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
Use
Node
Node
Node
Prob.
(End Node)
Branch?
Type
Start
0
1
0
0
Dec
Branch 1
1
2
0
0
Yes
Ch
Branch 2
1
3
0
0
Dec
Branch 3
2
4
0.7
0
Dec
Branch 4
2
5
0.3
0
Dec
Branch 5
3
8
0
0
Yes
Ch
Branch 6
3
17
0
0
Fin
Branch 7
4
6
0
0
Yes
Ch
Branch 8
4
11
0
–20,000
Fin
Branch 9
6
9
0.9
80,000
Fin
Branch 10
6
10
0.1
–100,000
Fin
Branch 11
5
7
0
0
Ch
Branch 12
5
14
0
–20,000
Yes
Fin
Branch 13
7
12
0.2
80,000
Fin
Branch 14
7
13
0.8
–100,000
Fin
Branch 15
8
15
0.6
Fin
Branch 16
8
16
0.4
–80,000
Fin
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 deci-
sion is $28,000. The results are presented below:
Results for 3-52. c.
Start
Ending
Branch
Profit
Use
Node
Node
Node
Prob.
(End Node)
Branch?
Type
Start
0
1
0
0
Dec
Branch 1
1
2
0
0
Ch
Branch 2
1
3
0
0
Yes
Dec
Branch 3
2
4
0.6
0
Branch 4
2
5
0.4
0
Dec
Branch 5
3
8
0
0
Yes
Ch
Branch 6
3
17
0
0
Fin
Branch 7
4
6
0
0
Yes
Ch
Branch 8
4
11
0
–20,000
Fin
Branch 9
6
9
0.8
80,000
Fin
Branch 10
6
10
0.2
–100,000
Fin
Branch 11
5
7
0
0
Ch
Branch 12
5
14
0
–20,000
Yes
Fin
Branch 13
7
12
0.2
80,000
Fin
Branch 14
7
13
0.8
–100,000
Fin
Branch 15
8
15
0.6
Fin
Branch 16
8
16
0.4
–80,000
Fin
d. Yes, Sue’s decision would change from her original decision. With the higher cost of infor-
mation, 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
Profit
Use
Node
Node
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 3
2
4
0.6
0
Decision
52,000
Branch 4
2
5
0.4
0
Decision
–30,000
Branch 5
3
8
0
0
Yes
Chance
28,000
Branch 6
3
17
0
0
Final
0
Branch 7
4
6
0
0
Yes
Chance
52,000
Branch 8
4
11
0
–30,000
Final
–30,000
Branch 9
6
9
0.9
70,000
Final
70,000
Branch 10
6
10
0.1
Final
Branch 11
5
7
0
0
Chance
–74,000
Branch 12
5
14
0
–30,000
Yes
Final
–30,000
Branch 13
7
12
0.2
70,000
Final
70,000
Branch 14
7
13
0.8
Final
Branch 15
8
15
0.6
Final
Branch 16
8
16
0.4
–80,000
Final
–80,000
e. The expected utility can be computed by replacing the monetary values with utility values.
Given the utility values in the problem, the expected utility is 0.62. The utility table represents a
risk seeker. The results are given below.
Results for 3-52. e.
Start
Ending
Branch
Profit
Use
Ending
Node
Node
Node
Node
Prob.
(End Node)
Branch?
Node
Type
Value
Start
0
1
0
0
1
Dec
0.62
Branch 1
1
2
0
0
2
Ch
0.256
Branch 3
2
4
0.6
0
4
Dec
0.36
Branch 4
2
5
0.4
0
5
Dec
0.1
Branch 5
3
8
0
0
Yes
8
Ch
0.62
Branch 6
3
17
0
0.2
17
Fin
0.20
Branch 7
4
6
0
0
Yes
6
Ch
0.36
Branch 8
4
11
0
0.1
11
Fin
0.1
Branch 9
6
9
0.9
0.4
9
Fin
0.4
Branch 10
6
10
0.1
0
10
Fin
0
Branch 11
5
7
0
0
7
Ch
0.08
Branch 12
5
14
0
0.1
Yes
14
Fin
0.1
Branch 13
7
12
0.2
0.4
12
Fin
0.4
Branch 14
7
13
0.8
0
13
Fin
0
Branch 15
8
15
0.6
1
15
Fin
1
Branch 16
8
16
0.4
0.05
16
Fin
0.05
f. This problem can be solved by replacing monetary values with utility values. The expected
utility is 0.80. The utility table given in the problem is representative of a risk avoider. The re-
sults 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 3
2
4
0.6
0
Dec
0.81
Branch 4
2
5
0.4
0
Dec
0.60
Branch 5
3
8
0
0
Yes
Ch
0.76
Branch 6
3
17
0
0.8
Fin
0.80
Branch 7
4
6
0
0
Yes
Ch
0.81
Branch 8
4
11
0
0.6
Fin
0.60
Branch 9
6
9
0.9
0.9
Fin
0.90
Branch 10
6
10
0.1
0
Fin
0.00
Branch 11
5
7
0
0
Ch
0.18
Branch 12
5
14
0
0.6
Yes
Fin
0.60
Branch 13
7
12
0.2
0.9
Fin
0.90
Branch 14
7
13
0.8
0
Fin
0.00
Branch 15
8
15
0.6
1
Fin
1.00
Branch 16
8
16
0.4
0.4
Fin
0.40
3-53. a. The decision table for Chris Dunphy along with the expected profits or expected mone-
tary 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.
Return in $1,000:
NO. OF WATCHES
EVENT 1
EVENT 2
EVENT 3
EVENT 4
EVENT 5
Probability
0.100
0.280
0.500
0.100
0.020
Expected Profit
100,000
100
110
120
135
140
117.1
150,000
90
120
140
155
170
131.5
200,000
85
110
135
160
175
126.3
250,000
80
120
155
170
180
139.7
350,000
50
100
160
190
210
136.2
400,000
45
95
170
200
230
140.7
450,000
30
90
165
230
245
138.6
500,000
20
85
160
270
295
138.7
3-54. a. Decision under uncertainty.
b.
Population
Population
Row
Same
Grows
Average
Large wing
–85,000
150,000
32,500
Small wing
–45,000
60,000
No wing
c. Best alternative: large wing.
3-55. a.
Weighted
Population
Population
Average with
Same
Grows
= 0.75
Large wing
–85,000
150,000
91,250
Small wing
–45,000
33,750
No wing
b. Best decision: large wing.
c. No.
3-56. a.
No
Mild
Severe
Expected
Congestion
Congestion
Congestion
Time
Tennessee
15
30
45
25
Back roads
20
25
35
24.17
Expressway
30
30
30
30
13
3-57. a. EMV can be used to determine the best strategy to minimize costs. The QM for
Windows solution is provided. The best decision is to go with the partial service
(maintenance) agreement.
Solution to 3-57a
Probabilities
0.2
0.8
Maint.
No Maint.
Expected
Row
Row
Cost ($)
Cost ($)
Value
Minimum
Maximum
($)
($)
($)
No Service Agreement
3,000
0
600
0
3,000
Partial Service Agreement
1,500
540
0
1,500
Agreement
Column best
500
0
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
1,260
Agreement
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
Solution to 3-58
Prob.
0.25
0.25
0.25
0.25
Judge
Trial
Court
Arbitration
Equally
Row
Row
Likely
Min.
Max.
Option 1
5,000
5,000
5,000
5,000
5,000
5,000
5,000
Option 2
10,000
5,000
2,000
4,250
10,000
Option 3
20,000
7,000
1,000
5,750
20,000
Option 4
30,000
3,750
30,000
Column best
5,750
5,000
30,000
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 4.
Payoff table
Laplace
Hurwicz
Event 1
Event 2
Average Value
Minimum
Maximum
Value
Alternative 1
0
0
0.0
0
0
0.00
Alternative 2
55,273
22,636.5
55,273
Alternative 3
120,000
52,500.0
120,000
Alternative 4
240,000
105,000.0
240,000
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.
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.
Success or failure of development effort:
Ok—Development effort ultimately a success
No—Development effort ultimately a failure
Column:
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)
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.
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
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
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
0 B
40/3
A 2/3 chance or B 1/3 chance
Study 1,3
20 A
0 B
20 A
40/3
A 2/3 chance or B 1/3 chance
Study 2,3
0 B
20 A
20 A
40/3
A 2/3 chance or B 1/3 chance
Study 1
25 A
0 B
0 B
25/3
A 1/3 chance or B 2/3 chance
Study 2
0 B
25 A
0 B
25/3
A 1/3 chance or B 2/3 chance