b. P(successful store | unfavorable research) = P(S1 | I2)
c. Now P(S1) = 0.6 and P(S2) = 0.4
( )
( )
( ) ( )
11
0.8 0.6 0.8
0.8 0.6 0.3 0.4
P S I
==
+
3-40. I1: favorable survey or information
I2: unfavorable survey
S1: facility successful
S2: facility unsuccessful
P(S1) = 0.3; P(S2) = 0.7
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.
3-42. a.
b.
S1: survey favorable
S2: survey unfavorable
S3: study favorable
( )
( )
( ) ( )
51
0.7 0.5 0.78
0.7 0.5 0.2 0.5
P S S ==
+
P(S6 | S2) = 1 – 0.27 = 0.73
P(S6 | S4) = 1 – 0.18 = 0.82
c. EMV(node 3) = 95,000(0.78) + (–65,000)(0.22)
= 59,800
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
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
3-44. 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
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
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
3-46.
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 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 max-
imizes expected utility.
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
a. A decision tree is provided:
b. The computations are shown in the following table. These numbers are entered into the tree
above. The best decision is to stock 1,500 gallons.
Table for Problem 3-49
Demand
Stock
500
1,000
1,500
2,000
EMV
500
2,000
–500
–3,000
–5,500
–$1,500
1,000
4,000
1,500
–1,000
1,500
–1,000
2,500
6,000
3,500
Maximum
2,000
4,000
6,000
8,000
$4,800 = EVwPI
Probabilities
0.2
0.3
0.4
0.1
c. EVwPI = (0.2)(2,000) + (0.3)(4,000) + (0.4)(6,000)
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
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.
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
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
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
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.
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:
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
SC
SD
SM
SC
Example computations:
( ) ( )
( )
( )
( )
( )
( )
P SRP SM P SM
P SM SRP
P SRP SM P SM P SRP SM P SM
=
+
These calculations are for the tree that follows:
EMV(2) = $171,600 – $28,600 = $143,000
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(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
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
(8–16) is an unsuccessful retail store given no information is obtained.
Results for 3-52. a.
Start
Ending
Branch
Profit
Use
Node
Node
Node
Node
Prob.
(End Node)
Branch?
Type
Value
Start
0
1
0
0
Dec
29,200
Branch 1
1
2
0
0
Yes
Ch
29,200
Branch 2
1
3
0
0
Dec
28,000
Branch 3
2
4
0.6
0
Dec
62,000
Branch 5
3
8
0
0
Yes
Ch
28,000
Branch 6
3
17
0
0
Fin
0
Branch 7
4
6
0
0
Yes
Ch
62,000
Branch 8
4
11
0
–20,000
Fin
–20,000
Branch 9
6
9
0.9
80,000
Fin
80,000
Branch 10
6
10
0.1
Fin
Branch 11
5
7
0
0
Ch
–64,000
Branch 12
5
14
0
–20,000
Yes
Fin
–20,000
Branch 13
7
12
0.2
80,000
Fin
80,000
Branch 14
7
13
0.8
Fin
Branch 15
8
15
0.6
Fin
Branch 16
8
16
0.4
–80,000
Fin
–80,000