Final Exam Solutions Summary
1. Answer the following questions using the following problem and its managerial summary
solution. MAX 8X1 + 9X2 + 10X3,
S.T.,
3X1+2X2+2X3 < 2250, 4X1+ 4X2+ 6X3 < 3600, X1+ X2+2X3 < 950, X1+X2 < 800,
all variables are non-negative, and the following managerial information:
Solution
Variable
Total
Allowable
Allowable
Profit Profit
Min. c(j)
Max.c(j)
-M
9.0
X1
0
8.0
0
X2
800.
9.0
7200.0
8.00
M
X3
66.67 10.0
666.67
0
13.5
Objective
Function
(Max.) =
Left Hand
(40%)
7866.67
Shadow
Allowable
Constraint
Slack/Surplus
Price
Min. RHS Max. RHS
C1
1733.33
516.67
0
1733.33
C2
3600
0
1.67
3200
3650
M
C3
933.33
16.67
0
933.33
M
C4
800
0
2.33
750
900
1.1 What are the optimal solution and the optimal value?
Optimal solution is: X1 = 0, X2 = 800, X3 = 66.67, with optimal value of 7866.67
1.2 What is the allowable increase for each of the coefficients of the decision variables can reach
before changing the optimal strategy.
For cost coefficient c1 = 8 it is 1. For cost coefficient c2 = 9 it is unlimited. While for cost coefficient
c3 = 10 it is 3.5
1.3 What is the value of the unused portion of resource C1?
It is 516.67. However, its monetary value is zero.
1.4 How much are you willing to pay for additional units of resource number four?
$2.33 up to 100 units.
1.5 What is the impact on the optimal solution if the objective function is changed to
7X1 + 9X2 + 10X3 . Why?
The cost coefficient c1 in the new objective function is within its allowable range. Therefore, there
is no impact on current optimal solution.
1.6 Is it profitable to produce a new product with net profit of $11 requiring 2 units of every
resource you have? Why?
Considering the shadow prices, It take 2(0) + 2(1.67) + 2(0) + 2(2.33) = $8 to produce a unit of the
new product. Since 8 <11, therefore it is profitable to produce the new product.
1.7 What is the impact of deleting the first constraint on the optimal solution? Why?
The first constraint is not a binding constraint (it has a non-zero slack); therefore deleting this
constraint does not have any impact on optimal solution.
1.8 What is the solution to the dual problem?
The solution to the dual is the shadow prices
2. Pick up a year of your choice and then write the maximum number of cars manufactured during that
year in USA.
(5%)
The year of your choice could be, say 1875, and the number of cars = 0.
3. The ABC wants to market a new series program, whose success probability was estimated to be 0.90.
Suppose a marketing research agency is hired to evaluate the program. The past experience indicates that
the agency has 80% of accuracy in predicting a success and 90% of accuracy in predicting a failure. What
is the probability that the program is successful given a favorable prediction? What is the probability that
the program is a failure given an unfavorable prediction?
(15%)
Positive
States of Nature Prior
Probability
Successful
.90
Unsuccessful
.10
Conditional
Probability
.80
.0.1
Joint
Probability
.72
.01
.73
Posterior
Probability
.99
.01
Negative
States of Nature Prior
Probability
Successful
.90
Unsuccessful
.10
Conditional
Probability
.20
.90
Joint
Probability
.18
.09
.27
Posterior
Probability
.67
.33
4. A manager is considering marketing (M) a product or not marketing (NM). But he is not sure whether
the market is good (g) or nor good (ng). His estimated payoff matrix is shown as follows.
M
NM
g
34,000
0
ng
–9,000
0
p(.)
0.4
a.
b.
According to the Min-expected regret criterion, what is the manager's decision? (10%)
Now the manager wants to have additional information for his decision making. How
much should he pay at most for the additional information?
(5%)
0.6
Payoff table
M
M1
G
34000
0
G1
-9000
0
G
0
34000
G1
9000
0
Regret
M
M1
ER(M) = 0 (.40) + 9000 (.60) = 5,400
ER(M1) = 34000 (.40) + 0 (.60) = 13,600
a. M = 5,400 Min-expected regret
b. Since in general EVPI = EOL, the EVPI is $5,400.
Alternatively, EVPI can be computed directly, as follow:
EREV 34000(.40) + -9000 (.60) = 8,2000
EREV 0 (.40) + 0 (.60) = 0
b. 13,600 – 8,200 = 5,400
3. Buzzy-B Toys must decide the course of action to follow in promoting a new whistling yo-yo.
Initially, management must decide whether to market the yo-yo or to conduct a test marketing program.
After test marketing the yo-yo, management must decide whether to abandon it or nationally distribute it.
A national success will increase profits by $500,000, and a failure will reduce profits by $100,000.
Abandoning the product will not affect profits. The test marketing will cost Buzzy-B a further $10,000.
If no test marketing is conducted, the probability for a national success is judged to be 0.45. The assumed
probability for a favorable test marketing result is 0.50. The conditional probability for national success
given favorable test marketing, is 0.80, for national success given unfavorable test results, it is 0.10.
Construct the decision tree diagram and perform backward induction analysis to determine the optimal
course of action.
(25%)
Net Payoffs
Success, 0.80
$490,000
Market
Failure, 0.20
-$110,000
Favorable, 0.50
Abandon
Test market
-$10,000
Success, 0.10
Unfavorable, 0.50
$490,000
Market
Failure, 0.90
-$110,000
Abandon
Don't test market
Success, 0.45
-$10,000
$500,000
Market
Failure, 0.55
Abandon
$-100,000
$0
Net Payoffs
$370,000
$370,000
Success, 0.80
$490,000
Market
Failure, 0.20
//
Favorable, 0.50
$180,000
Test market
$180,000
-$110,000
Abandon
-$10,000
-$50,000
Unfavorable, 0.50
//
-$10,000
Market
\\
Success, 0.10
Failure, 0.90
-$110,000
Abandon
Don't test market
$170,000
Market
$490,000
Success, 0.45
-$10,000
$500,000
Failure, 0.55
$170,000
Abandon
$-100,000
$0
Optimal Strategy:
Test the market
If Favorable, Market
Otherwise Abandon.