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Problem 4-17:
Minimize 38.00 x f 11.50 xu 6.50 xt 51.00 y f 15.00 yu 7.50 yt
Demand
Capacity
x f y f 5000
3.5 x f 1.3xu 0.8 xt ≤ 350 キ 60
x y 10000
2.2x 1.7 x 0.0x ≤ 420 キ 60
x y 5000
3.1x 2.6x 1.7 x ≤ 680 キ 60
Problem 7-21:
a) Consider
1, if a camera is located at opening i
xi
0, if not
min x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13
such that
x1 x4 x6 1
Room 1
x6 x8 x12 1
Room 2
x1 x2 x3 1
Room 3
x3 x4 x5 x7 1
Room 4
x7 x8 x9 x10 1
Room 5
x10 x12 x13 1
Room 6
x2 x5 x9 x11 11 Room 7
x11 x13 1
Room 8
b) Since x1 x5 x8 x13 1, it implies that the cameras should be located at the openings 1, 5, 8, and
13. Moreover, an alternative optimal solution is x1 x7 x11 x12 1 .
c) Change the constraint for room 7 to the following:
x2 x5 x9 x11 2
d) Since x3 x6 x9 x11 x12 1 , it implies that the cameras should be located at the openings 3, 6,
9, 11, and 12. Moreover, an alternative optimal solution is x2 x4 x6 x10 x11 1. Hence, the
optimal value = 5.
Problem 13-4:
Using P s1 , P s2 , P s3 , and the opportunity loss values, we can compute the expected opportunity
loss (EOL) for each of the decision alternatives. With P s1 0.65, P s2 0.15 , and P s3 0.20 ,
the expected opportunity loss for each of the two decision alternatives is
EOL d1 0.65 250 0.15 100 0.20 25 182.5
EOL d 2 0.65 100 0.15 100 0.20 75 95
No matter of whether the decision analysis involves minimization or maximization, the minimum
expected opportunity loss often provides the best decision alternative. Hence, with EOL d2 95 , d 2
is the recommended decision. Moreover, the calculated EOL is equal to the expected value of perfect
information. Thus, EOL (best decision) = EVPI; and this value is 95 .
Problem 13-18:
a) The payoffs of $2, 650, 000 and $650, 000 for first two incomes are calculated as below:
Outcome 1
Bid
$200, 000
Contract
$2, 000, 000
Market Research
$150, 000
High Demand
$5, 000, 000
$2, 650, 000
Outcome 2
Bid
$200, 000
Contract
$2, 000, 000
Market Research
$150, 000
Moderate Demand
$3, 000, 000
$650, 000
b) We have the following calculations:
EV(node8) =
0.85 2650 0.15 650 $2,350,000
EV (node 5) =
Max 2350, 1150 $2,350,000
EV(node9) =
0.225 2650 0.775 650 $1,100,000
EV(node6) =
Max 1100, 1150 $1,150,000
EV(node 10) =
0.6 2800 0.4 800 $2,000,000
EV(node 7) =
Max 2000, 1300 $2,000,000
EV(node 4) =
0.6 2350 0.4 1150 $1,870,000
EV(node3) =
Max 1870, 2000 $2,000,000
EV(node 2) =
0.8 2000 0.2 200 $1,560,000
EV(node 1) =
Max 1560, 0 $1,560,000
Hence, the expected profit is $1,560, 000 . The decision would be to bid on...