chapter 03 answers(1)

17

TEACHING SUGGESTIONS

Teaching Suggestion 3.1: Using the Steps of the Decision-Making Process.The six steps used in decision theory are discussed in this chapter.Students can be asked to describe a decision they made in the lastsemester, such as buying a car or selecting an apartment, and de-scribe the steps that they took. This will help in getting stu-dents involved in decision theory. It will also help them realizehow this material can be useful to them in making important per-sonal decisions.

Teaching Suggestion 3.2: Importance of Defining the Problemand Listing All Possible Alternatives.Clearly defining the problem and listing the possible alternatives canbe difficult. Students can be asked to do this for a typical decision-making problem, such as constructing a new manufacturing plant.Role-playing can be used to make this exercise more interesting.

Many students get too involved in the mathematical ap-proaches and do not pay enough attention to the importance ofcarefully defining the problem and considering all possible alter-natives. These initial steps are important. Students need to realizethat if they do not carefully define the problem and list all alterna-tives, most likely their analyses will be wrong.

Teaching Suggestion 3.3: Categorizing Decision-Making Types.Decision-making types are discussed in this chapter; decisionmaking under certainty, risk, and uncertainty are included. Stu-dents can be asked to describe an important decision they had tomake in the past year and categorize the decision type. A good ex-ample can be a financial investment of $1,000. In-class discussioncan help students realize the importance of decision theory and itspotential use.

Teaching Suggestion 3.4: Starting the EVPI Concept.The material on the expected value of perfect information (EVPI)can be started with a discussion of how to place a value on infor-mation and whether or not new information should be acquired.The use of EVPI to place an upper limit on what you should payfor information is a good way to start the section on this topic.

Teaching Suggestion 3.5: Starting the Decision-Making Under Uncertainty Material.The section on decision-making under uncertainty can be startedwith a discussion of optimistic versus pessimistic decision makers.Students can be shown how maximax is an optimistic approach,while maximin is a pessimistic decision technique. While few peo-ple use these techniques to solve real problems, the concepts andgeneral approaches are useful.

Teaching Suggestion 3.6: Decision Theory and Life-Time Decisions.This chapter investigates large and complex decisions. Duringone’s life, there are a few very important decisions that have amajor impact. Some call these “life-time decisions.” Students canbe asked to carefully consider these life-time decisions and howdecision theory can be used to assist them. Life-time decisions in-clude decisions about what school to attend, marriage, and the first job.

Teaching Suggestion 3.7: Popularity of Decision Trees Among Business Executives.Stress that decision trees are not just an academic subject; they area technique widely used by top-level managers. Everyone appreci-ates a graphical display of a tough problem. It clarifies issues andmakes a great discussion base. Harvard business students regularlyuse decision trees in case analysis.

Teaching Suggestion 3.8: Importance of Accurate Tree Diagrams.Developing accurate decision trees is an important part of thischapter. Students can be asked to diagram several decision situa-tions. The decisions can come from the end-of-chapter problems,the instructor, or from student experiences.

Teaching Suggestion 3.9: Diagramming a Large DecisionProblem Using Branches.Some students are intimidated by large and complex decisiontrees. To avoid this situation, students can be shown that a largedecision tree is like having a number of smaller trees or decisionsthat can be solved separately, starting at the end branches of thetree. This can help students use decision-making techniques onlarger and more complex problems.

Teaching Suggestion 3.10: Using Tables to Perform Bayesian Analysis.Bayesian analysis can be difficult; the formulas can be hard to remember and use. For many, using tables is the most effectiveway to learn how to revise probability values. Once students un-derstand how the tables are used, they can be shown that the for-mulas are making exactly the same calculations.

ALTERNATIVE EXAMPLES

Alternative Example 3.1: Goleb TransportGeorge Goleb is considering the purchase of two types of industrialrobots. The Rob1 (alternative 1) is a large robot capable of perform-ing a variety of tasks, including welding and painting. The Rob2 (al-ternative 2) is a smaller and slower robot, but it has all the capabilities

3C H A P T E R

Decision Analysis

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18 CHAPTER 3 DECIS ION ANALYS IS

of Rob1. The robots will be used to perform a variety of repair opera-tions on large industrial equipment. Of course, George can always donothing and not buy any robots (alternative 3). The market for the re-pair could be either favorable (event 1) or unfavorable (event 2).George has constructed a payoff matrix showing the expected returnsof each alternative and the probability of a favorable or unfavorablemarket. The data are presented below:

This problem can be solved using expected monetary value. Theequations are presented below:

EMV (alternative 1) � ($50,000)(0.6) � (�$40,000)(0.4)� $14,000

EMV (alternative 2) � ($30,000)(0.6) � (�$20,000)(0.4)� $10,000

EMV (alternative 3) � 0

The best solution is to purchase Rob1, the large robot.

Alternative Example 3.2: George Goleb is not confident aboutthe probability of a favorable or unfavorable market. (See Alterna-tive Example 3.1.) He would like to determine the equally likely(Laplace), maximax, maximin, coefficient of realism (Hurwicz), andminimax regret decisions. The Hurwicz coefficient should be 0.7.The problem data are summarized below:

The Laplace (equally likely) solution is computed averaging thepayoffs for each alternative and choosing the best. The results areshown below. Alternatives 1 and 2 both give the highest averagereturn of $5,000.

Average (alternative 1) � [$50,000 � (�$40,000)]/2 � $5,000

Average (alternative 2) � [$30,000 � (�$20,000)]/2 � $5,000

Average (alternative 3) � 0

The maximin decision (pessimistic) maximizes the minimum pay-off outcome for every alternative: these are �40,000; �20,000;and 0. Therefore, the decision is to do nothing.

The maximax decision (optimistic) maximizes the maximumpayoff for any alternative: these maximums are 50,000; 30,000;and 0. Therefore, the decision is to purchase the large robot (alternative 1).

Using the values above and the fact that P(FM) � 0.6 andP(UM) � 0.4, we can compute the conditional probability valuesof a favorable or unfavorable market given a positive or negative

EVENT 1 EVENT 2

Probability 0.6 0.4

Alternative 1 50,000 �40,000Alternative 2 30,000 �20,000Alternative 3 0 0

EVENT 1 EVENT 2

Probability 0.6 0.4

Alternative 1 50,000 �40,000Alternative 2 30,000 �20,000Alternative 3 0 0

Results of Favorable Market Unfavorable Market Survey (FM) (UM)

Positive (P) P(P | FM) � 0.9 P(P | UM) � 0.2Negative (N) P(N | FM) � 0.1 P(N | UM) � 0.8

The Hurwicz approach uses a coefficient of realism value of0.7, and a weighted average of the best and the worst payoffs foreach alternative is computed. The results are as follows:

Weighted average (alternative 1) � ($50,000)(0.7) � (�$40,000)(0.3)

� $23,000

Weighted average (alternative 2) � ($30,000)(0.7) � (�$20,000)(0.3)

� $15,000

Weighted average (alternative 3) � 0

The decision would be alternative 1.The minimax regret decision minimizes the maximum oppor-

tunity loss. The opportunity loss table for Goleb is as follows:

Favorable Unfavorable MaximumAlternatives Market Market in Row

Rob1 0 40,000 40,000Rob2 20,000 20,000 20,000Nothing 50,000 0 50,000

The alternative that minimizes the maximum opportunity loss isthe Rob2. This is due to the $20,000 in the last column in the tableabove. Rob1 has a maximum opportunity loss of $40,000, anddoing nothing has a maximum opportunity loss of $50,000.

Alternative Example 3.3: George Goleb is considering the pos-sibility of conducting a survey on the market potential for indus-trial equipment repair using robots. The cost of the survey is$5,000. George has developed a decision tree that shows the over-all decision, as in the figure on the next page.

This problem can be solved using EMV calculations. Westart with the end of the tree and work toward the beginning com-puting EMV values. The results of the calculations are shown inthe tree. The conditional payoff of the solution is $18,802.

Alternative Example 3.4: George (in Alternative Example 3.3)would like to determine the expected value of sample information(EVSI). EVSI is equal to the expected value of the best decisionwith sample information, assuming no cost to gather it, minus theexpected value of the best decision without sample information.Because the cost of the survey is $5,000, the expected value of thebest decision with sample information, assuming no cost to gatherit, is $23,802. The expected value of the best decision withoutsample information is found on the lower branch of the decisiontree to be $14,000. Thus, EVSI is $9,802.

Alternative Example 3.5: This example reveals how the condi-tional probability values for the George Goleb examples (above)have been determined. The probability values about the survey aresummarized in the following table:

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CHAPTER 3 DECIS ION ANALYS IS 19

survey result. The calculations are presented in the followingtwo tables.Probability revision given a positive survey result

State of Conditional Prior Joint Posterior Nature Probability Prob. Prob. Probability

FM 0.9 0.6 0.54 0.54/0.62 � 0.871UM 0.2 0.4 0.08 0.08/0.62 � 0.129Total 0.62 1.00

State of Conditional Prior Joint Posterior Nature Probability Prob. Prob. Probability

FM 0.1 0.6 0.06 0.06/0.38 � 0.158UM 0.8 0.4 0.32 0.32/0.38 � 0.842Total 0.38 1.00

Alternative Example 3.6: In the section on utility theory, MarkSimkin used utility theory to determine his best decision. What decision would Mark make if he had the following utility values?Is Mark still a risk seeker?

U(�$10,000) � 0.8

U($0) � 0.9

U($10,000) � 1Using the data above, we can determine the expected utility ofeach alternative as follows:

U(Mark plays the game) � 0.45(1) � 0.55(0.8) � 0.89

U(Mark doesn’t play the game) � 0.9

Thus, the best decision for Mark is not to play the game with anexpected utility of 0.9. Given these data, Mark is a risk avoider.

FirstDecision

Point

SecondDecision

Point

Result

s

Favor

able

Results

NegativeCondu

ct

Mar

ket S

urve

y

Do Not Conduct Survey

Rob1

Rob2

Favorable Market (0.871)

Unfavorable Market (0.129)



$45,000

–$45,000

–$5,000

$25,000

–$25,000

Rob1

Rob2





$45,000

–$45,000

–$5,000

$25,000

–$25,000

Rob1

Rob2





$50,000

–$40,000

$30,000

–$20,000

$0

$–5,000

$18,802

$14,000

$33,390

1

2

3

4

5

6

7

Survey (0.38)

Surve

y (0.6

2)

Figure for Alternative Example 3.3

Probability given a negative survey result



3-10. The purpose of Bayesian analysis is to determine poste-rior probabilities based on prior probabilities and new information.Bayesian analysis can be used in the decision-making processwhenever additional information is gathered. This information canthen be combined with prior probabilities in arriving at posteriorprobabilities. Once these posterior probabilities are computed,they can be used in the decision-making process as any other prob-ability value.

3-11. The expected value of sample information (EVSI) is theincrease in expected value that results from having sample infor-mation. It is computed as follows:

EVSI � (expected value with sample information) � (cost of information) � (expected value withoutsample information)

3-12. The overall purpose of utility theory is to incorporate a de-cision maker’s preference for risk in the decision-making process.

3-13. A utility function can be assessed in a number of differentways. A common way is to use a standard gamble. With a standardgamble, the best outcome is assigned a utility of 1, and the worstoutcome is assigned a utility of 0. Then, intermediate outcomes areselected and the decision maker is given a choice between havingthe intermediate outcome for sure and a gamble involving the bestand worst outcomes. The probability that makes the decision makerindifferent between having the intermediate outcome for sure and agamble involving the best and worst outcomes is determined. Thisprobability then becomes the utility of the intermediate value. Thisprocess is continued until utility values for all economic conse-quences are determined. These utility values are then placed on autility curve.

3-14. When a utility curve is to be used in the decision-makingprocess, utility values from the utility curve replace all monetaryvalues at the terminal branches in a decision tree or in the body ofa decision table. Then, expected utilities are determined in thesame way as expected monetary values. The alternative with thehighest expected utility is selected as the best decision.

3-15. A risk seeker is a decision maker who enjoys and seeksout risk. A risk avoider is a decision maker who avoids risk even ifthe potential economic payoff is higher. The utility curve for a riskseeker increases at an increasing rate. The utility curve for a riskavoider increases at a decreasing rate.

3-16. a. Decision making under uncertainty.b. Maximax criterion.c. Sub 100 because the maximum payoff for this is$300,000.

Row RowEquipment Favorable Unfavorable Maximum Minimum

Sub 100 300,000 �200,000 300,000 �200,000Oiler J 250,000 �100,000 250,000 �100,000Texan 75,000 �18,000 75,000 �18,000

3-17. Using the maximin criterion, the best alternative is theTexan (see table above) because the worst payoff for this ($�18,000) is better than the worst payoffs for the other decisions.

3-18. a. Decision making under risk—maximize expectedmonetary value.

SOLUTIONS TO DISCUSSION QUESTIONS

AND PROBLEMS

3-1. The purpose of this question is to make students use a per-sonal experience to distinguish between good and bad decisions.A good decision is based on logic and all of the available informa-tion. A bad decision is one that is not based on logic and the avail-able information. It is possible for an unfortunate or undesirableoutcome to occur after a good decision has been made. It is alsopossible to have a favorable or desirable outcome occur after a baddecision.

3-2. The decision-making process includes the following steps:(1) define the problem, (2) list the alternatives, (3) identify the pos-sible outcomes, (4) evaluate the consequences, (5) select an evalua-tion criterion, and (6) make the appropriate decision. The first foursteps or procedures are common for all decision-making problems.Steps 5 and 6, however, depend on the decision-making model.

3-3. An alternative is a course of action over which we havecomplete control. A state of nature is an event or occurrence inwhich we have no control. An example of an alternative is decid-ing whether or not to take an umbrella to school or work on a par-ticular day. An example of a state of nature is whether or not itwill rain on a particular day.

3-4. The basic differences between decision-making modelsunder certainty, risk, and uncertainty depend on the amount ofchance or risk that is involved in the decision. A decision-makingmodel under certainty assumes that we know with complete confi-dence the future outcomes. Decision-making-under-risk modelsassume that we do not know the outcomes for a particular decisionbut that we do know the probability of occurrence of those out-comes. With decision making under uncertainty, it is assumed thatwe do not know the outcomes that will occur, and furthermore, wedo not know the probabilities that these outcomes will occur.

3-5. The techniques discussed in this chapter used to solve deci-sion problems under uncertainty include maximax, maximin, equallylikely, coefficient of realism, and minimax regret. The maximaxdecision-making criterion is an optimistic decision-making criterion,while the maximin is a pessimistic decision-making criterion.

3-6. For a given state of nature, opportunity loss is the differencebetween the payoff for a decision and the best possible payoff forthat state of nature. It indicates how much better the payoff couldhave been for that state of nature. The minimax regret and the mini-mum expected opportunity loss are the criteria used with this.

3-7. Alternatives, states of nature, probabilities for all states ofnature and all monetary outcomes (payoffs) are placed on the deci-sion tree. In addition, intermediate results, such as EMVs for mid-dle branches, can be placed on the decision tree.

3-8. Using the EMV criterion with a decision tree involvesstarting at the terminal branches of the tree and working towardthe origin, computing expected monetary values and selecting thebest alternatives. The EMVs are found by multiplying the proba-bilities of the states of nature times the economic consequencesand summing the results for each alternative. At each decisionpoint, the best alternative is selected.

3-9. A prior probability is one that exists before additional in-formation is gathered. A posterior probability is one that can becomputed using Bayes Theorem based on prior probabilities andadditional information.



b. EMV (Sub 100) � 0.7(300,000) � 0.3(–200,000) � 150,000

EMV (Oiler J) � 0.7(250,000) � 0.3(–100,000) � 145,000

EMV (Texan) � 0.7(75,000) � 0.3(–18,000) � 47,100

Optimal decision: Sub 100.c. Ken would change decision if EMV(Sub 100) is lessthan the next best EMV, which is $145,000. Let X �payoff for Sub 100 in favorable market.

(0.7)(X) � (0.3)(�200,000) � 145,000

0.7X � 145,000 � 60,000 � 205,000

X � (205,000)/0.7 � 292,857.14

The decision would change if this payoff were less than 292,857.14,so it would have to decrease by about $7,143.

3-19. a. The expected value (EV) is computed for eachalternative.

EV(stock market) � 0.5(80,000) � 0.5(�20,000) � 30,000

EV(Bonds) � 0.5(30,000) � 0.5(20,000) � 25,000

EV(CDs) � 0.5(23,000) � 0.5(23,000) � 23,000

Therefore, he should invest in the stock market.

b. EVPI � EV(with perfect information) � (Maximum EV without P, I)

� [0.5(80,000) � 0.5(23,000)] � 30,000 � 51,500 � 30,000 � 21,500

Thus, the most that should be paid is $21,500.

3-20. The opportunity loss table is

Alternative Good Economy Poor Economy

Stock Market 0 43,000Bonds 50,000 3,000CDs 57,000 0

EOL(Stock Market) � 0.5(0) � 0.5(43,000) � 21,500*This minimizes EOL.

EOL(Bonds) � 0.5(50,000) � 0.5(3,000) � 26,500

EOL(CDs) � 0.5(57,000) � 0.5(0) � 28,500

3-21. a.

MarketAlternative Condition Good Fair Poor EMV

Stock market 1,400 800 0 880

Bank deposit 900 900 900 900

Probabilities of 0.4 0.4 0.2market conditions

b. Best decision: deposit $10,000 in bank.

3-22. a. Expected value with perfect information is1,400(0.4) � 900(0.4) � 900(0.2) � 1,100; the maxi-mum EMV without the information is 900. Therefore,Allen should pay at most EVPI � 1,100 – 900 � $200.

b. Yes, Allen should pay [1,100(0.4) � 900(0.4) �900(0.2)] � 900 � $80.

3-23. a. Opportunity loss table

Strong Fair Poor Max.Market Market Market Regret

Large 0 19,000 310,000 310,000Medium 250,000 0 100,000 250,000Small 350,000 29,000 32,000 350,000None 550,000 129,000 0 550,000

b. Minimax regret decision is to build medium.

3-24. a.

Stock Demand(Cases) (Cases) 11 12 13 EMV

11 385 385 385 38512.12 329 420 420 379.0513 273 364 455 341.25

Probabilities 0.45 0.35 0.20

b. Stock 11 cases.

c. If no loss is involved in excess stock, the recom-mended course of action is to stock 13 cases and to re-plenish stock to this level each week. This follows fromthe following decision table.

Stock Demand(Cases) (Cases) 11 12 13 EMV

11 385 385 385 38512 385 420 420 404.2513 385 420 455 411.25

3-25.

Manu- Demandfacture (Cases)(Cases) 6 7 8 9 EMV

6 300 300 300 300 3007 255 350 350 350 340.58 210 305 400 400 352.59 165 260 355 450 317

Probabilities 0.1 0.3 0.5 0.1

John should manufacture 8 cases of cheese spread.

3-26. Cost of produced case � $5.

Cost of purchased case � $16.

Selling price � $15.



b. Produce 300 cases each day.

3-27. a. The table presented is a decision table. The basis forthe decisions in the following questions is shown in thetable below.

Supply Demand(Cases) (Cases) 100 200 300 EMV

100 100(15) �100(5) � 1000 200(15) � 100(5) � 300(15) � 100(5) � 900100(16) � 900 200(16) � 800

200 100(15) � 100(3) � 200(15) � 200(5) � 2000 300(15) � 200(5) � 1610200(5) � 800 100(16) � 1900

300 100(15) � 200(3) � 200(15) � 100(3) � 300(15) � 300(5) � 3000 1800300(5) � 600 300(5) � 1800

Probabilities 0.3 0.4 0.3

b. Maximax decision: Very large station.

c. Maximin decision: Small station.

d. Equally likely decision: Very large station.

e. Criterion of realism decision: Very large station.

f. Opportunity loss table:

MARKET MINIMAX

Decision Good Fair Poor Row Alternatives Market Market Market Maximum

Small 250,000 10,000 0 250,000Medium 220,000 0 10,000 220,000Large 200,000 0 30,000 200,000Very Large 0 5,000 150,000 150,000

Minimax regret decision: Very large station.

Construct

Clinic

Do Nothing

$30,000



EMV for no clinic is $0

$100,000

–$40,000

$0

Payoff

1

3-28. EMV for node 1 � 0.5(100,000) � 0.5(�40,000) �$30,000. Choose the highest EMV, therefore construct the clinic.

Money recovered from each unsold case � $3.

EQUALLY CRIT. OF

MARKET MAXIMAX MAXIMIN LIKELY REALISM

Decision Row Row Row Weighted Alternatives Good Fair Poor Maximum Minimum Average Average

Small 50,000 20,000 �10,000 50,000 �10,000 20,000 38,000Medium 80,000 30,000 �20,000 80,000 �20,000 30,000 60,000Large 100,000 30,000 �40,000 100,000 �40,000 30,000 72,000Very Large 300,000 25,000 �160,000 300,000 �160,000 55,000 208,000



3-29. a.


Unfavorable Market (0.18)$95,000

–$45,000

–$5,000

Payoff

CONSTRUCT

DO NOT CONSTRUCT

2

1

Surve

y

Favor

able

(0.55

)

Survey

Negative (0.45)

Condu

ct

Marke

t Sur

vey

$69,800



–$45,000

–$5,000

CONSTRUCT

DO NOT CONSTRUCT

Do Not Conduct Survey

3

–$5,000$36,140

$36,140



–$40,000

$0

CONSTRUCT CLINIC

DO NOT CONSTRUCT

4

$30,000

b. EMV(node 2) � (0.82)($95,000) � (0.18)(–$45,000)� 77,900 � 8,100 � $69,800

EMV(node 3) � (0.11)($95,000) � (0.89)(–$45,000)� 10,450 � $40,050 � –$29,600

EMV(node 4) � $30,000

EMV(node 1) � (0.55)($69,800) � (0.45)(–$5,000)� 38,390 � 2,250 � $36,140

The EMV for using the survey � $36,140.

EMV(no survey) � (0.5)($100,000) � (0.5)(–$40,000)� $30,000

The survey should be used.

c. EVSI � ($36,140 � $5,000) � $30,000 � $11,140.

Thus, the physicians would pay up to $11,140 for the survey.



3-30.

Favorable Market

Unfavorable Market2

Favorable Market

Unfavorable Market3

Large Shop

No Shop

Small Shop

Favorable Market

Unfavorable Market4

Favorable Market

Unfavorable Market5

Large Shop

No Shop

Small Shop

Favorable Market

Unfavorable Market6

Favorable Market

Unfavorable Market7

Large Shop

No Shop

Small Shop

Favor

able

Surve

y

Unfavorable

Survey

No Survey

Marke

t

Surve

y

1

3-31.

a. EMV(node 2) � (0.9)(55,000) � (0.1)(–$45,000)� 49,500 � 4,500 � $45,000

EMV(node 3) � (0.9)(25,000) � (0.1)(–15,000)� 22,500 � 1,500 � $21,000

EMV(node 4) � (0.12)(55,000) � (0.88)(–45,000)� 6,600 � 39,600 � –$33,000

EMV(node 5) � (0.12)(25,000) � (0.88)(–15,000)� 3,000 � 13,200 � –$10,200

EMV(node 6) � (0.5)(60,000) � (0.5)(–40,000)� 30,000 � 20,000 � $10,000

EMV(node 7) � (0.5)(30,000) � (0.5)(–10,000)� 15,000 � 5,000 � $10,000

EMV(node 1) � (0.6)(45,000) � (0.4)(–5,000)� 27,000 � 2,000 � $25,000

Since EMV(market survey) > EMV(no survey), Jerry should con-duct the survey. Since EMV(large shop | favorable survey) islarger than both EMV(small shop | favorable survey) and EMV(noshop | favorable survey), Jerry should build a large shop if the sur-vey is favorable. If the survey is unfavorable, Jerry should buildnothing since EMV(no shop | unfavorable survey) is larger thanboth EMV(large shop | unfavorable survey) and EMV(small shop |unfavorable survey).




Unfavorable Market (0.1)2



Large Shop

No Shop

Small Shop





Large Shop

No Shop

Small Shop





Large Shop

No Shop

Small Shop

Favor

able

Surve

y (0.6

)

Unfavorable

Survey (0.4)

No Survey

Marke

t

Surve

y

1

$25,000

$45,000

–$5,000

$10,000

$45,000

$21,000

–$33,000

–$10,200

$10,000

$10,000

Payoff

$55,000

–$45,000

–$5,000

–$5,000

$25,000

$25,000

–$15,000

$55,000

–$45,000

–$15,000

$0

$30,000

$60,000

–$40,000

–$10,000

b. If no survey, EMV � 0.5(30,000) � 0.5(–10,000) �$10,000. To keep Jerry from changing decisions, the follow-ing must be true:

EMV(survey) ≥ EMV(no survey)

Let P � probability of a favorable survey. Then,

P[EMV(favorable survey)] � (1 � P) [EMV(unfavor-able survey)] ≥ EMV(no survey)

This becomes:

P(45,000) � (1 � P)(–5,000) ≥ $10,000

Solving gives

45,000P � 5,000 � 5,000P ≥ 10,00050,000P ≥ 15,000P ≥ 0.3

Thus, the probability of a favorable survey could be as low as0.3. Since the marketing professor estimated the probabilityat 0.6, the value can decrease by 0.3 without causing Jerry tochange his decision. Jerry’s decision is not very sensitive tothis probability value.



3-32.

2A3

A4

A5Infor

mation

Favor

able

(0.5)

Information

Unfavorable (0.5)

A 1

Gather

Mor

e

Infor

mation

A

2Do Not Gather

More Inform

ation

1

$2,750

$8,500

$8,500

$500

(0.9)

(0.1)

(0.1)

(0.9)

Payoff

$12,000

–$23,000

$2,000

–$13,000

–$3,000

3

4A3

A4

A5

–$3,000

–$9,000

–$7,000

(0.4)

(0.6)

(0.6)

(0.4)

$12,000

–$23,000

$2,000

–$13,000

–$3,000

5

6A3

A4

A5

$4,500

$4,500

$500

(0.7)

(0.3)

(0.3)

(0.7)

$15,000

–$20,000

$5,000

–$10,000

$0

7

A1: gather more information

A2: do not gather more information

A3: build quadplex

A4: build duplex

A5: do nothing

EMV(node 2) � 0.9(12,000) � 0.1(�23,000) � 8,500

EMV(node 3) � 0.9(2,000) � 0.1(�13,000) � 500

EMV(get information and then do nothing) � �3,000

EMV(node 4) � 0.4(12,000) � 0.6(�23,000) � �9,000

EMV(node 5) � 0.4(2,000) � 0.6(�13,000) � �7,000

EMV(get information and then do nothing) � �3,000

EMV(node 1) � 0.5(8,500) � 0.5(-3,000) � 2,750

EMV(build quadplex) � 0.7(15,000) � 0.3(�20,000) � 4,500

EMV(build duplex) � 0.7(5,000) � 0.3(�10,000) � 500

EMV(do nothing) � 0

Decisions: do not gather information; build quadplex.

3-33. I1: favorable research or information

I2: unfavorable research

S1: store successful

S2: store unsuccessful

P(S1) � 0.5; P(S2) � 0.5

P(I1 | S1) � 0.8; P(I2 | S1) � 0.2

P(I1 | S2) � 0.3; P(I2 | S2) � 0.7

a. P(successful store | favorable research) � P(S1 | I1)

b. P(successful store | unfavorable research) � P(S1 | I2)

c. Now P(S1) � 0.6 and P(S2) � 0.4

P S I( | ). ( . )

. ( . ) . ( . ).1 2

0 2 0 6

0 2 0 6 0 7 0 40 3=

+=

P S I( | ). ( . )

. ( . ) . ( . ).1 1

0 8 0 6

0 8 0 6 0 3 0 40 8=

+=

P S I( | ). ( . )

. ( . ) . ( . ).1 2

0 2 0 5

0 2 0 5 0 7 0 50 22=

+=

P S IP I S P S

P I S P S P I S( | )

( | ) ( )

( | ) ( ) ( |1 22 1 1

2 1 1 2 2

=+ )) ( )P S2

P S I( | ). ( . )

. ( . ) . ( . ).1 1

0 8 0 5

0 8 0 5 0 3 0 50 73=

+=

P S IP I S P S


( | ) ( )

( | ) ( ) ( |1 11 1 1

1 1 1 1 2

=+ )) ( )P S2



3-34. I1: favorable survey or information

I2: unfavorable survey

S1: facility successful

S2: facility unsuccessful

P(S1) � 0.3; P(S2) � 0.7

P(I1 | S1) � 0.8; P(I2 | S1) � 0.2

P(I1 | S2) � 0.3; P(I2 | S2) � 0.7

P(successful facility | favorable survey) � P(S1 | I1)

P(successful facility | unfavorable survey) � P(S1 | I2)

P S I( | ). ( . )

. ( . ) . ( . ).1 2

0 2 0 3

0 2 0 3 0 7 0 70 109=

+=

P S IP I S P S


( | ) ( )

( | ) ( ) ( |1 22 1 1

2 1 1 2 2

=+ )) ( )P S2

P S I( | ). ( . )

. ( . ) . ( . ).1 1

0 8 0 3

0 8 0 3 0 3 0 70 533=

+=

P S IP I S P S


( | ) ( )

( | ) ( ) ( |1 11 1 1

1 1 1 1 2

=+ )) ( )P S2

Fund A

Fund B

Good economy 0.2

Fair economy 0.3

Poor economy 0.5

Good economy 0.2

Fair economy 0.3

Poor economy 0.5

10,000

2,000

–5,000

6,000

4,000

0

b. EMV(A) � 10,000(0.2) � 2,000(0.3) � (�5,000)(0.5) � 100

EMV(B) � 6,000(0.2) � 4,000(0.3) � 0(0.5) � 2,400

Fund B should be selected.

c. Let X � payout for Fund A in a good economy.

EMV(A) � EMV(B)

X(0.2) � 2,000(0.3) � (–5,000)(0.5) � 2,400

0.2X � 4,300

X � 4,300/0.2 � 21,500Therefore, the return would have to be $21,500 for Fund A in agood economy for the two alternatives to be equally desirablebased on the expected values.

3-35. a.



3-36. a.

3

1

Survey

Favorable

Survey

Unfavorable

ProduceRazor

Do Not Produce Razor

Favorable Market

Unfavorable Market

4ProduceRazor


Favorable Market

Unfavorable Market

5

2

Study

Favorable

Study

Unfavorable

ProduceRazor


Favorable Market

Unfavorable Market

6ProduceRazor


Favorable Market

Unfavorable Market

7ProduceRazor


Favorable Market

Unfavorable Market

Con

duct

Sur

vey

Neither Test

ConductPilot

Payoff

$95,000

–$65,000

–$5,000

$95,000

–$65,000

–$5,000

$80,000

–$80,000

–$20,000

$80,000

–$80,000

–$20,000

$100,000

–$60,000

$0

Study

b. S1: survey favorable

S2: survey unfavorable

S3: study favorable

S4: study unfavorable

S5: market favorable

S6: market unfavorable

P(S6 | S1) � 1 – 0.778 � 0.222

P(S6 | S2) � 1 – 0.27 � 0.73

P(S6 | S3) � 1 – 0.89 � 0.11

P(S6 | S4) � 1 – 0.18 � 0.82

c. EMV(node 3) � 95,000(0.78) � (�65,000)(0.22) � 59,800

EMV(node 4) � 95,000(0.27) � (�65,000)(0.73)� �21,800

P S S( | ). ( . )

. ( . ) . ( . ).5 4

0 2 0 5

0 2 0 5 0 9 0 50 18=

+=

P S S( | ). ( . )

. ( . ) . ( . ).5 3

0 8 0 5

0 8 0 5 0 1 0 50 89=

+=

P S S( | ). ( . )

. ( . ) . ( . ).5 2

0 3 0 5

0 3 0 5 0 8 0 50 27=

+=

P S S( | ). ( . )

. ( . ) . ( . ).5 1

0 7 0 5

0 7 0 5 0 2 0 50 78=

+=

EMV(node 5) � 80,000(0.89) � (�80,000)(0.11) � 62,400

EMV(node 6) � 80,000(0.18) � (�80,000)(0.82)� �51,200

EMV(node 7) � 100,000(0.5) � (�60,000)(0.5) � 20,000

EMV(conduct survey) � 59,800(0.45) � (–5,000)(0.55)� 24,160

EMV(conduct pilot study) � 62,400(0.45) � (�20,000)(0.55)� 17,080

EMV(neither) � 20,000

Therefore, the best decision is to conduct the survey. If it is favor-able, produce the razor. If it is unfavorable, do not produce the razor.

3-37. The following computations are for the decision tree thatfollows.

EU(node 3) � 0.95(0.78) � 0.5(0.22) � 0.85

EU(node 4) � 0.95(0.27) � 0.5(0.73) � 0.62

EU(node 5) � 0.9(0.89) � 0(0.11) � 0.80

EU(node 6) � 0.9(0.18) � 0(0.82) � 0.16

EU(node 7) � 1(0.5) � 0.55(0.5) � 0.78

EU(conduct survey) � 0.85(0.45) � 0.8(0.55) � 0.823

EU(conduct pilot study) � 0.80(0.45) � 0.7(0.55) � 0.745

EU(neither test) � 0.81

Therefore, the best decision is to conduct the survey. Jim is a riskavoider.



3

1

SurveyProduceRazor


Market Favorable (0.78)

Market Unfavorable (0.22)

4ProduceRazor




5

2

ProduceRazor




6ProduceRazor




7ProduceRazor




Con

duct

Sur

vey

Neither Test

ConductPilot

Utility

0.95

0.5

0.8

0.95

0.5

0.8

0.9

0

0.7

0.9

0

0.7

1

0.55

0.81

Favorable(0.45)

Survey

Unfavorable(0.55)

Study

Favorable(0.45)

Study

Unfavorable(0.55)

0.82

0.745

0.85

0.62

0.80

0.16

0.78

Study

3-38. a. P(good economy | prediction of

good economy) �

P(poor economy | prediction of

good economy) �

P(good economy | prediction of

poor economy) �


poor economy) �

b. P(good economy | prediction of

good economy) �


good economy) �

P(good economy | prediction of

poor economy) �0 2 0 7

0 2 0 7 0 9 0 30 341

. ( . )

. ( . ) . ( . ).

+=

0 1 0 3

0 8 0 7 0 1 0 30 051

. ( . )

. ( . ) . ( . ).

+=

0 8 0 7

0 8 0 7 0 1 0 30 949

. ( . )

. ( . ) . ( . ).

+=

0 9 0 6

0 2 0 6 0 9 0 40 75

. ( . )

. ( . ) . ( . ).

+=

0 2 0 6

0 2 0 6 0 9 0 40 25

. ( . )

. ( . ) . ( . ).

+=

0 1 0 4

0 8 0 6 0 1 0 40 077

. ( . )

. ( . ) . ( . ).

+=

0 8 0 6

0 8 0 6 0 1 0 40 923

. ( . )

. ( . ) . ( . ).

+=


poor economy) �

3-39. The expected value of the payout by the insurance com-pany is

EV � 0(0.999) � 100,000(0.001) � 100

The expected payout by the insurance company is $100, but thepolicy costs $200, so the net gain for the individual buying thispolicy is negative (–$100). Thus, buying the policy does not maxi-mize EMV since not buying this policy would have an EMV of 0,which is better than –$100. However, a person who buys this pol-icy would be maximizing the expected utility. The peace of mindthat goes along with the insurance policy has a relatively high util-ity. A person who buys insurance would be a risk avoider.

0 9 0 3

0 2 0 7 0 9 0 30 659

. ( . )

. ( . ) . ( . ).

+=



3-40.

2

1

SurveyConstructClinic

Do Not Construct Clinic



3ConstructClinic



Unfavorable Market (0.89)Conduct

Market

Do Not Conduct

Survey

0.99

0

0.7

0.99

0

0.7

Favorable(0.55)

Survey

Unfavorable(0.45)

U = 0.76

U = 0.8118

U = 0.1089

4ConstructClinic




1.0

0.1

0.9

U = 0.55

Utility

$95,000

–$45,000

–$5,000

$95,000

–$45,000

–$5,000

$100,000

–$40,000

$0

Payoff

EU(node 2) � (0.82)(0.99) � (0.18)(0) � 0.8118

EU(node 3) � (0.11)(0.99) � (0.89)(0) � 0.1089

EU(node 4) � 0.5(1) � 0.5(0.1) � 0.55

EU(node 1) � (0.55)(0.8118) � (0.45)(0.7000) � 0.7615

EU(no survey) � 0.9

The expected utility with no survey (0.9) is higher than the ex-pected utility with a survey (0.7615), so the survey should be notused. The medical professionals are risk avoiders.

3-41. EU(large plant | survey favorable) � 0.78(0.95) � 0.22(0) � 0.741

EU(small plant | survey favorable) � 0.78(0.5) � 0.22(0.1) � 0.412

EU(no plant | survey favorable) � 0.2

EU(large plant | survey negative) � 0.27(0.95) � 0.73(0) � 0.2565

EU(small plant | survey negative) � 0.27(0.5) � 0.73(0.10) � 0.208

EU(no plant | survey negative) � 0.2

EU(large plant | no survey) � 0.5(1) � 0.5(0.05) � 0.525

EU(small plant | no survey) � 0.5(0.6) � 0.5(0.15) � 0.375

EU(no plant | no survey) � 0.3

EU(conduct survey) � 0.45(0.741) � 0.55(0.2565) � 0.4745

EU(no survey) � 0.525

John’s decision would change. He would not conduct the surveyand build the large plant.

3-42. a. Expected travel time on Broad Street � 40(0.5) �15(0.5) � 27.5 minutes. Broad Street has a lower ex-pected travel time.

Congestion (0.5)

NoCongestion (0.5)

1

Expressway

BroadStreet

30 Minutes,U = 0.7

40 Minutes,U = 0.2

15 Minutes,U = 0.9

b. Expected utility on Broad Street � 0.2(0.5) �0.9(0.5) � 0.55. Therefore, the expressway maximizesexpected utility.

c. Lynn is a risk avoider.

1.0

0.8

0.6

0.4

0.2

0

Util

ity

0 10 20 30 40

Time (minutes)

3-43. Selling price � $20 per gallon; manufacturing cost �$12 per gallon; salvage value � $13; handling costs � $1 pergallon; and advertising costs � $3 per gallon. From this informa-tion, we get:

marginal profit � selling price minus the manufacturing, handling,and advertising costs

marginal profit � $20 � $12 � $1 � $3 � $4 per gallon

If more is produced than is needed, a marginal loss is incurred.

marginal loss � $13 � $12 � $1 � $3 � $3 per gallon

In addition, there is also a shortage cost. Coren has agreed to fulfillany demand that cannot be met internally. This requires that Corenpurchase chemicals from an outside company. Because the cost ofobtaining the chemical from the outside company is $25 and theprice charged by Coren is $20, this results in

shortage cost � $5 per gallon

In other words, Coren will lose $5 for every gallon that is sold thathas to be purchased from an outside company due to a shortage.



a. A decision tree is shown below:

b. The computations are shown in the following table. Thesenumbers are entered into the tree above. The best decision is tostock 1,500 gallons.

Table for Problem 3-43

Demand

Stock 500 1,000 1,500 2,000 EMV

500 2,000 �500 �3,000 �5,500 �$1,5001,000 500 4,000 1,500 1,000 $1,8001,500 �1,000 2,500 6,000 3,500 $3,3002,000 �2,500 1,000 4,500 8,000 $2,400

Maximum 2,000 4,000 6,000 8,000 $4,800 � EVwPI


c. EVwPI � (0.2)(2,000) � (0.3)(4,000) � (0.4)(6,000) � (0.1)(8,000) � $4,800

EVPI � EVwPI � EMV � $4,800 � $3,300 � $1,5003-44. If no survey is to be conducted, the decision tree is fairly

straightforward. There are three main decisions, which are build-ing a small, medium, or large facility. Extending from thesedecision branches are three possible demands, representing thepossible states of nature. The demand for this type of facility couldbe either low (L), medium (M), or high (H). It was given in theproblem that the probability for a low demand is 0.15. The proba-bilities for a medium and a high demand are 0.40 and 0.45, respec-tively. The problem also gave monetary consequences for buildinga small, medium, or large facility when the demand could be low,medium, or high for the facility. These data are reflected in the fol-lowing decision tree.

Stock500

Stock1,000

Stock2,000

Stock1,500

Decision Tree

–$1,500

$2,400

$3,300

$1,800

(0.2) Demand

(0.3)

(0.4)

(0.1)

(0.2)

(0.3)

(0.4)

(0.1)

(0.2)

(0.3)

(0.4)

(0.1)

(0.2)

(0.3)

(0.4)

(0.1)

500

1,000

1,500

2,000

500

1,000

1,500

2,000

500

1,000

1,500

2,000

500

1,000

1,500

2,000

$2,000 = (500)(4)

–$500 = (500)(4) – (500)(5)

$500 = (500)(4) – (500)(3)

$4,000 = (1,000)(4)

$1,500 = (1,000)(4) – (5)(500)

–$1,000 = (1,000)(4) – (5)(1,000)

–$1,000 = (500)(4) – (3)(1,000)

$2,500 = (1,000)(4) – (3)(500)

–$2,500 = (500)(4) – (3)(1,500)

$1,000 = (1,000)(4) – (3)(1,000)

$4,500 = (1,500)(4) – (3)(500)

$3,500 = (1,500)(4) – (5)(500)

$6,000 = (1,500)(4)

$8,000 = (2,000)(4)

–$3,000 = (500)(4) – (1,000)(5)

–$5,500 = (500)(4) – (1,500)(5)

Decision Tree–No Survey

Small $500,000

Large $580,000

Medium $670,000

(0.15)

(0.40)

(0.45)

(0.15)

(0.40)

(0.45)

(0.15)

(0.40)

(0.45)

$500,000

$500,000

$500,000

$200,000

$700,000

$800,000

–$200,000

$400,000

$1,000,000

With no survey, we have: EMV(Small) � 500,000;EMV(Medium) � 670,000; and EMV(Large) � 580,000.The medium facility, with an expected monetary value of$670,000, is selected because it represents the highest ex-pected monetary value.

If the survey is used, we must compute the revised probabili-ties using Bayes’ theorem. For each alternative facility, threerevised probabilities must be computed, representing low,medium, and high demand for a facility. These probabilitiescan be computed using tables. One table is used to compute theprobabilities for low survey results, another table is used for



Small

Medium

Large

L

M

H

L

M

H

L

M

H

Small

Medium

Large

L

M

H

L

M

H

L

M

H

Small

Medium

Large

L

M

H

L

M

H

L

M

H

450,000

450,000

450,000

150,000

650,000

750,000

–250,000

350,000

950,000

450,000

450,000

650,000

750,000

150,000

350,000

950,000

–250,000

450,000

450,000

450,000

650,000

750,000

150,000

350,000

950,000

–250,000

450,000

Decision Tree–Survey

$495

,000

Low

(0.

310)

$821,000

High (0.325)

$646,000

Medium(0.365)

medium survey results, and a final table is used for high survey re-sults. These tables are shown below. These probabilities will be usedin the decision tree that follows.

For low survey results—A1:

State of Nature P(Bi) P(Ai | Bj) P(Bj and Ai) P(Bj | Ai)

B1 0.150 0.700 0.105 0.339B2 0.400 0.400 0.160 0.516B3 0.450 0.100 0.045 0.145

P(A1) � 0.310

For medium survey results—A2:


B1 0.150 0.200 0.030 0.082B2 0.400 0.500 0.200 0.548B3 0.450 0.300 0.135 0.370

P(A2) � 0.365

For high survey results—A3:


B1 0.150 0.100 0.015 0.046B2 0.400 0.100 0.040 0.123B3 0.450 0.600 0.270 0.831

P(A3) � 0.325

When survey results are low, the probabilities are P(L) �0.339; P(M) � 0.516; and P(H) � 0.145. This results inEMV(Small) � 450,000; EMV(Medium) � 495,000; andEMV(Large) � 233,600.

When survey results are medium, the probabilities are P(L) �0.082; P(M) � 0.548; and P(H) � 0.378. This results in EMV(Small) � 450,000; EMV(Medium) � 646,000; and EMV(Large) �522,800.

When survey results are high, the probabilities are P(L) �0.046; P(M) � 0.123; and P(H) � 0.831. This results inEMV(Small) � 450,000; EMV(Medium) � 710,100; andEMV(Large) � 821,000.

If the survey results are low, the best decision is to build themedium facility with an expected return of $495,000. If the surveyresults are medium, the best decision is also to build the mediumplant with an expected return of $646,000. On the other hand, ifthe survey results are high, the best decision is to build the largefacility with an expected monetary value of $821,000. The ex-pected value of using the survey is computed as follows:

EMV(with Survey) � 0.310(495,000) � 0.365(646,000) � 0.325(821,000) � 656,065

Because the expected monetary value for not conducting the sur-vey is greater (670,000), the decision is not to conduct the surveyand to build the medium-sized facility.



3-45. a.

Mary should select the traffic circle location (EMV � $250,000).

b. Use Bayes’ Theorem to compute posterior probabilities.

P(SD | SRP) = 0.78; P(SD¯̄ ¯̄ | SRP) = 0.22

P(SM | SRP) = 0.84; P(SM¯̄ ¯̄ ¯ | SRP) = 0.16

P(SC | SRP) = 0.91; P(SC¯̄ ¯̄ | SRP) = 0.09

P(SD | SRN) = 0.27; P(SD¯̄ ¯̄ | SRN) = 0.73

P(SM | SRN) = 0.36; P(SM¯̄ ¯̄ ¯ | SRN) = 0.64

P(SC | SRN) = 0.53; P(SC¯̄ ¯̄ | SRN) = 0.47

Example computations:

These calculations are for the tree that follows:

EMV(2) � $171,600 � $28,600 � $143,000

EMV(3) � $226,800 � $20,800 � $206,000

EMV(4) � $336,700 � $20,700 � $316,000

EMV(no grocery � A) � –$30,000

EMV(5) � $59,400 � $94,900 � –$35,500

EMV(6) � $97,200 � $83,200 � $14,000

EMV(7) � $196,100 � $108,100 � $88,000

EMV(no grocery � B) � –$30,000

P SC SRN( | ). ( . )

. ( . ) . ( . ).=

+=0 3 0 75

0 3 0 75 0 8 0 250 53

P SM SRP( | ). ( . )

. ( . ) . ( . ).=

+=0 7 0 6

0 7 0 6 0 2 0 40 84

P SM SRPP SRP SM P SM

P SRP SM P SM P SR( | )

( | ) ( )

( | ) ( ) (=

+ PP SM P SM| ) ( )

1

2

3

Succeed (0.5)

Don’t Succeed (0.5)

Succeed (0.6)


Succeed (0.75)


$75,000

$140,000

$250,000

Mall

No Grocery Store

Downtown

Traffic Circle

Payoff

$250,000

–$100,000

$300,000

–$100,000

$400,000

–$200,000

$0

EMV(8) � $75,000

EMV(9) � $140,000

EMV(10) � $250,000

EMV(no grocery � C) � $0

EMV(A) � (best of four alternatives) � $316,000

EMV(B) � (best of four alternatives) � $88,000

EMV(C) � (best of four alternatives) � $250,000

EMV(1) � (0.6)($316,000) � (0.4)($88,000)� $224,800

EMV(D) � (best of two alternatives)� $250,000

c. EVSI � [EMV(1) � cost] � (best EMV without sample information)

� $254,800 – $250,000 � $4,800.



FirstDecision

Point

SecondDecision

Point

Purch

ase

Mar

ket

Surve

y

Surve

y Res

ults

Positiv

e (0

.6)

Survey Results

Negative (0.4)

Do Not Purchase Market Survey

$220,000

–$130,000

–$230,000

$270,000

–$130,000

$370,000

–$30,000

$220,000

–$130,000

–$230,000

$270,000

–$130,000

$370,000

–$30,000

$250,000

–$100,000

–$200,000

$300,000

–$100,000

$400,000

$0

1

2

3

4

D

Downtown

Mall

Circle

No Grocery Store

SD (0.78)

SM (0.84)

SC (0.91)

SD (0.22)

SM (0.16)

SC (0.09)

SC (0.47)

5

6

7

Downtown

Mall

Circle

No Grocery Store

SD (0.27)

SM (0.36)

SC (0.53)

SD (0.73)

SM (0.64)

8

9

10

Downtown

Mall

Circle

No Grocery Store

SD (0.5)

SM (0.6)

SC (0.75)

SC (0.25)

SD (0.5)

SM (0.4)

A

B

C

Payoff

3-46. a. Sue can use decision tree analysis to find the best solu-tion. The results are presented below. In this case, the best decisionis to get information. If the information is favorable, she shouldbuild the retail store. If the information is not favorable, she shouldnot build the retail store. The EMV for this decision is $29,200.

In the following results (using QM for Windows), Branch 1(1–2) is to get information, Branch 2 (1–3) is the decision to not getinformation, Branch 3 (2–4) is favorable information, Branch 4(2–5) is unfavorable information, Branch 5 (3–8) is the decision tobuild the retail store and get no information, Branch 6 (3–17) is thedecision to not build the retail store and to get no information,Branch 7 (4–6) is the decision to build the retail store given favorableinformation, Branch 8 (4–11) is the decision to not build given favor-able information, Branch 9 (6–9) is a good market given favorable

information, Branch 10 (6–10) is a bad market given favorable in-formation, Branch 11 (5–7) is the decision to build the retail storegiven unfavorable information, Branch 12 (5–14) is the decisionnot to build the retail store given unfavorable information, Branch13 (7–12) is a successful retail store given unfavorable information,Branch 14 (7–13) is an unsuccessful retail store given unfavorableinformation, Branch 15 (8–15) is a successful retail store given thatno information is obtained, and Branch 16 (8–16) is an unsuccess-ful retail store given no information is obtained.



b. The suggested changes would be reflected in Branches 4 and 5. The decision stays the same, but the EMVincreases to $46,000. The results are provided in the tables that follow:

Results for 3-46. a.

Start Ending Branch Profit Use Node NodeNode Node Probability (End Node) Branch? Type Value

Start 0 1 0 0 Decision 29,200Branch 1 1 2 0 0 Yes Chance 29,200Branch 2 1 3 0 0 Decision 28,000Branch 3 2 4 0.6 0 Decision 62,000Branch 4 2 5 0.4 0 Decision �20,000Branch 5 3 8 0 0 Yes Chance 28,000Branch 6 3 17 0 0 Final 0Branch 7 4 6 0 0 Yes Chance 62,000Branch 8 4 11 0 �20,000 Final �20,000Branch 9 6 9 0.9 80,000 Final 80,000Branch 10 6 10 0.1 �100,000 Final �100,000Branch 11 5 7 0 0 Chance �64,000Branch 12 5 14 0 �20,000 Yes Final �20,000Branch 13 7 12 0.2 80,000 Final 80,000Branch 14 7 13 0.8 �100,000 Final �100,000Branch 15 8 15 0.6 100,000 Final 100,000Branch 16 8 16 0.4 �80,000 Final �80,000

Results for 3-46. b.


Start 0 1 0 0 Decision 37,400Branch 1 1 2 0 0 Yes Chance 37,400Branch 2 1 3 0 0 Decision 28,000Branch 3 2 4 0.7 0 Decision 62,000Branch 4 2 5 0.3 0 Decision �20,000Branch 5 3 8 0 0 Yes Chance 28,000Branch 6 3 17 0 0 Final 0Branch 7 4 6 0 0 Yes Chance 62,000Branch 8 4 11 0 �20,000 Final �20,000Branch 9 6 9 0.9 80,000 Final 80,000Branch 10 6 10 0.1 �100,000 Final �100,000Branch 11 5 7 0 0 Chance �64,000Branch 12 5 14 0 �20,000 Yes Final �20,000Branch 13 7 12 0.2 80,000 Final 80,000Branch 14 7 13 0.8 �100,000 Final �100,000Branch 15 8 15 0.6 100,000 Final 100,000Branch 16 8 16 0.4 �80,000 Final �80,000



c. Sue can determine the impact of the change by changing the probabilities and recomputing EMVs. This analysisshows the decision changes. Given the new probability values, Sue’s best decision is build the retail store withoutgetting additional information. The EMV for this decision is $28,000. The results are presented below:

Results for 3-46. c.


Start 0 1 0 0 Decision 28,000Branch 1 1 2 0 0 Chance 18,400Branch 2 1 3 0 0 Yes Decision 28,000Branch 3 2 4 0.6 0 Decision 44,000Branch 4 2 5 0.4 0 Decision �20,000Branch 5 3 8 0 0 Yes Chance 28,000Branch 6 3 17 0 0 Final 0Branch 7 4 6 0 0 Yes Chance 44,000Branch 8 4 11 0 �20,000 Final �20,000Branch 9 6 9 0.8 80,000 Final 80,000Branch 10 6 10 0.2 �100,000 Final �100,000Branch 11 5 7 0 0 Chance �64,000Branch 12 5 14 0 �20,000 Yes Final �20,000Branch 13 7 12 0.2 80,000 Final 80,000Branch 14 7 13 0.8 �100,000 Final �100,000Branch 15 8 15 0.6 100,000 Final 100,000Branch 16 8 16 0.4 �80,000 Final �80,000

d. Yes, Sue’s decision would change from her original decision. With the higher cost of information, Sue’s decisionis to not get the information and build the retail store. The EMV of this decision is $28,000. The results are givenbelow:

Results for 3-46. d.


Start 0 1 0 0 Decision 28,000Branch 1 1 2 0 0 Chance 19,200Branch 2 1 3 0 0 Yes Decision 28,000Branch 3 2 4 0.6 0 Decision 52,000Branch 4 2 5 0.4 0 Decision �30,000Branch 5 3 8 0 0 Yes Chance 28,000Branch 6 3 17 0 0 Final 0Branch 7 4 6 0 0 Yes Chance 52,000Branch 8 4 11 0 �30,000 Final �30,000Branch 9 6 9 0.9 70,000 Final 70,000Branch 10 6 10 0.1 �110,000 Final �110,000Branch 11 5 7 0 0 Chance �74,000Branch 12 5 14 0 �30,000 Yes Final �30,000Branch 13 7 12 0.2 70,000 Final 70,000Branch 14 7 13 0.8 �110,000 Final �110,000Branch 15 8 15 0.6 100,000 Final 100,000Branch 16 8 16 0.4 �80,000 Final �80,000



e. The expected utility can be computed by replacing the monetary values with utility values. Given the utility values in the prob-lem, the expected utility is 0.62. The utility table represents a risk seeker. The results are given below:

Results for 3-46. e.

Start Ending Branch Profit Use Ending Node NodeNode Node Probability (End Node) Branch? Node Type Value

Start 0 1 0 0 1 Decision 0.62Branch 1 1 2 0 0 2 Chance 0.256Branch 2 1 3 0 0 Yes 3 Decision 0.62Branch 3 2 4 0.6 0 4 Decision 0.36Branch 4 2 5 0.4 0 5 Decision 0.1Branch 5 3 8 0 0 Yes 8 Chance 0.62Branch 6 3 17 0 0.2 17 Final 0.20Branch 7 4 6 0 0 Yes 6 Chance 0.36Branch 8 4 11 0 0.1 11 Final 0.1Branch 9 6 9 0.9 0.4 9 Final 0.4Branch 10 6 10 0.1 0 10 Final 0Branch 11 5 7 0 0 7 Chance 0.08Branch 12 5 14 0 0.1 Yes 14 Final 0.1Branch 13 7 12 0.2 0.4 12 Final 0.4Branch 14 7 13 0.8 0 13 Final 0Branch 15 8 15 0.6 1 15 Final 1Branch 16 8 16 0.4 0.05 16 Final 0.05

f. This problem can be solved by replacing monetary values with utility values. The expected utility is 0.80. The utility tablegiven in the problem is representative of a risk avoider. The results are presented below:

Results for 3-46. f.


Start 0 1 0 0 Decision 0.80Branch 1 1 2 0 0 Chance 0.726Branch 2 1 3 0 0 Yes Decision 0.80Branch 3 2 4 0.6 0 Decision 0.81Branch 4 2 5 0.4 0 Decision 0.60Branch 5 3 8 0 0 Yes Chance 0.76Branch 6 3 17 0 0.8 Final 0.80Branch 7 4 6 0 0 Yes Chance 0.81Branch 8 4 11 0 0.6 Final 0.60Branch 9 6 9 0.9 0.9 Final 0.90Branch 10 6 10 0.1 0 Final 0.00Branch 11 5 7 0 0 Chance 0.18Branch 12 5 14 0 0.6 Yes Final 0.60Branch 13 7 12 0.2 0.9 Final 0.90Branch 14 7 13 0.8 0 Final 0.00Branch 15 8 15 0.6 1 Final 1.00Branch 16 8 16 0.4 0.4 Final 0.40

3-47. a. The decision table for Chris Dunphy along with the ex-pected profits or expected monetary values (EMVs) for each alter-native are shown on the next page.



Table for Problem 3-47a

Return in $1,000:

Event 1 Event 2 Event 3 Event 4 Event 5

Number of Watches Probability 0.100 0.200 0.500 0.100 0.100

100,000 Alternative 1 100,000 110,000 120,000 135,000 140,000150,000 Alternative 2 90,000 120,000 140,000 155,000 170,000200,000 Alternative 3 85,000 110,000 135,000 160,000 175,000250,000 Alternative 4 80,000 120,000 155,000 170,000 180,000300,000 Alternative 5 65,000 100,000 155,000 180,000 195,000350,000 Alternative 6 50,000 100,000 160,000 190,000 210,000400,000 Alternative 7 45,000 95,000 170,000 200,000 230,000450,000 Alternative 8 30,000 90,000 165,000 230,000 245,000500,000 Alternative 9 20,000 85,000 160,000 270,000 295,000

Expected profit:

Alternative Expected Profit

1 119,5002 135,5003 131,5004 144,5005 141,5006 145,0007 151,5008 151,0009 155,500 ← best alternative

For this decision problem, Alternative 9 gives the highest ex-pected profit of $155,500.

b. The expected value with perfect information is $175,500, andthe expected value of perfect information (EVPI) is $20,000.

c. The new probability estimates will give more emphasis toevent 2 and less to event 5. The overall impact is shownbelow. As you can see, stocking 400,000 watches is now thebest decision with an expected value of $140,700.

Return in $1,000:

EVENT 1 EVENT 2 EVENT 3 EVENT 4 EVENT 5

Probability 0.100 0.280 0.500 0.100 0.020

Alternative 1 100,000 110,000 120,000 135,000 140,000Alternative 2 90,000 120,000 140,000 155,000 170,000Alternative 3 85,000 110,000 135,000 160,000 175,000Alternative 4 80,000 120,000 155,000 170,000 180,000Alternative 5 65,000 100,000 155,000 180,000 195,000Alternative 6 50,000 100,000 160,000 190,000 210,000Alternative 7 45,000 95,000 170,000 200,000 230,000Alternative 8 30,000 90,000 165,000 230,000 245,000Alternative 9 20,000 85,000 160,000 270,000 295,000

Expected profit:


1 117.1002 131,5003 126,3004 139,7005 133,9006 136,2007 140,7008 138,6009 138,700

← best alternative: stock 400,000 watches



d. Stocking 400,000 is still the best alternative. The resultsare shown below.

Return in $1,000:

Event 1 Event 2 Event 3 Event 4 Event 5

Probability 0.100 0.280 0.500 0.100 0.020

Alternative 1 100,000 110,000 120,000 135,000 140,000Alternative 2 90,000 120,000 140,000 155,000 170,000Alternative 3 85,000 110,000 135,000 160,000 175,000Alternative 4 80,000 120,000 155,000 170,000 180,000Alternative 5 65,000 100,000 155,000 180,000 195,000Alternative 6 50,000 100,000 160,000 190,000 210,000Alternative 7 45,000 95,000 170,000 200,000 230,000Alternative 8 30,000 90,000 165,000 230,000 245,000Alternative 9 20,000 85,000 160,000 270,000 340,000

Expected profit


1 117,1002 131,5003 126,3004 139,7005 133,9006 136,2007 140,7008 138,6009 139,600

3-48. a. Decision under uncertainty.b.

Population Population Row Same Grows Average

Large wing �85,000 150,000 32,500Small wing �45,000 60,000 7,500No wing 0 0 0

c. Best alternative: large wing.

3-49. a. Note: This problem can also be solved using marginalanalysis.

Weighted Population Population Average with

Same Grows � = 0.75

Large wing �85,000 150,000 91,250Small wing �45,000 60,000 33,750No wing 0 0 0

b. Best decision: large wing.c. No.

3-50. a.

No Mild Severe Expected Congestion Congestion Congestion Time

Tennessee 15 30 45 25Back roads 20 25 35 24.17Expressway 30 30 30 30

Probabilities (30 days)/ (20 days)/ (10 days)/(60 days) = 1/2 (60 days) = 1/3 (60 days) = 1/6

← best alternative: stock 400,000 watches

b. Back roads (minimum time used).

c. Expected time with perfect information:15 � 1/2 + 25 � 1/3 + 30 � 1/6 = 20.83 minutes

Time saved is 31⁄3; minutes.

3-51. a. EMV can be used to determine the best strategy to min-imize costs. The QM for Windows solution is shown onthe next page. The best decision is to go with the partialservice (maintenance) agreement.



Solution to 3-51a

Expected Row Row Value Minimum Maximum

($) ($) ($)

Probabilities 0.2 0.8

Maint. No Maint. Cost ($) Cost ($)

No Service Agreement 3,000 0 600 0 3,000Partial Service Agreement 1,500 300 540 0 1,500Complete Service Agreement 500 500 500 500 500

Column best 500 0 500

The minimum expected monetary value is 500 given by CompleteService Agreement

b. The new probability estimates dramatically changeSim’s decision. The best decision given this new informa-tion is to still go with the complete service or maintenancepolicy with an expected cost of $500. The results areshown below.

Solution to 3-51b

Does Not Expected Needs Repair Need Repair Value

($) ($) ($)

Probabilities 0.8 0.2

No Service 3,000 0 2,400Agreement

Partial Service 1,500 300 1,260Agreement

Complete Service 500 500 500Agreement

Column best 500

3-52. We can use QM for Windows to solve this decision mak-ing under uncertainty problem. We have made up probability val-ues, which will be ignored in the analysis. As you can see, themaximax decision is Option 4, and the maximum decision is Op-tion 1. To compute the equality likely decision, we used equalprobability values of 0.25 for each of the four scenarios. As seenbelow, the equally likely decision, which is the same as the EMVdecision in this case, is Option 3.

Solution to 3-52

Expected Row Row Value ($) Minimum ($) Miximum ($)

Probabilities 0.25 0.25 0.25 0.25

Judge ($) Trial ($) Court ($) Arbitration ($)

Option 1 5,000 5,000 5,000 5,000 5,000 5,000 5,000Option 2 10,000 5,000 2,000 0 4,250 0 10,000Option 3 20,000 7,000 1,000 �5,000 5,750 �5,000 20,000Option 4 30,000 15,000 �10,000 �20,000 3,750 �20,000 30,000

Column best 5,750 5,000 30,000

The maximum expected monetary value is 5,750 given by Option 3.The maximum is 5,000 given by Option 1. The maximax is 30,000given by Option 4.



Payoff table

Laplace HurwiczEvent 1 Event 2 Average Value Minimum Maximum Value

Alternative 1 0 0 0.0 0 0 0.00Alternative 2 55,273 �10,000 22,636.5 �10,000 55,273 �2,819.97Alternative 3 120,000 �15,000 152,500.0 �15,000 120,000 �150.00Alternative 4 240,000 �30,000 105,000.0 �30,000 240,000 �300.00

SOLUTION TO STARTING RIGHT CASE

This is a decision-making-under-uncertainty case. There are twoevents: a favorable market (event 1) and an unfavorable market(event 2). There are four alternatives, which include do nothing(alternative 1), invest in corporate bonds (alternative 2), invest inpreferred stock (alternative 3), and invest in common stock (alter-native 4). The decision table is presented below. Note that for al-ternative 2, the return in a good market is $30,000 (1 � 0.13)5 �$55,273. The return in a good market is $120,000 (4 x $30,000)for alternative 3, and $240,000 (8 x $30,000) for alternative 4.

Regret table

Maximum Alternative Event 1 Event 2 Regret

Alternative 1 240,000 0 240,000Alternative 2 184,727 10,000 184,727Alternative 3 120,000 15,000 120,000Alternative 4 0 30,000 30,000

a. Sue Pansky is a risk avoider and should use the maximindecision approach. She should do nothing and not make aninvestment in Starting Right.b. Ray Cahn should use a coefficient of realism of 0.11.The best decision is to do nothing.c. Lila Battle should eliminate alternative 1 of doing noth-ing and apply the maximin criterion. The result is to invest inthe corporate bonds.d. George Yates should use the equally likely decision cri-terion. The best decision for George is to invest in commonstock.e. Pete Metarko is a risk seeker. He should invest in com-mon stock.f. Julia Day can eliminate the preferred stock alternativeand still offer alternatives to risk seekers (common stock) andrisk avoiders (doing nothing or investing in corporate bonds).

SOLUTIONS TO INTERNET CASES

Drink-at-Home, Inc. CaseAbbreviations and values used in the following decision trees:

Normal—proceed with research and development at a normalpace.

6 Month—Adopt the 6-month program: if a competitor’s productis available at the end of 6 months, then copy; otherwise proceedwith research and development.

8 Month—Adopt the 6-month program: proceed for 8 months; ifno competition at 8 months, proceed; otherwise stop development.

Success or failure of development effort:Ok—Development effort ultimately a successNo—Development effort ultimately a failure

Column:S—Sales revenueR—Research and development expendituresE—Equipment costsI—Introduction to market costs

Market size and Revenues:

Without WithCompetition Competition

S—Substantial (P � 0.1) $800,000 $400,000M—Moderate (P � 0.6) $600,000 $300,000L—Low (P � 0.3) $500,000 $250,000

Competition:C6—Competition at end of 6 months (P � .5)

No C6—No competition at end of 6 months (P � .5)C8—Competition at end of 8 months (P � .6)

No C8—No competition at end of 8 months (P � .4)C12—Competition at end of 12 months (P � .8)

No C12—No competition at end of 12 months (P � .2)



Drink-at-Home. Inc. Case (continued)

Normal

8 Month

6 Month

No C6 (.5)

No C8 (.4)

C12 (.8)

No C12 (.2)

No (.1) (Stop)

C8 (.6)

No C12 (.2)

C6 (.5)

C12 (.8)

Ok (.9)

Ok (.9)

Ok (.9)

No (.1)

No (.1)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

Mkt S R E I

400 – 100 – 100 – 150 = 50

300 – 100 – 100 – 150 = –50

250 – 100 – 100 – 150 = –100

800 – 100 – 100 – 150 = 450

600 – 100 – 100 – 150 = 250

500 – 100 – 100 – 150 = 150

– 100

– 80

= –100

= – 80

– 100 = –100

400 – 140 – 100 – 150 = 10

300 – 140 – 100 – 150 = –90

250 – 140 – 100 – 150 = –140

800 – 140 – 100 – 150 = 410

600 – 140 – 100 – 150 = 210

500 – 140 – 100 – 150 = 110

400 – 90 – 100 – 150 = 60

300 – 90 – 100 – 150 = –40

250 – 90 – 100 – 150 = –90

400 – 100 – 100 – 150 = 50

300 – 100 – 100 – 150 = –50

250 – 100 – 100 – 150 = –100

800 – 100 – 100 – 150 = 450

600 – 100 – 100 – 150 = 250

500 – 100 – 100 – 150 = 150

Normal

8 Month

6 Month

No C6 (.5)

No C8 (.4)

C12 (.8)

No C12 (.2)

No (.1) (Stop)

C8 (.6)

No C12 (.2)

C6 (.5)

C12 (.8)

Ok (.9)

Ok (.9)

Ok (.9)

No (.1)

No (.1)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

Mkt

50

–50

–100

450

250

150

–100

–80

10

–90

–140

410

210

110

60

–40

–90

50

–50

–100

450

–100

150250



Normal –6.4

8 Month –74.2

6 Month

No C6 (.5)

No C8 (.4)

C12 (.8)

No C12 (.2)No (.1) (Stop)

C8 (.6)

No C12 (.2)

C6 (.5)

C12 (.8)

–55

200

–45

–95

–55

(–74.2)

240

(4)

240

Ok (.9)

(19.3)

Ok (.9)

Ok (.9)

No (.1)

No (.1)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

S (.1)

M (.6)

L (.3)

Mkt

50

–50

–100

450

250

150

–100

–80

10

–90

–140

410

210

110

60

–40

–90

50

–50

–100

450

–100

150250

Drink-at-Home, Inc. Case (continued)

Ruth Jones’ Heart By-Pass Operation Case

One Year

Two Years

Five Years

Eight Years

0 Years

One Year

Five Years

Ten Years

Fifteen Years

Twenty Years

Twenty-fiveYears

.50

.20

.20

.10

.05

.45

.20

.13

.08

.05

.04

1

2

5

8

0

1

5

10

15

20

25

.50

.40

1.00

.80

0.0

.45

1.00

1.30

1.20

1.00

1.00

Prob. Years Expected Rate

2.7 years

5.95 years

No By-

pass

Surge

ry

Surgery

The optimal program is to adopt the 6-month program



Expected survival rate with surgery (5.95 years) exceeds thenonsurgical survival rate of 2.70 years. Surgery is favorable.

Ski Right Casea. Bob can solve this case using decision analysis. As you cansee, the best decision is to have Leadville Barts make the hel-mets and have Progressive Products do the rest with an ex-pected value of $2,600. The final option of not using Progres-sive, however, was very close with an expected value of $2,500.

EXPECTED

POOR AVERAGE GOOD EXCELLENT VALUE


Option 1—PP �5,000 �2,000 2,000 5,000 700Option 2—LB and PP �10,000 �4,000 6,000 12,000 2,600Option 3—TR and PP �15,000 �10,000 7,000 13,000 900Option 4—CC and PP �30,000 �20,000 10,000 30,000 1,000Option 5—LB, CC, and TR �60,000 �35,000 20,000 55,000 2,500

With Perfect Information �5,000 �2,000 25,000 55,000 17,900

The maximum expected monetary value is 2,600 given byOption 2 � LB and PP.b and c. The opportunity loss and the expected value of per-fect information is presented below. The EVPI is $15,300.Expected value with perfect information � 17,900Expected monetary value � 2,600Expected value of perfect information � 15,300

Opportunity loss table

POOR MARKET AVERAGE GOOD EXCELLENT


Option 1—PP 0 0 18,000 50,000Option 2—LB and PP 5,000 2,000 14,000 43,000Option 3—TR and PP 10,000 8,000 13,000 42,000Option 4—CC and PP 25,000 18,000 10,000 25,000Option 5—LB, CC, and TR 55,000 33,000 0 0

d. Bob was logical in approaching this problem. However,there are other alternatives that might be considered. Onepossibility is to sell the idea and the rights to produce thisproduct to Progressive Products for a fixed amount.

STUDY TIME CASE

Raquel must decide which of the three cases (1, 2, or 3) to study, andhow much time to devote to each. We will assume that it is equallylikely (a 1/3 chance) that each case is chosen. If she misses at most 8points (let’s assume she is correct in thinking that) on the other partsof the exam, scoring 20 points or more on this part will give her an Afor the course. Scoring 0 or 12 points on this portion of the exam will



Case 1 Case 2 Case 3on Exam on Exam on Exam EV Grade in Course

Study 1, 2, 3 12 B 12 B 12 B 12 BStudy 1,2 20 A 20 A 0 B 40/3 A 2/3 chance or B 1/3 chanceStudy 1,3 20 A 0 B 20 A 40/3 A 2/3 chance or B 1/3 chanceStudy 2,3 0 B 20 A 20 A 40/3 A 2/3 chance or B 1/3 chanceStudy 1 25 A 0 B 0 B 25/3 A 1/3 chance or B 2/3 chanceStudy 2 0 B 25 A 0 B 25/3 A 1/3 chance or B 2/3 chanceStudy 3 0 B 0 B 25 A 25/3 A 1/3 chance or B 2/3 chance

Thus, Raquel should study 2 cases since this will give her a2/3 chance of an A in the course. Notice that this also has the high-est expected value. This is a situation in which the values (points)are not always indicative of the importance of the result since 0 or12 results in a B for the course, and 20 or 25 results in an A for thecourse.

result in a grade of B for the course. The table below gives the differ-ent possibilities – points and grade in the course.


chapter 03 answers(1)

Documents

importance of decision

decision type

important decisions

lifetime decisions

pessimistic decision

popularity of decision

pessimistic decision

clude decisions