ADM 2302 Midterm Exam
ADM 2302 Midterm Exam (sample 1)
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ADM 2302
MIDTERM EXAM
Problem 1 (6 points) There are four marketing research firms (MR1, MR2, MR3, MR4) that HairWays has faith in to advertise
its products. HairWays has just come out with a new hairspray and they wish to have 30 newspaper ads, 15
television ads, and 25 radio ads available within three months. Given the size of the firms it is expected that
MR1 can produce 20 total ads, MR2 can produce 25 total ads, MR3 can produce 15 total ads, and MR4 can
produce 20 total ads. The bids submitted by the marketing research firms (in thousands of dollars per ad)
are:
MR1 MR2 MR3 MR4
Newspaper 16 10 12 12
Television 26 20 30 21
Radio 22 15 23 14
a) Formulate this problem as a linear program. Define the decision variables, objective function, constraints. DO NOT SOLVE (5 points)
b) Show that this LP fits the structure of a transportation problem. (1 point)
ADM 2302 Midterm Exam (sample 1)
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Problem 2: (6 points)
You have been hired to survey cable television customers to determine their attitudes toward a
restructuring of the basic cable service and rate schedule. The cable company wants to classify
participants according to whether or not they have children under the age of 16 in the household
and as to whether or not the household subscribes to any premium channels.
The basic cost of conducting an interview is $5. Those households with children under 16 are
asked a second series of questions which adds an additional cost of $4 to their interviews.
Households with premium channels are questioned about their viewing habits, and this adds an
additional $3 to the cost of their interviews. Cost is to be minimized.
The cable company wants responses from at least 50 customers. Premium subscribers should
constitute at least half of the customers surveyed. At least 60% of households surveyed should
have children under 16. Of those households with children under 16, no more than half should be
premium subscribers.
Formulate the corresponding linear programming model. (Define decision variables the objective
function and the constraints). DO NOT SOLVE
(Hint: use the cost information to assist in finding the decision variables)
ADM 2302 Midterm Exam (sample 1)
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Problem 3 (10 points)
Buster Sod operates an 800-acre irrigated farm in the Red River Valley of Arizona. Sod’s principal
activities are raising wheat, alfalfa and beef. The Red River Valley Water Authority has just given its
water allotments for the next year (Sod was allotted 1000 acre-feet), and Sod is busy preparing his
production plan for next year.
He figures that beef prices will hold at around $500 per ton and that wheat will sell at $2 per bushel.
Best guesses are that he will be able to sell alfalfa at $22 per ton, but if he needs more alfalfa to feed
his beef than he can raise, he will have to pay $28 per ton to get the alfalfa to his beef feedlot.
The wheat yield is 70 bushels per acre; alfalfa yield is 4 tons per acre. Other features of Sod’s
operation are given in the table below:
Water Land Alfalfa
Labor, Machinery Requirements Requirements Requirements
Activity And other Costs ($) (Acre-feet) (Acres) (Tons)
1 acre of Wheat 20 2 1
1 ton of beef 50 0.05 0.1 5
1 acre of alfalfa 28 3 1
The problem when formulated as an LP and solved is as follows:
Definition of the Decision Variables
W: the number of acres of wheat raised and sold
B: the number of tons of beef raised and sold
A: the number of tons of Alfalfa raised
Ab: the number of tons of Alfalfa bought
As: the number of tons of Alfalfa sold
Maximize Z = $120W + $450B – $7A – $28Ab + $22As (refer to the following page for calculations
that explain the of the objective function coefficients)
Subject to
W + 0.1B + 1/4A <= 800 (Acres Limit Constraints)
2W + 0.05B + 3As <= 1000 (Water Limit Constraints)
A + Ab = 5B + As (Alfalfa Requirements Constraints)
W, B, A, Ab, and As >= 0 (Non-negativity Constraints)
ADM 2302 Midterm Exam (sample 1)
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Please note that the objective function coefficients were computed as follow:
• $2 / 70 / $140 /bushel bushels acre acre = is the revenue from 1 acre of wheat. But 1 acre of
wheat costs $20 in labor, machinery, and other costs. So the net profit from planting l acre of wheat
is $120. (cW= 120) • $500 is the revenue of 1 ton of beef. But 1 ton of beef costs $50 in labor, machinery and other costs.
So the net profit from raising and selling one ton of beef is $450. (cB = 450)
• 1 ton of alfalfa takes up 1/4 acre of land. And 1 acre of alfalfa costs $28 in labor, machinery and
other costs. So the cost of 1 ton of alfalfa is: 7$28 4
1 = . A cost of $7 is equivalent to a profit of –
$7/ton of Alfalfa. (cA = -7).
a) Does Soy Buy or Sell Alfalfa? Justify. (0.5 points)
b) How much beef is being produced? And how much profit will Sod receive from the optimal operations of his farm? (1 points)
c) How much should Sod pay to acquire another acre-feet of water? Justify. (0.5 points)
d) Sod is considering buying an additional 1400 acres of land at a total price of $2,940,000. What would you recommend? Justify. (1.5 points)
e) Red River Valley Authority decided to decrease its water allotment to 500 acre-feet, what will happen to the optimal solution (not the shadow prices)? Justify. (1 points)
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f) What will happen to the optimal solution if the price of wheat triples (e.g. it increases from $2 per bushel to $6 per bushel)? What happens to the optimal value of the objective function? (1.5 points)
g) What will happen to the optimal solution and the optimal value of the objective function if the cost of alfalfa purchased increases form $28 to $29? (1 points)
h) How much can the cost of buying alfalfa decrease before the current optimal solution will change? (1 points)
i) Sod is considering raising and selling barely. The profit contribution of barely is $320/ton. Barely yield 10 tons per acre. For 1 acre of barely the water requirement per acre-feet is 6. Should Sod
start raising barely? Justify. (2 points)
