IS 320 Problems from Chapter 12
12.7
a) Max(A1) = $220 thousand, Max(A2) = $200 thousand.
Maximax = $220 thousand when choosing alternative A1.
b) Min(A1) = $110 thousand, Min(A2) = $150 thousand.
Maximin = $150 thousand when choosing alternative A2.
c) S1 is the most likely outcome. For this state, the maximum payoff of $220
thousand occurs with alternative A1.
d) Alternative A1 has the highest expected payoff of $194 thousand.
1
2
3
4
5
6
7
A
B
Payoff Table ($thousand)
C
D
Alternative
A1
A2
S1
220
200
State of Nature
S2
170
180
S3
110
150
Prior Probability
0.6
0.3
0.1
E
F
Expected
Payoff
($thousand)
194
189
e & f)
0.6
State 1
220
220
220
0.3
State 2
Alternative 1
170
0
194
170
170
0.1
State 3
110
110
110
1
194
0.6
State 1
200
200
200
0.3
State 2
Alternative 2
180
0
189
180
180
0.1
State 3
150
150
150
IS 320 Problems from Chapter 12
g)
16
17
18
19
20
21
22
23
24
25
26
27
28
M
Prior
Probability
of S1
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
N
Best
Alternative
1
2
2
2
1
1
1
1
1
1
O
Expected
Payoff
($thousand)
194
183
184
185
186.5
189
191.5
194
196.5
199
Let p = prior probability of S1.
For A1:
EP
= p(220) + (1 – 0.1 – p)(170) + (0.1)(110)
= 220p + 153 – 170p + 11
= 50p + 164
For A2:
EP
= p(200) + (1 – 0.1 – p)(180) + (0.1)(150)
= 200p + 162 – 180p + 15
= 20p + 177
A1 and A2 cross when 50p + 164 = 20p + 177 or 30p = 13 or p=0.433.
They should choose A2 when p ≤ 0.433, A1 when p > 0.433.
IS 320 Problems from Chapter 12
h)
16
17
18
19
20
21
22
23
24
25
26
27
28
M
Prior
Probability
of S1
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
N
Best
Alternative
1
2
2
2
2
2
1
1
1
1
O
Expected
Payoff
($thousand)
194
174
176.5
179
181.5
184
188.5
194
199.5
205
Let p = prior probability of S1.
For A1:
EP
= p(220) + (0.3)(170) + (1 – 0.3 – p)(110)
= 220p + 51 + 77 – 110p
= 110p + 128
For A2:
EP
= p(200) + (0.3)(180) + (1 – 0.3 – p)(150)
= 200p + 54 + 105 – 150p
= 50p + 159
A1 and A2 cross when 110p + 128 = 50p + 159 or 60p = 31 or p = 0.517.
They should choose A2 when p ≤ 0.517, A1 when p > 0.517.
IS 320 Problems from Chapter 12
i)
16
17
18
19
20
21
22
23
24
25
26
27
28
M
Prior
Probability
of S2
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
N
Best
Alternative
1
2
2
2
1
1
1
1
1
1
O
Expected
Payoff
($thousand)
194
180
181.5
183
185
188
191
194
197
200
Let p = prior probability of S2.
For A1:
EP
= (0.6)(220) + p(170) + (1 – 0.6 – p)(110)
= 132 + 170p + 44 – 110p
= 60p + 176
For A2:
EP
= (0.6)(200) + p(180) + (1 – 0.6 – p)(150)
= 120 + 180p + 60 – 150p
= 30p + 180
A1 and A2 cross when 60p + 176 = 30p + 180 or 30p = 4 or p = 0.133.
They should choose A2 when p ≤ 0.133, A1 when p > 0.133.
j) Alternative A1 should be chosen.
12.9
When x = 50, alternative A3 has the highest expected payoff of $5,600.
A
1
x=
2
3 Payoff Table ($hundred)
4
Alternative
5
A1
6
A2
7
A3
8
9
Prior Probability
10
B
50
C
D
S1
100
25
35
State of Nature
S2
50
40
150
S3
10
90
30
0.4
0.2
0.4
E
F
Expected
Payoff
($hundred)
54
54
56
IS 320 Problems from Chapter 12
When x = 75, alternative A1 has the highest expected payoff of $7,400.
A
1
x=
2
3 Payoff Table ($hundred)
4
Alternative
5
A1
6
A2
7
A3
8
9
Prior Probability
10
B
75
C
D
S1
150
25
35
State of Nature
S2
50
40
225
S3
10
90
30
0.4
0.2
0.4
E
F
Expected
Payoff
($hundred)
74
54
71
Barbara Miller should pay a maximum of $1,800 to increase x to 75.
12.17 a through d)
Prior
Probabilities
P (state)
0.1
user
nonuser
0.9
Conditional
Probabilities
P(finding | state)
Joint
Probabilities
P(state and finding)
0.095
user and positive
0.95
positive, given user
negative, given user
0.05
0.005
user and negative
Posterior
Probabilities
P(state | finding)
0.6786
user, given positive
0.0058
user, given negative
0.045
0.3214
nonuser and positive nonuser, given positive
0.05
positive, given nonuser
negative given nonuser
0.95
0.855
0.9942
nonuser and negative nonuser, given negative
IS 320 Problems from Chapter 12
e)
B
C
3 Data:
4
State of
Prior
Nature
Probability
5
0.1
User
6
0.9
Nonuser
7
8
9
10
11
12 Posterior
13 Probabilities:
P(Finding)
Finding
14
Positive
0.14
15
0.86
Negative
16
17
18
19
D
E
Positive
0.95
0.05
F
P(Finding | State)
Finding
Negative
0.05
0.95
User
0.679
0.006
P(State | Finding)
State of Nature
Nonuser
0.321
0.994
G
H