Operations Management

OPR 320: LINEAR MODELS IN DECISION MAKING Fall 2016-2017

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ASSIGNMENT # 1 (due before class on Tuesday, October 4th)

1) Every LP must fall into one of the following cases:

Case 1: The LP has unique optimal solution.

Case 2: The LP has alternative or multiple optimal solutions.

Case 3: The LP is infeasible.

Case 4: The LP is unbounded.

Using graphical method, please identify which of Cases 1-4 apply to each of the

following LPs, and provide the solution, if it exists.

For each part, circle the correct option, and show work on a separate page.

(a) max x1 + x2

s.t. x1 + x2 ≤ 4

x1 – x2 ≥ 5

x1 , x2 ≥ 0

(b) max 4×1 + x2

s.t. 8×1 + 2×2 ≤ 16

5×1 + 2×2 ≤ 12

x1 , x2 ≥ 0

(c) max -x1 + 3×2

s.t. x1 – x2 ≤ 4

x1 + 2×2 ≥ 4

x1 , x2 ≥ 0

(d) max 3×1 + x2

s.t. 2×1 + x2 ≤ 6

x1 + 3×2 ≤ 9

x1 , x2 ≥ 0

Case 1 / Case 2 / Case 3 / or Case 4 If exists, the optimal solution is: .

Case 1 / Case 2 / Case 3 / or Case 4 If exists, the optimal solution is: .

Case 1 / Case 2 / Case 3 / or Case 4 If exists, the optimal solution is: .

Case 1 / Case 2 / Case 3 / or Case 4 If exists, the optimal solution is: .

OPR 320: LINEAR MODELS IN DECISION MAKING Fall 2016-2017

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2) The marketing manager for Mountain Mist soda needs to decide how many TV spots

and magazine ads to run during the next quarter. Each TV spot costs $5,000 and is

expected to increase sales by 300,000 cans. Each magazine ad costs $2,000 and is

expected to increase sales by 500,000 cans. A total of $100,000 may be spent on TV

and magazine ads; however, Mountain Mist wants to spend no more than $70,000 on

TV spots and no more than $50,000 on magazine ads. Mountain Mist earns a profit of

$0.05 on each can it sells. a) Formulate an LP model for this problem. Clearly label and define the decision

variables, the objective function and each of the constraints.

b) Sketch the feasible region for this model using the graph paper available in Learn.

What is the optimal solution?

3) Jim solved a linear program involving profit maximization, subject to various

constraints, and obtained $120.2 as the optimal objective value. After solving the

problem, he noticed that he forgot to include the most important constraint in his

formulation. Then, he added that constraint and re-solved the problem. The new

optimal objective value is $160.7. Do you trust in the new objective value Jim has

come up with? Please explain your reasoning.

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