. If you let the pitch go by, and the pitcher throws a strike, you’re down to a 63 percent chance of getting on. If you instead put that ball in play, you only have a one-third chance of getting on base. Your choice is to put the ball in play and have a one-third chance of getting on base, or take a strike and still have a 63 percent chance of getting on base. That’s why you take it.”

Bickel has used decision analysis to determine a batter’s optimal strategies for all ball/strike counts. His decision analysis and his lucid explanation of the result- ing optimal strategies for various ball/strike counts have helped shape the way Gattis approaches each pitch when he is at bat.

DECISION ANALYSIS AT BAT*

Q.M. in ACtion

*Based on Joe Lemire, “this Photo is Just one Good Reason You need to Know the Story of Evan Gattis,” Sports Illustrated (June 10, 2013) and “Mastering the numbers Game—Sports illustrated Coverage.” Petroleum and Geosystems Engineering News, University of texas at Austin, http://www.pge.utexas.edu/news/136-eric-bickel.

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Figure 4.14 provides a graphical representation of the risk profile. Comparing Figures 4.5 and 4.14, we see that the PDC risk profile is changed by the strategy to conduct the market research study. In fact, the use of the market research study lowered the prob- ability of the $9 million loss from 0.20 to 0.05. PDC’s management would most likely view that change as a considerable reduction in the risk associated with the condominium project.

Expected Value of Sample Information In the PDC problem, the market research study is the sample information used to determine the optimal decision strategy. The expected value associated with the market research study is $15.93. In Section 4.3 we showed that the best expected value if the market research study

Strong (s1)

Weak (s2)

Unfavorable Report 0.23

Market Research Study

Large (d3) 20

Medium (d2)

0.94

0.06

0.35

0.65

Favorable Report 0.77

FIGURE 4.13 PDC DECISION TREE SHOWING ONLY BRANCHES ASSOCIATED WITH OPTIMAL DECISION STRATEGY

payoff ($ millions) probability 29 0.05 5 0.15 14 0.08 20 0.72

1.00

for each of the other payoffs are obtained by multiplying the probabilities for the branches from the chance nodes leading to the payoffs. By doing so, we find the probability of the 2$9 million payoff is (0.77)(0.06) 5 0.05; the probability of the $14 million payoff is (0.23) (0.35) 5 0.08; and the probability of the $5 million payoff is (0.23)(0.65) 5 0.15. The fol- lowing table showing the probability distribution for the payoffs for the PDC optimal deci- sion strategy is the tabular representation of the risk profile for the optimal decision strategy.

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0.8

0.6

0.4

0.2

–10 0 10 20 Pro�t ($ millions)

P ro

b ab

il it

FIGURE 4.14 RISK PROFILE FOR PDC CONDOMINIUM PROJECT WITH SAMPLE INFORMATION SHOWING PAYOFFS ASSOCIATED WITH OPTIMAL DECISION STRATEGY

is not undertaken is $14.20. Thus, we can conclude that the difference, $15.93 2 $14.20 5 $1.73, is the expected value of sample information (EVSi). In other words, conducting the market research study adds $1.73 million to the PDC expected value. In general, the expected value of sample information is as follows:

EVSI 5 uEVwSI 2 EVwoSIu (4.13)

where

EVSI 5 expected value of sample information

EVwSI 5 expected value with sample information about the states of nature

EVwoSI 5 expected value without sample information about the states of nature

Note the role of the absolute value in equation (4.13). For minimization problems, the expected value with sample information is always less than or equal to the expected value without sample information. In this case, EVSI is the magnitude of the difference between EVwSI and EVwoSI; thus, by taking the absolute value of the difference as shown in equation (4.13), we can handle both the maximization and minimization cases with one equation.

Efficiency of Sample Information In Section 4.3 we showed that the expected value of perfect information (EVPI) for the PDC problem is $3.2 million. We never anticipated that the market research report would obtain perfect information, but we can use an efficiency measure to express the value of the

The EVSI 5 $1.73 million suggests PDC should be willing to pay up to $1.73 million to conduct the market research study.

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4.6 Computing Branch Probabilities with Bayes’ theorem 133

market research information. With perfect information having an efficiency rating of 100%, the efficiency rating E for sample information is computed as follows:

E 5 EVSI

EVPI 3 100 (4.14)

For the PDC problem,

E 5 1.73

3.2 3 100 5 54.1%

In other words, the information from the market research study is 54.1% as efficient as perfect information.

Low efficiency ratings for sample information might lead the decision maker to look for other types of information. However, high efficiency ratings indicate that the sample information is almost as good as perfect information and that additional sources of informa- tion would not yield substantially better results.

4.6 Computing Branch Probabilities with Bayes’ Theorem In Section 4.5 the branch probabilities for the PDC decision tree chance nodes were speci- fied in the problem description. No computations were required to determine these prob- abilities. In this section we show how Bayes’ theorem can be used to compute branch probabilities for decision trees.

The PDC decision tree is shown again in Figure 4.15. Let

F 5 Favorable market research report

U 5 Unfavorable market research report

s1 5 Strong demand (state of nature 1)

s2 5 Weak demand (state of nature 2)

At chance node 2, we need to know the branch probabilities P(F) and P(U). At chance nodes 6, 7, and 8, we need to know the branch probabilities P(s1 | F), the probability of state of nature 1 given a favorable market research report, and P(s2 | F), the probability of state of nature 2 given a favorable market research report. P(s1 | F) and P(s2 | F) are referred to as posterior probabilities because they are conditional probabilities based on the outcome of the sample information. At chance nodes 9, 10, and 11, we need to know the branch probabilities P(s1 | U) and P(s2 | U); note that these are also posterior probabilities, denoting the probabilities of the two states of nature given that the market research report is unfavorable. Finally, at chance nodes 12, 13, and 14, we need the probabilities for the states of nature, P(s1) and P(s2), if the market research study is not undertaken.

In performing the probability computations, we need to know PDC’s assessment of the probabilities for the two states of nature, P(s1) and P(s2), which are the prior probabilities as discussed earlier. In addition, we must know the conditional probability of the market

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Strong (s1)

Weak (s2)

Unfavorable Report P(U )

Market Research Study

No Market Research Study

Favorable Report P(F )

Small (d1)

Large (d3)

Medium (d2)

P(s1| F)

P(s2| F)

P(s1| F)

P(s1| U)

P(s2| F)

P(s2| U)

P(s1| U)

P(s2| U)

P(s1| U)

P(s2| U)

P(s1)

P(s2)

P(s1)

P(s2)

P(s1)

FIGURE 4.15 THE PDC DECISION TREE

research outcomes (the sample information) given each state of nature. For example, we need to know the conditional probability of a favorable market research report given that the state of nature is strong demand for the PDC project; note that this conditional prob- ability of F given state of nature s1 is written P(F | s1). To carry out the probability calcu- lations, we will need conditional probabilities for all sample outcomes given all states of

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4.6 Computing Branch Probabilities with Bayes’ theorem 135

Note that the preceding probability assessments provide a reasonable degree of con- fidence in the market research study. If the true state of nature is s1, the probability of a favorable market research report is 0.90, and the probability of an unfavorable market research report is 0.10. If the true state of nature is s2, the probability of a favorable market research report is 0.25, and the probability of an unfavorable market research report is 0.75. The reason for a 0.25 probability of a potentially misleading favorable market research report for state of nature s2 is that when some potential buyers first hear about the new condominium project, their enthusiasm may lead them to overstate their real interest in it. A potential buyer’s initial favorable response can change quickly to a “no thank you” when later faced with the reality of signing a purchase contract and making a down payment.

In the following discussion we present a tabular approach as a convenient method for carrying out the probability computations. The computations for the PDC problem based on a favorable market research report (F) are summarized in Table 4.7. The steps used to develop this table are as follows:

Step 1. In column 1 enter the states of nature. In column 2 enter the prior probabilities for the states of nature. In column 3 enter the conditional probabilities of a favorable market research report (F) given each state of nature.

Step 2. In column 4 compute the joint probabilities by multiplying the prior prob- ability values in column 2 by the corresponding conditional probability values in column 3.

Step 3. Sum the joint probabilities in column 4 to obtain the probability of a favorable market research report, P(F).

Step 4. Divide each joint probability in column 4 by P(F) 5 0.77 to obtain the revised or posterior probabilities, P(s1 | F) and P(s2 | F).

Table 4.7 shows that the probability of obtaining a favorable market research report is P(F) 5 0.77. In addition, P(s1 | F) 5 0.94 and P(s2 | F) 5 0.06. In particular, note that a

A favorable market research report given that the state of nature is weak demand is often referred to as a “false positive,” while the converse (an unfavorable market research report given that the state of nature is strong demand) is referred to as a “false negative.”

Market Research

State of Nature Favorable, F Unfavorable, U Strong demand, s1 P(F Z s1) 5 0.90 P(U Z s1) 5 0.10 Weak demand, s2 P(F Z s2) 5 0.25 P(U Z s2) 5 0.75

nature, that is, P(F | s1), P(F | s2), P(U | s1), and P(U | s2). In the PDC problem we assume that the following assessments are available for these conditional probabilities:

States of prior Conditional Joint posterior Nature probabilities probabilities probabilities probabilities sj P(sj) P(F Z sj) P(F > sj) P(sj Z F) s1 0.8 0.90 0.72 0.94 s2 0.2 0.25 0.05 0.06

1.0 P(F ) 5 0.77 1.00

TABLE 4.7 BRANCH PROBABILITIES FOR THE PDC CONDOMINIUM PROJECT BASED ON A FAVORABLE MARKET RESEARCH REPORT

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136 Chapter 4 Decision Analysis

States of prior Conditional Joint posterior Nature probabilities probabilities probabilities probabilities sj P(sj) P(U Z sj) P(U > sj) P(sj Z U ) s1 0.8 0.10 0.08 0.35 s2 0.2 0.75 0.15 0.65

1.0 P(U ) 5 0.23 1.00

TABLE 4.8 BRANCH PROBABILITIES FOR THE PDC CONDOMINIUM PROJECT BASED ON AN UNFAVORABLE MARKET RESEARCH REPORT

favorable market research report will prompt a revised or posterior probability of 0.94 that the market demand of the condominium will be strong, s1.

The tabular probability computation procedure must be repeated for each possible sam- ple information outcome. Table 4.8 shows the computations of the branch probabilities of the PDC problem based on an unfavorable market research report. Note that the probability of obtaining an unfavorable market research report is P(U) 5 0.23. If an unfavorable report is obtained, the posterior probability of a strong market demand, s1, is 0.35 and of a weak market demand, s2, is 0.65. The branch probabilities from Tables 4.7 and 4.8 were shown on the PDC decision tree in Figure 4.9.

The tabular method can be used directly to compute the branch probabilities in the deci- sion tree. Alternatively, equation (4.15) provides a general formula for Bayes’ theorem for computing posterior probabilities.

BAYES’ THEOREM

PsAi u Bd 5 PsB u Aid PsAid

oj PsB u Ajd PsAjd (4.15)

To perform the Bayes’ theorem calculations for P(s1uU) with equation (4.15), we replace B with U (unfavorable report) and Ai with s1 in (4.15) so that we have

Pss1 uUd 5 PsU us1d Pss1d

oj PsU usjd Pssjd

5 0.10 3 0.80

s0.10 3 0.80d 1 s0.20 3 0.75d 5 0.35

which provides the same value as the tabular approach used to generate the values in Table 4.7.

The discussion in this section shows an underlying relationship between the probabilities on the various branches in a decision tree. It would be inappropriate to assume different prior probabilities, P(s1) and P(s2) without determining how these changes would alter P(F) and P(U), as well as the posterior probabilities P(s1uF), P(s2uF), P(s1uU), and P(s2uU).

Problem 23 asks you to compute the posterior probabilities.

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Summary 137

The Q.M. in Action, Decision Analysis Helps Treat and Prevent Hepatitis B, discusses how medical researchers use posterior probability information and decision analysis to understand the risks and costs associated with treatment and screening procedures.

Hepatitis B is a viral disease that left untreated can lead to fatal liver conditions such as cirrhosis and cancer. The hepatitis B virus can be treated, and there exists a vaccine to prevent it. However, in order to make economically prudent allocations of their limited health care budgets, public health officials require analysis on the cost ef- fectiveness (health benefit per dollar investment) of any potential health program. Unfortunately, since hepatitis B is a slow-progressing condition whose victims are often unaware of their potentially fatal infection, gathering data on the benefits of any public health policy addressing hepatitis B would take decades.

A multidisciplinary team consisting of management science researchers and a liver transplant surgeon from Stanford University applied decision analysis techniques to determine which combination of hepatitis B screen- ing, treatment, and vaccination would be appropriate in the United States. Their decision tree contained the

sequential decisions of: (1) whether or not to perform a blood test to screen an individual for a hepatitis B infection, (2) whether or not to treat infected individu- als, and (3) whether or not to vaccinate a noninfected (or nonscreened) individual.

For each policy, composed of a sequence of screen- ing, treatment, and vaccination decisions, the research- ers utilized existing infection and treatment knowledge to model future disease progression. Implementing their decision model in an Excel spreadsheet, the research- ers concluded that it is cost effective to screen adult Asian and Pacific Islanders so that infected individu- als can be treated (these individuals are genetically at a high risk for hepatitis B infection). Although it is not cost effective to universally vaccinate all U.S. adult Asian and Pacific Islanders, it proves to be cost effec- tive to vaccinate people in close contact with infected individuals. Influenced by these findings, the Centers for Disease Control updated its official policy in 2008 to recommend screening all adult Asian and Pacific Islanders and all adults in areas of intermediate (2 to 7%) hepatitis B prevalence.

DECISION ANALYSIS HELPS TREAT AND PREVENT HEPATITIS B*

Q.M. in ACtion

*David W. Hutton, Margaret L. Brandeau, and Samuel K. So, “Doing Good with Good oR: Supporting Cost-Effective Hepatitis B interven- tions.” Interfaces 41 (May/June 2011): 289–300.

Summary

Decision analysis can be used to determine a recommended decision alternative or an op- timal decision strategy when a decision maker is faced with an uncertain and risk-filled pattern of future events. The goal of decision analysis is to identify the best decision alterna- tive or the optimal decision strategy given information about the uncertain events and the possible consequences or payoffs. The uncertain future events are called chance events, and the outcomes of the chance events are called states of nature.

We showed how influence diagrams, payoff tables, and decision trees could be used to structure a decision problem and describe the relationships among the decisions, the chance events, and the consequences. We presented three approaches to decision making without probabilities: the optimistic approach, the conservative approach, and the mini- max regret approach. When probability assessments are provided for the states of nature, the expected value approach can be used to identify the recommended decision alternative or decision strategy.

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138 Chapter 4 Decision Analysis

In cases where sample information about the chance events is available, a sequence of decisions has to be made. First we must decide whether to obtain the sample information. If the answer to this decision is yes, an optimal decision strategy based on the specific sample information must be developed. In this situation, decision trees and the expected value ap- proach can be used to determine the optimal decision strategy.

Even though the expected value approach can be used to obtain a recommended deci- sion alternative or optimal decision strategy, the payoff that actually occurs will usually have a value different from the expected value. A risk profile provides a probability distri- bution for the possible payoffs and can assist the decision maker in assessing the risks as- sociated with different decision alternatives. Finally, sensitivity analysis can be conducted to determine the effect changes in the probabilities for the states of nature and changes in the values of the payoffs have on the recommended decision alternative.

Glossary

Bayes’ theorem A theorem that enables the use of sample information to revise prior probabilities. Branch Lines showing the alternatives from decision nodes and the outcomes from chance nodes. Chance event An uncertain future event affecting the consequence, or payoff, associated with a decision. Chance nodes Nodes indicating points where an uncertain event will occur. Conditional probabilities The probability of one event given the known outcome of a (possibly) related event. Consequence The result obtained when a decision alternative is chosen and a chance event occurs. A measure of the consequence is often called a payoff. Consequence nodes Nodes of an influence diagram indicating points where a payoff will occur. Conservative approach An approach to choosing a decision alternative without using probabilities. For a maximization problem, it leads to choosing the decision alternative that maximizes the minimum payoff; for a minimization problem, it leads to choosing the decision alternative that minimizes the maximum payoff. decision alternatives Options available to the decision maker. decision nodes Nodes indicating points where a decision is made. decision strategy A strategy involving a sequence of decisions and chance outcomes to provide the optimal solution to a decision problem. decision tree A graphical representation of the decision problem that shows the sequen- tial nature of the decision-making process. Efficiency The ratio of EVSI to EVPI as a percentage; perfect information is 100% efficient. Expected value (EV) For a chance node, it is the weighted average of the payoffs. The weights are the state-of-nature probabilities. Expected value approach An approach to choosing a decision alternative based on the expected value of each decision alternative. The recommended decision alternative is the one that provides the best expected value. Expected value of perfect information (EVpi) The expected value of information that would tell the decision maker exactly which state of nature is going to occur (i.e., perfect information).

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Problems 139

Expected value of sample information (EVSi) The difference between the expected value of an optimal strategy based on sample information and the “best” expected value without any sample information. influence diagram A graphical device that shows the relationship among decisions, chance events, and consequences for a decision problem. Joint probabilities The probabilities of both sample information and a particular state of nature occurring simultaneously. Minimax regret approach An approach to choosing a decision alternative without using probabilities. For each alternative, the maximum regret is computed, which leads to choos- ing the decision alternative that minimizes the maximum regret. Node An intersection or junction point of an influence diagram or a decision tree. Opportunity loss, or regret The amount of loss (lower profit or higher cost) from not making the best decision for each state of nature. Optimistic approach An approach to choosing a decision alternative without using prob- abilities. For a maximization problem, it leads to choosing the decision alternative corre- sponding to the largest payoff; for a minimization problem, it leads to choosing the deci- sion alternative corresponding to the smallest payoff. payoff A measure of the consequence of a decision such as profit, cost, or time. Each combi- nation of a decision alternative and a state of nature has an associated payoff (consequence). payoff table A tabular representation of the payoffs for a decision problem. posterior (revised) probabilities The probabilities of the states of nature after revising the prior probabilities based on sample information. prior probabilities The probabilities of the states of nature prior to obtaining sample information. Risk analysis The study of the possible payoffs and probabilities associated with a deci- sion alternative or a decision strategy. Risk profile The probability distribution of the possible payoffs associated with a deci- sion alternative or decision strategy. Sample information New information obtained through research or experimentation that enables an updating or revision of the state-of-nature probabilities. Sensitivity analysis The study of how changes in the probability assessments for the states of nature or changes in the payoffs affect the recommended decision alternative. States of nature The possible outcomes for chance events that affect the payoff associ- ated with a decision alternative.

Problems

1. The following payoff table shows profit for a decision analysis problem with two decision alternatives and three states of nature:testSELF

State of Nature

decision Alternative s1 s2 s3 d1 250 100 25 d2 100 100 75

a. Construct a decision tree for this problem. b. If the decision maker knows nothing about the probabilities of the three states of

nature, what is the recommended decision using the optimistic, conservative, and minimax regret approaches?

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140 Chapter 4 Decision Analysis

State of Nature

decision Alternative s1 s2 s3 s4 d1 14 9 10 5 d2 11 10 8 7 d3 9 10 10 11 d4 8 10 11 13

2. Suppose that a decision maker faced with four decision alternatives and four states of nature develops the following profit payoff table:

a. If the decision maker knows nothing about the probabilities of the four states of nature, what is the recommended decision using the optimistic, conservative, and minimax regret approaches?

b. Which approach do you prefer? Explain. Is establishing the most appropriate approach before analyzing the problem important for the decision maker? Explain.

c. Assume that the payoff table provides cost rather than profit payoffs. What is the recommended decision using the optimistic, conservative, and minimax regret approaches?

3. Southland Corporation’s decision to produce a new line of recreational products resulted in the need to construct either a small plant or a large plant. The best selection of plant size depends on how the marketplace reacts to the new product line. To conduct an analysis, marketing management has decided to view the possible long-run demand as low, me- dium, or high. The following payoff table shows the projected profit in millions of dollars:

testSELF

Long-Run demand

plant Size Low Medium High

Small 150 200 200 Large 50 200 500

a. What is the decision to be made, and what is the chance event for Southland’s problem? b. Construct an influence diagram. c. Construct a decision tree. d. Recommend a decision based on the use of the optimistic, conservative, and minimax

regret approaches.

4. Amy Lloyd is interested in leasing a new Honda and has contacted three automobile deal- ers for pricing information. Each dealer offered Amy a closed-end 36-month lease with no down payment due at the time of signing. Each lease includes a monthly charge and a mileage allowance. Additional miles receive a surcharge on a per-mile basis. The monthly lease cost, the mileage allowance, and the cost for additional miles follow:

Cost per dealer Monthly Cost Mileage Allowance Additional Mile

Hepburn Honda $299 36,000 $0.15 Midtown Motors $310 45,000 $0.20 Hopkins Automotive $325 54,000 $0.15

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Problems 141

Amy decided to choose the lease option that will minimize her total 36-month cost. The difficulty is that Amy is not sure how many miles she will drive over the next three years. For purposes of this decision, she believes it is reasonable to assume that she will drive 12,000 miles per year, 15,000 miles per year, or 18,000 miles per year. With this assump- tion Amy estimated her total costs for the three lease options. For example, she figures that the Hepburn Honda lease will cost her $10,764 if she drives 12,000 miles per year, $12,114 if she drives 15,000 miles per year, or $13,464 if she drives 18,000 miles per year. a. What is the decision, and what is the chance event? b. Construct a payoff table for Amy’s problem. c. If Amy has no idea which of the three mileage assumptions is most appropriate, what

is the recommended decision (leasing option) using the optimistic, conservative, and minimax regret approaches?

d. Suppose that the probabilities that Amy drives 12,000, 15,000, and 18,000 miles per year are 0.5, 0.4, and 0.1, respectively. What option should Amy choose using the expected value approach?

e. Develop a risk profile for the decision selected in part (d). What is the most likely cost, and what is its probability?

f. Suppose that after further consideration Amy concludes that the probabilities that she will drive 12,000, 15,000, and 18,000 miles per year are 0.3, 0.4, and 0.3, respectively. What decision should Amy make using the expected value approach?

5. In American football, touchdowns are worth 6 points. After scoring a touchdown, the scor- ing team may subsequently attempt to score 1 or 2 additional points. Going for 1 point is virtually an assured success, while going for 2 points is successful only with probability p. Consider the following game situation. The Temple Wildcats are losing by 14 points to the Killeen Tigers near the end of regulation time. The only way for Temple to win (or tie) this game is to score two touchdowns while not allowing Killeen to score again. The Temple coach must decide whether to attempt a 1-point or 2-point conversion after each touchdown. If the score is tied at the end of regulation time, the game goes into overtime. The Temple coach believes that there is a 50% chance that Temple will win if the game goes into over- time. The probability of successfully converting a 1-point conversion is 1.0. The probability of successfully converting a 2-point conversion is p. a. Assume Temple will score two touchdowns and Killeen will not score. Create a deci-

sion tree for the decision of whether Temple’s coach should go for a 1-point conver- sion or a 2-point conversion after each touchdown. The terminal nodes in the decision tree should be either WIN or LOSE for Temple.

b. Assume that a WIN results in a value of 1.0 and LOSE results in a value of 0. Further, as- sume that the probability of converting a 2-point conversion is p 5 0.4. Should Temple’s coach go for a 1-point conversion or 2-point conversion after scoring the first touchdown?

6. Investment advisors estimated the stock market returns for four market segments: com- puters, financial, manufacturing, and pharmaceuticals. Annual return projections vary de- pending on whether the general economic conditions are improving, stable, or declining. The anticipated annual return percentages for each market segment under each economic condition are as follows:

Economic Condition

Market Segment improving Stable declining Computers 10 2 24 Financial 8 5 23 Manufacturing 6 4 22 Pharmaceuticals 6 5 21

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142 Chapter 4 Decision Analysis

a. Assume that an individual investor wants to select one market segment for a new investment. A forecast shows stable to declining economic conditions with the fol- lowing probabilities: improving (0.2), stable (0.5), and declining (0.3). What is the preferred market segment for the investor, and what is the expected return percentage?

b. At a later date, a revised forecast shows a potential for an improvement in economic conditions. New probabilities are as follows: improving (0.4), stable (0.4), and declin- ing (0.2). What is the preferred market segment for the investor based on these new probabilities? What is the expected return percentage?

7. Hudson Corporation is considering three options for managing its data processing opera- tion: continuing with its own staff, hiring an outside vendor to do the managing (referred to as outsourcing), or using a combination of its own staff and an outside vendor. The cost of the operation depends on future demand. The annual cost of each option (in thousands of dollars) depends on demand as follows:

testSELF

a. If the demand probabilities are 0.2, 0.5, and 0.3, which decision alternative will mini- mize the expected cost of the data processing operation? What is the expected annual cost associated with that recommendation?

b. Construct a risk profile for the optimal decision in part (a). What is the probability of the cost exceeding $700,000?

8. The following payoff table shows the profit for a decision problem with two states of nature and two decision alternatives:testSELF

demand

Staffing Options High Medium Low Own staff 650 650 600 Outside vendor 900 600 300 Combination 800 650 500

State of Nature

decision Alternative s1 s2 d1 10 1 d2 4 3

a. Use graphical sensitivity analysis to determine the range of probabilities of state of nature s1 for which each of the decision alternatives has the largest expected value.

b. Suppose P(s1) 5 0.2 and P(s2) 5 0.8. What is the best decision using the expected value approach?

c. Perform sensitivity analysis on the payoffs for decision alternative d1. Assume the probabilities are as given in part (b), and find the range of payoffs under states of nature s1 and s2 that will keep the solution found in part (b) optimal. Is the solution more sensitive to the payoff under state of nature s1 or s2?

9. Myrtle Air Express decided to offer direct service from Cleveland to Myrtle Beach. Management must decide between a full-price service using the company’s new fleet of jet aircraft and a discount service using smaller capacity commuter planes. It is clear that the best choice depends on the market reaction to the service Myrtle Air offers. Manage- ment developed estimates of the contribution to profit for each type of service based upon

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Problems 143

a. What is the decision to be made, what is the chance event, and what is the consequence for this problem? How many decision alternatives are there? How many outcomes are there for the chance event?

b. If nothing is known about the probabilities of the chance outcomes, what is the recom mended decision using the optimistic, conservative, and minimax regret approaches?

c. Suppose that management of Myrtle Air Express believes that the probability of strong demand is 0.7 and the probability of weak demand is 0.3. Use the expected value approach to determine an optimal decision.

d. Suppose that the probability of strong demand is 0.8 and the probability of weak demand is 0.2. What is the optimal decision using the expected value approach?

e. Use graphical sensitivity analysis to determine the range of demand probabilities for which each of the decision alternatives has the largest expected value.

10. Video Tech is considering marketing one of two new video games for the coming holiday season: Battle Pacific or Space Pirates. Battle Pacific is a unique game and appears to have no competition. Estimated profits (in thousands of dollars) under high, medium, and low demand are as follows:

two possible levels of demand for service to Myrtle Beach: strong and weak. The follow- ing table shows the estimated quarterly profits (in thousands of dollars):

demand for Service

Service Strong Weak

Full price $960 2$490 Discount $670 $320

Video Tech is optimistic about its Space Pirates game. However, the concern is that profit- ability will be affected by a competitor’s introduction of a video game viewed as similar to Space Pirates. Estimated profits (in thousands of dollars) with and without competition are as follows:

demand

Battle pacific High Medium Low

Profit $1000 $700 $300 Probability 0.2 0.5 0.3

Space pirates demand

with Competition High Medium Low

Profit $800 $400 $200 Probability 0.3 0.4 0.3

Space pirates demand

without Competition High Medium Low

Profit $1600 $800 $400 Probability 0.5 0.3 0.2

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144 Chapter 4 Decision Analysis

a. Develop a decision tree for the Video Tech problem. b. For planning purposes, Video Tech believes there is a 0.6 probability that its com-

petitor will produce a new game similar to Space Pirates. Given this probability of competition, the director of planning recommends marketing the Battle Pacific video game. Using expected value, what is your recommended decision?

c. Show a risk profile for your recommended decision. d. Use sensitivity analysis to determine what the probability of competition for Space

Pirates would have to be for you to change your recommended decision alternative.

11. For the Pittsburgh Development Corporation problem in Section 4.3, the decision alterna- tive to build the large condominium complex was found to be optimal using the expected value approach. In Section 4.4 we conducted a sensitivity analysis for the payoffs associ- ated with this decision alternative. We found that the large complex remained optimal as long as the payoff for the strong demand was greater than or equal to $17.5 million and as long as the payoff for the weak demand was greater than or equal to 2$19 million. a. Consider the medium complex decision. How much could the payoff under strong

demand increase and still keep decision alternative d3 the optimal solution? b. Consider the small complex decision. How much could the payoff under strong

demand increase and still keep decision alternative d3 the optimal solution?

12. The distance from Potsdam to larger markets and limited air service have hindered the town in attracting new industry. Air Express, a major overnight delivery service, is considering establishing a regional distribution center in Potsdam. However, Air Express will not establish the center unless the length of the runway at the local air- port is increased. Another candidate for new development is Diagnostic Research, Inc. (DRI), a leading producer of medical testing equipment. DRI is considering building a new manufacturing plant. Increasing the length of the runway is not a requirement for DRI, but the planning commission feels that doing so will help convince DRI to locate its new plant in Potsdam. Assuming that the town lengthens the runway, the Potsdam planning commission believes that the probabilities shown in the following table are applicable.

dRi plant No dRi plant

Air Express Center 0.30 0.10 No Air Express Center 0.40 0.20

For instance, the probability that Air Express will establish a distribution center and DRI will build a plant is 0.30.

The estimated annual revenue to the town, after deducting the cost of lengthening the runway, is as follows:

dRi plant No dRi plant

Air Express Center $600,000 $150,000 No Air Express Center $250,000 2$200,000

If the runway expansion project is not conducted, the planning commission assesses the probability that DRI will locate its new plant in Potsdam at 0.6; in this case, the esti- mated annual revenue to the town will be $450,000. If the runway expansion project is not

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Problems 145

conducted and DRI does not locate in Potsdam, the annual revenue will be $0 because no cost will have been incurred and no revenues will be forthcoming. a. What is the decision to be made, what is the chance event, and what is the consequence? b. Compute the expected annual revenue associated with the decision alternative to

lengthen the runway. c. Compute the expected annual revenue associated with the decision alternative not to

lengthen the runway. d. Should the town elect to lengthen the runway? Explain. e. Suppose that the probabilities associated with lengthening the runway were as follows:

dRi plant No dRi plant

Air Express Center 0.40 0.10 No Air Express Center 0.30 0.20

What effect, if any, would this change in the probabilities have on the recommended decision?

13. Seneca Hill Winery recently purchased land for the purpose of establishing a new vineyard. Management is considering two varieties of white grapes for the new vineyard: Chardonnay and Riesling. The Chardonnay grapes would be used to produce a dry Chardonnay wine, and the Riesling grapes would be used to produce a semidry Riesling wine. It takes approximately four years from the time of planting before new grapes can be harvested. This length of time creates a great deal of uncertainty concerning future demand and makes the decision about the type of grapes to plant difficult. Three possibilities are being considered: Chardonnay grapes only; Riesling grapes only; and both Chardonnay and Riesling grapes. Seneca man- agement decided that for planning purposes it would be adequate to consider only two demand possibilities for each type of wine: strong or weak. With two possibilities for each type of wine, it was necessary to assess four probabilities. With the help of some forecasts in industry publications, management made the following probability assessments:

Riesling demand

Chardonnay demand Weak Strong

Weak 0.05 0.50 Strong 0.25 0.20

Riesling demand

Chardonnay demand Weak Strong

Weak $22,000 $40,000 Strong $26,000 $60,000

Revenue projections show an annual contribution to profit of $20,000 if Seneca Hill only plants Chardonnay grapes and demand is weak for Chardonnay wine, and $70,000 if Seneca only plants Chardonnay grapes and demand is strong for Chardonnay wine. If Seneca only plants Riesling grapes, the annual profit projection is $25,000 if demand is weak for Riesling grapes and $45,000 if demand is strong for Riesling grapes. If Seneca plants both types of grapes, the annual profit projections are shown in the following table:

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146 Chapter 4 Decision Analysis

State of Nature

decision Alternative s1 s2 s3 d1 250 100 25 d2 100 100 75

a. What is the decision to be made, what is the chance event, and what is the conse- quence? Identify the alternatives for the decisions and the possible outcomes for the chance events.

b. Develop a decision tree. c. Use the expected value approach to recommend which alternative Seneca Hill Winery

should follow in order to maximize expected annual profit. d. Suppose management is concerned about the probability assessments when demand

for Chardonnay wine is strong. Some believe it is likely for Riesling demand to also be strong in this case. Suppose the probability of strong demand for Chardonnay and weak demand for Riesling is 0.05 and that the probability of strong demand for Chardonnay and strong demand for Riesling is 0.40. How does this change the recom- mended decision? Assume that the probabilities when Chardonnay demand is weak are still 0.05 and 0.50.

e. Other members of the management team expect the Chardonnay market to become saturated at some point in the future, causing a fall in prices. Suppose that the an- nual profit projections fall to $50,000 when demand for Chardonnay is strong and Chardonnay grapes only are planted. Using the original probability assessments, determine how this change would affect the optimal decision.

14. The following profit payoff table was presented in Problem 1:

testSELF

The probabilities for the states of nature are P(s1) 5 0.65, P(s2) 5 0.15, and P(s3) 5 0.20. a. What is the optimal decision strategy if perfect information were available? b. What is the expected value for the decision strategy developed in part (a)? c. Using the expected value approach, what is the recommended decision without perfect

information? What is its expected value? d. What is the expected value of perfect information?

15. The Lake Placid Town Council decided to build a new community center to be used for conventions, concerts, and other public events, but considerable controversy surrounds the appropriate size. Many influential citizens want a large center that would be a show- case for the area. But the mayor feels that if demand does not support such a center, the community will lose a large amount of money. To provide structure for the decision process, the council narrowed the building alternatives to three sizes: small, medium, and large. Everybody agreed that the critical factor in choosing the best size is the number of people who will want to use the new facility. A regional planning consultant provided demand estimates under three scenarios: worst case, base case, and best case. The worst- case scenario corresponds to a situation in which tourism drops substantially; the base-case scenario corresponds to a situation in which Lake Placid continues to attract visitors at cur- rent levels; and the best-case scenario corresponds to a substantial increase in tourism. The consultant has provided probability assessments of 0.10, 0.60, and 0.30 for the worst-case, base-case, and best-case scenarios, respectively.

The town council suggested using net cash flow over a 5-year planning horizon as the criterion for deciding on the best size. The following projections of net cash flow (in thou- sands of dollars) for a 5-year planning horizon have been developed. All costs, including the consultant’s fee, have been included.

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Problems 147

demand Scenario

Worst Base Best Center Size Case Case Case

Small 400 500 660 Medium 2250 650 800 Large 2400 580 990

a. What decision should Lake Placid make using the expected value approach? b. Construct risk profiles for the medium and large alternatives. Given the mayor’s con-

cern over the possibility of losing money and the result of part (a), which alternative would you recommend?

c. Compute the expected value of perfect information. Do you think it would be worth trying to obtain additional information concerning which scenario is likely to occur?

d. Suppose the probability of the worst-case scenario increases to 0.2, the probability of the base-case scenario decreases to 0.5, and the probability of the best-case sce- nario remains at 0.3. What effect, if any, would these changes have on the decision recommendation?

e. The consultant has suggested that an expenditure of $150,000 on a promotional cam- paign over the planning horizon will effectively reduce the probability of the worst- case scenario to zero. If the campaign can be expected to also increase the probability of the best-case scenario to 0.4, is it a good investment?

16. Consider a variation of the PDC decision tree shown in Figure 4.9. The company must first decide whether to undertake the market research study. If the market research study is conducted, the outcome will either be favorable (F ) or unfavorable (U ). Assume there are only two decision alternatives, d1 and d2, and two states of nature, s1 and s2. The payoff table showing profit is as follows:

testSELF

State of Nature

decision Alternative s1 s2 d1 100 300 d2 400 200

a. Show the decision tree. b. Using the following probabilities, what is the optimal decision strategy?

P(F) 5 0.56 P(s1 Z F) 5 0.57 P(s1 Z U) 5 0.18 P(s1) 5 0.40 P(U) 5 0.44 P(s2 Z F) 5 0.43 P(s2 Z U) 5 0.82 P(s2) 5 0.60

17. Hemmingway, Inc., is considering a $5 million research and development (R&D) project. Profit projections appear promising, but Hemmingway’s president is concerned because the probability that the R&D project will be successful is only 0.50. Furthermore, the president knows that even if the project is successful, it will require that the company build a new production facility at a cost of $20 million in order to manufacture the prod- uct. If the facility is built, uncertainty remains about the demand and thus uncertainty about the profit that will be realized. Another option is that if the R&D project is success- ful, the company could sell the rights to the product for an estimated $25 million. Under this option, the company would not build the $20 million production facility.

The decision tree is shown in Figure 4.16. The profit projection for each outcome is shown at the end of the branches. For example, the revenue projection for the high demand outcome is $59 million. However, the cost of the R&D project ($5 million) and the cost

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148 Chapter 4 Decision Analysis

of the production facility ($20 million) show the profit of this outcome to be $59 2 $5 2 $20 5 $34 million. Branch probabilities are also shown for the chance events. a. Analyze the decision tree to determine whether the company should undertake the

R&D project. If it does, and if the R&D project is successful, what should the company do? What is the expected value of your strategy?

b. What must the selling price be for the company to consider selling the rights to the product?

c. Develop a risk profile for the optimal strategy.

18. Dante Development Corporation is considering bidding on a contract for a new office building complex. Figure 4.17 shows the decision tree prepared by one of Dante’s ana- lysts. At node 1, the company must decide whether to bid on the contract. The cost of preparing the bid is $200,000. The upper branch from node 2 shows that the company has a 0.8 probability of winning the contract if it submits a bid. If the company wins the bid, it will have to pay $2,000,000 to become a partner in the project. Node 3 shows that the company will then consider doing a market research study to forecast demand for the of- fice units prior to beginning construction. The cost of this study is $150,000. Node 4 is a chance node showing the possible outcomes of the market research study.

Nodes 5, 6, and 7 are similar in that they are the decision nodes for Dante to either build the office complex or sell the rights in the project to another developer. The decision to build the complex will result in an income of $5,000,000 if demand is high and $3,000,000 if demand is moderate. If Dante chooses to sell its rights in the project to another developer, income from the sale is estimated to be $3,500,000. The probabilities shown at nodes 4, 8, and 9 are based on the projected outcomes of the market research study. a. Verify Dante’s profit projections shown at the ending branches of the decision tree by

calculating the payoffs of $2,650,000 and $650,000 for first two outcomes. b. What is the optimal decision strategy for Dante, and what is the expected profit for

this project?

Not Successful 0.5

Start R&D Project ($5 million)

Do Not Start the R&D Project

Successful 0.5

Building Facility ($20 million)

Sell Rights

Pro�t ($ millions)

High Demand 0.5

Medium Demand 0.3

Low Demand 0.2

FIGURE 4.16 DECISION TREE FOR HEMMINGWAY, INC.

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Problems 149

c. What would the cost of the market research study have to be before Dante would change its decision about the market research study?

d. Develop a risk profile for Dante.

19. Hale’s TV Productions is considering producing a pilot for a comedy series in the hope of selling it to a major television network. The network may decide to reject the series, but it may also decide to purchase the rights to the series for either one or two years. At this point in time, Hale may either produce the pilot and wait for the network’s decision or transfer the rights for the pilot and series to a competitor for $100,000. Hale’s decision alternatives and profits (in thousands of dollars) are as follows:

Lose Contract 0.2

Bid

Do Not Bid

Win Contract 0.8

Market Research

No Market Research

Build Complex

Sell 6

1150

2650

650

1150

Build Complex

Sell 7

2800

800

1300

2200

Build Complex

Sell 5

2650

Pro�t ($1000s)

650Forecast High 0.6

Forecast Moderate 0.4

High Demand 0.85

Moderate Demand 0.15

High Demand 0.225

Moderate Demand 0.775

High Demand 0.6

Moderate Demand 0.4

FIGURE 4.17 DECISION TREE FOR THE DANTE DEVELOPMENT CORPORATION

State of Nature

decision Alternative Reject, s1 1 Year, s2 2 Years, s3 Produce pilot, d1 2100 50 150 Sell to competitor, d2 100 100 100

The probabilities for the states of nature are P(s1) 5 0.20, P(s2) 5 0.30, and P(s3) 5 0.50. For a consulting fee of $5000, an agency will review the plans for the comedy series and indicate the overall chances of a favorable network reaction to the series. Assume that the agency review will result in a favorable (F) or an unfavorable (U) review and that the fol- lowing probabilities are relevant:

P(F) 5 0.69 P(s1 Z F) 5 0.09 P(s1 Z U) 5 0.45 P(U) 5 0.31 P(s2 Z F) 5 0.26 P(s2 Z U) 5 0.39 P(s3 Z F) 5 0.65 P(s3 Z U) 5 0.16

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a. Construct a decision tree for this problem. b. What is the recommended decision if the agency opinion is not used? What is the

expected value? c. What is the expected value of perfect information? d. What is Hale’s optimal decision strategy assuming the agency’s information is used? e. What is the expected value of the agency’s information? f. Is the agency’s information worth the $5000 fee? What is the maximum that Hale

should be willing to pay for the information? g. What is the recommended decision?

20. Embassy Publishing Company received a six-chapter manuscript for a new college text- book. The editor of the college division is familiar with the manuscript and estimated a 0.65 probability that the textbook will be successful. If successful, a profit of $750,000 will be realized. If the company decides to publish the textbook and it is unsuccessful, a loss of $250,000 will occur.

Before making the decision to accept or reject the manuscript, the editor is consider- ing sending the manuscript out for review. A review process provides either a favorable (F) or unfavorable (U) evaluation of the manuscript. Past experience with the review pro- cess suggests that probabilities P(F) 5 0.7 and P(U) 5 0.3 apply. Let s1 5 the textbook is successful, and s2 5 the textbook is unsuccessful. The editor’s initial probabilities of s1 and s2 will be revised based on whether the review is favorable or unfavorable. The revised probabilities are as follows:

P(s1 Z F) 5 0.75 P(s1 Z U) 5 0.417 P(s2 Z F) 5 0.25 P(s2 Z U) 5 0.583

a. Construct a decision tree assuming that the company will first make the decision of whether to send the manuscript out for review and then make the decision to accept or reject the manuscript.

b. Analyze the decision tree to determine the optimal decision strategy for the publishing company.

c. If the manuscript review costs $5000, what is your recommendation? d. What is the expected value of perfect information? What does this EVPI suggest for

the company?

21. A real estate investor has the opportunity to purchase land currently zoned residential. If the county board approves a request to rezone the property as commercial within the next year, the investor will be able to lease the land to a large discount firm that wants to open a new store on the property. However, if the zoning change is not approved, the investor will have to sell the property at a loss. Profits (in thousands of dollars) are shown in the following payoff table:

State of Nature

Rezoning Approved Rezoning Not Approved decision Alternative s1 s2 Purchase, d1 600 2200 Do not purchase, d2 0 0

a. If the probability that the rezoning will be approved is 0.5, what decision is recom- mended? What is the expected profit?

b. The investor can purchase an option to buy the land. Under the option, the investor maintains the rights to purchase the land anytime during the next three months while

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learning more about possible resistance to the rezoning proposal from area residents. Probabilities are as follows:

Let H 5 High resistance to rezoning

L 5 Low resistance to rezoning

P(H) 5 0.55 P(s1 Z H) 5 0.18 P(s2 Z H) 5 0.82 P(L) 5 0.45 P(s1 Z L) 5 0.89 P(s2 Z L) 5 0.11

What is the optimal decision strategy if the investor uses the option period to learn more about the resistance from area residents before making the purchase decision?

c. If the option will cost the investor an additional $10,000, should the investor purchase the option? Why or why not? What is the maximum that the investor should be willing to pay for the option?

22. Lawson’s Department Store faces a buying decision for a seasonal product for which demand can be high, medium, or low. The purchaser for Lawson’s can order one, two, or three lots of the product before the season begins but cannot reorder later. Profit projec- tions (in thousands of dollars) are shown.

a. If the prior probabilities for the three states of nature are 0.3, 0.3, and 0.4, respectively, what is the recommended order quantity?

b. At each preseason sales meeting, the vice president of sales provides a personal opinion regarding potential demand for this product. Because of the vice president’s enthusiasm and optimistic nature, the predictions of market conditions have always been either “excellent” (E ) or “very good” (V ). Probabilities are as follows:

P(E ) 5 0.70 P(s1 Z E) 5 0.34 P(s1 Z V) 5 0.20 P(V) 5 0.30 P(s2 Z E) 5 0.32 P(s2 Z V) 5 0.26 P(s3 Z E) 5 0.34 P(s3 Z V) 5 0.54

What is the optimal decision strategy? c. Use the efficiency of sample information and discuss whether the firm should consider

a consulting expert who could provide independent forecasts of market conditions for the product.

23. Suppose that you are given a decision situation with three possible states of nature: s1, s2, and s3. The prior probabilities are P(s1) 5 0.2, P(s2) 5 0.5, and P(s3) 5 0.3. With sample information I, P(I Z s1) 5 0.1, P(I Z s2) 5 0.05, and P(I Z s3) 5 0.2. Compute the revised or posterior probabilities: P(s1 Z I), P(s2 Z I), and P(s3 Z I).

24. To save on expenses, Rona and Jerry agreed to form a carpool for traveling to and from work. Rona preferred to use the somewhat longer but more consistent Queen City Avenue. Although Jerry preferred the quicker expressway, he agreed with Rona that they should

testSELF

State of Nature

High demand Medium demand Low demand decision Alternative s1 s2 s3 Order 1 lot, d1 60 60 50 Order 2 lots, d2 80 80 30 Order 3 lots, d3 100 70 10

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take Queen City Avenue if the expressway had a traffic jam. The following payoff table provides the one-way time estimate in minutes for traveling to or from work:

State of Nature

Expressway Expressway Open Jammed decision Alternative s1 s2 Queen City Avenue, d1 30 30 Expressway, d2 25 45

Based on their experience with traffic problems, Rona and Jerry agreed on a 0.15 prob- ability that the expressway would be jammed.

In addition, they agreed that weather seemed to affect the traffic conditions on the expressway. Let

C 5 clear

O 5 overcast

R 5 rain

The following conditional probabilities apply:

P(C Z s1) 5 0.8 P(O Z s1) 5 0.2 P(R Z s1) 5 0.0 P(C Z s2) 5 0.1 P(O Z s2) 5 0.3 P(R Z s2) 5 0.6

a. Use Bayes’ theorem for probability revision to compute the probability of each weather condition and the conditional probability of the expressway open, s1, or jammed, s2, given each weather condition.

b. Show the decision tree for this problem. c. What is the optimal decision strategy, and what is the expected travel time?

25. The Gorman Manufacturing Company must decide whether to manufacture a component part at its Milan, Michigan, plant or purchase the component part from a supplier. The resulting profit is dependent upon the demand for the product. The following payoff table shows the projected profit (in thousands of dollars):

State of Nature

Low demand Medium demand High demand decision Alternative s1 s2 s3 Manufacture, d1 220 40 100 Purchase, d2 10 45 70

The state-of-nature probabilities are P(s1) 5 0.35, P(s2) 5 0.35, and P(s3) 5 0.30. a. Use a decision tree to recommend a decision. b. Use EVPI to determine whether Gorman should attempt to obtain a better estimate of

demand.

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Case Problem 1 Property Purchase Strategy 153

c. A test market study of the potential demand for the product is expected to report either a favorable (F) or unfavorable (U) condition. The relevant conditional probabilities are as follows:

P(F Z s1) 5 0.10 P(U Z s1) 5 0.90 P(F Z s2) 5 0.40 P(U Z s2) 5 0.60 P(F Z s3) 5 0.60 P(U Z s3) 5 0.40

What is the probability that the market research report will be favorable? d. What is Gorman’s optimal decision strategy? e. What is the expected value of the market research information? f. What is the efficiency of the information?

Case problem 1 Property Purchase Strategy Glenn Foreman, president of Oceanview Development Corporation, is considering submit- ting a bid to purchase property that will be sold by sealed bid at a county tax foreclosure. Glenn’s initial judgment is to submit a bid of $5 million. Based on his experience, Glenn estimates that a bid of $5 million will have a 0.2 probability of being the highest bid and securing the property for Oceanview. The current date is June 1. Sealed bids for the property must be submitted by August 15. The winning bid will be announced on September 1.

If Oceanview submits the highest bid and obtains the property, the firm plans to build and sell a complex of luxury condominiums. However, a complicating factor is that the property is currently zoned for single-family residences only. Glenn believes that a referendum could be placed on the voting ballot in time for the November election. Passage of the referendum would change the zoning of the property and permit construction of the condominiums.

The sealed-bid procedure requires the bid to be submitted with a certified check for 10% of the amount bid. If the bid is rejected, the deposit is refunded. If the bid is accepted, the deposit is the down payment for the property. However, if the bid is accepted and the bidder does not follow through with the purchase and meet the remainder of the financial obli gation within six months, the deposit will be forfeited. In this case, the county will offer the property to the next highest bidder.

To determine whether Oceanview should submit the $5 million bid, Glenn conducted some preliminary analysis. This preliminary work provided an assessment of 0.3 for the probability that the referendum for a zoning change will be approved and resulted in the following estimates of the costs and revenues that will be incurred if the condominiums are built:

Cost and Revenue Estimates Revenue from condominium sales $15,000,000

Cost Property $5,000,000 Construction expenses $8,000,000

If Oceanview obtains the property and the zoning change is rejected in November, Glenn believes that the best option would be for the firm not to complete the purchase of the property. In this case, Oceanview would forfeit the 10% deposit that accompanied the bid.

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Because the likelihood that the zoning referendum will be approved is such an impor- tant factor in the decision process, Glenn suggested that the firm hire a market research service to conduct a survey of voters. The survey would provide a better estimate of the likelihood that the referendum for a zoning change would be approved. The market research firm that Oceanview Development has worked with in the past has agreed to do the study for $15,000. The results of the study will be available August 1, so that Oceanview will have this information before the August 15 bid deadline. The results of the survey will be either a prediction that the zoning change will be approved or a prediction that the zoning change will be rejected. After considering the record of the market research service in previ- ous studies conducted for Oceanview, Glenn developed the following probability estimates concerning the accuracy of the market research information:

P(A Z s1) 5 0.9 P(N Z s1) 5 0.1 P(A Z s2) 5 0.2 P(N Z s2) 5 0.8

where

A 5 prediction of zoning change approval

N 5 prediction that zoning change will not be approved

s1 5 the zoning change is approved by the voters

s2 5 the zoning change is rejected by the voters

Managerial Report Perform an analysis of the problem facing the Oceanview Development Corporation, and prepare a report that summarizes your findings and recommendations. Include the follow- ing items in your report:

1. A decision tree that shows the logical sequence of the decision problem 2. A recommendation regarding what Oceanview should do if the market research

information is not available 3. A decision strategy that Oceanview should follow if the market research is conducted 4. A recommendation as to whether Oceanview should employ the market research

firm, along with the value of the information provided by the market research firm

Include the details of your analysis as an appendix to your report.

Case problem 2 Lawsuit Defense Strategy John Campbell, an employee of Manhattan Construction Company, claims to have injured his back as a result of a fall while repairing the roof at one of the Eastview apartment buildings. He filed a lawsuit against Doug Reynolds, the owner of Eastview Apartments, asking for damages of $1,500,000. John claims that the roof had rotten sections and that his fall could have been prevented if Mr. Reynolds had told Manhattan Construction about the problem. Mr. Reynolds notified his insurance company, Allied Insurance, of the lawsuit. Allied must defend Mr. Reynolds and decide what action to take regarding the lawsuit.

Some depositions and a series of discussions took place between both sides. As a result, John Campbell offered to accept a settlement of $750,000. Thus, one option is for Allied to pay John $750,000 to settle the claim. Allied is also considering making John a counter- offer of $400,000 in the hope that he will accept a lesser amount to avoid the time and cost of going to trial. Allied’s preliminary investigation shows that John’s case is strong; Allied

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Appendix 4.1 Using Analytic Solver Platform to Create Decision trees 155

is concerned that John may reject its counteroffer and request a jury trial. Allied’s lawyers spent some time exploring John’s likely reaction if they make a counteroffer of $400,000.

The lawyers concluded that it is adequate to consider three possible outcomes to rep- resent John’s possible reaction to a counteroffer of $400,000: (1) John will accept the counteroffer and the case will be closed; (2) John will reject the counteroffer and elect to have a jury decide the settlement amount; or (3) John will make a counteroffer to Allied of $600,000. If John does make a counteroffer, Allied decided that it will not make additional counteroffers. It will either accept John’s counteroffer of $600,000 or go to trial.

If the case goes to a jury trial, Allied considers three outcomes possible: (1) the jury may reject John’s claim and Allied will not be required to pay any damages; (2) the jury will find in favor of John and award him $750,000 in damages; or (3) the jury will conclude that John has a strong case and award him the full amount of $1,500,000.

Key considerations as Allied develops its strategy for disposing of the case are the probabilities associated with John’s response to an Allied counteroffer of $400,000 and the probabilities associated with the three possible trial outcomes. Allied’s lawyers believe that the probability that John will accept a counteroffer of $400,000 is 0.10, the probability that John will reject a counteroffer of $400,000 is 0.40, and the probability that John will, himself, make a counteroffer to Allied of $600,000 is 0.50. If the case goes to court, they believe that the probability that the jury will award John damages of $1,500,000 is 0.30, the probability that the jury will award John damages of $750,000 is 0.50, and the probability that the jury will award John nothing is 0.20.

Managerial Report Perform an analysis of the problem facing Allied Insurance and prepare a report that sum- marizes your findings and recommendations. Be sure to include the following items:

1. A decision tree 2. A recommendation regarding whether Allied should accept John’s initial offer to

settle the claim for $750,000 3. A decision strategy that Allied should follow if they decide to make John a coun-

teroffer of $400,000 4. A risk profile for your recommended strategy

Appendix 4.1 Using Analytic Solver Platform to Create Decision Trees In this appendix, we describe how Analytic Solver Platform can be used to develop a de- cision tree for the PDC problem presented in Section 4.3. The decision tree for the PDC problem is shown in Figure 4.18.

Getting Started: An Initial Decision Tree To build a decision tree for the PDC problem using Analytic Solver Platform, follow these steps in a blank workbook in Excel:

Step 1. Select cell A1 Step 2. Click the ANALYTiC SOLVER pLATFORM tab on the Ribbon Step 3. Click decision Tree in the Tools group

Select Node, and click Add Node

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Step 4. When the decision Tree dialog box appears, verify that decision is selected for Node Type, and click OK

A decision tree with one decision node and two branches (initially labeled as “Decision 1” and “Decision 2”) appears, as shown in Figure 4.19.

Adding a Branch The PDC problem has three decision alternatives (small, medium, and large condominium complexes), so we must add another decision branch to the tree.

Step 1. Select cell B5 Step 2. Click the ANALYTiC SOLVER pLATFORM tab in the Ribbon

FIGURE 4.18 DECISION TREE FOR THE PDC CONDOMINIUM PROJECT (PAYOFFS IN MILLIONS $)

– 9 Weak (s2)

Strong (s1)

Weak (s2)

Strong (s1)

Weak (s2)

Strong (s1)

Small (d1)

Medium (d2 )

Large (d3)

FIGURE 4.19 DECISION TREE WITH ONE DECISION NODE AND TWO BRANCHES CREATED WITH ANALYTIC SOLVER PLATFORM

A B C D E F G H

1 2 3 4 5 6 7 8 9

Decision 1

Decision 2

0 00

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Step 3. Click decision Tree in the Tools group Select Branch, and click Add Branch

Step 4. When the decision Tree dialog box appears, verify that decision is selected for Node Type, and click OK

A revised tree with three decision branches now appears in the Excel worksheet as shown in Figure 4.20.

Naming the Decision Alternatives The decision alternatives can be named by selecting the cells containing the labels “Deci- sion 1,” “Decision 2,” and “New Branch,” and then entering the corresponding PDC names Small, Medium, and Large (cells D2, D7, and D12). After naming the alternatives, the PDC tree with three decision branches appears as shown in Figure 4.21.

Adding Chance Nodes The chance event for the PDC problem is the demand for the condominiums, which may be either strong or weak. Thus, a chance node with two branches must be added at the end of each decision alternative branch. To add a chance node with two branches to the top decision alternative branch:

Step 1. Select cell F3 Step 2. Click the ANALYTiC SOLVER pLATFORM tab in the Ribbon Step 3. Select decision Tree from the Tools group

Select Node, and click Add Node Step 4. When the decision Tree dialog box appears, select Event/Chance in the Node

Type area Click OK

The tree now appears as shown in Figure 4.22.

FIGURE 4.20 DECISION TREE FOR THE PDC PROBLEM WITH THREE BRANCHES CREATED WITH ANALYTIC SOLVER PLATFORM

Decision 1

Decision 2

New Branch

0 00

A B C D E F G H

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

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158 Chapter 4 Decision Analysis

FIGURE 4.21 DECISION TREE FOR THE PDC PROBLEM WITH RENAMED BRANCHES CREATED WITH ANALYTIC SOLVER PLATFORM

0 00

A B C E F G H

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

Small

Medium

Large

FIGURE 4.22 DECISION TREE FOR THE PDC PROBLEM WITH AN ADDED CHANCE NODE CREATED WITH ANALYTIC SOLVER PLATFORM

Small

Medium

Large

0 00

50% Event 1

50% Event 2

A B C D E F G H 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

I J K L

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We next select the cells containing “Event 1” and “Event 2” (cells H2 and H7) and re- name them Strong and Weak to provide the proper names for the PDC states of nature. After doing so, we can copy the subtree for the chance node in cell F5 to the other two decision branches to complete the structure of the PDC decision tree as follows:

Step 1. Select cell F5 Step 2. Click the ANALYTiC SOLVER pLATFORM tab in the Ribbon Step 3. Click decision Tree in the Tools group

Select Node, and click Copy Node Step 4. Select cell F13 Step 5. Click the ANALYTiC SOLVER pLATFORM tab in the Ribbon Step 6. Click decision Tree in the Tools group

Select Node, and click paste Node

This copy-and-paste procedure places a chance node at the end of the Medium deci- sion branch. Repeating the same copy-and-paste procedure for the Large decision branch completes the structure of the PDC decision tree, as shown in Figure 4.23.

FIGURE 4.23 PDC DECISION TREE CREATED WITH ANALYTIC SOLVER PLATFORM

Small

Medium

Large

1 0

50% Strong

50% Weak

50% Strong

50% Weak

50% Strong

50% Weak

A B C D E F G H

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

I J K L

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Inserting Probabilities and Payoffs We now insert probabilities and payoffs into the decision tree. In Figure 4.23, we see that an equal probability of 0.5 is assigned automatically to each of the chance outcomes. For PDC, the probability of strong demand is 0.8 and the probability of weak demand is 1 minus the probability of strong demand, 5 1 2 0.8 5 0.2. We can enter 0.8 into cell H1 and the formula 51 2 H1 into cell H6. We enter the formula 5H1 into cells H11 and H21, and we enter the formula 5H6 into cells H16 and H26. In this way, all probabilities will be updated correctly if we change the value in cell H1.

To insert the payoffs, we enter 8 in H4, 7 in cell H9, 14 in cell H14, 5 in cell H19, 20 in cell H24, and 29 in cell H29. Note in Figure 4.24 that the payoffs also appear in the right-hand margin of the decision tree. The payoffs in the right margin are computed by a formula that adds the payoffs on all of the branches leading to the associated terminal node. For the PDC problem, no payoffs are associated with the decision alternatives branches, so we leave the default values of zero in cells D6, D16, and D26. The PDC decision tree is now complete. After inserting the PDC probabilities and payoffs, the PDC decision tree appears as shown in Figure 4.24.

FIGURE 4.24 PDC DECISION TREE WITH BRANCH PROBABILITIES AND PAYOFFS CREATED WITH ANALYTIC SOLVER PLATFORM

Small

Medium

Large

3 14.2

80% Strong

20% Weak

80% Strong

20% Weak

80% Strong

20% Weak

7.80

–9

12.20

14.2

–9

A B C D E F G H

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

I J K L

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Appendix 4.1 Using Analytic Solver Platform to Create Decision trees 161

Interpreting the Result When probabilities and payoffs are inserted, Analytic Solver Platform automatically makes the rollback computations necessary to determine the optimal solution. Optimal decisions are identified by the number in the corresponding decision node. In the PDC decision tree in Figure 4.24, cell B15 contains the decision node. Note that a “3” appears in this node, which tells us that decision alternative branch 3 provides the optimal decision. We can also easily identify the best decision using the Highlight function in Analytic Solver Platform. To highlight the best decision follow these steps:

Step 1. Click the ANALYTiC SOLVER pLATFORM tab in the Ribbon Step 2. Click decision Tree in the Tools group

Select Highlight, and click Highlight Best

Analytic Solver Platform highlights the best decision for the PDC problem. From Figure 4.25, we see that decision analysis recommends that PDC construct the Large con- dominium complex. The expected value of this decision appears at the beginning of the tree

FIGURE 4.25 PDC DECISION TREE WITH BEST DECISION HIGHLIGHTED CREATED WITH ANALYTIC SOLVER PLATFORM

Small

Medium

Large

14.2

8 88

7.80

80% Strong

20% Weak

2020

–9–9

14.20

80% Strong

20% Weak

1414

12.20

80% Strong

20% Weak

A B C D E F G H 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

I J K

–9

20 21 22 23 24

25 26 27 28

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162 Chapter 4 Decision Analysis

in cell A16. Thus, we see that the optimal expected value is $14.2 million. The expected values of the other decision alternatives are displayed at the end of the corresponding deci- sion branch. Thus, referring to cells E6 and E16, we see that the expected value of the Small complex is $7.8 million and the expected value of the Medium complex is $12.2 million.

Using software such as Analytic Solver Platform to develop decision trees allows for quick and easy sensitivity analysis. We can easily analyze the impact of changing branch probabilities and payoffs by simply changing these values in Excel and observing the im- pact on the optimal decision using Analytic Solver Platform. For instance, if we want to examine the impact of different values of Strong demand on our decision, we can change the value of cell H1 and see whether this changes the optimal decision.

A convenient way to summarize the sensitivity of a decision to a particular parameter is to combine the decision tree from XLMiner with a Data Table in Excel. Suppose we want to evaluate the impact of different probabilities of strong demand over a wide range of possibilities. The Excel worksheet shown in Figure 4.26 demonstrates the use of a Data Table to perform this sensitivity analysis. To create this Data Table, we follow these steps once we have created the decision tree for this problem in XLMiner:

FIGURE 4.26 PDC DECISION TREE AND DATA TABLE ILLUSTRATING SENSITIVITY ANALYSIS CREATED WITH ANALYTIC SOLVER PLATFORM

A B C D E F G H 1

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

I J K L M N O

Small

Medium

Large

3 14.2

80% Strong

20% Weak

80% Strong

20% Weak

80% Strong

20% Weak

7.80

8 0.0 Large Small Small Small Medium Medium Medium Medium Medium Large Large Large

14.2 7

7.1 7.2 7.7 8.6 9.5

10.4 11.3 14.2 17.1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

1414

2020

–9–9

12.20

14.2

–9

Probability Strong Demand

Expected Value

Best Decision

Data Table

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Appendix 4.1 Using Analytic Solver Platform to Create Decision trees 163

Step 1. Enter the values 0.0, 0.1, 0.2, etc. into cells M4 to M14, as shown in Figure 4.25, to represent the different scenarios for the probability of Strong demand

Step 2. Enter 5A16 into cell N3 to keep track of the optimal expected value in each scenario

Step 3. Enter the formula 5CHOOSE(B15,“Small”,“Medium”,“Large”) into cell O3. This will return the best decision in each scenario

Step 4. Select cells M3:O14 Step 5. Click the dATA tab in the Ribbon Step 6. Click What-if Analysis from the data Tools group

Select data Table… Step 7. When the data Table dialog box opens, enter 5H1 into the Column input

cell: box Click OK

Figure 4.26 shows the completed decision tree and Data Table. From Figure 4.26, we see that the best decision is to construct the Small complex if the probability of strong demand is 0, 0.1, or 0.2, the Medium complex if the probability is any value shown between 0.3 and 0.7, and the Large complex if the probability of strong demand is 0.8 or greater. The Data Table also provides the expected values for these decisions in each scenario. Such sensitiv- ity analysis can be greatly beneficial in demonstrating which values should be clarified, if possible, by procuring additional information.

The Excel function CHOOSE chooses a value from a list of possibilities based on the index in the referenced cell. Here, the CHOOSE function chooses “Small” in cell O3 if the value in cell B15 is 1; it enters “Medium” if the value in cell B15 is 2; it enters “Large” if the value in cell B15 is 3.

By entering 5H1 in Step 7, we tell Excel to substitute the values of 0, 0.1, 0.2, and so on for the probability of Strong demand and then return the related outputs.

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Utility and Game Theory

CONTENTS

5.1 THE MEANING OF UTILITY

5.2 UTILITY AND DECISION MAKING The Expected Utility Approach Summary of Steps for

Determining the Utility of Money

5.3 UTILITY: OTHER CONSIDERATIONS Risk Avoiders Versus Risk Takers

5.4 INTRODUCTION TO GAME THEORY Competing for Market Share Identifying a Pure Strategy

5.5 MIXED STRATEGY GAMES A Larger Mixed Strategy Game Summary of Steps for Solving

Two-Person, Zero-Sum Games Extensions

CHAPTER 5

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5.1 The Meaning of Utility 165

The decision analysis situations presented in Chapter 4 often expressed consequences or payoffs in terms of monetary values. With probability information available about the out- comes of the chance events, we defined the optimal decision alternative as the one that provided the best expected monetary value. However, in some situations the decision al- ternative with the best expected monetary value may not be the preferred alternative. A decision maker may also wish to consider intangible factors such as risk, image, or other nonmonetary criteria in order to evaluate the decision alternatives. When monetary value does not necessarily lead to the most preferred decision, expressing the value (or worth) of a consequence in terms of its utility will permit the use of expected utility to identify the most desirable decision alternative. The discussion of utility and its application in decision analysis is presented in the first part of this chapter.

In the last part of this chapter we introduce the topic of game theory. Game theory is the study of developing optimal strategies where two or more decision makers, usually called players, compete as adversaries. Game theory can be viewed as a relative of deci- sion analysis. A key difference, however, is that each player selects a decision strategy not by considering the possible outcomes of a chance event, but by considering the possible strategies selected by one or more competing players.

We note here that utility and game theory are separable topics. Either or both may be studied, and it is not required that you cover one topic before the other. The Q.M. in Action, Game Theory Used in 700-MHz Auction, describes how participants in an FCC auction used game theory to develop bidding strategies.

On January 24, 2008, the Federal Communications Com- mission (FCC) auctioned the rights to operate the 700- MHz frequency band in the United States. This bandwidth became available due to the switch of over-the-air tele- vision broadcasts from analog to digital transmission. The 700-MHz frequency bandwidth is highly desirable to companies because the high frequency can penetrate walls and other obstacles. Companies including Google, AT&T, Verizon Wireless, and many others placed bids on the rights to operate in this frequency band.

Game theory was central to this auction, as it was used by the FCC to establish the overall rules and procedures for the auction. To promote competition, the FCC used a “blind auction” format in which each

bid was anonymous. A blind auction assures that each bidder does not know which competitor(s) they are bidding against. Thus, large firms could not use their market dominance and deep pockets to intimidate smaller firms from placing additional bids. Further, bidding was allowed to continue until no new bids were received in order to prevent last-second winning bids (a practice known as auction sniping).

Most participants hired game theorists to de- vise bid strategies. Economists, mathematicians, engineers, and many others assisted companies in developing optimal bid plans. The auction lasted 261 rounds over 38 days and resulted in 101 winning bid- ders. The auction generated over $19 billion in rev- enue for the FCC.

game theory used in 700-mhz auction*

Q.M. in AcTion

*From E. Woyke, “Peeking into the Spectrum Auction.” Forbes, 2007.

5.1 The Meaning of Utility Utility is a measure of the total worth or relative desirability of a particular outcome; it reflects the decision maker’s attitude toward a collection of factors such as profit, loss, and risk. Researchers have found that as long as the monetary value of payoffs stays within a

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166 chapter 5 Utility and Game Theory

range that the decision maker considers reasonable, selecting the decision alternative with the best expected monetary value usually leads to selection of the most preferred decision. However, when the payoffs are extreme, decision makers are often unsatisfied or uneasy with the decision that simply provides the best expected monetary value.

As an example of a situation in which utility can help in selecting the best decision al- ternative, let us consider the problem faced by Swofford, Inc., a relatively small real estate investment firm located in Atlanta, Georgia. Swofford currently has two investment oppor- tunities that require approximately the same cash outlay. The cash requirements necessary prohibit Swofford from making more than one investment at this time. Consequently, three possible decision alternatives may be considered.

The three decision alternatives, denoted d1, d2, and d3, are

d1 5 make investment A

d2 5 make investment B

d3 5 do not invest

The monetary payoffs associated with the investment opportunities depend on the invest- ment decision and on the direction of the real estate market during the next six months (the chance event). Real estate prices will go up, remain stable, or go down. Thus the states of nature, denoted s1, s2, and s3, are

s1 5 real estate prices go up

s2 5 real estate prices remain stable

s3 5 real estate prices go down

Using the best information available, Swofford has estimated the profits, or payoffs, associ- ated with each decision alternative and state-of-nature combination. The resulting payoff table is shown in Table 5.1.

The best estimate of the probability that real estate prices will go up is 0.3; the best estimate of the probability that prices will remain stable is 0.5; and the best estimate of the probability that prices will go down is 0.2. Thus the expected values for the three decision alternatives are

EV(d1) 5 0.3($30,000) 1 0.5($20,000) 1 0.2(2$50,000) 5 $9000

EV(d2) 5 0.3($50,000) 1 0.5(2$20,000) 1 0.2(2$30,000) 5 2$1000

EV(d3) 5 0.3($0) 1 0.5($0) 1 0.2($0) 5 $0

Using the expected value approach, the optimal decision is to select investment A with an expected monetary value of $9000. Is it really the best decision alternative? Let us consider

State of Nature

Decision Alternative Prices Up s1 Prices Stable s2 Prices Down s3 Investment A, d1 $30,000 $20,000 2$50,000 Investment B, d2 $50,000 2$20,000 2$30,000 Do not invest, d3 $0 $0 $0

TABLE 5.1 PAYOFF TABLE FOR SWOFFORD, INC.

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5.2 Utility and Decision Making 167

some other factors that relate to Swofford’s capability for absorbing the loss of $50,000 if investment A is made and prices actually go down.

Actually, Swofford’s current financial position is weak. This condition is partly re- flected in Swofford’s ability to make only one investment. More important, however, the firm’s president believes that, if the next investment results in a substantial loss, Swofford’s future will be in jeopardy. Although the expected value approach leads to a recommenda- tion for d1, do you think the firm’s president would prefer this decision? We suspect that the president would select d2 or d3 to avoid the possibility of incurring a $50,000 loss. In fact, a reasonable conclusion is that, if a loss of even $30,000 could drive Swofford out of business, the president would select d3, believing that both investments A and B are too risky for Swofford’s current financial position.

The way we resolve Swofford’s dilemma is first to determine Swofford’s utility for the various monetary outcomes. Recall that the utility of any outcome is the total worth of that outcome, taking into account all risks and consequences involved. If the utilities for the various consequences are assessed correctly, the decision alternative with the highest expected utility is the most preferred, or best, alternative. In the next section we show how to determine the utility of the monetary outcomes so that the alternative with the highest expected utility can be identified.

5.2 Utility and Decision Making The procedure we use to establish utility values for the payoffs in Swofford’s situation requires that we first assign a utility value to the best and worst possible payoffs. Any values will work as long as the utility assigned to the best payoff is greater than the utility assigned to the worst payoff. In this case, $50,000 is the best payoff and 2$50,000 is the worst. Suppose, then, that we arbitrarily make assignments to these two payoffs as follows:

Utility of 2$50,000 5 u(2$50,000) 5 0

Utility of $50,000 5 u($50,000) 5 10

Let us now determine the utility associated with every other payoff. Consider the process of establishing the utility of a payoff of $30,000. First we ask

Swofford’s president to state a preference between a guaranteed $30,000 payoff and an opportunity to engage in the following lottery, or bet, for some probability of p that we select:

Lottery: Swofford obtains a payoff of $50,000 with probability p

and a payoff of 2$50,000 with probability (1 2 p)

If p is very close to 1, Swofford’s president would obviously prefer the lottery to the guaranteed payoff of $30,000 because the firm would virtually ensure itself a payoff of $50,000. If p is very close to 0, Swofford’s president would clearly prefer the guarantee of $30,000. In any event, as p increases continuously from 0 to 1, the preference for the guaranteed payoff of $30,000 decreases and at some point is equal to the preference for the lottery. At this value of p, Swofford’s president would have equal preference for the guaranteed payoff of $30,000 and the lottery; at greater values of p, Swofford’s president would prefer the lottery to the guaranteed $30,000 payoff. For example, let us assume that when p = 0.95, Swofford’s president is indifferent between the guaranteed payoff

utility values of 0 and 1 could have been selected here; we selected 0 and 10 in order to avoid any possible confusion between the utility value for a payoff and the probability p.

p is often referred to as the indifference probability.

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168 chapter 5 Utility and Game Theory

of $30,000 and the lottery. For this value of p, we can compute the utility of a $30,000 payoff as follows:

u($30,000) 5 pu($50,000) 1 (1 2 p)u(2$50,000)

5 0.95(10) 1 (0.05)(0)

5 9.5

Obviously, if we had started with a different assignment of utilities for a payoff of $50,000 and 2$50,000, the result would have been a different utility for $30,000. For ex- ample, if we had started with an assignment of 100 for $50,000 and 10 for 2$50,000, the utility of a $30,000 payoff would be

u($30,000) 5 0.95(100) 1 0.05(10)

5 0.95 1 0.05

5 95.5

Hence, we must conclude that the utility assigned to each payoff is not unique but merely depends on the initial choice of utilities for the best and worst payoffs. We will discuss utility choice further at the end of the section. For now, however, we will continue to use a value of 10 for the utility of $50,000 and a value of 0 for the utility of 2$50,000.

Before computing the utility for the other payoffs, let us consider the implication of Swofford’s president assigning a utility of 9.5 to a payoff of $30,000. Clearly, when p 5 0.95, the expected value of the lottery is

EV(lottery) 5 0.95($50,000) 1 0.05(2$50,000)

5 $47,500 2 $2,500

5 $45,000

Although the expected value of the lottery when p 5 0.95 is $45,000, Swofford’s presi- dent is indifferent between the lottery (and its associated risk) and a guaranteed payoff of $30,000. Thus, Swofford’s president is taking a conservative, or risk-avoiding, viewpoint. A decision maker who would choose a guaranteed payoff over a lottery with a superior expected payoff is a risk avoider (or is said to be risk averse). The president would rather have $30,000 for certain than risk anything greater than a 5% chance of incurring a loss of $50,000. In other words, the difference between the EV of $45,000 and the guaranteed payoff of $30,000 is the risk premium that Swofford’s president would be willing to pay to avoid the 5% chance of losing $50,000.

To compute the utility associated with a payoff of 2$20,000, we must ask Swofford’s president to state a preference between a guaranteed 2$20,000 payoff and an opportunity to engage again in the following lottery:

Lottery: Swofford obtains a payoff of $50,000 with probability p

and a payoff of 2$50,000 with probability (1 2 p)

Note that this lottery is exactly the same as the one we used to establish the utility of a payoff of $30,000 (in fact, we can use this lottery to establish the utility for any monetary value in the Swofford payoff table). We need to determine the value of p that would make the president indifferent between a guaranteed payoff of 2$20,000 and the lottery. For

the difference between the expected value of the lottery and the guaranteed payoff can be viewed as the risk premium the decision maker is willing to pay.

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5.2 Utility and Decision Making 169

example, we might begin by asking the president to choose between a certain loss of $20,000 and the lottery with a payoff of $50,000 with probability p 5 0.90 and a payoff of 2$50,000 with probability (1 2 p) 5 0.10. What answer do you think we would get? Surely, with this high probability of obtaining a payoff of $50,000, the president would elect the lottery. Next, we might ask whether p 5 0.85 would result in indifference between the loss of $20,000 for certain and the lottery. Again the president might prefer the lottery. Suppose that we continue until we get to p 5 0.55, at which point the president is indifferent between the payoff of 2$20,000 and the lottery. That is, for any value of p less than 0.55, the president would take a loss of $20,000 for certain rather than risk the potential loss of $50,000 with the lottery; and for any value of p above 0.55, the president would choose the lottery. Thus, the utility assigned to a payoff of 2$20,000 is

u(2$20,000) 5 pu($50,000) 1 (1 2 p)u(2$50,000)

5 0.55(10) 1 0.45(0)

5 5.5

Again let us assess the implication of this assignment by comparing it to the expected value approach. When p 5 0.55, the expected value of the lottery is

EV(lottery) 5 0.55($50,000) 1 0.45(2$50,000)

5 $27,500 2 $22,500

5 $5000

Thus, Swofford’s president would just as soon absorb a certain loss of $20,000 as take the lottery and its associated risk, even though the expected value of the lottery is $5000. Once again this preference demonstrates the conservative, or risk-avoiding, point of view of Swofford’s president.

In these two examples we computed the utility for the monetary payoffs of $30,000 and 2$20,000. We can determine the utility for any monetary payoff m in a similar fashion. First, we must find the probability p for which the decision maker is indifferent between a guaranteed payoff of m and a lottery with a payoff of $50,000 with probability p and 2$50,000 with probability (1 2 p). The utility of m is then computed as follows:

u(m) 5 pu($50,000) 1 (1 2 p)u(2$50,000)

5 p(10) 1 (1 2 p)0

5 10p

Using this procedure, we developed utility values for the rest of the payoffs in Swofford’s problem. The results are presented in Table 5.2.

Now that we have determined the utility value of each of the possible monetary values, we can write the original payoff table in terms of utility values. Table 5.3 shows the utility for the various outcomes in the Swofford problem. The notation we use for the entries in the utility table is uij, which denotes the utility associated with decision alternative di and state of nature sj. Using this notation, we see that u23 5 4.0.

The Expected Utility Approach We can now apply the expected value computations introduced in Chapter 4 to the utili- ties in Table 5.3 in order to select an optimal decision alternative for Swofford, Inc. How- ever, because utility values represent such a special case of expected value, we will refer to

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170 chapter 5 Utility and Game Theory

Monetary Indifference Utility Value Value of p Value $50,000 Does not apply 10.0 $30,000 0.95 9.5 $20,000 0.90 9.0 $ 0 0.75 7.5 2$20,000 0.55 5.5 2$30,000 0.40 4.0 2$50,000 Does not apply 0.0

TABLE 5.2 UTILITY OF MONETARY PAYOFFS FOR SWOFFORD, INC.

State of Nature

Decision Alternative Prices Up s1 Prices Stable s2 Prices Down s3 Investment A, d1 9.5 9.0 0.0 Investment B, d2 10.0 5.5 4.0 Do not invest, d3 7.5 7.5 7.5

TABLE 5.3 UTILITY TABLE FOR SWOFFORD, INC.

the expected value when applied to utility values as the expected utility (EU). Thus, the expected utility approach requires the analyst to compute the expected utility for each decision alternative and then select the alternative yielding the highest expected utility. With n possible states of nature, the expected utility of a decision alternative di is given by

EUsdid 5 o n

j5 1

Pssjduij (5.1)

The expected utility for each of the decision alternatives in the Swofford problem is

EU(d1) 5 0.3(9.5) 1 0.5(9.0) 1 0.2(0.0) 5 7.35

EU(d2) 5 0.3(10.0) 1 0.5(5.5) 1 0.2(4.0) 5 6.55

EU(d3) 5 0.3(7.5) 1 0.5(7.5) 1 0.2(7.5) 5 7.50

Note that the optimal decision using the expected utility approach is d3, do not invest. The ranking of alternatives according to the president’s utility assignments and the associated monetary values is as follows:

can you use the expected utility approach to determine the optimal decision? try Problem 1.

Ranking of Expected Decision Expected Monetary

Alternatives Utility Value Do not invest 7.50 $ 0 Investment A 7.35 $9000 Investment B 6.55 2$1000

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5.2 Utility and Decision Making 171

Note also that although investment A had the highest expected monetary value of $9000, the analysis indicates that Swofford should decline this investment. The rationale behind not selecting investment A is that the 0.20 probability of a $50,000 loss was considered to involve a serious risk by Swofford’s president. The seriousness of this risk and its associated impact on the company were not adequately reflected by the expected monetary value of investment A. We assessed the utility for each payoff to assess this risk adequately.

NOTES AND COMMENTS

1. In the Swofford problem we have been using a utility of 10 for the best payoff and 0 for the worst. We could have chosen any values so long as the utility associated with the best payoff ex- ceeds the utility associated with the worst payoff. In practice, frequently a utility of 1 is associated with the best payoff and a utility of 0 is associated with the worst payoff. Had we made this choice, the utility for any monetary value m would have been the value of p at which the decision maker was indifferent between a guaranteed payoff of

m and a lottery in which the best payoff is ob- tained with probability p and the worst payoff is obtained with probability (1 2 p). Thus, the utility for any monetary value would have been equal to the probability of earning the best pay- off. Often this choice is made because of the ease in computation. We chose not to do so to empha- size the distinction between the utility values and the indifference probabilities for the lottery.

2. Other approaches to decision analysis, such as the optimistic, conservative, and minimax regret

Summary of Steps for Determining the Utility of Money Before considering other aspects of utility, let us summarize the steps involved in determin- ing the utility for a monetary value and using it within the decision analysis framework. The following steps state in general terms the procedure used to solve the Swofford, Inc., investment problem:

Step 1. Develop a payoff table using monetary values. Step 2. Identify the best and worst payoff values in the table and assign each a utility

value, with u(best payoff) . u(worst payoff). Step 3. For every other monetary value m in the original payoff table, do the following

to determine its utility value: a. Define the lottery: The best payoff is obtained with probability p and the

worst payoff is obtained with probability (1 2 p). b. Determine the value of p such that the decision maker is indifferent between

a guaranteed payoff of m and the lottery defined in step 3(a). c. Calculate the utility of m as follows:

u(m) 5 pu(best payoff) 1 (1 2 p)u(worst payoff)

Step 4. Convert the payoff table from monetary values to utility values. Step 5. Apply the expected utility approach to the utility table developed in step 4 and

select the decision alternative with the highest expected utility.

(continued)

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172 chapter 5 Utility and Game Theory

5.3 Utility: Other Considerations In this section we describe how a risk-avoiding decision maker and a risk-taking decision maker differ in their assessment of utility. Expected utility is then used to show how a risk-avoiding decision maker and a risk-taking decision maker may prefer different decision alternatives for the same decision problem. We close this section by comparing expected monetary value and expected utility as criteria for decision making.

Risk Avoiders Versus Risk Takers The financial position of Swofford, Inc., was such that the firm’s president evaluated investment opportunities from a conservative, or risk-avoiding, point of view. How- ever, if the firm had a surplus of cash and a stable future, Swofford’s president might have been looking for investment alternatives that, although perhaps risky, contained a potential for substantial profit. That type of behavior would demonstrate that the president is a risk taker with respect to this decision.

A risk taker is a decision maker who would choose a lottery over a guaranteed payoff when the expected value of the lottery is inferior to the guaranteed payoff. In this section we analyze the decision problem faced by Swofford from the point of view of a decision maker who would be classified as a risk taker. We then compare the con- servative, or risk-avoiding, point of view of Swofford’s president with the behavior of a decision maker who is a risk taker.

For the decision problem facing Swofford, Inc., using the general procedure for de- veloping utilities as discussed in Section 5.2, a risk taker might express the utility for the various payoffs shown in Table 5.4. As before, u($50,000) 5 10 and u(2$50,000) 5 0. Note the difference in behavior reflected in Table 5.4 and Table 5.2. That is, in de- termining the value of p at which the decision maker is indifferent between a guar- anteed payoff of m and a lottery in which $50,000 is obtained with probability p and 2$50,000 with probability (1 2 p), the risk taker is willing to accept a greater risk of incurring a loss of $50,000 in order to gain the opportunity to realize a profit of $50,000.

To help develop the utility table for the risk taker, we have reproduced the Swof- ford, Inc., payoff table in Table 5.5. Using these payoffs and the risk taker’s utility val- ues given in Table 5.4, we can write the risk taker’s utility table as shown in Table 5.6.

approaches covered in Chapter 4, can be applied to utilities.

3. The procedure we described for determining the utility of monetary consequences can also be used to develop a utility measure for non- monetary consequences. Assign the best con- sequence a utility of 10 and the worst a utility of 0. Then create a lottery with a probability of p for the best consequence and (1 2 p) for the

worst consequence. For each of the other conse- quences, find the value of p that makes the de- cision maker indifferent between the lottery and the consequence. Then calculate the utility of the consequence in question as follows:

u(consequence) 5 pu(best consequence) 1 (1 2 p)u(worst consequence)

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5.3 Utility: other considerations 173

Using the state-of-nature probabilities P(s1) 5 0.3, P(s2) 5 0.5, and P(s3) 5 0.2, the expected utility for each decision alternative is

EU(d1) 5 0.3(5.0) 1 0.5(4.0) 1 0.2(0.0) 5 3.50

EU(d2) 5 0.3(10.0) 1 0.5(1.5) 1 0.2(1.0) 5 3.95

EU(d3) 5 0.3(2.5) 1 0.5(2.5) 1 0.2(2.5) 5 2.50

What is the recommended decision? Perhaps somewhat to your surprise, the analy- sis recommends investment B, with the highest expected utility of 3.95. Recall that this investment has a 2$1000 expected monetary value. Why is it now the recommended decision? Remember that the decision maker in this revised problem is a risk taker. Thus, although the expected value of investment B is negative, utility analysis has shown that this decision maker is enough of a risk taker to prefer investment B and its potential for the $50,000 profit.

Monetary Indifference Utility Value Value of p Value $50,000 Does not apply 10.0 $30,000 0.50 5.0 $20,000 0.40 4.0 $ 0 0.25 2.5 2$20,000 0.15 1.5 2$30,000 0.10 1.0 2$50,000 Does not apply 0.0

TABLE 5.4 REVISED UTILITY VALUES FOR SWOFFORD, INC., ASSUMING A RISK TAKER

State of Nature

Decision Alternative Prices Up s1 Prices Stable s2 Prices Down s3 Investment A, d1 $30,000 $20,000 2$50,000 Investment B, d2 $50,000 2$20,000 2$30,000 Do not invest, d3 $0 $0 $0

TABLE 5.5 PAYOFF TABLE FOR SWOFFORD, INC.

State of Nature

Decision Alternative Prices Up s1 Prices Stable s2 Prices Down s3 Investment A, d1 5.0 4.0 0.0 Investment B, d2 10.0 1.5 1.0 Do not invest, d3 2.5 2.5 2.5

TABLE 5.6 UTILITY TABLE OF A RISK TAKER FOR SWOFFORD, INC.

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174 chapter 5 Utility and Game Theory

The expected utility values give the following order of preference of the decision alter- natives for the risk taker and the associated expected monetary values:

Ranking of Expected Decision Expected Monetary

Alternatives Utility Value Investment B 3.95 2$1000 Investment A 3.50 $9000 Do not invest 2.50 $0

Comparing the utility analysis for a risk taker with the more conservative preferences of the president of Swofford, Inc., who is a risk avoider, we see that, even with the same decision problem, different attitudes toward risk can lead to different recommended deci- sions. The utility values established by Swofford’s president indicated that the firm should not invest at this time, whereas the utilities established by the risk taker showed a prefer- ence for investment B. Note that both of these decisions differ from the best expected monetary value decision, which was investment A.

We can obtain another perspective of the difference between behaviors of a risk avoider and a risk taker by developing a graph that depicts the relationship between monetary value and utility. We use the horizontal axis of the graph to represent monetary values and the vertical axis to represent the utility associated with each monetary value. Now, consider the data in Table 5.2, with a utility value corresponding to each monetary value for the original Swofford, Inc., problem. These values can be plotted on a graph such as that in Figure 5.1, and a curve can be drawn through the observed points. The resulting curve is the utility

0 -50

U ti

li ty

-40 -30 -20 -10 0 10 20 30 40 50 Monetary Value ($1000s)

FIGURE 5.1 UTILITY FUNCTION FOR MONEY FOR THE RISK AVOIDER

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5.3 Utility: other considerations 175

function for money for Swofford’s president. Recall that these points reflected the conservative, or risk-avoiding, nature of Swofford’s president. Hence, we refer to the curve in Figure 5.1 as a utility function for a risk avoider. Using the data in Table 5.4 developed for a risk taker, we can plot these points on a graph such as that in Figure 5.2. The resulting curve depicts the utility function for a risk taker.

By looking at the utility functions of Figures 5.1 and 5.2, we can begin to generalize about the utility functions for risk avoiders and risk takers. Although the exact shape of the utility function will vary from one decision maker to another, we can see the general shape of these two types of utility functions. The utility function for a risk avoider shows a diminishing marginal return for money. For example, the increase in utility going from a monetary value of 2$30,000 to $0 is 7.5 2 4.0 5 3.5, whereas the increase in utility in going from $0 to $30,000 is only 9.5 2 7.5 5 2.0.

However, the utility function for a risk taker shows an increasing marginal return for money. For example, in Figure 5.2, the increase in utility in going from 2$30,000 to $0 is 2.5 2 1.0 5 1.5, whereas the increase in utility in going from $0 to $30,000 is 5.0 – 2.5 5 2.5. Note also that in either case the utility function is always increasing; that is, more money leads to more utility. All utility functions possess this property.

We have concluded that the utility function for a risk avoider shows a diminishing marginal return for money and that the utility function for a risk taker shows an increasing marginal return. When the marginal return for money is neither decreasing nor increasing but remains constant, the corresponding utility function describes the behavior of a decision maker who is neutral to risk. The following characteristics are associated with a risk-neutral decision maker:

1. The utility function can be drawn as a straight line connecting the “best” and the “worst” points.

try Problem 5 for practice in plotting the utility function for risk-avoider, risk-taker, and risk-neutral decision makers.

-50 Monetary Value ($1000s)

U ti

li ty

-40 -30 -20 -10 0 10 20 30 40 50

FIGURE 5.2 UTILITY FUNCTION FOR MONEY FOR THE RISK TAKER

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176 chapter 5 Utility and Game Theory

2. The expected utility approach and the expected value approach applied to monetary payoffs result in the same action.

Figure 5.3 depicts the utility function of a risk-neutral decision maker using the Swofford, Inc., problem data. For comparison purposes, we also show the utility functions for the cases where the decision maker is either a risk taker or a risk avoider.

Generally, when the payoffs for a particular decision-making problem fall into a reason- able range—the best is not too good and the worst is not too bad—decision makers tend to express preferences in agreement with the expected monetary value approach. Thus, we suggest asking the decision maker to consider the best and worst possible payoffs for a problem and assess their reasonableness. If the decision maker believes that they are in the reasonable range, the decision alternative with the best expected monetary value can be used. However, if the payoffs appear unreasonably large or unreasonably small (for example, a huge loss) and if the decision maker believes that monetary values do not ad- equately reflect her or his true preferences for the payoffs, a utility analysis of the problem should be considered.

Unfortunately, determination of the appropriate utilities is not a trivial task. As we have shown, measuring utility requires a degree of subjectivity on the part of the decision maker, and different decision makers will have different utility functions. This aspect of utility often causes decision makers to feel uncomfortable about using the expected utility approach. However, if you encounter a decision situation in which you are convinced that monetary value is not the sole measure of performance, and if you agree that a quantitative analysis of the decision problem is desirable, you should recommend that utility analysis be considered.

-50 Monetary Value ($1000s)

U ti

li ty

-40 -30 -20 -10 0 10 20 30 40 50

Risk Neutral

Risk Avoider

Risk Taker

FIGURE 5.3 UTILITY FUNCTION FOR RISK-AVOIDER, RISK-TAKER, AND RISK- NEUTRAL DECISION MAKERS

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5.4 introduction to Game Theory 177

5.4 Introduction to Game Theory In decision analysis, a single decision maker seeks to select an optimal decision alternative after considering the possible outcomes of one or more chance events. In game theory, two or more decision makers are called players, and they compete as adversaries against each other. Each player selects a strategy independently without knowing in advance the strategy of the other player or players. The combination of the competing strategies pro- vides the value of the game to the players. Game theory applications have been developed for situations in which the competing players are teams, companies, political candidates, armies, and contract bidders.

In this section we describe two-person, zero-sum games. two-person means that two competing players take part in the game. Zero-sum means that the gain (or loss) for one player is equal to the corresponding loss (or gain) for the other player (what one player wins, the other player loses). As a result, the gain and loss balance out so that the game results

until 1944, when Von neumann and morgenstern published the book Theory of Games and Economic Behavior, the literature on decisions involving risk consisted primarily of applications involving the use of probability in gambling.

NOTES AND COMMENTS

1. Circumstances often dictate whether one acts as a risk avoider or a risk taker when making a decision. You may think of yourself as a risk avoider when faced with financial decisions, but if you have ever purchased a lottery ticket you have actually acted as a risk taker. For ex- ample, suppose you purchase a $1.00 lottery ticket for a simple lottery in which the object is to pick the six numbers that will be drawn from 50 potential numbers. Also suppose that if you win (correctly choose all six numbers that are drawn), you will receive $1,000,000. There are 15,890,700 possible winning combinations, so your probability of winning is 1y15,890,700 < 0.00000006293 (i.e., very low) and the ex- pected monetary value of your ticket is

15,890,700 s$1,000,000 2 1$d

111 2 115,890,7002s2$1d 5

15,890,700 s$1,000,000 2 $1d

1 15,890,699

15,890,700 s2$1d

< 2$0.94

(Why is the payoff for winning equal to $1,000,000 2 $1? You had to pay for your ticket!)

If a lottery ticket has a negative expected value, why does anyone play? The answer is in utility—most people who play lotteries associ- ate great utility with the possiblity of winning the $1,000,000 prize and relatively little util- ity with the $1.00 cost for a ticket, and so the expected value of the utility of the lottery ticket is positive even though the expected monetary value of the ticket is negative.

2. In many decision-making problems, expected monetary value and expected utility will lead to identical recommendations. In fact, this re- sult will always be true if the decision maker is risk neutral. In general, if the decision maker is almost risk neutral over the range of payoffs (from lowest to highest) for a particular deci- sion problem, the decision alternative with the best expected monetary value leads to selection of the most preferred decision alternative. The trick lies in recognizing the range of monetary values over which a decision maker’s utility function is risk neutral.

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178 chapter 5 Utility and Game Theory

in the sum of zero. Let us demonstrate a two-person, zero-sum game and its solution by considering two companies competing for market share.

Competing for Market Share Suppose that two companies are the only manufacturers of a particular product; they com- pete against each other for market share. In planning a marketing strategy for the coming year, each company is considering three strategies designed to take market share from the other company. The three strategies, assumed to be the same for both companies, are as follows:

Strategy 1 Increase advertising

Strategy 2 Provide quantity discounts

Strategy 3 Extend product warranty

A payoff table showing the percentage gain in the market share for Company A ex- pected for each combination of strategies follows. The notations a1, a2, and a3 identify the three strategies for Company A; the notations b1, b2, and b3 identify the three strategies for Company B. It is a zero-sum game because any gain in market share for Company A is a loss in market share for Company B.

Company B

Increase Quantity Extend Advertising Discounts Warranty b1 b2 b3 Increase Advertising, a1 4 3 2 Company A Quantity Discounts, a2 21 4 1 Extend Warranty, a3 5 22 0

In interpreting the entries in the table we see that if Company A increases advertising (a1) and Company B increases advertising (b1), Company A will come out ahead with an increase in market share of 4%; because this is a zero-sum game, Company B will suffer a corresponding 4% decrease in market share. On the other hand, if Company A provides quantity discounts (a2) and Company B increases advertising (b1), Company A is projected to lose 1% of market share to Company B. Company A is seeking payoff values that show relatively large increases in its market share. Company B is seeking payoff values that show decreases or small increases in Company A’s market share, and thus better results for Company B.

This game involving market share meets the requirements of a two-person, zero-sum game. The two companies are the two players and the zero-sum occurs because the gain (or loss) in market share for Company A is the same as the loss (or gain) in market share for Company B. Due to the planning horizon, each company must select a strategy be- fore knowing the other company’s strategy. What are the optimal strategies for the two companies?

The logic of game theory assumes that each company or player has the same infor- mation and will select a strategy that provides the best possible outcome from its point of view. Suppose Company A selects strategy a1. Market share increases of 4%, 3%,

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5.4 introduction to Game Theory 179

or 2% are possible depending upon Company B’s strategy. If Company B believes that Company A will use strategy a1, then Company B will employ strategy b3. Under the as- sumption that Company B will select the strategy that is best for it, Company A analyzes the game by protecting itself against the actions of Company B. In doing so, Company A identifies the minimum possible payoff for each of its actions. This payoff is the mini- mum value in each row of the payoff matrix. These row minimums are computed in the payoff table as follows:

Company B

Increase Quantity Extend Advertising Discounts Warranty b1 b2 b3 Minimum

Increase Advertising, a1 4 3 2 2

Company A Quantity Discounts, a2 21 4 1 21 Extend Warranty, a3 5 22 0 22

Maximum of row minimums

Considering the entries in the Minimum column, we see that Company A can be guaranteed an increase in market share of at least 2% by selecting the strategy that provides the maxi- mum of the row minimums (strategy a1). Thus, Company A follows a maximin procedure and selects strategy a1 as its best strategy.

Let us now look at the payoff table from the point of view of the other player, Company B. The entries in the payoff table represent losses in market share. Consider what happens to Company B if strategy b1 is selected. Market share decreases of 4%, 21%, and 5% are possible. Under the assumption that Company A will select the strategy that is best for it, Company B knows that if it selects strategy b1, a loss in market share of as much as 5% could be incurred. Thus, Company B analyzes the game by considering the maximum value in each column, which provides the maximum decrease in its market share for each Company A strategy. These column maximums are computed as follows:

By considering the entries in the Maximum row, Company B can be guaranteed a decrease in market share of no more than 2% by selecting the strategy that provides the minimum of the column maximums (strategy b3). Thus, Company B follows a minimax procedure and

Company B

Increase Quantity Extend Advertising Discounts Warranty b1 b2 b3 Minimum

Increase Advertising, a1 4 3 2 2 Company A Quantity Discounts a2 21 4 1 21 Extend Warranty, a3 5 22 0 22

Maximum 5 4 2

Minimum of column maximums

Maximum of row minimums

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180 chapter 5 Utility and Game Theory

selects strategy b3 as its best strategy. Under strategy b3, Company B knows that Company A cannot gain more than 2% in market share.

Identifying a Pure Strategy Whenever the maximum of the row minimums equals the minimum of the column maxi- mums, the players cannot improve their outcomes by changing strategies. The game is said to have a saddle point. With a saddle point, the optimal strategies and the value of the game cannot be improved by either player changing strategies. Thus, a pure strategy has been identified as being optimal for both players. The requirement for a pure strategy is as follows:

Maximum(Row minimums) 5 Minimum(Column maximums)

That is, the maximin value for Player A equals the minimax value for Player B, and so Player A would not alter his strategy even if he knew Player B’s strategy and Player B would not alter her strategy even if she knew Player A’s strategy. In our example, the solution to the game is a pure strategy. If Company A increases its advertising (strategy a1), Company B’s optimal strategy is to extend its product warranty (strategy b3); if Company B chooses to extend its product warranty (strategy b3), Company A’s optimal strategy is to increase its advertising (strategy a1). The value of the game shows that this optimal solution will increase Company A’s market share by 2% and decrease Company B’s market share by 2%.

With a pure strategy, neither player can improve its position by changing to a different strategy. In our marketing example, the pure strategy for Company A is a1. When Com- pany B selects its pure strategy b3, the value of the game shows an increase in Company A’s market share of 2%. Note that if Company B tries to change its pure strategy from b3, Company A’s market share will increase 4% if b1 is selected or will increase 3% if b2 is selected. Company B must stay with its pure strategy b3 to obtain its best result. Similarly, note that if Company A tries to change its pure strategy from a1, Company A’s market share will increase only 1% if a2 is selected or will not increase at all if a3 is selected. Company A must stay with its pure strategy a1 in order to keep its 2% increase in market share. Thus, even if one of the players discovered in advance the opponent’s strategy, no advantage could be gained by switching to a different strategy.

When a pure strategy is optimal for a two-person, zero-sum game, the following steps will find the optimal strategy for each player:

Step 1. Compute the minimum payoff for each row (Player A). Step 2. For Player A, select the strategy that provides the maximum of the row

minimums. Step 3. Compute the maximum payoff for each column (Player B). Step 4. For Player B, select the strategy that provides the minimum of the column

maximums. Step 5. If the maximin value (step 2) equals the minimax value (step 4), an optimal

pure strategy exists for both players. The optimal strategy for Player A is iden- tified in step 2, and the optimal strategy for Player B is identified in step 4. The value of the game is given by the value at the saddle point where the optimal strategies for both players intersect.

If in step 5 the maximin value for Player A does not equal the minimax value for Player B, a pure strategy is not optimal for the two-person, zero-sum game. In this case, a mixed strat- egy is best. In the next section we show when it is necessary to employ a mixed strategy.

if a pure strategy exists, it is the optimal solution for the game.

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5.5 Mixed Strategy Games 181

5.5 Mixed Strategy Games Consider the two-person, zero-sum game that occurs in a football game. The two competing players are the two football teams. On each play, the game is zero-sum because the yardage gained by one team is equal to the yardage lost by the other team. As usual in game theory, each team must select its strategy before knowing the strategy selected by the other team. In this example, let Team A be the team on offense trying to gain yardage and Team B be the team on defense trying to keep the yardage gained by Team A to a minimum. We define the offensive strategies for Team A as follows:

a1 5 running play

a2 5 passing play

The defensive strategies for Team B are as follows:

b1 5 run defense

b2 5 pass defense

The payoff table shows the yardage gained by Team A depending upon the strategies selected by the two teams.

Team B

Run Defense Pass Defense b1 b2

Team A Run, a1 1 6

Pass, a2 15 0

Team B

Run Defense Pass Defense b1 b2 Minimum

Run, a1 1 6 1

Team A Pass, a2 15 0 0

Maximum 15 6

Applying the five-step procedure used to identify a pure strategy, the row minimums and the column maximums are as follows:

The maximum of the row minimums is 1 and the minimum of the column maximums is 6. Because these values are not equal, the two-person, zero-sum game does not have an opti- mal pure strategy. In this case, a mixed strategy solution is best. With a mixed strategy the optimal solution for each player is to randomly select among the alternative strategies. In the football example, then, the offensive Team A will mix up or vary its selection of running (a1) and passing (a2) plays, while the defensive Team B will mix up or vary its selection of a run defense (b1) and a pass defense (b2).

When you think about a football game, it becomes clear that a pure strategy such as Team A always selecting a running play would not work. Team B would recognize Team

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182 chapter 5 Utility and Game Theory

A’s pure strategy and would always be prepared with a run defense. Thus, a Team A mixed strategy of sometimes running and sometimes passing would make sense. When a mixed strategy solution is needed, game theory will determine the optimal probabilities for each strategy for each player. That is, the game theory solution of the football example will tell the offensive team the optimal probabilities for a running play and a passing play. At the same time, the solution will tell the defensive team the optimal probabilities for a run defense and a pass defense. The following discussion shows how to calculate these mixed strategy probabilities.

Let

p 5 the probability Team A selects a running play

(l 2 p) 5 the probability Team A selects a passing play

When a mixed strategy solution exists, we seek to determine the probability p for Team A such that Team B cannot improve its result by changing its defensive strategy. First assume that Team B selects a run defense as shown in column b1. If Team A selects a running play with probability p and a passing play with probability (1 2 p), the expected value of the yardage gain for Team A is computed as follows:

If Team B selects b1:

EV(Yardage) 5 1p 1 15(1 2 p)

If Team B selects its pass defense as shown in column b2, the expected value of the yardage gain for Team A will be as follows:

If Team B selects b2:

EV(Yardage) 5 6p 1 0(1 2 p) 5 6p

To guarantee that Team B cannot change its strategy and decrease the expected value of the yardage gained by Team A, we set the two expected values equal and solve for the value of p.

1p 1 15(1 2 p) 5 6p

1p 1 15 2 15p 5 6p

20p 5 15

p 5 15y20 5 0.75

With p 5 0.75, (1 2 p) 5 1 2 0.75 5 0.25. This result tells Team A it should select a running play with a 0.75 probability and a passing play with a 0.25 probability. The expected value of the yardage gained, which is the value of the game, is

EV(Yardage) 5 1p 1 15(1 2 p) 5 1(0.75) 1 15(0.25) 5 4.5 yards per play

Now let us consider the optimal probabilities for Team B. Let

q 5 the probability Team B selects a run defense

(1 2 q) 5 the probability Team B selects a pass defense

Using the same logic we used for computing Team A’s optimal probabilities, we want to determine the value of q such that Team A cannot increase the expected value of the yardage

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5.5 Mixed Strategy Games 183

gained by changing its offensive strategy. We first compute the expected value of the yardage for Team B for the following two cases:

If Team A selects a1:

EV(Yardage) 5 lq 1 6(1 2 q)

If Team A selects a2:

EV(Yardage) 5 15q 1 0(1 2 q) 5 15q

To guarantee that Team A cannot change its strategy and affect the expected value of the yardage for Team B, we set the two expected values equal and solve for the value of q as follows:

1q 1 6(1 2 q) 5 15q

1q 1 6 2 6q 5 15q

20q 5 6

q 5 6y20 5 0.30

With q 5 0.30, (1 2 q) 5 1 2 0.30 5 0.70. This result tells Team B that it should select a run defense with a 0.30 probability and a pass defense with a 0.70 probability. The expected yardage gained, which is the value of the game, will remain 4.5 yards per play.

Thus, we have the optimal mixed strategy solution for the football game example. Any 2 3 2 two-person, zero-sum mixed strategy game can be solved algebraically as shown in this example. If a larger two-person, zero-sum game involves a mixed strategy, solving it is a bit more complicated.

A Larger Mixed Strategy Game Consider the following two-person, zero-sum game:

Player B

b1 b2 b3

a1 0 21 2 Player A a2 5 4 23 a3 2 3 24

Player B

b1 b2 b3 Minimum

a1 0 21 2 21 Player A a2 5 4 23 23 a3 2 3 24 24

Maximum 5 4 2

Following the usual procedure for identifying a pure strategy, we compute the row mini- mums and the column maximums:

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184 chapter 5 Utility and Game Theory

The maximum of the row minimums is 21 and the minimum of the column maximums is 2. Because the maximin and minimax values are not equal, the two-person, zero-sum game does not have an optimal pure strategy. However, with a problem larger than 2 3 2, we cannot use the algebraic solution for the mixed strategy probabilities as we did in the previous example.

If a game larger than 2 3 2 requires a mixed strategy, we first look for dominated strategies in order to reduce the size of the game. A dominated strategy exists if another strategy is at least as good regardless of what the opponent does. For example, consider strategies a2 and a3. The payoff table shows that in column b1, 5 . 2; in column b2, 4 . 3; and in column b3, 23 . 24. Thus, regardless of what Player B does, Player A will always prefer the higher values of strategy a2 compared to strategy a3. Thus, strategy a3 is dominated by strategy a2 and can be dropped from consideration by Player A. Eliminat- ing dominated strategies from the game reduces its size. After eliminating a3, the reduced game becomes

Player B

b1 b3

Player A a1 21 2

a2 4 23

Player B

b1 b2 b3

Player A a1 0 21 2

a2 5 4 23

Next, we look for more dominated strategies. Player A finds no other dominated strat- egies. However, consider strategies b1 and b2 for Player B. Remember that Player B is interested in smaller values. The payoff table shows that in row a1, 21 , 0, and in row a2, 4 , 5. Thus, regardless of what Player A does, Player B would always prefer the smaller values of strategy b2 compared to strategy b1. Thus, strategy b1 is dominated by strategy b2 and can be eliminated from the game. With this dominated strategy eliminated, the reduced game becomes

By successively eliminating dominated strategies, we reduce the game to a 2 3 2 game. The algebraic solution procedure described earlier in this section can now be used to identify the optimal probabilities for the mixed strategy solution.

Finally, it is important to realize that no hard-and-fast rule identifies dominated strategies. Basically, the analyst must make pairwise comparisons of the decision strat- egies in an attempt to identify dominated strategies. The goal is to identify and elimi- nate dominated strategies sequentially in order to reduce the game to a 2 3 2 game so that an algebraic solution procedure can be used to solve for the mixed strategy probabilities.

Problem 14 asks you to find the optimal probabilities for this example.

identifying and eliminating dominated strategies may reduce the game to a 2 3 2 game. if so, an algebraic procedure may be used to determine the mixed strategy solution.

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Summary 185

Summary of Steps for Solving Two-Person, Zero-Sum Games The following summary shows the steps used to solve two-person, zero-sum games:

1. Use the maximin procedure for Player A and the minimax procedure for Player B to determine whether a pure strategy solution exists. (See previous steps for identify- ing a pure strategy.) If a pure strategy exists, it is the optimal solution.

2. If a pure strategy does not exist and the game is larger than 2 3 2, identify a domi- nated strategy to eliminate a row or column. Develop the reduced payoff table and continue to use dominance to eliminate as many additional rows and columns as possible.

3. If the reduced game is 2 3 2, solve for the optimal mixed strategy probabilities algebraically.

If the game cannot be reduced to a 2 3 2 game, a linear programming model can be used to solve for the optimal mixed strategy probabilities. The formulation of a linear program- ming model to solve these larger game theory problems is beyond the scope of this text.

Extensions We presented the basic model for two-person, zero-sum games. However, game theory models extend beyond two-person, zero-sum games. One extension is a two-person, constant-sum game that occurs when the payoffs for the strategies chosen sum to a constant other than zero. In addition, game theory can be extended to include more general n-person games. Coopera- tive games where players are allowed preplay communications are another variation. Finally, some game theory models allow an infinite number of strategies to be available for the players.

Summary

In this chapter we showed how utility could be used in decision-making situations in which monetary value did not provide an adequate measure of the payoffs. Utility is a measure of the total worth of a consequence. As such, utility takes into account the decision maker’s assessment of all aspects of a consequence, including profit, loss, risk, and perhaps addi- tional nonmonetary factors. The examples showed how the use of expected utility can lead to decision recommendations that differ from those based on expected monetary value.

A decision maker’s judgment must be used to establish the utility for each consequence. We presented a step-by-step procedure to determine a decision maker’s utility for monetary payoffs. We also discussed how conservative, risk-avoiding decision makers assess utility differently from more aggressive, risk-taking decision makers. If the decision maker is risk neutral, we showed that the solution using expected utility is identical to the solution using expected monetary value.

We presented an introduction to game theory by describing how to solve two-person, zero-sum games. In these games, the two players end up with the sum of the gain (loss) to one player and the loss (gain) to the other player always equal to zero. We described the steps that can be used to determine whether a two-person, zero-sum game results in an optimal pure strategy. If a pure strategy is optimal, a saddle point determines the value of the game. If an optimal pure strategy does not exist for a two-person, zero-sum 2 3 2 game, we showed how to identify an optimal mixed strategy. With a mixed strategy, each player uses probability to select a strategy for each play of the game. We showed how dominance could be used to reduce the size of mixed strategy games. If the elimination of dominated

in 1994, John harsanui. John nash, and reinhard selten received the nobel Prize in economics for their work on noncooperative game theory.

Problem 17 demonstrates a simple constant-sum game theory problem.

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186 chapter 5 Utility and Game Theory

strategies can reduce a larger game to a 2 3 2 game, an algebraic solution procedure can be used to find a solution. If the game cannot be reduced to a 2 3 2 game, a linear program- ming model is needed to determine the optimal mixed strategy solution.

Glossary

Dominated strategy A strategy is dominated if another strategy is at least as good for every strategy that the opposing player may employ. A dominated strategy will never be selected by the player and as such can be eliminated in order to reduce the size of the game. Expected utility (EU) The weighted average of the utilities associated with a decision alternative. The weights are the state-of-nature probabilities. Game theory The study of decision situations in which two or more players compete as adversaries. The combination of strategies chosen by the players determines the value of the game to each player. Lottery A hypothetical investment alternative with a probability p of obtaining the best payoff and a probability of (1 2 p) of obtaining the worst payoff. Mixed strategy A game solution in which the player randomly selects the strategy to play from among several strategies with positive probabilities. The solution to the mixed strategy game identifies the probabilities that each player should use to randomly select the strategy to play. Pure strategy A game solution that provides a single best strategy for each player. Risk avoider A decision maker who would choose a guaranteed payoff over a lottery with a better expected payoff. Risk taker A decision maker who would choose a lottery over a better guaranteed payoff. Risk-neutral decision maker A decision maker who is neutral to risk. For this decision maker the decision alternative with the best expected monetary value is identical to the alternative with the highest expected utility. Saddle point A condition that exists when pure strategies are optimal for both players in a two-person, zero-sum game. The saddle point occurs at the intersection of the optimal strategies for the players, and the value of the saddle point is the value of the game. Two-person, zero-sum game A game with two players in which the gain to one player is equal to the loss to the other player. Utility A measure of the total worth of a consequence reflecting a decision maker’s attitude toward considerations such as profit, loss, and risk. Utility function for money A curve that depicts the relationship between monetary value and utility.

Problems

1. A firm has three investment alternatives. Payoffs are in thousands of dollars.

testSELF Economic Conditions

Decision Alternative Up, s1 Stable, s2 Down, s3 Investment A, d1 100 25 0 Investment B, d2 75 50 25 Investment C, d3 50 50 50 Probabilities 0.40 0.30 0.30

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Problems 187

a. Using the expected value approach, which decision is preferred? b. For the lottery having a payoff of $100,000 with probability p and $0 with probability

(1 2 p), two decision makers expressed the following indifference probabilities. Find the most preferred decision for each decision maker using the expected utility approach.

Indifference Probability (p)

Profit Decision Maker A Decision Maker B

$75,000 0.80 0.60 $50,000 0.60 0.30 $25,000 0.30 0.15

Damage

Decision Alternative None, s1 Minor, s2 Major, s3 Purchase insurance, d1 10,000 10,000 10,000 Do not purchase insurance, d2 0 100,000 200,000 Probabilities 0.96 0.03 0.01

c. Why don’t decision makers A and B select the same decision alternative?

2. Alexander Industries is considering purchasing an insurance policy for its new office building in St. Louis, Missouri. The policy has an annual cost of $10,000. If Alexander Industries doesn’t purchase the insurance and minor fire damage occurs, a cost of $100,000 is anticipated; the cost if major or total destruction occurs is $200,000. The costs, including the state-of-nature probabilities, are as follows:

a. Using the expected value approach, what decision do you recommend? b. What lottery would you use to assess utilities? (note: Because the data are costs, the

best payoff is $0.) c. Assume that you found the following indifference probabilities for the lottery defined

in part (b). What decision would you recommend?

Cost Indifference Probability

10,000 p 5 0.99 100,000 p 5 0.60

d. Do you favor using expected value or expected utility for this decision problem? Why?

3. In a certain state lottery, a lottery ticket costs $2. In terms of the decision to purchase or not to purchase a lottery ticket, suppose that the following payoff table applies:

State of Nature

Decision Alternatives Win, s1 Lose, s2 Purchase lottery ticket, d1 300,000 22 Do not purchase lottery ticket, d2 0 0

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188 chapter 5 Utility and Game Theory

a. A realistic estimate of the chances of winning is 1 in 250,000. Use the expected value approach to recommend a decision.

b. If a particular decision maker assigns an indifference probability of 0.000001 to the $0 payoff, would this individual purchase a lottery ticket? Use expected utility to justify your answer.

4. Two different routes accommodate travel between two cities. Route A normally takes 60 minutes and route B normally takes 45 minutes. If traffic problems are encountered on route A, the travel time increases to 70 minutes; traffic problems on route B increase travel time to 90 minutes. The probability of a delay is 0.20 for route A and 0.30 for route B. a. Using the expected value approach, what is the recommended route? b. If utilities are to be assigned to the travel times, what is the appropriate lottery? (note:

The smaller times should reflect higher utilities.) c. Use the lottery of part (b) and assume that the decision maker expresses indifference

probabilities of

p 5 0.80 for 60 minutes

p 5 0.60 for 70 minutes

What route should this decision maker select? Is the decision maker a risk taker or a risk avoider?

5. Three decision makers have assessed utilities for the following decision problem (payoff in dollars):testSELF

State of Nature

Decision Alternative s1 s2 s3 d1 20 50 220 d2 80 100 2100

The indifference probabilities are as follows:

Indifference Probability ( p)

Payoff Decision Maker A Decision Maker B Decision Maker C

100 1.00 1.00 1.00 80 0.95 0.70 0.90 50 0.90 0.60 0.75 20 0.70 0.45 0.60 220 0.50 0.25 0.40 2100 0.00 0.00 0.00

a. Plot the utility function for money for each decision maker. b. Classify each decision maker as a risk avoider, a risk taker, or risk neutral. c. For the payoff of 20, what is the premium that the risk avoider will pay to avoid risk?

What is the premium that the risk taker will pay to have the opportunity of the high payoff?

6. In Problem 5, if P(s1) 5 0.25, P(s2) 5 0.50, and P(s3) 5 0.25, find a recommended deci- sion for each of the three decision makers. (note: For the same decision problem, different utilities can lead to different decisions.)

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Problems 189

7. Suppose that the point spread for a particular sporting event is 10 points and that with this spread you are convinced you would have a 0.60 probability of winning a bet on your team. However, the local bookie will accept only a $1000 bet. Assuming that such bets are legal, would you bet on your team? (Disregard any commission charged by the bookie.) Remember that you must pay losses out of your own pocket. Your payoff table is as follows:

State of Nature

Decision Alternatives You Win You Lose

Bet $1000 2$1000 Don’t bet $0 $0

a. What decision does the expected value approach recommend? b. What is your indifference probability for the $0 payoff? (Although this choice isn’t

easy, be as realistic as possible. It is required for an analysis that reflects your attitude toward risk.)

c. What decision would you make based on the expected utility approach? In this case are you a risk taker or a risk avoider?

d. Would other individuals assess the same utility values you do? Explain. e. If your decision in part (c) was to place the bet, repeat the analysis assuming a mini-

mum bet of $10,000.

8. A Las Vegas roulette wheel has 38 different numerical values. If an individual bets on one number and wins, the payoff is 35 to 1. a. Show a payoff table for a $10 bet on one number for decision alternatives of bet and

do not bet. b. What is the recommended decision using the expected value approach? c. Do the Las Vegas casinos want risk-taking or risk-avoiding customers? Explain. d. What range of utility values would a decision maker have to assign to the $0 payoff

in order to have expected utility justify a decision to place the $10 bet?

9. A new product has the following profit projections and associated probabilities:

Profit Probability

$150,000 0.10 $100,000 0.25 $ 50,000 0.20 $0 0.15

2$ 50,000 0.20 2$100,000 0.10

a. Use the expected value approach to decide whether to market the new product. b. Because of the high dollar values involved, especially the possibility of a $100,000

loss, the marketing vice president has expressed some concern about the use of the expected value approach. As a consequence, if a utility analysis is performed, what is the appropriate lottery?

c. Assume that the following indifference probabilities are assigned. Do the utilities reflect the behavior of a risk taker or a risk avoider?

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190 chapter 5 Utility and Game Theory

Indifference Profit Probability( p)

$100,000 0.95 $ 50,000 0.70 $0 0.50

2$ 50,000 0.25

d. Use expected utility to make a recommended decision. e. Should the decision maker feel comfortable with the final decision recommended by

the analysis?

10. A television network has been receiving low ratings for its programs. Currently, manage- ment is considering two alternatives for the Monday night 8:00 p.m.–9:00 p.m. time slot: a comedy with a well-known star or a reality show. The percentages of viewing audience estimates depend on the degree of program acceptance. The relevant data are as follows:

Percentage of Viewing Audience

Program Acceptance Comedy Reality Show

High 30% 40% Moderate 25% 20% Poor 20% 15%

The probabilities associated with program acceptance levels are as follows:

Probability

Program Acceptance Comedy Reality Show

High 0.30 0.30 Moderate 0.60 0.40 Poor 0.10 0.30

a. Using the expected value approach, which program should the network choose? b. For a utility analysis, what is the appropriate lottery? c. Based on the lottery in part (b), assume that the network’s program manager has as-

signed the following indifference probabilities. Based on the use of utility measures, which program would you recommend? Is the manager a risk taker or a risk avoider?

Percentage of Audience Indifference Probability ( p)

30% 0.40 25% 0.30 20% 0.10

11. Consider the following two-person, zero-sum game. Identify the pure strategy. What is the value of the game?testSELF

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Problems 191

12. Two opposing armies, Red and Blue, must each decide whether to attack or defend. These decisions are made without knowledge of the opposing army’s decision. The payoff table, in terms of value of property gained or lost for the Red Army, appears below. Any gains for the Red Army are losses for the Blue Army.

Player B

b1 b2 b3

Player A a1 8 5 7

a2 2 4 10

Blue Army

Attack Defend

Red Army Attack 30 50

Defend 40 0

a. What is the optimal mixed strategy for the Red Army? b. What is the optimal mixed strategy for the Blue Army?

13. Two Indiana state senate candidates must decide which city to visit the day before the No- vember election. The same four cities—Indianapolis, Evansville, Fort Wayne, and South Bend—are available for both candidates. These cities are listed as strategies 1 to 4 for each candidate. Travel plans must be made in advance, so the candidates must decide which city to visit prior to knowing the other candidate’s plans. Values in the following table show thousands of voters for the Republican candidate based on the strategies selected by the two candidates. Which city should each candidate visit, and what is the value of the game?

Democrat Candidate

Fort South Indianapolis Evansville Wayne Bend b1 b2 b3 b4 Indianapolis, a1 0 215 28 20 Republican Evansville, a2 30 25 5 210 Candidate Fort Wayne, a3 10 225 0 20 South Bend, a4 20 20 10 15

14. In Section 5.5, we showed the following two-person, zero-sum game had a mixed strategy:

testSELF Player B

b1 b2 b3 a1 0 21 2 Player A a2 5 4 23 a3 2 3 24

a. Use dominance to reduce the game to a 2 3 2 game. Which strategies are dominated? b. Determine the optimal mixed strategy solution. c. What is the value of the game?

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192 chapter 5 Utility and Game Theory

15. In a gambling game, Player A and Player B both have a $1 and a $5 bill. Each player selects one of the bills without the other player knowing the bill selected. Simultaneously they both reveal the bills selected. If the bills do not match, Player A wins Player B’s bill. If the bills match, Player B wins Player A’s bill. a. Develop the game theory table for this game. The values should be expressed as the

gains (or losses) for Player A. b. Is there a pure strategy? Why or why not? c. Determine the optimal strategies and the value of this game. Does the game favor one

player over the other? d. Suppose Player B decides to deviate from the optimal strategy and begins playing

each bill 50% of the time. What should Player A do to improve Player A’s winnings? Comment on why it is important to follow an optimal game theory strategy.

16. Two companies compete for a share of the soft drink market. Each worked with an ad- vertising agency in order to develop alternative advertising strategies for the coming year. A variety of television advertisements, product promotions, in-store displays, and so on provides four different strategies for each company. The following table summarizes the projected change in market share for Company A once the two companies select their advertising strategy for the coming year. What is the optimal solution to this game for each of the players? What is the value of the game?

Company B

b1 b2 b3 b4 a1 3 0 2 4

Company A a2 2 22 1 0

a3 4 2 5 6 a4 22 6 21 0

17. If a soccer game ends in a tie, it goes into a penalty-kick shootout in which each team chooses five players to take penalty kicks. The team that makes the most subsequent penalty kicks wins the game. In a penalty-kick shootout, the shooter and the keeper each decide simultaneously on a direction to move. They can choose left, right, or mid- dle. These strategies yield the following game theory table, where the first value is the shooter’s probability of scoring and the second value is the keeper’s probability of stop- ping the shot:

Keeper

Left Center Right Left 0.35, 0.65 0.90, 0.10 0.85. 0.15 Shooter Center 0.30, 0.70 0.25, 0.75 0.45, 0.55 Right 0.95, 0.05 0.90, 0.10 0.30, 0.70

This is an example of a constant-sum game since each pair of entries in the game theory table sums to 1. This can be analyzed in the same manner as a zero-sum game. a. Use dominance to reduce the game to a 2 3 2 game. Which strategies are dominated? b. What is the solution to this game for the shooter and for the keeper? c. What is the shooter’s expected probability of scoring?

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case Problem Utility, Game Theory, and Product Line Extension Decisions 193

Case Problem Utility, Game Theory, and Product Line Extension Decisions ABC, Inc. and XYZ, Co. are the two dominant companies providing chargers, adapt- ers, and other accessories for cell phones. Each of these Silicon Valley companies is developing a new line of smartphone accessories, and each has a choice of technologies to use for these accessories. Each company can choose to focus on older (cheaper) tech- nology, recent (more expensive) technology, or cutting-edge (very expensive) technol- ogy. The share each company will gain or lose in the cell-phone accessories market depends on its technology choice and the technology choice of its competitor. The choice in technology investment must be made by each company before its competi- tor’s choice is revealed.

ABC, Inc. and XYZ, Co. are both led by young, aggressive CEOs; ABC’s CEO is Jack Webster and XYZ is run by Curtis Madsen. The CEOs of ABC and XYZ are each trying to determine the best technology in which to invest. The following tables pro- vide market share values and indifference probability values, p, for both Jack Webster (ABC) and Curtis Madsen (XYZ). Jack and Curtis have identical indifference prob- ability values.

Market Share Gain (Loss) Indifference Probability, p 25 — 20 0.85 15 0.70 10 0.58 5 0.45 0 0.35 –5 0.27 –10 0.20 –15 0.12 –20 0.05 –25 —

XYZ, Co.

Old Recent Cutting Edge Old 5 210 215

ABC, Inc. Recent 10 25 210 Cutting Edge 20 5 220

The following table shows the three possible technology investments (old, recent, and cutting edge) for ABC and XYZ and the resulting market share gain (or loss) for ABC, Inc. Because ABC and XYZ are the dominant companies in this market, whatever market share is gained by ABC is lost by XYZ and vice versa.

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194 chapter 5 Utility and Game Theory

Managerial Report Perform an analysis of the best decision of technology investment for ABC Inc. Prepare a report you would provide to Jack Webster that summarizes your analysis and findings. Include the following:

1. A graph of the utility function for ABC, Inc. Explain whether you would character- ize Jack Webster as a risk taker or a risk avoider and why.

2. Payoff tables for ABC, Inc. and XYZ, Co. using expected utilities for ABC and XYZ.

3. Recommendation for the best decision for ABC, Inc. and XYZ, Co. Is this a zero- sum game?

4. Detailed calculations and analysis to support your recommendation. 5. A discussion of how the expected utilities would change if ABC is taken over by a

more conservative, risk-avoiding firm.

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Time Series Analysis and Forecasting

CONTENTS

6.1 Time SerieS PaTTernS Horizontal Pattern Trend Pattern Seasonal Pattern Trend and Seasonal Pattern Cyclical Pattern Selecting a Forecasting method

6.2 ForeCaST aCCuraCy

6.3 moving averageS and exPonenTial SmooTHing moving averages Weighted moving averages exponential Smoothing

6.4 linear Trend ProjeCTion

6.5 SeaSonaliTy Seasonality Without Trend Seasonality With Trend models Based on monthly data

AppENdix 6.1 ForeCaSTing WiTH exCel daTa analySiS ToolS

CHAPTER 6

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196 Chapter 6 Time Series Analysis and Forecasting

The purpose of this chapter is to provide an introduction to time series analysis and forecast- ing. Suppose we are asked to provide quarterly forecasts of sales for one of our company’s products over the coming one-year period. Production schedules, raw materials purchas- ing, inventory policies, and sales quotas will all be affected by the quarterly forecasts we provide. Consequently, poor forecasts may result in poor planning and increased costs for the company. How should we go about providing the quarterly sales forecasts? good judgment, intuition, and an awareness of the state of the economy may give us a rough idea or “feeling” of what is likely to happen in the future, but converting that feeling into a number that can be used as next year’s sales forecast is challenging. The Q.m. in action, Forecasting energy needs in the utility industry, describes the role that forecasting plays in the utility industry.

Forecasting methods can be classified as qualitative or quantitative. Qualitative meth- ods generally involve the use of expert judgment to develop forecasts. Such methods are appropriate when historical data on the variable being forecast are either unavailable or not applicable. Quantitative forecasting methods can be used when (1) past information about the variable being forecast is available, (2) the information can be quantified, and (3) it is reasonable to assume that past is prologue (i.e., the pattern of the past will continue into the future). We will focus exclusively on quantitative forecasting methods in this chapter.

if the historical data are restricted to past values of the variable to be forecast, the fore- casting procedure is called a time series method and the historical data are referred to as a

A forecast is simply a prediction of what will happen in the future. Managers must accept that regardless of the technique used, they will not be able to develop perfect forecasts.

duke energy is a diversified energy company with a portfolio of natural gas and electric businesses and an affiliated real estate company. in 2006, duke energy merged with Cinergy of Cincinnati, ohio, to create one of north america’s largest energy companies, with as- sets totaling more than $70 billion. as a result of this merger the Cincinnati gas & electric Company became part of duke energy. Today, duke energy services over 5.5 million retail electric and gas customers in north Carolina, South Carolina, ohio, Kentucky, indiana, and ontario, Canada.

Forecasting in the utility industry offers some unique perspectives. Because energy is difficult to store, this product must be generated to meet the instantaneous re- quirements of the customers. electrical shortages are not just lost sales, but “brownouts” or “blackouts.” This situ- ation places an unusual burden on the utility forecaster. on the positive side, the demand for energy and the sale of energy are more predictable than for many other products. also, unlike the situation in a multiproduct firm, a great

amount of forecasting effort and expertise can be concen- trated on the two products: gas and electricity.

The largest observed electric demand for any given period, such as an hour, a day, a month, or a year, is defined as the peak load. The forecast of the annual electric peak load guides the timing decision for constructing future generating units, and the financial impact of this decision is great. obviously, a timing decision that leads to having the unit available no sooner than necessary is crucial.

The energy forecasts are important in other ways also. For example, purchases of coal as fuel for the gen- erating units are based on the forecast levels of energy needed. The revenue from the electric operations of the company is determined from forecasted sales, which in turn enters into the planning of rate changes and external financing. These planning and decision-making processes are among the most important managerial activities in the company. it is imperative that the decision makers have the best forecast information available to assist them in arriving at these decisions.

FORECASTING ENERGY NEEDS IN THE UTILITY INDUSTRY*

Q.M. in ACTion

*Based on information provided by Dr. Richard Evans of Duke Energy.

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6.1 Time Series Patterns 197

time series. The objective of time series analysis is to uncover a pattern in the historical data or time series and then extrapolate the pattern into the future; the forecast is based solely on past values of the variable and/or on past forecast errors.

in Section 6.1 we discuss the various kinds of time series that a forecaster might be faced with in practice. These include a constant or horizontal pattern, a trend, a seasonal pattern, both a trend and a seasonal pattern, and a cyclical pattern. in order to build a quantitative forecasting model it is also necessary to have a measurement of forecast accuracy. different measurements of forecast accuracy, and their respective advantages and disadvantages, are discussed in Section 6.2. in Section 6.3 we consider the simplest case, which is a horizontal or constant pattern. For this pattern, we develop the classical moving average, weighted moving average, and exponential smoothing models. many time series have a trend, and taking this trend into account is important; in Section 6.4 we provide regression models for finding the best model parameters when a linear trend is present. Finally, in Section 6.5 we show how to incorporate both a trend and seasonality into a forecasting model.

6.1 Time Series Patterns a time series is a sequence of observations on a variable measured at successive points in time or over successive periods of time. The measurements may be taken every hour, day, week, month, or year, or at any other regular interval.1 The pattern of the data is an important factor in understanding how the time series has behaved in the past. if such be- havior can be expected to continue in the future, we can use it to guide us in selecting an appropriate forecasting method.

To identify the underlying pattern in the data, a useful first step is to construct a time series plot. a time series plot is a graphical presentation of the relationship between time and the time series variable; time is represented on the horizontal axis and values of the time series variable are shown on the vertical axis. let us first review some of the common types of data patterns that can be identified when examining a time series plot.

Horizontal Pattern a horizontal pattern exists when the data fluctuate randomly around a constant mean over time. To illustrate a time series with a horizontal pattern, consider the 12 weeks of data in Table 6.1. These data show the number of gallons of gasoline (in 1000s) sold by a gasoline distributor in Bennington, vermont, over the past 12 weeks. The average value or mean for this time series is 19.25 or 19,250 gallons per week. Figure 6.1 shows a time series plot for these data. note how the data fluctuate around the sample mean of 19,250 gallons. although random variability is present, we would say that these data follow a horizontal pattern.

The term stationary time series2 is used to denote a time series whose statistical proper- ties are independent of time. in particular this means that

1. The process generating the data has a constant mean. 2. The variability of the time series is constant over time.

a time series plot for a stationary time series will always exhibit a horizontal pattern with random fluctuations. However, simply observing a horizontal pattern is not sufficient

1We limit our discussion to time series for which the values of the series are recorded at equal intervals. Cases in which the observations are made at unequal intervals are beyond the scope of this text. 2For a formal definition of stationarity, see K. ord and R. Fildes (2012), Principles of Business Forecasting. Mason, oH: Cengage Learning, p. 155.

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198 Chapter 6 Time Series Analysis and Forecasting

S al

es (

10 00

s of

g al

lo n

Week

0 1 2 3 4 5 6 7 8 9 10 11 12

FIGURE 6.1 gaSoline SaleS Time SerieS PloT

Week Sales (1000s of gallons) 1 17 2 21 3 19 4 23 5 18 6 16 7 20 8 18 9 22 10 20 11 15 12 22

TABLE 6.1 gaSoline SaleS Time SerieS

fileWEB Gasoline

evidence to conclude that the time series is stationary. more advanced texts on forecasting discuss procedures for determining if a time series is stationary and provide methods for transforming a time series that is nonstationary into a stationary series.

Changes in business conditions often result in a time series with a horizontal pattern that shifts to a new level at some point in time. For instance, suppose the gasoline distribu- tor signs a contract with the vermont Sate Police to provide gasoline for state police cars located in southern vermont beginning in week 13. With this new contract, the distributor naturally expects to see a substantial increase in weekly sales starting in week 13. Table 6.2 shows the number of gallons of gasoline sold for the original time series and the 10 weeks

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6.1 Time Series Patterns 199

after signing the new contract. Figure 6.2 shows the corresponding time series plot. note the increased level of the time series beginning in week 13. This change in the level of the time series makes it more difficult to choose an appropriate forecasting method. Selecting a forecasting method that adapts well to changes in the level of a time series is an important consideration in many practical applications.

Trend Pattern although time series data generally exhibit random fluctuations, a time series may also show gradual shifts or movements to relatively higher or lower values over a longer period

Week Sales (1000s of gallons) Week Sales (1000s of gallons) 1 17 12 22 2 21 13 31 3 19 14 34 4 23 15 31 5 18 16 33 6 16 17 28 7 20 18 32 8 18 19 30 9 22 20 29 10 20 21 34 11 15 22 33

TABLE 6.2 gaSoline SaleS Time SerieS aFTer oBTaining THe ConTraCT WiTH THe vermonT STaTe PoliCe

fileWEB GasolineRevised

0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

S al

es (

10 00

s of

g al

lo n

Week

FIGURE 6.2 gaSoline SaleS Time SerieS PloT aFTer oBTaining THe ConTraCT WiTH THe vermonT STaTe PoliCe

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200 Chapter 6 Time Series Analysis and Forecasting

of time. if a time series plot exhibits this type of behavior, we say that a trend pattern exists. a trend is usually the result of long-term factors such as population increases or decreases, shifting demographic characteristics of the population, improving technology, and/or changes in consumer preferences.

To illustrate a time series with a linear trend pattern, consider the time series of bicycle sales for a particular manufacturer over the past 10 years, as shown in Table 6.3 and Figure 6.3. note that 21,600 bicycles were sold in year 1, 22,900 were sold in year 2, and so on. in year 10, the most recent year, 31,400 bicycles were sold. visual inspection of the time series plot shows some up and down movement over the past 10 years, but the time series seems also to have a systematically increasing or upward trend.

Year Sales (1000s) 1 21.6 2 22.9 3 25.5 4 21.9 5 23.9 6 27.5 7 31.5 8 29.7 9 28.6 10 31.4

TABLE 6.3 BiCyCle SaleS Time SerieS

fileWEB Bicycle

S al

es (

10 00

Year

0 1 2 3 4 5 6 7 8 9

10 20

FIGURE 6.3 BiCyCle SaleS Time SerieS PloT

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6.1 Time Series Patterns 201

The trend for the bicycle sales time series appears to be linear and increasing over time, but sometimes a trend can be described better by other types of patterns. For in- stance, the data in Table 6.4 and the corresponding time series plot in Figure 6.4 show the sales revenue for a cholesterol drug since the company won Fda approval for the drug 10 years ago. The time series increases in a nonlinear fashion; that is, the rate of change of revenue does not increase by a constant amount from one year to the next. in fact, the revenue appears to be growing in an exponential fashion. exponential relation- ships such as this are appropriate when the percentage change from one period to the next is relatively constant.

Seasonal Pattern The trend of a time series can be identified by analyzing movements in historical data over multiple years. Seasonal patterns are recognized by observing recurring patterns

Year Revenue

1 23.1 2 21.3 3 27.4 4 34.6 5 33.8 6 43.2 7 59.5 8 64.4 9 74.2 10 99.3

TABLE 6.4 CHoleSTerol drug revenue Time SerieS ($ millionS)

fileWEB Cholesterol

R ev

en u

Year

0 1 2 3 4 5 6 7 8 9

120

100

FIGURE 6.4 CHoleSTerol drug revenue TimeS SerieS PloT ($ millionS)

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202 Chapter 6 Time Series Analysis and Forecasting

over successive periods of time. For example, a manufacturer of swimming pools expects low sales activity in the fall and winter months, with peak sales in the spring and summer months to occur each year. manufacturers of snow removal equipment and heavy clothing, however, expect the opposite yearly pattern. not surprisingly, the pattern for a time series plot that exhibits a recurring pattern over a one-year period due to seasonal influences is called a seasonal pattern. While we generally think of seasonal movement in a time series as occurring within one year, time series data can also exhibit seasonal patterns of less than one year in duration. For example, daily traffic volume shows within-the-day “sea- sonal” behavior, with peak levels occurring during rush hours, moderate flow during the rest of the day and early evening, and light flow from midnight to early morning. another example of an industry with sales that exhibit easily discernible seasonal patterns within a day is the restaurant industry.

as an example of a seasonal pattern, consider the number of umbrellas sold at a cloth- ing store over the past five years. Table 6.5 shows the time series and Figure 6.5 shows the corresponding time series plot. The time series plot does not indicate a long-term trend in sales. in fact, unless you look carefully at the data, you might conclude that the data follow a horizontal pattern with random fluctuation. However, closer inspection of the fluctuations in the time series plot reveals a systematic pattern in the data that occurs within each year. That is, the first and third quarters have moderate sales, the second quarter has the highest sales, and the fourth quarter tends to have the lowest sales volume. Thus, we would con- clude that a quarterly seasonal pattern is present.

Year Quarter Sales

1 1 125 2 153 3 106 4 88

2 1 118 2 161 3 133 4 102

3 1 138 2 144 3 113 4 80

4 1 109 2 137 3 125 4 109

5 1 130 2 165 3 128 4 96

TABLE 6.5 umBrella SaleS Time SerieS

fileWEB Umbrella

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6.1 Time Series Patterns 203

fileWEB TVSales

0 2 4 6 81 3 5 7 9 10 11 12 13 14 15 16 17 18 19 20

Time Period

180

100

120

140

160

S al

FIGURE 6.5 umBrella SaleS Time SerieS PloT

Year Quarter Sales (1000s) 1 1 4.8 2 4.1 3 6.0 4 6.5

2 1 5.8 2 5.2 3 6.8 4 7.4

3 1 6.0 2 5.6 3 7.5 4 7.8

4 1 6.3 2 5.9 3 8.0 4 8.4

TABLE 6.6 QuarTerly TeleviSion SeT SaleS Time SerieS

Trend and Seasonal Pattern Some time series include both a trend and a seasonal pattern. For instance, the data in Table 6.6 and the corresponding time series plot in Figure 6.6 show quarterly television set sales for a particular manufacturer over the past four years. Clearly an increasing trend is

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204 Chapter 6 Time Series Analysis and Forecasting

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Period

Q u

ar te

rl y

T el

ev is

io n

S et

S al

es (

10 00

FIGURE 6.6 QuarTerly TeleviSion SeT SaleS Time SerieS PloT

present. However, Figure 6.6 also indicates that sales are lowest in the second quarter of each year and highest in quarters 3 and 4. Thus, we conclude that a seasonal pattern also exists for television sales. in such cases we need to use a forecasting method that is capable of dealing with both trend and seasonality.

Cyclical Pattern a cyclical pattern exists if the time series plot shows an alternating sequence of points below and above the trend line that lasts for more than one year. many economic time se- ries exhibit cyclical behavior with regular runs of observations below and above the trend line. often the cyclical component of a time series is due to multiyear business cycles. For example, periods of moderate inflation followed by periods of rapid inflation can lead to a time series that alternates below and above a generally increasing trend line (e.g., a time series for housing costs). Business cycles are extremely difficult, if not impossible, to forecast. as a result, cyclical effects are often combined with long-term trend effects and referred to as trend-cycle effects. in this chapter we do not deal with cyclical effects that may be present in the time series.

Selecting a Forecasting Method The underlying pattern in the time series is an important factor in selecting a forecasting method. Thus, a time series plot should be one of the first analytic tools employed when trying to determine which forecasting method to use. if we see a horizontal pattern, then we need to select a method appropriate for this type of pattern. Similarly, if we observe a trend in the data, then we need to use a forecasting method that is capable of handling a trend effectively. in the next two sections we illustrate methods for assessing forecast accuracy

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6.1 Time Series Patterns 205

aCCo Brands Corporation is one of the world’s largest suppliers of branded office and consumer products and print finishing solutions. The company’s widely recog- nized brands include aT-a-glanCe®, day-Timer®, Five Star®, gBC®, Hilroy®, Kensington®, marbig®, mead®, noBo, Quartet®, rexel, Swingline®, Tilibra®, Wilson jones®, and many others.

Because it produces and markets a wide array of prod- ucts with a myriad of demand characteristics, aCCo Brands relies heavily on sales forecasts in planning its manufactur- ing, distribution, and marketing activities. By viewing its relationship in terms of a supply chain, aCCo Brands and its customers (which are generally retail chains) establish close collaborative relationships and consider each other to be valued partners. as a result, aCCo Brands’ customers share valuable information and data that serve as inputs into aCCo Brands’ forecasting process.

in her role as a forecasting manager for aCCo Brands, vanessa Baker appreciates the importance of this additional information. “We do separate forecasts of demand for each major customer,” said Baker, “and we generally use twenty- four to thirty-six months of history to generate monthly fore- casts twelve to eighteen months into the future. While trends are important, several of our major product lines, including school, planning and organizing, and decorative calendars, are heavily seasonal, and seasonal sales make up the bulk of our annual volume.”

daniel marks, one of several account-level strategic forecast managers for aCCo Brands, adds,

The supply chain process includes the total lead time

from identifying opportunities to making or procuring

the product to getting the product on the shelves to align

with the forecasted demand; this can potentially take

several months, so the accuracy of our forecasts is criti-

cal throughout each step of the supply chain. adding to

this challenge is the risk of obsolescence. We sell many

dated items, such as planners and calendars, which have

a natural, built-in obsolescence. in addition, many of

our products feature designs that are fashion-conscious

or contain pop culture images, and these products can

also become obsolete very quickly as tastes and popu-

larity change. an overly optimistic forecast for these

products can be very costly, but an overly pessimis-

tic forecast can result in lost sales potential and give

our competitors an opportunity to take market share

from us.

in addition to looking at trends, seasonal components, and cyclical patterns, Baker and marks must contend with several other factors. Baker notes, “We have to adjust our forecasts for upcoming promotions by our customers.” marks agrees and adds:

We also have to go beyond just forecasting consumer

demand; we must consider the retailer’s specific needs

in our order forecasts, such as what type of display will

be used and how many units of a product must be on

display to satisfy their presentation requirements. Cur-

rent inventory is another factor—if a customer is car-

rying either too much or too little inventory, that will

affect their future orders, and we need to reflect that

in our forecasts. Will the product have a short life be-

cause it is tied to a cultural fad? What are the retailer’s

marketing and markdown strategies? our knowledge

of the environments in which our supply chain partners

are competing helps us to forecast demand more accu-

rately, and that reduces waste and makes our custom-

ers, as well as aCCo Brands, far more profitable.

FORECASTING DEMAND FOR A BROAD LINE OF OFFICE PRODUCTS*

Q.M. in ACTion

*The authors are indebted to Vanessa Baker and Daniel Marks of ACCo Brands for providing input for this Q.M. in Action.

and consider forecasting models that can be used in situations for which the underlying pattern is horizontal; in other words, no trend or seasonal effects are present. We then consider methods appropriate when trend and/or seasonality are present in the data. The Q.m. in action, Forecasting demand for a Broad Product line of office Products, describes the considerations made by aCCo Brands when forecasting demand for its consumer and office products.

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206 Chapter 6 Time Series Analysis and Forecasting

6.2 Forecast Accuracy in this section we begin by developing forecasts for the gasoline time series shown in Table 6.1 using the simplest of all the forecasting methods, an approach that uses the most recent week’s sales volume as the forecast for the next week. For instance, the distributor sold 17,000 gallons of gasoline in week 1; this value is used as the forecast for week 2. next, we use 21, the actual value of sales in week 2, as the forecast for week 3, and so on. The forecasts obtained for the historical data using this method are shown in Table 6.7 in the column labeled Forecast. Because of its simplicity, this method is often referred to as a naïve forecasting method.

How accurate are the forecasts obtained using this naïve forecasting method? To an- swer this question we will introduce several measures of forecast accuracy. These measures are used to determine how well a particular forecasting method is able to reproduce the time series data that are already available. By selecting the method that is most accurate for the data already known, we hope to increase the likelihood that we will obtain more accurate forecasts for future time periods.

The key concept associated with measuring forecast accuracy is forecast error. if we denote Yt and Y

⁄ t as the actual and forecasted values of the time series for period t, respec-

tively, the forecasting error for period t is

et 5 Yt 2 Y ⁄

t (6.1)

That is, the forecast error for time period t is the difference between the actual and the forecasted values for period t.

Absolute Absolute Time Value of Squared Value of Series Forecast Forecast Forecast percentage percentage Week Value Forecast Error Error Error Error Error

1 17 2 21 17 4 4 16 19.05 19.05 3 19 21 22 2 4 210.53 10.53 4 23 19 4 4 16 17.39 17.39 5 18 23 25 5 25 227.78 27.78 6 16 18 22 2 4 212.50 12.50 7 20 16 4 4 16 20.00 20.00 8 18 20 22 2 4 211.11 11.11 9 22 18 4 4 16 18.18 18.18 10 20 22 22 2 4 210.00 10.00 11 15 20 25 5 25 233.33 33.33 12 22 15 7 7 49 31.82 31.82

Totals 5 41 179 1.19 211.69

TABLE 6.7 ComPuTing ForeCaSTS and meaSureS oF ForeCaST aCCuraCy uSing THe moST reCenT value aS THe ForeCaST For THe nexT Period

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6.2 Forecast Accuracy 207

For instance, because the distributor actually sold 21,000 gallons of gasoline in week 2 and the forecast, using the sales volume in week 1, was 17,000 gallons, the forecast error in week 2 is

Forecast error in week 2 5 e2 5 Y2 2 Y ⁄

2 5 21 2 17 5 4

The fact that the forecast error is positive indicates that in week 2 the forecasting method underestimated the actual value of sales. next we use 21, the actual value of sales in week 2, as the forecast for week 3. Since the actual value of sales in week 3 is 19, the forecast error for week 3 is e3 5 19 2 21 5 22. in this case, the negative forecast error indicates the forecast overestimated the actual value for week 3. Thus, the forecast error may be positive or negative, depending on whether the forecast is too low or too high. a complete summary of the forecast errors for this naïve forecasting method is shown in Table 6.7 in the column labeled Forecast error. it is important to note that because we are using a past value of the time series to produce a forecast for period t, we do not have sufficient data to produce a naïve forecast for the first week of this time series.

a simple measure of forecast accuracy is the mean or average of the forecast errors. if we have n periods in our time series and k is the number of periods at the beginning of the time series for which we cannot produce a naïve forecast, the mean forecast error (mFe) is

mFe 5 o

t5 k 1 1

n 2 k (6.2)

Table 6.7 shows that the sum of the forecast errors for the gasoline sales time series is 5; thus, the mean or average error is 5/11 5 0.45. Because we do not have sufficient data to produce a naïve forecast for the first week of this time series, we must adjust our calculations in both the numerator and denominator accordingly. This is common in forecasting; we often use k past periods from the time series to produce forecasts, and so we frequently cannot produce forecasts for the first k periods. in those instances the summation in the numerator starts at the first value of t for which we have produced a forecast (so we begin the summation at t 5 k 1 1), and the denominator (which is the number of periods in our time series for which we are able to produce a forecast) will also reflect these circumstances. in the gasoline example, although the time series consists of 12 values, to compute the mean error we divided the sum of the forecast errors by 11 because there are only 11 forecast errors (we cannot generate forecast sales for the first week using this naïve forecasting method).

also note that in the gasoline time series, the mean forecast error is positive, which implies that the method is generally underforecasting; in other words, the observed values tend to be greater than the forecasted values. Because positive and negative forecast errors tend to offset one another, the mean error is likely to be small; thus, the mean error is not a very useful measure of forecast accuracy.

The mean absolute error, denoted mae, is a measure of forecast accuracy that avoids the problem of positive and negative forecast errors offsetting one another. as you might expect given its name, mae is the average of the absolute values of the forecast errors:

mae 5 o

t5 k 1 1

uet u

n 2 k (6.3)

This is also referred to as the mean absolute deviation or mad. Table 6.7 shows that the sum of the absolute values of the forecast errors is 41; thus

mae 5 average of the absolute value of forecast errors 5 41

11 5 3.73

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208 Chapter 6 Time Series Analysis and Forecasting

another measure that avoids the problem of positive and negative errors offsetting each other is obtained by computing the average of the squared forecast errors. This measure of forecast accuracy, referred to as the mean squared error, is denoted mSe:

mSe 5 o

t5 k 1 1

e2t

n 2 k (6.4)

From Table 6.7, the sum of the squared errors is 179; hence,

mSe 5 average of the sum of squared forecast errors 5 179

11 5 16.27

The size of mae and mSe depends upon the scale of the data. as a result, it is difficult to make comparisons for different time intervals (such as comparing a method of forecasting monthly gasoline sales to a method of forecasting weekly sales) or to make comparisons across different time series (such as monthly sales of gasoline and monthly sales of oil filters). To make comparisons such as these we need to work with relative or percentage error measures. The mean absolute percentage error, denoted maPe, is such a measure. To compute maPe we must first compute the percentage error for each forecast:

1etYt2100

For example, the percentage error corresponding to the forecast of 17 in week 2 is computed by dividing the forecast error in week 2 by the actual value in week 2 and multiplying the result by 100. For week 2 the percentage error is computed as follows:

Percentage error for week 2 5 1e2Y22100 5 1 4

212100 5 19.05% Thus, the forecast error for week 2 is 19.05% of the observed value in week 2. a complete summary of the percentage errors is shown in Table 6.7 in the column labeled Percentage error. in the next column, we show the absolute value of the percentage error. Finally, we find the maPe, which is calculated as:

maPe 5 o

t5 k 1 1

`1etYt2100 ` n 2 k

(6.5)

Table 6.7 shows that the sum of the absolute values of the percentage errors is 211.69; thus

maPe 5 average of the absolute value of percentage forecast errors

5 211.69

11 5 19.24%

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6.2 Forecast Accuracy 209

in summary, using the naïve (most recent observation) forecasting method, we obtained the following measures of forecast accuracy:

mae 5 3.73

mSe 5 16.27

maPe 5 19.24%

These measures of forecast accuracy simply measure how well the forecasting method is able to forecast historical values of the time series. now, suppose we want to forecast sales for a future time period, such as week 13. in this case the forecast for week 13 is 22, the actual value of the time series in week 12. is this an accurate estimate of sales for week 13? unfortunately there is no way to address the issue of accuracy associated with forecasts for future time periods. However, if we select a forecasting method that works well for the historical data, and we have reason to believe the historical pattern will continue into the future, we should obtain forecasts that will ultimately be shown to be accurate.

Before closing this section, let us consider another method for forecasting the gasoline sales time series in Table 6.1. Suppose we use the average of all the historical data available as the forecast for the next period. We begin by developing a forecast for week 2. Since there is only one historical value available prior to week 2, the forecast for week 2 is just the time series value in week 1; thus, the forecast for week 2 is 17,000 gallons of gasoline. To compute the forecast for week 3, we take the average of the sales values in weeks 1 and 2. Thus,

Y ⁄

3 5 17 1 21

2 5 19

Similarly, the forecast for week 4 is

Y ⁄

4 5 17 1 21 1 19

3 5 19

The forecasts obtained using this method for the gasoline time series are shown in Table 6.8 in the column labeled Forecast. using the results shown in Table 6.8, we obtained the following values of mae, mSe, and maPe:

mae 5 26.81

11 5 2.44

mSe 5 89.07

11 5 8.10

maPe 5 141.34

11 5 12.85%

We can now compare the accuracy of the two forecasting methods we have considered in this section by comparing the values of mae, mSe, and maPe for each method.

Naïve Method Average of past Values

mae 3.73 2.44 mSe 16.27 8.10 maPe 19.24% 12.85%

Try Problem 1 for practice in computing measures of forecast accuracy.

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210 Chapter 6 Time Series Analysis and Forecasting

For each of these measures, the average of past values provides more accurate forecasts than using the most recent observation as the forecast for the next period. in general, if the underlying time series is stationary, the average of all the historical data will provide the most accurate forecasts.

evaluating different forecasts based on historical accuracy is only helpful if histori- cal patterns continue into the future. as we noted in Section 6.1, the 12 observations of Table 6.1 comprise a stationary time series. in Section 6.1 we mentioned that changes in business conditions often result in a time series that is not stationary. We discussed a situ- ation in which the gasoline distributor signed a contract with the vermont State Police to provide gasoline for state police cars located in southern vermont. Table 6.2 shows the number of gallons of gasoline sold for the original time series and the 10 weeks after signing the new contract, and Figure 6.2 shows the corresponding time series plot. note the change in level in week 13 for the resulting time series. When a shift to a new level such as this occurs, it takes several periods for the forecasting method that uses the average of all the historical data to adjust to the new level of the time series. However, in this case the simple naïve method adjusts very rapidly to the change in level because it uses only the most recent observation available as the forecast.

measures of forecast accuracy are important factors in comparing different forecast- ing methods, but we have to be careful to not rely too heavily upon them. good judgment and knowledge about business conditions that might affect the value of the variable to be forecast also have to be considered carefully when selecting a method. Historical forecast accuracy is not the sole consideration, especially if the pattern exhibited by the time series is likely to change in the future.

in the next section we will introduce more sophisticated methods for developing forecasts for a time series that exhibits a horizontal pattern. using the measures of forecast accuracy de- veloped here, we will be able to assess whether such methods provide more accurate forecasts

Absolute Absolute Time Value of Squared Value of Series Forecast Forecast Forecast percentage percentage Week Value Forecast Error Error Error Error Error

1 17 2 21 17.00 4.00 4.00 16.00 19.05 19.05 3 19 19.00 0.00 0.00 0.00 0.00 0.00 4 23 19.00 4.00 4.00 16.00 17.39 17.39 5 18 20.00 22.00 2.00 4.00 211.11 11.11 6 16 19.60 23.60 3.60 12.96 222.50 22.50 7 20 19.00 1.00 1.00 1.00 5.00 5.00 8 18 19.14 21.14 1.14 1.31 26.35 6.35 9 22 19.00 3.00 3.00 9.00 13.64 13.64 10 20 19.33 0.67 0.67 0.44 3.33 3.33 11 15 19.40 24.40 4.40 19.36 229.33 29.33 12 22 19.00 3.00 3.00 9.00 13.64 13.64

Totals 4.52 26.81 89.07 2.75 141.34

TABLE 6.8 ComPuTing ForeCaSTS and meaSureS oF ForeCaST aCCuraCy uSing THe average oF all THe HiSToriCal daTa aS THe ForeCaST For THe nexT Period

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6.3 Moving Averages and Exponential Smoothing 211

than we obtained using the simple approaches illustrated in this section. The methods that we will introduce also have the advantage that they adapt well to situations in which the time series changes to a new level. The ability of a forecasting method to adapt quickly to changes in level is an important consideration, especially in short-term forecasting situations.

6.3 Moving Averages and Exponential Smoothing in this section we discuss three forecasting methods that are appropriate for a time series with a horizontal pattern: moving averages, weighted moving averages, and exponential smoothing. These methods are also capable of adapting well to changes in the level of a horizontal pattern such as what we saw with the extended gasoline sales time series (Table 6.2 and Figure 6.2). However, without modification they are not appropriate when consider- able trend, cyclical, or seasonal effects are present. Because the objective of each of these methods is to “smooth out” random fluctuations in the time series, they are referred to as smoothing methods. These methods are easy to use and generally provide a high level of accuracy for short-range forecasts, such as a forecast for the next time period.

Moving Averages The moving averages method uses the average of the most recent k data values in the time series as the forecast for the next period. mathematically, a moving average forecast of order k is as follows:

Y ⁄

t1 1 5 osmost recent k data valuesd

k 5

o t

i5 t2 k 1 1

5 Yt2 k 1 1 1 · · · 1 Yt2 1 1 Yt

k (6.6)

where

Y ⁄

t11 5 forecast of the time series for period t 1 1

Yi 5 actual value of the time series in period i

k 5 number of periods of time series data used to generate the forecast

The term moving is used because every time a new observation becomes available for the time series, it replaces the oldest observation in the equation and a new average is com- puted. Thus, the periods over which the average is calculated change, or move, with each ensuing period.

To illustrate the moving averages method, let us return to the original 12 weeks of gasoline sales data in Table 6.1 and Figure 6.1. The time series plot in Figure 6.1 indicates that the gasoline sales time series has a horizontal pattern. Thus, the smoothing methods of this section are applicable.

To use moving averages to forecast a time series, we must first select the order k, or number of time series values to be included in the moving average. if only the most recent values of the time series are considered relevant, a small value of k is preferred. if a greater number of past values are considered relevant, then we generally opt for a

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212 Chapter 6 Time Series Analysis and Forecasting

larger value of k. As mentioned earlier, a time series with a horizontal pattern can shift to a new level over time. A moving average will adapt to the new level of the series and resume providing good forecasts in k periods. Thus a smaller value of k will track shifts in a time series more quickly (the naïve approach discussed earlier is actually a moving average for k 5 1). On the other hand, larger values of k will be more effective in smoothing out random fluctuations. Thus, managerial judgment based on an understanding of the behavior of a time series is helpful in choosing an appropriate value of k.

To illustrate how moving averages can be used to forecast gasoline sales, we will use a three-week moving average (k 5 3). We begin by computing the forecast of sales in week 4 using the average of the time series values in weeks 1 to 3.

Y ⁄

4 5 average of weeks 1 to 3 5 17 1 21 1 19

3 5 19

Thus, the moving average forecast of sales in week 4 is 19 or 19,000 gallons of gaso- line. Because the actual value observed in week 4 is 23, the forecast error in week 4 is e4 5 23 2 19 5 4.

We next compute the forecast of sales in week 5 by averaging the time series values in weeks 2–4.

Y ⁄

5 5 average of weeks 2 to 4 5 21 1 19 1 23

3 5 21

Hence, the forecast of sales in week 5 is 21 and the error associated with this forecast is e5 5 18 2 21 5 23. A complete summary of the three-week moving average forecasts for the gasoline sales time series is provided in Table 6.9. Figure 6.7 shows the original time series plot and the three-week moving average forecasts. Note how the graph of the mov- ing average forecasts has tended to smooth out the random fluctuations in the time series.

Absolute Absolute Time Value of Squared Value of Series Forecast Forecast Forecast Percentage Percentage Week Value Forecast Error Error Error Error Error

1 17 2 21 3 19 4 23 19 4 4 16 17.39 17.39 5 18 21 23 3 9 216.67 16.67 6 16 20 24 4 16 225.00 25.00 7 20 19 1 1 1 5.00 5.00 8 18 18 0 0 0 0.00 0.00 9 22 18 4 4 16 18.18 18.18 10 20 20 0 0 0 0.00 0.00 11 15 20 25 5 25 233.33 33.33 12 22 19 3 3 9 13.64 13.64

Totals 0 24 92 220.79 129.21

TABLE 6.9 SummAry OF THree-Week mOviNg AverAge CAlCulATiONS

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6.3 Moving Averages and Exponential Smoothing 213

S al

es (

10 00

s of

g al

lo n

Week

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Three-week moving average forecasts

FIGURE 6.7 gaSoline SaleS Time SerieS PloT and THree-WeeK moving aerage ForeCaSTS

To forecast sales in week 13, the next time period in the future, we simply compute the average of the time series values in weeks 10, 11, and 12.

Y ⁄

13 5 average of weeks 10 to 12 5 20 1 15 1 22

3 5 19

Thus, the forecast for week 13 is 19 or 19,000 gallons of gasoline.

Forecast Accuracy in Section 6.2 we discussed three measures of forecast accuracy: mean absolute error (mae); mean squared error (mSe); and mean absolute percentage error (maPe). using the three-week moving average calculations in Table 6.9, the values for these three measures of forecast accuracy are

mae 5 o 12

t5 4

uet u

12 2 3 5

9 5 2.67

mSe 5 o 12

t5 4

e2t

12 2 3 5

9 5 10.22

maPe 5 o 12

t5 4

`1 et Yt 2100 `

12 2 3 5

129.21

9 5 14.36%

Can you now use moving averages to develop forecasts? Try Problem 7.

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214 Chapter 6 Time Series Analysis and Forecasting

in Section 6.2 we showed that using the most recent observation as the forecast for the next week (a moving average of order k 5 1) resulted in values of mae 5 3.73, mSe 5 16.27, and maPe 5 19.24%. Thus, in each case the three-week moving average approach has provided more accurate forecasts than simply using the most recent observation as the forecast. also note how the formulas for the mae, mSe, and maPe reflect that our use of a three-week moving average leaves us with insufficient data to generate forecasts for the first three weeks of our time series.

To determine if a moving average with a different order k can provide more accurate forecasts, we recommend using trial and error to determine the value of k that minimizes the mSe. For the gasoline sales time series, it can be shown that the minimum value of mSe corresponds to a moving average of order k 5 6 with mSe 5 6.79. if we are willing to assume that the order of the moving average that is best for the historical data will also be best for future values of the time series, the most accurate moving average forecasts of gasoline sales can be obtained using a moving average of order k 5 6.

Weighted Moving Averages in the moving averages method, each observation in the moving average calculation receives equal weight. one variation, known as weighted moving averages, involves selecting a dif- ferent weight for each data value in the moving average and then computing a weighted average of the most recent k values as the forecast.

Y ⁄

t1 1 5 wtYt 1 wt2 1Yt2 1 1 . . . 1 wt2 k 1 1Yt2 k 1 1 (6.7)

where

Y ⁄

t+1 5 forecast of the time series for period t 1 1

Yt 5 actual value of the time series in period t

wt 5 weight applied to the actual time series value for period t

k 5 number of periods of time series data used to generate the forecast

generally the most recent observation receives the largest weight, and the weight decreases with the relative age of the data values. let us use the gasoline sales time series in Table 6.1 to illustrate the computation of a weighted three-week moving average. We will assign a weight of wt 5

3 6 to the most recent observation, a weight of wt21 5

2 6 to the second most

In situations where you need to compare forecasting methods for different time periods, such as comparing a forecast of weekly sales to a forecast of monthly sales, relative measures such as MAPE are preferred.

A moving average forecast of order k is just a special case of the weighted moving averages method in which each weight is equal to 1/k; for example, a moving average forecast of order k 5 3 is just a special case of the weighted moving averages method in which each weight is equal to 13.

recent observation, and a weight of wt22 5 1 6 to the third most recent observation. using this

weighted average, our forecast for week 4 is computed as follows:

Forecast for week 4 5 1

6 (17) 1

6 (21) 5

6 (19) 5 19.33

note that the sum of the weights is equal to 1 for the weighted moving average method.

Forecast Accuracy To use the weighted moving averages method, we must first select the number of data values to be included in the weighted moving average and then choose weights for each of these data values. in general, if we believe that the recent past is a better predictor of the future than the distant past, larger weights should be given to the more recent observations. However, when the time series is highly variable, selecting approximately equal weights for the data values may be preferable. The only requirements in selecting the weights are that they be nonnegative and that their sum must equal 1. To determine whether

Use Problem 8 to practice using weighted moving averages to produce forecasts.

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6.3 Moving Averages and Exponential Smoothing 215

one particular combination of number of data values and weights provides a more accurate forecast than another combination, we recommend using mSe as the measure of forecast accuracy. That is, if we assume that the combination that is best for the past will also be best for the future, we would use the combination of number of data values and weights that minimized mSe for the historical time series to forecast the next value in the time series.

Exponential Smoothing Exponential smoothing also uses a weighted average of past time series values as a fore- cast; it is a special case of the weighted moving averages method in which we select only one weight—the weight for the most recent observation. The weights for the other data values are computed automatically and become smaller as the observations move farther into the past. The exponential smoothing model follows.

Y ⁄

t1 1 5 aYt 1 s1 2 adY ⁄

t (6.8)

where

Y ⁄

t+1 5 forecast of the time series for period t 1 1

Yt 5 actual value of the time series in period t

Y ⁄

t 5 forecast of the time series for period t

a 5 smoothing constant (0 # a # 1)

equation (6.8) shows that the forecast for period t + 1 is a weighted average of the actual value in period t and the forecast for period t. The weight given to the actual value in period t is the smoothing constant a and the weight given to the forecast in period t is 1 2 a. it turns out that the exponential smoothing forecast for any period is actually a weighted average of all the previous actual values of the time series. let us illustrate by working with a time series involving only three periods of data: Y1, Y2, and Y3.

To initiate the calculations, we let Y ⁄

1 equal the actual value of the time series in period 1; that is, Y

⁄ 1 5 Y1. Hence, the forecast for period 2 is

Y ⁄

2 5 aY1 1 (1 2 a)Y ⁄

5 aY1 1 (1 2 a)Y1 5 Y1

We see that the exponential smoothing forecast for period 2 is equal to the actual value of the time series in period 1.

The forecast for period 3 is

Y ⁄

3 5 aY2 1 (1 2 a)Y ⁄

2 5 aY2 1 (1 2 a)Y1

Finally, substituting this expression for Y ⁄

3 into the expression for Y ⁄

4, we obtain

Y ⁄

4 5 aY3 1 (1 2 a)Y ⁄

5 aY3 1 (1 2 a)[aY2 1 (1 2 a)Y1]

5 aY3 1 a(1 2 a)Y2 1 (1 2 a) 2Y1

We now see that Y ⁄

4 is a weighted average of the first three time series values. The sum of the coefficients, or weights, for Y1, Y2, and Y3 equals 1. a similar argument can be made to show that, in general, any forecast Y

⁄ t+1 is a weighted average of all the t previous time

series values.

There are several exponential smoothing procedures. Because it has a single smoothing constant a, the method presented here is often referred to as single exponential smoothing.

The term exponential smoothing comes from the exponential nature of the weighting scheme for the historical values.

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216 Chapter 6 Time Series Analysis and Forecasting

despite the fact that exponential smoothing provides a forecast that is a weighted aver- age of all past observations, all past data do not need to be retained to compute the forecast for the next period. in fact, equation (6.8) shows that once the value for the smoothing constant a is selected, only two pieces of information are needed to compute the forecast for period t 1 1: Yt , the actual value of the time series in period t; and Y

⁄ t , the forecast for

period t. To illustrate the exponential smoothing approach to forecasting, let us again consider

the gasoline sales time series in Table 6.1 and Figure 6.1. as indicated previously, to ini- tialize the calculations we set the exponential smoothing forecast for period 2 equal to

the actual value of the time series in period 1. Thus, with Y1 5 17, we set Y ⁄

2 5 17 to initiate the computations. referring to the time series data in Table 6.1, we find an actual time series value in period 2 of Y2 5 21. Thus, in period 2 we have a forecast error of e2 5 21 – 17 5 4.

Continuing with the exponential smoothing computations using a smoothing constant of a 5 0.2, we obtain the following forecast for period 3.

Y ⁄

3 5 0.2Y2 1 0.8Y ⁄

2 5 0.2(21) 1 0.8(17) 5 17.8

once the actual time series value in period 3, Y3 5 19, is known, we can generate a forecast for period 4 as follows.

Y ⁄

4 5 0.2Y3 1 0.8Y ⁄

3 5 0.2(19) 1 0.8(17.8) 5 18.04

Continuing the exponential smoothing calculations, we obtain the weekly forecast values shown in Table 6.10. note that we have not shown an exponential smoothing fore- cast or a forecast error for week 1 because no forecast was made (we used actual sales for week 1 as the forecasted sales for week 2 to initialize the exponential smoothing process). For week 12, we have Y12 5 22 and Y

⁄ 12 5 18.48. We can we use this information to gener-

ate a forecast for week 13.

Y ⁄

135 0.2Y12 1 0.8Y ⁄

125 0.2(22) 1 0.8(18.48) 5 19.18

Week Time Series Value Forecast Forecast Error Squared Forecast Error 1 17 2 21 17.00 4.00 16.00 3 19 17.80 1.20 1.44 4 23 18.04 4.96 24.60 5 18 19.03 21.03 1.06 6 16 18.83 22.83 8.01 7 20 18.26 1.74 3.03 8 18 18.61 20.61 0.37 9 22 18.49 3.51 12.32 10 20 19.19 0.81 0.66 11 15 19.35 24.35 18.92 12 22 18.48 3.52 12.39

Totals 10.92 98.80

TABLE 6.10 Summary oF THe exPonenTial SmooTHing ForeCaSTS and ForeCaST errorS For THe gaSoline SaleS Time SerieS WiTH SmooTHing ConSTanT a 5 0.2

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6.3 Moving Averages and Exponential Smoothing 217

Thus, the exponential smoothing forecast of the amount sold in week 13 is 19.18, or 19,180 gallons of gasoline. With this forecast, the firm can make plans and decisions accordingly.

Figure 6.8 shows the time series plot of the actual and forecast time series values. note in particular how the forecasts “smooth out” the irregular or random fluctuations in the time series.

Forecast Accuracy in the preceding exponential smoothing calculations, we used a smoothing constant of a 5 0.2. although any value of a between 0 and 1 is accept- able, some values will yield more accurate forecasts than others. insight into choosing a good value for a can be obtained by rewriting the basic exponential smoothing model as follows.

Y ⁄

t1 1 5

Y ⁄

t1 1 5

Y ⁄

t1 1 5

aYt 1 s1 2 adY ⁄

aYt 1 Y ⁄

t 2 aY ⁄

Y ⁄

t 1 asYt 2 Y ⁄

t d 5 Y ⁄

t 1 a et

(6.9)

Thus, the new forecast Y ⁄

t+1 is equal to the previous forecast Y ⁄

t plus an adjustment, which is the smoothing constant a times the most recent forecast error, et 5 Yt – Y

⁄ t. That is, the

forecast in period t 1 1 is obtained by adjusting the forecast in period t by a fraction of the forecast error from period t. if the time series contains substantial random variability, a small value of the smoothing constant is preferred. The reason for this choice is that if much of the forecast error is due to random variability, we do not want to overreact and adjust the forecasts too quickly. For a time series with relatively little random variability,

S al

es (

10 00

s of

g al

lo n

Week

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Forecast time series with a 5 0.2

Actual time series

FIGURE 6.8 aCTual and ForeCaST gaSoline Time SerieS WiTH SmooTHing ConSTanT a 5 0.2

Try Problem 9 for practice using exponential smoothing to produce forecasts.

66314_ch06_ptg01_hires.indd 217 23/12/14 1:59 PM

218 Chapter 6 Time Series Analysis and Forecasting

a forecast error is more likely to represent a real change in the level of the series. Thus, larger values of the smoothing constant provide the advantage of quickly adjusting the forecasts to changes in the time series; this allows the forecasts to react more quickly to changing conditions.

The criterion we will use to determine a desirable value for the smoothing con- stant a is the same as the criterion we proposed for determining the order or number of periods of data to include in the moving averages calculation. That is, we choose the value of a that minimizes the mSe. a summary of the mSe calculations for the exponential smoothing forecast of gasoline sales with a 5 0.2 is shown in Table 6.10. note that there is one less squared error term than the number of time periods; this is because we had no past values with which to make a forecast for period 1. The value of the sum of squared forecast errors is 98.80; hence mSe 5 98.80/11 5 8.98. Would a different value of a provide better results in terms of a lower mSe value? Trial and error is often used to determine if a different smoothing constant a can provide more accurate forecasts, but we can avoid trial and error and determine the value of a that minimizes mSe through the use of nonlinear optimization as discussed in Chapter 12 (see exercise 12.19).

NOTES AND COMMENTS

1. Spreadsheet packages are effective tools for implementing exponential smoothing. With the time series data and the forecasting formu- las in a spreadsheet as shown in Table 6.10, you can use the mae, mSe, and maPe to evaluate different values of the smoothing con- stant a.

2. We presented the moving average, weighted moving average, and exponential smoothing methods in the context of a stationary time se- ries. These methods can also be used to fore- cast a nonstationary time series that shifts in

level but exhibits no trend or seasonality. mov- ing averages with small values of k adapt more quickly than moving averages with larger val- ues of k. Weighted moving averages that place relatively large weights on the most recent val- ues adapt more quickly than weighted moving averages that place relatively equal weights on the k time series values used in calculating the forecast. exponential smoothing models with smoothing constants closer to 1 adapt more quickly than models with smaller values of the smoothing constant.

6.4 Linear Trend Projection in this section we present forecasting methods that are appropriate for time series exhibiting trend patterns. Here we show how regression analysis may be used to forecast a time series with a linear trend. in Section 6.1 we used the bicycle sales time series in Table 6.3 and Figure 6.3 to illustrate a time series with a trend pattern. let us now use this time series to illustrate how regression analysis can be used to forecast a time series with a linear trend. The data for the bicycle time series are repeated in Table 6.11 and Figure 6.9.

although the time series plot in Figure 6.9 shows some up and down movement over the past 10 years, we might agree that the linear trend line shown in Figure 6.10 provides a reasonable approximation of the long-run movement in the series. We can use regression analysis to develop such a linear trend line for the bicycle sales time series.

in regression analysis we use known values of variables to estimate the relation- ship between one variable (called the dependent variable) and one or more other related

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6.4 Linear Trend Projection 219

variables (called independent variables). This relationship is usually found in a manner that minimizes the sum of squared errors (and so also minimizes the mSe). With this rela- tionship we can then use values of the independent variables to estimate the associated value of the dependent variable. When we estimate a linear relationship between the dependent variable (which is usually denoted as y) and a single independent variable (which is usually denoted as x), this is referred to as simple linear regression. estimating the relationship between the dependent variable and a single independent variable requires that we find the values of parameters b0 and b1 for the straight line y 5 b0 1 b1x.

Because our use of simple linear regression analysis yields the linear relationship be- tween the independent variable and the dependent variable that minimizes the mSe, we can use this approach to find a best-fitting line to a set of data that exhibits a linear trend. in finding a linear trend, the variable to be forecasted (Yt, the actual value of the time series in

S al

es (

10 00

Year

0 2 4 6 8 10

FIGURE 6.9 BiCyCle SaleS Time SerieS PloT

Year Sales (1000s) 1 21.6 2 22.9 3 25.5 4 21.9 5 23.9 6 27.5 7 31.5 8 29.7 9 28.6 10 31.4

TABLE 6.11 BiCyCle SaleS Time SerieS

fileWEB Bicycle

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220 Chapter 6 Time Series Analysis and Forecasting

period t) is the dependent variable and the trend variable (time period t) is the independent variable. We will use the following notation for our linear trendline.

Y ⁄

t 5 b0 1 b1t (6.10)

where t 5 the time period

Y ⁄

t 5 linear trend forecast in period t (i.e., the estimated value of Yt in period t)

b0 5 the Y-intercept of the linear trendline

b1 5 the slope of the linear trendline

in equation (6.10) the time variable begins at t 5 1 corresponding to the first time series observation (year 1 for the bicycle sales time series) and continues until t 5 n corresponding to the most recent time series observation (year 10 for the bicycle sales time series). Thus, for the bicycle sales time series t 5 1 corresponds to the oldest time series value and t 5 10 corresponds to the most recent year. Calculus may be used to show that the equations given below for b0 and b1 yield the line that minimizes the mSe. The equations for computing the values of b0 and b1 are

b1 5 o

t5 1 tYt 2 o

t5 1

to n

t5 1

Yt nn

o n

t5 1 t 2 2 1o

t5 1 t2

nn (6.11)

b0 5 Y 2 b1t (6.12)

S al

es (

10 00

Year

0 2 4 6 8 10

FIGURE 6.10 Trend rePreSenTed By a linear FunCTion For THe BiCyCle SaleS Time SerieS

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6.4 Linear Trend Projection 221

where

t 5 the time period

Yt 5 actual value of the time series in period t

n 5 number of periods in the time series

Y 5 average value of the time series; that is, Y 5 o n

t5 1

Ytnn t 5 mean value of t; that is, t 5 o

t5 1

tnn

let us calculate b0 and b1 for the bicycle data in Table 6.11; the intermediate summary calculations necessary for computing the values of b0 and b1 are

t Yt tYt t 2

1 21.6 21.6 1 2 22.9 45.8 4 3 25.5 76.5 9 4 21.9 87.6 16 5 23.9 119.5 25 6 27.5 165.0 36 7 31.5 220.5 49 8 29.7 237.6 64 9 28.6 257.4 81 10 31.4 314.0 100

Totals 55 264.5 1545.5 385

and the final calculations of the values of b0 and b1 are

t 5 55

10 5 5.5

Y 5 264.5

10 5 26.45

b1 5 1545.5 2 s55ds264.5d/10

385 2 552/10 5 1.10

b0 5 26.45 2 1.10s5.5d 5 20.40

Therefore,

Y ⁄

t 5 20.4 1 1.1t (6.13)

is the regression equation for the linear trend component for the bicycle sales time series. The slope of 1.1 in this trend equation indicates that over the past 10 years, the firm

has experienced an average growth in sales of about 1100 units per year. if we assume that the past 10-year trend in sales is a good indicator for the future, we can use equation (6.13) to project the trend component of the time series. For example, substituting t 5 11 into equation (6.13) yields next year’s trend projection, Y

⁄ 11:

Y ⁄

11 5 20.4 1 1.1(11) 5 32.5

Thus, the linear trend model yields a sales forecast of 32,500 bicycles for the next year.

Problem 20 provides additional practice in using regression analysis to estimate the linear trend in a time series data set.

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222 Chapter 6 Time Series Analysis and Forecasting

Table 6.12 shows the computation of the minimized sum of squared errors for the bicycle sales time series. as previously noted, minimizing sum of squared errors also mini- mizes the commonly used measure of accuracy, mSe. For the bicycle sales time series,

mSe 5 o

t5 1

e2t

n 5

30.7

10 5 3.07

note that in this example we are not using past values of the time series to produce forecasts, and so k 5 0; that is, we can produce a forecast for each period of the time series and so do not have to adjust our calculations of the mae, mSe, or maPe for k.

We can also use the trendline to forecast sales farther into the future. For instance, using equation (6.13), we develop annual forecasts for two and three years into the future as follows:

Y ⁄

12 5 20.4 1 1.1(12) 5 33.6

Y ⁄

13 5 20.4 1 1.1(13) 5 34.7

note that the forecasted value increases by 1100 bicycles in each year.

Sales (1000s) Forecast Forecast Squared Forecast

Week Yt Y ⁄

t Error Error

1 21.6 21.5 0.1 0.01 2 22.9 22.6 0.3 0.09 3 25.5 23.7 1.8 3.24 4 21.9 24.8 22.9 8.41 5 23.9 25.9 22.0 4.00 6 27.5 27.0 0.5 0.25 7 31.5 28.1 3.4 11.56 8 29.7 29.2 0.5 0.25 9 28.6 30.3 21.7 2.89 10 31.4 31.4 0.0 0.00

Total 30.70

TABLE 6.12 Summary oF THe linear Trend ForeCaSTS and ForeCaST errorS For THe BiCyCle SaleS Time SerieS

NOTES AND COMMENTS

1. Statistical packages such as minitab and SaS, as well as excel, have routines to perform re- gression analysis. regression analysis mini- mizes the sum of squared error and under certain assumptions it also allows the analyst to make statistical statements about the parameters and the forecasts.

2. While the use of a linear function to model the trend is common, some time series exhibit a cur- vilinear (nonlinear) trend. more advanced texts discuss how to develop nonlinear models such as quadratic models and exponential models for these more complex relationships.

in this section we used simple linear regression to estimate the relationship between the dependent variable (Yt , the actual value of the time series in period t) and a single indepen- dent variable (the trend variable t). However, some regression models include several inde- pendent variables. When we estimate a linear relationship between the dependent variable

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6.5 Seasonality 223

and more than one independent variable, this is referred to as multiple linear regression. in the next section we will apply multiple linear regression to time series that include seasonal effects and to time series that include both seasonal effects and a linear trend.

6.5 Seasonality in this section we show how to develop forecasts for a time series that has a seasonal pattern. To the extent that seasonality exists, we need to incorporate it into our forecasting models to ensure accurate forecasts. We begin the section by considering a seasonal time series with no trend and then discuss how to model seasonality with a linear trend.

Seasonality Without Trend let us consider again the data from Table 6.5, the number of umbrellas sold at a clothing store over the past five years. We repeat the data here in Table 6.13, and Figure 6.11 again shows the corresponding time series plot. The time series plot does not indicate any long- term trend in sales. in fact, unless you look carefully at the data, you might conclude that the data follow a horizontal pattern with random fluctuation and that single exponential smoothing could be used to forecast sales. However, closer inspection of the time series plot reveals a pattern in the fluctuations. That is, the first and third quarters have moderate sales, the second quarter the highest sales, and the fourth quarter tends to be the lowest quarter in terms of sales volume. Thus, we conclude that a quarterly seasonal pattern is present.

We can model a time series with a seasonal pattern by treating the season as a categori- cal variable. Categorical variables are data used to categorize observations of data. When

Year Quarter Sales 1 1 125 2 153 3 106 4 88 2 1 118 2 161 3 133 4 102 3 1 138 2 144 3 113 4 80 4 1 109 2 137 3 125 4 109 5 1 130 2 165 3 128 4 96

TABLE 6.13 umBrella SaleS Time SerieS

fileWEB Umbrella

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224 Chapter 6 Time Series Analysis and Forecasting

0 2 4 6 810 3 5 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Period

180

100

120

140

160

S al

FIGURE 6.11 umBrella SaleS Time SerieS PloT

a categorical variable has k levels, k 2 1 dummy variables (sometimes called 0-1 variables) are required. So if there are four seasons, we need three dummy variables. For instance, in the umbrella sales time series, the quarter to which each observation corresponds is treated as a season; it is a categorical variable with four levels: Quarter 1, Quarter 2, Quarter 3, and Quarter 4. Thus, to model the seasonal effects in the umbrella time series we need 4 2 1 5 3 dummy variables. The three dummy variables can be coded as follows:

Qtr1t 5 51 if period t is a Quarter 10 otherwise Qtr2t 5 51 if period t is a Quarter 20 otherwise Qtr3t 5 51 if period t is a Quarter 30 otherwise

using Ft to denote the forecasted value of sales for period t, the general form of the equation relating the number of umbrellas sold to the quarter the sales take place follows.

Y ⁄

t 5 b0 1 b1 Qtr1t 1 b2 Qtr2t 1 b3 Qtr3t (6.14)

note that the fourth quarter will be denoted by a setting of all three dummy variables to 0. Table 6.14 shows the umbrella sales time series with the coded values of the dummy variables shown. We can use a multiple linear regression model to find the values of b0, b1, b2, and b3 that minimize the sum of squared errors. For this regression model Yt is the dependent variable and the quarterly dummy variables Qtr1t, Qtr2t, and Qtr3t are the inde- pendent variables.

Note that we have numbered the observations in Table 6.14 as periods 1 to 20. For example, year 3, quarter 3 is observation 11.

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6.5 Seasonality 225

using the data in Table 6.14 and regression analysis, we obtain the following equation:

Y ⁄

t 5 95.0 1 29.0 Qtr1t 1 57.0 Qtr2t 1 26.0 Qtr3t (6.15)

we can use equation (6.15) to forecast quarterly sales for next year.

Quarter 1: Sales 5 95.0 1 29.0(1) 1 57.0(0) 1 26.0(0) 5 124

Quarter 2: Sales 5 95.0 1 29.0(0) 1 57.0(1) 1 26.0(0) 5 152

Quarter 3: Sales 5 95.0 1 29.0(0) 1 57.0(0) 1 26.0(1) 5 121

Quarter 4: Sales 5 95.0 1 29.0(0) 1 57.0(0) 1 26.0(0) 5 95

it is interesting to note that we could have obtained the quarterly forecasts for next year by simply computing the average number of umbrellas sold in each quarter, as shown in the following table.

period Year Quarter Qtr1 Qtr2 Qtr3 Sales

1 1 1 1 0 0 125 2 2 0 1 0 153 3 3 0 0 1 106 4 4 0 0 0 88 5 2 1 1 0 0 118 6 2 0 1 0 161 7 3 0 0 1 133 8 4 0 0 0 102 9 3 1 1 0 0 138 10 2 0 1 0 144 11 3 0 0 1 113 12 4 0 0 0 80 13 4 1 1 0 0 109 14 2 0 1 0 137 15 3 0 0 1 125 16 4 0 0 0 109 17 5 1 1 0 0 130 18 2 0 1 0 165 19 3 0 0 1 128 20 4 0 0 0 96

TABLE 6.14 umBrella SaleS Time SerieS WiTH dummy variaBleS

Year Quarter 1 Quarter 2 Quarter 3 Quarter 4

1 125 153 106 88 2 118 161 133 102 3 138 144 113 80 4 109 137 125 109 5 130 165 128 96

average 124 152 121 95

For practice using categorical variables to estimate seasonal effects, try Problem 24.

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226 Chapter 6 Time Series Analysis and Forecasting

nonetheless, for more complex problem situations, such as dealing with a time series that has both trend and seasonal effects, this simple averaging approach will not work.

Seasonality with Trend We now consider situations for which the time series contains both a seasonal effect and a linear trend by showing how to forecast the quarterly television set sales time series intro- duced in Section 6.1. The data for the television set time series are shown in Table 6.15. The time series plot in Figure 6.12 indicates that sales are lowest in the second quarter of each year and increase in quarters 3 and 4. Thus, we conclude that a seasonal pattern exists for television set sales. However, the time series also has an upward linear trend that will need to be accounted for in order to develop accurate forecasts of quarterly sales. This is easily done by combining the dummy variable approach for handling seasonality with the approach we discussed in Section 6.4 for handling a linear trend.

The general form of the regression equation for modeling both the quarterly seasonal effects and the linear trend in the television set time series is:

Y ⁄

t 5 b0 1 b1Qtr1t 1 b2Qtr2t 1 b3Qtr3t 1 b4t (6.16)

where

Y ⁄

t 5 forecast of sales in period t

Qtr1t 5 1 if time period t corresponds to the first quarter of the year; 0, otherwise

Qtr2t 5 1 if time period t corresponds to the second quarter of the year; 0, otherwise

Qtr3t 5 1 if time period t corresponds to the third quarter of the year; 0, otherwise

t 5 time period

Year Quarter Sales (1000s) 1 1 4.8 2 4.1 3 6.0 4 6.5 2 1 5.8 2 5.2 3 6.8 4 7.4 3 1 6.0 2 5.6 3 7.5 4 7.8 4 1 6.3 2 5.9 3 8.0 4 8.4

TABLE 6.15 TeleviSion SeT SaleS Time SerieS

fileWEB TVSales

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6.5 Seasonality 227

3.0

4.0

2.0 10 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Period

9.0

8.0

7.0

6.0

5.0

Q u

ar te

rl y

T el

ev is

io n

S et

S al

es (

10 00

FIGURE 6.12 TeleviSion SeT SaleS Time SerieS PloT

For this regression model Yt is the dependent variable and the quarterly dummy variables Qtr1t, Qtr2t, and Qtr3t and the time period t are the independent variables.

Table 6.16 shows the revised television set sales time series that includes the coded values of the dummy variables and the time period t. using the data in Table 6.16 with the

period Year Quarter Qtr1 Qtr2 Qtr3 Sales (1000s)

1 1 1 1 0 0 4.8 2 2 0 1 0 4.1 3 3 0 0 1 6.0 4 4 0 0 0 6.5 5 2 1 1 0 0 5.8 6 2 0 1 0 5.2 7 3 0 0 1 6.8 8 4 0 0 0 7.4 9 3 1 1 0 0 6.0 10 2 0 1 0 5.6 11 3 0 0 1 7.5 12 4 0 0 0 7.8 13 4 1 1 0 0 6.3 14 2 0 1 0 5.9 15 3 0 0 1 8.0 16 4 0 0 0 8.4

TABLE 6.16 TeleviSion SeT SaleS Time SerieS WiTH dummy variaBleS and Time Period

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228 Chapter 6 Time Series Analysis and Forecasting

regression model that includes both the seasonal and trend components, we obtain the fol- lowing equation that minimizes our sum of squared errors:

Y ⁄

t 5 6.07 2 1.36 Qtr1t 2 2.03 Qtr2t 2 0.304 Qtr3t 1 0.146t (6.17)

We can now use equation (6.17) to forecast quarterly sales for next year. next year is year 5 for the television set sales time series; that is, time periods 17, 18, 19, and 20.

Forecast for Time Period 17 (Quarter 1 in year 5)

Y ⁄

17 5 6.07 2 1.36(1) 2 2.03(0) 2 0.304(0) 1 0.146(17) 5 7.19

Forecast for Time Period 18 (Quarter 2 in year 5)

Y ⁄

18 5 6.07 2 1.36(0) 2 2.03(1) 2 0.304(0) 1 0.146(18) 5 6.67

Forecast for Time Period 19 (Quarter 3 in year 5)

Y ⁄

19 5 6.07 2 1.36(0) 2 2.03(0) 2 0.304(1) 1 0.146(19) 5 8.54

Forecast for Time Period 20 (Quarter 4 in year 5)

Y ⁄

20 5 6.07 2 1.36(0) 2 2.03(0) 2 0.304(0) 1 0.146(20) 5 8.99

Thus, accounting for the seasonal effects and the linear trend in television set sales, the estimates of quarterly sales in year 5 are 7190, 6670, 8540, and 8990.

The dummy variables in the equation actually provide four equations, one for each quarter. For instance, if time period t corresponds to quarter 1, the estimate of quarterly sales is

Quarter 1: Sales 5 6.07 2 1.36(1) 2 2.03(0) 2 0.304(0) 1 0.146t 5 4.71 1 0.146t

Similarly, if time period t corresponds to quarters 2, 3, and 4, the estimates of quarterly sales are:

Quarter 2: Sales 5 6.07 2 1.36(0) 2 2.03(1) 2 0.304(0) 1 0.146t 5 4.04 1 0.146t

Quarter 3: Sales 5 6.07 2 1.36(0) 2 2.03(0) 2 0.304(1) 1 0.146t 5 5.77 1 0.146t

Quarter 4: Sales 5 6.07 2 1.36(0) 2 2.03(0) 2 0.304(0) 1 0.146t 5 6.07 1 0.146t

The slope of the trend line for each quarterly forecast equation is 0.146, indicating a consistent growth in sales of about 146 sets per quarter. The only difference in the four equations is that they have different intercepts.

Models Based on Monthly Data in the preceding television set sales example, we showed how dummy variables can be used to account for the quarterly seasonal effects in the time series. Because there were four levels for the categorical variable season, three dummy variables were required. However,

Problem 28 provides another example of using regression analysis to forecast time series data with both trend and seasonal effects.

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Summary 229

many businesses use monthly rather than quarterly forecasts. For monthly data, season is a categorical variable with 12 levels, and thus 12 2 1 5 11 dummy variables are required. For example, the 11 dummy variables could be coded as follows:

month1 5 51 if january0 otherwise month2 5 51 if February0 otherwise . . .

month1 5 51 if november0 otherwise other than this change, the approach for handling seasonality remains the same.

Summary

This chapter provided an introduction to basic methods of time series analysis and forecasting. We first showed that the underlying pattern in the time series can often be identified by constructing a time series plot. Several types of data patterns can be distinguished, including a horizontal pattern, a trend pattern, and a seasonal pattern. The forecasting methods we have discussed are based on which of these patterns are present in the time series.

We also discussed that the accuracy of the method is an important factor in deter- mining which forecasting method to use. We considered three measures of forecast accuracy: mean absolute error (mae), mean squared error (mSe), and mean absolute percentage error (maPe). each of these measures is designed to determine how well a particular forecasting method is able to reproduce the time series data that are already available. By selecting the method that is most accurate for the data already known, we hope to increase the likelihood that we will obtain more accurate forecasts for future time periods.

For a time series with a horizontal pattern, we showed how moving averages, weighted moving averages, and exponential smoothing can be used to develop a forecast. The moving averages method consists of computing an average of past data values and then using that average as the forecast for the next period. in the weighted moving average and exponen- tial smoothing methods, weighted averages of past time series values are used to compute forecasts. These methods also adapt well to a horizontal pattern that shifts to a different level and then resumes a horizontal pattern.

For time series that have only a long-term linear trend, we showed how regression analysis can be used to make trend projections. For a time series with a seasonal pattern, we showed how dummy variables and regression analysis can be used to develop an equa- tion with seasonal effects. We then extended the approach to include situations where the time series contains both a seasonal and a linear trend effect by showing how to combine the dummy variable approach for handling seasonality with the approach for handling a linear trend.

Whenever a categorical variable such as season has k levels, k 2 1 dummy variables are required.

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230 Chapter 6 Time Series Analysis and Forecasting

Glossary

Categorical (dummy) variable a variable used to categorize observations of data. used when modeling a time series with a seasonal pattern. Cyclical pattern a cyclical pattern exists if the time series plot shows an alternating sequence of points below and above the trend line lasting more than one year. dependent variable The variable that is being predicted or explained in a regression analysis. Exponential smoothing a forecasting method that uses a weighted average of past time series values as the forecast; it is a special case of the weighted moving averages method in which we select only one weight—the weight for the most recent observation. Forecast error The difference between the actual time series value and the forecast. independent variable a variable used to predict or explain values of the dependent vari- able in regression analysis. Mean absolute error (MAE) The average of the absolute values of the forecast errors. Mean absolute percentage error (MApE) The average of the absolute values of the percentage forecast errors. Mean squared error (MSE) The average of the sum of squared forecast errors. Moving averages a forecasting method that uses the average of the k most recent data values in the time series as the forecast for the next period. Regression analysis a procedure for estimating values of a dependent variable given the values of one or more independent variables in a manner that minimizes the sum of the squared errors. Seasonal pattern a seasonal pattern exists if the time series plot exhibits a repeating pattern over successive periods. Smoothing constant a parameter of the exponential smoothing model that provides the weight given to the most recent time series value in the calculation of the forecast value. Stationary time series a time series whose statistical properties are independent of time. For a stationary time series, the process generating the data has a constant mean and the variability of the time series is constant over time. Time series a sequence of observations on a variable measured at successive points in time or over successive periods of time. Time series plot a graphical presentation of the relationship between time and the time series variable. Time is shown on the horizontal axis and the time series values are shown on the verical axis. Trend pattern a trend pattern exists if the time series plot shows gradual shifts or move- ments to relatively higher or lower values over a longer period of time. Weighted moving averages a forecasting method that involves selecting a different weight for the k most recent data values values in the time series and then computing a weighted average of the of the values. The sum of the weights must equal one.

Problems

1. Consider the following time series data.

Week 1 2 3 4 5 6

Value 18 13 16 11 17 14

using the naïve method (most recent value) as the forecast for the next week, compute the following measures of forecast accuracy.

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Problems 231

a. mean absolute error b. mean squared error c. mean absolute percentage error d. What is the forecast for week 7?

2. refer to the time series data in exercise 1. using the average of all the historical data as a forecast for the next period, compute the following measures of forecast accuracy: a. mean absolute error b. mean squared error c. mean absolute percentage error d. What is the forecast for week 7?

3. exercises 1 and 2 used different forecasting methods. Which method appears to provide the more accurate forecasts for the historical data? explain.

4. Consider the following time series data.

Month 1 2 3 4 5 6 7

Value 24 13 20 12 19 23 15

a. Compute mSe using the most recent value as the forecast for the next period. What is the forecast for month 8?

b. Compute mSe using the average of all the data available as the forecast for the next period. What is the forecast for month 8?

c. Which method appears to provide the better forecast?

5. Consider the following time series data.

Week 1 2 3 4 5 6

Value 18 13 16 11 17 14

a. Construct a time series plot. What type of pattern exists in the data? b. develop a three-week moving average for this time series. Compute mSe and a fore-

cast for week 7. c. use a 5 0.2 to compute the exponential smoothing values for the time series. Com-

pute mSe and a forecast for week 7. d. Compare the three-week moving average forecast with the exponential smoothing

forecast using a 5 0.2. Which appears to provide the better forecast based on mSe? explain.

e. use trial and error to find a value of the exponential smoothing coefficient a that results in a smaller mSe than what you calculated for a 5 0.2.

6. Consider the following time series data.

Month 1 2 3 4 5 6 7

Value 24 13 20 12 19 23 15

a. Construct a time series plot. What type of pattern exists in the data? b. develop a three-week moving average for this time series. Compute mSe and a fore-

cast for week 8. c. use a 5 0.2 to compute the exponential smoothing values for the time series. Com-

pute mSe and a forecast for week 8. d. Compare the three-week moving average forecast with the exponential smoothing

forecast using a 5 0.2. Which appears to provide the better forecast based on mSe? e. use trial and error to find a value of the exponential smoothing coefficient a that

results in a smaller mSe than what you calculated for a 5 0.2.

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232 Chapter 6 Time Series Analysis and Forecasting

7. refer to the gasoline sales time series data in Table 6.1. a. Compute four-week and five-week moving averages for the time series. b. Compute the mSe for the four-week and five-week moving average forecasts. c. What appears to be the best number of weeks of past data (three, four, or five) to use in

the moving average computation? recall that mSe for the three-week moving average is 10.22.

8. refer again to the gasoline sales time series data in Table 6.1. a. using a weight of 1/2 for the most recent observation, 1/3 for the second most recent,

and 1/6 for third most recent, compute a three-week weighted moving average for the time series.

b. Compute the mSe for the weighted moving average in part (a). do you prefer this weighted moving average to the unweighted moving average? remember that the mSe for the unweighted moving average is 10.22.

c. Suppose you are allowed to choose any weights as long as they sum to 1. Could you always find a set of weights that would make the mSe smaller for a weighted moving average than for an unweighted moving average? Why or why not?

9. With the gasoline time series data from Table 6.1, show the exponential smoothing fore- casts using a 5 0.1. a. applying the mSe measure of forecast accuracy, would you prefer a smoothing

constant of a 5 0.1 or a 5 0.2 for the gasoline sales time series? b. are the results the same if you apply mae as the measure of accuracy? c. What are the results if maPe is used?

10. With a smoothing constant of a 5 0.2, equation (6.8) shows that the forecast for week 13 of the gasoline sales data from Table 6.1 is given by Y

⁄ 13 5 0.2Y12 1 0.8Y

⁄ 12. However, the

forecast for week 12 is given by Y ⁄

12 5 0.2Y11 1 0.8Y ⁄

11. Thus, we could combine these two results to show that the forecast for week 13 can be written

Y ⁄

13 5 0.2Y12 1 0.8(0.2Y11 1 0.8Y ⁄

11) 5 0.2Y12 1 0.16Y11 1 0.64Y ⁄

a. making use of the fact that Y ⁄

11 5 0.2Y10 1 0.8Y ⁄

10 (and similarly for Y ⁄

10 and Y ⁄

9), continue to expand the expression for Y

⁄ 13 until it is written in terms of the past data

values Y12, Y11, Y10, Y9, Y8, and the forecast for period 8. b. refer to the coefficients or weights for the past values Y12, Y11, Y10, Y9, and Y8. What

observation can you make about how exponential smoothing weights past data val- ues in arriving at new forecasts? Compare this weighting pattern with the weighting pattern of the moving averages method.

11. For the Hawkins Company, the monthly percentages of all shipments received on time over the past 12 months are 80, 82, 84, 83, 83, 84, 85, 84, 82, 83, 84, and 83. a. Construct a time series plot. What type of pattern exists in the data? b. Compare a three-month moving average forecast with an exponential smoothing fore-

cast for a 5 0.2. Which provides the better forecasts using mSe as the measure of model accuracy?

c. What is the forecast for next month?

12. Corporate triple a bond interest rates for 12 consecutive months follow.

9.5 9.3 9.4 9.6 9.8 9.7 9.8 10.5 9.9 9.7 9.6 9.6

a. Construct a time series plot. What type of pattern exists in the data? b. develop three-month and four-month moving averages for this time series. does the three-

month or four-month moving average provide the better forecasts based on mSe? explain. c. What is the moving average forecast for the next month?

fileWEB Gasoline

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Problems 233

13. The values of alabama building contracts (in millions of dollars) for a 12-month period follow.

240 350 230 260 280 320 220 310 240 310 240 230

a. Construct a time series plot. What type of pattern exists in the data? b. Compare a three-month moving average forecast with an exponential smoothing fore-

cast. use a 5 0.2. Which provides the better forecasts based on mSe? c. What is the forecast for the next month?

14. The following time series shows the sales of a particular product over the past 12 months.

Month Sales Month Sales

1 105 7 145 2 135 8 140 3 120 9 100 4 105 10 80 5 90 11 100 6 120 12 110

a. Construct a time series plot. What type of pattern exists in the data? b. use a 5 0.3 to compute the exponential smoothing values for the time series. c. use trial and error to find a value of the exponential smoothing coefficient a that

results in a relatively small mSe.

15. Ten weeks of data on the Commodity Futures index are 7.35, 7.40, 7.55, 7.56, 7.60, 7.52, 7.52, 7.70, 7.62, and 7.55. a. Construct a time series plot. What type of pattern exists in the data? b. use trial and error to find a value of the exponential smoothing coefficient a that

results in a relatively small mSe.

16. Since its inception in 1967, the Super Bowl has been one of the most watched events on u.S. television every year. The number of u.S. households that tuned in for each Super Bowl, reported by nielsen.com, is provided in the data set SuperBowlratings. a. Construct a time series plot for the data. What type of pattern exists in the data?

discuss some of the patterns that may have resulted in the pattern exhibited in the time series plot of the data.

b. given the pattern of the time series plot developed in part (a), do you think the fore- casting methods discussed in this chapter are appropriate to develop forecasts for this time series? explain.

c. use simple linear regression analysis to find the parameters for the line that minimizes mSe for this time series.

17. Consider the following time series.

t 1 2 3 4 5

Yt 6 11 9 14 15

a. Construct a time series plot. What type of pattern exists in the data? b. use simple linear regression analysis to find the parameters for the line that minimizes

mSe for this time series. c. What is the forecast for t 5 6?

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fileWEB SuperBowlRatings

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234 Chapter 6 Time Series Analysis and Forecasting

18. The following table reports the percentage of stocks in a portfolio for the nine previous quarters.

Year Quarter Stock%

1 1 29.8 1 2 31.0 1 3 29.9 1 4 30.1 2 1 32.2 2 2 31.5 2 3 32.0 2 4 31.9 3 1 30.0

a. Construct a time series plot. What type of pattern exists in the data? b. use trial and error to find a value of the exponential smoothing coefficient a that

results in a relatively small mSe. c. using the exponential smoothing model you developed in part (b), what is the forecast

of the percentage of stocks in a typical portfolio for the second quarter of year 3?

19. Consider the following time series.

t 1 2 3 4 5 6 7

Yt 120 110 100 96 94 92 88

a. Construct a time series plot. What type of pattern exists in the data? b. use simple linear regression analysis to find the parameters for the line that minimizes

mSe for this time series. c. What is the forecast for t 5 8?

20. Because of high tuition costs at state and private universities, enrollments at community colleges have increased dramatically in recent years. The following data show the enroll- ment (in thousands) for jefferson Community College for the nine most recent years.

Year Enrollment (1000s)

1 6.5 2 8.1 3 8.4 4 10.2 5 12.5 6 13.3 7 13.7 8 17.2 9 18.1

fileWEB JeffersonEnrollment

a. Construct a time series plot. What type of pattern exists in the data? b. use simple linear regression analysis to find the parameters for the line that minimizes

mSe for this time series. c. What is the forecast for next year?

21. The Centers for disease Control and Prevention office on Smoking and Health (oSH) is the lead federal agency responsible for comprehensive tobacco prevention and control. oSH was established in 1965 to reduce the death and disease caused by tobacco use and exposure to secondhand smoke. one of the many responsibilities of the oSH is to collect

fileWEB Portfolio

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Problems 235

data on tobacco use. The following data show the percentage of u.S. adults who were users of tobacco for a recent 11-year period (http://www.cdc.gov/tobacco/data_statistics /tables/trends/cig_smoking/index.htm).

Year percentage of Adults Who Smoke

1 22.8 2 22.5 3 21.6 4 20.9 5 20.9 6 20.8 7 19.8 8 20.6 9 20.6 10 19.3 11 18.9

a. Construct a time series plot. What type of pattern exists in the data? b. use simple linear regression to find the parameters for the line that minimizes mSe

for this time series. c. one of oSH’s goals is to cut the percentage of u.S. adults who were users of tobacco

to 12% or less within nine years of the last year of these data. does your regression model from part (b) suggest that oSH is on target to meet this goal? if not, use your model from part (b) to estimate the number of years that must pass after these data have been collected before oSH will achieve this goal.

22. The president of a small manufacturing firm is concerned about the continual increase in manufacturing costs over the past several years. The following figures provide a time series of the cost per unit for the firm’s leading product over the past eight years.

Year Cost / Unit ($) Year Cost / Unit ($)

1 20.00 5 26.60 2 24.50 6 30.00 3 28.20 7 31.00 4 27.50 8 36.00

a. Construct a time series plot. What type of pattern exists in the data? b. use simple linear regression analysis to find the parameters for the line that minimizes

mSe for this time series. c. What is the average cost increase that the firm has been realizing per year? d. Compute an estimate of the cost/unit for next year.

23. The medical community unanimously agrees on the health benefits of regular exercise, but are adults listening? during each of the past 15 years, a polling organization has sur- veyed americans about their exercise habits. in the most recent of these polls, slightly over half of all american adults reported that they exercise for 30 or more minutes at least three times per week. The following data show the percentages of adults who reported that they exercise for 30 or more minutes at least three times per week during each of the 15 years of this study.

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236 Chapter 6 Time Series Analysis and Forecasting

percentage of Adults Who Reported That They Exercise for 30 or More Year Minutes at Least Three Times per Week

1 41.0 2 44.9 3 47.1 4 45.7 5 46.6 6 44.5 7 47.6 8 49.8 9 48.1 10 48.9 11 49.9 12 52.1 13 50.6 14 54.6 15 52.4

a. Construct a time series plot. does a linear trend appear to be present? b. use simple linear regression to find the parameters for the line that minimizes mSe

for this time series. c. use the trend equation from part (b) to forecast the percentage of adults next year (year

16 of the study) who will report that they exercise for 30 or more minutes at least three times per week.

d. Would you feel comfortable using the trend equation from part (b) to forecast the percentage of adults three years from now (year 18 of the study) who will report that they exercise for 30 or more minutes at least three times per week?

24. Consider the following time series.

Quarter Year 1 Year 2 Year 3

1 71 68 62 2 49 41 51 3 58 60 53 4 78 81 72

a. Construct a time series plot. What type of pattern exists in the data? b. use a multiple linear regression model with dummy variables as follows to develop an

equation to account for seasonal effects in the data. Qtr1 5 1 if Quarter 1, 0 otherwise; Qtr2 5 1 if Quarter 2, 0 otherwise; Qtr3 5 1 if Quarter 3, 0 otherwise.

c. Compute the quarterly forecasts for next year.

25. Consider the following time series data.

Quarter Year 1 Year 2 Year 3

1 4 6 7 2 2 3 6 3 3 5 6 4 5 7 8

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Problems 237

a. Construct a time series plot. What type of pattern exists in the data? b. use a multiple regression model with dummy variables as follows to develop an

equation to account for seasonal effects in the data. Qtr1 5 1 if Quarter 1, 0 otherwise; Qtr2 5 1 if Quarter 2, 0 otherwise; Qtr3 5 1 if Quarter 3, 0 otherwise.

c. Compute the quarterly forecasts for next year.

26. The quarterly sales data (number of copies sold) for a college textbook over the past three years follow.

Quarter Year 1 Year 2 Year 3

1 1690 1800 1850 2 940 900 1100 3 2625 2900 2930 4 2500 2360 2615

a. Construct a time series plot. What type of pattern exists in the data? b. use a regression model with dummy variables as follows to develop an equation to

account for seasonal effects in the data. Qtr1 5 1 if Quarter 1, 0 otherwise; Qtr2 5 1 if Quarter 2, 0 otherwise; Qtr3 5 1 if Quarter 3, 0 otherwise.

c. Compute the quarterly forecasts for next year. d. let t 5 1 to refer to the observation in quarter 1 of year 1; t 5 2 to refer to the obser-

vation in quarter 2 of year 1; . . . ; and t 5 12 to refer to the observation in quarter 4 of year 3. using the dummy variables defined in part (b) and also using t, develop an equation to account for seasonal effects and any linear trend in the time series. Based upon the seasonal effects in the data and linear trend, compute the quarterly forecasts for next year.

27. air pollution control specialists in southern California monitor the amount of ozone, car- bon dioxide, and nitrogen dioxide in the air on an hourly basis. The hourly time series data exhibit seasonality, with the levels of pollutants showing patterns that vary over the hours in the day. on july 15, 16, and 17, the following levels of nitrogen dioxide were observed for the 12 hours from 6:00 a.m. to 6:00 p.m.

July 15: 25 28 35 50 60 60 40 35 30 25 25 20 July 16: 28 30 35 48 60 65 50 40 35 25 20 20 July 17: 35 42 45 70 72 75 60 45 40 25 25 25

a. Construct a time series plot. What type of pattern exists in the data? b. use a multiple linear regression model with dummy variables as follows to develop

an equation to account for seasonal effects in the data:

Hour1 5 1 if the reading was made between 6:00 a.m. and 7:00 a.m.; 0 otherwise

Hour2 5 1 if the reading was made between 7:00 a.m. and 8:00 a.m.; 0 otherwise . . .

Hour11 5 1 if the reading was made between 4:00 p.m. and 5:00 p.m.; 0 otherwise

note that when the values of the 11 dummy variables are equal to 0, the observation corresponds to the 5:00 p.m. to 6:00 p.m. hour.

c. using the equation developed in part (b), compute estimates of the levels of nitrogen dioxide for july 18.

d. let t 5 1 to refer to the observation in hour 1 on july 15; t 5 2 to refer to the obser- vation in hour 2 of july 15; . . . ; and t 5 36 to refer to the observation in hour 12 of july 17. using the dummy variables defined in part (b) and t, develop an equation to

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238 Chapter 6 Time Series Analysis and Forecasting

account for seasonal effects and any linear trend in the time series. Based upon the seasonal effects in the data and linear trend, compute estimates of the levels of nitrogen dioxide for july 18.

28. South Shore Construction builds permanent docks and seawalls along the southern shore of long island, new york. although the firm has been in business only five years, rev- enue has increased from $308,000 in the first year of operation to $1,084,000 in the most recent year. The following data show the quarterly sales revenue in thousands of dollars.

Quarter Year 1 Year 2 Year 3 Year 4 Year 5

1 20 37 75 92 176 2 100 136 155 202 282 3 175 245 326 384 445 4 13 26 48 82 181

a. Construct a time series plot. What type of pattern exists in the data? b. use a multiple regression model with dummy variables as follows to develop an

equation to account for seasonal effects in the data. Qtr1 5 1 if Quarter 1, 0 otherwise; Qtr2 5 1 if Quarter 2, 0 otherwise; Qtr3 5 1 if Quarter 3, 0 otherwise.

c. let Period 5 1 to refer to the observation in quarter 1 of year 1; Period 5 2 to refer to the observation in quarter 2 of year 1; . . . and Period 5 20 to refer to the observa- tion in quarter 4 of year 5. using the dummy variables defined in part (b) and Period, develop an equation to account for seasonal effects and any linear trend in the time series. Based upon the seasonal effects in the data and linear trend, compute estimates of quarterly sales for year 6.

Case problem 1 Forecasting Food and Beverage Sales The vintage restaurant, on Captiva island near Fort myers, Florida, is owned and oper- ated by Karen Payne. The restaurant just completed its third year of operation. during that time, Karen sought to establish a reputation for the restaurant as a high-quality dining establishment that specializes in fresh seafood. Through the efforts of Karen and her staff, her restaurant has become one of the best and fastest-growing restaurants on the island.

To better plan for future growth of the restaurant, Karen needs to develop a system that will enable her to forecast food and beverage sales by month for up to one year in advance. Table 6.17 shows the value of food and beverage sales ($1000s) for the first three years of operation.

Managerial Report Perform an analysis of the sales data for the vintage restaurant. Prepare a report for Karen that summarizes your findings, forecasts, and recommendations. include the following:

1. a time series plot. Comment on the underlying pattern in the time series. 2. using the dummy variable approach, forecast sales for january through december

of the fourth year.

assume that january sales for the fourth year turn out to be $295,000. What was your forecast error? if this error is large, Karen may be puzzled about the difference between your forecast and the actual sales value. What can you do to resolve her uncertainty in the forecasting procedure?

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Case Problem 2 Forecasting Lost Sales 239

Month First Year Second Year Third Year january 242 263 282 February 235 238 255 march 232 247 265 april 178 193 205 may 184 193 210 june 140 149 160 july 145 157 166 august 152 161 174 September 110 122 126 october 130 130 148 november 152 167 173 december 206 230 235

TABLE 6.17 Food and Beverage SaleS For THe vinTage reSTauranT ($1000s)

fileWEB Vintage

Case problem 2 Forecasting Lost Sales The Carlson department Store suffered heavy damage when a hurricane struck on august 31. The store was closed for four months (September through december), and Carlson is now involved in a dispute with its insurance company about the amount of lost sales during the time the store was closed. Two key issues must be resolved: (1) the amount of sales Carlson would have made if the hurricane had not struck and (2) whether Carlson is entitled to any compensation for excess sales due to increased business activity after the storm. more than $8 billion in federal disaster relief and insurance money came into the county, resulting in increased sales at department stores and numerous other businesses.

Table 6.18 gives Carlson’s sales data for the 48 months preceding the storm. Table 6.19 reports total sales for the 48 months preceding the storm for all department stores in the county, as well as the total sales in the county for the four months the Carlson depart- ment Store was closed. Carlson’s managers asked you to analyze these data and develop

Month Year 1 Year 2 Year 3 Year 4 Year 5 january 1.45 2.31 2.31 2.56 February 1.80 1.89 1.99 2.28 march 2.03 2.02 2.42 2.69 april 1.99 2.23 2.45 2.48 may 2.32 2.39 2.57 2.73 june 2.20 2.14 2.42 2.37 july 2.13 2.27 2.40 2.31 august 2.43 2.21 2.50 2.23 September 1.71 1.90 1.89 2.09 october 1.90 2.13 2.29 2.54 november 2.74 2.56 2.83 2.97 december 4.20 4.16 4.04 4.35

TABLE 6.18 SaleS For CarlSon deParTmenT STore ($ millionS)

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240 Chapter 6 Time Series Analysis and Forecasting

estimates of the lost sales at the Carlson department Store for the months of September through december. They also asked you to determine whether a case can be made for excess storm-related sales during the same period. if such a case can be made, Carlson is entitled to compensation for excess sales it would have earned in addition to ordinary sales.

Managerial Report Prepare a report for the managers of the Carlson department Store that summarizes your findings, forecasts, and recommendations. include the following:

1. an estimate of sales for Carlson department Store had there been no hurricane 2. an estimate of countywide department store sales had there been no hurricane 3. an estimate of lost sales for the Carlson department Store for September through

december

in addition, use the countywide actual department stores sales for September through december and the estimate in part (2) to make a case for or against excess storm-related sales.

Appendix 6.1 Forecasting with Excel Data Analysis Tools in this appendix we show how excel can be used to develop forecasts using three forecast- ing methods: moving averages, exponential smoothing, and trend projection. We also show how to use excel Solver for least-squares fitting of models to data.

Moving Averages To show how excel can be used to develop forecasts using the moving averages method, we develop a forecast for the gasoline sales time series in Table 6.1 and Figure 6.1. We assume that the user has entered the week in rows 2 through 13 of column a and the sales data for the 12 weeks into worksheet rows 2 through 13 of column B (as in Figure 6.13).

Month Year 1 Year 2 Year 3 Year 4 Year 5 january 46.80 46.80 43.80 48.00 February 48.00 48.60 45.60 51.60 march 60.00 59.40 57.60 57.60 april 57.60 58.20 53.40 58.20 may 61.80 60.60 56.40 60.00 june 58.20 55.20 52.80 57.00 july 56.40 51.00 54.00 57.60 august 63.00 58.80 60.60 61.80 September 55.80 57.60 49.80 47.40 69.00 october 56.40 53.40 54.60 54.60 75.00 november 71.40 71.40 65.40 67.80 85.20 december 117.60 114.00 102.00 100.20 121.80

TABLE 6.19 deParTmenT STore SaleS For THe CounTy ($ millionS)

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Appendix 6.1 Forecasting with Excel Data Analysis Tools 241

If the Data Analysis option does not appear in the Analysis group, you will have to include the Add-In in Excel. To do so, click on the File tab, then click Options, and then Add-Ins. Click Go next to the Excel Add-Ins drop-down box. Click the box next to Analysis ToolPak and click OK.

The following steps can be used to produce a three-week moving average.

Step 1. Select the data tab Step 2. From the Analysis group select the data Analysis option Step 3. When the data Analysis dialog box appears, choose Moving Average and

click OK Step 4. When the Moving Average dialog box appears:

enter B2:B13 in the input Range box enter 3 in the interval box enter C2 in the Output Range box Click OK

once you have completed this step (as shown in Figure 6.14), the three-week moving average forecasts will appear in column C of the worksheet as in Figure 6.15. note that forecasts for pe- riods of other lengths can be computed easily by entering a different value in the interval box.

FIGURE 6.14 exCel moving average dialogue Box For a 3-Period moving average

FIGURE 6.13 gaSoline SaleS daTa in exCel arranged To uSe THe moving averageS FunCTion To develoP ForeCaSTS

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242 Chapter 6 Time Series Analysis and Forecasting

Exponential Smoothing To show how excel can be used for exponential smoothing, we again develop a forecast for the gasoline sales time series in Table 6.1 and Figure 6.1. We assume that the user has entered the week in rows 2 through 13 of column a and the sales data for the 12 weeks into worksheet rows 2 through 13 of column B (as in Figure 6.13), and that the smoothing constant is a 5 0.2. The following steps can be used to produce a forecast.

Step 1. Select the data tab Step 2. From the Analysis group select the data Analysis option Step 3. When the data Analysis dialog box appears, choose Exponential Smooth-

ing and click OK Step 4. When the Exponential Smoothing dialog box appears:

enter B2:B13 in the input Range box enter 0.8 in the damping factor box enter C2 in the Output Range box Click OK

once you have completed this step (as shown in Figure 6.16), the exponential smoothing forecasts will appear in column C of the worksheet (as in Figure 6.17). note that the value we entered in the damping factor box is 1 2 a; forecasts for other smoothing constants can be computed easily by entering a different value for 1 2 a in the damping factor box.

Trend Projection To show how excel can be used for trend projection, we develop a forecast for the bicycle sales time series in Table 6.3 and Figure 6.3. We assume that the user has entered the year (1–10) for each observation into worksheet rows 2 through 11 of column a and the sales values into worksheet rows 2 through 11 of column B as shown in Figure 6.18. The following steps can be used to produce a forecast for year 11 by trend projection.

Step 1. Select the Formulas tab Step 2. Select two cells in the row where you want the regression coefficients b1 and

b0 to appear (for this example, choose d1 and e1)

FIGURE 6.15 gaSoline SaleS daTa and ouTPuT oF moving averageS FunCTion in exCel

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Appendix 6.1 Forecasting with Excel Data Analysis Tools 243

FIGURE 6.16 exCel exPonenTial SmooTHing dialogue Box For a 5 0.20

FIGURE 6.17 gaSoline SaleS daTa and ouTPuT oF exPonenTial SmooTHing FunCTion in exCel

FIGURE 6.18 BiCyCle SaleS daTa in exCel arranged To uSe THe lineST FunCTion To Find THe linear Trend

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244 Chapter 6 Time Series Analysis and Forecasting

FIGURE 6.19 exCel inSerT FunCTion dialogue Box For THe Finding THe Trend line uSing THe lineST FunCTion in exCel

Step 3. Click on the insert Function key Step 4. When the insert Function dialog box appears:

Choose Statistical in the Or select a category box Choose LiNEST in the Select a function box Click OK

See Figure 6.19 for an example of this step.

Step 5. When the Function Arguments dialog box appears: enter B2:B11 in the Known_y’s box enter a2:a11 in the Known_x’s box Click OK

See Figure 6.20 for an example of this step.

Step 6. Hit the F2 key and then simultaneously hit the Shift, Control, and enter keys (Shift 1 Control 1 enter) to create an array that contains the values of the regression coefficients b1 and b0

at this point you have generated the regression coefficients b1 and b0 in the two cells you originally selected in step 1. it is important to note that cell d1 contains b1 and cell e1 contains b0.

To generate a forecast, in a blank cell, multiply the value of the independent variable t by b1 and add the value of b0 to this product. For example, if you wish to use this linear trend model to generate a forecast for year 11 and the value of b1 is in cell d1 and the value of b0 is in cell e1, then enter 511*d11e1 in a blank cell. The forecast for year 11, in this case 32.5, will appear in the blank cell in which you enter this formula.

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Appendix 6.1 Forecasting with Excel Data Analysis Tools 245

FIGURE 6.20 exCel FunCTion argumenTS dialogue Box For THe Finding THe Trend line uSing THe lineST FunCTion in exCel

Models with Seasonality and No Trend To show how excel can be used to fit models with seasonality, we develop a forecast for the umbrella sales time series in Table 6.13 and Figure 6.11. We assume that the user has entered the year (1–5) for each observation into worksheet rows 3 through 22 of column a; the values for the quarter in worksheet rows 3 through 22 of column B; the values for the quarterly dummy variables Qtr1t, Qtr2t, and Qtr3t in worksheet rows 3 through 22 of columns C, d, and e, respectively; and the sales values into worksheet rows 3 through 22 of column F. The following steps can be used to produce a forecast for year 11 by trend projection as shown in Figure 6.21.

Step 1. Select the Formulas tab Step 2. Select four cells in the row where you want the regression coefficients b3, b2,

b1, and b0 to appear (for this example, choose g1:j1) Step 3. Click on the insert Function key Step 4. When the insert Function dialog box appears:

Choose Statistical in the Or select a category box Choose LiNEST in the Select a function box Click OK

Step 5. When the Function Arguments dialog box appears: enter F3:F22 in the Known_y’s box enter C3:e22 in the Known_x’s box Click OK

See Figure 6.22 for an example of this step.

Step 6. Hit the F2 key and then simultaneously hit the Shift, Control, and enter keys (Shift 1 Control 1 enter) to create an array that contains the values of the regression coefficients b3, b2, b1, and b0

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246 Chapter 6 Time Series Analysis and Forecasting

FIGURE 6.22 exCel FunCTion argumenTS dialogue Box For Finding THe SeaSonal ComPonenTS uSing THe lineST FunCTion in exCel

FIGURE 6.21 umBrella SaleS daTa in exCel arranged To uSe THe lineST FunCTion To Find THe SeaSonal ComPonenTS

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Appendix 6.1 Forecasting with Excel Data Analysis Tools 247

at this point you have generated the regression coefficients b3, b2, b1, and b0 in cells g1:j1 selected in step 1. it is important to note that the first cell you selected contains b3, the second cell you selected contains b2, the third cell you selected contains b1, and the fourth cell you selected contains b0 (i.e., if you selected cells g1:j1 in step 1, the value of b1 will be in cell g1, the value of b2 will be in H1, the value of b1 will be in i1, and the value of b0 will be in cell j1).

To generate a forecast, in a blank cell, add together b0 and the product of b1 and Qtr1t, the product of b2 and Qtr2t, and the product of b3 and Qtr3t. For example, if you wish to use this linear trend model to generate a forecast for the first quarter of next year and the value of b3 is in cell g1, the value of b2 is in cell H1, the value of b1 is in cell i1, and the value of b0 is in cell j1, then enter 51*g110*H110*i11j1 in a blank cell. The forecast for the first quarter of next year, in this case 124.0, will appear in the blank cell in which you enter this formula.

Models with Seasonality and Linear Trend To show how excel can be used to fit models with seasonality and a linear trend, we develop a forecast for the umbrella set time series in Table 6.13 and Figure 6.11. We assume that the user has entered the year (1–5) for each observation into worksheet rows 3 through 22 of column a; the values for the quarter in worksheet rows 3 through 22 of column B; the values for the quarterly dummy variables Qtr1t, Qtr2t, and Qtr3t into worksheet rows 3 through 22 of columns C, d, and e, respectively; the values of period t into worksheet rows 3 through 22 of column F; and the sales values into worksheet rows 3 through 22 of column g. The following steps can be used to produce a forecast for year 11 by trend projection as shown in Figure 6.23.

FIGURE 6.23 umBrella Time SerieS daTa in exCel arranged To uSe THe lineST FunCTion To Find BoTH THe SeaSonal ComPonenTS and Trend ComPonenT

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248 Chapter 6 Time Series Analysis and Forecasting

Step 1. Select the Formulas tab Step 2. Select five cells in the row where you want the regression coefficients b4, b3,

b2, b1, and b0 to appear for this example; choose H1:l1 Step 3. Click on the insert Function key Step 4. When the insert Function dialog box appears:

Choose Statistical in the Or select a category box Choose LiNEST in the Select a function box Click OK

Step 5. When the Function Arguments dialog box appears: enter g3:g22 in the Known_y’s box enter C3:F22 in the Known_x’s box Click OK

at this point you have generated the regression coefficients b4, b3, b2, b1, and b0 in cells H1:l1 selected in step 1. it is important to note that the first cell you selected contains b4, the second cell you selected contains b3, the third cell you selected contains b2, the fourth cell you selected contains b1, and the fifth cell you selected contains b0 (i.e., if you selected cells H1:l1 in step 1, the value of b4 will be in cell H1, the value of b1 will be in cell i1, the value of b2 will be in j1, the value of b1 will be in K1, and the value of b0 will be in cell l1).

To generate a forecast, in a blank cell, add together b0 and the product of b1 and Qtr1t, the product of b2 and Qtr2t, the product of b3 and Qtr3t, and the product of b4 and t. For example, if you wish to use this linear trend model to generate a forecast for the first quarter of year 5 and the value of b4 is in cell H1, the value of b3 is in cell i1, the value of b2 is in cell j1, the value of b1 is in cell K1, and the value of b0 is in cell l1, then enter 517*H11 1*i110*j110*K11l1 in a blank cell. The forecast for the first quarter of next year, in this case 7.19, will appear in the blank cell in which you enter this formula.

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Introduction to Linear Programming

CONTENTS

7.1 A SIMPLE MAXIMIZATION PROBLEM Problem Formulation Mathematical Model for the

RMC Problem

7.2 GRAPHICAL SOLUTION PROCEDURE A Note on Graphing Lines Summary of the Graphical

Solution Procedure for Maximization Problems

Slack Variables

7.3 EXTREME POINTS AND THE OPTIMAL SOLUTION

7.4 COMPUTER SOLUTION OF THE RMC PROBLEM Interpretation of Answer Report

7.5 A SIMPLE MINIMIZATION PROBLEM Summary of the Graphical

Solution Procedure for Minimization Problems

Surplus Variables Computer Solution of the M&D

Chemicals Problem

7.6 SPECIAL CASES Alternative Optimal Solutions Infeasibility Unbounded

7.7 GENERAL LINEAR PROGRAMMING NOTATION

AppENdix 7.1 SOLVING LINEAR PROGRAMS wITH EXCEL SOLVER

AppENdix 7.2 SOLVING LINEAR PROGRAMS wITH LINGO

CHAPTER 7

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250 Chapter 7 Introduction to Linear Programming

Linear programming is a problem-solving approach developed to help managers make deci- sions. Numerous applications of linear programming can be found in today’s competitive business environment. For instance, IBM uses linear programming to perform capacity planning and to make capacity investment decisions for its semiconductor manufacturing operations. GE Capital uses linear programming to help determine optimal lease structur- ing. Marathon Oil Company uses linear programming for gasoline blending and to evaluate the economics of a new terminal or pipeline. The Q.M. in Action, Timber Harvesting Model at Meadwestvaco Corporation, provides another example of the use of linear programming. Later in the chapter another Q.M. in Action illustrates how IBM uses linear programming and other quantitative methods to plan and operate its semiconductor supply chain.

To illustrate some of the properties that all linear programming problems have in com- mon, consider the following typical applications:

1. A manufacturer wants to develop a production schedule and an inventory policy that will satisfy sales demand in future periods. Ideally, the schedule and policy will enable the company to satisfy demand and at the same time minimize the total production and inventory costs.

2. A financial analyst must select an investment portfolio from a variety of stock and bond investment alternatives. The analyst would like to establish the portfolio that maximizes the return on investment.

3. A marketing manager wants to determine how best to allocate a fixed advertising budget among alternative advertising media such as radio, television, newspaper, and magazine. The manager would like to determine the media mix that maximizes advertising effectiveness.

4. A company has warehouses in a number of locations. Given specific customer demands, the company would like to determine how much each warehouse should ship to each customer so that total transportation costs are minimized.

These examples are only a few of the situations in which linear programming has been used successfully, but they illustrate the diversity of linear programming applications. A close scrutiny reveals one basic property they all have in common. In each example, we were concerned with maximizing or minimizing some quantity. In example 1, the manu- facturer wanted to minimize costs; in example 2, the financial analyst wanted to maximize return on investment; in example 3, the marketing manager wanted to maximize advertising effectiveness; and in example 4, the company wanted to minimize total transportation costs. In all linear programming problems, the maximization or minimization of some quantity is the objective.

Meadwestvaco Corporation is a major producer of pre- mium papers for periodicals, books, commercial print- ing, and business forms. The company also produces pulp and lumber, designs and manufactures packag- ing systems for beverage and other consumables mar-

kets, and is a world leader in the production of coated board and shipping containers. Quantitative analyses at Meadwestvaco are developed and implemented by the company’s Decision Analysis Department. The depart- ment assists decision makers by providing them with analytical tools of quantitative methods as well as per- sonal analysis and recommendations.

TIMBER HARVESTING MODEL AT MEADWESTVACO CORPORATION*

Q.M. in ACtIon

*Based on information provided by Dr. Edward P. Winkofsky of MeadWestvaco Corporation.

(continued)

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7.1 A Simple Maximization Problem 251

All linear programming problems also have a second property: restrictions or constraints that limit the degree to which the objective can be pursued. In the first example, the manufacturer is restricted by constraints requiring product demand to be satisfied and by the constraints limiting production capacity. The financial analyst’s port- folio problem is constrained by the total amount of investment funds available and the maximum amounts that can be invested in each stock or bond. The marketing manager’s media selection decision is constrained by a fixed advertising budget and the availability of the various media. In the transportation problem, the minimum-cost shipping schedule is constrained by the supply of product available at each warehouse. Thus, constraints are another general feature of every linear programming problem.

7.1 A Simple Maximization Problem RMC, Inc., is a small firm that produces a variety of chemical-based products. In a par- ticular production process, three raw materials are used to produce two products: a fuel additive and a solvent base. The fuel additive is sold to oil companies and is used in the production of gasoline and related fuels. The solvent base is sold to a variety of chemical firms and is used in both home and industrial cleaning products. The three raw materials are blended to form the fuel additive and solvent base as indicated in Table 7.1, which shows that a ton of fuel additive is a mixture of 0.4 tons of material 1 and 0.6 tons of mate- rial 3. A ton of solvent base is a mixture of 0.5 tons of material 1, 0.2 tons of material 2, and 0.3 tons of material 3.

Linear programming was initially referred to as “programming in a linear structure.” In 1948, Tjalling Koopmans suggested to George Dantzig that the name was much too long: Koopman’s suggestion was to shorten it to linear programming. George Dantzig agreed, and the field we now know as linear programming was named.

Meadwestvaco uses quantitative models to assist with the long-range management of the company’s timberland. Through the use of large-scale linear pro- grams, timber harvesting plans are developed to cover a substantial time horizon. These models consider wood market conditions, mill pulpwood requirements, harvesting capacities, and general forest management principles. within these constraints, the model arrives at an optimal harvesting and purchasing schedule based on discounted cash flow. Alternative schedules reflect

changes in the various assumptions concerning for- est growth, wood availability, and general economic conditions.

Quantitative methods are also used in the develop- ment of the inputs for the linear programming models. Timber prices and supplies as well as mill requirements must be forecast over the time horizon, and advanced sampling techniques are used to evaluate land holdings and to project forest growth. The harvest schedule is then developed using quantitative methods.

product

Fuel Additive Solvent Base Material 1 0.4 0.5 Material 2 0.2 Material 3 0.6 0.3

TABLE 7.1 MATERIAL REQUIREMENTS PER TON FOR THE RMC PROBLEM

0.6 tons of material 3 is used in each ton of fuel additive

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252 Chapter 7 Introduction to Linear Programming

RMC’s production is constrained by a limited availability of the three raw materials. For the current production period, RMC has available the following quantities of each raw material:

Amount Available Material for production Material 1 20 tons Material 2 5 tons Material 3 21 tons

It is important to understand that we are maximizing profit contribution, not profit. Overhead and other shared costs must be deducted before arriving at a profit figure.

Because of spoilage and the nature of the production process, any materials not used for current production are useless and must be discarded.

The accounting department analyzed the production figures, assigned all relevant costs, and arrived at prices for both products that will result in a profit contribution1 of $40 for every ton of fuel additive produced and $30 for every ton of solvent base produced. Let us now use linear programming to determine the number of tons of fuel additive and the number of tons of solvent base to produce in order to maximize total profit contribution.

Problem Formulation problem formulation is the process of translating a verbal statement of a problem into a mathematical statement. The mathematical statement of the problem is referred to as a mathematical model. Developing an appropriate mathematical model is an art that can only be mastered with practice and experience. Even though every problem has at least some unique features, most problems also have many common or similar features. As a result, some general guidelines for developing a mathematical model can be helpful. we will illustrate these guidelines by developing a mathematical model for the RMC problem.

Understand the Problem Thoroughly The RMC problem is relatively easy to under- stand. RMC wants to determine how much of each product to produce in order to maximize the total contribution to profit. The number of tons available for the three materials that are required to produce the two products will limit the number of tons of each product that can be produced. More complex problems will require more work in order to understand the problem. However, understanding the problem thoroughly is the first step in developing any mathematical model.

Describe the Objective RMC’s objective is to maximize the total contribution to profit.

Describe Each Constraint Three constraints limit the number of tons of fuel additive and the number of tons of solvent base that can be produced.

Constraint 1: The number of tons of material 1 used must be less than or equal to the 20 tons available.

Constraint 2: The number of tons of material 2 used must be less than or equal to the 5 tons available.

Constraint 3: The number of tons of material 3 used must be less than or equal to the 21 tons available.

1From an accounting perspective, profit contribution is more correctly described as the contribution margin per ton; overhead and other shared costs have not been allocated to the fuel additive and solvent base costs.

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7.1 A Simple Maximization Problem 253

Define the Decision Variables The decision variables are the controllable inputs in the problem. For the RMC problem the two decision variables are (1) the number of tons of fuel additive produced, and (2) the number of tons of solvent base produced. In developing the mathematical model for the RMC problem, we will use the following notation for the decision variables:

F 5 number of tons of fuel additive

S 5 number of tons of solvent base

Write the Objective in Terms of the Decision Variables RMC’s profit contribution comes from the production of F tons of fuel additive and S tons of solvent base. Because RMC makes $40 for every ton of fuel additive produced and $30 for every ton of solvent base produced, the company will make $40F from the production of the fuel additive and $30S from the production of the solvent base. Thus,

Total profit contribution 5 40F 1 30S

Because the objective—maximize total profit contribution—is a function of the decision variables F and S, we refer to 40F 1 30S as the objective function. Using “Max” as an abbreviation for maximize, we can write RMC’s objective as follows:

Max 40F 1 30S (7.1)

Write the Constraints in Terms of the Decision Variables

Constraint 1:

Tons of material 1 used # Tons of material 1 available

Every ton of fuel additive that RMC produces will use 0.4 tons of material 1. Thus, 0.4F tons of material 1 is used to produce F tons of fuel additive. Similarly, every ton of solvent base that RMC produces will use 0.5 tons of material 1. Thus, 0.5S tons of material 1 is used to produce S tons of solvent base. Therefore, the number of tons of material 1 used to produce F tons of fuel additive and S tons of solvent base is

Tons of material 1 used 5 0.4F 1 0.5S

Because 20 tons of material 1 are available for use in production, the mathematical statement of constraint 1 is

0.4F 1 0.5S # 20 (7.2)

Constraint 2:

Tons of material 2 used # Tons of material 2 available

Fuel additive does not use material 2. However, every ton of solvent base that RMC pro- duces will use 0.2 tons of material 2. Thus, 0.2S tons of material 2 is used to produce S tons of solvent base. Therefore, the number of tons of material 2 used to produce F tons of fuel additive and S tons of solvent base is

Tons of material 2 used 5 0.2S

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254 Chapter 7 Introduction to Linear Programming

Because 5 tons of material 2 are available for production, the mathematical statement of constraint 2 is

0.2S # 5 (7.3)

Constraint 3:

Tons of material 3 used # Tons of material 3 available

Every ton of fuel additive RMC produces will use 0.6 tons of material 3. Thus, 0.6F tons of material 1 is used to produce F tons of fuel additive. Similarly, every ton of solvent base RMC produces will use 0.3 tons of material 3. Thus, 0.3S tons of material 1 is used to produce S tons of solvent base. Therefore, the number of tons of material 3 used to produce F tons of fuel additive and S tons of solvent base is

Tons of material 3 used 5 0.6F 1 0.3S

Because 21 tons of material 3 are available for production, the mathematical statement of constraint 3 is

0.6F 1 0.3S # 21 (7.4)

Add the Nonnegativity Constraints RMC cannot produce a negative number of tons of fuel additive or a negative number of tons of solvent base. Therefore, nonnegativity constraints must be added to prevent the decision variables F and S from having negative values. These nonnegativity constraints are

F $ 0 and S $ 0

Nonnegativity constraints are a general feature of many linear programming problems and may be written in the abbreviated form:

F, S $ 0 (7.5)

Mathematical Model for the RMC Problem Problem formulation is now complete. we have succeeded in translating the verbal state- ment of the RMC problem into the following mathematical model:

Max 40F 1 30S

Subject to (s.t.)

0.4F 1 0.5S # 20 Material 1

0.2 S # 5 Material 2

0.6F 1 0.3S # 21 Material 3

F, S $ 0

Our job now is to find the product mix (i.e., the combination of F and S) that satisfies all the constraints and, at the same time, yields a maximum value for the objective function. Once these values of F and S are calculated, we will have found the optimal solution to the problem.

This mathematical model of the RMC problem is a linear program. The RMC problem has an objective and constraints that, as we said earlier, are common properties of all linear

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7.2 Graphical Solution Procedure 255

programs. But what is the special feature of this mathematical model that makes it a linear program? The special feature that makes it a linear program is that the objective function and all constraint functions (the left-hand sides of the constraint inequalities) are linear functions of the decision variables.

Mathematical functions in which each variable appears in a separate term and is raised to the first power are called linear functions. The objective function (40F 1 30S) is linear because each decision variable appears in a separate term and has an exponent of 1. The amount of material 1 used (0.4F 1 0.5S) is also a linear function of the decision variables for the same reason. Similarly, the functions on the left-hand side of the material 2 and material 3 constraint inequalities (the constraint functions) are also linear functions. Thus, the mathematical formulation is referred to as a linear program.

Linear programming has nothing to do with computer programming. The use of the word programming here means “choosing a course of action.” Linear programming in- volves choosing a course of action when the mathematical model of the problem contains only linear functions.

Try Problem 1 to test your ability to recognize the types of mathematical relationships that can be found in a linear program.

NOTES AND COMMENTS

1. The three assumptions necessary for a lin- ear programming model to be appropriate are proportionality, additivity, and divisibility. Proportionality means that the contribution to the objective function and the amount of re- sources used in each constraint are proportional to the value of each decision variable. Additiv- ity means that the value of the objective func- tion and the total resources used can be found by summing the objective function contribution and the resources used for all decision variables. Divisibility means that the decision variables are

continuous. The divisibility assumption plus the nonnegativity constraints mean that decision variables can take on any value greater than or equal to zero.

2. Quantitative analysts formulate and solve a variety of mathematical models that contain an objective function and a set of constraints. Models of this type are referred to as mathemati- cal programming models. Linear programming models are a special type of mathematical pro- gramming model in that the objective function and all constraint functions are linear.

7.2 Graphical Solution Procedure A linear programming problem involving only two decision variables can be solved using a graphical solution procedure. Let us begin the graphical solution procedure by developing a graph that displays the possible solutions (F and S values) for the RMC problem. The graph in Figure 7.1 has values of F on the horizontal axis and values of S on the vertical axis. Any point on the graph can be identified by its F and S values, which indicate the position of the point along the horizontal and vertical axes, respectively. Thus, every point on the graph corresponds to a possible solution. The solution of F 5 0 and S 5 0 is referred to as the origin. Because both F and S must be nonnegative, the graph in Figure 7.1 only displays solutions where F $ 0 and S $ 0.

Earlier we determined that the inequality representing the material 1 constraint was

0.4F 1 0.5S # 20

To show all solutions that satisfy this relationship, we start by graphing the line correspond- ing to the equation

0.4F 1 0.5S 5 20

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256 Chapter 7 Introduction to Linear Programming

we graph this equation by identifying two points that satisfy this equation and then draw- ing a line through the points. Setting F 5 0 and solving for S gives 0.5S 5 20, or S 5 40; hence the solution (F 5 0, S 5 40) satisfies the preceding equation. To find a second solu- tion satisfying this equation, we set S 5 0 and solve for F. Doing so, we obtain 0.4F 5 20, or F 5 50. Thus, a second solution satisfying the equation is (F 5 50, S 5 0). with these two points, we can now graph the line. This line, called the material 1 constraint line, is shown in Figure 7.2.

Recall that the inequality representing the material 1 constraint is

0.4F 1 0.5S # 20

Can you identify all the solutions that satisfy this constraint? First, note that any point on the line 0.4F 1 0.5S 5 20 must satisfy the constraint. But where are the solutions satisfy- ing 0.4F 1 0.5S , 20? Consider two solutions (F 5 10, S 5 10) and (F 5 40, S 5 30). Figure 7.2 shows that the first solution is on the same side of the constraint line as the origin while the second solution is on the side of the constraint line opposite of the origin. which of these solutions satisfies the material 1 constraint? For (F 5 10, S 5 10) we have

0.4F 1 0.5S 5 0.4(10) 1 0.5(10) 5 9

Because 9 tons is less than the 20 tons of material 1 available, the F 5 10, S 5 10 solution satisfies the constraint. For F 5 40 and S 5 30 we have

0.4F 1 0.5S 5 0.4(40) 1 0.5(30) 5 31

10 20 30 40 50

T on

s of

S ol

ve n

t B

as e

Tons of Fuel Additive

Solution with F = 10 and S = 40

(10, 40)

Solution with F = 20 and S = 15

(20, 15)

FIGURE 7.1 GRAPH SHOwING TwO SOLUTIONS FOR THE TwO-VARIABLE RMC PROBLEM

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7.2 Graphical Solution Procedure 257

The 31 tons is greater than the 20 tons available, so the F 5 40, S 5 30 solution does not satisfy the constraint.

If a particular solution satisfies the constraint, all other solutions on the same side of the constraint line will also satisfy the constraint. If a particular solution does not satisfy the constraint, all other solutions on the same side of the constraint line will not satisfy the constraint. Thus, you need to evaluate only one solution to determine which side of a constraint line provides solutions that will satisfy the constraint. The shaded area in Figure 7.3 shows all the solutions that satisfy the material 1 constraint.

Next let us identify all solutions that satisfy the material 2 constraint:

0.2S # 5

we start by drawing the constraint line corresponding to the equation 0.2S 5 5. Because this equation is equivalent to the equation S 5 25, we simply draw a line whose S value is 25 for every value of F; this line is parallel to and 25 units above the horizontal axis. Figure 7.4 shows the line corresponding to the material 2 constraint. Following the approach we used for the material 1 constraint, we realize that only solutions on or below the line will satisfy the material 2 constraint. Thus, in Figure 7.4 the shaded area corresponds to the solutions that satisfy the material 2 constraint.

Similarly, we can determine the solutions that satisfy the material 3 constraint. Figure 7.5 shows the result. For practice, try to graph the feasible solutions that satisfy the material 3 constraint and determine whether your result agrees with that shown in Figure 7.5.

we now have three separate graphs showing the solutions that satisfy each of the three constraints. In a linear programming problem, we need to identify the solutions that satisfy

You should now be able to graph a constraint line and find the solution points that satisfy the constraint. Try Problem 2.

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M aterial 1

(10, 10)

(40, 30)

(50, 0)

(0, 40)

0.4F + 0.5S = 20

FIGURE 7.2 MATERIAL 1 CONSTRAINT LINE

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258 Chapter 7 Introduction to Linear Programming

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Tons of Fuel Additive 0

0.4F + 0.5S = 20

M aterial 1

(10, 10)

(40, 30)

FIGURE 7.3 SOLUTIONS THAT SATISFY THE MATERIAL 1 CONSTRAINT

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0.2 S = 5 Material 2 (0, 25)

FIGURE 7.4 SOLUTIONS THAT SATISFY THE MATERIAL 2 CONSTRAINT

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7.2 Graphical Solution Procedure 259

all the constraints simultaneously. To find these solutions, we can draw the three constraints on one graph and observe the region containing the points that do in fact satisfy all the constraints simultaneously.

The graphs in Figures 7.3, 7.4, and 7.5 can be superimposed to obtain one graph with all three constraints. Figure 7.6 shows this combined constraint graph. The shaded region in this figure includes every solution point that satisfies all the constraints simultaneously. Because solutions that satisfy all the constraints simultaneously are termed feasible solu- tions, the shaded region is called the feasible solution region, or simply the feasible region. Any point on the boundary of the feasible region, or within the feasible region, is a feasible solution point for the linear programming problem.

Now that we have identified the feasible region, we are ready to proceed with the graphical solution method and find the optimal solution to the RMC problem. Recall that the optimal solution for a linear programming problem is the feasible solution that pro- vides the best possible value of the objective function. Let us start the optimizing step of the graphical solution procedure by redrawing the feasible region on a separate graph. Figure 7.7 shows the graph.

One approach to finding the optimal solution would be to evaluate the objective func- tion for each feasible solution; the optimal solution would then be the one yielding the

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0.6F

+ 0.3S = 21

aterial 3

(35, 0)

(0, 70) S

FIGURE 7.5 SOLUTIONS THAT SATISFY THE MATERIAL 3 CONSTRAINT

Can you now find the feasible region given several constraints? Try Problem 7.

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260 Chapter 7 Introduction to Linear Programming

largest value. The difficulty with this approach is that the infinite number of feasible solu- tions makes evaluating all feasible solutions impossible. Hence, this trial-and-error proce- dure cannot be used to identify the optimal solution.

Rather than trying to compute the profit contribution for each feasible solution, we select an arbitrary value for profit contribution and identify all the feasible solutions that yield the selected value. For example, what feasible solutions provide a profit contribution of $240? These solutions are given by the values of F and S in the feasible region that will make the objective function

40F 1 30S 5 240

This expression is simply the equation of a line. Thus all feasible solutions (F, S) yielding a profit contribution of $240 must be on the line. we learned earlier in this section how to graph a constraint line. The procedure for graphing the profit or objective function line is the same. Letting F 5 0, we see that S must be 8; thus the solution point (F 5 0, S 5 8) is on the line. Similarly, by letting S 5 0 we see that the solution point (F 5 6, S 5 0) is also on the line. Drawing the line through these two points identifies all the solutions that have a profit contribution of $240. A graph of this profit line is presented in Figure 7.8. The graph shows that an infinite number of feasible production combinations will provide a $240 profit contribution.

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Material 2

M aterial 1

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M aterial 3

Feasible Region

FIGURE 7.6 FEASIBLE REGION FOR THE RMC PROBLEM

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7.2 Graphical Solution Procedure 261

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Feasible Region

FIGURE 7.7 FEASIBLE REGION FOR THE RMC PROBLEM

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(0, 8) Pro�t Line 40F + 30S = 240

(6, 0)

FIGURE 7.8 $240 PROFIT LINE FOR THE RMC PROBLEM

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262 Chapter 7 Introduction to Linear Programming

The objective is to find the feasible solution yielding the highest profit contribution, so we proceed by selecting higher profit contributions and finding the solutions that yield the stated values. For example, what solutions provide a profit contribution of $720? what solutions provide a profit contribution of $1200? To answer these questions, we must find the F and S values that are on the profit lines:

40F 1 30S 5 720 and 40F 1 30S 5 1200

Using the previous procedure for graphing profit and constraint lines, we graphed the $720 and $1200 profit lines presented in Figure 7.9. Not all solution points on the $1200 profit line are in the feasible region, but at least some points on the line are; thus, we can obtain a feasible solution that provides a $1200 profit contribution.

Can we find a feasible solution yielding an even higher profit contribution? Look at Figure 7.9 and make some general observations about the profit lines. You should be able to identify the following properties: (1) The profit lines are parallel to each other, and (2) profit lines with higher profit contributions are farther from the origin.

Because the profit lines are parallel and higher profit lines are farther from the origin, we can obtain solutions that yield increasingly higher values for the objective function by continuing to move the profit line farther from the origin but keeping it parallel to the other profit lines. However, at some point any further outward movement will place the profit line entirely outside the feasible region. Because points outside the feasible region are un- acceptable, the point in the feasible region that lies on the highest profit line is an optimal solution to the linear program.

You should now be able to identify the optimal solution point for the RMC problem. Use a ruler and move the profit line as far from the origin as you can. what is the last point in the feasible region? This point, which is the optimal solution, is shown graphically in Fig ure 7.10. The optimal values for the decision variables are the F and S values at this point.

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40F + 30S = 1200

40F + 30S = 720

40F + 30S = 240

FIGURE 7.9 SELECTED PROFIT LINES FOR THE RMC PROBLEM

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7.2 Graphical Solution Procedure 263

Depending on the accuracy of your graph, you may or may not be able to determine the exact optimal values of F and S directly from the graph. However, refer to Figure 7.6 and note that the optimal solution point for the RMC example is at the intersection of the material 1 and material 3 constraint lines. That is, the optimal solution is on both the mate- rial 1 constraint line,

0.4F 1 0.5S 5 20 (7.6)

and the material 3 constraint line,

0.6F 1 0.3S 5 21 (7.7)

Thus, the values of the decision variables F and S must satisfy both equations (7.6) and (7.7) simultaneously. Using (7.6) and solving for F gives

0.4F 5 20 2 0.5S

F 5 50 2 1.25S (7.8)

Substituting this expression for F into equation (7.7) and solving for S yields

0.6(50 2 1.25S) 1 0.3S 5 21

30 2 0.75S 1 0.3S 5 21

20.45S 5 29

S 5 20

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0 M

axim um

Pro�t Line Optimal Solution Point

(25, 20)

FIGURE 7.10 OPTIMAL SOLUTION FOR THE RMC PROBLEM

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264 Chapter 7 Introduction to Linear Programming

Substituting S 5 20 in equation (7.8) and solving for F provides

F 5 50 2 1.25(20)

5 50 2 25 5 25

Thus, the exact location of the optimal solution point is F 5 25 and S 5 20. This solution point provides the optimal production quantities for RMC at 25 tons of fuel additive and 20 tons of solvent base and yields a profit contribution of 40(25) 1 30(20) 5 $1600.

For a linear programming problem with two decision variables, you can determine the exact values of the decision variables at the optimal solution by first using the graphical procedure to identify the optimal solution point and then solving the two simultaneous equations associated with this point.

A Note on Graphing Lines An important aspect of the graphical method is the ability to graph lines showing the con- straints and the objective function of the linear program. The procedure we used for graph- ing the equation of a line is to find any two points satisfying the equation and then draw the line through the two points. For the RMC constraints, the two points were easily found by setting F 5 0 and solving the constraint equation for S. Then we set S 5 0 and solved for F. For the material 1 constraint line

0.4F 1 0.5S 5 20

this procedure identified the two points (F 5 0, S 5 40) and (F 5 50, S 5 0). The material 1 constraint line was then graphed by drawing a line through these two points.

All constraints and objective function lines in two-variable linear programs can be graphed if two points on the line can be identified. However, finding the two points on the line is not always as easy as shown in the RMC problem. For example, suppose a company manufactures two models of a tablet computer: the Professional (P) and the Assistant (A). Management needs 50 units of the Professional model for its own sales force and expects sales of the remaining Professionals to be less than or equal to 50% of the sales of the As- sistant. A constraint enforcing this requirement is

P 2 50 # 0.5A

P 2 0.5A # 50

Using the equality form of the constraint and setting P 5 0, we find that the point (P 5 0, A 5 2100) is on the constraint line. Setting A 5 0, we find a second point (P 5 50, A 5 0) on the constraint line. If we have drawn only the nonnegative (P $ 0, A $ 0) portion of the graph, the first point (P 5 0, A 5 2100) cannot be plotted because A 5 2100 is not on the graph. whenever we have two points on the line, but one or both of the points can- not be plotted in the nonnegative portion of the graph, the simplest approach is to enlarge the graph. In this example, the point (P 5 0, A 5 2100) can be plotted by extending the graph to include the negative A axis. Once both points satisfying the constraint equation have been located, the line can be drawn. The constraint line and the solutions that satisfy the constraint P 2 0.5A # 50 are shown in Figure 7.11.

Although the optimal solution to the RMC problem consists of integer values for the decision variables, this result will not always be the case.

Try Problem 10 to test your ability to use the graphical solution procedure to identify the optimal solution and find the exact values of the decision variables at the optimal solution.

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7.2 Graphical Solution Procedure 265

As another example, consider a problem involving two decision variables, R and T. Suppose that the number of units of R produced has to be at least equal to the number of units of T produced. A constraint enforcing this requirement is

R $ T

R 2 T $ 0

To find all solutions satisfying the constraint as an equality, we first set R 5 0 and solve for T. This result shows that the origin (T 5 0, R 5 0) is on the constraint line. Setting T 5 0 and solving for R provides the same point. However, we can obtain a second point on the line by setting T equal to any value other than zero and then solving for R. For instance, setting T 5 100 and solving for R, we find that the point (T 5 100, R 5 100) is on the line. with the two points (R 5 0, T 5 0) and (R 5 100, T 5 100), the constraint line R 2 T 5 0 and the solutions that satisfy the constraint R 2 T $ 0 can be plotted as shown in Figure 7.12.

Summary of the Graphical Solution Procedure for Maximization Problems As we have seen, the graphical solution procedure is a method for solving two-variable linear programming problems such as the RMC problem. The steps of the graphical solution procedure for a maximization problem are summarized here:

1. Prepare a graph for each constraint that shows the solutions that satisfy the constraint.

2. Determine the feasible region by identifying the solutions that satisfy all the constraints simultaneously.

Can you graph a constraint line when the origin is on the constraint line? Try Problem 5.

For additional practice in using the graphical solution procedure, try Problem 24.

100 200 300

100

200

300

(50, 0)

–100

(0, –100)

P –

0 .5

A ≤

5 0

FIGURE 7.11 SOLUTIONS THAT SATISFY THE CONSTRAINT P 2 0.5A # 50

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266 Chapter 7 Introduction to Linear Programming

3. Draw an objective function line showing the values of the decision variables that yield a specified value of the objective function.

4. Move parallel objective function lines toward larger objective function values until further movement would take the line completely outside the feasible region.

5. Any feasible solution on the objective function line with the largest value is an optimal solution.

Slack Variables In addition to the optimal solution and its associated profit contribution, the RMC managers will want information about the production requirements for the three materials. we can determine this information by substituting the optimal solution values (F 5 25, S 5 20) into the constraints of the linear program.

100 200 300

100

200

300

R –

T =

(100, 100)

(0, 0)

FIGURE 7.12 FEASIBLE SOLUTIONS FOR THE CONSTRAINT R 2 T $ 0

Tons Required Tons Unused Constraint for F 5 25, S 5 20 Tons Available Tons Material 1 0.4(25) 1 0.5(20) 5 20 20 0 Material 2 0.2(20) 5 4 5 1 Material 3 0.6(25) 1 0.3(20) 5 21 21 0

Can you identify the slack associated with a constraint? Try Problem 24, part (e).

Thus, the optimal solution tells management that the production of 25 tons of fuel addi- tive and 20 tons of solvent base will require all available material 1 and material 3 but only 4 of the 5 tons of material 2. The 1 ton of unused material 2 is referred to as slack. In linear programming terminology, any unused or idle capacity for a # constraint is referred to as the slack associated with the constraint. Thus, the material 2 constraint has a slack of 1 ton.

Often variables, called slack variables, are added to the formulation of a linear pro- gramming problem to represent the slack, or unused capacity, associated with a constraint. Unused capacity makes no contribution to profit, so slack variables have coefficients of zero in the objective function. More generally, slack variables represent the difference between

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7.2 Graphical Solution Procedure 267

the right-hand side and the left-hand side of a # constraint. After the addition of three slack variables, denoted S1, S2, and S3, the mathematical model of the RMC problem becomes

Max

s.t.

40F 1 30S 1 0S1 1 0S2 1 0S3

0.4F 1 0.5S 1 1S1 5 20

0.2S 1 1S2 5 5

0.6F 1 0.3S 1 1S3 5 21 F, S, S1, S2, S3 $ 0

whenever a linear program is written in a form with all the constraints expressed as equali- ties, it is said to be written in standard form.

Referring to the standard form of the RMC problem, we see that at the optimal solution (F 5 25, S 5 20) the values for the slack variables are

Can you write a linear program in standard form? Try Problem 18.

Value of Constraint Slack Variable Material 1 S1 5 0 Material 2 S2 5 1 Material 3 S3 5 0

NOTES AND COMMENTS

1. In the standard form representation of a linear program, the objective function coefficients for the slack variables are zero. This condition im- plies that slack variables, which represent un- used resources, do not affect the value of the objective function. However, in some applica- tions, some or all of the unused resources can be sold and contribute to profit. In such cases the corresponding slack variables become decision variables representing the amount of resources to be sold. For each of these variables, a non- zero coefficient in the objective function would

reflect the profit associated with selling a unit of the corresponding resource.

2. Redundant constraints do not affect the feasible region; as a result they can be removed from a linear programming model without affecting the optimal solution. However, if the linear pro- gramming model is to be resolved later, changes in some of the data might change a previously redundant constraint into a binding constraint. Thus, we recommend keeping all constraints in the linear programming model even though one or more of the constraints may be redundant.

Could we have used the graphical analysis to provide some of the previous information? The answer is yes. By finding the optimal solution in Figure 7.6, we see that the material 1 constraint and the material 3 constraint restrict, or bind, the feasible region at this point. Thus, the optimal solution requires the use of all of these two resources. In other words, the graph shows that at the optimal solution material 1 and material 3 will have zero slack. But, because the material 2 constraint is not binding the feasible region at the optimal solution, we can expect some slack for this resource.

Finally, some linear programs may have one or more constraints that do not affect the feasible region; that is, the feasible region remains the same whether or not the constraint is included in the problem. Because such a constraint does not affect the feasible region, it is called a redundant constraint. Redundant constraints can be dropped from the problem without having any effect on the optimal solution. However, in most linear programming problems redundant constraints are not discarded because they are not immediately recog- nizable as being redundant. The RMC problem had no redundant constraints because each constraint had an effect on the feasible region.

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7.3 Extreme Points and the Optimal Solution Suppose that the profit contribution for 1 ton of solvent base increases from $30 to $60 while the profit contribution for 1 ton of fuel additive and all the constraints remain un- changed. The complete linear programming model of this new problem is identical to the mathematical model in Section 7.2, except for the revised objective function:

Max 40F 1 60S

How does this change in the objective function affect the optimal solution to the RMC problem? Figure 7.13 shows the graphical solution of the RMC problem with the revised objective function. Note that because the constraints do not change, the feasible region remains unchanged. However, the profit lines must be altered to reflect the new objective function.

By moving the profit line in a parallel manner away from the origin, we find the opti- mal solution as shown in Figure 7.13. The values of the decision variables at this point are F 5 18.75 and S 5 25. The increased profit for the solvent base caused a change in the optimal solution. In fact, as you might suspect, we cut back the production of the lower profit fuel additive and increase the production of the higher profit solvent base.

what do you notice about the location of the optimal solutions in the linear programming problems that we solved thus far? Look closely at the graphical solutions in Figures 7.10 and 7.13. An important observation that you should be able to make is that the optimal solutions occur at one of the vertices, or “corners,” of the feasible region. In linear programming terminology these vertices are referred to as the extreme points of the feasible region.

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Tons of Fuel Additive

M aximum Pro�t Line (40F + 60S = 2250)

40F + 60S = 1200

Optimal Solution (F = 18.75, S = 25)

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FIGURE 7.13 OPTIMAL SOLUTION FOR THE RMC PROBLEM wITH AN OBJECTIVE FUNCTION OF 40F 1 60S

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7.4 Computer Solution of the RMC Problem 269

Thus, the RMC has five vertices or five extreme points (Figure 7.14). we can now state our observation about the location of optimal solutions:2

The optimal solution to a linear programming problem can be found at an extreme point of the feasible region for the problem.

This property means that, if you are looking for the optimal solution to a linear pro- gramming problem, you do not have to evaluate all feasible solution points. In fact, you have to consider only the feasible solutions that occur at the extreme points of the feasible region. Thus, for the RMC problem, instead of computing and comparing the profit for all feasible solutions, we can find the optimal solution by evaluating the five extreme-point solutions and selecting the one that provides the highest profit. Actually, the graphical so- lution procedure is nothing more than a convenient way of identifying an optimal extreme point for two-variable problems.

7.4 Computer Solution of the RMC Problem Computer programs designed to solve linear programming problems are widely available. After a short period of familiarization with the specific features of the program, most users can solve linear programming problems with few difficulties. Problems involving thousands of variables and thousands of constraints are now routinely solved with computer pack- ages. Some of the leading commercial packages include CPLEX, LINGO, MOSEK, Gurobi,

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FIGURE 7.14 THE FIVE EXTREME POINTS OF THE FEASIBLE REGION FOR THE RMC PROBLEM

For additional practice in identifying the extreme points of the feasible region and determining the optimal solution by computing and comparing the objective function value at each extreme point, try Problem 13.

2In Section 7.6 we show that two special cases (infeasibility and unboundedness) in linear programming have no optimal solution. the observation stated does not apply to these cases.

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270 Chapter 7 Introduction to Linear Programming

Excel Solver, and Analytic Solver Platform for Excel. Packages are also available for free download. A good example is Clp (COIN-OR linear programming) available from the COIN-OR organization at http://www.coin-or.org.

Probably the most widely used tool is Solver, which is built into Microsoft Excel. Therefore, the computer output we discuss is based on the output provided by Excel Solver. The complete details for how to formulate the RMC problem in Excel and use Solver are contained in Appendix 7.1. Appendix 7.2 demonstrates the use of LINGO, a stand-alone software package for solving optimization problems.

Recall the RMC linear program:

Max

s.t.

40F 1 30S

0.4F 1 0.5S

0.2S

0.6F 1 0.3S

# 20 Material 1 # 5 Material 2 # 21 Material 3

F, S $ 0

Figure 7.15 shows the optimal solution to the RMC problem. This output is based on the Answer Report from Excel Solver, but includes the variable names we have used in our linear programming model. This allows you to easily link the answer report to the model under discussion. we will use this style to show the solutions to optimization problems throughout Chapters 7–12.

Interpretation of Answer Report Let us look more closely at the answer report in Figure 7.15 and interpret the computer solu- tion provided for the RMC problem. First, note the number 1600.000 in the Objective Cells (Max) section, which appears in the Final Value column to the right of objective function value, Maximize Total Profit. This number indicates that the optimal solution to this problem will provide a profit of $1600. Directly below the objective function value are the values of

Instructions on how to solve linear programs using Excel and LINGO are provided in appendixes at the end of the chapter.

FIGURE 7.15 ANSwER REPORT FOR THE RMC PROBLEM

Objective Cells (Max) Name Original Value Final Value Maximize Total Profit 0.000 1600.000

Variable Cells Model Variable Name Original Value Final Value integer F Tons Produced Fuel Additive 0.000 25.000 Contin S Tons Produced Solvent Base 0.000 20.000 Contin

Constraints Constraint Number Name Cell Value Status Slack 1 Material 1 Amount Used 20.000 Binding 0.000 2 Material 2 Amount Used 4.000 Not Binding 1.000 3 Material 3 Amount Used 21.000 Binding 0.000

fileWEB RMC

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7.5 A Simple Minimization Problem 271

Thus, we see that the binding constraints (the Material 1 Amount Used and Material 3 Amount Used constraints) have zero slack at the optimal solution. The Material 2 Amount Used constraint has 1 ton of slack, or unused capacity.

7.5 A Simple Minimization Problem M&D Chemicals produces two products that are sold as raw materials to companies manu- facturing bath soaps and laundry detergents. Based on an analysis of current inventory levels and potential demand for the coming month, M&D’s management has specified that the combined production for products A and B must total at least 350 gallons. Separately, a major customer’s order for 125 gallons of product A must also be satisfied. Product A requires 2 hours of processing time per gallon while product B requires 1 hour of process- ing time per gallon, and for the coming month, 600 hours of processing time are available. M&D’s objective is to satisfy these requirements at a minimum total production cost. Pro- duction costs are $2 per gallon for product A and $3 per gallon for product B.

To find the minimum-cost production schedule, we will formulate the M&D Chemicals problem as a linear program. Following a procedure similar to the one used for the RMC problem, we first define the decision variables and the objective function for the problem. Let

A 5 number of gallons of product A

B 5 number of gallons of product B

Because the production costs are $2 per gallon for product A and $3 per gallon for product B, the objective function that corresponds to the minimization of the total produc- tion cost can be written as

Min 2A 1 3B

Next, consider the constraints placed on the M&D Chemicals problem. To satisfy the major customer’s demand for 125 gallons of product A, we know A must be at least 125. Thus, we write the constraint

1A $ 125

Constraint Value of Number Constraint Name Slack Variable 1 Material 1 Amount Used 0 2 Material 2 Amount Used 1 3 Material 3 Amount Used 0

the decision variables at the optimal solution. These are shown as the Final Value column of the Variable Cells section in the answer report. Thus, we have F 5 25 tons of fuel additive and S 5 20 tons of solvent base as the optimal production quantities. we will discuss the meaning of the Integer column in the Variable Cells section in Chapter 11.

The Constraints section of the answer report provides information about the status of the constraints. Recall that the RMC problem had three less-than-or-equal-to constraints corresponding to the tons available for each of the three raw materials. The information shown in the Slack column provides the value of the slack variable for each of the three constraints. This information is summarized as follows:

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272 Chapter 7 Introduction to Linear Programming

Because the combined production for both products must total at least 350 gallons, we can write the constraint

1A 1 1B $ 350

Finally, the limitation on available processing time of 600 hours means that we need to add the constraint

2A 1 1B # 600

After adding the nonnegativity constraints (A, B $ 0), we have the followin