# MDS TEXTBOOK ANS

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C H A P T E R
Decision Analysis
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 last
semester, such as buying a car or selecting an apartment, and describe the steps that they took. This will help in getting students involved in decision theory. It will also help them realize
how this material can be useful to them in making important personal decisions.
Teaching Suggestion 3.2: Importance of Defining the Problem
and Listing All Possible Alternatives.
Clearly defining the problem and listing the possible alternatives can
be difficult. Students can be asked to do this for a typical decisionmaking 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 approaches and do not pay enough attention to the importance of
carefully defining the problem and considering all possible alternatives. These initial steps are important. Students need to realize
that if they do not carefully define the problem and list all alternatives, most likely their analyses will be wrong.
Teaching Suggestion 3.6: Decision Theory
and Life-Time Decisions.
This chapter investigates large and complex decisions. During
one’s life, there are a few very important decisions that have a
major impact. Some call these “life-time decisions.” Students can
be asked to carefully consider these life-time decisions and how
decision theory can be used to assist them. Life-time decisions include decisions about what school to attend, marriage, and the
first job.
Teaching Suggestion 3.7: Popularity of Decision Trees
Stress that decision trees are not just an academic subject; they are
a technique widely used by top-level managers. Everyone appreciates a graphical display of a tough problem. It clarifies issues and
makes a great discussion base. Harvard business students regularly
use decision trees in case analysis.
Teaching Suggestion 3.8: Importance of Accurate
Tree Diagrams.
Developing accurate decision trees is an important part of this
chapter. Students can be asked to diagram several decision situations. The decisions can come from the end-of-chapter problems,
the instructor, or from student experiences.
Teaching Suggestion 3.3: Categorizing Decision-Making Types.
Decision-making types are discussed in this chapter; decision
making under certainty, risk, and uncertainty are included. Students can be asked to describe an important decision they had to
make in the past year and categorize the decision type. A good example can be a financial investment of \$1,000. In-class discussion
can help students realize the importance of decision theory and its
potential use.
Teaching Suggestion 3.9: Diagramming a Large Decision
Problem Using Branches.
Some students are intimidated by large and complex decision
trees. To avoid this situation, students can be shown that a large
decision tree is like having a number of smaller trees or decisions
that can be solved separately, starting at the end branches of the
tree. This can help students use decision-making techniques on
larger and more complex problems.
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 information and whether or not new information should be acquired.
The use of EVPI to place an upper limit on what you should pay
for information is a good way to start the section on this topic.
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 effective
way to learn how to revise probability values. Once students understand how the tables are used, they can be shown that the formulas are making exactly the same calculations.
Teaching Suggestion 3.5: Starting the Decision-Making
Under Uncertainty Material.
The section on decision-making under uncertainty can be started
with 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 people use these techniques to solve real problems, the concepts and
general approaches are useful.
ALTERNATIVE EXAMPLES
Alternative Example 3.1: Goleb Transport
George Goleb is considering the purchase of two types of industrial
robots. The Rob1 (alternative 1) is a large robot capable of performing a variety of tasks, including welding and painting. The Rob2 (alternative 2) is a smaller and slower robot, but it has all the capabilities
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of Rob1. The robots will be used to perform a variety of repair operations on large industrial equipment. Of course, George can always do
nothing and not buy any robots (alternative 3). The market for the repair could be either favorable (event 1) or unfavorable (event 2).
George has constructed a payoff matrix showing the expected returns
of each alternative and the probability of a favorable or unfavorable
market. The data are presented below:
Probability
Alternative 1
Alternative 2
Alternative 3
EVENT 1
EVENT 2
0.6
0.4
50,000
30,000
0
40,000
20,000
0
This problem can be solved using expected monetary value. The
equations 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 about
the probability of a favorable or unfavorable market. (See Alternative Example 3.1.) He would like to determine the equally likely
(Laplace), maximax, maximin, coefficient of realism (Hurwicz), and
minimax regret decisions. The Hurwicz coefficient should be 0.7.
The problem data are summarized below:
EVENT 1
Probability
Alternative 1
Alternative 2
Alternative 3
EVENT 2
0.6
0.4
50,000
30,000
0
40,000
20,000
0
The Laplace (equally likely) solution is computed averaging the
payoffs for each alternative and choosing the best. The results are
shown below. Alternatives 1 and 2 both give the highest average
return 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 payoff 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 maximum
payoff for any alternative: these maximums are 50,000; 30,000;
and 0. Therefore, the decision is to purchase the large robot
(alternative 1).
The Hurwicz approach uses a coefficient of realism value of
0.7, and a weighted average of the best and the worst payoffs for
each 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 opportunity loss. The opportunity loss table for Goleb is as follows:
Alternatives
Rob1
Rob2
Nothing
Favorable
Market
Unfavorable
Market
Maximum
in Row
0
20,000
50,000
40,000
20,000
0
40,000
20,000
50,000
The alternative that minimizes the maximum opportunity loss is
the Rob2. This is due to the \$20,000 in the last column in the table
above. Rob1 has a maximum opportunity loss of \$40,000, and
doing nothing has a maximum opportunity loss of \$50,000.
Alternative Example 3.3: George Goleb is considering the possibility of conducting a survey on the market potential for industrial equipment repair using robots. The cost of the survey is
\$5,000. George has developed a decision tree that shows the overall decision, as in the figure on the next page.
This problem can be solved using EMV calculations. We
start with the end of the tree and work toward the beginning computing EMV values. The results of the calculations are shown in
the 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 decision
with sample information, assuming no cost to gather it, minus the
expected value of the best decision without sample information.
Because the cost of the survey is \$5,000, the expected value of the
best decision with sample information, assuming no cost to gather
it, is \$23,802. The expected value of the best decision without
sample information is found on the lower branch of the decision
tree to be \$14,000. Thus, EVSI is \$9,802.
Alternative Example 3.5: This example reveals how the conditional probability values for the George Goleb examples (above)
have been determined. The probability values about the survey are
summarized in the following table:
Results of
Survey
Positive (P)
Negative (N)
Favorable Market
(FM)
Unfavorable Market
(UM)
P(P | FM) 0.9
P(N | FM) 0.1
P(P | UM) 0.2
P(N | UM) 0.8
Using the values above and the fact that P(FM) 0.6 and
P(UM) 0.4, we can compute the conditional probability values
of a favorable or unfavorable market given a positive or negative
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First
Decision
Point
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DECISION ANALYSIS
Second
Decision
Point
\$
0
,39
33
Favorable Market (0.871)
2
b1
Ro
Favorable Market (0.871)
Rob2
2)
Su
1
(0
3
.6
Unfavorable Market (0.129)
lts le
su rab
e
R vo
Fa
–\$45,000
\$25,000
–\$25,000
–\$5,000
8)
.3
(0
ey
rv
ts
Su sul tive
Re ega
N
ey
ct urv
u
d tS
on e
C ark
M
\$45,000
r
Favorable Market (0.158)
2
80
,
8
\$1
y
ve
Unfavorable Market (0.129)
4
Unfavorable Market (0.842)
b1
Ro
Favorable Market (0.158)
Rob2
5
Unfavorable Market (0.842)
\$45,000
–\$45,000
\$25,000
–\$25,000
–\$5,000
Do
\$–5,000
ey
rv
Su
ct
du
on
tC
No
Favorable Market (0.60)
\$1
4
0
,00
6
Unfavorable Market (0.40)
b1
Ro
Rob2
Favorable Market (0.60)
7
Unfavorable Market (0.40)
\$50,000
–\$40,000
\$30,000
–\$20,000
\$0
Figure for Alternative Example 3.3
survey result. The calculations are presented in the following
two tables.
Probability revision given a positive survey result
State of
Nature
FM
UM
Total
Conditional
Probability
Prior
Prob.
Joint
Prob.
Posterior
Probability
0.9
0.2
0.6
0.4
0.54
0.08
0.62
0.54/0.62 0.871
0.08/0.62 0.129
1.00
Probability given a negative survey result
State of
Nature
FM
UM
Total
Alternative Example 3.6: In the section on utility theory, Mark
Simkin 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) 1
Using the data above, we can determine the expected utility of
each alternative as follows:
U(Mark plays the game) 0.45(1) 0.55(0.8) 0.89
Conditional
Probability
Prior
Prob.
Joint
Prob.
Posterior
Probability
0.1
0.8
0.6
0.4
0.06
0.32
0.38
0.06/0.38 0.158
0.32/0.38 0.842
1.00
U(Mark doesn’t play the game) 0.9
Thus, the best decision for Mark is not to play the game with an
expected utility of 0.9. Given these data, Mark is a risk avoider.
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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 times 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
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 overall purpose of utility theory is to incorporate a decision maker’s preference for risk in the decision-making process.
3-13. 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.
3-14. 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.
3-15. 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.
3-16.
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
Row
Unfavorable Maximum
200,000
100,000
18,000
300,000
250,000
75,000
Row
Minimum
200,000
100,000
18,000
3-17. 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.
3-18.
a. Decision making under risk—maximize expected
monetary value.
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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
3-22. 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.
3-23.
X (205,000)/0.7 292,857.14
3-19. a. The expected value (EV) is computed for each
alternative.
Stock
(Cases)
EV(CDs) 0.5(23,000) 0.5(23,000) 23,000
Strong
Market
Fair
Market
Poor
Market
Max.
Regret
0
250,000
350,000
550,000
19,000
0
29,000
129,000
310,000
100,000
32,000
0
310,000
250,000
350,000
550,000
b. Minimax regret decision is to build medium.
3-24.
EV(Bonds) 0.5(30,000) 0.5(20,000) 25,000
a. Opportunity loss table
Large
Medium
Small
None
The decision would change if this payoff were less than 292,857.14,
so it would have to decrease by about \$7,143.
EV(stock market) 0.5(80,000) 0.5(20,000) 30,000
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DECISION ANALYSIS
a.
Demand
(Cases)
11
12
13
EMV
11
12
13
385
329
273
385
420
364
385
420
455
38512.
379.05
341.25
Probabilities
0.45
0.35
0.20
Therefore, he should invest in the stock market.
b. Stock 11 cases.
b. EVPI EV(with perfect information)
(Maximum EV without P, I)
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.
[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.
Stock
(Cases)
The opportunity loss table is
Alternative
Stock Market
Bonds
CDs
Good Economy
Poor Economy
0
50,000
57,000
43,000
3,000
0
11
12
13
EOL(Bonds) 0.5(50,000) 0.5(3,000) 26,500
EOL(CDs) 0.5(57,000) 0.5(0) 28,500
a.
Alternative
Market
Condition
Good
Fair
Stock market
1,400
Bank deposit
900
Probabilities of
market conditions
0.4
Poor
EMV
800
0
880
900
900
900
0.4
0.2
b. Best decision: deposit \$10,000 in bank.
11
12
13
EMV
385
385
385
385
420
420
385
420
455
385
404.25
411.25
3-25.
EOL(Stock Market) 0.5(0) 0.5(43,000) 21,500*
This minimizes EOL.
3-21.
Demand
(Cases)
Manufacture
(Cases)
Demand
(Cases)
6
7
8
9
Probabilities
6
7
8
9
EMV
300
255
210
165
300
350
305
260
300
350
400
355
300
350
400
450
300
340.5
352.5
317
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.
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Money recovered from each unsold case \$3.
Supply
(Cases)
Demand
(Cases)
100
200
300
EMV
100
100(15) 100(5) 1000
100(15) 100(3) 200(5) 800
100(15) 200(3) 300(5) 600
300(15) 100(5) 200(16) 800
300(15) 200(5) 100(16) 1900
300(15) 300(5) 3000
900
200
200(15) 100(5) 100(16) 900
200(15) 200(5) 2000
300
Probabilities
200(15) 100(3) 300(5) 1800
0.3
0.4
0.3
b. Produce 300 cases each day.
3-27. a. The table presented is a decision table. The basis for
the decisions in the following questions is shown in the
table below.
MARKET
Decision
Alternatives
Small
Medium
Large
Very Large
Good
Fair
50,000
80,000
100,000
300,000
20,000
30,000
30,000
25,000
Poor
MAXIMIN
Row
Maximum
Row
Minimum
Row
Average
50,000
80,000
100,000
300,000
10,000
20,000
40,000
160,000
20,000
30,000
30,000
55,000
10,000
20,000
40,000
160,000
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
Alternatives
Good
Market
Fair
Market
Poor
Market
Row
Maximum
Small
Medium
Large
Very Large
250,000
220,000
200,000
0
10,000
0
0
5,000
0
10,000
30,000
150,000
250,000
220,000
200,000
150,000
Minimax regret decision: Very large station.
3-28. EMV for node 1 0.5(100,000) 0.5(40,000) \$30,000. Choose the highest EMV, therefore construct the clinic.
Payoff
Favorable Market (0.5)
ct
tru
ns nic
o
C Cli
\$100,000
1
Unfavorable Market (0.5)
–\$40,000
\$30,000
Do
N
oth
i
ng
EMV for no clinic is \$0
EQUALLY
LIKELY
MAXIMAX
\$0
CRIT. OF
REALISM
Weighted
Average
38,000
60,000
72,000
208,000
1610
1800
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3-29.
DECISION ANALYSIS
a.
Payoff
Favorable Market (0.82)
CONSTRUCT
2
Unfavorable Market (0.18)
5)
ey .5
rv e (0
\$69,800
u
S bl
a
r
vo
Fa
DO NOT CONSTRUCT
1
t
y
uc ve \$36,140
nd Sur
o
C et
k
ar
M
–\$45,000
–\$5,000
Favorable Market (0.11)
CONSTRUCT
\$95,000
5)
ey .4
rv (0
Su tive
ga
Ne
3
Unfavorable Market (0.89)
\$95,000
–\$45,000
–\$5,000
\$36,140
–\$5,000
DO NOT CONSTRUCT
Do
tC
No
on
du
ct
Favorable Market (0.5)
Su
rv
CONSTRUCT CLINIC
ey
4
Unfavorable Market (0.5)
\$100,000
–\$40,000
\$30,000
DO NOT CONSTRUCT
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.
\$0
23
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DECISION ANALYSIS
3-30.
Favorable Market
Large Shop
2
Unfavorable Market
No Shop
e
bl
ra
o
y
v e
Fa urv
S
Favorable Market
Small Shop
3
Unfavorable Market
1
et
k
ar ey
M urv
S
N
U
Su nfa
rv vo
ey ra
b
Favorable Market
Large Shop
4
le
Unfavorable Market
No Shop
Favorable Market
o
Small Shop
Su
5
rv
ey
Unfavorable Market
Favorable Market
Large Shop
6
Unfavorable Market
No Shop
Favorable Market
Small Shop
7
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 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).
Unfavorable Market
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\$45,000
DECISION ANALYSIS
Favorable Market (0.9)
Large Shop
2
Unfavorable Market (0.1)
\$45,000
Payoff
\$55,000
–\$45,000
No Shop
–\$5,000
le 6)
ab 0.
or y (
v
e
Fa urv
S
\$21,000
Favorable Market (0.9)
Small Shop
3
\$25,000
Unfavorable Market (0.1)
\$25,000
–\$15,000
1
t
ke
ar ey
M urv
S
U
Su nfa
rv vo
ey ra
(0 ble
.4
)
–\$33,000
Favorable Market (0.12)
\$55,000
Large Shop
4
Unfavorable Market (0.88)
–\$5,000
–\$45,000
No Shop
–\$5,000
–\$10,200
N
o
Favorable Market (0.12)
\$25,000
Small Shop
Su
5
rv
ey
\$10,000
Unfavorable Market (0.88)
Favorable Market (0.5)
Large Shop
6
Unfavorable Market (0.5)
\$10,000
–\$15,000
\$60,000
–\$40,000
No Shop
\$0
\$10,000
Favorable Market (0.5)
\$30,000
Small Shop
7
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.
Unfavorable Market (0.5)
–\$10,000
25
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DECISION ANALYSIS
3-32.
Payoff
\$8,500
(0.9)
2
(0.1)
\$500
(0.9)
3
(0.1)
A3
\$8,500
\$2,750
e
or
M on
r
i
e t
A 1 ath ma
G for
In
A4
)
n 5
i o 0.
at le (
rm b
fo ra
In avo
F
A5
\$12,000
–\$23,000
\$2,000
–\$13,000
–\$3,000
1
In
U for
nf m
av at
or ion
ab
le
(0
.
–\$9,000
(0.4)
4
(0.6)
–\$7,000
(0.4)
5
(0.6)
A3
–\$3,000
A4
5)
A5
r n
he io
at at
G rm
ot fo
A 2 o N e In
D or
M
A4
–\$23,000
\$2,000
–\$13,000
–\$3,000
\$4,500
(0.7)
6
(0.3)
\$500
(0.7)
7
(0.3)
A3
\$4,500
\$12,000
\$15,000
–\$20,000
\$5,000
–\$10,000
A5
\$0
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
A4: build duplex
a. P(successful store | favorable research) P(S1 | I1)
A5: do nothing
EMV(node 2) 0.9(12,000) 0.1(23,000) 8,500
P ( S1 | I1 ) =
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 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
P ( I1 | S1 ) P ( S1 )
P ( I1 | S1 ) P ( S1 ) + P ( I1 | S2 ) P ( S2 )
0.8(0.5)
= 0.73
0.8(0.5) + 0.3(0.5)
P ( S1 | I1 ) =
b. P(successful store | unfavorable research) P(S1 | I2)
P ( S1 | I 2 ) =
P ( I 2 | S1 ) P ( S1 )
P ( I 2 | S1 ) P ( S1 ) + P ( I 2 | S2 ) P ( S2 )
P ( S1 | I 2 ) =
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 )
I2: unfavorable research
S1: store successful
S2: store unsuccessful
0.2(0.5)
= 0.22
0.2(0.5) + 0.7(0.5)
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3-34.
I1: favorable survey or information
EMV(B) 6,000(0.2) 4,000(0.3) 0(0.5)
2,400
S1: facility successful
S2: facility unsuccessful
Fund B should be selected.
P(S1) 0.3; P(S2) 0.7
c. Let X payout for Fund A in a good economy.
P(I1 | S1) 0.8; P(I2 | S1) 0.2
EMV(A) EMV(B)
P(I1 | S2) 0.3; P(I2 | S2) 0.7
X(0.2) 2,000(0.3) (–5,000)(0.5) 2,400
P(successful facility | favorable survey) P(S1 | I1)
P ( S1 | I1 ) =
P ( I1 | S1 ) P ( S1 )
P ( I1 | S1 ) P ( S1 ) + P ( I1 | S2 ) P ( S2 )
P ( S1 | I1 ) =
0.8(0.3)
= 0.533
0.8(0.3) + 0.3(0.7)
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.
P(successful facility | unfavorable survey) P(S1 | I2)
P ( S1 | I 2 ) =
P ( I 2 | S1 ) P ( S1 )
P ( I 2 | S1 ) P ( S1 ) + P ( I 2 | S2 ) P ( S2 )
P ( S1 | I 2 ) =
0.2(0.3)
= 0.109
0.2(0.3) + 0.7(0.7)
a.
Good economy 0.2
Fair economy 0.3
10,000
2,000
Fund A
Poor economy 0.5
Good economy 0.2
Fund B
27
b. EMV(A) 10,000(0.2) 2,000(0.3)
(5,000)(0.5) 100
I2: unfavorable survey
3-35.
DECISION ANALYSIS
Fair economy 0.3
Poor economy 0.5
–5,000
6,000
4,000
0
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3-36.
Page 28
DECISION ANALYSIS
a.
Payoff
Favorable Market
Survey
Favorable
Produce
Razor
3
Do Not Produce Razor
Favorable Market
1
Survey
Unfavorable
Produce
Razor
4
Unfavorable Market
Do Not Produce Razor
tS
urv
ey
Unfavorable Market
Co
nd
uc
Favorable Market
Study
Favorable
Conduct
Pilot
Study
Produce
Razor
5
Unfavorable Market
Do Not Produce Razor
Favorable Market
2
Ne
Unfavorable
Do Not Produce Razor
6
Unfavorable Market
er
ith
Study
Produce
Razor
Te
st
Favorable Market
Produce
Razor
7
Unfavorable Market
Do Not Produce Razor
b.
\$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
S1: survey favorable
EMV(node 5) 80,000(0.89) (80,000)(0.11) 62,400
S2: survey unfavorable
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
S3: study favorable
S4: study unfavorable
S5: market favorable
S6: market unfavorable
P ( S5 | S1 ) =
0.7(0.5)
= 0.78
0.7(0.5) + 0.2(0.5)
P(S6 | S1) 1 – 0.778 0.222
P ( S5 | S2 ) =
0.3(0.5)
= 0.27
0.3(0.5) + 0.8(0.5)
P(S6 | S2) 1 – 0.27 0.73
P ( S5 | S3 ) =
0.8(0.5)
= 0.89
0.8(0.5) + 0.1(0.5)
P(S6 | S3) 1 – 0.89 0.11
P ( S5 | S 4 ) =
0.2(0.5)
= 0.18
0.2(0.5) + 0.9(0.5)
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
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.
3-37. 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.
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DECISION ANALYSIS
29
Utility
0.85
Survey
0.82
Favorable
(0.45)
Produce
Razor
0.62
ey
Survey
urv
Unfavorable
(0.55)
Produce
Razor
4
uc
nd
Co
Conduct 0.745
Pilot
2
Study
Favorable
(0.45)
Produce
Razor
5
Study
Ne
er
ith
Unfavorable
(0.55)
Market Unfavorable (0.73)
Produce
Razor
6
Market Favorable (0.89)
Market Unfavorable (0.11)
Te
st
0.78
7
Market Unfavorable (0.82)
a. P(good economy | prediction of
good economy) 0.8(0.6 )
= 0.923
0.8(0.6 ) + 0.1(0.4 )
P(poor economy | prediction of
good economy) 0.1(0.4 )
= 0.077
0.8(0.6 ) + 0.1(0.4 )
P(good economy | prediction of
poor economy) 0.2(0.6 )
= 0.25
0.2(0.6 ) + 0.9(0.4 )
P(poor economy | prediction of
poor economy) 0.9(0.6 )
= 0.75
0.2(0.6 ) + 0.9(0.4 )
b. P(good economy | prediction of
good economy) 0.8(0.7)
= 0.949
0.8(0.7) + 0.1(0.3)
P(poor economy | prediction of
good economy) 0.1(0.3)
= 0.051
0.8(0.7) + 0.1(0.3)
P(good economy | prediction of
poor economy) 0.2(0.7)
= 0.341
0.2(0.7) + 0.9(0.3)
0.5
0.9
0
0.9
0
0.7
Market Favorable (0.5)
Market Unfavorable (0.5)
Do Not Produce Razor
3-38.
0.95
0.7
Market Favorable (0.18)
Do Not Produce Razor
Produce
Razor
0.5
0.8
Do Not Produce Razor
0.16
0.95
0.8
Market Favorable (0.27)
Do Not Produce Razor
0.80
Study
Market Unfavorable (0.22)
Do Not Produce Razor
1
tS
3
Market Favorable (0.78)
1
0.55
0.81
P(poor economy | prediction of
poor economy) 0.9(0.3)
= 0.659
0.2(0.7) + 0.9(0.3)
3-39. 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 –\$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.
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DECISION ANALYSIS
3-40.
Survey
Favorable
(0.55)
U = 0.76
ct
du t 1
n
e
Co ark
M
Survey
o
D
Unfavorable
(0.45)
U = 0.8118
Construct
2
Clinic
Favorable Market (0.82)
Unfavorable Market (0.18)
Do Not Construct Clinic
U = 0.1089
Construct
3
Clinic
Favorable Market (0.11)
Unfavorable Market (0.89)
Do Not Construct Clinic
c
du
on
C y
ot rve
Su
N
U = 0.55
t
Construct
Clinic
4
Favorable Market (0.5)
Unfavorable Market (0.5)
Payoff
Utility
\$95,000
0.99
–\$45,000
0
–\$5,000
0.7
\$95,000
0.99
–\$45,000
0
–\$5,000
0.7
\$100,000
1.0
–\$40,000
0.1
\$0
0.9
Do Not Construct Clinic
EU(node 2) (0.82)(0.99) (0.18)(0) 0.8118
b. Expected utility on Broad Street 0.2(0.5) 0.9(0.5) 0.55. Therefore, the expressway maximizes
expected utility.
EU(node 3) (0.11)(0.99) (0.89)(0) 0.1089
EU(node 4) 0.5(1) 0.5(0.1) 0.55
c. Lynn is a risk avoider.
EU(node 1) (0.55)(0.8118) (0.45)(0.7000) 0.7615
EU(no survey) 0.9
1.0
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.
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
Utility
3-41.
0.8
0.6
0.4
0.2
0
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
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 per
gallon; and advertising costs \$3 per gallon. From this information, we get:
marginal profit selling price minus the manufacturing, handling,
EU(conduct survey) 0.45(0.741) 0.55(0.2565) 0.4745
marginal profit \$20 \$12 \$1 \$3 \$4 per gallon
EU(no survey) 0.525
If more is produced than is needed, a marginal loss is incurred.
John’s decision would change. He would not conduct the survey
and 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 expected travel time.
Expressway
Street
30 Minutes,
U = 0.7
1
Congestion (0.5)
40 Minutes,
U = 0.2
No
Congestion (0.5)
15 Minutes,
U = 0.9
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.
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31
DECISION ANALYSIS
a. A decision tree is shown below:
Decision Tree
(0.2) Demand
500
\$2,000 = (500)(4)
(0.3)
1,000
–\$1,500
(0.4)
1,500
–\$3,000 = (500)(4) – (1,000)(5)
Stock
500
(0.1)
2,000
–\$5,500 = (500)(4) – (1,500)(5)
(0.2)
500
(0.3)
1,000
\$4,000 = (1,000)(4)
(0.4)
1,500
\$1,500 = (1,000)(4) – (5)(500)
(0.1)
2,000
–\$1,000 = (1,000)(4) – (5)(1,000)
(0.2)
500
(0.3)
1,000
(0.4)
1,500
\$6,000 = (1,500)(4)
(0.1)
2,000
\$3,500 = (1,500)(4) – (5)(500)
(0.2)
500
(0.3)
1,000
(0.4)
1,500
(0.1)
2,000
\$1,800
Stock
1,000
\$3,300
Stock
1,500
Stock
2,000
\$2,400
–\$500 = (500)(4) – (500)(5)
\$500 = (500)(4) – (500)(3)
–\$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)
\$8,000 = (2,000)(4)
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.
Decision Tree–No Survey
(0.15)
(0.40)
Table for Problem 3-43
00
0,0
Demand
Stock
500
500
2,000
1,000
500
1,500 1,000
2,000 2,500
Maximum
2,000
Probabilities
0.2
1,000
1,500
2,000
500 3,000 5,500
4,000 1,500
1,000
2,500 6,000
3,500
1,000 4,500
8,000
4,000 6,000
8,000
0.3
0.4
EMV
\$1,500
\$1,800
\$3,300
\$2,400
\$4,800 EVwPI
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
3-44. 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.
50
ll \$
a
Sm
(0.45)
(0.15)
Medium \$670,000
(0.40)
La
(0.45)
rge
\$5
80
,00
0
(0.15)
(0.40)
\$500,000
\$500,000
\$500,000
\$200,000
\$700,000
\$800,000
–\$200,000
\$400,000
(0.45) \$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 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
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DECISION ANALYSIS
medium survey results, and a final table is used for high survey results. These tables are shown below. These probabilities will be used
in the decision tree that follows.
Decision Tree–Survey
For low survey results—A1:
State of Nature
B1
B2
B3
P(Bi)
0.150
0.400
0.450
Small
P(Ai | Bj)
P(Bj and Ai)
0.700
0.400
0.100
P(A1) 0.105
0.160
0.045
0.310
P(Bj | Ai)
0.339
0.516
0.145
0.150
0.400
0.450
P(Ai | Bj)
B1
B2
B3
P(Bi)
P(Bj and Ai)
0.200
0.500
0.300
P(A2) 0.150
0.400
0.450
P(Ai | Bj)
450,000
Medium
M
H
0.030
0.200
0.135
0.365
P(Bj and Ai)
0.100
0.100
0.600
P(A3) 0.015
0.040
0.270
0.325
L
P(Bj | Ai)
0.082
0.548
0.370
For high survey results—A3:
State of Nature
M
L
Large
\$49
Low 5,000
(0.3
10)
B1
B2
B3
P(Bi)
450,000
H
For medium survey results—A2:
State of Nature
L
H
L
Small
P(Bj | Ai)
0.046
0.123
0.831
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.
M
H
L
\$646,000
Medium
Medium
(0.365)
M
H
L
Large
,000
\$821 .325)
(0
High
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) 0.378. 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 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:
M
M
H
L
Small
M
450,000
150,000
650,000
750,000
–250,000
350,000
950,000
450,000
450,000
450,000
150,000
650,000
750,000
–250,000
350,000
950,000
450,000
450,000
H
450,000
L
Medium
M
H
L
Large
M
H
150,000
650,000
750,000
–250,000
350,000
950,000
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3-45.
DECISION ANALYSIS
a.
Payoff
\$75,000
Succeed (0.5)
\$250,000
1
Do
w
w
nto
Don’t Succeed (0.5)
n
\$140,000
Mall
Succeed (0.6)
–\$100,000
\$300,000
2
Don’t Succeed (0.4)
Tra
ffic
Cir
cle
No
Gr
oc
er
yS
to
\$250,000
re
Succeed (0.75)
–\$100,000
\$400,000
3
Don’t Succeed (0.25)
–\$200,000
\$0
Mary should select the traffic circle location (EMV \$250,000).
b. Use Bayes’ Theorem to compute posterior probabilities.
¯¯¯¯ | SRP) = 0.22
P(SD | SRP) = 0.78;
P(SD
P(SM | SRP) = 0.84;
P(SC | SRP) = 0.91;
¯¯¯¯¯ | SRP) = 0.16
P(SM
¯¯¯¯ | SRP) = 0.09
P(SC
P(SM | SRN) = 0.36;
¯¯¯¯ | SRN) = 0.73
P(SD
¯¯¯¯¯
P(SM | SRN) = 0.64
P(SC | SRN) = 0.53;
¯¯¯¯ | SRN) = 0.47
P(SC
P(SD | SRN) = 0.27;
Example computations:
P ( SM | SRP ) =
P ( SRP | SM ) P ( SM )
P ( SRP | SM ) P ( SM ) + P ( SRP | SM ) P ( SM )
0.7(0.6 )
P ( SM | SRP ) =
= 0.84
0.7(0.6 ) + 0.2(0.4 )
P ( SC | SRN ) =
0.3(0.75)
= 0.53
0.3(0.75) + 0.8(0.25)
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.
33
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DECISION ANALYSIS
Second
Decision
Point
First
Decision
Point
Payoff
SD (0.78)
Downtown
2
ts
ul
es .6)
R (0
ey e
rv itiv
u
S os
P
A
3
SM (0.16)
Circle
4
SC (0.09)
No Grocery Store
lts
su )
Re (0.4
ey e
rv tiv
Su ega
N
D
5
Mall
SD (0.73)
SM (0.36)
6
SM (0.64)
SC (0.53)
No
tP
Circle
ur
c
ha
se
7
SC (0.47)
No Grocery Store
M
ar
ke
tS
ur
v
8
Mall
C
SD (0.5)
SM (0.6)
9
SM (0.4)
SC (0.75)
Circle
10
SC (0.25)
No Grocery Store
3-46. a. Sue can use decision tree analysis to find the best solution. The results are presented below. 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 good market given favorable
–\$130,000
\$270,000
–\$130,000
\$370,000
–\$230,000
\$220,000
–\$130,000
\$270,000
–\$130,000
\$370,000
–\$230,000
–\$30,000
SD (0.5)
Downtown
ey
\$220,000
–\$30,000
SD (0.27)
Downtown
B
Do
SM (0.84)
SC (0.91)
1
t
ke
ar
M
se ey
a
h
rv
rc Su
u
P
Mall
SD (0.22)
\$250,000
–\$100,000
\$300,000
–\$100,000
\$400,000
–\$200,000
\$0
information, Branch 10 (6–10) is a bad market 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.
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b. The suggested changes would be reflected in Branches 4 and 5. The decision stays the same, but the EMV
increases to \$46,000. The results are provided in the tables that follow:
Results for 3-46. a.
Start
Node
Ending
Node
Branch
Probability
Profit
(End Node)
0
1
1
2
2
3
3
4
4
6
6
5
5
7
7
8
8
1
2
3
4
5
8
17
6
11
9
10
7
14
12
13
15
16
0
0
0
0.6
0.4
0
0
0
0
0.9
0.1
0
0
0.2
0.8
0.6
0.4
0
0
0
0
0
0
0
0
20,000
80,000
100,000
0
20,000
80,000
100,000
100,000
80,000
Start
Node
Ending
Node
Branch
Probability
Profit
(End Node)
0
1
1
2
2
3
3
4
4
6
6
5
5
7
7
8
8
1
2
3
4
5
8
17
6
11
9
10
7
14
12
13
15
16
0
0
0
0.7
0.3
0
0
0
0
0.9
0.1
0
0
0.2
0.8
0.6
0.4
0
0
0
0
0
0
0
0
20,000
80,000
100,000
0
20,000
80,000
100,000
100,000
80,000
Start
Branch 1
Branch 2
Branch 3
Branch 4
Branch 5
Branch 6
Branch 7
Branch 8
Branch 9
Branch 10
Branch 11
Branch 12
Branch 13
Branch 14
Branch 15
Branch 16
Use
Branch?
Yes
Yes
Yes
Yes
Node
Type
Decision
Chance
Decision
Decision
Decision
Chance
Final
Chance
Final
Final
Final
Chance
Final
Final
Final
Final
Final
Node
Value
29,200
29,200
28,000
62,000
20,000
28,000
0
62,000
20,000
80,000
100,000
64,000
20,000
80,000
100,000
100,000
80,000
Results for 3-46. b.
Start
Branch 1
Branch 2
Branch 3
Branch 4
Branch 5
Branch 6
Branch 7
Branch 8
Branch 9
Branch 10
Branch 11
Branch 12
Branch 13
Branch 14
Branch 15
Branch 16
Use
Branch?
Yes
Yes
Yes
Yes
Node
Type
Decision
Chance
Decision
Decision
Decision
Chance
Final
Chance
Final
Final
Final
Chance
Final
Final
Final
Final
Final
Node
Value
37,400
37,400
28,000
62,000
20,000
28,000
0
62,000
20,000
80,000
100,000
64,000
20,000
80,000
100,000
100,000
80,000
35
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DECISION ANALYSIS
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 3-46. c.
Start
Node
Ending
Node
Branch
Probability
Profit
(End Node)
0
1
1
2
2
3
3
4
4
6
6
5
5
7
7
8
8
1
2
3
4
5
8
17
6
11
9
10
7
14
12
13
15
16
0
0
0
0.6
0.4
0
0
0
0
0.8
0.2
0
0
0.2
0.8
0.6
0.4
0
0
0
0
0
0
0
0
20,000
80,000
100,000
0
20,000
80,000
100,000
100,000
80,000
Start
Branch 1
Branch 2
Branch 3
Branch 4
Branch 5
Branch 6
Branch 7
Branch 8
Branch 9
Branch 10
Branch 11
Branch 12
Branch 13
Branch 14
Branch 15
Branch 16
Use
Branch?
Yes
Yes
Yes
Yes
Node
Type
Decision
Chance
Decision
Decision
Decision
Chance
Final
Chance
Final
Final
Final
Chance
Final
Final
Final
Final
Final
Node
Value
28,000
18,400
28,000
44,000
20,000
28,000
0
44,000
20,000
80,000
100,000
64,000
20,000
80,000
100,000
100,000
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 3-46. d.
Start
Branch 1
Branch 2
Branch 3
Branch 4
Branch 5
Branch 6
Branch 7
Branch 8
Branch 9
Branch 10
Branch 11
Branch 12
Branch 13
Branch 14
Branch 15
Branch 16
Start
Node
Ending
Node
Branch
Probability
Profit
(End Node)
0
1
1
2
2
3
3
4
4
6
6
5
5
7
7
8
8
1
2
3
4
5
8
17
6
11
9
10
7
14
12
13
15
16
0
0
0
0.6
0.4
0
0
0
0
0.9
0.1
0
0
0.2
0.8
0.6
0.4
0
0
0
0
0
0
0
0
30,000
70,000
110,000
0
30,000
70,000
110,000
100,000
80,000
Use
Branch?
Yes
Yes
Yes
Yes
Node
Type
Decision
Chance
Decision
Decision
Decision
Chance
Final
Chance
Final
Final
Final
Chance
Final
Final
Final
Final
Final
Node
Value
28,000
19,200
28,000
52,000
30,000
28,000
0
52,000
30,000
70,000
110,000
74,000
30,000
70,000
110,000
100,000
80,000
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37
DECISION ANALYSIS
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 3-46. e.
Start
Node
Ending
Node
Branch
Probability
0
1
1
2
2
3
3
4
4
6
6
5
5
7
7
8
8
1
2
3
4
5
8
17
6
11
9
10
7
14
12
13
15
16
0
0
0
0.6
0.4
0
0
0
0
0.9
0.1
0
0
0.2
0.8
0.6
0.4
Start
Branch 1
Branch 2
Branch 3
Branch 4
Branch 5
Branch 6
Branch 7
Branch 8
Branch 9
Branch 10
Branch 11
Branch 12
Branch 13
Branch 14
Branch 15
Branch 16
Profit
(End Node)
0
0
0
0
0
0
0.2
0
0.1
0.4
0
0
0.1
0.4
0
1
0.05
Use
Branch?
Yes
Yes
Yes
Yes
Ending
Node
Node
Type
Node
Value
1
2
3
4
5
8
17
6
11
9
10
7
14
12
13
15
16
Decision
Chance
Decision
Decision
Decision
Chance
Final
Chance
Final
Final
Final
Chance
Final
Final
Final
Final
Final
0.62
0.256
0.62
0.36
0.1
0.62
0.20
0.36
0.1
0.4
0
0.08
0.1
0.4
0
1
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 3-46. f.
Start
Branch 1
Branch 2
Branch 3
Branch 4
Branch 5
Branch 6
Branch 7
Branch 8
Branch 9
Branch 10
Branch 11
Branch 12
Branch 13
Branch 14
Branch 15
Branch 16
Start
Node
Ending
Node
Branch
Probability
Profit
(End Node)
0
1
1
2
2
3
3
4
4
6
6
5
5
7
7
8
8
1
2
3
4
5
8
17
6
11
9
10
7
14
12
13
15
16
0
0
0
0.6
0.4
0
0
0
0
0.9
0.1
0
0
0.2
0.8
0.6
0.4
0
0
0
0
0
0
0.8
0
0.6
0.9
0
0
0.6
0.9
0
1
0.4
3-47. 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.
Use
Branch?
Yes
Yes
Yes
Yes
Node
Type
Decision
Chance
Decision
Decision
Decision
Chance
Final
Chance
Final
Final
Final
Chance
Final
Final
Final
Final
Final
Node
Value
0.80
0.726
0.80
0.81
0.60
0.76
0.80
0.81
0.60
0.90
0.00
0.18
0.60
0.90
0.00
1.00
0.40
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DECISION ANALYSIS
Table for Problem 3-47a
Return in \$1,000:
Number of
Watches
Probability
100,000
150,000
200,000
250,000
300,000
350,000
400,000
450,000
500,000
Alternative 1
Alternative 2
Alternative 3
Alternative 4
Alternative 5
Alternative 6
Alternative 7
Alternative 8
Alternative 9
Event 1
Event 2
Event 3
Event 4
Event 5
0.100
0.200
0.500
0.100
0.100
100,000
90,000
85,000
80,000
65,000
50,000
45,000
30,000
20,000
110,000
120,000
110,000
120,000
100,000
100,000
95,000
90,000
85,000
120,000
140,000
135,000
155,000
155,000
160,000
170,000
165,000
160,000
135,000
155,000
160,000
170,000
180,000
190,000
200,000
230,000
270,000
140,000
170,000
175,000
180,000
195,000
210,000
230,000
245,000
295,000
Expected profit:
Alternative
Expected Profit
1
2
3
4
5
6
7
8
9
119,500
135,500
131,500
144,500
141,500
145,000
151,500
151,000
155,500 ← best alternative
For this decision problem, Alternative 9 gives the highest expected profit of \$155,500.
b. The expected value with perfect information is \$175,500, and
the expected value of perfect information (EVPI) is \$20,000.
c. 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:
Probability
Alternative 1
Alternative 2
Alternative 3
Alternative 4
Alternative 5
Alternative 6
Alternative 7
Alternative 8
Alternative 9
EVENT 1
EVENT 2
EVENT 3
EVENT 4
EVENT 5
0.100
0.280
0.500
0.100
0.020
100,000
90,000
85,000
80,000
65,000
50,000
45,000
30,000
20,000
110,000
120,000
110,000
120,000
100,000
100,000
95,000
90,000
85,000
120,000
140,000
135,000
155,000
155,000
160,000
170,000
165,000
160,000
135,000
155,000
160,000
170,000
180,000
190,000
200,000
230,000
270,000
140,000
170,000
175,000
180,000
195,000
210,000
230,000
245,000
295,000
Expected profit:
Alternative
1
2
3
4
5
6
7
8
9
Expected Profit
117.100
131,500
126,300
139,700
133,900
136,200
140,700 ← best alternative:
stock 400,000 watches
138,600
138,700
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DECISION ANALYSIS
d. Stocking 400,000 is still the best alternative. The results
are shown below.
Return in \$1,000:
Event 1
Probability
Alternative 1
Alternative 2
Alternative 3
Alternative 4
Alternative 5
Alternative 6
Alternative 7
Alternative 8
Alternative 9
Event 2
Event 3
0.100
0.280
0.500
0.100
0.020
110,000
120,000
110,000
120,000
100,000
100,000
95,000
90,000
85,000
120,000
140,000
135,000
155,000
155,000
160,000
170,000
165,000
160,000
135,000
155,000
160,000
170,000
180,000
190,000
200,000
230,000
270,000
140,000
170,000
175,000
180,000
195,000
210,000
230,000
245,000
340,000
b. Back roads (minimum time used).
c. Expected time with perfect information:
15 1/2 + 25 1/3 + 30 1/6 = 20.83 minutes
Expected Profit
1
2
3
4
5
6
7
8
9
3-48.
117,100
131,500
126,300
139,700
133,900
136,200
140,700 ← best alternative:
stock 400,000 watches
138,600
139,600
Time saved is 31⁄3; minutes.
3-51.
a. Decision under uncertainty.
b.
Population
Same
Large wing
Small wing
No wing
Population
Grows
85,000
45,000
0
150,000
60,000
0
Row
Average
32,500
7,500
0
c. Best alternative: large wing.
3-49. a. Note: This problem can also be solved using marginal
analysis.
Large wing
Small wing
No wing
Population
Same
Population
Grows
85,000
45,000
0
150,000
60,000
0
Weighted
Average with
␣ = 0.75
91,250
33,750
0
b. Best decision: large wing.
c. No.
3-50.
a.
No
Mild
Severe
Expected
Congestion Congestion Congestion
Time
Tennessee
Expressway
Probabilities
Event 5
100,000
90,000
85,000
80,000
65,000
50,000
45,000
30,000
20,000
Expected profit
Alternative
Event 4
15
20
30
30
25
30
45
35
30
(30 days)/
(20 days)/
(60 days) = 1/2 (60 days) = 1/3
25
24.17
30
(10 days)/
(60 days) = 1/6
a. EMV can be used to determine the best strategy to minimize costs. The QM for Windows solution is shown on
the next page. The best decision is to go with the partial
service (maintenance) agreement.
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DECISION ANALYSIS
Solution to 3-51a
Probabilities
0.2
0.8
Maint.
Cost (\$)
No Maint.
Cost (\$)
3,000
1,500
500
0
300
500
No Service Agreement
Partial Service Agreement
Complete Service Agreement
Expected
Value
(\$)
Row
Minimum
(\$)
600
540
500
0
0
500
3,000
1,500
500
500
0
500
Column best
Row
Maximum
(\$)
The minimum expected monetary value is 500 given by Complete
Service Agreement
b. The new probability estimates dramatically change
Sim’s decision. 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 below.
Solution to 3-51b
Needs Repair
(\$)
Probabilities
No Service
Agreement
Partial Service
Agreement
Complete Service
Agreement
Does Not
Need Repair
(\$)
0.8
Expected
Value
(\$)
0.2
3,000
0
2,400
1,500
300
1,260
500
500
500
Column best
500
3-52. We can use QM for Windows to solve this decision making under uncertainty problem. We have made up probability values, which will be ignored in the analysis. As you can see, the
maximax decision is Option 4, and the maximum decision is Option 1. To compute the equality likely decision, we used equal
probability values of 0.25 for each of the four scenarios. As seen
below, the equally likely decision, which is the same as the EMV
decision in this case, is Option 3.
Solution to 3-52
Expected
Value (\$)
Probabilities
Option 1
Option 2
Option 3
Option 4
0.25
0.25
0.25
0.25
Judge (\$)
Trial (\$)
Court (\$)
Arbitration (\$)
5,000
10,000
20,000
30,000
5,000
5,000
7,000
15,000
5,000
2,000
1,000
10,000
5,000
0
5,000
20,000
Column best
The maximum expected monetary value is 5,750 given by Option 3.
The maximum is 5,000 given by Option 1. The maximax is 30,000
given by Option 4.
Row
Minimum (\$)
Row
Miximum (\$)
5,000
4,250
5,750
3,750
5,000
0
5,000
20,000
5,000
10,000
20,000
30,000
5,750
5,000
30,000
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DECISION ANALYSIS
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 below. 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
Alternative 1
Alternative 2
Alternative 3
Alternative 4
Event 1
Event 2
Laplace
Average Value
Minimum
Maximum
Hurwicz
Value
0
55,273
120,000
240,000
0
10,000
15,000
30,000
0.0
22,636.5
152,500.0
105,000.0
0
10,000
15,000
30,000
0
55,273
120,000
240,000
0.00
2,819.97
150.00
300.00
Regret table
Alternative
Event 1
Event 2
Maximum
Regret
Alternative 1
Alternative 2
Alternative 3
Alternative 4
240,000
184,727
120,000
0
0
10,000
15,000
30,000
240,000
184,727
120,000
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:
S—Substantial (P 0.1)
M—Moderate (P 0.6)
L—Low (P 0.3)
Without
Competition
With
Competition
\$800,000
\$600,000
\$500,000
\$400,000
\$300,000
\$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)
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DECISION ANALYSIS
Drink-at-Home. Inc. Case (continued)
Mkt
S (.1)
C12 (.8)
L (.3)
Ok (.9)
S (.1)
No C12 (.2)
Normal
M (.6)
M (.6)
L (.3)
No (.1) (Stop)
S (.1)
Ok (.9)
M (.6)
L (.3)
No C8 (.4)
S (.1)
No (.1)
M (.6)
nth
6 Mo
L (.3)
S (.1)
C6 (.5)
M (.6)
L (.3)
S (.1)
C12 (.8)
M (.6)
L (.3)
Ok (.9)
S (.1)
No C12 (.2)
No C6 (.5)
R
E
M (.6)
L (.3)
No (.1)
= –100
– 80
= – 80
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
– 100
= –100
Mkt
S (.1)
C12 (.8)
S (.1)
No C12 (.2)
Normal
M (.6)
L (.3)
Ok (.9)
M (.6)
L (.3)
No (.1) (Stop)
Ok (.9)
No C8 (.4)
No (.1)
nth
6 Mo
S (.1)
M (.6)
L (.3)
S (.1)
M (.6)
L (.3)
S (.1)
C6 (.5)
M (.6)
L (.3)
S (.1)
C12 (.8)
M (.6)
L (.3)
Ok (.9)
S (.1)
No C12 (.2)
No C6 (.5)
M (.6)
L (.3)
No (.1)
50
–50
–100
450
250
150
–100
C8 (.6)
8 Month
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
C8 (.6)
8 Month
S
–80
10
–90
–140
410
210
110
60
–40
–90
50
–50
–100
450
250
150
–100
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Drink-at-Home, Inc. Case (continued)
Mkt
S (.1)
Ok (.9)
C12 (.8)
M (.6)
–55
L (.3)
240
M (.6)
(4)
S (.1)
Normal –6.4
L (.3)
No C12 (.2)
No (.1) (Stop)
–100
C8 (.6)
S (.1)
8 Month –74.2
Ok (.9)
–95
(–74.2)
No C8 (.4)
M (.6)
L (.3)
S (.1)
M (.6)
200
L (.3)
6 Mo
No (.1)
nth
S (.1)
C6 (.5)
M (.6)
–45
L (.3)
S (.1)
C 12 (.8)
M (.6)
–55
Ok (.9)
No C6 (.5)
L (.3)
S (.1)
M (.6)
240
(19.3)
L (.3)
No C12 (.2)
No (.1)
–80
10
–90
–140
410
210
110
60
–40
–90
50
–50
–100
450
250
150
–100
The optimal program is to adopt the 6-month program
Ruth Jones’ Heart By-Pass Operation Case
N
o
Su Byrg pa
er ss
y
50
–50
–100
450
250
150
Prob.
Years
Expected Rate
One Year
.50
1
.50
Two Years
.20
2
.40
Five Years
.20
5
1.00
Eight Years
.10
8
.80
2.7 years
0.0
.05
0
One Year
.45
1
.45
Five Years
.20
5
1.00
Ten Years
.13
10
1.30
Fifteen Years
.08
15
1.20
Twenty Years
.05
20
1.00
Twenty-five
Years
.04
25
1.00
ry
e
rg
Su
0 Years
5.95 years
43
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Expected survival rate with surgery (5.95 years) exceeds the
nonsurgical survival rate of 2.70 years. Surgery is favorable.
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.
POOR
Probabilities
0.1
AVERAGE
GOOD
EXCELLENT
0.3
0.4
0.2
EXPECTED
VALUE
Option 1—PP
Option 2—LB and PP
Option 3—TR and PP
Option 4—CC and PP
Option 5—LB, CC, and TR
5,000
10,000
15,000
30,000
60,000
2,000
4,000
10,000
20,000
35,000
2,000
6,000
7,000
10,000
20,000
5,000
12,000
13,000
30,000
55,000
700
2,600
900
1,000
2,500
With Perfect Information
5,000
2,000
25,000
55,000
17,900
The maximum expected monetary value is 2,600 given by
Option 2 LB and PP.
b and c. The opportunity loss and the expected value of perfect information is presented below. 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
Probabilities
Option 1—PP
Option 2—LB and PP
Option 3—TR and PP
Option 4—CC and PP
Option 5—LB, CC, and TR
AVERAGE
GOOD
EXCELLENT
0.1
0.3
0.4
0.2
0
5,000
10,000
25,000
55,000
0
2,000
8,000
18,000
33,000
18,000
14,000
13,000
10,000
0
50,000
43,000
42,000
25,000
0
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
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result in a grade of B for the course. The table below gives the different possibilities – points and grade in the course.
Study 1, 2, 3
Study 1,2
Study 1,3
Study 2,3
Study 1
Study 2
Study 3
Case 1
on Exam
Case 2
on Exam
Case 3
on Exam
EV
12 B
20 A
20 A
0B
25 A
0B
0B
12 B
20 A
0B
20 A
0B
25 A
0B
12 B
0B
20 A
20 A
0B
0B
25 A
12
40/3
40/3
40/3
25/3
25/3
25/3
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.