CHAPTER 3
Decision Analysis
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 decisionmaking 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 decisionmaking 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 $300,000.
Equipment
Sub 100
Oiler J
Texan
Favorable
300,000
250,000
75,000
Unfavorable
Row
Row
Maximum
Minimum
300,000
250,000
75,000
–200,000
–100,000
–200,000
–100,000
–18,000
–18,000
3-18. Using the maximin criterion, the best alternative is the Texan (see table above) because the
worst payoff for this ($–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
Stock Market
Poor Economy
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
Good
Fair
Poor
EMV
Stock market
1,400
800
0
880
Bank deposit (CD)
900
900
900
900
0.4
0.4
0.2
Probabilities
conditions
Condition
of
market
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
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
Large
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
100
100(15) –100(5) = 200(15) – 100(5) –100(16) 300(15) – 100(5) –200(16)
1000
= 900
= 800
200
100(15) + 100(3) – 200(15) – 200(5) = 2000
200(5) = 800
300
100(15) + 200(3) – 200(15) + 100(3) –300(5) 300(15) – 300(5) = 3000
300(5) = 600
= 1800
Prob.
0.3
0.4
b. Produce 300 cases each day.
300(15) – 200(5) –100(16)
= 1900
0.3
EMV
900
1610
1800
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.
MARKET
Decision
Good
Fair
Poor
Alternatives
EQUALLY
LIKELY
OF
CRIT.
REALISM
MAXIMAX
MAXIMIN
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
Decision
Good
Alternatives Market
MINIMAX
Fair
Poor
Row
Market
Market
Maximum
Small
250
10
0
250
Medium
220
0
10
220
Large
200
0
30
200
0
5
150
150
Very Large
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
Option 2 total monthly payments: 36($380) = $13,680
Option 3 total monthly payments: 36($430) = $15,480
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
Option 1 excess mileage cost if 54,000 miles are driven = (54000 – 36000)(0.35) = 6300
Option 2 excess mileage cost if 54,000 miles are driven = (54000 – 45000)(0.25) = 2250
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
50,000P
P
10,000
15,000
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)
P ( S1 I1 ) =
P ( S1 I1 ) =
P ( I1 S1 ) P ( S1 )
P ( I1 S1 ) P ( S1 ) + P ( I1 S 2 ) P ( S 2 )
0.8 ( 0.5 )
= 0.73
0.8 ( 0.5 ) + 0.3 ( 0.5 )
b. P(successful store | unfavorable research) = P(S1 | I2)
P ( S1 I 2 ) =
P ( S1 I 2 ) =
P ( I 2 S1 ) P ( S1 )
P ( I 2 S1 ) P ( S1 ) + P ( I 2 S 2 ) P ( S 2 )
0.2 ( 0.5 )
= 0.22
0.2 ( 0.5 ) + 0.7 ( 0.5 )
c. Now P(S1) = 0.6 and P(S2) = 0.4
P ( S1 I1 ) =
0.8 ( 0.6 )
= 0.8
0.8 ( 0.6 ) + 0.3 ( 0.4 )
P ( S1 I 2 ) =
0.2 ( 0.6 )
= 0.3
0.2 ( 0.6 ) + 0.7 ( 0.4 )
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)
P ( S1 I1 ) =
P ( S1 I1 ) =
P ( I1 S1 ) P ( S1 )
P ( I1 S1 ) P ( S1 ) + P ( I1 S 2 ) P ( S 2 )
0.8 ( 0.3)
= 0.533
0.8 ( 0.3) + 0.3 ( 0.7 )
P(successful facility | unfavorable survey) = P(S1 | I2)
P ( S1 I 2 ) =
P ( S1 I 2 ) =
P ( I 2 S1 ) P ( S1 )
P ( I 2 S1 ) P ( S1 ) + P ( I 2 S 2 ) P ( S 2 )
0.2 ( 0.3)
= 0.109
0.2 ( 0.3) + 0.7 ( 0.7 )
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
P ( S5 S1 ) =
0.7 ( 0.5 )
0.7 ( 0.5 ) + 0.2 ( 0.5 )
= 0.78
P(S6 | S1) = 1 – 0.778 = 0.222
P ( S5 S 2 ) =
0.3 ( 0.5 )
0.3 ( 0.5 ) + 0.8 ( 0.5 )
= 0.27
P(S6 | S2) = 1 – 0.27 = 0.73
P ( S5 S 3 ) =
0.8 ( 0.5 )
0.8 ( 0.5 ) + 0.1( 0.5 )
P(S6 | S3) = 1 – 0.89 = 0.11
= 0.89
P ( S5 S 4 ) =
0.2 ( 0.5 )
0.2 ( 0.5 ) + 0.9 ( 0.5 )
= 0.18
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 ( 0.6 )
0.8 ( 0.6 ) + 0.1( 0.4 )
= 0.923
P(poor economy | prediction of
good economy) =
0.1( 0.4 )
0.8 ( 0.6 ) + 0.1( 0.4 )
= 0.077
P(good economy | prediction of
poor economy) =
0.2 ( 0.6 )
0.2 ( 0.6 ) + 0.9 ( 0.4 )
= 0.25
P(poor economy | prediction of
poor economy) =
0.9 ( 0.6 )
0.2 ( 0.6 ) + 0.9 ( 0.4 )
= 0.75
b. P(good economy | prediction of
0.8 ( 0.7 )
good economy) =
= 0.949
0.8 ( 0.7 ) + 0.1( 0.3)
P(poor economy | prediction of
good economy) =
0.1( 0.3)
0.8 ( 0.7 ) + 0.1( 0.3)
= 0.051
P(good economy | prediction of
poor economy) =
0.2 ( 0.7 )
0.2 ( 0.7 ) + 0.9 ( 0.3)
= 0.341
P(poor economy | prediction of
poor economy) =
0.9 ( 0.3)
0.2 ( 0.7 ) + 0.9 ( 0.3)
= 0.659
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
500
1,000
1,500
2,000
EMV
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
2,000
4,000
6,000
8,000
$4,800 = EVwPI
Maximum
Probabilities
0.2
0.3
0.4
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
0.1
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:
State of Nature
P(Bi)
P(Ai | Bj)
P(Bj and Ai)
P(Bj | Ai)
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:
State of Nature
P(Bi)
P(Ai | Bj)
P(Bj and Ai)
P(Bj | Ai)
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:
P ( SM SRP ) =
P ( SM SRP ) =
P ( SC SRN ) =
P ( SRP SM ) P ( SM )
(
) ( )
P ( SRP SM ) P ( SM ) + P SRP SM P SM
0.7 ( 0.6 )
= 0.84
0.7 ( 0.6 ) + 0.2 ( 0.4 )
0.3 ( 0.75 )
0.3 ( 0.75 ) + 0.8 ( 0.25 )
= 0.53
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
Node
Node
Node Node
(End Node) Branch? Type
Value
Prob.
Use
Start
0
1
0
0
Dec
29,200
Branch 1
1
2
0
0
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
Ch
28,000
Branch 6
3
17
0
0
Fin
0
Branch 7
4
6
0
0
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
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
Yes
Yes
Yes
Yes
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 Ending Branch
Profit
Node
Node
Node Node
(End Node) Branch? Type
Value
Prob.
Use
Start
0
1
0
0
Dec
37,400
Branch 1
1
2
0
0
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
Ch
28,000
Branch 6
3
17
0
0
Fin
0
Branch 7
4
6
0
0
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
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
Yes
Yes
Yes
Yes
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 Ending Branch
Profit
Node
Node
Node Node
(End Node) Branch? Type
Value
Prob.
Use
Start
0
1
0
0
Dec
28,000
Branch 1
1
2
0
0
Ch
18,400
Branch 2
1
3
0
0
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
Ch
28,000
Branch 6
3
17
0
0
Fin
0
Branch 7
4
6
0
0
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
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
Yes
Yes
Yes
Yes
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.
Start Ending Branch
Node Node
Profit
Use
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
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
Chance
28,000
Branch 6
3
17
0
0
Branch 7
4
6
0
0
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
Branch 12 5
14
0
–30,000
Branch 13 7
12
0.2
Branch 14 7
13
Branch 15 8
Branch 16 8
Yes
Yes
Final
Yes
Chance
0
52,000
Chance
–74,000
Final
–30,000
70,000
Final
70,000
0.8
–110,000
Final
–110,000
15
0.6
100,000
Final
100,000
16
0.4
–80,000
Final
–80,000
Yes
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
Node Node
Use
Ending Node
Prob.
(End Node) Branch? Node
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
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
8
Ch
0.62
Branch 6
3
17
0
0.2
17
Fin
0.20
Branch 7
4
6
0
0
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
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
Yes
Yes
Yes
Yes
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 Ending Branch
Profit
Use
Node
Node
Node Node
Prob.
(End Node) Branch? Type
Value
Start
0
1
0
0
Dec
0.80
Branch 1
1
2
0
0
Ch
0.726
Branch 2
1
3
0
0
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
Ch
0.76
Branch 6
3
17
0
0.8
Fin
0.80
Branch 7
4
6
0
0
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
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
Yes
Yes
Yes
Yes
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 $1,000:
NO. OF WATCHES
EVENT 1 EVENT 2 EVENT 3 EVENT 4 EVENT 5
Probability
0.100
0.280
0.500
0.100
0.020
Expected Profit
100,000
100
110
120
135
140
117.1
150,000
90
120
140
155
170
131.5
200,000
85
110
135
160
175
126.3
250,000
80
120
155
170
180
139.7
300,000
65
100
155
180
195
133.9
350,000
50
100
160
190
210
136.2
400,000
45
95
170
200
230
140.7
450,000
30
90
165
230
245
138.6
500,000
20
85
160
270
295
138.7
Population
Population
Row
Same
Grows
Average
3-54. a. Decision under uncertainty.
b.
Large wing
–85,000
150,000
32,500
Small wing
–45,000
60,000
7,500
0
0
0
No wing
c. Best alternative: large wing.
3-55. a.
Weighted
Population
Population
Average with
Same
Grows
α = 0.75
Large wing
–85,000
150,000
91,250
Small wing
–45,000
60,000
33,750
0
0
0
No wing
b. Best decision: large wing.
c. No.
3-56. a.
No
Mild
Severe
Expected
Congestion
Congestion
Congestion
Time
Tennessee
15
30
45
25
Back roads
20
25
35
24.17
Expressway
30
30
30
30
Probabilities
(30
days)/(60 (20
days)/(60 (10
days)/(60
days) = 1/2
days) = 1/3
days) = 1/6
b. Back roads (minimum time used).
c. Expected time with perfect information: 15 × 1/2 + 25 × 1/3 + 30 × 1/6 = 20.83 minutes
Time saved is 3
1
3
; minutes.
3-57. a. EMV can be used to determine the best strategy to minimize costs. The QM for
Windows solution is provided. The best decision is to go with the partial service
(maintenance) agreement.
Solution to 3-57a
Probabilities
0.2
0.8
Maint.
No Maint.
Expected
Row
Row
Cost ($)
Cost ($)
Value
Minimum
Maximum
($)
($)
($)
No Service Agreement
3,000
0
600
0
3,000
Partial Service Agreement
1,500
300
540
0
1,500
500
500
500
500
500
500
0
500
Complete
Agreement
Column best
Service
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
500
500
500
Complete
Agreement
Service
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 Equall
y
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
5,750
5,000
30,000
Column best
SOLUTION TO STARTING RIGHT CASE
This is a decision-making-under-uncertainty case. There are two events: 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 in preferred stock
(alternative 3), and invest in common stock (alternative 4). The decision table is presented. Note
that for alternative 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.
Payoff table
Laplace
Event 1
Hurwicz
Event 2
Average Value Minimum Maximum Value
Alternative 1 0
0
0.0
0
0
0.00
Alternative 2 55,273
–10,000
22,636.5
–10,000
55,273
–2,819.97
Alternative 3 120,000
–15,000
52,500.0
–15,000
120,000
–150.00
Alternative 4 240,000
–30,000
105,000.0
–30,000
240,000
–300.00
Regret table
Maximum
Alternative
Event 1
Event 2
Regret
Alternative 1
240,000
0
240,000
Alternative 2
184,727
10,000
184,727
Alternative 3
120,000
15,000
120,000
Alternative 4
0
30,000
30,000
a. Sue Pansky is a risk avoider and should use the maximin decision approach. She should
do nothing and not make an investment 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 nothing and apply the maximin
criterion. The result is to invest in the corporate bonds.
d. George Yates should use the equally likely decision criterion. The best decision for
George is to invest in common stock.
e. Pete Metarko is a risk seeker. He should invest in common stock.
f. Julia Day can eliminate the preferred stock alternative and still offer alternatives to risk
seekers (common stock) and risk avoiders (doing nothing or investing in corporate bonds).
SOLUTIONS TO INTERNET CASES
Drink-at-Home, Inc. Case
Abbreviations and values used in the following decision trees:
Normal—proceed with research and development at a normal pace.
6 Month—Adopt the 6-month program: if a competitor’s product is available at the end of 6
months, then copy; otherwise proceed with research and development.
8 Month—Adopt the 6-month program: proceed for 8 months; if no competition at 8 months,
proceed; otherwise stop development.
Success or failure of development effort:
Ok—Development effort ultimately a success
No—Development effort ultimately a failure
Column:
S— Sales revenue
R— Research and development expenditures
E— Equipment costs
I—Introduction to market costs
Market size and Revenues:
Without
With
Competition
Competition
S—Substantial (P = 0.1)
$800,000
$400,000
M—Moderate (P = 0.6)
$600,000
$300,000
L—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)
The optimal program is to adopt the 6-month program. However, as the expected value is
negative, perhaps another alternative of doing nothing should be considered.
Ruth Jones’ Heart By-Pass Operation Case
1. Expected survival rate with surgery (5.95 years) exceeds the nonsurgical survival rate of
2.70 years. Surgery is normal.
Ski Right Case
a. Bob can solve this case using decision analysis. As you can see, the best decision is to
have Leadville Barts make the helmets and have Progressive Products do the rest with an
expected value of $2,600. The final option of not using Progressive, however, was very close
with an expected value of $2,500.
EXPECTE
D
POOR
AVERAGE
GOOD
EXCELLENT VALUE
Probabilities
0.1
0.3
0.4
0.2
Option 1—PP
–5,000
–2,000
2,000
5,000
700
Option 2—LB and PP
–10,000
–4,000
6,000
12,000
2,600
Option 3—TR, LB, and PP
–15,000
–10,000
7,000
13,000
900
Option 4—CC and PP
–30,000
–20,000
10,000
30,000
1,000
Option 5—LB, CC, and TR
–60,000
–35,000
20,000
55,000
2,500
With Perfect Information
–5,000
–2,000
20,000
55,000
17,900
The maximum expected monetary value is $2,600 given by Option 2 – LB and PP.
b and c. The opportunity loss and the expected value of perfect information is presented. The
EVPI is $15,300.
Expected value with perfect information = $17,900
Expected monetary value = $2,600
Expected value of perfect information = $15,300
Opportunity loss table
POOR MARKET
AVERAGE
GOOD
EXCELLENT
EXPECTED
VALUE
Probabilities
0.1
0.3
0.4
0.2
Option 1
0
0
18,000
50,000
17,200,
Option 2
5,000
2,000
14,000
43,000
15,300,
Option 3
10,000
8,000
13,000
42,000
17,000
Option 4
25,000
18,000
10,000
25,000
88,000
Option 5
55,000
33,000
0
0
15,400
d. Bob was logical in approaching this problem. However, there are other alternatives that
might be considered. One possibility is to sell the idea and the rights to produce this product
to Progressive Products for a fixed amount.
STUDY TIME CASE
Raquel must decide which of the three cases (1, 2, or 3) to study, and how much time to devote to
each. We will assume that it is equally likely (a 1/3 chance) that each case is chosen. If she
misses at most 8 points (let’s assume she is correct in thinking that) on the other parts of the
exam, scoring 20 points or more on this part will give her an A for the course. Scoring 0 or 12
points on this portion of the exam will result in a grade of B for the course. The table gives the
different possibilities – points and grade in the course.
Case 1
Case 2
Case 3
on
Exam
on Exam
on
Exam
EV
Grade in Course
Study 1, 2, 3
12 B
12 B
12 B
12
B
Study 1,2
20 A
20 A
0B
40/3
A 2/3 chance or B 1/3 chance
Study 1,3
20 A
0B
20 A
40/3
A 2/3 chance or B 1/3 chance
Study 2,3
0B
20 A
20 A
40/3
A 2/3 chance or B 1/3 chance
Study 1
25 A
0B
0B
25/3
A 1/3 chance or B 2/3 chance
Study 2
0B
25 A
0B
25/3
A 1/3 chance or B 2/3 chance
Study 3
0B
0B
25 A
25/3
A 1/3 chance or B 2/3 chance
Thus, Raquel should study 2 cases since this will give her a 2/3 chance of an A in the course.
Notice that this also has the highest expected value. This is a situation in which the values
(points) are not always indicative of the importance of the result since 0 or 12 results in a B for
the course, and 20 or 25 results in an A for the course.