Project Management

BARRY RENDER • RALPH M. STAIR, JR. • MICHAEL E. HANNA • TREVOR S. HALE

T H I R T E E N T H E D I T I O N

QUANTITATIVE ANALYSIS for MANAGEMENT

New York, NY

BARRY RENDER Charles Harwood Professor Emeritus of Management Science

Crummer Graduate School of Business, Rollins College

MICHAEL E. HANNA Professor of Decision Sciences,

University of Houston–Clear Lake

TREVOR S. HALE Associate Professor of Management Sciences,

University of Houston–Downtown

T H I R T E E N T H E D I T I O N

QUANTITATIVE ANALYSIS for MANAGEMENT

RALPH M. STAIR, JR. Professor Emeritus of Information and Management Sciences,

Florida State University

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Library of Congress Cataloging-in-Publication Data Names: Render, Barry, author. | Stair, Ralph M., author. | Hanna, Michael E.,    author. | Hale, Trevor S., author. Title: Quantitative analysis for management / Barry Render, Charles Harwood    Professor of Management Science, Crummer Graduate School of Business,    Rollins College; Ralph M. Stair, Jr., Professor of Information and    Management Sciences, Florida State University; Michael E. Hanna, Professor    of Decision Sciences, University of Houston/Clear Lake; Trevor S. Hale,    Associate Professor of Management Sciences, University of Houston/Downtown. Description: Thirteenth edition. | Boston : Pearson, 2016. | Includes    bibliographical references and index. Identifiers: LCCN 2016037461| ISBN 9780134543161 (hardcover) | ISBN    0134543165 (hardcover) Subjects: LCSH: Management science—Case studies. | Operations research—Case    studies. Classification: LCC T56 .R543 2016 | DDC 658.4/03—dc23 LC record available at https://lccn.loc.gov/ 2016037461

ISBN 10: 0-13-454316-5 ISBN 13: 978-0-13-454316-1

10 9 8 7 6 5 4 3 2 1

To my wife and sons—BR

To Lila and Leslie—RMS

To Zoe and Gigi—MEH

To Valerie and Lauren—TSH

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iii

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

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

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

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

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

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

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

About the Authors

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iv  ABOUT THE AUTHORS

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

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

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

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

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

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

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

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v

CHAPTER 1 Introduction to Quantitative Analysis 1

CHAPTER 2 Probability Concepts and Applications 21

CHAPTER 3 Decision Analysis 63

CHAPTER 4 Regression Models 111

CHAPTER 5 Forecasting 147

CHAPTER 6 Inventory Control Models 185

CHAPTER 7 Linear Programming Models: Graphical and Computer Methods 237

CHAPTER 8 Linear Programming Applications 289

CHAPTER 9 Transportation, Assignment, and Network Models 319

CHAPTER 10 Integer Programming, Goal Programming, and Nonlinear Programming 357

CHAPTER 11 Project Management 387

CHAPTER 12 Waiting Lines and Queuing Theory Models 427

CHAPTER 13 Simulation Modeling 461

CHAPTER 14 Markov Analysis 501

CHAPTER 15 Statistical Quality Control 529

ONLINE MODULES

1 Analytic Hierarchy Process M1-1

2 Dynamic Programming M2-1

3 Decision Theory and the Normal Distribution M3-1

4 Game Theory M4-1

5 Mathematical Tools: Determinants and Matrices M5-1

6 Calculus-Based Optimization M6-1

7 Linear Programming: The Simplex Method M7-1

8 Transportation, Assignment, and Network Algorithms M8-1

Brief Contents

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vi

PREFACE xiii

CHAPTER 1 Introduction to Quantitative Analysis 1 1.1 What Is Quantitative Analysis? 2 1.2 Business Analytics 2 1.3 The Quantitative Analysis Approach 3

Defining the Problem 4

Developing a Model 4

Acquiring Input Data 4

Developing a Solution 5

Testing the Solution 5

Analyzing the Results and Sensitivity Analysis 6

Implementing the Results 6

The Quantitative Analysis Approach and Modeling in the Real World 6

1.4 How to Develop a Quantitative Analysis Model 6 The Advantages of Mathematical Modeling 9

Mathematical Models Categorized by Risk 9

1.5 The Role of Computers and Spreadsheet Models in the Quantitative Analysis Approach 9

1.6 Possible Problems in the Quantitative Analysis Approach 12 Defining the Problem 12

Developing a Model 13

Acquiring Input Data 14

Developing a Solution 14

Testing the Solution 14

Analyzing the Results 15

1.7 Implementation—Not Just the Final Step 15 Lack of Commitment and Resistance to

Change 16

Lack of Commitment by Quantitative Analysts 16

Summary 16 Glossary 16 Key Equations 17

Self-Test 17 Discussion Questions and Problems 18 Case Study: Food and Beverages at Southwestern University Football Games 19 Bibliography 20

CHAPTER 2 Probability Concepts and Applications 21

2.1 Fundamental Concepts 22 Two Basic Rules of Probability 22

Types of Probability 22

Mutually Exclusive and Collectively Exhaustive Events 23

Unions and Intersections of Events 25

Probability Rules for Unions, Intersections, and Conditional Probabilities 25

2.2 Revising Probabilities with Bayes’ Theorem 27 General Form of Bayes’ Theorem 28

2.3 Further Probability Revisions 29 2.4 Random Variables 30 2.5 Probability Distributions 32

Probability Distribution of a Discrete Random Variable 32

Expected Value of a Discrete Probability Distribution 32

Variance of a Discrete Probability Distribution 33

Probability Distribution of a Continuous Random Variable 34

2.6 The Binomial Distribution 35 Solving Problems with the Binomial

Formula 36

Solving Problems with Binomial Tables 37

2.7 The Normal Distribution 38 Area Under the Normal Curve 40

Using the Standard Normal Table 40

Haynes Construction Company Example 41

The Empirical Rule 44

2.8 The F Distribution 44 2.9 The Exponential Distribution 46

Arnold’s Muffler Example 47

2.10 The Poisson Distribution 48 Summary 50 Glossary 50 Key Equations 51 Solved Problems 52 Self-Test 54 Discussion Questions and Problems 55 Case Study: WTVX 61 Bibliography 61

Appendix 2.1: Derivation of Bayes’ Theorem 61

Contents

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CONTENTS  vii

CHAPTER 3 Decision Analysis 63 3.1 The Six Steps in Decision Making 63 3.2 Types of Decision-Making

Environments 65 3.3 Decision Making Under Uncertainty 65

Optimistic 66

Pessimistic 66

Criterion of Realism (Hurwicz Criterion) 67

Equally Likely (Laplace) 67

Minimax Regret 67

3.4 Decision Making Under Risk 69 Expected Monetary Value 69

Expected Value of Perfect Information 70

Expected Opportunity Loss 71

Sensitivity Analysis 72

A Minimization Example 73

3.5 Using Software for Payoff Table Problems 75 QM for Windows 75

Excel QM 75

3.6 Decision Trees 77 Efficiency of Sample Information 82

Sensitivity Analysis 82

3.7 How Probability Values Are Estimated by Bayesian Analysis 83 Calculating Revised Probabilities 83

Potential Problem in Using Survey Results 85

3.8 Utility Theory 86 Measuring Utility and Constructing a Utility

Curve 86

Utility as a Decision-Making Criterion 88

Summary 91 Glossary 91 Key Equations 92 Solved Problems 92 Self-Test 97 Discussion Questions and Problems 98 Case Study: Starting Right Corporation 107 Case Study: Toledo Leather Company 107 Case Study: Blake Electronics 108 Bibliography 110

CHAPTER 4 Regression Models 111 4.1 Scatter Diagrams 112 4.2 Simple Linear Regression 113 4.3 Measuring the Fit of the Regression

Model 114 Coefficient of Determination 116

Correlation Coefficient 116

4.4 Assumptions of the Regression Model 117 Estimating the Variance 119

4.5 Testing the Model for Significance 119 Triple A Construction Example 121

The Analysis of Variance (ANOVA) Table 122

Triple A Construction ANOVA Example 122

4.6 Using Computer Software for Regression 122 Excel 2016 122

Excel QM 123

QM for Windows 125

4.7 Multiple Regression Analysis 126 Evaluating the Multiple Regression

Model 127

Jenny Wilson Realty Example 128

4.8 Binary or Dummy Variables 129 4.9 Model Building 130

Stepwise Regression 131

Multicollinearity 131

4.10 Nonlinear Regression 131 4.11 Cautions and Pitfalls in Regression

Analysis 134 Summary 135 Glossary 135 Key Equations 136 Solved Problems 137 Self-Test 139 Discussion Questions and Problems 139 Case Study: North–South Airline 144 Bibliography 145

Appendix 4.1: Formulas for Regression Calculations 145

CHAPTER 5 Forecasting 147 5.1 Types of Forecasting Models 147

Qualitative Models 147

Causal Models 148

Time-Series Models 149

5.2 Components of a Time-Series 149 5.3 Measures of Forecast Accuracy 151 5.4 Forecasting Models—Random Variations

Only 154 Moving Averages 154

Weighted Moving Averages 154

Exponential Smoothing 156

Using Software for Forecasting Time Series 158

5.5 Forecasting Models—Trend and Random Variations 160 Exponential Smoothing with Trend 160

Trend Projections 163

5.6 Adjusting for Seasonal Variations 164 Seasonal Indices 165

Calculating Seasonal Indices with No Trend 165

Calculating Seasonal Indices with Trend 166

5.7 Forecasting Models—Trend, Seasonal, and Random Variations 167 The Decomposition Method 167

Software for Decomposition 170

Using Regression with Trend and Seasonal Components 170

5.8 Monitoring and Controlling Forecasts 172 Adaptive Smoothing 174

Summary 174 Glossary 174 Key Equations 175 Solved Problems 176 Self-Test 177 Discussion Questions and Problems 178 Case Study: Forecasting Attendance at SWU Football Games 182 Case Study: Forecasting Monthly Sales 183 Bibliography 184

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viii  CONTENTS

CHAPTER 6 Inventory Control Models 185 6.1 Importance of Inventory Control 186

Decoupling Function 186

Storing Resources 187

Irregular Supply and Demand 187

Quantity Discounts 187

Avoiding Stockouts and Shortages 187

6.2 Inventory Decisions 187 6.3 Economic Order Quantity: Determining

How Much to Order 189 Inventory Costs in the EOQ Situation 189

Finding the EOQ 191

Sumco Pump Company Example 192

Purchase Cost of Inventory Items 193

Sensitivity Analysis with the EOQ Model 194

6.4 Reorder Point: Determining When to Order 194

6.5 EOQ Without the Instantaneous Receipt Assumption 196 Annual Carrying Cost for Production Run

Model 196

Annual Setup Cost or Annual Ordering Cost 197

Determining the Optimal Production Quantity 197

Brown Manufacturing Example 198

6.6 Quantity Discount Models 200 Brass Department Store Example 202

6.7 Use of Safety Stock 203 6.8 Single-Period Inventory Models 209

Marginal Analysis with Discrete Distributions 210

Café du Donut Example 210

Marginal Analysis with the Normal Distribution 212

Newspaper Example 212

6.9 ABC Analysis 214 6.10 Dependent Demand: The Case for

Material Requirements Planning 214 Material Structure Tree 215

Gross and Net Material Requirements Plans 216

Two or More End Products 218

6.11 Just-In-Time Inventory Control 219 6.12 Enterprise Resource Planning 220

Summary 221 Glossary 221 Key Equations 222 Solved Problems 223 Self-Test 225 Discussion Questions and Problems 226 Case Study: Martin-Pullin Bicycle Corporation 234 Bibliography 235

Appendix 6.1: Inventory Control with QM for Windows 235

CHAPTER 7 Linear Programming Models: Graphical and Computer Methods 237

7.1 Requirements of a Linear Programming Problem 238

7.2 Formulating LP Problems 239 Flair Furniture Company 240

7.3 Graphical Solution to an LP Problem 241 Graphical Representation of Constraints 241

Isoprofit Line Solution Method 245

Corner Point Solution Method 248

Slack and Surplus 250

7.4 Solving Flair Furniture’s LP Problem Using QM for Windows, Excel 2016, and Excel QM 251 Using QM for Windows 251

Using Excel’s Solver Command to Solve LP Problems 252

Using Excel QM 255

7.5 Solving Minimization Problems 257 Holiday Meal Turkey Ranch 257

7.6 Four Special Cases in LP 261 No Feasible Solution 261

Unboundedness 261

Redundancy 262

Alternate Optimal Solutions 263

7.7 Sensitivity Analysis 264 High Note Sound Company 265

Changes in the Objective Function Coefficient 266

QM for Windows and Changes in Objective Function Coefficients 266

Excel Solver and Changes in Objective Function Coefficients 267

Changes in the Technological Coefficients 268

Changes in the Resources or Right-Hand-Side Values 269

QM for Windows and Changes in Right-Hand- Side Values 270

Excel Solver and Changes in Right-Hand-Side Values 270

Summary 272 Glossary 272 Solved Problems 273 Self-Test 277 Discussion Questions and Problems 278 Case Study: Mexicana Wire Winding, Inc. 286 Bibliography 288

CHAPTER 8 Linear Programming Applications 289 8.1 Marketing Applications 289

Media Selection 289

Marketing Research 291

8.2 Manufacturing Applications 293 Production Mix 293

Production Scheduling 295

8.3 Employee Scheduling Applications 299 Labor Planning 299

8.4 Financial Applications 300 Portfolio Selection 300

Truck Loading Problem 303

8.5 Ingredient Blending Applications 305

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CONTENTS  ix

Diet Problems 305

Ingredient Mix and Blending Problems 306

8.6 Other Linear Programming Applications 308 Summary 310 Self-Test 310 Problems 311 Case Study: Cable & Moore 318 Bibliography 318

CHAPTER 9 Transportation, Assignment, and Network Models 319

9.1 The Transportation Problem 320 Linear Program for the Transportation

Example 320

Solving Transportation Problems Using Computer Software 321

A General LP Model for Transportation Problems 322

Facility Location Analysis 323

9.2 The Assignment Problem 325 Linear Program for Assignment Example 325

9.3 The Transshipment Problem 327 Linear Program for Transshipment

Example 327

9.4 Maximal-Flow Problem 330 Example 330

9.5 Shortest-Route Problem 332 9.6 Minimal-Spanning Tree Problem 334

Summary 337 Glossary 338 Solved Problems 338 Self-Test 340 Discussion Questions and Problems 341 Case Study: Andrew–Carter, Inc. 352 Case Study: Northeastern Airlines 353 Case Study: Southwestern University Traffic Problems 354 Bibliography 355

Appendix 9.1: Using QM for Windows 355

CHAPTER 10 Integer Programming, Goal Programming, and Nonlinear Programming 357

10.1 Integer Programming 358 Harrison Electric Company Example of Integer

Programming 358

Using Software to Solve the Harrison Integer Programming Problem 360

Mixed-Integer Programming Problem Example 360

10.2 Modeling with 0–1 (Binary) Variables 363 Capital Budgeting Example 364

Limiting the Number of Alternatives Selected 365

Dependent Selections 365

Fixed-Charge Problem Example 366

Financial Investment Example 367

10.3 Goal Programming 368 Example of Goal Programming: Harrison Electric

Company Revisited 369

Extension to Equally Important Multiple Goals 370

Ranking Goals with Priority Levels 371

Goal Programming with Weighted Goals 371

10.4 Nonlinear Programming 372 Nonlinear Objective Function and Linear

Constraints 373

Both Nonlinear Objective Function and Nonlinear Constraints 373

Linear Objective Function with Nonlinear Constraints 374

Summary 375 Glossary 375 Solved Problems 376 Self-Test 378 Discussion Questions and Problems 379 Case Study: Schank Marketing Research 384 Case Study: Oakton River Bridge 385 Bibliography 385

CHAPTER 11 Project Management 387 11.1 PERT/CPM 389

General Foundry Example of PERT/CPM 389

Drawing the PERT/CPM Network 390

Activity Times 391

How to Find the Critical Path 392

Probability of Project Completion 395

What PERT Was Able to Provide 398

Using Excel QM for the General Foundry Example 398

Sensitivity Analysis and Project Management 399

11.2 PERT/Cost 400 Planning and Scheduling Project Costs:

Budgeting Process 400

Monitoring and Controlling Project Costs 403

11.3 Project Crashing 405 General Foundry Example 406

Project Crashing with Linear Programming 407

11.4 Other Topics in Project Management 410 Subprojects 410

Milestones 410

Resource Leveling 410

Software 410

Summary 410 Glossary 410 Key Equations 411 Solved Problems 412 Self-Test 414 Discussion Questions and Problems 415 Case Study: Southwestern University Stadium Construction 422 Case Study: Family Planning Research Center of Nigeria 423 Bibliography 424

Appendix 11.1: Project Management with QM for Windows 424

CHAPTER 12 Waiting Lines and Queuing Theory Models 427

12.1 Waiting Line Costs 428 Three Rivers Shipping Company Example 428

12.2 Characteristics of a Queuing System 429 Arrival Characteristics 429

Waiting Line Characteristics 430

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x  CONTENTS

Service Facility Characteristics 430

Identifying Models Using Kendall Notation 431

12.3 Single-Channel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M /1) 434 Assumptions of the Model 434

Queuing Equations 434

Arnold’s Muffler Shop Case 435

Enhancing the Queuing Environment 438

12.4 Multichannel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M/m) 439 Equations for the Multichannel Queuing

Model 439

Arnold’s Muffler Shop Revisited 440

12.5 Constant Service Time Model (M/D/1) 442 Equations for the Constant Service Time

Model 442

Garcia-Golding Recycling, Inc. 443

12.6 Finite Population Model (M/M/1 with Finite Source) 443 Equations for the Finite Population

Model 444

Department of Commerce Example 444

12.7 Some General Operating Characteristic Relationships 445

12.8 More Complex Queuing Models and the Use of Simulation 446 Summary 446 Glossary 447 Key Equations 447 Solved Problems 449 Self-Test 451 Discussion Questions and Problems 452 Case Study: New England Foundry 457 Case Study: Winter Park Hotel 459 Bibliography 459

Appendix 12.1: Using QM for Windows 460

CHAPTER 13 Simulation Modeling 461 13.1 Advantages and Disadvantages of

Simulation 462 13.2 Monte Carlo Simulation 463

Harry’s Auto Tire Example 464

Using QM for Windows for Simulation 468

Simulation with Excel Spreadsheets 469

13.3 Simulation and Inventory Analysis 471 Simkin’s Hardware Store 472

Analyzing Simkin’s Inventory Costs 475

13.4 Simulation of a Queuing Problem 476 Port of New Orleans 476

Using Excel to Simulate the Port of New Orleans Queuing Problem 478

13.5 Simulation Model for a Maintenance Policy 479 Three Hills Power Company 479

Cost Analysis of the Simulation 481

13.6 Other Simulation Issues 484 Two Other Types of Simulation Models 484

Verification and Validation 485

Role of Computers in Simulation 485

Summary 486 Glossary 486 Solved Problems 487 Self-Test 489 Discussion Questions and Problems 490 Case Study: Alabama Airlines 496 Case Study: Statewide Development Corporation 497 Case Study: FB Badpoore Aerospace 498 Bibliography 500

CHAPTER 14 Markov Analysis 501 14.1 States and State Probabilities 502

The Vector of State Probabilities for Grocery Store Example 503

14.2 Matrix of Transition Probabilities 504 Transition Probabilities for Grocery Store

Example 504

14.3 Predicting Future Market Shares 505 14.4 Markov Analysis of Machine

Operations 506 14.5 Equilibrium Conditions 507 14.6 Absorbing States and the Fundamental

Matrix: Accounts Receivable Application 510 Summary 514 Glossary 514 Key Equations 514 Solved Problems 515 Self-Test 518 Discussion Questions and Problems 519 Case Study: Rentall Trucks 523 Bibliography 525

Appendix 14.1: Markov Analysis with QM for Windows 525

Appendix 14.2: Markov Analysis with Excel 526

CHAPTER 15 Statistical Quality Control 529 15.1 Defining Quality and TQM 529 15.2 Statistical Process Control 531

Variability in the Process 531

15.3 Control Charts for Variables 532 The Central Limit Theorem 533

Setting x--Chart Limits 534

Setting Range Chart Limits 536

15.4 Control Charts for Attributes 537 p-Charts 537

c-Charts 539

Summary 541 Glossary 541 Key Equations 541 Solved Problems 542 Self-Test 543 Discussion Questions and Problems 543 Bibliography 546

Appendix 15.1: Using QM for Windows for SPC 547

APPENDICES 549

APPENDIX A Areas Under the Standard Normal Curve 550

APPENDIX B Binomial Probabilities 552

APPENDIX C Values of e−L for Use in the Poisson Distribution 557

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CONTENTS  xi

APPENDIX D F Distribution Values 558

APPENDIX E Using POM-QM for Windows 560

APPENDIX F Using Excel QM and Excel Add-Ins 563

APPENDIX G Solutions to Selected Problems 564

APPENDIX H Solutions to Self-Tests 568

INDEX 571

ONLINE MODULES

MODULE 1 Analytic Hierarchy Process M1-1 M1.1 Multifactor Evaluation Process M1-2 M1.2 Analytic Hierarchy Process M1-3

Judy Grim’s Computer Decision M1-3

Using Pairwise Comparisons M1-5

Evaluations for Hardware M1-5

Determining the Consistency Ratio M1-6

Evaluations for the Other Factors M1-7

Determining Factor Weights M1-8

Overall Ranking M1-9

Using the Computer to Solve Analytic Hierarchy Process Problems M1-9

M1.3 Comparison of Multifactor Evaluation and Analytic Hierarchy Processes M1-9 Summary M1-10 Glossary M1-10 Key Equations M1-10 Solved Problems M1-11 Self-Test M1-12 Discussion Questions and Problems M1-12 Bibliography M1-14

Appendix M1.1: Using Excel for the Analytic Hierarchy Process M1-14

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

Dynamic Programming M2-2 M2.2 Dynamic Programming

Terminology M2-5 M2.3 Dynamic Programming Notation M2-7 M2.4 Knapsack Problem M2-8

Types of Knapsack Problems M2-8

Roller’s Air Transport Service Problem M2-8

Summary M2-13 Glossary M2-14 Key Equations M2-14 Solved Problem M2-14 Self-Test M2-16 Discussion Questions and Problems M2-17 Case Study: United Trucking M2-19 Bibliography M2-20

MODULE 3 Decision Theory and the Normal Distribution M3-1

M3.1 Break-Even Analysis and the Normal Distribution M3-1

Barclay Brothers Company’s New Product Decision M3-1

Probability Distribution of Demand M3-2

Using Expected Monetary Value to Make a Decision M3-4

M3.2 Expected Value of Perfect Information and the Normal Distribution M3-5 Opportunity Loss Function M3-5

Expected Opportunity Loss M3-5

Summary M3-7 Glossary M3-7 Key Equations M3-7 Solved Problems M3-7 Self-Test M3-8 Discussion Questions and Problems M3-9 Bibliography M3-10

Appendix M3.1: Derivation of the Break-Even Point M3-10

Appendix M3.2: Unit Normal Loss Integral M3-11

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

Summary M4-7 Glossary M4-7 Solved Problems M4-14 Self-Test M4-8 Discussion Questions and Problems M4-9 Bibliography M4-10

MODULE 5 Mathematical Tools: Determinants and Matrices M5-1

M5.1 Matrices and Matrix Operations M5-1 Matrix Addition and Subtraction M5-2

Matrix Multiplication M5-2

Matrix Notation for Systems of Equations M5-5

Matrix Transpose M5-5

M5.2 Determinants, Cofactors, and Adjoints M5-5 Determinants M5-5

Matrix of Cofactors and Adjoint M5-7

M5.3 Finding the Inverse of a Matrix M5-9 Summary M5-10 Glossary M5-10 Key Equations M5-10 Self-Test M5-11 Discussion Questions and Problems M5-11 Bibliography M5-12

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

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

Second Derivatives M6-6

M6.4 Maximum and Minimum M6-6 M6.5 Applications M6-8

Economic Order Quantity M6-8

Total Revenue M6-8

Summary M6-9 Glossary M6-10 Key Equations M6-10 Solved Problems M6-10 Self-Test M6-11 Discussion Questions and Problems M6-11 Bibliography M6-12

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xii  CONTENTS

MODULE 7 Linear Programming: The Simplex Method M7-1

M7.1 How to Set Up the Initial Simplex Solution M7-2 Converting the Constraints to Equations M7-2

Finding an Initial Solution Algebraically M7-3

The First Simplex Tableau M7-4

M7.2 Simplex Solution Procedures M7-7 The Second Simplex Tableau M7-8

Interpreting the Second Tableau M7-11

The Third Simplex Tableau M7-12

Review of Procedures for Solving LP Maximization Problems M7-14

M7.3 Surplus and Artificial Variables M7-15 Surplus Variables M7-15

Artificial Variables M7-15

Surplus and Artificial Variables in the Objective Function M7-16

M7.4 Solving Minimization Problems M7-16

The Muddy River Chemical Corporation Example M7-16

Graphical Analysis M7-17

Converting the Constraints and Objective Function M7-18

Rules of the Simplex Method for Minimization Problems M7-18

First Simplex Tableau for the Muddy River Chemical Corporation Problem M7-19

Developing a Second Tableau M7-20

Developing a Third Tableau M7-22

Fourth Tableau for the Muddy River Chemical Corporation Problem M7-23

Review of Procedures for Solving LP Minimization Problems M7-24

M7.5 Special Cases M7-25 Infeasibility M7-25

Unbounded Solutions M7-25

Degeneracy M7-26

More Than One Optimal Solution M7-27

M7.6 Sensitivity Analysis with the Simplex Tableau M7-27 High Note Sound Company Revisited M7-27

Changes in the Objective Function Coefficients M7-28

Changes in Resources or RHS Values M7-30

M7.7 The Dual M7-32 Dual Formulation Procedures M7-33

Solving the Dual of the High Note Sound Company Problem M7-33

M7.8 Karmarkar’s Algorithm M7-34 Summary M7-35 Glossary M7-35 Key Equations M7-36 Solved Problems M7-36 Self-Test M7-40 Discussion Questions and Problems M7-41 Bibliography M7-50

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

M8.1 The Transportation Algorithm M8-1

Developing an Initial Solution: Northwest Corner Rule M8-2

Stepping-Stone Method: Finding a Least-Cost Solution M8-3

Special Situations with the Transportation Algorithm M8-9

Unbalanced Transportation Problems M8-9

Degeneracy in Transportation Problems M8-10

More Than One Optimal Solution M8-12

Maximization Transportation Problems M8-13

Unacceptable or Prohibited Routes M8-13

Other Transportation Methods M8-13

M8.2 The Assignment Algorithm M8-13 The Hungarian Method (Flood’s Technique) 14

Making the Final Assignment M8-18

Special Situations with the Assignment Algorithm M8-18

Unbalanced Assignment Problems M8-18

Maximization Assignment Problems M8-19

M8.3 Maximal-Flow Problem M8-20 Maximal-Flow Technique M8-20

M8.4 Shortest-Route Problem M8-24 Shortest-Route Technique M8-24

Summary M8-26 Glossary M8-26 Solved Problems M8-26 Self-Test M8-33 Discussion Questions and Problems M8-33 Bibliography M8-43

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xiii

Overview

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

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

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

Special Features

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

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

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

●● Margin notes highlight the important topics in the text.

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

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

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

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

Preface

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xiv  PREFACE

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

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

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

●● Case Studies, at the end of most chapters, provide additional challenging managerial applications.

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

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

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

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

●● Excel QM and Excel 2016 are used to solve problems throughout the book.

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

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

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

Significant Changes to the Thirteenth Edition

In the thirteenth edition, we have introduced Excel 2016 in all of the chapters. Updated screen- shots are integrated in the appropriate sections so that students can easily learn how to use Excel 2016 for the calculations. The Excel QM add-in is used with Excel 2016, allowing students with limited Excel experience to easily perform the necessary calculations. This also allows students to improve their Excel skills as they see the formulas automatically written in Excel QM.

From the Companion Website, students can access files for all of the examples used in the textbook in Excel 2016, QM for Windows, and Excel QM. Other files with all of the end-of- chapter problems involving these software tools are available to the instructors.

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

Examples and problems have been updated, and many new ones have been added. New screenshots are provided for almost all of the examples in the book. A brief summary of the changes in each chapter of the thirteenth edition is presented here.

Chapter 1 Introduction to Quantitative Analysis. The section on business analytics has been updated, and a new end-of-chapter problem has been added.

Chapter 2 Probability Concepts and Applications. The Modeling in the Real World box has been updated. New screenshots of Excel 2016 have been added throughout.

Chapter 3 Decision Analysis. A new QA in Action box has been added. New screenshots of Excel 2016 have been added throughout. A new case study has been added.

Chapter 4 Regression Models. New screenshots of Excel 2016 have been added throughout. A new end-of-chapter problem has been added.

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PREFACE  xv

Chapter 5 Forecasting. A new QA in Action box has been added. New screenshots of Excel 2016 have been added throughout. Two new end-of-chapter problems have been added.

Chapter 6 Inventory Control Models. A new QA in Action box has been added. New screenshots of Excel 2016 have been added throughout. Two new end-of-chapter problems have been added.

Chapter 7 Linear Programming Models: Graphical and Computer Methods. The Learning Objectives have been modified slightly. Screenshots have been updated to Excel 2016.

Chapter 8 Linear Programming Applications. Two new problems have been added to the Internet Homework Problems. Excel 2016 screenshots have been incorporated throughout.

Chapter 9 Transportation, Assignment, and Network Models. Two new problems have been added to the Internet Homework Problems. Excel 2016 screenshots have been incorporated throughout.

Chapter 10 Integer Programming, Goal Programming, and Nonlinear Programming. Two new problems have been added to the Internet Homework Problems. Excel 2016 screenshots have been incorporated throughout.

Chapter 11 Project Management. Four new end-of-chapter problems and a new Modeling in the Real World box have been added.

Chapter 12 Waiting Lines and Queuing Theory Models. Four new end-of-chapter problems have been added.

Chapter 13 Simulation Modeling. Two new end-of-chapter problems have been added.

Chapter 14 Markov Analysis. Two new end-of-chapter problems have been added.

Chapter 15 Statistical Quality Control. Two new end-of-chapter problems have been added. Excel 2016 screenshots have been updated throughout.

Modules 1–8 The only significant change to the modules is the update to Excel 2016 throughout.

Online Modules

To streamline the book, eight topics are contained in modules available on the Companion Website for the book.

1. Analytic Hierarchy Process

2. Dynamic Programming

3. Decision Theory and the Normal Distribution

4. Game Theory

5. Mathematical Tools: Determinants and Matrices

6. Calculus-Based Optimization

7. Linear Programming: The Simplex Method

8. Transportation, Assignment, and Network Algorithms

Software

Excel 2016 Excel 2016 instructions and screen captures are provided, throughout the book. Instructions for activating the Solver and Analysis ToolPak add-ins in Excel 2016 are provided in an appendix. The use of Excel is more prevalent in this edition of the book than in previous editions.

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xvi  PREFACE

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

POM-QM for Windows This software, developed by Professor Howard Weiss, is available to students at the Companion Website. This is very user-friendly and has proven to be a very popular software tool for users of this textbook. Modules are available for every major prob- lem type presented in the textbook. At press time, only version 4.0 of POM-QM for Windows was available. Updates for version 5.0 will be released on the Companion Website as they become available.

Companion Website

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

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

Files for Examples in Excel, Excel QM, and POM-QM for Windows Students can download the files that were used for examples throughout the book. This helps them become familiar with the software, and it helps them understand the input and formulas necessary for working the examples.

Internet Homework Problems In addition to the end-of-chapter problems in the textbook, there are additional problems that instructors may assign. These are available for download at the Companion Website.

Internet Case Studies Additional case studies are available for most chapters.

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

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

Instructor Resources

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

●● Register, Redeem, Login: At www.pearsonhighered.com/irc, instructors can access a variety of print, media, and presentation resources that are available with this text in downloadable, digital format. For most texts, resources are also available for course management platforms such as Blackboard, WebCT, and Course Compass.

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

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PREFACE  xvii

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

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

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

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

Acknowledgments

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

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

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

Stephen Achtenhagen, San Jose University M. Jill Austin, Middle Tennessee State University Raju Balakrishnan, University of Michigan–Dearborn Hooshang Beheshti, Radford University Jason Bergner, University of Nevada Bruce K. Blaylock, Radford University Rodney L. Carlson, Tennessee Technological University Edward Chu, California State University, Dominguez Hills John Cozzolino, Pace University–Pleasantville Ozgun C. Demirag, Penn State–Erie Shad Dowlatshahi, University of Missouri–Kansas City Ike Ehie, Kansas State University Richard Ehrhardt, University of North Carolina–Greensboro Sean Eom, Southeast Missouri State University Ephrem Eyob, Virginia State University Mira Ezvan, Lindenwood University Wade Ferguson, Western Kentucky University Robert Fiore, Springfield College Frank G. Forst, Loyola University of Chicago Stephen H. Goodman, University of Central Florida Irwin Greenberg, George Mason University Nicholas G. Hall, Ohio State University Robert R. Hill, University of Houston–Clear Lake Bharat Jain, Towson University Vassilios Karavas, Nomura International

Arun Khanal, Nobel College Kenneth D. Lawrence, New Jersey Institute of Technology Jooh Lee, Rowan College Richard D. Legault, University of Massachusetts–Dartmouth Douglas Lonnstrom, Siena College Daniel McNamara, University of St. Thomas Peter Miller, University of Windsor Ralph Miller, California State Polytechnic University Shahriar Mostashari, Campbell University David Murphy, Boston College Robert C. Myers, University of Louisville Barin Nag, Towson State University Nizam S. Najd, St. Gregory’s University Harvey Nye, Central State University Alan D. Olinsky, Bryant College Savas Ozatalay, Widener University Young Park, California University of Pennsylvania Yusheng Peng, Brooklyn College Dane K. Peterson, Missouri State University Sanjeev Phukan, Bemidji State University Ranga Ramasesh, Texas Christian University Bonnie Robeson, Johns Hopkins University Grover Rodich, Portland State University Vijay Shah, West Virginia University–Parkersburg L. Wayne Shell, Nicholls State University

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xviii  PREFACE

We are very grateful to all the people at Pearson who worked so hard to make this book a success. These include Donna Battista, Jeff Holcomb, Ashley Santora, Neeraj Bhalla, Vamanan Namboodiri, and Dan Tylman. We are also grateful to Angela Urquhart and Andrea Archer at Thistle Hill Publishing Services. Thank you all!

Barry Render [email protected]

Ralph Stair

Michael Hanna [email protected]

Trevor S. Hale [email protected]

Thomas Sloan, University of Massachusetts–Lowell Richard Slovacek, North Central College Alan D. Smith, Robert Morris University John Swearingen, Bryant College Jack Taylor, Portland State University Andrew Tiger, Union University

Chris Vertullo, Marist College James Vigen, California State University, Bakersfield Larry Weinstein, Wright State University Fred E. Williams, University of Michigan–Flint Mela Wyeth, Charleston Southern University Oliver Yu, San Jose State University

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 1

1.4 Prepare a quantitative analysis model.

1.5 Use computers and spreadsheet models to perform quantitative analysis.

1.6 Recognize possible problems in using quantitative analysis.

1.7 Recognize implementation concerns of quantitative analysis.

1.1 Describe the quantitative analysis approach and understand how to apply it to a real situation.

1.2 Describe the three categories of business analytics.

1.3 Describe the use of modeling in quantitative analysis.

After completing this chapter, students will be able to:

Introduction to Quantitative Analysis

LEARNING OBJECTIVES

1 CHAPTER

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

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

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

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2  CHAPTER 1 • InTRoDUCTIon To QUAnTITATIvE AnAlysIs

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

1.1 What Is Quantitative Analysis?

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

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

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

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

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

Quantitative analysis has been particularly important in many areas of management. The field of production management, which evolved into production/operations management (POM) as society became more service oriented, uses quantitative analysis extensively. While POM focuses on the internal operations of a company, the field of supply chain management takes a more complete view of the business and considers the entire process of obtaining materials from suppliers, using the materials to develop products, and distributing these products to the final consumers. Supply chain management makes extensive use of many management science mod- els. Another area of management that could not exist without the quantitative analysis methods presented in this book, and perhaps the hottest discipline in business today, is business analytics.

1.2 Business Analytics

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

Business analytics is often broken into three categories: descriptive, predictive, and prescrip- tive. Descriptive analytics involves the study and consolidation of historical data for a business and an industry. It helps a company measure how it has performed in the past and how it is per- forming now. Predictive analytics is aimed at forecasting future outcomes based on patterns in the past data. Statistical and mathematical models are used extensively for this purpose. Prescrip- tive analytics involves the use of optimization methods to provide new and better ways to operate

Quantitative analysis uses a scientific approach to decision making.

Both qualitative and quantitative factors must be considered.

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1.3 THE QUAnTITATIvE AnAlysIs APPRoACH  3

based on specific business objectives. The optimization models presented in this book are very important to prescriptive analytics. While there are only three business analytics categories, many business decisions are made based on information obtained from two or three of these categories.

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

1.3 The Quantitative Analysis Approach

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

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

Defining the problem can be the most important step.

Concentrate on only a few problems.

TABLE 1.1 Business Analytics and Quantitative Analysis Models

BUSINESS ANALYTICS CATEGORY QUANTITATIVE ANALYSIS TECHNIQUE

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

Statistical quality control (Chapter 15)

Predictive analytics Decision analysis and decision trees (Chapter 3)

Regression models (Chapter 4)

Forecasting (Chapter 5)

Project scheduling (Chapter 11)

Waiting line models (Chapter 12)

Simulation (Chapter 13)

Markov analysis (Chapter 14)

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

Linear programming (Chapters 7, 8)

Transportation and assignment models (Chapter 9)

Integer programming, goal programming, and nonlinear programming (Chapter 10)

Network models (Chapter 9)

Quantitative analysis has been in existence since the begin- ning of recorded history, but it was Frederick Winslow Taylor who in the late 1800s and early 1900s pioneered the appli- cation of the principles of the scientific approach to manage- ment. Dubbed the “Father of Industrial Engineering,” Taylor is credited with introducing many new scientific and quantitative techniques. These new developments were so successful that many companies still use his techniques in managerial deci- sion making and planning today. Indeed, many organizations employ a staff of operations research or management science

personnel or consultants to apply the principles of scien- tific management to the challenges and opportunities of the twenty-first century.

The origin of many of the techniques discussed in this book can be traced to individuals and organizations that have applied the principles of scientific management first developed by Taylor; they are discussed in History boxes throughout the book. Trivia: Taylor was also a world-class golfer and tennis player, finishing just off the medal stand in golf at the 1900 Olympics and winning the inaugural men’s doubles title (with Clarence Clark) at the U.S. Open Tennis Championships.

The origin of Quantitative AnalysisHISTORY

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4  CHAPTER 1 • InTRoDUCTIon To QUAnTITATIvE AnAlysIs

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

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

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

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

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

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

Of course, there are many other types of models. Architects sometimes make a physical model of a building that they will construct. Engineers develop scale models of chemical plants, called pilot plants. A schematic model is a picture, drawing, or chart of reality. Automobiles, lawn mowers, gears, fans, smartphones, and numerous other devices have schematic models (drawings and pictures) that reveal how these devices work. What sets quantitative analysis apart from other techniques is that the models that are used are mathematical. A mathematical model is a set of mathematical relationships. In most cases, these relationships are expressed in equa- tions and inequalities, as they are in a spreadsheet model that computes sums, averages, or stan- dard deviations.

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

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

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

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

Defining the Problem

Developing a Model

Acquiring Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

FIGURE 1.1 The Quantitative Analysis Approach

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1.3 THE QUAnTITATIvE AnAlysIs APPRoACH  5

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

Developing a Solution Developing a solution involves manipulating the model to arrive at the best (optimal) solution to the problem. In some cases, this requires that an equation be solved for the best decision. In other cases, you can use a trial-and-error method, trying various approaches and picking the one that results in the best decision. For some problems, you may wish to try all possible values for the variables in the model to arrive at the best decision. This is called complete enumera- tion. This book also shows you how to solve very difficult and complex problems by repeat- ing a few simple steps until you find the best solution. A series of steps or procedures that are repeated is called an algorithm, named after Algorismus (derived from Muhammad ibn Musa al-Khwarizmi), a Persian mathematician of the ninth century.

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

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

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

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

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

operations Research and oil spills

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

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

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

Source: Based on N. Altay and W. Green, “OR/MS Research in Disaster Op- erations Management,” European Journal of Operational Research 175, 1 (2006): 475–493, © Trevor S. Hale.

IN ACTION

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

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

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

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

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

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

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

1.4 How to Develop a Quantitative Analysis Model

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

Profit = Revenue - Expenses

In many cases, we can express revenue as the selling price per unit multiplied times the num- ber of units sold. Expenses can often be determined by summing fixed cost and variable cost.

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

Expenses include fixed and variable costs.

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Defining the Problem CSX Transportation, Inc., has 35,000 employees and annual revenue of $11 billion. It provides rail freight services to 23 states east of the Mississippi River, as well as parts of Canada. CSX receives orders for rail delivery service and must send empty railcars to customer locations. Moving these empty railcars results in hundreds of thousands of empty-car miles every day. If allocations of railcars to customers is not done properly, problems arise from excess costs, wear and tear on the system, and congestion on the tracks and at rail yards.

Developing a Model In order to provide a more efficient scheduling system, CSX spent 2 years and $5 million developing its Dynamic Car-Planning (DCP) system. This model will minimize costs, including car travel distance, car han- dling costs at the rail yards, car travel time, and costs for being early or late. It does this while at the same time filling all orders, making sure the right type of car is assigned to the job, and getting the car to the destination in the allowable time.

Acquiring Input Data In developing the model, the company used historical data for testing. In running the model, DCP uses three external sources to obtain information on the customer car orders, the available cars of the type needed, and the transit-time standards. In addition to these, two internal input sources provide informa- tion on customer priorities and preferences and on cost parameters.

Developing a Solution This model takes about 1 minute to load but only 10 seconds to solve. Because supply and demand are constantly changing, the model is run about every 15 minutes. This allows final decisions to be delayed until absolutely necessary.

Testing the Solution The model was validated and verified using existing data. The solutions found using DCP were determined to be very good compared to assignments made without DCP.

Analyzing the Results Since the implementation of DCP in 1997, more than $51 million has been saved annually. Due to the im- proved efficiency, it is estimated that CSX avoided spending another $1.4 billion to purchase an additional 18,000 railcars that would have been needed without DCP. Other benefits include reduced congestion in the rail yards and reduced congestion on the tracks, which are major concerns. This greater efficiency means that more freight can ship by rail rather than by truck, resulting in significant public benefits. These benefits include reduced pollution and greenhouse gases, improved highway safety, and reduced road maintenance costs.

Implementing the Results Both senior-level management who championed DCP and key car-distribution experts who supported the new approach were instrumental in gaining acceptance of the new system and overcoming problems dur- ing the implementation. The job description of the car distributors was changed from car allocators to cost technicians. They are responsible for seeing that accurate cost information is entered into DCP, and they also manage any exceptions that must be made. They were given extensive training on how DCP works so they could understand and better accept the new system. Due to the success of DCP, other railroads have implemented similar systems and achieved similar benefits. CSX continues to enhance DCP to make it even more customer friendly and to improve car-order forecasts.

Source: Based on M. F. Gorman et al., “CSX Railway Uses OR to Cash In on Optimized Equipment Distribution,” Interfaces 40, 1 (January–February 2010): 5–16, © Trevor S. Hale.

MODELING IN THE REAL WORLD

Railroad Uses optimization Models to save Millions

Defining the Problem

Developing a Model

Acquiring Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

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Variable cost is often expressed as the variable cost per unit multiplied times the number of units. Thus, we can also express profit in the following mathematical model:

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

where

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

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

EXAMPLE: PRITCHETT’S PRECIOUS TIME PIECES We will use the Bill Pritchett clock repair shop ex- ample to demonstrate the use of mathematical models. Bill’s company, Pritchett’s Precious Time Pieces, buys, sells, and repairs old clocks and clock parts. Bill sells rebuilt springs for a price per unit of $8. The fixed cost of the equipment to build the springs is $1,000. The variable cost per unit is $3 for spring material. In this example,

s = 8 f = 1,000 n = 3

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

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

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

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

0 = sX - f - nX

This can be written as

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

f = 1s - n2X X =

f

s - n

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

BEP = Fixed cost

1Selling price per unit2 - 1Variable cost per unit2

BEP = f

s - n (1-2)

The BEP results in $0 profits.

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

BEP = +1,000>1+8 - +32 = 200 units, or springs,

The Advantages of Mathematical Modeling There are a number of advantages of using mathematical models:

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

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

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

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

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

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

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

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

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

Developing a solution, testing the solution, and analyzing the results are important steps in the quantitative analysis approach. Because we will be using mathematical models, these steps re- quire mathematical calculations. Excel 2016 can be used to help with these calculations, and some spreadsheets developed in Excel will be shown in some chapters. However, some of the techniques presented in this book require sophisticated spreadsheets and are quite tedious to de- velop. Fortunately, there are two software programs available from the Companion Website for this book that make this much easier:

1. POM-QM for Windows is an easy-to-use decision support program that was developed for production and operations management (POM) and quantitative methods (QM) courses.

Deterministic means with complete certainty.

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POM for Windows and QM for Windows were originally separate software packages for each type of course. These are now combined into one program called POM-QM for Win- dows. As seen in Program 1.1, it is possible to display all the modules, only the POM mod- ules, or only the QM modules. The images shown in this textbook will typically display only the QM modules. Hence, in this book, reference will usually be made to QM for Windows. Appendix E at the end of the book provides more information about QM for Windows.

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

Select a module from the drop-down menu.

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

PROGRAM 1.1 The QM for windows Main Menu

Click Solve to run the program.

Enter the data.

PROGRAM 1.2A Entering the Data for Pritchett’s Precious Time Pieces Example into QM for windows

Additional output is available from the Window menu.

PROGRAM 1.2B QM for windows solution screen for Pritchett’s Precious Time Pieces Example

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

2. Excel QM, an add-in for Excel, can also be used to perform the mathematical calculations for the techniques discussed in each chapter. When installed in Excel 2016, Excel QM will appear as a tab on the ribbon. From this tab, the appropriate model can be selected from a menu, as shown in Program 1.3. Appendix F has more information about this. Excel files with the example problems shown can be downloaded from the Companion Website.

To use Excel QM in Excel 2016 to solve the break-even problem presented earlier, from the Alphabetical menu (see Program 1.3) select Breakeven Analysis. When this is done, a worksheet is prepared automatically, and the user simply inputs the fixed cost, variable cost, and revenue (selling price per unit), as shown in Program 1.4. The solution is calculated when all the inputs have been entered.

Excel 2016 contains some functions, special features, formulas, and tools that help with some of the questions that might be posed in analyzing a business problem. One such feature, Goal Seek, is shown in Program 1.5 as it is applied to the break-even example. Excel 2016 also has some add-ins that must be activated before using them the first time. These include the Data Analysis add-in and the Solver add-in, which will be discussed in later chapters.

Select the Excel QM tab.

Select the Alphabetical menu to see the techniques.

PROGRAM 1.3 Excel QM in Excel 2016 Ribbon and Menu of Techniques

The problem data are entered here.

The results are shown here.

PROGRAM 1.4 Entering the Data for Pritchett’s Precious Time Pieces Example into Excel QM in Excel 2016

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1.6 Possible Problems in the Quantitative Analysis Approach

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

Defining the Problem One view of decision makers is that they sit at a desk all day long, waiting until a problem arises, and then stand up and attack the problem until it is solved. Once it is solved, they sit down, relax, and wait for the next big problem. In the worlds of business, government, and education, prob- lems are, unfortunately, not easily identified. There are four potential roadblocks that quantita- tive analysts face in defining a problem. We use an application, inventory analysis, throughout this section as an example.

CONFLICTING VIEWPOINTS The first difficulty is that quantitative analysts must often consider conflicting viewpoints in defining the problem. For example, there are at least two views that

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

If the goal is $175 profit (B23) and this is obtained by changing the volume (B12), the Goal Seek window inputs are these.

PROGRAM 1.5 Using Goal seek in the Break-Even Problem to Achieve a specified Profit

Major league operations Research at the Department of Agriculture

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

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

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

Source: From “Ross Ohlendorf: From Major League Pitcher to Unpaid Intern,” by Rick Klein, © 2009, ABCnews.com.

IN ACTION

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

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

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

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

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

UNDERSTANDING THE MODEL A second major concern involves the trade-off between the complex- ity of the model and ease of understanding. Managers simply will not use the results of a model they do not understand. Complex problems, though, require complex models. One trade-off is to simplify assumptions in order to make the model easier to understand. The model loses some of its reality but gains some acceptance by management.

One simplifying assumption in inventory modeling is that demand is known and constant. This means that probability distributions are not needed and it allows us to build simple, easy- to-understand models. Demand, however, is rarely known and constant, so the model we build

All viewpoints should be considered before formally defining the problem.

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

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lacks some reality. Introducing probability distributions provides more realism but may put com- prehension beyond all but the most mathematically sophisticated managers. One approach is for the quantitative analyst to start with the simple model and make sure that it is completely understood. Later, more complex models can be introduced slowly as managers gain more confi- dence in using the new approach. Explaining the impact of the more sophisticated models (e.g., carrying extra inventory called safety stock) without going into complete mathematical details is sometimes helpful. Managers can understand and identify with this concept, even if the specific mathematics used to find the appropriate quantity of safety stock is not totally understood.

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

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

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

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

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

ONLY ONE ANSWER IS LIMITING The second problem is that quantitative models usually give just one answer to a problem. Most managers would like to have a range of options and not be put in a take-it-or-leave-it position. A more appropriate strategy is for an analyst to present a range of options, indicating the effect that each solution has on the objective function. This gives managers a choice, as well as information on how much it will cost to deviate from the optimal solution. It also allows problems to be viewed from a broader perspective, since nonquantitative factors can be considered.

Testing the Solution The results of quantitative analysis often take the form of predictions of how things will work in the future if certain changes are made now. To get a preview of how well solutions will really

Obtaining accurate input data can be very difficult.

Hard-to-understand mathematics and single solutions can become problematic in developing a solution.

1C. W. Churchman, “Reliability of Models in the Social Sciences,” Interfaces 4, 1 (November 1973): 1–12.

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1.7 IMPlEMEnTATIon—noT JUsT THE FInAl sTEP  15

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

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

1.7 Implementation—Not Just the Final Step

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

Assumptions should be reviewed.

PlATo Helps 2004 olympic Games in Athens

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

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

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

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

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

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

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

IN ACTION

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16  CHAPTER 1 • InTRoDUCTIon To QUAnTITATIvE AnAlysIs

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

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

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

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

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

Management support and user involvement are important.

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

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

Summary

Glossary

Algorithm A set of logical and mathematical operations per- formed in a specific sequence.

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

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

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

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

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

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

Model A representation of reality or of a real-life situation.

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sElF-TEsT  17

1. In analyzing a problem, you should normally study a. the qualitative aspects. b. the quantitative aspects. c. both a and b. d. neither a nor b. 2. Quantitative analysis is a. a logical approach to decision making. b. a rational approach to decision making. c. a scientific approach to decision making. d. all of the above. 3. Frederick Winslow Taylor a. was a world-class golfer and tennis player. b. pioneered the principles of scientific management. c. is known as the “Father of Industrial Engineering.” d. all of the above. 4. An input (such as variable cost per unit or fixed cost) for

a model is an example of a. a decision variable. b. a parameter. c. an algorithm. d. a stochastic variable. 5. The point at which total revenue equals total cost

(meaning zero profit) is called the

a. zero-profit solution. b. optimal-profit solution. c. break-even point. d. fixed-cost solution. 6. Quantitative analysis is typically associated with the use

of a. schematic models. b. physical models. c. mathematical models. d. scale models. 7. Sensitivity analysis is most often associated with which

step of the quantitative analysis approach? a. defining a problem b. acquiring input data c. implementing the results d. analyzing the results 8. A deterministic model is one in which a. there is some uncertainty about the parameters used in

the model. b. there is a measurable outcome. c. all parameters used in the model are known with com-

plete certainty. d. there is no available computer software.

Parameter A measurable input quantity that is inherent in a problem.

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

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

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

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

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

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

Variable A measurable quantity that is subject to change.

Key Equations

(1-1) Profit = sX - f - nX

where

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

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

(1-2) BEP = f

s - n

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

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

glossary at the end of the chapter. ●● Use the key at the back of the book (see Appendix H) to correct your answers. ●● Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

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18  CHAPTER 1 • InTRoDUCTIon To QUAnTITATIvE AnAlysIs

9. The term algorithm a. is named after Algorismus. b. is named after a ninth-century Persian mathematician. c. describes a series of steps or procedures to be repeated. d. all of the above. 10. An analysis to determine how much a solution would

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

a. sensitivity or postoptimality analysis. b. schematic or iconic analysis. c. futurama conditioning. d. both b and c.

11. Decision variables are a. controllable. b. uncontrollable. c. parameters. d. constant numerical values associated with any complex

problem. 12. Which of the following categories of business analytics

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

Discussion Questions and Problems

Discussion Questions 1-1 What is the difference between quantitative and

qualitative analysis? Give several examples. 1-2 Define quantitative analysis. What are some of the

organizations that support the use of the scientific approach?

1-3 What are the three categories of business analytics? 1-4 What is the quantitative analysis process? Give sev-

eral examples of this process. 1-5 Brief ly trace the history of quantitative analysis.

What happened to the development of quantitative analysis during World War II?

1-6 Give some examples of various types of models. What is a mathematical model? Develop two exam- ples of mathematical models.

1-7 List some sources of input data. 1-8 What is implementation, and why is it important? 1-9 Describe the use of sensitivity analysis and postopti-

mality analysis in analyzing the results. 1-10 Managers are quick to claim that quantitative ana-

lysts talk to them in a jargon that does not sound like English. List four terms that might not be un- derstood by a manager. Then explain in nontechnical language what each term means.

1-11 Why do you think many quantitative analysts don’t like to participate in the implementation process? What could be done to change this attitude?

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

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

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

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

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

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

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

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

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

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

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

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

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

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CAsE sTUDy  19

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

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

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

1-23 Golden Age Retirement Planners specializes in pro- viding financial advice for people planning for a comfortable retirement. The company offers semi- nars on the important topic of retirement planning. For a typical seminar, the room rental at a hotel is $1,000, and the cost of advertising and other inci- dentals is about $10,000 per seminar. The cost of the materials and special gifts for each attendee is $60 per person attending the seminar. The company charges $250 per person to attend the seminar, as this seems to be competitive with other companies in the same business. How many people must attend each seminar for Golden Age to break even?

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

cost $5 each. The tutor was paid $25 per hour, for a total of $75 for each tutoring session.

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

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

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

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

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

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

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

1-26 Bismarck Manufacturing intends to increase capac- ity through the addition of new equipment. Two vendors have presented proposals. The fixed cost for proposal A is $65,000 and for proposal B, $34,000. The variable cost for A is $10 and for B, $14. The revenue generated by each unit is $18.

(a) What is the break-even point for each proposal? (b) If the expected volume is 8,300 units, which al-

ternative should be chosen?

Southwestern University (SWU), a large state college in Ste- phenville, Texas, 30 miles southwest of the Dallas/Fort Worth metroplex, enrolls close to 20,000 students. The school is the dominant force in the small city, with more students during fall and spring than permanent residents.

A longtime football powerhouse, SWU is a member of the Big Eleven conference and is usually in the top 20 in college football rankings. To bolster its chances of reaching the elusive

and long-desired number-one ranking, in 2013 SWU hired the legendary Billy Bob Dillon as its head coach. Although the number-one ranking remained out of reach, attendance at the six Saturday home games each year increased. Prior to Dillon’s arrival, attendance generally averaged 25,000–29,000. Season ticket sales bumped up by 10,000 just with the announcement of the new coach’s arrival. Stephenville and SWU were ready to move to the big time!

Case Study

Food and Beverages at Southwestern University Football Games

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20  CHAPTER 1 • InTRoDUCTIon To QUAnTITATIvE AnAlysIs

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

Maddux’s fixed costs are interesting. He estimated that the prorated portion of the stadium cost would be as follows: salaries for food services at $300,000 ($60,000 for each of the six home games); 2,400 square feet of stadium space at $5 per square foot per game; and six people per booth in each of the six booths for 5 hours at $12 an hour. These fixed costs will be proportionately allocated to each of the products based on the percentages provided in the table. For example, the revenue from soft drinks would be expected to cover 25% of the total fixed costs.

ITEM SELLING

PRICE/UNIT VARIABLE COST/UNIT

PERCENT REVENUE

Soft drink $5.00 $1.50 25%

Coffee 4.00 1.00 25

Hot dogs 6.00 2.00 20

Hamburgers 7.50 3.00 20

Misc. snacks 2.00 1.00 10

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

Discussion Question 1. Prepare a brief report for Dr. Starr that covers the items

noted.

HEIZER, JAY; RENDER, BARRY, OPERATIONS MANAGEMENT, 6th ed., © 2001. Reprinted and Electronically reproduced by permission of Pearson Education, Inc., New York, NY.

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Churchman, C. W. “Reliability of Models in the Social Sciences,” Interfaces 4, 1 (November 1973): 1–12.

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 21

2.6 Understand the binomial distribution.

2.7 Understand the normal distribution and use the normal table.

2.8 Understand the F distribution. 2.9 Understand the exponential distribution and its

relation to queuing theory.

2.10 Understand the Poisson distribution and its relation to queuing theory.

2.1 Understand the basic foundations of probability analysis.

2.2 Use Bayes’ Theorem to establish posterior probabilities.

2.3 Use Bayes’ Theorem to establish further probability revisions.

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

2.5 Explain the difference between discrete and continuous probability distributions.

After completing this chapter, students will be able to:

LEARNING OBJECTIVES

Probability Concepts and Applications

2 CHAPTER

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

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

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

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

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22  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

2.1 Fundamental Concepts

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

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

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

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

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

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

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

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

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

Here is an example. Demand for white latex paint at Diversey Paint and Supply has always been 0, 1, 2, 3, or 4 gallons per day. Over the past 200 working days, the owner notes the frequencies of demand as shown in Table 2.2. If this past distribution is a good indicator of future sales, we can find the probability of each possible outcome occurring in the future by converting the data into percentages.

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

Objective probabilities can also be determined using what is called the classical or logi- cal method. Without performing a series of trials, we can often logically determine what the

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

CHAPTER TITLE

3 Decision Analysis

4 Regression Models

5 Forecasting

6 Inventory Control Models

11 Project Management

12 Waiting Lines and Queuing Theory Models

13 Simulation Modeling

14 Markov Analysis

15 Statistical Quality Control

Module 3 Decision Theory and the Normal Distribution

Module 4 Game Theory

TABLE 2.1 Chapters in This Book That Use Probability

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2.1 FUnDAmEnTAL ConCEPTs  23

probabilities of various events should be. For example, the probability of tossing a fair coin once and getting a head is

P1head2 = 1 2

d Number of ways of getting a head d Number of possible outcomes (head or tail)

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

P1spade2 = 13 52

d Number of chances of drawing a spade d Number of possible outcomes

= 1�4 = 0.25 = 25%

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

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

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

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

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

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

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

QUANTITY DEMANDED (GALLONS)

NUMBER OF DAYS

PROBABILITY

0 40 0.201= 40>2002 1 80 0.401= 80>2002 2 50 0.251= 50>2002 3 20 0.101= 20>2002 4 10 0.051= 10>2002

Total 200 1.001= 200>2002

TABLE 2.2 Relative Frequency Approach to Probability for Paint sales

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

BA

FIGURE 2.1 Venn Diagram for Events That Are mutually Exclusive

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24  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

BA

FIGURE 2.2 Venn Diagram for Events That Are not mutually Exclusive

Defining the Problem The scarcity of liver organs for transplants has reached critical levels in the United States. Every year about 1,500 people die waiting for a liver transplant (liverfoundation.org). There is a need to develop a model to evaluate policies for allocating livers to terminally ill patients who need them.

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

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

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

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

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

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

Source: Based on A. A. B. Pritsker, “Life and Death Decisions,” OR/MS Today 25, 4 (August 1998): 22–28, © Trevor S. Hale.

MODELING IN THE REAL WORLD

Liver Transplants in the United states

Defining the Problem

Developing a Model

Acquiring Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

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

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

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

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2.1 FUnDAmEnTAL ConCEPTs  25

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

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

The notation for the probability would be

P1intersection of event A and event B2 = P1A and B2 = P1A ¨ B2 = P1AB2

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

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

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

P1union of event A and event B2 = P1A or B2 = P1A ´ B2

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

(A and B) = the 7 of hearts is drawn

The notation for the probability would be

P1A and B2 = P1the 7 of hearts is drawn2 = 1�52 Also, the union of event A and event B would be

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

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

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

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

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

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26  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

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

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

P1A �B2 = P1AB2 P1B2 (2-3)

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

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

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

P1A �B2 = P1AB2 P1B2 =

1�52 13�52

= 1�13

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

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

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

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

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

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

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

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

P1A2 = 0.5

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2.2 REVising PRoBABiLiTiEs wiTH BAyEs’ THEoREm  27

and

P1B2 = 0.5 Because A and B are independent,

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

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

2.2 Revising Probabilities with Bayes’ Theorem

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

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

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

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

P1fair2 = 0.50 P1loaded2 = 0.50 We also know that

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

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

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

New Information

Bayes’ Process

Posterior Probabilities

Prior Probabilities

FIGURE 2.3 Using Bayes’ Process

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28  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

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

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

P1fair � 32 = P1fair and 32 P132 =

0.083

0.383 = 0.22

The probability that the die rolled was loaded is

P1loaded � 32 = P1loaded and 32 P132 =

0.300

0.383 = 0.78

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

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

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

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

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

where

A′ = the complement of the event A; for example, if A is the event “fair die,” then A′ is “loaded die”

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

P1A �B2 = P1AB2 P1B2

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

STATE OF NATURE

P(B �STATE OF NATURE)

PRIOR PROBABILITY

JOINT PROBABILITY

POSTERIOR PROBABILITY

A P(B �A) *P(A) =P(B and A) P(B and A)>P(B) = P(A �B) A′ P(B �A′) *P(A′) =P(B and A′)

P(B) P(B and A′)>P(B) = P(A′ �B)

TABLE 2.3 Tabular Form of Bayes’ Calculations given That Event B Has occurred

STATE OF NATURE

P(3 �STATE OF NATURE)

PRIOR PROBABILITY

JOINT PROBABILITY

POSTERIOR PROBABILITY

Fair die 0.166 *0.5 = 0.083 0.083>0.383 = 0.22 Loaded die 0.600 *0.5 = 0.300 0.300>0.383 = 0.78

P(3) = 0.383

TABLE 2.4 Bayes’ Calculations given That a 3 is Rolled in This Example

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

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

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2.3 FURTHER PRoBABiLiTy REVisions  29

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

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

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

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

P1fair die � 3 rolled2 = P13 � fair2P1fair2 P13 � fair2P1fair2 + P13 � loaded2P1loaded2

= 10.166210.502

10.166210.502 + 10.60210.502

= 0.083

0.383 = 0.22

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

2.3 Further Probability Revisions

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

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

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

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

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

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

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

= 0.013

0.193 = 0.067

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

= 0.18

0.193 = 0.933

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30  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

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

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

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

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

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

2.4 Random Variables

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

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

X = number of refrigerators sold during the day

Flight safety and Probability Analysis

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

Determining airline safety is a matter of applying the con- cepts of objective probability analysis. The chance of getting killed in a scheduled domestic flight is about 1 in 5 million. This is a probability of about 0.0000002. Another measure is the number of deaths per passenger mile flown. The number is about 1 passenger per billion passenger miles flown, or a prob- ability of about 0.000000001. Without question, flying is safer than many other forms of transportation, including driving. For a typical weekend, more people are killed in car accidents than a typical air disaster.

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

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

In addition, some proposed safety issues are not completely certain. For example, a Thermal Neutron Analysis device used to detect explosives at airports had a probability of 0.15 of giving a false alarm, resulting in a high cost of inspection and long flight delays. This would indicate that money should be spent on de- veloping more reliable equipment for detecting explosives. The result would be safer air travel with fewer unnecessary delays.

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

Sources: Based on Robert Machol, “Flying Scared,” OR/MS Today 23, 5 (October 1996): 32–37; and Arnold Barnett, “The Worst Day Ever,” OR/MS Today 28, 6 (December 2001): 28–31, © Trevor S. Hale.

IN ACTION

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2.4 RAnDom VARiABLEs  31

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

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

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

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

A continuous random variable is a random variable that has an infinite or an unlimited set of values. Are there any examples of continuous random variables in Table 2.5 or 2.6? Looking at Table 2.5, we can see that testing the lifetime of a lightbulb is an experiment whose results can be described with a continuous random variable. In this case, the random variable, S, is the time the bulb burns. It can last for 3,206 minutes, 6,500.7 minutes, 251.726 minutes, or any other value between 0 and 80,000 minutes. In most cases, the range of a continuous random variable is stated as lower value … S … upper value, such as 0 … S … 80,000. The random variable R in Table 2.5 is also continuous. Can you explain why?

TABLE 2.5 Examples of Random Variables

EXPERIMENT

OUTCOME

RANDOM VARIABLE

RANGE OF RANDOM VARIABLES

Stock 50 Christmas trees Number of Christmas trees sold X = number of Christmas trees sold 0, 1, 2, . . . , 50

Inspect 600 items Number of acceptable items Y = number of acceptable items 0, 1, 2, . . . , 600

Send out 5,000 sales letters Number of people responding to the letters

Z = number of people responding to the letters

0, 1, 2, . . . , 5,000

Build an apartment building Percent of building completed after 4 months

R = percent of building completed after 4 months

0 … R … 100

Test the lifetime of a lightbulb (minutes)

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

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

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

EXPERIMENT

OUTCOME

RANGE OF RANDOM VARIABLES

RANDOM VARIABLES

Students respond to a questionnaire

Strongly agree (SA)

Agree (A)

Neutral (N)

Disagree (D)

Strongly disagree (SD)

X = e 5 if SA4 if A3 if N 2 if D

1 if SD

1, 2, 3, 4, 5

One machine is inspected

Defective

Not defective Y = b 0 if defective1 if not defective 0, 1 Consumers respond to how they like a product

Good

Average

Poor Z = c 3 if good2 if average

1 if poor

1, 2, 3

TABLE 2.6 Random Variables for outcomes That Are not numbers

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32  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

2.5 Probability Distributions

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

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

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

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

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

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

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

E1X2 = a n

i=1 Xi P1Xi2

= X1P1X12 + X2P1X22 + Á + XnP1Xn2 (2-6) where

Xi = random variable’s possible values

P1Xi2 = probability of each possible value of the random variable

a n

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

E1X2 = expected value or mean of the random variable

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

RANDOM VARIABLE X (SCORE)

NUMBER

PROBABILITY P(X)

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

Total 100 1.0 = 100>100

TABLE 2.7 Probability Distribution for Quiz scores

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2.5 PRoBABiLiTy DisTRiBUTions  33

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

E1X2 = a 5

i=1 XiP1Xi2

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

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

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

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

s2 = Variance = a n

i=1 3Xi - E1X242P1Xi2 (2-7)

where

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

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

To compute the variance, each value of the random variable is subtracted from the expected value, squared, and multiplied times the probability of occurrence of that value. The results

0.4

0.3

0.2

0.1

0 1 2 3

X

P (X

)

4 5

FIGURE 2.4 Probability Distribution for Dr. shannon’s Class

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

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34  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

are then summed to obtain the variance. Here is how this procedure is done for Dr. Shannon’s quiz scores:

Variance = a 5

i=1 3Xi - E1X242P1Xi2

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

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

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

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

1 = square root s = standard deviation

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

s = 1Variance = 11.29 = 1.14

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

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

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

PROGRAM 2.1A Excel 2016 output for the Dr. shannon Example

PROGRAM 2.1B Formulas in an Excel spreadsheet for the Dr. shannon Example

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2.6 THE BinomiAL DisTRiBUTion  35

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

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

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

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

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

The inclusion or exclusion of either or both endpoints (5.22 or 5.26) has no impact on the probability.

In this section, we have investigated the fundamental characteristics and properties of prob- ability distributions in general. In the next five sections, we introduce three important continuous distributions—the normal distribution, the F distribution, and the exponential distribution—and two discrete distributions—the Poisson distribution and the binomial distribution.

2.6 The Binomial Distribution

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

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

2. The probability stays the same from one trial to the next.

P ro

ba bi

lit y

5.06 5.10 5.14 5.18 5.22 5.26 5.30 Weight (grams)

FIGURE 2.5 graph of sample Density Function

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

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36  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

3. The trials are statistically independent. 4. The number of trials is a positive integer.

A common example of this process is tossing a coin. The binomial distribution is used to find the probability of a specific number of successes

out of n trials of a Bernoulli process. To find this probability, it is necessary to know the following:

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

We let

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

The binomial formula is

Probability of r successes in n trials = n!

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

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

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

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

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

Thus,

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

40.55 - 4

= 5142132122112 413212211211!2 10.0625210.52 = 0.15625

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

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

NUMBER OF HEADS (r)

PROBABILITY = 5!

r!(5 − r)! (0.5)r(0.5)5− r

0 0.03125 =

5!

0!15 - 02!10.52 010.525 - 0

1 0.15625 =

5!

1!15 - 12!10.52 110.525 - 1

2 0.31250 =

5!

2!15 - 22!10.52 210.525 - 2

3 0.31250 =

5!

3!15 - 32!10.52 310.525 - 3

4 0.15625 =

5!

4!15 - 42!10.52 410.525 - 4

5 0.03125 =

5!

5!15 - 52!10.52 510.525 - 5

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

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2.6 THE BinomiAL DisTRiBUTion  37

TABLE 2.9 A sample Table for the Binomial Distribution

P

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

1 0 0.9500 0.9000 0.8500 0.8000 0.7500 0.7000 0.6500 0.6000 0.5500 0.5000 1 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000

2 0 0.9025 0.8100 0.7225 0.6400 0.5625 0.4900 0.4225 0.3600 0.3025 0.2500 1 0.0950 0.1800 0.2500 0.3200 0.3750 0.4200 0.4550 0.4800 0.4950 0.5000 2 0.0025 0.0100 0.0225 0.0400 0.0625 0.0900 0.1225 0.1600 0.2025 0.2500

3 0 0.8574 0.7290 0.6141 0.5120 0.4219 0.3430 0.2746 0.2160 0.1664 0.1250 1 0.1354 0.2430 0.3251 0.3840 0.4219 0.4410 0.4436 0.4320 0.4084 0.3750 2 0.0071 0.0270 0.0574 0.0960 0.1406 0.1890 0.2389 0.2880 0.3341 0.3750 3 0.0001 0.0010 0.0034 0.0080 0.0156 0.0270 0.0429 0.0640 0.0911 0.1250

4 0 0.8145 0.6561 0.5220 0.4096 0.3164 0.2401 0.1785 0.1296 0.0915 0.0625 1 0.1715 0.2916 0.3685 0.4096 0.4219 0.4116 0.3845 0.3456 0.2995 0.2500 2 0.0135 0.0486 0.0975 0.1536 0.2109 0.2646 0.3105 0.3456 0.3675 0.3750 3 0.0005 0.0036 0.0115 0.0256 0.0469 0.0756 0.1115 0.1536 0.2005 0.2500 4 0.0000 0.0001 0.0005 0.0016 0.0039 0.0081 0.0150 0.0256 0.0410 0.0625

5 0 0.7738 0.5905 0.4437 0.3277 0.2373 0.1681 0.1160 0.0778 0.0503 0.0313 1 0.2036 0.3281 0.3915 0.4096 0.3955 0.3602 0.3124 0.2592 0.2059 0.1563 2 0.0214 0.0729 0.1382 0.2048 0.2637 0.3087 0.3364 0.3456 0.3369 0.3125 3 0.0011 0.0081 0.0244 0.0512 0.0879 0.1323 0.1811 0.2304 0.2757 0.3125 4 0.0000 0.0005 0.0022 0.0064 0.0146 0.0284 0.0488 0.0768 0.1128 0.1563 5 0.0000 0.0000 0.0001 0.0003 0.0010 0.0024 0.0053 0.0102 0.0185 0.0313

6 0 0.7351 0.5314 0.3771 0.2621 0.1780 0.1176 0.0754 0.0467 0.0277 0.0156 1 0.2321 0.3543 0.3993 0.3932 0.3560 0.3025 0.2437 0.1866 0.1359 0.0938 2 0.0305 0.0984 0.1762 0.2458 0.2966 0.3241 0.3280 0.3110 0.2780 0.2344 3 0.0021 0.0146 0.0415 0.0819 0.1318 0.1852 0.2355 0.2765 0.3032 0.3125 4 0.0001 0.0012 0.0055 0.0154 0.0330 0.0595 0.0951 0.1382 0.1861 0.2344 5 0.0000 0.0001 0.0004 0.0015 0.0044 0.0102 0.0205 0.0369 0.0609 0.0938 6 0.0000 0.0000 0.0000 0.0001 0.0002 0.0007 0.0018 0.0041 0.0083 0.0156

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

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

0.4

0.3

0.2

0.1

0 1 2 3 4 5 6

P ro

ba bi

lit y,

P (r

)

Value of r (number of successes)

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

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38  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

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

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

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

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

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

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

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

2.7 The Normal Distribution

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

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

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

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

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

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

PROGRAM 2.2A Excel output for the Binomial Example

PROGRAM 2.2B Function in an Excel 2016 spreadsheet for Binomial Probabilities

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

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2.7 THE noRmAL DisTRiBUTion  39

Smaller m, same s

Larger m, same s

m 5 50 6040

50 60m 5 40

50 m 5 6040

FIGURE 2.7 normal Distribution with Different Values for m

Same m, smaller s

Same m, larger s

m

FIGURE 2.8 normal Distribution with Different Values for s

Probability Assessments of Curling Champions

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

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

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

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

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

IN ACTION

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40  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

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

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

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

Z = X - m

s (2-13)

where

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

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

Z = X - m

s =

130 - 100 15

= 30

15 = 2 standard deviations

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

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

55 70 85 100 115 130 145

–3 –2 –1 0 1 2 3 Z 5

X 5 IQ

x – m s

m

FIGURE 2.9 normal Distribution showing the Relationship Between Z Values and X Values

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2.7 THE noRmAL DisTRiBUTion  41

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

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

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

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

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

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

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

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

Z = X - m

s

= 125 - 100

20

= 25

20 = 1.25

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

To be sure you understand the concept of symmetry in Table 2.10, try to find a probability such as P(X < 85). Note that the standard normal table shows only positive Z values.

s 5 20 days

X 5 125 days

Z 5 1.25

m 5 100 days

P(X # 125)

0.89435

FIGURE 2.10 normal Distribution for Haynes Construction

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42  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

TABLE 2.10 standardized normal Distribution Function

AREA UNDER THE NORMAL CURVE

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

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

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

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

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

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

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

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

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

0.8 .78814 .79103 .79389 .79673 .79955 .80234 .80511 .80785 .81057 .81327

0.9 .81594 .81859 .82121 .82381 .82639 .82894 .83147 .83398 .83646 .83891

1.0 .84134 .84375 .84614 .84849 .85083 .85314 .85543 .85769 .85993 .86214

1.1 .86433 .86650 .86864 .87076 .87286 .87493 .87698 .87900 .88100 .88298

1.2 .88493 .88686 .88877 .89065 .89251 .89435 .89617 .89796 .89973 .90147

1.3 .90320 .90490 .90658 .90824 .90988 .91149 .91309 .91466 .91621 .91774

1.4 .91924 .92073 .92220 .92364 .92507 .92647 .92785 .92922 .93056 .93189

1.5 .93319 .93448 .93574 .93699 .93822 .93943 .94062 .94179 .94295 .94408

1.6 .94520 .94630 .94738 .94845 .94950 .95053 .95154 .95254 .95352 .95449

1.7 .95543 .95637 .95728 .95818 .95907 .95994 .96080 .96164 .96246 .96327

1.8 .96407 .96485 .96562 .96638 .96712 .96784 .96856 .96926 .96995 .97062

1.9 .97128 .97193 .97257 .97320 .97381 .97441 .97500 .97558 .97615 .97670

2.0 .97725 .97778 .97831 .97882 .97932 .97982 .98030 .98077 .98124 .98169

2.1 .98214 .98257 .98300 .98341 .98382 .98422 .98461 .98500 .98537 .98574

2.2 .98610 .98645 .98679 .98713 .98745 .98778 .98809 .98840 .98870 .98899

2.3 .98928 .98956 .98983 .99010 .99036 .99061 .99086 .99111 .99134 .99158

2.4 .99180 .99202 .99224 .99245 .99266 .99286 .99305 .99324 .99343 .99361

2.5 .99379 .99396 .99413 .99430 .99446 .99461 .99477 .99492 .99506 .99520

2.6 .99534 .99547 .99560 .99573 .99585 .99598 .99609 .99621 .99632 .99643

2.7 .99653 .99664 .99674 .99683 .99693 .99702 .99711 .99720 .99728 .99736

2.8 .99744 .99752 .99760 .99767 .99774 .99781 .99788 .99795 .99801 .99807

2.9 .99813 .99819 .99825 .99831 .99836 .99841 .99846 .99851 .99856 .99861

3.0 .99865 .99869 .99874 .99878 .99882 .99886 .99889 .99893 .99896 .99900

3.1 .99903 .99906 .99910 .99913 .99916 .99918 .99921 .99924 .99926 .99929

3.2 .99931 .99934 .99936 .99938 .99940 .99942 .99944 .99946 .99948 .99950

3.3 .99952 .99953 .99955 .99957 .99958 .99960 .99961 .99962 .99964 .99965

3.4 .99966 .99968 .99969 .99970 .99971 .99972 .99973 .99974 .99975 .99976

3.5 .99977 .99978 .99978 .99979 .99980 .99981 .99981 .99982 .99983 .99983

3.6 .99984 .99985 .99985 .99986 .99986 .99987 .99987 .99988 .99988 .99989

3.7 .99989 .99990 .99990 .99990 .99991 .99991 .99992 .99992 .99992 .99992

3.8 .99993 .99993 .99993 .99994 .99994 .99994 .99994 .99995 .99995 .99995

3.9 .99995 .99995 .99996 .99996 .99996 .99996 .99996 .99996 .99997 .99997

Source: Based on Richard I. Levin and Charles A. Kirkpatrick, Quantitative Approaches to Management, 4th ed. 1978, 1975, 1971, 1965 by McGraw-Hill, Inc., © Trevor S. Hale.

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2.7 THE noRmAL DisTRiBUTion  43

Now let us look at the Haynes problem from another perspective. If the firm finishes this triplex in 75 days or less, it will be awarded a bonus payment of $5,000. What is the probability that Haynes will receive the bonus?

Figure 2.11 illustrates the probability we are looking for in the shaded area. The first step is again to compute the Z value:

Z = X - m

s

= 75 - 100

20

= -25 20

= -1.25

This Z value indicates that 75 days is -1.25 standard deviations to the left of the mean. But the standard normal table is structured to handle only positive Z values. To solve this problem, we observe that the curve is symmetric. The probability that Haynes will finish in 75 days or less is equivalent to the probability that it will finish in more than 125 days. A moment ago (in Fig- ure 2.10) we found the probability that Haynes will finish in less than 125 days. That value is 0.89435. So the probability it takes more than 125 days is

P1X 7 1252 = 1.0 - P1X … 1252 = 1.0 - 0.89435 = 0.10565

Thus, the probability of completing the triplex in 75 days or less is 0.10565, or about 11%. One final example: What is the probability that the triplex will take between 110 and 125

days? We see in Figure 2.12 that

P1110 6 X 6 1252 = P1X … 1252 - P1X 6 1102 That is, the shaded area in the graph can be computed by finding the probability of completing the building in 125 days or less minus the probability of completing it in 110 days or less.

P(X # 75 days) Area of interest

X 5 75 days m 5 100 days

Z 5 21.25

0.89435

FIGURE 2.11 Probability That Haynes will Receive the Bonus by Finishing in 75 Days or Less

s 5 20 days

m 5 100 days

110 days

125 days

FIGURE 2.12 Probability That Haynes will Complete in 110 to 125 Days

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44  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

Recall that P1X … 125 days2 is equal to 0.89435. To find P1X 6 110 days2, we follow the two steps developed earlier:

1. Z = X - m

s =

110 - 100 20

= 10

20

= 0.5 standard deviations

2. From Table 2.10, the area for Z = 0.50 is 0.69146. So the probability the triplex can be completed in less than 110 days is 0.69146. Finally,

P1110 … X … 1252 = 0.89435 - 0.69146 = 0.20289 The probability that it will take between 110 and 125 days is about 20%. Programs 2.3A and 2.3B show how Excel can be used for this.

The Empirical Rule While the probability tables for the normal distribution can provide precise probabilities, many situations require less precision. The empirical rule was derived from the normal distribution and is an easy way to remember some basic information about normal distributions. The empiri- cal rule states that for a normal distribution

approximately 68% of the values will be within {1 standard deviation of the mean approximately 95% of the values will be within {2 standard deviations of the mean almost all (about 99.7%) of the values will be within {3 standard deviations of the mean

Figure 2.13 illustrates the empirical rule. The area from point a to point b in the first drawing represents the probability, approximately 68%, that the random variable will be within {1 stan- dard deviation of the mean. The middle drawing illustrates the probability, approximately 95%, that the random variable will be within {2 standard deviations of the mean. The last drawing illustrates the probability, about 99.7%, that the random variable will be within {3 standard deviations of the mean.

2.8 The F Distribution The F distribution is a continuous probability distribution that is helpful in testing hypotheses about variances. The F distribution will be used in Chapter 4 when regression models are tested for significance. Figure 2.14 provides a graph of the F distribution. As with a graph for any con- tinuous distribution, the area underneath the curve represents probability. Note that for a large value of F, the probability is very small.

The F statistic is the ratio of two sample variances from independent normal distributions. Every F distribution has two sets of degrees of freedom associated with it. One of the degrees of freedom is associated with the numerator of the ratio, and the other is associated with the de- nominator of the ratio. The degrees of freedom are based on the sample sizes used in calculating the numerator and denominator.

PROGRAM 2.3A Excel 2016 output for the normal Distribution Example

PROGRAM 2.3B Function in an Excel 2016 spreadsheet for the normal Distribution Example

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2.8 THE F DisTRiBUTion  45

Appendix D provides values of F associated with the upper tail of the distribution for cer- tain probabilities (denoted by a) and degrees of freedom for the numerator 1df12 and degrees of freedom for the denominator 1df22.

To find the F value that is associated with a particular probability and degrees of freedom, refer to Appendix D. The following notation will be used:

df1 = degrees of freedom for the numerator df2 = degrees of freedom for the denominator

Consider the following example:

df1 = 5 df2 = 6

a = 0.05

16% 16%

2.5% 2.5%

68%

a m b 2 1s 1 1s

95%

a m b

22s 12s

99.7%

a m b

23s 13s

0.15% 0.15%

FIGURE 2.13 Approximate Probabilities from the Empirical Rule

Fa

FIGURE 2.14 The F Distribution

Figure 2.13 is very important, and you should comprehend the meanings of ± 1, 2, and 3 standard deviation symmetrical areas.

Managers often speak of 95% and 99% confidence intervals, which roughly refer to ± 2 and 3 standard deviation graphs.

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46  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

From Appendix D, we get

Fa, df1, df2 = F0.05, 5, 6 = 4.39

This means

P1F 7 4.392 = 0.05 The probability is very low (only 5%) that the F value will exceed 4.39. There is a 95% prob- ability that it will not exceed 4.39. This is illustrated in Figure 2.15. Appendix D also provides F values associated with a = 0.01. Programs 2.4A and 2.4B illustrate Excel functions for the F distribution.

2.9 The Exponential Distribution

The exponential distribution, also called the negative exponential distribution, is used in deal- ing with queuing problems. The exponential distribution often describes the time required to service a customer. The exponential distribution is a continuous distribution. Its probability function is given by

f 1X2 = me-mx (2-14) where

X = random variable (service times) m = average number of units the service facility can handle in a specific period of time e = 2.718 (the base of the natural logarithm)

F = 4.39 0.05

FIGURE 2.15 F Value for 0.05 Probability with 5 and 6 Degrees of Freedom

PROGRAM 2.4A Excel 2016 output for the F Distribution

PROGRAM 2.4B Functions in an Excel 2016 spreadsheet for the F Distribution

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2.9 THE ExPonEnTiAL DisTRiBUTion  47

The general shape of the exponential distribution is shown in Figure 2.16. Its expected value and variance can be shown to be

Expected value = 1 m

= Average service time (2-15)

Variance = 1

m2 (2-16)

As with any other continuous distribution, probabilities are found by determining the area under the curve. For the normal distribution, we found the area by using a table of probabilities. For the exponential distribution, the probabilities can be found using the exponent key on a calcula- tor with the formula below. The probability that an exponentially distributed time, X, required to serve a customer is less than or equal to time t is given by the formula

P1X … t2 = 1 - e-mt (2-17) The time period used in describing m determines the units for the time t. For example, if m is the average number served per hour, the time t must be given in hours. If m is the average number served per minute, the time t must be given in minutes.

Arnold’s Muffler Example Arnold’s Muffler Shop installs new mufflers on automobiles and small trucks. The mechanic can install new mufflers at a rate of about three per hour, and this service time is exponentially distributed. What is the probability that the time to install a new muffler will be 1�2 hour or less? Using Equation 2-17, we have

X = exponentially distributed service time m = average number that can be served per time period = 3 per hour t = 1�2 hour = 0.5 hour

P1X … 0.52 = 1 - e-310.52 = 1 - e-1.5 = 1 - 0.2231 = 0.7769 Figure 2.17 shows the area under the curve from 0 to 0.5 to be 0.7769. Thus, there is about a 78% chance the time will be no more than 0.5 hour and about a 22% chance that the time will be longer than this. Similarly, we could find the probability that the service time is no more 1>3 hour or 2>3 hour, as follows:

P aX … 1 3 b = 1 - e-31132 = 1 - e-1 = 1 - 0.3679 = 0.6321

P aX … 2 3 b = 1 - e-31232 = 1 - e-2 = 1 - 0.1353 = 0.8647

f(X )

X

FIGURE 2.16 Exponential Distribution

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48  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

While Equation 2-17 provides the probability that the time (X) is less than or equal to a particu- lar value t, the probability that the time is greater than a particular value t is found by observing that these two events are complementary. For example, to find the probability that the mechanic at Arnold’s Muffler Shop would take longer than 0.5 hour, we have

P1X 7 0.52 = 1 - P1X … 0.52 = 1 - 0.7769 = 0.2231 Programs 2.5A and 2.5B illustrate how a function in Excel can find exponential probabilities.

2.10 The Poisson Distribution

An important discrete probability distribution is the Poisson distribution.1 We examine it be- cause of its key role in complementing the exponential distribution in queuing theory in Chap- ter 12. The distribution describes situations in which customers arrive independently during a certain time interval, and the number of arrivals depends on the length of the time interval. Ex- amples are patients arriving at a health clinic, customers arriving at a bank window, passengers arriving at an airport, and telephone calls going through a central exchange.

The formula for the Poisson distribution is

P1X2 = l xe-l

X! (2-18)

FIGURE 2.17 Probability That the mechanic will install a muffler in 0.5 Hour

PROGRAM 2.5B Function in an Excel 2016 spreadsheet for the Exponential Distribution

0 0

0.5

1

1.5

0.7769

P(service time # 0.5) 5 0.7769

2

2.5

0.5 1 1.5 2 2.5

PROGRAM 2.5A Excel 2016 output for the Exponential Distribution

The Poisson probability distribution is used in many queuing models to represent arrival patterns.

1This distribution, derived by Siméon Denis Poisson in 1837, is pronounced “pwah-sahn.”

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2.10 THE Poisson DisTRiBUTion  49

where

P1X2 = probability of exactly X arrivals or occurrences l = average number of arrivals per unit of time (the mean arrival rate), pronounced “lambda”

e = 2.718, the base of the natural logarithm X = number of occurrences 10, 1, 2, c2

The mean and variance of the Poisson distribution are equal and are computed simply as

Expected value = l (2-19)

Variance = l (2-20)

With the help of the table in Appendix C, the values of e-l are easy to find. We can use these in the formula to find probabilities. For example, if l = 2, from Appendix C we find e-2 = 0.1353. The Poisson probabilities that X is 0, 1, and 2 when l = 2 are as follows:

P1X2 = e -llx

X!

P102 = e -220

0! =

10.135321 1

= 0.1353 ≈ 14%

P112 = e -221

1! =

e-22

1 =

0.1353122 1

= 0.2706 ≈ 27%

P122 = e -222

2! =

e-24

2112 = 0.1353142

2 = 0.2706 ≈ 27%

These probabilities, as well as others for l = 2 and l = 4, are shown in Figure 2.18. No- tice that the chances that 9 or more customers will arrive in a particular time period are virtually nil. Programs 2.6A and 2.6B illustrate how Excel can be used to find Poisson probabilities.

It should be noted that the exponential and Poisson distributions are related. If the number of occurrences per time period follows a Poisson distribution, then the time between occurrences follows an exponential distribution. For example, if the number of phone calls arriving at a cus- tomer service center followed a Poisson distribution with a mean of 10 calls per hour, the time between each phone call would be exponentially distributed with a mean time between calls of 1>10 hour (6 minutes).

0.30

0.25

0.20

0.15

0.10

0.05

0.00 0 1 2 3 4 5

X

l 5 2 Distribution l 5 4 Distribution

P ro

ba bi

lit y

6 7 8 9

0.25

0.20

0.15

0.10

0.05

0.00 0 1 2 3 4 5

X

P ro

ba bi

lit y

6 7 8 9

FIGURE 2.18 sample Poisson Distributions with l = 2 and l = 4

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50  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

PROGRAM 2.6A Excel 2016 output for the Poisson Distribution

PROGRAM 2.6B Functions in an Excel 2016 spreadsheet for the Poisson Distribution

This chapter presents the fundamental concepts of probability and probability distributions. Probability values can be ob- tained objectively or subjectively. A single probability value must be between 0 and 1, and the sum of all probability val- ues for all possible outcomes must be equal to 1. In addition, probability values and events can have a number of properties. These properties include mutually exclusive, collectively ex- haustive, statistically independent, and statistically dependent events. Rules for computing probability values depend on these fundamental properties. It is also possible to revise probability values when new information becomes available. This can be done using Bayes’ Theorem.

We also covered the topics of random variables, discrete probability distributions (such as Poisson and binomial), and continuous probability distributions (such as normal, F, and exponential). A probability distribution is any statement of a probability function having a set of collectively exhaustive and mutually exclusive events. All probability distributions follow the basic probability rules mentioned previously.

The topics presented here will be very important in many of the chapters to come. Basic probability concepts and distri- butions are used for decision theory, inventory control, Markov analysis, project management, simulation, and statistical qual- ity control.

Summary

Glossary

Bayes’ Theorem A formula that is used to revise probabili- ties based on new information.

Bernoulli Process A process with two outcomes in each of a series of independent trials in which the probabilities of the outcomes do not change.

Binomial Distribution A discrete distribution that describes the number of successes in independent trials of a Bernoulli process.

Classical or Logical Method An objective way of assessing probabilities based on logic.

Collectively Exhaustive Events A collection of all possible outcomes of an experiment.

Conditional Probability The probability of one event occur- ring given that another has taken place.

Continuous Probability Distribution A probability distri- bution with a continuous random variable.

Continuous Random Variable A random variable that can assume an infinite or unlimited set of values.

Discrete Probability Distribution A probability distribution with a discrete random variable.

Discrete Random Variable A random variable that can only assume a finite or limited set of values.

Expected Value The (weighted) average of a probability distribution.

F Distribution A continuous probability distribution that is the ratio of the variances of samples from two independent normal distributions.

Independent Events The situation in which the occurrence of one event has no effect on the probability of occurrence of a second event.

Intersection The set of all outcomes that are common to both events.

Joint Probability The probability of events occurring to- gether (or one after the other).

Mutually Exclusive Events A situation in which only one of two or more events can occur on any given trial or experiment.

Negative Exponential Distribution A continuous probabil- ity distribution that describes the time between customer arrivals in a queuing situation.

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Normal Distribution A continuous bell-shaped distribution that is a function of two parameters, the mean and standard deviation of the distribution.

Objective Approach A method of determining probability values based on historical data or logic.

Poisson Distribution A discrete probability distribution used in queuing theory.

Prior Probability A probability value determined before new or additional information is obtained. It is sometimes called an a priori probability estimate.

Probability A statement about the likelihood of an event oc- curring. It is expressed as a numerical value between 0 and 1, inclusive.

Probability Density Function The mathematical function that describes a continuous probability distribution. It is represented by f(X).

Probability Distribution The set of all possible values of a random variable and their associated probabilities.

Random Variable A variable that assigns a number to every possible outcome of an experiment.

Relative Frequency Approach An objective way of deter- mining probabilities based on observing frequencies over a number of trials.

Revised or Posterior Probability A probability value that re- sults from new or revised information and prior probabilities.

Standard Deviation The square root of the variance. Subjective Approach A method of determining probability

values based on experience or judgment. Union The set of all outcomes that are contained in either of

these two events. Variance A measure of dispersion or spread of the probabil-

ity distribution.

Key Equations

(2-1) 0 … P1event2 … 1 A basic statement of probability.

(2-2) P1A or B2 = P1A2 + P1B2 - P1A and B2 Probability of the union of two events.

(2-3) P1A �B2 = P1AB2 P1B2

Conditional probability.

(2-4) P1AB2 = P1A �B2P1B2 Probability of the intersection of two events.

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

Bayes’ Theorem in general form.

(2-6) E1X2 = a n

i=1 XiP1Xi2

An equation that computes the expected value (mean) of a discrete probability distribution.

(2-7) s2 = Variance = a n

i=1 3Xi - E1X242P1Xi2

An equation that computes the variance of a discrete probability distribution.

(2-8) s = 1Variance = 2s2 An equation that computes the standard deviation from the variance.

(2-9) Probability of r successes in n trials = n!

r!1n - r2! p rqn - r

A formula that computes probabilities for the binomial probability distribution.

(2-10) Expected value 1mean2 = np The expected value of the binomial distribution.

(2-11) Variance = np11 - p2 The variance of the binomial distribution.

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

The density function for the normal probability distribution.

(2-13) Z = X - m

s

An equation that computes the number of standard deviations, Z, the point X is from the mean m.

(2-14) f 1X ) = me-mx

The exponential distribution.

(2-15) Expected value = 1 m

The expected value of an exponential distribution.

(2-16) Variance = 1

m2

The variance of an exponential distribution.

(2-17) P1X … t2 = 1 - e-mt

Formula to find the probability that an exponential ran- dom variable, X, is less than or equal to time t.

(2-18) P1X2 = l xe-l

X!

The Poisson distribution.

(2-19) Expected value = l

The mean of a Poisson distribution.

(2-20) Variance = l

The variance of a Poisson distribution.

KEy EQUATions  51

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52  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

Solved Problems

Solved Problem 2-1 In the past 30 days, Roger’s Rural Roundup has sold 8, 9, 10, or 11 lottery tickets. It never sold fewer than 8 or more than 11. Assuming that the past is similar to the future, find the probabilities for the number of tickets sold if sales were 8 tickets on 10 days, 9 tickets on 12 days, 10 tickets on 6 days, and 11 tickets on 2 days.

Solution

SALES NO. DAYS PROBABILITY

8 10 0.333

9 12 0.400

10 6 0.200

11 2 0.067

Total 30 1.000

Solved Problem 2-2 A class contains 30 students. Ten are female (F) and U.S. citizens (U); 12 are male (M) and U.S. citizens; 6 are female and non-U.S. citizens (N); 2 are male and non-U.S. citizens.

A name is randomly selected from the class roster, and it is female. What is the probability that the student is a U.S. citizen?

Solution

P1FU2 = 10>30 = 0.333 P1FN2 = 6>30 = 0.200

P1MU2 = 12>30 = 0.400 P1MN2 = 2>30 = 0.067

P1F2 = P1FU2 + P1FN2 = 0.333 + 0.200 = 0.533 P1M2 = P1MU2 + P1MN2 = 0.400 + 0.067 = 0.467 P1U2 = P1FU2 + P1MU2 = 0.333 + 0.400 = 0.733 P1N2 = P1FN2 + P1MN2 = 0.200 + 0.067 = 0.267

P1U �F2 = P1FU2 P1F2 =

0.333

0.533 = 0.625

Solved Problem 2-3 Your professor tells you that if you score an 85 or better on your midterm exam, then you have a 90% chance of getting an A for the course. You think you have only a 50% chance of scoring 85 or better. Find the probability that both your score is 85 or better and you receive an A in the course.

Solution P1A and 852 = P1A � 852 * P1852 = 10.90210.502

= 45%

Solved Problem 2-4 A statistics class was asked if it believed that all tests on the Monday following the football game win over their archrival should be postponed automatically. The results were as follows:

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soLVED PRoBLEms  53

Strongly agree 40

Agree 30

Neutral 20

Disagree 10

Strongly disagree 0

100

Transform this into a numeric score, using the following random variable scale, and find a prob- ability distribution for the results:

Strongly agree 5

Agree 4

Neutral 3

Disagree 2

Strongly disagree 1

Solution

OUTCOME PROBABILITY, P (X)

Strongly agree (5) 0.4 = 40>100 Agree (4) 0.3 = 30>100 Neutral (3) 0.2 = 20>100 Disagree (2) 0.1 = 10>100 Strongly disagree (1) 0.0 = 0>100 Total 1.0 = 100>100

Solved Problem 2-5 For Solved Problem 2-4, let X be the numeric score. Compute the expected value of X.

Solution

E1X2 = a 5

i=1 XiP1Xi2 = X1P1X12 + X2P1X22

+ X3P1X32 + X4P1X42 + X5P1X52 = 510.42 + 410.32 + 310.22 + 210.12 + 1102 = 4.0

Solved Problem 2-6 Compute the variance and standard deviation for the random variable X in Solved Problems 2-4 and 2-5.

Solution

Variance = a 5

i=1 1Xi - E1X222P1Xi2

= 15 - 42210.42 + 14 - 42210.32 + 13 - 42210.22 + 12 - 42210.12 + 11 - 42210.02 = 112210.42 + 102210.32 + 1-12210.22 + 1-22210.12 + 1-32210.02 = 0.4 + 0.0 + 0.2 + 0.4 + 0.0 = 1.0

The standard deviation is

s = 1Variance = 11 = 1 M02_REND3161_13_AIE_C02.indd 53 28/10/16 10:17 AM

Solved Problem 2-7 A candidate for public office has claimed that 60% of voters will vote for her. If 5 registered voters were sampled, what is the probability that exactly 3 would say they favor this candidate?

Solution We use the binomial distribution with n = 5, p = 0.6, and r = 3:

P1exactly 3 successes in 5 trials2 = n! r!1n - r2! p

rqn - r = 5!

3!15 - 32!10.62 310.425 - 3 = 0.3456

Solved Problem 2-8 The length of the rods coming out of our new cutting machine can be said to approximate a normal distribution with a mean of 10 inches and a standard deviation of 0.2 inch. Find the probability that a rod selected randomly will have a length

(a) of less than 10.0 inches. (b) between 10.0 and 10.4 inches. (c) between 10.0 and 10.1 inches. (d) between 10.1 and 10.4 inches. (e) between 9.6 and 9.9 inches. (f) between 9.9 and 10.4 inches. (g) between 9.886 and 10.406 inches.

Solution First, compute the standard normal distribution, the Z value:

Z = X - m

s

Next, find the area under the curve for the given Z value by using a standard normal distribution table.

(a) P1X 6 10.02 = 0.50000 (b) P110.0 6 X 6 10.42 = 0.97725 - 0.50000 = 0.47725 (c) P110.0 6 X 6 10.12 = 0.69146 - 0.50000 = 0.19146 (d) P110.1 6 X 6 10.42 = 0.97725 - 0.69146 = 0.28579 (e) P19.6 6 X 6 9.92 = 0.97725 - 0.69146 = 0.28579 (f) P19.9 6 X 6 10.42 = 0.19146 + 0.47725 = 0.66871 (g) P19.886 6 X 6 10.4062 = 0.47882 + 0.21566 = 0.69448

54  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

Self-Test

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

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

1. If only one event may occur on any one trial, then the events are said to be a. independent. b. exhaustive. c. mutually exclusive. d. continuous.

2. New probabilities that have been found using Bayes’ Theorem are called a. prior probabilities. b. posterior probabilities. c. Bayesian probabilities. d. joint probabilities.

3. A measure of central tendency is a. expected value. b. variance. c. standard deviation. d. all of the above.

4. To compute the variance of a discrete random variable, you need to know a. the variable’s possible values. b. the expected value of the variable. c. the probability of each possible value of the

variable. d. all of the above.

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DisCUssion QUEsTions AnD PRoBLEms  55

5. The square root of the variance is a. the expected value. b. the standard deviation. c. the area under the normal curve. d. all of the above.

6. Which of the following is an example of a discrete distribution? a. the normal distribution b. the exponential distribution c. the Poisson distribution d. the Z distribution

7. The total area under the curve for any continuous distribution must equal a. 1. b. 0. c. 0.5. d. none of the above.

8. Probabilities for all the possible values of a discrete random variable a. may be greater than 1. b. may be negative on some occasions. c. must sum to 1. d. are represented by the area underneath the curve.

9. In a standard normal distribution, the mean is equal to a. 1. b. 0. c. the variance. d. the standard deviation.

10. The probability of two or more independent events occurring is a. the marginal probability. b. the simple probability. c. the conditional probability. d. the joint probability. e. all of the above.

11. In the normal distribution, 95.45% of the population lies within a. 1 standard deviation of the mean. b. 2 standard deviations of the mean. c. 3 standard deviations of the mean. d. 4 standard deviations of the mean.

12. If a normal distribution has a mean of 200 and a standard deviation of 10, 99.7% of the population falls within what range of values? a. 170–230 b. 180–220 c. 190–210 d. 175–225 e. 170–220

13. If two events are mutually exclusive, then the probability of the intersection of these two events a. is 0. b. is 0.5. c. is 1.0. d. cannot be determined without more information.

14. If P1A2 = 0.4 and P1B2 = 0.5 and P1A and B2 = 0.2, then P1B �A2 = a. 0.80. b. 0.50. c. 0.10. d. 0.40. e. none of the above.

15. If P1A2 = 0.4 and P1B2 = 0.5 and P1A and B2 = 0.2, then P1A or B2 = a. 0.7. b. 0.9. c. 1.1. d. 0.2. e. none of the above.

Discussion Questions and Problems

Discussion Questions 2-1 What are the two basic laws of probability? 2-2 What is the meaning of mutually exclusive events?

What is meant by collectively exhaustive? Give an example of each.

2-3 Describe the various approaches used in determining probability values.

2-4 Why is the probability of the intersection of two events subtracted in the sum of the probability of two events?

2-5 Describe what it means for two events to be independent.

2-6 What is Bayes’ Theorem, and when can it be used? 2-7 Describe the characteristics of a Bernoulli process.

How is a Bernoulli process associated with the bino- mial distribution?

2-8 What is a random variable? What are the various types of random variables?

2-9 What is the difference between a discrete probability distribution and a continuous probability distribu- tion? Give your own example of each.

2-10 What is the expected value, and what does it mea- sure? How is it computed for a discrete probability distribution?

2-11 What is the variance, and what does it measure? How is it computed for a discrete probability distribution?

2-12 Name three business processes that can be described by the normal distribution.

2-13 A card is drawn from a standard deck of playing cards. For each of the following pairs of events, in- dicate if the events are mutually exclusive, and indi- cate if the events are exhaustive.

(a) Draw a spade and draw a club. (b) Draw a face card and draw a number card. (c) Draw an ace and draw a three. (d) Draw a red card and draw a black card.

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56  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

(e) Draw a five and draw a diamond. (f) Draw a red card and draw a diamond.

Problems 2-14 A student taking Management Science 301 at East

Haven University will receive one of the five pos- sible grades for the course: A, B, C, D, or F. The distribution of grades over the past 2 years is as follows:

GRADE NUMBER OF STUDENTS

A 80

B 75

C 90

D 30

F 25

Total 300

If this past distribution is a good indicator of future grades, what is the probability of a student receiving a C in the course?

2-15 A silver dollar is flipped twice. Calculate the prob- ability of each of the following occurring:

(a) a head on the first flip (b) a tail on the second flip given that the first toss

was a head (c) two tails (d) a tail on the first and a head on the second (e) a tail on the first and a head on the second or a

head on the first and a tail on the second (f) at least one head on the two flips

2-16 An urn contains 8 red chips, 10 green chips, and 2 white chips. A chip is drawn and replaced, and then a second chip drawn. What is the probability of

(a) a white chip on the first draw? (b) a white chip on the first draw and a red on the

second? (c) two green chips being drawn? (d) a red chip on the second given that a white chip

was drawn on the first?

2-17 Evertight, a leading manufacturer of quality nails, produces 1-, 2-, 3-, 4-, and 5-inch nails for various uses. In the production process, if there is an over- run or the nails are slightly defective, they are placed in a common bin. Yesterday, 651 of the 1-inch nails, 243 of the 2-inch nails, 41 of the 3-inch nails, 451 of the 4-inch nails, and 333 of the 5-inch nails were placed in the bin.

(a) What is the probability of reaching into the bin and getting a 4-inch nail?

(b) What is the probability of getting a 5-inch nail?

(c) If a particular application requires a nail that is 3 inches or shorter, what is the probability of get- ting a nail that will satisfy the requirements of the application?

2-18 Last year, at Northern Manufacturing Company, 200 people had colds during the year. One hundred fifty- five people who did no exercising had colds, and the remainder of the people with colds were involved in a weekly exercise program. Half of the 1,000 em- ployees were involved in some type of exercise.

(a) What is the probability that an employee will have a cold next year?

(b) Given that an employee is involved in an exer- cise program, what is the probability that he or she will get a cold next year?

(c) What is the probability that an employee who is not involved in an exercise program will get a cold next year?

(d) Are exercising and getting a cold independent events? Explain your answer.

2-19 The Springfield Kings, a professional basketball team, has won 12 of its last 20 games and is expected to con- tinue winning at the same percentage rate. The team’s ticket manager is anxious to attract a large crowd to tomorrow’s game but believes that depends on how well the Kings perform tonight against the Galveston Comets. He assesses the probability of drawing a large crowd to be 0.90 should the team win tonight. What is the probability that the team wins tonight and that there will be a large crowd at tomorrow’s game?

2-20 David Mashley teaches two undergraduate statistics courses at Kansas College. The class for Statistics 201 consists of 7 sophomores and 3 juniors. The more advanced course, Statistics 301, has 2 sopho- mores and 8 juniors enrolled. As an example of a business sampling technique, Professor Mashley ran- domly selects, from the stack of Statistics 201 regis- tration cards, the class card of one student and then places that card back in the stack. If that student was a sophomore, Mashley draws another card from the Statistics 201 stack; if not, he randomly draws a card from the Statistics 301 group. Are these two draws independent events? What is the probability of

(a) a junior’s name on the first draw? (b) a junior’s name on the second draw given that a

sophomore’s name was drawn first? (c) a junior’s name on the second draw given that a

junior’s name was drawn first? (d) a sophomore’s name on both draws? (e) a junior’s name on both draws? (f) one sophomore’s name and one junior’s name on

the two draws, regardless of order drawn?

2-21 The oasis outpost of Abu Ilan, in the heart of the Negev desert, has a population of 20 Bedouin

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

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

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DisCUssion QUEsTions AnD PRoBLEms  57

tribesmen and 20 Farima tribesmen. El Kamin, a nearby oasis, has a population of 32 Bedouins and 8 Farima. A lost Israeli soldier, accidentally sepa- rated from his army unit, is wandering through the desert and arrives at the edge of one of the oases. The soldier has no idea which oasis he has found, but the first person he spots at a distance is a Bed- ouin. What is the probability that he has wandered into Abu Ilan? What is the probability that he is in El Kamin?

2-22 The lost Israeli soldier mentioned in Problem 2-21 decides to rest for a few minutes before en- tering the desert oasis he has just found. Closing his eyes, he dozes off for 15 minutes, wakes, and walks toward the center of the oasis. The first per- son he spots this time he again recognizes as a Bedouin. What is the posterior probability that he is in El Kamin?

2-23 Ace Machine Works estimates that the probability its lathe tool is properly adjusted is 0.8. When the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. If the lathe is out of adjustment, however, the probability of a good part being produced is only 0.2. A part randomly chosen is inspected and found to be acceptable. At this point, what is the posterior probability that the lathe tool is properly adjusted?

2-24 The Boston South Fifth Street Softball League consists of three teams: Mama’s Boys, team 1; the Killers, team 2; and the Machos, team 3. Each team plays the other teams just once during the sea- son. The win–loss record for the past 5 years is as follows:

WINNER (1) (2) (3)

Mama’s Boys (1) X 3 4

The Killers (2) 2 X 1

The Machos (3) 1 4 X

Each row represents the number of wins over the past 5 years. Mama’s Boys beat the Killers 3 times, beat the Machos 4 times, and so on.

(a) What is the probability that the Killers will win every game next year?

(b) What is the probability that the Machos will win at least one game next year?

(c) What is the probability that Mama’s Boys will win exactly one game next year?

(d) What is the probability that the Killers will win fewer than two games next year?

2-25 The schedule for the Killers next year is as follows (refer to Problem 2-24):

Game 1: The Machos Game 2: Mama’s Boys

(a) What is the probability that the Killers will win their first game?

(b) What is the probability that the Killers will win their last game?

(c) What is the probability that the Killers will break even—win exactly one game?

(d) What is the probability that the Killers will win every game?

(e) What is the probability that the Killers will lose every game?

(f) Would you want to be the coach of the Killers?

2-26 The Northside rif le team has two markspersons, Dick and Sally. Dick hits a bull’s-eye 90% of the time, and Sally hits a bull’s-eye 95% of the time.

(a) What is the probability that either Dick or Sally or both will hit the bull’s-eye if each takes one shot?

(b) What is the probability that Dick and Sally will both hit the bull’s-eye?

(c) Did you make any assumptions in answering the preceding questions? If you answered yes, do you think that you are justified in making the assumption(s)?

2-27 In a sample of 1,000 representing a survey from the entire population, 650 people were from Laketown, and the rest of the people were from River City. Out of the sample, 19 people had some form of cancer. Thirteen of these people were from Laketown.

(a) Are the events of living in Laketown and having some sort of cancer independent?

(b) Which city would you prefer to live in, assum- ing that your main objective was to avoid having cancer?

2-28 Compute the probability of “loaded die given that a 3 was rolled,” as shown in the example in Section 2.3, this time using the general form of Bayes’ Theorem from Equation 2-5.

2-29 Which of the following are probability distributions? Why?

(a)

RANDOM VARIABLE X PROBABILITY

2 0.1

-1 0.2 0 0.3

1 0.25

2 0.15

(b)

RANDOM VARIABLE Y PROBABILITY

1 1.1

1.5 0.2

2 0.3

2.5 0.25

3 -1.25

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58  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

(c)

RANDOM VARIABLE Z PROBABILITY

1 0.1

2 0.2

3 0.3

4 0.4

5 0.0

2-30 Harrington Health Food stocks 5 loaves of Neutro- Bread. The probability distribution for the sales of Neutro-Bread is listed in the following table. How many loaves will Harrington sell on average?

NUMBER OF LOAVES SOLD PROBABILITY

0 0.05

1 0.15

2 0.20

3 0.25

4 0.20

5 0.15

2-31 What are the expected value and variance of the fol- lowing probability distribution?

RANDOM VARIABLE X PROBABILITY

1 0.05

2 0.05

3 0.10

4 0.10

5 0.15

6 0.15

7 0.25

8 0.15

2-32 There are 10 questions on a true–false test. A student feels unprepared for this test and randomly guesses the answer for each of these.

(a) What is the probability that the student gets exactly 7 correct?

(b) What is the probability that the student gets exactly 8 correct?

(c) What is the probability that the student gets exactly 9 correct?

(d) What is the probability that the student gets exactly 10 correct?

(e) What is the probability that the student gets more than 6 correct?

2-33 Gary Schwartz is the top salesman for his company. Records indicate that he makes a sale on 70% of his sales calls. If he calls on four potential clients, what is the probability that he makes exactly 3 sales? What is the probability that he makes exactly 4 sales?

2-34 If 10% of all disk drives produced on an assembly line are defective, what is the probability that there will be exactly one defect in a random sample of 5

of these? What is the probability that there will be no defects in a random sample of 5?

2-35 Trowbridge Manufacturing produces cases for per- sonal computers and other electronic equipment. The quality control inspector for this company be- lieves that a particular process is out of control. Nor- mally, only 5% of all cases are deemed defective due to discolorations. If 6 such cases are sampled, what is the probability that there will be 0 defective cases if the process is operating correctly? What is the probability that there will be exactly 1 defective case?

2-36 Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection pro- cedure is to select 6 items, and if there are 0 or 1 defective cases in the group of 6, the process is said to be in control. If the number of defects is more than 1, the process is out of control. Suppose that the true proportion of defective items is 0.15. What is the probability that there will be 0 or 1 defects in a sample of 6 if the true proportion of defects is 0.15?

2-37 An industrial oven used to cure sand cores for a fac- tory manufacturing engine blocks for small cars is able to maintain fairly constant temperatures. The temperature range of the oven follows a normal dis- tribution with a mean of 450°F and a standard devia- tion of 25°F. Leslie Larsen, president of the factory, is concerned about the large number of defective cores that have been produced in the past several months. If the oven gets hotter than 475°F, the core is defective. What is the probability that the oven will cause a core to be defective? What is the prob- ability that the temperature of the oven will range from 460° to 470°F?

2-38 Steve Goodman, production foreman for the Florida Gold Fruit Company, estimates that the average sale of oranges is 4,700 and the standard deviation is 500 oranges. Sales follow a normal distribution.

(a) What is the probability that sales will be greater than 5,500 oranges?

(b) What is the probability that sales will be greater than 4,500 oranges?

(c) What is the probability that sales will be less than 4,900 oranges?

(d) What is the probability that sales will be less than 4,300 oranges?

2-39 Susan Williams has been the production manager of Medical Suppliers, Inc., for the past 17 years. Medi- cal Suppliers, Inc., is a producer of bandages and arm slings. During the past 5 years, the demand for No-Stick bandages has been fairly constant. On aver- age, sales have been about 87,000 packages of No- Stick. Susan has reason to believe that the distribution of No-Stick follows a normal curve, with a standard de- viation of 4,000 packages. What is the probability that sales will be less than 81,000 packages?

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DisCUssion QUEsTions AnD PRoBLEms  59

2-40 Armstrong Faber produces a standard number-two pencil called Ultra-Lite. Since Chuck Armstrong started Armstrong Faber, sales have grown steadily. With the increase in the price of wood products, however, Chuck has been forced to increase the price of the Ultra-Lite pencils. As a result, the demand for Ultra-Lite has been fairly stable over the past 6 years. On average, Armstrong Faber has sold 457,000 pencils each year. Furthermore, 90% of the time sales have been between 454,000 and 460,000 pencils. It is expected that the sales follow a normal distribution with a mean of 457,000 pencils. Estimate the standard deviation of this distribution. (Hint: Work backward from the normal table to find Z. Then apply Equation 2-13.)

2-41 The time to complete a construction project is nor- mally distributed with a mean of 60 weeks and a standard deviation of 4 weeks.

(a) What is the probability the project will be fin- ished in 62 weeks or less?

(b) What is the probability the project will be fin- ished in 66 weeks or less?

(c) What is the probability the project will take lon- ger than 65 weeks?

2-42 A new integrated computer system is to be installed worldwide for a major corporation. Bids on this project are being solicited, and the contract will be awarded to one of the bidders. As a part of the pro- posal for this project, bidders must specify how long the project will take. There will be a significant pen- alty for finishing late. One potential contractor de- termines that the average time to complete a project of this type is 40 weeks with a standard deviation of 5 weeks. The time required to complete this project is assumed to be normally distributed.

(a) If the due date of this project is set at 40 weeks, what is the probability that the contractor will have to pay a penalty (i.e., the project will not be finished on schedule)?

(b) If the due date of this project is set at 43 weeks, what is the probability that the contractor will have to pay a penalty (i.e., the project will not be finished on schedule)?

(c) If the bidder wishes to set the due date in the pro- posal so that there is only a 5% chance of being late (and consequently only a 5% chance of having to pay a penalty), what due date should be set?

2-43 Patients arrive at the emergency room of Costa Val- ley Hospital at an average of 5 per day. The demand for emergency room treatment at Costa Valley fol- lows a Poisson distribution.

(a) Using Appendix C, compute the probability of exactly 0, 1, 2, 3, 4, and 5 arrivals per day.

(b) What is the sum of these probabilities, and why is the number less than 1?

2-44 Using the data in Problem 2-43, determine the prob- ability of more than 3 visits for emergency room ser- vice on any given day.

2-45 Cars arrive at Carla’s Muffler Shop for repair work at an average of 3 per hour, following an exponential distribution.

(a) What is the expected time between arrivals? (b) What is the variance of the time between

arrivals?

2-46 A particular test for the presence of steroids is to be used after a professional track meet. If steroids are present, the test will accurately indicate this 95% of the time. However, if steroids are not present, the test will indicate this 90% of the time (so it is wrong 10% of the time and predicts the presence of steroids). Based on past data, it is believed that 2% of the ath- letes do use steroids. This test is administered to one athlete, and the test is positive for steroids. What is the probability that this person actually used steroids?

2-47 Market Researchers, Inc., has been hired to perform a study to determine if the market for a new product will be good or poor. In similar studies performed in the past, whenever the market actually was good, the market research study indicated that it would be good 85% of the time. On the other hand, when- ever the market actually was poor, the market study incorrectly predicted it would be good 20% of the time. Before the study is performed, it is believed there is a 70% chance the market will be good. When Market Researchers, Inc., performs the study for this product, the results predict the market will be good. Given the results of this study, what is the probability that the market actually will be good?

2-48 Policy Pollsters is a market research firm specializ- ing in political polls. Records indicate in past elec- tions, when a candidate was elected, Policy Pollsters had accurately predicted this 80% of the time and was wrong 20% of the time. Records also show, for losing candidates, Policy Pollsters accurately pre- dicted they would lose 90% of the time and was wrong only 10% of the time. Before the poll is taken, there is a 50% chance of winning the election. If Policy Pollsters predicts a candidate will win the election, what is the probability that the candidate will actually win? If Policy Pollsters predicts that a candidate will lose the election, what is the probabil- ity that the candidate will actually lose?

2-49 Burger City is a large chain of fast-food restaurants specializing in gourmet hamburgers. A mathe- matical model is now used to predict the success of new restaurants based on location and demo- graphic information for that area. In the past, 70% of all restaurants that were opened were success- ful. The mathematical model has been tested in the existing restaurants to determine how effective it is. For the restaurants that were successful, 90%

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60  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

of the time the model predicted they would be, while 10% of the time the model predicted a fail- ure. For the restaurants that were not successful, when the mathematical model was applied, 20% of the time it incorrectly predicted a successful restaurant, while 80% of the time it was accurate and predicted an unsuccessful restaurant. If the model is used on a new location and predicts the restaurant will be successful, what is the probabil- ity that it actually is successful?

2-50 A mortgage lender attempted to increase its business by marketing its subprime mortgage. This mort- gage is designed for people with a less-than-perfect credit rating, and the interest rate is higher to offset the extra risk. In the past year, 20% of these mort- gages resulted in foreclosure as customers defaulted on their loans. A new screening system has been de- veloped to determine whether to approve customers for the subprime loans. When the system is applied to a credit application, the system will classify the application as “Approve for loan” or “Reject for loan.” When this new system was applied to recent customers who had defaulted on their loans, 90% of these customers were classified as “Reject.” When this same system was applied to recent loan custom- ers who had not defaulted on their loan payments, 70% of these customers were classified as “Approve for loan.”

(a) If a customer did not default on a loan, what is the probability that the rating system would have classified the applicant in the reject category?

(b) If the rating system had classified the applicant in the reject category, what is the probability that the customer would not default on a loan?

2-51 Use the F table in Appendix D to find the value of F for the upper 5% of the F distribution with

(a) df1 = 5, df2 = 10 (b) df1 = 8, df2 = 7 (c) df1 = 3, df2 = 5 (d) df1 = 10, df2 = 4

2-52 Use the F table in Appendix D to find the value of F for the upper 1% of the F distribution with

(a) df1 = 15, df2 = 6 (b) df1 = 12, df2 = 8 (c) df1 = 3, df2 = 5 (d) df1 = 9, df2 = 7

2-53 For each of the following F values, determine whether the probability indicated is greater than or less than 5%:

(a) P1F3,4 7 6.82 (b) P1F7,3 7 3.62 (c) P1F20,20 7 2.62 (d) P1F7,5 7 5.12 (e) P1F7,5 6 5.12

2-54 For each of the following F values, determine whether the probability indicated is greater than or less than 1%:

(a) P1F5,4 7 142 (b) P1F6,3 7 302 (c) P1F10,12 7 4.22 (d) P1F2,3 7 352 (e) P1F2,3 6 352

2-55 Nite Time Inn has a toll-free telephone number so that customers can call at any time to make a reser- vation. A typical call takes about 4 minutes to com- plete, and the time required follows an exponential distribution. Find the probability that a call takes

(a) 3 minutes or less. (b) 4 minutes or less. (c) 5 minutes or less. (d) longer than 5 minutes.

2-56 During normal business hours on the east coast, calls to the toll-free reservation number of the Nite Time Inn arrive at a rate of 5 per minute. It has been de- termined that the number of calls per minute can be described by the Poisson distribution. Find the prob- ability that in the next minute, the number of calls arriving will be

(a) exactly 5. (b) exactly 4. (c) exactly 3. (d) exactly 6. (e) less than 2.

2-57 In the Arnold’s Muffler example for the exponen- tial distribution in this chapter, the average rate of service was given as 3 per hour, and the times were expressed in hours. Convert the average service rate to the number per minute and convert the times to minutes. Find the probabilities that the service times will be less than 1>2 hour, 1>3 hour, and 2>3 hour. Com- pare these probabilities to the probabilities found in the example.

See our Internet home page, at www.pearsonhighered.com/render, for additional homework problems, Problems 2-58 to 2-65.

Internet Homework Problems

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APPEnDix 2.1: DERiVATion oF BAyEs’ THEoREm  61

WTVX, Channel 6, is located in Eugene, Oregon, home of the University of Oregon. The station was owned and oper- ated by George Wilcox, a former Duck (University of Or- egon football player). Although there were other television stations in Eugene, WTVX was the only station that had a weatherperson who was a member of the American Meteoro- logical Society (AMS). Every night, Joe Hummel would be introduced as the only weatherperson in Eugene who was a member of the AMS. This was George’s idea, and he believed that this gave his station the mark of quality and helped with market share.

In addition to being a member of the AMS, Joe was the most popular person on any of the local news programs. Joe was always trying to find innovative ways to make the weather interesting, and this was especially difficult during the winter months, when the weather seemed to remain the same over long periods of time. Joe’s forecast for next month, for example, was that there would be a 70% chance of rain every day and that what happens on one day (rain or shine) was not in any way dependent on what happened the day before.

One of Joe’s most popular features of the weather re- port was to invite questions during the actual broadcast.

Questions would be phoned in, and they were answered on the spot by Joe. Once a 10-year-old boy asked what caused fog, and Joe did an excellent job of describing some of the various causes.

Occasionally, Joe would make a mistake. For example, a high school senior asked Joe what the chances were of getting 15 days of rain in the next month (30 days). Joe made a quick calcu- lation: 170%2 * 115 days>30 days2 = 170%211>22 = 35,. Joe quickly found out what it was like being wrong in a univer- sity town. He had over 50 phone calls from scientists, mathema- ticians, and other university professors, telling him that he had made a big mistake in computing the chances of getting 15 days of rain during the next 30 days. Although Joe didn’t understand all of the formulas the professors mentioned, he was determined to find the correct answer and make a correction during a future broadcast.

Discussion Question 1. What are the chances of getting 15 days of rain during the

next 30 days? 2. What do you think about Joe’s assumptions concerning

the weather for the next 30 days?

Case Study

WTVX

Bibliography

Berenson, Mark, David Levine, and Timothy Krehbiel. Basic Business Statis- tics, 10th ed. Upper Saddle River, NJ: Prentice Hall, 2006.

Campbell, S. Flaws and Fallacies in Statistical Thinking. Upper Saddle River, NJ: Prentice Hall, 1974.

Feller, W. An Introduction to Probability Theory and Its Applications, Vols. 1 and 2. New York: John Wiley & Sons, Inc., 1957 and 1968.

Groebner, David, Patrick Shannon, Phillip Fry, and Kent Smith. Business Sta- tistics, 8th ed. Upper Saddle River, NJ: Prentice Hall, 2011.

Hanke, J. E., A. G. Reitsch, and D. W. Wichern. Business Forecasting, 9th ed. Upper Saddle River, NJ: Prentice Hall, 2008.

Huff, D. How to Lie with Statistics. New York: W. W. Norton & Company, Inc., 1954.

Newbold, Paul, William Carlson, and Betty Thorne. Statistics for Business and Economics, 6th ed. Upper Saddle River, NJ: Prentice Hall, 2007.

Appendix 2.1: Derivation of Bayes’ Theorem

We know that the following formulas are correct:

P1A �B2 = P1AB2 P1B2 (1)

P1B �A2 = P1AB2 P1A2 (2)

3which can be rewritten as P1AB2 = P1B �A2P1A24 and

P1B �A′2 = P1A′B2 P1A′2

3which can be rewritten as P1A′B2 = P1B �A′2P1A′24. (3)

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62  CHAPTER 2 • PRoBABiLiTy ConCEPTs AnD APPLiCATions

Furthermore, by definition, we know that

P1B2 = P1AB2 + P1A′B2 = P1B �A2P1A2 + P1B �A′2P1A′2 (4)

from (2) from (3)

Substituting Equations 2 and 4 into Equation 1, we have

P1A �B2 = P1AB2 P1B2

= P1B �A2P1A2

P1B �A2P1A2 + P1B �A′2P1A′2 (5)

This is the general form of Bayes’ Theorem, shown as Equation 2-5 in this chapter.

i from (2) from (4)

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 63

3.5 Use computers to solve basic decision-making problems.

3.6 Develop accurate and useful decision trees.

3.7 Revise probability estimates using Bayesian analysis.

3.8 Understand the importance and use of utility theory in decision making.

3.1 List the steps of the decision-making process.

3.2 Describe the types of decision-making environments.

3.3 Make decisions under uncertainty.

3.4 Use probability values to make decisions under risk.

After completing this chapter, students will be able to:

Decision Analysis

LEARNING OBJECTIVES

3 CHAPTER

To a great extent, the successes or failures that a person experiences in life depend on the decisions that he or she makes. The person who managed the ill-fated space shuttle Chal-lenger is no longer working for NASA. The person who designed the top-selling Mustang became president of Ford. Why and how did these people make their respective decisions? In general, what is involved in making good decisions? One decision may make the difference between a successful career and an unsuccessful one. Decision theory is an analytic and sys- tematic approach to the study of decision making. In this chapter, we present the mathematical models useful in helping managers make the best possible decisions.

What makes the difference between good and bad decisions? A good decision is one that is based on logic, considers all available data and possible alternatives, and applies the quantita- tive approach we are about to describe. Occasionally, a good decision results in an unexpected or unfavorable outcome. But if it is made properly, it is still a good decision. A bad decision is one that is not based on logic, does not use all available information, does not consider all al- ternatives, and does not employ appropriate quantitative techniques. If you make a bad decision but are lucky and a favorable outcome occurs, you have still made a bad decision. Although occasionally good decisions yield bad results, in the long run, using decision theory will result in successful outcomes.

3.1 The Six Steps in Decision Making

Whether you are deciding about getting a haircut today, building a multimillion-dollar plant, or buying a new camera, the steps in making a good decision are basically the same:

Decision theory is an analytic and systematic way to tackle problems.

A good decision is based on logic.

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64  CHAPTER 3 • DECision AnALysis

Six Steps in Decision Making

1. Clearly define the problem at hand. 2. List the possible alternatives. 3. Identify the possible outcomes or states of nature. 4. List the payoff (typically profit) of each combination of alternatives and outcomes. 5. Select one of the mathematical decision theory models. 6. Apply the model and make your decision.

We use the Thompson Lumber Company case as an example to illustrate these decision theory steps. John Thompson is the founder and president of Thompson Lumber Company, a profitable firm located in Portland, Oregon.

Step 1. The problem that John Thompson identifies is whether to expand his product line by manufacturing and marketing a new product, backyard storage sheds.

Step 2. Thompson’s second step is to generate the alternatives that are available to him. In deci- sion theory, an alternative is defined as a course of action or a strategy that the decision maker can choose. John decides that his alternatives are to construct (1) a large new plant to manufac- ture the storage sheds, (2) a small plant, or (3) no plant at all (i.e., he has the option of not devel- oping the new product line).

One of the biggest mistakes that decision makers make is to leave out some important alter- natives. Although a particular alternative may seem to be inappropriate or of little value, it might turn out to be the best choice.

The next step involves identifying the possible outcomes of the various alternatives. A com- mon mistake is to forget about some of the possible outcomes. Optimistic decision makers tend to ignore bad outcomes, whereas pessimistic managers may discount a favorable outcome. If you don’t consider all possibilities, you will not be making a logical decision, and the results may be undesirable. If you do not think the worst can happen, you may design another Edsel automobile. In decision theory, those outcomes over which the decision maker has little or no control are called states of nature.

Step 3. Thompson determines that there are only two possible outcomes: the market for the storage sheds could be favorable, meaning that there is a high demand for the product, or it could be unfavorable, meaning that there is a low demand for the sheds.

Once the alternatives and states of nature have been identified, the next step is to express the payoff resulting from each possible combination of alternatives and outcomes. In decision theory, we call such payoffs or profits conditional values. Not every decision, of course, can be based on money alone–-any appropriate means of measuring benefit is acceptable.

Step 4. Because Thompson wants to maximize his profits, he can use profit to evaluate each consequence.

John Thompson has already evaluated the potential profits associated with the various outcomes. With a favorable market, he thinks a large facility would result in a net profit of $200,000 to his firm. This $200,000 is a conditional value because Thompson’s receiving the money is conditional upon both his building a large factory and having a good market. The conditional value if the market is unfavorable would be a $180,000 net loss. A small plant would result in a net profit of $100,000 in a favorable market, but a net loss of $20,000 would occur if the market was unfavorable. Finally, doing nothing would result in $0 profit in either market. The easiest way to present these values is by constructing a decision table, sometimes called a payoff table. A decision table for Thompson’s conditional values is shown in Table 3.1. All of the alternatives are listed down the left side of the table, and all of the possible outcomes or states of nature are listed across the top. The body of the table contains the actual payoffs.

Steps 5 and 6. The last two steps are to select a decision theory model and apply it to the data to help make the decision. Selecting the model depends on the environment in which you’re oper- ating and the amount of risk and uncertainty involved.

The first step is to define the problem.

The second step is to list alternatives.

The third step is to identify possible outcomes.

The fourth step is to list payoffs.

During the fourth step, the decision maker can construct decision or payoff tables.

The last two steps are to select and apply the decision theory model.

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3.3 DECision MAKinG UnDER UnCERTAinTy  65

3.2 Types of Decision-Making Environments

The types of decisions people make depend on how much knowledge or information they have about the situation. There are three decision-making environments:

●● Decision making under certainty

●● Decision making under uncertainty

●● Decision making under risk

In the environment of decision making under certainty, decision makers know with cer- tainty the consequence of every alternative or decision choice. Naturally, they will choose the alternative that will maximize their well-being or will result in the best outcome. For example, let’s say that you have $1,000 to invest for a 1-year period. One alternative is to open a savings account paying 4% interest, and another is to invest in a government Treasury bond paying 6% interest. If both investments are secure and guaranteed, there is a certainty that the Treasury bond will pay a higher return. The return after 1 year will be $60 in interest.

In decision making under uncertainty, there are several possible outcomes for each al- ternative, and the decision maker does not know the probabilities of the various outcomes. As an example, the probability that a Democrat will be president of the United States 25 years from now is not known. Sometimes it is impossible to assess the probability of success of a new undertaking or product. The criteria for decision making under uncertainty are explained in Section 3.4.

In decision making under risk, there are several possible outcomes for each alternative, and the decision maker knows the probability of occurrence of each outcome. We know, for example, that when playing cards using a standard deck, the probability of being dealt a club is 0.25. The probability of rolling a 5 on a die is 1/6. In decision making under risk, the decision maker usually attempts to maximize his or her expected well-being. Decision theory models for business problems in this environment typically employ two equivalent criteria: maximization of expected monetary value and minimization of expected opportunity loss.

In the Thompson Lumber example, John Thompson is faced with decision making under uncertainty. If either a large plant or a small plant is constructed, the actual payoff depends on the state of nature, and probabilities are not known. If probabilities for a favorable market and for an unfavorable market were known, the environment would change from uncertainty to risk. For the third alternative, do nothing, the payoff does not depend on the state of nature and is known with certainty.

3.3 Decision Making Under Uncertainty

The presentation in this section of the criteria for decision making under uncertainty (and also for decision making under risk) is based on the assumption that the payoff is something in which larger values are better and high values are desirable. For payoffs such as profit, total sales, total return on investment, and interest earned, the best decision would be one that resulted in some type of maximum payoff. However, there are situations in which lower payoff values (e.g., cost) are better, and these payoffs would be minimized rather than maximized. The statement of the decision criteria would be modified slightly for such minimization problems. These differences will be mentioned in this section, and an example will be provided in a later section.

Probabilities are not known.

Probabilities are known.

TABLE 3.1 Decision Table with Conditional Values for Thompson Lumber

STATE OF NATURE

ALTERNATIVE FAVORABLE MARKET

($) UNFAVORABLE MARKET

($)

Construct a large plant 200,000 -180,000 Construct a small plant 100,000 -20,000 Do nothing 0 0

Note: It is important to include all alternatives, including “do nothing.”

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66  CHAPTER 3 • DECision AnALysis

STATE OF NATURE

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($)

MINIMUM IN A ROW

($)

Construct a large plant 200,000 -180,000 -180,000 Construct a small plant 100,000 -20,000 -20,000 Do nothing 0 0 0

Maximin

Several criteria exist for making decisions under conditions of uncertainty. The ones that we cover in this section are as follows:

1. Optimistic 2. Pessimistic 3. Criterion of realism (Hurwicz) 4. Equally likely (Laplace) 5. Minimax regret

The first four criteria can be computed directly from the decision (payoff) table, whereas the minimax regret criterion requires use of an opportunity loss table. Let’s take a look at each of the five models and apply them to the Thompson Lumber example.

Optimistic In using the optimistic criterion, the best (maximum) payoff for each alternative is considered, and the alternative with the best (maximum) of these is selected. Hence, the optimistic criterion is sometimes called the maximax criterion. In Table 3.2, we see that Thompson’s optimistic choice is the first alternative, “construct a large plant.” By using this criterion, the highest of all possible payoffs ($200,000 in this example) may be achieved, while if any other alternative were selected, it would be impossible to achieve a payoff this high.

In using the optimistic criterion for minimization problems in which lower payoffs (e.g., cost) are better, you would look at the best (minimum) payoff for each alternative and choose the alternative with the best (minimum) of these.

Pessimistic In using the pessimistic criterion, the worst (minimum) payoff for each alternative is considered, and the alternative with the best (maximum) of these is selected. Hence, the pessimistic criterion is sometimes called the maximin criterion. This criterion guarantees the payoff will be at least the maximin value (the best of the worst values). Choosing any other alternative may allow a worse (lower) payoff to occur.

Thompson’s maximin choice, “do nothing,” is shown in Table 3.3. This decision is associ- ated with the maximum of the minimum number within each row or alternative.

In using the pessimistic criterion for minimization problems in which lower payoffs (e.g., cost) are better, you would look at the worst (maximum) payoff for each alternative and choose the alternative with the best (minimum) of these.

Maximax is an optimistic approach.

Maximin is a pessimistic approach.

TABLE 3.2 Thompson’s Maximax Decision

STATE OF NATURE

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($)

MAXIMUM IN A ROW

($)

Construct a large plant 200,000 -180,000 200,000 Maximax

Construct a small plant 100,000 -20,000 100,000 Do nothing 0 0 0

TABLE 3.3 Thompson’s Maximin Decision

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3.3 Decision Making UnDer Uncertainty  67

Both the maximax and maximin criteria consider only one extreme payoff for each alterna- tive, while all other payoffs are ignored. The next criterion considers both of these extremes.

Criterion of Realism (Hurwicz Criterion) Often called the weighted average, the criterion of realism (the Hurwicz criterion) is a com- promise between an optimistic and a pessimistic decision. To begin, a coefficient of realism, A, is selected. This measures the degree of optimism of the decision maker and is between 0 and 1. When a is 1, the decision maker is 100% optimistic about the future. When a is 0, the decision maker is 100% pessimistic about the future. The advantage of this approach is that it allows the decision maker to build in personal feelings about relative optimism and pessimism. The weighted average is computed as follows:

Weighted average = a1Best in row2 + 11 - a21Worst in row2

For a maximization problem, the best payoff for an alternative is the highest value, and the worst payoff is the lowest value. Note that when a = 1, this is the same as the optimistic criterion, and when a = 0, this is the same as the pessimistic criterion. This value is computed for each alter- native, and the alternative with the highest weighted average is then chosen.

If we assume that John Thompson sets his coefficient of realism, a, to be 0.80, the best de- cision would be to construct a large plant. As seen in Table 3.4, this alternative has the highest weighted average: $124,000 = 10.8021$200,0002 + 10.2021-$180,0002.

In using the criterion of realism for minimization problems, the best payoff for an alterna- tive would be the lowest payoff in the row, and the worst would be the highest payoff in the row. The alternative with the lowest weighted average is then chosen.

Because there are only two states of nature in the Thompson Lumber example, only two payoffs for each alternative are present and both are considered. However, if there are more than two states of nature, this criterion will ignore all payoffs except the best and the worst. The next criterion will consider all possible payoffs for each decision.

Equally Likely (Laplace) One criterion that uses all the payoffs for each alternative is the equally likely, also called Laplace, decision criterion. This involves finding the average payoff for each alternative and selecting the alternative with the best or highest average. The equally likely approach assumes that all probabilities of occurrence for the states of nature are equal, and thus each state of nature is equally likely.

The equally likely choice for Thompson Lumber is the second alternative, “construct a small plant.” This strategy, shown in Table 3.5, is the one with the maximum average payoff.

In using the equally likely criterion for minimization problems, the calculations are exactly the same, but the best alternative is the one with the lowest average payoff.

Minimax Regret The next decision criterion that we discuss is based on opportunity loss or regret. Opportunity loss refers to the difference between the optimal profit or payoff for a given state of nature and the actual payoff received for a particular decision for that state of nature. In other words, it’s the amount lost by not picking the best alternative in a given state of nature.

Criterion of realism uses the weighted average approach.

Equally likely criterion uses the average outcome.

Minimax regret criterion is based on opportunity loss.

TABLE 3.4 thompson’s criterion of realism Decision

STATE OF NATURE

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($)

CRITERION OF REALISM OR WEIGHTED AVERAGE

(A = 0.8) ($)

Construct a large plant 200,000 -180,000 124,000 Realism

Construct a small plant 100,000 -20,000 76,000 Do nothing 0 0 0

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68  CHAPTER 3 • DECision AnALysis

The first step is to create the opportunity loss table by determining the opportunity loss for not choosing the best alternative for each state of nature. Opportunity loss for any state of nature, or any column, is calculated by subtracting each payoff in the column from the best payoff in the same column. For a favorable market, the best payoff is $200,000 as a result of the first alterna- tive, “construct a large plant.” The opportunity loss is 0, meaning that it is impossible to achieve a higher payoff in this state of nature. If the second alternative is selected, a profit of $100,000 would be realized in a favorable market, and this is compared to the best payoff of $200,000. Thus, the opportunity loss is 200,000 - 100,000 = 100,000. Similarly, if “do nothing” is se- lected, the opportunity loss would be 200,000 - 0 = 200,000.

For an unfavorable market, the best payoff is $0 as a result of the third alternative, “do noth- ing,” so this has 0 opportunity loss. The opportunity losses for the other alternatives are found by subtracting the payoffs from this best payoff ($0) in this state of nature, as shown in Table 3.6. Thompson’s opportunity loss table is shown as Table 3.7.

Using the opportunity loss (regret) table, the minimax regret criterion first considers the maximum (worst) opportunity loss for each alternative. Next, looking at these maximum values, pick the alternative with the minimum (or best) number. By doing this, the opportunity loss actually realized is guaranteed to be no more than this minimax value. In Table 3.8, we can see that the minimax regret choice is the second alternative, “construct a small plant.” When this al- ternative is selected, we know the maximum opportunity loss cannot be more than 100,000 (the minimum of the maximum regrets).

In calculating the opportunity loss for minimization problems such as those involving costs, the best (lowest) payoff or cost in a column is subtracted from each payoff in that column. Once the opportunity loss table has been constructed, the minimax regret criterion is applied in exactly the same way as just described. The maximum opportunity loss for each alternative is found, and the alternative with the minimum of these maximums is selected. As with maximiza- tion problems, the opportunity loss can never be negative.

We have considered several decision-making criteria to be used when probabilities of the states of nature are not known and cannot be estimated. Now we will see what to do if the prob- abilities are available.

TABLE 3.7 opportunity Loss Table for Thompson Lumber

STATE OF NATURE

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($)

Construct a large plant 0 180,000

Construct a small plant 100,000 20,000

Do nothing 200,000 0

TABLE 3.6 Determining opportunity Losses for Thompson Lumber

STATE OF NATURE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($)

200,000 – 200,000 0 – (–180,000)

200,000 – 100,000 0 – (–20,000)

200,000 – 0 0 – 0

TABLE 3.5 Thompson’s Equally Likely Decision

STATE OF NATURE

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($) ROW AVERAGE

($)

Construct a large plant 200,000 -180,000 10,000 Construct a small plant 100,000 -20,000 40,000

Equally likely

Do nothing 0 0 0

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3.4 DECision MAKinG UnDER RisK  69

3.4 Decision Making Under Risk

Decision making under risk is a decision situation in which several possible states of nature may occur and the probabilities of these states of nature are known. In this section, we consider one of the most popular methods of making decisions under risk: selecting the alternative with the highest expected monetary value (or simply expected value). We also use the probabilities with the opportunity loss table to minimize the expected opportunity loss.

Expected Monetary Value Given a decision table with conditional values (payoffs) that are monetary values and prob- ability assessments for all states of nature, it is possible to determine the expected monetary value (EMV) for each alternative. The expected value, or the mean value, is the long-run aver- age value of that decision. The EMV for an alternative is just the sum of possible payoffs of the alternative, each weighted by the probability of that payoff occurring.

This could also be expressed simply as the expected value of X, or E(X), which was discussed in Section 2.6 of Chapter 2.

EMV1alternative i2 = gXiP1Xi2 (3-1) where

Xi = payoff for the alternative in state of nature i P1Xi2 = probability of achieving payoff Xi (i.e., probability of state of nature i)

g = summation symbol If this were expanded, it would become

EMV 1alternative2 = 1Payoff in first state of nature2 * 1Probability of first state of nature2

+ 1Payoff in second state of nature2 * 1Probability of second state of nature2 + Á + 1Payoff in last state of nature2 * 1Probability of last state of nature2

EMV is the weighted sum of possible payoffs for each alternative.

Ford Uses Decision Theory to Choose Parts suppliers

Ford Motor Company manufactures about 5 million cars and trucks annually and employs more than 200,000 people at about 100 facilities around the globe. Such a large company often needs to make large supplier decisions under tight deadlines.

This was the situation when researchers at MIT teamed up with Ford management and developed a data-driven supplier se- lection tool. This computer program aids in decision making by applying some of the decision-making criteria presented in this

chapter. Decision makers at Ford are asked to input data about their suppliers (part costs, distances, lead times, supplier reliabil- ity, etc.), as well as the type of decision criterion they want to use. Once these are entered, the model outputs the best set of suppli- ers to meet the specified needs. The result is a system that is now saving Ford Motor Company over $40 million annually.

Source: Based on E. Klampfl, Y. Fradkin, C. McDaniel, and M. Wolcott, “Ford Uses OR to Make Urgent Sourcing Decisions in a Distressed Supplier Environ- ment,” Interfaces 39, 5 (2009): 428–442, © Trevor S. Hale.

IN ACTION

TABLE 3.8 Thompson’s Minimax Decision Using opportunity Loss

STATE OF NATURE

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($)

MAXIMUM IN A ROW

($)

Construct a large plant 0 180,000 180,000

Construct a small plant 100,000 20,000 100,000

Minimax

Do nothing 200,000 0 200,000

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70  CHAPTER 3 • DECision AnALysis

The alternative with the maximum EMV is then chosen. Suppose that John Thompson now believes that the probability of a favorable market is

exactly the same as the probability of an unfavorable market; that is, each state of nature has a 0.50 probability. Which alternative would give the greatest EMV? To determine this, John has expanded the decision table, as shown in Table 3.9. His calculations follow:

EMV 1large plant2 = 1$200,000210.502 + 1-$180,000210.502 = $10,000 EMV 1small plant2 = 1$100,000210.502 + 1-$20,000210.502 = $40,000 EMV 1do nothing2 = 1$0210.502 + 1$0210.502 = $0

The largest expected value ($40,000) results from the second alternative, “construct a small plant.” Thus, Thompson should proceed with the project and put up a small plant to manufacture storage sheds. The EMVs for constructing the large plant and for doing nothing are $10,000 and $0, respectively.

When using the EMV criterion with minimization problems, the calculations are the same, but the alternative with the smallest EMV is selected.

Expected Value of Perfect Information John Thompson has been approached by Scientific Marketing, Inc., a firm that proposes to help John make the decision about whether to build the plant to produce storage sheds. Scientific Marketing claims that its technical analysis will tell John with certainty whether the market is favorable for his proposed product. In other words, it will change his environment from one of decision making under risk to one of decision making under certainty. This information could prevent John from making a very expensive mistake. Scientific Marketing would charge Thomp- son $65,000 for the information. What would you recommend to John? Should he hire the firm to make the marketing study? Even if the information from the study is perfectly accurate, is it worth $65,000? What would it be worth? Although some of these questions are difficult to answer, determining the value of such perfect information can be very useful. It places an upper bound on what you should be willing to spend on information such as that being sold by Scien- tific Marketing. In this section, two related terms are investigated: the expected value of perfect information (EVPI) and the expected value with perfect information (EVwPI). These tech- niques can help John make his decision about hiring the marketing firm.

The expected value with perfect information is the expected or average return, in the long run, if we have perfect information before a decision has to be made. To calculate this value, we choose the best alternative for each state of nature and multiply its payoff times the probability of occurrence of that state of nature.

EVwPI = g1Best payoff in state of nature i21probability of state of nature i2 (3-2) If this were expanded, it would become

EVwPI

= 1Best payoff in first state of nature2 * 1Probability of first state of nature2 + 1Best payoff in second state of nature2 * 1Probability of second state of nature2 + Á + 1Best payoff in last state of nature2 * 1Probability of last state of nature2

EVPI places an upper bound on what to pay for information.

TABLE 3.9 Decision Table with Probabilities and EMVs for Thompson Lumber

STATE OF NATURE

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($) EMV ($)

Construct a large plant 200,000 –180,000 10,000

Construct a small plant 100,000 –20,000 40,000

Best EMV

Do nothing 0 0 0

Probabilities 0.50 0.50

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3.4 DECision MAKinG UnDER RisK  71

The EVPI is the expected value with perfect information minus the expected value without perfect information (i.e., the best or maximum EMV). Thus, the EVPI is the improvement in EMV that results from having perfect information.

EVPI = EVwPI - Best EMV (3-3)

By referring to Table 3.9, Thompson can calculate the maximum that he would pay for in- formation, that is, the EVPI. He follows a three-stage process. First, the best payoff in each state of nature is found. If the perfect information says the market will be favorable, the large plant will be constructed, and the profit will be $200,000. If the perfect information says the market will be unfavorable, the “do nothing” alternative is selected, and the profit will be 0. These val- ues are shown in the “with perfect information” row in Table 3.10. Second, the expected value with perfect information is computed. Then, using this result, EVPI is calculated.

The expected value with perfect information is

EVwPI = 1$200,000210.502 + 1$0210.502 = $100,000 Thus, if we had perfect information, the payoff would average $100,000.

The maximum EMV without additional information is $40,000 (from Table 3.9 ). Therefore, the increase in EMV is

EVPI = EVwPI - Maximum EMV = $100,000 - $40,000 = $60,000

Thus, the most Thompson would be willing to pay for perfect information is $60,000. This, of course, is again based on the assumption that the probability of each state of nature is 0.50.

This EVPI also tells us that the most we would pay for any information (perfect or im- perfect) is $60,000. In a later section, we’ll see how to place a value on imperfect or sample information.

In finding the EVPI for minimization problems, the approach is similar. The best payoff in each state of nature is found, but this is the lowest payoff for that state of nature rather than the highest. The EVwPI is calculated from these lowest payoffs, and this is compared to the best (lowest) EMV without perfect information. The EVPI is the improvement that results, and this is the best EMV - EVwPI.

Expected Opportunity Loss An alternative approach to maximizing EMV is to minimize expected opportunity loss (EOL). First, an opportunity loss table is constructed. Then the EOL is computed for each alternative by multiplying the opportunity loss by the probability and adding these together. In Table 3.7, we presented the opportunity loss table for the Thompson Lumber example. Using these opportu- nity losses, we compute the EOL for each alternative by multiplying the probability of each state of nature times the appropriate opportunity loss value and adding these together:

EVPI is the expected value with perfect information minus the maximum EMV.

EOL is the cost of not picking the best solution.

TABLE 3.10 Decision Table with Perfect information

STATE OF NATURE

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($) EMV ($)

Construct a large plant 200,000 –180,000 10,000

Construct a small plant 100,000 –20,000 40,000

Do nothing 0 0 0

With perfect information 200,000 0 100,000

EVwPI

Probabilities 0.50 0.50

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72  CHAPTER 3 • DECision AnALysis

EOL1construct large plant2 = 10.521$02 + 10.521$180,0002 = $90,000

EOL1construct small plant2 = 10.521$100,0002 + 10.521$20,0002 = $60,000

EOL1do nothing2 = 10.521$200,0002 + 10.521$02 = $100,000

Table 3.11 gives these results. Using minimum EOL as the decision criterion, the best decision would be the second alternative, “construct a small plant.”

It is important to note that minimum EOL will always result in the same decision as maxi- mum EMV and that the EVPI will always equal the minimum EOL. Referring to the Thompson case, we used the payoff table to compute the EVPI to be $60,000. Note that this is the minimum EOL we just computed.

Sensitivity Analysis In previous sections, we determined that the best decision (with the probabilities known) for Thompson Lumber was to construct the small plant, with an expected value of $40,000. This conclusion depends on the values of the economic consequences and the two probability val- ues of a favorable and an unfavorable market. Sensitivity analysis investigates how our decision might change given a change in the problem data. In this section, we investigate the impact that a change in the probability values would have on the decision facing Thompson Lumber. We first define the following variable:

P = probability of a favorable market

Because there are only two states of nature, the probability of an unfavorable market must be 1 - P.

We can now express the EMVs in terms of P, as shown in the following equations. A graph of these EMV values is shown in Figure 3.1.

EMV1large plant2 = $200,000P - $180,00011 - P2 = $200,000P - $180,000 + 180,000P = $380,000P - $180,000

EMV1small plant2 = $100,000P - $20,00011 - P2 = $100,000P - $20,000 + 20,000P = $120,000P - $20,000

EMV1do nothing2 = $0P + $011 - P2 = $0 As you can see in Figure 3.1, the best decision is to do nothing as long as P is between 0 and

the probability associated with point 1, where the EMV for doing nothing is equal to the EMV for the small plant. When P is between the probabilities for points 1 and 2, the best decision is to build the small plant. Point 2 is where the EMV for the small plant is equal to the EMV for the large plant. When P is greater than the probability for point 2, the best decision is to construct

EOL will always result in the same decision as the maximum EMV.

Sensitivity analysis investigates how our decision might change with different input data.

TABLE 3.11 EoL Table for Thompson Lumber

STATE OF NATURE

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($) EOL ($)

Construct a large plant 0 180,000 90,000

Construct a small plant 100,000 20,000 60,000

Best EOL

Do nothing 200,000 0 100,000

Probabilities 0.50 0.50

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3.4 DECision MAKinG UnDER RisK  73

the large plant. Of course, this is what you would expect as P increases. The value of P at points 1 and 2 can be computed as follows:

Point 1: EMV 1do nothing2 = EMV 1small plant2 0 = $120,000P - $20,000 P =

20,000

120,000 = 0.167

Point 2: EMV 1small plant2 = EMV 1large plant2 $120,000P - $20,000 = $380,000P - $180,000

260,000P = 160,000 P = 160,000

260,000 = 0.615

The results of this sensitivity analysis are displayed in the following table:

BEST ALTERNATIVE RANGE OF P VALUES

Do nothing Less than 0.167

Construct a small plant 0.167 – 0.615

Construct a large plant Greater than 0.615

A Minimization Example The previous examples have illustrated how to apply the decision-making criterion when the payoffs are to be maximized. The following example illustrates how the criteria are applied to problems in which the payoffs are costs that should be minimized.

The Business Analytics department at State University will be signing a 3-year lease for a new copy machine, and three different machines are being considered. For each of these, there is a monthly fee, which includes service on the machine, plus a charge for each copy. The number of copies that would be made each month is uncertain, but the department has estimated that the number of copies per month could be 10,000 or 20,000 or 30,000 per month. The monthly cost for each machine based on each of the three levels of activity is shown in Table 3.12.

Which machine should be selected? To determine the best alternative, a specific criterion must be chosen. If the decision maker is optimistic, only the best (minimum) payoff for each decision is considered. These are shown in Table 3.13, and the best (minimum) of these is 700. Thus, Machine C would be selected, allowing the possibility of achieving this best cost of $700.

If the decision maker is pessimistic, only the worst (maximum) payoff for each decision is considered. These are also shown in Table 3.13, and the best (minimum) of these is 1,150. Thus, Machine A would be selected based on the pessimistic criterion. This would guarantee that the cost would be no more than 1,150, regardless of which state of nature occurred.

0

EMV (large plant)

EMV (small plant)

EMV (do nothing)

Value of P

Point 2

Point 1

EMV Value

$300,000

$200,000

$100,000

2$100,000

2$200,000

1.615.167

FIGURE 3.1 sensitivity Analysis

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74  CHAPTER 3 • DECision AnALysis

Using the Hurwicz criterion, if we assume that the decision maker is 70% optimistic (the coefficient of realism is 0.7), the weighted average of the best and the worst payoff for each alternative would be calculated using the formula

Weighted average = 0.71Best payoff2 + 11 - 0.721Worst payoff2 For each of the three copy machines, we would get the following values:

Machine A: 0.719502 + 0.311,1502 = 1,010 Machine B: 0.718502 + 0.311,3502 = 1,000 Machine C: 0.717002 + 0.311,3002 = 880

The decision would be to select Machine C based on this criterion because it has the lowest weighted average cost.

For the equally likely criterion, the average payoff for each machine would be calculated.

Machine A: 1950 + 1,050 + 1,1502>3 = 1,050 Machine B: 1850 + 1,100 + 1,3502>3 = 1,100 Machine C: 1700 + 1,000 + 1,3002>3 = 1,000

Based on the equally likely criterion, Machine C would be selected because it has the lowest average cost.

To apply the EMV criterion, probabilities must be known for each state of nature. Past re- cords indicate that 40% of the time the number of copies made in a month was 10,000, while 30% of the time it was 20,000 and 30% of the time it was 30,000. The probabilities for the three states of nature would be 0.4, 0.3, and 0.3. We can use these to calculate the EMVs, and the results are shown in Table 3.14. Machine C would be selected because it has the lowest EMV. The monthly cost would average $970 with this machine, while the other machines would aver- age a higher cost.

To find the EVPI, we first find the payoffs (costs) that would be experienced with perfect information. The best payoff in each state of nature is the lowest value (cost) in that state of nature, as shown in the bottom row of Table 3.14. These values are used to calculate the EVwPI. With these costs, we find

EVwPI = $925 Best EMV without perfect information = $970

EVPI = 970 - 925 = $45

Perfect information would lower the expected value by $45.

TABLE 3.12 Payoff Table with Monthly Copy Costs for Business Analytics Department

10,000 COPIES PER

MONTH

20,000 COPIES PER

MONTH

30,000 COPIES PER

MONTH

Machine A 950 1,050 1,150

Machine B 850 1,100 1,350

Machine C 700 1,000 1,300

TABLE 3.13 Best and Worst Payoffs (Costs) for Business Analytics Department

10,000 COPIES PER

MONTH

20,000 COPIES PER

MONTH

30,000 COPIES PER

MONTH

BEST PAYOFF

(MINIMUM)

WORST PAYOFF

(MAXIMUM)

Machine A 950 1,050 1,150 950 1,150

Machine B 850 1,100 1,350 850 1,350

Machine C 700 1,000 1,300 700 1,300

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3.5 UsinG soFTWARE FoR PAyoFF TABLE PRoBLEMs  75

To apply criteria based on opportunity loss, we must first develop the opportunity loss table. In each state of nature, the opportunity loss indicates how much worse each payoff is than the best possible payoff in that state of nature. The best payoff (cost) would be the lowest cost. Thus, to get the opportunity loss in this case, we subtract the lowest value in each column from all the values in that column, and we obtain the opportunity loss table.

Once the opportunity loss table has been developed, the minimax regret criterion is applied exactly as it was for the Thompson Lumber example. The maximum regret for each alternative is found, and the alternative with the minimum of these maximums is selected. As seen in Table 3.15, the minimum of these maximums is 150, so Machine C would be selected based on this criterion.

The probabilities are used to compute the expected opportunity losses as shown in Table 3.15. Machine C has the lowest EOL of $45, so it would be selected based on the minimum EOL criterion. As previously noted, the minimum EOL is equal to the expected value of perfect information.

3.5 Using Software for Payoff Table Problems

It is easy to use QM for Windows or Excel QM or even Excel 2016 without any add-ins to per- form the calculations associated with payoff tables. The Thompson Lumber example is used in the following examples for illustration.

QM for Windows To use QM for Windows for the Thompson Lumber example, select Module – Decision Theory. Then enter New-Payoff Tables, and a window appears allowing you to set up the problem. Enter a title, the number of options (alternatives), and the number of scenarios (states of nature), and specify if it is to be maximized or minimized. In this example, there are three alternatives or options and two states of nature or scenarios. When these are entered, click OK and a window appears (Program 3.1A ) allowing you to enter the probabilities and payoffs. You can also change the names of the alternatives and states of nature by simply typing over the default names. When you click Solve, the output in Program 3.1B appears. Additional output such as EVPI and oppor- tunity loss results are available by clicking Window on the screen where the output is displayed.

Excel QM To use Excel QM for the Thompson Lumber example, from the Excel QM ribbon in Excel 2016, click the Alphabetical menu and select Decision Analysis – Decision Tables. When the window

TABLE 3.14 Expected Monetary Values and Expected Values with Perfect information for Business Analytics Department

10,000 COPIES PER

MONTH

20,000 COPIES PER

MONTH

30,000 COPIES PER

MONTH EMV

Machine A 950 1,050 1,150 1,040

Machine B 850 1,100 1,350 1,075

Machine C 700 1,000 1,300 970

With perfect information 700 1,000 1,150 925

Probability 0.4 0.3 0.3

TABLE 3.15 opportunity Loss Table for Business Analytics Department

10,000 COPIES PER

MONTH

20,000 COPIES PER

MONTH

30,000 COPIES PER

MONTH MAXIMUM EOL

Machine A 250 50 0 250 115

Machine B 150 100 200 200 150

Machine C 0 0 150 150 45

Probability 0.4 0.3 0.3

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76  CHAPTER 3 • DECision AnALysis

opens to enter the parameters of the problem, you should enter a title, the number of alternatives, and the number of states of nature and specify if it is to be maximized or minimized. Click OK and an empty payoff table is created. You simply enter the probabilities, payoffs, and names for the rows and columns. The calculations are performed automatically, and the results are shown in Program 3.2A. While the formulas are automatically developed in Excel QM, Program 3.2B shows the important formulas in this example. To see the formulas in Excel, simply hold down the control key and press the grave accent ( ̀ ) key (usually found above the Tab key). To return the Excel 2016 spreadsheet display to the numbers instead of the formulas, simply press the keys again.

Enter data into table and type the row and column names.

PROGRAM 3.1A QM for Windows input for Thompson Lumber Example

Click Window to see more information.

PROGRAM 3.1B QM for Windows output screen for Thompson Lumber Example

To see the formulas, hold down the control key (Ctrl) and press the ` (grave accent) key, which is usually found above the Tab key.

PROGRAM 3.2A Excel QM Results for Thompson Lumber Example

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3.6 DECision TREEs  77

3.6 Decision Trees

Any problem that can be presented in a decision table can also be graphically illustrated in a decision tree. All decision trees are similar in that they contain decision nodes or decision points and state-of-nature nodes or state-of-nature points:

●● A decision node from which one of several alternatives may be chosen

●● A state-of-nature node out of which one state of nature will occur

In drawing the tree, we begin at the left and move to the right. Thus, the tree presents the decisions and outcomes in sequential order. Lines or branches from the squares (decision nodes) represent alternatives, and branches from the circles represent the states of nature. Figure 3.2 gives the basic decision tree for the Thompson Lumber example. First, John decides whether to construct a large plant, a small plant, or no plant. Then, once that decision is made, the possible states of nature or outcomes (favorable or unfavorable market) will occur. The next step is to put the payoffs and probabilities on the tree and begin the analysis.

Analyzing problems with decision trees involves five steps:

Five Steps of Decision Tree Analysis

1. Define the problem. 2. Structure or draw the decision tree. 3. Assign probabilities to the states of nature. 4. Estimate payoffs for each possible combination of alternatives and states of nature. 5. Solve the problem by computing EMVs for each state-of-nature node. This is done by

working backward, that is, starting at the right of the tree and working back to decision nodes on the left. Also, at each decision node, the alternative with the best EMV is selected.

PROGRAM 3.2B Key Formulas in Excel QM for Thompson Lumber Example

A Decision Node A State-of-Nature Node

Favorable Market

Unfavorable Market

Favorable Market

Unfavorable Market

Construct Small Plant

Co nst

ruc t

Lar ge

Pla nt

Do Nothing

1

2

FIGURE 3.2 Thompson’s Decision Tree

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78  CHAPTER 3 • DECision AnALysis

from the market survey. Of course, you would expect to find a high probability of a favorable market given that the research indicated that the market was good. Don’t forget, though, there is a chance that John’s $10,000 market survey didn’t result in perfect or even reliable information. Any market research study is subject to error. In this case, there is a 22% chance that the market for sheds will be unfavorable given that the survey results are positive.

We note that there is a 27% chance that the market for sheds will be favorable given that John’s survey results are negative. The probability is much higher, 0.73, that the market will actually be unfavorable given that the survey was negative.

Finally, when we look to the payoff column in Figure 3.4, we see that $10,000, the cost of the marketing study, had to be subtracted from each of the top 10 tree branches. Thus, a large plant with a favorable market would normally net a $200,000 profit. But because the market study was conducted, this figure is reduced by $10,000 to $190,000. In the unfavorable case, the loss of $180,000 would increase to a greater loss of $190,000. Similarly, conducting the survey and building no plant now results in a -$10,000 payoff.

With all probabilities and payoffs specified, we can start calculating the EMV at each state- of-nature node. We begin at the end, or right side of the decision tree, and work back toward the origin. When we finish, the best decision will be known.

Most of the probabilities are conditional probabilities.

The cost of the survey had to be subtracted from the original payoffs.

We start by computing the EMV of each branch.

The final decision tree with the payoffs and probabilities for John Thompson’s decision situation is shown in Figure 3.3. Note that the payoffs are placed at the right side of each of the tree’s branches. The probabilities are shown in parentheses next to each state of nature. Begin- ning with the payoffs on the right of the figure, the EMVs for each state-of-nature node are then calculated and placed by their respective nodes. The EMV of the first node is $10,000. This represents the branch from the decision node to construct a large plant. The EMV for node 2, to construct a small plant, is $40,000. Building no plant or doing nothing has, of course, a payoff of $0. The branch leaving the decision node leading to the state-of-nature node with the highest EMV should be chosen. In Thompson’s case, a small plant should be built.

A MORE COMPLEX DECISION FOR THOMPSON LUMBER—SAMPLE INFORMATION When sequential deci- sions need to be made, decision trees are much more powerful tools than decision tables. Let’s say that John Thompson has two decisions to make, with the second decision dependent on the outcome of the first. Before deciding about building a new plant, John has the option of conduct- ing his own marketing research survey, at a cost of $10,000. The information from his survey could help him decide whether to construct a large plant or a small plant or not to build at all. John recognizes that such a market survey will not provide him with perfect information, but it may help quite a bit nevertheless.

John’s new decision tree is represented in Figure 3.4. Let’s take a careful look at this more complex tree. Note that all possible outcomes and alternatives are included in their logical sequence. This is one of the strengths of using decision trees in making decisions. The user is forced to examine all possible outcomes, including unfavorable ones. He or she is also forced to make decisions in a logical, sequential manner.

Examining the tree, we see that Thompson’s first decision point is whether to conduct the $10,000 market survey. If he chooses not to do the study (the lower part of the tree), he can con- struct a large plant, a small plant, or no plant. This is John’s second decision point. The market will be either favorable (0.50 probability) or unfavorable (also 0.50 probability) if he builds. The payoffs for each of the possible consequences are listed along the right side. As a matter of fact, the lower portion of John’s tree is identical to the simpler decision tree shown in Figure 3.3. Why is this so?

The upper part of Figure 3.4 reflects the decision to conduct the market survey. State-of- nature node 1 has two branches. There is a 45% chance that the survey results will indicate a fa- vorable market for storage sheds. We also note that the probability is 0.55 that the survey results will be negative. The derivation of this probability will be discussed in the next section.

The rest of the probabilities shown in parentheses in Figure 3.4 are all conditional proba- bilities or posterior probabilities (these probabilities will also be discussed in the next section). For example, 0.78 is the probability of a favorable market for the sheds given a favorable result

All outcomes and alternatives must be considered.

Favorable Market (0.5)

Unfavorable Market (0.5)

Favorable Market (0.5)

Unfavorable Market (0.5)

Construct Small Plant Con

stru ct L

arg e P

lant

Do Nothing

= (0.5)($200,000) + (0.5)( –$180,000)

Payo�s

$200,000

–$180,000

$100,000

–$20,000

$0

1

2

EMV for Node 1 = $10,000

EMV for Node 2 = $40,000

= (0.5)($100,000) + (0.5)( –$20,000)

Alternative with Best EMV Is Selected

FIGURE 3.3 Completed and solved Decision Tree for Thompson Lumber

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3.6 DECision TREEs  79

from the market survey. Of course, you would expect to find a high probability of a favorable market given that the research indicated that the market was good. Don’t forget, though, there is a chance that John’s $10,000 market survey didn’t result in perfect or even reliable information. Any market research study is subject to error. In this case, there is a 22% chance that the market for sheds will be unfavorable given that the survey results are positive.

We note that there is a 27% chance that the market for sheds will be favorable given that John’s survey results are negative. The probability is much higher, 0.73, that the market will actually be unfavorable given that the survey was negative.

Finally, when we look to the payoff column in Figure 3.4, we see that $10,000, the cost of the marketing study, had to be subtracted from each of the top 10 tree branches. Thus, a large plant with a favorable market would normally net a $200,000 profit. But because the market study was conducted, this figure is reduced by $10,000 to $190,000. In the unfavorable case, the loss of $180,000 would increase to a greater loss of $190,000. Similarly, conducting the survey and building no plant now results in a -$10,000 payoff.

With all probabilities and payoffs specified, we can start calculating the EMV at each state- of-nature node. We begin at the end, or right side of the decision tree, and work back toward the origin. When we finish, the best decision will be known.

Most of the probabilities are conditional probabilities.

The cost of the survey had to be subtracted from the original payoffs.

We start by computing the EMV of each branch.

Favorable Market (0.78) $190,000

Unfavorable Market (0.22) –$190,000

Favorable Market (0.78) $90,000

Unfavorable Market (0.22) –$30,000

No Plant

Small Plant

Lar ge

Pla nt

–$10,000

Favorable Market (0.27) $190,000

Unfavorable Market (0.73) –$190,000

Favorable Market (0.27) $90,000

Unfavorable Market (0.73) –$30,000

No Plant

Small Plant

Lar ge

Pla nt

–$10,000

Favorable Market (0.50) $200,000

Unfavorable Market (0.50) –$180,000

Favorable Market (0.50) $100,000

Unfavorable Market (0.50) –$20,000

No Plant

Small Plant

Lar ge

Pla nt

$0

Su rv

ey Re

su lts

Fa vo

ra ble

(0

.4 5) Survey

Results

Negative

(0.55)

C on

du ct

M ar

ke t S

ur ve

y

Do Not Conduct Survey

First Decision Point

Second Decision Point

Payo�s

2

3

5

6

7

4

1

FIGURE 3.4 Larger Decision Tree with Payoffs and Probabilities for Thompson Lumber

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80  CHAPTER 3 • DECision AnALysis

1. Given favorable survey results,

EMV1node 22 = EMV1large plant � positive survey2 = 10.7821$190,0002 + 10.2221-$190,0002 = $106,400

EMV1node 32 = EMV1small plant � positive survey2 = 10.7821$90,0002 + 10.2221-$30,0002 = $63,600

The EMV of no plant in this case is -$10,000. Thus, if the survey results are favorable, a large plant should be built. Note that we bring the expected value of this decision ($106,400) to the decision node to indicate that if the survey results are positive, our expected value will be $106,400. This is shown in Figure 3.5.

2. Given negative survey results,

EMV1node 42 = EMV1large plant � negative survey2 = 10.2721$190,0002 + 10.7321-$190,0002 = -$87,400

EMV1node 52 = EMV1small plant � negative survey2 = 10.2721$90,0002 + 10.7321-$30,0002 = $2,400

EMV calculations for favorable survey results are made first.

Su rv

ey Re

su lts

Fa vo

ra ble

(0. 45

)

Favorable Market (0.78) $190,000

Unfavorable Market (0.22) –$190,000

Favorable Market (0.78) $90,000

Unfavorable Market (0.22) –$30,000

No Plant

Small Plant

Lar ge

Pla nt

–$10,000

Favorable Market (0.27) $190,000

Unfavorable Market (0.73) –$190,000

Favorable Market (0.27) $90,000

Unfavorable Market (0.73) –$30,000

No Plant

Small Plant

–$10,000

Favorable Market (0.50) $200,000

Unfavorable Market (0.50) –$180,000

Favorable Market (0.50) $100,000

Unfavorable Market (0.50) –$20,000

No Plant

Small Plant

$0

Survey

Results

Negative

(0.55)

C on

du ct

M ar

ke t S

ur ve

y

Do Not Conduct Survey

First Decision Point

Second Decision Point

Payo�s

$106,400

$63,600

–$87,400

$2,400

$10,000

Lar ge

Pla nt

Lar ge

Pla nt

$40,000

49,200

2

3

5

6

7

1

4

$4 0,

00 0

$2 ,4

00 $1

06 ,4

00

$4 9,

20 0

FIGURE 3.5 Thompson’s Decision Tree with EMVs shown

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3.6 DECision TREEs  81

The EMV of no plant is again -$10,000 for this branch. Thus, given a negative survey result, John should build a small plant with an expected value of $2,400, and this figure is indicated at the decision node.

3. Continuing on the upper part of the tree and moving backward, we compute the expected value of conducting the market survey:

EMV1node 12 = EMV1conduct survey2 = 10.4521$106,4002 + 10.5521$2,4002 = $47,880 + $1,320 = $49,200

4. If the market survey is not conducted,

EMV1node 62 = EMV1large plant2 = 10.5021$200,0002 + 10.5021-$180,0002 = $10,000

EMV1node 72 = EMV1small plant2 = 10.5021$100,0002 + 10.5021-$20,0002 = $40,000

The EMV of no plant is $0. Thus, building a small plant is the best choice, given that the marketing research is not

performed, as we saw earlier. 5. We move back to the first decision node and choose the best alternative. The EMV of con-

ducting the survey is $49,200, versus an EMV of $40,000 for not conducting the study, so the best choice is to seek marketing information. If the survey results are favorable, John should construct a large plant, but if the research is negative, John should construct a small plant.

In Figure 3.5, these expected values are placed on the decision tree. Notice on the tree that a pair of slash lines / / through a decision branch indicates that particular alternative is dropped from further consideration. This is because its EMV is lower than the EMV for the best alterna- tive. After you have solved several decision tree problems, you may find it easier to do all of your computations on the tree diagram.

EXPECTED VALUE OF SAMPLE INFORMATION With the market survey he intends to conduct, John Thomp- son knows that his best decision will be to build a large plant if the survey is favorable or a small plant if the survey results are negative. But John also realizes that conducting the market research is not free. He would like to know what the actual value of doing a survey is. One way of measuring the value of market information is to compute the expected value of sample information (EVSI), which is the increase in expected value resulting from the sample information.

The expected value with sample information (EV with SI) is found from the decision tree, and the cost of the sample information is added to this, since this was subtracted from all the payoffs before the EV with SI was calculated. The expected value without sample information (EV without SI) is then subtracted from this to find the value of the sample information.

EVSI = 1EV with SI + cost2 - 1EV without SI2 (3-4) where

EVSI = expected value of sample information EV with SI = expected value with sample information EV without SI = expected value without sample information

In John’s case, his EMV would be $59,200 if he hadn’t already subtracted the $10,000 study cost from each payoff. (Do you see why this is so? If not, add $10,000 back into each payoff, as in the original Thompson problem, and recompute the EMV of conducting the market study.) From the lower branch of Figure 3.5, we see that the EMV of not gathering the sample informa- tion is $40,000. Thus,

EVSI = 1$49,200 + $10,0002 - $40,000 = $59,200 - $40,000 = $19,200

EMV calculations for unfavorable survey results are done next.

We continue working backward to the origin, computing EMV values.

EVSI measures the value of sample information.

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82  CHAPTER 3 • DECision AnALysis

This means that John could have paid up to $19,200 for a market study and still come out ahead. Since it costs only $10,000, the survey is indeed worthwhile.

Efficiency of Sample Information There may be many types of sample information available to a decision maker. In developing a new product, information could be obtained from a survey, from a focus group, from other market research techniques, or from actual use of a test market to see how sales will be. While none of these sources of information are perfect, they can be evaluated by comparing the EVSI with the EVPI. If the sample information was perfect, then the efficiency would be 100%. The efficiency of sample information is

Efficiency of sample information = EVSI

EVPI 100% (3-5)

In the Thompson Lumber example,

Efficiency of sample information = 19,200

60,000 100% = 32%

Thus, the market survey is only 32% as efficient as perfect information.

Sensitivity Analysis As with payoff tables, sensitivity analysis can be applied to decision trees as well. The overall approach is the same. Consider the decision tree for the expanded Thompson Lumber problem shown in Figure 3.5. How sensitive is our decision (to conduct the marketing survey) to the probability of favorable survey results?

Let p be the probability of favorable survey results. Then 11 - p2 is the probability of nega- tive survey results. Given this information, we can develop an expression for the EMV of con- ducting the survey, which is node 1:

EMV1node 12 = 1$106,4002p + 1$2,400211 - p2 = $104,000p + $2,400

We are indifferent when the EMV of conducting the marketing survey, node 1, is the same as the EMV of not conducting the survey, which is $40,000. We can find the indifference point by equating EMV(node 1) to $40,000:

$104,000p + $2,400 = $40,000 $104,000p = $37,600

p = $37,600

$104,000 = 0.36

As long as the probability of favorable survey results, p, is greater than 0.36, our decision will stay the same. When p is less than 0.36, our decision will be not to conduct the survey.

We could also perform sensitivity analysis for other problem parameters. For example, we could find how sensitive our decision is to the probability of a favorable market given favorable survey results. At this time, this probability is 0.78. If this value goes up, the large plant becomes more attractive. In this case, our decision would not change. What happens when this probabil- ity goes down? The analysis becomes more complex. As the probability of a favorable market (given favorable survey) result goes down, the small plant becomes more attractive. At some point, the small plant will result in a higher EMV (given favorable survey results) than the large plant. This, however, does not conclude our analysis. As the probability of a favorable market (given favorable survey results) continues to fall, there will be a point where not conducting the survey, with an EMV of $40,000, will be more attractive than conducting the marketing survey. We leave the actual calculations to you. It is important to note that sensitivity analysis should consider all possible consequences.

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3.7 HoW PRoBABiLiTy VALUEs ARE EsTiMATED By BAyEsiAn AnALysis  83

3.7 How Probability Values Are Estimated by Bayesian Analysis

There are many ways of getting probability data for a problem such as Thompson’s. The num- bers (such as 0.78, 0.22, 0.27, and 0.73 in Figure 3.4) can be assessed by a manager based on experience and intuition. They can be derived from historical data, or they can be computed from other available data using Bayes’ Theorem. The advantage of Bayes’ Theorem is that it incorporates both our initial estimates of the probabilities and information about the accuracy of the information source (e.g., market research survey).

The Bayes’ Theorem approach recognizes that a decision maker does not know with cer- tainty what state of nature will occur. It allows the manager to revise his or her initial or prior probability assessments based on new information. The revised probabilities are called posterior probabilities. (Before continuing, you may wish to review Bayes’ Theorem in Chapter 2.)

Calculating Revised Probabilities In the Thompson Lumber case solved in Section 3.7, we made the assumption that the following four conditional probabilities were known:

P1favorable market1FM2 � survey results positive2 = 0.78 P1unfavorable market1UM2 � survey results positive2 = 0.22

P1favorable market1FM2 � survey results negative2 = 0.27 P1unfavorable market1UM2 � survey results negative2 = 0.73

We now show how John Thompson was able to derive these values with Bayes’ Theorem. From discussions with market research specialists at the local university, John knows that special surveys such as his can either be positive (i.e., predict a favorable market) or be negative (i.e., predict an unfavorable market). The experts have told John that, statistically, of all new products with a favorable market (FM), market surveys were positive and predicted success correctly 70% of the time. Thirty percent of the time the surveys falsely predicted negative results or an unfavorable market (UM). On the other hand, when there was actually an unfavorable market for a new product, 80% of the surveys correctly predicted negative results. The surveys incor- rectly predicted positive results the remaining 20% of the time. These conditional probabilities are summarized in Table 3.16. They are an indication of the accuracy of the survey that John is thinking of undertaking.

Recall that without any market survey information, John’s best estimates of a favorable and unfavorable market are

P1FM2 = 0.50 P1UM2 = 0.50

These are referred to as the prior probabilities. We are now ready to compute Thompson’s revised or posterior probabilities. These desired

probabilities are the reverse of the probabilities in Table 3.16. We need the probability of a favor- able or unfavorable market given a positive or negative result from the market study. The general form of Bayes’ Theorem presented in Chapter 2 is

P1A �B2 = P1B �A2P1A2 P1B �A2P1A2 + P1B �A′2P1A′2 (3-6)

Bayes’ Theorem allows decision makers to revise probability values.

TABLE 3.16 Market survey Reliability in Predicting states of nature

STATE OF NATURE

RESULT OF SURVEY FAVORABLE MARKET

(FM) UNFAVORABLE MARKET

(UM)

Positive (predicts favorable market for product)

P(survey positive �FM) = 0.70 P(survey positive �UM) = 0.20

Negative (predicts unfavor- able market for product)

P(survey negative �FM) = 0.30 P(survey negative �UM) = 0.80

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84  CHAPTER 3 • DECision AnALysis

success drop to 27% if the survey report is negative. This is valuable management information, as we saw in the earlier decision tree analysis.

Potential Problem in Using Survey Results In many decision-making problems, survey results or pilot studies are done before an actual decision (such as building a new plant or taking a particular course of action) is made. As dis- cussed earlier in this section, Bayes’ analysis is used to help determine the correct conditional probabilities that are needed to solve these types of decision theory problems. In computing these conditional probabilities, we need to have data about the surveys and their accuracies. If a decision to build a plant or to take another course of action is actually made, we can determine the accuracy of our surveys. Unfortunately, we cannot always get data about those situations in which the decision was not to build a plant or not to take some course of action. Thus, some- times when we use survey results, we are basing our probabilities only on those cases in which a decision to build a plant or take some course of action is actually made. This means that, in some situations, conditional probability information may not be not quite as accurate as we would like. Even so, calculating conditional probabilities helps to refine the decision-making process and, in general, to make better decisions.

New probabilities provide valuable information.

where

A, B = any two events A′ = complement of A

We can let A represent a favorable market and B represent a positive survey. Then, substitut- ing the appropriate numbers into this equation, we obtain the conditional probabilities given that the market survey is positive:

P1FM � survey positive2 = P1survey positive �FM2P1FM2 P1survey positive �FM2P1FM2 + P1survey positive �UM2P1UM2

= 10.70210.502

10.70210.502 + 10.20210.502 = 0.35

0.45 = 0.78

P1UM � survey positive2 = P1survey positive �UM2P1UM2 P1survey positive �UM2P1UM2 + P1survey positive �FM2P1FM2

= 10.20210.502

10.20210.502 + 10.70210.502 = 0.10

0.45 = 0.22

Note that the denominator (0.45) in these calculations is the probability of a positive survey. An alternative method for these calculations is to use a probability table as shown in Table 3.17.

The conditional probabilities, given that the market survey is negative, are

P1FM � survey negative2 = P1survey negative �FM2P1FM2 P1survey negative �FM2P1FM2 + P1survey negative �UM2P1UM2

= 10.30210.502

10.30210.502 + 10.80210.502 = 0.15

0.55 = 0.27

P1UM � survey negative2 = P1survey negative �UM2P1UM2 P1survey negative �UM2P1UM2 + P1survey negative �FM2P1FM2

= 10.80210.502

10.80210.502 + 10.30210.502 = 0.40

0.55 = 0.73

Note that the denominator (0.55) in these calculations is the probability of a negative survey. These computations could also have been performed in a table instead, as in Table 3.18.

The calculations shown in Tables 3.17 and 3.18 can easily be performed in Excel spread- sheets. Programs 3.3A and 3.3B show the final output for this example and the formulas.

The posterior probabilities now provide John Thompson with estimates for each state of nature if the survey results are positive or negative. As you know, John’s prior probability of success without a market survey was only 0.50. Now he is aware that the probability of success- fully marketing storage sheds will be 0.78 if his survey shows positive results. His chances of

TABLE 3.17 Probability Revisions Given a Positive survey

POSTERIOR PROBABILITY

STATE OF NATURE

CONDITIONAL PROBABILITY

P(SURVEY POSITIVE � STATE OF NATURE) PRIOR PROBABILITY

JOINT PROBABILITY

P(STATE OF NATURE � SURVEY

POSITIVE)

FM 0.70 * 0.50 = 0.35 0.35/0.45 = 0.78 UM 0.20 * 0.50 = 0.10 0.10/0.45 = 0.22

P(survey results positive) = 0.45 1.00

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3.7 HoW PRoBABiLiTy VALUEs ARE EsTiMATED By BAyEsiAn AnALysis  85

success drop to 27% if the survey report is negative. This is valuable management information, as we saw in the earlier decision tree analysis.

Potential Problem in Using Survey Results In many decision-making problems, survey results or pilot studies are done before an actual decision (such as building a new plant or taking a particular course of action) is made. As dis- cussed earlier in this section, Bayes’ analysis is used to help determine the correct conditional probabilities that are needed to solve these types of decision theory problems. In computing these conditional probabilities, we need to have data about the surveys and their accuracies. If a decision to build a plant or to take another course of action is actually made, we can determine the accuracy of our surveys. Unfortunately, we cannot always get data about those situations in which the decision was not to build a plant or not to take some course of action. Thus, some- times when we use survey results, we are basing our probabilities only on those cases in which a decision to build a plant or take some course of action is actually made. This means that, in some situations, conditional probability information may not be not quite as accurate as we would like. Even so, calculating conditional probabilities helps to refine the decision-making process and, in general, to make better decisions.

New probabilities provide valuable information.

PROGRAM 3.3A Results of Bayes’ Calculations in Excel 2016

PROGRAM 3.3B Formulas Used for Bayes’ Calculations in Excel 2016

TABLE 3.18 Probability Revisions Given a negative survey

POSTERIOR PROBABILITY

STATE OF NATURE

CONDITIONAL PROBABILITY

P(SURVEY NEGATIVE � STATE OF NATURE) PRIOR PROBABILITY

JOINT PROBABILITY

P(STATE OF NATURE � SURVEY

NEGATIVE)

FM 0.30 * 0.50 = 0.15 0.15/0.55 = 0.27 UM 0.80 * 0.50 = 0.40 0.40/0.55 = 0.73

P(survey results negative) = 0.55 1.00

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86  CHAPTER 3 • DECision AnALysis

for the best outcome is greater than the utility for the worst outcome, using 0 and 1 has some benefits. Because we have chosen to use 0 and 1, all other outcomes will have a utility value be- tween 0 and 1. In determining the utilities of all outcomes, other than the best or worst outcome, a standard gamble is considered. This gamble is shown in Figure 3.7.

In Figure 3.7, p is the probability of obtaining the best outcome, and 11 - p2 is the prob- ability of obtaining the worst outcome. Assessing the utility of any other outcome involves deter- mining the probability (p) that makes you indifferent between alternative 1, which is the gamble between the best and worst outcomes, and alternative 2, which is obtaining the other outcome for sure. When you are indifferent between alternatives 1 and 2, the expected utilities for these two alternatives must be equal. This relationship is shown as

Expected utility of alternative 2 = Expected utility of alternative 1 Utility of other outcome = 1p21Utility of best outcome, which is 12

+ 11 - p21Utility of the worst outcome, which is 02 Utility of other outcome = 1p2112 + 11 - p2102 = p (3-7)

Now all you have to do is to determine the value of the probability (p) that makes you indif- ferent between alternatives 1 and 2. In setting the probability, you should be aware that utility assessment is completely subjective. It’s a value set by the decision maker that can’t be mea- sured on an objective scale. Let’s take a look at an example.

Jane Dickson would like to construct a utility curve revealing her preference for money be- tween $0 and $10,000. A utility curve is a graph that plots utility value versus monetary value. She can invest her money either in a bank savings account or in a real estate deal.

If the money is invested in the bank, in 3 years Jane would have $5,000. If she invested in the real estate, after 3 years she could either have nothing or have $10,000. Jane, however, is very conservative. Unless there is an 80% chance of getting $10,000 from the real estate deal, Jane would prefer to have her money in the bank, where it is safe. What Jane has done here is to assess her utility for $5,000. When there is an 80% chance (this means that p is 0.8) of getting $10,000, Jane is indifferent between putting her money in real estate and putting it in the bank. Jane’s utility for $5,000 is thus equal to 0.8, which is the same as the value for p. This utility assessment is shown in Figure 3.8.

Other utility values can be assessed in the same way. For example, what is Jane’s utility for $7,000? What value of p would make Jane indifferent between $7,000 and the gamble that would result in either $10,000 or $0? For Jane, there must be a 90% chance of getting the $10,000. Otherwise, she would prefer the $7,000 for sure. Thus, her utility for $7,000 is 0.90. Jane’s utility for $3,000 can be determined in the same way. If there were a 50% chance of obtaining the $10,000, Jane would be indifferent between having $3,000 for sure and taking the gamble of either winning the $10,000 or getting nothing. Thus, the utility of $3,000 for Jane is 0.5. Of course, this process can be continued until Jane has assessed her utility for as many monetary values as she wants. These assessments, however, are enough to get an idea of Jane’s feelings toward risk. In fact, we can plot these points in a utility curve, as is done in Figure 3.9. In the figure, the assessed utility points of $3,000, $5,000, and $7,000 are shown by dots, and the rest of the curve is inferred from these.

Utility assessment assigns the worst outcome a utility of 0 and the best outcome a utility of 1.

When you are indifferent, the expected utilities are equal.

Once utility values have been determined, a utility curve can be constructed.

3.8 Utility Theory

We have focused on the EMV criterion for making decisions under risk. However, there are many occasions in which people make decisions that would appear to be inconsistent with the EMV criterion. When people buy insurance, the amount of the premium is greater than the ex- pected payout to them from the insurance company because the premium includes the expected payout, the overhead cost, and the profit for the insurance company. A person involved in a lawsuit may choose to settle out of court rather than go to trial even if the expected value of go- ing to trial is greater than the proposed settlement. A person buys a lottery ticket even though the expected return is negative. Casino games of all types have negative expected returns for the player, and yet millions of people play these games. A businessperson may rule out one potential decision because it could bankrupt the firm if things go bad, even though the expected return for this decision is better than that of all other alternatives.

Why do people make decisions that don’t maximize their EMV? They do this because the monetary value is not always a true indicator of the overall value of the result of the decision. The overall worth of a particular outcome is called utility, and rational people make decisions that maximize the expected utility. Although at times the monetary value is a good indicator of utility, there are other times when it is not. This is particularly true when some of the values involve an extremely large payoff or an extremely large loss. Suppose that you are the lucky holder of a lottery ticket. Five minutes from now a fair coin could be flipped, and if it comes up tails, you would win $5 million. If it comes up heads, you would win nothing. Just a moment ago a wealthy person offered you $2 million for your ticket. Let’s assume that you have no doubts about the validity of the offer. The person will give you a certified check for the full amount, and you are absolutely sure the check would be good.

A decision tree for this situation is shown in Figure 3.6. The EMV for rejecting the offer indicates that you should hold on to your ticket, but what would you do? Just think, $2 million for sure instead of a 50% chance at nothing. Suppose you were greedy enough to hold on to the ticket, and then lost. How would you explain that to your friends? Wouldn’t $2 million be enough to be comfortable for a while?

Most people would choose to sell the ticket for $2 million. Most of us, in fact, would proba- bly be willing to settle for a lot less. Just how low we would go is, of course, a matter of personal preference. People have different feelings about seeking or avoiding risk. Using the EMV alone is not always a good way to make these types of decisions.

One way to incorporate your own attitudes toward risk is through utility theory. In the next section, we explore first how to measure utility and then how to use utility measures in decision making.

Measuring Utility and Constructing a Utility Curve The first step in using utility theory is to assign utility values to each monetary value in a given situation. It is convenient to begin utility assessment by assigning the worst outcome a utility of 0 and the best outcome a utility of 1. Although any values may be used as long as the utility

The overall value of the result of a decision is called utility.

EMV is not always the best approach.

Accept O�er

Reject O�er

$2,000,000

$0

Heads (0.5)

Tails (0.5)

$5,000,000EMV = $2,500,000

FIGURE 3.6 your Decision Tree for the Lottery Ticket

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3.8 UTiLiTy THEoRy  87

for the best outcome is greater than the utility for the worst outcome, using 0 and 1 has some benefits. Because we have chosen to use 0 and 1, all other outcomes will have a utility value be- tween 0 and 1. In determining the utilities of all outcomes, other than the best or worst outcome, a standard gamble is considered. This gamble is shown in Figure 3.7.

In Figure 3.7, p is the probability of obtaining the best outcome, and 11 - p2 is the prob- ability of obtaining the worst outcome. Assessing the utility of any other outcome involves deter- mining the probability (p) that makes you indifferent between alternative 1, which is the gamble between the best and worst outcomes, and alternative 2, which is obtaining the other outcome for sure. When you are indifferent between alternatives 1 and 2, the expected utilities for these two alternatives must be equal. This relationship is shown as

Expected utility of alternative 2 = Expected utility of alternative 1 Utility of other outcome = 1p21Utility of best outcome, which is 12

+ 11 - p21Utility of the worst outcome, which is 02 Utility of other outcome = 1p2112 + 11 - p2102 = p (3-7)

Now all you have to do is to determine the value of the probability (p) that makes you indif- ferent between alternatives 1 and 2. In setting the probability, you should be aware that utility assessment is completely subjective. It’s a value set by the decision maker that can’t be mea- sured on an objective scale. Let’s take a look at an example.

Jane Dickson would like to construct a utility curve revealing her preference for money be- tween $0 and $10,000. A utility curve is a graph that plots utility value versus monetary value. She can invest her money either in a bank savings account or in a real estate deal.

If the money is invested in the bank, in 3 years Jane would have $5,000. If she invested in the real estate, after 3 years she could either have nothing or have $10,000. Jane, however, is very conservative. Unless there is an 80% chance of getting $10,000 from the real estate deal, Jane would prefer to have her money in the bank, where it is safe. What Jane has done here is to assess her utility for $5,000. When there is an 80% chance (this means that p is 0.8) of getting $10,000, Jane is indifferent between putting her money in real estate and putting it in the bank. Jane’s utility for $5,000 is thus equal to 0.8, which is the same as the value for p. This utility assessment is shown in Figure 3.8.

Other utility values can be assessed in the same way. For example, what is Jane’s utility for $7,000? What value of p would make Jane indifferent between $7,000 and the gamble that would result in either $10,000 or $0? For Jane, there must be a 90% chance of getting the $10,000. Otherwise, she would prefer the $7,000 for sure. Thus, her utility for $7,000 is 0.90. Jane’s utility for $3,000 can be determined in the same way. If there were a 50% chance of obtaining the $10,000, Jane would be indifferent between having $3,000 for sure and taking the gamble of either winning the $10,000 or getting nothing. Thus, the utility of $3,000 for Jane is 0.5. Of course, this process can be continued until Jane has assessed her utility for as many monetary values as she wants. These assessments, however, are enough to get an idea of Jane’s feelings toward risk. In fact, we can plot these points in a utility curve, as is done in Figure 3.9. In the figure, the assessed utility points of $3,000, $5,000, and $7,000 are shown by dots, and the rest of the curve is inferred from these.

Utility assessment assigns the worst outcome a utility of 0 and the best outcome a utility of 1.

When you are indifferent, the expected utilities are equal.

Once utility values have been determined, a utility curve can be constructed.

Other Outcome Utility = ?

(p)

Alte rna

tive 1

Alternative 2

(1 – p) Worst Outcome Utility = 0

Best Outcome Utility = 1

FIGURE 3.7 standard Gamble for Utility Assessment

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88  CHAPTER 3 • DECision AnALysis

computed instead of the EMV. Let’s take a look at an example in which a decision tree is used and expected utility values are computed in selecting the best alternative.

Mark Simkin loves to gamble. He decides to play a game that involves tossing thumbtacks in the air. If the point on the thumbtack is facing up after it lands, Mark wins $10,000. If the point on the thumbtack is down, Mark loses $10,000. Should Mark play the game (alternative 1), or should he not play the game (alternative 2)?

Alternatives 1 and 2 are displayed in the tree shown in Figure 3.11. As can be seen, alterna- tive 1 is to play the game. Mark believes that there is a 45% chance of winning $10,000 and a 55% chance of suffering the $10,000 loss. Alternative 2 is not to gamble. What should Mark do? Of course, this depends on Mark’s utility for money. As stated previously, he likes to gamble. Using the procedure just outlined, Mark was able to construct a utility curve showing his prefer- ence for money. Mark has a total of $20,000 to gamble, so he has constructed the utility curve based on a best payoff of $20,000 and a worst payoff of a $20,000 loss. This curve appears in Figure 3.12.

We see that Mark’s utility for –$10,000 is 0.05, his utility for not playing ($0) is 0.15, and his utility for $10,000 is 0.30. These values can now be used in the decision tree. Mark’s objec- tive is to maximize his expected utility, which can be done as follows:

Step 1. U1-$10,0002 = 0.05 U1$02 = 0.15

U1$10,0002 = 0.30

Utility values replace monetary values.

Mark’s objective is to maximize expected utility.

Jane’s utility curve is typical of a risk avoider. A risk avoider is a decision maker who gets less utility or pleasure from a greater risk and tends to avoid situations in which high losses might occur. As monetary value increases on her utility curve, the utility increases at a slower rate.

Figure 3.10 illustrates that a person who is a risk seeker has an opposite-shaped utility curve. This decision maker gets more utility from a greater risk and higher potential payoff. As monetary value increases on his or her utility curve, the utility increases at an increasing rate. A person who is indifferent to risk has a utility curve that is a straight line. The shape of a person’s utility curve depends on the specific decision being considered, the monetary values involved in the situation, the person’s psychological frame of mind, and how the person feels about the fu- ture. It may well be that you have one utility curve for some situations you face and completely different curves for others.

Utility as a Decision-Making Criterion After a utility curve has been determined, the utility values from the curve are used in mak- ing decisions. Monetary outcomes or values are replaced with the appropriate utility values, and then decision analysis is performed as usual. The expected utility for each alternative is

The shape of a person’s utility curve depends on many factors.

$5,000 U($5,000) = p = 0.80

p = 0.80

Inv est

in Rea

l Es tate

Invest in Bank

$0 U($0.00) = 0.0

$10,000 U($10,000) = 1.0

Utility for $5,000 = U($5,000) = pU($10,000) + (1 – p) U($0) = (0.8)(1) + (0.2)(0) = 0.8

(1 – p) = 0.20

FIGURE 3.8 Utility of $5,000

$0 $1,000 $3,000 $5,000 $7,000 $10,000

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

U til

ity

Monetary Value

U($10,000) = 1.0

U($7,000) = 0.90

U($5,000) = 0.80

U($3,000) = 0.50

U($0) = 0

FIGURE 3.9 Utility Curve for Jane Dickson

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3.8 UTiLiTy THEoRy  89

computed instead of the EMV. Let’s take a look at an example in which a decision tree is used and expected utility values are computed in selecting the best alternative.

Mark Simkin loves to gamble. He decides to play a game that involves tossing thumbtacks in the air. If the point on the thumbtack is facing up after it lands, Mark wins $10,000. If the point on the thumbtack is down, Mark loses $10,000. Should Mark play the game (alternative 1), or should he not play the game (alternative 2)?

Alternatives 1 and 2 are displayed in the tree shown in Figure 3.11. As can be seen, alterna- tive 1 is to play the game. Mark believes that there is a 45% chance of winning $10,000 and a 55% chance of suffering the $10,000 loss. Alternative 2 is not to gamble. What should Mark do? Of course, this depends on Mark’s utility for money. As stated previously, he likes to gamble. Using the procedure just outlined, Mark was able to construct a utility curve showing his prefer- ence for money. Mark has a total of $20,000 to gamble, so he has constructed the utility curve based on a best payoff of $20,000 and a worst payoff of a $20,000 loss. This curve appears in Figure 3.12.

We see that Mark’s utility for –$10,000 is 0.05, his utility for not playing ($0) is 0.15, and his utility for $10,000 is 0.30. These values can now be used in the decision tree. Mark’s objec- tive is to maximize his expected utility, which can be done as follows:

Step 1. U1-$10,0002 = 0.05 U1$02 = 0.15

U1$10,0002 = 0.30

Utility values replace monetary values.

Mark’s objective is to maximize expected utility.

U til

ity

Monetary Outcome

Risk Avoider

Risk Seeker

Ri sk

In di�

er en

ce

FIGURE 3.10 Preferences for Risk

$0

Tack Lands Point Up (0.45)

Alte rna

tive 1

Ma rk P

lay s th

e G am

e

Alternative 2

–$10,000

$10,000 Tack Lands Point Down (0.55)

Mark Does Not Play the Game

FIGURE 3.11 Decision Facing Mark simkin

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90  CHAPTER 3 • DECision AnALysis

–$20,000

U til

ity

Monetary Outcome

0.75

1.00

0.50

0.30 0.25

0.15

0.05 0

–$10,000 $0 $10,000 $20,000

FIGURE 3.12 Utility Curve for Mark simkin

Decision Analysis Helps American Airlines Assess Uncertainty of Bid Quotes

American Airlines, Inc. (AA), is the world’s largest airline in pas- senger miles transported with annual revenue of over $21 billion. Although its primary goal is to transport passengers, AA also has to manage ancillary functions such as full-truckload (FTL) freight shipment of maintenance equipment and in-flight service items. The inventory value of these goods as they move point to point worldwide can be over $1 billion at any given time.

Each year AA issues approximately 500 requests for quotes (RFQs) in the bid process for its FTL point-to-point freight ship- ment routes. AA needed a should-cost model to assess these quotes to ensure that it does not overpay its FTL suppliers.

Working with researchers at North Carolina State University, AA developed a decision tree-based analysis that estimates reason- able costs for these shipments. The fully expanded decision tree for this problem has nearly 60,000 end points.

AA has now used this decision tree model on more than 20 RFQs to prioritize its contractual opportunities and obtain accurate assessments of the FTL costs, thus minimizing the risk of overpaying its FTL suppliers.

Source: Based on M. J. Bailey et al., “American Airlines Uses Should-Cost Modeling to Assess the Uncertainty of Bids for Its Full-Truckload Shipment Routes,” Interfaces 41, 2 (March–April 2011): 194–196, © Trevor S. Hale.

IN ACTION

0.15

Tack Lands Point Up (0.45)

Alte rna

tive 1

Pla y th

e G am

e

Alternative 2

0.05

0.30 Tack Lands Point Down (0.55)

Don’t Play

Utility

FIGURE 3.13 Using Expected Utilities in Decision Making

Step 2. Replace monetary values with utility values. Refer to Figure 3.13. Here are the expected utilities for alternatives 1 and 2:

E1alternative 1: play the game2 = 10.45210.302 + 10.55210.052 = 0.135 + 0.027 = 0.162

E1alternative 2: don>t play the game2 = 0.15

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GLossARy  91

Therefore, alternative 1 is the best strategy using utility as the decision criterion. If EMV had been used, alternative 2 would have been the best strategy. The utility curve is a risk-seeker util- ity curve, and the choice of playing the game certainly reflects this preference for risk.

Decision theory is an analytic and systematic approach to studying decision making. Six steps are usually involved in making decisions in three environments: decision making un- der certainty, uncertainty, and risk. In decision making under uncertainty, decision tables are constructed to compute criteria such as maximax, maximin, criterion of realism, equally likely, and minimax regret. Methods such as determining expected monetary value (EMV), expected value of perfect information (EVPI), expected opportunity loss (EOL), and sensitivity anal- ysis are used in decision making under risk.

Decision trees are another option, particularly for larger decision problems, when one decision must be made before other decisions can be made. For example, a decision to take a sample or to perform market research is made before we decide to construct a large plant, a small one, or no plant. In this case, we can also compute the ex- pected value of sample information (EVSI) to determine the value of the market research. The efficiency of sample information compares the EVSI to the EVPI. Bayesian analysis can be used to revise or update probability values using both the prior probabilities and other probabilities related to the accuracy of the information source.

Summary

Glossary

Alternative A course of action or a strategy that may be chosen by a decision maker.

Coefficient of Realism (A) A number from 0 to 1. When the coefficient is close to 1, the decision criterion is opti- mistic. When the coefficient is close to 0, the decision crite- rion is pessimistic.

Conditional Probability A posterior probability. Conditional Value or Payoff A consequence, normally

expressed in a monetary value, that occurs as a result of a particular alternative and state of nature.

Criterion of Realism A decision-making criterion that uses a weighted average of the best and worst possible payoffs for each alternative.

Decision Making Under Certainty A decision-making environment in which the future outcomes or states of na- ture are known.

Decision Making Under Risk A decision-making environ- ment in which several outcomes or states of nature may occur as a result of a decision or alternative. The probabili- ties of the outcomes or states of nature are known.

Decision Making Under Uncertainty A decision-making environment in which several outcomes or states of nature may occur. The probabilities of these outcomes, however, are not known.

Decision Node (Point) In a decision tree, this is a point where the best of the available alternatives is chosen. The branches represent the alternatives.

Decision Table A payoff table. Decision Theory An analytic and systematic approach to

decision making. Decision Tree A graphical representation of a decision-

making situation. Efficiency of Sample Information A measure of how good

the sample information is relative to perfect information. Equally Likely A decision criterion that places an equal

weight on all states of nature.

Expected Monetary Value (EMV) The average value of a decision if it can be repeated many times. This is determined by multiplying the monetary values by their respective prob- abilities. The results are then added to arrive at the EMV.

Expected Value of Perfect Information (EVPI) The aver- age or expected value of information if it were completely accurate. The increase in EMV that results from having perfect information.

Expected Value of Sample Information (EVSI) The in- crease in EMV that results from having sample or imperfect information.

Expected Value with Perfect Information (EVwPI) The average or expected value of a decision if perfect knowl- edge of the future is available.

Hurwicz Criterion The criterion of realism. Laplace Criterion The equally likely criterion. Maximax An optimistic decision-making criterion. This

selects the alternative with the highest possible return. Maximin A pessimistic decision-making criterion. This

alternative maximizes the minimum payoff. It selects the alternative with the best of the worst possible payoffs.

Minimax Regret A criterion that minimizes the maximum opportunity loss.

Opportunity Loss The amount you would lose by not pick- ing the best alternative. For any state of nature, this is the difference between the consequences of any alternative and the best possible alternative.

Optimistic Criterion The maximax criterion. Payoff Table A table that lists the alternatives, states of

nature, and payoffs in a decision-making situation. Posterior Probability A conditional probability of a state of

nature that has been adjusted based on sample information. This is found using Bayes’ Theorem.

Prior Probability The initial probability of a state of nature before sample information is used with Bayes’ Theorem to obtain the posterior probability.

Regret Opportunity loss.

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92  CHAPTER 3 • DECision AnALysis

Risk Avoider A person who avoids risk. On the utility curve, as the monetary value increases, the utility increases at a decreasing rate. This decision maker gets less utility for a greater risk and higher potential returns.

Risk Seeker A person who seeks risk. On the utility curve, as the monetary value increases, the utility increases at an increasing rate. This decision maker gets more utility for a greater risk and higher potential returns.

Sequential Decisions Decisions in which the outcome of one decision influences other decisions.

Standard Gamble The process used to determine utility values.

State of Nature An outcome or occurrence over which the decision maker has little or no control.

State-of-Nature Node (Point) In a decision tree, a point where the EMV is computed. The branches coming from this node represent states of nature.

Utility The overall value or worth of a particular outcome. Utility Assessment The process of determining the utility of

various outcomes. This is normally done using a standard gamble between any outcome for sure and a gamble be- tween the worst and best outcomes.

Utility Curve A graph or curve that reveals the relationship between utility and monetary values. When this curve has been constructed, utility values from the curve can be used in the decision-making process.

Utility Theory A theory that allows decision makers to incorporate their risk preference and other factors into the decision-making process.

Weighted Average Criterion Another name for the criterion of realism.

Key Equations

(3-1) EMV1alternative i2 = gXiP1Xi2 An equation that computes expected monetary value.

(3-2) EVwPI = g1Best payoff in state of nature i2 * 1Probability of state of nature i2

An equation that computes the expected value with perfect information.

(3-3) EVPI = EVwPI - Best EMV

An equation that computes the expected value of perfect information.

(3-4) EVSI = 1EV with SI + cost2 - 1EV without SI2

An equation that computes the expected value (EV) of sample information (SI).

(3-5) Efficiency of sample information = EVSI

EVPI 100%

An equation that compares sample information to perfect information.

(3-6) P1A �B2 = P1B �A2P1A2 P1B �A2P1A2 + P1B �A′2P1A′2

Bayes’ Theorem—the conditional probability of event A given that event B has occurred.

(3-7) Utility of other outcome = 1p2112 + 11 - p2102 = p

An equation that determines the utility of an intermediate outcome.

Solved Problems

Solved Problem 3-1 Maria Rojas is considering the possibility of opening a small dress shop on Fairbanks Avenue, a few blocks from the university. She has located a good mall that attracts students. Her options are to open a small shop, a medium-sized shop, or no shop at all. The market for a dress shop can be good, aver- age, or bad. The probabilities for these three possibilities are 0.2 for a good market, 0.5 for an average market, and 0.3 for a bad market. The net profit or loss figures for the medium-sized and small shops for the various market conditions are given in the following table. Building no shop at all yields no loss and no gain.

a. What do you recommend? b. Calculate the EVPI. c. Develop the opportunity loss table for this situation. What decisions would be made using the

minimax regret criterion and the minimum EOL criterion?

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soLVED PRoBLEMs  93

ALTERNATIVE

GOOD MARKET

($)

AVERAGE MARKET

($)

BAD MARKET

($)

Small shop 75,000 25,000 –40,000

Medium-sized shop 100,000 35,000 –60,000

No shop 0 0 0

Solution

a. Since the decision-making environment is risk (probabilities are known), it is appropriate to use the EMV criterion. The problem can be solved by developing a payoff table that contains all alternatives, states of nature, and probability values. The EMV for each alternative is also com- puted, as in the following table:

STATE OF NATURE

ALTERNATIVE GOOD

MARKET ($) AVERAGE

MARKET ($) BAD

MARKET ($) EMV ($)

Small shop 75,000 25,000 –40,000 15,500

Medium-sized shop 100,000 35,000 –60,000 19,500

No shop 0 0 0 0

Probabilities 0.20 0.50 0.30

EMV1small shop2 = 10.221$75,0002 + 10.521$25,0002 + 10.321-$40,0002 = $15,500 EMV1medium shop2 = 10.221$100,0002 + 10.521$35,0002 + 10.321-$60,0002 = $19,500

EMV1no shop2 = 10.221$02 + 10.521$02 + 10.321$02 = $0 As can be seen, the best decision is to build the medium-sized shop. The EMV for this alternative is $19,500.

b. EVwPI = 10.22$100,000 + 10.52$35,000 + 10.32$0 = $37,500 EVPI = $37,500 - $19,500 = $18,000

c. The opportunity loss table is shown here.

STATE OF NATURE

ALTERNATIVE

GOOD MARKET

($)

AVERAGE MARKET

($)

BAD MARKET

($) MAXIMUM

($) EOL ($)

Small shop 25,000 10,000 40,000 40,000 22,000

Medium-sized shop 0 0 60,000 60,000 18,000

No shop 100,000 35,000 0 100,000 37,500

Probabilities 0.20 0.50 0.30

The best payoff in a good market is 100,000, so the opportunity losses in the first column indicate how much worse each payoff is than 100,000. The best payoff in an average market is 35,000, so the opportunity losses in the second column indicate how much worse each payoff is than 35,000. The best payoff in a bad market is 0, so the opportunity losses in the third column indicate how much worse each payoff is than 0.

The minimax regret criterion considers the maximum regret for each decision, and the decision corresponding to the minimum of these is selected. The decision would be to build a small shop, since

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94  CHAPTER 3 • DECision AnALysis

the maximum regret for this is 40,000, while the maximum regret for each of the other two alterna- tives is higher, as shown in the opportunity loss table.

The decision based on the EOL criterion would be to build the medium shop. Note that the mini- mum EOL ($18,000) is the same as the EVPI computed in part (b). The calculations are

EOL1small2 = 10.2225,000 + 10.5210,000 + 10.3240,000 = 22,000 EOL1medium2 = 10.220 + 10.520 + 10.3260,000 = 18,000 EOL1no shop2 = 10.22100,000 + 10.5235,000 + 10.320 = 37,500

Solved Problem 3-2 Cal Bender and Becky Addison have known each other since high school. Two years ago they entered the same university and today they are taking undergraduate courses in the business school. Both hope to graduate with degrees in finance. In an attempt to make extra money and to use some of the knowledge gained from their business courses, Cal and Becky have decided to look into the possibil- ity of starting a small company that would provide word processing services to students who needed term papers or other reports prepared in a professional manner. Using a systems approach, Cal and Becky have identified three strategies. Strategy 1 is to invest in a fairly expensive microcomputer sys- tem with a high-quality laser printer. In a favorable market, they should be able to obtain a net profit of $10,000 over the next 2 years. If the market is unfavorable, they can lose $8,000. Strategy 2 is to purchase a less expensive system. With a favorable market, they could get a return during the next 2 years of $8,000. With an unfavorable market, they would incur a loss of $4,000. Their final strategy, strategy 3, is to do nothing. Cal is basically a risk taker, whereas Becky tries to avoid risk.

a. What type of decision procedure should Cal use? What would Cal’s decision be? b. What type of decision maker is Becky? What decision would Becky make? c. If Cal and Becky were indifferent to risk, what type of decision approach should they use? What

would you recommend if this were the case?

Solution The problem is one of decision making under uncertainty. Before answering the specific ques- tions, a decision table should be developed showing the alternatives, states of nature, and related consequences.

ALTERNATIVE

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($)

Strategy 1 10,000 –8,000

Strategy 2 8,000 –4,000

Strategy 3 0 0

a. Since Cal is a risk taker, he should use the maximax decision criteria. This approach selects the row that has the highest or maximum value. The $10,000 value, which is the maximum value from the table, is in row 1. Thus, Cal’s decision is to select strategy 1, which is an optimistic decision approach.

b. Becky should use the maximin decision criteria because she wants to avoid risk. The minimum or worst outcome for each row, or strategy, is identified. These outcomes are -$8,000 for strategy 1, -$4,000 for strategy 2, and $0 for strategy 3. The maximum of these values is selected. Thus, Becky would select strategy 3, which reflects a pessimistic decision approach.

c. If Cal and Becky are indifferent to risk, they could use the equally likely approach. This approach selects the alternative that maximizes the row averages. The row average for strategy 1 is $1,000 3i.e., $1,000 = 1$10,000 - $8,0002>24. The row average for strategy 2 is $2,000, and the row average for strategy 3 is $0. Thus, using the equally likely approach, the decision is to select strategy 2, which maximizes the row averages.

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soLVED PRoBLEMs  95

Solved Problem 3-3 Monica Britt has enjoyed sailing small boats since she was 7 years old, when her mother started sail- ing with her. Today, Monica is considering the possibility of starting a company to produce small sail- boats for the recreational market. Unlike other mass-produced sailboats, however, these boats will be made specifically for children between the ages of 10 and 15. The boats will be of the highest quality and extremely stable, and the sail size will be reduced to prevent problems of capsizing.

Her basic decision is whether to build a large manufacturing facility, a small manufacturing facil- ity, or no facility at all. With a favorable market, Monica can expect to make $90,000 from the large facility or $60,000 from the smaller facility. If the market is unfavorable, however, Monica estimates that she would lose $30,000 with a large facility and she would lose only $20,000 with the small facility. Because of the expense involved in developing the initial molds and acquiring the neces- sary equipment to produce fiberglass sailboats for young children, Monica has decided to conduct a pilot study to make sure that the market for the sailboats will be adequate. She estimates that the pilot study will cost her $10,000. Furthermore, the pilot study can be either favorable or unfavorable. Monica estimates that the probability of a favorable market, given a favorable pilot study, is 0.8. The probability of an unfavorable market, given an unfavorable pilot study, is estimated to be 0.9. Monica feels that there is a 0.65 chance that the pilot study will be favorable. Of course, Monica could bypass the pilot study and simply make the decision as to whether to build a large plant, small plant, or no facility at all. Without doing any testing in a pilot study, she estimates that the probability of a favor- able market is 0.6. What do you recommend? Compute the EVSI.

Solution Before Monica starts to solve this problem, she should develop a decision tree that shows all alterna- tives, states of nature, probability values, and economic consequences. This decision tree is shown in Figure 3.14.

(0.6) Market Favorable

(0.6) Market Favorable

Do Not Conduct Study

Conduct Study

(0.35)

U nfavorable

Study

(0 .6

5) F

av or

ab le

St

ud y

(0.4) Market Unfavorable

(0.4) Market Unfavorable

$60,000

–$20,000

$90,000

–$30,000

No Facility

(0.8) Market Favorable

(0.8) Market Favorable

(0.2) Market Unfavorable

(0.2) Market Unfavorable

No Facility

(0.1) Market Favorable

(0.1) Market Favorable

(0.9) Market Unfavorable

(0.9) Market Unfavorable

No Facility

Sma ll Fa

cility

Large Facility

Sma ll Fa

cility

Large Facility

Sma ll Fa

cility

Large Facility

$0

2

3

$50,000

–$30,000

$80,000

–$40,000

–$10,000

4

5

$50,000

–$30,000

$80,000

–$40,000

–$10,000

6

1

7

A

B

C

D

FIGURE 3.14 Monica’s Decision Tree, Listing Alternatives, states of nature, Probability Values, and Financial outcomes for solved Problem 3-3

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96  CHAPTER 3 • DECision AnALysis

The EMV at each of the numbered nodes is calculated as follows:

EMV1node 22 = 60,00010.62 + 1-20,00020.4 = 28,000 EMV1node 32 = 90,00010.62 + 1-30,00020.4 = 42,000 EMV1node 42 = 50,00010.82 + 1-30,00020.2 = 34,000 EMV1node 52 = 80,00010.82 + 1-40,00020.2 = 56,000 EMV1node 62 = 50,00010.12 + 1-30,00020.9 = -22,000 EMV1node 72 = 80,00010.12 + 1-40,00020.9 = -28,000 EMV1node 12 = 56,00010.652 + 1-10,00020.35 = 32,900

At each of the square nodes with letters, the decisions would be:

Node B: Choose Large Facility, since the EMV = $42,000.

Node C: Choose Large Facility, since the EMV = $56,000.

Node D: Choose No Facility, since the EMV = -$10,000.

Node A: Choose Do Not Conduct Study, since the EMV 1$42,0002 for this is higher than the EMV ($32,000) for node 1.

Based on the EMV criterion, Monica would select Do Not Conduct Study and then select Large Facility. The EMV of this decision is $42,000. Choosing to conduct the study would result in an EMV of only $32,900. Thus, the expected value of sample information is

EVSI = $32,900 + $10,000 - $42,000 = $900

Solved Problem 3-4 Developing a small driving range for golfers of all abilities has long been a desire of John Jenkins. John, however, believes that the chance of a successful driving range is only about 40%. A friend of John’s has suggested that he conduct a survey in the community to get a better feeling of the demand for such a facility. There is a 0.9 probability that the research will be favorable if the driving range facility will be successful. Furthermore, it is estimated that there is a 0.8 probability that the market- ing research will be unfavorable if indeed the facility will be unsuccessful. John would like to deter- mine the chances of a successful driving range given a favorable result from the marketing survey.

Solution This problem requires the use of Bayes’ Theorem. Before we start to solve the problem, we will define the following terms:

P1SF2 = probability of successful driving range facility P1UF2 = probability of unsuccessful driving range facility

P1RF �SF2 = probability that the research will be favorable given a successful driving range facility P1RU �SF2 = probability that the research will be unfavorable given a successful driving range facility P1RU �UF2 = probability that the research will be unfavorable given an unsuccessful driving range facility P1RF �UF2 = probability that the research will be favorable given an unsuccessful driving range facility

Now, we can summarize what we know:

P1SF2 = 0.4 P1RF �SF2 = 0.9

P1RU �UF2 = 0.8

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From this information we can compute the three additional probabilities that we need to solve the problem:

P1UF2 = 1 - P1SF2 = 1 - 0.4 = 0.6 P1RU �SF2 = 1 - P1RF �SF2 = 1 - 0.9 = 0.1 P1RF �UF2 = 1 - P1RU �UF2 = 1 - 0.8 = 0.2

Now we can put these values into Bayes’ Theorem to compute the desired probability:

P1SF �RF2 = P1RF �SF2 * P1SF2 P1RF �SF2 * P1SF2 + P1RF �UF2 * P1UF2

= 10.9210.42

10.9210.42 + 10.2210.62

= 0.36

10.36 + 0.122 = 0.36

0.48 = 0.75

In addition to using formulas to solve John’s problem, it is possible to perform all calculations in a table:

Revised Probabilities Given a Favorable Research Result

STATE OF NATURE

CONDITIONAL PROBABILITY

PRIOR PROBABILITY

JOINT PROBABILITY

POSTERIOR PROBABILITY

Favorable market 0.9 * 0.4 = 0.36 0.36>0.48 = 0.75

Unfavorable market 0.2 * 0.6 = 0.12 0.12>0.48 = 0.25 0.48

As you can see from the table, the results are the same. The probability of a successful driving range, given a favorable research result, is 0.36>0.48, or 0.75.

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

glossary at the end of the chapter. ●● Use the key at the back of the book (see Appendix H) to correct your answers. ●● Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

1. In decision theory terminology, a course of action or a strategy that may be chosen by a decision maker is called

a. a payoff. b. an alternative. c. a state of nature. d. none of the above. 2. In decision theory, probabilities are associated with a. payoffs. b. alternatives. c. states of nature. d. none of the above. 3. If probabilities are available to the decision maker, then

the decision-making environment is called a. certainty. b. uncertainty. c. risk. d. none of the above.

4. Which of the following is a decision-making criterion that is used for decision making under risk?

a. expected monetary value criterion b. Hurwicz criterion (criterion of realism) c. optimistic (maximax) criterion d. equally likely criterion 5. The minimum expected opportunity loss is a. equal to the highest expected payoff. b. greater than the expected value with perfect informa-

tion. c. equal to the expected value of perfect information. d. computed when finding the minimax regret decision. 6. In using the criterion of realism (Hurwicz criterion), the

coefficient of realism (a) a. is the probability of a good state of nature. b. describes the degree of optimism of the decision

maker.

sELF-TEsT  97

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98  CHAPTER 3 • DECision AnALysis

c. describes the degree of pessimism of the decision maker.

d. is usually less than zero. 7. The most that a person should pay for perfect information

is a. the EVPI. b. the maximum EMV minus the minimum EMV. c. the maximum EOL. d. the maximum EMV. 8. The minimum EOL criterion will always result in the

same decision as a. the maximax criterion. b. the minimax regret criterion. c. the maximum EMV criterion. d. the equally likely criterion. 9. A decision tree is preferable to a decision table when a. a number of sequential decisions are to be made. b. probabilities are available. c. the maximax criterion is used. d. the objective is to maximize regret. 10. Bayes’ Theorem is used to revise probabilities. The new

(revised) probabilities are called a. prior probabilities. b. sample probabilities. c. survey probabilities. d. posterior probabilities. 11. On a decision tree, at each state-of-nature node, a. the alternative with the greatest EMV is selected. b. an EMV is calculated. c. all probabilities are added together. d. the branch with the highest probability is selected. 12. The EVSI a. is found by subtracting the EMV without sample infor-

mation from the EMV with sample information.

b. is always equal to the expected value of perfect infor- mation.

c. equals the EMV with sample information assuming no cost for the information minus the EMV without sam- ple information.

d. is usually negative. 13. The efficiency of sample information a. is the EVSI/(maximum EMV without SI) expressed as

a percentage. b. is the EVPI/EVSI expressed as a percentage. c. would be 100% if the sample information were perfect. d. is computed using only the EVPI and the maximum

EMV. 14. On a decision tree, once the tree has been drawn and the

payoffs and probabilities have been placed on the tree, the analysis (computing EMVs and selecting the best alternative) is done by

a. working backward (starting on the right and moving to the left).

b. working forward (starting on the left and moving to the right).

c. starting at the top of the tree and moving down. d. starting at the bottom of the tree and moving up. 15. In assessing utility values, a. the worst outcome is given a utility of -1. b. the best outcome is given a utility of 0. c. the worst outcome is given a utility of 0. d. the best outcome is given a value of -1. 16. If a rational person selects an alternative that does not

maximize the EMV, we would expect that this alternative a. minimizes the EMV. b. maximizes the expected utility. c. minimizes the expected utility. d. has zero utility associated with each possible payoff.

Discussion Questions and Problems

Discussion Questions 3-1 Give an example of a good decision that you made

that resulted in a bad outcome. Also give an example of a bad decision that you made that had a good out- come. Why was each decision good or bad?

3-2 Describe what is involved in the decision process. 3-3 What is an alternative? What is a state of nature? 3-4 Discuss the differences among decision making

under certainty, decision making under risk, and de- cision making under uncertainty.

3-5 What techniques are used to solve decision-mak- ing problems under uncertainty? Which technique results in an optimistic decision? Which technique results in a pessimistic decision?

3-6 Define opportunity loss. What decision-making cri- teria are used with an opportunity loss table?

3-7 What information should be placed on a decision tree?

3-8 Describe how you would determine the best decision using the EMV criterion with a decision tree.

3-9 What is the difference between prior and posterior probabilities?

3-10 What is the purpose of Bayesian analysis? De- scribe how you would use Bayesian analysis in the decision-making process.

3-11 What is the EVSI? How is this computed? 3-12 How is the efficiency of sample information

computed? 3-13 What is the overall purpose of utility theory? 3-14 Brief ly discuss how a utility function can be

assessed. What is a standard gamble, and how is it used in determining utility values?

3-15 How is a utility curve used in selecting the best deci- sion for a particular problem?

3-16 What is a risk seeker? What is a risk avoider? How do the utility curves for these types of decision mak- ers differ?

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DisCUssion QUEsTions AnD PRoBLEMs  99

Problems 3-17 Kenneth Brown is the principal owner of Brown Oil,

Inc. After quitting his university teaching job, Ken has been able to increase his annual salary by a fac- tor of over 100. At the present time, Ken is forced to consider purchasing some more equipment for Brown Oil because of competition. His alternatives are shown in the following table:

EQUIPMENT

FAVORABLE MARKET

($)

UNFAVORABLE MARKET

($)

Sub 100 300,000 –200,000

Oiler J 250,000 –100,000

Texan 75,000 –18,000

For example, if Ken purchases a Sub 100 and if there is a favorable market, he will realize a profit of $300,000. On the other hand, if the market is unfa- vorable, Ken will suffer a loss of $200,000. But Ken has always been a very optimistic decision maker.

(a) What type of decision is Ken facing? (b) What decision criterion should he use? (c) What alternative is best?

3-18 Although Ken Brown (discussed in Problem 3-17) is the principal owner of Brown Oil, his brother Bob is credited with making the company a financial suc- cess. Bob is vice president of finance. Bob attributes his success to his pessimistic attitude about business and the oil industry. Given the information from Problem 3-17, it is likely that Bob will arrive at a dif- ferent decision. What decision criterion should Bob use, and what alternative will he select?

3-19 The Lubricant is an expensive oil newsletter to which many oil giants subscribe, including Ken Brown (see Problem 3-17 for details). In the last issue, the letter described how the demand for oil products would be extremely high. Apparently, the American consumer will continue to use oil products even if the price of these products doubles. Indeed, one of the articles in the Lubricant states that the chance of a favorable market for oil products was 70%, while the chance of an unfavorable market was only 30%. Ken would like to use these probabilities in determining the best decision.

(a) What decision model should be used? (b) What is the optimal decision? (c) Ken believes that the $300,000 figure for the Sub

100 with a favorable market is too high. How much lower would this figure have to be for Ken to change his decision made in part (b)?

3-20 Mickey Lawson is considering investing some money that he inherited. The following payoff ta- ble gives the profits that would be realized during the next year for each of three investment alterna- tives Mickey is considering:

STATE OF NATURE

DECISION ALTERNATIVE

GOOD ECONOMY

POOR ECONOMY

Stock market 80,000 –20,000

Bonds 30,000 20,000

CDs 23,000 23,000

Probability 0.5 0.5

(a) What decision would maximize expected profits?

(b) What is the maximum amount that should be paid for a perfect forecast of the economy?

3-21 Develop an opportunity loss table for the investment problem that Mickey Lawson faces in Problem 3-20. What decision would minimize the expected oppor- tunity loss? What is the minimum EOL?

3-22 Allen Young has always been proud of his personal investment strategies and has done very well over the past several years. He invests primarily in the stock market. Over the past several months, how- ever, Allen has become very concerned about the stock market as a good investment. In some cases, it would have been better for Allen to have his money in a bank than in the market. During the next year, Allen must decide whether to invest $10,000 in the stock market or in a certificate of deposit (CD) at an interest rate of 9%. If the market is good, Allen be- lieves that he could get a 14% return on his money. With a fair market, he expects to get an 8% return. If the market is bad, he will most likely get no re- turn at all—in other words, the return would be 0%. Allen estimates that the probability of a good market is 0.4, the probability of a fair market is 0.4, and the probability of a bad market is 0.2, and he wishes to maximize his long-run average return.

(a) Develop a decision table for this problem. (b) What is the best decision?

3-23 In Problem 3-22, you helped Allen Young determine the best investment strategy. Now, Allen is thinking about paying for a stock market newsletter. A friend of Allen said that these types of letters could predict very accurately whether the market would be good, fair, or poor. Then, based on these predictions, Allen could make better investment decisions.

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

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

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100  CHAPTER 3 • DECision AnALysis

(a) What is the most that Allen would be willing to pay for a newsletter?

(b) Allen now believes that a good market will give a return of only 11% instead of 14%. Will this information change the amount that Allen would be willing to pay for the newsletter? If your an- swer is yes, determine the most that Allen would be willing to pay, given this new information.

3-24 Today’s Electronics specializes in manufacturing modern electronic components. It also builds the equipment that produces the components. Phyl- lis Weinberger, who is responsible for advising the president of Today’s Electronics on electronic manu- facturing equipment, has developed the following table concerning a proposed facility:

PROFIT ($)

STRONG MARKET

FAIR MARKET

POOR MARKET

Large facility 550,000 110,000 –310,000

Medium-sized facility

300,000 129,000 –100,000

Small facility 200,000 100,000 –32,000

No facility 0 0 0

(a) Develop an opportunity loss table. (b) What is the minimax regret decision?

3-25 Brilliant Color is a small supplier of chemicals and equipment that are used by some photographic stores to process 35mm film. One product that Bril- liant Color supplies is BC-6. John Kubick, president of Brilliant Color, normally stocks 11, 12, or 13 cases of BC-6 each week. For each case that John sells, he receives a profit of $35. Like many photo- graphic chemicals, BC-6 has a very short shelf life, so if a case is not sold by the end of the week, John must discard it. Since each case costs John $56, he loses $56 for every case that is not sold by the end of the week. There is a probability of 0.45 of selling 11 cases, a probability of 0.35 of selling 12 cases, and a probability of 0.2 of selling 13 cases.

(a) Construct a decision table for this problem. Include all conditional values and probabilities in the table.

(b) What is your recommended course of action? (c) If John is able to develop BC-6 with an ingredi-

ent that stabilizes it so that it no longer has to be discarded, how would this change your recom- mended course of action?

3-26 Megley Cheese Company is a small manufacturer of several different cheese products. One of the prod- ucts is a cheese spread that is sold to retail outlets. Jason Megley must decide how many cases of cheese

spread to manufacture each month. The probability that the demand will be six cases is 0.1, seven cases is 0.3, eight cases is 0.5, and nine cases is 0.1. The cost of every case is $45, and the price that Jason gets for each case is $95. Unfortunately, any cases not sold by the end of the month are of no value, due to spoilage. How many cases of cheese should Jason manufacture each month?

3-27 Farm Grown, Inc., produces cases of perishable food products. Each case contains an assortment of veg- etables and other farm products. Each case costs $5 and sells for $15. If there are any cases not sold by the end of the day, they are sold to a large food pro- cessing company for $3 a case. The probability that daily demand will be 100 cases is 0.3, the probabil- ity that daily demand will be 200 cases is 0.4, and the probability that daily demand will be 300 cases is 0.3. Farm Grown has a policy of always satisfying customer demands. If its own supply of cases is less than the demand, it buys the necessary vegetables from a competitor. The estimated cost of doing this is $16 per case.

(a) Draw a decision table for this problem. (b) What do you recommend?

3-28 In Problem 3-27, Farm Grown, Inc., has reason to believe the probabilities may not be reliable due to changing conditions. If these probabilities are ignored, what decision would be made using the op- timistic criterion? What decision would be made us- ing the pessimistic criterion?

3-29 Mick Karra is the manager of MCZ Drilling Prod- ucts, which produces a variety of specialty valves for oil field equipment. Recent activity in the oil fields has caused demand to increase drastically, and a decision has been made to open a new manufactur- ing facility. Three locations are being considered, and the size of the facility would not be the same in each location. Thus, overtime might be necessary at times. The following table gives the total monthly cost (in $1,000s) for each possible location under each demand possibility. The probabilities for the demand levels have been determined to be 20% for low demand, 30% for medium demand, and 50% for high demand.

DEMAND IS LOW

DEMAND IS MEDIUM

DEMAND IS HIGH

Ardmore, OK 85 110 150

Sweetwater, TX 90 100 140

Lake Charles, LA 110 120 130

(a) Which location would be selected based on the optimistic criterion?

(b) Which location would be selected based on the pessimistic criterion?

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DisCUssion QUEsTions AnD PRoBLEMs  101

(c) Which location would be selected based on the minimax regret criterion?

(d) Which location should be selected to minimize the expected cost of operation?

(e) How much is a perfect forecast of the demand worth?

(f) Which location would minimize the expected opportunity loss?

(g) What is the expected value of perfect informa- tion in this situation?

3-30 Even though independent gasoline stations have been having a difficult time, Susan Solomon has been thinking about starting her own independent gasoline station. Susan’s problem is to decide how large her station should be. The annual returns will depend on both the size of her station and a number of marketing factors related to the oil industry and demand for gasoline. After a careful analysis, Susan developed the following table:

SIZE OF FIRST STATION

GOOD MARKET

($)

FAIR MARKET

($)

POOR MARKET

($)

Small 50,000 20,000 –10,000

Medium 80,000 30,000 –20,000

Large 100,000 30,000 –40,000

Very large 300,000 25,000 –160,000

For example, if Susan constructs a small station and the market is good, she will realize a profit of $50,000.

(a) Develop a decision table for this decision. (b) What is the maximax decision? (c) What is the maximin decision? (d) What is the equally likely decision? (e) What is the criterion of realism decision? Use an

a value of 0.8. (f) Develop an opportunity loss table. (g) What is the minimax regret decision?

3-31 Beverly Mills has decided to lease a hybrid car to save on gasoline expenses and to do her part to help keep the environment clean. The car she selected is available from only one dealer in the local area, but that dealer has several leasing options to accom- modate a variety of driving patterns. All the leases are for 3 years and require no money at the time of signing the lease. The first option has a monthly cost of $330, a total mileage allowance of 36,000 miles (an average of 12,000 miles per year), and a cost of $0.35 per mile for any miles over 36,000. The fol- lowing table summarizes each of the three lease options:

3-YEAR LEASE

MONTHLY COST

MILEAGE ALLOWANCE

COST PER EXCESS MILE

Option 1 $330 36,000 $0.35

Option 2 $380 45,000 $0.25

Option 3 $430 54,000 $0.15

Beverly has estimated that, during the 3 years of the lease, there is a 40% chance she will drive an aver- age of 12,000 miles per year, a 30% chance she will drive an average of 15,000 miles per year, and a 30% chance that she will drive 18,000 miles per year. In evaluating these lease options, Beverly would like to keep her costs as low as possible.

(a) Develop a payoff (cost) table for this situation. (b) What decision would Beverly make if she were

optimistic? (c) What decision would Beverly make if she were

pessimistic? (d) What decision would Beverly make if she

wanted to minimize her expected cost (monetary value)?

(e) Calculate the expected value of perfect informa- tion for this problem.

3-32 Refer to the leasing decision facing Beverly Mills in Problem 3-31. Develop the opportunity loss table for this situation. Which option would be chosen based on the minimax regret criterion? Which alternative would result in the lowest expected opportunity loss?

3-33 The game of roulette is popular in many casinos around the world. In Las Vegas, a typical roulette wheel has the numbers 1–36 in slots on the wheel. Half of these slots are red, and the other half are black. In the United States, the roulette wheel typi- cally also has the numbers 0 (zero) and 00 (double zero), and both of these are on the wheel in green slots. Thus, there are 38 slots on the wheel. The dealer spins the wheel and sends a small ball in the opposite direction of the spinning wheel. As the wheel slows, the ball falls into one of the slots, and that is the winning number and color. One of the bets available is simply red or black, for which the odds are 1 to 1. If the player bets on either red or black and that happens to be the winning color, the player wins the amount of her bet. For example, if the player bets $5 on red and wins, she is paid $5 and she still has her original bet. On the other hand, if the winning color is black or green when the player bets red, the player loses the entire bet.

(a) What is the probability that a player who bets red will win the bet?

(b) If a player bets $10 on red every time the game is played, what is the expected monetary value (expected win)?

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102  CHAPTER 3 • DECision AnALysis

(c) In Europe, there is usually no 00 on the wheel, just the 0. With this type of game, what is the probability that a player who bets red will win the bet? If a player bets $10 on red every time in this game (with no 00), what is the expected monetary value?

(d) Since the expected profit (win) in a roulette game is negative, why would a rational person play the game?

3-34 Refer to Problem 3-33 for details about the game of roulette. Another bet in a roulette game is called a “straight up” bet, which means that the player is bet- ting that the winning number will be the number that she chose. In a game with 0 and 00, there is a total of 38 possible outcomes (the numbers 1 to 36 plus 0 and 00), and each of these has the same chance of occurring. The payout on this type of bet is 35 to 1, which means the player is paid 35 and gets to keep the original bet. If a player bets $10 on the number 7 (or any single number), what is the expected mon- etary value (expected win)?

3-35 The Technically Techno company has several pat- ents for a variety of different flash memory devices that are used in computers, cell phones, and a va- riety of other things. A competitor has recently in- troduced a product based on technology very similar to something patented by Technically Techno last year. Consequently, Technically Techno has sued the other company for patent infringement. Based on the facts in the case as well as the record of the lawyers involved, Technically Techno believes there is a 40% chance that it will be awarded $300,000 if the lawsuit goes to court. There is a 30% chance that Technically Techno will be awarded only $50,000 if it goes to court and wins, and there is a 30% chance that Technically Techno will lose the case and be awarded nothing. The estimated cost of legal fees if Technically Techno goes to court is $50,000. However, the other company has offered to pay Technically Techno $75,000 to settle the dispute without going to court. The estimated legal cost of this would be only $10,000. If Technically Techno wished to maximize the expected gain, should it ac- cept the settlement offer?

3-36 A group of medical professionals is considering the construction of a private clinic. If the medical de- mand is high (i.e., there is a favorable market for the clinic), the physicians could realize a net profit of $100,000. If the market is not favorable, they could lose $40,000. Of course, they don’t have to proceed at all, in which case there is no cost. In the absence of any market data, the physicians’ best guess is that there is a 50–50 chance the clinic will be successful. Construct a decision tree to help analyze this prob- lem. What should the medical professionals do?

3-37 The physicians in Problem 3-36 have been ap- proached by a market research firm that offers to perform a study of the market at a fee of $5,000. The market researchers claim their experience enables them to use Bayes’ Theorem to make the following statements of probability:

Probability of a favorable market given

a favorable study = 0.82 Probability of an unfavorable market given

a favorable study = 0.18 Probability of a favorable market given

an unfavorable study = 0.11 Probability of an unfavorable market given

an unfavorable study = 0.89 Probability of a favorable research study = 0.55

Probability of an unfavorable research study = 0.45

(a) Develop a new decision tree for the medical pro- fessionals to reflect the options now open with the market study.

(b) Use the EMV approach to recommend a strategy. (c) What is the expected value of sample informa-

tion? How much might the physicians be willing to pay for a market study?

(d) Calculate the efficiency of this sample information.

3-38 Jerry Smith is thinking about opening a bicycle shop in his hometown. Jerry loves to take his own bike on 50-mile trips with his friends, but he believes that any small business should be started only if there is a good chance of making a profit. Jerry can open a small shop, a large shop, or no shop at all. The prof- its will depend on the size of the shop and whether the market is favorable or unfavorable for his prod- ucts. Because there will be a 5-year lease on the building that Jerry is thinking about using, he wants to make sure that he makes the correct decision. Jerry is also thinking about hiring his old marketing professor to conduct a marketing research study. If the study is conducted, the study could be favorable (i.e., predicting a favorable market) or unfavorable (i.e., predicting an unfavorable market). Develop a decision tree for Jerry.

3-39 Jerry Smith (see Problem 3-38) has done some analysis about the profitability of the bicycle shop. If Jerry builds the large bicycle shop, he will earn $60,000 if the market is favorable, but he will lose $40,000 if the market is unfavorable. The small shop will return a $30,000 profit in a favorable market and a $10,000 loss in an unfavorable market. At the present time, he believes that there is a 50–50 chance that the market will be favorable. His old marketing professor will charge him $5,000 for the marketing research. It is estimated that there is a 0.6 probability that the survey will be favorable. Furthermore, there is a 0.9 probability that the market will be favorable

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DisCUssion QUEsTions AnD PRoBLEMs  103

given a favorable outcome from the study. However, the marketing professor has warned Jerry that there is only a probability of 0.12 of a favorable market if the marketing research results are not favorable. Jerry is confused.

(a) Should Jerry use the marketing research? (b) Jerry, however, is unsure the 0.6 probability of

a favorable marketing research study is correct. How sensitive is Jerry’s decision to this prob- ability value? How far can this probability value deviate from 0.6 without causing Jerry to change his decision?

3-40 Bill Holliday is not sure what he should do. He can build a quadplex (i.e., a building with four apart- ments), build a duplex, gather additional informa- tion, or simply do nothing. If he gathers additional information, the results could be either favorable or unfavorable, but it would cost him $3,000 to gather the information. Bill believes that there is a 50–50 chance that the information will be favorable. If the rental market is favorable, Bill will earn $15,000 with the quadplex or $5,000 with the duplex. Bill doesn’t have the financial resources to do both. With an unfavorable rental market, however, Bill could lose $20,000 with the quadplex or $10,000 with the duplex. Without gathering additional informa- tion, Bill estimates that the probability of a favor- able rental market is 0.7. A favorable report from the study would increase the probability of a favorable rental market to 0.9. Furthermore, an unfavorable re- port from the additional information would decrease the probability of a favorable rental market to 0.4. Of course, Bill could forget all of these numbers and do nothing. What is your advice to Bill?

3-41 Peter Martin is going to help his brother who wants to open a food store. Peter initially believes that there is a 50–50 chance that his brother’s food store would be a success. Peter is considering doing a market re- search study. Based on historical data, there is a 0.8 probability that the marketing research will be favor- able given a successful food store. Moreover, there is a 0.7 probability that the marketing research will be unfavorable given an unsuccessful food store.

(a) If the marketing research is favorable, what is Peter’s revised probability of a successful food store for his brother?

(b) If the marketing research is unfavorable, what is Peter’s revised probability of a successful food store for his brother?

(c) If the initial probability of a successful food store is 0.60 (instead of 0.50), find the probabilities in parts (a) and (b).

3-42 Mark Martinko has been a class A racquetball player for the past 5 years, and one of his biggest goals is to own and operate a racquetball facility. Unfortunately,

Mark thinks that the chance of a successful racquet- ball facility is only 30%. Mark’s lawyer has recom- mended that he employ one of the local marketing research groups to conduct a survey concerning the success or failure of a racquetball facility. There is a 0.8 probability that the research will be favorable given a successful racquetball facility. In addition, there is a 0.7 probability that the research will be unfavorable given an unsuccessful facility. Com- pute revised probabilities of a successful racquetball facility given a favorable and given an unfavorable survey.

3-43 A financial advisor has recommended two possible mutual funds for investment: Fund A and Fund B. The return that will be achieved by each of these de- pends on whether the economy is good, fair, or poor. A payoff table has been constructed to illustrate this situation:

STATE OF NATURE

INVESTMENT GOOD

ECONOMY FAIR

ECONOMY POOR

ECONOMY

Fund A $10,000 $2,000 –$5,000

Fund B $6,000 $4,000 0

Probability 0.2 0.3 0.5

(a) Draw the decision tree to represent this situation. (b) Perform the necessary calculations to determine

which of the two mutual funds is better. Which one should you choose to maximize the expected value?

(c) Suppose there is a question about the return of Fund A in a good economy. It could be higher or lower than $10,000. What value for this would cause a person to be indifferent between Fund A and Fund B (i.e., the EMVs would be the same)?

3-44 Jim Sellers is thinking about producing a new type of electric razor for men. If the market were favor- able, he would get a return of $100,000, but if the market for this new type of razor were unfavorable, he would lose $60,000. Since Ron Bush is a good friend of Jim Sellers, Jim is considering the possibil- ity of using Bush Marketing Research to gather ad- ditional information about the market for the razor. Ron has suggested that Jim use either a survey or a pilot study to test the market. The survey would be a sophisticated questionnaire administered to a test market. It will cost $5,000. Another alternative is to run a pilot study. This would involve producing a limited number of the new razors and trying to sell them in two cities that are typical of American cit- ies. The pilot study is more accurate but is also more expensive. It will cost $20,000. Ron Bush has sug- gested that it would be a good idea for Jim to con- duct either the survey or the pilot before Jim makes the decision concerning whether to produce the new

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104  CHAPTER 3 • DECision AnALysis

razor. But Jim is not sure if the value of the survey or the pilot is worth the cost.

Jim estimates that the probability of a successful market without performing a survey or pilot study is 0.5. Furthermore, the probability of a favorable sur- vey result given a favorable market for razors is 0.7, and the probability of a favorable survey result given an unsuccessful market for razors is 0.2. In addition, the probability of an unfavorable pilot study given an unfavorable market is 0.9, and the probability of an unsuccessful pilot study result given a favorable market for razors is 0.2.

(a) Draw the decision tree for this problem without the probability values.

(b) Compute the revised probabilities needed to complete the decision, and place these values in the decision tree.

(c) What is the best decision for Jim? Use EMV as the decision criterion.

3-45 Jim Sellers has been able to estimate his utility for a number of different values. He would like to use these utility values in making the decision in Prob- lem 3-44: U1-$80,0002 = 0, U1-$65,0002 = 0.5,

U1-$60,0002 = 0.55, U1-$20,0002 = 0.7, U1-$5,0002 = 0.8, U1$02 = 0.81, U1$80,0002 =

0.9, U1$95,0002 = 0.95, and U1$100,0002 = 1. Resolve Problem 3-44 using utility values. Is Jim a

risk avoider?

3-46 Two states of nature exist for a particular situation: a good economy and a poor economy. An economic study may be performed to obtain more informa- tion about which of these will actually occur in the coming year. The study may forecast either a good economy or a poor economy. Currently there is a 60% chance that the economy will be good and a 40% chance that it will be poor. In the past, when- ever the economy was good, the economic study predicted it would be good 80% of the time. (The other 20% of the time the prediction was wrong.) In the past, whenever the economy was poor, the eco- nomic study predicted it would be poor 90% of the time. (The other 10% of the time the prediction was wrong.)

(a) Use Bayes’ Theorem and find the following: P1good economy � prediction of good economy2 P1poor economy � prediction of good economy2 P1good economy � prediction of poor economy2 P1poor economy � prediction of poor economy2

(b) Suppose the initial (prior) probability of a good economy is 70% (instead of 60%) and the initial probability of a poor economy is 30% (instead of

40%). Find the posterior probabilities in part (a) based on these new values.

3-47 The Long Island Life Insurance Company sells a term life insurance policy. If the policy holder dies during the term of the policy, the company pays $100,000. If the person does not die, the company pays out nothing, and there is no further value to the policy. The company uses actuarial tables to de- termine the probability that a person with certain characteristics will die during the coming year. For a particular individual, it is determined that there is a 0.001 chance that the person will die in the next year and a 0.999 chance that the person will live and the company will pay out nothing. The cost of this policy is $200 per year. Based on the EMV criterion, should the individual buy this insurance policy? How would utility theory help explain why a person would buy this insurance policy?

3-48 In Problem 3-37, you helped the medical profession- als analyze their decision using expected monetary value as the decision criterion. This group has also assessed its utility for money: U1-$45,0002 = 0, U1-$40,0002 = 0.1, U1-$5,0002 = 0.7, U1$02= 0.9, U1$95,0002 = 0.99, a n d U1$100,0002 = 1. Use expected utility as the decision criterion, and determine the best decision for the medical profes- sionals. Are the medical professionals risk seekers or risk avoiders?

3-49 In this chapter, a decision tree was developed for John Thompson (see Figure 3.5 for the complete de- cision tree analysis). After completing the analysis, John was not completely sure that he is indifferent to risk. After going through a number of standard gam- bles, John was able to assess his utility for money. Here are some of the utility assessments: U1-$190,0002 = 0, U1-$180,0002 = 0.05, U1-$30,0002 = 0.10, U1-$20,0002 = 0.15, U1-$10,0002 = 0.2, U1$02 = 0.3, U1$90,0002 = 0.15, U1$100,0002 = 0.6, U1$190,0002 = 0.95,

and U1$200,0002 = 1.0. If John maximizes his expected utility, does his decision change?

3-50 In the past few years, the traffic problems in Lynn McKell’s hometown have gotten worse. Now, Broad Street is congested about half the time. The normal travel time to work for Lynn is only 15 minutes when Broad Street is used and there is no con- gestion. With congestion, however, it takes Lynn 40 minutes to get to work. If Lynn decides to take the expressway, it will take 30 minutes, regardless of the traffic conditions. Lynn’s utility for travel time is U115 minutes2 = 0.9, U130 minutes2 = 0.7, a n d U140 minutes2 = 0.2.

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DisCUssion QUEsTions AnD PRoBLEMs  105

(a) Which route will minimize Lynn’s expected travel time?

(b) Which route will maximize Lynn’s utility? (c) When it comes to travel time, is Lynn a risk

seeker or a risk avoider?

3-51 Coren Chemical, Inc., develops industrial chemicals that are used by other manufacturers to produce pho- tographic chemicals, preservatives, and lubricants. One of their products, K-1000, is used by several photographic companies to make a chemical that is used in the film-developing process. To produce K-1000 efficiently, Coren Chemical uses the batch approach, in which a certain number of gallons is produced at one time. This reduces setup costs and allows Coren Chemical to produce K-1000 at a competitive price. Unfortunately, K-1000 has a very short shelf life of about 1 month.

Coren Chemical produces K-1000 in batches of 500 gallons, 1,000 gallons, 1,500 gallons, and 2,000 gallons. Using historical data, David Coren was able to determine that the probability of selling 500 gal- lons of K-1000 is 0.2. The probabilities of selling 1,000, 1,500, and 2,000 gallons are 0.3, 0.4, and 0.1, respectively. The question facing David is how many gallons to produce of K-1000 in the next batch run. K-1000 sells for $20 per gallon. Manufacturing cost is $12 per gallon, and handling costs and warehous- ing costs are estimated to be $1 per gallon. In the past, David has allocated advertising costs to K-1000 at $3 per gallon. If K-1000 is not sold after the batch run, the chemical loses much of its important prop- erties as a developer. It can, however, be sold at a salvage value of $13 per gallon. Furthermore, David has guaranteed to his suppliers that there will always be an adequate supply of K-1000. If David does run out, he has agreed to purchase a comparable chemi- cal from a competitor at $25 per gallon. David sells all of the chemical at $20 per gallon, so his shortage means that David loses the $5 to buy the more ex- pensive chemical.

(a) Develop a decision tree of this problem. (b) What is the best solution? (c) Determine the expected value of perfect

information.

3-52 The Jamis Corporation is involved with waste man- agement. During the past 10 years, it has become one of the largest waste disposal companies in the Midwest, serving primarily Wisconsin, Illinois, and Michigan. Bob Jamis, president of the company, is considering the possibility of establishing a waste treatment plant in Mississippi. From past experience, Bob believes that a small plant in northern Missis- sippi would yield a $500,000 profit, regardless of the market for the facility. The success of a medium- sized waste treatment plant would depend on the market. With a low demand for waste treatment, Bob

expects a $200,000 return. A medium demand would yield a $700,000 return in Bob’s estimation, and a high demand would return $800,000. Although a large facility is much riskier, the potential return is much greater. With a high demand for waste treat- ment in Mississippi, the large facility should return a million dollars. With a medium demand, the large facility will return only $400,000. Bob estimates that the large facility would be a big loser if there were a low demand for waste treatment. He estimates that he would lose approximately $200,000 with a large treatment facility if demand were indeed low. Look- ing at the economic conditions for the upper part of the state of Mississippi and using his experience in the field, Bob estimates that the probability of a low demand for treatment plants is 0.15. The probability of a medium demand for a waste treatment facility is approximately 0.40, and the probability of a high demand for a waste treatment facility is 0.45.

Because of the large potential investment and the possibility of a loss, Bob has decided to hire a market research team that is based in Jackson, Mis- sissippi. This team will perform a survey to get a better feeling for the probability of a low, medium, or high demand for a waste treatment facility. The cost of the survey is $50,000. To help Bob determine whether to go ahead with the survey, the marketing research firm has provided Bob with the following information:

P(survey results ∣ possible outcomes)

SURVEY RESULTS

POSSIBLE OUTCOME

LOW SURVEY RESULTS

MEDIUM SURVEY RESULTS

HIGH SURVEY RESULTS

Low demand 0.7 0.2 0.1

Medium demand 0.4 0.5 0.1

High demand 0.1 0.3 0.6

As you see, the survey could result in three pos- sible outcomes. Low survey results would mean that a low demand is likely. In a similar fashion, medium survey results or high survey results would mean a medium or a high demand, respectively. What should Bob do?

3-53 Mary is considering opening a new grocery store in town. She is evaluating three sites: downtown, the mall, and out at the busy traffic circle. Mary cal- culated the value of successful stores at these loca- tions as follows: downtown, $250,000; the mall, $300,000; the circle, $400,000. Mary calculated the losses if unsuccessful to be $100,000 at either down- town or the mall and $200,000 at the circle. Mary figures her chance of success to be 50% downtown, 60% at the mall, and 75% at the traffic circle.

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106  CHAPTER 3 • DECision AnALysis

(a) Draw a decision tree for Mary and select her best alternative.

(b) Mary has been approached by a marketing re- search firm that offers to study the area to de- termine if another grocery store is needed. The cost of this study is $30,000. Mary believes there is a 60% chance that the survey results will be positive (show a need for another grocery store). SRP = survey results positive, SRN = sur- vey results negative, SD = success downtown, SM = success at mall, SC = success at circle, SD′ = don’t succeed downtown, and so on. For studies of this nature: P1SRP � success2 = 0.7, P1SRN � success2 = 0.3, P1SRP � not success2 = 0.2, and P1SRN � not success2 = 0.8.

Calculate the revised probabilities for success (and not success) for each location, depending on survey results. (c) How much is the marketing research worth to

Mary? Calculate the EVSI.

3-54 Sue Reynolds has to decide if she should get in- formation (at a cost of $20,000) to help her decide whether to invest in a retail store. If she gets the in- formation, there is a 0.6 probability that the infor- mation will be favorable and a 0.4 probability that the information will not be favorable. If the informa- tion is favorable, there is a 0.9 probability that the store will be a success. If the information is not fa- vorable, the probability of a successful store is only 0.2. Without any information, Sue estimates that the probability of a successful store will be 0.6. A suc- cessful store will give a return of $100,000. If the store is built but is not successful, Sue will see a loss of $80,000. Of course, she could always decide not to build the retail store.

(a) What do you recommend?

(b) What impact would a 0.7 probability of obtain- ing favorable information have on Sue’s deci- sion? The probability of obtaining unfavorable information would be 0.3.

(c) Sue believes that the probabilities of a successful and an unsuccessful retail store, given favorable information, might be 0.8 and 0.2, respectively, instead of 0.9 and 0.1, respectively. What im- pact, if any, would this have on Sue’s decision and the best EMV?

(d) Sue had to pay $20,000 to get information. Would her decision change if the cost of the in- formation increased to $30,000?

(e) Using the data in this problem and the following utility table, compute the expected utility. Is this the curve of a risk seeker or a risk avoider?

MONETARY VALUE UTILITY

$100,000 1

$80,000 0.4

$0 0.2

–$20,000 0.1

–$80,000 0.05

–$100,000 0

(f) Compute the expected utility given the follow- ing utility table. Does this utility table represent a risk seeker or a risk avoider?

MONETARY VALUE UTILITY

$100,000 1

$80,000 0.9

$0 0.8

–$20,000 0.6

–$80,000 0.4

–$100,000 0

See our Internet home page, at www.pearsonhighered.com/render, for additional homework problems, Problems 3-55 to 3-60.

Internet Homework Problems

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CAsE sTUDy  107

After watching a movie about a young woman who quit a suc- cessful corporate career to start her own baby food company, Julia Day decided that she wanted to do the same. In the movie, the baby food company was very successful. Julia knew, how- ever, that it is much easier to make a movie about a successful woman starting her own company than to actually do it. The product had to be of the highest quality, and Julia had to get the best people involved to launch the new company. Julia resigned from her job and launched her new company–-Starting Right.

Julia decided to target the upper end of the baby food mar- ket by producing baby food that contained no preservatives but had a great taste. Although the price would be slightly higher than for existing baby food, Julia believed that parents would be willing to pay more for a high-quality baby food. Instead of putting baby food in jars, which would require preservatives to stabilize the food, Julia decided to try a new approach. The baby food would be frozen. This would allow for natural ingredients, no preservatives, and outstanding nutrition.

Getting good people to work for the new company was also important. Julia decided to find people with experience in fi- nance, marketing, and production to get involved with Starting Right. With her enthusiasm and charisma, Julia was able to find such a group. Their first step was to develop prototypes of the new frozen baby food and to perform a small pilot test of the new product. The pilot test received rave reviews.

The final key to getting the young company off to a good start was to raise funds. Three options were considered: corpo- rate bonds, preferred stock, and common stock. Julia decided that each investment should be in blocks of $30,000. Further- more, each investor should have an annual income of at least $40,000 and a net worth of $100,000 to be eligible to invest in Starting Right. Corporate bonds would return 13% per year for the next 5 years. Julia furthermore guaranteed that investors in the corporate bonds would get at least $20,000 back at the end

of 5 years. Investors in preferred stock should see their initial investment increase by a factor of 4 with a good market or see the investment worth only half of the initial investment with an unfavorable market. The common stock had the greatest poten- tial. The initial investment was expected to increase by a factor of 8 with a good market, but investors would lose everything if the market was unfavorable. During the next 5 years, it was expected that inflation would increase by a factor of 4.5% each year.

Discussion Question 1. Sue Pansky, a retired elementary school teacher, is con-

sidering investing in Starting Right. She is very conserva- tive and is a risk avoider. What do you recommend?

2. Ray Cahn, who is currently a commodities broker, is also considering an investment, although he believes that there is only an 11% chance of success. What do you recommend?

3. Lila Battle has decided to invest in Starting Right. While she believes that Julia has a good chance of being success- ful, Lila is a risk avoider and very conservative. What is your advice to Lila?

4. George Yates believes that there is an equally likely chance for success or failure. What is your recommendation?

5. Peter Metarko is extremely optimistic about the market for the new baby food. What is your advice for Pete?

6. Julia Day has been told that developing the legal docu- ments for each fund-raising alternative is expensive. Julia would like to offer alternatives for both risk-averse and risk-seeking investors. Can Julia delete one of the finan- cial alternatives and still offer investment choices for risk seekers and risk avoiders?

Case Study

Starting Right Corporation

The Toledo Leather Company has been producing leather goods for more than 30 years. It purchases prepared hides from tanners and produces leather clothing accessories such as wal- lets, belts, and handbags. The firm has just developed a new leather product and has prepared a 1-year production and sales plan for it. The new product is best described as a combination billfold, key case, and credit card carrier. As company presi- dent Peggy Lane has noted, “It is a super carryall for small this-and-that.” Lane has placed her administrative assistant, Harold Hamilton, in charge of the project.

Hamilton has established that material and variable over- head for the carryall should be about $1.50 per case over the next year given a 5-day week and no overtime. Unit labor and machining costs, however, depend on the choice of machine that will be used for production. Hamilton has narrowed the choice

down to two specialized pieces of equipment. Machine 1 is a semi-automated machine that will cut the material to the size needed for one unit and also will sew it, install the rings and snaps, and emboss it with two types of designs. This machine costs $250,000 and will add $2.50 per case to the average vari- able cost for labor and other machine-related costs. This piece of equipment has a production capacity of 640 units per day. However, estimated downtime for maintenance and repairs is 12.5% (1/8 of the total time).

Machine 2 is fully automated. It cuts, sews, and installs rings and snaps and is capable of embossing the case with three types of designs. This machine costs $350,000 and will add $1.75 per case to the average variable cost for labor and other machine-related costs. Machine 2 has a higher production ca- pacity (estimated at 800 units per day) than the semi-automated

Case Study

Toledo Leather Company

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108  CHAPTER 3 • DECision AnALysis

machine. However, estimated downtime is 25% (1/4 of the total time), consistent with its great complexity.

Marketing estimates for the next year have been more dif- ficult to project than production costs and capacity estimates. However, $6.00 seems the likeliest selling price for the carryall. The price brings it in line with somewhat comparable products on the market, but because the carryall offers more features than these other products, it has the potential to outsell them. Sales volume estimates center on 140,000 units for the year, but anal- ysis of the potential market has been difficult because this new product is so different from the products now being sold. Ham- ilton’s best estimates of sales at $6.00 per unit and the probabili- ties attached to these volumes are as follows:

SALES VOLUME PROBABILITY

120,000 units 0.15

130,000 units 0.25

140,000 units 0.40

150,000 units 0.15

160,000 units 0.05

Given these marketing estimates and the machine capaci- ties, the company will have to decide either to modify the ma- chines to increase capacity or to work overtime if demand is at the higher levels. Management can make this decision based on

the first week’s worth of sales, which are expected to be a good indicator of the annual sales level. Overtime premiums would raise the costs by $1.20 per case on the semi-automated ma- chine and by $0.90 on the fully automated machine. Modifica- tion of Machine 1, the semi-automated one, would cost $15,000 to meet the highest level of sales. Modification of Machine 2 would cost $20,000.

Lane has directed Hamilton to make a decision based on first-year sales, since demand for a product such as this is un- certain after its initial popularity passes. Toledo operates on a 50-week year because the company usually closes down for the winter holidays.

Discussion Questions 1. Using a decision tree based on maximizing expected

profit, decide which machine Toledo Leather should se- lect. Should overtime be scheduled? Or should a machine be modified and, if so, under what circumstances?

2. Set up a payoff matrix for the sales volumes given (as- sume the machines cannot be modified and overtime is used), and assume that the probabilities for the five lev- els of sales are not known. Then decide which machine should be purchased using the maximax criterion, the maximin criterion, and the equally likely criterion.

Source: Professor Emeritus Michael Ballot, ESB, University of the Pacific.

In 1979, Steve Blake founded Blake Electronics in Long Beach, California, to manufacture resistors, capacitors, inductors, and other electronic components. During the Vietnam War, Steve was a radio operator, and it was during this time that he became proficient at repairing radios and other communications equip- ment. Steve viewed his 4-year experience with the army with mixed feelings. He hated army life, but this experience gave him the confidence and the initiative to start his own electron- ics firm.

Over the years, Steve kept the business relatively un- changed. By 1992, total annual sales were in excess of $2 million. In 1996, Steve’s son, Jim, joined the company after finishing high school and 2 years of courses in electronics at Long Beach Community College. Jim was always aggressive in high school athletics, and he became even more aggressive as general sales manager of Blake Electronics. This aggressiveness bothered Steve, who was more conservative. Jim would make deals to supply companies with electronic components before he bothered to find out if Blake Electronics had the ability or capacity to produce the components. On several occasions, this behavior caused the company some embarrassing moments when Blake Electronics was unable to produce the electronic components for companies with which Jim had made deals.

In 2000, Jim started to go after government contracts for electronic components. By 2002, total annual sales had in- creased to more than $10 million, and the number of employees

exceeded 200. Many of these employees were electronic spe- cialists and graduates of electrical engineering programs from top colleges and universities. But Jim’s tendency to stretch Blake Electronics to take on additional contracts continued as well, and by 2007, Blake Electronics had a reputation with gov- ernment agencies as a company that could not deliver what it promised. Almost overnight, government contracts stopped, and Blake Electronics was left with an idle workforce and unused manufacturing equipment. This high overhead started to melt away profits, and in 2009, Blake Electronics was faced with the possibility of sustaining a loss for the first time in its history.

In 2010, Steve decided to look at the possibility of manu- facturing electronic components for home use. Although this was a totally new market for Blake Electronics, Steve was con- vinced that this was the only way to keep Blake Electronics from dipping into the red. The research team at Blake Electron- ics was given the task of developing new electronic devices for home use. The first idea from the research team was the Mas- ter Control Center. The basic components for this system are shown in Figure 3.15.

The heart of the system is the master control box. This unit, which would have a retail price of $250, has two rows of five buttons. Each button controls one light or appliance and can be set as either a switch or a rheostat. When set as a switch, a light finger touch on the button turns a light or appliance on or off. When set as a rheostat, a finger touching the button controls the

Case Study

Blake Electronics

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CAsE sTUDy  109

Master Control Box

Outlet Adapter

Light Switch Adapter

Lightbulb Disk

B LA

K E

TABLE 3.19 success Figures for MAi

SURVEY RESULTS

OUTCOME FAVORABLE UNFAVORABLE TOTAL

Successful venture

35 20 55

Unsuccessful venture

15 30 45

FIGURE 3.15 Master Control Center

intensity of the light. Leaving your finger on the button makes the light go through a complete cycle ranging from off to bright and back to off again.

To allow for maximum flexibility, each master control box is powered by two D-sized batteries that can last up to a year, depending on usage. In addition, the research team has devel- oped three versions of the master control box—versions A, B, and C. If a family wants to control more than 10 lights or appli- ances, another master control box can be purchased.

The lightbulb disk, which would have a retail price of $2.50, is controlled by the master control box and is used to control the intensity of any light. A different disk is available for each button position for all three master control boxes. By in- serting the lightbulb disk between the lightbulb and the socket, the appropriate button on the master control box can completely control the intensity of the light. If a standard light switch is used, it must be on at all times for the master control box to work.

One disadvantage of using a standard light switch is that only the master control box can be used to control the particular light. To avoid this problem, the research team developed a spe- cial light switch adapter that would sell for $15. When this de- vice is installed, either the master control box or the light switch adapter can be used to control the light.

When used to control appliances other than lights, the mas- ter control box must be used in conjunction with one or more outlet adapters. The adapters are plugged into a standard wall outlet, and the appliance is then plugged into the adapter. Each outlet adapter has a switch on top that allows the appliance to be controlled from the master control box or the outlet adapter. The price of each outlet adapter would be $25.

The research team estimated that it would cost $500,000 to develop the equipment and procedures needed to manufacture the master control box and accessories. If successful, this ven- ture could increase sales by approximately $2 million. But will the master control boxes be a successful venture? With a 60%

chance of success estimated by the research team, Steve had serious doubts about trying to market the master control boxes even though he liked the basic idea. Because of his reservations, Steve decided to send requests for proposals (RFPs) for addi- tional marketing research to 30 marketing research companies in southern California.

The first RFP to come back was from a small company called Marketing Associates, Inc. (MAI), which would charge $100,000 for the survey. According to its proposal, MAI has been in business for about 3 years and has conducted about 100 marketing research projects. MAI’s major strengths appeared to be individual attention to each account, experienced staff, and fast work. Steve was particularly interested in one part of the proposal, which revealed MAI’s success record with previous accounts. This is shown in Table 3.19.

The only other proposal to be returned was by a branch office of Iverstine and Walker, one of the largest marketing research firms in the country. The cost for a complete survey would be $300,000. While the proposal did not contain the same success record as MAI, the proposal from Iverstine and Walker did contain some interesting information. The chance of get- ting a favorable survey result, given a successful venture, was 90%. On the other hand, the chance of getting an unfavorable survey result, given an unsuccessful venture, was 80%. Thus, it appeared to Steve that Iverstine and Walker would be able to predict the success or failure of the master control boxes with a great amount of certainty.

Steve pondered the situation. Unfortunately, both market- ing research teams gave different types of information in their proposals. Steve concluded that there would be no way that the two proposals could be compared unless he got additional information from Iverstine and Walker. Furthermore, Steve wasn’t sure what he would do with the information and whether it would be worth the expense of hiring one of the marketing research firms.

Discussion Questions 1. Does Steve need additional information from Iverstine

and Walker? 2. What would you recommend?

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110  CHAPTER 3 • DECision AnALysis

See our Internet home page, at www.pearsonhighered.com/render, for these additional case studies: (1) Drink-At-Home, Inc.: This case involves the development and marketing of a new beverage. (2) Ruth Jones’ Heart Bypass Operation: This case deals with a medical decision regarding

surgery. (3) Ski Right: This case involves the development and marketing of a new ski helmet. (4) Study Time: This case is about a student who must budget time while studying for a final exam.

Internet Case Studies

Bibliography

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Carassus, Laurence, and Miklós Rásonyi. “Optimal Strategies and Utility- Based Prices Converge When Agents’ Preferences Do,” Mathematics of Operations Research 32, 1 (February 2007): 102–117.

Congdon, Peter. Bayesian Statistical Modeling, 2nd ed., New York: John Wiley & Sons, Inc., 2007.

Ewing, Paul L., Jr. “Use of Decision Analysis in the Army Base Realignment and Closure (BRAC) 2005 Military Value Analysis,” Decision Analysis 3 (March 2006): 33–49.

Hammond, J. S., R. L. Keeney, and H. Raiffa. “The Hidden Traps in Decision Making,” Harvard Business Review (September–October 1998): 47–60.

Hurley, William J. “The 2002 Ryder Cup: Was Strange’s Decision to Put Tiger Woods in the Anchor Match a Good One?” Decision Analysis 4, 1 (March 2007): 41–45.

Kirkwood, C. W. “An Overview of Methods for Applied Decision Analysis,” Interfaces 22, 6 (November–December 1992): 28–39.

Kirkwood, C. W. “Approximating Risk Aversion in Decision Analysis Appli- cations,” Decision Analysis 1 (March 2004): 51–67.

Luce, R. and H. Raiffa. Games and Decisions: Introduction and Critical Survey. New York: Wiley, 1957 (Reprinted in 1989 by Dover Publications).

Maxwell, Daniel T. “Improving Hard Decisions,” OR/MS Today 33, 6 (December 2006): 51–61.

Patchak, William M. “Software Survey: Decision Analysis.” OR/MS Today 39, 5 (October 2012).

Paté-Cornell, M. Elisabeth, and Robin L. Dillon. “The Respective Roles of Risk and Decision Analyses in Decision Support,” Decision Analysis 3 (December 2006): 220–232.

Pennings, Joost M. E., and Ale Smidts. “The Shape of Utility Functions and Organizational Behavior,” Management Science 49, 9 (September 2003): 1251–1263.

Raiffa, Howard, John W. Pratt, and Robert Schlaifer. Introduction to Statistical Decision Theory. Boston: MIT Press, 1995.

Raiffa, Howard, and Robert Schlaifer. Applied Statistical Decision Theory. New York: John Wiley & Sons, Inc., 2000.

Render, B., and R. M. Stair. Cases and Readings in Management Science, 2nd ed. Boston: Allyn & Bacon, Inc., 1988.

Schlaifer, R. Analysis of Decisions Under Uncertainty. New York: McGraw- Hill Book Company, 1969.

Smith, James E., and Robert L. Winkler. “The Optimizer’s Curse: Skepticism and Postdecision Surprise in Decision Analysis,” Management Science 52 (March 2006): 311–322.

van Binsbergen, Jules H., and Leslie M. Marx. “Exploring Relations Between Decision Analysis and Game Theory,” Decision Analysis 4, 1 (March 2007): 32–40.

Wallace, Stein W. “Decision Making Under Uncertainty: Is Sensitivity Analy- sis of Any Use?” Operations Research 48, 1 (2000): 20–25.

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 111

4.6 Use computer software for regression analysis.

4.7 Develop a multiple regression model and use it for prediction purposes.

4.8 Use dummy variables to model categorical data.

4.9 Determine which variables should be included in a multiple regression model.

4.10 Transform a nonlinear function into a linear one for use in regression.

4.11 Understand and avoid mistakes commonly made in the use of regression analysis.

4.1 Identify variables, visualize them in a scatter diagram, and use them in a regression model.

4.2 Develop simple linear regression equations from sample data and interpret the slope and intercept.

4.3 Calculate the coefficient of determination and the coefficient of correlation and interpret their meanings.

4.4 List the assumptions used in regression and use residual plots to identify problems.

4.5 Interpret the F test in a linear regression model.

After completing this chapter, students will be able to:

LEARNING OBJECTIVES

Regression Models

4 CHAPTER

Regression analysis is a very valuable tool for today’s manager. Regression has been used to model things such as the relationship between level of education and income, the price of a house and the square footage, and the sales volume for a company relative to the dollars spent on advertising. When businesses are trying to decide which location is best for a new store or branch office, regression models are often used. Cost estimation models are often regression models. The applicability of regression analysis is virtually limitless.

There are generally two purposes for regression analysis. The first is to understand the relationship between variables such as advertising expenditures and sales. The second purpose is to predict the value of one variable based on the value of the other. Because of this, regression is a very important forecasting technique and will be addressed again in Chapter 5.

In this chapter, the simple linear regression model will first be developed, and then a more complex multiple regression model will be used to incorporate even more variables into our model. In any regression model, the variable to be predicted is called the depen- dent variable or response variable. The value of this is said to be dependent upon the value of an independent variable, which is sometimes called an explanatory variable or a predictor variable.

Two purposes of regression analysis are to understand the relationship between variables and to predict the value of one based on the other.

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112  CHAPTER 4 • REgREssIon MoDELs

4.1 Scatter Diagrams

To investigate the relationship between variables, it is helpful to look at a graph of the data. Such a graph is often called a scatter diagram or a scatter plot. Normally, the independent variable is plotted on the horizontal axis and the dependent variable is plotted on the vertical axis. The following example will illustrate this.

Triple A Construction Company renovates old homes in Albany. Over time, the company has found that its dollar volume of renovation work is dependent on the Albany area payroll. The figures for Triple A’s revenues and the amount of money earned by wage earners in Albany for the past 6 years are presented in Table 4.1. Economists have predicted the local area payroll to be $600 million next year, and Triple A wants to plan accordingly.

Figure 4.1 provides a scatter diagram for the Triple A Construction data given in Table 4.1. This graph indicates that higher values for the local payroll seem to result in higher sales for the company. There is not a perfect relationship because not all the points lie in a straight line, but there is a relationship. A line has been drawn through the data to help show the relationship that exists between the payroll and sales. The points do not all lie on the line, so there would be some error involved if we tried to predict sales based on payroll using this or any other line. Many lines could be drawn through these points, but which one best represents the true relationship? Regression analysis provides the answer to this question.

A scatter diagram is a graph of the data.

TRIPLE A'S SALES ($100,000s)

LOCAL PAYROLL

($100,000,000s)

6 3

8 4

9 6

5 4

4.5 2

9.5 5

TABLE 4.1 Triple A Construction Company sales and Local Payroll

FIGURE 4.1 scatter Diagram of Triple A Construction Company Data

0 1 2 0

2

4

6

8

10

12

3 4 5 6 7 8 Payroll ($100,000,000s)

S al

es ($

10 0,

00 0s

)

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4.2 sIMPLE LInEAR REgREssIon  113

4.2 Simple Linear Regression

In any regression model, there is an implicit assumption (which can be tested) that a relationship exists between the variables. There is also some random error that cannot be predicted. The un- derlying simple linear regression model is

Y = b0 + b1X + P (4-1)

where

Y = dependent variable (response variable)

X = independent variable (predictor variable or explanatory variable)

b0 = intercept (value of Y when X = 0)

b1 = slope of regression line

P = random error

The true values for the intercept and slope are not known, and therefore they are estimated using sample data. The regression equation based on sample data is given as

Yn = b0 + b1X (4-2)

where

Yn = predicted value of Y

b0 = estimate of b0, based on sample results

b1 = estimate of b1, based on sample results

In the Triple A Construction example, we are trying to predict the sales, so the dependent variable (Y) would be sales. The variable we use to help predict sales is the Albany area payroll, so this is the independent variable (X). Although any number of lines can be drawn through these points to show a relationship between X and Y in Figure 4.1, the line that will be chosen is the one that in some way minimizes the errors. Error is defined as

Error = 1Actual value2 - 1Predicted value2 e = Y - Yn (4-3)

Since errors may be positive or negative, the average error could be zero even though there are extremely large errors—both positive and negative. To eliminate the difficulty of negative errors canceling positive errors, the errors can be squared. The best regression line will be defined as the one with the minimum sum of the squared errors. For this reason, regression analysis is sometimes called least squares regression.

Statisticians have developed formulas that we can use to find the equation of a straight line that would minimize the sum of the squared errors. The simple linear regression equation is

Yn = b0 + b1X

The following formulas can be used to compute the slope and the intercept:

X = gX n

= Average 1mean2 of X values

Y = gY n

= Average 1mean2 of Y values

b1 = g1X - X21Y - Y2

g1X - X22 (4-4) b0 = Y - b1X (4-5)

The dependent variable is Y and the independent variable is X.

Estimates of the slope and intercept are found from sample data.

The regression line minimizes the sum of the squared errors.

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114  CHAPTER 4 • REgREssIon MoDELs

The preliminary calculations are shown in Table 4.2. There are other “shortcut” formulas that are helpful when doing the computations on a calculator, and these are presented in Appendix 4.1. They will not be shown here, as computer software will be used for most of the other examples in this chapter.

Computing the slope and the intercept of the regression equation for the Triple A Construc- tion Company example, we have

X = gX 6

= 24

6 = 4

Y = gX 6

= 42

6 = 7

b1 = g1X - X21Y - Y2

g1X - X22 = 12.5

10 = 1.25

b0 = Y - b1X = 7 - 11.252142 = 2 The estimated regression equation therefore is

Yn = 2 + 1.25X

or

Sales = 2 + 1.251Payroll2 If the payroll next year is $600 million 1X = 62, then the predicted value would be

Yn = 2 + 1.25162 = 9.5 or $950,000.

One of the purposes of regression is to understand the relationship among variables. This model tells us that each time the payroll increases by $100 million (represented by X), we would expect the sales to increase by $125,000, since b1 = 1.25 ($100,000). This model helps Triple A Construction see how the local economy and company sales are related.

4.3 Measuring the Fit of the Regression Model

A regression equation can be developed for any variables X and Y, even random numbers. We certainly would not have any confidence in the ability of one random number to predict the value of another random number. How do we know that the model is actually helpful in predicting Y based on X? Should we have confidence in this model? Does the model provide better predic- tions (smaller errors) than simply using the average of the Y values?

In the Triple A Construction example, sales figures (Y) varied from a low of 4.5 to a high of 9.5, and the mean was 7. If each sales value is compared with the mean, we see how far they deviate from the mean, and we could compute a measure of the total variability in sales. Because

Deviations (errors) may be positive or negative.

Y X 1X − X 22 1X − X 2 1Y − Y 2 6 3 13 - 422 = 1 13 - 4216 - 72 = 1 8 4 14 - 422 = 0 14 - 4218 - 72 = 0 9 6 16 - 422 = 4 16 - 4219 - 72 = 4 5 4 14 - 422 = 0 14 - 4215 - 72 = 0 4.5 2 12 - 422 = 4 12 - 4214.5 - 72 = 5 9.5 5 15 - 422 = 1 15 - 4219.5 - 72 = 2.5

gY = 42 Y = 42>6 = 7

gX = 24 X = 24>6 = 4

g1X - X22 = 10 g1X - X21Y - Y2 = 12.5

TABLE 4.2 Regression Calculations for Triple A Construction

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4.3 MEAsURIng THE FIT oF THE REgREssIon MoDEL  115

Y is sometimes higher and sometimes lower than the mean, there may be both positive and nega- tive deviations. Simply summing these values would be misleading because the negatives would cancel out the positives, making it appear that the numbers are closer to the mean than they ac- tually are. To prevent this problem, we will use the sum of squares total (SST) to measure the total variability in Y:

SST = g1Y - Y22 (4-6) If we did not use X to predict Y, we would simply use the mean of Y as the prediction, and

the SST would measure the accuracy of our predictions. However, a regression line may be used to predict the value of Y, and while there are still errors involved, the sum of these squared errors will be less than the total sum of squares just computed. The sum of squares error (SSE) is

SSE = ge2 = g1Y - Yn22 (4-7) Table 4.3 provides the calculations for the Triple A Construction example. The mean

1Y = 72 is compared to each value, and we get SST = 22.5

The prediction 1Yn2 for each observation is computed and compared to the actual value. This results in

SSE = 6.875

The SSE is much lower than the SST. Using the regression line has reduced the variability in the sum of squares by 22.5 - 6.875 = 15.625. This is called the sum of squares regression (SSR) and indicates how much of the total variability in Y is explained by the regression model. Mathematically, this can be calculated as

SSR = g1Yn - Y22 (4-8) Table 4.3 indicates

SSR = 15.625

There is a very important relationship among the sums of squares that we have computed:

1Sum of squares total2 = 1Sum of squares due to regression2 + 1Sum of squares error2 SST = SSR + SSE (4-9)

Figure 4.2 displays the data for Triple A Construction. The regression line is shown, as is a line representing the mean of the Y values. The errors used in computing the sums of squares are shown on this graph. Notice how the sample points are closer to the regression line than they are to the mean.

The SST measures the total variability in Y about the mean.

The SSE measures the variability in Y about the regression line.

TABLE 4.3 sum of squares for Triple A Construction

Y X 1Y − Y 22 Yn 1Y - Yn 22 1Yn - Y 22

6 3 16 - 722 = 1 2 + 1.25132 = 5.75 0.0625 1.563 8 4 18 - 722 = 1 2 + 1.25142 = 7.00 1 0 9 6 19 - 722 = 4 2 + 1.25162 = 9.50 0.25 6.25 5 4 15 - 722 = 4 2 + 1.25142 = 7.00 4 0

4.5 2 14.5 - 722 = 6.25 2 + 1.25122 = 4.50 0 6.25 9.5 5 19.5 - 722 = 6.25 2 + 1.25152 = 8.25 1.5625 1.563

Y = 7 g1Y - Y22 = 22.5 SST = 22.5

g1Y - Yn22 = 6.875 SSE = 6.875

g1Yn - Y22 = 15.625 SSR = 15.625

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116  CHAPTER 4 • REgREssIon MoDELs

Coefficient of Determination The SSR is sometimes called the explained variability in Y, while the SSE is the unexplained variability in Y. The proportion of the variability in Y that is explained by the regression equation is called the coefficient of determination and is denoted by r2. Thus,

r2 = SSR

SST = 1 -

SSE

SST (4-10)

Either the SSR or the SSE can be used to find r2. For Triple A Construction, we have

r2 = 15.625

22.5 = 0.6944

This means that about 69% of the variability in sales (Y) is explained by the regression equation based on payroll (X).

If every point in the sample were on the regression line (meaning all errors are 0), then 100% of the variability in Y could be explained by the regression equation, so r2 = 1 and SSE = 0. The lowest possible value of r2 is 0, indicating that X explains 0% of the variability in Y. Thus, r2 can range from a low of 0 to a high of 1. In developing regression equations, a good model will have an r2 value close to 1.

Correlation Coefficient Another measure related to the coefficient of determination is the coefficient of correla- tion. This measure also expresses the degree or strength of the linear relationship. It is usu- ally expressed as r and can be any number between and including +1 and -1. Figure 4.3 illustrates possible scatter diagrams for different values of r. The value of r is the square root of r2. It is negative if the slope is negative, and it is positive if the slope is positive. Thus,

r = ±2r2 (4-11) For the Triple A Construction example with r2 = 0.6944,

r = 10.6944 = 0.8333 We know it is positive because the slope is +1.25.

r2 is the proportion of variability in Y that is explained by the regression equation.

FIGURE 4.2 Deviations from the Regression Line and from the Mean

0 1 2 0

2

4

6

8

10

12

3 4 5 6 7 8 Payroll ($100,000,000s)

S al

es ($

10 0,

00 0s

)

Y – Y ^ Y

Y – Y Y – Y

^

Y = 2 1 1.25 X ^

Y

X

The correlation coefficient ranges from -1 to +1.

If every point lies on the regression line, r2 = 1 and SSE = 0.

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4.4 AssUMPTIons oF THE REgREssIon MoDEL  117

(a) Perfect Positive Correlation: r 5 11

Y

X (b) Positive Correlation: 0 , r ,1

Y

X

(c) No Correlation: r 5 0

Y

X (d ) Perfect Negative Correlation: r 5 21

Y

X

FIGURE 4.3 Four Values of the Correlation Coefficient

4.4 Assumptions of the Regression Model

If we can make certain assumptions about the errors in a regression model, we can perform statistical tests to determine if the model is useful. The following assumptions are made about the errors:

1. The errors are independent. 2. The errors are normally distributed. 3. The errors have a mean of zero. 4. The errors have a constant variance (regardless of the value of X).

Using Regression as Part of Improvement Initiative at Trane/Ingersoll Rand

Trane U.S. Inc. is a brand of Ingersoll Rand and is a leader in both home and commercial air conditioning. In an effort to im- prove quality and efficiency, the company consolidated its man- ufacturing of evaporator and condenser coils into one location in South Carolina. A key part of the coil unit is a header, which controls the flow of water or refrigerant through the coil. With about 120 unique headers produced on each shift, with increased demand for their air conditioners due to federal refrigerant stan- dards, and with growing pressure to reduce costs, Trane began an improvement initiative in 2010.

Value stream mapping, a lean manufacturing technique, was used to visualize the entire process and identify waste in the system. A new layout was suggested to improve processing and reduce the overall time of production. To predict how well the new layout would perform in the manufacture of the headers, data were collected about the time to move materials into posi- tion, time to set up a job, time to set up a machine, and time to process a job. Several regression models were developed to

determine which job characteristics were most significant in the various times involved in processing and to measure the variabil- ity in times. These were used in simulation models that were developed to determine the overall impact of the new layout and to determine where bottlenecks might occur. One of the bottle- necks, at a robotic machine for welding, was eliminated by re- tooling the robot to allow the machine operator to operate both sides of the robot. This enables the robot to process two headers simultaneously.

The improvement initiatives from this project are estimated to save over $700,000 per year. This project highlighted the im- portance of data-based analytics and lean manufacturing tools in identifying the improvement tactics for complex value streams. The company plans similar improvement initiatives for other value streams in the future.

Source: Based on John B. Jensen, Sanjay L. Ahire, and Manoj K. Malhotra, “Trane/Ingersoll Rand Combines Lean and Operations Research Tools to Redesign Feeder Manufacturing Operations,” Interfaces 43, 4 (July–August 2013): 325–340, © Trevor S. Hale.

IN ACTION

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118  CHAPTER 4 • REgREssIon MoDELs

It is possible to check the data to see if these assumptions are met. Often a plot of the residu- als will highlight any glaring violations of the assumptions. When the errors (residuals) are plot- ted against the independent variable, the pattern should appear random.

Figure 4.4 presents some typical error patterns, with Figure 4.4A displaying a pattern that is expected when the assumptions are met and the model is appropriate. The errors are random and no discernible pattern is present. Figure 4.4B demonstrates an error pattern in which the errors increase as X increases, violating the constant variance assumption. Figure 4.4C shows errors consistently increasing at first and then consistently decreasing. A pattern such as this would indicate that the model is not linear and some other form (perhaps quadratic) should be used. In general, patterns in the plot of the errors indicate problems with the assumptions or the model specification.

A plot of the errors may highlight problems with the model.

E rr

or

X

E rr

or

X

E rr

or

X

FIGURE 4.4A Pattern of Errors Indicating Randomness

FIGURE 4.4B nonconstant Error Variance

FIGURE 4.4C Pattern of Errors Indicating Relationship Is not Linear

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4.5 TEsTIng THE MoDEL FoR sIgnIFICAnCE  119

Estimating the Variance While the errors are assumed to have constant variance 1s22, this is usually not known. It can be estimated from the sample results. The estimate of s2 is the mean squared error (MSE) and is denoted by s2. The MSE is the sum of squares due to error divided by the degrees of freedom:1

s2 = MSE = SSE

n - k - 1 (4-12)

where

n = number of observations in the sample k = number of independent variables

In the Triple A Construction example, n = 6 and k = 1, so

s2 = MSE = SSE

n - k - 1 =

6.8750

6 - 1 - 1 =

6.8750

4 = 1.7188

From this, we can estimate the standard deviation as

s = 1MSE (4-13) This is called the standard error of the estimate or the standard deviation of the regres-

sion. In this example,

s = 1MSE = 11.7188 = 1.31 This is used in many of the statistical tests about the model. It is also used to find interval

estimates for both Y and regression coefficients.2

4.5 Testing the Model for Significance

Both the MSE and r2 provide a measure of accuracy in a regression model. However, when the sample size is too small, it is possible to get good values for both of these even if there is no re- lationship between the variables in the regression model. To determine whether these values are meaningful, it is necessary to test the model for significance.

To see if there is a linear relationship between X and Y, a statistical hypothesis test is per- formed. The underlying linear model was given in Equation 4-1 as

Y = b0 + b1X + P

If b1 = 0, then Y does not depend on X in any way. The null hypothesis says there is no linear relationship between the two variables (i.e., b1 = 0). The alternate hypothesis is that there is a linear relationship (i.e., b1 ≠ 0). If the null hypothesis can be rejected, then we have proven that a linear relationship does exist, so X is helpful in predicting Y. The F distribution is used for testing this hypothesis. Appendix D contains values for the F distribution that can be used when calculations are performed by hand. See Chapter 2 for a review of the F distribution. The results of the test can also be obtained from both Excel and QM for Windows.

The F statistic used in the hypothesis test is based on the MSE (seen in the previous section) and the mean squared regression (MSR). The MSR is calculated as

MSR = SSR

k (4-14)

The error variance is estimated by the MSE.

2The MSE is a common measure of accuracy in forecasting. When used with techniques besides regression, it is com- mon to divide the SSE by n rather than n - k - 1.

1See the bibliography at the end of this chapter for books with further details.

An F test is used to determine if there is a relationship between X and Y.

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120  CHAPTER 4 • REgREssIon MoDELs

where

k = number of independent variables in the model

The F statistic is

F = MSR

MSE (4-15)

Based on the assumptions regarding the errors in a regression model, this calculated F statistic is described by the F distribution with

Degrees of freedom for the numerator = df1 = k

Degrees of freedom for the denominator = df2 = n - k - 1

where

k = the number of independent 1X2 variables If there is very little error, the denominator (MSE) of the F statistic is very small relative

to the numerator (MSR), and the resulting F statistic will be large. This is an indication that the model is useful. A significance level related to the value of the F statistic is then found. When- ever the F value is large, the observed significance level (p-value) will be low, indicating that it is extremely unlikely that this could have occurred by chance. When the F value is large (with a resulting small significance level), we can reject the null hypothesis that there is no linear relationship. This means that there is a linear relationship and the values of MSE and r2 are meaningful.

The hypothesis test just described is summarized here:

Steps in Hypothesis Test for a Significant Regression Model

1. Specify null and alternative hypotheses:

H0:b1 = 0

H1:b1 ≠ 0

2. Select the level of significance 1a). Common values are 0.01 and 0.05. 3. Calculate the value of the test statistic using the formula

F = MSR

MSE

4. Make a decision using one of the following methods:

(a) Reject the null hypothesis if the test statistic is greater than the F value from the table in Appendix D. Otherwise, do not reject the null hypothesis:

Reject if Fcalculated 7 Fa, df1, df2 df1 = k

df2 = n - k - 1

(b) Reject the null hypothesis if the observed significance level, or p-value, is less than the level of significance (a). Otherwise, do not reject the null hypothesis:

p@value = P1F 7 calculated test statistic2 Reject if p@value 6 a

If the significance level for the F test is low, there is a relationship between X and Y.

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4.5 TEsTIng THE MoDEL FoR sIgnIFICAnCE  121

Triple A Construction Example To illustrate the process of testing the hypothesis about a significant relationship, consider the Triple A Construction example. Appendix D will be used to provide values for the F distribution.

STEP 1.

H0:b1 = 0 1no linear relationship between X and Y2 H1:b1 ≠ 0 1linear relationship exists between X and Y2

STEP 2.

Select a = 0.05

STEP 3. Calculate the value of the test statistic. The MSE was already calculated to be 1.7188. The MSR is then calculated so that F can be found:

MSR = SSR

k =

15.6250

1 = 15.6250

F = MSR

MSE =

15.6250

1.7188 = 9.09

STEP 4. (a) Reject the null hypothesis if the test statistic is greater than the F value from the table in Appendix D:

df1 = k = 1 df2 = n - k - 1 = 6 - 1 - 1 = 4

The value of F associated with a 5% level of significance and with degrees of freedom 1 and 4 is found in Appendix D. Figure 4.5 illustrates this:

F0.05, 1,4 = 7.71 Fcalculated = 9.09 Reject H0 because 9.09 7 7.71

Thus, there is sufficient data to conclude that there is a statistically significant relationship between X and Y, so the model is helpful. The strength of this relationship is measured by r2 = 0.69. Thus, we can conclude that about 69% of the variability in sales (Y) is explained by the regression model based on local payroll (X).

F 5 7.71 9.09

0.05

FIGURE 4.5 F Distribution for Triple A Construction Test for significance

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122  CHAPTER 4 • REgREssIon MoDELs

The Analysis of Variance (ANOVA) Table When software such as Excel or QM for Windows is used to develop regression models, the out- put provides the observed significance level, or p-value, for the calculated F value. This is then compared to the level of significance 1a) to make the decision.

Table 4.4 provides summary information about the ANOVA table. This shows how the num- bers in the last three columns of the table are computed. The last column of this table, labeled Significance F, is the p-value, or observed significance level, which can be used in the hypoth- esis test about the regression model.

Triple A Construction ANOVA Example The Excel output that includes the ANOVA table for the Triple A Construction data is shown in the next section. The observed significance level for F = 9.0909 is given to be 0.0394. This means

P1F 7 9.09092 = 0.0394 Because this probability is less than 0.05 (a), we would reject the hypothesis of no linear rela- tionship and conclude that there is a linear relationship between X and Y. Note in Figure 4.5 that the area under the curve to the right of 9.09 is clearly less than 0.05, which is the area to the right of the F value associated with a 0.05 level of significance.

4.6 Using Computer Software for Regression

Software such as Excel 2016, Excel QM, and QM for Windows is usually used for the regression calculations. All of these will be illustrated using the Triple A Construction example.

Excel 2016 To use Excel 2016 for regression, enter the data in columns. Click the Data tab to access the data ribbon. Then select Data Analysis, as shown in Program 4.1A. If Data Analysis does not appear on the Data ribbon, the Excel add-in for this has not yet been activated. See Appendix F at the end of the book for instructions on how to activate this. Once this add-in is activated, it will re- main on the Data ribbon for future use.

When the Data Analysis window opens, scroll down to and highlight Regression and click OK, as illustrated in Program 4.1A. The Regression window will open, as shown in Program 4.1B, and you can input the X and Y ranges. Check the Labels box because the cells in the first row of the X and Y ranges include the variable names. To have the output presented on this page rather than on a new worksheet, select Output Range and give a cell address for the start of the output. Click the OK button, and the output appears in the output range specified. Program 4.1C shows the intercept (2), slope (1.25), and other information that was previously calculated for the Triple A Construction example.

The sums of squares are shown in the column headed by SS. Another name for error is re- sidual. In Excel, the sum of squares error is shown as the sum of squares residual. The values in this output are the same values shown in Table 4.3:

Sum of squares regression = SSR = 15.625 Sum of squares error (residual) = 6.8750 Sum of squares total = SST = 22.5

The coefficient of determination 1r2) is shown to be 0.6944. The coefficient of correlation (r) is called Multiple R in the Excel output, and this is 0.8333.

DF SS MS F SIGNIFICANCE F

Regression k SSR MSR = SSR>k MSR>MSE P(F 7 MSR>MSE) Residual n - k - 1 SSE MSE = SSE>(n - k - 1) Total n - 1 SST

TABLE 4.4 Analysis of Variance Table for Regression

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4.6 UsIng CoMPUTER soFTwARE FoR REgREssIon  123

Go to the Data tab.Click OK.

Select Data Analysis.

When Data Analysis window opens, scroll down to highlight Regression.

Excel QM To use Excel QM for regression, from the Excel QM ribbon select the module by clicking Al- phabetical - Forecasting - Multiple Regression, as shown in Program 4.2A. A window will open to allow you to specify the size of the problem, as shown in Program 4.2B. Enter a name or title for the problem, the number of past observations, and the number of independent (X) variables. For the Triple A Construction example, there are 6 past periods of data and 1 independent vari- able. When you click OK, a spreadsheet will appear for you to enter the X and Y data in the shaded area. The calculations are automatically performed, so there is nothing to do after enter- ing the data. The results are seen in Program 4.2C.

Specify the X and Y ranges.

Click OK to have Excel develop the regression model.

Check the Labels box if the first row in the X and Y ranges includes the variable names.

Specify location for the output. To put this on the current worksheet, click Output Range and give a cell location for this to begin.

The regression coefficients are given here.

A high r2 (close to 1) is desirable.

A low (e.g., less than 0.05) Significance F (p-value for overall model) indicates a significant relationship between X and Y.The SSR (regression), SSE

(residual or error), and SST (total) are shown in the SS column of the ANOVA table.

PROGRAM 4.1A Accessing the Regression option in Excel 2016

PROGRAM 4.1B Data Input for Regression in Excel 2016

PROGRAM 4.1C Excel 2016 output for Triple A Construction Example

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124  CHAPTER 4 • REgREssIon MoDELs

PROGRAM 4.2A Using Excel QM for Regression

Go to the Excel QM tab in Excel 2016.

When options appear, click Multiple Regression.

Left-click on the Alphabetical menu.

Point the cursor at Forecasting.

PROGRAM 4.2B Initializing the spreadsheet in Excel QM Input a title.

Click OK.

Input the number of past observations.

Input the number of independent (X) variables.

PROGRAM 4.2C Input and Results for Regression in Excel QM

Enter the past observations of Y and X. Results appear automatically.

To forecast Y based on any value of X, simply input the value of X here.

Correlation coefficient is given here.

The intercept and slope are shown here.

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4.6 UsIng CoMPUTER soFTwARE FoR REgREssIon  125

QM for Windows To develop a regression equation using QM for Windows for the Triple A Construction Com- pany data, select the Forecasting module and select New. Then choose Least Squares—Simple and Multiple Regression, as illustrated in Program 4.3A. This opens the window shown in Program 4.3B. Enter the number of observations (6 in this example) and specify the number of independent variables (1 in this example). Left-click OK and a window opens allowing you to input the data, as shown in Program 4.3C. After entering the data, click Solve, and the forecasting results shown in Program 4.3D are displayed. The equation as well as other infor- mation is provided on this screen. Note the regression equation is displayed across two lines. Additional output, including the ANOVA table, is available by clicking the Window option on the toolbar.

After selecting the Forecasting module, click New and select Least Squares – Simple and Multiple Regression.

PROGRAM 4.3A QM for windows Regression option in Forecasting Module

PROGRAM 4.3B QM for windows screen to Initialize the Problem

PROGRAM 4.3C Data Input for Triple A Construction Example

Specify the number of observations.

Specify the number of variables.

Click OK.

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126  CHAPTER 4 • REgREssIon MoDELs

4.7 Multiple Regression Analysis

The multiple regression model is a practical extension of the model we just observed. It allows us to build a model with several independent variables. The underlying model is

Y = b0 + b1X1 + b2X2 + Á + bkXk + P (4-16)

where

Y = dependent variable (response variable)

Xi = ith independent variable (predictor variable or explanatory variable)

b0 = intercept (value of Y when all Xi = 0)

bi = coefficient of the ith independent variable

k = number of independent variables

P= random error

To estimate the values of these coefficients, a sample is taken and the following equation is developed:

Yn = b0 + b1X1 + b2X2 + Á + bkXk (4-17)

where

Yn = predicted value of Y

b0 = sample intercept (and is an estimate of b0)

bi = sample coefficient of ith variable (and is an estimate of bi)

Consider the case of Jenny Wilson Realty, a real estate company in Montgomery, Alabama. Jenny Wilson, owner and broker for this company, wants to develop a model to determine a suggested listing price for houses based on the size of the house and the age of the house. She selects a sample of houses that have sold recently in a particular area, and she records the sell- ing price, the square footage of the house, the age of the house, and also the condition (good, excellent, or mint) of each house, as shown in Table 4.5. Initially Jenny plans to use only the square footage and age to develop a model, although she wants to save the information on

Additional information is available by clicking the Window menu.

Regression equation is across two lines.

PROGRAM 4.3D QM for windows output for Triple A Construction Example

A multiple regression model has more than one independent variable.

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4.7 MULTIPLE REgREssIon AnALysIs  127

condition of the house to use later. She wants to find the coefficients for the following multiple regression model:

Yn = b0 + b1X1 + b2X2

where

Yn = predicted value of dependent variable (selling price)

b0 = Y intercept

X1 and X2 = value of the two independent variables (square footage and age), respectively

b1 and b2 = slopes for X1 and X2, respectively

The mathematics of multiple regression becomes quite complex, so we leave formulas for b0, b1, and b2 to regression textbooks.

3 Excel can be used to develop a multiple regression model just as it was used for a simple linear regression model. When entering the data in Excel, it is im- portant that all of the independent variables are in adjoining columns to facilitate the input. From the Data tab in Excel, select Data Analysis and then Regression, as shown earlier, in Program 4.1A. This opens the Regression window to allow the input, as shown in Program 4.4A. Note that the X Range includes the data in two columns (B and C) because there are two independent variables. The Excel output that Jenny Wilson obtains is shown in Program 4.4B, and it provides the following equation:

Yn = b0 + b1X1 + b2X2 = 146,630.89 + 43.82 X1 - 2898.69 X2

Evaluating the Multiple Regression Model A multiple regression model can be evaluated in a manner similar to the way a simple linear regression model is evaluated. Both the p-value for the F test and r2 can be interpreted the same

SELLING PRICE ($)

SQUARE FOOTAGE

AGE

CONDITION

95,000 1,926 30 Good

119,000 2,069 40 Excellent

124,800 1,720 30 Excellent

135,000 1,396 15 Good

142,800 1,706 32 Mint

145,000 1,847 38 Mint

159,000 1,950 27 Mint

165,000 2,323 30 Excellent

182,000 2,285 26 Mint

183,000 3,752 35 Good

200,000 2,300 18 Good

211,000 2,525 17 Good

215,000 3,800 40 Excellent

219,000 1,740 12 Mint

TABLE 4.5 Jenny wilson Real Estate Data

Excel can be used to develop multiple regression models.

3See, for example, Norman R. Draper and Harry Smith, Applied Regression Analysis, 3rd ed. (New York: John Wiley & Sons, Inc., 1998).

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128  CHAPTER 4 • REgREssIon MoDELs

with multiple regression models as they are with simple linear regression models. However, as there is more than one independent variable, the hypothesis that is being tested with the F test is that all the coefficients are equal to 0. If all these are 0, then none of the independent variables in the model is helpful in predicting the dependent variable.

To determine which of the independent variables in a multiple regression model is significant, a significance test on the coefficient for each variable is performed. While sta- tistics textbooks can provide the details of these tests, the results of these tests are au- tomatically displayed in the Excel output. The null hypothesis is that the coefficient is 0 1H0:bi = 02, and the alternate hypothesis is that it is not 0 1H1:bi ≠ 02. The test statistic is calculated in Excel, and the p-value is given. If the p-value is lower than the level of sig- nificance 1a), then the null hypothesis is rejected, and it can be concluded that the variable is significant.

Jenny Wilson Realty Example In the Jenny Wilson Realty example in Program 4.4B, the overall model is statistically significant and useful in predicting the selling price of the house because the p-value for the F test is 0.002. The r2 value is 0.6719, so 67% of the variability in selling price for these houses can be explained by the regression model. However, there were two inde- pendent variables in the model–-square footage and age. It is possible that one of these is significant and the other is not. The F test simply indicates that the model as a whole is significant.

PROGRAM 4.4A Input screen for Jenny wilson Realty Multiple Regression in Excel 2016

PROGRAM 4.4B Excel 2016 output screen for Jenny wilson Realty Multiple Regression Example

The columns containing data for the X variables should be adjoining.

The X range is from the start of the first column to the bottom of the second column.

The coefficients are here.

The coefficient of determination is 0.67.

A low Significance F value indicates that the model is statistically significant.

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4.8 BInARy oR DUMMy VARIABLEs  129

Two significance tests can be performed to determine if square footage or age (or both) is significant. In Program 4.4B, the results of two hypothesis tests are provided. The first test for variable X1 (square footage) is

H0:b1 = 0 H1:b1 ≠ 0

Using a 5% level of significance 1a = 0.052, the null hypothesis is rejected because the p-value for this is 0.0013. Thus, square footage is helpful in predicting the price of a house.

Similarly, the variable X2 (age) is tested using the Excel output, and the p-value is 0.0039. The null hypothesis is rejected because this is less than 0.05. Thus, age is also helpful in predict- ing the price of a house.

4.8 Binary or Dummy Variables

All of the variables we have used in regression examples have been quantitative variables such as sales figures, payroll numbers, square footage, and age. These have all been easily measurable and have had numbers associated with them. There are many times when we believe a qualitative variable rather than a quantitative variable would be helpful in predicting the dependent vari- able Y. For example, regression may be used to find a relationship between annual income and certain characteristics of the employees. Years of experience at a particular job would be a quan- titative variable. However, information regarding whether or not a person has a college degree might also be important. This would not be a measurable value or quantity, so a special variable called a dummy variable (or a binary variable or an indicator variable) would be used. A dummy variable is assigned a value of 1 if a particular condition is met (e.g., a person has a col- lege degree) and a value of 0 otherwise.

Return to the Jenny Wilson Realty example. Jenny believes that a better model can be de- veloped if the condition of the property is included. To incorporate the condition of the house into the model, Jenny looks at the information available (see Table 4.5) and sees that the three categories are good condition, excellent condition, and mint condition. Since these are not quan- titative variables, she must use dummy variables. These are defined as

X3 = 1 if house is in excellent condition = 0 otherwise

X4 = 1 if house is in mint condition = 0 otherwise

Notice there is no separate variable for “good” condition. If X3 and X4 are both 0, then the house cannot be in excellent or mint condition, so it must be in good condition. When using dummy variables, the number of variables must be 1 less than the number of categories. In this problem, there were three categories (good, excellent, and mint condition), so we must have two dummy variables. If we had mistakenly used too many variables and the number of dummy variables equaled the number of categories, then the mathematical computations could not be performed or would not give reliable values.

These dummy variables will be used with the two previous variables (X1 - square footage, and X2 - age) to try to predict the selling prices of houses for Jenny Wilson. Programs 4.5A and 4.5B provide the Excel input and output for these new data, and this shows how the dummy vari- ables were coded. The significance level for the F test is 0.00017, so this model is statistically significant. The coefficient of determination 1r22 is 0.898, so this is a much better model than the previous one. The regression equation is

Yn = 121,658 + 56.43X1 - 3,962X2 + 33,162X3 + 47,369X4

This indicates that a house in excellent condition 1X3 = 1, X4 = 02 would sell for about $33,162 more than a house in good condition 1X3 = 0, X4 = 02. A house in mint condition 1X3 = 0, X4 = 12 would sell for about $47,369 more than a house in good condition.

A dummy variable is also called an indicator variable or a binary variable.

The number of dummy variables must equal 1 less than the number of categories of a qualitative variable.

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130  CHAPTER 4 • REgREssIon MoDELs

4.9 Model Building

In developing a good regression model, possible independent variables are identified and the best ones are selected to be used in the model. The best model is a statistically significant model with a high r2 and few variables.

As more variables are added to a regression model, r2 will usually increase, and it cannot decrease. It is tempting to keep adding variables to a model to try to increase r2. However, if too many independent variables are included in the model, problems can arise. For this reason, the adjusted r2 value is often used (rather than r2) to determine if an ad- ditional independent variable is beneficial. The adjusted r2 takes into account the number of independent variables in the model, and it is possible for the adjusted r2 to decrease. The formula for r2 is

r2 = SSR

SST = 1 -

SSE

SST

The adjusted r2 is

Adjusted r2 = 1 - SSE>1n - k - 12

SST>1n - 12 (4-18)

Notice that as the number of variables (k) increases, n - k - 1 will decrease. This causes SSE>1n - k - 12 to increase, and consequently the adjusted r2 will decrease unless the extra

The X range includes columns B, C, D, and E but not F.

PROGRAM 4.5A Input screen for Jenny wilson Realty Example with Dummy Variables in Excel 2016

PROGRAM 4.5B output screen for Jenny wilson Realty Example with Dummy Variables in Excel 2016

The overall model is helpful, since the Significance F probability is low (much less than 5%).

Each of the variables individually is helpful, since the p-value for each of these is low (much less than 5%).

The coefficient of age is negative, indicating that the price decreases as the house gets older.

The adjusted r2 may decrease when more variables are added to the model.

The value of r2 can never decrease when more variables are added to the model.

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4.10 nonLInEAR REgREssIon  131

variable in the model causes a significant decrease in the SSE. Thus, the reduction in error (and SSE) must be sufficient to offset the change in k.

As a general rule of thumb, if the adjusted r2 increases when a new variable is added to the model, the variable should probably remain in the model. If the adjusted r2 decreases when a new variable is added, the variable should not remain in the model. Other factors should also be considered when trying to build the model, but they are beyond the introductory level of this chapter.

Stepwise Regression While the process of model building may be tedious, there are many statistical software pack- ages that include stepwise regression procedures to do this. Stepwise regression is an automated process to systematically add independent variables to or delete them from a regression model. A forward stepwise procedure puts the most significant variable in the model first and then adds the next variable that will improve the model the most given that the first variable is already in the model. Variables continue to be added in this fashion until all the variables are in the model or until any remaining variables do not significantly improve the model. A backward stepwise procedure begins with all independent variables in the model, and one by one the least helpful variables are deleted. This continues until only significant variables remain. Many variations of these stepwise models exist.

Multicollinearity In the Jenny Wilson Realty example illustrated in Program 4.5B, we saw an r2 of about 0.90 and an adjusted r2 of 0.85. While other variables such as the size of the lot, the number of bedrooms, and the number of bathrooms might be related to the selling price of a house, we may not want to include these in the model. It is likely that these variables would be correlated with the square footage of the house (e.g., more bedrooms usually means a larger house), which is already in- cluded in the model. Thus, the information provided by these additional variables might be du- plication of information already in the model.

When an independent variable is correlated with one other independent variable, the vari- ables are said to share collinearity. If an independent variable is correlated with a combination of other independent variables, the condition of multicollinearity exists. This can create prob- lems in interpreting the coefficients of the variables, as several variables are providing duplicate information. For example, if two independent variables were monthly salary expenses for a com- pany and annual salary expenses for a company, the information provided in one is also provided in the other. Several sets of regression coefficients for these two variables would yield exactly the same results. Thus, individual interpretation of these variables would be questionable, al- though the model itself is still good for prediction purposes. When multicollinearity exists, the overall F test is still valid, but the hypothesis tests related to the individual coefficients are not. A variable may appear to be significant when it is insignificant, or a variable may appear to be insignificant when it is significant.

4.10 Nonlinear Regression

The regression models we have seen are linear models. However, at times there exist nonlinear relationships between variables. Some simple variable transformations can be used to create an apparently linear model from a nonlinear relationship. This allows us to use Excel and other linear regression programs to perform the calculations. We will demonstrate this in the following example.

On every new automobile sold in the United States, the fuel efficiency (as measured by miles per gallon [MPG] of gasoline) of the automobile is prominently displayed on the window sticker. The MPG is related to several factors, one of which is the weight of the au- tomobile. Engineers at Colonel Motors, in an attempt to improve fuel efficiency, have been asked to study the impact of weight on MPG. They have decided that a regression model should be used to do this.

Multicollinearity exists when a variable is correlated to other variables.

A variable should not be added to the model if it causes the adjusted r2 to decrease.

Transformations may be used to turn a nonlinear model into a linear model.

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132  CHAPTER 4 • REgREssIon MoDELs

A sample of 12 new automobiles was selected, and the weight and MPG rating were re- corded. Table 4.6 provides these data. A scatter diagram of the data in Figure 4.6A shows the weight and MPG. A linear regression line is drawn through the points. Excel was used to de- velop a simple linear regression equation to relate the MPG (Y) to the weight in thousands of pounds 1X12 in the form

Yn = b0 + b1X1

The Excel output is shown in Program 4.6. From this, we get the equation

Yn = 47.6 - 8.2X1

or

MPG = 47.6 - 8.21Weight in thousands of pounds)

The model is useful since the significance level for the F test is small and r2 = 0.7446. However, further examination of the graph in Figure 4.6A brings into question the use of a linear model. Perhaps a nonlinear relationship exists, and maybe the model should be modi- fied to account for this. A quadratic model is illustrated in Figure 4.6B. This model would be of the form

MPG = b0 + b11Weight2 + b21Weight22

MPG

WEIGHT (1,000s LB.)

MPG

WEIGHT (1,000s LB.)

12 4.58 20 3.18

13 4.66 23 2.68

15 4.02 24 2.65

18 2.53 33 1.70

19 3.09 36 1.95

19 3.11 42 1.92

TABLE 4.6 Automobile weight Versus MPg

1.00 2.00 3.00 4.00 5.00 0

5

10

25

15

20

30

35

40

45

Weight (1,000s lb.)

M P

G

FIGURE 4.6A Linear Model for MPg Data

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4.10 nonLInEAR REgREssIon  133

The easiest way to develop this model is to define a new variable

X2 = 1Weight22

This gives us the model

Yn = b0 + b1X1 + b2X2

We can create another column in Excel and again run the regression tool. The output is shown in Program 4.7. The new equation is

Yn = 79.8 - 30.2X1 + 3.4X2

The significance level for F is low (0.0002), so the model is useful, and r2 = 0.8478. The adjusted r2 increased from 0.719 to 0.814, so this new variable definitely improved the model.

This model is good for prediction purposes. However, we should not try to interpret the coefficients of the variables due to the correlation between X1 (weight) and X2 (weight squared). Normally, we would interpret the coefficient for X1 as the change in Y that results from a 1-unit change in X1, while holding all other variables constant. Obviously, holding one

1.00 2.00 3.00 4.00 5.00 0

5

10

25

15

20

30

35

40

45

Weight (1,000s lb.)

M P

G

FIGURE 4.6B nonlinear Model for MPg Data

A low significance value for F and a high r2 are indications of a good model.

PROGRAM 4.6 Excel 2016 output with Linear Regression Model for MPg Data

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134  CHAPTER 4 • REgREssIon MoDELs

variable constant while changing the other is impossible in this example, since X2 = X21. If X1 changes, then X2 must change also. This is an example of a problem that exists when multicol- linearity is present.

Other types of nonlinearities can be handled using a similar approach. A number of transformations exist that may help to develop a linear model from variables with nonlinear relationships.

4.11 Cautions and Pitfalls in Regression Analysis

This chapter has provided a brief introduction to regression analysis, one of the most widely used quantitative techniques in business. However, some common errors are made with regres- sion models, so caution should be observed when using them.

If the assumptions are not met, the statistical tests may not be valid. Any interval estimates are also invalid, although the model can still be used for prediction purposes.

Correlation does not necessarily mean causation. Two variables (such as the price of auto- mobiles and your annual salary) may be highly correlated to one another, but one is not causing the other to change. They may both be changing due to other factors such as the economy in general or the inflation rate.

If multicollinearity is present in a multiple regression model, the model is still good for pre- diction, but interpretation of individual coefficients is questionable. The individual tests on the regression coefficients are not valid.

Using a regression equation beyond the range of X is very questionable. A linear relation- ship may exist within the range of values of X in the sample. What happens beyond this range is unknown; the linear relationship may become nonlinear at some point. For example, there is usually a linear relationship between advertising and sales within a limited range. As more money is spent on advertising, sales tend to increase even if everything else is held constant. However, at some point, increasing advertising expenditures will have less impact on sales un- less the company does other things to help, such as opening new markets or expanding the prod- uct offerings. If advertising is increased and nothing else changes, the sales will probably level off at some point.

Related to the limitation regarding the range of X is the interpretation of the intercept 1b02. Since the lowest value for X in a sample is often much greater than 0, the intercept is a point on the regression line beyond the range of X. Therefore, we should not be concerned if the t-test for this coefficient is not significant, as we should not be using the regression equation to predict a value of Y when X = 0. This intercept is merely used in defining the line that fits the sample points the best.

Using the F test and concluding a linear regression model is helpful in predicting Y does not mean that this is the best relationship. While this model may explain much of the variability in Y, it is possible that a nonlinear relationship might explain even more.

PROGRAM 4.7 Excel 2016 output with nonlinear Regression Model for MPg Data

A high correlation does not mean one variable is causing a change in the other.

The regression equation should not be used with values of X that are below the lowest value of X or above the highest value of X found in the sample.

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 135

Similarly, if it is concluded that no linear relationship exists, another type of relationship could exist.

A statistically significant relationship does not mean it has any practical value. With large enough samples, it is possible to have a statistically significant relationship, but r2 might be 0.01. This would normally be of little use to a manager. Similarly, a high r2 could be found due to random chance if the sample is small. The F test must also show significance to place any value in r2.

gLossARy  135

A significant F value may occur even when the relationship is not strong.

Regression analysis is an extremely valuable quantitative tool. Using scatter diagrams helps to see relationships between vari- ables. The F test is used to determine if the results can be con- sidered useful. The coefficient of determination 1r22 is used to measure the proportion of variability in Y that is explained by the regression model. The correlation coefficient measures the relationship between two variables.

Multiple regression involves the use of more than one in- dependent variable. Dummy variables (binary or indicator vari- ables) are used with qualitative or categorical data. Nonlinear models can be transformed into linear models.

We saw how to use Excel to develop regression models. Interpretation of computer output was presented, and several examples were provided.

Summary

Glossary

Adjusted r2 A measure of the explanatory power of a regression model that takes into consideration the number of independent variables in the model.

Binary Variable See Dummy Variable. Coefficient of Correlation 1r 2 A measure of the strength of

the relationship between two variables. Coefficient of Determination 1r2 2 The percent of the

variability in the dependent variable 1Y2 that is explained by the regression equation.

Collinearity A condition that exists when one independent variable is correlated with another independent variable.

Dependent Variable The Y variable in a regression model. This is what is being predicted.

Dummy Variable A variable used to represent a qualitative factor or condition. Dummy variables have values of 0 or 1. This is also called a binary variable or an indicator variable.

Error The difference between the actual value (Y) and the predicted value 1Yn2.

Explanatory Variable The independent variable in a regres- sion equation.

Independent Variable The X variable in a regression equation. This is used to help predict the dependent variable.

Indicator Variable (Binary Variable) A type of indepen- dent variable that is either zero or one to indicate the absence (zero) or presence (one) of some condition or force.

Least Squares A reference to the criterion used to select the regression line, to minimize the squared distances between the estimated straight line and the observed values.

Mean Squared Error (MSE) An estimate of the error variance.

Multicollinearity A condition that exists when one independent variable is correlated with other independent variables.

Multiple Regression Model A regression model that has more than one independent variable.

Observed Significance Level Another name for p-value. p-Value A probability value that is used when testing

a hypothesis. The hypothesis is rejected when this is low.

Predictor Variable Another name for explanatory variable. Regression Analysis A forecasting procedure that uses

the least-squares approach on one or more independent variables to develop a forecasting model.

Residual Another term for error. Response Variable The dependent variable in a regression

equation. Scatter Diagrams or Scatter Plots Diagrams of the variable

to be forecasted, plotted against another variable, such as time. Also called scatter plots.

Standard Error of the Estimate An estimate of the stan- dard deviation of the errors; sometimes called the standard deviation of the regression.

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136  CHAPTER 4 • REgREssIon MoDELs

Key Equations

(4-1) Y = b0 + b1X + P

Underlying linear model for simple linear regression.

(4-2) Yn = b0 + b1X

Simple linear regression model computed from a sample.

(4-3) e = Y - Yn

Error in regression model.

(4-4) b1 = g1X - X21Y - Y2

g1X - X22 Slope in the regression line.

(4-5) b0 = Y - b1X

Intercept in the regression line.

(4-6) SST = g1Y - Y22 Total sums of squares.

(4-7) SSE = ge2 = g1Y - Yn22 Sum of squares due to error.

(4-8) SSR = g1Yn - Y22 Sum of squares due to regression.

(4-9) SST = SSR + SSE

Relationship among sums of squares in regression.

(4-10) r2 = SSR

SST = 1 -

SSE

SST

Coefficient of determination.

(4-11) r = {2r2 Coefficient of correlation. This has the same sign as the slope.

(4-12) s2 = MSE = SSE

n - k - 1

An estimate of the variance of the errors in regression; n is the sample size and k is the number of independent variables.

(4-13) s = 1MSE An estimate of the standard deviation of the regres- sion. Also called the standard error of the estimate.

(4-14) MSR = SSR

k

Mean square regression. k is the number of independent variables.

(4-15) F = MSR

MSE

F statistic used to test significance of overall regression model.

(4-16) Y = b0 + b1X1 + b2X2 + Á + bkXk + P

Underlying model for multiple regression model.

(4-17) Yn = b0 + b1X1 + b2X2 + Á + bkXk

Multiple regression model computed from a sample.

(4-18) Adjusted r2 = 1 - SSE>1n - k - 12

SST>1n - 12 Adjusted r2 used in building multiple regression models.

Stepwise Regression An automated process to systemati- cally add or delete independent variables from a regression model.

Sum of Squares Error (SSE) The total sum of the squared differences between each observation (Y) and the predicted value 1Yn2.

Sum of Squares Regression (SSR) The total sum of the squared differences between each predicted value 1Yn2 and the mean 1Y2.

Sum of Squares Total (SST) The total sum of the squared differences between each observation (Y) and the mean 1Y2.

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soLVED PRoBLEMs  137

Solved Problems

Solved Problem 4-1 Judith Thompson runs a florist shop on the Gulf Coast of Texas, specializing in floral arrangements for weddings and other special events. She advertises weekly in the local newspapers and is consider- ing increasing her advertising budget. Before doing so, she decides to evaluate the past effectiveness of these ads. Five weeks are sampled, and the advertising dollars and sales volume for each of these are shown in the following table. Develop a regression equation that would help Judith evaluate her advertising. Find the coefficient of determination for this model.

SALES ($1,000) ADVERTISING ($100)

11 5

6 3

10 7

6 2

12 8

Solution

SALES Y ADVERTISING X 1X − X 22 1X − X 2 1Y − Y 2 11 5 15 - 522 = 0 15 - 52111 - 92 = 0 6 3 13 - 522 = 4 13 - 5216 - 92 = 6 10 7 17 - 522 = 4 17 - 52110 - 92 = 2 6 2 12 - 522 = 9 12 - 5216 - 92 = 9 12 8 18 - 522 = 9 18 - 52112 - 92 = 9

gY = 45 Y = 45>5

= 9

gX = 25 X = 25>5

= 5

g1X - X22 = 26 g1X - X21Y - Y2 = 26

b1 = g1X - X21Y - Y2

g1X - X22 = 26

26 = 1

b0 = Y - b1X = 9 - 112152 = 4 The regression equation is

Yn = 4 + 1X

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138  CHAPTER 4 • REgREssIon MoDELs

To compute r2, we use the following table:

Y X Yn = 4 + 1X 1Yn − Y 22 1Y − Y 22

11 5 9 111 - 922 = 4 111 - 922 = 4 6 3 7 16 - 722 = 1 16 - 922 = 9 10 7 11 110 - 1122 = 1 110 - 922 = 1 6 2 6 16 - 622 = 0 16 - 922 = 9 12 8 12 112 - 1222 = 0 112 - 922 = 9

gY = 45 gX = 25 g1Y - Yn22 = 6 g1Y - Y22 = 32 Y = 9 X = 5 SSE SST

The slope 1b1 = 12 tells us that for each 1-unit increase in X (or $100 in advertising), sales increase by 1 unit (or $1,000). Also, r2 = 0.8125, indicating that about 81% of the variability in sales can be explained by the regression model with advertising as the independent variable.

Solved Problem 4-2 Use Excel with the data in Solved Problem 4-1 to find the regression model. What does the F test say about this model?

Solution Program 4.8 provides the Excel output for this problem. We see the equation is

Yn = 4 + 1X

The coefficient of determination 1r22 is shown to be 0.8125. The significance level for the F test is 0.0366, which is less than 0.05. This indicates the model is statistically significant. Thus, there is suf- ficient evidence in the data to conclude that the model is useful and there is a relationship between X (advertising) and Y (sales).

PROGRAM 4.8 Excel 2016 output for solved Problem 4-2

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DIsCUssIon QUEsTIons AnD PRoBLEMs  139

Self-Test

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

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

1. One of the assumptions in regression analysis is that a. the errors have a mean of 1. b. the errors have a mean of 0. c. the observations (Y) have a mean of 1. d. the observations (Y) have a mean of 0.

2. A graph of the sample points that will be used to develop a regression line is called a. a sample graph. b. a regression diagram. c. a scatter diagram. d. a regression plot.

3. When using regression, an error is also called a. an intercept. b. a prediction. c. a coefficient. d. a residual.

4. In a regression model, Y is called a. the independent variable. b. the dependent variable. c. the regression variable. d. the predictor variable.

5. A quantity that provides a measure of how far each sample point is from the regression line is a. the SSR. b. the SSE. c. the SST. d. the MSR.

6. The percentage of the variation in the dependent variable that is explained by a regression equation is measured by a. the coefficient of correlation. b. the MSE. c. the coefficient of determination. d. the slope.

7. In a regression model, if every sample point is on the regression line (all errors are 0), then a. the correlation coefficient would be 0. b. the correlation coefficient would be -1 or 1.

c. the coefficient of determination would be -1. d. the coefficient of determination would be 0.

8. When using dummy variables in a regression equation to model a qualitative or categorical variable, the number of dummy variables should equal a. the number of categories. b. 1 more than the number of categories. c. 1 less than the number of categories. d. the number of other independent variables in the

model. 9. A multiple regression model differs from a simple linear

regression model because the multiple regression model has more than one a. independent variable. b. dependent variable. c. intercept. d. error.

10. The overall significance of a regression model is tested using an F test. The model is significant if a. the F value is low. b. the significance level of the F value is low. c. the r2 value is low. d. the slope is lower than the intercept.

11. A new variable should not be added to a multiple regression model if that variable causes a. r2 to decrease. b. the adjusted r2 to decrease. c. the SST to decrease. d. the intercept to decrease.

12. A good regression model should have a. a low r2 and a low significance level for the F test. b. a high r2 and a high significance level for the F test. c. a high r2 and a low significance level for the F test. d. a low r2 and a high significance level for the F test.

Discussion Questions and Problems

Discussion Questions 4-1 What is the meaning of least squares in a regression

model? 4-2 Discuss the use of dummy variables in regression

analysis. 4-3 Discuss how the coefficient of determination and the

coefficient of correlation are related and how they are used in regression analysis.

4-4 Explain how a scatter diagram can be used to iden- tify the type of regression to use.

4-5 Explain how the adjusted r2 value is used in devel- oping a regression model.

4-6 Explain what information is provided by the F test. 4-7 What is the SSE? How is this related to the SST and

the SSR? 4-8 Explain how a plot of the residuals can be used in

developing a regression model.

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140  CHAPTER 4 • REgREssIon MoDELs

Problems 4-9 John Smith has developed the following forecasting

model: Yn = 36 + 4.3X1

where Yn = demand for K10 air conditioners X1 = the outside temperature 1°F2

(a) Forecast the demand for K10 when the tempera- ture is 70°F.

(b) What is the demand for a temperature of 80°F? (c) What is the demand for a temperature of 90°F?

4-10 The operations manager of a musical instrument distributor feels that demand for a particular type of guitar may be related to the number of YouTube views for a music video by the popular rock group Marble Pumpkins during the preceding month. The manager has collected the data shown in the follow- ing table:

YouTube VIEWS (1,000s)

GUITAR SALES

30 8

40 11

70 12

60 10

80 15

50 13

(a) Graph these data to see whether a linear equa- tion might describe the relationship between the views on YouTube and guitar sales.

(b) Using the equations presented in this chapter, compute the SST, SSE, and SSR. Find the least- squares regression line for these data.

(c) Using the regression equation, predict guitar sales if there were 40,000 views last month.

4-11 Using the data in Problem 4-10, test to see if there is a statistically significant relationship between sales and YouTube views at the 0.05 level of sig- nificance. Use the formulas in this chapter and Appendix D.

4-12 Using computer software, find the least-squares re- gression line for the data in Problem 4-10. Based on the F test, is there a statistically significant relation- ship between the demand for guitars and the number of YouTube views?

4-13 Students in a management science class have just re- ceived their grades on the first test. The instructor has provided information about the first test grades in some previous classes, as well as the final aver- ages for the same students. Some of these grades have been sampled and are as follows:

STUDENT 1 2 3 4 5 6 7 8 9

1st test grade 98 77 88 80 96 61 66 95 69

Final average 93 78 84 73 84 64 64 95 76

(a) Develop a regression model that could be used to predict the final average in the course based on the first test grade.

(b) Predict the final average of a student who made an 83 on the first test.

(c) Give the values of r and r2 for this model. Inter- pret the value of r2 in the context of this problem.

4-14 Using the data in Problem 4-13, test to see if there is a statistically significant relationship between the grade on the first test and the final average at the 0.05 level of significance. Use the formulas in this chapter and Appendix D.

4-15 Using computer software, find the least-squares re- gression line for the data in Problem 4-13. Based on the F test, is there a statistically significant relation- ship between the first test grade and the final average in the course?

4-16 Steve Caples, a real estate appraiser in Lake Charles, Louisiana, has developed a regression model to help appraise residential housing in the Lake Charles area. The model was developed using recently sold homes in a particular neighborhood. The price (Y) of the house is based on the square footage (X) of the house. The model is

Yn = 33,478 + 62.4X

The coefficient of correlation for the model is 0.63.

(a) Use the model to predict the selling price of a house that is 1,860 square feet.

(b) A house with 1,860 square feet recently sold for $165,000. Explain why this is not what the model predicted.

(c) If you were going to use multiple regression to develop an appraisal model, what other quantita- tive variables might be included in the model?

(d) What is the coefficient of determination for this model?

4-17 Accountants at the firm Walker and Walker be- lieved that several traveling executives submit unusually high travel vouchers when they return from business trips. The accountants took a sam- ple of 200 vouchers submitted from the past year; they then developed the following multiple regres- sion equation relating expected travel cost (Y) to number of days on the road 1X12 and distance traveled 1X22 in miles:

Yn = $90.00 + $48.50X1 + $0.40X2

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

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

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DIsCUssIon QUEsTIons AnD PRoBLEMs  141

The coefficient of correlation computed was 0.68.

(a) If Thomas Williams returns from a 300-mile trip that took him out of town for 5 days, what is the expected amount that he should claim as expenses?

(b) Williams submitted a reimbursement request for $685; what should the accountant do?

(c) Comment on the validity of this model. Should any other variables be included? Which ones? Why?

4-18 Thirteen students entered the undergraduate busi- ness program at Rollins College 2 years ago. The following table indicates what their grade-point av- erages (GPAs) were after being in the program for 2 years and what each student scored on the SAT exam (maximum 2400) when he or she was in high school. Is there a meaningful relationship between grades and SAT scores? If a student scores a 1200 on the SAT, what do you think his or her GPA will be? What about a student who scores 2400?

STUDENT SAT SCORE GPA STUDENT SAT SCORE GPA

A 1263 2.90 H 1443 2.53

B 1131 2.93 I 2187 3.22

C 1755 3.00 J 1503 1.99

D 2070 3.45 K 1839 2.75

E 1824 3.66 L 2127 3.90

F 1170 2.88 M 1098 1.60

G 1245 2.15

4-19 Bus and subway ridership in Washington, D.C., dur- ing the summer months is believed to be heavily tied to the number of tourists visiting the city. During the past 12 years, the following data have been obtained:

YEAR

NUMBER OF TOURISTS (1,000,000s)

RIDERSHIP

(100,000s)

1 7 15

2 2 10

3 6 13

4 4 15

5 14 25

6 15 27

7 16 24

8 12 20

9 14 27

10 20 44

11 15 34

12 7 17

(a) Plot these data and determine whether a linear model is reasonable.

(b) Develop a regression model. (c) What is expected ridership if 10 million tourists

visit the city?

(d) If there are no tourists at all, explain the pre- dicted ridership.

4-20 Use computer software to develop a regression model for the data in Problem 4-19. Explain what this output indicates about the usefulness of this model.

4-21 The following data give the starting salary for stu- dents who recently graduated from a local university and accepted jobs soon after graduation. The starting salary, grade-point average (GPA), and major (busi- ness or other) are provided.

SALARY $29,500 $46,000 $39,800 $36,500

GPA 3.1 3.5 3.8 2.9

Major Other Business Business Other

SALARY $42,000 $31,500 $36,200

GPA 3.4 2.1 2.5 Major Business Other Business

(a) Using a computer, develop a regression model that could be used to predict starting salary based on GPA and major.

(b) Use this model to predict the starting salary for a business major with a GPA of 3.0.

(c) What does the model say about the starting sal- ary for a business major compared to a nonbusi- ness major?

(d) Do you believe this model is useful in predicting the starting salary? Justify your answer, using in- formation provided in the computer output.

4-22 The following data give the selling price, square footage, number of bedrooms, and age of houses that have sold in a neighborhood in the past 6 months. Develop three regression models to predict the sell- ing price based upon each of the other factors indi- vidually. Which of these is best?

SELLING PRICE ($)

SQUARE FOOTAGE

BEDROOMS

AGE (YEARS)

84,000 1,670 2 30

79,000 1,339 2 25

91,500 1,712 3 30

120,000 1,840 3 40

127,500 2,300 3 18

132,500 2,234 3 30

145,000 2,311 3 19

164,000 2,377 3 7

155,000 2,736 4 10

168,000 2,500 3 1

172,500 2,500 4 3

174,000 2,479 3 3

175,000 2,400 3 1

177,500 3,124 4 0

184,000 2,500 3 2

195,500 4,062 4 10

195,000 2,854 3 3

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142  CHAPTER 4 • REgREssIon MoDELs

4-23 Use the data in Problem 4-22 and develop a regres- sion model to predict selling price based on the square footage and number of bedrooms. Use this to predict the selling price of a 2,000-square-foot house with three bedrooms. Compare this model with the models in Problem 4-22. Should the num- ber of bedrooms be included in the model? Why or why not?

4-24 Use the data in Problem 4-22 and develop a regres- sion model to predict selling price based on the square footage, number of bedrooms, and age. Use this to predict the selling price of a 10-year-old, 2,000-square-foot house with three bedrooms.

4-25 The total expenses of a hospital are related to many factors. Two of these factors are the number of beds in the hospital and the number of admissions. Data were collected on 14 hospitals, as shown in the fol- lowing table:

HOSPITAL

NUMBER OF BEDS

ADMISSIONS (100s)

TOTAL EXPENSES ($1,000,000s)

1 215 77 57

2 336 160 127

3 520 230 157

4 135 43 24

5 35 9 14

6 210 155 93

7 140 53 45

8 90 6 6

9 410 159 99

10 50 18 12

11 65 16 11

12 42 29 15

13 110 28 21

14 305 98 63

Find the best regression model to predict the total expenses of a hospital. Discuss the accuracy of this model. Should both variables be included in the model? Why or why not?

4-26 A sample of 20 automobiles was taken, and the miles per gallon (MPG), horsepower, and total weight were recorded. Develop a linear regression model to predict MPG, using horsepower as the only indepen- dent variable. Develop another model with weight as the independent variable. Which of these two mod- els is better? Explain.

MPG HORSEPOWER WEIGHT

44 67 1,844

44 50 1,998

40 62 1,752

MPG HORSEPOWER WEIGHT

37 69 1,980

37 66 1,797

34 63 2,199

35 90 2,404

32 99 2,611

30 63 3,236

28 91 2,606

26 94 2,580

26 88 2,507

25 124 2,922

22 97 2,434

20 114 3,248

21 102 2,812

18 114 3,382

18 142 3,197

16 153 4,380

16 139 4,036

4-27 Use the data in Problem 4-26 to develop a multiple linear regression model. How does this compare with each of the models in Problem 4-26?

4-28 Use the data in Problem 4-26 to find the best qua- dratic regression model. (There is more than one to consider.) How does this compare to the models in Problems 4-26 and 4-27?

4-29 A sample of nine public universities and nine private universities was taken. The total cost for the year (including room and board) and the median SAT score (maximum total is 2400) at each school were recorded. It was felt that schools with higher median SAT scores would have a better reputation and would charge more tuition as a result of that. The data are in the following table. Use regression to help answer the following questions based on the sample data. Do schools with higher SAT scores charge more in tuition and fees? Are private schools more expensive than public schools when SAT scores are taken into consideration? Discuss how accurate you believe these results are, using information related to the re- gression models.

CATEGORY TOTAL COST ($) MEDIAN SAT

Public 21,700 1990

Public 15,600 1620

Public 16,900 1810

Public 15,400 1540

Public 23,100 1540

Public 21,400 1600

Public 16,500 1560

Public 23,500 1890

(Continued on next page)

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DIsCUssIon QUEsTIons AnD PRoBLEMs  143

CATEGORY TOTAL COST ($) MEDIAN SAT

Public 20,200 1620

Private 30,400 1630

Private 41,500 1840

Private 36,100 1980

Private 42,100 1930

Private 27,100 2130

Private 34,800 2010

Private 32,100 1590

Private 31,800 1720

Private 32,100 1770

4-30 In 2012, the total payroll for the New York Yankees was almost $200 million, while the total payroll for the Oakland Athletics (a team known for using baseball analytics or sabermetrics) was about $55 million, less than one-third of the Yankees’ payroll. In the following table, you will see the payrolls (in millions) and the total number of victories for the baseball teams in the American League in the 2012 season. Develop a regression model to predict the to- tal number of victories based on the payroll. Use the model to predict the number of victories for a team with a payroll of $79 million. Based on the results of the computer output, discuss the relationship be- tween payroll and victories.

TEAM

PAYROLL ($1,000,000s)

NUMBER OF VICTORIES

Baltimore Orioles 81.4 93

Boston Red Sox 173.2 69

Chicago White Sox 96.9 85

Cleveland Indians 78.4 68

Detroit Tigers 132.3 88

Kansas City Royals 60.9 72

Los Angeles Angels 154.5 89

Minnesota Twins 94.1 66

New York Yankees 198.0 95

Oakland Athletics 55.4 94

Seattle Mariners 82.0 75

Tampa Bay Rays 64.2 90

Texas Rangers 120.5 93

Toronto Blue Jays 75.5 73

4-31 The number of victories (W), earned run average (ERA), runs scored (R), batting average (AVG), and on-base percentage (OBP) for each team in the American League in the 2012 season are provided in the following table. The ERA is one measure of the effectiveness of the pitching staff, and a lower number is better. The other statistics are measures of the effectiveness of the hitters, and a higher number is better for each of these.

TEAM W ERA R AVG OBP

Baltimore Orioles 93 3.90 712 0.247 0.311

Boston Red Sox 69 4.70 734 0.260 0.315

Chicago White Sox 85 4.02 748 0.255 0.318

Cleveland Indians 68 4.78 667 0.251 0.324

Detroit Tigers 88 3.75 726 0.268 0.335

Kansas City Royals 72 4.30 676 0.265 0.317

Los Angeles Angels 89 4.02 767 0.274 0.332

Minnesota Twins 66 4.77 701 0.260 0.325

New York Yankees 95 3.85 804 0.265 0.337

Oakland Athletics 94 3.48 713 0.238 0.310

Seattle Mariners 75 3.76 619 0.234 0.296

Tampa Bay Rays 90 3.19 697 0.240 0.317

Texas Rangers 93 3.99 808 0.273 0.334

Toronto Blue Jays 73 4.64 716 0.245 0.309

(a) Develop a regression model that could be used to predict the number of victories based on the ERA.

(b) Develop a regression model that could be used to predict the number of victories based on the runs scored.

(c) Develop a regression model that could be used to predict the number of victories based on the batting average.

(d) Develop a regression model that could be used to predict the number of victories based on the on-base percentage.

(e) Which of the four models is better for predicting the number of victories?

(f) Find the best multiple regression model to pre- dict the number of wins. Use any combination of the variables to find the best model.

4-32 The closing stock price for each of two stocks was recorded over a 12-month period. The closing price for the Dow Jones Industrial Average (DJIA) was also recorded over this same time period. These val- ues are shown in the following table:

MONTH DJIA STOCK 1 STOCK 2

1 11,168 48.5 32.4

2 11,150 48.2 31.7

3 11,186 44.5 31.9

4 11,381 44.7 36.6

5 11,679 49.3 36.7

6 12,081 49.3 38.7

7 12,222 46.1 39.5

8 12,463 46.2 41.2

9 12,622 47.7 43.3

10 12,269 48.3 39.4

11 12,354 47.0 40.1

12 13,063 47.9 42.1

13 13,326 47.8 45.2

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144  CHAPTER 4 • REgREssIon MoDELs

(a) Develop a regression model to predict the price of stock 1 based on the Dow Jones Industrial Average.

(b) Develop a regression model to predict the price of stock 2 based on the Dow Jones Industrial Average.

(c) Which of the two stocks is most highly corre- lated to the Dow Jones Industrial Average over this time period?

4-33 Annual rainfall plays an important role in corn agri- culture. The drought of 2011 literally affected corn prices for years. Given the following data, build a

model and predict the harvest for 2016 given that the total rainfall was 6.45 inches. Critique your prediction.

YEAR RAIN (INCHES) REAP (TONS)

2011 2.06 325

2012 5.11 408

2013 7.43 609

2014 6.12 512

2015 7.14 544

In January 2012, Northern Airlines merged with Southeast Air- lines to create the fourth largest U.S. carrier. The new North– South Airline inherited both an aging fleet of Boeing 727-300 aircraft and Stephen Ruth. Stephen was a tough former Secre- tary of the Navy who stepped in as new president and chairman of the board.

Stephen’s first concern in creating a financially solid com- pany was maintenance costs. It was commonly surmised in the airline industry that maintenance costs rise with the age of the aircraft. He quickly noticed that historically there had been a significant difference in the reported B727-300 maintenance costs (from ATA Form 41s) in both the airframe and the engine areas between Northern Airlines and Southeast Airlines, with Southeast having the newer fleet.

On February 12, 2012, Peg Jones, vice president for op- erations and maintenance, was called into Stephen’s office and asked to study the issue. Specifically, Stephen wanted to know whether the average fleet age was correlated to direct airframe maintenance costs and whether there was a relationship be- tween average fleet age and direct engine maintenance costs. Peg was to report back by February 26 with the answer, along with quantitative and graphical descriptions of the relationship.

Peg’s first step was to have her staff construct the average age of the Northern and Southeast B727-300 fleets, by quarter,

since the introduction of that aircraft to service by each airline in late 1993 and early 1994. The average age of each fleet was calculated by first multiplying the total number of calendar days each aircraft had been in service at the pertinent point in time by the average daily utilization of the respective fleet to determine the total fleet hours flown. The total fleet hours flown was then divided by the number of aircraft in service at that time, giving the age of the “average” aircraft in the fleet.

The average utilization was found by taking the actual total fleet hours flown on September 30, 2011, from Northern and Southeast data, and dividing by the total days in service for all aircraft at that time. The average utilization for Southeast was 8.3 hours per day, and the average utilization for Northern was 8.7 hours per day. Because the available cost data were calcu- lated for each yearly period ending at the end of the first quarter, average fleet age was calculated at the same points in time. The fleet data are shown in the following table. Airframe cost data and engine cost data are both shown paired with fleet average age in that table.

Discussion Questions 1. Prepare Peg Jones’s response to Stephen Ruth.

Note: Dates and names of airlines and individuals have been changed in this case to maintain confidentiality. The data and issues described here are real.

Case Study

North–South Airline

north–south Airline Data for Boeing 727-300 Jets

NORTHERN AIRLINES DATA SOUTHEAST AIRLINES DATA

YEAR

AIRFRAME COST PER

AIRCRAFT ($)

ENGINE COST PER

AIRCRAFT ($)

AVERAGE

AGE (HOURS)

AIRFRAME COST PER

AIRCRAFT ($)

ENGINE COST PER

AIRCRAFT ($)

AVERAGE

AGE (HOURS)

2001 51.80 43.49 6,512 13.29 18.86 5,107

2002 54.92 38.58 8,404 25.15 31.55 8,145

2003 69.70 51.48 11,077 32.18 40.43 7,360

2004 68.90 58.72 11,717 31.78 22.10 5,773

2005 63.72 45.47 13,275 25.34 19.69 7,150

2006 84.73 50.26 15,215 32.78 32.58 9,364

2007 78.74 79.60 18,390 35.56 38.07 8,259

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APPEnDIx 4.1: FoRMULAs FoR REgREssIon CALCULATIons  145

Bibliography

Berenson, Mark L., David M. Levine, and Timothy C. Krehbiel. Basic Business Statistics, 12th ed. Upper Saddle River, NJ: Pearson, 2012.

Black, Ken. Business Statistics: For Contemporary Decision Making, 8th ed. John Wiley & Sons, Inc., 2014.

Draper, Norman R., and Harry Smith. Applied Regression Analysis, 3rd ed. New York: John Wiley & Sons, Inc., 1998.

Kuiper, S., and J. Sklar. Practicing Statistics: Guided Investigations for the Second Course. Upper Saddle River, NJ: Pearson, 2013.

Kutner, Michael, John Neter, and Chris J. Nachtsheim. Applied Linear Regres- sion Models, 4th ed. Boston; New York: McGraw-Hill/Irwin, 2004.

Mendenhall, William, and Terry L. Sincich. A Second Course in Statistics: Regression Analysis, 7th ed. Upper Saddle River, NJ: Pearson, 2012.

Y X Y2 X2 XY

6 3 62 = 36 32 = 9 3162 = 18 8 4 82 = 64 42 = 16 4182 = 32 9 6 92 = 81 62 = 36 6192 = 54 5 4 52 = 25 42 = 16 4152 = 20 4.5 2 4.52 = 20.25 22 = 4 214.52 = 9 9.5 5 9.52 = 90.25 52 = 25 519.52 = 47.5

gY = 42 gX = 24 gY2 = 316.5 gX2 = 106 gXY = 180.5 Y = 42>6 = 7 X = 24>6 = 4

TABLE 4.7 Preliminary Calculations for Triple A Construction

Appendix 4.1: Formulas for Regression Calculations

When performing regression calculations by hand, there are other formulas that can make the task easier and are math- ematically equivalent to the ones presented in the chapter. These, however, make it more difficult to see the logic be- hind the formulas and to understand what the results actu- ally mean.

When using these formulas, it helps to set up a table with the columns shown in Table 4.7, which has the Triple A Con- struction Company data that were used earlier in the chapter. The sample size (n) is 6. The totals for all columns are shown, and the averages for X and Y are calculated. Once this is done, we can use the following formulas for computations in a simple linear regression model (one independent variable). The simple linear regression equation is again given as

Yn = b0 + b1X

Slope of regression equation:

b1 = gXY - nXY gX2 - nX2

b1 = 180.5 - 6142172

106 - 61422 = 1.25

Intercept of regression equation:

b0 = Y - b1X b0 = 7 - 1.25142 = 2

Sum of squares of the error:

SSE = gY2 - b0gY - b1gXY SSE = 316.5 - 21422 - 1.251180.52 = 6.875

Estimate of the error variance:

s2 = MSE = SSE

n - 2

s2 = 6.875

6 - 2 = 1.71875

Estimate of the error standard deviation:

s = 1MSE s = 21.71875 = 1.311

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146  CHAPTER 4 • REgREssIon MoDELs

Coefficient of determination:

r2 = 1 - SSE

gY2 - nY 2 r2 = 1 -

6.875

316.5 - 61722 = 0.6944

This formula for the correlation coefficient automatically determines the sign of r. This could also be found by

taking the square root of r2 and giving it the same sign as the slope:

r = ngXY - gXgY23ngX2 - 1gX2243ngY2 - 1gY224

r = 61180.52 - 1242142223611062 - 2424361316.52 - 4224 = 0.833

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 147

5.5 Apply forecast models for trends and random variations.

5.6 Manipulate data to account for seasonal variations.

5.7 Apply forecast models for trends, seasonal variations, and random variations.

5.8 Explain how to monitor and control forecasts.

5.1 Understand and know when to use various families of forecasting models.

5.2 Compare moving averages, exponential smoothing, and other time-series models.

5.3 Calculate measures of forecast accuracy.

5.4 Apply forecast models for random variations.

After completing this chapter, students will be able to:

Forecasting

LEARNING OBJECTIVES

5 CHAPTER

Every day, managers make decisions without knowing what will happen in the future. In-ventory is ordered though no one knows what sales will be, new equipment is purchased though no one knows the demand for products, and investments are made though no one knows what profits will be. Managers are always trying to reduce this uncertainty and to make better estimates of what will happen in the future. Accomplishing this is the main purpose of forecasting.

There are many ways to forecast the future. In numerous firms (especially smaller ones), the entire process is subjective, involving seat-of-the-pants methods, intuition, and years of experience.

There are also more formal forecasting techniques, both quantitative and qualitative in na- ture. The primary focus of this chapter will be understanding time-series models and determin- ing which model works best with a particular set of data.

5.1 Types of Forecasting Models

Figure 5.1 lists some of the more common forecasting models, and these are categorized by the type of model. The first category involves qualitative models, while the others are quantitative in nature and use mathematical models to develop better forecasts.

Qualitative Models Qualitative models are forecasting techniques based on judgmental or subjective factors. When a totally new product such as the iPad is introduced, forecasting demand is very difficult due to

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148  CHAPTER 5 • FoRECAsTing

the lack of any historical sales data on that particular product or on similar products. The com- pany must rely on expert opinion, individual experiences and judgment, and other subjective factors.

Here is a brief overview of four different qualitative forecasting techniques:

1. Delphi method. This iterative group process allows experts, who may be located in differ- ent places, to make forecasts. There are three different types of participants in the Delphi process: decision makers, staff personnel, and respondents. The decision-making group usually consists of 5 to 10 experts who will be making the actual forecast. The staff person- nel assist the decision makers by preparing, distributing, collecting, and summarizing a series of questionnaires and survey results. The respondents are a group of people whose judgments are valued and are being sought. This group provides inputs to the decision makers before the forecast is made. In the Delphi method, when the results of the first questionnaire are obtained, the re- sults are summarized, and the questionnaire is modified. Both the summary of the results and the new questionnaire are then sent to the same respondents for a new round of re- sponses. The respondents, upon seeing the results from the first questionnaire, may view things differently and may modify their original responses. This process is repeated with the hope that a consensus is reached.

2. Jury of executive opinion. This method takes the opinions of a small group of high-level managers, often in combination with statistical models, and results in a group estimate of demand.

3. Sales force composite. In this approach, each salesperson estimates what sales will be in his or her region; these forecasts are reviewed to ensure that they are realistic and are then combined at the district and national levels to reach an overall forecast.

4. Consumer market survey. This method solicits input from customers or potential customers regarding their future purchasing plans. It can help not only in preparing a forecast but also in improving product design and planning for new products.

Causal Models A variety of quantitative forecasting models are available when past numerical data are available. Forecasting models are identified as causal models if the variable to be forecast is influenced by or correlated with other variables included in the model. For example, daily sales of bottled wa- ter might depend on the average temperature, the average humidity, and so on. A causal model would include factors such as these in the mathematical model. Regression models (see Chapter 4) and other more complex models would be classified as causal models.

Overview of four qualitative or judgmental approaches: Delphi method, jury of executive opinion, sales force composite, and consumer market survey.

Forecasting Techniques

Qualitative Models

Causal Methods

Delphi Method

Moving Averages

Regression Analysis

Exponential Smoothing

Multiple Regression

Trend Projections

Jury of Executive Opinion

Sales Force Composite

Consumer Market Survey

Decomposition

Time-Series Methods

FIGURE 5.1 Forecasting Models

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5.2 CoMPonEnTs oF A TiME-sERiEs  149

Time-Series Models Time series are also quantitative models, and many time-series methods are available. A time- series model is a forecasting technique that attempts to predict the future values of a variable by using only historical data on that one variable. These models are extrapolations of past values of that series. While other factors may have influenced these past values, the impact of those other factors is captured in the previous values of the variable being predicted. Thus, if we are fore- casting weekly sales for lawn mowers, we use the past weekly sales for lawn mowers in making the forecast for future sales, ignoring other factors such as the economy, competition, and even the selling price of the lawn mowers.

5.2 Components of a Time-Series

A time series is sequence of values recorded at successive intervals of time. The intervals can be days, weeks, months, years, or other time units. Examples include weekly sales of HP personal computers, quarterly earnings reports of Microsoft Corporation, daily shipments of Eveready batteries, and annual U.S. consumer price indices. A time series may consist of four possible components—trend, seasonal, cyclical, and random.

The trend (T) component is the general upward or downward movement of the data over a relatively long period of time. For example, total sales for a company may be increasing con- sistently over time. The consumer price index, one measure of inflation, is increasing over time. While consistent upward (or downward) movements are indicative of a trend, there may be a positive (or negative) trend present in a time series and yet the values do not increase (or de- crease) in every time period. There may be an occasional movement that seems inconsistent with the trend due to random or other fluctuations.

Figure 5.2 shows scatter diagrams for several time series. The data for all the series are quar- terly, and there are four years of data. Series 3 has both trend and random components present.

The seasonal (S) component is a pattern of fluctuations above or below an average value that repeats at regular intervals. For example, with monthly sales data on snowblowers, sales tend to be high in December and January and lower in summer months, and this pattern is expected to

Series 4: Trend, Seasonal, and Random Variations

Series 3: Trend and Random Variations

Series 2: Seasonal Variations Only

Series 1: Random Variations Only

1 2 3 4 5 6 7 8 Time Period (Quarters)

S al

es

9 10 11 12 13 14 15 16

FIGURE 5.2 scatter Diagram for Four Time series of Quarterly Data

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150  CHAPTER 5 • FoRECAsTing

repeat every year. Quarterly sales for a consumer electronics store may be higher in the fourth quarter of every year and lower in other quarters. Daily sales in a retail store may be higher on Saturdays than on other days of the week. Hourly sales in a fast-food restaurant are usually ex- pected to be higher around the lunch hour and the dinner hour, while other times are not as busy. Series 2 in Figure 5.2 illustrates seasonal variations for quarterly data.

If a seasonal component is present in a set of data, the number of seasons depends on the type of data. With quarterly data, there are four seasons because there are four quarters in a year. With monthly data, there would be 12 seasons because there are 12 months in a year. With daily sales data for a retail store that is open seven days a week, there would be seven seasons. With hourly data, there could be 24 seasons if the business is open 24 hours a day.

A cyclical (C) component of a time series is a pattern in annual data that tends to repeat ev- ery several years. The cyclical component of a time series is used only when making very long- range forecasts, and it is usually associated with the business cycle. Sales or economic activity might reach a peak and then begin to recede and contract, reaching a bottom or trough. At some point after the trough is reached, activity would pick up as recovery and expansion take place. A new peak would be reached, and the cycle would begin again. Figure 5.3 shows 16 years of annual data for a time series that has a cyclical as well as a random component.

While a time series with cyclical variations and a time series with seasonal variations may look similar on a graph, cyclical variations tend to be irregular in length (from a few years to 10 or more years) and magnitude, and they are very difficult to predict. Seasonal variations are very consistent in length and magnitude; are much shorter, as the seasons repeat in a year or less; and are much easier to predict. The forecasting models presented in this chapter will be for short- range or medium-range forecasts and the cyclical component will not be included in these models.

The random (R) component consists of irregular, unpredictable variations in a time series. Any variation in a times series that cannot be attributed to trend, seasonal, or cyclical variations would fall into this category. If data for a time series tend to be level with no discernible trend or seasonal pattern, random variations would be the cause for any changes from one time period to the next. Series 1 in Figure 5.2 is an example of a time series with only a random component.

There are two general forms of time-series models in statistics. The first is a multiplicative model, which assumes that demand is the product of the four components. It is stated as follows:

Demand = T * S * C * R

An additive model adds the components together to provide an estimate. Multiple regres- sion is often used to develop additive models. This additive relationship is stated as follows:

Demand = T + S + C + R

There are other models that may be a combination of these. For example, one of the com- ponents (such as trend) might be additive, while another (such as seasonality) could be multi- plicative. Understanding the components of a time series will help in selecting an appropriate forecasting technique to use. The methods presented in this chapter will be grouped according to the components considered when the forecast is developed.

0 2 4 6 8 10 Time Period (Years)

S al

es

12 14 16 18

FIGURE 5.3 scatter Diagram of a Time series with Cyclical and Random Components

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5.3 MEAsUREs oF FoRECAsT ACCURACy  151

5.3 Measures of Forecast Accuracy

We discuss several different forecasting models in this chapter. To see how well one model works or to compare that model with other models, the forecast values are compared with the actual or observed values. The forecast error (or deviation) is defined as follows:

Forecast error = Actual value - Forecast value

One measure of accuracy is the mean absolute deviation (MAD). This is computed by tak- ing the sum of the absolute values of the individual forecast errors and dividing by the number of errors (n):

MAD = g �Forecast error �

n (5-1)

To illustrate how to compute the measures of accuracy, the naïve model will be used to forecast one period into the future. The naïve model does not attempt to address any of the components of a time series, but it is very quick and easy to use. A naïve forecast for the next time period is the actual value that was observed in the current time period. For example, suppose someone asked you to forecast the high temperature in your city for tomorrow. If you forecast the tem- perature to be whatever the high temperature is today, your forecast would be a naïve forecast. This technique is often used if a forecast is needed quickly and accuracy is not a major concern.

Consider Walker Distributors’ sales of wireless speakers shown in Table 5.1. Suppose that in the past, Walker had forecast sales for each month to be the sales that were actually achieved in the previous month. Table 5.1 gives these forecasts as well as the absolute value of the errors. In forecasting for the next time period (month 11), the forecast would be 190. Notice that there is no error computed for month 1, since there was no forecast for this month, and there is no error for month 11, since the actual value of this is not yet known. Thus, the number of errors (n) is 9.

From this, we see the following:

MAD = g �Forecast error �

n =

160

9 = 17.8

This means that, on the average, each forecast missed the actual value by 17.8 units.

The naïve forecast for the next period is the actual value observed in the current period.

TABLE 5.1 Computing the Mean Absolute Deviation (MAD) MONTH

ACTUAL SALES OF WIRELESS

SPEAKERS FORECAST

SALES

ABSOLUTE VALUE OF ERRORS (DEVIATION). |ACTUAL–FORECAST|

1 110 — —

2 100 110 �100 - 110 � = 10 3 120 100 �120 - 100 � = 20 4 140 120 �140 - 120 � = 20 5 170 140 �170 - 140 � = 30 6 150 170 �150 - 170 � = 20 7 160 150 �160 - 150 � = 10 8 190 160 �190 - 160 � = 30 9 200 190 �200 - 190 � = 10

10 190 200 �190 - 200 � = 10 11 — 190 —

Sum of � errors � = 160

MAD = 160>9 = 17.8

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152  CHAPTER 5 • FoRECAsTing

Other measures of the accuracy of historical errors in forecasting are sometimes used besides the MAD. One of the most common is the mean squared error (MSE), which is the average of the squared errors:1

MSE = g1Error22

n (5-2)

1 In regression analysis, the MSE formula is usually adjusted to provide an unbiased estimator of the error variance. Throughout this chapter, we will use the formula provided here.

Hurricane Landfall Location Forecasts and the Mean Absolute Deviation

Scientists at the National Hurricane Center (NHC) of the National Weather Service have the very difficult job of pre- dicting where the eye of a hurricane will hit land. Accurate forecasts are extremely important to coastal businesses and resi- dents who need to prepare for a storm or perhaps even evacu- ate. They are also important to local government officials, law enforcement agencies, and other emergency responders who will provide help once a storm has passed. Over the years, the NHC has tremendously improved the forecast accuracy (mea- sured by the mean absolute deviation [MAD]) in predicting the actual landfall location for hurricanes that originate in the At- lantic Ocean.

The NHC provides forecasts and periodic updates of where the hurricane eye will hit land. Such landfall location predictions

are recorded when a hurricane is 72 hours, 48 hours, 36 hours, 24 hours, and 12 hours away from actually reaching land. Once the hurricane has come ashore, these forecasts are compared to the actual landfall location, and the error (in miles) is recorded. At the end of the hurricane season, the errors for all the hurri- canes in that year are used to calculate the MAD for each type of forecast (12 hours away, 24 hours away, etc.). The graph below shows how the landfall location forecast has improved since 1989. During the early 1990s, the landfall forecast when the hur- ricane was 48 hours away had an MAD close to 200 miles; in 2009, this number was down to about 75 miles. Clearly, there has been vast improvement in forecast accuracy, and this trend is continuing.

Source: Based on National Hurricane Center, © Trevor S. Hale.

IN ACTION

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Besides the MAD and MSE, the mean absolute percent error (MAPE) is sometimes used. The MAPE is the average of the absolute values of the errors expressed as percentages of the actual values. This is computed as follows:

MAPE = g ` Error

Actual `

n 100% (5-3)

There is another common term associated with error in forecasting. Bias is the average error and tells whether the forecast tends to be too high or too low and by how much. Thus, bias may be negative or positive. It is not a good measure of the actual size of the errors because the nega- tive errors can cancel out the positive errors.

Three common measures of error are MAD, MSE, and MAPE. Bias gives the average error and may be positive or negative.

Defining the Problem To drive production at each of Tupperware’s 15 plants in the United States, Latin America, Africa, Europe, and Asia, the firm needs accurate forecasts of demand for its products.

Developing a Model A variety of statistical models are used, including moving averages, exponential smoothing, and regression analysis. Qualitative analysis is also employed in the process.

Acquiring Input Data At world headquarters in Orlando, Florida, huge databases are maintained that map the sales of each product, the test market results of each new product (since 20% of the firm’s sales come from products less than 2 years old), and where each product falls in its own life cycle.

Developing a Solution Each of Tupperware’s 50 profit centers worldwide develops computerized monthly, quarterly, and 12-month sales projections. These are aggregated by region and then globally.

Testing the Solution Reviews of these forecasts take place in the sales, marketing, finance, and production departments.

Analyzing the Results Participating managers analyze forecasts with Tupperware’s version of a “jury of executive opinion.”

Implementing the Results Forecasts are used to schedule materials, equipment, and personnel at each plant.

Source: Based on interviews by the authors with Tupperware executives, © Trevor S. Hale.

MODELING IN THE REAL WORLD

Forecasting at Tupperware international

Defining the Problem

Developing a Model

Acquiring Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

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5.4 Forecasting Models—Random Variations Only

If all variations in a time series are due to random variations, with no trend, seasonal, or cycli- cal component, some type of averaging or smoothing model would be appropriate. The averag- ing techniques in this chapter are moving average, weighted moving average, and exponential smoothing. These methods will smooth out the forecasts and not be too heavily influenced by random variations. However, if there is a trend or seasonal pattern present in the data, then a technique that incorporates that particular component into the forecast should be used.

Moving Averages Moving averages are useful if we can assume that market demands will stay fairly steady over time. For example, a 4-month moving average is found simply by summing the demand during the past 4 months and dividing by 4. With each passing month, the most recent month’s data are added to the sum of the previous 3 months’ data, and the earliest month is dropped. This tends to smooth out short-term irregularities in the data series.

An n-period moving average forecast, which serves as an estimate of the next period’s demand, is expressed as follows:

Moving average forecast = Sum of demands in previous n periods

n (5-4)

Mathematically, this is written as

Ft+ 1 = Yt + Yt- 1 + Á + Yt- n + 1

n (5-5)

where

Ft+ 1 = forecast for time period t + 1 Yt = actual value in time period t n = number of periods to average

A 4-month moving average has n = 4; a 5-month moving average has n = 5.

WALLACE GARDEN SUPPLY EXAMPLE Storage shed sales at Wallace Garden Supply are shown in the middle column of Table 5.2. A 3-month moving average is indicated on the right. The forecast for the next January, using this technique, is 16. Were we simply asked to find a forecast for next January, we would have to make only this one calculation. The other forecasts are necessary only if we wish to compute the MAD or another measure of accuracy.

Weighted Moving Averages A simple moving average gives the same weight 11>n2 to each of the past observations being used to develop the forecast. On the other hand, a weighted moving average allows different weights to be assigned to the previous observations. As the weighted moving average method typically assigns greater weight to more recent observations, this forecast is more responsive to changes in the pattern of the data that occur. However, this is also a potential drawback to this method because the heavier weight would also respond just as quickly to random fluctuations.

A weighted moving average may be expressed as

Ft+ 1 = g1Weight in period i21Actual value in period i2

g1Weights2 (5-6) Mathematically, this is

Ft+ 1 = w1Yt + w2Yt- 1 + Á + wnYt- n + 1

w1 + w2 + Á + wn (5-7)

where

wi = weight for ith observation

Moving averages smooth out variations when forecasting demands that are fairly steady.

Weights can be used to put more emphasis on recent periods.

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Wallace Garden Supply decides to use a 3-month weighted moving average forecast with weights of 3 for the most recent observation, 2 for the next observation, and 1 for the most dis- tant observation. This would be implemented as follows:

WEIGHT APPLIED PERIOD

3 Last month

2 2 months ago

1 3 months ago

3 * Sales last month + 2 * Sales 2 months ago + 1 * Sales 3 months ago 6

Sum of the weights

The results of the Wallace Garden Supply weighted average forecast are shown in Table 5.3. In this particular forecasting situation, you can see that weighting the latest month more heavily provides a much more accurate projection, and calculating the MAD for each of these would verify this.

Choosing the weights obviously has an important impact on the forecasts. One way to choose weights is to try various combinations of weights, calculate the MAD for each, and select the set of weights that results in the lowest MAD. Some forecasting software has an option to search for the best set of weights, and forecasts using these weights are then provided. The best set of weights can also be found by using nonlinear programming, as will be seen in a later chapter.

Some software packages require that the weights add to 1, and this would simplify Equation 5-7 because the denominator would be 1. Forcing the weights to sum to 1 is easily achieved by di- viding each of the weights by the sum of the weights. In the Wallace Garden Supply example in Table 5.3, the weights are 3, 2, and 1, which add to 6. These weights could be revised to the new weights 3/6, 2/6, and 1/6, which add to 1. Using these weights gives the same forecasts shown in Table 5.3.

Both simple and weighted moving averages are effective in smoothing out sudden fluctua- tions in the demand pattern in order to provide stable estimates. Moving averages do, however, have two problems. First, increasing the size of n (the number of periods averaged) does smooth out fluctuations better, but it makes the method less sensitive to real changes in the data, should they occur. Second, moving averages cannot pick up trends very well. Because they are aver- ages, they will always stay within past levels and will not predict a change to either a higher or a lower level.

Moving averages have two problems: the larger number of periods may smooth out real changes, and they don’t pick up trends.

TABLE 5.2 Wallace garden supply shed sales

MONTH ACTUAL SHED SALES 3-MONTH MOVING AVERAGE

January 10

February 12

March 13

April 16 110 + 12 + 132>3 = 11.67 May 19 112 + 13 + 162>3 = 13.67 June 23 113 + 16 + 192>3 = 16.00 July 26 116 + 19 + 232>3 = 19.33 August 30 119 + 23 + 262>3 = 22.67 September 28 123 + 26 + 302>3 = 26.33 October 18 126 + 30 + 282>3 = 28.00 November 16 130 + 28 + 182>3 = 25.33 December 14 128 + 18 + 162>3 = 20.67 January — 118 + 16 + 142>3 = 16.00

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Exponential Smoothing Exponential smoothing is a forecasting method that is easy to use and is handled efficiently by computers. Although it is a type of moving average technique, it involves little record keeping of past data. The basic exponential smoothing formula can be shown as follows:

New forecast = Last period’s forecast (5-8) + a1Last period’s actual demand - Last period>s forecast2

where a is a weight (or smoothing constant) that has a value between 0 and 1, inclusive. Equation 5-8 can also be written mathematically as

Ft+ 1 = Ft + a1Yt - Ft2 (5-9) where

Ft+ 1 = new forecast 1for time period t + 12 Ft = previous forecast (for time period t) a = smoothing constant 10 … a … 12 Yt = previous period’s actual demand

The concept here is not complex. The latest estimate of demand is equal to the old estimate adjusted by a fraction of the error (last period’s actual demand minus the old estimate).

The smoothing constant, a, can be changed to give more weight to recent data when the value is high or more weight to past data when it is low. For example, when a = 0.5, it can be shown mathematically that the new forecast is based almost entirely on demand in the past three periods. When a = 0.1, the forecast places little weight on any single period, even the most re- cent, and it takes many periods (about 19) of historical values into account.2

For example, in January, a demand for 142 of a certain car model for February was predicted by a dealer. Actual February demand was 153 autos, so the forecast error was 153 - 142 = 11. Using a smoothing constant a = 0.20, the exponential smoothing forecast for March demand would be found by adding 20% of this error to the previous forecast of 142. Substituting into the formula, we obtain

New forecast 1for March demand2 = 142 + 0.21153 - 1422 = 144.2

TABLE 5.3 Weighted Moving Average Forecast for Wallace garden supply

MONTH ACTUAL SHED SALES 3-MONTH WEIGHTED MOVING AVERAGE

January 10

February 12

March 13

April 16 313 * 132 + 12 * 122 + 11024>6 = 12.17 May 19 313 * 162 + 12 * 132 + 11224>6 = 14.33 June 23 313 * 192 + 12 * 162 + 11324>6 = 17.00 July 26 313 * 232 + 12 * 192 + 11624>6 = 20.5 August 30 313 * 262 + 12 * 232 + 11924>6 = 23.83 September 28 313 * 302 + 12 * 262 + 12324>6 = 27.5 October 18 313 * 282 + 12 * 302 + 12624>6 = 28.33 November 16 313 * 182 + 12 * 282 + 13024>6 = 23.33 December 14 313 * 162 + 12 * 182 + 12824>6 = 18.67 January — 313 * 142 + 12 * 162 + 11824>6 = 15.33

2 The term exponential smoothing is used because the weight of any one period’s demand in a forecast decreases expo- nentially over time. See an advanced forecasting book for algebraic proof.

The smoothing constant, A, allows managers to assign weight to recent data.

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Thus, the forecast for the demand for cars in March would be 144 after being rounded by the dealer.

Suppose that actual demand for the cars in March was 136. A forecast for the demand in April, using the exponential smoothing model with a constant of a = 0.20, can be made:

New forecast 1for April demand2 = 144.2 + 0.21136 - 144.22 = 142.6, or 143 autos

GETTING STARTED The exponential smoothing approach is easy to use and has been applied suc- cessfully by banks, manufacturing companies, wholesalers, and other organizations. To develop a forecast with this method, a smoothing constant must be selected, and a previous forecast must be known or assumed. If a company has been routinely using this method, there is no problem. However, if a company is just beginning to use the method, these issues must be addressed.

The appropriate value of the smoothing constant, a, can make the difference between an accurate forecast and an inaccurate forecast. In picking a value for the smoothing constant, the objective is to obtain the most accurate forecast. Several values of the smoothing constant may be tried, and the one with the lowest MAD could be selected. This is analogous to how weights are selected for a weighted moving average forecast. Some forecasting software will automati- cally select the best smoothing constant. QM for Windows will display the MAD that would be obtained with values of a ranging from 0 to 1 in increments of 0.01.

When the exponential smoothing model is used for the first time, there is no prior forecast for the current time period to use in developing a forecast for the next time period. Therefore, a previous value for a forecast must be assumed. It is common practice to assume a forecast for time period 1 and to use this to develop a forecast for time period 2. The time period 2 forecast is then used to forecast time period 3, and this continues until the forecast for the current time period is found. Unless there is reason to assume another value, the forecast for time period 1 is assumed to be equal to the actual value in time period 1 (i.e., the error is zero). The value of the forecast for time period 1 usually has very little impact on the value of forecasts many time periods into the future.

PORT OF BALTIMORE EXAMPLE Let us apply this concept with a trial-and-error testing of two values of a in an example. The port of Baltimore has unloaded large quantities of grain from ships dur- ing the past eight quarters. The port’s operations manager wants to test the use of exponential smoothing to see how well the technique works in predicting tonnage unloaded. He assumes that the forecast of grain unloaded in the first quarter was 175 tons. Two values of a are examined: a = 0.10 and a = 0.50. Table 5.4 shows the detailed calculations for a = 0.10 only.

To evaluate the accuracy of each smoothing constant, we can compute the absolute devia- tions and MADs (see Table 5.5). Based on this analysis, a smoothing constant of a = 0.10 is preferred to a = 0.50 because its MAD is smaller.

TABLE 5.4 Port of Baltimore Exponen- tial smoothing Forecasts for a = 0.10 and a = 0.50

QUARTER

ACTUAL TONNAGE

UNLOADED FORECAST

USING A = 0.10 FORECAST

USING A = 0.50

1 180 175 175

2 168 175.5 = 175.00 + 0.101180 - 1752 177.5 3 159 174.75 = 175.50 + 0.101168 - 175.502 172.75 4 175 173.18 = 174.75 + 0.101159 - 174.752 165.88 5 190 173.36 = 173.18 + 0.101175 - 173.182 170.44 6 205 175.02 = 173.36 + 0.101190 - 173.362 180.22 7 180 178.02 = 175.02 + 0.101205 - 175.022 192.61 8 182 178.22 = 178.02 + 0.101180 - 178.022 186.30 9 ? 178.60 = 178.22 + 0.101182 - 178.222 184.15

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Using Software for Forecasting Time Series The two software packages available with this book, Excel QM and QM for Windows, are very easy to use for forecasting. The previous examples will be solved using these.

In using Excel QM with Excel 2016 for forecasting, from the Excel QM ribbon, select Alphabetical as shown in Program 5.1A. Then simply select whichever method you wish to use, and an initialization window opens to allow you to set the size of the problem. Program 5.1B shows this window for the Wallace Garden Supply weighted moving average example with the number of past periods and the number of periods in the weighted average already entered. Clicking OK results in the screen shown in Program 5.1C, allowing you to enter the data and the weights. Once all numbers are entered, the calculations are automatically performed, as you see in Program 5.1C.

To use QM for Windows to forecast, under Modules select Forecasting. Then select New and Time Series Analysis, as shown in Program 5.2A. This would be the same for all of the models presented in this chapter. In the window that opens (Program 5.2B), specify the number

TABLE 5.5 Absolute Deviations and MADs for the Port of Baltimore Example

QUARTER

ACTUAL TONNAGE

UNLOADED

FORECAST WITH

A = 0.10

ABSOLUTE DEVIATIONS FOR A = 0.10

FORECAST WITH

A = 0.50

ABSOLUTE DEVIATIONS FOR A = 0.50

1 180 175 5 175 5

2 168 175.5 7.5 177.5 9.5

3 159 174.75 15.75 172.75 13.75

4 175 173.18 1.82 165.88 9.12

5 190 173.36 16.64 170.44 19.56

6 205 175.02 29.98 180.22 24.78

7 180 178.02 1.98 192.61 12.61

8 182 178.22 3.78 186.30 4.3

Sum of absolute deviations 82.45 98.63

MAD = g �Deviation �

n = 10.31

MAD = 12.33

Click the Excel QM tab.

With cursor on Forecasting, a menu appears to the right. Select the model you want and an initialization window will open.

Click Alphabetical and scroll down to Forecasting.

PROGRAM 5.1A selecting the Forecasting Model in Wallace garden supply Problem

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Enter the number of periods to be averaged.

Click OK.

Enter a title. Enter the number of past periods of data. PROGRAM 5.1B initializing Excel QM spreadsheet for Wallace garden supply Problem

Enter the demand data and the weights. The calculations will automatically be performed.

The measures of accuracy are shown here.

The forecast for the next period is here.

PROGRAM 5.1C Excel QM output for Wallace garden supply Problem

In the Forecasting module, click New and Time-Series Analysis.

PROGRAM 5.2A selecting Time-series Analysis in QM for Windows in the Forecasting Module

Enter a title.

Enter the number of past periods of data.

Click OK.

PROGRAM 5.2B Entering Data for Port of Baltimore Example in QM for Windows

of past periods of data. For the Port of Baltimore example, this is 8. Click OK, and an input screen appears, allowing you to enter the data and select the forecasting technique, as shown in Program 5.2C. Based on the method you choose, other windows may open to allow you to specify the necessary parameters for that model. In this example, the exponential smoothing model is selected, so the smoothing constant must be specified. After entering the data, click Solve, and

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the output shown in Program 5.2D appears. From this output screen, you can select Window to see additional information such as the details of the calculations and a graph of the time series and the forecasts.

5.5 Forecasting Models—Trend and Random Variations

If there is trend present in a time series, the forecasting model must account for this and cannot simply average past values. Two very common techniques will be presented here. The first is ex- ponential smoothing with trend, and the second is called trend projection or simply a trend line.

Exponential Smoothing with Trend An extension of the exponential smoothing model that will explicitly adjust for trend is called the exponential smoothing with trend model. The idea is to develop an exponential smooth- ing forecast and then adjust this for trend. Two smoothing constants, a and b, are used in this model, and both of these values must be between 0 and 1. The level of the forecast is adjusted by multiplying the first smoothing constant, a, by the most recent forecast error and adding it to the previous forecast. The trend is adjusted by multiplying the second smoothing constant, b, by the most recent error or excess amount in the trend. A higher value gives more weight to recent observations and thus responds more quickly to changes in the patterns.

As with simple exponential smoothing, the first time a forecast is developed, a previous forecast 1Ft2 must be given or estimated. If none is available, often the initial forecast is assumed to be perfect. In addition, a previous trend 1Tt2 must be given or estimated. This is often estimated by using other past data, if available; by using subjective means; or by calculating the increase (or decrease) observed during the first few time periods of the data available. Without such an estimate available, the trend is sometimes assumed to be 0 initially, although this may lead to poor forecasts if the trend is large and b is small. Once these initial conditions have been set, the exponential smoothing forecast including trend 1FITt2 is developed using three steps:

Two smoothing constants are used.

Estimate or assume initial values for Ft and Tt.

Enter the value for the smoothing constant.

Enter the data. Then click Solve at the top of the page.

Click here to see the models. Other input areas appear based on the model. Select Exponential Smoothing and a window opens for you to enter the smoothing constant.

PROGRAM 5.2C selecting the Model and Entering Data for Port of Baltimore Example in QM for Windows

Additional output is available under Window.

The measures of accuracy are shown here.

The forecast for next period is here.

PROGRAM 5.2D output for Port of Baltimore Example in QM for Windows

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STEP1. Compute the smoothed forecast 1Ft+ 12 for time period t + 1 using the equation Smoothed forecast = Previous forecast including trend + a1Last error2

Ft+ 1 = FITt + a1Yt - FITt2 (5-10)

STEP 2. Update the trend 1Tt+ 12 using the equation Smoothed trend = Previous trend + b1Error or excess in trend2

Tt+ 1 = Tt + b1Ft+ 1 - FITt2 (5-11)

STEP 3. Calculate the trend-adjusted exponential smoothing forecast 1FITt+ 12 using the equation Forecast including trend 1FITt+ 12 = Smoothed forecast 1Ft+ 12 + Smoothed trend 1Tt+ 12

FITt+ 1 = Ft+ 1 + Tt+ 1 (5-12)

where

Tt = smoothed trend for time period t Ft = smoothed forecast for time period t

FITt = forecast including trend for time period t a = smoothing constant for forecasts b = smoothing constant for trend

Consider the case of Midwestern Manufacturing Company, which has a demand for electri- cal generators over the period 2007 to 2013 as shown in Table 5.6. To use the trend-adjusted ex- ponential smoothing method, first set initial conditions (previous values for F and T) and choose a and b. Assuming that F1 is perfect and T1 is 0 and picking 0.3 and 0.4 for the smoothing con- stants, we have

F1 = 74 T1 = 0 a = 0.3 b = 0.4

This results in

FIT1 = F1 + T1 = 74 + 0 = 74

Following the three steps to get the forecast for 2008 (time period 2), we have

STEP 1. Compute Ft+ 1 using the equation

Ft+ 1 = FITt + a1Yt - FITt2 F2 = FIT1 + 0.31Y1 - FIT12 = 74 + 0.3174 - 742 = 74

STEP 2. Update the trend 1Tt+ 12 using the equation Tt+ 1 = Tt + b1Ft+ 1 - FITt2

T2 = T1 + 0.41F2 - FIT12 = 0 + 0.4174 - 742 = 0

TABLE 5.6 Midwestern Manufacturing’s Demand

YEAR ELECTRICAL GENERATORS SOLD

2007 74

2008 79

2009 80

2010 90

2011 105

2012 142

2013 122

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STEP 3. Calculate the trend-adjusted exponential smoothing forecast 1FITt+ 12 using the equation FIT2 = F2 + T2 = 74 + 0 = 74

For 2009 (time period 3), we have

STEP 1. F3 = FIT2 + 0.31Y2 - FIT22 = 74 + 0.3179 - 742 = 75.5

STEP 2. T3 = T2 + 0.41F3 - FIT22 = 0 + 0.4175.5 - 742 = 0.6

STEP 3. FIT3 = F3 + T3 = 75.5 + 0.6 = 76.1

The other results are shown in Table 5.7. The forecast for 2014 would be about 131.35. To have Excel QM perform the calculations in Excel 2016, from the Excel QM ribbon, select

the alphabetical list of techniques and choose Forecasting and then Trend Adjusted Exponential Smoothing. After specifying the number of past observations, enter the data and the values for a and b, as shown in Program 5.3.

TABLE 5.7 Midwestern Manufacturing Exponential smoothing with Trend Forecasts

TIME (t)

DEMAND (Yt) Ft+1 = FITt + 0.3 1Yt − FITt 2 Tt+1 = Tt + 0.4 1Ft+1 − FITt 2 FITt+1 = Ft+1 + Tt+1

1 74 74 0 74

2 79 74 = 74 + 0.3174 - 742 0 = 0 + 0.4174 - 742 74 = 74 + 0 3 80 75.5 = 74 + 0.3179 - 742 0.6 = 0 + 0.4175.5 - 742 76.1 = 75.5 + 0.6 4 90 77.270 1.068 78.338 = 77.270 + 1.068

= 76.1 + 0.3180 - 76.12 = 0.6 + 0.4177.27 - 76.12 5 105 81.837 2.468 84.305 = 81.837 + 2.468

= 78.338 + 0.3190 - 78.3382 = 1.068 + 0.4181.837 - 78.3382 6 142 90.514 4.952 95.466 = 90.514 + 4.952

= 84.305 + 0.31105 - 84.3052 = 2.468 + 0.4190.514 - 84.3052 7 122 109.426 10.536 119.962 = 109.426 + 10.536

= 95.466 + 0.31142 - 95.4662 = 4.952 + 0.41109.426 - 95.4662 8 120.573 10.780 131.353 = 120.573 + 10.780

= 119.962 + 0.31122 - 119.9622 = 10.536 + 0.41120.573 - 119.9622

The forecast for next period is here.

PROGRAM 5.3 output from Excel QM in Excel 2016 for Trend- Adjusted Exponential smoothing Example

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Trend Projections Another method for forecasting time series with trend is called trend projection. This technique fits a trend line to a series of historical data points and then projects the line into the future for medium- to long-range forecasts. There are several mathematical trend equations that can be developed (e.g., exponential and quadratic), but in this section we look at linear (straight line) trends only. A trend line is simply a linear regression equation in which the independent variable (X) is the time period. The first time period will be time period 1. The second time period will be time period 2, and so forth. The last time period will be time period n. The form of this is

Yn = b0 + b1X

where

Yn = predicted value b0 = intercept b1 = slope of the line X = time period (i.e., X = 1, 2, 3, c, n)

The least squares regression method may be applied to find the coefficients that minimize the sum of the squared errors, thereby also minimizing the mean squared error (MSE). Chapter 4 provides a detailed explanation of least squares regression and the formulas to calculate the coef- ficients by hand. In this section, we use computer software to perform the calculations.

Let us consider the case of Midwestern Manufacturing’s demand for generators that was presented in Table 5.6. A trend line can be used to predict demand (Y) based on the time period using a regression equation. For the first time period, which was 2007, we let X = 1. For 2008, we let X = 2, and so forth. Using computer software to develop a regression equation as we did in Chapter 4, we get the following equation:

Yn = 56.71 + 10.54X

To project demand in 2014, we first denote the year 2014 in our new coding system as X = 8:

1Sales in 20142 = 56.71 + 10.54182 = 141.03, or 141 generators

We can estimate demand for 2015 by inserting X = 9 in the same equation:

1Sales in 20152 = 56.71 + 10.54192 = 151.57, or 152 generators

Program 5.4 provides output from Excel QM in Excel 2016. To run this model, from the Excel QM ribbon, select Alphabetical to see the techniques. Then select Forecasting and

A trend line is a regression equation with time as the independent variable.

To forecast other time periods, enter the time period here.

The forecast for next period is here.

PROGRAM 5.4 output from Excel QM in Excel 2016 for Trend Line Example

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Regression / Trend Analysis. When the input window opens, enter the number of past observa- tions (7), and the spreadsheet is initialized, allowing you to input the seven Y values and the seven X values, as shown in Program 5.4.

This problem could also be solved using QM for Windows. To do this, select the Forecast- ing module, and then enter a new problem by selecting New – Time Series Analysis. When the next window opens, enter the number of observations (7) and press OK. Enter the values (Y) for the seven past observations when the input window opens. It is not necessary to enter the values for X (1, 2, 3, . . . , 7) because QM for Windows will automatically use these numbers. Then click Solve to see the results shown in Program 5.5.

Figure 5.4 provides a scatter diagram and a trend line for these data. The projected demand in each of the next three years is also shown on the trend line.

5.6 Adjusting for Seasonal Variations

Time-series forecasting such as that in the example of Midwestern Manufacturing involves looking at the trend of data over a series of time observations. Sometimes, however, recurring variations at certain seasons of the year make a seasonal adjustment in the trend line forecast necessary. Demand for coal and fuel oil, for example, usually peaks during cold winter months. Demand for golf clubs or suntan lotion may be highest in summer. Analyzing data in monthly or quarterly terms usually makes it easy to spot seasonal patterns. A seasonal index is often used

Forecasts for future time periods are shown here.

The trend line is shown over two lines.

PROGRAM 5.5 output from QM for Windows for Trend Line Example

0 0

20

40

60

80

100

120

140

160

180

1 2 3 4 5 6 7 8 9 10 11 Time Period

G en

er at

or D

em an

d

Trend Line Y 5 56.71 1 10.54X ^

Projected demand for next 3 years is shown on the trend line.

FIGURE 5.4 generator Demand and Projections for next Three years Based on Trend Line

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5.6 ADjUsTing FoR sEAsonAL VARiATions  165

in multiplicative time-series forecasting models to make an adjustment in the forecast when a seasonal component exists. An alternative is to use an additive model such as a regression model that will be introduced in a later section.

Seasonal Indices A seasonal index indicates how a particular season (e.g., month or quarter) compares with an average season. An index of 1 for a season would indicate that the season is average. If the index is higher than 1, the values of the time series in that season tend to be higher than average. If the index is lower than 1, the values of the time series in that season tend to be lower than average.

Seasonal indices are used with multiplicative forecasting models in two ways. First, each observation in a time series is divided by the appropriate seasonal index to remove the impact of seasonality. The resulting values are called deseasonalized data. Using these deseasonalized values, forecasts for future values can be developed using a variety of forecasting techniques. Once the forecasts of future deseasonalized values have been developed, they are multi- plied by the seasonal indices to develop the final forecasts, which now include variations due to seasonality.

There are two ways to compute seasonal indices. The first method, based on an overall average, is easy and may be used when there is no trend present in the data. The second method, based on a centered-moving-average approach, is more difficult, but it must be used when there is trend present.

Calculating Seasonal Indices with No Trend When no trend is present, the index can be found by dividing the average value for a particular season by the average of all the data. Thus, an index of 1 means the season is average. For ex- ample, if the average sales in January were 120 and the average sales in all months were 200, the seasonal index for January would be 120>200 = 0.60, so January is below average. The next example illustrates how to compute seasonal indices from historical data and to use these in forecasting future values.

Monthly sales of one brand of telephone answering machine at Eichler Supplies are shown in Table 5.8 for the two most recent years. The average demand in each month is computed, and these values are divided by the overall average (94) to find the seasonal index for each month. We then use the seasonal indices from Table 5.8 to adjust future forecasts. For example, suppose we expected the third year’s annual demand for answering machines to be 1,200 units, which is

An average season has an index of 1.

TABLE 5.8 Answering Machine sales and seasonal indices

MONTH

SALES DEMAND AVERAGE 2-YEAR

DEMAND MONTHLY DEMANDa

AVERAGE SEASONAL

INDEXbYEAR 1 YEAR 2

January 80 100 90 94 0.957

February 85 75 80 94 0.851

March 80 90 85 94 0.904

April 110 90 100 94 1.064

May 115 131 123 94 1.309

June 120 110 115 94 1.223

July 100 110 105 94 1.117

August 110 90 100 94 1.064

September 85 95 90 94 0.957

October 75 85 80 94 0.851

November 85 75 80 94 0.851

December 80 80 80 94 0.851

Total average demand = 1,128

a Average monthly demand = 1,128

12 months = 94 b Seasonal index =

Average 2@year demand

Average monthly demand

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166  CHAPTER 5 • FoRECAsTing

100 per month. We would not forecast each month to have a demand of 100, but we would adjust these based on the seasonal indices as follows:

Jan. 1,200

12 * 0.957 = 96 July

1,200

12 * 1.117 = 112

Feb. 1,200

12 * 0.851 = 85 Aug.

1,200

12 * 1.064 = 106

Mar. 1,200

12 * 0.904 = 90 Sept.

1,200

12 * 0.957 = 96

Apr. 1,200

12 * 1.064 = 106 Oct.

1,200

12 * 0.851 = 85

May 1,200

12 * 1.309 = 131 Nov.

1,200

12 * 0.851 = 85

June 1,200

12 * 1.223 = 122 Dec.

1,200

12 * 0.851 = 85

Calculating Seasonal Indices with Trend When both trend and seasonal components are present in a time series, a change from one month to the next could be due to a trend, to a seasonal variation, or simply to random fluctuations. To help with this problem, the seasonal indices should be computed using a centered moving aver- age (CMA) approach whenever trend is present. Using this approach prevents a variation due to trend from being incorrectly interpreted as a variation due to the season. Consider the following example.

Quarterly sales figures for Turner Industries are shown in Table 5.9. Notice that there is a definite trend, as the total is increasing each year, and there is an increase for each quarter from one year to the next as well. The seasonal component is obvious, as there is a definite drop from the fourth quarter of one year to the first quarter of the next. A similar pattern is observed in comparing the third quarters to the fourth quarters immediately following.

If a seasonal index for quarter 1 were computed using the overall average, the index would be too low and misleading, since this quarter has less trend than any of the others in the sample. If the first quarter of year 1 were omitted and replaced by the first quarter of year 4 (if it were available), the average for quarter 1 (and consequently the seasonal index for quarter 1) would be considerably higher. To derive an accurate seasonal index, we should use a CMA.

Consider quarter 3 of year 1 for the Turner Industries example. The actual sales in that quar- ter were 150. To determine the magnitude of the seasonal variation, we should compare this with an average quarter centered at that time period. Thus, we should have a total of four quarters (1 year of data) with an equal number of quarters before and after quarter 3 so the trend is av- eraged out. Thus, we need 1.5 quarters before quarter 3 and 1.5 quarters after it. To obtain the CMA, we take quarters 2, 3, and 4 of year 1, plus one-half of quarter 1 for year 1 and one-half of quarter 1 for year 2. The average will be

CMA 1quarter 3 of year 12 = 0.511082 + 125 + 150 + 141 + 0.511162 4

= 132.00

We compare the actual sales in this quarter to the CMA, and we have the following seasonal ratio:

Seasonal ratio = Sales in quarter 3

CMA =

150

132.00 = 1.136

Centered moving averages are used to compute seasonal indices when there is trend.

TABLE 5.9 Quarterly sales ($1,000,000s) for Turner industries

QUARTER YEAR 1 YEAR 2 YEAR 3 AVERAGE

1 108 116 123 115.67

2 125 134 142 133.67

3 150 159 168 159.00

4 141 152 165 152.67

Average 131.00 140.25 149.50 140.25

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Thus, sales in quarter 3 of year 1 are about 13.6% higher than an average quarter at this time. All of the CMAs and the seasonal ratios are shown in Table 5.10.

Since there are two seasonal ratios for each quarter, we average these to get the seasonal index. Thus,

Index for quarter 1 = I1 = 10.851 + 0.8482>2 = 0.85 Index for quarter 2 = I2 = 10.965 + 0.9602>2 = 0.96 Index for quarter 3 = I3 = 11.136 + 1.1272>2 = 1.13 Index for quarter 4 = I4 = 11.051 + 1.0632>2 = 1.06

The sum of these indices should be the number of seasons (4), since an average season should have an index of 1. In this example, the sum is 4. If the sum were not 4, an adjustment would be made. We would multiply each index by 4 and divide this by the sum of the indices.

Steps Used to Compute Seasonal Indices Based on CMAs

1. Compute a CMA for each observation (where possible). 2. Compute Seasonal ratio = Observation/CMA for that observation. 3. Average seasonal ratios to get seasonal indices. 4. If seasonal indices do not add to the number of seasons, multiply each index by (Number

of seasons)/(Sum of the indices).

The seasonal indices found using the CMAs will be used to remove the impact of seasons so that the trend is easier to identify. Later, the seasonal indices will be used again to adjust the trend forecast based on the seasonal variations.

5.7 Forecasting Models—Trend, Seasonal, and Random Variations

When both trend and seasonal components are present in a time series, the forecasting model selected must address these. The decomposition method, which uses seasonal indices, is a very common approach. Multiple regression models are also common, with dummy variables used to adjust for seasonal variations in an additive time-series model.

The Decomposition Method The process of isolating linear trend and seasonal factors to develop more accurate forecasts is called decomposition. The first step is to compute seasonal indices for each season as we have done with the Turner Industries data. Then, the data are deseasonalized by dividing each number

TABLE 5.10 Centered Moving Averages and seasonal Ratios for Turner industries

YEAR QUARTER SALES

($1,000,000s) CMA SEASONAL

RATIO

1 1 108

2 125

3 150 132.000 1.136

4 141 134.125 1.051

2 1 116 136.375 0.851

2 134 138.875 0.965

3 159 141.125 1.127

4 152 143.000 1.063

3 1 123 145.125 0.848

2 142 147.875 0.960

3 168

4 165

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168  CHAPTER 5 • FoRECAsTing

by its seasonal index, as shown in Table 5.11. Figure 5.5 provides a graph of the original data, as well as a graph of the deseasonalized data. Notice how smooth the deseasonalized data are.

A trend line is then found using the deseasonalized data. Using computer software with these data, we have3

b1 = 2.34 b0 = 124.78

The trend equation is

Yn = 124.78 + 2.34X

where

X = time

This equation is used to develop the forecast based on trend, and the result is multiplied by the appropriate seasonal index to make a seasonal adjustment. For the Turner Industries data, the

TABLE 5.11 Deseasonalized Data for Turner industries

SALES ($1,000,000s) SEASONAL INDEX DESEASONALIZED SALES

($1,000,000s)

108 0.85 127.059

125 0.96 130.208

150 1.13 132.743

141 1.06 133.019

116 0.85 136.471

134 0.96 139.583

159 1.13 140.708

152 1.06 143.396

123 0.85 144.706

142 0.96 147.917

168 1.13 148.673

165 1.06 155.660

3 If you do the calculations by hand, the numbers may differ slightly from these due to rounding.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 0

50

100

150

200

Time Period (Quarters)

S al

es ($

1, 00

0, 00

0s )

Sales Data

Deseasonalized Sales Data

FIGURE 5.5 scatterplot of Turner industries original sales Data and Deseasonalized Data

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5.7 FoRECAsTing MoDELs—TREnD, sEAsonAL, AnD RAnDoM VARiATions  169

forecast for the first quarter of year 4 (time period X = 13 and seasonal index I1 = 0.85) would be found as follows:

Yn = 124.78 + 2.34X = 124.78 + 2.341132 = 155.2 1forecast before adjustment for seasonality2

We multiply this by the seasonal index for quarter 1, and we get

Yn * I1 = 155.2 * 0.85 = 131.92

Using the same procedure, we find the forecasts for quarters 2, 3, and 4 of the next year. Table 5.12 shows the unadjusted forecast based on the trend line for each of the four quarters of year 4. The last column of this table gives the final forecasts that were found by multiplying each trend forecast by the appropriate seasonal index. A scatter diagram showing the original data and the deseasonalized data, as well as the unadjusted forecast and the final forecast, for these next four quarters is provided in Figure 5.6.

TABLE 5.12 Turner industry Forecasts for Four Quarters of year 4

YEAR QUARTER TIME PERIOD (X) TREND FORECAST Yn = 124.78 + 2.34X

SEASONAL INDEX

FINAL (ADJUSTED) FORECAST

4 1 13 155.20 0.85 131.92

2 14 157.54 0.96 151.24

3 15 159.88 1.13 180.66

4 16 162.22 1.06 171.95

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0

50

100

150

200

Time Period (Quarters)

S al

es ($

1, 00

0, 00

0)

Sales Data Deseasonalized Sales Data Unadjusted Trend Forecast Forecast Adjusted for Season

FIGURE 5.6 scatterplot of Turner industries’ original sales Data and Deseasonalized Data with Unadjusted and Adjusted Forecasts

Steps to Develop a Forecast Using the Decomposition Method

1. Compute seasonal indices using CMAs. 2. Deseasonalize the data by dividing each number by its seasonal index. 3. Find the equation of a trend line using the deseasonalized data. 4. Forecast for future periods using the trend line. 5. Multiply the trend line forecast by the appropriate seasonal index.

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170  CHAPTER 5 • FoRECAsTing

Most forecasting software, including Excel QM and QM for Windows, includes the decompo- sition method as one of the available techniques. This will automatically compute the CMAs, deseasonalize the data, develop the trend line, make the forecast using the trend equation, and adjust the final forecast for seasonality.

Software for Decomposition Program 5.6A shows the QM for Windows input screen that appears after selecting the Forecast- ing module, selecting New-Time Series, and specifying 12 past observations. The multiplicative decomposition method is selected, and you can see the inputs that are used for this problem. When this is solved, the output screen in Program 5.6B appears. Additional output is available by selecting details or graph from the Windows drop-down menu (not shown in Program 5.6B) that appears after the problem is solved. The forecasts are slightly different from the ones shown in Table 5.12 due to round-off error.

Excel QM can be used in Excel 2016 for this problem also. From the QM Ribbon, select the menus By Chapter and select Chapter 5-Forecasting-Decomposition. Specify there are 12 past observations and 4 seasons, and use a centered moving average.

Using Regression with Trend and Seasonal Components Multiple regression may be used to forecast with both trend and seasonal components present in a time series. One independent variable is time, and other independent variables are dummy variables to indicate the season. If we forecast quarterly data, there are four categories (quarters),

Multiple regression can be used to develop an additive decomposition model.

Select Rescale: set average to 1.

Specify Centered Moving Average approach.

Specify the number of seasons (4 for quarterly data).

Select Multiplicative Decomposition from the drop-down menu.

Input the data.

PROGRAM 5.6A QM for Windows input screen for Turner industries Example

The final forecast is obtained by multiplying the trend (unadjusted) forecast by the seasonal indices (factors).

The seasonal indices (factors) are shown here.

The trend line is shown here over two lines.

PROGRAM 5.6B QM for Windows output screen for Turner industries Example

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so we would use three dummy variables. The basic model is an additive decomposition model and is expressed as follows:

Yn = a + b1X1 + b2X2 + b3X3 + b4X4

where

X1 = time period X2 = 1 if quarter 2

= 0 otherwise X3 = 1 if quarter 3

= 0 otherwise X4 = 1 if quarter 4

= 0 otherwise

If X2 = X3 = X4 = 0, then the quarter would be quarter 1. It is an arbitrary choice as to which of the quarters would not have a specific dummy variable associated with it. The forecasts will be the same, regardless of which quarter does not have a specific dummy variable.

Using Excel QM in Excel 2016 to develop the regression model, from the Excel QM rib- bon, we select the menu for Chapter 5—Forecasting, and then choose the Multiple Regression method. The input screen in Program 5.7A appears, and the number of observations (12) and the number of independent variables (4) are entered. An initialized spreadsheet appears, and the values for all the variables are entered in columns B through F, as shown in Program 5.7B. Once

Enter the number of past periods of data.

Enter the number of independent variables.

PROGRAM 5.7A Excel QM Multiple Regression initialization screen for Turner industries

Enter the values of Y and X1–X4 as shown.

The regression coefficients are shown here.

Enter the values of the variables to obtain any forecast.

PROGRAM 5.7B Excel QM Multiple Regression output screen for Turner industries

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172  CHAPTER 5 • FoRECAsTing

this has been done, the regression coefficients appear. The regression equation (with rounded coefficients) is

Yn = 104.1 + 2.3X1 + 15.7X2 + 38.7X3 + 30.1X4

If this is used to forecast sales in the first quarter of the next year, we get

Yn = 104.1 + 2.31132 + 15.7102 + 38.7102 + 30.1102 = 134 For quarter 2 of the next year, we get

Yn = 104.1 + 2.31142 + 15.7112 + 38.7102 + 30.1102 = 152 Notice these are not the same values we obtained using the multiplicative decomposition method. We could compare the MAD or MSE for each method and choose the one that is better.

5.8 Monitoring and Controlling Forecasts

After a forecast has been completed, it is important that it not be forgotten. No manager wants to be reminded when his or her forecast is horribly inaccurate, but a firm needs to determine why the actual demand (or whatever variable is being examined) differed significantly from that projected.4

One way to monitor forecasts to ensure that they are performing well is to employ a track- ing signal. A tracking signal is a measurement of how well the forecast is predicting actual values. As forecasts are updated every week, month, or quarter, the newly available demand data are compared to the forecast values.

The tracking signal is computed as the running sum of the forecast errors (RSFE) divided by the mean absolute deviation:

Tracking signal = RSFE

MAD (5-13)

= g1Forecast error2

MAD

A tracking signal measures how well predictions fit actual data.

Forecasting at nBC Universal

NBC Universal (NBCU), a world leader in the production, distri- bution, and marketing of entertainment, news, and information, had revenues of over $14 billion in 2005. NBCU owns a television network and several stations in the United States, an impressive portfolio of cable networks, a major motion picture company, and very popular theme parks. Over 60 percent of NBCU’s rev- enues are from the sales of on-air advertising time on its television networks and stations.

Each year, the up-front market is a brief period in late May when the television networks sell a majority of their on-air ad- vertising inventory, right after announcing their program sched- ules for the upcoming broadcast year. To address the challenging problem of forecasting up-front market demand, NBCU initially relied primarily on judgment models and then tried time-series

forecasting models. These models proved, however, to be rather unsatisfactory due to the unique nature of the demand popu- lation. NBCU now estimates up-front demand using a novel procedure that combines the Delphi forecasting method with grassroots forecasting.

The system, which has been in place since 2004, has been used to support sales decisions each year worth over $4.5 bil- lion. The system enables NBCU to easily analyze pricing scenarios across all of its television properties, while predicting demand with a high level of accuracy. NBCU’s sales leaders have credited the forecast system with giving the company a unique competi- tive advantage over its competitors.

Source: Based on S. Bollapragada, S. Gupta, B. Hurwitz, P. Miles, and R. Tyagi, “NBC-Universal Uses a Novel Qualitative Forecasting Technique to Predict Advertising Demand,” Interfaces 38, 2 (2008): 103–111, © Trevor S. Hale.

IN ACTION

4 If the forecaster is accurate, he or she usually makes sure that everyone is aware of his or her talents. Very seldom does one read articles in Fortune, Forbes, or the Wall Street Journal, however, about money managers who are consis- tently off by 25% in their stock market forecasts.

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5.8 MoniToRing AnD ConTRoLLing FoRECAsTs  173

where

MAD = g (Forecast error)

n

as seen earlier in Equation 5-1. Positive tracking signals indicate that demand is greater than the forecast. Negative signals

mean that demand is less than the forecast. A good tracking signal—that is, one with a low RSFE—has about as much positive error as it has negative error. In other words, small devia- tions are okay, but the positive and negative deviations should balance so that the tracking signal centers closely around zero.

When tracking signals are calculated, they are compared with predetermined control limits. When a tracking signal exceeds an upper or lower limit, a signal is tripped. This means that there is a problem with the forecasting method, and management may want to reevaluate the way it forecasts demand. Figure 5.7 shows the graph of a tracking signal that is exceeding the range of acceptable variation. If the model being used is exponential smoothing, perhaps the smoothing constant needs to be readjusted.

How do firms decide what the upper and lower tracking limits should be? There is no single answer, but they try to find reasonable values—in other words, limits not so low as to be triggered with every small forecast error and not so high as to allow bad forecasts to be regu- larly overlooked. George Plossl and Oliver Wight, two inventory control experts, suggested using maximums of {4 MADs for high-volume stock items and {8 MADs for lower-volume items.5

Other forecasters suggest slightly lower ranges. One MAD is equivalent to approximately 0.8 standard deviation, so that {2 MADs = 1.6 standard deviations, {3 MADs = 2.4 stan- dard deviations, and {4 MADs = 3.2 standard deviations. This suggests that for a fore- cast to be “in control,” 89% of the errors are expected to fall within {2 MADs, 98, within {3 MADs, or 99.9% within {4 MADs whenever the errors are approximately normally distributed.6

Here is an example that shows how the tracking signal and RSFE can be computed. Kim- ball’s Bakery’s quarterly sales of croissants (in thousands), as well as forecast demand and error computations, are in the following table. The objective is to compute the tracking signal and determine whether forecasts are performing adequately.

Setting tracking limits is a matter of setting reasonable values for upper and lower limits.

1

2

0 MADs

Upper Control Limit

Lower Control Limit

Signal Tripped

Tracking Signal

Acceptable Range

Time

FIGURE 5.7 Plot of Tracking signals

6 To prove these three percentages to yourself, just set up a normal curve for standard deviations (Z values). Using the normal table in Appendix D, you find that the area under the curve is 0.89. This represents MADs. Similarly, standard deviations encompasses 98% of the area, and so on for MADs.

5 See G. W. Plossl and O. W. Wight, Production and Inventory Control (Upper Saddle River, NJ: Prentice Hall, 1967).

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174  CHAPTER 5 • FoRECAsTing

TIME PERIOD

FORECAST DEMAND

ACTUAL DEMAND ERROR RSFE `

FORECAST ` ERROR CUMULATIVE

ERROR MAD TRACKING

SIGNAL

1 100 90 -10 -10 10 10 10.0 -1 2 100 95 -5 -15 5 15 7.5 -2 3 100 115 +15 0 15 30 10.0 0 4 110 100 -10 -10 10 40 10.0 -1 5 110 125 +15 +5 15 55 11.0 +0.5 6 110 140 +30 +35 30 85 14.2 +2.5

In period 6, the calculations are

MAD = g �Forecast error �

n =

85

6

= 14.2

Tracking signal = RSFE

MAD =

35

14.2

= 2.5 MADs

This tracking signal is within acceptable limits. We see that it drifted from -2.0 MADs to +2.5 MADs.

Adaptive Smoothing A lot of research has been published on the subject of adaptive forecasting. This refers to com- puter monitoring of tracking signals and self-adjustment if a signal passes its preset limit. In exponential smoothing, the a and b coefficients are first selected based on values that minimize error forecasts and are then adjusted accordingly whenever the computer notes an errant tracking signal. This is called adaptive smoothing.

Forecasts are a critical part of a manager’s function. Demand forecasts drive the production, capacity, and scheduling sys- tems in a firm and affect the financial, marketing, and person- nel planning functions.

In this chapter, we introduced three types of forecasting models: time series, causal, and qualitative. Moving averages, exponential smoothing, trend projection, and decomposition time-series models were developed. Regression and multiple regression models were recognized as causal models. Four qualitative models were brief ly discussed. In addition, we

explained the use of scatter diagrams and measures of fore- casting accuracy. In future chapters, you will see the useful- ness of these techniques in determining values for the various decision-making models.

As we learned in this chapter, no forecasting method is perfect under all conditions. Even when management has found a satisfactory approach, it must still monitor and control its forecasts to make sure that errors do not get out of hand. Forecasting can often be a very challenging but rewarding part of managing.

Summary

Glossary

Adaptive Smoothing The process of automatically monitor- ing and adjusting the smoothing constants in an exponential smoothing model.

Bias A technique for determining the accuracy of a forecast- ing model by measuring the average error and its direction.

Causal Models Models that forecast using variables and fac- tors in addition to time.

Centered Moving Average An average of the values cen- tered at a particular point in time. This is used to compute seasonal indices when trend is present.

Cyclical Component A pattern in which the annual data in a time series tend to repeat every several years.

Decision-Making Group A group of experts in a Delphi technique that has the responsibility of making the forecast.

Decomposition A forecasting model that decomposes a time series into its seasonal and trend components.

Delphi A judgmental forecasting technique that uses deci- sion makers, staff personnel, and respondents to determine a forecast.

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KEy EQUATions  175

Deseasonalized Data Time-series data in which each value has been divided by its seasonal index to remove the effect of the seasonal component.

Deviation A term used in forecasting for error. Error The difference between the actual value and the fore-

cast value. Exponential Smoothing A forecasting method that is a

combination of the last forecast and the last observed value. Least Squares A procedure used in trend projection and

regression analysis to minimize the squared distances be- tween the estimated straight line and the observed values.

Mean Absolute Deviation (MAD) A technique for deter- mining the accuracy of a forecasting model by taking the average of the absolute deviations.

Mean Absolute Percent Error (MAPE) A technique for determining the accuracy of a forecasting model by taking the average of the absolute errors as a percentage of the observed values.

Mean Squared Error (MSE) A technique for determining the accuracy of a forecasting model by taking the average of the squared error terms for a forecasting model.

Moving Average A forecasting technique that averages past values in computing the forecast.

Naïve Model A time-series forecasting model in which the forecast for next period is the actual value for the current period.

Qualitative Models Models that forecast using judgments, experience, and qualitative and subjective data.

Random Component The irregular, unpredictable variations in a time series.

Running Sum of the Forecast Errors (RSFE) Used to de- velop a tracking signal for time-series forecasting models, this is a running total of the errors and may be positive or negative.

Seasonal Component A pattern of fluctuations in a time se- ries above or below an average value that repeats at regular intervals.

Seasonal Index An index number that indicates how a par- ticular season compares with an average time period (with an index of 1 indicating an average season).

Smoothing Constant A value between 0 and 1 that is used in an exponential smoothing forecast.

Time-series Model A forecasting technique that predicts the future values of a variable by using only historical data on that one variable.

Tracking Signal A measure of how well the forecast is pre- dicting actual values.

Trend Component The general upward or downward move- ment of the data in a time series over a relatively long pe- riod of time.

Trend Projection The use of a trend line to forecast a time series with trend present. A linear trend line is a regression line with time as the independent variable.

Weighted Moving Average A moving average forecasting method that places different weights on past values.

Key Equations

(5-1) MAD = g �Forecast error �

n

A measure of overall forecast error called mean absolute deviation.

(5-2) MSE = g1Error22

n

A measure of forecast accuracy called mean squared error.

(5-3) MAPE = g ` Error

Actual `

n 100%

A measure of forecast accuracy called mean absolute percent error.

(5-4) Moving average forcast

= Sum of demands in previous n periods

n

An equation for computing a moving average forecast.

(5-5) Ft+ 1 = Yt + Yt- 1 + Á + Yt- n + 1

n

A mathematical expression for a moving average forecast.

(5-6) Ft+ 1= g1Weight in period i21Actual value in period i2

g1Weights2 An equation for computing a weighted moving average forecast.

(5-7) Ft+ 1 = w1Yt + w2Yt- 1 + Á + wnYt- n + 1

w1 + w2 + Á + wn A mathematical expression for a weighted moving aver- age forecast.

(5-8) New forecast = Last period’s forecast + a(Last period’s actual demand - Last period’s forecast)

An equation for computing an exponential smoothing forecast.

(5-9) Ft+ 1 = Ft + a1Yt - Ft2 Equation 5-8 rewritten mathematically.

(5-10) Ft+ 1 = FITt + a1Yt - FITt2 Equation to update the smoothed forecast 1Ft+ 12 used in the trend-adjusted exponential smoothing model.

(5-11) Tt+ 1 = Tt + b1Ft+ 1 - FITt2 Equation to update the smoothed trend value 1Tt+ 12 used in the trend-adjusted exponential smoothing model.

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176  CHAPTER 5 • FoRECAsTing

(5-12) FITt+ 1 = Ft+ 1 + Tt+ 1 Equation to develop forecast including trend (FIT) in the trend-adjusted exponential smoothing model.

(5-13) Tracking signal = RSFE

MAD

= g1Forecast error2

MAD

An equation for monitoring forecasts with a tracking signal.

Solved Problems

Solved Problem 5-1 Demand for patient surgery at Washington General Hospital has increased steadily in the past few years, as seen in the following table:

YEAR OUTPATIENT SURGERIES PERFORMED

1 45

2 50

3 52

4 56

5 58

6 —

The director of medical services predicted 6 years ago that demand in year 1 would be 42 surger- ies. Using exponential smoothing with a weight of a = 0.20, develop forecasts for years 2 through 6. What is the MAD?

Solution

YEAR ACTUAL FORECAST (SMOOTHED) ERROR �ERROR �

1 45 42 +3 3 2 50 42.6 = 42 + 0.2145 - 422 +7.4 7.4 3 52 44.1 = 42.6 + 0.2150 - 42.62 +7.9 7.9 4 56 45.7 = 44.1 + 0.2152 - 44.12 +10.3 10.3 5 58 47.7 = 45.7 + 0.2156 - 45.72 +10.3 10.3 6 — 49.8 = 47.7 + 0.2158 - 47.72 — —

MAD = g �Errors �

n =

38.9

5 = 7.78

38.9

Solved Problem 5-2 Quarterly demand for the Jaguar XJ8 at a New York auto dealership is forecast with the equation

Yn = 10 + 3X

where

X = time period (quarter): quarter 1 of last year = 0 quarter 2 of last year = 1 quarter 3 of last year = 2 quarter 4 of last year = 3 quarter 1 of this year = 4, and so on

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and

Yn = predicted quarterly demand

The demand for luxury sedans is seasonal, and the indices for quarters 1, 2, 3, and 4 are 0.80, 1.00, 1.30, and 0.90, respectively. Using the trend equation, forecast the demand for each quarter of next year. Then adjust each forecast for seasonal (quarterly) variations.

Solution Quarter 2 of this year is coded X = 5; quarter 3 of this year, X = 6; and quarter 4 of this year, X = 7. Hence, quarter 1 of next year is coded X = 8; quarter 2, X = 9; and so on.

Yn 1next year quarter 12 = 10 + 132182 = 34 Adjusted forecast = 10.8021342 = 27.2 Yn 1next year quarter 22 = 10 + 132192 = 37 Adjusted forecast = 11.0021372 = 37 Yn 1next year quarter 32 = 10 + 1321102 = 40 Adjusted forecast = 11.3021402 = 52 Yn 1next year quarter 42 = 10 + 1321112 = 43 Adjusted forecast = 10.9021432 = 38.7

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

glossary at the end of the chapter. ●● Use the key at the back of the book (see Appendix H) to correct your answers. ●● Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

1. Qualitative forecasting models include a. regression analysis. b. Delphi. c. time-series models. d. trend lines. 2. A forecasting model that uses only historical data for the

variable being forecast is called a a. time-series model. b. causal model. c. Delphi model. d. variable model. 3. One example of a causal model is a. exponential smoothing. b. trend projections. c. moving averages. d. regression analysis. 4. Which of the following is a time-series model? a. the Delphi model b. regression analysis c. exponential smoothing d. multiple regression 5. Which of the following is not a component of a time

series? a. seasonality b. causal variations c. trend d. random variations 6. Which of the following may be negative? a. MAD b. bias c. MAPE d. MSE

7. When comparing several forecasting models to determine which one best fits a particular set of data, the model that should be selected is the one

a. with the highest MSE. b. with the MAD closest to 1. c. with a bias of 0. d. with the lowest MAD. 8. In exponential smoothing, if you wish to give a

significant weight to the most recent observations, then the smoothing constant should be

a. close to 0. b. close to 1. c. close to 0.5. d. less than the error. 9. A trend equation is a regression equation in which a. there are multiple independent variables. b. the intercept and the slope are the same. c. the dependent variable is time. d. the independent variable is time. 10. Sales for a company are typically higher in the summer

months than in the winter months. This variation would be called a

a. trend. b. seasonal factor. c. random factor. d. cyclical factor. 11. An exponential smoothing forecast with a smoothing

constant of 1 would result in the same forecast as a. a trend line forecast. b. an exponential smoothing forecast with a smoothing

constant of 0. c. a naïve forecast. d. a 2-period moving average forecast.

sELF-TEsT  177

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178  CHAPTER 5 • FoRECAsTing

12. If the seasonal index for January is 0.80, then a. January sales tend to be 80% higher than an average

month. b. January sales tend to be 20% higher than an average

month. c. January sales tend to be 80% lower than an average

month. d. January sales tend to be 20% lower than an average

month. 13. If both trend and seasonal components are present in a

time series, then the seasonal indices a. should be computed based on an overall average. b. should be computed based on CMAs. c. will all be greater than 1. d. should be ignored in developing the forecast.

14. Which of the following is used to alert the user of a forecasting model that a significant error occurred in one of the periods?

a. a seasonal index b. a smoothing constant c. a tracking signal d. a regression coefficient 15. If the multiplicative decomposition model is used to

forecast daily sales for a retail store, how many seasons will there be?

a. 4 b. 7 c. 12 d. 365

Discussion Questions and Problems

Discussion Questions 5-1 Describe briefly the steps used to develop a forecast-

ing system. 5-2 What is a time-series forecasting model? 5-3 What is the difference between a causal model and a

time-series model? 5-4 What is a qualitative forecasting model, and when is

it appropriate? 5-5 What are some of the problems and drawbacks of

the moving average forecasting model? 5-6 What effect does the value of the smoothing con-

stant have on the weight given to the past forecast and the past observed value?

5-7 Describe briefly the Delphi technique. 5-8 What is MAD, and why is it important in the selec-

tion and use of forecasting models? 5-9 Explain how the number of seasons is determined

when forecasting with a seasonal component. 5-10 A seasonal index may be less than one, equal to one,

or greater than one. Explain what each of these val- ues would mean.

5-11 How is the impact of seasonality removed from a time series?

5-12 In using the decomposition method, the forecast based on trend is found using the trend line. How is the seasonal index used to adjust this forecast based on trend?

5-13 Explain what would happen if the smoothing con- stant in an exponential smoothing model was equal to zero. Explain what would happen if the smooth- ing constant was equal to one.

5-14 Explain when a CMA (rather than an overall aver- age) should be used in computing a seasonal index. Explain why this is necessary.

Problems 5-15 Develop a 4-month moving average forecast for

Wallace Garden Supply, and compute the MAD. A 3-month moving average forecast was developed in the section on moving averages in Table 5.2.

5-16 Using MAD, determine whether the forecast in Problem 5-15 or the forecast in the section concern- ing Wallace Garden Supply is more accurate.

5-17 Data collected on the yearly demand for 50-pound bags of fertilizer at Wallace Garden Supply are shown in the following table. Develop a 3-year mov- ing average to forecast sales. Then estimate demand again with a weighted moving average in which sales in the most recent year are given a weight of 2 and sales in the other 2 years are each given a weight of 1. Which method do you think is better?

YEAR DEMAND FOR FERTILIZER

(1,000s OF BAGS)

1 4

2 6

3 4

4 5

5 10

6 8

7 7

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

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

(Continued on next page)

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DisCUssion QUEsTions AnD PRoBLEMs  179

YEAR DEMAND FOR FERTILIZER

(1,000s OF BAGS)

8 9

9 12

10 14

11 15

5-18 Develop a trend line for the demand for fertilizer in Problem 5-17, using any computer software.

5-19 In Problems 5-17 and 5-18, three different fore- casts were developed for the demand for fertilizer. These three forecasts are a 3-year moving average, a weighted moving average, and a trend line. Which one would you use? Explain your answer.

5-20 Use exponential smoothing with a smoothing con- stant of 0.3 to forecast the demand for fertilizer given in Problem 5-17. Assume that last period’s forecast for year 1 is 5,000 bags to begin the procedure. Would you prefer to use the exponential smoothing model or the weighted average model developed in Problem 5-17? Explain your answer.

5-21 A college student has just completed her junior year. The following table summarizes her grade-point av- erage (GPA) for each of the past nine quarters:

YEAR SEMESTER GPA

Freshman Fall 2.4

Winter 2.9

Spring 3.1

Sophomore Fall 3.2

Winter 3.0

Spring 2.9

Junior Fall 2.8

Winter 3.6

Spring 3.2

(a) Forecast the student’s GPA for the fall semester of her senior year by using a three-period moving average.

(b) Forecast the student’s GPA for the fall semester of her senior year by using exponential smooth- ing with a = 0.2.

(c) Which of the two methods provides a more ac- curate forecast? Justify your answer.

5-22 Sales of Cool-Man air conditioners have grown steadily during the past 5 years:

The sales manager had predicted, before the business started, that year 1’s sales would be 410 air condi- tioners. Using exponential smoothing with a weight of a = 0.30, develop forecasts for years 2 through 6.

5-23 Using smoothing constants of 0.6 and 0.9, develop forecasts for the sales of Cool-Man air conditioners (see Problem 5-22).

5-24 What effect did the smoothing constant have on the forecast for Cool-Man air conditioners? (See Prob- lems 5-22 and 5-23.) Which smoothing constant gives the more accurate forecast?

5-25 Use a 3-year moving average forecasting model to forecast the sales of Cool-Man air conditioners (see Problem 5-22).

5-26 Using the trend projection method, develop a fore- casting model for the sales of Cool-Man air condi- tioners (see Problem 5-22).

5-27 Would you use exponential smoothing with a smooth- ing constant of 0.3, a 3-year moving average, or a trend line to predict the sales of Cool-Man air condi- tioners? Refer to Problems 5-22, 5-25, and 5-26.

5-28 Sales of industrial vacuum cleaners at R. Lowenthal Supply Co. over the past 13 months are as follows:

SALES ($1,000s) MONTH

SALES ($1,000s) MONTH

11 January 14 August

14 February 17 September

16 March 12 October

10 April 14 November

15 May 16 December

17 June 11 January

11 July

(a) Using a moving average with three periods, de- termine the demand for vacuum cleaners for next February.

(b) Using a weighted moving average with three pe- riods, determine the demand for vacuum cleaners for February. Use 3, 2, and 1 for the weights of the most recent, second most recent, and third most recent periods, respectively. For example, if you were forecasting the demand for February, November would have a weight of 1, December would have a weight of 2, and January would have a weight of 3.

(c) Evaluate the accuracy of each of these methods. (d) What other factors might R. Lowenthal consider

in forecasting sales?

5-29 Passenger miles flown on Northeast Airlines, a com- muter firm serving the Boston hub, are as follows for the past 12 weeks:

YEAR SALES

1 450

2 495

3 518

YEAR SALES

4 563

5 584

6 ?

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180  CHAPTER 5 • FoRECAsTing

ACTUAL PASSENGER

ACTUAL PASSENGER

WEEK MILES (1,000s) WEEK

MILES (1,000s)

1 17 7 20

2 21 8 18

3 19 9 22

4 23 10 20

5 18 11 15

6 16 12 22

(a) Assuming an initial forecast for week 1 of 17,000 miles, use exponential smoothing to compute miles for weeks 2 through 12. Use a = 0.2.

(b) What is the MAD for this model? (c) Compute the RSFE and tracking signals. Are they

within acceptable limits?

5-30 Emergency calls to Winter Park, Florida’s 911 sys- tem for the past 24 weeks are as follows:

WEEK CALLS WEEK CALLS WEEK CALLS

1 50 9 35 17 55

2 35 10 20 18 40

3 25 11 15 19 35

4 40 12 40 20 60

5 45 13 55 21 75

6 35 14 35 22 50

7 20 15 25 23 40

8 30 16 55 24 65

(a) Compute the exponentially smoothed forecast of calls for each week. Assume an initial forecast of 50 calls in the first week, and use a = 0.1. What is the forecast for week 25?

(b) Reforecast each period using a = 0.6. (c) Actual calls during week 25 were 85. Which

smoothing constant provides a superior forecast?

5-31 How would the forecast for week 25 of the previous problem change if the initial forecast was 40 instead of 50? How would the forecast for week 25 change if the forecast for week 1 was assumed to be 60?

5-32 Sales of vacuum cleaners over the past 13 months were as follows:

MONTH SALES MONTH SALES

January 9 July 9

February 12 August 12

March 14 September 15

April 8 October 10

May 13 November 12

June 15 December 14

(a) Using a moving average with three periods, pre- dict the demand for vacuum cleaners for next February.

(b) Using a three-period weighted moving average with weights 3, 2, and 1, predict the demand for vacuum cleaners for February.

(c) Evaluate and comment on the accuracy of each of these models.

5-33 Consulting income at Kate Walsh Associates for the period February–July has been as follows:

MONTH INCOME ($1,000s)

February 70.0

March 68.5

April 64.8

May 71.7

June 71.3

July 72.8

Use exponential smoothing to forecast August’s in- come. Assume that the initial forecast for Febru- ary is $65,000. The smoothing constant selected is a = 0.1.

5-34 Resolve Problem 5-33 with a = 0.3. Using MAD, which smoothing constant provides a better forecast?

5-35 A major source of revenue in Texas is a state sales tax on certain types of goods and services. Data are com- piled, and the state comptroller uses them to project future revenues for the state budget. One particular category of goods is classified as Retail Trade. Four years of quarterly data (in $1,000,000s) for one par- ticular area of southeast Texas follow:

QUARTER YEAR 1 YEAR 2 YEAR 3 YEAR 4

1 218 225 234 250

2 247 254 265 283

3 243 255 264 289

4 292 299 327 356

(a) Compute a seasonal index for each quarter based on a CMA.

(b) Deseasonalize the data, and develop a trend line on the deseasonalized data.

(c) Use the trend line to forecast the sales for each quarter of year 5.

(d) Use the seasonal indices to adjust the forecasts found in part (c) to obtain the final forecasts.

5-36 Using the data in Problem 5-35, develop a multiple regression model to predict sales (with both trend and seasonal components), using dummy variables to incorporate the seasonal factor into the model. Use this model to predict sales for each quarter of the next year. Comment on the accuracy of this model.

5-37 Trevor Harty, an avid mountain biker, always wanted to start a business selling top-of-the-line mountain

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DisCUssion QUEsTions AnD PRoBLEMs  181

bikes and other outdoor supplies. A little over 6 years ago, he and a silent partner opened a store called Hale and Harty Trail Bikes and Supplies. Growth was rapid in the first 2 years, but since that time, growth in sales has slowed a bit, as expected. The quarterly sales (in $1,000s) for the past 4 years are shown in the table below:

QUARTER YEAR 1 YEAR 2 YEAR 3 YEAR 4

1 274 282 282 296

2 172 178 182 210

3 130 136 134 158

4 162 168 170 182

(a) Develop a trend line using the data in the table. Use this to forecast sales for each quarter of year 5. What does the slope of this line indicate?

(b) Use the multiplicative decomposition model to incorporate both trend and seasonal components into the forecast. What does the slope of this line indicate?

(c) Compare the slope of the trend line in part (a) to the slope in the trend line for the decomposi- tion model that was based on the deseasonalized sales figures. Discuss why these are so different and explain which is the better one to use.

5-38 The unemployment rates in the United States dur- ing a 10-year period are given in the following table. Use exponential smoothing to find the best forecast for next year. Use smoothing constants of 0.2, 0.4, 0.6, and 0.8. Which one had the lowest MAD?

YEAR 1 2 3 4 5 6 7 8 9 10

Unemployment rate (%)

7.2 7.0 6.2 5.5 5.3 5.5 6.7 7.4 6.8 6.1

5-39 Management of Davis’s Department Store has used time-series extrapolation to forecast retail sales for the next four quarters. The sales estimates are $100,000, $120,000, $140,000, and $160,000 for the respective quarters before adjusting for seasonal- ity. Seasonal indices for the four quarters have been found to be 1.30, 0.90, 0.70, and 1.10, respectively. Compute a seasonalized or adjusted sales forecast.

5-40 In the past, Judy Holmes’s tire dealership sold an av- erage of 1,000 radials each year. In the past 2 years, 200 and 250, respectively, were sold in fall, 350 and 300 in winter, 150 and 165 in spring, and 300 and 285 in summer. With a major expansion planned, Judy projects sales next year to increase to 1,200 ra- dials. What will the demand be each season?

5-41 The following table provides the Dow Jones Industrial Average (DJIA) opening index value on the first work- ing day of 1994–2013. Develop a trend line and use it to predict the opening DJIA index value for years 2014, 2015, and 2016. Find the MSE for this model.

YEAR DJIA YEAR DJIA

2013 13,104 2003 8,342

2012 12,392 2002 10,022

2011 11,577 2001 10,791

2010 10,431 2000 11,502

2009 8,772 1999 9,213

2008 13,262 1998 7,908

2007 12,460 1997 6,448

2006 10,718 1996 5,117

2005 10,784 1995 3,834

2004 10,453 1994 3,754

5-42 Using the DJIA data in Problem 5-41 and expo- nential smoothing with trend adjustment, forecast the opening DJIA value for 2014. Use a = 0.8 and b = 0.2. Compare the MSE for this technique with the MSE for the trend line.

5-43 Refer to the DJIA data in Problem 5-41.

(a) Use an exponential smoothing model with a smoothing constant of 0.4 to predict the open- ing DJIA index value for 2014. Find the MSE for this.

(b) Use QM for Windows or Excel to find the smoothing constant that would provide the low- est MSE.

5-44 The following table gives the average monthly ex- change rate between the U.S. dollar and the euro for 2009. It shows that 1 euro was equivalent to 1.289 U.S. dollars in January 2009. Develop a trend line that could be used to predict the exchange rate for 2010. Use this model to predict the exchange rate for January 2010 and February 2010.

MONTH EXCHANGE RATE

January 1.289

February 1.324

March 1.321

April 1.317

May 1.280

June 1.254

July 1.230

August 1.240

September 1.287

October 1.298

November 1.283

December 1.311

5-45 For the data in Problem 5-44, develop an exponen- tial smoothing model with a smoothing constant of 0.3. Using the MSE, compare this with the model in Problem 5-44.

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182  CHAPTER 5 • FoRECAsTing

Southwestern University Football Game Attendance, 2008–2013

2008 2009 2010

GAME ATTENDEES OPPONENT ATTENDEES OPPONENT ATTENDEES OPPONENT

1 34,200 Baylor 36,100 Oklahoma 35,900 TCU

2a 39,800 Texas 40,200 Nebraska 46,500 Texas Tech

3 38,200 LSU 39,100 UCLA 43,100 Alaska

4b 26,900 Arkansas 25,300 Nevada 27,900 Arizona

5 35,100 USC 36,200 Ohio State 39,200 Rice

2011 2012 2013

GAME ATTENDEES OPPONENT ATTENDEES OPPONENT ATTENDEES OPPONENT

1 41,900 Arkansas 42,500 Indiana 46,900 LSU

2a 46,100 Missouri 48,200 North Texas 50,100 Texas

3 43,900 Florida 44,200 Texas A&M 45,900 Prairie View A&M

4b 30,100 Miami 33,900 Southern 36,300 Montana

5 40,500 Duke 47,800 Oklahoma 49,900 Arizona State

aHomecoming games. bDuring the fourth week of each season, Stephenville hosted a hugely popular southwestern crafts festival. This event brought tens of thousands of tourists to the town, especially on weekends, and had an obvious negative impact on game attendance.

Source: J. Heizer and B. Render, Operations Management, 11th ed., © 2014. Reprinted and electronically reproduced by permission of Pearson Education, Inc., New York, NY.

See our Internet home page, at www.pearsonhighered.com/render, for additional homework problems, Problems 5-46 to 5-54.

Internet Homework Problems

Southwestern University (SWU), a large state college in Ste- phenville, Texas, 30 miles southwest of the Dallas/Fort Worth metroplex, enrolls close to 20,000 students. In a typical town– gown relationship, the school is a dominant force in the small city, with more students during fall and spring than permanent residents.

A longtime football powerhouse, SWU is a member of the Big Eleven conference and is usually in the top 20 in col- lege football rankings. To bolster its chances of reaching the elu- sive and long-desired number-one ranking, in 2008 SWU hired the legendary Billy Bob Taylor as its head coach. Although the number-one ranking remained out of reach, attendance at the five Saturday home games each year increased. Prior to Tay- lor’s arrival, attendance generally averaged 25,000 to 29,000 per game. Season ticket sales bumped up by 10,000 just with the announcement of the new coach’s arrival. Stephenville and SWU were ready to move to the big time!

The immediate issue facing SWU, however, was not NCAA ranking. It was capacity. The existing SWU stadium,

built in 1953, has seating for 54,000 fans. The following table indicates attendance at each game for the past 6 years.

One of Taylor’s demands upon joining SWU had been a stadium expansion, or possibly even a new stadium. With atten- dance increasing, SWU administrators began to face the issue head-on. Taylor had wanted dormitories solely for his athletes in the stadium as an additional feature of any expansion.

SWU’s president, Dr. Marty Starr, decided it was time for his vice president of development to forecast when the existing stadium would “max out.” He also sought a revenue projection, assuming an average ticket price of $20 in 2014 and a 5% in- crease each year in future prices.

Discussion Questions 1. Develop a forecasting model, justify its selection over

other techniques, and project attendance through 2015. 2. What revenues are to be expected in 2014 and 2015? 3. Discuss the school’s options.

Case Study

Forecasting Attendance at SWU Football Games

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 183 CAsE sTUDy  183

For years The Glass Slipper restaurant has operated in a resort community near a popular ski area of New Mexico. The restau- rant is busiest during the first 3 months of the year, when the ski slopes are crowded and tourists flock to the area.

When James and Deena Weltee built The Glass Slipper, they had a vision of the ultimate dining experience. As the view of surrounding mountains was breathtaking, a high priority was placed on having large windows and providing a spectacular view from anywhere inside the restaurant. Special attention was also given to the lighting, colors, and overall ambiance, result- ing in a truly magnificent experience for all who came to enjoy gourmet dining. Since its opening, The Glass Slipper has de- veloped and maintained a reputation as one of the “must visit” places in that region of New Mexico.

While James loves to ski and truly appreciates the moun- tains and all that they have to offer, he also shares Deena’s dream of retiring to a tropical paradise and enjoying a more re- laxed lifestyle on the beach. After some careful analysis of their financial condition, they knew that retirement was many years away. Nevertheless, they were hatching a plan to bring them closer to their dream. They decided to sell The Glass Slipper and open a bed and breakfast on a beautiful beach in

TABLE 5.13 Monthly Revenue ($1,000s)

MONTH 2011 2012 2013

January 438 444 450

February 420 425 438

March 414 423 434

April 318 331 338

May 306 318 331

June 240 245 254

July 240 255 264

August 216 223 231

September 198 210 224

October 225 233 243

November 270 278 289

December 315 322 335

Mexico. While this would mean that work was still in their future, they could wake up in the morning to the sight of the palm trees blowing in the wind and the waves lapping at the shore. They also knew that hiring the right manager would al- low James and Deena the time to begin a semi-retirement in a corner of paradise.

To make this happen, James and Deena would have to sell The Glass Slipper for the right price. The price of the business would be based on the value of the property and equipment as well as projections of future income. A forecast of sales for the next year is needed to help in the determination of the value of the restaurant. Monthly sales for each of the past 3 years are provided in Table 5.13.

Discussion Questions 1. Prepare a graph of the data. On this same graph, plot a

12-month moving average forecast. Discuss any apparent trend and seasonal patterns.

2. Use regression to develop a trend line that could be used to forecast monthly sales for the next year. Is the slope of this line consistent with what you observed in question 1? If not, discuss a possible explanation.

3. Use the multiplicative decomposition model and these data to forecast sales for each month of the next year. Dis- cuss why the slope of the trend equation with this model is so different from that of the trend equation in question 2.

Case Study

Forecasting Monthly Sales

See our Internet home page, at www.pearsonhighered.com/render, for the additional case study on Akron Zoological Park. This case involves forecasting attendance at Akron’s zoo.

Internet Case Study

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184  CHAPTER 5 • FoRECAsTing

Bibliography

Billah, Baki, Maxwell L. King, Ralph D. Snyder, and Anne B. Koehler. “Ex- ponential Smoothing Model Selection for Forecasting,” International Journal of Forecasting 22, 2 (April–June 2006): 239–247.

Black, Ken. Business Statistics: For Contemporary Decision Making, 8th ed. New York: John Wiley & Sons, Inc., 2014.

Diebold, F. X. Elements of Forecasting, 4th ed. Cincinnati: Cengage College Publishing, 2007.

Gardner, Everette, Jr. “Exponential Smoothing: The State of the Art—Part II,” International Journal of Forecasting 22, 4 (October 2006): 637–666.

Granger, Clive W., and J. M. Hashem Pesaran. “Economic and Statisti- cal Measures of Forecast Accuracy,” Journal of Forecasting 19, 7 (December 2000): 537–560.

Hanke, J. E., and D. W. Wichern. Business Forecasting, 9th ed. Upper Saddle River, NJ: Pearson, 2009.

Heizer, J., B. Render, and C. Munson. Operations Management: Sustainability and Supply Chain Management, 12th ed. Upper Saddle River, NJ: Pearson, 2017.

Hyndman, Rob J. “The Interaction Between Trend and Seasonality,” In- ternational Journal of Forecasting 20, 4 (October–December 2004): 561–563.

Hyndman, Rob J., and Anne B. Koehler. “Another Look at Measures of Fore- cast Accuracy,” International Journal of Forecasting 22, 4 (October 2006): 679–688.

Levine, David M., David F. Stephan, and Kathryn A. Szabat. Statistics for Managers Using Microsoft Excel, 7th ed. Upper Saddle River, NJ: Pearson, 2014.

Meade, Nigel. “Evidence for the Selection of Forecasting Methods,” Journal of Forecasting 19, 6 (November 2000): 515–535.

Snyder, Ralph D., and Roland G. Shami. “Exponential Smoothing of Sea- sonal Data: A Comparison,” Journal of Forecasting 20, 3 (April 2001): 197–202.

Yurkiewicz, J. “Forecasting Software Survey,” OR/MS Today 39, 3 (June 2012): 52–61.

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 185

6.7 Understand the use of safety stock.

6.8 Compute single period inventory quantities using marginal analysis.

6.9 Understand the importance of ABC analysis.

6.10 Describe the use of material requirements planning in solving dependent-demand inventory problems.

6.11 Discuss just-in-time inventory concepts to reduce inventory levels and costs.

6.12 Discuss enterprise resource planning systems.

6.1 Understand the importance of inventory control.

6.2 Understand the various types of inventory related decisions.

6.3 Use the economic order quantity (EOQ) to determine how much to order.

6.4 Compute the reorder point (ROP) in determining when to order more inventory.

6.5 Handle inventory problems that allow noninstantaneous receipt.

6.6 Handle inventory problems that allow quantity discounts.

After completing this chapter, students will be able to:

Inventory Control Models

LEARNING OBJECTIVES

6 CHAPTER

Inventory is one of the most expensive and important assets to many companies, representing as much as 50% of total invested capital. Managers have long recognized that good inventory control is crucial. On one hand, a firm can try to reduce costs by reducing on-hand inventory levels. On the other hand, customers become dissatisfied when frequent inventory outages, called stockouts, occur. Thus, companies must find the balance between low and high inventory levels. As you would expect, cost minimization is the major factor in achieving this delicate balance.

Inventory is any stored resource that is used to satisfy a current or a future need. Raw ma- terials, work in process, and finished goods are examples of inventory. Inventory levels for fin- ished goods are a direct function of demand. When we determine the demand for completed clothes dryers, for example, it is possible to use this information to determine the amounts of sheet metal, paint, electric motors, switches, and other raw materials and work in process that are needed to produce the finished product.

All organizations have some type of inventory planning and control system. A bank has methods to control its inventory of cash. A hospital has methods to control blood supplies and other important items. State and federal governments, schools, and virtually every manufac- turing and production organization are concerned with inventory planning and control. Study- ing how organizations control their inventory is equivalent to studying how they achieve their

Inventory is any stored resource that is used to satisfy a current or future need.

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186  CHAPTER 6 • InvEnTORy COnTROl MODEls

objectives by supplying goods and services to their customers. Inventory is the common thread that ties all the functions and departments of the organization together.

Figure 6.1 illustrates the basic components of an inventory planning and control sys- tem. The planning phase is concerned primarily with what inventory is to be stocked and how it is to be acquired (whether it is to be manufactured or purchased). This information is then used in forecasting demand for the inventory and in controlling inventory levels. The feedback loop in Figure 6.1 provides a way of revising the plan and forecast based on expe- riences and observation.

Through inventory planning, an organization determines what goods and/or services are to be produced. In cases of physical products, the organization must also determine whether to produce these goods or to purchase them from another manufacturer. When this has been determined, the next step is to forecast the demand. As discussed in Chapter 5, there are many mathematical techniques that can be used in forecasting demand for a particular product. The emphasis in this chapter is on inventory control—that is, how to maintain adequate inventory levels within an organization.

6.1 Importance of Inventory Control

Inventory control serves several important functions and adds a great deal of flexibility to the operation of the firm. Consider the following five uses of inventory:

1. The decoupling function 2. Storing resources 3. Irregular supply and demand 4. Quantity discounts 5. Avoiding stockouts and shortages

Decoupling Function One of the major functions of inventory is to decouple manufacturing processes within the or- ganization. If you did not store inventory, there could be many delays and inefficiencies. For example, when one manufacturing activity has to be completed before a second activity can be started, it could stop the entire process. If, however, you have some stored inventory between processes, it could act as a buffer.

FIGURE 6.1 Inventory Planning and Control

Planning on what to stock and how to

get it

Forecasting parts/product

demand

Controlling inventory

levels

Feedback metrics to

revise plans and forecasts

Inventory can act as a buffer.

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6.2 InvEnTORy DECIsIOns  187

Storing Resources Agricultural and seafood products often have definite seasons over which they can be harvested or caught, but the demand for these products is somewhat constant during the year. In these and similar cases, inventory can be used to store these resources.

In a manufacturing process, raw materials can be stored by themselves, in work in process, or in the finished product. Thus, if your company makes lawn mowers, you might obtain lawn mower tires from another manufacturer. If you have 400 finished lawn mowers and 300 tires in inventory, you actually have 1,900 tires stored in inventory. Three hundred tires are stored by themselves, and 1,600 (1,600 = 4 tires per lawn mower * 400 lawn mowers) tires are stored in the finished lawn mowers. In the same sense, labor can be stored in inventory. If you have 500 subassemblies and it takes 50 hours of labor to produce each assembly, you actually have 25,000 labor hours stored in inventory in the subassemblies. In general, any resource, physical or other- wise, can be stored in inventory.

Irregular Supply and Demand When the supply of or demand for an item is irregular, storing certain amounts in inventory can be important. If the greatest demand for Diet-Delight beverage is during the summer, you will have to make sure that there is enough supply to meet this irregular demand. This might require that you produce more of the soft drink in the winter than is actually needed to meet the winter demand. The inventory levels of Diet-Delight will gradually build up over the winter, but this inventory will be needed in the summer. The same is true for ir- regular supplies.

Quantity Discounts Another use of inventory is to take advantage of quantity discounts. Many suppliers offer dis- counts for large orders. For example, an electric jigsaw might normally cost $20 per unit. If you order 300 or more saws in one order, your supplier may lower the cost to $18.75. Purchas- ing in larger quantities can substantially reduce the cost of products. There are, however, some disadvantages of buying in larger quantities. You will have higher storage costs and higher costs due to spoilage, damaged stock, theft, insurance, and so on. Furthermore, by investing in more inventory, you will have less cash to invest elsewhere.

Avoiding Stockouts and Shortages Another important function of inventory is to avoid shortages or stockouts. If you are repeatedly out of stock, customers are likely to go elsewhere to satisfy their needs. Lost goodwill can be an expensive price to pay for not having the right item at the right time.

6.2 Inventory Decisions

Even though there are literally millions of different types of products produced in our society, there are only two fundamental decisions that you have to make when controlling inventory:

1. How much to order 2. When to order

The purpose of all inventory models and techniques is to determine rationally how much to order and when to order. As you know, inventory fulfills many important functions within an organization. But as the inventory levels go up to provide these functions, the cost of storing and holding inventory also increases. Thus, you must reach a fine balance in establishing inventory levels. A major objective in controlling inventory is to minimize total inventory costs. Some of the most significant inventory costs follow:

1. Cost of the items (purchase cost or material cost) 2. Cost of ordering 3. Cost of carrying, or holding, inventory 4. Cost of stockouts

A major objective of all inventory models is to minimize inventory costs.

Resources can be stored in work in process.

Inventory can alleviate irregular demand and supply patterns.

Buying in bulk can be an economic advantage.

Carrying safety stock can increase customer satisfaction.

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Defining the Problem Managing today’s energy supply chain is an ever-increasing complex problem. Yet, to survive in today’s marketplace, oil and gas companies must optimize their supply chains.

Developing a Model Researchers from Portugal developed a unidirectional model of oil and gas products from refineries to dis- tribution center. The model incorporated many different parameters, including number of products, maxi- mum storage capacities for each product, and pipeline capacities, as well as the demand for each product.

Acquiring the Input Data The input data consisted of the current inventory levels of specific oil and gas products at different points along the supply chain.

Developing a Solution A large mixed-integer linear program (see Chapter 10 in this book) was developed as part of the solution procedure to optimize the supply chain.

Testing the Solution All the problems that were solved by the Portuguese researchers were based on real data from Companhia Logistica Combustiveis, a large Portuguese oil and gas distribution company. Testing was done, and the results were compared to real-world scenarios.

Analyzing the Results Results obtained from the model were compared to the real-world operation. The results identified oppor- tunities for marked improvements in flow rates and reductions in pumping costs.

Implementing the Results As a result, oil and gas companies can better plan and manage their supply chains with fewer disruptions and lower overall costs.

Source: Based on S. Relvas, S. N. B. Magatão, A. P. F. D. Barbosa-Póvoa, and F. Neves, “Integrated Scheduling and Inventory Management of an Oil Products Distribution System,” Omega 41 (2013): 955–968, © Trevor S. Hale.

MODELING IN THE REAL WORLD

Managing Today’s Energy supply Chain

ORDERING COST FACTORS CARRYING COST FACTORS

Developing and sending purchase orders Cost of capital

Processing and inspecting incoming inventory Taxes

Bill paying Insurance

Inventory inquiries Spoilage

Utilities, phone bills, and so on for the purchasing department Theft

Salaries and wages for purchasing department employees Obsolescence

Supplies such as forms and paper for the purchasing department Salaries and wages for warehouse employees

Utilities and building costs for the warehouse

Supplies such as forms and paper for the warehouse

TABLE 6.1 Inventory Cost Factors

The most common factors associated with ordering cost and holding cost are shown in Ta- ble 6.1. Notice that the ordering cost is generally independent of the size of the order and often involves personnel time. An ordering cost is incurred each time an order is placed, whether the order is for 1 unit or 1,000 units. The time to process the paperwork, pay the bill, and so forth does not depend on the number of units ordered.

Defining the Problem

Developing a Model

Acquiring the Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

Defining the Problem

Developing a Model

Acquiring the Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

Defining the Problem

Developing a Model

Acquiring the Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

Defining the Problem

Developing a Model

Acquiring the Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

Defining the Problem

Developing a Model

Acquiring the Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

Defining the Problem

Developing a Model

Acquiring the Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

Defining the Problem

Developing a Model

Acquiring the Input Data

Developing a Solution

Testing the Solution

Analyzing the Results

Implementing the Results

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6.3 ECOnOMIC ORDER QUAnTITy: DETERMInIng HOw MUCH TO ORDER  189

On the other hand, the holding cost varies as the size of the inventory varies. If 1,000 units are placed into inventory, the taxes, insurance, cost of capital, and other factors in the holding cost will be higher than if only 1 unit was put into inventory. Similarly, if the inventory level is low, there is little chance of spoilage and obsolescence.

The cost of the items, or the purchase cost, is what is paid to acquire the inventory. The stockout cost indicates the lost sales and goodwill (future sales) that result from not having the items available for the customers. This is discussed later in the chapter.

6.3 Economic Order Quantity: Determining How Much to Order

The economic order quantity (EOQ) is one of the oldest and most widely known inventory control techniques. Research on its use dates back to a 1915 publication by Ford W. Harris. This technique is still used by a large number of organizations today. It is relatively easy to use, but it does make a number of assumptions. Some of the most important assumptions follow:

1. Demand is known and constant over time. 2. The lead time—that is, the time between the placement of the order and the receipt of the

order—is known and constant. 3. The receipt of inventory is instantaneous. In other words, the inventory from an order

arrives in one batch, at one point in time. 4. The purchase cost per unit is constant throughout the year. Quantity discounts are not

possible. 5. The only variable costs are the cost of placing an order, ordering cost, and the cost of hold-

ing or storing inventory over time, holding or carrying cost. The holding cost per unit per year and the ordering cost per order are constant throughout the year.

6. Orders are placed so that stockouts or shortages are avoided completely.

When these assumptions are not met, adjustments must be made to the EOQ model. These are discussed later in this chapter.

With these assumptions, inventory usage has a sawtooth shape, as in Figure 6.2. In Figure 6.2, Q represents the amount that is ordered. If this amount is 500 dresses, all 500 dresses arrive at one time when an order is received. Thus, the inventory level jumps from 0 to 500 dresses. In general, an inventory level increases from 0 to Q units when an order arrives.

Because demand is constant over time, inventory drops at a uniform rate over time. (Refer to the sloped lines in Figure 6.2.) Another order is placed such that when the inventory level reaches 0, the new order is received, and the inventory level again jumps to Q units, represented by the vertical lines. This process continues indefinitely over time.

Inventory Costs in the EOQ Situation The objective of most inventory models is to minimize the total costs. With the assumptions just given, the relevant costs are the ordering cost and the carrying or holding cost. All other costs,

The inventory usage curve has a sawtooth shape.

0

Inventory Level

Minimum Inventory

Time

Order Quantity = Q = Maximum Inventory Level

FIGURE 6.2 Inventory Usage over Time

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190  CHAPTER 6 • InvEnTORy COnTROl MODEls

such as the cost of the inventory itself (the purchase cost), are constant. Thus, if we minimize the sum of the ordering and carrying costs, we are also minimizing the total costs.

The annual ordering cost is simply the number of orders per year times the cost of plac- ing each order. Since the inventory level changes daily, it is appropriate to use the average inventory level to determine annual holding or carrying cost. The annual carrying cost will equal the average inventory times the inventory carrying cost per unit per year. Again looking at Figure 6.2, we see that the maximum inventory is the order quantity (Q), and the average inventory will be one-half of that. Table 6.2 provides a numerical example to illustrate this. Notice that for this situation, if the order quantity is 10, the average inventory will be 5, or one- half of Q. Thus:

Average inventory level = Q

2 (6-1)

Using the following variables, we can develop mathematical expressions for the annual ordering and carrying costs:

Q = number of units in an order EOQ = Q* = optimal number of units to order

D = annual demand in units for the inventory item Co = ordering cost of each order Ch = holding or carrying cost per unit per year

Annual ordering cost = 1Number of orders placed per year2 * 1Ordering cost per order2

= Annual demand

Number of units in each order * 1Ordering cost per order2

= D

Q Co

Annual holding or carrying cost = 1Average inventory2 * 1Carrying cost per unit per year2

= Order quantity

2 * 1Carrying cost per unit per year2

= Q

2 Ch

A graph of the holding cost, the ordering cost, and the total of these two is shown in Figure 6.3. The lowest point on the total cost curves occurs where the ordering cost is equal to the carrying cost. Thus, to minimize total costs, given this situation, the order quantity should occur where these two costs are equal.

The objective of the simple EOQ model is to minimize total inventory cost. The relevant costs are the ordering and holding costs.

The average inventory level is one-half the maximum level.

INVENTORY LEVEL

DAY BEGINNING ENDING AVERAGE

April 1 (order received) 10 8 9

April 2 8 6 7

April 3 6 4 5

April 4 4 2 3

April 5 2 0 1

Maximum level April 1 = 10 units Total of daily averages = 9 + 7 + 5 + 3 + 1 = 25 Number of days = 5 Average inventory level = 25>5 = 5 units

TABLE 6.2 Computing Average Inventory

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6.3 ECOnOMIC ORDER QUAnTITy: DETERMInIng HOw MUCH TO ORDER  191

Finding the EOQ When the EOQ assumptions are met, total cost is minimized when:

Annual holding cost = Annual ordering cost

Q

2 Ch =

D

Q Co

Solving this for Q gives the optimal order quantity:

Q2 Ch = 2DCo

Q2 = 2DCo

Ch

Q = A2DCoCh

We derive the EOQ equation by setting ordering cost equal to carrying cost.

global Fashion Firm Fashions Inventory Management system

Founded in 1975, the Spanish retailer Zara currently has more than 1,600 stores worldwide, launches more than 10,000 new designs each year, and is recognized as one of the world’s princi- pal fashion retailers. Goods were shipped from two central ware- houses to each of the stores, based on requests from individual store managers. These local decisions inevitably led to inefficient warehouse, shipping, and logistics operations when assessed on a global scale. Recent production overruns, inefficient sup- ply chains, and an ever-changing marketplace (to say the least) caused Zara to tackle this problem.

A variety of operations research models were used in rede- signing and implementing an entirely new inventory management

system. The new centralized decision-making system replaced all store-level inventory decisions, thus providing results that were more globally optimal. Having the right products in the right places at the right time for customers has increased sales from 3% to 4% since implementation. This translated into an increase in revenue of over $230 million in 2007 and over $350 million in 2008. Talk about fashionistas!

Source: Based on F. Caro, J. Gallien, M. Díaz, J. García, J. M. Corredoira, M. Montes, J.A. Ramos, and J. Correa, “Zara Uses Operations Research to Re- engineer Its Global Distribution Process,” Interfaces 40, 1 (January–February 2010): 71–84, © Trevor S. Hale.

IN ACTION

Cost

Minimum Total Cost

Curve for Total Cost of Carrying and Ordering

Carrying Cost Curve

Ordering Cost Curve

Order QuantityOptimal Order Quantity

FIGURE 6.3 Total Cost as a Function of Order Quantity

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192  CHAPTER 6 • InvEnTORy COnTROl MODEls

This optimal order quantity is often denoted by Q*. Thus, the economic order quantity is given by the following formula:

EOQ = Q* = A2DCoCh This EOQ formula is the basis for many more advanced models, and some of these are discussed later in this chapter.

Economic Order Quantity (EOQ) Model

Annual ordering cost = D

Q Co (6-2)

Annual holding cost = Q

2 Ch (6-3)

EOQ = Q* = A2DCoCh (6-4) Sumco Pump Company Example Sumco, a company that sells pump housings to other manufacturers, would like to reduce its inventory cost by determining the optimal number of pump housings to obtain per order. The an- nual demand is 1,000 units, the ordering cost is $10 per order, and the average carrying cost per unit per year is $0.50. Using these figures, if the EOQ assumptions are met, we can calculate the optimal number of units per order:

Q* = A2DCoCh = B 211,000211020.50 = 140,000 = 200 units

The relevant total annual inventory cost is the sum of the ordering cost and the carrying cost:

Total annual cost = Ordering cost + Holding cost

In terms of the variables in the model, the total cost (TC) can now be expressed as

TC = D

Q Co +

Q

2 Ch (6-5)

The total annual inventory cost for Sumco is computed as follows:

TC = D

Q Co +

Q

2 Ch

= 1,000

200 1102 + 200

2 10.52

= $50 + $50 = $100

The number of orders per year 1D>Q2 is 5, and the average inventory 1Q>22 is 100. As you might expect, the ordering cost is equal to the carrying cost. You may wish to try dif-

ferent values for Q, such as 100 or 300 pumps. You will find that the minimum total cost occurs when Q is 200 units. The EOQ, Q*, is 200 pumps.

The total annual inventory cost is equal to ordering cost plus the holding cost for the simple EOQ model.

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6.3 ECOnOMIC ORDER QUAnTITy: DETERMInIng HOw MUCH TO ORDER  193

USING EXCEL QM FOR BASIC EOQ INVENTORY PROBLEMS The Sumco Pump Company example, and a variety of other inventory problems we address in this chapter, can be easily solved using Excel QM. Program 6.1A shows the input data for Sumco and the Excel formulas needed for the EOQ model. Program 6.1B contains the solution for this example, including the op- timal order quantity, maximum inventory level, average inventory level, and the number of orders.

Purchase Cost of Inventory Items Sometimes the total inventory cost expression is written to include the actual cost of the mate- rial purchased. With the EOQ assumptions, the purchase cost does not depend on the particular order policy found to be optimal because, regardless of how many orders are placed each year, we still incur the same annual purchase cost of D * C, where C is the purchase cost per unit and D is the annual demand in units.1

It is useful to know how to calculate the average inventory level in dollar terms when the price per unit is given. This can be done as follows. Using the variable Q to represent the

1Later in this chapter, we discuss the case in which price can affect order policy—that is, when quantity discounts are offered.

Enter the demand rate, the ordering cost, the holding cost, and, if it is available, the unit price here.

Total unit cost is given here.

PROGRAM 6.1B Excel QM solution for the sumco Pump Company Example

Total cost—including holding, ordering, and, if it is given, unit cost—is computed here.

PROGRAM 6.1A Input Data and Excel QM Formulas for the sumco Pump Company Example

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194  CHAPTER 6 • InvEnTORy COnTROl MODEls

quantity of units ordered and assuming a unit cost of C, we can determine the average dollar value of inventory:

Average dollar level = 1CQ2

2 (6-6)

This formula is analogous to Equation 6-1. Many businesses and industries often express the inventory carrying cost as an annual

percentage of the unit cost or price. When this is the case, a new variable is introduced. Let I be the annual inventory holding charge as a percent of unit price or cost. Then the cost of storing one unit of inventory for the year, Ch, is given by Ch = IC, where C is the unit price or cost of an inventory item. Q* can be expressed, in this case, as

Q* = B 2DCoIC (6-7) Sensitivity Analysis with the EOQ Model The EOQ model assumes that all input values are fixed and known with certainty. However, since these values are often estimated or may change over time, it is important to understand how the order quantity might change if different input values are used. Determining the effects of these changes is called sensitivity analysis.

The EOQ formula is given as follows:

EOQ = B 2DCoCh Because of the square root in the formula, any changes in the inputs 1D, Co, Ch2 will result in relatively minor changes in the optimal order quantity. For example, if Co were to increase by a factor of 4, the EOQ would increase by only a factor of 2. Consider the Sumco example just presented. The EOQ for this company is as follows:

EOQ = B 211,000211020.50 = 200 If we increased Co from $10 to $40,

EOQ = B 211,000214020.50 = 400 In general, the EOQ changes by the square root of a change in any of the inputs.

6.4 Reorder Point: Determining When to Order

Now that we have decided how much to order, we look at the second inventory question: when to order. The time between placing and receiving an order, called the lead time or delivery time, is often a few days or even a few weeks. Inventory must be available to meet the demand during this time, and this inventory can either be on hand now or on order but not yet received. The total of these is called the inventory position. Thus, the when to order decision is usually expressed in terms of a reorder point (ROP), the inventory position at which an order should be placed. The ROP is given as

ROP = 1Demand per day2 * 1Lead time for a new order in days2 = d * L (6-8)

Figure 6.4 has two graphs showing the ROP. One of these has a relatively small lead time, while the other has a relatively large lead time. When the inventory position reaches the ROP, a new order should be placed. While waiting for that order to arrive, the demand will be met either with

I is the annual carrying cost as a percentage of the cost per unit.

The reorder point (ROP) determines when to order inventory. It is found by multiplying the daily demand times the lead time in days.

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6.4 REORDER POInT: DETERMInIng wHEn TO ORDER  195

inventory currently on hand or with inventory that already has been ordered but will arrive when the on-hand inventory falls to zero. Let’s look at an example.

PROCOMP’S COMPUTER CHIP EXAMPLE Procomp’s demand for computer chips is 8,000 per year. The firm has a daily demand of 40 units, and the order quantity is 400 units. Delivery of an order takes three working days. The reorder point for chips is calculated as follows:

ROP = d * L = 40 units per day * 3 days = 120 units

Hence, when the inventory stock of chips drops to 120, an order should be placed. The order will arrive three days later, just as the firm’s stock is depleted to 0. Since the order quantity is 400 units, the ROP is simply the on-hand inventory. This is the situation in the first graph in Figure 6.4.

Suppose the lead time for Procomp Computer Chips was 12 days instead of 3 days. The reorder point would be

ROP = 40 units per day * 12 days = 480 units

Since the maximum on-hand inventory level is the order quantity of 400, an inventory position of 480 would be

Inventory position = 1Inventory on hand2 + 1Inventory on order2 480 = 80 + 400

FIGURE 6.4 Reorder Point graphs

0

Inventory Level

ROP

Lead time = L ROP < Q

Q

Time

0

Inventory Level

On hand

Lead time = L ROP > Q

Q

Time

On Order

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196  CHAPTER 6 • InvEnTORy COnTROl MODEls

Thus, a new order would have to be placed when the on-hand inventory fell to 80 while there was one other order in-transit. The second graph in Figure 6.4 illustrates this type of situation.

6.5 EOQ Without the Instantaneous Receipt Assumption

When a firm receives its inventory over a period of time, a new model is needed that does not require the instantaneous inventory receipt assumption. This new model is applicable when inventory continuously flows or builds up over a period of time after an order has been placed or when units are produced and sold simultaneously. Under these circumstances, the daily demand rate must be taken into account. Figure 6.5 shows inventory levels as a function of time. Because this model is especially suited to the production environment, it is commonly called the produc- tion run model.

In this model, instead of having an ordering cost, there will be a setup cost. This is the cost of setting up the production facility to manufacture the desired product. It nor- mally includes the salaries and wages of employees who are responsible for setting up the equipment, engineering and design costs of making the setup, paperwork, supplies, utilities, and so on. The carrying cost per unit is composed of the same factors as in the traditional EOQ model, although the annual carrying cost equation changes due to a change in average inventory.

The optimal production quantity can be derived by setting the setup cost equal to the holding or carrying cost and solving for the order quantity. Let’s start by developing the ex- pression for carrying cost. You should note, however, that making setup cost equal to carrying cost does not always guarantee optimal solutions for models more complex than the produc- tion run model.

Annual Carrying Cost for Production Run Model As with the EOQ model, the carrying cost of the production run model is based on the average inventory, and the average inventory is one-half the maximum inventory level. However, since the replenishment of inventory occurs over a period of time and demand continues during this time, the maximum inventory will be less than the order quantity Q. We can develop the annual carrying, or holding, cost expression using the following variables:

Q = number of units per order, or production run Cs = setup cost Ch = holding or carrying cost per unit per year p = daily production rate d = daily demand rate t = length of production run in days

The production run model eliminates the instantaneous receipt assumption.

Solving the production run model involves setting the setup cost equal to the holding cost and solving for Q.

Inventory Level

Time

Maximum Inventory

t

Part of Inventory Cycle During Which Production Is Taking Place

There Is No Production During This Part of the Inventory Cycle

FIGURE 6.5 Inventory Control and the Production Process

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6.5 EOQ wITHOUT THE InsTAnTAnEOUs RECEIPT AssUMPTIOn  197

The maximum inventory level is as follows:

1Total produced during the production run2 - 1Total used during production run2 = 1Daily production rate21Number of days of production2 - 1Daily demand21Number of days of production2 = 1pt2 - 1dt2

Since

Total produced = Q = pt,

we know that

t = Q p

Maximum inventory level = pt - dt = p Q p

- d Q p

= Q a1 - d p b

Since the average inventory is one-half of the maximum, we have

Average inventory = Q

2 a1 - d

p b (6-9)

and

Annual holding cost = Q

2 a1 - d

p bCh (6-10)

Annual Setup Cost or Annual Ordering Cost When a product is produced over time, setup cost replaces ordering cost. Both of these are inde- pendent of the size of the order and the size of the production run. This cost is simply the num- ber of orders (or production runs) times the ordering cost (setup cost). Thus,

Annual setup cost = D

Q Cs (6-11)

and

Annual ordering cost = D

Q Co (6-12)

Determining the Optimal Production Quantity When the assumptions of the production run model are met, costs are minimized if the setup cost equals the holding cost. We can find the optimal quantity by setting these costs to be equal and solving for Q. Thus,

Annual holding cost = Annual setup cost

Q

2 a1 - d

p bCh = DQ Cs

Solving this for Q, we get the optimal production quantity 1Q*2:

Q* = H 2DCsCh a1 - dp b (6-13) It should be noted that if the situation does not involve production but rather involves the receipt of inventory gradually over a period of time, this same model is appropriate, but Co replaces Cs in the formula.

The maximum inventory level in the production model is less than Q.

Here is the formula for the optimal production quantity. Notice the similarity to the basic EOQ model.

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198  CHAPTER 6 • InvEnTORy COnTROl MODEls

Production Run Model

Annual holding cost = Q

2 a1 - d

p bCh

Annual setup cost = D

Q Cs

Optimal production quantity Q* = H 2DCsCh a1 - dp b Brown Manufacturing Example Brown Manufacturing produces commercial refrigeration units in batches. The firm’s estimated demand for the year is 10,000 units. It costs about $100 to set up the manufacturing process, and the carrying cost is about 50 cents per unit per year. When the production process has been set up, 80 refrigeration units can be manufactured daily. The demand during the production period has traditionally been 60 units each day. Brown operates its refrigeration unit production area 167 days per year. How many refrigeration units should Brown Manufacturing produce in each batch? How long should the production part of the cycle shown in Figure 6.5 last? Here is the solution:

Annual demand = D = 10,000 units

Setup cost = Cs = $100

Carrying cost = Ch = $0.50 per unit per year

Daily production rate = p = 80 units daily

Daily demand rate = d = 60 units daily

Q* = R 2DCsCh a1 - dp b = R2 * 10,000 * 1000.5 a1 - 6080 b = B 2,000,0000.511>42 = 116,000,000 = 4,000 units

If Q* = 4,000 units and we know that 80 units can be produced daily, the length of each produc- tion cycle will be Q>p = 4,000>80 = 50 days. Thus, when Brown decides to produce refrig- eration units, the equipment will be set up to manufacture the units for a 50-day time span. The number of production runs per year will be D>Q = 10,000>4,000 = 2.5. This means that the average number of production runs per year is 2.5. There will be three production runs in one year with some inventory carried to the next year, so only two production runs are needed in the second year.

USING EXCEL QM FOR PRODUCTION RUN MODELS The Brown Manufacturing production run model can also be solved using Excel QM. Program 6.2A contains the input data and the Excel formulas for this problem. Program 6.2B provides the solution results, including the optimal production quantity, maximum inventory level, average inventory level, and number of setups.

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Enter the demand rate, setup cost, and holding cost. Notice that the holding cost is a fixed dollar amount rather than a percentage of the unit price.

Enter the daily production rate and the daily demand rate.

Calculates the optimal production quantity.

Calculates the maximum inventory level. Calculates the average number of setups.

Calculates the annual holding cost and setup cost.

PROGRAM 6.2A Excel QM Formulas and Input Data for the Brown Manufacturing Problem

PROGRAM 6.2B Excel QM solutions for the Brown Manufacturing Problem

U.s. Defense logistics Agency saves on Inventory Costs Through Risk Management system

The U.S. Defense Logistics Agency (DLA; www.dla.mil) had cost overruns in its inventory supply chain. Some items had stable but infrequent demand, while others had highly variable demand pat- terns. The Logistics Management Institute (LMI; www.lmi.org) de- veloped an inventory control solution named PNG that contained two modules: Peak Policy and Next Gen. The system utilized a risk management approach to control inventory.

Since the DLA implemented the two modules in January 2013, customer service metrics have improved, buyer workload has decreased, and inventory cost savings have totaled nearly $400 million per year.

Source: Based on Tovey C. Bachman, Pamela J. Williams, Kristen M. Cheman, Jeffrey Curtis, and Robert Carroll, “PNG: Effective Inventory Control for Items with Highly Variable Demand,” Interfaces 46, 1 (January–February 2016): 18–32, © Trevor S. Hale.

IN ACTION

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6.6 Quantity Discount Models

In developing the EOQ model, we assumed that quantity discounts were not available. However, many companies do offer quantity discounts. If such a discount is possible, and all of the other EOQ assumptions are met, it is possible to find the quantity that minimizes the total inventory cost by using the EOQ model and making some adjustments.

When quantity discounts are available, the purchase cost or material cost becomes a rel- evant cost, as it changes based on the order quantity. The total relevant costs are as follows:

Total cost = Material cost + Ordering cost + Carrying cost

Total cost = DC + D

Q Co +

Q

2 Ch (6-14)

where

D = annual demand in units Co = ordering cost of each order C = cost per unit

Ch = holding or carrying cost per unit per year

Since the holding cost per unit per year is based on the cost of the item, it is convenient to ex- press this as

Ch = IC

where

I = holding cost as a percentage of the unit cost 1C2 For a specific purchase cost (C), given the assumptions we have made, ordering the EOQ will minimize total inventory costs. However, in the discount situation, this quantity may not be large enough to qualify for the discount, so we must also consider ordering this minimum quantity for the discount. A typical quantity discount schedule is shown in Table 6.3.

As can be seen in the table, the normal cost for the item is $5. When 1,000 to 1,999 units are ordered at one time, the cost per unit drops to $4.80, and when the quantity ordered at one time is 2,000 or more units, the cost is $4.75 per unit. As always, management must decide when and how much to order. But with quantity discounts, how does the manager make these decisions?

As with other inventory models discussed so far, the overall objective will be to mini- mize the total cost. Because the unit cost for the third discount in Table 6.3 is lowest, you might be tempted to order 2,000 units or more to take advantage of the lower material cost. Placing an order for that quantity with the greatest discount cost, however, might not mini- mize the total inventory cost. As the discount quantity goes up, the material cost goes down, but the carrying cost increases because the orders are large. Thus, the major trade-off when considering quantity discounts is between the reduced material cost and the increased car- rying cost.

Figure 6.6 provides a graphical representation of the total cost for this situation. Notice the cost curve drops considerably when the order quantity reaches the minimum for each dis- count. With the specific costs in this example, we see that the EOQ for the second price category

The overall objective of the quantity discount model is to minimize total inventory costs, which now include actual material costs.

DISCOUNT NUMBER

DISCOUNT QUANTITY

DISCOUNT (%)

DISCOUNT COST ($)

1 0 to 999 0 5.00

2 1,000 to 1,999 4 4.80

3 2,000 and over 5 4.75

TABLE 6.3 Quantity Discount schedule

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11,000 … Q … 1,9992 is less than 1,000 units. Although the total cost for this EOQ is less than the total cost for the EOQ with the cost in category 1, the EOQ is not large enough to obtain this discount. Therefore, the lowest possible total cost for this discount price occurs at the minimum quantity required to obtain the discount 1Q = 1,0002. The process for determining the mini- mum cost quantity in this situation is summarized in the following box.

Quantity Discount Model

1. For each discount price (C), compute EOQ = B 2DCoIC . 2. If EOQ 6 Minimum for discount, adjust the quantity to Q = Minimum for discount.

3. For each EOQ or adjusted Q, compute Total cost = DC + D

Q Co +

Q

2 Ch.

4. Choose the lowest-cost quantity.

Total Cost

$ TC Curve for Discount 1

TC Curve for Discount 2

EOQ for Discount 2

TC Curve for Discount 3

0 1,000 2,000

Order Quantity

FIGURE 6.6 Total Cost (TC) Curve for the Quantity Discount Model

Intel Improves Inventory Operations

Faced with skyrocketing distribution costs, Intel turned to quantitative analysis to design, construct, and implement a multi–echelon inventory optimization (MEIO) system. The system spanned all of Intel’s global supply chain and included suppliers, third-party logistics providers, dealers, and retailers, among oth- ers. Because of the inherent complexities, designing and con- structing the system took several years by itself.

However, as a result, Intel was able to optimize inventory levels worldwide. After one year of operation, the MEIO system

reduced global inventory levels by approximately 11 percent without sacrificing delivery schedules, service levels improved by 8 per- centage points, and an order of magnitude reduction in the number of costly expedites was realized.

Source: Based on B. Wieland, P. Mastrantonio, S. Willems, and K.G. Kempf, “Optimizing Inventory Levels Within Intel ’s Channel Supply Demand Operations,” Interfaces 42, 6 (November–December 2012): 517–527, © Trevor S. Hale.

IN ACTION

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Brass Department Store Example Let’s see how this procedure can be applied by showing an example. Brass Department Store stocks toy race cars. Recently, the store was given a quantity discount schedule for the cars; this quantity discount schedule is shown in Table 6.3. Thus, the normal cost for the toy race cars is $5. For orders between 1,000 and 1,999 units, the unit cost is $4.80, and for orders of 2,000 or more units, the unit cost is $4.75. Furthermore, the ordering cost is $49 per order, the annual demand is 5,000 race cars, and the inventory carrying charge as a percentage of cost, I, is 20% or 0.2. What order quantity will minimize the total inventory cost?

The first step is to compute EOQ for every discount in Table 6.3. This is done as follows:

EOQ1 = B 12215,0002149210.2215.002 = 700 cars per order EOQ2 = B 12215,0002149210.2214.802 = 714 cars per order EOQ3 = B 12215,0002149210.2214.752 = 718 cars per order

The second step is to adjust those quantities that are below the allowable discount range. Since EOQ1 is between 0 and 999, it does not have to be adjusted. EOQ2 is below the allowable range of 1,000 to 1,999, and therefore it must be adjusted to 1,000 units. The same is true for EOQ3; it must be adjusted to 2,000 units. After this step, the following order quantities must be tested in the total cost equation:

Q1 = 700 Q2 = 1,000 Q3 = 2,000

The third step is to use Equation 6-14 to compute a total cost for each of the order quantities. This is accomplished with the aid of Table 6.4.

The fourth step is to select the order quantity with the lowest total cost. Looking at Table 6.4, you can see that an order quantity of 1,000 toy race cars minimizes the total cost. It should be recognized, however, that the total cost for ordering 2,000 cars is only slightly greater than the total cost for ordering 1,000 cars. Thus, if the third discount cost is lowered to $4.65, for example, this order quantity might be the one that minimizes the total inven- tory cost.

USING EXCEL QM FOR QUANTITY DISCOUNT PROBLEMS As seen in the previous analysis, the quantity discount model is more complex than the inventory models discussed so far in this chapter. For- tunately, we can use the computer to simplify the calculations. Program 6.3A shows the Excel formulas and input data needed for Excel QM for the Brass Department Store problem. Program 6.3B provides the solution to this problem, including adjusted order quantity and total cost for each price break.

EOQ values are adjusted.

EOQ values are computed.

The total cost is computed.

Q* is selected.

DISCOUNT NUMBER

UNIT PRICE (C)

ORDER

QUANTITY (Q)

ANNUAL

MATERIAL COST 1$ 2 = DC

ANNUAL ORDERING

COST 1$ 2 = D Q

Co

ANNUAL CARRYING

COST 1$ 2 = Q 2

Ch

TOTAL ($)

1 $5.00 700 25,000 350.00 350.00 25,700.00

2 4.80 1,000 24,000 245.00 480.00 24,725.00

3 4.75 2,000 23,750 122.50 950.00 24,822.50

TABLE 6.4 Total Cost Computations for Brass Department store

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6.7 Use of Safety Stock

When the EOQ assumptions are met, it is possible to schedule orders to arrive so that stock- outs are completely avoided. However, if the demand or the lead time is uncertain, the exact demand during the lead time (which is the ROP in the EOQ situation) will not be known with certainty. Therefore, to prevent stockouts, it is necessary to carry additional inventory called safety stock.

When demand is unusually high during the lead time, you dip into the safety stock in- stead of encountering a stockout. Thus, the main purpose of safety stock is to avoid stockouts when the demand is higher than expected. Its use is shown in Figure 6.7. Note that although stockouts can often be avoided by using safety stock, there is still a chance that they may occur. The demand may be so high that all the safety stock is used up, and thus there is still a stockout.

One of the best ways to implement a safety stock policy is to adjust the reorder point. In the EOQ situation where the demand and lead time are constant, the reorder point is simply

Enter the quantity discount schedule with the quantity and unit price for each price break.

Computes the order quantity for each price break and adjusts it upward if necessary.

Enter the demand rate, setup cost, and holding cost.

Computes the holding and setup costs.

Computes the total cost of each price break.

Determines the optimal order quantity, Q*.

PROGRAM 6.3A Excel QM Formulas and Input Data for the Brass Department store Quantity Discount Problem

PROGRAM 6.3B Excel QM solutions for the Brass Department store Quantity Discount Problem

Safety stock is extra stock kept on hand. It helps in avoiding stockouts.

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the amount of inventory that would be used during the lead time (i.e., the daily demand times the lead time in days). This is assumed to be known with certainty, so there is no need to place an order when the inventory position is more than this. However, when the daily demand or the lead time fluctuates and is uncertain, the exact amount of inventory that will be used during the lead time is uncertain. The average inventory usage during the lead time should be computed, and some safety stock should be added to this to avoid stockouts. The reorder point becomes

ROP = 1Average demand during lead time2 + 1Safety stock2 ROP = 1Average demand during lead time2 + SS (6-15) where

SS = safety stock

How to determine the correct amount of safety stock is the only remaining question. Two important factors in this decision are the stockout cost and the holding cost. The stockout cost usually involves lost sales and lost goodwill, which result in loss of future sales. If the holding cost is low but the stockout cost is high, a large amount of safety stock should be carried to avoid stockouts because it costs little to carry this and stockouts are expensive. On the other hand, if the stockout cost is low but the holding cost is high, a lower amount of safety stock would be preferred because having a stockout would cost very little but having too much safety stock would result in a much higher annual holding cost.

Inventory on Hand

Time

Inventory on Hand

Time

Safety Stock, SS

0 Units

0 Units

Stockout Is Avoided

Stockout

FIGURE 6.7 Use of safety stock

Safety stock is included in the ROP.

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How is the optimum stock level determined? If demand f luctuates, the lead time is constant, and both the stockout cost per unit and the holding cost per unit are known, the use of a payoff/cost table might be considered. With only a small number of possible de- mand values during the lead time, a cost table could be constructed in which the different possible demand levels would be the states of nature and the different amounts of safety stock would be the alternatives. Using the techniques discussed in Chapter 3, the expected cost could be calculated for each safety stock level, and the minimum cost solution could be found.

However, a more general approach is to determine what service level is desired and then find the safety stock level that would accomplish this. A prudent manager will look at the hold- ing cost and the stockout cost to help determine an appropriate service level. A service level indicates what percentage of the time customer demand is met. In other words, the service level is the percentage of time that stockouts are avoided. Thus,

Service level = 1 - Probability of a stockout

or

Probability of a stockout = 1 - Service level

Once the desired service level is established, the amount of safety stock to carry can be found using the probability distribution of demand during the lead time.

SAFETY STOCK WITH THE NORMAL DISTRIBUTION Equation 6-15 provides the general formula for de- termining the reorder point. When demand during the lead time is normally distributed, the reor- der point becomes

ROP = 1Average demand during lead time2 + Zsd LT (6-16) where

Z = number of standard deviations for a given service level sd LT = standard deviation of demand during the lead time

Thus, the amount of safety stock is simply Zsd LT. The following example looks at how to deter- mine the appropriate safety stock level when demand during the lead time is normally distrib- uted and the mean and standard deviation are known.

HINSDALE COMPANY EXAMPLE The Hinsdale Company carries a variety of electronic inventory items, and these are typically identified by SKU. One particular item, SKU A3378, has a de- mand that is normally distributed during the lead time, with a mean of 350 units and a standard deviation of 10. Hinsdale wants to follow a policy that results in stockouts occurring only 5% of the time on any order. How much safety stock should be maintained, and what is the reorder point? Figure 6.8 helps visualize this example.

From the normal distribution table (Appendix A), we have Z = 1.65:

ROP = 1Average demand during lead time2 + ZsdLT = 350 + 1.651102 = 350 + 16.5 = 366.5 units 1or about 367 units2

So the reorder point is 366.5, and the safety stock is 16.5 units.

CALCULATING LEAD TIME DEMAND AND STANDARD DEVIATION If the mean and standard deviation of demand during the lead time are not known, they must be calculated from historical demand and lead time data. Once these are found, Equation 6-16 can be used to find the safety stock and reorder point. Throughout this section, we assume that lead time is in days, although the same procedure can be applied to weeks, months, or any other time period. We will also assume that if demand fluctuates, the distribution of demand each day is identical to and independent of de- mand on other days. If both daily demand and lead time fluctuate, they are also assumed to be independent.

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There are three situations to consider. In each of the following ROP formulas, the average demand during the lead time is the first term and the safety stock 1ZsdLT2 is the second term.

1. Demand is variable but lead time is constant:

ROP = dL + Z1sd2L2 (6-17) where

d = average daily demand

sd = standard deviation of daily demand

L = lead time in days

2. Demand is constant, but lead time is variable:

ROP = dL + Z1dsL2 (6-18) where

L = average lead time

sL = standard deviation of lead time

d = daily demand

3. Both demand and lead time are variable:

ROP = d L + Z12Ls2d + d2s2L2 (6-19) Notice that the third situation is the most general case and the others can be derived from that. If either demand or lead time is constant, the standard deviation and variance for that would be 0, and the average would just equal the constant amount. Thus, the formula for ROP in situation 3 can be simplified to the ROP formula given for that situation.

HINSDALE COMPANY EXAMPLE, CONTINUED Hinsdale has decided to determine the safety stock and ROP for three other items: SKU F5402, SKU B7319, and SKU F9004.

For SKU F5402, the daily demand is normally distributed, with a mean of 15 units and a standard deviation of 3. Lead time is exactly 4 days. Hinsdale wants to maintain a 97% service level. What is the reorder point, and how much safety stock should be carried?

SS

5% Area of Normal Curve

m = 350 X = ?

m = Mean demand = 350

s = Standard deviation = 10

= Mean demand + Safety stockX

SS = Safety stock = X – m 5 Zs

X – m Z =

s

FIGURE 6.8 safety stock and the normal Distribution

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From Appendix A, for a 97% service level Z = 1.88. Since demand is variable but lead time is constant, we find

ROP = dL + Z1sd1L2 = 15142 + 1.8813142 = 15142 + 1.88162 = 60 + 11.28 = 71.28

So the average demand during the lead time is 60, and the safety stock is 11.28 units. For SKU B7319, the daily demand is constant at 25 units per day, and the lead time is nor-

mally distributed, with a mean of 6 days and a standard deviation of 3. Hinsdale wants to main- tain a 98% service level on this particular product. What is the reorder point?

From Appendix A, for a 98% service level Z = 2.05. Since demand is constant but lead time is variable, we find

ROP = dL + Z1dsL2 = 25162 + 2.051252132 = 150 + 2.051752 = 150 + 153.75 = 303.75

So the average demand during the lead time is 150, and the safety stock is 154.03 units. For SKU F9004, the daily demand is normally distributed, with a mean of 20 units and a

standard deviation of 4, and the lead time is normally distributed, with a mean of 5 days and a standard deviation of 2. Hinsdale wants to maintain a 94% service level on this particular prod- uct. What is the reorder point?

From Appendix A, for a 94% service level Z = 1.55. Since both demand and lead time are variable, we find

ROP = d L + Z12Ls2d + d2s2L2 = 1202152 + 1.551251422 + 1202212222 = 100 + 1.5511680 = 100 + 1.55140.992 = 100 + 63.53 = 163.53

So the average demand during the lead time is 100, and the safety stock is 63.53 units. As the service level increases, the safety stock increases at an increasing rate. Table 6.5

illustrates how the safety stock level would change in the Hinsdale Company (SKU A3378) example for changes in the service level. As the amount of safety stock increases, the annual holding cost increases as well.

CALCULATING ANNUAL HOLDING COST WITH SAFETY STOCK When the EOQ assumptions of constant demand and constant lead time are met, the average inventory is Q>2, and the annual hold- ing cost is 1Q>22Ch. When safety stock is carried because demand fluctuates, the holding cost

A safety stock level is determined for each service level.

SERVICE LEVEL (%)

Z VALUE FROM NORMAL CURVE TABLE

SAFETY STOCK (UNITS)

90 1.28 12.8

91 1.34 13.4

92 1.41 14.1

93 1.48 14.8

94 1.55 15.5

95 1.65 16.5

96 1.75 17.5

97 1.88 18.8

98 2.05 20.5

99 2.33 23.3

99.99 3.72 37.2

TABLE 6.5 safety stock for skU A3378 at Different service levels

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208  CHAPTER 6 • InvEnTORy COnTROl MODEls

for this safety stock is added to the holding cost of the regular inventory to get the total annual holding cost. If demand during the lead time is normally distributed and safety stock is used, the average inventory on the order quantity (Q) is still Q>2, but the average amount of safety stock carried is simply the amount of the safety stock (SS) and not one-half this amount. Since demand during the lead time is normally distributed, there would be times when inventory usage during the lead time exceeded the expected amount and some safety stock would be used. But it is just as likely that the inventory usage during the lead time would be less than the expected amount and the order would arrive while some regular inventory remained in addition to all the safety stock. Thus, on the average, the company would always have this full amount of safety stock in inventory, and a holding cost would apply to all of this. From this, we have

Total annual holding cost = Holding cost of regular inventory + Holding cost of safety stock

THC = Q

2 Ch + 1SS2Ch (6-20)

where

THC = total annual holding cost Q = order quantity

Ch = holding cost per unit per year SS = safety stock

In the Hinsdale example for SKU A3378, let’s assume that the holding cost is $2 per unit per year. The amount of safety stock needed to achieve various service levels is shown in Table 6.5. The holding cost for the safety stock would be these amounts times $2 per unit. As illustrated in Figure 6.9, this holding cost would increase extremely rapidly once the service level reached 98%.

USING EXCEL QM FOR SAFETY STOCK PROBLEMS To use Excel QM to determine the safety stock and reorder point, select Excel QM from the Add-Ins tab, and select Inventory—Reorder Point/Safety Stock (normal distribution). Enter a title when the input window appears, and click OK. Program 6.4A shows the input screen and formulas for the first three Hinsdale Company examples. Pro- gram 6.4B shows the solution.

FIGURE 6.9 service level versus Annual Carrying Costs

Carrying cost increases at an increasing rate as the service level increases.

($) 80

70

60

50

40

30

20

90 91 92 93 94 95 96

Service Level (%)

In ve

nt or

y C

ar ry

in g

C os

ts ($

)

97 98 99 99.99 (%)

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6.8 sInglE-PERIOD InvEnTORy MODEls  209

6.8 Single-Period Inventory Models

So far, we have considered inventory decisions in which demand continues in the future and future orders will be placed for the same product. There are some products for which a decision to meet the demand for a single time period is made, and items that do not sell during this time period are of no value or have a greatly reduced value in the future. For example, a daily news- paper is worthless after the next paper is available. Other examples include weekly magazines, programs printed for athletic events, certain prepared foods that have a short life, and some sea- sonal clothes that have greatly reduced value at the end of the season. This type of problem is often called the news vendor problem or a single-period inventory model.

For example, a large restaurant might be able to stock from 20 to 100 cartons of doughnuts to meet a demand that ranges from 20 to 100 cartons per day. While this could be modeled us- ing a payoff table (see Chapter 3), we would have to analyze 101 possible alternatives and states of nature, which would be quite tedious. A simpler approach for this type of decision is to use marginal, or incremental, analysis.

The standard deviation of demand during the lead time is calculated here.

The standard deviation of demand during the lead time is calculated here.

The average demand and standard deviation during the lead time are entered here, if available.

If daily demand is normally distributed but lead time is constant, the data are entered here.

If daily demand and/or lead time is normally distributed, the data are entered here. If one or the other is constant, enter zero for the standard deviation of that quantity.

Solution to the first Hinsdale example where standard deviation of demand during lead time was given.

Solution to second Hinsdale example where daily demand was normally distributed.

Solution to third Hinsdale example where daily demand was constant but lead time was normally distributed.

PROGRAM 6.4A Excel QM Formulas and Input Data for the Hinsdale safety stock Problem

PROGRAM 6.4B Excel QM solutions for the Hinsdale safety stock Problem

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A decision-making approach using marginal profit and marginal loss is called marginal analysis. Marginal profit (MP) is the additional profit achieved if one additional unit is stocked and sold. Marginal loss (ML) is the loss that occurs when an additional unit is stocked but can- not be sold.

When there are a manageable number of alternatives and states of nature and we know the probabilities for each state of nature, marginal analysis with discrete distributions can be used. When there are a very large number of possible alternatives and states of nature and the prob- ability distribution of the states of nature can be described with a normal distribution, marginal analysis with the normal distribution is appropriate.

Marginal Analysis with Discrete Distributions Finding the inventory level with the lowest cost is not difficult when we follow the marginal analysis procedure. This approach says that we would stock an additional unit only if the ex- pected marginal profit for that unit equals or exceeds the expected marginal loss. This relation- ship is expressed symbolically as follows:

P = probability that demand will be greater than or equal to a given supply (or the probability of selling at least one additional unit)

1 - P = probability that demand will be less than a given supply (or the probability that one additional unit will not sell)

The expected marginal profit is found by multiplying the probability that a given unit will be sold by the marginal profit, P1MP2. Similarly, the expected marginal loss is the probability of not selling the unit multiplied by the marginal loss, or 11 - P21ML2.

The optimal decision rule is to stock the additional unit if

P1MP2 Ú 11 - P2ML With some basic mathematical manipulations, we can determine the level of P for which

this relationship holds:

P1MP2 Ú ML - P1ML2 P1MP2 + P1ML2 Ú ML P1MP + ML2 Ú ML

or

P Ú ML

ML + MP (6-21)

In other words, as long as the probability of selling one more unit (P) is greater than or equal to ML>1MP + ML2, we would stock the additional unit.

Steps of Marginal Analysis with Discrete Distributions

1. Determine the value of ML

ML + MP for the problem.

2. Construct a probability table, and add a cumulative probability column. 3. Keep ordering inventory as long as the probability (P) of selling at least one additional unit

is greater than ML

ML + MP .

Café du Donut Example Café du Donut is a popular New Orleans dining spot on the edge of the French Quarter. Its spe- cialty is coffee and doughnuts; it buys the doughnuts fresh daily from a large industrial bakery. The café pays $4 for each carton (containing two dozen doughnuts) delivered each morning. Any cartons not sold at the end of the day are thrown away, for they would not be fresh enough

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to meet the café’s standards. If a carton of doughnuts is sold, the total revenue is $6. Hence, the marginal profit per carton of doughnuts is

MP = Marginal profit = $6 - $4 = $2

The marginal loss is ML = $4, since the doughnuts cannot be returned or salvaged at day’s end. From past sales, the café’s manager estimates that the daily demand will follow the prob-

ability distribution shown in Table 6.6. The manager then follows the three steps to find the opti- mal number of cartons of doughnuts to order each day.

Step 1. Determine the value of ML

ML + MP for the decision rule

P Ú ML

ML + MP =

$4

$4 + $2 =

4

6 = 0.67

P Ú 0.67

So the inventory stocking decision rule is to stock an additional unit if P Ú 0.67.

Step 2. Add a new column to the table to reflect the probability that doughnut sales will be at each level or greater. This is shown in the right-hand column of Table 6.7. For example, the probability that demand will be 4 cartons or greater is 1.00 1= 0.05 + 0.15 + 0.15 + 0.20 + 0.25 + 0.10 + 0.102. Similarly, the probability that sales will be 8 cartons or greater is 0.45 1= 0.25 + 0.10 + 0.102—that is, the sum of the probabilities for sales of 8, 9, and 10 cartons.

Step 3. Keep ordering additional cartons as long as the probability of selling at least one ad- ditional carton is greater than P, which is the indifference probability. If Café du Donut orders 6 cartons, marginal profits will still be greater than marginal loss, since

P at 6 cartons = 0.80 7 0.67

DAILY SALES (CARTONS OF DOUGHNUTS)

PROBABILITY (P) THAT DEMAND WILL BE AT THIS LEVEL

4 0.05

5 0.15

6 0.15

7 0.20

8 0.25

9 0.10

10 0.10

Total 1.00

TABLE 6.6 Café du Donut’s Probability Distribution

DAILY SALES (CARTONS OF DOUGHNUTS)

PROBABILITY (P) THAT DEMAND WILL BE AT

THIS LEVEL

PROBABILITY (P) THAT DEMAND WILL BE AT THIS

LEVEL OR GREATER

4 0.05 1.00 Ú 0.66 5 0.15 0.95 Ú 0.66 6 0.15 0.80 Ú 0.66 7 0.20 0.65

8 0.25 0.45

9 0.10 0.20

10 0.10 0.10

Total 1.00

TABLE 6.7 Marginal Analysis for Café du Donut

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212  CHAPTER 6 • InvEnTORy COnTROl MODEls

Marginal Analysis with the Normal Distribution When product demand or sales follow a normal distribution, which is a common business situ- ation, marginal analysis with the normal distribution can be applied. First, we need to find four values:

1. The average or mean sales for the product, m 2. The standard deviation of sales, s 3. The marginal profit for the product, MP 4. The marginal loss for the product, ML

Once these quantities are known, the process of finding the best stocking policy is somewhat similar to marginal analysis with discrete distributions. We let X* = optimal stocking level.

Steps of Marginal Analysis with the Normal Distribution

1. Determine the value of ML

ML + MP for the problem.

2. Locate P on the normal distribution (Appendix A), and find the associated Z value. 3. Find X* using the relationship

Z = X* - m

s

to solve for the resulting stocking policy:

X* = m + Zs

Newspaper Example Demand for copies of the Chicago Tribune newspaper at Joe’s Newsstand is normally distrib- uted and has averaged 60 papers per day, with a standard deviation of 10 papers. With a marginal loss of 20 cents and a marginal profit of 30 cents, what daily stocking policy should Joe follow?

Step 1. Joe should stock the Chicago Tribune as long as the probability of selling the last unit is at least ML>1ML + MP2:

ML

ML + MP =

20 cents

20 cents + 30 cents =

20

50 = 0.40

Let P = 0.40.

Step 2. Figure 6.10 shows the normal distribution. Since the normal table has cumulative areas under the curve between the left side and any point, we look for 0.60 1= 1.0 - 0.402 in order to get the corresponding Z value:

Z = 0.25 standard deviations from the mean

Step 3. In this problem, m = 60 and s = 10, so

0.25 = X* - 60

10

or

X* = 60 + 0.251102 = 62.5, or 62 newspapers.

Thus, Joe should order 62 copies of the Chicago Tribune daily, since the probability of selling 63 is slightly less than 0.40.

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6.8 sInglE-PERIOD InvEnTORy MODEls  213

When P is greater than 0.50, the same basic procedure is used, although caution should be used when looking up the Z value. Let’s say that Joe’s Newsstand also stocks the Chicago Sun- Times, which has a marginal loss of 40 cents and a marginal profit of 10 cents. The daily sales have averaged 100 copies of the Chicago Sun-Times, with a standard deviation of 10 papers. The optimal stocking policy is as follows:

ML

ML + MP =

40 cents

40 cents + 10 cents =

40

50 = 0.80

The normal curve is shown in Figure 6.11. Since the normal curve is symmetrical, we find Z for an area under the curve of 0.80 and multiply this number by -1 because all values below the mean are associated with a negative Z value:

Z = -0.84 standard deviations from the mean for an area of 0.80

With m = 100 and s = 10,

-0.84 = X* - 100

10

or

X* = 100 - 0.841102 = 91.6, or 91 newspapers.

So Joe should order 91 copies of the Chicago Sun-Times every day.

FIGURE 6.10 Joe’s stocking Decision for the Chicago Tribune Mean Daily Sales

Area Under the Curve Is 0.40

Area Under the Curve Is 1 – 0.40 = 0.60 (Z = 0.25)

X *

Optimal Stocking Policy (62 Newspapers)

m 5 60 X = Demand

Area Under the Curve Is 0.80 (Z = –0.84)

X *

Optimal Stocking Policy (91 Newspapers)

m 5 100 X = Demand

FIGURE 6.11 Joe’s stocking Decision for the Chicago Sun-Times

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214  CHAPTER 6 • InvEnTORy COnTROl MODEls

The optimal stocking policies in these two examples are intuitively consistent. When mar- ginal profit is greater than marginal loss, we would expect X* to be greater than the average demand, m, and when marginal profit is less than marginal loss, we would expect the optimal stocking policy, X*, to be less than m.

6.9 ABC Analysis

Earlier, we showed how to develop inventory policies using quantitative techniques. There are also some very practical considerations that should be incorporated into the implementation of inventory decisions, such as ABC analysis.

The purpose of ABC analysis is to divide all of a company’s inventory items into three groups (group A, group B, and group C) based on the overall inventory value of the items. A prudent manager should spend more time managing those items representing the greatest dol- lar inventory cost because this is where the greatest potential savings are. A brief description of each group follows, with general guidelines as to how to categorize items.

The inventory items in the A group account for a major portion of the inventory costs of the organization. As a result, their inventory levels must be monitored carefully. These items typically make up more than 70% of the company’s business in dollars but may consist of only 10% of all inventory items. In other words, a few inventory items are very costly to the com- pany. Thus, great care should be taken in forecasting the demand and developing good inventory management policies for this group of items (refer to Table 6.8). Since there are relatively few of these, the time involved would not be excessive.

The items in the B group are typically moderately priced items and represent much less in- vestment than the A items. It may not be appropriate to spend as much time developing optimal inventory policies for this group as with the A group, since these inventory costs are much lower. Typically, the group B items represent about 20% of the company’s business in dollars and about 20% of the items in inventory.

The items in the C group are the very low-cost items that represent very little in terms of the total dollars invested in inventory. These items may constitute only 10% of the company’s busi- ness in dollars, but they may consist of 70% of the items in inventory. From a cost–benefit per- spective, it would not be good to spend as much time managing these items as the A and B items.

For the group C items, the company should develop a very simple inventory policy, and this may include a relatively large safety stock. Since the items cost very little, the holding cost as- sociated with a large safety stock will also be very low. More care should be taken in determin- ing the safety stock with the higher priced group B items. For the very expensive group A items, the cost of carrying the inventory is so high that it is beneficial to carefully analyze the demand for these and set the safety stock at an appropriate level. Otherwise, the company may have an exceedingly high holding cost for the group A items.

6.10 Dependent Demand: The Case for Material Requirements Planning

In all the inventory models discussed earlier, we assume that the demand for one item is inde- pendent of the demand for other items. For example, the demand for refrigerators is usually in- dependent of the demand for toaster ovens. Many inventory problems, however, are interrelated; the demand for one item is dependent on the demand for another item. Consider a manufacturer of small power lawn mowers. The demand for lawn mower wheels and spark plugs is depen- dent on the demand for lawn mowers. Four wheels and one spark plug are needed for each fin- ished lawn mower. Usually, when the demand for different items is dependent, the relationship

INVENTORY GROUP

DOLLAR USAGE (%)

INVENTORY ITEMS (%)

ARE QUANTITATIVE CONTROL TECHNIQUES USED?

A 70 10 Yes

B 20 20 In some cases

C 10 70 No

TABLE 6.8 summary of ABC Analysis

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6.10 DEPEnDEnT DEMAnD: THE CAsE FOR MATERIAl REQUIREMEnTs PlAnnIng  215

between the items is known and constant. Thus, you should forecast the demand for the final products and compute the requirements for component parts.

As with the inventory models discussed previously, the major questions that must be an- swered are how much to order and when to order. But with dependent demand, inventory sched- uling and planning can be very complex indeed. In these situations, material requirements planning (MRP) can be employed effectively. Some of the benefits of MRP follow:

1. Increased customer service and satisfaction 2. Reduced inventory costs 3. Better inventory planning and scheduling 4. Higher total sales 5. Faster response to market changes and shifts 6. Reduced inventory levels without reduced customer service

Although most MRP systems are computerized, the analysis is straightforward and similar from one computerized system to the next. Here is the typical procedure.

Material Structure Tree We begin by developing a bill of materials (BOM). The BOM identifies the components, de- scribes them, and indicates the number required in the production of one unit of the final product. From the BOM, we develop a material structure tree. Let’s say that demand for product A is 50 units. Each unit of A requires 2 units of B and 3 units of C. Now, each unit of B requires 2 units of D and 3 units of E. Furthermore, each unit of C requires 1 unit of E and 2 units of F. Thus, the de- mand for B, C, D, E, and F is completely dependent on the demand for A. Given this information, a material structure tree can be developed for the related inventory items (see Figure 6.12).

The structure tree has three levels: 0, 1, and 2. Items above any level are called parents, and items below any level are called components. There are three parents: A, B, and C. Each parent item has at least one level below it. Items B, C, D, E, and F are components because each item has at least one level above it. In this structure tree, B and C are both parents and components.

Note that the number in the parentheses in Figure 6.12 indicates how many units of that par- ticular item are needed to make the item immediately above it. Thus, B(2) means that it takes 2 units of B for every unit of A, and F(2) means that it takes 2 units of F for every unit of C.

After the material structure tree has been developed, the number of units of each item re- quired to satisfy demand can be determined. This information can be displayed as follows:

Part B: 2 * number of A>s = 2 * 50 = 100 Part C: 3 * number of A>s = 3 * 50 = 150 Part D: 2 * number of B>s = 2 * 100 = 200 Part E: 3 * number of B>s + 1 * number of C>s = 3 * 100 + 1 * 150 = 450 Part F: 2 * number of C>s = 2 * 150 = 300

Parents and components are identified in the material structure tree.

The material structure tree shows how many units are needed at every level of production.

B(2) C(3)

Material Structure Tree for Item A

D(2) E(3) E(1) F(2)

Level 0

1

2

A

FIGURE 6.12 Material structure Tree for Item A

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216  CHAPTER 6 • InvEnTORy COnTROl MODEls

Thus, for 50 units of A we need 100 units of B, 150 units of C, 200 units of D, 450 units of E, and 300 units of F. Of course, the numbers in this table could have been determined directly from the material structure tree by multiplying the numbers along the branches times the de- mand for A, which is 50 units for this problem. For example, the number of units of D needed is simply 2 * 2 * 50 = 200 units.

Gross and Net Material Requirements Plans Once the materials structure tree has been developed, we construct a gross material require- ments plan. This is a time schedule that shows when an item must be ordered from suppliers when there is no inventory on hand or when the production of an item must be started in order to satisfy the demand for the finished product at a particular date. Let’s assume that all of the items are produced or manufactured by the same company. It takes one week to make A, two weeks to make B, one week to make C, one week to make D, two weeks to make E, and three weeks to make F. With this information, the gross material requirements plan can be constructed to reveal the production schedule needed to satisfy the demand of 50 units of A at a future date. (Refer to Figure 6.13.)

The interpretation of the material in Figure 6.13 is as follows: If you want 50 units of A at week 6, you must start the manufacturing process in week 5. Thus, in week 5 you need 100 units of B and 150 units of C. These two items take 2 weeks and 1 week to produce. (See the lead times.) Production of B should be started in week 3, and production of C should be started in week 4. (See the order release for these items.) Working backward, the same compu- tations can be made for all the other items. The material requirements plan graphically reveals when each item should be started and completed in order to have 50 units of A at week 6. Now, a net requirements plan can be developed given the on-hand inventory in Table 6.9; here is how it is done.

Using these data, we can develop a net material requirements plan that includes the gross requirements, on-hand inventory, net requirements, planned order receipts, and planned order release for each item. It is developed by beginning with A and working backward through the other items. Figure 6.14 shows a net material requirements plan for product A.

Use on-hand inventory to compute net requirements.

FIGURE 6.13 gross Material Requirements Plan for 50 Units of A Lead Time 5 1 Week

Lead Time 5 2 Weeks

Lead Time 5 1 Week

Lead Time 5 1 Week

Lead Time 5 2 Weeks

Lead Time 5 3 Weeks

50

50

100

100

150

150

200

200

300 150

300 150

300

300

A

B

C

D

E

F

Week

1 2 4 5 6

Required Date

Order Release

Every part A requires 2 part B’s in week 5

Required Date

Order Release

Required Date

Order Release

Required Date

Order Release

Required Date

Order Release

Required Date

Order Release

3

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6.10 DEPEnDEnT DEMAnD: THE CAsE FOR MATERIAl REQUIREMEnTs PlAnnIng  217

The net requirements plan is constructed like the gross requirements plan. Starting with item A, we work backward determining net requirements for all items. These computations are done by referring constantly to the structure tree and lead times. The gross requirements for A are 50 units in week 6. Ten items are on hand, and thus the net requirements and planned-order receipt are both 40 items in week 6. Because of the one-week lead time, the planned order re- lease is 40 items in week 5. (See the arrow connecting the order receipt and order release.) Look down column 5 and refer to the structure tree in Figure 6.13. Eighty 12 * 402 items of B and 120 = 3 * 40 items of C are required in week 5 in order to have a total of 50 items of A in

ITEM ON-HAND INVENTORY

A 10

B 15

C 20

D 10

E 10

F 5

TABLE 6.9 On-Hand Inventory

Item 1 2 3 4 5 6 Lead Time

Week

A

B

C

D

E

F

Gross On-Hand 10 Net Order Receipt Order Release

Gross On-Hand 15 Net Order Receipt Order Release

Gross On-Hand 20 Net Order Receipt Order Release

Gross On-Hand 10 Net Order Receipt Order Release

Gross On-Hand 10 Net Order Receipt Order Release

Gross On-Hand 5 Net Order Receipt Order Release

1 Week

2 Weeks

1 Week

1 Week

2 Weeks

3 Weeks

50 10 40 40

40

80 15 65 65

65

100

120

100185

195

120 20

100 100

195 10

185 185

100 0

100 100

200 5

195 195

A

A

B C

C

130 10

120 120

B

FIGURE 6.14 net Material Requirements Plan for 50 Units of A

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218  CHAPTER 6 • InvEnTORy COnTROl MODEls

week 6. The letter A in the upper-right corner for items B and C means that this demand for B and C was generated as a result of the demand for the parent, A. Now the same type of analysis is done for B and C to determine the net requirements for D, E, and F.

Two or More End Products So far, we have considered only one end product. For most manufacturing companies, there are normally two or more end products that use some of the same parts or compo- nents. All of the end products must be incorporated into a single net material requirements plan.

In the MRP example just discussed, we developed a net material requirements plan for product A. Now, we show how to modify the net material requirements plan when a second end product is introduced. Let’s call the second end product AA. The material structure tree for prod- uct AA is as follows:

AA

D(3) F(2)

Let’s assume that we need 10 units of AA. With this information, we can compute the gross re- quirements for AA:

Part D: 3 * number of AA>s = 3 * 10 = 30 Part F: 2 * number of AA>s = 2 * 10 = 20

To develop a net material requirements plan, we need to know the lead time for AA. Let’s assume that it is one week. We also assume that we need 10 units of AA in week 6 and that we have no units of AA on hand.

Now, we are in a position to modify the net material requirements plan for product A to in- clude AA. This is done in Figure 6.15.

Look at the top row of the figure. As you can see, we have a gross requirement of 10 units of AA in week 6. We don’t have any units of AA on hand, so the net requirement is also 10 units of AA. Because it takes one week to make AA, the order release of 10 units of AA is in week 5. This means that we start making AA in week 5 and have the finished units in week 6.

Because we start making AA in week 5, we must have 30 units of D and 20 units of F in week 5. See the rows for D and F in Figure 6.15. The lead time for D is one week. Thus, we must give the order release in week 4 to have the finished units of D in week 5. Note that there was no inventory on hand for D in week 5. The original 10 units of inventory of D were used in week 5 to make B, which was subsequently used to make A. We also need to have 20 units of F in week 5 to produce 10 units of AA by week 6. Again, we have no on-hand inventory of F in week 5. The original 5 units were used in week 4 to make C, which was subsequently used to make A. The lead time for F is three weeks. Thus, the order release for 20 units of F must be in week 2. (See the F row in Figure 6.15.)

This example shows how the inventory requirements of two products can be reflected in the same net material requirements plan. Some manufacturing companies can have more than 100 end products that must be coordinated in the same net material requirements plan. Although such a situation can be very complicated, the same principles we used in this example are em- ployed. Remember that computer programs have been developed to handle large and complex manufacturing operations.

In addition to using MRP to handle end products and finished goods, MRP can be used to handle spare parts and components. This is important because most manufacturing companies sell spare parts and components for maintenance. A net material requirements plan should also reflect these spare parts and components.

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6.11 JUsT-In-TIME InvEnTORy COnTROl  219

6.11 Just-In-Time Inventory Control

During the past several decades, there has been a trend to make the manufacturing process more efficient. One objective is to have less in-process inventory on hand. This is known as just-in- time (JIT) inventory. With this approach, inventory arrives just in time to be used during the manufacturing process to produce subparts, assemblies, or finished goods. One technique of im- plementing JIT is a manual procedure called kanban. Kanban in Japanese means “card.” With a dual-card kanban system, there is a conveyance kanban, or C-kanban, and a production kanban, or P-kanban. The kanban system is very simple. Here is how it works:

Four Steps of Kanban

Arrow 1: As shown in Figure 6.16, full containers along with their C-kanban card are taken from the storage area to a user area, typically on a manufacturing line (see arrow 1).

Arrow 2: During the manufacturing process, parts in the container are used up by the user. When the container is empty, the empty container along with the same C-kanban card is taken back to the storage area (depicted by arrow number 2). Here the user picks up a new full container, detaches the P-kanban card from it, attaches his or her C-kanban card to it, and returns with it to the user area (depicted by arrow number 1 again).

Item 1 2 3 4 5 6 Lead Time

Week

AA

A

C

D

E

F

Gross On-Hand: 0 Net Order Receipt Order Release

Gross On-Hand: 10 Net Order Receipt Order Release

Gross On-Hand: 20 Net Order Receipt Order Release

Gross On-Hand: 10 Net Order Receipt Order Release

Gross On-Hand: 10 Net Order Receipt Order Release

Gross On-Hand: 5 Net Order Receipt Order Release

B Gross On-Hand: 15 Net Order Receipt Order Release

Inventory

10 0

10 10

1 Week

1 Week

1 Week

1 Week

2 Weeks

3 Weeks

10

50 10 40 40

40

100

120

100185

195

120 20

100 100

130 10

120 120

195 10

185 185

100 0

100 100

200 5

195 195

2 Weeks

65

80 15 65 65

30

30 0

30 30

20

20 0

20 20

C AA

AA

A

A

B

B C

FIGURE 6.15 net Material Requirements Plan, Including AA

With JIT, inventory arrives just before it is needed.

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220  CHAPTER 6 • InvEnTORy COnTROl MODEls

Arrow 3: This detached P-kanban card is attached to an empty container in the storage area and then—and only then—is the empty container taken back to the upstream pro- ducer area (depicted by arrow number 3).

Arrow 4: This empty container is then refilled with parts and taken with its P-kanban card back to the storage area (depicted by arrow number 4). This kanban process con- tinuously cycles throughout the day. Kanban is sometimes known as a “pull” pro- duction system.

At a minimum, two containers are required using the kanban system. One container is used at the user area, and another container is being refilled for future use. In reality, there are usually more than two containers. This is how inventory control is accomplished. Inventory managers can introduce additional containers and their associated P-kanbans into the system. In a similar fashion, the inventory manager can remove containers and their P-kanbans to have tighter con- trol over inventory buildups.

In addition to being a simple, easy-to-implement system, the kanban system can be very effective in controlling inventory costs and in uncovering production bottlenecks. Inventory ar- rives at the user area or on the manufacturing line just before it is needed. Inventory does not build up unnecessarily, cluttering the production line or adding to unnecessary inventory ex- pense. The kanban system reduces inventory levels and makes for a more effective operation. It is like putting the production line on an inventory diet. Like any diet, the inventory diet imposed by the kanban system makes the production operation more streamlined. Furthermore, produc- tion bottlenecks and problems can be uncovered. Many production managers remove containers and their associated P-kanbans from the kanban system in order to “starve” the production line to uncover bottlenecks and potential problems.

In implementing a kanban system, a number of work rules or kanban rules are normally implemented. One typical kanban rule is that no containers are filled without the appropriate P-kanban. Another rule is that each container must hold exactly the specified number of parts or inventory items. These and similar rules make the production process more efficient. Only those parts that are actually needed are produced. The production department does not produce inven- tory just to keep busy. It produces inventory or parts only when they are needed in the user area or on an actual manufacturing line.

6.12 Enterprise Resource Planning

Over the years, MRP has evolved to include not only the materials required in production but also the labor hours, material cost, and other resources related to production. When approached in this fashion, the term MRP II is often used, and the word resource replaces the word require- ments. As this concept evolved and sophisticated computer software programs were developed, these systems were called enterprise resource planning (ERP) systems.

The objective of an ERP system is to reduce costs by integrating all of the operations of a firm. This starts with the supplier of the materials needed and flows through the organization to include invoicing the customer for the final product. Data are entered once into a database, and then these data can be quickly and easily accessed by anyone in the organization. This benefits

P-kanban and

Container

C-kanban and

Container

Producer Area

Storage Area

User Area

4 1

23

FIGURE 6.16 The kanban system

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glOssARy  221

not only the functions related to planning and managing inventory but also other business pro- cesses such as accounting, finance, and human resources.

The benefits of a well-developed ERP system are reduced transaction costs and increased speed and accuracy of information. However, there are drawbacks as well. The software is ex- pensive to buy and costly to customize. The implementation of an ERP system may require a company to change its normal operations, and employees are often resistant to change. Also, training employees on the use of the new software can be expensive.

There are many ERP systems available. The most common ones include SAP, Oracle, and PeopleSoft. Even small systems can cost hundreds of thousands of dollars. The larger systems can cost hundreds of millions of dollars.

This chapter introduces the fundamentals of inventory control theory. We showed that the two most important problems are (1) how much to order and (2) when to order.

We investigated the economic order quantity, which deter- mines how much to order, and the reorder point, which determines when to order. In addition, we explored the use of sensitivity analysis to determine what happens to computations when one or more of the values used in one of the equations change.

The basic EOQ inventory model presented in this chap- ter makes a number of assumptions: (1) known and constant demand and lead times, (2) instantaneous receipt of inventory, (3) no quantity discounts, (4) no stockouts or shortage, and (5) the only variable costs are ordering cost and carrying cost. If these assumptions are valid, the EOQ inventory model pro- vides optimal solutions. On the other hand, if these assump- tions do not hold, the basic EOQ model does not apply. In these cases, more complex models are needed, including the

production run, quantity discount, and safety stock models. When the inventory item is for use in a single time period, the marginal analysis approach is used. ABC analysis is used to determine which items represent the greatest potential inven- tory cost so these items can be more carefully managed.

When the demand for inventory is not independent of the demand for another product, a technique such as MRP is needed. MRP can be used to determine the gross and net ma- terial requirements for products. Computer software is neces- sary to implement major inventory systems, including MRP systems, successfully. Today many companies are using ERP software to integrate all of the operations within a firm, includ- ing inventory, accounting, finance, and human resources.

JIT can lower inventory levels, reduce costs, and make a manufacturing process more efficient. Kanban, a Japa- nese word meaning “card,” is one way to implement the JIT approach.

Summary

Glossary

ABC Analysis An analysis that divides inventory into three groups. Group A is more important than group B, which is more important than group C.

Average Inventory The average inventory on hand. In this chapter, the average inventory is Q>2 for the EOQ model.

Bill of Materials (BOM) A material structure tree of the components in a product, with a description and the quantity required to make one unit of that product.

Carrying Cost The cost of holding inventory over time. Economic Order Quantity (EOQ) The amount of inventory

ordered that will minimize the total inventory cost. It is also called the optimal order quantity, or Q*.

Enterprise Resource Planning (ERP) A computerized information system that integrates and coordinates the operations of a firm.

Instantaneous Inventory Receipt A system in which inventory is received or obtained at one point in time and not over a period of time.

Inventory Position The amount of inventory on hand plus the amount in any orders that have been placed but not yet received.

Just-in-Time (JIT) Inventory An approach whereby inven- tory arrives just in time to be used in the manufacturing process.

Kanban A manual JIT system developed by the Japanese. Kanban means “card” in Japanese.

Lead Time The time it takes to receive an order after it is placed (called L in the chapter).

Marginal Analysis A decision-making technique that uses marginal profit and marginal loss in determining optimal decision policies. Marginal analysis is used when the num- ber of alternatives and states of nature is large.

Marginal Loss (ML) The loss that would be incurred by stocking and not selling an additional unit.

Marginal Profit (MP) The additional profit that would be realized by stocking and selling one more unit.

Material Requirements Planning (MRP) An inventory model that can handle dependent demand.

Production Run Model An inventory model in which inven- tory is produced or manufactured over time instead of being ordered or purchased. This model eliminates the instanta- neous receipt assumption.

Ordering Cost The cost of placing an order with a supplier.

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Quantity Discount The cost per unit when large orders of an inventory item are placed.

Reorder Point (ROP) The number of units on hand when an order for more inventory is placed.

Safety Stock Extra inventory that is used to help avoid stockouts.

Setup Cost The cost to set up the manufacturing or produc- tion process for the production run model.

Sensitivity Analysis The process of determining how sensi- tive the optimal solution is to changes in the values used in the equations.

Service Level The chance, expressed as a percent, that there will not be a stockout. Service level = 1 - Probability of a stockout.

Stockout A situation that occurs when there is no inventory on hand.

Key Equations

Equations 6-1 through 6-8 are associated with the economic order quantity (EOQ).

(6-1) Average inventory level = Q

2

(6-2) Annual ordering cost = D

Q Co

(6-3) Annual holding cost = Q

2 Ch

(6-4) EOQ = Q* = A2DCoCh (6-5) TC =

D

Q Co +

Q

2 Ch

Total relevant inventory cost.

(6-6) Average dollar level = 1CQ2

2

(6-7) Q* = B 2DCoIC EOQ with Ch expressed as percentage of unit cost.

(6-8) ROP = d * L Reorder point: d is the daily demand and L is the lead time in days.

Equations 6-9 through 6-13 are associated with the production run model.

(6-9) Average inventory = Q

2 a1 - d

p b

(6-10) Annual holding cost = Q

2 a1 - d

p bCh

(6-11) Annual setup cost = D

Q Cs

(6-12) Annual ordering cost = D

Q Co

(6-13) Q* = H 2DCsCh a1 - dp b Optimal production quantity.

Equation 6-14 is used for the quantity discount model.

(6-14) Total cost = DC + D

Q Co +

Q

2 Ch

Total inventory cost (including purchase cost).

Equations 6-15 to 6-20 are used when safety stock is required.

(6-15) ROP = 1Average demand during lead time2 + SS General formula for determining the reorder point when safety stock (SS) is carried.

(6-16) ROP = 1Average demand during lead time2 + Zsd LT Reorder point formula when demand during lead time is normally distributed with a standard deviation of sdLT.

(6-17) ROP = dL + Z1sd2L2 Formula for determining the reorder point when daily demand is normally distributed but lead time is constant, where d is the average daily demand, L is the constant lead time in days, and sd is the standard deviation of daily demand.

(6-18) ROP = dL + Z1dsL2 Formula for determining the reorder point when daily demand is constant but lead time is normally distributed, where L is the average lead time in days, d is the constant daily demand, and sL is the standard deviation of lead time.

(6-19) ROP = d L + Z12Ls2d + d2s2L2 Formula for determining the reorder point when both daily demand and lead time are normally distributed, where d is the average daily demand, L is the average lead time in days, sL is the standard deviation of lead time, and sd is the standard deviation of daily demand.

(6-20) THC = Q

2 Ch + 1SS2Ch

Total annual holding cost formula when safety stock is carried.

Equation 6-21 is used for marginal analysis.

(6-21) P Ú ML

ML + MP Decision rule in marginal analysis for stocking ad- ditional units.

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Solved Problems

Solved Problem 6-1 Patterson Electronics supplies microcomputer circuitry to a company that incorporates microproces- sors into refrigerators and other home appliances. One of the components has an annual demand of 250 units, and this is constant throughout the year. The carrying cost is estimated to be $1 per unit per year, and the ordering cost is $20 per order.

a. To minimize cost, how many units should be ordered each time an order is placed? b. How many orders per year are needed with the optimal policy? c. What is the average inventory if costs are minimized? d. Suppose the ordering cost is not $20 and Patterson has been ordering 150 units each time an or-

der is placed. For this order policy to be optimal, what would the ordering cost have to be?

Solution a. The EOQ assumptions are met, so the optimal order quantity is

EOQ = Q* = B 2DCoCh = B 212502201 = 100 units b. Number of orders per year =

D

Q =

250

100 = 2.5 orders per year

Note that this would mean that in one year the company places three orders and in the next year it needs to place only two orders, since some inventory would be carried over from the previous year. It averages 2.5 orders per year.

c. Average inventory = Q

2 =

100

2 = 50 units

d. Given an annual demand of 250, a carrying cost of $1, and an order quantity of 150, Patterson Electronics must determine what the ordering cost would have to be for the order policy of 150 units to be optimal. To find the answer to this problem, we must solve the traditional EOQ equa- tion for the ordering cost. As you can see in the calculations that follow, an ordering cost of $45 is needed for the order quantity of 150 units to be optimal.

Q = A2DCoCh Co = Q2

Ch 2D

= 115022112

212502

= 22,500

500 = $45

Solved Problem 6-2 Flemming Accessories produces paper slicers used in offices and in art stores. The minislicer has been one of its most popular items: annual demand is 6,750 units and is constant throughout the year. Kristen Flemming, owner of the firm, produces the minislicers in batches. On average, Kristen can manufacture 125 minislicers per day. Demand for these slicers during the production process is 30 per day. The setup cost for the equipment necessary to produce the minislicers is $150. The carrying cost is $1 per minislicer per year. How many minislicers should Kristen manufacture in each batch?

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Solution The data for Flemming Accessories are summarized as follows:

D = 6,750 units

Cs = $150

Ch = $1

d = 30 units p = 125 units

This is a production run problem that involves a daily production rate and a daily demand rate. The appropriate calculations are shown here:

Q* = B 2DCsCh11 - d>p2 = B 216,750211502111 - 30>1252 = 1,632

Solved Problem 6-3 Dorsey Distributors has an annual demand for 1400 units of a metal detector. The cost of a typical detector to Dorsey is $400. The carrying cost is estimated to be 20% of the unit cost, and the order- ing cost is $25 per order. If Dorsey orders in quantities of 300 or more, it can get a 5% discount on the cost of the detectors. Should Dorsey take the quantity discount? Assume the demand is constant.

Solution The solution to any quantity discount model involves determining the total cost of each alternative af- ter quantities have been computed and adjusted for the original problem and every discount. We start the analysis with no discount:

EOQ 1no discount2 = B 211,400212520.214002 = 29.6 units

Total cost 1no discount2 = Material cost + Ordering cost + Carrying cost

= $40011,4002 + 1,4001$252 29.6

+ 29.61$400210.22

2

= $560,000 + $1,183 + $1,183 = $562,366

The next step is to compute the total cost for the discount:

EOQ 1with discount2 = B 211,400212520.21$3802 = 30.3 units

Q1adjusted2 = 300 units Because this last economic order quantity is below the discounted price, we must adjust the order quantity to 300 units. The next step is to compute total cost:

Total cost 1with discount2 = Material cost + Ordering cost + Carrying cost

= $38011,4002 + 1,4001252 300

+ 3001$380210.22

2

= $532,000 + $117 + $11,400 = $543,517

The optimal strategy is to order 300 units at a total cost of $543,517.

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Solved Problem 6-4 The F. W. Harris Company sells an industrial cleaner to a large number of manufacturing plants in the Houston area. An analysis of the demand and costs has resulted in a policy of ordering 300 units of this product every time an order is placed. The demand is constant, at 25 units per day. In an agreement with the supplier, F. W. Harris is willing to accept a lead time of 20 days, since the supplier has provided an excellent price. What is the reorder point? How many units are actually in inventory when an order should be placed?

Solution The reorder point is

ROP = d * L = 251202 = 500 units This means that an order should be placed when the inventory position is 500. Since the ROP is greater than the order quantity, Q = 300, an order must have been placed already but not yet deliv- ered. So the inventory position must be

Inventory position = 1Inventory on hand2 + 1Inventory on order2 500 = 200 + 300

There would be 200 units on hand and an order of 300 units in transit.

Solved Problem 6-5 The B. N. Thayer and D. N. Thaht Computer Company sells a desktop computer that is popular among gaming enthusiasts. In the past few months, demand has been relatively consistent, although it does fluctu- ate from day to day. The company orders the computer cases from a supplier. It places an order for 5,000 cases at the appropriate time to avoid stockouts. The demand during the lead time is normally distributed, with a mean of 1,000 units and a standard deviation of 200 units. The holding cost per unit per year is esti- mated to be $4. How much safety stock should the company carry to maintain a 96% service level? What is the reorder point? What will the total annual holding cost be if this policy is followed?

Solution Using the table for the normal distribution, the Z value for a 96% service level is about 1.75. The stan- dard deviation is 200. The safety stock is calculated as

SS = ZsdLT = 1.7512002 = 375 units For a normal distribution with a mean of 1,000, the reorder point is

ROP = 1Average demand during lead time2 + SS = 1,000 + 350 = 1,350 units

The total annual holding cost is

THC = Q

2 Ch + 1SS2Ch = 5,0002 4 + 137524 = $11,500

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

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

1. Which of the following is a basic component of an inventory control system? a. planning what inventory to stock and how to acquire it b. forecasting the demand for parts and products c. controlling inventory levels d. developing and implementing feedback measurements

for revising plans and forecasts e. all of the above are components of an inventory control

system

2. Which of the following is a valid use of inventory? a. the decoupling function b. to take advantage of quantity discounts c. to avoid shortages and stockouts d. to smooth out irregular supply and demand e. all of the above are valid uses of inventory

sElF-TEsT  225

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226  CHAPTER 6 • InvEnTORy COnTROl MODEls

3. One assumption necessary for the EOQ model is instantaneous replenishment. This means a. the lead time is zero. b. the production time is assumed to be zero. c. the entire order is delivered at one time. d. replenishment cannot occur until the on-hand inven-

tory is zero. 4. If the EOQ assumptions are met and a company orders

the EOQ each time an order is placed, then the a. total annual holding cost is minimized. b. total annual ordering cost is minimized. c. total of all inventory cost is minimized. d. order quantity will always be less than the average

inventory. 5. If the EOQ assumptions are met and a company orders

more than the EOQ, then the a. total annual holding cost will be greater than the total

annual ordering cost. b. total annual holding cost will be less than the total an-

nual ordering cost. c. total annual holding cost will be equal to the total an-

nual ordering cost. d. total annual holding cost will be equal to the total an-

nual purchase cost. 6. The reorder point is

a. the quantity that is reordered each time an order is placed.

b. the amount of inventory that would be needed to meet demand during the lead time.

c. equal to the average inventory when the EOQ assump- tions are met.

d. assumed to be zero if there is instantaneous replenishment. 7. If the EOQ assumptions are met, then

a. the annual stockout cost will be zero. b. the total annual holding cost will equal the total annual

ordering cost. c. the average inventory will be one-half the order quantity. d. all of the above are true.

8. In the production run model, the maximum inventory level will be a. greater than the production quantity. b. equal to the production quantity. c. less than the production quantity. d. equal to the daily production rate plus the daily demand.

9. Why is the annual purchase (material) cost not considered to be a relevant inventory cost if the EOQ assumptions are met? a. This cost will be zero. b. This cost is constant and not affected by the order

quantity. c. This cost is insignificant compared with the other in-

ventory costs. d. This cost is never considered to be an inventory cost.

10. A JIT system will usually result in a. a low annual holding cost. b. very few orders per year. c. frequent shutdowns in an assembly line. d. high levels of safety stock.

11. Manufacturers use MRP when a. the demand for one product is dependent on the de-

mand for other products. b. the demand for each product is independent of the de-

mand for other products. c. the demand is totally unpredictable. d. the purchase cost is extremely high.

12. In using marginal analysis, an additional unit should be stocked if a. MP = ML. b. the probability of selling that unit is greater than or

equal to MP>1MP + ML2. c. the probability of selling that unit is less than or equal

to ML>1MP + ML2. d. the probability of selling that unit is greater than or

equal to ML>1MP + ML2. 13. In using marginal analysis with the normal distribution, if

marginal profit is less than marginal loss, we expect the optimal stocking quantity to be a. greater than the standard deviation. b. less than the standard deviation. c. greater than the mean. d. less than the mean.

14. The inventory position is defined as a. the amount of inventory needed to meet the demand

during the lead time. b. the amount of inventory on hand. c. the amount of inventory on order. d. the total of the on-hand inventory plus the on-order

inventory.

Discussion Questions and Problems

Discussion Questions 6-1 Why is inventory an important consideration for

managers? 6-2 What is the purpose of inventory control? 6-3 Under what circumstances can inventory be used as

a hedge against inflation? 6-4 Why wouldn’t a company always store large quanti-

ties of inventory to eliminate shortages and stock outs?

6-5 What are some of the assumptions made in using the EOQ?

6-6 Discuss the major inventory costs that are used in determining the EOQ.

6-7 What is the ROP? How is it determined? 6-8 If the ROP is greater than the order quantity, explain

how the ROP is implemented. Can the ROP be more than twice the order quantity, and, if so, how is such a situation handled?

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DIsCUssIOn QUEsTIOns AnD PROBlEMs  227

6-9 Let the annual demand for an arbitrary commodity be 1,000 units per year and the associated EOQ be 400 units per order. Under this circumstance, the number of orders per year would be 1D>Q2 = 2.5 orders per year. How can this be so?

6-10 What is the purpose of sensitivity analysis? 6-11 What assumptions are made in the production run

model? 6-12 What happens to the production run model when the

daily production rate becomes very large? 6-13 Briefly describe what is involved in solving a quan-

tity discount model. 6-14 When using safety stock, how is the standard devia-

tion of demand during the lead time calculated if daily demand is normally distributed but lead time is constant? How is it calculated if daily demand is constant but lead time is normally distributed? How is it calculated if both daily demand and lead time are normally distributed?

6-15 Briefly explain the marginal analysis approach to the single-period inventory problem.

6-16 Briefly describe what is meant by ABC analysis. What is the purpose of this inventory technique?

6-17 What is the overall purpose of MRP? 6-18 What is the difference between the gross and the net

material requirements plans? 6-19 What is the objective of JIT?

Problems 6-20 Lila Battle has determined that the annual demand

for number 6 screws is 100,000 screws. Lila, who works in her brother’s hardware store, is in charge of purchasing. She estimates that it costs $10 ev- ery time an order is placed. This cost includes her wages, the cost of the forms used in placing the or- der, and so on. Furthermore, she estimates that the cost of carrying one screw in inventory for a year is one-half of 1 cent. Assume that the demand is con- stant throughout the year.

(a) How many number 6 screws should Lila order at a time if she wishes to minimize total inventory cost?

(b) How many orders per year would be placed? What would the annual ordering cost be?

(c) What would the average inventory be? What would the annual holding cost be?

6-21 It takes approximately eight working days for an or- der of number 6 screws to arrive once the order has been placed. (Refer to Problem 6-20.) The demand for number 6 screws is fairly constant, and on aver- age, Lila has observed that her brother’s hardware

store sells 500 of these screws each day. Because the demand is fairly constant, Lila believes that she can avoid stockouts completely if she orders the number 6 screws only at the correct time. What is the ROP?

6-22 Lila’s brother believes that she places too many or- ders for screws per year. He believes that an order should be placed only twice per year. If Lila follows her brother’s policy, how much more would this cost every year over the ordering policy that she de- veloped in Problem 6-20? If only two orders were placed each year, what effect would this have on the ROP?

6-23 Barbara Bright is the purchasing agent for West Valve Company. West Valve sells industrial valves and fluid control devices. One of the most popular valves is the Western, which has an annual demand of 4,000 units. The cost of each valve is $90, and the inventory carrying cost is estimated to be 10% of the cost of each valve. Barbara has made a study of the costs involved in placing an order for any of the valves that West Valve stocks, and she has concluded that the average ordering cost is $25 per order. Fur- thermore, it takes about two weeks for an order to arrive from the supplier, and during this time, the de- mand per week for West valves is approximately 80.

(a) What is the EOQ? (b) What is the ROP? (c) Is the ROP greater than the EOQ? If so, how is

this situation handled? (d) What is the average inventory? What is the an-

nual holding cost? (e) How many orders per year would be placed?

What is the annual ordering cost?

6-24 Ken Ramsing has been in the lumber business for most of his life. Ken’s biggest competitor is Pacific Woods. Through many years of experience, Ken knows that the ordering cost for an order of ply- wood is $25 and that the carrying cost is 25% of the unit cost. Both Ken and Pacific Woods receive plywood in loads that cost $100 per load. Further- more, Ken and Pacific Woods use the same supplier of plywood, and Ken was able to find out that Pacific Woods orders in quantities of 4,000 loads at a time. Ken also knows that 4,000 loads is the EOQ for Pa- cific Woods. What is the annual demand in loads of plywood for Pacific Woods?

6-25 Shoe Shine is a local retail shoe store located on the north side of Centerville. Annual demand for a pop- ular sandal is 500 pairs, and John Dirk, the owner of Shoe Shine, has been in the habit of ordering 100 pairs at a time. John estimates that the ordering cost is $10 per order. The cost of the sandal is $5 per pair. For John’s ordering policy to be correct, what would

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

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

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228  CHAPTER 6 • InvEnTORy COnTROl MODEls

the carrying cost as a percentage of the unit cost have to be? If the carrying cost was 10% of the cost, what would the optimal order quantity be?

6-26 In Problem 6-20, you helped Lila Battle determine the optimal order quantity for number 6 screws. She had estimated that the ordering cost was $10 per order. At this time, though, she believes that this estimate was too low. Although she does not know the exact order- ing cost, she believes that it could be as high as $40 per order. How would the optimal order quantity change if the ordering cost were $20, $30, and $40?

6-27 Ross White’s machine shop uses 2,500 brackets dur- ing the course of a year, and this usage is relatively constant throughout the year. These brackets are pur- chased from a supplier 100 miles away for $15 each, and the lead time is 2 days. The holding cost per bracket per year is $1.50 (or 10% of the unit cost), and the ordering cost per order is $18.75. There are 250 working days per year.

(a) What is the EOQ? (b) Given the EOQ, what is the average inventory?

What is the annual inventory holding cost? (c) In minimizing cost, how many orders would be

placed each year? What would be the annual ordering cost?

(d) Given the EOQ, what is the total annual inven- tory cost (including purchase cost)?

(e) What is the time between orders? (f) What is the ROP?

6-28 Ross White (see Problem 6-27) wants to reconsider his decision of buying the brackets and is consider- ing making the brackets in-house. He has determined that setup cost would be $25 in machinist time and lost production time and that 50 brackets could be produced in a day once the machine has been set up. Ross estimates that the cost (including labor time and materials) of producing one bracket would be $14.80. The holding cost would be 10% of this cost.

(a) What is the daily demand rate? (b) What is the optimal production quantity? (c) How long will it take to produce the optimal

quantity? How much inventory is sold during this time?

(d) If Ross uses the optimal production quantity, what would be the maximum inventory level? What would be the average inventory level? What is the annual holding cost?

(e) How many production runs would there be each year? What would be the annual setup cost?

(f) Given the optimal production run size, what is the total annual inventory cost?

(g) If the lead time is one-half day, what is the ROP?

6-29 Upon hearing that Ross White (see Problems 6-27 and 6-28) is considering producing the brackets in- house, the vendor has notified Ross that the purchase

price would drop from $15 per bracket to $14.50 per bracket if Ross would purchase the brackets in lots of 1,000. Lead times, however, would increase to 3 days for this larger quantity.

(a) What is the total annual inventory cost plus pur- chase cost if Ross buys the brackets in lots of 1,000 at $14.50 each?

(b) If Ross does buy in lots of 1,000 brackets, what is the new ROP?

(c) Given the options of purchasing the brackets at $15 each, producing them in-house at $14.80, and taking advantage of the discount, what is your recommendation to Ross White?

6-30 After analyzing the costs of various options for ob- taining brackets, Ross White (see Problems 6-27 through 6-29) recognizes that although he knows that the lead time is 2 days and the demand per day averages 10 units, the demand during the lead time often varies. Ross has kept very careful records and has determined that lead time demand is normally distributed with a standard deviation of 1.5 units.

(a) What Z value would be appropriate for a 98% service level?

(b) What safety stock should Ross maintain if he wants a 98% service level?

(c) What is the adjusted ROP for the brackets? (d) What is the annual holding cost for the safety

stock if the annual holding cost per unit is $1.50?

6-31 Annual demand for the Dobbs model airplane kit is 80,000 units. Albert Dobbs, president of Dobbs’ Ter- rific Toys, controls one of the largest toy companies in Nevada. He estimates that the ordering cost is $40 per order. The carrying cost is $7 per unit per year. It is 25 days from the time that Albert places an order for the model airplane kits until they are received at his warehouse. During this time, the daily demand is estimated to be 450 units.

(a) Compute the EOQ, ROP, and optimal number of orders per year.

(b) Albert now believes that the carrying cost may be as high as $14 per unit per year. Furthermore, he estimates that the lead time may be 35 days instead of 25 days. Redo part (a), using these re- vised estimates.

6-32 Morgan Arthur has spent the past few weeks deter- mining inventory costs for Armstrong, a toy manu- facturer located near Cincinnati, Ohio. She knows that annual demand will be 30,000 units per year and that the carrying cost will be $1.50 per unit per year. The ordering cost, on the other hand, can vary from $45 per order to $50 per order. During the past 450 working days, Morgan has observed the following frequency distribution for the ordering cost:

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ORDERING COST FREQUENCY

$45 85

$46 95

$47 90

$48 80

$49 55

$50 45

Morgan’s boss would like Morgan to determine an EOQ value for each possible ordering cost and to determine an EOQ value for the expected ordering cost.

6-33 Douglas Boats is a supplier of boating equipment for the states of Oregon and Washington. It sells 5,000 White Marine WM-4 diesel engines every year. These engines are shipped to Douglas in a ship- ping container of 100 cubic feet, and Douglas Boats keeps the warehouse full of these WM-4 motors. The warehouse can hold 5,000 cubic feet of boating supplies. Douglas estimates that the ordering cost is $10 per order, and the carrying cost is estimated to be $10 per motor per year. Douglas Boats is consid- ering the possibility of expanding the warehouse for the WM-4 motors. How much should Douglas Boats expand, and how much would it be worth for the company to make the expansion? Assume demand is constant throughout the year.

6-34 Northern Distributors is a wholesale organization that supplies retail stores with lawn care and house- hold products. One building is used to store Neverfail lawn mowers. The building is 25 feet wide by 40 feet deep by 8 feet high. Anna Oldham, manager of the warehouse, estimates that about 60% of the ware- house can be used to store the Neverfail lawn mow- ers. The remaining 40% is used for walkways and a small office. Each Neverfail lawn mower comes in a box that is 5 feet by 4 feet by 2 feet high. The annual demand for these lawn mowers is 12,000, and the ordering cost for Northern Distributors is $30 per or- der. It is estimated that it costs Northern $2 per lawn mower per year for storage. Northern Distributors is thinking about increasing the size of the warehouse. The company can do this only by making the ware- house deeper. At the present time, the warehouse is 40 feet deep. How many feet of depth should be added onto the warehouse to minimize the annual inventory costs? How much should the company be willing to pay for this addition? Remember that only 60% of the total area can be used to store Neverfail lawn mowers. Assume all EOQ conditions are met.

6-35 Lisa Surowsky was asked to help in determining the best ordering policy for a new product. Currently, the demand for the new product has been projected to be about 1,000 units annually. To get a handle on the carrying and ordering costs, Lisa prepared a

series of average inventory costs. Lisa thought that these costs would be appropriate for the new prod- uct. The results are summarized in the following ta- ble. These data were compiled for 10,000 inventory items that were carried or held during the year and were ordered 100 times during the past year. Help Lisa determine the EOQ.

COST FACTOR COST ($)

Taxes 2,000

Processing and inspection 1,500

New product development 2,500

Bill paying 500

Ordering supplies 50

Inventory insurance 600

Product advertising 800

Spoilage 750

Sending purchasing orders 800

Inventory inquiries 450

Warehouse supplies 280

Research and development 2,750

Purchasing salaries 3,000

Warehouse salaries 2,800

Inventory theft 800

Purchase order supplies 500

Inventory obsolescence 300

6-36 Jan Gentry is the owner of a small company that produces electric scissors used to cut fabric. The annual demand is 8,000 scissors, and Jan produces the scissors in batches. On average, Jan can produce 150 scissors per day, and during the production pro- cess, demand for scissors has been about 40 scissors per day. The cost to set up the production process is $100, and it costs Jan 30 cents to carry one pair of scissors for one year. How many scissors should Jan produce in each batch?

6-37 Jim Overstreet, inventory control manager for Itex, receives wheel bearings from Wheel-Rite, a small producer of metal parts. Unfortunately, Wheel-Rite can produce only 500 wheel bearings per day. Itex receives 10,000 wheel bearings from Wheel-Rite each year. Since Itex operates 200 working days each year, its average daily demand for wheel bear- ings is 50. The ordering cost for Itex is $40 per order, and the carrying cost is 60 cents per wheel bearing per year. How many wheel bearings should Itex order from Wheel-Rite at one time? Wheel-Rite has agreed to ship the maximum number of wheel bearings that it produces each day to Itex when an order has been received.

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6-38 North Manufacturing has a demand for 1,000 pumps each year. The cost of a pump is $50. It costs North Manufacturing $40 to place an order, and the carry- ing cost is 25% of the unit cost. If pumps are ordered in quantities of 200, North Manufacturing can get a 3% discount on the cost of the pumps. Should North Manufacturing order 200 pumps at a time and take the 3% discount?

6-39 Linda Lechner is in charge of maintaining hospital supplies at General Hospital. During the past year, the mean lead time demand for bandage BX-5 was 60. Furthermore, the standard deviation for BX-5 was 7. Linda would like to maintain a 90% service level. What safety stock level do you recommend for BX-5?

6-40 Linda Lechner has just been severely chastised for her inventory policy. (See Problem 6-39.) Sue Sur- rowski, her boss, believes that the service level should be either 95% or 98%. Compute the safety stock levels for a 95% and a 98% service level. Linda knows that the carrying cost of BX-5 is 50 cents per unit per year. Compute the carrying cost that is associated with a 90%, a 95%, and a 98% ser- vice level.

6-41 Ralph Janaro simply does not have time to ana- lyze all of the items in his company’s inventory. As a young manager, he has more important things to do. The following is a table of six items in inventory along with the unit cost and the demand in units.

(a) Find the total amount spent on each item during the year. What is the total investment for all of these?

(b) Find the percentage of the total investment in in- ventory that is spent on each item.

IDENTIFICATION CODE

UNIT COST ($)

DEMAND IN UNITS

XX1 5.84 1,200

B66 5.40 1,110

3CPO 1.12 896

33CP 74.54 1,104

R2D2 2.00 1,110

RMS 2.08 961

(c) Based on the percentages in part (b), which item(s) would be classified in categories A, B, and C using ABC analysis?

(d) Which item(s) should Ralph most carefully con- trol using quantitative techniques?

6-42 Thaarugo, Inc., produces a GPS device that is be- coming popular in parts of Scandinavia. When Thaarugo produces one of these, a printed circuit board (PCB) is used, and it is populated with several electronic components. Thaarugo determines that it

needs about 16,000 of this type of PCB each year. Demand is relatively constant throughout the year, and the ordering cost is about $25 per order; the holding cost is 20% of the price of each PCB. Two companies are competing to become the dominant supplier of the PCBs, and both have now offered discounts, as shown in the following table. Which of the two suppliers should be selected if Thaarugo wishes to minimize total annual inventory cost? What would be the annual inventory cost?

SUPPLIER A SUPPLIER B

QUANTITY PRICE QUANTITY PRICE

1–199 38.40 1–299 39.50

200–499 35.80 300–999 35.40

500 or more 34.70 1,000 or more 34.60

6-43 Dillard Travey receives 5,000 tripods annually from Quality Suppliers to meet his annual demand. Dil- lard runs a large photographic outlet, and the tripods are used primarily with 35-mm cameras. The order- ing cost is $15 per order, and the carrying cost is 50 cents per unit per year. Quality is starting a new option for its customers. When an order is placed, Quality will ship one-third of the order every week for three weeks instead of shipping the entire order at one time. Weekly demand over the lead time is 100 tripods.

(a) What is the order quantity if Dillard has the en- tire order shipped at one time?

(b) What is the order quantity if Dillard has the or- der shipped over three weeks using the new op- tion from Quality Suppliers, Inc.? To simplify your calculations, assume that the average inven- tory is equal to one-half of the maximum inven- tory level for Quality’s new option.

(c) Calculate the total cost for each option. What do you recommend?

6-44 Quality Suppliers, Inc., has decided to extend its shipping option. (Refer to Problem 6-43 for de- tails.) Now, Quality Suppliers is offering to ship the amount ordered in five equal shipments, one each week. It will take five weeks for this entire order to be received. What are the order quantity and total cost for this new shipping option?

6-45 The Hardware Warehouse is evaluating the safety stock policy for all its items, as identified by the SKU code. For SKU M4389, the company always orders 80 units each time an order is placed. The daily demand is constant, at 5 units per day; the lead time is normally distributed, with a mean of three days and a standard deviation of two days. Holding cost is $3 per unit per year. A 95% service level is to be maintained.

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DIsCUssIOn QUEsTIOns AnD PROBlEMs  231

(a) What is the standard deviation of demand during the lead time?

(b) How much safety stock should be carried, and what should be the reorder point?

(c) What is the total annual holding cost?

6-46 For SKU A3510 at the Hardware Warehouse, the order quantity has been set at 150 units each time an order is placed. The daily demand is normally distributed, with a mean of 12 units and a standard deviation of 4. It always takes exactly five days for an order of this item to arrive. The holding cost has been determined to be $10 per unit per year. Due to the large sale volume of this item, management wants to maintain a 99% service level.

(a) What is the standard deviation of demand during the lead time?

(b) How much safety stock should be carried, and what should be the reorder point?

(c) What is the total annual holding cost?

6-47 H & K Electronic Warehouse sells a 12-pack of AAA batteries, and this is a very popular item. De- mand for this is normally distributed, with an aver- age of 50 packs per day and a standard deviation of 16. The average delivery time is five days, with a standard deviation of two days. Delivery time has been found to be normally distributed. A 96% ser- vice level is desired.

(a) What is the standard deviation of demand during the lead time?

(b) How much safety stock should be carried, and what should be the reorder point?

6-48 Xemex has collected the following inventory data for the six items that it stocks:

ITEM CODE

UNIT COST

($)

ANNUAL DEMAND (UNITS)

ORDERING COST ($)

CARRYING COST AS A

PERCENTAGE OF UNIT COST

1 10.60 600 40 20

2 11.00 450 30 25

3 2.25 500 50 15

4 150.00 560 40 15

5 4.00 540 35 16

6 4.10 490 40 17

Lynn Robinson, Xemex’s inventory manager, does not feel that all of the items can be controlled. What order quantities do you recommend for which inven- tory product(s)?

6-49 Georgia Products offers the following discount schedule for its 4- by 8-foot sheets of good-quality plywood:

ORDER UNIT COST ($)

9 sheets or less 18.00

10 to 50 sheets 17.50

More than 50 sheets 17.25

Home Sweet Home Company orders plywood from Georgia Products. Home Sweet Home has an or- dering cost of $45. The carrying cost is 20%, and the annual demand is 100 sheets. What do you recommend?

6-50 Sunbright Citrus Products produces orange juice, grapefruit juice, and other citrus-related items. Sun- bright obtains fruit concentrate from a cooperative in Orlando consisting of approximately 50 citrus grow- ers. The cooperative will sell a minimum of 100 cans of fruit concentrate to citrus processors such as Sun- bright. The cost per can is $9.90.

Last year, the cooperative developed the Incen- tive Bonus Program to give an incentive to their large customers to buy in quantity. It works like this: if 200 cans of concentrate are purchased, 10 cans of free concentrate are included in the deal. In addition, the names of the companies purchasing the concen- trate are added to a drawing for a new personal com- puter. The personal computer has a value of about $3,000, and currently about 1,000 companies are eligible for this drawing. At 300 cans of concentrate, the cooperative will give away 30 free cans and will also place the company name in the drawing for the personal computer. When the quantity goes up to 400 cans of concentrate, 40 cans of concentrate will be given away free with the order. In addition, the company is also placed in a drawing for the per- sonal computer and a free trip for two. The value of the trip for two is approximately $5,000. About 800 companies are expected to qualify and to be in the running for this trip.

Sunbright estimates that its annual demand for fruit concentrate is 1,000 cans. In addition, the or- dering cost is estimated to be $10, while the carrying cost is estimated to be 10%, or about $1 per unit. The firm is intrigued with the Incentive Bonus Pro- gram. If the company decides that it will keep the free cans, the trip, or the computer if they are won, what should it do?

6-51 John Lindsay sells CDs that contain 25 software packages that perform a variety of financial func- tions, including net present value, internal rate of re- turn, and other financial programs typically used by business students majoring in finance. Depending on the quantity ordered, John offers the following price discounts. The annual demand is 2,000 units on av- erage. His setup cost to produce the CDs is $250. He estimates the holding cost to be 10% of the price, or about $1 per unit per year.

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232  CHAPTER 6 • InvEnTORy COnTROl MODEls

QUANTITY ORDERED

FROM TO PRICE

1 500 $10.00

501 1,000 9.95

1,001 1,500 9.90

1,501 2,000 9.85

(a) What is the optimal number of CDs to produce at a time?

(b) What is the impact of the following quantity– price schedule on the optimal order quantity?

QUANTITY ORDERED

FROM TO PRICE

1 500 $10.00

501 1,000 9.99

1,001 1,500 9.98

1,501 2,000 9.97

6-52 Teresa Granger is the manager of Chicago Cheese, which produces cheese spreads and other cheese-re- lated products. E-Z Spread Cheese is a product that has always been popular. The probability of sales, in cases, is as follows:

DEMAND (CASES) PROBABILITY

10 0.2

11 0.3

12 0.2

13 0.2

14 0.1

A case of E-Z Spread Cheese sells for $100 and has a cost of $75. Any cheese that is not sold by the end of the week is sold to a local food processor for $50. Teresa never sells cheese that is more than a week old. Use marginal analysis to determine how many cases of E-Z Spread Cheese to produce each week to maximize average profit.

6-53 Harry’s Hardware does a brisk business during the year. During Christmas, Harry’s Hardware sells Christmas trees for a substantial profit. Unfortu- nately, any trees not sold at the end of the season are totally worthless. Thus, the number of trees to stock for a given season is a very important decision. The following table reveals the demand for Christmas trees:

DEMAND FOR CHRISTMAS TREES

PROBABILITY

50 0.05

75 0.1

100 0.2

125 0.3

150 0.2

175 0.1

200 0.05

Harry sells trees for $80 each, but his cost is only $20.

(a) Use marginal analysis to determine how many trees Harry should stock at his hardware store.

(b) If the cost increases to $35 per tree and Harry continues to sell trees for $80 each, how many trees should Harry stock?

(c) Harry is thinking about increasing the price to $100 per tree. Assume that the cost per tree is $20. With the new price, it is expected that the probability of selling 50, 75, 100, or 125 trees will be 0.25 each. Harry does not expect to sell more than 125 trees with this price increase. What do you recommend?

6-54 In addition to selling Christmas trees during the Christmas holidays, Harry’s Hardware sells all the ordinary hardware items (see Problem 6-53). One of the most popular items is Great Glue HH, a glue that is made just for Harry’s Hardware. The selling price is $4 per bottle, but unfortunately, the glue gets hard and unusable after one month. The cost of the glue is $1.20. During the past several months, the mean sales of glue have been 60 units, and the standard deviation is 7. How many bottles of glue should Harry’s Hardware stock? Assume that sales follow a normal distribution.

6-55 The marginal loss on Washington Reds, a brand of apples from the state of Washington, is $35 per case. The marginal profit is $15 per case. During the past year, the mean sales of Washington Reds in cases was 45,000 cases, and the standard deviation was 4,450. How many cases of Washington Reds should be brought to market? Assume that sales follow a normal distribution.

6-56 Linda Stanyon has been the production manager for Plano Produce for over eight years. Plano Produce is a small company located near Plano, Illinois. One produce item, tomatoes, is sold in cases, with daily sales averaging 400 cases. Daily sales are assumed to be normally distributed. In addition, 85% of the time the sales are between 350 and 450 cases. Each case costs $10 and sells for $15. All cases that are not sold must be discarded.

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DIsCUssIOn QUEsTIOns AnD PROBlEMs  233

(a) Using the information provided, estimate the standard deviation of sales.

(b) Using the standard deviation in part (a), deter- mine how many cases of tomatoes Linda should stock.

6-57 Paula Shoemaker produces a weekly stock market report for an exclusive readership. She normally sells 3,000 reports per week, and 70% of the time her sales range from 2,900 to 3,100. The report costs Paula $15 to produce, but Paula is able to sell reports for $350 each. Of course, any reports not sold by the end of the week have no value. How many reports should Paula produce each week?

6-58 Emarpy Appliance produces all kinds of major ap- pliances. Richard Feehan, the president of Emarpy, is concerned about the production policy for the company’s best-selling refrigerator. The demand for this has been relatively constant at about 8,000 units each year. The production capacity for this product is 200 units per day. Each time production starts, it costs the company $120 to move materials into place, reset the assembly line, and clean the equip- ment. The holding cost of a refrigerator is $50 per year. The current production plan calls for 400 re- frigerators to be produced in each production run. Assume there are 250 working days per year.

(a) What is the daily demand of this product? (b) If the company were to continue to produce 400

units each time production starts, how many days would production continue?

(c) Under the current policy, how many production runs per year would be required? What would the annual setup cost be?

(d) If the current policy continues, how many refrig- erators would be in inventory when production stops? What would the average inventory level be?

(e) If the company produces 400 refrigerators at a time, what would the total annual setup cost and holding cost be?

6-59 Consider the Emarpy Appliance situation in Prob- lem 6-58. If Richard Feehan wants to minimize the total annual inventory cost, how many refrigerators should be produced in each production run? How much would this save the company in inventory costs compared with the current policy of producing 400 in each production run?

6-60 This chapter presents a material structure tree for item A in Figure 6.12. Assume that it now takes 1 unit of item B to make every unit of item A. What impact does this have on the material structure tree and the number of items D and E that are needed?

6-61 Given the information in Problem 6-60, develop a gross material requirements plan for 50 units of item A.

6-62 Using the data from Figures 6.12–6.14, develop a net material requirements plan for 50 units of item A assuming that it takes only 1 unit of item B for each unit of item A.

6-63 The demand for product S is 100 units. Each unit of S requires 1 unit of T and 0.5 ounce of U. Each unit of T requires 1 unit of V, 2 units of W, and 1 unit of X. Finally, each unit of U requires 0.5 ounce of Y and 3 units of Z. All items are manufactured by the same firm. It takes two weeks to make S, one week to make T, two weeks to make U, two weeks to make V, three weeks to make W, one week to make X, two weeks to make Y, and one week to make Z.

(a) Construct a material structure tree and a gross material requirements plan for the dependent in- ventory items.

(b) Identify all levels, parents, and components. (a) Construct a net material requirements plan using

the following on-hand inventory data:

ITEM S T U V W X Y Z

On-Hand Inventory

20

20

10

30

30

25

15

10

6-64 The Webster Manufacturing Company produces a popular type of serving cart. This product, the SL72, is made from the following parts: 1 unit of part A, 1 unit of part B, and 1 unit of subassembly C. Each subassembly C is made up of 2 units of part D, 4 units of part E, and 2 units of part F. Develop a material structure tree for this.

6-65 Blair H. Dodds III runs a medium- to large-sized home eBay business dealing in vintage photographs. The annual demand for his photos is approximately 50,000. The annual overhead cost (excluding the purchase price) to buy the photographs is $4,000 per year. Given that this cost represents the optimal an- nual ordering cost and the optimal ordering quantity is 400 photographs at a time, determine the ordering cost and the average inventory level.

6-66 The lead time for each of the parts in the SL72 (Prob- lem 6-64) is one week, except for part B, which has a lead time of two weeks. Develop a net material re- quirements plan for an order of 800 SL72s. Assume that currently there are no parts in inventory.

6-67 Refer to Problem 6-66. Develop a net material re- quirements plan assuming that there are cur- rently 150 units of part A, 40 units of part B, 50 units of subassembly C, and 100 units of part F in inventory.

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234  CHAPTER 6 • InvEnTORy COnTROl MODEls

See our Internet home page, at www.pearsonhighered.com/render, for additional homework problems, Problems 6-68 to 6-75.

Internet Homework Problems

Martin-Pullin Bicycle Corp. (MPBC), located in Dallas, is a wholesale distributor of bicycles and bicycle parts. Formed in 1981 by cousins Ray Martin and Jim Pullin, the firm’s primary retail outlets are located within a 400-mile radius of the dis- tribution center. These retail outlets receive their orders from Martin-Pullin within two days after notifying the distribution center, provided that the stock is available. However, if an or- der is not fulfilled by the company, no backorder is placed; the retailers arrange to get their shipment from other distributors, and MPBC loses that amount of business.

Demands for AirWing Model

MONTH 2014 2015 FORECAST FOR 2016

January 6 7 8

February 12 14 15

March 24 27 31

April 46 53 59

May 75 86 97

June 47 54 60

July 30 34 39

August 18 21 24

September 13 15 16

October 12 13 15

November 22 25 28

December 38 42 47

Total 343 391 439

The company distributes a wide variety of bicycles. The most popular model, and the major source of revenue for the company, is the AirWing. MPBC receives all the models from a single manufacturer overseas, and shipment takes as long as four weeks from the time an order is placed. With the cost of communication, paperwork, and customs clearance included, MPBC estimates that each time an order is placed, it incurs a cost of $65. The purchase price paid by MPBC, per bicycle, is roughly 60% of the suggested retail price for all the styles available, and the inventory carrying cost is 1% per month (12% per year) of the purchase price paid by MPBC. The retail price (paid by the customers) for the AirWing is $170 per bicycle.

MPBC is interested in making an inventory plan for 2016. The firm wants to maintain a 95% service level with its cus- tomers to minimize the losses on the lost orders. The data collected for the past two years are summarized in the accom- panying table. A forecast for AirWing model sales in 2016 has been developed and will be used to make an inventory plan for MPBC.

Discussion Questions 1. Develop an inventory plan to help MPBC. 2. Discuss ROPs and total costs. 3. How can you address demand that is not at the level of the

planning horizon?

Source: Professor Kala Chand Seal, Loyola Marymount University.

Case Study

Martin-Pullin Bicycle Corporation

See our Internet home page, at www.pearsonhighered.com/render, for these additional case studies: (1) LaPlace Power and Light: This case involves a public utility in Louisiana and its use of elec- tric cables for connecting houses to power lines. (2) Western Ranchman Outfitters: This case involves managing the inventory of a popular style of jeans when the delivery date is sometimes unpredictable. (3) Professional Video Management: This case involves a video system in which discounts from suppliers are possible. (4) Drake Radio: This case involves ordering FM tuners.

Internet Case Studies

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APPEnDIx 6.1: InvEnTORy COnTROl wITH QM FOR wInDOws  235

Bibliography

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Bradley, James R., and Richard W. Conway. “Managing Cyclic Inventories,” Production and Operations Management 12, 4 (Winter 2003): 464–479.

Chan, Lap Mui Ann, David Simchi-Levi, and Julie Swann. “Pricing, Produc- tion, and Inventory Policies for Manufacturing with Stochastic Demand and Discretionary Sales,” Manufacturing & Service Operations Manage- ment 8, 2 (Spring 2006): 149–168.

Chickering, David Maxwell, and David Heckerman. “Targeted Advertising on the Web with Inventory Management,” Interfaces 33, 5 (September– October 2003): 71–77.

Chopra, Sunil, Gilles Reinhardt, and Maqbool Dada. “The Effect of Lead Time Uncertainty on Safety Stocks,” Decision Sciences 35, 1 (Winter 2004): 1–24.

Desai, Preyas S., and Oded Koenigsberg. “The Role of Production Lead Time and Demand Uncertainty in Marketing Durable Goods,” Management Science 53, 1 (January 2007): 150–158.

Jayaraman, Vaidyanathan, Cindy Burnett, and Derek Frank. “Separating In- ventory Flows in the Materials Management Department of Hancock Medical Center,” Interfaces 30, 4 (July–August 2000): 56–64.

Kapuscinski, Roman, Rachel Q. Zhang, Paul Carbonneau, Robert Moore, and Bill Reeves. “Inventory Decisions in Dell’s Supply Chain,” Interfaces 34, 3 (May–June 2004): 191–205.

Karabakal, Nejat, Ali Gunal, and Warren Witchie. “Supply-Chain Analysis at Volkswagen of America,” Interfaces 30, 4 (July–August 2000): 46–55.

Kök, A. Gürhan, and Kevin H. Shang. “Inspection and Replenishment Policies for Systems with Inventory Record Inaccuracy,” Manufacturing & Ser- vice Operations Management 9, 2 (Spring 2007): 185–205.

Ramasesh, Ranga V., and Ram Rachmadugu. “Lot-Sizing Decision Under Limited-Time Price Reduction,” Decision Sciences 32, 1 (Winter 2001): 125–143.

Ramdas, Kamalini, and Robert E. Spekman. “Chain or Shackles: Understanding What Drives Supply-Chain Performance,” Interfaces 30, 4 (July– August 2000): 3–21.

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Shin, Hojung, and W. C. Benton. “Quantity Discount-Based Inventory Coor- dination: Effectiveness and Critical Environmental Factors,” Production and Operations Management 13, 1 (Spring 2004): 63–76.

Appendix 6.1: Inventory Control with QM for Windows

A variety of inventory control models are covered in this chapter. Each model makes different assumptions and uses slightly differ- ent approaches. The use of QM for Windows is similar for these different types of inventory problems. As you can see in the Inven- tory menu for QM for Windows, most of the inventory problems discussed in this chapter can be solved using your computer.

To demonstrate QM for Windows, we start with the basic EOQ model. Sumco, a manufacturing company discussed in the chapter, has an annual demand of 1,000 units, an ordering cost of $10 per order, and a carrying cost of $0.50 per unit per year. With these data, we can use QM for Windows to determine the economic order quantity. The results are shown in Program 6.5.

The production run inventory problem—which requires the daily production and demand rates in addition to the an- nual demand, the ordering cost per order, and the carrying cost per unit per year—is also covered in this chapter. The Brown Manufacturing example is used in this chapter to show how the