Decision Trees Solution Render

SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS

3-1. The purpose of this question is to make students use a personal experience to distinguish

between good and bad decisions. A good decision is based on logic and all of the available

information. A bad decision is one that is not based on logic and the available information. It is

possible for an unfortunate or undesirable outcome to occur after a good decision has been made.

It is also possible to have a favorable or desirable outcome occur after a bad decision.

3-2. The decision-making process includes the following steps: (1) define the problem, (2) list

the alternatives, (3) identify the possible outcomes, (4) evaluate the consequences, (5) select an

evaluation criterion, and (6) make the appropriate decision. The first four steps 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 have complete control. A state of nature

is an event or occurrence in which we have no control. An example of an alternative is deciding

whether or not to take an umbrella to school or work on a particular day. An example of a state

of nature is whether or not it will rain on a particular day.

3-4. The basic differences between decision-making models under certainty, risk, and

uncertainty depend on the amount of chance or risk that is involved in the decision. A decision-

making model under certainty assumes that we know with complete confidence the future

outcomes. Decision-making-under-risk models assume that we do not know the outcomes for a

particular decision but that we do know the probability of occurrence of those outcomes. With

decision making under uncertainty, it is assumed that we do not know the outcomes that will

occur, and furthermore, we do not know the probabilities that these outcomes will occur.

3-5. The techniques discussed in this chapter used to solve decision problems under uncertainty

include maximax, maximin, equally likely, coefficient of realism, and minimax regret. The

maximax decision-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 difference between the payoff for a

decision and the best possible payoff for that state of nature. It indicates how much better the

payoff could have been for that state of nature. The minimax regret and the minimum expected

opportunity loss are the criteria used with this.

3-7. Alternatives, states of nature, probabilities for all states of nature and all monetary

outcomes (payoffs) are placed on the decision tree. In addition, intermediate results, such as

EMVs for middle branches, can be placed on the decision tree.

3-8. Using the EMV criterion with a decision tree involves starting at the terminal branches of

the tree and working toward the origin, computing expected monetary values and selecting the

best alternatives. The EMVs are found by multiplying the probabilities of the states of nature by

the economic consequences and summing the results for each alternative. At each decision point,

the best alternative is selected.

3-9. A prior probability is one that exists before additional information is gathered. A posterior

probability is one that can be computed using Bayes’ Theorem based on prior probabilities and

additional information.

3-10. The purpose of Bayesian analysis is to determine posterior probabilities based on prior

probabilities and new information. Bayesian analysis can be used in the decision-making process

whenever additional information is gathered. This information can then be combined with prior

probabilities in arriving at posterior probabilities. Once these posterior probabilities are

computed, they can be used in the decision-making process as any other probability value.

3-11. The expected value of sample information (EVSI) is the increase in expected value that

results from having sample information. It is computed as follows:

EVSI = (expected value with sample information)

+ (cost of information) – (expected value without

sample information)

3-12. The expect value of sample information (EVSI) and the expected value of perfect

information (EVPI) are calculated. The ratio EVSI/EVPI is calculated and multiplied by 100%

to get the efficiency of sample information.

3-13. The overall purpose of utility theory is to incorporate a decision maker’s preference for

risk in the decision-making process.

3-14. A utility function can be assessed in a number of different ways. A common way is to use

a standard gamble. With a standard gamble, the best outcome is assigned a utility of 1, and the

worst outcome is assigned a utility of 0. Then, intermediate outcomes are selected and the

decision maker is given a choice between having the intermediate outcome for sure and a gamble

involving the best and worst outcomes. The probability that makes the decision maker indifferent

between having the intermediate outcome for sure and a gamble involving the best and worst

outcomes is determined. This probability then becomes the utility of the intermediate value. This

process is continued until utility values for all economic consequences are determined. These

utility values are then placed on a utility curve.

3-15. When a utility curve is to be used in the decision-making process, utility values from the

utility curve replace all monetary values at the terminal branches in a decision tree or in the body

of a decision table. Then, expected utilities are determined in the same way as expected

monetary values. The alternative with the highest expected utility is selected as the best decision.

3-16. A risk seeker is a decision maker who enjoys and seeks out risk. A risk avoider is a

decision maker who avoids risk even if the potential economic payoff is higher. The utility curve

for a risk seeker increases at an increasing rate. The utility curve for a risk avoider increases at a

decreasing rate.

3-17. a. Decision making under uncertainty.

b. Maximax criterion.

c. Sub 100 because the maximum payoff for this is $300,000.

Row Row

Equipment Favorable Unfavorable Maximum Minimum

Sub 100 300,000 –200,000 300,000 –200,000

Oiler J 250,000 –100,000 250,000 –100,000

Texan 75,000 –18,000 75,000 –18,000

3-18. Using the maximin criterion, the best alternative is the Texan (see table above) because the

worst payoff for this ($–18,000) is better than the worst payoffs for the other decisions.

3-19. a. Decision making under risk—maximize expected monetary value.

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 less than 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.

3-20. a. The expected value (EV) is computed for each alternative.

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.

3-21. The opportunity loss table is

Alternative Good Economy Poor Economy

Stock Market 0 43,000

Bonds 50,000 3,000

CDs 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

3-22. a.

Market

Alternative Condition Good Fair Poor EMV

Stock market 1,400 800 0 880

Bank deposit (CD) 900 900 900 900

Probabilities of market

conditions

0.4 0.4 0.2

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

3-23. a. Expected value with perfect information is 1,400(0.4) + 900(0.4) + 900(0.2) = 1,100;

the maximum 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.

3-24. a. Opportunity loss table

Strong Fair Poor Max.

Market Market Market Regret

Large 0 19,000 310,000 310,000

Medium 250,000 0 100,000 250,000

Small 350,000 29,000 32,000 350,000

None 550,000 129,000 0 550,000

b. Minimax regret decision is to build medium.

3-25. a.

Stock Demand

(Cases) (Cases) 11 12 13 EMV

11 385 385 385 385

12 329 420 420 379.05

13 273 364 455 341.25

Probabilities 0.45 0.35 0.20

b. Stock 11 cases.

c. If no loss is involved in excess stock, the recommended course of action is to stock 13

cases and to replenish stock to this level each week. This follows from the following

decision table.

Stock Demand

(Cases) (Cases) 11 12 13 EMV

11 385 385 385 385

12 385 420 420 404.25

13 385 420 455 411.25

3-26.

Manufacture

(Cases)

Demand

(Cases) 6 7 8 9 EMV

6 300 300 300 300 300

7 255 350 350 350 340.5

8 210 305 400 400 352.5

9 165 260 355 450 317

Probabilities 0.1 0.3 0.5 0.1

John should manufacture 8 cases of cheese spread.

3-27. Cost of produced case = $5.

Cost of purchased case = $16.

Selling price = $15.

Money recovered from each unsold case = $3.

Supply Demand(Cases)

(Cases) 100 200 300 EMV

100 100(15) –100(5) =

1000

200(15) – 100(5) –100(16)

= 900

300(15) – 100(5) –200(16)

= 800

900

200 100(15) + 100(3) –

200(5) = 800

200(15) – 200(5) = 2000 300(15) – 200(5) –100(16)

= 1900

1610

300 100(15) + 200(3) –

300(5) = 600

200(15) + 100(3) –300(5)

= 1800

300(15) – 300(5) = 3000 1800

Prob. 0.3 0.4 0.3

b. Produce 300 cases each day.

3-28. a. The table presented is a decision table. The basis for the decisions in the following

questions is shown in the table. The values in the table are in 1,000s.

Decision

MARKET

MAXIMAX

MAXIMIN

EQUALLY

LIKELY

CRIT. OF

REALISM

Alternatives

Good Fair Poor Row

Max.

Row

Min.

Row Ave. Weighted

Ave.

Small 50 20 –10 50 –10 20 38

Medium 80 30 –20 80 –20 30 60

Large 100 30 –40 100 –40 30 72

Very Large 300 25 –160 300 –160 55 208

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 (values in the table are in 1,000s):

MARKET MINIMAX

Decision Good Fair Poor Row

Alternatives Market Market Market Maximum

Small 250 10 0 250

Medium 220 0 10 220

Large 200 0 30 200

Very Large 0 5 150 150

3-29. Note this problem is based on costs, so the minimum values are the best.

a. For a 3-year lease, there are 36 months of payments.

Option 1 total monthly payments: 36($330) = $11,880



Excess mileage costs based on 36,000 mileage allowance for Option 1, 45,000 for Option 2, and

54,000 for option 3.

Option 1 excess mileage cost if 45,000 miles are driven = (45000 – 36000)(0.35) = 3150



The total cost for each option in each state of nature is obtained by adding the total monthly

payment cost to the excess mileage cost.

Total cost table

Lease option 36000 miles driven 45000 miles driven 54000 miles driven

Option 1 11,880 15030 18180

Option 2 13,680 13680 15930

Option 3 15,480 15480 15480

b. Optimistic decision: Option 1 because the best (minimum) payoff (cost) for this is 11,800

which is better (lower) than the best payoff for each of the others.

c. Pessimistic decision: Option 3 because the worst (maximum) payoff (cost) for this is 15,480 is

better (lower) than the worst payoff for each of the others.

d. Select Option 2.

EMV(Option 1) = 11,880(0.4) + 15,030(0.3) + 18,180(0.3) = 14,715

EMV(Option 2) = 13,680(0.4) + 13,680(0.3) + 15,930(0.3) = 14,355

EMV(Option 3) = 15,480(0.4) + 15,480 (0.3) + 15,480(0.3) = 15,480

(e) EVPI for a minimization problem = (Best EMV without PI) - (EV with PI)

EV with PI = 11,880(0.4) + 13,680(0.3) + 15,480(0.3) = 13,500

EVPI = 14,355 – 13,500 = 855

3-30. Note that this is a minimization problem, so the opportunity loss is based on the lowest

(best) cost in each state of nature.

Opportunity loss table

Lease option 36000 miles driven 45000 miles driven 54000 miles driven

Option 1 11880 – 11880 = 0 15030 - 13680 = 1350 18180 - 15480 = 2700

Option 2 13680 – 11880 = 1800 13680 - 13680 = 0 15930 - 15480 = 450

Option 3 15480 – 11880 = 3600 15480 - 13680 = 1800 15480 - 15480 = 0

The maximum regrets are 2700 for option 1, 1800 for option 2, and 3600 for option 3. Option 2

is selected because 1800 is lower than the other maximums.

EOL(option 1) = 0(0.4) + 1350(0.3) + 2700(0.3) = 1215

EOL(option 2) = 1800(0.4) + 0(0.3) + 450(0.3) = 855

EOL(option 3) = 3600(0.4) + 1800(0.3) + 0(0.3) = 1980

Option 2 has the lowest EOL, so this alternative is selected based on the EOL criterion.

3-31. a. P(red) = 18/38; P(not red) = 20/38

b. EMV = Expected win = 10(18/38) + (-10)(20/38) = -0.526

c. P(red) = 18/37; P(not red) = 19/37

EMV = Expected win = 10(18/37) + (-10)(19/37) = -0.270

d. The enjoyment of playing the game and possibly winning adds utility to the game. A person

would play this game because the expected utility of playing the game is positive even though

the expected monetary value is negative.

3-32. A $10 bet on number 7 would pay 35($10) = 350 if the number 7 is the winner.

P(number 7) = 1/38; P(not seven) = 37/38

EMV = Expected win = 350(1/38) + (-10)(37/38) = -0.526

3-33. Payoff table with cost of $50,000 in legal fees deducted if suit goes to court and $10,000 in

legal fees if settle.

Win big Win small Lose EMV

Go to court 250000 0 -50000 85000

Settle 65000 65000 65000 65000

Prob. 0.4 0.3 0.3

Decision based on EMV: Go to court

3-34. EMV for node 1 = 0.5(100,000) + 0.5(–40,000) = $30,000. Choose the highest EMV,

therefore construct the clinic.

3-35. a.

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.

3-36.

3-37.

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 conduct the survey. Since EMV(large

shop | favorable survey) is larger than both EMV(small shop | favorable survey) and EMV(no shop

| favorable survey), Jerry should build a large shop if the survey is favorable. If the survey is

unfavorable, Jerry should build nothing since EMV(no shop | unfavorable survey) is larger than

both EMV(large shop | unfavorable survey) and EMV(small shop | unfavorable survey).

b. If no survey, EMV = 0.5(30,000) + 0.5(–10,000) = $10,000. To keep Jerry from

changing decisions, the following must be true:

EMV(survey) ≥ EMV(no survey)

Let P = probability of a favorable survey. Then,

P[EMV(favorable survey)] + (1 – P) [EMV(unfavorable survey)] ≥ EMV(no survey)

This becomes:

P(45,000) + (1 – P)(–5,000) ≥ $10,000

Solving gives

45,000P + 5,000 – 5,000P ≥ 10,000

50,000P ≥ 15,000

P ≥ 0.3

Thus, the probability of a favorable survey could be as low as 0.3. Since the marketing

professor estimated the probability at 0.6, the value can decrease by 0.3 without causing Jerry

to

change his decision. Jerry’s decision is not very sensitive to this probability value.

3-38.

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.

3-39. 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)

1 1 1

1 1

1 1 1 1 2 2

1 1

0.8 0.50.73

0.8 0.5 0.3 0.5

P I S P SP S I

P I S P S P I S P S

P S I

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

2 1 1

1 2

2 1 1 2 2 2

1 2

0.2 0.50.22

0.2 0.5 0.7 0.5

P I S P SP S I

P I S P S P I S P S

P S I

c. Now P(S1) = 0.6 and P(S2) = 0.4

1 1

1 2

0.8 0.60.8

0.8 0.6 0.3 0.4

0.2 0.60.3

0.2 0.6 0.7 0.4

P S I

P S I

3-40. 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)

1 1 1

1 1

1 1 1 1 2 2

1 1

0.8 0.30.533

0.8 0.3 0.3 0.7

P I S P SP S I

P I S P S P I S P S

P S I

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

2 1 1

1 2

2 1 1 2 2 2

1 2

0.2 0.30.109

0.2 0.3 0.7 0.7

P I S P SP S I

P I S P S P I S P S

P S I

3-41. a.

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 = 21,500

Therefore, the return would have to be $21,500 for Fund A in a good economy for the two

alternatives to be equally desirable based on the expected values.

3-42. a.

b.

S1: survey favorable

S2: survey unfavorable

S3: study favorable

S4: study unfavorable

S5: market favorable

S6: market unfavorable

5 1

0.7 0.50.78

0.7 0.5 0.2 0.5P S S

P(S6 | S1) = 1 – 0.778 = 0.222

5 2

0.3 0.50.27

0.3 0.5 0.8 0.5P S S

P(S6 | S2) = 1 – 0.27 = 0.73

5 3

0.8 0.50.89

0.8 0.5 0.1 0.5P S S

P(S6 | S3) = 1 – 0.89 = 0.11

5 4

0.2 0.50.18

0.2 0.5 0.9 0.5P S S

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

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 favorable, produce the razor. If it is

unfavorable, do not produce the razor.

3-43. The following computations are for the decision tree that follows.

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 risk avoider.

3-44. a. P(good economy | prediction of

good economy) =

0.8 0.60.923

0.8 0.6 0.1 0.4

P(poor economy | prediction of

good economy) =

0.1 0.40.077

0.8 0.6 0.1 0.4

P(good economy | prediction of

poor economy) =

0.2 0.60.25

0.2 0.6 0.9 0.4


poor economy) =

0.9 0.60.75

0.2 0.6 0.9 0.4

b. P(good economy | prediction of

good economy) =

0.8 0.70.949

0.8 0.7 0.1 0.3


good economy) =

0.1 0.30.051

0.8 0.7 0.1 0.3

P(good economy | prediction of

poor economy) =

0.2 0.70.341

0.2 0.7 0.9 0.3


poor economy) =

0.9 0.30.659

0.2 0.7 0.9 0.3

3-45. The expected value of the payout by the insurance company is

EV = 0(0.999) + 100,000(0.001) = 100

The expected payout by the insurance company is $100, but the policy costs $200, so the net

gain for the individual buying this policy is negative (–$100). Thus, buying the policy does not

maximize EMV since not buying this policy would have an EMV of 0, which is better than a

negative $100. However, a person who buys this policy would be maximizing the expected

utility. The peace of mind that goes along with the insurance policy has a relatively high utility.

A person who buys insurance would be a risk avoider.

3-46.

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 expected utility with a survey

(0.7615), so the survey should be not used. The medical professionals are risk avoiders.

3-47. 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 survey and build the large plant.

3-48. a. Expected travel time on Broad Street = 40(0.5) + 15(0.5) = 27.5 minutes. Broad Street

has a lower expected travel time.

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

maximizes expected utility.

c. Lynn is a risk avoider.

3-49. Selling price = $20 per gallon; manufacturing cost = $12 per gallon; salvage value = $13;

handling costs = $1 per gallon; and advertising costs = $3 per gallon. From this information, 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 fulfill any demand that cannot be

met internally. This requires that Coren purchase chemicals from an outside company. Because

the cost of obtaining the chemical from the outside company is $25 and the price 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 that has to be purchased from an

outside company due to a shortage.

a. A decision tree is provided:

b. The computations are shown in the following table. These numbers are entered into the tree

above. The best decision is to stock 1,500 gallons.

Table for Problem 3-49

Demand

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

500 2,000 –500 –3,000 –5,500 –$1,500

1,000 500 4,000 1,500 –1,000 $1,800

1,500 –1,000 2,500 6,000 3,500 $3,300

2,000 –2,500 1,000 4,500 8,000 $2,400

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

Probabilities 0.2 0.3 0.4 0.1

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 = $1,500

3-50. If no survey is to be conducted, the decision tree is fairly straightforward. There are three

main decisions, which are building a small, medium, or large facility. Extending from these

decision branches are three possible demands, representing the possible states of nature. The

demand for this type of facility could be either low (L), medium (M), or high (H). It was given in

the problem that the probability for a low demand is 0.15. The probabilities for a medium and a

high demand are 0.40 and 0.45, respectively. The problem also gave monetary consequences for

building a small, medium, or large facility when the demand could be low, medium, or high for

the facility. These data are reflected in the following decision tree.

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 expected monetary value.

If the survey is used, we must compute the revised probabilities using Bayes’ theorem. For

each alternative facility, three revised probabilities must be computed, representing low,

medium, and high demand for a facility. These probabilities can be computed using tables. One

table is used to compute the probabilities for low survey results, another table is used for medium

survey results, and a final table is used for high survey results. These probabilities will be used in

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 0.339

B2 0.400 0.400 0.160 0.516

B3 0.450 0.100 0.045 0.145

P(A1) = 0.310

For medium survey results—A2:


B1 0.150 0.200 0.030 0.082

B2 0.400 0.500 0.200 0.548

B3 0.450 0.300 0.135 0.370

P(A2) = 0.365

For high survey results—A3:


B1 0.150 0.100 0.015 0.046

B2 0.400 0.100 0.040 0.123

B3 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 in EMV(Small) = 450,000; EMV(Medium) = 495,000; and EMV(Large) =

233,600.

When survey results are medium, the probabilities are P(L) = 0.082; P(M) = 0.548; and P(H)

= 0.370. 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 in EMV(Small) = 450,000; EMV(Medium) = 710,100; and EMV(Large) =

821,000.

If the survey results are low, the best decision is to build the medium facility with an

expected return of $495,000. If the survey results are medium, the best decision is also to build

the medium plant with an expected return of $646,000. On the other hand, if the survey results

are high, the best decision is to build the large facility with an expected monetary value of

$821,000. The expected 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 survey is greater (670,000), the

decision is not to conduct the survey and to build the medium-sized facility.

3-51. 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:

0.7 0.60.84

0.7 0.6 0.2 0.4

0.3 0.750.53

0.3 0.75 0.8 0.25

P SRP SM P SMP SM SRP

P SRP SM P SM P SRP SM P SM

P SM SRP

P SC SRN

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

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.

3-52. a. Sue can use decision tree analysis to find the best solution. In this case, the best

decision is to get information. If the information is favorable, she should build the retail store. If

the information is not favorable, she should not 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 get information, Branch 3 (2–4) is favorable information,

Branch 4 (2–5) is unfavorable information, Branch 5 (3–8) is the decision to build the retail store

and get no information, Branch 6 (3–17) is the decision to not build the retail store and to get no

information, Branch 7 (4–6) is the decision to build the retail store given favorable information,

Branch 8 (4–11) is the decision to not build given favorable information, Branch 9 (6–9) is a

successful retail store given favorable information, Branch 10 (6–10) is an unsuccessful retail

store given favorable information, Branch 11 (5–7) is the decision to build the retail store given

unfavorable information, Branch 12 (5–14) is the decision not to build the retail store given

unfavorable information, Branch 13 (7–12) is a successful retail store given unfavorable

information, Branch 14 (7–13) is an unsuccessful retail store given unfavorable information,

Branch 15 (8–15) is a successful retail store given that no information is obtained, and Branch 16

(8–16) is an unsuccessful retail store given no information is obtained.

Results for 3-52. a.

Start Ending Branch Profit Use Node Node

Node Node Prob. (End Node) Branch? Type Value

Start 0 1 0 0 Dec 29,200

Branch 1 1 2 0 0 Yes Ch 29,200

Branch 2 1 3 0 0 Dec 28,000

Branch 3 2 4 0.6 0 Dec 62,000

Branch 4 2 5 0.4 0 Dec –20,000

Branch 5 3 8 0 0 Yes Ch 28,000

Branch 6 3 17 0 0 Fin 0

Branch 7 4 6 0 0 Yes Ch 62,000

Branch 8 4 11 0 –20,000 Fin –20,000

Branch 9 6 9 0.9 80,000 Fin 80,000

Branch 10 6 10 0.1 –100,000 Fin –100,000

Branch 11 5 7 0 0 Ch –64,000

Branch 12 5 14 0 –20,000 Yes Fin –20,000

Branch 13 7 12 0.2 80,000 Fin 80,000

Branch 14 7 13 0.8 –100,000 Fin –100,000

Branch 15 8 15 0.6 100,000 Fin 100,000

Branch 16 8 16 0.4 –80,000 Fin –80,000

b. The suggested changes would be reflected in Branches 3 and 4. The decision stays the same,

but the EMV increases to $37,400. The results are provided in the tables that follow. In these

tables, BR = Branch; Prob. = Probability; and for Node Type, Dec = Decision, Ch = Chance, and

Fin = Final.

Results for 3-52. b.



Start 0 1 0 0 Dec 37,400

Branch 1 1 2 0 0 Yes Ch 37,400

Branch 2 1 3 0 0 Dec 28,000

Branch 3 2 4 0.7 0 Dec 62,000

Branch 4 2 5 0.3 0 Dec –20,000

Branch 5 3 8 0 0 Yes Ch 28,000

Branch 6 3 17 0 0 Fin 0

Branch 7 4 6 0 0 Yes Ch 62,000

Branch 8 4 11 0 –20,000 Fin –20,000

Branch 9 6 9 0.9 80,000 Fin 80,000

Branch 10 6 10 0.1 –100,000 Fin –100,000

Branch 11 5 7 0 0 Ch –64,000

Branch 12 5 14 0 –20,000 Yes Fin –20,000

Branch 13 7 12 0.2 80,000 Fin 80,000

Branch 14 7 13 0.8 –100,000 Fin –100,000

Branch 15 8 15 0.6 100,000 Fin 100,000

Branch 16 8 16 0.4 –80,000 Fin –80,000

c. Sue can determine the impact of the change by changing the probabilities and recomputing

EMVs. This analysis shows the decision changes. Given the new probability values, Sue’s best

decision is build the retail store without getting additional information. The EMV for this

decision is $28,000. The results are presented below:

Results for 3-52. c.



Start 0 1 0 0 Dec 28,000

Branch 1 1 2 0 0 Ch 18,400

Branch 2 1 3 0 0 Yes Dec 28,000

Branch 3 2 4 0.6 0 Dec 44,000

Branch 4 2 5 0.4 0 Dec –20,000

Branch 5 3 8 0 0 Yes Ch 28,000

Branch 6 3 17 0 0 Fin 0

Branch 7 4 6 0 0 Yes Ch 44,000

Branch 8 4 11 0 –20,000 Fin –20,000

Branch 9 6 9 0.8 80,000 Fin 80,000

Branch 10 6 10 0.2 –100,000 Fin –100,000

Branch 11 5 7 0 0 Ch –64,000

Branch 12 5 14 0 –20,000 Yes Fin –20,000

Branch 13 7 12 0.2 80,000 Fin 80,000

Branch 14 7 13 0.8 –100,000 Fin –100,000

Branch 15 8 15 0.6 100,000 Fin 100,000

Branch 16 8 16 0.4 –80,000 Fin –80,000

d. Yes, Sue’s decision would change from her original decision. With the higher cost of

information, Sue’s decision is to not get the information and build the retail store. The EMV of

this decision is $28,000. The results are given below:

Results for 35-2. d.


Node Node Probability (End Node) Branch? Type Value

Start 0 1 0 0 Decision 28,000

Branch 1 1 2 0 0 Chance 19,200

Branch 2 1 3 0 0 Yes Decision 28,000

Branch 3 2 4 0.6 0 Decision 52,000

Branch 4 2 5 0.4 0 Decision –30,000

Branch 5 3 8 0 0 Yes Chance 28,000

Branch 6 3 17 0 0 Final 0

Branch 7 4 6 0 0 Yes Chance 52,000

Branch 8 4 11 0 –30,000 Final –30,000

Branch 9 6 9 0.9 70,000 Final 70,000

Branch 10 6 10 0.1 –110,000 Final –110,000

Branch 11 5 7 0 0 Chance –74,000

Branch 12 5 14 0 –30,000 Yes Final –30,000

Branch 13 7 12 0.2 70,000 Final 70,000

Branch 14 7 13 0.8 –110,000 Final –110,000

Branch 15 8 15 0.6 100,000 Final 100,000

Branch 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 problem, the expected utility is 0.62. The utility table represents a

risk seeker. The results are given below.

Results for 3-52. e.

Start Ending Branch Profit Use Ending Node Node

Node Node Prob. (End Node) Branch? Node Type Value

Start 0 1 0 0 1 Dec 0.62

Branch 1 1 2 0 0 2 Ch 0.256

Branch 2 1 3 0 0 Yes 3 Dec 0.62

Branch 3 2 4 0.6 0 4 Dec 0.36

Branch 4 2 5 0.4 0 5 Dec 0.1

Branch 5 3 8 0 0 Yes 8 Ch 0.62

Branch 6 3 17 0 0.2 17 Fin 0.20

Branch 7 4 6 0 0 Yes 6 Ch 0.36

Branch 8 4 11 0 0.1 11 Fin 0.1

Branch 9 6 9 0.9 0.4 9 Fin 0.4

Branch 10 6 10 0.1 0 10 Fin 0

Branch 11 5 7 0 0 7 Ch 0.08

Branch 12 5 14 0 0.1 Yes 14 Fin 0.1

Branch 13 7 12 0.2 0.4 12 Fin 0.4

Branch 14 7 13 0.8 0 13 Fin 0

Branch 15 8 15 0.6 1 15 Fin 1

Branch 16 8 16 0.4 0.05 16 Fin 0.05

f. This problem can be solved by replacing monetary values with utility values. The expected

utility is 0.80. The utility table given in the problem is representative of a risk avoider. The

results are presented below:

Results for 3-52. f.



Start 0 1 0 0 Dec 0.80

Branch 1 1 2 0 0 Ch 0.726

Branch 2 1 3 0 0 Yes Dec 0.80

Branch 3 2 4 0.6 0 Dec 0.81

Branch 4 2 5 0.4 0 Dec 0.60

Branch 5 3 8 0 0 Yes Ch 0.76

Branch 6 3 17 0 0.8 Fin 0.80

Branch 7 4 6 0 0 Yes Ch 0.81

Branch 8 4 11 0 0.6 Fin 0.60

Branch 9 6 9 0.9 0.9 Fin 0.90

Branch 10 6 10 0.1 0 Fin 0.00

Branch 11 5 7 0 0 Ch 0.18

Branch 12 5 14 0 0.6 Yes Fin 0.60

Branch 13 7 12 0.2 0.9 Fin 0.90

Branch 14 7 13 0.8 0 Fin 0.00

Branch 15 8 15 0.6 1 Fin 1.00

Branch 16 8 16 0.4 0.4 Fin 0.40

3-53. a. The decision table for Chris Dunphy along with the expected profits or expected

monetary values (EMVs) for each alternative are shown on the next page.

Table for Problem 3-53a

Return in $1,000

NO. OF WATCHES

EVENT 1

EVENT 2

EVENT 3

EVENT 4

EVENT 5

Probability 0.10 0.20 0.50 0.10 0.10 Expected

Profit

100,000 100 110 120 135 140 119.5

150,000 90 120 140 155 170 135.5

200,000 85 110 135 160 175 131.5

250,000 80 120 155 170 180 144.5

300,000 65 100 155 180 195 141.5

350,000 50 100 160 190 210 145

400,000 45 95 170 200 230 151.5

450,000 30 90 165 230 245 151

500,000 20 85 160 270 295 155.5

b. For this decision problem, Alternative 9, stocking 500,000, gives the highest expected

profit of $155,500.

c. The expected value with perfect information is $175,500, and the expected value of perfect

information (EVPI) is $20,000.

d. The new probability estimates will give more emphasis to event 2 and less to event 5. The

overall impact is shown below. As you can see, stocking 400,000 watches is now the best

decision with an expected value of $140,700.

Return in $1,000:

NO. OF WATCHES

EVENT 1

EVENT 2

EVENT 3

EVENT 4

EVENT 5

Probability 0.100 0.280 0.500 0.100 0.020 Expected Profit

100,000 100 110 120 135 140 117.1

150,000 90 120 140 155 170 131.5

200,000 85 110 135 160 175 126.3

250,000 80 120 155 170 180 139.7

300,000 65 100 155 180 195 133.9

350,000 50 100 160 190 210 136.2

400,000 45 95 170 200 230 140.7

450,000 30 90 165 230 245 138.6

500,000 20 85 160 270 295 138.7

3-54. a. Decision under uncertainty.

b.

Population Population Row

Same Grows Average

Large wing –85,000 150,000 32,500

Small wing –45,000 60,000 7,500

No wing 0 0 0

c. Best alternative: large wing.

3-55. a.

Weighted

Population Population Average with

Same Grows = 0.75

Large wing –85,000 150,000 91,250

Small wing –45,000 60,000 33,750

No wing 0 0 0

b. Best decision: large wing.

c. No.

3-56. a.

No Mild Severe Expected

Congestion Congestion Congestion Time

Tennessee 15 30 45 25

Back roads 20 25 35 24.17

Expressway 30 30 30 30

Probabilities (30 days)/(60

days) = 1/2

(20 days)/(60

days) = 1/3

(10 days)/(60

days) = 1/6

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 3 1

3; minutes.

3-57. a. EMV can be used to determine the best strategy to minimize costs. The QM for

Windows solution is provided. The best decision is to go with the partial service

(maintenance) agreement.

Solution to 3-57a

Probabilities 0.2 0.8

Maint. No Maint. Expected Row Row

Cost ($) Cost ($) Value Minimum Maximum

($) ($) ($)

No Service Agreement 3,000 0 600 0 3,000

Partial Service Agreement 1,500 300 540 0 1,500

Complete Service

Agreement

500 500 500 500 500

Column best 500 0 500

The minimum expected monetary value is $500 given by Complete Service Agreement

b. The new probability estimates dramatically change Sim’s expected values (costs). The

best decision given this new information is to still go with the complete service or

maintenance policy with an expected cost of $500. The results are shown in the table.

Solution to 3-57b

Probabilities 0.8 0.2

Does Not Expected

Needs Repair Need Repair Value

($) ($) ($)

No Service Agreement 3,000 0 2,400

Partial Service Agreement 1,500 300 1,260

Complete Service

Agreement

500 500 500

Column best 500

3-58. We can use QM for Windows to solve this decision making under uncertainty problem.

We have shown probability values for the equally likely calculations. As you can see, the

maximax decision is Option 4 based on the $30,000, and the maximin decision is Option 1 based

on the 5,000. As seen in the table, the equally likely decision is Option 3 because the average

value for this is $5750.

Solution to 3-58

Prob. 0.25 0.25 0.25 0.25

Judge Trial Court Arbitration Equally Row Row

Likely Min. Max.

Option 1 5,000 5,000 5,000 5,000 5,000 5,000 5,000

Option 2 10,000 5,000 2,000 0 4,250 0 10,000

Option 3 20,000 7,000 1,000 –5,000 5,750 –5,000 20,000

Option 4 30,000 15,000 –10,000 –20,000 3,750 –20,000 30,000

Column best 5,750 5,000 30,000

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