Instructor s Manual for Making Hard Decisions With DecisionTools

INSTRUCTOR’S MANUAL for

Making Hard Decisions with DecisionTools, 3rd Ed.

Revised 2013 Samuel E. Bodily University of Virginia Robert T. Clemen Duke University Robin Dillon Georgetown University Terence Reilly Babson College

Table of Contents GENERAL INFORMATION Introduction Influence Diagrams Decision Analysis Software

1 2 2

CHAPTER NOTES AND PROBLEM SOLUTIONS Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Section 1 Cases Athens Glass Works Integrated Siting Systems, Inc. International Guidance and Controls George’s T-Shirts Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Section 2 Cases Lac Leman Festival de la Musique Sprigg Lane Appshop, Inc. Calambra Olive Oil Scor-eStore.com Chapter 14 Chapter 15 Chapter 16 Chapter 17 Section 3 Cases John Carter Sleepmore Mattress Manufacturing Susan Jones TOPICAL INDEX TO PROBLEMS

4 8 14 39 66 85 90 98 116 130 145 165 176 201 223 240 267 283 297 330 339 353 371 395 403 416 423 438 447 461

GENERAL INFORMATION INTRODUCTION  Making Hard Decisions with DecisionTools, DecisionTools, 3rd  Edition presents the basic basic techniques of modern decision analysis. The emphasis of the text is on the development of models to represent decision situations and the use of probability and utility theory to represent uncertainties and preferences, respectively, in those models. This is a new edition of the text. New examples and problems have been added throughout the text and some chapters have either been completely rewritten (Chapters 5 & 11) or are entirely new (Chapters 6 & 13). In addition, we have added 15 cases from Darden Business Publishing. The Darden cases are grouped together at the end of each of the three sections.

The first section of the book deals with structuring decision models. This part of the process is undoubtedly the most critical. It is in the structuring phase that one comes to terms with the decision situation, clarifies clarifies one’s objectives in the context of that situation, and confronts questions regarding the problem’s essential elements. One must decide exactly what aspects of a p roblem are to be included in a model and make fundamental modeling choices regarding how to represent each facet. Discussions with decision analysts confirm that in most real-world applications, the majority of the time is spent in structuring the problem, and this is where most of the important insights are found and creative new alternatives invented. The discussion of model structuring integrates notions of tradeoffs and multiple objectives, something new to the second edition of the book. Although complete discussion of modeling and analysis techniques are put off until later, students should have enough information so that they can analyze simple multiattribute models after finishing Chapter 4. This early introduction to this material has proven to be an excellent motivator for students. Give them an interesting problem, ask them to discuss the objectives and tradeoffs, and you will have trouble getting them to be quiet!  Making Hard Decisions with DecisionTools DecisionTools provides a one-semester one-semester introduction introduction to the tools and concepts of decision analysis. The text can be reasonably well adapted to different curricula; additional material (readings, cases, problems from other sources) can be included easily at many different points. For example, Chapters 8 and 15 discuss judgmental aspects of probability assessment assessment and decision making, and an instructor can introduce more behavioral material at these points. Likewise, Chapter 16 delves into the additive utility function for decision making. Some instructors may wish to present goal programming or the analytic hierarchy process here.

The Darden cases are grouped at the end of each of the 3 sections. Instead of tying each case to a particular chapter, a group of cases are associated with a group of chapters. The goal is to show that the various concepts and tools covered throughout the section can be applied to the cases for that section. For example, to solve the cases at the end of Section One, Modeling Decisions, students will need to understand the objectives of the decision maker (Chapter 2), structure and solve the decision model (Chapters 3 and 4),  perform a sensitivity sensitivity analysis analysis (Chapter 5), and, perhaps, incorporate incorporate organizational organizational decision making making concepts (Chapter 6). Instructors can either assign a case analysis after covering a set of chapters asking the students to incorporate all the relevant material or can assign a case after each chapter highlighting that chapter’s material. Students need to understand that a complete and insightful analysis is based on investigating the case using more than one or two concepts. Incorporating Keeney’s (1992) value-focused thinking was challenging because some colleagues preferred to have all of the multiple-objective material material put in the same place (Chapters 15 and 16), whereas others  preferred to integrate integrate the material material throughout the text. Ultimately Ultimately the latter latter was chosen chosen especially stressing stressing the role of values in structuring decision models. In particular, students must read about structuring values at the beginning of Chapter 3 before going on to structuring influence diagrams diagrams or decision trees. The reason is simply that it makes sense to understand what one wants before trying to structure the decision. In order for an instructor to locate problems on specific topics or concepts without having to read through all the problems, a topical cross-reference for the problems is included in each chapter and a topical index for all of the problems and case studies is provided at the end of the manual.

1

INFLUENCE DIAGRAMS The most important innovation in the first edition of  Making Hard Decisions was the integration of influence diagrams throughout the book. Indeed, in Chapter 3 influence diagrams are presented before decision trees as structuring tools. The presentation and use of influence diagrams reflects their current  position in the decision-analysis toolkit. toolkit. They appear to be most most useful for (1) (1) structuring problems problems and (2)  presenting overviews overviews to an audience with little technical background. background. In certain situations, influence influence diagrams can be used to great advantage. For example, understanding value-of-information value-of-information analysis is a  breeze with influence influence diagrams, diagrams, but tortuous with decision trees. trees. On the other other hand, decision decision trees still  provide the best best medium for understanding many basic decision-analysis decision-analysis concepts, such as risk-return trade-offs or subjective probability assessment.

Some instructors may want to read more about influence diagrams prior to teaching a course using  Making  Hard Decisions with DecisionTools DecisionTools. The basic reference is Howard and Matheson (1981, reprinted in ). This first paper offers a very general overview, but relatively little in the way of nitty-gritty, hands-on help. Aside from Chapters 3 and 4 of Making Hard Decisions with DecisionTools DecisionTools, introductory discussions of influence diagrams can be found in Oliver and Smith (1990) and McGovern, Samson, and Wirth (1993). In the field of artificial intelligence, intelligence, belief nets (which can be thought of as influence diagrams that contain only uncertainty nodes) are used to represent probabilistic knowledge structures. For introductions to belief nets, consult Morawski (1989a, b) as well as articles in Oliver and Smith (1990). Matzkevich and Abramson (1995) provides an excellent recent review of network models, including influence diagrams and  belief nets. The conference on Uncertainty in Artificial Intelligence has been held annually since 1985, and the conference always publishes a book of proceedings. For individuals who wish to survey the field broadly, these volumes provide up-to-date information on the representation and use of network models. Selected Bibliography for Influence Diagrams Howard, R. A. (1989). Knowledge maps.  Management Science, 35, 903-922. Howard, R. A., and J. E. Matheson (1981). “Influence Diagrams.” R. Howard and J. Matheson (Eds.),The Principles and Applications of Decision Analysis, Vol II , Palo Alto: Strategic Decisions Group, (1984), 719-762. Reprinted in  Decision Analysis, Vol 2 (2005), 127-147. Matzkevich, I., and B. Abramson (1995). “Decision Analytic Networks in Artificial Intelligence.”  Management Science, 41, 1-22. McGovern, J., D. Samson, and A. Wirth (1993). “Influence Diagrams for Decision Analysis.” In S. Nagel (Ed.), Computer-Aided Decision Analysis. Westport, CT: Quorum, 107-122. Morawski, P. (1989a). “Understanding Bayesian Belief Networks.” AI Expert  (May),  (May), 44-48. Morawski, P. (1989b). “Programming Bayesian Belief Networks.” AI Expert  (August),  (August), 74-79.  Neapolitan, R. R. E. (1990). Probabilistic Reasoning in Expert Systems. New York: Wiley. Oliver, R. M., and J. Q. Smith (1989).  Influence Diagrams, Belief Nets Nets and Decision Analysis (Proceedings of an International Conference 1988, Berkeley). New York: John Wiley. Pearl, J. (1988). Probabilistic Reasoning in Intelligent Systems . San Mateo, CA: Morgan Kaufman. Shachter, R. D. (1986). “Evaluating Influence Diagrams,” Operations Research, 34, 871-882. Shachter, R. D. (1988). Probabilistic inference and influence diagrams. Operations Research, 36, 389-604. Shachter, R. D., and C. R. Kenley (1989). Gaussian Influence Diagrams.  Management Science, 35, 527550. DECISION ANALYSIS SOFTWARE  Making Hard Decisions with DecisionTools DecisionTools integrates Palisade Corporation’s DecisionTools, version 6.0 throughout the text. DecisionTools consists of six programs (PrecisionTree, TopRank, @RISK, StatTools, StatTools,  NeuralTools, and Evolver), each each designed to help with different different aspects of modeling and solving solving decision  problems. Instructions Instructions are given given on how to use PrecisionTree and @RISK, @RISK, typically, at the end of the chapter. PrecisionTree PrecisionTree is a versatile program that allows the user to construct and solve both decision trees and influence diagrams. @RISK allows the user to insert probability distributions into a spreadsheet and run a Monte Carlo simulation. . Each of these programs are Excel add-ons, which means that they run within Excel by adding their ribbon of commands to Excel’s toolbar.

2

In the textbook, instructions have been included at the ends of appropriate chapters for using the programs that correspond to the chapter topic. The instructions provide step-by-step guides through the important features of the programs. They have been written to be a self-contained tutorial. Some supplemental information is contained in this manual especially related to the implementation implementation of specific problem solutions. Some general guidelines: To run an add-in within Excel, it is necessary to have the “Ignore other applications” option turned off. Choose Tools on the menu bar, then Options, and click on the General tab in the resulting Options dialog box. Be sure that the box by  Ignore other applications is not checked. Macros in the add-in program become disabled automatically if the security level is set to High. To change the security level to Medium, in the Tools menu, point to  Macros and then click Security. When the program crashes, restart the computer. It may appear as if the program has closed  properly and can be reopened, but it probably has not, and it is best to restart the the computer. The student version of PrecisionTree may limit the tree to 50 nodes. Some of the problems that examine the value of information in Chapter 12 can easily exceed this limit. When running @RISK simulations in the student version, make sure that only one worksheet is open at a time. Otherwise, the program will display that error message “Model Extends Beyond Allowed Region of Worksheet.” •

•

•

•

•

More tips are provided throughout this manual as they relate to implementing specific problem solutions.

JOIN THE DECISION ANALYSIS SOCIETY OF INFORMS

Instructors and students both are encouraged to join the Decision Analysis Society of INFORMS (Institute for Operations Research and Management Science). This organization provides a wide array of services for decision analysts, including a newsletter, Internet list server, a site on the World Wide Web (https://www https://www.informs.org/ .informs.org/Community/DA Community/DAS S), annual meetings, and information on job openings and candidates for d ecision-analysis ecision-analysis positions. For information on how to join, visit the web site.

3

CHAPTER 1 Introduction to Decision Analysis Notes This chapter serves as an introduction to the book and the course. It sets the tone and presents the basic approach that will be used. The ideas of subjective judgment and modeling are stressed. Also, we mention some basic aspects of decisions: uncertainty, preferences, decision structure, and sensitivity analysis.

In teaching decision analysis courses, it is critical to distinguish at the outset between good decisions and good outcomes. Improving decisions mostly means improving the decision-making process. Students should make decisions with their eyes open, having carefully considered the important issues at hand. This is not to say that a good decision analysis foresees every possible outcome; indeed, many possible outcomes are so unlikely that they may have no bearing whatsoever on the decision to be made. Often it is helpful to imagine yourself in the future, looking back at your decision now. Will you be able to say, regardless of the outcome: “Given everything I knew at the time — and I d id a pretty good job of digging out the important issues — I made the appropriate decision. If I were put back in the same situation, I would go through the process pretty much in the same way and would probably make the same decision.” If your decision making lets you say this, then you are making good decisions. The issue is not whether you can foresee some unusual outcome that really is unforeseen, even by the experts. The issue is whether you carefully consider the aspects of the decision that are important and meaningful to you. Chapter 1 emphasizes a modeling approach and the idea of a requisite model. If the notion of a requisite model seems a bit slippery, useful references are the articles by Phillips. (Specific references can be found in Making Hard Decisions with DecisionTools.) The concept is simple: A decision model is requisite if it incorporates all of the essential elements of the decision situation. The cyclical process of modeling, solution, sensitivity analysis, and then modeling again, provides the mechanism for identifying areas that require more elaboration in the model and portions where no more modeling is needed (or even where certain aspects can be ignored altogether). After going through the decision analysis cycle a few times, the model should provide a reasonable representation of the situation and should provide insight regarding the situation and available options. Note that the process, being a human one, is not guaranteed to converge in any technical sense. Convergence to a requisite model must arise from 1) technical modeling expertise on the part of the analyst, and 2) desire on the part of the decision maker to avoid the cognitive dissonance associated with an incomplete or inappropriate model. Also important is that the modeling approach presented throughout the book emphasizes value-focused thinking (Keeney, 1992), especially the notion that values should be considered at the earliest phases of the decision-making process. This concept is initially introduced on pages 5-6. To show that that decision analysis really is used very broadly, we have included the section “Where is Decision Analysis Used?” Two references are given. The Harvard Business Review article by Ulvila and Brown is particularly useful for students to read any time during the course to get a feel for real-world applications of decision analysis. Finally, we have included the section “Where Does the Software Fit In?” to introduce the DecisionTools suite of programs. Topical cross-reference for problems Constructionist view Creativity Rice football Requisite models Subjective judgments

1.12,Du Pont and Chlorofluorocarbons 1.8 1.7 1.2 1.3, 1.5

4

Solutions 1.1. Answers will be based on personal experience. It is important here to be sure the distinction is made  between good decisions on one hand (or a good decision-making process) and lucky outcomes on the other. 1.2. We will have models to represent the decision structure as well as uncertainty and preferences. The whole point of using models is to create simplifications of the real world in such a way that analysis of the model yields insight regarding the real-world situation. A requisite model is one that includes all essential elements of the problem. Alternatively, a requisite model is one which, when subjected to sensitivity analysis, yields no new intuitions. Not only are all essential elements included, but also all extraneous elements are excluded. 1.3. Subjective judgments will play large roles in the modeling of uncertainty and preferences. Essentially we will build representations of personal beliefs (probabilities) and preferences (utilities). In a more subtle  — and perhaps more important — way, subjective judgments also direct the modeling process. Subjective  judgments are necessary for determining the appropriateness of a model’s structure, what should be included in the model, and so on. Thus, subjective judgments play central roles in decision analysis. Good decision analysis cannot be done without subjective judgments. 1.4. An appropriate answer would be that decision analysis can improve your decisions — the way you make decisions — by providing a framework for dealing with difficult decisions in a systematic way. Along with the analytical framework, decision analysis provides a set of tools for constructing and analyzing decision models, the purpose of which is to obtain insight regarding difficult decision problems. 1.5. You require her subjective judgments on a number of matters. First is the problem of identifying important aspects of the problem. Her input also will be required for the development of models of her uncertainty and her preferences. Thus, her judgments will be critical to the analysis.

This question may also lead students to consider the implications of delegating decisions to agents. How can you ensure that the agent will see things the way you do? Will the same aspects of the problem be important? Does the agent agree with you regarding the uncertainty inherent in the situation (which outcomes are more or less likely)? Does the agent have the same feeling regarding trade-offs that must be made? In many cases it may be appropriate to obtain and use an expert’s information. Can you identify some specific decision situations where you would be willing to accept an agent’s recommendation? Does it matter who the agent is? Can you identify other situations in which some of the agent ’s input can be taken at face value (a forecast, say), but must be incorporated into a model based primarily on your own  judgments? 1.6. Answers will be based on personal experience. 1.7. Some of the issues are 1) the monetary costs of staying in Division 1-A and of moving to Division III, 2) impact on both alumni and local businesses of moving to Division III, 3) political and social impact on campus of changing divisions.

Alternatives include 1) stay in Division 1-A, 2) move to Division III, 3) move to Division II, 4) delay the decision for a year or more to gather information, 5) investigate other sources of funding to cover the deficit, 6) drop out from the NCAA altogether ... There is considerable uncertainty around the impact on the school of switching divisions. What will the fallout be from the faculty, students, alumni, and local businesses if Rice went to Division III? Will it impact recruiting? If so, how? What are the financial consequences? Is the d eficit due to mismanagement or is it structural? What are the long-term consequences versus the immediate uproar? Sources of information could be surveys given to each constituency and/or interviews with leaders of the constituencies. Perhaps other schools have changed divisions, and information can be found from their experience. The objectives that different groups want to work toward include 1) minimize short-term and long-term deficit, 2) minimize social upheaval, 3) maximize enjoyment of collegiate sports, 4) maximize student 5

opportunity to participate in sports, 5) maximize quality of sports programs. Some students may identify still other objectives. Trading off these objectives may mean trying to balance the issues that are important to different constituent groups. 1.8. This is a creativity question. The Friends of Rice Athletics could fund raise, tuition and/or ticket prices could be increased, the stadium’s name can be sold, the athletic staff could all take a pay cut, etc. 1.9. Answers will be based on personal experience. 1.10. Instead of thinking only about risk versus return, the socially responsible investor also must consider how to trade off risk and return for ethical integrity. It would not be unreasonable to suspect that to obtain a higher level of ethical integrity in the portfolio, the investor must accept a lower expected return, higher level of risk, or both. 1.11. For the most part, decision analysis is most appropriate for strategic, or one-time, decisions. These are situations that we have not thought about before and “don’t know what to do.” Hence, it is worthwhile to engage in some “decision making,” or decision analysis, to figure out what would be an appropriate action.

This is not to say that decision analysis is inappropriate for repetitive decisions. In fact, if a decision is repeated many times, the savings that can be achieved over time by improving the decision-making process can be substantial. In fact, this is the basis of much of management science. However, the reliance on subjective judgments for the construction of tailored decision models in each decision situation may render decision analysis, as portrayed here, unsuitable for dealing with repetitive situations. The point, though, is that if one anticipates a long string of repetitive decisions in the future, and an optimal decision strategy has not been previously developed, then the situation is indeed one of “not knowing what to do.” A decisionmodeling approach would indeed be appropriate in that case. 1.12. Beliefs and values do appear to change and develop over time as we think about new issues. Decision analysis implicitly provides a framework for such changes through the identification and modeling of decision problems, beliefs regarding uncertainty, and preferences. Case Study: Commercial Space Travel A student’s answer to being an early adopter or waiting until the industry matures is a personal choice and depends on many factors. Some of these are: track record of industry, affordability, health of student vis-àvis demands of space travel, interest level, etc.

It certainly is true that new firms can come along and change an industry with leaner production or management systems. Often, these firms do not have to contend with the legacy of older systems in more established firms. In addition, the savings of a younger workforce and less established pension program can  be quite significant. Thus, it is reasonable that the new furry animals can be competitive with a massive governmental organization. On the other hand, the lack of experience of extreme situations might turn into a disaster for a newly established firm. The cost savings of the newer firms could come from more efficient operations or it could come from not having the equipment and policies in place to handle unusual situations. A space-flight disaster would make headlines across the world and probably doom the responsible for-profit company. To continue the survival-of-the-fittest analogy, it is not that every for-profit company will survive by avoiding life-threatening situations; it is that a subgroup will survive. Would you want to put your life or the life of a loved one on the line given the uncertainties surrounding early adopters in space travel? Case Study: Du Pont and Chlorofluorocarbons The major issues include shareholder welfare, social and environmental responsibility, and ethics. Of course, all of these might be thought of as means for ensuring the long-run profitability or survivability of the firm. The major sources of uncertainty involve research and development. Will substitute products work? Will they be accepted? The CEO might wonder whether the ozone problem really is a problem, or

6

whether the observed recent changes are part of a normal cycle. Finally, could Du Pont ’s efforts really have an effect, and how much? It is undoubtedly the case that Du Pont’s views of the situation have changed over time. Early on, the chlorofluorocarbon issue was essentially ignored; no one knew that a problem existed. In the 1970s and 1980s, it became apparent that a problem did exist, and as scientific evidence accumulated, the problem appeared to become more serious. Finally, we have arrived at a position where the ozone issue clearly matters. (In fact, it matters mostly because of consumers ’ views and preferences rather than because of the scientific evidence, which appears to be less than conclusive.) Du Pont appears to be asking “Can we do anything to help?” Many companies have developed a kind of “social awareness” in the past two decades as a way to maintain a high-integrity profile. Case Study: Choosing a Vice-Presidential Candidate A vice president tends not to have an important role in American politics except in gaining electoral votes during the election. A running mate is often chosen to balance the ticket geographically and ideologically. For example, choosing a conservative, women from Alaska helped McCain appeal to the conservative base of the Republican Party and to women. Alaska, however, has the minimum number of possible electoral votes at 3. While McCain could reasonably count on winning Alaska’s 3 electoral votes, he could have chosen someone else from a more p opulous state for the electoral votes. McCain must have thought that Ms. Palin would provide a ticket with a wide appeal and that she could help pick up votes across the whole country.

It is hard to know how McCain’s health affected his choice of Ms. Palin. Clearly, he knew how he felt, and given that he is still in office eight years later, it is reasonable to assume that his health was not a major concern when choosing Ms. Palin. A portion of the population, however, did find his age coupled with her inexperience troubling. If he personally was not concerned, he might at least have considered how the voters would perceive Ms. Palin being one heartbeat away from the presidency of the U.S.A. The president is constantly gathering information, from the daily threat-assessment reports to meetings with his cabinet, congressional members, and world leaders. However, even with all of these intelligence reports, much uncertainty still remains, often requiring the president to make a judgment call. One of the more famous examples of this is President Obama’s decision to send U.S. forces into Pakistan after Osama  bin Laden. Although it was thought that bin Laden was hiding inside a residence, there was not definitive  proof. Moreover, Obama also had to make judgment calls concerning the size of the force to send in and whether to alert Pakistani officials. Generally, the president’s decisions are based (hopefully) on both facts and judgments. McCain’s choice of Sarah Palin led many voters to question his judgment. Choosing Sarah Palin might have turned out to be a very good choice for the United States, but it certainly had many political overtones. In all fairness, the choice of a vice-presidential running mate is a very  political decision, one specially aimed at winning the election – a political event. On the other hand, appearances are of utmost importance in elections, and even an unsubstantiated rumor can completely derail a candidate. Thus, in choosing his running mate, McCain probably should have weighed the pros and cons of each candidate using his fundamental objectives, the fundamental objectives of his party, and, of course, the fundamental objectives of the United States as a whole.

7

CHAPTER 2 Elements of Decision Problems Notes This chapter is intended to start the reader thinking about decision problems in decision-analysis terms. Thus, we talk about decisions to make, uncertain events, and valuing consequences. To make sure that the richness of the terrain is understood, we introduce the concepts of dynamic decision making, a planning horizon, and trade-offs.

In our definition of terms, we refer to a decision maker ’s objectives where the term values is used to refer to the decision maker ’s set of objectives and their structure. The terms decision and alternative are adopted, and are used throughout the book rather than similar terms such as “choice” and “option.” Likewise, we have adopted the term uncertain event  (and sometimes chance event ), which then has outcomes. Finally, and perhaps most significant, we have adopted Savage’s term consequence to refer to what the decision maker experiences as a result of a combination of alternative(s) chosen and chance outcome(s). Another term that we use that comes from Keeney’s value-focused thinking is the notion of decision context . This term is discussed in the text and still more thoroughly in Keeney’s book. Briefly, it refers to the specific identification of the problem (from which we might suspect that when one solves the wrong problem, one has used the wrong decision context). It also can be used as a way to identify the class of alternatives that one is willing to consider; a broader context (safety in auto travel as compared to specific traffic laws, for example) leads a decision maker to consider a broader class of alternatives. The time value of money appears in Chapter 2 and may seem out of place in some ways. It is here because it is a fundamental way that streams of cash flows are valued, and because it provides a nice example of a  basic trade-off. Also, we have found that since most students have already been exposed to discounting, we have been able to incorporate NPV calculations into problems and case studies throughout t he book. For the few students who have not encountered the topic, the early introduction to discounting in Chapter 2  provides enough information for them to proceed. Of course, the section on NPV may be skipped and used as a reference later for problems that require discounting or for the discussion of trade-offs in Chapter 15. Topical cross-reference for problems Requisite models “Secretary” problem Sequential decisions Time value of money

2.13 2.6 2.2, 2.6, Early Bird, Inc. 2.9-2.12, The Value of Patience

Solutions 2.1. a. Some objectives might be to minimize cost, maximize safety, maximize comfort, maximize reliability, maximize cargo capacity (for shopping or vacationing), maximize maneuverability (in city traffic). Students will undoubtedly come up with others as well. b. In this new context, appropriate objectives might be minimize travel time, maximize exercise, minimize total transportation cost, minimize use of fossil fuels, maximize ease (suitably defined) of visiting friends and shopping. New alternatives to consider include using a bicycle or public transportation, walking, rollerblading, skateboarding, motorcycle or scooter, renting a car, such as Zipcar. One might even consider moving in order to live in a more convenient location. 2.2. Future alternatives can affect the eventual value of the consequence. For example, a university faculty member, when accepting a position at a different institution, may not immediately resign his or her position at the first university. Instead, a leave of absence may be taken. The leave of absence provides the opportunity to decide in the future whether to stay at the new institution or return to the old one. A faculty member would most likely think about the two different situations — resigning the current position immediately versus taking a leave and postponing a permanent decision — in very different ways.

8

Another good example is purchasing a house. For many people in our mobile society, it is important to think about the potential for selling the house in the future. Many purchasers might buy an unusual h ouse that suits them fine. However, if the house is too unusual, would-be purchasers might be afraid that, if they decide to sell the house in the near future , it may be difficult to find a buyer and the sales price might be lower than it would be for a more conventional house. Finally, the current choice might eliminate a future valuable option. For example, our policy of powering cars with fossil fuels reduces our options for using oil for potentially more valuable and less destructive future activities. 2.3. In the first case, the planning horizon may be tied directly to the solution of the specific problem at hand. If the problem is an isolated one not expected to repeat, this is a reasonable horizon. If more similar  problems are anticipated, the planning horizon might change to look forward in time far enough to anticipate future such situations. If the firm is considering hiring a permanent employee or training existing employees, then a planning horizon should be long enough to accommodate employee-related issues (training, reviews, career advancement, and so on). In thi s broader context, the firm must consider objectives related to hiring a new person (or training), which might include maximizing the welfare of current employees, minimizing long-term costs of dealing with the class of problems, satisfying affirmative-action requirements, or equity in treatment of employees. 2.4. In making any decision, it is important to 1) use all currently available information and 2) think carefully about future uncertainty. Thus it i s necessary to keep track of exactly what i nformation is available at each point in time. If information is lost or forgotten, then it will either be treated as an uncertainty or simply not used when deciding. Clearly, the farmer would want to keep up to date on the weather and incorporate any change to the forecast. 2.5. Some possibilities: insurance, hire another firm to manage the protection operation, press for regulatory decisions and evaluations (i.e., get public policy makers to do the necessary analysis), do nothing, develop a “cleanup cooperative” with other firms, or design and develop equipment th at can serve a day-to-day purpose but be converted easily to cleanup equipment. Students may come up with a wide variety of ideas. 2.6. The employer should think about qualifications of the applicants. The qualifications that he seeks should be intimately related to what the employer wants to accomplish (objectives — e.g., increase market share) and hence to the way the successful applicant will be evaluated (attributes — e.g., sales). The  planning horizon may be critical. Is the employer interested in long-term or short-term performance? The uncertainty that the employer faces, of course, i s the uncertainty regarding the applicant ’s future  performance on the specified attributes.

If the decision maker must decide whether to make a job offer at the end of each interview, then the  problem becomes a dynamic one. That is, after each interview the decision maker must decide whether to make the offer (and end the search) or to continue the search for at least one more interview, at which time the same decision arises. In this version of the problem, the decision maker faces an added uncertainty: the qualifications of the applicants still to come. (This dynamic problem is sometimes known as the “Secretary Problem,” and has been analyzed extensively and in many different forms in the operations-research literature. For example, see DeGroot (2004) Optimal Statistical Decisions , Hoboken, NJ: Wiley & Sons. P. 325.) 2.7. Decisions to make: How to invest current funds. Possible alternatives include do nothing, purchase specific properties, purchase options, etc. Other decisions might include how to finance the purchase, wh en to resell, how much rent to charge, and so on. Note that the situation is a dynamic one if we consider future investment opportunities that may be limited by current investments.

Uncertain events: Future market conditions (for resale or rentin g), occupancy rates, costs (management, maintenance, insurance), and rental income.

9

Possible outcomes: Most likely such an investor will be interested in future cash flows. Important trade-offs include time value of money and current versus future investment opportunities. 2.8. Answers depend on personal experience and will vary widely. Be sure to consider current and future decisions and uncertain events, the planning horizon, and important trade-offs. 2.9. NPV

=

-2500 1500 1700 + + 1.130 1.131 1.132 = -2500 + 1327.43 + 1331.35 = $158.78.

Or use Excel’s function NPV: =-2500+NPV(0.13,1500,1700) = $158.78 The Excel file, “Problem 2.9.xls” has the equation set-up as a reference to cells that contain the cash flows. 2.10. NPV

=

-12000 5000 5000 -2000 6000 6000 + + + + + 1.12 1.122 1.123 1.124 1.125 1.126

= -10,714.29 + 3985.97 + 3558.90 - 1271.04 + 3404.56 + 3039.79 = $2003.90 Using Excel’s NPV function: =NPV(0.12,-12000,5000, 5000,-2000,6000,6000) = $2,003.90 The internal rate of return (IRR) for this cash flow is approximately 19.2%. The Excel file, “Problem 2.10.xls” has the equation set-up as a reference to cells that contain the cash flows. 2.11.

If the annual rate = 10%, then the monthly (periodic) rate r  = 10% / 12 = 0.83%.

 NPV(0.83%)

= -1000 +

90 90 90 + + ... + 2 1.0083 1.008312 1.0083

= $23.71. Or use Excel’s NPV function, assume the 12 payments of $90 appear in cells B13:B24: =-1000+NPV(0.1/12,B13:B24)= $23.71 (As shown in the Excel file “Problem 2.11.xls”) If the annual rate = 20%, then the monthly (periodic) rate r  = 20% / 12 = 1.67%.  NPV(1.67%)

= -1000 +

90 90 90 + + ... + = $-28.44. 1.0167 1.01672 1.016712

10

Or use Excel’s NPV function, assume the 12 payments of $90 appear in cells B13:B24: =-1000+NPV(0.2/12,B13:B24)= $-28.44 (As shown in the Excel file “Problem 2.11.xls”) The annual interest rate (IRR) that gives NPV=0 is approximately 14.45%. You can verify this result by substituting 14.45% / 12 = 1.20% for r  in the calculations above. Or with Excel’s IRR function, IRR(Values, Guess), assume the series of payments (the initial $1000  payment and the series of 12 payments of $90) are in cells B12:B24: =IRR(B12:B24,0) = 1.20% (As shown in the Excel file “Problem 2.11.xls”) 2.12. a. If the annual rate = 10%, then the monthly rate r  = 10%/12 = 0.83%. Always match the periodicity of the rate to that of the payments or cash flows.

 NPV(Terry) = 600 +

-55 -55 -55 + + ... + 1.0083 2 1.0083 1.008312

= $-25.60. Be sure to get the orientation correct. For Terry, the loan is a positive cash flow, and the payments are negative cash flows (outflows). Thus, the NPV is negative. Because of the negative NPV, Terry should know that this deal is not in his favor and that the actual interest rate being charged is not 10% annually. If it were, then NPV should equal zero. The actual annual interest being charged must be greater than 10% as  NPV is less than zero. Or with Excel’s NPV function, assume the series of 12 payments of $55 are in cells B12:B23. =NPV(0.1/12,B12:B23)+600 = -$25.60 These calculations and those associated with the remaining parts of the question are shown in the Excel file “Problem 2.12.xls”. b. For the manager, the $600 loan is a negative cash flow, and the payments are positive cash flows. Hence,

 NPV(Mgr)

= -600 +

55 55 55 + + ... + 1.0083 1.00832 1.008312

= $25.60.

Or with Excel’s NPV function, assume the series of 12 receipts of $55 are in cells B12:B23. =NPV(0.1/12,B12:B23)-600 = $25.60 c. If the annual rate is 18%, then NPV is about $-0.08. In other words, the actual rate on this loan (the internal rate of return or IRR) is just under 18%.

Using Excel’s IRR function, and assuming the cash flows are in cells B11:B23:

11

=IRR(B11:B23,0)*12 = 17.97% annually 2.13. Should future decisions ever be treated as u ncertain events? Under some circumstances, this may not  be unreasonable.

If the node for selling the car is included at all, then possible consequences must be considered. For example, the consequence would be the price obtained if he decides to sell, whereas if he keeps the car, the consequence would be the length of the car’s life and cost to maintain and repair it. If the node is a decision node, the requisite model would have to identify the essential events and information prior to the decision. If the node is a chance event, this amounts to collapsing the model, and hence may be useful in a first-cut analysis of a complicated problem. It w ould be necessary to think about scenarios that would lead to selling the car or not, and to evaluate the uncertainty surrounding each scenario. 2.14. Vijay’s objectives include maximizing profit, minimizing unsavory behavior, minimizing legal costs, and maximizing Rising Moon’s appeal. Students will think of other objectives. Vijay’s decision is to apply for a liquor license, and if granted, then he could decide on how to manage drinking at Rising Moon. For example, he might be able to create a separate area of his place, such as a beer garden, where drinking alcohol is allowed. Vijay could also decide to broaden his menu in other ways than serving alcohol. The uncertainties include future sales and profit for Risin g Moon, market reaction to offering alcohol, amount of disruption occurring from serving alcohol, and legal liabilities. Consequence measures for sales, profit, and legal costs are clear. He could simply count the number of disruptions to the business due to alcohol or he could try to associate a cost figure to the unsavory behavior. Rising Moon’s appeal could be measured  by the change in sales volume due to introducing alcohol.

Vijay will certainly, as law requires, hedge by carrying insurance, and he will want to think carefully about the level of insurance. As mentioned, he might be able to have a designated area for drinking alcohol. He could gather information now via surveys or speaking to other local merchants. And he can always change his mind later and stop serving alcohol. Case Study: The Value of Patience The Excel solution for this case is provided in the file “Value of Patience case.xlsx”. 1.

NPV

= -385,000 +

100,000 1.18

+

100,000 1.18 2

+ ... +

100,000 1.18 7

= $-3847.

Thus, Union should not accept the project because the NPV is negative. Using Excel’s NPV function and assuming the series of 7 payments of $100,000 are in cells B12:B18: =-385000+NPV(0.18,B12:B18) = -$3847 2.

NPV

= -231,000 +

50,000 1.10

+

50,000 1.10

2

+ ... +

50,000 1.10 7

= $12,421.

This portion of the project is acceptable to Briggs because it has a positive NPV. Using Excel’s NPV function and assuming the series of 7 payments of $50,000 are in cells E12:E18: = -231,000+NPV(0.1,E12:E18) = $12,421

12

3.

NPV

= -154,000 +

50,000 1.18

+

50,000 1.18

2

+ ... +

50,000 1.18

7

= $36,576.

Thus, this portion of the project is profitable to Union. Using Excel’s NPV function and assuming the series of 7 payments of $50,000 are in cells H12:H18: = -154,000+NPV(0.18,H12:H18) = $36,576 Some students will want to consider the other $231,000 that Union was considering investing as part of the entire project. Note, however, that if Union invests this money at their 18% rate, the NPV for that particular investment would be zero. Thus the NPV for the entire $385,000 would be the sum of the two NPVs, or $36,576. 4. Patience usually refers to a willingness to wait. Briggs, with the lower interest rate, is willing to wait longer than Union to be paid back. The higher interest rate for Union can be thought of as an indication of impatience; Union needs to be paid back sooner than Briggs.

The uneven split they have engineered exploits this difference between the two parties. For Briggs, a  payment of $50,000 per year is adequate for the initial investment of $231,000. On the other hand, the less  patient Union invests less ($154,000) and so the $50,000 per year is satisfactory. As an alternative arrangement, suppose that the two parties arrange to split the annual payments in such a way that Union gets more money early, and Briggs gets more later. For example, suppose each invests half, or $192,500. Union gets $100,000 per year for years 1-3, and Briggs gets $100,000 per year for years 4-7. This arrangement provides a positive NPV for each side: NPV(Union) = $24,927, NPV(Briggs) = $45,657. Briggs really is more patient than Union! Case Study: Early Bird, Inc. 1. The stated objective is to gain market share by the end of this time. Other objectives might be to maximize profit (perhaps appropriate in a broader strategic context) or to enhance its public image. 2. Early Bird’s planning horizon must be at least through the end of the current promotion. In a first-cut analysis, the planning horizon might be set at the end of the promotion plus two months (to evaluate how sales, profits, and market share stabilize after th e promotion is over). If another promotion is being planned, it may be appropriate to consider how the outcome of the current situation could affect the next promotion decision. 3, 4. Customer response

 New Morning’s reaction

Move up promotion start date?

Reaction to  New Morning

 Now

13

Market Share or Profits

CHAPTER 3 Structuring Decisions Notes Chapters 3 and 4 might be considered the heart of Making Hard Decisions with DecisionTools. Here is where most of the action happens. Chapter 3 describes the process of structuring objectives and buildi ng influence diagrams and decision trees. Chapter 4 di scusses analysis. The chapter begins with a comprehensive discussion of value structuring and incorporates value-focused thinking throughout. Constructing influence diagrams and decision trees to reflect multiple objectives is demonstrated, and the chapter contains a discussion of scales for measuring achievement of fundamental objectives, including how to construct scales for objectives wi th no natural measures.

Understanding one’s objectives in a decision context is a crucial step in modeling the decision. This section of the chapter shows how to identify and structure values, with an important emphasis on distinguishing  between fundamental and means objectives and creating hierarchies and networks, respectively, to represent these. The fundamental objectives are the main reasons for caring about a decision in the first  place, and so they play a large role in subsequent modeling with influence diagrams or decision trees. Students can generally grasp the concepts of influence diagrams and how to interpret them. Creating influence diagrams, on the other hand, seems to be much more difficult. Thus, in teaching students how to create an influence diagram for a specific situation, we stress basic influence diagrams, in particular the  basic risky decision and imperfect information. Students should be able to identify these basic forms and modify them to match specific problems. The problems at the end of the chapter range from simple identification of basic forms to construction of diagrams that are fairly complex. The discussion of decision trees is straightforward, and many students have already been exposed to decision trees somewhere in their academic careers. Again, a useful strategy seems to be to stress some of the basic forms. Also discussed in Chapter 3 is the matter of including in the decision model appropriate details. One issue is the inclusion of probabilities and payoffs. More crucial is the clarity test and the development of scales for measuring fundamental objectives. The matter of clarifying definitions of alternatives, outcomes, and consequences is absolutely crucial in real-world problems. Th e clarity test forces us to define all aspects of a problem with great care. The advantage in the classroom of religiously applying the clarity test is that the  problems one addresses obtain much more realism and relevance for the students. It is very easy to be lazy and gloss over definitional issues in working through a problem (e.g., “Let’s assume that the market could go either up or down”). If care is taken to define events to pass the clarity test (“The market goes u p means that the Standard & Poor’s 500 Index rises”), problems become more realistic and engaging. The last section in Chapter 3 describes in detail how to use PrecisionTree for structuring decisions. The instructions are intended to be a self-contained tutorial on constructing decision t rees and influence diagrams. PrecisionTree does have an interactive video tutorial along with video tutorials on the basics and videos from experts. These videos along with example spreadsheets and the manual all can be found in the PrecisionTree menu ribbon under Help, then choosing W elcome to PrecisionTree. Please note that if you enter probability values that do not sum to 100% for a chance node, then the  program uses normalized probability values. For example, if a chance node has two branches and the corresponding probabilities entered are (10%, 80%), then the model will use (11.11%, 88.89%) – i.e., 0.1/(0.1+0.8) and 0.8/(0.1+0.8). Because this chapter focuses on structuring the decision, many of the problems do not have all of the numbers required to complete the model. In some cases, the spreadsheet solution provides the structure of the problem only, and the formulas were deleted (for example, problem 3.10). In other cases, the model is

14

completed with representative numbers (for example, problem 3.5). In the completed model, you will see expected values and standard deviations in the d ecision trees and influence diagrams. These topics are discussed in Chapter 4. Topical cross-reference for problems Branch pay-off formula Calculation nodes Clarity test Constructed scales Convert to tree Decision trees

3.25 – 3.28 3.14, Prescribed Fire 3.4, 3.5, 3.12, 3.21, Prescribed Fire 3.3, 3.14, 3.20, 3.21 3.11, 3.26, 3.28, Prescribed Fire 3.5 - 3.7, 3.6 - 3.11, 3.13, 3.20 - 3.28, Prescribed Fire, S.S. Kuniang, Hillblom Estate 3.9, 3.11 3.4, 3.6 - 3.9, 3.11, 3.14, 3.16, 3.20, 3.21, 3.26, 3.28, Prescribed Fire, S.S. Kuniang 3.24, 3.25 3.1 - 3.3, 3.7, 3.10, 3.14 - 3.19, 3.21, 3.23, Prescribed Fire 3.5, 3.9, 3.24 - 3.28, Prescribed Fire, S.S. Kuniang 3.20 3.9

Imperfect information Influence diagrams  Net present value Objectives PrecisionTree Sensitivity analysis Umbrella problem

Solutions 3.1. Fundamental objectives are the essential reasons we care about a decision, whereas means objectives are things we care about because they help us achieve the fundamental objectives. In the automotive safety example, maximizing seat-belt use is a means objective because it helps to achieve the fundamental objectives of minimizing lives lost and injuries. We try to measure achievement of fundamental objectives  because we want to know how a consequence “stacks up” in terms of the things we care about.

Separating means objectives from fundamental objectives is important in Chapter 3 if only to be sure that we are clear on the fundamental objectives, so that we know what to measure. In Chapter 6 we will see that the means-objectives network is fertile ground for creating new alternatives. 3.2. Answers will vary because different indivi duals have different objectives. Here is one possibilit y. (Means objectives are indicated by italics.)

Best Apartment

Minimize Travel time

To School To Shopping

Maximize Ambiance

To Leisure-time Activities

Maximize Use of leisure time

Alone Friends Neighbors

 Maximize  Discretionary $$ Centrally located 

 Maximize  features (e.g.,  pool, sauna, laundry)

Parking at apartment   Maximize windows, light 

15

 Minimize  Rent 

3.3. A constructed scale for “ambiance” might be the following:

Best

Many large windows. Unit is like new. Entrance and landscape are clean and inviting with many plants and open areas. -Unit has excellent light into living areas, but bedrooms are poorly lit. Unit is clean and maintained, but there is some evidence of wear. Entrance and landscaping includes some  plants and usable open areas but is not luxurious. -Unit has one large window that admits sufficient light to living room. Unit is reasonably clean; a few defects in walls, woodwork, floors. Entrance is not inviting but does appear safe. Landscaping is adequate with a few plants. Minimal open areas. -Unit has at least one window per room, but the windows are small. Considerable wear. Entrance is dark. Landscaping is poor; few plants, and small open areas are not inviting. Worst Unit has few windows, is not especially clean. Carpet has stains, woodwork and walls are marred. Entrance is dark and dreary, appears unsafe. Landscaping is poor or nonexistent; no plants, no usable open areas. 3.4. It is reasonable in this situation to assume that the bank’s objective is to maximize its profit on the loan, although there could be other objectives such as serving a particular clientele or gaining market share. The main risk is whether the borrower will default on the loan, and the credit report serves as imperfect information. Assuming that profit is the only objective, a simple influence diagram would be:

Credit report

Default?

Make Loan?

Profit

 Note the node labeled “Default” Some students may be tempted to call this node something like “Credit worthy?” In fact, though, what matters to the bank is whether the money is paid back or not. A more  precise analysis would require the banker to consider the probability distribution for the amount paid back (perhaps calculated as NPV for various possible cash flows). Another question is whether the arrow from “Default” to “Credit Report” might not be better going the other way. On one hand, it might be easier to think about the probability of default given a particular credit report. But it might be more difficult to make judgments about the likelihood of a particular report without conditioning first on whether the borrower defaults. Also, note that the “Credit Report” node will probably have as its outcome some kind of summary measure  based on many credit characteristics reported by the credit bureau. It might have something like ratings that  bonds receive (AAA, AA, A, and so on). Arriving at a summary measure that passes the clarity test could  be difficult and certainly would be an important aspect of the problem. If the diagram above seems incomplete, a “Credit worthiness” node could be included and connected to  both “Credit report” and “Default”:

16

Credit worthiness

Credit report

Default?

Make Loan?

Profit

Both of these alternative influence diagrams are shown in the Excel file “Problem 3.4.xlsx.” Two different types of arcs are used in the diagrams: 1) value only and 2) value and timing, and these are explained in the text. A value influence type influences the payoff calculation and a timing type exists if the outcome  precedes that calculation chronologically (or is known prior to the event). 3.5. This is a range-of-risk dilemma. Important components of profit include all of the different costs and revenue, especially box-office receipts, royalties, licensing fees, foreign rights, and so on. Furthermore, the definition of profits to pass the clarity test would require specification of a planning horizon. At the specified time in the future, all costs and revenues would be combined to calculate the movie’s profits. In its simplest form, the decision tree would be as drawn below. Of course, other pertinent chance nodes could  be included.

Revenue

Make movie

Don't make m ovie

Profit = Revenu e-Cost

Valu e of best alternative

The revenue for the movie is drawn as a continuous uncertainty node in the above decision tree. Continuous distributions can be handled two ways in PrecisionTree either with a discrete approximation (see Chapter 8 in the text) or with simulation (see Chapter 11 in the text). This decision tree with a discrete approximation of some sample revenue values is shown in the Excel file “Problem 3.5.xlsx.” A potentially useful exercise is to have the students alter the sample values to see the effect on the model and specifically the preferred alternative. 3.6. Strengths

Weaknesses

 Influence Diagrams Compact Good for communication Good for overview of large problems

 Decision Trees Display details Flexible representation

Details suppressed

Become very messy for large problems

17

Which should he use? I would certainly use influence diagrams first to present an overview. If details must  be discussed, a decision tree may work well for that. 3.7. This problem can be handled well with a simple decision tree and consequence matrix. See Chapter 4 for a discussion of symmetry in decision problems.

Best Representation Max Communication Overview Influence Diagram

Decision Tree

Details Details hidden

Excellent for large problems

Poor, due to complexity

Details displayed 

Max flexibility Best for symmetric decision problems

Very flexible for assymetric decisions

3.8.

Win Senate election?

Run for Senate? Run for Senate Run for House

Yes  No

Outcome Run for Senate House

Win Senate? Yes  No Yes  No

Outcome US Senator Lawyer US Representative US Representative

 Note that the outcome of the “Win Senate” event is vacuous if the decision is made to run for the house. Some students will want to include an arc from the decision to the chance node on the grounds that the chance of winning the election depends on the choice made:

18

Win election?

Run for Senate?

Win Lose

Outcome Run for Senate

Run for Senate Run for House

Win Election? Outcome Yes US Senator  No Lawyer Yes US Representative

House

 Note that it is not possible to lose the House election.

The arc is only to capture the asymmetry of the problem. To model asymmetries in an influence diagram, PrecisionTree uses structure arcs. When a structural influence is desired, it is necessary to specify how the  predecessor node will affect the structure of the outcomes from the successor node. By using a structure arc, if the decision is made to run for the house, the “Win election?” node is skipped. This influence diagram is shown in the Excel file “Problem 3.8.xlsx.” 3.9. (Thanks to David Braden for this solution.) The following answers are based on the interpretation that the suit will be ruined if it rains. They are a good first pass at the problem structure (but see below).

suit not ruin ed, plus a sense of relief  rain

no rain take umbrella suit not ruin ed, but inconvenience incurred  suit ruined  do n ot take umbrella rain

no rain suit not ruined 

(A) Decision tree

19

Weather  Forecast

Rain

Take Umbrella?

Take Umbrella?

Satisfaction

(B) Basic Risky Decision

Rain

Satisfaction

(C) Imperfect Information

The Excel solution “Problem 3.9.xlsx” shows a realization of this problem assuming the cost of the suit is $200, the cost of the inconvenience of carrying an umbrella when it is not raining is $20, the probability of rain is 0.25, and the weather forecaster is 90% accurate.  Note that the wording of the problem indicates that the suit may be ruined if it rains. For example, the degree of damage probably depends on the amount  of rain that hits the suit, which is itself uncertain! The following diagrams capture this uncertainty. suit no t ruined,  plus a sen se of relief  rain no rain suit not ruined, but inconvenience incurred 

take umbrella suit ruined 

suit ruined 

do not take umbrella suit not ruined  rain no rain

suit not ruined, but s ome effort sp ent to avoid ruin ing the suit

suit not ruined 

(A) Decision tree

20

Weather  Forecast

Rain

Take Umbrella?

Take Umbrella?

Ruin Suit

Satisfaction

Rain

Satisfaction

Ruin Suit

(C) Imperfect Information

(B) Basic Risky Decision 3.10. (Thanks to David Braden for this solutio n.)

The decision alternatives are (1) use the low -sodium saline solution, and (2) don’t use the low-sodium saline solution. The uncertain variables are: (1) The effect of the saline solution, consequences for which are patient survival or death; (2) Possibility of court-martial if the saline solution is used and the patent dies. The possible consequences are court-martial or no court-martial. The decision tree:  pat ient dead 

do not use saline solution  pat ien t saved 

 pat ient su rvi ves an d th e use of saline solu tion justi fied for ot her patients

use saline solution

 pat ient dead an d do ctors suffer 

 pat ient dead 

court-martial no cou rt-martial

 pat ient dead 

This decision tree is drawn in the Excel file “Problem 3.10.xlsx.” 3.11. a.

Sunny

Best

Weather  Rainy

Outdoors

Indoors Party decision

Terrible

Good 

Satisfaction  No party

21

Bad 

This influence diagram is drawn in the Excel file “Problem 3.11.xlsx” with some sample values assumed (on a utility scale, a sunny party outside is worth 100, an indoors party is worth 80, no party is worth 20, a  party outside in the rain is worth 0, and the probability of rain is 0.3). A structure only arc is added in the file between party decision and weather to include the asymmetries to skip the weather uncertainty if the decision is made to have no party or have one indoors. The second worksheet in the file shows the default decision tree created by the “Convert to Tree” button on the influence diagram settings dialog box. (Click on the name of the influence diagram “Problem 3.11a” to access the influence diagram settings. The Convert to Decision Tree button creates a decision tree from the current influence diagram. This can be used to check the model specified by an influence diagram to insure that the specified relationships and chronological ordering of nodes are correct. Conversion to a decision tree also shows the impacts of any Bayesian revisions made between nodes in the influence diagram. Once a model described with an influence diagram is converted to decision tree, it may be further edited and enhanced in decision tree format. However, any edits made to the model in decision tree format will not be reflected in the original influence diagram. b. The arrow points from “Weather” to “Forecast” because we can easily think about the chances associated with the weather and then the chances associated with the forecast, given the weather. That is, if the weather really will be sunny, what are the chances that the forecaster will predict sunny weather? (Of course, it is also possible to draw the arrow in the other direction. However, doing so suggests that it is easy to assess the chances associated with the different forecasts, regardless of the weather. Such an assessment can be hard to make, though; most people find the former approach easier to deal with.)

Forecast

Weather 

Party decision

Satisfaction

22

Sunny Rainy

Outdoors

Indoors Forecast = “Sunny”

No party

Rainy

Outdoors

Terrible

Good

Sunny

Forecast = “Rainy”

Best

Indoors

No party

Bad

Best Terrible

Good

Bad

The influence diagram including the weather forecast is shown in the third worksheet and the associated default decision tree created by the “Convert to Tree” function is shown in the fourth worksheet. Additionally, we assumed that the weather forecaster is 90% accurate. 3.12. The outcome “Cloudy,” defined as fully overcast and no blue sky, might be a useful distinction,  because such an evening outdoors would not be as nice for most parties as a partly-cloudy sky. Actually, defining “Cloudy” to pass the clarity test is a difficult task. A possible definition is “At least 90% of the sky is cloud-covered for at least 90% of the time.” The NWS definition of rain is probably not as useful as one which would focus on whether the guests are forced indoors. Rain could come as a dreary drizzle, thunderstorm, or a light shower, for instance. The drizzle and the thunderstorm would no doubt force the guests inside, but the shower might not. One possibility would be to create a constructed scale that measures the quality of the weather in terms that are appropriate for the party context. Here are some p ossible levels: (Best) ---(Worst)

Clear or partly cloudy. Light breeze. No precipitation. Cloudy and humid. No precipitation. Thunderclouds. Heavy downpour just before the party. Cloudy and light winds (gusts to 15mph). Showers off and on. Overcast. Heavy continual rain.

3.13.

23

Change product Engineer says “Fix #3 .”

Replace #3

Behind schedule

#3 Defective #3 not defective

Change product

Behin d scedule ,costly

Behind schedule

Engi nee r says “#3 OK.” Replace #3

On schedule, costly

#3 Defective #3 not defective

On schedule, costly Behi nd scedule,costly

This decision tree is drawn in the Excel file “Problem 3.13.xlsx.” 3.14. Forecast

Hurricane Path

Safety Decision

Evacuation Cost

Consequence

 Note that Evacuation Cost is high or low depending only on the evacuation decision. Thus, there is no arc from Hurricane Path to Evacuation Cost. This influence diagram is drawn in the Excel file “Problem 3.14.xlsx.” Because PrecisionTree allows only one payoff node per influence diagram, the “Safety” and “Evaluation Cost” nodes are represented by calculation nodes. A calculation node (represented by a rounded blue rectangle) takes the results from  predecessor nodes and combines them using calculations to generate new values. These nodes can be used to score how each decision either maximizes safety or minimizes cost. The constructed scale for safety is meant to describe conditions during a hurricane. Issues that should be considered are winds, rain, waves due to the hurricane’s storm surge, and especially the damage to  buildings that these conditions can create. Here is a possible constructed scale: (Best) --

Windy, heavy rain, and high waves, but little or no damage to property or infrastructure. After the storm passes there is little to do beyond cleaning up a little debris. Rain causes minor flooding. Isolated instances of property damage, primarily due to window breakage. For people who stay inside a strong building during the storm, risk is minimal. Brief interruption of power service.

24

--

Flooding due to rain and storm surge. Buildings within 100 feet of shore sustain heavy damage. Wind strong enough to break many windows, but structural collapse rarely occurs. Power service interrupted for at least a day following the storm. -Flooding of roads and neighborhoods in the storm’s path causes areas with high property damage. Many roofs are severely damaged, and several poorly constructed buildings collapse altogether. Both electrical power and water service are interrupted for at least a day following the storm. (Worst) Winds destroy many roofs and buildings, putting occupants at high risk of injury or death. Extensive flooding in the storm’s path. Water and electrical service are interrupted for several days after the storm. Structural collapse of older wood-frame buildings occurs throughout the region, putting occupants at high risk. 3.15. Answers to this question will depend largely on individual preferences, although there are some typical responses. Some fundamental objectives: improve one’s quality of life by making better decisions, help others make better decisions, improve career, graduate from college, improve one’s GPA (for one w ho is motivated by grades). Some means objectives: satisfy a requirement for a major, improve a GPA (to have  better job opportunities), satisfy a prerequisite for another course or for graduate school. Note that “making  better decisions” is itself best viewed as a means objective because it can provide ways to improve one’s life. Only a very few people (academics and textbook writers, for example), would find the study of decision analysis to be its own reward!

The second set of questions relates to the design of the course and whether it is possible to modify the course so that it can better suit the student’s objectives. Although I do not want to promote classroom chaos, this is a valuable exercise for both students and instructor to go through together. (And the earlier in the term, the better!) Look at the means objectives, and try to elaborate the means objectives as much as  possible. For example, if a class is primarily taking the course to satisfy a major requirement, it might make sense to find ways to make the course relate as much as possible to the students’ major fields of study. 3.16. While individual students will have their own fundamental objectives, we based our hierarchy on a study titled  "Why do people use Facebook?" (Personality and Individual Differences, Volume 52, Issue 3, February 2012, Pages 243–249). The authors, Ashwini Nadkarni and Stefan G. Hofmann from Boston University, propose that Facebook meets two primary human needs: (1) the need to belong and (2) the need for self-presentation. Thus, a particular student’s objectives may be a variation on these themes. Students may find it interesting to see how their fundamental o bjectives for their own Facebook page compare and contrast to the study.

The fact that one’s private area can sometimes be seen by outsiders, particularly employers or future employers, interferes with presenting your private self. If the private area were truly private, then a user could be more truthful discussing their revelries and celebrations or even their private thoughts. If you  believe someone is eavesdropping on your conversation, then you are naturally more guarded with your speech. Thus, Facebook does not provide a good means to expressing your private self.

Maximize Socialization

Maximize Need for Belonging

Maximize Self Esteem

Maximize Self Worth

Maximize Need for Self Presentation

Minimize Loneliness

25

Maximize Public Persona

Maximize Private Persona

Clearly, each individual will have their own FOH, but there are some facts that students may want to take into consideration. First, it is naïve to think that future employers will not “Google” or “Facebook” you. Such concerns are not usually on the mind of a college freshman or particularly a high-school student, but there are a host of problems that can arise from indiscriminately sharing private information. Even if there is no illegal activity being shown (underage drinking, drug use, etc.), different audiences will have different norms, and pictures of drinking, dancing, and partying could be considered compromising or unprofessional. Second, a future employer or even your current employer may be interested in your postings. They may want to know what religion you practice, what your interests are, what organizations you belong to, such as the NRA. All of these could bias them, good or bad, towards you. Also, discretion might be an important aspect of your position, and employers might view your postings to determine if you can be trusted with  proprietary information. Posted Information can also be used by competing firms either to gain a direct benefit or more nefariously to befriend you, and thereby learn more about their competitor. Your personal profile may include job details and thus provide an opening by unscrupulous ‘cyber sharks’, or by competing businesses hoping to learn from the eyes and ears of the opposition. You may even have posted confidential information without realizing its sensitive nature. Facebook, Inc. itself has rights to your private information and has wide latitude on how it can use your info. Facebook’s privacy agreement states "We may use information about you that we collect from other sources, including but not limited to newspapers and Internet sources such as blogs, instant messaging services, Facebook Platform developers and other users of Facebook, to supplement your profile.” Facebook can also sell a user's data to private companies, stating: "We may share your information with third parties, including responsible companies with which we have a relationship." There are also many data mining and identify theft issues that could result from even the public areas of your Facebook page. Searches can be performed to discover what movi es you like, what music you listen to, what organizations you belong to, etc. This information can be used for marketing, recruiting, or even identity theft. Finally, a student’s short-term and long-term objectives may differ. Short term, the focus will most likely  be on being connected and being cool, i.e., partying, sexual prowess, being wild-n-crazy, etc. Many highschool and college students compete using Facebook t o see who can have the most friends. Having 500 to 1000 Facebook friends is actually common. F or these users, their objectives are to attract their friend’s friends and present a profile attractive to large swaths of the population. Long term, the focus will most likely be on being professional and staying connected to their college friends. 3.17. Here are my (Clemen’s) fundamental-objectives hierarchy and means-objective network (itali cs) in the context of purchasing or building a telescope. The diagram does provide insight! For example, many astronomers focus (so to speak) on image quality, and so there is a tendency to overemphasize aperture and quality of eyepieces. But for me, an important issue is enjoying myself as much as possible, and that can mean taking the telescope out of the city. All of the means objectives related to “Maximize enjoyment of viewing sessions” (which is intended to capture aspects other than enjoying the quality of the images viewed) push me toward a smaller, lighter, less expensive telescope. Thus, it is arguable that the basic question I must ask is whether I just want to get out at night and enjoy seeing a few interesting sights, or whether my interest really is in seeing very faint objects with as much clarity as possible.

Of course, the creative solution would be to find an inexpensive and highly transportable telescope with large aperture, excellent optics, and very stable mount. Unfortunately, such a telescope doesn’t exist; all of the desired physical features would lead to a very expensive telescope!

26

Best T elescope

Maximize image clarity Maximize  brightness

 Max aperture

Maximize image defnition

 Max quality of optics

Maximize enjoyment of viewing sessions

Maximize quality of astrophotography

Maximize stability

 Max quality of tracking device

 Max stability of mount 

 Max additional viewing accessories  Min light  pollution in sky

 Max visits to dark-sky site

 Min cost of telescope  Min total weight 

 Max transportability

 Minimize telescope size

27

3.18. Objectives are, of course, a matter of personal preferences, and so answers will vary considerably. a. Here is an objectives hierarchy for the decision context of going out to dinner: Time

Driv ing time

Minimize cost

Ordering/serv ing

Dollar cost  Atmosphere Maximize experience

Location Menu Quality of f ood

b. A simple hierarchy for deciding from among d ifferent trips: Company (f riends) Maximize experience

Learning Relax

Preparation Minimize cost

Time Expense

Trav el cost

Some means objectives might be going to a particular kind of resort; maximizing time spent shopping, golfing, or on the beach; maximizing nights spent in youth hostels; using a travel agent (to reduce time spent in preparation); maximizing time in a foreign country (to maximize language learning, for example). c. Here is a possible fundamental-objectives hierarchy for choosing a child’s name: Similarity of name Maximize f amily ties

Closeness of namesake by child

Ease of use/learning

by child’s play mates by relativ es Nicknames

Minimize negativ e potential Teasing by “f riends”

3.19. Responding to this question requires considerable introspection and can be very troubling for many students. At the same time, it can be very enlightening. I have had students face up to these issues in

28

analyzing important personal decisions such as where to relocate after graduation, whether to make (or accept) a marriage proposal, or whether to have child ren. The question, “What is important in my life?” must be asked and, if answered clearly, can provide the individual with important insight and guidance. 3.20. a. These influence diagrams and decision trees are drawn in the Excel file “Problem 3.20.xlsx.”

Surgery results

Recover fully

Have surgery

Die Have surgery?

Quality of life

Don’t have surgery

Long, healthy life

Death

Progressive debilitation

b.

Surgery results

Quality of life

Have surgery?

Recover fully Have surgery

Complications

Long, healthy life

Complications Die

Don’t have surgery

Death

Full recovery

Long healthy life afte difficult treatment

Partial recovery

Invalid after difficult treatment

Death

Progressive debilitation

Death after difficult treatment

Given the possibility of complications and eventual consequences, the surgery looks considerably l ess appealing. c. Defining this scale is a personal matter, but it must capture important aspects of what life would be like in case complications arise. Here is one possible scale:

29

(Best) --

No complications. Normal, healthy life. Slight complications lead to minor health annoyances, need for medication, frequent visits. Little or no pain experienced. Able to engage in most age-appropriate activities. -Recovery from surgery requires more than two weeks of convalescence. Pain is intense  but intermittent. Need for medication is constant after recovery. Unable to engage in all age-appropriate activities. -Recovery requires over a month. Chronic pain and constant need for medication. Confined to wheelchair 50% of the time. (Worst) Complete invalid for remainder of life. Restricted to bed and wheelchair. Constant pain, sometimes intense. Medication schedule complicated and occasionally overwhelming. 3.21. This question follows up on the personal decision situation that was identified in problem 1.9. 3.22 This decision tree is drawn in the Excel file “Problem 3.22.xlsx.”

“Not to be” (comm it suicid e)

“What dream s may come ” (What comes after death?)

“To be” “Bear fardels” (continue to li ve) (burdens of li fe)

3.23. This decision tree is drawn in the Excel file “Problem 3.23.xlsx.”

 Ai rcraft hostile

Crew safe

Shoot  Ai rcraft not hostile  Ai rcraft ho sti le Don't shoo t  Ai rcraft no t hostile

Crew safe Civilia ns kil led Harm to crew Crew safe Civilians safe

Rogers’s most crucial objectives in this situation are to save lives, those of his crew and of any civilians who are not involved. It is not unreasonable to consider objectives of saving his ship or improving the relationship with Iran, but in the heat of action, these were probably not high on Rogers’ list. The risk that Rogers faces is that the blip on the radar screen may not represent a hostile aircraft. The main trade-off, of course, is the risk to his crew versus possibly killing innocent civilians. As usual, there are a lot of ways this decision tree could be made more elaborate. For example, a “Wait” alternative might be included. The tree above assumes that if the decision is to shoot, the incoming aircraft would be hit (generally a safe assumption these days), but one might want to include the possibility of missing.

30

3.24. This is a straightforward calculation of NPV. Assuming that all the cash flows happen at the end of the year, the following table shows the cash flows: Cash Flows

Year 0 1 2 3 4 5 6 7 8 9

Stop

Continue  No Patent

Continue Patent License

0 0 0 0 0 0 0 0 0 0

0 -2 0 0 0 0 0 0 0 0

0 -2 0 5 5 5 5 5 0 0

Continue Continue Continue Patent Patent Patent Develop Develop Develop Dem. High Dem. Med Dem. Low 0 0 0 -2 -2 -2 0 0 0 -5 -5 -5 -5 -5 -5 11 6.6 3 11 6.6 3 11 6.6 3 11 6.6 3 11 6.6 3

Present values are calculated by applying the appropriate discount rate to each cash flow; the discount rate 1 is for the cash flows in year i. Finally, NPV is the sum of the present values. Also, the NPV function 1.15i in Excel can be used for the calculations as shown in the Excel file “Problem 3.24.xls.”

Present Values

Year 0 1 2 3 4 5 6 7 8 9 NPV

Discount Factor 1 0.8696 0.7561 0.6575 0.5718 0.4972 0.4323 0.3759 0.3269 0.2843

Stop

Continue  No Patent

Continue Patent License

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 -1.74 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 -1.74 0.00 3.29 2.86 2.49 2.16 1.88 0.00 0.00

0.00

-1.74

10.93

Continue Continue Patent Patent Develop Develop Dem. High Dem. Med 0.00 0.00 -1.74 -1.74 0.00 0.00 -3.29 -3.29 -2.86 -2.86 5.47 3.28 4.76 2.85 4.14 2.48 3.60 2.16 3.13 1.88 13.20

4.76

Continue Patent Develop Dem. Low 0.00 -1.74 0.00 -3.29 -2.86 1.49 1.30 1.13 0.98 0.85 -2.14

In file “Problem 3.24.xlsx”, the decision tree references the NPV calculations to d emonstrate the process of choosing to continue or stop development. The ability to build these trees in Excel and reference cells as done in this problem makes this a powerful program. The payoff for each branch of the tree is a formula that corresponds to the correct cell in the NPV calculations worksheet. Alternative assumptions can be made about the timing of the cash flows. For example, it would not be unreasonable to believe that the expenses must be paid at the beginning of the year and that revenue arrives at the end of the year. The most realistic scenario, however, is that all cash flows are evenly spread out over the year for which they are specified.

31

3.25. This decision tree is relatively complex compared to the ones that we have seen so far. Buying the new car does not involve any risk. However, the used car has an uncertainty each year for t he next three years. The decision tree is shown below. Note that it is also possible to calculate the NPVs for the ends of the branches; the natural interest rate to use would be 10%, although it would be best to use a rate that reflects what you could earn in another investment. This decision tree representation does not discount the values.

Down Payment New Car  -5500

Purchase

Maintenance and loan payments Year 1 Year 2 Year 3 -2522.20

-2722.20

Repairs Year 1

-650 (0.2)

Used Car  -5500

-1650 (0.6)

-2650 (0.2)

32

-2722.20

Net Salvage Value 3626

-9841

Repairs Year 2

Repairs Year 3

Net Salvage Value

-700 (0.2)

-500 (0.2) -1500 (0.6) -2500 (0.2)

2000 2000 2000

-5350 -6350 -7350

-1700 (0.6)

-500 (0.2) 2000 -1500 (0.6) 2000 -2500 (0.2) 2000

-6350 -7350 -8350

-2700 (0.2)

-500 (0.2) 2000 -1500 (0.6) 2000 -2500 (0.2) 2000

-7350 -8350 -9350

-700 (0.2)

-500 (0.2) 2000 -1500 (0.6) 2000 -2500 (0.2) 2000

-6350 -7350 -8350

-1700 (0.6)

-500 (0.2) 2000 -1500 (0.6) 2000 -2500 (0.2) 2000

-7350 -8350 -9350

-2700 (0.2)

-500 (0.2) 2000 -1500 (0.6) 2000 -2500 (0.2) 2000

-8350 -9350 -10350

-700 (0.2)

-500 (0.2) 2000 -1500 (0.6) 2000 -2500 (0.2) 2000

-7350 -8350 -9350

-1700 (0.6)

-500 (0.2) 2000 -1500 (0.6) 2000 -2500 (0.2) 2000

-8350 -9350 -10350

-2700 (0.2)

-500 (0.2) 2000 -1500 (0.6) 2000 -2500 (0.2) 2000

-9350 -10350 -11350

The Excel file “Problem 3.25.xlsx” contains two solutions for this problem. The workbook consists of 3 worksheets. The first worksheet is titled Data & Formulas, and contains the input data for the problem (car costs, loan value, etc.) along with some formulas. The formulas in cells G29 and H29 calculate the 3-year net cost of the new and used car respectively. Remember to include the initial $5,500 payment. The formulas in M12 and L12 also calculate the 3-year net cost but incorporate the time value of money. We used 10% as the interest rate. The next two worksheets show the decision trees, with and without incorporating the time value of money. Both of these trees are linked trees, which we introduce in Chapter 4. There is a method for students to solve the problem not using linked trees, and this is explained in the text box as Option 1. Essentially, students will need to create 27 formulas for 3-year net cost, one for each unique combination of maintenance costs across the 3 years. Once these 27 formulas have been created, it is a simple matter to use Excel referencing to reference the end node with it s corresponding formula. Assigning this problem i n Chapter 3 will help the students realize the flexibility of linked decision trees when they encounter them in Chapter 4. Additional hints might be needed. Finally, this problem can also be used to exhibit good-modeling techniques to the students via separating out the data inputs and placing all of them into one area. Not only does this help with constructing the model, but it also facilitates running a sensitivity analysis. 3.26a. The influence diagram from Exercise 3.11(a) is shown here drawn in PrecisionTree.

Weather 

Party decision Satisfaction

In order for the “Convert To Tree” button to automatically adjust for the asymmetries, a structure arc is needed from “Party Decision” to “Weather” (represented by the dotted arrow). This influence diagram and the corresponding converted decision tree are shown in the first two worksheets in the Excel file “Problem 3.26.xlsx.” Some assumed values for outcomes and probabilities are shown in the converted decision tree. 70.0%

Sunny

0

FALSE

Outdoors

100

Weather   70

0 Rainy Converted Problem 3.26 (a)

0

30.0%

0

0

0

Party decision 80 Indoors

TRUE

1

0

80

FALSE

No party

0

0 20

b. Adding the arrow from “Weather” to “Party Decision” means that the information regarding the weather is known before the time of the decision. Therefore, in the Converted Decision Tree, the “Weather” chance events will appear in the tree prior to the “Party Decision.”

33

Weather 

Party Decision

Satisfaction

Outdoors

0.7

100

100

Party Decision

70.0%

Sunny

TRUE

100

0

FALSE

Indoors

80 No party

FALSE 20

0 80 0 20

Weather 

Converted Problem 3.26 (b)

94 Outdoors

0

0

0

Party Decision

30.0%

Rainy

FALSE

80

0 Indoors

No party

TRUE

0.3

80

80

FALSE 20

0 20

c.

Forecast

Weather 

Party decision

Satisfaction

If the arrow went from "Party" t o "Forecast", then you would have to make the party decision before you got the forecast. If an arrow started at "Forecast" and went to "Weather", we would be stating that somehow the forecast influences the weather.

34

95.5%

Sunny

 

0 Outdoors

TRUE 0

100

Weather    

95.455 4.5%

Rainy

 

0 "Sunny" forecast 66.0% 0

0.03 0

Party decision  

95.455 FALSE

Indoors

0

0 No party

80

FALSE

0

0 Converted Problem 3.26 (c)

0.63

20

Forecast 90.2 20.6%

Sunny

0 Outdoors

FALSE 0

Weather    

20.588 Rainy

"Rainy" forecast

34.0%

0 100

79.4%

0

0

0

Party decision

0

80 Indoors

TRUE

0.34

0 No party

FALSE 0

80 0 20

3.27. This problem is more challenging than some previous ones as it pushes the students to incorporate numerical values into the decision model, and to do so, using formulas. Also, the data were provided in two different formats (annual and monthly) requiring the students t o pick one format for the consequence measure. The solution in the spreadsheet “Problem 3.27.xlsx” uses monthly values. A portion of the tree is shown below.

The formula for Jameson’s monthly take home if he stays the course is $2,000 + (Actual Revenue – VJ’s Cost) x (1 – Tax Rate) Subtracting the required $2,400 is the surplus/deficit Jameson faces, and is the consequence measure we used. If the student uses the revenue values, then he or she will need to subtract $2,400 after the analysis. This only works because the same value is being subtracted from all end nodes. Also, students will run into PrecisionTree adding the values along the branches (cumulative method). For example, using monthly revenue values of $2,300 in year 1 and $2,400 in year 2 has a cumulative value of $4,700. In this case, $4,800 would need to be subtracted. Better is to use the consequence measure that fits the problem. The formula for Jameson’s monthly take home if he pursues an MSW in 2 years is (Annual salary/12) x (1 – Tax Rate) – Loan Payment Again, we subtracted $2,400 for the consequence measure.

35

So what should Jameson do? The future looks bleak for him, but bleaker if he stays the course. Over the next 3 years, we see him having a monthly deficit of $570 on average if he stays the course, but only a $75 deficit if he gets his MSW. Either way, he is falling behind every month, and expenses associated with raising children grow almost as fast as the children do. 3.28. From an income maximization point of view, Susan should not quit. There are no financial advantages of her quitting. As a matter of fact, the only financial ramification is the loss of 6 months of salary or $12,000. It could be that Susan had set her mind on quitting and cannot process not quitting based on something that might only happen. She is anchored to quitting, and is probably excited to start a new life.

Susan is not as concerned with income as she is with running through their savings and being evicted. To model this, we consider all the different scenarios that were presented for Susan and Denzel, and for each calculate what would be left in their savings account. If this ever drops below zero, then they run the risk of eviction. The timeframe for them is clearly 6 months because both Susan and Denzel will be on their feet  by then with their new careers. The solution, shown below and in the file “Problem 3.28.xlsx,” shows the structure of the problem and the values we used for the end nodes. To help understand the formulas, the spreadsheet cells have been named and thus the formulas are of the form: =Savings+6*(Denzel’s Contr if Laid Off + Assistance) - Six_Months_Req. This formula reports their end savings account balance when Denzel is laid off, when Susan is not contributing, and they do receive assistance. See the file for complete details. Please note that the above equals a negative value (-$2,000), but we are not saying the savings account balance can go negative. Rather, this measures by amount of their deficit. Susan should definitely not quit her job at the coffee shop. If she does, there is a 70% chance they will have at least a $2,000 deficit. Any deficit is to be avoided. If she stays working at Joes, then no matter what happens to Denzel, they will have at least $4,000 left in their savings account. To exhibit the iterative natures of modeling, we extended the model by adding a third alternative, namely, staying at the coffee shop for a few months. Cell E7 allows you to enter any value between 0 and 6 for the

36

number of Months Susan stays at Joe’s Coffee. By making this a cell reference, she can dynamically change it value to view intermediate alternatives. For example, if she stays at Joe’s for 3 months, then the  probability of a deficit drops to 8%. Case Study: Prescribed Fire 1. Appropriate objectives for the agency in the context of this decision relate to costs. In particular, they would like to minimize the cost of disposing of the material, and they would like to minimize the cost of containing fires that go out of control. The latter could include damages if a fire escapes. 2. Many influence-diagram representations are possible. Here is one popular alternative: Fire  behavior 

Burn or YUM & Burn

Problem cost

Total cost of containment

Treatment cost

Total cost

A possible decision tree: Total cost

Success High cost Burn

Problems

Med cost Low cost

Escape Success High cost YUM & Burn

Problems

Med cost Low cost

Escape

To pass the clarity test, the costs must be precisely defined. Also, fire behavior must be defined. For example, it would be necessary to distinguish between and escaped fire and a fire that develops problems that can be brought under control. Excel file “Prescribed fire case study.xlsx” shows the above influence diagram with one additional structural arc. In order for the “Problem Cost” uncertainty to occur only if the “Fire behavior” results in  problems as shown in the above decision tree, an additional structural arc was added between these two nodes. The file also contains the resulting decision tree from automatically converting the influence diagram. While the above decision tree treats the resulting multiple objectives at the outcome of each

37

 branch, we used calculation nodes for the “Treatment cost” and “Total cost of containment” and are thus represented as additional nodes in the converted decision tree. Case Study: The S.S. Kuniang 1. Again, many different influence-diagram representations are possible here. A popular alternative is:

Highest competitor bid

Bid amount

Winning bid Coast Guard  judgment

Cost

Bid 6 5 5 -

Winning CG Bid  judgment 6 H 6 H 6 L -

Cost 6 + refit cost Cost of next best alternative -

}

This representation shows that the decision maker has to think about chances associated with the highest  bid from a competitor and the Coast Guard’s judgment. An arc might be included from “Winning bid” to “Coast Guard judgment” if the decision maker feels that different winning bids would affect the chances associated with the coast guard’s judgment (which is a distinct possibility). Also, it would be reasonable to include a separate decision node for purchasing alternative transportation i f the bid fails. The representation above includes that consideration implicitly in the cost table. A possible decision tree: Coast Guard  judgment

Win bid 

Cost of  S.S. Kuniang

Bid amount Lose bid 

Cost of next  best alternative

Additional nodes might be added. As with the influence diagram, a decision regarding what to do if the  bid is lost could be included. Also, a decision to keep or sell Kuniang might be included in the event of a high Coast Guard judgment. The file “SSKuniang I.xlsx” shows representations of both the above influence diagram and the decision tree. Case Study: The Hillblom Estate, Part I  No solution is provided for this case. We recommend that the instructor consult the article by Lippman and McCardle, (2004), “Sex, Lies and the Hillblom Estate,” Decision Analysis, 1, pp 149–166.

38

CHAPTER 4 Making Choices Notes

After having structured a number of decisions in Chapter 3, it seems natural to talk about analyzing models. Thus, Chapter 4 is titled “Making Choices,” and discusses the analysis of decision models. We cover decision trees, influence diagrams, risk profiles and first-order stochastic dominance, and the analysis of a two-attribute decision model. The last section in Chapter 4 provides additional instructions for using some of the advanced features in PrecisionTree. The opening example concerning the Texaco-Pennzoil negotiations is quite useful because it includes multiple decisions and multiple chance nodes. Furthermore, the problem could be quite complex if analyzed fully. Our analysis here is a simplification, and yet the modeling simplifications made are reasonable and would provide an excellent springboard for a more complete analysis. In fact, we will revisit Texaco-Pennzoil in later chapters. The discussion of the algorithm for solving decision trees is straightforward and takes at most part of one class period. Solving influence diagrams, however, is somewhat more involved, partly because of the symmetric representation that influence diagrams use. While PrecisionTree solves both of these models with a click of a button, it is important to make sure the students still understand the steps involved. The discussion of risk profiles and dominance are placed in Chapter 4 so that students get an early idea about using dominance to screen alternatives. The discussion is not technically deep because probability has not yet been introduced; we revisit the idea of stochastic dominance later in Chapter 7 after discussing  probability and CDFs. Also, note that the text only discusses first-order stochastic dominance. Two  problems in Chapter 14 (14.19 and 14.20) can be used as introductions to second-order stochastic dominance, but only after the students have begun to think about risk aversion. The chapter continues the topic of multiple attributes by leading the students through the analysis of a simple two-attribute job choice. Much of the discussion is about how to construct an appropriate twoattribute value function. With the value function in place, we see how to calculate expected value, create risk profiles, and determine stochastic dominance when there are multiple objectives. The last section provides additional instructions (beyond those in Chapter 3) for using PrecisionTree’s advanced features including reference nodes to graphically prune the tree without leaving out any of the mathematical details and linked trees (decision trees linked to a spreadsheet model). An example is also  provided that demonstrates a multiple-attribute model using formulas set-up in a spreadsheet. Finally, instructions are provided for analyzing the model and generating risk profiles. Topical cross-reference for problems

Branch pay-off formula Calculation nodes Certainty equivalent Convert to tree Decision criterion Decision trees

4.19, Southern Electronics Southern Electronics 4.3 Southern Electronics 4.20, 4.23 4.1, 4.4, 4.6-4.8, 4.14, 4.15, 4.19, Southern Electronics, Strenlar, SS Kuniang Part II, Marketing Specialists 4.15, Marketing Specialists 4.19, Marketing Specialists 4.1, 4.5, 4.10, 4.11, 4.13, Southern Electronics 4.12, 4.18, Job Offers, Strenlar

Goal Seek Linked decision trees Influence diagrams Multiple objectives

39

 Negotiations PrecisionTree

Southern Electronics 4.4-4.8, 4.10, 4.15, 4.16, 4.19, Job Offers, Strenlar, Southern Electronics, S.S. Kuniang Part II, Marketing Specialists 4.21 4.3, Strenlar 4.9, 4.14-4.17, Job Offers, Marketing Specialists 4.11, 4.15, Strenlar S.S. Kuniang, Part II 4.2, 4.6, 4.7, 4.9, 4.11, 4.14-4.17, 4.22, Job Offers, Marketing Specialists 4.15 4.3 4.14, Strenlar, Marketing Specialists 4.12, 4.18, Job Offers 4.13

Probability Risk aversion Risk profiles Sensitivity analysis Solver Stochastic dominance Stock options Texaco-Pennzoil Time value of money Trade-off weights Umbrella problem

Solutions 4.1. No. At least, not if the decision tree and influence diagram each represent the same problem (identical

details and definitions). Decision trees and influence diagrams are called “isomorphic,” meaning that they are equivalent representations. The solution to any given problem should not depend on the representation. Thus, as long as the decision tree and the influence diagram represent the same problem, their solutions should be the same. 4.2. There are many ways to express the idea of stochastic dominance. Any acceptable answer must capture

the notion that a stochastically dominant alternative is a better gamble. In Chapter 4 we have discussed first-order stochastic dominance; a dominant alternative in this sense is a lottery or gamble that can be viewed as another (dominated) alternative with extra value included in some circumstances. 4.3. A variety of reasonable answers exist. For example, it could be argued that the least Liedtke should

accept is $4.63 billion, the expected value of his “Counteroffer $5 billion” alternative. However, this amount depends on the fact that the counteroffer is for $5 billion, not some other amount. Hence, another reasonable answer is $4.56 billion, the expected value of going to court. If Liedtke is risk-averse, though, he might want to settle for less than $4.56 billion. If he is very risk-averse, he might accept Texaco ’s $2 billion counteroffer instead of taking the risk of going to court and coming away with nothing. The least that a risk-averse Liedtke would accept would be his certainty equivalent  (see Chapter 13), the sure amount that is equivalent, in his mind, to the risky situation of making the best counteroffer he could come up with. What would such a certainty equivalent be? See the epilogue to the chapter to find out the final settlement. The following problems (4.4 – 4.9) are straightforward and designed to reinforce the computations used in decision trees. Because PrecisionTree does all the calculations, the instructor may want the students to do some of these problems by hand to reinforce the methodology of calculating expected value or creating a risk profile. 4.4. The Excel file “Problem 4.4.xlsx” contains this decision tree. The results of the run Decision Analysis

 button (fifth button from the left on the PrecisionTree toolbar) are shown in the worksheets labeled Probability Chart, Cumulative Chart, and Statistical Summary. EMV(A)

= 0.1(20) + 0.2(10) + 0.6(0) + 0.1(-10) = 3.0

EMV(B)

= 0.7(5) + 0.3(-1) = 3.2

40

Alternative B has a larger EMV than A and a lower standard deviation or narrower range. While B’s maximum payoff is only 5, B’s payoff will be larger than A’s payoff 70% of the time. Neither A nor B dominates. 4.5. The Excel file “Problem 4.5.xlsx” contains this influence diagram. The most challenging part of

implementing the influence diagram is to enter the payoff values. The payoff values reference the outcome values listed in the value table. The value table is a standard Excel spreadsheet with values of influencing nodes. In order for the influence diagram to calculate the expected value of the model, it is necessary to fill in the value tables for all diagram nodes. is the summary statistics are shown in the upper left of the worksheet. Change a value or probability in the diagram, and you will immediately see the impact on the statistics of the model. It is possible to use formulas that combine values for influence nodes to calculate the payoff node vales. The results of the run Decision Analysis button (fifth button from the left on the PrecisionTree toolbar) are shown in the worksheets labeled Statistics, RiskProfile, CumulativeRiskProfile, and ScatterProfile. The results of an influence diagram will only show the Optimal Policy. An analysis based on a decision tree will  produce the risk profiles and summary statistics for all the alternatives, not just the optimal one. The following figure shows how to solve the influence diagram by hand. See the online supplement for more explanation.

Event A

Choice

A

20

.1

10

.2

0

.6

-10

.1

Event B

5

.7

-1

.3

B Value Event

Event

Choice

A

B

Value

A

20

5

20

-1

20

10 0 -10 B

20 10 0 -10

41

5

10

-1

10

5

0

-1

0

5

-10

-1

-10

5

5

-1

-1

5

5

-1

-1

5

5

-1

-1

5

5

-1

-1

Solution: 1.

2.

3.

Reduce Event B: Choice Event A EMV A 20 10 0 -10 B 20 10 0 -10 Reduce Event A: Choice EMV A 3.0 B 3.2 Reduce Choice: B 3.2

20 10 0 -10 3.2 3.2 3.2 3.2

4.6. The Excel file “Problem 4.6.xlsx” contains this decision tree. The results of the run Decision Analysis

 button are shown in the worksheets labeled Probability Chart, Cumulative Chart, and Summary Statistics. EMV(A) = 0.1(20) + 0.2(10) + 0.6(6) + 0.1(5) = 8.1 EMV(B) = 0.7(5) + 0.3(-1) = 3.2 This problem demonstrates deterministic dominance. All of the outcomes in A are at least as good as the outcomes in B. 4.7. Choose B, because it costs less for exactly the same risky prospect. Choosing B is like choosing A but  paying one less dollar.

EMV(A) EMV(B)

= 0.1(18) + 0.2(8) + 0.6(-2) + 0.1(-12) = 1.0 = 0.1(19) + 0.2(9) + 0.6(-1) + 0.1(-11) = 1 + 0.1(18) + 0.2(8) + 0.6(-2) + 0.1(-12) = 1 + EMV(A) = 2.0

The Excel file “Problem 4.7.xlsx” contains this decision tree. The dominance of alternative B over alternative A is easily seen in the Risk Profile and theCumulative Risk Profile. 4.8. The Excel file “Problem 4.8.xlsx” contains this decision tree. The results of the run Decision Analysis

 button are shown in the worksheets labeled Probability Chart, Cumulative Chart, and Summary Statistics.

42

 A1

$8

8 EMV(A) = 0.27(8) + 0.73(4) = 5.08

.27

EMV = 7.5

 A2

0.5 0.5

 A

.73

$15 $4

0.45

$10

B EMV(B) = 0.45(10) + 0.55(0) = 4.5

$0

0.55

$0

4.9. The risk profiles and cumulative risk profiles are shown in the associated tabs in the Excel file

“Problem 4.8.xlsx.” The following risk profiles were generated by hand. The profiles generated by PrecisionTree only include the two primary alternatives defined by the original decision “A” or “B”. To also include the A-A1 and A-A2 distinction, the decision tree would need to be restructured so that there was only one decision node with three primary alternatives, “A-A1”, “A-A2”, and “B”. Risk profiles: 0.8

 A-A1

0.7

 A-A2

0.6

B

0.5 0.4 0.3 0.2 0.1 0 0

2

4

6

8

43

10

12

14

Cumulative risk profiles:

1  A-A2  A-A1

0.75

B 0.5

0.25 0 -2

0

2

4

6

8

10

12

14

 None of the alternatives is stochastically dominated (first-order) because the cumulative risk-profile lines cross. 4.10. The Excel file “Problem 4.10.xlsx” contains two versions of this influence diagram. The first version

is as shown below. The second version (in Worksheet “Alt. Influence Diagram”) takes advantage of PrecisonTree’s structural arcs to incorporate the asymmetries associated with the model. For example, if you choose A for Choice 1, you don’t care about the result of Event B. In the basic influence diagram, you need to consider all combinations, but in the alternative diagram, structural arcs are used to skip the irrelevant nodes created by the asymmetries. Compare the value tables for the two alternatives – incorporating structure arcs makes the value table much simpler because the asymmetries are captured by the structure arcs. As mentioned previously, these results generated by PrecisionTree when analyzing an influence diagram only show the optimal alternative, in this case Choice B. To solve the influence diagram manually, see the diagrams and tables below. The online supplement for Chapter 4 provides a detailed explanation.

44

16

Choice 1

Event A

A B

Choice 2

8 A' Event A'

Choice 2 4

0 15 Event B

Value Choice 1 A

Event A Choice 2

Choice 2 8

Event A' 0 15 0

A'

15 4

8

0 15

A'

0 15

B

Choice 2

0

8

15 0

A'

15 4

8

0 15 0

A'

15

45

Event B 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10

Value 8 8 8 8 0 0 15 15 4 4 4 4 4 4 4 4 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10

0 10

Reduce Event B: Choice Event 1  A  A Choice 2

Choice 2 8  A'

4

8  A'

B

Choice 2

8  A'

4

8  A'

Event  A' 0 15 0 15 0 15 0 15 0 15 0 15 0 15 0 15

EMV 8 8 0 15 4 4 4 4 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5

Reducing Event A': Choice Event 1  A  A Choice 2 4 B

Choice 2 4

Choice 2 8  A' 8  A' 8  A' 8  A'

EMV 8 7.5 4 4 4.5 4.5 4.5 4.5

Reducing Choice 2: Choice Event 1  A EMV  A Choice 2 8 4 4 B Choice 2 4.5 4 4.5

Reducing Event A: Choice 1 EMV  A 5.08 B 4.5

Finally, reducing Choice 1 amounts to choosing A because it has the higher EMV. This problem shows how poorly an influence diagram works for an asymmetric decision problem!

46

4.11. If A deterministically dominates B, then for every possible consequence level  x, A must  have a  probability of achieving  x or more that is at least as great as B ’s probability of achieving  x or more:

P(A ≥ x) ≥ P(B ≥ x) .

The only time these two probabilities will be the same is 1) for  x less than the minimum possible under B, when P(A ≥ x) = P(B ≥  x) = 1.0 — both are bound to do better than such a low  x; and 2) when  x is greater than the greatest possible consequence under A, in which case P(A ≥ x) = P(B ≥ x) = 0. Here neither one could possibly be greater than  x. 4.12. It is important to consider the ranges of consequences because smaller ranges represent less

opportunity to make a meaningful change in utility. The less opportunity to make a real change, the less important is that objective, and so the lower the weight. 4.13. Reduce “Weather”:

Take umbrella? Take it Don’t take it

EMV 80  p(100)

Reducing “Take Umbrella?” means that “Take it” would be chosen if  p ≤ 0.8, and “Don’t take it” would be chosen if p > 0.8. The Excel file “Problem 4.13.xlsx” contains the influence diagram for this problem. PrecisionTree allows you to link the probability of weather to a cell location for variable  p. Thus, to consider different  probability values, you simply need to change the value for the probability in cell J6, the location we chose for p. 4.14. a. There is not really enough information here for a full analysis. However, we do know that the

expected net revenue is $6000 per month. This is a lot more than the sure $3166.67 = $400,000 × (0.095/12) in interest per month that the investor would earn in the money market. b. If the investor waits, someone else might buy the complex, or the seller might withdraw it from the

market. But the investor might also find out whether the electronics plant rezoning is approved. He still will not know the ultimate effect on the apartment complex, but his beliefs about future income from and value of the complex will depend on what happens. He has to decide whether the risk of losing the complex to someone else if he waits is offset by the potential opportunity to make a more informed choice later. c. Note that the probability on the rezoning event is missing. Thus, we do not have all the information for a

full analysis. We can draw some conclusions, though. For all intents and purposes, purchasing the option dominates the money-market alternative, because it appears that with the option the investor can do virtually as well as the money-market consequence, no matter what happens. Comparing the option with the immediate purchase, however, is more difficult because we do no t know the precise meaning of “substantial long-term negative effect” on the apartment complex ’s value. That is, this phrase does not pass the clarity test! The point of this problem is that, even with the relatively obscure information we have, we can suggest that the option is worth considering because it will allow him to make an informed decision. With full information we could mount a full-scale attack and determine which alternative has the greatest EMV. The structure of the tree is drawn in the Excel file “Problem 4.14.xlsx.” All the numbers necessary to do a complete analysis are not provided.

47

Buy C omplex Rezone denied

Buy option

Expect $6000 per month + higher v alue

Inv est $399,000 in $3158.75/month = $399,000 (0.095/12) money market Buy C omplex

Rezone approv ed Inv est $399,000 in money market Rezone denied

Expect $6000 per month + lower v alue $3158.75/month = $399,000 (0.095/12) Expect $6000 per month + higher v alu

Buy c omplex Rezone approv ed

Do nothing. Inv est $400,000 in money market.

Expect $6000 per month + lower v alue

$3166.67/month = $400,000 (0.095/12)

4.15. a. 1.00

.50

Option

1000

-500

2000

3000

$

2000

3000

$

1.00

Buy stock

-500 -46.65

1000 85.10

48

1.00

Do nothing

2000

1000

-500

3000

$

3.35 3.33

 No immediate conclusions can be drawn. No one alternative dominates another. b. The Excel file “Problem 4.15.xlsx” contains this decision tree and the risk profiles generated by

PrecisionTree. Net Contribution  Apricot wins (0. 25) Option worth $3.5 per share f or 1000 shares. Buy option on 1000 shares

3000 = 3500 - 500

 Apricot loses (0.75) Option worthless .

-500

 Apricot wins (0. 25) Buy 17 shares f or $484.50. Invest $15.50 in money market.

 Apricot loses (0.75)

Do nothing. Inv est $500 in money market f or 1 month.

85.10 = 33.5 (17) + 15.5 (1.0067) - 500

-46.65= 25.75 (17) + 15.5 (1.0067) - 500 3.35 = 500(1.0067) - 500 3.33

Analyzing the decision tree with  p = 0.25 gives EMV(Option) = 0.25(3000) + 0.75(-500) = $375 EMV(Buy stock) = 0.25(85.10) + 0.75(-46.65) = -$13.71 EMV(Do nothing) = $3.33. Thus, the optimal choice would be to purchase the option. If  p = 0.10, though, we have EMV(Option) = 0.10(3000) + 0.90(-500) = $-150 EMV(Buy stock) = 0.10(85.10) + 0.90(-46.65) = -$33.48 EMV(Do nothing) = $3.33.  Now it would be best to do nothing. To find the breakeven value of  p, set up the equation  p (3000) + (1 -  p)(-500) = 3.33

Solve this for  p:

49

 p (3000) +  p (500) = 503.33  p (3500) = 503.33  p = 503.33/3500 = 0.1438.

When p = 0.1438, EMV(Option) = EMV(Do nothing). The break-even analysis can also be found using the built-in Excel tool: Goal Seek. Because we are altering a probability value, it is necessary to use formulas to guarantee that the probabilities always add to one. These formulas are incorporated in the decision tree model. Then, use Excel’s Goal Seek (from the Tools menu). Select the outcome of the Buy option branch (cell $C$ 13) as the “Set Cell”, the value of the money market branch (3.33) as the “To value” (you can’t enter a cell reference here with goal seek), and the  probability of winning the lawsuit (cell: $F$5) as the “By changing cell.” Goal Seek will find the  probability 14.4%. 4.16. The decision tree for this problem is shown in the Excel file “Problem 4.16.xlsx.” The cumulative risk  profile generated by PrecisionTree is shown in the second worksheet. Cumulative risk profiles:

1

0.75 Buy 0.5

Make

0.25 0 30

35

40

45

50

55

Johnson Marketing should make the processor because the cumulative risk profile for “Make” lies to the left of the cumulative risk profile for “Buy.” (Recall that the objective is to minimize cost , and so the leftmost distribution is preferred.) Making the processor stochastically dominates the “Buy” alternative. 4.17. Analysis of personal decision from problems 1.9 and 3.21. 4.18. a. Stacy has three objectives: minimize distance, minimize cost, and maximize variety. Because she

has been on vacation for two weeks, we can assume that she has not been out to lunch in the past week, so on Monday, all of the restaurants would score the same in terms of variety. Thus, for this problem, we can analyze the problem in terms of cost and distance. The following table gives the calculations for part a:

Sam’s Sy’s Bubba’s Blue China Eating Excel-Soaring

Distance 10 9 7 2 2 5

Distance Score 0 13 38 100 100 63

Cost $3.50 $2.85 $6.50 $5.00 $7.50 $9.00

50

Cost Score 89 100 41 65 24 0

Overall Score 45 56 39 83 62 31

In the table, “Distance Score” and “Cost Score” are calculated as in the text. For example, Sam ’s cost score is calculated as 100(3.50 - 9.00)/(2.85 - 9.00) = 89. The overall score is calculated by equally weighting the cost and distance scores. Thus, S(Sam ’s) = 0.5(0) + 0.5(89) = 45. The overall scores in the table are rounded to integer values. Blue China has the highest score and would be the recommended choice for Monday ’s lunch. b. Let’s assume that Stacy does not go out for lunch on Tuesday or Wednesday. For Thursday ’s selection, we now must consider all three attributes, because now variety plays a role. Here are Stacy ’s calculations

for Thursday:

Sam’s Sy’s Bubba’s Blue China Eating Excel-Soaring

Distance 10 9 7 2 2 5

Distance Score 0 13 38 100 100 63

Cost $3.50 $2.85 $6.50 $5.00 $7.50 $9.00

Cost Score 89 100 41 65 24 0

Variety Score 100 100 100 0 100 100

Overall Score 63 71 59 55 75 54

The score for variety shows Blue China with a zero and all others with 100, reflecting Monday ’s choice. The overall score is calculated by giving a weight of 1/3 to each of the individual scores. Now the recommended alternative is The Eating Place with an overall score of 75. If we assume that Stacy has been out to eat twice before making Thursday ’s choice, then the table would have zeroes under variety for both Blue China and The Eating Place, and the recommended choice would  be Sy’s.  Note that it is necessary to do the calculations for part b; we cannot assume that Stacy would automatically go to the next best place based on the calculations in part a. The reason is that a previous choice could be so much better than all of the others on price and distance that even though Stacy has already been there once this week, it would still be the preferred alternative. 4.19. The linked-tree is in the Excel file “Problem 4.19.xlsx.” The spreadsheet model defines the profit  based on the problem parameters: cost per mug and revenue per mug, the order decision, and the demand uncertainty. The value of the order decision node is linked to the “Order Quantity” cell in the spreadsheet model ($B$5). The value for the demand chance nodes are linked to the “Demand” cell in the spreadsheet model ($B$8), and the outcome node values are linked to the Profit (no resale) in the spreadsheet model ($B$12) for part a, and to the Profit (resale) in the spreadsheet model ($B$13) for part b.

The results of the model show that they should order 15,000 if they can't sell the extras at a discount, and they should order 18,000 if they can.

51

Spreadsheet Model

Order Quantity Cost per mug Cost Demand  Number Sold Revenue per mug Revenue Profit (no resale) Profit (resale)

12000 6.75 =Order_Quantity*Cost_per_mug 10000 =IF(Demand
25.0% 10000

Order 12,000

FALSE 12000

 

158500

Demand  

194425

Demand is 15,000

Demand is 20,000

0

50.0% 15000

 

206400

25.0% 20000

Problem 4.19

0

0  

206400

Order Decision 228062.5

Demand is 10,000

25.0% 10000

Order 15,000

TRUE 15000

 

138250

228062.5

Demand is 20,000

258000  

 

0.25 258000 0

25.0% 10000

FALSE

 

25.0% 20000

Demand is 10,000

0.5

50.0% 15000

18000

 

0.25

Demand

Demand is 15,000

Order 18,000

 

 

118000

Demand  

225775

Demand is 15,000

15000

Demand is 20,000

0

50.0%

 

0

25.0% 20000

237750

 

309600

4.20. The most compelling argument for using EMV when deciding among alternatives is presented later in

the book when we discuss utility theory (Chapters 14 & 15). There we show that if you agree with a few  postulates and if you are risk neutral, then you are most satisfied when choosing the alternative that maximizes EMV. As an example of one of the postulates, consider transitivity, which states: if you prefer A to B and prefer B to C, then you prefer A to C. As most people readily agree to the transitivity and other  postulates, most people are served well by choosing the alternative that maximizes EMV.

52

The cons of using EMV as a decision criterion is that the decision maker may not be risk neutral, that is, the amount of risk associated to the different alternatives does matter to him or her. The EMV still serves as the single best summary of what could happen, but in this case, the decision maker is interested in more than a single number summary. To choose the alternative they like the most, the risk level of the alternative needs to be taken into consideration. Typically, we do this by examining the range of possible payouts via the risk  profile. Not only does the risk profile report the different payouts, but it also reports the probability of those  payouts occurring. The pros of using EMV is that it makes logical sense and that it is straightforward. Incorporating risk into the decision criterion involves making tradeoffs. Comparing risk profiles requires the decision maker to somehow balance to rewards and risks, whereas comparing EMVs requires no tradeoffs. It is quite easy to view a list of EMVs and choose the largest one. Even in one-off cases, the EMV summarizes all the different payoffs using both the monetary values and their probabilities into a coherent, easy to understand single value useful for making comparisons. 4.21. This is an algebra problem. We are given:

1 + 2 + 3 = 1; 1 + 2 + 3 = 1; 1 + 2 + 3 = 1;  1 + 2 + 3 = 1. We want to show that the nine end-node probability values sum to one:

11 + 12 + 13 + 21 + 22 + 23 + 31 + 32 + 33 = 1. We do this by grouping terms.

11 + 12 + 13 + 21 + 22 + 23 + 31 + 32 + 33 = 1 (1 + 2 + 3 ) + 2 (1 + 2 + 3 ) + 3 (1 + 2 + 3) = 1 + 2 + 3 = 1 4.22. Yes, if A dominates B, then the EV(A) is always better than the EV(B). For example, assume our

consequence measure is profit, and thus the more profit the better. If alternative A dominates alternative B, 1, A’s associated payoff or profit is always as good as and at then for any probability value , 0 times better than B’s associated profit. Because EV(A) is a weighted average of the profit values, weighted  by the probabilities, EV(A) must be greater than EV(B).

 ≤≤

   ∗� >  ∗ � = (), where  �  is the profit value of A associated to the probability . EV(A) =

4.23. Students often mix up the mode and mean, and this problem is to help them distinguish these

statistics. The Excel solution is given in file “Problem 4.23.xlsx,” and w e copied the summary statistics  below. Note that the values (in model and below) are given in $1000s. Thus, bidding $750,000 has an EMV of $35,480 and bidding $700,000 has an EMV of $40,950. Because the probability of the higher bid amount has a 30% chance of being accepted versus the 60% chance of the lower bid amount being accepted, we see that bidding the lower amount produces the higher average profit. The modal value for both alternatives is zero. This is because 0 is the most likely outcome for either bid amount. As a matter of fact, bidding $750,000 results in a $0 profit 70% of the time and bidding $700,000 results in a $0 profit 40% of the time. These probabilities can be read from either the probability or cumulative charts. Clearly, the mean or EMV is different from the mode and tells us a different story. Both statistics give us different insights, and both are useful.

53

Statistics Mean

Bid = $750,000 $35.48

Bid =$700,000 $40.95

Minimum

$0

-$20

$205

$155

$0

$0

Maximum Mode Std. Deviation Skewness

$58

$46

1.3130

0.7538

Kurtosis

3.2847

2.4924

Case Study: Southern Electronics, Part I 1. Steve’s reservation price is $10 Million, the value of the offer from Big Red. This is the least he should

accept from Banana. 2. This influence diagram is modeled in the Excel file “Southern Case Parts 1 & 2.xlsx.” An additional

structure arc is included between “Choice” and “EYF Success” to model the fact that if Steve chooses Big Red, the probability of EYF Success is irrelevant.

EYF Success

Choice

Banana Share Price

Yes (0.4)  No (0.6)

EYF Success Yes  No

Share Price $40 $25

Big Red Banana Value

Choice Big Red Banana

54

Banana Share Price 40 25 40 25

Value 10 10 11 8.75

3.

Share price EYF succeeds (0.4)

EV | succ ess = $40

11 M = 5 M + (40 x 150,000)

EMV = $9.65 Million

Banana

EYF Fails (0.6)

EV | f ailure = $25

Big Red

8.75 M = 5 M + (25 x 150,000)

10M = 5 M + (50 x 100,000)

The decision tree shown in the second worksheet of the Excel file “Southern Case.xlsx” is the version of the tree that is automatically generated by the “Convert to Tree” function for the associated influence diagram in the first worksheet. It shows calculation nodes for the Banana share price as represented in the influence diagram but not explicitly in the above decision tree. Also, formulas are used in the calculation of the output values for the Banana decision where the formula includes the certain $5M p lus 150000 shares * BranchVal("Banana Share Price"). 4. Obviously, Steve cannot accept the Banana offer because for him it has a lower EMV than Big Red  ’s offer. Furthermore, it is riskier than Big Red ’s.

 Note that Banana calculates the expected value of its offer as EMV

= 0.6[$5 M + (50 × 150,000)] + 0.4[$5 M + (25 × 150,000)] = 0.6 (12.5 M) + 0.4 (8.75 M) = 11 M.

So naturally Banana thinks it is making Steve a good offer! Case Study: Southern Electronics, Part II 1. The second part of the Southern Electronics case now includes an option to hedge against the risk of the

share price. This decision tree is shown in the third worksheet in the Excel file “Southern Electronics Parts 1 & 2.xlsx.”

55

Share price Expected share price = $40

EYF success (0.4)

$11.73 M = $530,000 + (40 x 280,000)

EMV = $10.05 M

Banana $8.93 M = $530,000 + (30 x 280,000). Stev e exercises option and sells shares f or $30 each.

EYF failure (0.6)

Big Red

$10 M

Steve’s expected value is EMV(Steve) = 0.4($11.73 M) + 0.6($8.93 M) = $10.05 M. 2. Banana’s expected cost is calculated by noting that

• Banana pays $530,000 in cash. • Banana must hand over 280,000 shares worth $30 per share. Thus, 280,000 × $30 = $8.4 M • If the EYF fails (which they assess as a 40% chance), Banana must buy back the shares at an expected price of $5 per share over the market value ($30 - $25). In this case Banana would incur a cost of $5 × 280,000 = 1.4 M. Thus, Banana’s expected cost is E(cost) = $0.53 M + $8.4 M + 0.4 (1.4 M) = $9.49 M. The expected cost of this offer thus is less than the $9.5 M cost for their original offer. Of course, Steve ’s offer is somewhat riskier than the original offer. This problem demonstrates how two parties in a negotiation can exploit differences in probability  judgments to arrive at an agreement. The essence of such exploitation is to set up some sort of a bet. In this case, the bet involves whether the EYF fails. Accepting this offer would mean that Banana is “betting” that the EYF succeeds, and Steve is “betting” (or is protected) if the EYF fails. (Side note from Clemen: EYF stands for Elongated Yellow Fruit. I stole the abbreviation from James Thurber, whose editor once complained that Thurber used fancy phrases (elongated yellow fruit) instead of simple ones (banana)). Case Study: Strenlar

Strenlar is a very messy case. Here is the structure in an influence diagram. This model is shown in the Excel file “Strenlar.xlsx.” This influence diagram is shown in the first worksheet.

56

Court outcome?

Decision?

Mfg process success?

Payoff 

In order to do a good job with the analysis, a number of assumptions must be made. Here is a reasonable set: • $8 million in profits is the present value of all profits to be realized over time. • $35 million in sales is also the present value of all future sales. • Fred Wallace ’s interest rate is 7.5% per year. • Fred’s time frame is 10 years. We will use 10 years to evaluate the job offer and the lump sum. • If Fred goes to court, he continues to be liable for the $500,000 he owes. The $8 million in  profits is net after repaying the $500,000. As indicated in the case, If Fred and PI reach an agreement, then PI repays the debt. • If Fred accepts the lump sum and options, and if the manufacturing process works, then the options pay 70,000 × $12 at a point 18 months in the future. Thus, the present value of the options is

12 × 70,000 1.0375

3

= $752,168. That is, we figure three periods at 3.75% per period. The

 purpose of this is simply to put a value on the options. • The options are non-transferable. Thus, there is no need to figure out the “market value” of the options. These assumptions are set-up in a spreadsheet model on the second worksheet in the Excel file “Strenlar.xlsx.” These assumptions are then used in the decision tree model on the same worksheet. For example, the outcome of going to court, winning, and the manufacturing process is the NPV of all profits from Strenlar over time (cell $B$3). Also, in cell H3 the NPV of 10 years making $100,000 per year is calculated at $686,408 and in H4 the NPV of the option is calculated at $752,168.

57

Fred ’s tree looks like this:

An explanation of the end-node values: If Fred wins in court, but cannot fix the manufacturing process, then he must pay back the • $500,000. • If he loses in court, then he must also pay the $60,000 in court fees. If he accepts the job (present value of $686,408) and fixes the manufacturing process, he also • earns 6% of $35 million, which is $2.1 million. • If he accepts the job and the process cannot be fixed, he has $686,408. If he takes the lump sum, then either he earns $500,000 or an additional $752,168 as the present • value of the options. Going to court is clearly Fred ’s best choice if he wants to maximize EMV. In fact, this conclusion is remarkably robust to the assumptions that are made (as long as the assumptions are reasonable). If he is risk averse—and the case indicates that he may be substantially so—one of the other alternatives may be  preferable. Furthermore, it would be appropriate to consider alternatives he might pursue over the next ten years if he were to take the lump sum. The analysis above is OK in financial terms, but that is as far as it goes. It does ignore other objectives that Fred might have. It may be helpful to characterize Fred ’s three alternatives: 1. Stick with Strenlar; don ’t succumb to PI. This alternative has by far the greatest “upside potential” of $8 Million. Such a consequence also would be quite satisfying and his future would be secure. He could also fail, however, or suffer from hypertension. This clearly is the riskiest choice. 2. Play it safe. Accept the PI job offer. After all, there is a good chance Fred will earn a lot of money in royalties. But will things be the same at PI as they were in the past? 3. Accept the lump sum. Fred can have $500,000 immediately and ten years to do something else,  plus a chance at an additional $840,000 18 months in the future. What could he do with resources like this?

58

From this characterization, it is clear that Fred has to think about whether the potential wealth to be gained from Strenlar is worth substantial risk, both financially and in terms of his health. If he decides that it is not, then he should consider whether the security of the PI job outweighs his potential for alternative pursuits if he were to take the lump sum. Wallace’s risk profiles show clearly that refusing PI is by far the riskiest choice. The cumulative risk profile shows, at least from the financial perspective, that taking the job dominates the lump sum offer.

Cumulative Probabilities for Decision Tree 'Strenlar' Choice Comparison for Node 'Decision' 1

0.8    y    t    i     l    i     b    a     b    o    r    P    e    v    i    t    a     l    u    m    u    C

0.6 Go to Court 0.4

Accept job Accept lum sum

0.2

0 -$1,000,000

$1,000,000

$3,000,000

$5,000,000

$7,000,000

$9,000,000

Case Study: Job Offers

Robin’s decision tree is shown in the Excel file: “Job Offers.xlsx” as shown below. The consequences in the tree use formula in the cells under each branch to link each path to the associated overall score in the spreadsheet model.

59

Weights:

100 $1500 $1500

50.0%

 

400

0

48.6

70.0%

 

0.35

56.9

 

56.9

15.0%

 

0.075

73.6

 

73.6

15.0%

 

0.075

40.3

 

40.3

70.0%

 

0.35

48.6

 

48.6

15.0%

 

0.075

65.3

 

65.3

75

25

56

48.6

75

50

56

56.9

75

100

56

73.6

25

25

56

40.3

25

50

56

48.6

25

100

56

65.3

100

37.5

0

29.2

100

57.5

0

35.8

100

80

0

43.3

0

0

100

50.0

Disposable Income  

54.0

100 $1300 $1300

50.0% 0

   

Snowfall 49.9

200 400 Job Case

0.075

 

0.50

Magazine Overall Rating Score

58.2

200

Madison Publishing TRUE

 

48.6

0.33

Snowfall Rating

Snowfall  

0

15.0%

0.17

Income Rating

Decision 54.0

29.2

MPR Manufacturing F AL SE 0

0

15.0%

150

 

29.2

 

35.8

Snow fa ll  

36.0 70.0%

230

35.8 15.0%

320

43.3

Pandemonium Pizza FALSE 50.0

0

0  

43.3

0 50

1. The student must calculate the proportional scores in the consequence matrix. For example, the snowfall

rating when it snows 100 cm is 100(100 cm - 0)/(400 cm - 0) = 25. These formulas are shown in the spreadsheet model. 2. Annual snowfall obviously is a proxy in this situation. It may not be perfect, depending on other climatic

conditions. for example, 200 cm of snowfall in Flagstaff may be very different than 200 cm in Minnesota; snow in Flagstaff typically melts after each storm, but not so in Minnesota. Other possible proxies might be  proximity to mountains, average snowpack during February (for example), or the average high temperature during winter months. Another possibility is to consider a non-weather index such as the amount of wintersports activity in the area. 3. The weights are k m = 1/2, k s = 1/3, and k i = 1/6. Calculate these by solving the equations k m = 1.5 k s, k m = 3k i, and k m +k s + k i = 1.

Using these weights, we can calculate the overall scores at the end of each branch of the tree:

Weights: Ratings:

Madison

MPR Pandemonium

Income 0.17 75 75 75 25 25 25 100 100 100 0

Snowfall 0.33 25 50 100 25 50 100 37.5 57.5 80 0

Magazine 0.50 56 56 56 56 56 56 0 0 0 100

60

Overall Score 49 57 74 41 49 66 29 36 43 50

4. Expected values for the three individual attributes and the overall score are given in the following table:

EXPECTED VALUES Income $1200 $1600 $1420

Pandemonium MPR Madison

Snowfall 0 cm 231.5 cm 215 cm

Magazine Score 95 50 75

Overall Score 50 36 55

5, 6. The cumulative risk profiles are shown below. Note that different alternatives are stochastically

dominant for different attributes. Looking at the overall score, MPR is clearly dominated. The Excel file “Job Offers Case.xlsx” shows the cumulative risk profile for all three measures and separate risk profiles for each measure individually. In expected value, Madison Publishing comes in second on each attribute but is a clear winner overall. Likewise, Madison looks good in the overall score risk profile, clearly dominating MPR. A good strategy for Robin might be to go to Madison Publishing in St. Paul and look for a good deal on an apartment.  Nevertheless, Robin would probably want to perform some sensitivity analysis on the weights used to determine how robust the choice of Madison is (in terms of the calculated expected values and risk profiles) to variations in those weights. Legend: Madison MPR Pandemonium

1.00 0.80 0.60 0.40 0.20

Income

0.00 $1,000

$1,200

$1,400

$1,600

$1,800

1.00 0.80 0.60 0.40 0.20

Snowfall

0.00 -100

0

100

200

300

400

61

500

1.00 0.80 0.60 0.40 0.20

Magazine Score

0.00 0

20

40

60

80

100

1.00 0.80 0.60 0.40 0.20

Overall score

0.00 0

20

40

60

80

100

Case Study: SS Kuniang, Part II 1. The problem is how does one sort through all of the possible bids between $3 and $10 million to find the

one with the lowest expected cost. One possibility is to construct a model that calculates the expected cost for a given bid and then search for the optimum bid by trial and error. A better approach would be to construct a model and use an optimization routine to find the optimum bid; for example, a spreadsheet model can be constructed in Microsoft Excel, and then Excel ’s Solver can be used to find the optimum. This model is constructed in the Excel file “SSKuniang II.xlsx.” 2. The details provided lead to the following decision tree:

Coast Guard  judgm ent $9 M (0.185) $4 M (0.630) Win bid (p | Bid)

$1.5 M (0.185)

Bid amount Lose bid (1-p | Bid )

Max (Bid, $13.5 M Max (Bid, $6 M ) Max (Bid, $2.25 M

$15 M (Tug-barge

Two things are notable in the decision tree. First, the probability of submitting the winning bid, p | Bid, is calculated according to the formula given in the problem. (Incidentally, this way of calculating the  probability is consistent with a belief that the highest competitive bid is uniformly distributed between $3 and $10 million; see Problems 9.27 and 9.28). Second, there is, strictly speaking, a decision following the Coast Guard judgment, and that would be whether to complete fitting out the Kuniang or to go with the tug and barge. Because the bid will never be more than $10 million, however, the final cost after the Coast Guard judgment will never be more than $13.5 million, less than the $15 million for the tug and barge. To run the model constructed in the spreadsheet, use the built-in Solver tool to minimize the cost of the Decision (cell $B$6) while constraining the bid amount (cell $A$22) between $3 and $10. To run the Solver tool, select Solver from the Tools menu. Using this model, the optimal bid is $9.17 million with an expected cost of $10.57 million.

62

Case Study: Marketing Specialists, Ltd. 1. See the file “Marketing Specialists.xlsx” for the full analysis.

2. With the commission rate set to 15%, we can use PrecisisonTree to calculate the following:

Cost Plus: Commission:

EMV = €291,391 EMV = €56,675

Std Deviation = €20,040 Std Deviation = €958,957

P(Cost Plus better than Commission) = 51.5% The risk profiles (below) show how risky the Commission option is compared to Cost Plus. In addition, the cumulative risk profile shows that neither option stochastically dominates the other. At a commission rate of 15%, it would be difficult to justify the Commission Option.

63

3. Using Goal Seek, it is straightforward to find that the breakeven commission rate is about 16.96%.

Although this sets the EMVs equal, the risk profiles are still vastly different (see below). As a result, it is still difficult to argue for the Commission Option. Why take the Commission when it gives the same EMV  but is so much riskier?

64

So the problem Grace Choi faces is to find a commission rate that gives a high enough EMV to offset the increased risk. For example, when the commission rate is 25%, the statistics and risk profiles are: EMV = €850,136; Std Deviation = €518,868; P(Cost Plus better than Commission) = 12.25%

This looks a lot better for Commission! Note that some students will want a commission rate of 30% or more to be sure that the Commission generates at least as much as Cost Plus. That might be a good opening offer from which to negotiate, but probably there is a lower commission rate that would work for Grace and Marketing Specialists. A good follow-up exercise would be to incorporate the idea of risk tolerance (Chapter 14) to find the commission rate that makes the expected utilities of the two options the same.

65

CHAPTER 5 Sensitivity Analysis Notes

Because no optimal procedure exists for performing sensitivity analysis, this chapter is somewhat “looser” than the preceding. An effort has been made to present some of the more basic sensitivity analysis approaches and tools. It is important to keep in mind that the purpose of sensitivity analysis is to refine the decision model, with the ultimate objective of obtaining a requisite model. For those instructors who enjoy basing lectures on problems and cases, an appealing way to introduce sensitivity analysis is through the Southern Electronics cases from Chapter 4. If this case has just been discussed, it can be used as a platform for launching into sensitivity analysis. Start by constructing a simple (one bar) tornado diagram for the value to Steve of Banana ’s offer. The endpoints of the bar are based on the value of Banana ’s stock ($20 to $60, say). Because the bar crosses the $10 million mark, the value of Big Red ’s offer, it would appear that the uncertainty about the stock price should be modeled. A second tornado diagram can be created which considers the sensitivity of the expected value to 1) the probability of success (from 0.35 up to 0.5, say); 2) the value of Banana ’s stock if the EYF succeeds ($35 to $60); and 3) the value of Banana ’s stock if the EYF fails ($20 to $30). An advantage of this is to show that tornado diagrams can be used on EMVs as well as sure consequences. After showing tornado diagrams for Southern Electronics, it is natural to construct a two-way sensitivity graph for the problem. One possibility is to construct a three-point probability distribution for the value of the stock if the EYF succeeds. Denote two of the three probabilities by  p and q, and construct the graph to show the region for which Banana ’s offer is preferred. Sensitivity analysis is one of those topics in decision analysis that can be tedious and boring if done by hand but quite exciting when a computer can be used. PrecisionTree, along with Goal Seek and Data Tables in Excel provide some very powerful tools for sensitivity analysis. The software provides the capability, to determine which inputs have the largest effect on a particular output, how much change you can expect in an output when a given input changes by a defined amount, and which variables in the model change the rank ordering of the alternatives. The chapter provides step-by-step instructions for setting up a sensitivity analysis in PrecisionTree to create tornado diagrams, sensitivity graphs, and spider graphs. Instructions are also provided on how to use Goal Seek and Data Tables as sensitivity analysis tools. PrecisionTree saves the entries for the sensitivity analysis dialog box, which is helpful upon returning to the model. When creating student handout worksheets, however, you may want your students to make their own entries, in which case, make sure the dialog box is empty. In the Excel solution files, the variables or cells used in the sensitivity analysis have been shaded green. This should help in reading and understanding the analysis a little better. Topical cross-reference for problems

Cost-to-loss ratio problem Linked Decision Trees Multiple objectives  Negotiations  Net present value PrecisionTree Requisite models Sensitivity analysis Texaco-Pennzoil

66

5.6, 5.7 5.9, 5.12 5.10, 5.12, Job Offers Part II, MANPADS DuMond International 5.10 5.8 - 5.12, Dumond International, Strenlar Part II, Job Offers Part II, MANPADS 5.4 5.1-5.12, DuMond International, Strenlar Part II, Job Offers Part II, MANPADS 5.11

Tornado diagrams Trade-off weights Two-way sensitivity analysis Umbrella problem

5.12, Strenlar Part II 5.9, Job Offers Part II 5.5, 5.10, 5.11, , Job Offers Part II, Strenlar Part II, MANPADS 5.6

Solutions 5.1. Sensitivity analysis answers the question “What matters in this decision?” Or, “How do the results

change if one or more inputs change?” To ask it still another way, “How much do the inputs have to change  before the decision changes?” Or “At what point does the most preferred alternative become the second most preferred and which alternative moves into the number one spot?” We have framed the main issue in sensitivity analysis as “What matters” because of our focus on constructing a requisite model. Clearly, if a decision is insensitive to an input—the decision does not change as the input is varied over a reasonable range—then variation in that input does not matter. An adequate model will fix such an input at a “best guess” level and proceed. By answering the question, “What matters in this decision,” sensitivity analysis helps identify elements of the decision situation that must be included in the model. If an input can vary widely without affecting the decision, then there is no need to model variation in that input. On the other hand, if the variation matters, then the input ’s uncertainty should be modeled carefully, as an uncertain variable or as a decision if it is under the decision maker ’s control. 5.2. This is a rather amorphous question. The decision apparently is, “In which house shall w e live?”

Important variables are the value of the current home, costs of moving, purchase price, financing arrangements (for the current house as well as for a new one), date of the move, transportation costs, and so on. What role would sensitivity analysis play? The couple might ask whether the decision would change if they took steps to reduce driving time in the future (e.g., by deciding to have no more children). How does the attractiveness of the different alternatives vary as the family ’s size and the nature of future outings are varied? (For example, how much more skiing would the family have to do before living out of town is  preferred?) Can the family put a price tag on moving into town; that is, is there an amount of money (price, monthly payment, etc.) such that if a house in town costs more than this amount, the family would prefer to stay in the country? 5.3. Another rather amorphous question. Students may raise such issues as:

• Is a retail business the right thing to pursue? (Is the right problem being addressed?) • Does the father really want to be a retailer? • Operating costs and revenues may vary considerably. These categories cover many possible inputs that might be subjected to sensitivity analysis. To use sensitivity analysis in this problem, the first step would be to determine some kind of performance measure (NPV, cash flow, payback, profit). Then a tornado diagram could be constructed showing how the selected performance measure varies over the range of values for the inputs. The tornado diagram will suggest further modeling steps. 5.4. From the discussion in the text, the basic issue is whether some relevant uncertainty could be resolved

during the life of the option. Some possibilities include: • Obtaining a new job, promotion, or raise, • Obtaining cash for a down payment, • Learning about one ’s preferences. “Is this house right for me?” • Are there hidden defects in the house that will require repairs? • Are zoning decisions or other developments hanging in the balance? If such an uncertainty exists, then the option may have value. If not, it may be a dominated alternative. 67

5.5. Each line is a line of indifference where two of the alternatives have the same EMV. Now imagine a  point where two lines cross. At that point, all three of the alternatives must have the same EMV.

Point D is at t = 0.4565 and v = 0.3261 and is the location when all 3 EMVs equal. Thus, at D, the EMV(High-Risk Stock) = EMV(Low-Risk Stock) = $500. The exact location of D can be found using algebra, using Solver, or using a combination of algebra and Goal Seek. Because there are two unknowns, Goal Seek needs additional information. For example, to use Goal Seek: Substitute   = (9 − 14)⁄8, which corresponds to Line CDE, into the expression for EMV(High-Risk Stock). Now you have one equation and one unknown, and you can use Goal Seek to find the value of t for which the new expression equals 500. Point D is unique in that there is no other point at which all 3 alternatives have equal expected monetary values. 5.6.

Cost of protective action = C  Expected loss if no action taken =  pL C 

Set C  = pL, and solve for  p: p =  L . C 

Thus, if  p ≥  L , take protective action. C 

The only information needed is  p and  L . Note that the specific values of C  and L are not required, only their relative values. 5.7. The best way to see if it is necessary to model the uncertainty about  D is to revert to the regular costto-loss problem. If  pL < C  < C  + D, then one would not take action, and if  pL > C  + D , then the optimal choice would be to take action. However, if C  < pL < C  + D then the occurrence of damage  D does matter. The choice between taking action and not is unclear, and one would want to include  D in the decision tree. 5.8. This problem can either be set up to maximize expected crop value (first diagram below) or minimize

expected loss (second diagram below). The solution is the same either way; the difference is the perspective (crop value vs. loss). The decision tree that maximizes expected crop value is modeled in the Excel file “Problem 5.8.xlsx” along with the sensitivity reports. The expected loss from doing nothing is much greater than for either of the two measures, and so it is certainly appropriate to take some action. The expected loss for burners is almost entirely below that for sprinklers, the only overlap being between $14.5K and $15K. It would be reasonable to set the burners without pursuing the analysis further. Another argument in favor of this is that most likely the same factors lead to more or less damage for both  burners and sprinklers. With this reasoning, there would be a negligible chance that the burners would  produce a high loss and the sprinklers a low loss. A final note: Some students may solve this problem without calculating the expected loss, comparing the range of losses from burners or sprinklers if damage occurs with the $50K loss from doing nothing. However, if uncertainty about the weather is ignored altogether, the appropriate analysis has the loss ranging from $0 to $50K for no action, $5 to $25K for the burners, and $2 to $32K for the sprinklers. Because the three ranges overlap so much, no obvious choice can be made. It is, therefore, appropriate and necessary to include the probability of adverse weather and calculate the expected losses.

68

Decision tree for maximizing expected crop value.

Loss ($ 000’s) Freeze (0.5)

Exp Loss = $25K

Decision tree for minimizing expected loss.

50

Do nothing No f reeze (0.5) Freeze (0.5) Set burners Exp Loss = $12.5K to $15K

No f reeze (0.5) Freeze (0.5)

0 20 to 25

5 27 to 32

Use sprinklers Exp Loss = $14.5K to $17K

No f reeze (0.5)

2

5.9. This decision tree (shown in Figure 4.40 in the text) is modeled in the Excel file “Problem.5.9.xlsx.”

The model is a linked tree where the uncertainty node for the amount of fun is linked to cell $F$6 in the spreadsheet model (“Fun Level for Forest Job”), and the uncertainty node for the amount of work is linked to cell $G$7 in the spreadsheet model (“Salary Level for In-town Job”). The outcome nodes for the Forest Job are linked to cell $F$8 and the outcome nodes for the In-Town Job to cell $G$8. The user can then vary the weights to see that Sam will still prefer the forest job. The sensitivity analysis gives the following results: Expected Overall Score k s Forest Job In-Town Job 0.50 71.25 57.50 0.75 76.125 56.25 Thus, regardless of the precise value of k s, the optimal choice is the forest job. In fact, a much stronger statement can be made; it turns out that for no value of k s between zero and one is the in-town job

69

 preferred. Smaller values of k s favor the in-town job, but even setting k s = 0 leaves the expected overall scores equal to 60 and 61.5 for the in-town and forest jobs, respectively. Another way to show the same result is to realize that the expected overall scores are linear in the weights and in the expected scores for the individual attributes. Because the forest job has higher expected scores on both attributes, there cannot exist a set of weights that makes the in-town job have the higher overall expected score. 5.10. Using the base values of $5000 for the cash flows and 11% for the interest rate,

 NPV

5000 5000 5000 = -14000 + 1.11 + + ... + 2 1.11 1.116 = $7153.

When the cash flows are varied from $2500 to $7000, and the interest rate is varied from 9.5% to 12%, the following tornado diagram results: Cash f lows $2500

$7000

Interest rate

NPV

12%

-$4000

0

$4000

9.5%

$8000

$12,000

$16,000

This graph assumes that the cash flows vary between $2500 and $7000 and the amount is the same across all 6 years. The range of possible interest rates appears not to pose a problem; NPV remains positive within a relatively narrow range. On the other hand, NPV is more sensitive to the range of cash flows. It would be appropriate to set the interest rate at 11% for the analysis but to model the uncertainty about the cash flows with some care. The tornado diagram generated by PrecisionTree is shown in the Excel file “Problem 55.10.xlsx.” The solution file also shows a two-way data table to calculate the swing of NPV as the annual payment and the interest rate vary. Note that the table reports many more values than the tornado diagram, as the diagram only incorporates one column and one row of the table.

Tornado Graph of Decision Tree 'Problem 5.10' Expected Value of Entire Model Annual Payment (B8)

Interest Rate (B13)

-$4,000 -$2,0 00

$0

$2,000 $4,000 $6,000 $8,000 $10,0 00 $12,000 $14 ,000 $16,000 Expected Value

The problem can also be modeled by varying each year between $2500 and $7000, one at a time. This model is shown in the Excel file “Problem 5.10.xlsx.” Because of the discounting factor, the early year  payments are more influential than the later years. Said differently, NPV is more sensitive to the early year  payments than to later years. 70

Tornado Graph of Decision Tree 'Problem 5.10 by Year' Expected Value of Entire Model Year 1 (B8) Year 2 (C8) Year 3 (D8) Year 4 (E8) Year 5 (F8) Year 6 (G8) $4,500

$5,000

$5,500

$6,000

$6,500

$7,000

$7,500

$8,000

$8,500

$9,000

$9,500

Expected Value

Some students may see the error message: “Model Extends Beyond Allowed Region of Worksheet,” which means that their version of the software is limited to one decision model per workbook. . 5.11. First, sensitivity analysis by hand: Let ’s establish some notation for convenience:

Strategy A = Accept $2 billion. Strategy B = Counteroffer $5 billion, then refuse if Texaco offers $3 billion. Strategy C = Counteroffer $5 billion, then accept if Texaco offers $3 billion. EMV(A) = 2 EMV(B) = 0.17 (5) + 0.5 [ p 10.3 + q 5 + (1- p - q) 0] + 0.33 [ p 10.3 + q 5 + (1- p - q) 0] = 0.85 + 8.549  p + 4.15 q. EMV(C) = 0.17 (5) + 0.5 [ p 10.3 + q 5 + (1- p - q) 0] + 0.33 (3) = 1.85 + 5.15  p + 2.5 q.  Now construct three inequalities: • EMV(A) > EMV(B) 2 > 0.85 + 8.549  p + 4.15 q 0.135 - 0.485 q > p .

(1)

• EMV(A) > EMV(C) 2 > 1.85 + 5.15  p + 2.5 q 0.03 - 0.485 q > p .

(2)

• EMV(B) > EMV(C) 0.85 + 8.549  p + 4.15 q > 1.85 + 5.15  p + 2.5 q  p > 0.294 - 0.485 q.

(3)

Plot these three inequalities as lines on a graph with  p on the vertical axis and q on the horizontal axis. Note that only the region below the line  p + q = 1 is feasible because  p + q must be less than or equal to one.

71

p

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3

I

0.2 II

0.1 II I

IV 0

q

0

0. 1 0 .2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 1. 0

These three lines divide the graph into four separate regions, labeled I, II, III, and IV. Inequality (3) divides regions I and II. For points above this line,  p > 0.294 - 0.485 q, and so EMV(B) > EMV (C). Inequality (1) divides regions II and III. For points above this line,  p > 0.135 - 0.485 q, and EMV(B) > EMV(A). As a result of this, we know that B is the preferred choice in region I and that C is the preferred choice in region II [where EMV(C) > EMV (B) > EMV(A)]. Inequality (2) divides regions III and IV. For points above this line,  p > 0.03 - 0.485 q, and EMV(C) > EMV (A). Thus, we now know that C is the preferred choice in region III [where EMV(C) > EMV(A) and EMV(C) > EMV(B)], and A is preferred in region IV. Thus, we can redraw the graph, eliminating the line  between regions II and III: p

1.0 0.9 0.8 0.7 p > 0.15

0.6

q > 0.35

0.5 0.4 B

0.3 0.2  A

0.1

C

0

q

0

0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 1. 0

The shaded area in the figure represents those points for which  p > 0.15 and q > 0.35. Note that all of these  points fall in the “Choose B” region. Thus, Liedtke should adopt strategy B: Counteroffer $5 billion, then refuse if Texaco offers $3 billion.

72

The two-way SA feature of PrecisionTree can be used here, but with limited results. The widest any one  probability can vary is up to the point that the remaining probabilities sum to 0 or sum to 1. In this case, the most we can vary “Probability of Large Award” is between 0 and 0.5. The model is in the Excel file “Problem 5.11.xlsx.” Because cell D18 (“Probability of No Award”) contains the formula “=1 – (D12 + D16)”, “Probability of Large Award” can vary at most from 0 and 0.5. The same is true for “Probability of Medium Award”. In deciding how much you can vary the variables in a two-way analysis, the complete rectangle defined by the range of values must fit inside the large triangular region shown above. The two-way sensitivity graph is on the second tab, and is hard to read. Where there is a kink in the 3D graph, then the optimal alternative has changed. In other words, the alternative with the maximum EMV has changed. Easier to read, butperhaps not available in the student version, is the strategy graph. The workbook contains two strategy graphs. The second one uses 50 steps, and thus computed a combination of 2500 values. From these graphs, we clearly see which alternative has the maximum EMV. Remember, the PrecisionTree is limited to working with rectangular regions, and the whole region needs to fit within the triangle. Therefore, the PrecisionTree two-way SA provides a limited and incomplete analysis compared to the algebraic solution previously given. If the student version does not allow strategy graphs, then by calculating the differences from row to row, we can see where the value changes (i.e., where the kink occurs). As shown below, the differences show which alternative is optimal. Taking the differences, as we have done, only works for linear models.

Prob Large A 0 0.05 0 2 2 0. 05 2 .0975 2 .2225 0.1 2.355 2.48 0.15 2.6125 2.7375 0.2 2.87 2.995 0.25 3.1275 3.2525 0. 3 3 .4147 3 .6222 0.35 3.84215 4.04965 0. 4 4 .2696 4 .4771 0.45 4.69705 4.90455 0.5 5.1245 5.332 Successive differences: 0.0975 0.2225 Accept $2B 0.2575 0.2575  preferred. 0.2575 0.2575 0.2575 0.2575 Counter and 0.2575 0.2575 accept $3B 0.2872 0.3697 0.42745 0.42745 0.42745 0.42745 Counter and 0.42745 0.42745 refuse $3B 0.42745 0.42745

0.1 2.09 2 .3475 2.605 2.8625 3.12 3.40225 3 .8297 4.25715 4 .6846 5.11205 5.5395 0.2575 0.2575 0.2575 0.2575 0.28225 0.42745 0.42745 0.42745 0.42745 0.42745

0.15 2.215 2 .4725 2.73 2.9875 3.245 3.60975 4 .0372 4.46465 4 .8921 5.31955 5.747 0.2575 0.2575 0.2575 0.2575 0.36475 0.42745 0.42745 0.42745 0.42745 0.42745

Prob 0.2 2.34 2 .5975 2.855 3.1125 3.3898 3.81725 4 .2447 4.67215 5 .0996 5.52705 5.9545

Medium Award 0.25 0.3 2.465 2.59 2 .722 5 2 .847 5 2.98 3.105 3.2375 3.37735 3.5973 3.8048 4.02475 4.23225 4 .452 2 4 .659 7 4.87965 5.08715 5 .307 1 5 .514 6 5.73455 5.94205 6.162 6.3695

0.2575 0.2575 0.2575 0.2773 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745

0.35 2.715 2. 972 5 3.23 3.58485 4.0123 4.43975 4. 867 2 5.29465 5. 722 1 6.14955 6.577

0.4 2.84 3. 097 5 3.3649 3.79235 4.2198 4.64725 5. 074 7 5.50215 5. 929 6 6.35705 6.7845

0.45 2.965 3. 222 5 3.5724 3.99985 4.4273 4.85475 5. 282 2 5.70965 6. 137 1 6.56455 6.992

0.5 3.09 3 .3524 5 3.7799 4.20735 4.6348 5.06225 5. 489 7 5.91715 6. 344 6 6.77205 7.1995

0.2575 0.2575 0.2575 0.2575 0.2575 0.26245 0.2575 0.2575 0.2575 0.2674 0.3499 0.42745 0.2575 0.27235 0.35485 0.42745 0.42745 0.42745 0.3598 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745 0.42745

5.12. The first thing to notice is that the net annual cost values have changed for Barnard and Charlotte, but

not for Astoria, as Astoria has no rental. Specifically, when incorporating the taxes on rental income and depreciation, Barnard’s net annual cost has dropped by almost $5,000 and Charlotte’s by over $3,000. Thus, instead of there be an $11,000 difference in annual costs among the 3 properties, there is only a $6,000 difference. This helps improve the attractiveness of Barnard and Charlotte. The Excel file for the model is “Problem 5.12.xlsx” and the sensitivity graphs are found in “Problem 5.12 Sensitivity Graphs.” Remember: what we call sensitivity graphs, PrecisionTree calls strategy graphs. Many of the sensitivity analysis insights discussed in the chapter also hold here. Astoria is still the cheapest no matter the value of the input variable for all the 8 input variables, except monthly rent. The main difference, as mentioned above, is that the properties are now closer in value. For example, when varying the interest rate between 4% and 8%, Astoria’s net annual cost is always less than the other 2 properties net annual cost values. At an interest rate of 4%, Charlotte is within $700 of Astoria’s net annual cost, but

73

 previously, when we did not incorporate depreciation and taxes on the rental income, then the closest Charlotte came to Astoria was $4,000. A subtle difference between the analyses is the influence of the tax rate. Previously, taxes could swing Charlotte’s and Barnard’s net annual cost by $4,000 to $5,000; whereas now it only swings it by only $2,000 to $2,500. Remember, that tax rate’s impact is counterintuitive in that costs go down as the tax rate increases. When incorporating a more complete tax analysis, the effect of tax rate overall is muted by taxes on the rental income. While the months-occupied variable never changes the rank ordering of the properties, the analysis does show a significant impact on net annual cost. Previously, when “Months Occupied” = 0 months, then Barnard’s net annual cost was nearly $60,000; whereas now it is only $48,218. The monthly rent value again can make Charlotte the least expensive property. Previously, when Charlotte’s rent was $2,100 or more per month, than it was cheaper than Astoria. Now, Charlotte’s rent needs only be above $1,900 per month for it to have a smaller net annual cost. Barnard is always the most expensive no matter the monthly rent (up to $2,900/month), but now it is within $200 at $2,900/month. The Excel file “Problem 5.12 Tornado Diagrmas.xlsx” report the 3 tornado diagrams and now show that the loan’s interest rate is more influential than months occupied for both Barnard and Charlotte. For example, previously months occupied could vary net annual cost by $24,000 for Barnard ($2,000 monthly rent), and now varies it by approximately $16,000. This is the buffer we mentioned. Because the interest rate can vary Barnard’s net annual cost by $19,276, interest rate has become the most influential variable for Barnard. The same holds true for Charlotte; interest rate is now more influential than months occupied. We choose to run a two-way SA on “Monthly Rent” and “Interest Rate.” Remember that Ch arlotte’s rent is set at $500 less than Barnard’s. The Excel file is “Problem 5.12 Two-Way SA.xlsx.” As expected, the results show a larger region when Charlotte is the cheapest than the previous analysis, and, again, Barnard is never the cheapest within this rectangle. The two-way graphs show that when the interest rate is 4%, then Charlotte has a lower net annual cost than Astoria when Charlotte’s rent is $1,629 or more.

Sensitivity of Decision Tree 'Problem 5.12' Expected Value of Node 'Housing Decision' (B34)

$36,000 $34,000    e $32,000    u     l    a $30,000    V     d    e $28,000    t    c    e    p $26,000    x    E

8.0% 7.4% 6.9% 6.3% Interest Rate (B4) 5.7% 5.1% 4.6%

$24,000 $22,000 $20,000

   0   6   1    0   8   7   7   3    5    5   4   9  ,    5  ,    6    2    4  ,    7    1    0    1   1  ,    8    0    6  ,    9    1   1    8    1     $    $  ,    0    7    7  ,    1 4.0%    1   1    5    3     $    $    $  ,    1    4    9  ,    2    2    2    4  ,    3    2    2    6    1    1    0     $    $    $  ,    4    0    5  ,    2    8    6  ,    2     $    $    7    2    2  ,    7  ,    7     $    $  ,    8    2    2  ,     $    $    2    2     $    $     $

Monthly Rent (Barnard) (C10)

74

Strategy Region for Node 'Housing Decision' 8.0% 7.5% 7.0%     )    46.5%    B     (    e    t    a    R6.0%    t    s    e    r    e    t    n    I 5.5%

Astoria St Charlotte St

5.0% 4.5% 4.0%    0    0    4  ,    1     $

   0    0    6  ,    1     $

   0    0    8  ,    1     $

   0    0    0  ,    2     $

   0    0    2  ,    2     $

   0    0    4  ,    2     $

   0    0    6  ,    2     $

Monthly Rent (Barnard) (C10)

   0    0    8  ,    2     $

   0    0    0  ,    3     $

5.13. The consequence measure “Appreciation + Equity – Net Annual Cost” substantially changes the

 preference order. Barnard St is now ranked first at $15,645 and Astoria St. is ranked last at $5,582. With this measure, larger values are more preferred. The Excel file “Problem 5.13.xlsx” contains the spreadsheet model and note that the values given in the text for Barnard and Charlotte are off. Charlotte should be $9,584.

Strategy Region of Decision Tree 'Problem 5.13' Expected Value of Node 'Decision' (B34) With Variation of Interest Rate (B4)

$25,000

$20,000    e    u     l    a$15,000    V     d    e    t    c    e$10,000    p    x    E

Astoria St Barnard St Charlotte St

$5,000

$0    %    5  .    3

   %    0  .    4

   %    5  .    4

   %    0  .    5

   %    5  .    5

   %    0  .    6

   %    5  .    6

   %    0  .    7

   %    5  .    7

   %    0  .    8

   %    5  .    8

Interest Rate (B4)

The Barnard Street house is the most preferred property for all the variables listed in Table 5.2, except “Appreciation” and “Months Occupied,” for every value in the range. For example, the sensitivity graph for interest rate with between 4% and 8% is shown above. For each interest rate in this range, Barnard always has the maximum value. The only time the rank ordering of the alternatives change is for the variables: “Appreciation” and “Months Occupied.”

75

Strategy Region of Decision Tree 'Problem 5.13' Expected Value of Node 'Decision' (B34) With Variation of Months Occupied (C18) $25,000 $20,000 $15,000    e    u     l    a $10,000    V     d    e    t    c    e $5,000    p    x    E

Astoria St Barnard St Charlotte St

$0 -$5,000 -$10,000

   2   -

0

2

4

6

8

   0    1

   2    1

   4    1

Months Occupied (C18)

Strategy Region of Decision Tree 'Problem 5.13' Expected Value of Node 'Decision' (B34) With Variation of Appreciation Percentage (B26) $50,000 $40,000 $30,000 $20,000

   e    u     l    a $10,000    V     d $0    e    t    c    e -$10,000    p    x    E

Astoria St Barnard St

-$20,000

Charlotte St

-$30,000 -$40,000 -$50,000    %    0    0  .    4   -

   %    0    0  .    2   -

   %    0    0  .    0

   %    0    0  .    2

   %    0    0  .    4

   %    0    0  .    6

Appreciation Percentage (B26)

   %    0    0  .    8

   %    0    0  .    0    1

   %    0    0  .    2    1

From 0 to 12 months, Barnard’s “Appreciation + Equity – Net Annual Cost” value is always larger than Charlotte’s, but is only larger than Astoria’s when months occupied greater than or equal to 5 months. “Appreciation” is now the most influential variable. Notice that the y-axis now goes from -$50,000 to $50,000, showing that as the appreciation percentage changes, the “Appreciation + Equity – Net A nnual Cost” value can vary up to $100,000. For annual appreciation rates less than 2.8%, Astoria is the better choice, then, after 2.8%, Barnard is more preferred. Part c asks the students to append a chance node to Figure 5.14 to account for the uncertainty in the appreciation percentage. You can either let the students define their own chance node (number of

76

outcomes, probabilities, etc.), or you can give them specific instructions. The file “Problem 5.13 part c.xlsx” contains our solution, and, as shown below, it has 3 outcomes. Note also this is a linked tree. You can tell this because the branch values are not payoffs, but the input values used to calculate the payoffs.

Running a two-way SA on the probabilities “p 1” and “q 1” in Figure 5.14 shows the Barnard is optimal for all probability values. We set the steps equal to 50 for each “p 1” and “q 1” resulting in 2500 evaluation  points. On this measure, Barnard dominates the other two properties. In part d, It is clear that “Net Annual Cost” is attempting to capture the real, or out-of-pocket, costs experienced by home ownership. These are items that Sanjay and Sarah will be paying cash for. The other consequence (“Appreciation + Equity – Net Annual Cost”) is harder to understand. In some ways, it is measuring the annual value Sanjay and Sarah are earning in their home. When they sell their home, the will realize both the appreciation and equity. This, however, does not happen incrementally as this measure indicates. Either it happens when they sell their house or it comes into play when they take out a homeequity loan. In the screenshot of the model, we see that Appreciation + Equity – Net Annual Cost = $35,470 for Barnard when there is full occupancy, $8,000 in repair and upkeep, and a 9% appreciation rate. This does not mean that Sanjay and Sarah have an additional $35,000 in their hands nor can they take a loan out for that amount.

Strategy Region for Node 'Housing Decision' 100%

    ) 80%    2    1    J     (    t    s    o    C 60%    U    &    R     h    g    i 40%    H     f    o     b    o    r    P

Barnard

20%

0%

   %    0

   %    5

   %    0    1

   %    5    1

   %    0    2

Prob of 12 Months Occup (J11)

77

   %    5    2

   %    0    3

   %    5    3

Case Study: DuMond International 1. If the changes suggested by Dilts and Lillovich are incorporated, EMV(Current product) increases to

$930,000, but EMV(New product) stays the same at $1,100,000. Thus, the new product would be preferred. If the changes suggested by Jenkins and Kellogg are incorporated, EMV(New product) drops to $925,000. Recall, though, that Jenkins and Kellogg were satisfied with Milnor ’s analysis of the current product, so their changes still leave the new product as the preferred choice. If all of the suggested changes are incorporated, EMV(New product) = $925,000 and EMV(Current  product) = $930,000. Thus, the combination of optimism about the current product and pessimism about the new leads to a scenario in which the current product barely has a better EMV than the new one. Because no one embraced all of the changes, though, all board members should be convinced that the new  product is the preferred choice. This case is set-up as a spreadsheet model in the Excel file “Dumond Case.xlsx.” The decision tree is structured so that it references cells in the spreadsheet, so that the user can vary the parameters of the model, and see how the preferred decision changes.

Case Study: Strenlar, Part II 1. The solution to this case depends to a great extent on how the decision was modeled in the first place.

The sensitivity analysis that follows is based on the model discussed above in the solution for Strenlar, Part I, in Chapter 4. The Excel solution file is “Strenlar Case Part II.xlsx.” The table shows the parameters for which we wish to perform a sensitivity analysis. For the Reject PI option, this includes P(Win Lawsuit), P(Manufacturing Process Works), legal fees, and profits. For the Accept Job option, the parameters are the interest rate, gross sales, and P(Manufacturing Process Works). For the Accept Lump Sum and Options alternative, the analysis focuses on the interest rate, stock price if Strenlar succeeds, and P(Manufacturing Process Works).

P(Win Lawsuit) P(Mfg Process Works) Gross Sales Legal Fees Fixed Cost Variable Cost (% of Gross Sales) PI Stock Price Interest Rate

Pessimistic 50% 70% $25,000,000 $90,000 $8,000,000 80.0% $48.00 5.0%

Base 60% 80% $35,000,000 $60,000 $5,000,000 62.9% $52.00 7.5%

Sensitivity Analysis Table for Strenlar Model.

78

Optimistic 75% 90% $45,000,000 $20,000 $2,000,000 60.0% $57.00 12.5%

Because Refuse PI depends on profits, which in turn depends on gross sales, we have chosen to expand the model slightly. We have assumed as base values that fixed costs would equal $5 million and that, for sales of $35 million, the variable cost would be $22 million. This leaves profits of $8 million as specified in the case. Specifically, we assumed: Profit = Gross Sales – Variable Cost – Fixed Cost, where Fixed Costs = $5 million and Variable Cost = (35/22) * Gross Sales.  Now it is possible to run sensitivity analysis using PrecisionTree on all three variables (Gross Sales, Variable Costs, Fixed Costs) and obtain comparable results for the Refuse PI and Accept Job alternatives. The tornado graph for the Refuse PI alternative shows that Variable Cost and Gross Sales are the two most influential variables. These two variables are also the ones to cause Fred’s payoff to drop below $3.9 million.

Tornado Graph of Refuse PI Expected Value of Entire Model Variable Costs (B14) Gross Sales (B5) Fixed Costs (B13) Prob of Winning Case (B4) Prob(Mfg Process Works) (B3) Legal Fees (B8) Interest Rate (B9) PI Stock Price (B10)    0    0    0  ,    0    0    0  ,    1     $

   0    0    0  ,    0    0    0  ,    2     $

   0    0    0  ,    0    0    0  ,    3     $

Expected Value

   0    0    0  ,    0    0    0  ,    4     $

   0    0    0  ,    0    0    0  ,    5     $

   0    0    0  ,    0    0    0  ,    6     $

   0    0    0  ,    0    0    0  ,    7     $

The tornado diagram for Accept Job shows that Fred’s payoffs are always below $3.2 million and that Gross Sales is the most influential variable pushing his payoff below $2 million. The sensitivity graph shows that Refusing PI has the better payoff until gross sales are less than $27 million.

Tornado Graph of Accept Job Expected Value of Entire Model Gross Sales (B5) Prob(Mfg Process Works) (B3) Interest Rate (B9) Legal Fees (B8) Prob of Winning Case (B4) PI Stock Price (B10) Fixed Costs (B13) Variable Costs (B14)    0    0    0  ,    0    0    8  ,    1     $

   0    0    0  ,    0    0    0  ,    2     $

   0    0    0  ,    0    0    2  ,    2     $

   0    0    0  ,    0    0    4  ,    2     $

Expected Value

79

   0    0    0  ,    0    0    6  ,    2     $

   0    0    0  ,    0    0    8  ,    2     $

   0    0    0  ,    0    0    0  ,    3     $

   0    0    0  ,    0    0    2  ,    3     $

Strategy Region of Decision Tree 'Strenlar' Expected Value of Node 'Decision' (B23) With Variation of Gross Sales (B5) $5,500,000 $5,000,000 $4,500,000    e$4,000,000    u     l    a    V$3,500,000     d    e    t    c$3,000,000    e    p    x    E$2,500,000

Refuse PI Accept job

$2,000,000

Accept lum sum

$1,500,000 $1,000,000    0    0    0  ,    0    0    0  ,    0    2     $

   0    0    0  ,    0    0    0  ,    5    2     $

   0    0    0  ,    0    0    0  ,    0    3     $

   0    0    0  ,    0    0    0  ,    5    3     $

   0    0    0  ,    0    0    0  ,    0    4     $

   0    0    0  ,    0    0    0  ,    5    4     $

   0    0    0  ,    0    0    0  ,    0    5     $

Gross Sales (B5)

The only other variable to cause a change to the rank ordering of the alternatives is Variable Costs. The sensitivity graph shows the variable cost only affects the Refuse PI alternative. If he takes the job at PI, then his royalties are tied to gross sales and not profit.

Strategy Region of Decision Tree 'Strenlar' Expected Value of Node 'Decision' (B23) With Variation of Variable Costs (B14)

$7,000,000 $6,000,000    e$5,000,000    u     l    a    V     d$4,000,000    e    t    c    e    p    x    E$3,000,000

Refuse PI Accept job Accept lum sum

$2,000,000 $1,000,000

   %    0    0  .    5    5

   %    0    0  .    0    6

   %    0    0  .    5    6

   %    0    0  .    0    7

   %    0    0  .    5    7

   %    0    0  .    0    8

   %    0    0  .    5    8

Variable Costs (B14)

. Another intriguing possibility is to see how pessimistic Fred could be before Refuse PI is no longer optimal. Consider Scenario 1 in the table. In this pessimistic scenario, all the variables are at their base value (best guess value) or they are less, and four of them are at their lowest value. Even in this case, refusing PI has a larger EMV than accepting the job (compare $1.77 million to $1.74 million). In Scenario 2, we kept all the values from Scenario 1, except we increased the stock price to its maximum upper bound of $57. Doing so only changed the lump-sum payoff, from $0.9 million to $1.4 million, a $500,000 gain, but not enough to overtake the other alternatives. Scenario 3 keeps all the same values as in

80

Scenario 1, except for increasing the fixed cost value only by $89,505. The EMV of Refuse PI and of Accept Job are equal in Scenario 3. In other words, Fred would have to be relatively pessimistic overall  before the Reject PI option is not optimal Scenario 1

Scenario 2

Scenario 3

P(Win Lawsuit)

60%

(Base Value)

60%

60%

P(Mfg Process Works)

70%

(Pessimistic)

70%

70%

$25,000,000

(Pessimistic)

$25,000,000

$25,000,000

$90,000

(Pessimistic)

$90,000

$90,000

$5,000,000

$5,000,000

$5,089,505

60%

60%

$48.00

(Base Value) (Slightly Pessimistic) (Pessimistic)

$57.00

$48.00

7.5%

(Base Value)

7.5%

7.5%

Gross Sales Legal Fees Fixed Cost Variable Cost (% of Gross Sales) PI Stock Price

60 %

Interest Rate

Case Study: Job Offers, Part II 1. The sensitivity analysis gives the following results:

P($1500) 0 1

Expected Overall Score Madison MPR 50 36 58 36

Pandemonium 50 50

Thus, there is no question about the preference for Madison Publishing; with the probability set at the most  pessimistic value of 0, Madison and Pandemonium are equivalent. MPR, of course, is never in the running at all. A one-way sensitivity plot for the probability of disposable income being $1500 for the Madison job generated by PrecisionTree is shown in the second worksheet. The one-way plot does not show the alternative policies, only the range of possible outcome results. 2. The algebraically generated two-way sensitivity graph is shown below. The labels indicate the optimal

choice in each region. The graph makes good sense! When the weights on snowfall and income are small (and hence the weight for the magazine score is high), Pandemonium is the best, reflecting its strong showing in the magazine dimension. Likewise, when the magazine weight is low, MPR is best, reflecting its strength in the income and snowfall dimensions, but poor showing with the magazine score.

1.0

ki

0.8 0.6

MPR

0.4 0.2

Madison Pandemonium

0.0 0.0

0.2

0.4

0.6

81

0.8

ks

1.0

Sensitivity analysis using PrecisionTree. The decision model is in the Excel file “Job Offers Case II.xlsx.” Question 1: To vary only the probability of $1,500 from 0 to 1, you must choose Spreadsheet Cell in the SA dialog box, for Type of Value. Question 2: You can use PrecisionTree to create the same optimal regions as w e did above algebraically. To do so, use the model and run a two-way SA, where you vary the weights for snowfall on the x-axis and income on the y-axis. The weight for magazine rating is the formula 1 – (snowfall wt + income wt). Vary  both these weights from 0 to 1. This will create a rectangular region not the desired the triangular region. See figure below. However, we can simply ignore the region above the main diagonal. We can do this for weights, but not for probabilities. With the weights, if one of this is negative; the model still calculates a value. The value it calculates is meaningless, which is w hy we ignore the upper triangular region of the two-way analysis. We cannot do this for probabilities, because probabilities can never be negative. Once PrecisionTree encounters a negative probability, it will not calculate any results.

Strategy Region for Node 'Decision' 1.00

0.80

    ) 0.60    4    F     (    t    W    e    m    o    c    n0.40    I

Madison Publishing MPR Manufacturing Pandemonium Pizza

0.20

0.00

   0    0  .    0

   0    2  .    0

   0    4  .    0

   0    6  .    0

   0    8  .    0

   0    0  .    1

Snowfall Wt (G4)

Case Study: MANPADS

See file “MANPADS.xlsx” for the decision tree and the full analysis. 1. Decision tree shown below. Using the values given in the case,

E(Cost | Countermeasures) = $14,574 million E(Cost | No Countermeasures) = $17,703 million Putting the tree together is relatively straightforward. The only odd aspect of the model is working out P(Interdiction | Attempt, Countermeasures). The information in the case is not entirely clear; in this solution f  is interpreted as the extent to which P(Interdiction | Attempt) is increased. The wording in the case suggests that  f  should be interpreted as the proportional decrease in P(Interdiction | Attempt), but that 82

does not necessarily make sense. In the original article, the based value for  f was set to 0, and the range was 0-0.25. You may want to be lenient with the students on this point.

83

2. Risk profiles shown below. The real difference is that adopting countermeasures greatly reduces the

chance of a very large loss. So the policy question is whether the cost is worth that reduction in risk. The cumulative graph (see the Excel file) shows that neither alternative stochastically stochastically dominates the other.

Probabilities for Decision Tree 'MANPADS' Choice Comparison for Node 'Decision' 80% 70% 60% 50%

   y    t    i     l    i     b    a40%     b    o    r    P

Countermeasures

30%

No Countermeasures

20% 10% 0%

   0    0    0    0    2   -

0

   0    0    0    0    2

   0    0    0    0    4

   0    0    0    0    6

   0    0    0    0    8

   0    0    0    0    0    1

   0    0    0    0    2    1

3. See the Excel file for both one-way and two-way sensitivity analyses. Changing the inputs in this model

can result in changes to both alternatives, so this sensitivity analysis has been run on the difference between the expected costs (B45 on "MANPADS Tree"). When the difference is positive, positive, Countermeasures Countermeasures has the lower expected cost. Also, without any real guidance on reasonable ranges, I've varied each input by plus or minus 25%. The tornado and spider charts in the Excel file show that the top three variables in terms of sensitivity are P(Attempt), Economic Loss, and P(Hit | Attack). Moreover, only the first two could result in No Countermeasures being preferred. But remember that this is a one-way analysis. The two-way analysis in the file shows how the strategy changes as we vary P(Attempt), Economic Loss, and P(Hit | Attack). (So it is actually a three-way analysis.) When P(Hit | Attack) = 0.80, the region for No Countermeasures is fairly small. small. When P(Hit | Attack) decreases to 0.60, though, the No Countermeasures region is much larger. So it would not take too much movement of these three variables together to result in  No Countermeasures having having the lower expected expected cost.

84

CHAPTER 6 Organizational Use of Decision Analysis Notes This chapter puts the activity of decision analysis directly into an organizational context. This context is very different from that of an individual thinking alone about a personal decision. Generally, organizational context is given short shrift in textbooks about the methodology of decision analysis. Yet it is where decision analysts will earn their keep. And it is where the creation of value from better decisions will be most strongly demonstrated. Efforts to create organizational buy-in and to create change will have huge rewards.

This doesn’t mean that individual ability and ideas don’t matter. Organizational success depends strongly on individual creativity. This is another area that may not be treated in textbooks on decision analysis. How to foster that creativity and how to capture the value it creates is an integral part of this chapter and can be a huge component of value creation in the organization. A six-step process for the organizational use of decision analysis is described in the chapter. This approach laces together a decision board and a strategy team as the work of the organizational decision analysis  proceeds through the six steps. Not only does it help help in the selection selection of the path down which to proceed, proceed, but it also looks towards the final desired results and focuses effort on the change necessary to achieve those results. Creativity comes into play in several of the steps. Creativity is needed to identify issues, determine alternatives and develop a vision statement. A key in the organization is to establish a corporate culture and tradition that stay out of the way of creativity. We simply had more to say about creativity than would fit into the b ook pages. So there is an online Supplement to Chapter 6. This has many more ideas on reducing blocks to creativity and additional creativity techniques. The supplement also contains questions and problems of its own. The solutions to those problems are included in a separate section at the end of this document. Topical cross-reference for problems (S prefix indicates problems in the Online Supplement). Analytical complexity 6.3 Barriers to implementation implementation 6.1, Eastman Kodak Brainstorming 6.10 Commitment to action 6.2 Creativity 6.4, 6.8, 6.10, 6.12, Eastman Kodak, S6.2, S6.3, S6.4 Creativity blocks S6.1 Decision board 6.2 Decision quality 6.13 Implementation 6.1 Incubation 6.11 Means objectives 6.6, 6.9 Organizational characteristics characteristics 6.3, 6.7 Six-step process 6.4 Spider diagram 6.13 Strategy table 6.5

This technical note was written by Samuel E. Bodily, John Tyler Professor at the University of Virginia Darden Graduate School of Business Administration. Administration. Copyright  2013 by Samuel E. Bodily. It rd  accompanies Making Hard Decisions Decisions with Decision Tools, Tools, 3  Ed , by Robert T. Clemen and Terrence Reilly, Mason, OH: South-Western. Some answers were contributed by Robert T. Clemen from a previous edition of the book . 85

Solutions 6.1. Here we are looking to tie the concepts from the chapter to the students’ personal experiences. Students may mention barriers such as turf wars, lack of people skills, inadequate selling of the proposal, inadequate attempts to get buy-in from stakeholders, missed attributes, lack of transparency, incorrect understanding of who will make the decision, to list just a few. Students can explain how ideas from the chapter can reduce these barriers. For example, bringing together a carefully composed decision board and analysis team through the lacing of the six-step process can improve buy-in from stakeholders, reduce missed attributes, improve transparency and improve identification of the decision maker(s). 6.2. First the decision-board participants must be wisely chosen. They must be those individuals who have a stake in the decision, such as project managers or engineers on the project that the decision most affects. The participants must also be compatible in that they work well together and can communicate freely among themselves.

Second, a simple vision statement must provide the decision-board with a definitive purpose to the project and set specific goals for the project. This will give incentive for each board member to commit to the  project and follow follow through until until the end. 6.3. a. If you are single and your retirement only affects yourself, this decision would have a low organizational complexity. complexity. However if you are a married person with children and your retirement affected all of these people, then the decision could have a higher organizational complexity. complexity.

This decision is analytically complex due to the uncertainty of the different investments and the usually large number of investment options to choose between. Therefore this decision may be in the upper right quadrant or the lower right quadrant. b. This would generally have a high level of organizational complexity, unless the organization was very small, because of the large number of people with different personalities and because there would most likely be conflicts and power struggles among the players. It may also have a fairly high degree of analytical complexity due to the fact that you would be uncertain of the reactions of the many choices that could be made for each position and the group dynamics involved at each level of the newly created organizational chart. Therefore Therefore it might be in the upper right quadrant, possibly the upper left quadrant. c. The organizational complexity depends on the size and shape of this organization. If this were a small organization with very few divisions and/or employees, it would have a low complexity. However if this was a company of the size, say, of Johnson & Johnson, this decision would have a high level of organizational complexity. This would in most cases have a high degree of analytical complexity. complexity. The uncertainties, the dynamics and the many interrelated and numerous factors would make this analytically complex. Therefore, probably upper right. 6.4. Here students bring in personal experiences and relate the many positives of the six-step decisionmaking process to real life problems. 6.5. In the following diagram we show a set of choices using brackets that fits with the “Broad-based Expansion” designation. It is a consistent strategy. It might be hard to argue that it is necessarily combining the best strategies of that figure without more information about the context. Each alternative strategy comprises a consistent sent of choices, including one option in each column.

86

Each alternative strategy comprises a consistent set of choices  —one  — option under each decision. Strategy Short-Term Profit Improvement Current Focused Reallocation Joint Venture and License Focus Broad-Based Expansion

International Focus Worldwide presence Build critical sales force mass in top 12 countries

Domestic Marketing Maintain current sales force levels

Expand sales force to maintain current officeBuild, acquire based companies, or  coverage and  joint venture to increased build critical hospital and mass in key physician Germany, coverage Japan, etc. Maintain  Acquire current sales company in force levels Germany; and increase out-license in advertising Japan, etc.

Licensing and Joint Ventures Out-licensing limited to lowpotential drugs and focused in-licensing with joint research agreements

Generics Stay out and promote trademark to prevent share erosion

R&D Strategy Current

Concentrate on key product classes pro duct Selectively pro enter to use duct  Aggressive manufacturing  Aggressively pro out-licensing to capacity license to duct achieve foreign exploit potential and strengths in-licensing to Cautious add domestic entry strategy products based on Expand entire marketing program  Aggressive strength licensing and key joint venture

Bodily & Allen, Interfaces 29:6

6.6. Here we are looking for the students to experience the process of identifying fundamental objectives and listing means objectives, from which they may generate new alternatives. Suppose a student picks the decision about a job to take, in other words to “choose an employer.” One of their objectives might be to “have the opportunity to work with leaders in my area of expertise.” An example means objective might be to meet and spend time with leaders in that area of expertise. The student might then imagine some alternatives such as “go to a professional meeting with leaders in my area of expertise” or “send a piece of my best writing on new ideas” to a leader in the area of expertise. 6.7. Students may reference examples of young companies in Silicon Valley where the culture is one of enjoying oneself creatively, with few rules and walls (both figuratively and literally). Students’ answers would reflect their own views about freedom to create. An organization that will thrive on chaos will most likely be relatively small, with managers that are highly flexible and creative themselves. In fact, the idea of a “shell” or “virtual” organization may be appropriate; such a firm may provide a minimum amount of structure required legally and for the sake of outsiders. However, within the shell the organization may be extremely flexible, or even constantly in flux. Imagine a holding company which permits its subsidiaries to metamorphose, segment, and recombine at will. Managers in such a situation must be the antithesis of  bureaucrats. They must eschew traditional empire-building because such behavior attempts to lock in certain organizational structures. Success and excellence must be defined in new ways; meeting next quarter’s sales quota will probably not be a universally appropriate objective. Other issues include the kinds of people one hires and nurturing a corporate culture that values change and evolution while disdaining the status quo. 6.8. Those of us who attended formal schooling need to take a step back and determine whether we have had the curiosity and creativity schooled out of us. And if we are less curious and creative we need to take steps to get that curiosity and creativity back into our everyday lives. Today’s educators need to be aware of the importance of curiosity and creativity in today’s schoolchildren. Instead of simply correcting a child for an incorrect answer, the educator should try to see in what ways that answer was creative and compliment the student for being creative before revealing the correct answer.

87

6.9. Here are some possible alternatives linked to specific objectives: Require safety features – Require the improvement of current safety features such as seatbelts and • airbags. Educate the public about safety – Require that those about to obtain a license receive a minimum • of car safety education. Enforce traffic laws – Require adults to take refresher driving courses (education) if they • accumulate too many traffic tickets? Have reasonable traffic laws – Have a citizen/police panel that reviews traffic laws. • Minimize driving under the influence of alcohol – Provide rides home from social gatherings that • involve heavy use of alcohol; establish programs that remove keys from those who may have consumed too much alcohol. 6.10. This is a good exercise to show how brainstorming and value-focused thinking can be combined. A good list of alternatives will include suggestions that target all aspects of PeachTree’s fundamental objectives. Other than the typical suggestions of distributing flyers, posters, and placing advertisements in local college papers, PeachTree might sponsor a contest, have an open house, have managers be guest speakers at local schools, and so on. 6.11. This should get the students to think about unconscious incubation from their own lives, maybe about the idea that came to them in the shower or while taking a run. It can be informative about how to set up opportunities for incubating ideas in the future. You can help students realize that while it is fantastic when it works, it won’t always produce break-through ideas. It is said that the inventor of television first had the idea for sending electronic pictures as a stream of rows of dots when, as a kid on the farm, he saw rows and rows of individual plants in a field forming a mosaic picture. 6.12. Some additional examples could be the following: Internal Management of firm: Inventory reduction, operational improvements, reduction of SKUs, • External Arrangements: Mergers and acquisitions, data mining and sharing. • 6.13. Answers will be very individual. Case Study: Eastman Kodak There are a couple of points to this case. One is that the kind of decision making needed early in the typical  business life-cycle life-cycle may not be the same as those those needed late in in the typical life-cycle. Early Early on the kind of creativity needed would be largely related to the technological challenges, relating relating to how to be the world’s leader in putting chemical emulsions onto plastic film. The mindset may be that of the chemical engineer.

Late in its business life-cycle Kodak found itself making decisions regarding two different paths. One path choice was about how to contract successfully a chemical film business that was being disrupted by digital technology. Another path choice related to developing potential business(es) that might renew the company. By that time the management of Eastman Kodak was primarily chemical engineers who had grown to their  positions by being being very good at laying laying down chemical chemical emulsions onto onto plastic film. film. 1. Virtually any of the creativity-enhancing techniques could be used to generate alternatives for Kodak. You can ask students what ideas these techniques produced and which were more useful. Ask also how these techniques might help someone get out o f the mindset that so successfully brought the company success, but which might not now lead to success. 2. You can imagine how difficult it would be for Kodak to come by organizational change. It could have  been helpful perhaps perhaps in this situation situation to add individuals individuals to the the decision board that that were completely completely outside the company and even the industry. For the decision about “renewing” it might be helpful to set up a board that was largely independent of existing management. This isn’t easy for most organizations to do well.

88

Chapter 6 Online Supplement: Solutions to Questions and Problems S6.1. Most students will imagine the engineer as male mainly because women rarely wear neckties and their hair is usually longer. Therefore, you would not necessarily mention these factors if it were a female. S6.2. This is a personal question and the questions generated should be different for each student. This question is included so that students will think about what is important to them. If they come up with a good list of questions, they should be better prepared for decisions like the job choice p roblem. Lists of questions should be complete but not redundant. A list of questions regarding potential jobs might include: Does this job increase my salary? • Could I live in a nicer house or neighborhood? • Does it improve the opportunities available for my children? • Does it increase my opportunities for outdoor recreation? • Does it enhance my professional development? •

Asking such questions up front can help one to focus on exactly what matters, and the exercise may enhance one’s creativity in generating alternatives. Note that the lists described in the chapter tend to focus on characteristics of alternatives. Here the purpose of the list is to be sure that alternatives are examined in ways that are particularly meaningful to the decision maker. S6.3. Many creative solutions will be generated by the students such as: chopped up for insulating material, melted into art projects, used as buoys to identify crab traps, sterilized for reuse as containers for homemade wine. If the student’s list is short with a broad range o f possibilities (s)he was most likely using flexible thinking. On the other hand if the student’s list is fairly long but more narrowly focused, (s)he was most likely using fluid thinking. S6.4. This question works well because students generally have a lot of good ideas about good and bad aspects of their programs. If left to their own devices, most students try brainstorming. As an instructor, your job is to be as open-minded as possible when you hear their ideas!

89

CHAPTER 7 Probability Basics Notes rd   Making Hard Decisions with DecisionTools, DecisionTools, 3  Ed. assumes that students have had some introduction to

 probability. In In Chapter 7, probability probability basics are are presented and examples examples worked out in order to strengthen strengthen the student’s understanding of probability and their ability to manipulate probabilities. The focus is on manipulations that are useful in decision analysis. Bayes’ theorem of course is useful for dealing with  problems involving involving information, and the “law of total total probability” probability” is pertinent when we discuss probability probability decomposition in Chapter 8. The use of PrecisionTree to perform Bayesian calculations (“flipping the tree”) is discussed in the last section of this chapter. Of particular note in Chapter 7 is the topic of conditional independence. This concept plays a central role in the development of probability models in decision analysis. The nature of the role is particularly obvious in the construction of influence diagrams; the absence of an arc between two chance nodes that may be connected through other nodes is an indication of conditional independence. Care should be taken in the construction of influence diagrams to identify conditional independence. Each statement of conditional independence means one less arc, which means less probability assessment or modeling. At the same time that conditional independence is important for modeling in decision analysis, it is most likely a new concept for students. Probability and statistics courses teach about marginal independence, a special case of conditional independence in which the condition is the sure outcome. However, students sometimes have difficulty with the idea that a common conditioning event can be carried through all  probabilities in a calculation calculation as is the case case in conditional independence. In In addition, the intuition intuition behind conditional independence often is new to students; if I already know C, then knowing B will not tell me any more about A. This chapter also includes an online supplement on covariance and correlation. The solutions to the  problems in the the online supplement supplement are included at the end of this this chapter. Topical cross-reference for problems

Bayes’ theorem Conditional independence Conjunction effect Linear transformations transformations PrecisionTree Sensitivity analysis Simpson’s paradox Two-way sensitivity analysis Texaco-Pennzoil

7.14, 7.21, 7.28, 7.30, AIDS 7.23, 7.23,7.28, 7.34 7.24, 7.25 7.18, 7.19, 7.20 7.14, 7.21, 7.30 7.32, 7.33, AIDS  Decision Analysis Monthly Monthly , Discrimination and the Death Penalty 7.33 7.29

Solutions 7.1. We often have to make decisions in the face of uncertainty. Probability is a formal way to cope with

and model that uncertainty. 7.2. An uncertain quantity or random variable is an event that is uncertain and has a quantitative outcome

(time, age, $, temperature, weight, . . . ). Often a non-quantitative event can be the basis for defining an uncertain quantity; specific non-quantitative outcomes (colors, names, categories) correspond to quantitative outcomes of the uncertain quantity (light wavelength, number of letters, classification classification number). Uncertain quantities are important in decision analysis because they permit us to build b uild models that may be subjected to quantitative analysis.

145

7.3.

7.4.

P(A and B) = 0.12

 —  P( B  ) = 0.35

 —  P(A and B ) = 0.29

P(B | A) =

0.12 = 0.293 0.41

P(A) = 0.41

P(A | B) =

0.12 = 0.185 0.65

P(B) = 0.65

 –   —  0.06 P(A | B ) = 0.35 = 0.171

 —  P(A or B) = P(A and B) + P(A and B  ) + P( –  A and B) = 0.12 + 0.53 + 0.29 = 0.94 or P(A or B)

or P(A or B)

= P(A) + P(B) - P(A and B) = 0.41 + 0.65 - 0.12 = 0.94  –   and  —  = 1 - P(A B ) = 1 - 0.06 = 0.94

7.5.

 A a nd B

 A a nd B

 A a nd B

 A B

From the diagram, it is clear that  —  P(A) = P(A and B) + P(A and B  ) and  –  P(B) = P(A and B) + P(A and B).  —  But P(A or B) clearly equals P(A and B) + P(A and B  ) + P( –  A and B) because because of property property 2. Thus,  —  P(A or B) = P(A and B) + P(A and B  ) + P( –  A and B) = P(A) + P(B) - P(A and B).

146

Joint. P(left-handed and red-haired) = 0.08 Conditional P(red-haired | left-handed) = 0.20 Conditional P(Cubs win | Orioles lose) = 0.90 Conditional P(Disease | positive) = 0.59 Joint P(success and no cancer) = 0.78 Conditional P(cancer | success) Conditional P(food prices up | drought) Conditional P(bankrupt | lose crop) = 0.50 Conditional, but with a joint condition: P(lose crop | temperature high and no rain) Conditional P(arrest | trading on insider information) Joint P(trade on insider information and get caught)

7.6.a. b. c. d. e. f. g. h. i.  j. k.

7.7. For stock AB, E(Return of AB) = 0.15(-2%) + 0.50(5%) + 0.35(11%) = 6.1%. Similarly, E(Return of

CD) = 5.0%, and E(Return of EF) = 9.0%. These are needed to calculate the variances and standard deviations. Var(Return of AB) = 0.15(-2% - 6.1%) 2 + 0.50(5% - 6.1%) 2 + 0.35(11% - 6.1%) 2 = 0.15(-0.02 – 0.061) 2 + 0.50(0.05 – 0.061) 2 + 0.35(0.11 – 0.061) 2 = 0.0019 Thus, the standard deviation of the return on AB is

0.0019 = 0.043 = 4.3%. √ 0.0019

Similarly, Var(Return of CD) = 0.0003 and the standard deviation of the return on CD is 1.8%. Also, Var(Return of EF) = 0.0243 and the standard deviation of the return on EF is 15.6%. 7.8.

 A

A

B

0.2772 0.1450 0.4222

B

0.1428 0.4350 0.5778 0 .42

0.58

1

 –  ) = 1 - P(A) = 1 - 0.42 = 0.58 P(A) = 0.42 is given, so P(A  —  P( B | A) = 1- P(B | A) = 1 - 0.66 = 0.34 0.34  —  P( B |  –  A ) = 1- P(B |  –  A  ) = 1 - 0.25 0 .25 = 0.75 P(B) = P(B | A) P(A) + P(B |  –  A  ) P( –  A ) = 0.66(0.42) + 0.25(0.58) = 0.4222  —  P( B  ) = 1 - P(B) = 1- 0.4222 = 0.5778 P(A | B) =

P(A and B) P(B|A)P(A) 0.66(0.42) = = 0.4222 = 0.6566 P(B) P(B)

 –  | B) = 1 - P(A | B) = 1 - 0.6566 = 0.3434 P(A

147

 —   —  P( B |A)P(A) 0.34(0.42)  —  P(A and B ) P(A | B  ) =  =  = 0.5778  = 0.2471  —   —  P( B ) P( B )  —   –   —  P(A | B ) = 1 - P(A | B  ) = 1 - 0.2471 = 0.7529  – 

7.9. P(A  ) = 1 - P(A) = 1 - 0.10) = 0.90

 —  P( B | A) = 1 - P(B | A) = 1 - 0.39 = 0.61 0.61  —  P( B |  –  A ) = 1 - P(B |  –  A  ) = 1 - 0.39 = 0.61 P(B) = P(B | A) P(A) + P(B |  –  A  ) P( –  A ) = 0.39(0.10) + 0.39(0.90) = 0.39  —  P( B  ) = 1- P(B) = 1 - 0.39 = 0.61  –  ) = 0.39. At this point, it should be clear that A and B are independent because P(B) = P(B | A) = P(B | A  —   –   –   –   —  Thus, P(A) = P(A | B) B) = P(A | B ) = 0.10, and P(A  ) = P(A | B) = P(A | B ) = 0.90. (Actually, the fact that A and B are independent can be seen in the statement of the problem.) 7.10. a. P(Y  =  = 2) = 0.4, but P( Y  =  = 2 |  X  =  = -2) = P( Y  =  = 2 |  X  =  = 2) = 1.0 and P( Y  =  = 2 | X  =  = 0) = 0  X = -2) = 0.2, but P( X  =  = -2 | Y  =  = 2) = 0.5 and P( X  =  = -2 | Y  =  = 0) = 0 b. P( X   |X |.|. But it is not a linear relationship, and the covariance relationship c. X  and  and Y  are  are dependent. In fact, Y  =  =  X 

does not capture this nonlinear relationship. 7.11. The influence diagram would show conditional independence between hemlines and stock prices,

given adventuresomeness: adventuresomeness:

 Adve ntu resom resom eness en ess

Hemlines

Stock Pri Pri ces

Thus (blatantly ignoring the clarity test), the probability statements would be P(Adventuresomeness), P(Hemlines | Adventuresomeness), and P(Stock prices | Adventuresomeness). 7.12. In many cases, it is not feasible to use a discrete model because of the large number of possible

outcomes. The continuous model is a “convenient fiction” that allows us to construct a model and analyze it.

148

7.13.

B

B

 A

0.204

0.476

0.68

 A

0.006

0.314

0.32

0 .2 1

0.79

1

P(B and A) = P(B | A) P(A) = 0.30 (0.68) = 0.204  –  ) = 0.02 (0.32) = 0.006 P(B and  –  A ) = P(B |  –  A ) P(A  —  P( B  and A) 0.476  —  P( B | A) =  = 0.68 = 0.70 P(A)  —  P( B  and  –  A) 0.314  —   –  P( B | A ) = = 0.32 = 0.98  –  P(A) P(B) = P(B and A) + P(B and  –  A  ) = 0.204 + 0.006 = 0.21  —  P( B  ) = 1 - P(B) = 1 - 0.21 = 0.79 P(A | B) =

P(A and B) 0.204 = 0.21 = 0.970 P(B)

 –  | B) = 1 - P(A | B) = 1 - 0.970 = 0.030 P(A 0.030  —   —  P(A and B ) 0.476 P(A | B  ) =  = 0.79 = 0.603  —  P( B )  —   –   —  P(A | B ) = 1 - P(A | B  ) = 1 - 0.603 = 0.397 7.14.

P(offer) = 0.50 P(good interview | offer) = 0.95 P(good interview | no offer) = 0.75 P(offer | good interview) = P(offer | good) P(good | offer) P(offer) = P(good | offer) P(offer) + P(good | no offer) P(no offer) 0.95 (0.50) =  0.95 (0.50) + 0.75 (0.50) = 0.5588

See also “Problem 7.14.xlsx” for a solution using PrecisionTree.

149

7.15. a. E( X )

= 0.05 (1) + 0.45 (2) + 0.30(3) + 0.20(4) = 0.05 + 0.90 + 0.90 + 0.80 = 2.65

Var( X ) = 0.05 (1-2.65)2 + 0.45 (2-2.65) 2 + 0.30(3-2.65) 2 + 0.20(4-2.65) 2 = 0.05 (2.72) + 0.45 (0.42) + 0.30(0.12) + 0.20(1.82) = 0.728 σ X  b.

E( X )

= 0.728 = 0.853

= 0.13 (-20) + 0.58 (0) + 0.29(100) = -2.60 + 0 + 29 = 26.40

Var( X ) = 0.13 (-20 - 26.40) 2 + 0.58 (0 - 26.40) 2 + 0.29(100 - 26.40) 2 = 0.13 (2152.96) + 0.58 (696.96) + 0.29(5416.96) = 2255.04 σ X  c.

E( X )

= 2255.04 = 47.49

= 0.368 (0) + 0.632 (1) = 0.632

Var( X ) = 0.368 (0 - 0.632) 2 + 0.632 (1 - 0.632) 2 = 0.368 (0.632) 2 + 0.632 (0.368) 2 = 0.368 (0.632) [0.632 + 0.368] = 0.368 (0.632) = 0.233 σ X  = 0.233 = 0.482 7.16.

E( X )

= (1 - p) (0) +  p (1) = p

Var( X ) = (1 -  p) (0 -  p)2 + p (1 - p)2 = (1 -  p) p2 + p (1 -  p)2 = (1 -  p) p[ p + (1 -  p)] = (1 -  p) p  – 

7.17. It is true that P(A | B) + P(A | B) = 1, because the condition (B) is the same in each probability. Thus,

 —  these two probabilities are complements. However, the question is about P(A | B) + P(A | B ), which can equal anything between 0 and 2. There is no requirement that these two probabilities add up, because the  —  conditions (B and B  ) are different. 7.18. a.

E(Revenue from A) = $3.50 E(Unit sales) = $3.50 (2000) = $7000 Var(Revenue from A) = 3.502 Var(Unit sales) = 3.502 (1000) = 12,250 “dollars squared”

150

E(Total revenue) = $3.50 (2000) + $2.00 (10,000) + $1.87 (8500) = $42,895

b.

Var(Total revenue) = 3.502 (1000) + 2.00 2 (6400) + 1.87 2 (1150) = 41,871 “dollars squared” 7.19. Let  X 1 = random number of breakdowns for Computer 1, and  X 2 = random number of breakdowns

for Computer 2. Cost = $200 ( X 1) + $165 ( X 2) E(Cost) = $200 E( X 1) + $165 E( X 2) = $200 (5) + $165 (3.6) = $1594 If X 1 and X 2 are independent, then Var(Cost) = 2002 Var( X 1) + 1652 Var( X 2) = 200 2 (6) + 165 2 (7) = 430,575 “dollars squared” σCost =

430,575 “dollars squared”

= $656.18

The assumption made for the variance computation is that the computers break down independently of one another. Given that they are in separate buildings and operated separately, this seems like a reasonable assumption. 7.20. The possible values for revenue are 100 ($3) = $300 and 300 ($2) = $600, each with probability 0.5.

thus, the expected revenue is 0.5 ($300) + 0.5 ($600) = $450. The manager’s mistake is in thinking that the expected value of the product is equal to the product of the expected values, which is true only if the two variables are independent, which is not true in this case. 7.21. Notation:

“Pos” = positive “Neg” = negative “D” = disease  —  “ D  ” = no disease

 —   —  P(Pos) = P(Pos | D) P(D) + P(Pos | D  ) P( D ) = 0.95 (0.02) + 0.005 (0.98) = 0.0239 P(Neg) = 1 - P(Pos) = 1 - 0.0239 = 0.9761 P(D | Pos) =

P(Pos | D) P(D) 0.95 (0.02) = 0.0239 = 0.795 P(Pos)

P(D | Neg) =

P(Neg | D) P(D) 0.05 (0.02) = = 0.0010 P(Neg) 0.9761

151

Test positive (0.0239)

Test negative (0.9761)

Disease (0.795)

Probability Table

No disease (0.205)

Pos D

0.0190 0.0010

0.02

D

0.0049 0.9751

0.98

Disease (0.0010) No disease (0.9990)

Neg

0.0239 0.9761

1

7.22. Test results and field results are conditionally independent given the level of carcinogenic risk.

Alternatively, given the level of carcinogenic risk, knowing the test results will not help specify the field results. 7.23.

P(TR+ and FR+ | CP high) = P(TR+ | FR+ and CP high) P(FR+ | CP high) = P(TR+ | CP high) P(FR+ | CP high)

The second equality follows because FR and TR are conditionally independent given CP. In other words, we just multiply the probabilities together. This is true for all four of the probabilities required: P(TR+ and FR+ | CP high) = 0.82 (0.95) = 0.779 P(TR+ and FR- | CP high) = 0.82 (0.05) = 0.041 P(TR- and FR- | CP low) = 0.79 (0.83) = 0.6557 P(TR- and FR+ | CP low) = 0.79 (0.17) = 0.1343 7.24. Students’ answers will vary considerably here, depending on their opinions. However, most will rate

h as more likely than f. Tversky and Kahneman (1982) (see reference in text) found that as many as 85% of experimental subjects ranked the statements in this way, which is inconsistent with the idea of joint  probability (see the next question). Moreover, this phenomenon was found to occur consistently regardless of the degree of statistical sophistication of the subject. 7.25. a. The students’ explanations will vary, but many of them argue on the basis of the degree to which

Linda’s description is consistent with the possible classifications. Her description makes her sound not much like a bank teller and a lot like an active feminist. Thus, statement h (bank teller and feminist) is more consistent with the description than f (bank teller). Tversky and Kahneman claim that the conjunction effect observed in the responses to problem 7.25 stem from the representativeness heuristic. This heuristic is discussed in Chapter 8 of  Making Hard Decisions with DecisionTools.

152

b.

Bank teller  Feminist

Bank teller and feminist P(Bank teller and feminist) = P(Feminist | Bank teller) P(Bank teller). Since P(Feminist | Bank teller) must  be less than or equal to one, P(Bank teller and feminist) must be less than or equal to P(Bank teller). The area for the intersection of the two outcomes cannot be larger than the area for Bank teller. c. The friend is interpreting h as a conditional outcome instead of a joint outcome. Statement h clearly is a  joint outcome, because both outcomes (bank teller and feminist) occur. 7.26. To start, we need some labels. Let us say that we have chosen Door A. The host has opened Door B,

revealing the goat, and Door C remains closed. The question is whether we should switch to C. The decision rule is simple: switch if the probability of the car being behind Door C is greater than the  probability that5 it is behind A. Let “Car C” denote the outcome that the car is behind C, and likewise with the goats and the other doors. We want to calculate P(Car C | Goat B). Use Bayes theorem: P(Car C | Goat B) = P(Goat B | Car C) P(Car C)  P(Goat B | Car A) P(Car A) + P(Goat B | Car B) P(Car B) + P(Goat B | Car C) P(Car C)

The prior probabilities P(Car A), P(Car B), and P(Car C) are all equal to 1/3. For the conditional  probabilities, the key is to think about the host’s behavior. The host would never open a door to reveal the car. Thus, P(Goat B | Car C) = 1 and P(Goat B | Car B) = 0. Finally, what if the car is behind A? What is P(Goat B | Car A)? In this case, we assume that the host would randomly choose B or C, so P(Goat B | Car A) = 0.5. Plug these numbers into the formula to get: 1 (1/3) 2 P(Car C | Goat B) =  0.5 (1/3) + 0 (1/3) + 1 (1/3) = 3 Thus, you should always switch when the host reveals the goat! Here’s another way to think about it: You had a one-third chance of getting the correct door in the first  place. Thus, there is a two-thirds chance that the goat is behind B or C. By showing the goat behind B, the host has effectively shifted the entire two-thirds probability over to Door C. Still another way to think about the problem: If you played this game over and over, one-third of the time the car would be behind A, and two-thirds of the time it would be behind one of the other doors. Thus, twothirds of the time, the host shows you which door the car is not  behind. If you always switch to the door that the host did not open, you will find the car 2/3 of the time. The other 1/3 the car is behind the door you chose in the first place.

153

This question was asked of Marilyn Vos Savant, the person with the highest recorded I.Q. Her published answer was correct, but it created quite a stir because many people (including PhDs) did not understand how to solve the problem. 7.27. The host is proposing a decision tree that looks like this:

Keep

Switch

x

0.5

x/2

0.5

2x

But this is not correct. Suppose that x is equal to $100. Then the host is saying that if you swap, y ou have equally likely chances at an envelope with $200 and an envelope with $50. But that’s not the case! (If it were true, you would definitely want to switch.) Labeling the two envelopes A and B, the contestant correctly understands that the decision tree is as follows:

Keep A

Switch to B

A has x

(0.5)

B has x

(0.5)

A has x

(0.5)

B has x

(0.5)

x x/2 x/2 x

The two decision branches are equivalent from the point of view of the decision maker. 7.28. The solution is a straightforward application of Bayes’ theorem. For any 1 ≤  ≤ , we are given that  �   = (  ). By Bayes’ theorem:

  

�   = This is what we were to show.

   and  , for 1 ≤  ≤   and

 �  ( )  (  )( ) = = � . (  ) (  )

7.29

E(Payoff)

= $4.56 billion (calculated previously)

Var(Payoff)

= 0.2 (10.3 - 4.56) 2 + 0.5 (5 - 4.56) 2 + 0.3 (0 - 4.56) 2 = 12.9244 billion-dollars-squared

σPayoff  =

12.9244 = $3.5951 billion

7.30 . Let “+” indicate positive results, and “-” indicate negative results.

P(+)

= P(+ | Dome) P(Dome) + P(+ | No dome) P(No Dome) = 0.99 (0.6) + 0.15 (0.4) = 0.654 154

P(+ | Dome) P(Dome) P(Dome | +) = P(+ | Dome) P(Dome) + P(+ | No dome) P(No Dome) =

0.99 (0.6) 0.99 (0.6) + 0.15 (0.4)

= 0.908 P(No Dome | +) = 1 - 0.908 = 0.092 We can now calculate the EMV for Site 1, given test results are positive: EMV(Site 1 | +) = (EMV | Dome) P(Dome | +) + (EMV | No dome) P(No dome | +) = ($52.50 K) 0.908 + (-$53.75 K) 0.092 = $42.725 K [EMV|Dome and EMV|No dome have been calculated and appear in Figure 7.15.] EMV(Site 1 | +) is greater than EMV(Site 2 | +). If the test gives a positive result, choose Site 1. If the results are negative: P(-) = 1- P(+) = 1 - 0.654 = 0.346 P(Dome | -) =

P(- | Dome) P(Dome) P(- | Dome) P(Dome) + P(- | No dome) P(No Dome)

0.01 (0.6) = 0.01 (0.6) + 0.85 (0.4) = 0.017 P(No Dome | -) = 1 - 0.017 = 0.983 We can now calculate the EMV for Site 1, given test results are negative: EMV(Site 1 | -) = (EMV | Dome) P(Dome | -) + (EMV | No dome) P(No dome | -) = ($52.50 K) 0.017 + (-$53.75 K) 0.983 = -$51.944 K EMV(Site 1 | -) is less than the EMV(Site 2 | -). If the test gives a negative result, choose Site 2. 7.31.

P(+ and Dome) = P(+ | Dome) P(Dome) = 0.99 (0.60) = 0.594. P(+ and Dome and Dry) = P(Dry | + and Dome) P(+ and Dome) But P(Dry | + and Dome) = P(Dry | Dome) = 0.60. That is, the presence or absence of the dome is what matters, not the test results themselves. Therefore: P(+ and Dome and Dry) = 0.60 (0.594) = 0.356

155

Finally, P(Dome | + and Dry) =

P(Dome and + and Dry) P(+ and Dry)

But P(+ and Dry) = P(+ and Dry | Dome) P(Dome) + P(+ and Dry | No dome) P(No dome) and P(+ and Dry | Dome)

= P(Dry | + and Dome) P(+ | Dome) = P(Dry | Dome) P(+ | Dome) = 0.6 (0.99)

P(+ and Dry | No dome)

= P(Dry | + and No dome) P(+ | No dome) = P(Dry | No dome) P(+ | No dome) = 0.85 (0.15)

 Now we can substitute back in: P(+ and Dry) = 0.6 (0.99) (0.6) + 0.85 (0.15) (0.4) = 0.407 and P(Dome | + and Dry)

0.356 = 0.6 (0.99) (0.6) + 0.85 (0.15) (0.4) = 0.874.

7.32. EMV(Site 1) =  p(52.50) + (1 -  p)(-53.75)

Set this equal to EMV( Site 2) = 0, and solve for  p:  p(52.50) + (1 -  p)(-53.75) = 0

52.50 p + 53.75  p = 53.75  p =

53.75  52.50 + 53.75

= 0.5059. If 0.55 < P(Dome) < 0.65, then the optimal choice for the entire region is to drill at Site #1. 7.33. Choose Site 1 if

EMV(Site 1) > EMV(Site 2) q(52.50) + (1 - q) (-53.75) >  p(-200) + (1- p) (50) q > -2.3529  p + 0.9765

156

q

= P(Dome at Site 1)

0.9765

1.00

0.75 Site 1

0.50

0.25

Site 2

p

0 0.25

0.50 0.4150

0.75

= P(Dry at Site 2)

1.00

7.34. P(FR pos) = P(FR pos | CP High) P(CP High) + P(FR pos | CP Low) P(CP Low)

= 0.95 (0.27) + 0.17 (0.73) = 0.3806 P(FR + | TR +) = P(FR + | CP High and TR +) P(CP High | TR +) + P(FR + | CP Low and TR +) P(CP Low | TR +) But FR and TR are conditionally independent given CP, so P(FR + | CP High and TR +)

= P(FR + | CP High) = 0.95

P(FR + | CP Low and TR +)

= P(FR + | CP Low) = 0.17

and

We can calculate P(CP High | TR +) using Bayes’ theorem: P(TR + | CP High) P(CP High) P(CP High | TR +) = P(TR + | CP High) P(CP High) + P(TR + | CP Low) P(CP Low) 0.82 (0.27) = 0.82 (0.27) + 0.21 (0.73) = 0.5909. Therefore P(CP Low | TR +) = 1 - P(CP High | TR +) = 1 - 0.5909 = 0.4091.

157

Substitute back to obtain P(FR + | TR +) = 0.95 (0.5909) + 0.17 (0.4091) = 0.6309. It is important to note that P(FR + | TR +) ≠ P(FR +) = 0.3806. Thus, the two are not fully independent, even though they are conditionally independent given CP. Another way to say it is that conditional independence does not necessarily imply regular (marginal) independence. 7.35 Students have a hard time understanding what they are to show here, as it seems obvious, but there is

some algebra required to show the return of portfolio is the weighted average of the individual returns. The trick for this problem is that you need to break the portfolio weights down into the number of shares and stock prices. Specifically, using the notation below, the weight,  , satisfies:

 =

 0  0 + 0 +  0

Let “Port0” be the initial price of the portfolio and “Port1” be the value of the portfolio after one period. To calculate the ending value of the portfolio, we need to know how many shares of each stock we own  because the ending value is the sum of number of shares times ending stock value. Let  ,  , and   be the numbers of shares of sticks AB, CD, and EF in the portfolio. Thus, “Port0” and “Port1” satisfy:

0 =   0 +  0 +  0 1 =   1 +  1 +  1 Therefore, return of the portfolio,  , is

  =

1 − 0 0

=

  1 + 1 +  1 − (  0 + 0 +  0)   0 + 0 + 0

=

  ( 1 − 0) + (1 −0) +  (1 −0)   0 + 0 +  0

=

=

+ (1 − 0 )  ( 1 − 0) +  0 +  0 +  0  0 +  0 +  0  (1 − 0 ) +  0 + 0 +  0 ( 1 − 0 )  0  0 +  0 +  0  0 (1 − 0 ) + 0 +  0 + 0 +  0 0 (1 − 0 )  0 +  0 + 0 +  0 0

=   +   +  

158

Case Study: Decision Analysis Monthly

May:

P(Renew) = P(Renew | Gift) P(Gift) + P(Renew | Promo) P(Promo) + P(Renew | Previous) P(Previous) = 0.75 (0.70) + 0.50 (0.20) + 0.10 (0.10) = 0.6350

June:

P(Renew) = 0.85 (0.45) + 0.60 (0.10) + 0.20 (0.45) = 0.5325

There is good news because the proportion of renewals in each category increased from May to June. However, as indicated by Calloway, the overall proportion has indeed decreased because the mix of gift,  promotional, and previous subscriptions has changed along with the increase in proportion renewed. The overall decrease should indeed be looked at as bad news in a sense. If the editors can project the future mix of expiring subscriptions, they will be able to make an educated guess as to whether the trend is toward an overall increase or decrease in subscriptions. (This problem is an example of Simpson’s paradox. ) Case Study: Screening for Colorectal Cancer 1. We know that P(Blood) = 0.10 from the text: “...10% had blood in their stools.” These 10% underwent

colonoscopy, and 2.5% of those actually had cancer, which gives us P(Cancer | Blood) = 0.025. Thus, P(Cancer and Blood) = 0.025 (0.10) = 0.0025. Finally, we know that “... approximately 5 out of 1000” had no blood but did develop cancer. This gives P(No Blood and Cancer) = 0.005. With these numbers, we can construct the complete probability table. The probabilities that we already have are indicated in bold: Blood  No Blood

Cancer 0.0025  0.0050

0.0075

No Cancer 0.0975 0.8950 0.9925

0.10

0.90 1.00

P(Cancer | Blood) = 0.025 as indicated above. We can calculate P(Cancer | No Blood) = 0.005/0.90 = .0056. 2. The expected cost of the policy is 60 million ($10) + 0.10 (60 million) ($750) = $5.1 billion. the

expected number of people who must undergo colonoscopy is 6 million. And the number of people who have colonoscopy done needlessly is 0.975 (6 million) = 58.5 million. 3. If we save 3 lives per thousand, then out of 60 million people tested we would expect to save 60 million

times 3/1000 or 180,000 lives. The total cost, ignoring the time value of money, would be $5.1 billion times 13, or $66.3 billion. Divide cost by number of lives to get the cost per life: $368,333 per life saved. 4. This is a tough question, but we face such questions constantly. On one hand, a lot of people are

inconvenienced needlessly by this screening procedure. On the other hand, the cost of $368,333 is a relatively low figure. Economic analyses of the value of a life typically give a figure in the neighborhood of $4 million. However, this analysis is not complete. The real issue is that the colonoscopy procedure itself can lead to complications and misinterpretations, and hence it is itself a risky procedure. A complete analysis would have to take this into account. Doing so will increase the overall cost of the screening policy. Moreover, we have put no dollar figures on the inconvenience, concern, and worry that so many people must go through needlessly.

159

Case Study: AIDS 1. P(Inf | ELISA+)

P(ELISA+ | Inf) P(Inf) =P(ELISA+ | Inf) P(Inf) + P(ELISA+ | Not Inf) P(Not Inf) 0.997 (0.0038) = 0.997 (0.0038) + 0.015 (1 - 0.0038) = 0.20. 2.

P(Inf|ELISA+)  NY drug users 0.98

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

RI gays 0.73

 NJ recruits 0.13

0

0.05

0.1

0.15

0.2 P(Inf)

3. P(Inf | ELISA-)

P(ELISA- | Inf) P(Inf) =P(ELISA- | Inf) P(Inf) + P(ELISA- | Not Inf) P(Not Inf) 0.003 (0.0038) = 0.003 (0.0038) + 0.985 (1 - 0.0038) = 0.0000116. 1

0.8

0.6

0.4

0.2 P(Inf | ELISA 0 0.50

0.60

0.70

0.80

0.90

160

1.00

0.25

0.3

0.35

0.4

4. P(Inf | WB+, ELISA+)

P(WB+ | Inf, ELISA+) P(Inf | ELISA+) = P(WB+ | Inf, ELISA+) P(Inf | ELISA+) + P(WB+ | Not Inf, ELISA+) P(Not Inf | ELISA+)

We have the values for specificity, sensitivity, false positive rate, and false negative rate for the Western Blot from the text of the case. As indicated in the problem, all of these figures are conditioned on a positive ELISA result: P(WB+ | Inf, ELISA+) = 0.993 P(WB- | Inf, ELISA+) = 0.007 P(WB+ | Not Inf, ELISA+) = 0.084 P(WB- | Not Inf, ELISA+) = 0.916. Likewise, we calculated P(Inf | ELISA+) in question 1: P(Inf | ELISA+) = 0.20 P(Not Inf | ELISA+) = 0.80 Thus, we have P(Inf | WB+, ELISA+) 0.993 (0.20) = 0.993 (0.20) + 0.084 (0.80) = 0.75. We can also calculate P(Inf | WB-, ELISA+) P(WB- | Inf, ELISA+) P(Inf | ELISA+) = P(WB- | Inf, ELISA+) P(Inf | ELISA+) + P(WB- | Not Inf, ELISA+) P(Not Inf | ELISA+) 0.007 (0.20) = 0.007 (0.20) + 0.916 (0.80) = 0.00193.  Note that in using P(Inf | ELISA+) = 0.20, we are implicitly using as a prior P(Inf) = 0.0038.

161

5.

1.00 P(Inf | ELISA+, WB+ 0.80

0.60

0.40

0.20

0.00 0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

1.00

0.80

0.60 P(Inf | ELISA+, WB-

0.40

0.20

0.00 0.00

0.20

0.40

0.60

0.80

1.00

6. Students’ answers will vary considerably here. The question itself provides some guidance for

identifying the costs associated with false negatives and false positives. Clearly, a false positive could lead an individual to unnecessary psychological distress. At a societal level, a high false positive rate could lead to an unduly high level of expenditures on the disease. On the other hand, false negatives could have severe social impact as infected individuals could spread the disease unknowingly.

162

One of society’s fundamental choices is to balance the rates of false positives and false negatives. If false negatives are deemed much more serious than false positives, for example, then it would be appropriate to develop tests that have a very high level of sensitivity. It is appropriate to draw an analogy here with Type I and Type II errors in statistical hypothesis testing. The  probabilities of these errors (often labeled α and β, respectively), correspond to false positive and false negative rates. When the probability of a Type I error is reduced by changing the decision rule (everything else being equal), the probability of a Type II error increases. Case Study: Discrimination and the Death Penalty 1. Let DP denote “Death Penalty,” DW “Defendant White,” and DB “Defendant Black.” From Table 7.5,

P(DP | DW) = 19/160 = 0.119 P(DP | DB) = 17/166 = 0.102 Based on these data, there appears to be little difference in the rate at which black defendants get the death  penalty. 2. Let VW and VB denote “Victim White” and “Victim Black,” respectively. Based on Table 7.6, for white

victims: P(DP | DW, VW) = 19/151 = 0.126 P(DP | DB, VW) = 11/63 = 0.175 For black victims: P(DP | DW, VB) = 0/9 = 0 P(DP | DB, VB) = 6/103 = 0.058  Now the interpretation is different. After disaggregating the data on the basis of victim race, blacks appear to get the death penalty more frequently (by about 5 percentage points) than whites, regardless of the race of the victim. 3. How do we resolve the apparent paradox? How could there be no difference between the overall rate of

death penalties (or even a slightly lower rate for blacks) with the aggregate data, but a clear difference — in the opposite direction — with the disaggregate data? This is an example of Simpson’s paradox. The problem is that the mix of victim races differs considerably from white defendants to black defendants. There are so few black victims of whites that the low death penalty rate in this case (0) plays a very small role in calculating the overall death-penalty rate for whites. Likewise, there are so many black victims of black defendants, so the relatively low death-penalty rate for  black defendant/black victim brings down the overall death-penalty rate for black victims. The Decision Analysis Monthly case is another example of Simpson’s paradox.

163

Chapter 7 Online Supplement: Solutions to Problems 7S.1. P( X =2, Y =10) = P( Y =10 |  X =2) P( X =2) = 0.9 (0.3) = 0.27. Likewise,

P( X =2, Y =20) = P(Y =20 | X =2) P( X =2) = 0.1 (0.3) = 0.03 P( X =4, Y =10) = P(Y =10 | X =4) P( X =4) = 0.25 (0.7) = 0.175 P( X =4, Y =20) = P( Y =20 | X =4) P( X =4) = 0.75 (0.7) = 0.525 E( X ) = 0.3(2) + 0.7(4) = 3.4 P(Y  = 10) = P( Y =10 | X =2) P( X =2) + P( Y =10 | X =4) P( X =4) = 0.27 + 0.175 = 0.445 P(Y  = 20) = 1-0.445 = 0.555 E(Y ) = 0.445 (10) + 0.555 (20) = 15.55  Now calculate:  X 

Y 

 X  - E( X )

Y  - E(Y )

2 2 4 4

10 20 10 20

-1.4 -1.4 0.6 0.6

-5.55 4.45 -5.55 4.45

( X -E( X ))(Y -E(Y )) 7.77 -6.23 -3.33 2.67

P( X , Y ) 0.27 0.03 0.175 0.525

The covariance is the expected value of the cross products in the next-to-last column. To calculate it, use the joint probabilities in the last column: Cov( X , Y ) = 0.27 (7.77) + 0.03 (-6.23) + 0.175 (-3.33) + 0.525 (2.67) = 2.73 Calculate the standard deviations by squaring the deviations in the third and fourth columns (for  X  and Y , respectively), finding the expected value of the squared deviations, and finding the square root: σ   X =

0.3 (-1.4 2) + 0.7 (0.6 2) = 0.917

σ Y =

0.445 (-5.552) + 0.555 (4.45 2) = 4.970

2.73 Thus, the correlation is  ρ   XY  =  0.917 (4.970) = 0.60. 7S.2. The basic setup for this problem is the same as it is for the previous problem. We already have the  joint probabilities, so we can start by calculating the expected values of  X  and Y :

E( X ) =

(-2) + (-1) + 0 + 1 + 2 = 0  5

E(Y ) = 0.2 (0) + 0.4 (1) + 0.4 (2) = 1.2 Thus, we have the following table:  X Y X  - E( X ) Y  - E(Y ) -2 2 -2 0.8 -1 1 -1 -0.2 0 0 0 -1.2 1 1 1 -0.2 2 2 2 0.8

( X -E( X ))(Y -E(Y )) -1.6 0.2 0 -0.2 1.6

The covariance is the expected value of the numbers in the last column, each of which can occur with  probability 1/5. Calculating this expected value gives a covariance of zero. Likewise, the correlation equals zero. 164

CHAPTER 8 Subjective Probability Notes

Chapter 8 is the key chapter in Section 2 of  Making Hard Decisions with DecisionTools, 3rd  Ed . While  probability models can, and often are, created in other ways, it is the subjective interpretation of probability and the use of personal judgments that make decision analysis unique. At the same time, subjective probability has been a source of considerable criticism. The only counter to this criticism is to understand the nature of the subjective judgments that must be made and to make those  judgments with great care. And the only way to do so is to spend the time required to think hard about what is known regarding any given uncertain event. The probability-assessment framework presented in the chapter is just the framework; the decision maker or assessor must provide the hard thinking and work to generate good subjective probability models. Question 8.4 is an excellent in-class exercise for motivating quantitative modeling of subjective  probabilities. Have the students answer the questions, and tabulate their results. A wide variety of interpretations exist for many of these verbal phrases. Two books are particularly good sources for probabilities of everyday events for which you might want to have students assess probabilities. These books can be used to compare the assessments with data-based risk estimates: Laudan, L. (1994) The Book of Risks. New York: Wiley Siskin, B., J. Staller, and D. Rorvik (1989) What are the Chances. New York: Crown The online supplement introduces the idea of a Dutch Book and shows why subjective assessments should follow the laws of probability. Solutions to the two online problems (S8.1 and S8.2) are included here. Topical cross-reference for problems

Ambiguity Bayes’ theorem Conjunction effect CDFs Decomposition Ellsberg paradox Odds Discrete approximations Probability assessment Probability assessment heuristics Requisite models Scientific information Sensitivity analysis Subjective judgments

8.23, S8.1, S8.2, Assessing Cancer Risk, Space Shuttle Challenger 8.10 8.21 8.11 - 8.16 8.5, 8.7, 8.17 - 8.19 8.25 8.9, 8.10, 8.26 8.11, 8.15, 8.16 8.3, 8.6, 8.11 - 8.14 8.20 - 8.22 8.23 Assessing Cancer Risk, Breast Implants, Space Shuttle Challenger  8.24 8.2

Solutions 8.1. Answers will vary considerably, but students might think about subjective probability as a degree of

 belief, uncertainty in one’s mind, or a willingness to bet or accept lotteries. They may appropriately contrast the subjective interpretation of probability with a frequency interpretation. 8.2. The model under discussion relates a number of financial ratios to the probability of d efault. First,

there is a good deal of subjective judgment involved in deciding which financial ratios to include in such a

165

model. Second, in using past data your friend has made the implicit assumption that no fundamental changes in causes of default will occur, or that the data from the past are appropriate for understanding which firms may default in the future. A bank officer using this model is making an implicit subjective  judgment that the model and data used are adequate for estimating the probability of default of the  particular firms that have applied for loans. Finally, the bank officer implicitly judges that your friend has done a good job! 8.3. Assessing a discrete probability requires only one judgment. Assessing a continuous probability

distribution can require many subjective judgments of interval probabilities, cumulative probabilities, or fractiles in order to sketch out the CDF. Even so, the fundamental probability assessments required in the continuous case are essentially the same as in the discrete case. 8.4. Answers will, of course, vary a lot. As a motivation for careful quantitative modeling of probability, it

is instructive to collect responses from a number of people in the class and show the ranges of their responses. Thus, it is clear that different people interpret these verbal phrases in different ways. Answers can be checked for consistency, as well. In particular, a > 0.5, g > 0.5, l < j, e < j, p < m < i, and o < f < k. 8.5. Answers will vary here, too, but many students will decompose the assessment into how well they will

do on homework (for which they have a good deal of information) and how well they will do on a final exam or project.

Final

Homework

Exam

Course Grade

8.6. The students’ assessments may vary considerably. However, they should be reasonable and indicate in

some way that some effort went into the assessment process. Answers should include some discussion of thought processes that led to the different assessments. 8.7. It is possible to assess probabilities regarding one’s own performance, but such assessments are

complicated because the outcome depends on effort expended. For example, an individual might assess a relatively high probability for an A in a course, thus creating something of a personal commitment to work hard in the course. 8.8. Considering the assessments in 8.7, it might be appropriate to d ecompose the event into uncertain

factors (homework, professor’s style, exams, and so on) and then think about how much effort to put into the course. It would then be possible to construct risk profiles for the final grade, given different levels of effort. 8.9. This problem calls for the subjective assessment of odds. Unfortunately, no formal method for

assessing odds directly has been provided in the text. Such a formal approach could be constructed in terms of bets or lotteries as in the case of probabilities, but with the uncertainty stated in odds form. 8.10. First, let us adopt some notation. Let NW and NL denote “Napoleon wins” and “Napoleon loses,”

respectively. Also, let “P&E” denote that the Prussians and English have joined forces. The best way to handle this problem is to express Bayes’ theorem in odds form. Show that P(NW | P&E) P(P&E | NW) P(NW) P(NL | P&E) = P(P&E | NL) P(NL) . 166

We have that P(NW) = 0.90 and P(NW | P&E) = 0.60, and so P(NL) = 0.10 and P(NL | P&E) = 0.40. Thus, we can substitute: 0.60 P(P&E | NW) 0.90 0.40 = P(P&E | NL) 0.10 or P(P&E | NW) 1 P(P&E | NL) = 6 . Thus, Napoleon would have had to judge the probability of the Prussian and English joining forces as six time more likely if he is to lose than if he is to win. 8.11. This problem requires a student to assess a subjective CDF for his or her score in the decision analysis

course. Thus, answers will vary considerably, depending on personal assessments. For illustrative purposes, assume that the student assesses the 0.05, 0.25, 0.50, 0.75, and 0.95 fractiles:  x0.05 = 71

 x0.25 = 81

 x0.50 = 0.87

 x0.75 = 89

 x0.95 = 93

These assessments can be used to create a subjective CDF: 1.00

0.75

0.50

0.25

60

70

80

90

100

To use these judgments in deciding whether to drop the course, we can use either Swanson-Megill or the Pearson-Tukey method. The Pearson-Tukey method approximates the expected DA score as: EP-T(DA Score) ≈ 0.185 (71) + 0.63 (87) + 0.185 (93) = 85.15. Assume that the student has a GPA of 2.7. Using E P-T(DA Score) = 85.15 to calculate expected salary, E(Salary | Drop Course) = $4000 (2.7) + $24,000 = $34,800 E(Salary | Don’t drop) = 0.6 ($4000 x 2.7) + 0.4 ($170 × EP-T(DA Score)) + $24,000 = 0.6 ($4000 × 2.7) + 0.4 ($170 × 85.15) + $24,000 = $36,270. Thus, the optimal choice is not to drop the course.

167

8.12. Again, the assessments will be based on personal judgments and will vary among students. As an

example, suppose the following assessments are made: P(DJIA ≤ 2000) = 0.05 P(DJIA > 3000) = 0.05 P(DJIA ≤ 2600) = 0.50 P(DJIA ≤ 2350) = 0.25 P(DJIA ≤ 2800) = 0.75 These assessments result in the following graph:

8.13. If you have worked this problem and looked in the back of the book for the answers, you have most

likely found that relatively few of the answers fell within the ranges you stated, indicating that you, like most people are very overconfident in your judgments. Given this, it would make sense to return to Problem 8.12 and broaden the assessed distributions. (Note from Dr. Clemen: I have tried to do exercises like this myself, knowing about the overconfidence phenomenon and how to make subjective probability  judgments, and I still make the same mistake!) 8.14. The cumulative distribution function provides a “picture” of what the forecaster sees as reasonable

outcomes for the uncertain quantity. If the CDF is translated into a probability density function, it is still easier to see how the forecaster thinks about the relative chances of the possible outcomes. The key advantages of probabilistic forecasting are 1) that it provides a complete picture of the uncertainty, as opposed to a point forecast which may give no indication of how accurate it is likely to be or how large the error might be; and 2) the decision maker can use the probability distribution in a decision analysis if desired. The disadvantage is that making the necessary assessments for the probabilistic forecast may take some time. Some decision makers (the uninitiated) may have difficulty interpreting a probabilistic forecast. 8.15. a. This question requires students to make personal judgments. As an example, suppose the following

assessments are made: P(S ≤ 65) = 0.05 P(S > 99) = 0.05 P(S ≤ 78) = 0.25 P(S ≤ 85) = 0.50 P(S ≤ 96) = 0.75

168

b. Now the student would ask whether she would be willing to place a 50- 50 bet in which she wins if 78 <

S ≤ 85 and loses if 85 < S ≤ 96. Is there a problem with betting on an event over which you have some control? See problems 8.8, 8.9. c. Three-point Extended Pearson-Tukey approximation:

EP-T(Score) ≈ 0.185 (65) + 0.63 (85) + 0.185 (99) = 83.89. d. Three-point Extended Swanson-Megill approximation:

ES-M(Score) ≈ 1/6 (68.25) + 2/3 (85) + 1/6 (98.25) = 84.42. Because the assessments did not include the tenth and ninetieth percentiles, we use a straight-line interpolation. The tenth percentile is calculated as: 65 + (0.10 – 0.05)*(13/0.2) = 68.25. The ninetieth percentile is calculated as: 96 + (0.90 – 0.75)*(3/0.2) = 98.25. The Extended P-T and S-M are quite close to each other, within 0.53 percentage points. 8.16. We continue with the example given above in 8.12, using the Dow Jones Industrial Average. The

CDF is:

169

a.

EP-T(DJIA) = 0.185 (2000) + 0.63 (2600) + 0.185 (3000) = 2563

b.

ES-M(DJIA) = 1/6 (2087.5) + 2/3 (2600) + 1/6 (2950) = 2573.

Because the assessments did not include the tenth and ninetieth percentiles, we use a straight-line interpolation. The tenth percentile is calculated as: 2000 + (0.10 – 0.05)*(350/0.2) = 2087.5. The ninetieth percentile is calculated as: 2800 + (0.90 – 0.75)*(200/0.2) = 2950. 8.17. The issue in this problem is whether an assessment made in one way is better than another. The assessments will probably not be perfectly consistent. That is, P(Mets win series) =  p will probably not be

exactly equal to P(Mets win series | Mets win Pennant) × P(Mets win pennant). For example, P(Mets win series) may be holistically assessed as 0.02, while P(Mets win series | Mets win Pennant) = 0.6 and P(Mets win pennant) = 0.1, giving P(Mets win series) = 0.06. For many individuals, particularly those with some knowledge of the Mets’ current team, the decomposed assessment may be easier, and they may have more confidence in the final result. 8.18. Students must assess P(Hospitalized | Accident), P(Hospitalized | No Accident), and P(Accident).

With these assessments, they could calculate P(Hospitalized) = P(Hospitalized | Accident) P(Accident) + P(Hospitalized | No Accident) P(No accident) It would be possible to decompose the assessment further or in other ways. For example, one might consider the possibility of contracting a serious disease or requiring hospitalization for mental illness. 8.19. This problem is actually a full-scale project requiring considerable research. The students might

consider assessing a distribution for industry sales and a distribution for the firm’s market share. Together, these two quantities give the distribution of revenue for the firm: Market Share

Industry Sales

Revenue

8.20. Tversky and Kahneman (1971) attribute the gambler’s fallacy to the representativeness heuristic and a

misunderstanding of random processes. People tend to think that small segments of a random process will  be highly representative of the overall process. Hence, after a string of red on a roulette wheel, it is thought that black must occur to balance the sample and make it more representative of the overall process (50% red, 50% black). Source: Tversky, A., and Kahneman, D. (1971) “Belief in the Law of Small Numbers,” Psychological Bulletin , 76, 105-110. 8.21. “Linda” (Problem 7.25) is a classic example of the representativeness heuristic. Many people judge

that Linda’s description is less representative of a bank teller than of a feminist bank teller. 8.22. The “regression to the mean” phenomenon could easily be at work here. The D is most likely an

“outlier” and, hence, is likely to be followed by an improvement. Another argument is that your parents should not use the D as a basis to compare you to other D students (representativeness heuristic). In contrast, they should consider your “base rate” (as a B student) and not overweight the poor exam  performance. 8.23. In principle, the notion of a requisite model is appropriate in answering this question, but the

application of the concept is delicate. It is possible to perform sensitivity analysis on, say, a three-point 170

discrete approximation by wiggling the representative points. Do small changes in the 0.05, 0.5, or 0.95 fractiles affect the choice? If not, then further assessments are probably not necessary. If they do, then more assessments, possibly decomposed assessments, and a clearer picture of the CDF are needed to obtain an unequivocal decision. 8.24. There are a variety of different possibilities here. Perhaps the most straightforward is to obtain

“upper” and “lower” probabilities and perform a sensitivity analysis. That is, solve the decision tree or influence diagram with each probability. Does the optimal choice change? If not, there is really no problem. If the decision is sensitive to the range of probabilities, it may be necessary to assess the probabilities more carefully. Another simple solution is to assess a continuous distribution for the probability in question (a “second-order” probability distribution). Now estimate the expected value of this distribution using bracket medians or the Pearson-Tukey method. Finally, use the expected value as the probability in the decision  problem. 8.25. a, b. Most people choose A and D because these two are the options for which the probability of

winning is known. c. Choosing A and D may appear to be consistent because both of these involve known probabilities.

However, consider the EMVs for the lotteries and the implied values for P(Blue). If A is preferred to B, then EMV(A) > EMV(B) 1 3 (1000) > P(Blue) (1000) 1 P(Blue) < 3 . However, if D is preferred to C, then EMV(D) > EMV(C) 1 P(Blue) (1000) + P(Yellow) (1000) > 3 (1000) + P(Yellow) (1000) 1 P(Blue) > 3 . 1 1 The inconsistency arises because it clearly is not possible to have both P(Blue) < 3 and P(Blue) > 3 . (Exactly the same result obtains if we use the utility of $1000, instead of the dollar value.) Case Study: Assessing Cancer Risk — From Mouse to Man 1. Some assumptions and judgments that must be made: •

•

•

Organisms react in a specified way to both high and low doses of the substance. Researchers have developed dosage response models, but validating those models has been difficult. Threshold effects may exist; at a low dosage level, an organism may be able to process a particular toxin effectively, although at a higher level (beyond the threshold) reactions to the toxin may appear. The test species is judged to react to the toxin in the same way that humans do. Some evidence indicates, though, that human livers are better at processing toxins than mouse or rat livers. Thus, dosage responses may not be the same across species. Lab and field exposures are similar in nature. However, in the field many more complicating and  possibly interactive effects exist.

The first two assumptions take shape in what is called a “dose-response” model. This is a mathematical relationship that estimates the magnitude of the physiological response to different doses of the chemical.

171

What kinds of evidence would help to nail down the effects of toxic substances? Long term studies of toxins in the field and in the lab would be most useful. We need to know effects of low doses on humans in the field, in order to refine the human dose-response model, but this information may be very difficult to gather. Certainly, no controlled studies could be performed! 2. The question is whether one bans substances on the grounds that they have not been demonstrated to be

safe, or does one permit their use on the grounds that they have not been demonstrated to be dangerous. The choice depends on how the decision maker values the potential economic benefits relative to the  potential (but unknown) risks. 3. The issue of credibility of information sources is one with which scientists are beginning to wrestle, and

it is a complicated one. Intuitively, one would give more weight to those information sources that are more credible. However, systematic ways of assessing credibility are not yet available. Furthermore, the overall impact of differences in source credibility on the decision maker’s posterior beliefs is unclear. Case Study: Breast Implants 1. There clearly are differences in the quality of information that is available in most situations. Science

teaches us to beware of inferences based on small samples, yet anecdotes can be used to paint compelling scenarios. Are judges prepared to make judgments regarding the quality of information presented as “scientific”? How can a judge, not trained in the science himself, be expected to make reasonable  judgments in this respect? And if a judge is ill prepared, what about jurors? 2. The questions asked in the last paragraph of the quoted passage clearly relate primarily to preferences. In

a democratic, capitalistic society, we generally assume that individual consumers should get to make their own decisions, based on their own preferences. In this case, however, the issue of preference deals with how much risk is acceptable. And that question presumes that the decision maker knows what the risk is. The level of risk, however, is a matter of uncertainty (“facts,” in contrast to “values”), and it takes experts to measure that risk. In situations where individual consumers cannot realistically be expected to understand fully the risks, we often expect the government to step in to regulate the consumer risk. It is not so much a matter of protecting the consumer from himself as it is being sure that the risks are appropriately measured, the information disseminated to the consumers, and , where appropriate, appropriate standards set. Case Study: The Space Shuttle Challenger 1. With little or no information, does one refrain from launching the spacecraft on the grounds that no proof

exists that the launch would be safe, or does one launch on the grounds that there is no proof that doing so is unsafe. Since the Challenger  accident, NASA has implemented a system whereby the policy is clearly stated: Do not launch if there are doubts as to the safety. 2. These subjective estimates made by different people are based on different information and different  perspectives, and are used for different purposes. It is important for a decision maker to look beyond the  biases, try to judge the “credibility” of the judgments, and take these into account in developing his or her own probabilities or beliefs. The same caveats as in question 3 of the Cancer Risk case apply here, however. That is, even though it seems appropriate to weight more credible sources more heavily, neither  precise methods nor an understanding of the impact of doing so exist at this time. 3. The overall effect of slight optimism in making each individual assessment would be a very

overoptimistic probability of failure. That is, P(failure) would wind up being much lower than it should be. 4. Reichhardt’s editorial raises the question of what is an “acceptable risk.” How should society determine

what an acceptable risk would be? How society should choose an acceptable risk level for enterprises such as nuclear power generation, genetic research, and so on, has been a hotly debated topic. Furthermore, different people are willing to accept different levels of risk. For example, an astronaut, thrilled with the  prospect of actually being in space, may be more willing to accept a high level of risk than a NASA administrator who may be subject to social and political repercussions in the event of an accident. Because 172

of the diversity of preferences, there is no obvious way to determine a single level of acceptable risk that would be satisfactory for everyone. Further reading: Fischhoff, B., S. Lichtenstein, P. Slovic, S. Derby, & R. Keeney (1981)  Acceptable Risk . Cambridge: Cambridge University Press. ****************

Chapter 8 Online Supplement: Solutions to Questions and Problems S8.1. a. For each project, the investor appears to believe that

E(profit) = 0.5 (150,000) + 0.5 (-100,000) = 75,000 - 50,000 = 25,000 b. However, since only one of the projects will succeed, he will gain $150,000 for the successful project,  but lose $100,000 for each of the other two. Thus, he is guaranteed to lose $50,000 no matter what happens. c. For a set of mutually exclusive and collectively exhaustive outcomes, he appears to have assessed

 probabilities that add up to 1.5. d. “Knowing nothing” does not necessarily imply a probability of 0.5. In this case, “knowing nothing” is

really not appropriate, because the investor does know something: Only one project will succeed. If, on top of that, he wants to invoke equally likely outcomes, then he should use P(Success) = 1/3 for each project.  Note that it is also possible to work parts a and b in terms of final wealth. Assume that he starts with $300,000. For part a, expected final wealth, given that he invests in one project, would be: E(wealth) = 0.5 (450,000) + 0.5 (200,000) = 325,000. Because $325,000 is an improvement on the initial wealth, the project looks good. For part b, though, if he starts with $300,000 and invests in all three projects, he will end up with only $250,000 for the project that succeeds. As before, he is guaranteed to lose $50,000. S8.2. a. If he will accept Bet 1, then it must have non-negative expected value:

P(Cubs win) ($20) + [1 - P(Cubs win)] (-$30) ≥ 0. Increasing the “win” amount to something more than $20, or decreasing the amount he must pay if he loses ($30) will increase the EMV of the bet. However, reducing the “win” amount or increasing the “lose” amount may result in a negative EMV, in which case he would not bet. The same argument holds true for Bet 2. b. Because he is willing to accept Bet 1, we know that

P(Cubs win) ($20) + [1 - P(Cubs win)] (-$30) ≥ 0, which can be reduced algebraically to P(Cubs win) ≥ 0.60. Likewise, for Bet 2, we know that

173

P(Cubs win) (-20) + [1 - P(Cubs win)] ($40) ≥ 0. This can be reduced to P(Cubs win) ≤ 0.67. Thus, we have 0.60 ≤ P(Cubs win) ≤ 0.67. c. Set up a pair of bets using the strategy from Chapter 8. From Bet 1 we infer that P(Cubs win) = 0.60, and

from Bet 2 P(Yankees win) = 0.33. Use these to make up Bets A and B: A: He wins 0.4 X if Cubs win He loses 0.6 X if Yankees win B: He wins 0.67 Y if the Yankees win He loses 0.33 Y if the Cubs win We can easily verify that the EMV of each bet is equal to 0: EMV(A) = P(Cubs win) (0.4 X) + [1 - P(Cubs win)] (- 0.6 X) = 0.6 (0.4 X) - 0.4 (0.6 X) =0 EMV(B) = P(Yankees win) (0.67 Y) + [1- P(Yankees win)] (-0.33 Y) = 0.33 (0.67 Y) - 0.67 (0.33 Y) =0 If the Cubs win, his position is: 0.4 X - 0.33 Y = W If the Yankees win: -0.6 X + 0.67 Y = Z Following the strategy in the book, set W = Z = -$100 to be sure that he pays us $100 net, regardless of what happens: 0.4 X - 0.33 Y = -$100 -0.6 X + 0.67 Y = -$100  Now solve these two equations for X and Y to obtain X = Y = -$1500. Thus, the original bets A and B  become: A: He wins -$600 if Cubs win He loses -$900 if Yankees win B: He wins -$1000 if the Yankees win He loses -$500 if the Cubs win The minus sign means that he is taking the “other side” of the bet, though (i.e. winning -$600 is the same as losing $600). Thus, these two bets really are: A: He loses $600 if Cubs win He wins $900 if Yankees win B: He loses $1000 if the Yankees win He wins $500 if the Cubs win

174

Finally, compare these bets to Bets 1 and 2 in the book. He said he w ould bet on the Cubs at odds of 3:2 or  better, but we have him betting betting on the Cubs (in Bet B) at odds of 2:1, which is worse. worse. (That is, he has to put up 2 dollars to win 1, rather than 1.5 to win 1.) The same problem exists with bet A: he is betting on the Yankees at odds of 2:3, which is worse than 1:2. As a result, he will not accept either of these bets! The reason for this result is that the solutions for X and Y are negative. In fact, it is possible to show algebraically that if A and B are both negative, then X and Y will both be negative. n egative. This has the effect of reversing the bets in such a way that your friend will accept neither. The conclusion is that, even though his  probabilities appear appear incoherent, you cannot set up a Dutch book against him.

175

CHAPTER 9 Theoretical Probability Models Notes Chapter 9 is a straightforward treatment treatment of probability modeling using six standard distributions. The six distributions (binomial, (binomial, Poisson, exponential, normal, triangular, and beta) were chosen because they cover a variety of different probability-modeling probability-modeling situations. To calculate probabilities, probabilities, students can use Excel, @RISK, or online probability calculators.

In addition to the six distributions treated in the main part of the chapter, the uniform distribution is developed in problems 9.27 - 9.29 and the lognormal distribution in problem 9.36 and the Municipal Solid Waste case study. Depending on the nature of the course and the level of the students, instructors may wish to introduce other distributions. One note of caution: Chapter 9 provides an introduction to probability distributions that are used in Chapter 10 (fitting model parameters and natural conjugate priors) as well as Chapter 11 (creating random variates in a simulation). In particular, if the course is intended to move on to Chapter 11, it is important to expose students to the uniform distribution, which is fundamental to the simulation process, and some of the other distributions as well. Although @RISK can run a full simulation, in this chapter, we use @RISK only to view a d istribution’s istribution’s graph and to calculate probability values from the distribution. Step-by-step instructions for @RISK are  provided in the chapter. The Define Distribution button button is used to choose choose a distribution, distribution, and the desired desired  probabilities can be determined determined by sliding the left left and right delimiters delimiters to the the appropriate values. The values can also be typed into the text box over each delimiter and probabilities can be typed into the probability over the graph. Topical cross-reference for problems Bayes’ theorem Beta distribution Binomial distribution

9.17, 9.21, 9.35 9.4, 9.5, 9.6, 9.15, 9.24 9.1, 9.5, 9.16, 9.18, 9.19, 9.29, 9.30, 9.32, 9.36, Overbooking 9.37, Municipal Solid Waste 9.11 9.5, 9.6, 9.7, 9.12, 9.14, 9.15, Earthquake Prediction 9.14, 9.22, 9.25, 9.26 9.37, Municipal Solid Waste 9.12, 9.13, 9.28 9.2, 9.5 - 9.7, 9.11, 9.25, 9.2 9.26, 6, 9.31, 9.32, Municipal Solid Waste 9.18 9.3, 9.5, 9.8, 9.9, 9.13, 9.15, 9.19, 9.20, 9.33, 9.34, Earthquake Earthquake Prediction 9.36 9.23 9.1-9.9, 9.11 – 9.16, 9.18, 9.19, 9.21, 9.24, 9.30, 9.35, 9.36, 9.17, 9.24, Earthquake Prediction 9.27 - 9.29

Central limit theorem Empirical rule Exponential distribution Linear transformations transformations Lognormal distribution Memoryless property  Normal distribution distribution Pascal sampling Poisson distribution PrecisionTree Requisite models @RISK Sensitivity analysis Uniform distribution

176

Solutions 9.1. Find P(Gone 6 or more weekends out of 12)  R ≥ 6 | n = 12, p = 0.65) = PB( R  R' ≤ 6 | n = 12,  p = 0.35) = 0.915. = PB( R

Being gone 6 or more weekends out of 12 is the same as staying home on 6 or fewer. Using @RISK: select the binomial as the distribution type, 12 for the n parameter, and 0.65 for the p  parameter. Move the left delimiter delimiter to 5.5, and the right delimiter delimiter to 12. The probability bar then shows the desired probability: 91.54%. See “Problem 9.1.xlsx.”  X  < 9.2. P(Loss) = P N ( X   < 0 | µ = 2000, σ = 1500) = P( Z  <  <

0 - 2000  < -1.33) = 0.0918. 1500 ) = P( Z  <

P(Gain greater than 4000)  X  > = P N ( X   > 4000 | µ = 2000, σ = 1500) = P( Z  >  >

4000 - 2000 ) 1500

 Z  > = P( Z   > 1.33) = 1 - P( Z ≤ 1.33) = 1 - 0.9082 = 0.0918.  Z ≤ -1.33) = P( Z ≥ 1.33) because of symmetry of the normal distribution.  Note that P( Z 

Using @RISK: select the normal distribution, 2000 for the µ parameter, and 1500 for the σ parameter. To determine the probability of a loss, move the left delimiter to 0. The left-hand side of the probability bar then shows the desired probability: 9.1%. To determine d etermine the probability that the return will be greater than 4000, move the right delimiter 4000 and the right-hand side of the probability bar shows the desired  probability: 9.1%. 9.1%. See “Problem 9.2.xlsx”  X  = 9.3. P(No chocolate chips) = P P( X   = 0 | m = 3.6) = 0.027  X  < P(Fewer than 5 chocolate chips) = P P( X   < 5 | m = 3.6)  X ≤ 4 | m = 3.6) = 0.706 = PP( X   X  > P(More than 10 chocolate chips) = P P( X   > 10 | m = 3.6)  X ≤ 10 | m = 3.6) = 1 - 0.999 = 0.001. = 1 - P P( X 

Using @RISK: select the Poisson distribution and 3.6 for   =   . To determine the probability of no chocolate chips, move the left delimiter to 0.5. The probability bar then shows the probability of 2.7%. To determine the probability of fewer than 5 chips, move the left delimiter bar to 4.5. The probability of more than 10 can be found by moving the right delimiter to 10.5, and the probability shown is 99.9%. See “Problem 9.3.xlsx” 9.4. P(Net Contribution > $600000) = P(1,680,000Q – 300000 > $600000) = 0.536|1  = 1.1, 2  = 3.48) 3.48) = 8.0%.  ( > 0.536|

P(Net Contribution < $100000) = P(1,680,000Q – 300000 < $100000) =  ( < 0.238| 0.238|1  = 1.1, 2  = 3.48) 3.48) = 57.3%.

177

Using @RISK: Select the beta distribution, d istribution, 1.1 for the 1  parameter, and 3.48 for the 2  parameter. To determine the probability of net contribution being greater than $600,000, move the right delimiter to 0.536. The right-hand side of the probability bar then shows the probability is 8.0%. To determine the  probability of net contribution being less than $100,000, move the the right delimiter delimiter to 0.238. The left-hand left-hand side of the probability bar then shows the probability is 57.3%. See “Problem 9.4.xlsx” 9.5. This problem requires significant trial and error with a dose of calculus if you are going to solve it algebraically. @RISK makes the problem easier to solve and the solution can be found in “Problem 9.5.xlsx.” To use @RISK, first enter the parameter values for the desired theoretical distribution. For example, to solve Part a, pull up the binomial distribution and enter 12 for n and 0.85 for p. Next, enter the desired probability value into the probability bar of the Define D efine Distribution window of @RISK. Note that you can actually type in the value and that the values are stated in percentages. Thus, to finish solving Part a, type 2.4 (for 2.4%) into the leftmost section of the probability bar as shown in the figure below. We choose the left hand side of the probability bar because Part a is asking for the area to the left of r :  ( ≤  |  = 12,   = .85) = 0.024. The answer is the value above the appropriate delimiter, the left delimiter in this case.

See “Problem 9.5.xlsx” for the answers to the remaining parts. You need to be careful when working with discrete distributions because probability probability values do not change until an outcome occurs. Thus, for Part a, r  could be any value between 7 (inclusive) to 8 (exclusive).

Type 2.4 into this section of the probability bar.

9.6. This problem requires @RISK. First, using the alternative- parameter method, enter the percentiles given for each part. Note the for Part b, the 25th percentile should be 25 and not 125. After defining the distribution using the percentiles, you can determine the parameter values by undoing the alternative parameter method. method. After unchecking unchecking the Alternative Alternative Parameters checkbox checkbox and clicking OK, the standard  parameter values values are reported. See “Problem 9.6.xlsx.” 9.6.xlsx.”

178

Warning: Do not use cell references for the stated percentiles in this problem. The alternative-parameter method will not report the parameter values if the percentiles are cell locations. The beta distribution does not allow the alternative- parameter method, but the betageneral does. By setting the minimum to zero and the maximum to one, the betageneral is the same as a beta distribution. ∞ 9.7. If PE(T ≥ 5 | m) = 0.24, then ∫5  −  = e-5m = 0.24. Take logs of both sides:

ln[ e-5] = ln(0.24) -5m = -1.4271 m = 0.2854.

Alternatively, the file “Problem 9.7.xlsx” contains the @RISK solution where the alternative- parameter method is used. Remember that @RISK uses  for the exponential parameter and we used  in the text. The relationship is   = 1⁄. 9.8. One can either use tables of Poisson values to find m or use Excel’s Goal Seek feature. The Excel file “Problem 9.8.xlsx” contains the solution. Note that the problem reports two conditions that are desired, but the Poisson has only one parameter. Therefore, you need only use one of the conditions to solve for m, then check to make sure the other condition also holds. Because the Poisson has no upper limit for x, it is much easier to use: (  ≤ 2|  2|) = 0.095. Solution m = 5.394.  X  = 9.9. If PP( X   = 0 | m) = 0.175, then

e- m m 0 = 0.175 0! e-m = 0.175 because m0 = 0! = 1. m = -ln(0.175) = 1.743.

Alternatively, Excel’s Goal Seek can be used as explained in Problem 9.8 above and in the solution file “Problem 9.9.xlsx.” 9.10. Having more information rather than less is always better. Here we are assuming that the quality of the information remains constant. If we think about gathering data one value at a time, then we w e can see that as data are gathered, each successive data point helps refine the choice of distribution. d istribution. If the data were completely accurate (no biases, no incorrectly calibrated values, etc.), then eventually a unique distribution will be identified. Additional data values will not disconfirm the distribution. That is, each additional data value will match the distribution. If, however, the data have hav e inaccuracies, then there may not be any one theoretical distribution that matches all the data.

Even in cases where additional data show that there is no one theoretical distribution matching all the data, having that data can be helpful. For example, it tells us that either the underlying uncertainty does not match a theoretical distribution and/or the data are not completely accurate. Suppose you felt a normal distribution was a good match to the uncertain in question. The first two data values then uniquely define a normal. The third value may or may not confirm the choice. If it doesn’t, then this tells us valuable information. The more data we gather, the better our estimates of the mean and standard deviation. 9.11. P N (µ - σ < Y ≤ µ + σ | µ, σ) = P N (Y ≤ µ + σ | µ, σ) - P N (Y ≤ µ - σ | µ, σ)

179

= P( Z ≤

(µ + σ) - µ σ

) - P( Z ≤

(µ - σ) - µ σ

)

= P( Z ≤ 1) - P( Z ≤ -1) = 0.8413 - 0.1587 = 0.6826. P N (µ - 2 σ < Y ≤ µ + 2σ | µ, σ) = P N (Y ≤ µ + 2σ | µ, σ) - P N (Y ≤ µ - 2σ | µ, σ) = P( Z ≤

(µ + 2σ) - µ σ

) - P( Z ≤

(µ - 2 σ) - µ σ

)

= P( Z ≤ 2) - P( Z ≤ -2) = 0.9772 - 0.0228 = 0.9544. Using @RISK, select the normal distribution, enter 0 for µ and 1 for σ. Moving the left and right delimiters to -1 and 1 respectively shows the probability of 68.3%. Similarly, setting the delimiters at -2 and 2 shows a probability value of 95.4%. See “Problem 9.11.xlsx.” 9.12. If the mean is 10 days, then m = 1/10 = 0.1. a.

PE(T ≤ 1 | m = 0.1) = 1 - e -1(0.1) = 0.0952.

Using @RISK, select the exponential distribution and enter 10 for the β parameter. Sliding the left delimiter to 1 shows the desired probability: 9.5%. b.

PE(T ≤ 6 | m = 0.1) = 1 - e -6(0.1) = 0.4512.

Move the left delimiter to 6 and the probability -is: 45.1%. c.

PE(6 ≤ T ≤ 7 | m = 0.1) = e -6(0.1) - e-7(0.1) = 0.0522.

Set the left delimiter to 6 and the right to 7. The probability in the middle of the bar is 5.2%. If you cannot read the probability from the bar, you can from the Statistics along the right in the “Dif. P” row. d.

P(T ≤ 7 | T ≥ 6, m = 0.1) =

P(T ≤ 7 and T ≥ 6 | m = 0.1) P(T ≥ 6 | m = 0.1)

=

PE(6 ≤ T ≤ 7 | m = 0.1) PE(T ≥ 6 | m = 0.1)

0.0522 = 1-P (T  E ≤ 6 | m = 0.1) = 0.0952. This problem demonstrates the “memoryless” property of the exponential distribution; the probability of time T  lasting another day is the same no matter how much time has already elapsed. See also problem 9.28. The solution is in “Problem 9.12.xlsx.” 9.13. Because of independence among arrivals, the probability distribution for arrivals over the next 15 minutes is independent of how many arrived previously. Thus, for both questions,

PP( X  = 1 in 15 minutes | m = 6 per hour) 180

= PP( X  = 1 in 15 minutes | m = 1.5 per 15 minutes) = 0.335 Using @RISK, select the Poisson distribution and enter 1.5 for the λ parameter. Set the left delimiter to 0.5 and the right to 1.5. The middle of the bar shows the desired probability: 33.5%. See “Problem 9.13.xlsx.” 9.14. a. E(T A) = 5 years and E( T B) = 10 years, choose B. b.

P(T A ≥ 5 | m = 0.2) = e -5(0.2) = 0.368 P(T B ≥ 10 | m = 0.1) = e -10(0.1) = 0.368.

For exponential random variables, the probability is 0.368 that the random variable will exceed its expected value. Using @RISK, select the exponential distribution and enter   = 1/ = 1/.2 = 5. Repeat for the second exponential, this time setting   = 10. See file “Problem 9.14.xlsx.” c.

 —  i. Average lifetime = 0.5 ( T A) + 0.5 ( T B) = T  .  —  E( T  ) = 0.5 (5) + 0.5 (10) = 7.5.  —  Var( T  ) = 0.5 2 (25) + 0.5 2 (100) = 31.25. ii. Difference ∆T  = T B - T A

E(∆T  ) = E( T B) - E(T A) = 10 - 5 = 5 Var(∆T  ) = 12 Var(T B) + (-1) 2 Var(T A) = 25 +100 = 125. 9.15. a. PE(T ≥ 2 hours | m = 8.5 cars per 10 hours) = PE(T ≥ 2 hours | m = 0.85 cars per hour) = e-2(0.85) = 0.183.

Using @RISK, select the exponential distribution and enter   = 1/  = 1/.85 = 1.176. Moving the right delimiter to 2 gives the probability of 18.3%. See file “Problem 9.15.xlsx.” b. PP( X  = 0 in 2 hours | m = 8.5 cars per 10 hours) = PP( X  = 0 in 2 hours | m = 1.70 cars per 2 hours) = 0.183.

Using @RISK, select the Poisson distribution and enter    = 8.5/5 = 1.70. Moving the left delimiter to 0 gives the probability of 18.3%. See file “Problem 9.15.xlsx.” This probability must be the same as the answer to part a because the outcome is the same. The first sale happening after 2 hours pass ( T ≥ 2) is the same as no sales ( X  = 0) in the first two hours. c. E(Bonus) = $200 P P( X  = 13 | m = 8.5) + $300 P P( X  = 14 | m = 8.5)

+ $500 PP( X  = 15 | m = 8.5) + $700 P P( X ≥ 16 | m = 8.5) = $200 (0.0395) + $300 (0.024) + $500 (0.0145) + $700 (0.0138) = $32.01 181

@RISK can be used to calculate the probabilities reported above, and then inserted into the expected bonus formula. See file “Problem 9.15.xlsx.” d. P(Pay $200 bonus) = P P( X  = 13 | m = 8.5) = 0.0395. But this is the probability for any given day. There are 8 such days, for each of which P(Pay $20 bonus) = 0.0395. Now we have a binomial problem. Find the  probability of 2 out of 8 “successes” when P(success) = 0.0395:

PB( R = 2 | n = 8, p = 0.0395) = 0.033. Using @RISK, select the binomial distribution and n = 8 and p = 0.035. Moving the left delimiter to 1.5 and the right delimiter to 2.5 gives the probability of 3.4%. See file “Problem 9.15.xlsx.” 9.16. a. Assuming independence from one questionnaire item to the next, we can calculate the probability using the binomial distribution: P B( R ≤ 2 | n = 10,  p = 0.90). Using Excel’s BinomDist() function, we can calculate BinomDist(2, 10, 0.90, TRUE) to obtain 0.000000374.

Using @RISK, select the binomial distribution and n = 10 and p = 0.90. Moving the left delimiter to 1.5 and the right delimiter to 2.5 gives a probability of 0.0%. See file “Problem 9.16.xlsx.” b. Why might the occurrences not  be independent? Imagine the following: You have just learned that the first answers fell outside of the ranges that you specified. Would you want to modify the ranges of the remaining five? Most of us would. And that would mean that, given data regarding hits or misses on some, you might change the probability of a hit or miss on the others. Thus, the data — whether each item is a hit or a miss — may not be independent, but only because of uncertainty about the level of calibration of the  probability assessor! For a more complete discussion, see Harrison, J.M. (1977), “Independence and Calibration in Decision Analysis,”  Management Science, 24, 320-328. 9.17. a. A hit is defined as market share =  p = 0.3. A flop is market share =  p = 0.1. Given n = 20 and  x = 4, we can find, for example,

P( x = 4 | n = 20, p = 0.3) = P B( R = 4 | n = 20, p = 0.3) = 0.130 and P( x = 4 | n = 20, p = 0.1) = P B( R = 4 | n = 20, p = 0.1) = 0.090. Thus, P( X  = 4 | Hit) P(Hit) P(Hit | X  = 4) =  X  P(  = 4 | Hit) P(Hit) + P( X  = 4 | Flop) P(Flop) 0.13 (0.2) = 0.13 (0.2) + 0.09 (0.8) = 0.265. b. For P(Hit) = 0.4,

0.13 (0.4) P(Hit | X  = 4) = 0.13 (0.4) + 0.09 (0.6) = 0.491. c. For P(Hit) = 0.5,

0.13 (0.5) P(Hit | X  = 4) = 0.13 (0.5) + 0.09 (0.5) = 0.591. d. For P(Hit) = 0.75,

0.13 (0.75) P(Hit | X  = 4) = 0.13 (0.75) + 0.09 (0.25) = 0.813.

182

e. For P(Hit) = 0.9,

P(Hit | X  = 4) =

0.13 (0.9) = 0.929. 0.13 (0.9) + 0.09 (0.9)

f. For P(Hit) = 1.0,

0.13 (1.0) P(Hit | X  = 4) = 0.13 (1.0) + 0.09 (0) = 1.0.

Posterior probability

1 0.8 0.6 0.4 0.2

Prior probability 0 0

0.2

0.4

0.6

0.8

1

The very slight bow relative to the 45° line indicates that 4 favorable responses out of 20 people lead to a slight but undramatic revision of the prior probability. 9.18. a. PB( R > 15 | n = 20,  p = 0.6) = P B( R' = 15 -  R < 5 | n = 20, p = 0.4) = PB( R' ≤ 4 | n = 20,  p = 0.4) = 0.051.

To use @RISK, select the binomial distribution with parameters n = 20 and p = 0.60. Set the right delimiter to 15.5. The right-hand side of the bar shows the desired probability of 5.1%. See “Problem 9.18.xlsx.” b.

P(First 12 in favor) = (0.6) 12 = 0.0022

To use @RISK, select the binomial distribution with parameters n = 12 and p = 0.60. Move the right delimiter to 11.5, and the right-hand side shows a probability of 0.02%. See “Problem 9.18.xlsx.” P(Stop after 13th) = P(11 out of 12 in favor and 13th in favor) = PB( R = 11 | n = 12, p = 0.6) × 0.6 = PB( R' = 12 -  R = 1 | n = 12,  p = 0.4) × 0.6 = 0.017 (0.6) = 0.0102. To use @RISK, select the binomial distribution with parameters n = 12 and p = 0.60. Move the left delimiter to 10.5 and the right to 11.5. The probability of  R = 11 is shown in the middle of the bar or to the right in the “Dif. P” row of the statistics table. See “Problem 9.18.xlsx.” P(Stop after 18th) = P(11 out of 17 in favor and 18th in favor) = PB( R = 11 | n = 17, p = 0.6) × 0.6 = PB( R' = 17 - R = 1 | n = 17, p = 0.4) × 0.6 = 0.002 (0.6) = 0.0012.

183

To use @RISK, select the binomial distribution with parameters n = 17 and p = 0.60. Move the left delimiter to 10.5 and the right to 11.5. The probability of  R = 11 is shown in the middle of the bar or to the right in the “Dif. P” row of the statistics table. See “Problem 9.18.xlsx.” 9.19. a. PP( X  > 2 | m = 1.1) = 1 - P P( X ≤ 2 | m = 1.1) = 1.00 - 0.90 = 0.1.

To use @RISK, select the Poisson distribution with the parameter λ = 1.1. Set the right delimiter to 2.5 and the desired probability value is shown in the right-hand side of the bar to be 10%.See “Problem 9.19.xlsx.” b. Now we have to obtain 18 or more that conform out of 20, given that the probability of conformance equals 0.90 and the probability that a box is rectified is 0.1. This is a binomial model:

P(18 or more pass) = P(2 or fewer are rectified) = PB( R ≤ 2 | n = 20,  p = 0.1) = 0.677. To use @RISK, select the binomial distribution with parameters n = 20 and p = 10%. Set the delimiter to 2.5, and the desired probability of 67.7% is shown in the left-hand side. See “Problem 9.19.xlsx.” c. Let X  be the random number of cases out of 20 that are rectified. Then  X  cases will have no nonconforming bottles. Also, (20 -  X ) may have up to 11 non-conforming bottles; we know that at least one  bottle in each case is OK. Given  X  cases are rectified, the expected number of nonconforming bottles out of 20 × 11 bottles is

0 ( X ) + 11 (0.1) (20 -  X ) = 1.1 (20 -  X ). The expected number of cases rectified are np = 20(0.1) = 2. Therefore E( X ) = 2, and E(# of nonconforming bottles) = 1.1 (20 - 2) = 1.1 (18) = 19.8. 9.20. The Poisson process is often used to model situations involving arrivals, and here we are using it to model the arrivals of the Definitely-Purchase responses. The issue here is whether the expected arrival rate (m) is constant over time. This rate may vary during the day. For example, people who arrive early to the conference might have a different acceptance rate than those arriving at day’s end. It might be possible to  break the day into a few different periods, each with its own rate, and use a separate Poisson process for each period. Cost

 New E(Cost) = $382.5 $3,825

m=3 (0.5)

X (170) X($1,700)

m = 1.5 (0.5)

X (170)

Old  E(Cost) = $375

Poisson, m = 2.5

184

X(150)

9.21. a. The older machines appear to have the edge. They have both lower expected cost and less uncertainty. See Excel file “Problem 9.21.xlsx.” b. Find P(m = 1.5 per month |  X  = 6 in 3 months) = P( m = 4.5 per 3 months |  X  = 6 in 3 months):

Using Bayes theorem, P( m = 4.5 per 3 months |  X  = 6 in 3 months) P( X  = 6 | m = 4.5) P( m = 4.5 per 3 months) = P( X  = 6 | m = 4.5) P( m = 4.5 ) + P( X  = 6 | m =9) P(m = 9)

X($1,700)

PP( X  = 6 | m = 4.5) 0.5 = P ( X  = 6 | m = 4.5) 0.5 + P ( X  = 6 | m = 9) 0.5 P P X($1,500)

0.128 (0.50) $3,750 = 0.128 (0.50) + 0.091= (0.50) = 0.5845. c. Now the expected cost of the new machines would be

E(Cost) = 0.5845 (1.5 x $1700) + 0.4155 (3.0 x $1700) = $3,610. This is less than $3,750, so the new information suggests that the new machines would be the better choice. d. The source of the information often provides us with some indication of credibility or biasedness. We might discount information from the distributor because of incentives to sell the product. A trade magazine report might be viewed as more credible or less biased. Does this mean Bayes’ theorem is inappropriate?  Not entirely, but very simple applications that do not account for bias or differential credibility may be.

One straightforward way to incorporate the idea of biasedness or credibility would be to think in terms of “equivalent independent information.” For example, a positive report from a source you think has a positive  bias might be adjusted downward. If the distributor claimed “6 out of 10” preferred the product, say, this might be adjusted down to 5 of 10 or 4 of 10. In other cases, it might be judged that the information  presented is to some extent redundant with information already obtained, in which case the reported sample size might be adjusted downward. Thus “6 out of 10” from a redundant sample might be judged to be equivalent to an independent test in which the results were “3 of 5.” The “adjusted” or “equivalent independent” sample results could then be used in Bayes’ theorem. 9.22. This is a straightforward application of the formula on page 284 : if  =    + , then  () =    + ( ). 9.23. All we need to do is to create a model or representation of the uncertainty that adequately captures the essential features of that uncertainty. A requisite model of uncertainty will not always be a perfect model,  but the aspects that are important will be well modeled. For example, if a normal distribution is used to represent the weights of lab animals, the mean may be 10 grams with standard deviation 0.3 grams. In fact, all of the animals may weigh between 9 and 11 grams. According to the normal distribution, 99.92% of the animals should fall in this range. 9.24. a. The assessments give P( Q < 0.08) = P( Q > 0.22) = 0.10 and P( Q < 0.14) = 0.50. Therefore,

P(0.08 < Q < 0.14) = P( Q < 0.14) - P( Q < 0.08) = 0.40. Likewise, P(0.14 < Q < 0.22) = P( Q < 0.22) - P( Q < 0.14) = 0.40. b. Answer: 1  = 5.97 and 2  = 39.68. See “Problem 9.24.xlsx.”

185

Using the alternative-parameter method, we choose the betageneral distribution (as the beta does not allow alternative parameters). The betageneral has 4 parameters and we have 3 assessments: P(Q < 0.08) = 0.10, P( Q < 0.14) = 0.50, and P( Q < 0.22) = 0.90. We used these 3 assessments and set the minimum equal to 0 for the fourth input. To find the parameter values for 1  and 2, undo the alternative parameter method. The resulting distribution matches the assessments perfectly, but as Problem 9.23 indicates, it is not an exact representation. For example, the maximum is 110%, which is clearly impossible for Q. Effectively, however, the maximum is 40% as there is a 99.9% chance that Q is less than 40%. c. E(Q) = 14.6%,. There is a 54% chance that market share will be less than 14.6% and a 57% chance that it will be less than 15% 9.25. a. No, because the second investment is substantially riskier as indicated by the higher standard deviation. b. It makes some sense. We need a single peak (mode) and a reasonably symmetric distribution for the normal distribution to provide a good fit. Returns can be positive or negative, and we might expect deviations around a central most-likely value to be evenly balanced between positive and negative deviations. c. P( R1 < 0%) = P N ( R1 ≤ 0 | µ = 10, σ = 3) 0 - 10 = P( Z ≤ ) = P( Z  < -3.33) = 0.0004. 3

To use @RISK, select the normal distribution; enter 0.1 for µ and 0.03 for σ. Set the left delimiter to 0, and the probability in the left-hand side is 0.04%. @RISK may not show the necessary 4 decimal places. P( R2 < 0%) = P N ( R2 ≤ 0 | µ = 20, σ = 12) 0 - 20 = P( Z ≤ 12 ) = P( Z  < -1.67) = 0.0475. To use @RISK, select the normal distribution; enter 0.2 for µ and 0.12 for σ. Set the left delimiter to 0, and the probability shown is 4.8%. P( R1 > 20%) = P N ( R1 > 20 | µ = 10, σ = 3) 20 - 10 = P( Z  > ) = P( Z  > 3.33) = 0.0004. 3 To use @RISK, select the normal distribution; enter 0.1 for  µ ανd 0.03 for σ. Set the right delimiter to 0.20, and the probability in the right-hand side is 0.04%. @RISK may not show the necessary 4 decimal  places P( R2 < 10%) = P N ( R2 <10 | µ = 20, σ = 12) 10 - 20 = P( Z ≤ 12 ) = P( Z  < -0.83) = 0.2033. To use @RISK, select the normal distribution; enter 0.2 for µ and 0.12 for σ. Set the left delimiter to 0.10, and the probability in the left-hand side is 20.2%. d. Find P( R1 > R2) = P( R1 - R2 > 0) = P( ∆ R > 0)

= P N (∆ R > 0 | µ = 10 - 20, σ =

32 + 122 - 0.5 (3) (12) )

= P N (∆ R > 0 | µ = -10, σ = 11.62) = P( Z ≤

0 - (-10) ) = P( Z  > 0.86) = 1 - P( Z  < 0.86) = 0.1949. 11.62

186

Part d cannot be calculated using the Define Distribution window as we have done in the previous  problems. Using the simulation feature of @RISK would allow one to complete this problem, but we wait until Chapter 11 for running simulations. e. The probability distributions could be used as the basis for a model about the uncertainty regarding their returns. This uncertainty can be included in a decision tree with appropriate portfolio alternatives. 9.26. Let X  denote return (in percent),  M  = McDonalds, and S  = US Steel. We have prior probability P( M ) = 0.80. a.

P(6 < X  < 18 | M ) = P N (6 < X  < 18 | µ = 14, σ = 4) 6 - 14 18 - 14 = P( 4 < Z  < 4 ) = P(-2 < Z  < 1) = 0.8185.

To use @RISK, select the normal distribution; enter 14 for µ and 4 for σ. Set the left delimiter to 6 and the right 18. The desired probability is shown in the middle of the bar to be 81.9%. P(6 < X  < 18 | S ) = P N (6 < X  < 18 | µ = 12, σ = 3) 6 - 12 18 - 12 = P( 3 < Z  < 3 ) = P(-2 < Z  < 2) = 0.9544. To use @RISK, select the normal distribution; enter 12 for µ and 3 for σ. Set the left delimiter to 6 and the right 18. The desired probability is shown in the middle of the bar to be 95.5%. b.

P(6 < X  < 18) = P(6 <  X  < 18 |  M ) P( M ) + P(6 <  X  < 18 | S ) P(S ) = 0.8185 (0.8) + 0.9544 (0.2) = 0.84568.

12 - 14 ) = P( Z  > -0.5) = 0.6915. 4 To use @RISK, select the normal distribution; enter 14 for  µ and 4 for σ. Set the right delimiter to 18. The desired probability is shown in right-hand side of the bar to be 69.1%. c.

P( X  > 12 |  M ) = P N ( X  > 12 | µ = 14, σ = 4) = P( Z  >

12 - 12 ) = P( Z  >0) = 0.5. 3 To use @RISK, select the normal distribution; enter 12 for  µ and 3 for σ. Set the right delimiter to 12. The desired probability is shown in right-hand side of the bar to be 50.0%. P( X  > 12 | S ) = P N ( X  > 12 | µ = 12, σ = 3) = P( Z  >

P( X  > 12 |  M ) P( M ) P( M  | X  > 12) = P( X  > 12 | M ) P( M ) + P( X  > 12 | S ) P(S ) 0.6915 (0.8) = 0.6915 (0.8) + 0.5 (0.2) = 0.847. d.

E(Return) = 0.5 E( X  | M ) + 0.5 E( X  | S ) = 0.5 (14) + 0.5 (12) = 13%. Var(Return) = 0.5 2 Var( X  | M ) + 0.5 2 Var( X  | S ) = 0.25 (4 2) + 0.25 (3 2) = 6.25.

9.27. a. This problem introduces the uniform probability distribution. The density function is  f U( x | b, a) =

1 b - a

  when a ≤ x ≤ b, and 0 otherwise:

187

f(x)

1 b- a

0

x a

b

1 The area under the density function is just the area of the rectangle: b - a (b - a) = 1. b.

c - a PU( X  < c | a, b) = b - a . When a = 3 and b = 5, then

4.5 - 3 1.5 PU( X  < 4.5 | a = 3, b = 5) = 5 - 3 = 2.0 = 0.75. To use @RISK, select the uniform distribution; enter 3 for the minimum parameter and 5 for the maximum  parameter. Set the left delimiter to 4.5. The desired probability is shown in the left-hand side of the bar to  be 75.0%. c.

4.3 - 3 1.3 P( X  < 4.3 | a = 3, b = 5) = 5 - 3 = 2.0 = 0.65.

Moving the left delimiter to 4.3 results in left-hand side of the bar showing a probability to be 65.0%. 0.75 - 0.25 P(0.25 <  X  < 0.75 | a = 0, b = 1) = = 0.5. 1-0 To use @RISK, select the uniform distribution; enter 0 for the minimum parameter and 1 for the maximum  parameter. Set the left delimiter to 0.25 and the right delimiter to 0.75. The desired probability is shown in the middle of the bar to be 50.0%. 10 - 3.4 P( X  > 3.4 | a = 0, b = 10) = 10 - 0 = 0.66. To use @RISK, select the uniform distribution; enter 0 for the minimum parameter and 10 for the maximum parameter. Set the right delimiter to 3.4. The desired probability is shown in the right-hand side of the bar to be 66.0%. 0 - (-1) 1 P( X  < 0 | a = -1, b = 4) = 4 - (-1) = 5 = 0.2. To use @RISK, select the uniform distribution; enter -1 for the minimum parameter and 4 for the maximum parameter. Set the left delimiter to 0.0. The desired probability is shown in left-hand side of the  bar to be 20.0%.

188

d. P(X
)

1

0

x a

e. E( X ) =

a + b

2

=

b

3+5 (b - a)2 (5 - 3)2 4 = 4. Var( X ) = = = = 0.33. 2 12 12 12

9.28. a.

0

s

1.0

10.5

1 P(S  < 1) = (height × width) = 10.5 x 1.0 = 0.0952. To use @RISK, select the uniform distribution; enter 0 for the minimum parameter and 10.5 for the maximum parameter. Set the left delimiter to 1.0. The desired probability is shown in left-hand side of the  bar to be 9.5%. b. P(S < 6) = 6 ×

1 10.5 = 0.5714.

With the setup in Part a, move the left delimiter to 6 and the left-hand side of the bar shows 57.1%. c. P(6 < S  < 7) = 1 ×

1 10.5 = 0.0952.

With the setup in Part a, move the left delimiter to 6 and the right delimiter to 7. The middle of the bar P(6 < S  < 7) 0.0952 shows 9.5%.d. P(S ≤ 7 | S  > 6) = P(S  > 6) = 1 - 0.5714 = 0.2222.  Note that this result is different from the probability in part a. The uniform distribution does not have a “memoryless” property like the exponential does (problem 9.12).

189

9.29. a.

0

36 min

1.5 hr 

t

36 P(T ≤ 36) = 90 = 0.40. To use @RISK, select the uniform distribution; enter 0 for the minimum parameter and 90 for the maximum parameter. We are using minutes as the base unit. Set the left delimiter to 36. The desired  probability is shown in left-hand side of the bar to be 40.0%. b. Use a binomial model with  =   ( ≤ 36) = 0.40. Each of 18 customers has a 0.40 chance of being a “success” (shopping for 36 minutes or less). We require the probability that 10 or more are “successes” out of 18:

PB( R ≥ 10 | n = 18,  p = 0.4) = 1 - P B( R ≤ 9 | n = 18,  p = 0.4) = 1 - 0.865 = 0.135. To use @RISK, select the binomial distribution; enter 18 for n and 0.4 for  p. Set the right delimiter to 9.5. The desired probability is shown in right-hand side of the bar to be 13.5%. 9.30. a. We define  to be the number of working engines on the plane. Thus, for a two-engine plane,  ≥ 1for the plane to land safely and for a four-engine plane,  ≥ 2. Does the binomial distribution fit the uncertainty? We have dichotomous outcomes (engine working or not). Each engine may or may not have the same probability of working. Because all airplane engines undergo routine safety checks and follow a schedule for replacing parts, even if the part is working, it is likely that both independence and constant probability holds true. b. For a given p,  ( ≥ 1|  = 2, ) = 2()(1  ) + 2 and  ( ≥ 2 |  = 4, ) = 6(2 )(1  )2 + 4(3 )(1  ) + 4. See the file “Problem 9.30.xlsx” for the complete solution. For probability values above 67%, the four-engine plane is safer and for probability values below 67%, the two-engine plane is safer. Counterintuitively, as the engines become less reliable, the safety of the two-engine plane surpasses the safety of the four-engine plane. Overall, the safety drops for both planes, but does so more quickly for the four-engine plane. Probability of Engine Working

Prob of 2-engine landing safely

Prob of 4-engine landing safely

0.95

99.750%

99.952%

0.8

96.000%

97.280%

0.67

89.110%

89.183%

0.1

19.000%

5.230%

190

9.31. Let X  be the score of a problem drinker and Y  be the score of a non-problem drinker.

 (  > 75|  = 80,  = 5) = 84.1% and

 (  > 75| = 60,  = 10) = 6.7% 9.32 Let L denote the uncertain length of an envelope. a. P N ( L > 5.975 | µ = 5.9, σ = 0.0365) = P( Z  >

5.975 - 5.9 ) = P( Z  > 2.055) = 0.02. 0.0365

To use @RISK, select the normal distribution; enter 5.9 for µ and 0.0365 for σ. Set the right delimiter to 5.975 and the desired probability is shown in the right-hand side to be 2.0%. b. We will use a binomial model, in which  p = P(Envelope fits) = 0.98. We need

PB( R ≤ 18 | n = 20,  p = 0.98) = P B( R ≥ 2 | n = 20,  p = 0.02) = 1 - P B( R ≤ 1 | n = 20, p = 0.02) = 1 - 0.94 = 0.06. That is, about 6% of the boxes will contain 2 or more cards that do not fit in the envelopes. To use @RISK, select the binomial distribution; enter 20 for n and 0.98 for  p. Set the left delimiter to 18.5 and the desired probability is shown in the left-hand side to be 6.0%. 9.33. a. Reasons for using the Poisson distribution include: - Machines often break down independently of one another. - The probability of a breakdown is small. - The rate of breakdowns is roughly constant over time. - Machines can break down at any point in time. b. There are a few ways to solve this problem. The brute-force way is to alter the λ parameter until the Poisson distribution closely matches the desired probabilities. At λ = 0.7, we have a close fit:

P(X = 0) = 0.50 P(X = 1) = 0.35 P(X = 2 or 3) = 0.15 P(X ≥ 4) = 0

P P( X  = 0 | m = 0.70) = 0.497 P P( X  = 1 | m = 0.70) = 0.348 P P( X  = 2 or 3 | m = 0.70) = 0.150 PP( X ≥ 4 | m = 0.70) = 0.006

Another approach is to use the Poisson formula on one of the probabilities. The probability that  X  = 0 is the easy one: e-m m0 = e-m. 0! Solve for m to get m = 0.693. P( X  = 0) = 0.5 =

c. The expected number of breakdowns is m, whatever value was chosen.

191

9.34. Cost 0 or 1 f ailure (0.736)

Leav plant open

0

E(Cost) = $3960 2 or more f ailures $15,000 (0.264)

Close plant

$10,000

P(0 or 1 failure) = P P( X  = 0 | m = 1) + P P( X  = 1 | m = 1) = 0.368 + 0.368 = 0.736. P(2 or more failures) = 1 - P(0 or 1 failure) = 1 - 0.736 = 0.264. Because E(Cost) for leaving the plant open is $3960 < $10,000, the choice would be to leave the plant open. 9.35. This problem is designed to show that Bayesian updating either done sequentially or all at once  produces the same posterior probabilities. a. The posterior probabilities after the first hour are: P(Failure) = 0.04, P(Potential) = 0.44, and P(Success) = 0.52. These become the prior probabilities for the second hour. Given 17 Definitely-Purchase responses in the second hour, we update as before: P(Failure|  = 17) =

P(  = 17|Failure)P(Failure) P(  = 17|Failure)P(Failure) + P(  = 17|Potential)P(Potential) + P(  = 17|Success)P(Success) =

=

PP (  = 17|m = 10)(0.04) PP(  = 17|m = 10)(0.04) + PP (  = 17|m = 15)(0.44) + PP (  = 17|m = 20)(0.52) 0.013(0.04) 0.0128(0.04) + .0847(0.44) + 0.0760(0.52)

= 0.0073.

Similarly, P(Potential|  = 17) = .4793 and P(Success|  = 17) = .5135 b. The prior probabilities are: P(Failure) = P(Potential) = P(Success) = 1/3. Given 35 Definitely-Purchase responses in the first two hours, we update as before: P(Failure|  = 35 in 2 hrs) =

P(  = 35|Failure)P(Failure) P(  = 35|Failure)P(Failure) + P(  = 35|Potential)P(Potential) + P(  = 35|Success)P(Success) =

PP (  = 35|m = 20)(1/3) PP(  = 35|m = 20)(1/3) + PP (  = 35|m = 30)(1/3) + PP (  = 35|m = 40)(1/3)

192

=

0.00069(1/3) 0.00069(1/3) + .04531(1/3) + 0.04854(1/3)

= 0.0073.

Similarly, P(Potential|  = 17) = .4793 and P(Success|  = 17) = .5135. You need be very careful with rounding in this problem, and thus should not use @RISK as it only reports the first decimal place. See “Problem 9.35.xlsx” for a complete solution. 9.36. a, b. Reasons for using the binomial distribution are the following: The number of machines to break down can be between 0 and 50. The probability of any machine breaking down is 0.004. The machines appear to break down independently of each other. Thus, a binomial distribution with  p = 0.004 and n = 50 would be appropriate.

Reasons for using the Poisson distribution: The probability of an individual machine breaking down is small and can occur any time during the day. Breakdowns seem to occur independently of one another. The expected number of breakdowns in one day is 50 × 0.004 = 0.2, so a Poisson distribution with m = 0.2 would be appropriate. # parts required (breakdowns) 0 (0.819) 1 (0.164) Stock 0 parts E(Cost) = $13.00

2 (0.016) 3 (0.001) . . .

0 (0.819) 1 (0.164) Stock 1 part E(Cost) = $11.24

2 (0.016) 3 (0.001) . . .

0 (0.819) 1 (0.164) Stock 2 parts E(Cost) = $20.13

2 (0.016) 3 (0.001) . . .

c. Calculations for E(Cost):

E(Cost | Stock 0) = E(Breakdowns) × $65.00 = 0.2 × $65.00 = $13.00 50

E(Cost | Stock 1) = $10 + $65 × ∑ x=1 [ x - 1]P( X = x)

193

Cost ($) 0 65 130 195

10 10 75 140

20 20 20 85

50

50

= $10 + $65 × [∑ x=1  x P( X = x) - ∑ x=1 P( X = x)] = $10 + $65 × [E(# breakdowns) - {1 - P( X  = 0)}] = $10 + $65 × [0.2 - (1 - 0.819)] = $11.235. 50

E(Cost | Stock 2) = $20 + $65 × ∑ x=2 [ x - 2]P( X = x) 50

50

= $20 + $65 × [∑ x=2  x P( X = x) - 2 ∑ x=2 P( X = x)] 50

= $20 + $65 × [∑ x=1  x P( X = x) - P( X = 1) - 2 {1 - P( X = 0) - P( X = 1)}] = $20 + $65 × [E(# breakdowns) - P( X = 1) - 2 {1 - P( X = 0) - P( X = 1)}] = $20 + $65 × [0.2 - 0.164 - 2 {1 - 0.819 - 0.164}] = $20.13. This decision tree is modeled in the Excel file “Problem 9.36.xlsx.” The Binomial distribution is included in the spreadsheet model using the Excel function BINOMDIST. For example, to find the probability of no  breakdowns use the formula “=BINOMDIST(0,50,0.004,FALSE)”. More possible outcomes are included in the spreadsheet model (up to 7 parts required due to breakdowns) therefore there is some round-off difference between the solution above and the answers in the spreadsheet. The preferred alternative is to stock 1 part with an expected cost of $11.20. 9.37. a. E( X ) = e[10 + 0.5 (0.09)]  = $23,040.

Var( X ) = e2(10) (e0.09 - 1) (e 0.09) = 49,992,916 σ X  = $7,071.

b. PL ( X  > 50,000 | µ = 10, σ = 0.3) = P N (Y = ln( X ) > ln(50,000) | µ = 10, σ = 0.3)

= P N (Y  > 10.8198 | µ = 10, σ = 0.3) = P( Z  >

10.8198 - 10 ) = P( Z  > 2.73) = 0.0032. 0.3

c. First, find the probability distribution for Q = 200 X . According to the hint, Q will be approximately normal with mean µ = 200 (23, 0404) = $4.608 million, variance σ2 = 200 (49,992,915.95) =

9,998,583,190, and standard deviation σ = $99,992.91, or about $0.1 million.  Now find the 0.95 fractile of this normal distribution. That is, find the value q such that P N (Q ≤ q | = 4.608, σ = 0.1) = 0.95. We know that P( Z ≤ 1.645) = 0.95. Therefore, P( Z ≤

q - 4.608

0.1

) = 0.95

194

only if q - 4.608

0.1

= 1.645

or q = 4.608 + 1.645 (0.1) = $4.772 million.

Thus, if the company has on hand $4.772 million, they should be able to satisfy all claims with 95%  probability. Case Study: Overbooking 1. Use a binomial distribution, and let  Res = n = number of reservations sold.

PB( R > 16 | Res = 17,  p = 0.96) = P B( R = 0 | n = 17,  p = 0.04) = 0.4996. PB( R > 16 |  Res = 18, p = 0.96) = P B( R ≤ 1 | n = 18,  p = 0.04) = 0.8393. PB( R > 16 |  Res = 19, p = 0.96) = P B( R ≤ 2 | n = 19,  p = 0.04) = 0.9616. This model is shown in the Excel file “Overbooking I.xlsx.” The probabilities for possible arrivals given a certain number of reservations are shown in a table in the second worksheet in the file. For example, the  probability that 17 people show-up given 17 reservations are taken (i.e., 0 no-shows) can be found using the Excel formula “ =BINOMDIST(0,17,0.04,TRUE)”. @RISK can be used to determine this value, but you would then need to manually transfer the values from @RISK into the spreadsheet model, while Excel  provides useful formulas for many of the distributions. This case and the following cases use the built-in Excel functions rather than finding the value in @RISK and typing it into the spreadsheet. 2.

E(R | Res = 16) = $225 (16) = $3600 16

E(C1 | Res = 16) = $900 + $100 ∑ x=0  x PB( X = x | n = 16,  p = 0.96) = $900 + $100 E( X ) = $900 + $100 (15.36) = $2436. E(C2 | Res = 16) = 0 (No extra passengers can arrive.) E(Profit |  Res = 16) = $3600 - $2436 - $0 = $1164. These calculations are shown in the first worksheet in the Excel file. 3.

E(R | Res = 17) = $225 (17) = $3825 16

E(C1 | Res = 17) = $900 + $100 ∑ x=0  x PB( X = x | n = 17,  p = 0.96) + $1600 P B( X = 17 | n = 17,  p = 0.96) = $900 + $782.70 + $799.34 = $2482.04 E(C2 | Res = 17) = $325 P B( X = 17 | n = 17,  p = 0.96) = $325 (0.4996) = $162.37

195

E(Profit |  Res = 17) = $3825.00 - $2482.04 - $162.37 = $1180.59. E(R | Res = 18) = $225 (18) = $4050 16

E(C1 | Res = 18) = $900 + $100 ∑ x=0  x PB( X = x | n = 18,  p = 0.96) + $1600 P B( X ≥ 17 | n = 18, p = 0.96) = $900 + $253.22 + $1342.89 = $2496.11 E(C2 | Res = 18) = $325 P B( X = 17 | n = 18,  p = 0.96) + $650 PB( X = 18 | n = 18, p = 0.96) = $325 (0.3597) + $650 (0.4796) = $428.65 E(Profit |  Res = 18) = $4050.00 - $2496.11 - $428.65 = $1125.24. E(R | Res = 19) = $225 (19) = $4275 16

E(C1 | Res = 19) = $900 + $100 ∑ x=0  x PB( X = x | n = 19,  p = 0.96) + $1600 P B( X ≥ 17 | n = 19, p = 0.96) = $900 + $60.74 + $1538.56 = $2499.30 E(C2 | Res = 19) = $325 P B( X = 17 | n = 19,  p = 0.96) + $650 PB( X = 18 | n = 19, p = 0.96) + $975 PB( X = 19 | n = 19, p = 0.96) = $325 (0.1367) + $650 (0.3645) + $975 (0.4604) = $730.26 E(Profit |  Res = 19) = $4275.00 - $2499.30 - $730.26 = $1045.44. The optimum amount to overbook is by one seat. By selling 17 reservations, Mockingbird obtains the highest expected profit. Case Study: Earthquake Prediction Excel provides many useful functions to solve for the probabilities. The model is shown in the Excel file “Earthquake Prediction.xlsx.”

If you want to do the raw calculations in a spreadsheet for this problem, a recursive formula for calculating Poisson probabilities is helpful: e-mmk +1 e-mmk  m PP( X  = k  + 1 | m ) = (k + 1)! = k! k+1

(

)  = PP( X  = k  | m )(k m+ 1 ).

A recursive formula like this can be entered and calculated easily in a spreadsheet. It saves laborious calculations of many factorial terms.

196

1.

PP( X ≤ 10 in next year | m = 24.93/year) = P P( X ≤ 10 | m = 24.93) = 0.000613

Or this value can be found using the Excel formula “=POISSON(10,24.93,TRUE)” PP( X ≤ 7 in next 6 months | m = 24.93/year) = P P( X ≤ 7 | m = 12.465) = 0.071069 This value can be found using the Excel formula “=POISSON(7,12.465,TRUE)” PP( X  > 3 in next month | m = 24.93/year) = 1 - P P( X ≤ 3 | m = 2.0775) = 0.157125 This value can be found using the Excel formula “=1-POISSON(3,2.0775,TRUE)” These formulas are shown in the first worksheet of the Excel file. 2. f(x)

Density function for earthquake magnitude

Magnitude

4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

PE( M ≤ 6.0 | m = 1.96) = 1 - e -1.96(6.0-4.0) = 0.980 PE(5.0 ≤ M ≤ 7.5 | m = 1.96) = e -1.96(5.0-4.0) - e-1.96(7.5-4.0) = 0.140 PE( M ≥ 6.4 | m = 1.96) = e -1.96(6.4-4.0) = 0.009 Range 4.0 - 5.0 5.0 - 6.0 6.0 - 7.0 7.0 - 8.0 8.0 +

Probability 0.859 0.121 0.017 0.002 0.001

197

8.0

8.5

Sensitivity analysis: Find m that corresponds to the data in Table 9.1: Magnitude ( x) 8.0+

Empirical P( M ≥ x) 1 = 0.0004 2493

Exponential rate  m 1.96

7.0

13 2493 = 0.0052

1.75

6.0

93 = 0.0373 2493

1.64

5.0

493 2493 = 0.1978

1.62

Thus, m could be less than 1.96. What value should be chosen for the model? A value of 1.96 makes sense if the concern is primarily with large earthquakes. Furthermore, this choice will result in larger probabilities of large earthquakes, a kind of “worst case” scenario. These calculations are shown in the second worksheet of the Excel file “Earthquake Prediction.xls”. Also, the probability as a function of m is implemented such that m can be varied to see the impact on the  probability. 3. Performing the calculations as described in the case gives

P(At least 1 quake with  M > 8 in next year) ≈ 0.01 P(At least 1 quake with  M > 8 in next 5 years) ≈ 0.124. These calculations are shown in the third worksheet of the Excel file “Earthquake Prediction.xls”. 4. This question asks students to think about preparation for serious earthquakes. Clearly, the potential exists for substantial damage. Other than the issues raised in the case, one would want to look at training for residents, protection for utilities, emergency measures for providing power and water to hospitals, and distributing water, food, shelter, and other necessities in the event of a severe quake. This question may stimulate an interesting discussion, especially if some of the students have experienced a large quake. Case Study: Municipal Solid Waste 1. The pollutant levels follow lognormal distributions, and so the log-pollutant levels follow normal distributions with corresponding parameters. Thus, we need to find the probability that the log-pollutant level exceeds the logarithm of the corresponding established level from Table 9.3. This can be done using the Normal Distribution Probability Tables or with Excel functions. This case is modeled in the spreadsheet “Municipal Solid Waste.xlsx.”

For a small plant: Dioxin/Furan (DF) permit level = 500 ng/Nm 3 P(DF > 500 ng/Nm3) = P N (ln(DF) > ln(500) | µ = 3.13, σ = 1.2) = P( Z ≤

ln(500) - 3.13 ) = P( Z  > 2.57) = 0.0051. 1.2

198

Or with the Excel function: =1-NORMDIST(LN(500),3.13, 1.2, TRUE) = 0.0051 Particulate Matter (PM) permit level = 69 mg/dscm P(PM > 69 mg/dscm) = P N (ln(PM) > ln(69) | µ = 3.43, σ = 0.44) = P( Z ≤

ln(69) - 3.43 ) = P( Z  > 1.83) = 0.0338. 0.44

Or with the Excel function: =1-NORMDIST(LN(69),3.43, 0.44, TRUE) = 0.0338 For a medium plant: DF permit level = 125 ng/Nm 3 P(DF > 125 ng/Nm3) = P N (ln(DF) > ln(125) | µ = 3.13, σ = 1.2) = P( Z ≤

ln(125) - 3.13 ) = P( Z  > 1.42) = 0.0785. 1.2

Or with the Excel function: =1-NORMDIST(LN(125),3.13, 1.2, TRUE) = 0.0785 Particulate Matter: Analysis is the same as for the small plant. 2. SO2 permit level = 30 ppmdv

For a single observation, the analysis follows question 1: P(SO2 > 30 ppmdv) = P N (ln(SO 2) > ln(30) | µ = 3.2, σ = 0.39) = P( Z ≤

ln(30) - 3.2 ) = P( Z  > 0.52) = 0.3030. 0.39

Or with the Excel function: =1-NORMDIST(LN(30),3.2, 0.39, TRUE) = 0.3030  —  For a geometric average of 24 independent observations of SO 2, denoted by S  ,  —   —  0.39 P( S > 30 ppmdv) = P N (ln( S ) > ln(30) | µ = 3.2, σ = ) 24 = P( Z ≤

ln(30) - 3.2 ) = P( Z  > 2.53) = 0.0057. 0.39 / 24 199

Or with the Excel function: =1-NORMDIST(LN(30),3.2, 0.0796, TRUE) = 0.0057  Naturally, this is a much smaller probability than for the single-observation case. It is due to the effect of the Central Limit Theorem on the distribution of averages. 3. They appear to be able to satisfy the less strict requirements for a small plant more easily. However, for a medium plant they also appear to be in good shape; the highest probability of noncompliance is for Dioxins and Furans (0.0785). A larger plant might prove to be better in the long run because of extra capacity and the stricter incineration standards. In addition, the officials should consider the effects of recycling program growth, measures that would require recyclable packaging, and other potential impacts on waste disposal needs.

200

CHAPTER 10 Using Data Notes

This chapter is a combination of elementary data analysis and regression analysis. It is clear that data can  be used as a basis for developing probability models in decision analysis. The first part of the chapter does this in a straightforward way, using data to generate histograms and CDFs, and fitting theoretical distributions to data. For most students, the only new element of this early section is the development of data-based CDFs. Additional reading on this topic can be found in Vatter, P., S. Bradley, S. Frey, & B. Jackson (1978) Quantitative Methods in Management: Text and Cases, Homewood, IL: Richard D. Irwin, Inc. The chapter then provides instructions for using @RISK to fit distributions to data. @RISK allows one to analyze data to find a best-fitting distribution from among a set of candidate families. It will run a fully automated fitting procedure and generate a report that ranks a set of distributions according to how well they fit the input data. The rankings are based on one of six goodness-of-fit measures. The second part of the chapter considers the use of data to understand relationships via regression analysis. For instructors, it is important to realize that we do not intend for this section to be a stand-alone treatment of regression. In fact, we can practically guarantee that classical statisticians who teach basic statistics courses will disapprove of the treatment of regression in this chapter. The chapter is to relate regression to decision analysis so that students understand how this statistical tool can be used in the decision-modeling  process, and thus does not cover the typical statistical-inference procedures. Nevertheless, this section does  provide a review and a decision-analysis perspective for students who have already covered regression in a  basic statistics course. The chapter concludes by reinforcing the distinction between decision-analysis modeling and conventional statistical inference. Case studies that use regression in a decision-making context make a good addition to the problems in the  book. For example, in a bidding context it is sometimes possible to construct a regression model of the highest competitive bids in previous auctions. For an upcoming auction, this model can serve as the basis for developing a probability distribution for the highest competitive bid. From this distribution, the bidder can derive the probability of winning the auction with a specific bid amount, which in turn provides a framework for choosing an optimal (expected-value-maximizing) bid. For an example of such a case, see “J.L. Hayes and Sons” in Vatter, et al (1978). A brief introduction to Bayesian inference using natural conjugate priors has been placed online (cengagebrain.com). Conceptually, this is very difficult material. In fact, this material was included only after a number of reviewers of  Making Hard Decisions specifically requested it! Of course, this online section has as a strict prerequisite the sections in Chapter 9 on the binomial, normal, and beta distributions. Additional reading at a level that students can understand is in  An Introduction to Bayesian Inference and  Decision, by R. Winkler, New York: Holt, Rinehart, Winston (1972), and Statistics: A Bayesian Perspective, by D. Berry, Belmont, CA: Duxbury (1995). Topical cross-reference for problems

@RISK Bayesian Analysis

10.4, 10.8 – 10.12, Taco Shells 10S.1 – 10S.5, Forecasting Sales, Overbooking: Bayesian Analysis 10.12, 10.17, 10.18 10.10 10.4, 10.6, 10.8, 10.9, 10.11, 10.13, 10.14, Taco Shells 10.11 10.5, 10.7, 10.10, 10.12

Beta distribution Binomial distribution CDFs Exponential distribution Histograms

201

 Natural conjugate conjugate prior  Normal distribution distribution Pearson-Tukey approximation Poisson distribution PrecisionTree Regression Subjective judgments

10S.1 – 10S.5, Forecasting Forecasting Sales, Overbooking: Bayesian Analysis 10.8, 10.9 Taco Shells 10.10 Taco Shells 10.3, 10.13, 10.14 10.2

Solutions 10.1. As long as what happened in the past can be reasonably expected to extend into the future, then

 patterns of past outcomes can be used to estimate estimate probability distributions. distributions. Even if such extrapolation extrapolation is not fully justified, an empirically derived probability distribution distribution may provide a basis that can be adjusted subjectively. 10.2. Many decision makers trust “objective” data more than “subjective” judgments. However, implicit implicit in such a position is the subjective judgment that the data and analysis are appropriate for constructing a model of the uncertainty for the current situation. In many cases, the data may not be adequate in this regard. For example, estimating a failure rate among machines implicitly assumes that the failure rate from the past (old machines) will be the same in the future (new machines). In the case of economic forecasting, the assumption is that the economy has not changed in any material way. 10.3. a. Important subjective judgments include the specification of the particular explanatory variables

(and hence exclusion of others), the form of the regression function, and that past data will be appropriate for understanding property damage due to future storms. Another important modeling decision is the use of  property damage as as the response variable; variable; presumably, presumably, this variable variable is useful for the insurance company. Because the population density is not constant along the coastline, however, it may be better to use something like property damage per unit of population density. b. The interpretation of  β 1 would be the expected change in property damage for every additional mile of

diameter holding constant the other variables in the model, specifically, holding constant the barometric  pressure, the wind speed, and the time of year. The sign of 1  indicates the direction of change; positive 1 means an expected increase in property damage while a negative 1  means an expected decrease in  property damage. c. The interpretation of  β 5 would be the expected change in property damage due to the hurricane hurricane hitting a

city. (Including such a categorical variable is one way the insurance company could account for variable  population density. density. Redefining the response variable variable as described in in the answer to part part a is another.) 10.4. One alternative is to estimate the fractiles by drawing lines on Figure 10.7: 1   y 0.8    t    i    l    i    b   a0.6    b   o   r    P0.4   e   v    i    t 0.2   a    l   u   m 0   u    C $600

Empirical Distribution of Costs

$700

$800

$900

$1,000 $1,100 $1,200 $1,300 $1,400 $1,500

Operating Costs

From this, we estimate 0.65  ≈  $1,100 and 0.35  ≈  $900. 202

To obtain a more precise p recise estimate, use the Distribution-Fitting Distribution-Fitting button in @RISK to fit the data. Simply S imply highlight the data and click the button. The input distribution (in blue) in the results window is the empirical CDF. Typing 35% in the left-hand and right-hand sides of the probability bar shows that 0.65  = $1,070 and 0.35  = $903. See figure below.

10.5. Answers to this question will vary depending on the intervals chosen. A histogram with six bins or

intervals is shown below. The bin widths are each $116.55 and were chosen using the Auto (automatic) feature of the DecisionTools program StatTools. The numerical details of the bins is given below the histogram. Histogram of A random sample of 20 observations for the annual operating costs of Big Berthas. / Data Set #1 4.5 4 3.5 3    y    c    n    e    u    q    e    r    F

2.5 2 1.5 1 0.5 0    5    7  .    4    1    7     $

   0    3  .    1    3    8     $

   5    8  .    7    4    9     $

   0    4  .    4    6    0    1     $

203

   6    9  .    0    8    1    1     $

   1    5  .    7    9    2    1     $

A random sample of 20 observations for the annual operating costs of Big Berthas. / Data Set #1 Bin Min

Bin Max

Bin Midpoint

Freq.

Rel. Freq.

Prb. Density

Bin #1

$656.47

$773.03

$714.75

3

0.1500

0.00129

Bin #2

$773.03

$889.58

$831.30

3

0.1500

0.00129

Bin #3

$889.58

$1006.13

$947.85

4

0.2000

0.00172

Bin #4

$1006.13

$1122.68

$1064.40

4

0.2000

0.00172

Bin #5

$1122.68

$1239.23

$1180.96

3

0.1500

0.00129

Bin #6

$1239.23

$1355.78

$1297.51

3

0.1500

0.00129

Histogram

10.6. Either draw lines as was done for Exercise 10.4 or use @RISK’s distribution fitting fitting procedure on the

residuals. For the regression using only Ads, the 0.2 fractile is -$1,600 and the 0.80 fractile is $1,632. For the regression using only all three variables, the 0.20 fractile is -$361 and the 0.80 fractile is $369. When predicting sales using only Ads, we would be 60% confident that actual sales will be within -$1,600 and $1,632 of the predicted value. When predicting sales using all three variables, we would be 60% confident that actual sales will be within -$361 and $369 of the predicted value. 10.7. Answers will vary considerably here. There is no really good way to resolve this problem in the

context of using data alone. Here are some possibilities: a.  b. c. d.

Fit a theoretical theoretical distribution distribution to the data. Use the theoretical theoretical distribution distribution as a basis for for determining small probabilities. Use a combination combination of subjective assessments assessments and the the data to come come up with probabilities. probabilities. Collapse the categories. That is, look at situations situations where two or more more failures occurred, being sure that the newly defined categories all have at least five observations. Do nothing. Estimate the probability probability of three failures as 1/260.

10.8.Using the distribution fitting feature of @RISK, we have the following comparisons.

Operating Cost $700 $800 $900 $1,000 $1,200 $1,400

Cumulative Empirical Prob 10% 20% 30% 50% 20% 0%

Cumulative  Normal Prob 6.8% 15.8% 30.4% 49.1 17% 2.7%

The figure below shows the data-based empirical distribution distribution overlaying the normal distribution with   = $1,004.71 and   = $204.39. From the cumulative probabilities above, the normal is underestimating the left-hand tail (compare 10% to 6.8% for $700 operating costs) and overestimating the right-hand tail (compare 0% to 2.7% for $1,400 operating costs). Overall, the normal is a very good fit. The figure shows that it is the best fitting for the chi-square, K-S, and A-D fit measures and is second best for the AIC and BIC measures of fit.

204

10.9. a. The figure below shows the empirical CDF and superimposed on the best-fitting normal

distribution, which has a mean of 9.7620 grams and a standard deviation of 1.2008 grams. Weight (grams) 8 9.5 11.5

Cumulative Empirical Prob 6.7% 33.3% 6.7%

Cumulative  Normal Prob 7.1% 43.3% 7.4%

205

b. The normal distribution probability is not terribly close and neither is it terribly far off. The only serious

departure from normality is that the distribution is slightly skewed. @RISK  @RISK shows shows that the Laplace distribution is the best fitting distribution for all five measures of fit. The Laplace has an unusual shape as shown below. The peak in the middle results from two exponential functions being placed back to back. Is the Laplace better than the normal? To answer this, it helps know what the Laplace is used to model. Researchers have recently being using the Laplace instead of the normal if they believe the tails need more weight. In this case, the scientist has to ask if the weights of the lab animals can be extreme. If so, the Laplace could be better than the normal.

10.10. a, @RISK recommends a Poisson distribution to model the data with m = 0.39. The Poisson is a

natural choice here. In Chapter 9, we w e discussed the four conditions necessary to use the Poisson. The independence condition (#3 in Chapter 9) is always tricky to confirm. From the fit results, the Poisson is a very good choice. The parameter m = 0.39 is interpreted as expected number of defective bulbs per box.

206

b, c. The parameter m = 0.39 and did not change for the Poisson distribution. @RISK reports a binomial with n = 3 and  p = 0.12333. See below. It is not necessary to use @RISK’s results. For example, a binomial distribution with n = 12 and  p = 39/1200 would be a reasonable distribution. The total number of bulbs

checked is 1200, and 39 were found defective.

10.11. a. An exponential distribution often is used to model times between arrivals, especially when the

arrivals are independent (in this case, no groups of shoppers arriving at once). b. No, @RISK reports that the triangular distribution is the #1 fit for four of the five fit measures. The

exponential is certainly close enough to be used here. c. The fit for the exponential distribution improved only slightly. For example, the K-S distance went from

0.17 to 0.15. The fit in Part b had a shift of 0.00115 and a mean of 1.77. The fit now has no shift and a mean of 1.87. This is based on the sample mean equaling 1.87 minutes. Thus, we will use an exponential 1 distribution with   = 1.87 or m = 1.87 = 0.53. 10.12 a. A Poisson distribution might be appropriate for these data. The problem involves the occurrence of

outcomes (nests) across a continuum (area). Certainly, the probability of a nest at any given location is quite small. The occurrence of nests may not be perfectly independent due to territorial behavior by birds, however; the probability of a nest may be reduced if another nest is nearby. b. The Poisson has the second or third best fit. The sample mean and sample standard deviation is

compared to the means and standard deviations estimated from the theoretical fitted distributions in the table below.

Sample IntUniform  NegBin Poisson Geomet

Mean 6.375 7.000 6.375 6.375 6.375

St. Dev. 3.294 3.742 3.385 2.525 6.857

207

c. The figure below shows the spike at 9 testing sites. The data shows nearly a 21% chance of finding 9

nesting sites per five acres and the Poisson reports 8.2%. Because none of the fitted distributions do a good  job of modeling this uncertainty, even though it naturally fits a Poisson, we would use the empirical distribution to model this uncertainty.

10.13. a. The regression equation is:

E(Sales Price | House Size, Lot Size, Attractiveness) = 15.04 + 0.0854 (House Size) + 20.82 (Lot Size) + 2.83 (Attractiveness). b. Creating this graph requires calculation of the residuals from the regression (which is done automatically

with Excel’s regression procedure or with StatTools). Then use the residuals in exactly the same way that we used the halfway-house data in Table 10.9 in the text to create a CDF. To use @RISK to create the CDF of the residuals, use either the RiskCumul distribution or use @RISK’s fitting procedure, which constructs the empirical distribution.. The file “Problem 10.13.xlsx” uses the RiskCumul formula. Even though we did not ask, notice in the Fit-Results window below that the normal distribution fits the residuals quite well.

208

c. The expected Sales Prices for the two properties are

1.

E(Sales Price | House Size = 2700, Lot Size = 1.6, Attractiveness = 75) = 15.04 + 0.0854 (2700) + 20.82 (1.6) + 2.83 (75) = 491 ($1000s).

2.

E(Sales Price | House Size = 2000, Lot Size = 2.0, Attractiveness = 80) = 15.04 + 0.0854 (2000) + 20.82 (2.0) + 2.83 (80) = 453.7 ($1000s).

d.

In the graph, House 2 is the leftmost or blue line.From the graph, it is easy to see that House #2 is reasonably priced, according to the model. At $480K, its list price falls just below the 0.80 fractile of the distribution. Presuming that the owners have built in some negotiating room, it looks as though Sandy may  be able to make a reasonable offer and obtain the property for a price close to the price suggested by the model. House #1 is a different story. Its list price falls well above the probability distribution given by the model. It is either way overpriced (which suggests that Sandy may have to offer a very low price), or there is something about the house that increases its value but is not reflected in the model.

209

10.14. a. Calculating the regression coefficients gives the expression

E(Sales ($1000s) | Index, Price, Invest, Ad) = 3275.89 + 56.95 (Index) - 15.18 (Price) + 1.55 (Invest) + 7.57 (Ad) The interpretations are as follows: • A one-point increase in the Spending Index leads to an expected increase of $56,950 in Sales when

holding constant the System Price, Capital Investment, and Advertising and Marketing variables. • A $1,000 increase in System Price leads to an expected decrease of $15,180 in Sales when holding constant the Spending Index, Capital Investment, and Advertising and Marketing variables. • A $1,000 increase in Capital Investment leads to an expected increase of $1,550 in Sales when holding constant the Spending Index, System Price, and Advertising and Marketing variables. • A $1,000 increase in Advertising and Marketing leads to and expected increase of $7,570 in Sales when holding constant the Spending Index, System Price, and Capital Investment variables. b.

E(Sales ($1000s) | Index = 45.2, Price = 70, Invest = 145, Ad = 90) = 3275.89 + 56.95 (45.2) - 15.18 (70) + 1.55 (145) + 7.57 (90) = $5,695 ($1000s)

The residuals from the regression can be used to estimate the probability that Sales will exceed $6 million. This can be done by creating the CDF graph for sales and reading off P(Sales > $6 Million | Index, Price, Invest, and Ad). Or we can go straight to the calculations themselves. To exceed $6 million, the error would have to be greater than $305,000. This amount falls at about the 0.90 fractile of the error distribution. Thus, we estimate a probability of about 10% that Sales will exceed $6 million. c. Using the conditions and estimate from part b, the 0.10 fractile (in $1,000s) is about $5,390. If we  blithely use the model as it is, we can estimate how far the price must be dropped for the 0.10 fractile to be 6000. The difference is 610. Every $1,000 decrease in price leads to an expected decrease of $15,180 in Sales, and the same incremental change applies to the fractiles as well. Thus, we divide 610 by 15.18 to arrive at about 40. That is, the price must be dropped by about $40,000 to obtain a 90% probability that Sales will exceed $6 million. This implies a System Price in the neighborhood of $30,000.  However, it is important to realize that the model was never meant to be used under these conditions!  The lowest System Price in the data set is $56.2, and that was with a much lower Spending Index, Capital Investment, and Advertising. So the model is not applicable when talking about a system price of $30,000.

The best advice to give Thomas is that the goal of $6 million in sales in the first half of next year will be difficult to achieve. Although he might drop the price further or spend more in advertising, achieving the target will take a substantial amount of luck! Case Study: Taco Shells 1. Cost per unbroken shell

New supplier $25.00/case

Current supplier $23.75/case

# of unbroken shells (X)

# of unbroken shells (Y)

=25.00 / X

= 23.75 / Y

210

Discrete chance nodes really would be correct, because only integer numbers of shells will be unbroken. However, the chance node would have to have 500 branches, so a continuous fan is more practical. 2. The new supplier certainly has the higher expected number of usable shells per case. The two CDFs are

about the same in terms of riskiness. The data and a description how to create these CDFs in @RISK are in the file “Taco Shells Case.xlsx.”

211

3. The statistics in the figure above allows us to easily construct either the ES-M or EP-T discrete

approximations discussed in Chapter 8. For the EP-T approximation, the median is assigned a probability of 0.63, and the 0.05 and 0.95 fractiles are assigned probability values of 0.185. Cost per usable shell Cost per case

 

465

463 New Supplier

TRUE

$25

 

0.185 $0.0540

Usable shells  

$0.0531 472

483

 

63.0% 470

 

0.63 $0.0532

 

18.5% 482.8

Taco Supplier

 

18.5%

0.185

 

$0.0518

 

$0.0556

 

$0.0539

 

$0.0528

Supplier? $0.0531 429

427 Current Supplier

FALSE

$23.75

0

18.5%

Usable shells  

$0.0540 442

441 450

0

63.0%

0

18.5% 450.2

Expected cost per usable shell for new supplier = $0.0531 Expected cost per usable shell for current supplier = $0.0540 This decision tree is modeled in the Excel file “Taco Shells Case.xlsx.” In the decision tree, the consequences are determined using a Branch pay-off formula that divides the cost per case by the number of usable shells. Click on the setting for the decision tree to view the default formula or the outcome nodes to view the node specific formula. A linked was also created. 4. Average number of unbroken shells for new supplier = 473. (Add up all 12 observations, and divide by

12. Thus, average cost per usable shell for new supplier = $25.00 / 473 = $0.0529. Similarly, for the current supplier the average number of unbroken shells for current supplier = 441, and the average cost per usable shell = $23.75 / 441 = $0.0539.  Note that these results are very close to those found in question 3. Two advantages of using the CDFs: a. The variability in the distributions is clear from the CDFs, but not when using the sample average directly.  b. Question 4 requires us to approximate $25.00 $25.00 E(  X  ) ≈ E( X ) and E(

$23.75 Y 

)≈

$23.75 E(Y ) .

1 1 However, this is not generally appropriate. That is, E( X ) ≠ E( X ) in general. The use of the CDF and the costs of the shells directly is more appropriate. 5. Ortiz should certainly consider taste, reliability of the supplier, reputation, whether he can obtain other  products from the same supplier, etc.

212

Chapter 10 Online Supplement: Solutions to Problems and Case Studies 10S.1. a. The natural conjugate prior distribution for µ is normal with mean m0 = 9.4 grams and σ0

0.8 =  1.96 = 0.4082. Note that m0 must be halfway between 9.0 and 9.8, and halfway between 8.6 and 10.2. The standard deviation σ0 is found by realizing that the distance 10.2 - 9.4 must be equal to 1.96 σ0. Thus, P N(µ ≥ 10 grams | m0 = 9.4, σ0 = 0.4082) 10 - 9.4 = P( Z ≥  0.4082 ) = P( Z ≥ 1.47) = 0.0708. b. The posterior distribution for µ is normal with

1.52 9.4 ( 15 ) + 9.76 (0.4082 2) m* = = 9.59 1.52 2  + 0.4082 15 and

σ* =

1.52 2 15  0.4082 = 0.2810 1.52 2 15  + 0.4082

 Now we have P N(µ ≥ 10 grams | m* = 9.59, σ* = 0.2810) 10 - 9.59 = P( Z ≥  0.2810 ) = P( Z ≥ 1.46) = 0.0721.  Note that the probability has changed only slightly. The mean has shifted up from 9.4 to 9.59 grams, but the standard deviation is less. The net effect is that the probability that µ is greater than 10 grams is still about the same. 10S.2. a. The predictive probability that a single animal weighs more than 11 grams is equal to

P( X ≥ 11 | m0 = 9.4, σ0 = 0.4082) = P N( X ≥ 11 grams | m0 = 9.4, σ p =

1.52 + 0.4082 2 = 1.5546)

213

11 - 9.4 = P( Z ≥ 1.5546 ) = P( Z ≥ 1.03) = 0.1515 b.

P( X ≥ 11 | m* = 9.59, σ * = 0.2810) = P N( X ≥ 11 grams | m* = 9.59, σ p =

1.52 + 0.2810 2 = 1.5261)

11 - 9.59 = P( Z ≥ 1.5261 ) = P( Z ≥ 0.92) = 0.1788 10.S.3. a. The graph for  f β(q | r 0 = 1, n0 = 20):

Density

q 0

0.1

0.2

0.3

0.4

0.5

b. The posterior distribution is  f β(q | r * = 40, n* = 1220), which is essentially a spike at q = 40/1220 =

0.0328.

214

10S.4. a. The prior distribution is  f β(c | r 0 = 3, n0 = 6): Density

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1 c

b. The predictive probability is P( R = 3 | n = 4, r 0 = 3, n0 = 6)

=

5! 3! 4! 5! = 0.2381 3! 2! 1! 2! 9!

Likewise, P( R = 4 | n = 4, r 0 = 3, n0 = 6) = 0.1190, and so P( R > 2 | n = 4, r 0 = 3, n0 = 6) = 0.2381 + 0.1190 = 0.3571. c. The posterior distribution is  f β(c | r * = 3 + 3 = 6, n* = 6 + 4 = 10): Density

Posterior 

Prior 

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1 c

d. The predictive probabilities are:

P( R = r  | n = 10, r * = 6, n* = 10) 6 0.1750 7 0.1715 8 0.1393 9 0.0867 10 0.0325 r 

Thus, P( R > 5 | n = 10, r * = 6, n* = 10) is equal to the sum of these five probabilities, or 0.6050. 215

e. Now the posterior distribution is  f β(c | r ** = 6 + 6 = 12, n** = 10 + 10 = 20). His probability that the

measure will pass must be the expected value of this distribution, or E(C) = 12/20 = 0.60. 10S.5. a. For the comptroller,

P N(µ > 11,000 | m0 = 10,000, σ0 = 800) = P( Z  >

11,000 − 10,000 800

)

= P( Z  > 1.25) = 0.1056. For her friend, P N(µ > 11,000 | m0 = 12,000, σ0 = 750) = P( Z  >

11,000 − 12,000 750

)

= P( Z  > -1.33) = 0.9082. b. The posterior distribution for the comptroller is normal with

10,000(

1500 2

m* =

9

) + 11,003(800 2 )

1500 2 9

= 10,721 +

800 2

and

σ* =

Thus,

15002 2 9  (800 ) = 424. 15002 2 9  + 800

P N(µ > 11,000 | m* = 10,721, σ* = 424) = P( Z  >

11,000 − 10,721 ) 424

= P( Z  > 0.66) = 0.2546. For her friend, the posterior distribution is normal with

216

12,000(

1500 2

m* =

9

) + 11,003(750 2 )

1500 2 9

= 11,310 +

750 2

and 15002 2 9  (750 ) = 416. 15002 2 9  + 750

σ* =

Thus,

P N(µ > 11,000 | m* = 11,310, σ* = 416) = P( Z  >

11,000 − 11,310 416

)

= P( Z  > -0.75) = 0.7734. c. For the comptroller, the posterior distribution is normal with

10,000( m** =

1500 2 144

2

) + 11,254(800 )

1500 2 144

= 11,224 +

800

2

and

σ** =

Thus,

15002  (8002) 144 = 123.5. 15002 2 144  + 800

P N(µ > 11,000 | m** = 11,224, σ** = 123.5) = P( Z  > = P( Z  > -1.81) = 0.9649.

For her friend, the posterior distribution is normal with 12,000( m** =

1500 2 144

) + 11,254(750 2 )

1500 2 144

= 11,274 +

750

2

and

217

11,000 − 11,224 123.5

)

σ** =

Thus,

15002 2 144  (750 ) = 123.3. 15002  + 7502 144

P N(µ > 11,000 | m** = 11,274, σ** = 123.3)= P( Z  >

11,000 − 11,274 123.3

)

= P( Z  > -2.22) = 0.9868. d. Eventually the data overwhelm any prior information. In the limit, as more data are collected, the

comptroller and her friend will end up with the same posterior distribution. Case Study: Forecasting Sales 1. Average error = 2416. Standard deviation = 3555. Bill Maught’s opinion is reasonable; Morley does

appear to underestimate, and by more than Maught suspected. These data and calculations are shown in the first worksheet of the Excel file “Forecasting Sales Case.xlsx.” 2.

P(Forecast too low by 1700 or more) = P(Error ≥ 1700) = 0.5. Note that we have defined Error as (Sales Forecast) so that Error is what must be added to the forecast in order to get sales. If the forecast is too low  by 1700 (or more), we would need to add 1700 (or more) to the forecast to obtain actual sales. Thus, we want the probability that Error is 1700 or more. 3. Maught says, “... I’d bet even money that his average forecast error is above 1700 units.” This means that 1700 is Maught’s median, which is also the mean m0 for a normal prior distribution.

“...About a 95% chance that on average he underforecasts by 1000 units or more.” This statement indicates that 1000 is the 0.05 fractile of the distribution.

218

We know that P( Z ≤ -1.645) = 0.05. Thus, 1000 - 1700 = -1.645. σ0 Solving for σ0: σ0 =

1000 - 1700 = 426. -1.645

Thus, Maught’s prior distribution for µ, Morley’s average error, is a normal distribution with parameters m0 = 1700 and σ0 = 426.  —  Given the data with  X  = 2416 and n = 14, Maught’s posterior distribution is normal with 30002 1700 ( 14 ) + 2416 (426 2) m* = = 1857 30002 2 14  + 426 and

σ* =

30002 2 14  (426 ) = 376. 30002 2 14  + 426

Thus, P(µ > 1700 | m* = 1857, σ* = 376) = P( Z  >

1700 - 1857 ) = P( Z  > -0.42) = 0.6628. 376

219

You can also use @RISK to model the normal distribution. To determine the desired probability that µ > 1700 by setting the left delimiter to 1700 and the right delimiter to 8300. Then graph will show the desired  probability: 66.19%. 4. The probability distribution for Sales will be based on a predictive distribution of the upcoming year’s

error. The predictive distribution for the error is normal with m* = 1857 and σ p = 30002 + 3762 = 3023. Because Sales = Forecast + Error, and we know that Forecast = 187,000, then we know that the distribution for Sales is normal with mean E(Sales) = 187,000 + 1857 = 188,857, and standard deviation 3023. Thus,

P(Sales > 190,000) = P( Z  >

190,000 − 188,857 3023

) = P( Z  > 0.38) = 0.3520.

You can use @RISK to determine the probability that the Sales > 190,000 by setting the left delimiter to 190,000 and the right delimiter to 210,000. The graph shows that the desired probability is 35.27%. Case Study: Overbooking: Bayesian Analysis 1, 2. See the Excel file “Overbooking Bayesian.xlsx” for the model and analysis.

In the original Overbooking case study at the end of Chapter 9, we used a binomial distribution to model the uncertainty about how many people arrive. In this case study, we include prior information about the no-show rate in the form of a beta distribution. Question 1 asks for the predictive distribution for the number of arrivals, given that Mockingbird sells 17 reservations. Let r  denote the number of no-shows and n the number of reservations sold. The predictive distribution for the number of no-shows is a beta binomial distribution; substituting r 0 = 1 and n0 = 15 into the beta-binomial formula and simplifying gives P(No-shows = r  | Reservations sold = n, r 0 = 1, n0 = 15)

220

(n - r  +13)! n! 14! = (n-r)! 13! ( n + 14)! . The following table shows the predictive probability distribution for no-shows given that Mockingbird sells 16, 17, 18, and 19 reservations. Note that the number of arrivals equals ( n - no-shows). For example, P(Arrivals = 17 | Reservations sold = 17) = P(No-shows = 0 | n = 17) = 0.452. The table also includes calculations of expected values for revenue, costs, and profits, allowing us to determine the number of reservations n that maximizes expected profit. The formulas for calculating these quantities are given in the original case in Chapter 9. You can see that the optimum number of reservations to sell is 17. Parameters: n r 0 n0  No-shows 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Expected Values: n: Revenue E(C1) E(C2) E(Profit)

16 1 15

17 1 15

18 1 15

19 1 15

0.467 0.257 0.138 0.072 0.036 0.017 0.008 0.003 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Predictive 0.452 0.256 0.141 0.076 0.039 0.020 0.009 0.004 0.002 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Probability 0.438 0.254 0.144 0.079 0.043 0.022 0.011 0.005 0.002 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

16 $3,600 $2,393 $0 $1,207

17 $3,825 $2,442 $147 $1,237

18 $4,050 $2,467 $367 $1,216

0.424 0.252 0.146 0.083 0.046 0.024 0.013 0.006 0.003 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 19 $4,275 $2,481 $625 $1,169

3. If all 17 passengers arrive, the manager’s posterior distribution for the no-show rate becomes a beta

distribution with parameters r * = r 0 + r  = 1 + 0 = 1 n* = n0 + n = 15 + 17 = 32.

221

4. The following table shows the predictive distributions for no-shows as in question 1, but now with r * = 1 and n* = 32. The expected value calculations indicate that the optimum action is not to overbook, but to sell

exactly 16 reservations. Parameters: n r* n*  No-shows 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Expected Values: n: Revenue E(C1) E(C2) E(Profit)

16 1 32

17 1 32

18 1 32

19 1 32

0.660 0.229 0.076 0.024 0.007 0.002 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Predictive 0.646 0.234 0.081 0.027 0.009 0.003 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Probability 0.633 0.237 0.086 0.030 0.010 0.003 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

16 $3,600 $2,450 $0 $1,150

17 $3,825 $2,482 $210 $1,133

18 $4,050 $2,493 $488 $1,068

0.620 0.240 0.090 0.033 0.011 0.004 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

19 $4,275 $2,498 $790 $987

222

CHAPTER 11 Monte Carlo Simulation Notes Simulation, sometimes referred to as “Risk Analysis” when applied in decision-analysis settings, is a very important and powerful tool. Many of my students have not been exposed to the idea of simulation modeling before reading about it in Making Hard Decisions with DecisionTools, 3rd  ed., and the opportunity to work on relatively complex decision problems proves to be enlightening and engaging, while those students that have little experience using a spreadsheet may find it very challenging. Students typically come away from Chapter 11 feeling good about the tools they have learned.

The first portion of the chapter discusses the general procedure for simulating random variables, including the general CDF-inversion method. Following this, we turn to the implementation of simulation using @RISK with detailed step-by-step instructions and several illustrations. When entering the probability distributions in a spreadsheet model, it is often helpful at first to use the Define Distribution window to better understand how to assign values to function arguments. Then, once you better understand the syntax of the distribution function arguments, you can enter the arguments yourself directly in Excel, bypassing the Define Distribution window. Students should have used the Define Distribution window in Chapters 8, 9 & 10, and hence should be comfortable with it. Topical cross-reference for problems @RISK

Beta distribution Binomial distribution Decision trees vs simulation Decision variables Discrete distribution EP-T vs ES-M Cumulative risk profiles  Normal distribution Monty Hall Objectives Opportunity cost Poisson distribution PrecisionTree Requisite model Sensitivity analysis Sequential simulations Stochastic dominance Subjective judgments Triangular distribution Uniform distribution

11.8-11.16, Manufacturing Process, La Hacienda Musa, Overbooking Part III 11.8 – 11.10 Overbooking, Part II 11.5 11.7, La Hacienda Musa 11.16 11.11, 11.12 Choosing a Manufacturing Process 11.5, 11.6, Choosing a Manufacturing Process, La Hacienda Musa 11.15 11.9 11.8 Choosing a Manufacturing Process 11.11, 11.12 11.1, 11.10 11.13, 11.14, Overbooking, Part II 11.8 – 11.10, La Hacienda Musa Choosing a Manufacturing Process 11.3, 11.4 11.13, 11.13, 11.14

Solutions 11.1. Constructing a simulation model can help a decision maker specify clearly what is uncertain in the environment and the nature of the uncertainty. This is part of the structuring phase of decision analysis, and insight can be gained in the process of iterative modeling and analysis with the goal of reaching a requisite decision model. The analysis itself (running the simulation) can produce simulation-generated probability distributions for those uncertain quantities in which the decision maker is interested (e.g., NPV, profit,  payoff, cost).

223

11.2. (Note: the difference between this question and 11.1 is that this one is meant to focus on the nuts and  bolts of the simulation process.) After constructing a model of the uncertainty, the computer generates random occurrences (e.g., revenue and cost) according to the specified distributions, and then aggregates these occurrences according to the model (e.g., profit = revenue - cost). Finally, these results are tracked for all iterations of the model to produce a simulation-generated distribution of the variable (profit) in which the decision maker is interested. 11.3. Answers will depend on a student’s attitudes toward risk. In particular, how much additional risk is the student willing to take for an increase in expected profit? Clearly, 700 calendars is the maximum almost every student will order because the expected profit drops after that. See Figure 11.10. A risk-neutral student will order 700 calendars and the more risk averse the student is, the fewer calendars they will order. Figure 11.11 illustrates the midd le 90% of profit values, showing an increase in the range as the order size increases. 11.4. Her subjective judgments regarding important sources of uncertainty and the nature of that uncertainty are required for building the model. She may also be needed to understand the relationships among the variables and how they impact the consequence measure. She may also be interested in more than one consequence, and thus her input is critical as to what should be measured. For example, Leah may  be interested in profit as well as the probability of selling out. 11.5. Simulation works by brute force, in that it samples the input distributions thousands of times. Decision trees do not sample the distributions at all. Rather, they use a discrete distribution to approximate a continuous distribution. It is the judicious choice of the discrete distribution that results in the two methods producing similar results. The ES -M and EP-T approximations were engineered so that th ey closely match the mean and the standard deviation of the distribution they are replacing. Generally, we are mainly concerned with the mean and the standard deviation of the consequence measure. If our approximations match the mean and standard deviation of the inputs, then the decision tree model should closely match the mean and standard deviatio n of the consequence measure. 11.6. The problem with saying that the simulation models choose values randomly is that it sounds as if simulation models are capricious or erratic in their operation. A better statement would be along the lines of saying that the values chosen by a simulation model for an uncertainty are governed by the properties of the uncertainty and controlled by the probability distribution being used. This is why we stress that the  probability distribution should have the same properties of the uncertainty it is modeling. Much of the text has been devoted to eliciting the most accurate probability distribution possible. In developing decision trees, a considerable amount of time is spent to get the distribution just right. 11.7. Many students believe that using a discrete distribution to model a decision variable is appropriate and have trouble answering this question.

There are two problems with modeling a decision variable as an uncertainty. First, a decision variable represents a value that is under the control of the decision maker and an uncertainty represents a value that is not. Second, analyzing a model with a decision variable modeled as an uncertainty is difficult at best. Suppose Leah model her quantity-ordered decision using a di screte distribution, say 30% chance of ordering 680 calendars and 70% of ordering 700 calendars. After running the model, how would Leah analyze the results to determine the best order quantity? Because she must order one and only one amount, she would have to sort all the simulation results by order quantity. She would need to gather all the 680 calendar orders into one spreadsheet and determine the expected profit, the risk, etc. She would also need to repeat all this for the 700 calendar order. Much simpler would be to use RiskSimTable(680, 700), in which case the program does all the sorting and analysis for us.

224

11.8. See file “Problem 11.8.xlsx” for the simulation model. The new consequence measure is labeled “Profit*” and includes the opportunity cost of missing sales due to running out of calendars.

Because Profit* includes an extra cost, expected Profit* is less than expected Profit. Also, the order quantity that maximizes expected Profit* is larger than the order quantity that maximized expected Profit. The results of running 9 simulations at 10,000 iterations are given below. The largest two EMVs are highlighted showing that ordering 710 calendars maximizes expected Profit*.

Order Quantities

Expected Profit*

650 660 670 680 690 700 710 720 730

Expected Profit

$5,653.21 $5,712.56 $5,758.32 $5,791.47 $5,813.22 $5,824.86 $5,827.65 $5,822.83 $5,811.53

$5,763.23 $5,803.96 $5,833.63 $5,853.07 $5,863.25 $5,865.20 $5,859.97 $5,848.55 $5,831.87

The graph below shows the expected Profit and expected Profit* for the different order quantities. We see that for order quantities less than 710, expected Profit* drops more quickly than does expected Profit. This tells us that Profit* is more affected by running out of stock than is Profit. For order quantities above 710, expected Profit* is less affected by running out of stock than is Profit. Thus, the statement that Leah should err on the high side when she orders holds even more for Profit*. We labeled the new consequence Profit* because it is similar to profit, but not equal to profit. It is a proxy measure for long-run profit that is designed to capture the disappointment of shoppers when they discover there are no more calendars. Being a proxy measure, it does not exactly capture long-run profit and should  be treated with care. In other words, Leah needs to be careful when interpreting the simulation results on Profit*. $5,900.00 $5,850.00 $5,800.00 Expected Profit*

$5,750.00

Expected Profit

$5,700.00 $5,650.00 $5,600.00 640

660

680

700

720

225

740

11.9 This problem shows that Leah may have other objectives, and by changing the consequence measure the simulation model can provide insights into her decision problem. The file “Problem 11.9.xlsx” has the simulation model with the new consequence measures. The table below reports the results.

If Leah wants to minimize leftovers, then she should order no calendars. That would guarantee no unsold calendars, but clearly ordering 0 calendars is not what she wants. In this case, the objective of minimizing leftovers is not very well thought out. A little more helpful is for Leah to order the minimum number of calendars, which according her assessments is 600 calendars. However, ordering the minimum that one expects to sell is not likely to grow the business. The simulation model shows, via the table below, that the expected number of unsold calendars increases as does her order quantity. If she were to order 680 calendars (expected demand), then she can expect to have 21 unsold calendars. If she were to order 700 calendars (maximizes expected profit), then she can expect to have 33 leftovers. We used the RiskMean function to calculate the values in the Expected Leftovers column. If her objective is to maximize the probability P(Profit > $5,200), then, according to the model, she should order no more than 660 calendars. This may sound counterintuitive, but as her order size increases, so does the standard deviation of profit. The larger the standard deviation is, the more weight there is in the tails, and thus, there is a higher probability of seeing extreme profits. Specifically, as the order size increases, P(Profit < $5,200) also increases. The @RISK output window below shows the profit distributions when ordering 650 and 710 calendars. The standard deviation of profit when ordering 710 calendars is nearly 300% higher than when ordering 650 calendars, pushing the tails of the distribution further out. We used the function RiskTarget to calculate the cumulative probability of Profit being less than $5,200. For the third objective, Leah wants to meet the hurdle of the 5th percentile being at least $5,200. The table  below shows that this hurdle is met for order quantities of 690 or less. The reason the 5th percentile increases as the order size increases is again because the standard deviation is also increasing. For this objective, Leah should consider ordering 690 calendars or less. We used the function RiskPercentile to calculate the 5th percentile values.

Order Quantities

Expected Leftovers

650 660 670 680 690 700 710 720 730

7 10 15 21 27 33 41 49 57

P(Profit > $5,200) 100% 100% 99% 98% 97% 95% 93% 91% 89%

226

5th Percentile $ $ $ $ $ $ $ $ $

5,397.88 5,357.88 5,317.88 5,277.88 5,237.88 5,197.88 5,157.88 5,117.88 5,077.88

11.10 Students often question how much detail to include and how do you determine if you have a requisite model. This question shows that going through the work to guarantee that only whole calendars are sold is not worth the effort. The maximum difference in expected profit between the model that allows fractional calendar sales and one that allows only whole number of calendars to be sold is one penny. See “Problem 11.10.xlsx.” Clearly, fractional parts are not realistic, but including them does not materially affect the results of the analysis. th

th

11.11 We now need the 10  and 90  percentiles of the demand distribution. As we did in Chapter 9, we can use the Define Distribution feature to pull up the demand distribution and read off the desired percentiles. The 10th percentile is 623 calendars, the 90th percentile is 752 calendars, and median is still 670 calendars. The file “Problem 11.11.xlsx” contains the simulation model and the decision tree models with all 6 alternatives. Both the ES-M and EP-T trees are in the spreadsheet and they are linked to the profit simulation model. The results of both the EP-T and ES-M trees are displayed in the table below.

The EP-T did a better job in this problem being a maximum of $16.57 from the simulation results. Three of the six ES-M estimates were over $28 away from the simulation results.

Order Quantity 600 650 700 750 800 850

Sim Expected Profit $5,400.00 $5,763.23 $5,865.20 $5,785.92 $5,625.71 $5,436.79

ES-M Expected Profit $5,400.00 $5,791.50 $5,873.17 $5,781.50 $5,585.83 $5,385.83

227

Difference $0.00 -$28.27 -$7.97 $4.42 $39.87 $50.95

Order Quantity 600 650 700 750 800 850

Sim Expected Profit $5,400.00 $5,763.23 $5,865.20 $5,785.92 $5,625.71 $5,436.79

EP-T Expected Profit $5,400.00 $5,765.83 $5,849.88 $5,770.13 $5,642.28 $5,442.28

Difference $0.00 -$2.59 $15.32 $15.79 -$16.57 -$5.49

11.12. Again, we see that the EP-T approximation estimated the simulation model more closely than the ES-M approximation. See table below and file “Problem 11.12.xlsx.” The file contains the simulation model and the decision tree models with all 6 alternatives. Both the ES-M and EP-T trees are in the spreadsheet and they are linked to the profit simulation model.

Order Quantity 600 650 700 750 800 850

Sim St. Dev. Profit $0.00 $154.06 $393.66 $557.32 $637.60 $668.52

ES-M St. Dev. Profit $0.00 $130.81 $293.62 $487.21 $495.84 $495.84

Difference $0.00 $23.25 $100.04 $70.11 $141.76 $172.68

Order Quantity 600 650 700 750 800 850

Sim St. Dev. Profit $0.00 $154.06 $393.66 $557.32 $637.60 $668.52

EP-T St. Dev. Profit $0.00 $176.68 $345.11 $539.50 $674.82 $674.82

Difference $0.00 -$22.61 $48.55 $17.83 -$37.22 -$6.30

In the second part of the problem, we are asking to compare the EP-T approximation using 780 as the 95th  percentile to the EP-T approximation using 781 as the 95th percentile. When comparing the results of the simulation model to that of the decision tree, we should be using 781 as it is the 95th of the input distribution. In the text, we used 780 as it was Leah’s assessed value. For comparison purposes, 781 is a slightly more accurate value. The table below shows that there is at most a $2.40 difference between the two different approximations. However, the second table shows that the using the slightly more accurate value actually produced higher, albeit slightly, errors.

Order Quantity 600 650 700 750 800 850

Using 781: EP-T Expected Profit $5,400.00 $5,765.83 $5,849.88 $5,770.13 $5,644.68 $5,444.68

Using 780: EP-T Expected Profit $5,400.00 $5,765.83 $5,849.88 $5,770.13 $5,642.28 $5,442.28 228

Difference $0.00 $0.00 $0.00 $0.00 $2.40 $2.40

Difference: Sim – EP-T(781) $0.00 -$2.59 $15.32 $15.80 -$18.97 -$7.89

Difference: Sim – EP-T(780) $0.00 -$2.59 $15.32 $15.79 -$16.57 -$5.49

11.13. a. Each probability can be a random variable itself. For example, for Project 1, first generate p1, its  probability of success, from a uniform distribution that ranges from 0.45 to 0.55. Then, once p1 is determined, determine whether Project 1 succeeds using  p1 as the probability of success. In influence diagram form:  No uncertainty or vagueness about  probabilities:

Project 1

Project 2

Project 3

Project 4

Project 5

Payoff 

With uncertainty or vagueness about probabilities: p

1

Project 1

p

2

Project 2

p

3

Project 3

p

4

Project 4

p

5

Project 5

Payoff 

Although there are many ways to solve this problem, any solution should comply with the above influence diagram. In particular, students could choose almost any distribution for each  ,   = 1, …, 5, as long as the values are constrained between   − 0.05 to   + 0.05,   = 1, … , 5. Possible distribution choices are the  beta, the triangular, and the uniform distributions. The file “Problem 11.13a.xlsx” shows that we chose the

229

uniform. We also supplied some monetary values for each project to illustrate the monetary impact on the  portfolio of projects when the probability of success of each project itself is uncertain. b. In part b, the probabilities are dependent. We can link them through an “optimism” node to represent the decision maker’s unknown level of optimism/pessimism r egarding the chance of success of the projects: Optimism level

p

1

Project 1

p

p

2

p

4

3

Project 3

Project 2

Project 4

p

5

Project 5

Payoff 

How could such a model be implemented? The file “Problem 11.13b.xslx” has two possible solutions. Both solutions are based on choosing an “optimism parameter” (OPT) according to a distribution. In the solution file, OPT is based on a uniform distribution between 0 and 1. Of course, this distribution should be assessed  by the decision maker. For Model 1 in the spreadsheet, the OPT value determines a value on the line segment that goes from (0,   − 0.05) to (1,   + 0.05). Specifically, given a value for OPT, the probability of success is determined by the equation for a line: 2 × (0.05) ×  +   − 0.05 2 × (0.05) ×  +   − 0.05 (1,   + 0.05) (0,   − 0.05) OPT = 0

OPT

OPT = 1

For Model 2 in the spreadsheet, the OPT value determines the modal (most likely) value for a triangular distribution with minimum set to   − 0.05 and maximum set to  + 0.05.

230

Density f unction for p

1

0.45

0.55

Mode = 0.45 + OPT * (0.55 - 0.45)

For either model, OPT values close to zero imply the boss is pessimistic about the probability of success for all the projects, that is, probability of success is close to   − 0.05. For OPT values close to one half, the  boss is neither pessimistic nor optimistic, that is, probability of success is close to  . For OPT values close to one, the boss is optimistic about all the probability of success for all the projects, that is, probability of success is close to   + 0.05. The difference between the models is that in Model 1 we use a uniform distribution between zero and one to choose the OPT value, and then this directly determines the probability of success via 2 × (0.05) ×  +   − 0.05. In Model 2, we again use the same uniform to choose the OPT value, but then the  probability value is drawn from a triangular distribution whose mode is 2 × (0.05) ×  +   − 0.05.

11.14. a. A straightforward sensitivity analysis would calculate expected values with the probabilities set at various values. The results would be a triangular grid that could be plotted in terms of p1 = P($10,000) and  p2 = P($5000): p

1

0.5

0.5

p

2

To use simulation to incorporate the uncertainty about the probabilities, it is possible to sample the  probabilities from uniform distributions. Care must be taken when sampling the values as each probability, 1 , 2 , and 3  can only range from 0 to 0.5. To do this, we sample 1from a uniform distribution between 0 and 0.5, and we sample 2 from a uniform distribution between (0.5 − 1 ) and 0.5. To ensure the  probabilities sum to one 3  is not sampled, but computed as 1 − (1 + 2 ). If we had sampled 2  from a uniform distribution between 0 and 0.5, then 3  could be larger than 0.5. The Excel solution can be found in “Problem 11.14.xlsx.” The expected payoff when each investment is equally likely is $5,333. The expected payoff when incorporating the uncertainty is $5,876, based on 10,000 iterations. b. No, it would not be possible to have each of the three probabilities chosen from a uniform distribution  between zero and one, because the three probabilities would never sum to one. Some kind of building-up  process, like that described in part a, must be followed.

231

11.15 This is a logic problem and the key to solving it correctly is that Monty always reveals the unattractive prize, which we said was a donkey. Thus, if you pick Curtain #1, and the prize is behind Curtain #2, then Monte must open Curtain #3. Monty never opens the curtain you choose; if he did, you would see that you had chosen a donkey and would obviously switch. If you choose Curtain #1 and the  prize is behind it, then Monty can open either of the other curtains, as they both contain donkeys.

The solution file “Problem 11.15.xlsx” contains one possible solution. See the file and formulas for the logic. The answer is that you double your chances of winning by switching. Many people believe that, regardless of which curtain Monty opens, it is still equally likely that the prize is  behind the remaining two. This would be true, but only if Monte randomly opened one of the two oth er doors. However, he always reveals the donkey, and this strategic move essentially provides information you can use. It is possible to use Bayes theorem to work out the posterior probability of winning given switch vs don’t switch, but we will leave that to the reader. Many different solutions can be found online, including decision tree solutions that clarify the logic behind which curtain Monty opens. 11.16. Although this problem may appear difficult, you need only replace one discrete distribution with another in the example model from the text. Specifically, replace the discrete distribution that chooses a day from 1 to 365 with equal probability with a discrete distribution that chooses a day according to observed births. In the file “Problem 11.16.xlsx,” we placed the data (observed births) in the AK and AL columns. The discrete distribution we used to choose a birthday is then:

=RiskDiscrete($AK$5:$AK$370,$AL$5:$AL$370) This discrete distribution is repeated 30 times in cells B4 to B33. Case Study: Choosing a Manufacturing Process 1. D

0

V

0

P

0

Z

0

D

1

V

1

P

1

Z

1

D

2

V

2

Z

2

P

2

NPV

This looks workable as an influence diagram. However, to do it all in one fell swoop like this, the problem must be simplified substantially. Even if Z 0, Z 1, and Z 2 each had only three outcomes, and V 0, V 1, and V 2 each were modeled with three-point d iscrete approximations, then there would be 27 outcomes each for P0, P1, and P2. These would combine to produce 273 = 19,683 outcomes under NPV. If all that is required is the expected value E(NPV) and standard deviation of NPV, one could find E(P0), E(P1), and E(P2), along with the standard deviations, and combine these to get E(NPV) and the standard deviation. If the entire distribution is needed, though, some other method would be required. 232

2. Assume the profit for each year arrives at the end of next year, and thus must be discounted. Let P1i denote the profit for Process 1 in year i. Then P1i = $8 Di - Di V i - $8000 Z i - $12,000,

and  NPV1 =

P10 P11 P12 + + . 1.10 1.102 1.103

For Process 2, P2i = $8 Di - Di V i - $6000 Z i - $12,000,

and  NPV2 = -$60,000 +

P20 P21 P22 + + . 1.10 1.102 1.103

3, 4. This problem is modeled in the Excel file “Manufacturing Case.xlsx.” For one run of 10,000 iterations, we obtained the following results:

Process 1: E(NPV1) = $92,084.

Standard deviation = $48,200

233

P(NPV < 0) = 0.031

Process 2: E(NPV2) = $114,337. Standard deviation = $72,808

P(NPV < 0) ≈ 0.049

5. Process 2’s expected NPV is $22,261 larger than process 1’s expected NPV. The st andard deviation of Process 2 is also larger, by nearly $25,000 . While process 2 does not quite stochastically dominate Process 1, as shown in the crossing CDFs below, Process 2 does provide the opportunity to make a lot more money than Process 1 with approximately the same downside risks as Process 1.

234

Case Study: La Hacienda Musa For the model and full analysis, see “La Hacienda Musa.xlsx.” The results below were obtained from 10,000 iterations. 1a. The graph below shows Maria's profit (C26) risk profile if 100 hectares are planted organic. Expected  profit = $83,080, and P(Profit < 0) = 15.4%

1b. The graph below calculates the risk profile for the difference between the profit for conventional profit versus organic (B19 in the Excel file). When negative, organic has the higher profit. You can see that P(Organic more profitable than conventional) = 77.3%.

235

1c. This graph shows the profit risk profile when 50 hectares are organic and 50 conventional. Expected  profit = $60,976. 10th percentile = -$10,581; 50th percentile (median) = $59,613 90th percentile = $135,257.

1d. This overlay shows that, as more organic is planted, the distribution moves higher, having a higher mean (see table below). Standard deviation also increases. In particular, though, the lower tails stay about the same, but the longer upper tails indicate greater chance of high profit.

Sim #5, 0 hectares

Sim #1, 100 hectares

Hectares planted organic

Sim #1

Sim #2

Sim #3

Sim #4

Sim #5

100 hectares

75 hectares

50 he ctares

25 hectares

0 hectares

236

2a. NOTE: See Chapter 14 for information about risk tolerance and the exponential utility function. Looking only at risk profiles and statistics for total profit, it looks as though Maria would prefer to go with 100 hectares organic. There is, however, a bit of a risk-return trade-off. To incorporate this trade-off, we use her risk tolerance and calculate her utility (C40 in the Excel file). Her expected utility (EU) is the mean of this output cell after the simulation completes. The table below reports the EUs for each of the five levels of organic, showing that 100 hectares organic does indeed have the highest EU = 0.359.

Sim #5, 0 hectares

Sim #1, 100 hectares

Hectares planted organic

Sim #1

Sim #2

Sim #3

Sim #4

Sim #5

100 hectares

75 hectares

50 he ctares

25 hectares

0 hectares

237

Hectares organic

EU

100

0.359

75

0.333

50

0.298

25

0.255

0

0.202

(Results from @RISK Excel report)

2bi. The figure below shows that the expected incremental cost to the company is $4,763. Note, though, that there is about a 76% chance that the incremental cost is zero. This occurs when the organic price falls above the GMP, when the company does not pay anything more than the market price.

2bii. This is a relatively easy question to answer. Keller did not need a GMP at all in order to decide to  plant all 100 hectares in organic bananas. Any GMP will be an improvement for her.

238

2biii. Assuming there are other farmers out there that could increase their organic area in response to the company's incentive, there are plenty of approaches the company could take. Perhaps the most straightforward would be to sign a fixed-price contract whereby the farmer agrees to sell the bananas for a specified price. (Essentially, this amounts to a futures contract.) This would remove all price uncertainty from both sides. The company's challenge would be to find a fixed price that would be equivalent, in terms of the farmer's expected utility, so that the farmer was just indifferent between the fixed price arrangement and the uncertain profit associated with waiting to sell the bananas at the current market price. Case Study: Overbooking, Part II 1. The results for 1000 iterations of the simulation model are shown below. The simulation model is saved in the Excel file: “Overbooking II part 1.xls”.

Reservations sold: Expected profit: Std Deviation: # Iterations:

16 $1,164 $78 1000

17 $1,180 $186 1000

18 $1,126 $247 1000

19 $1,045 $272 1000

2. Uncertainty about the no-show rate and the cost can be included in the spreadsheet by introducing random variables for these quantities. For 1000 iterations, the results are shown below. The simulation model is saved in the Excel file: “Overbooking II part 2.xls”.

Reservations sold: Expected profit: Std Deviation: # Iterations:

16 $1,165 $80 1000

17 $1,192 $177 1000

18 $1,155 $237 1000

19 $1,101 $274 1000

In both questions 1 and 2, selling 17 reservations — overbooking by one seat — comes out being slightly  better than selling 16. However, it is important to realize that most of th e time all 17 ticket holders will show up. It may be important to consider the cost of lost goodwill. 3. The uncertainty about the no-show rate and costs could be addressed in a straight sensitivity analysis of the problem as solved in Chapter 9.

239

CHAPTER 12 Value of Information Notes

Chapter 12 is where influence diagrams meet decision trees head to head. And influence diagrams definitely come out on top! Teaching value of information using influence diagrams is a snap. On the other hand, using decision trees to solve value-of-information problems is complicated at best. At worst, they require painstaking flipping of probabilities via Bayes’ theorem (a step handled automatically by the influence-diagram programs). And the trees become large and bushy. The chapter discusses both the Expected Value of Information and the Expected Value of Imperfect Information (referred to as the EVSI (expected value of sample  information) in a previous edition and in Raiffa and Schlaifer’s (1961)  Applied Statistical Decision Theory). We provide instructions for calculating  both the EVPI and EVII with PrecisionTree by adding additional arcs to influence diagrams and by rearranging nodes of the decision tree. The student version of PrecisionTree may limit the model size to 50 nodes, therefore some of the value of information trees are too large to develop completely. We have either summarized parts of the tree or referenced separate sub-trees to overcome this limitation. Topical cross-reference for problems

Bayes’ theorem Deterministic dominance Oil wildcatting PrecisionTree Texaco-Pennzoil Wizardry

12.7, 12.14, DuMond International 12.3, 12.10 12.7 12.2, 12.4, 12.5, 12.7, 12.9 - 12.14 12.12, 12.13, Texaco-Pennzoil Revisited Texaco-Pennzoil Revisited

Solutions 12.1. The issue typically is whether to obtain information about some uncertain factor, and this decision

must be made up front: Should we hire the expert? Should we conduct a survey? The decision must be made in anticipation of what the information will be and its possible impact and value with regard to the decision itself. As a result, the focus is on what the value of the information is expected to be. Because there is uncertainty as to what the value of the information is going to be, we need a summary measure of its worth. Sometimes the information will have no value, sometimes it will have some value, and at time it will have great value; the expected value tells us on average its worth. The expected value averages the value across all the possible information outcomes.

240

12.2. 0.1 20 0.2 10

EMV(A) = 7.00  A

0.6 5 0.1 0

B

EMV(B) = 6.00 6

Perfect Inf ormation

 A  A = 20

0.1

EMV(Inf o) = 8.20  A = 10

0.2

B  A B  A

 A = 5

0.6

B  A

 A = 0

0.1

B

20 6 10 6 5 6 0 6

EVPI = EMV(Info) - EMV(A) = 8.20 - 7.00 = 1.20. This decision tree model is saved in the Excel file “Problem 12.2.xlsx.” 12.3. EVPI = 0 because one would still choose A regardless of knowing which outcome for the chance

node were to occur. To state it slightly differently, no matter what you found out about the outcome for the chance node following A, your decision would still be the same: Choose A. Because the information never changes the optimal alternative, the expected value of the information equals zero.

241

12.4. a. The following decision trees are saved in the Excel file “Problem 12.4.xlsx.” Each part is shown in

a separate worksheet. 0.1 0.2 EMV(A) = 3.00  A

0.6

E

0.1

20 10 0 -10

EMV(B) = 3.20 B F

0.7

5

0.3

-1

Perf ect Inf ormation about E

 A  A = 20 (0.1)

EMV(Inf o about E) = 6.24

B

20 0.7 5 0.3

 A  A = 10 (0.2)

10 0.7 5

B

E

0.3  A  A = 0 (0.6)

B

 A B

242

-1 -10

0.7 5 0.3

EVPI(E) = EMV(Info) - EMV(B) = 6.24 - 3.20 = 3.04

-1

0 0.7 5 0.3

 A = -10 (0.1)

-1

-1

b. 0.1 0.2 EMV(A) = 3.0  A

0.6

E

0.1

20 10 0 -10

EMV(B) = 3.2 B

0.7

5

0.3

-1

F Perf ect Information about F

0.1 0.2

EMV(Inf o about F ) = 4.4

0.6

E

0.1

B=5 (0.7) B

F

0.2 0.6

E B = -1 (0.3)

0.1 B

243

10 0 -10

5 0.1

EVPI(F) = EMV(Info) - EMV(B) = 4.4 - 3.2 = 1.2

20

20 10 0 -10 -1

c. 0.1 0.2 EMV(A) = 3.0  A

0.6

E

0.1 EMV(B) = 3.2 B F

0 -10 5

0.3

-1

 A = 20 (0.1)

 A = 10 (0.2)

E

10

0.7

Perfect Information about E and F EMV(Inf o about E and F) = 6.42

20

 A = 0 (0.6)

 A = -10 (0.1)

B=5 (0.7) F B = -1 (0.3) B=5 (0.7) F B = -1 (0.3) B=5 (0.7) F B = -1 (0.3) B=5 (0.7) F B = -1 (0.3)

EVPI(E and F) = EMV(Info) - EMV(B) = 6.42 - 3.2 = 3.22. 12.5. The basic influence diagram is:

Decision A or B

Chance F Chance E

EV = 3.2 Payoff 

244

 A

20

B

5

 A

20

B

-1

 A

10

B  A B  A

5 10 -1 0

B  A B  A B  A B

5 0 -1 -10 5 -10 -1