IE 335-Exam 2 Version A - Version A Industrial Engineering. Solutions to Exam 2

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Version A Industrial Engineering 335 Operations Research – Optimization

Spring 2012

Solutions to Exam 2 Problem 1. a. False. See Lecture 27. b. False. See Lecture 17. c. True. See Lecture 24. d. True. See Lecture 20. e. False. See Lecture 22.

Problem 2. a. The decision variables of the given LP can be interpreted as follows: x i = amount of Product i produced

for i = A, B, C.

The objective function maximizes the total profit. Constraint (1) models the constraint on Resource X. Constraint (2) models the constraint on Resource Y. Constraint (3) models the fact that the amount of Product C that the company produces must be greater than or equal to twice the amount of Product B it produces. The last constraint models the fact that negative quantities of any product cannot be produced. b. The dual of the given linear program is: min s.t.

23v1 + 36v2

2v1 + 3v2 ≥ 3

3v1 + v2 − 2v3 ≥ 2

v1 + 4v2 + v3 ≥ 5 v1 , v2 ≥ 0, v3 ≤ 0 c. The optimal profit is (3 × 0) + (2 × 4) + (5 × 8) = 48.

d. The optimal solution to the dual linear program is v1∗ = 0, v2∗ = 1.333, v3∗ = −0.333. e. No. Buying 3 extra units of Resource Y corresponds to increasing the RHS of constraint (2) to 39, which is in the allowable range for this constraint. Therefore, buying 3 extra units of Resource Y will increase the profit by $1.333 × 3 = $3.999 < $6. f. Yes. Buying 6 extra units of Resource Y corresponds to increasing the RHS of constraint (2) to 42, which is outside the allowable range for this constraint. Since constraint (2) is a ≤-type constraint, we know that buying 6 extra units of Resource Y will increase the profit at least by $1.333 × (41.4 − 36) = $7.1982 > $6. g. Increasing the unit profit of Product C by 2 changes the coefficient of xC to 7, which is in the allowable range for xC . Therefore, the profit will change by 8 × ($7 − $5) = 8 × $2 = $16.

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Version A h. Decreasing the unit profit of Product A by 1 corresponds to decreasing the coefficient of xA to 2, which is in the allowable range for xA . Therefore, the current optimal solution remains valid under the new circumstances, i.e., xA∗ = 0, xB∗ = 4, xC∗ = 8 is still an optimal solution. i. The optimal value of dual variable v3 for constraint (3) represents how much the optimal profit will change if the RHS of constraint (3) is increased by 1. More specifically, v3 is how much the optimal profit will change if the amount of Product C that the company must now produce is at least 1 more than twice the amount of Product B it produces.

Problem 3. This problem is similar to the problem in Studio 2, and can also be seen as a transportation problem as described in Lecture 25. Below is a graphical representation of a valid minimum cost flow model for this problem. −5

A1

(0, +∞) (1 , + ∞) (9 ,+ ∞ )

(3 ,

(7

(1

∞ ,+

(0, +∞) ) +∞ (3 ,

)

−9

,

) +∞ ) +∞

B1

A2

7

B2

3

C2

3

D2

5

(9 ,

)

C1

+∞

−1

+∞ ) (0, +∞) (9

(9 ,

) ∞ ,+ 1 ( ∞) (1 , + (0, +∞)

−3

D1

(7 ,

,+ ∞ ) +∞ )

Nodes A1 , B1 , C1 , D1 represent the four stations today, and nodes A2 , B2 , C2 , D2 represent the four stations tomorrow. The number next to each node represents the net demand of that node. The arc labels represent (unit cost, capacity). The flow on each arc (i, j) represents the number of ambulances moved from i to j: for example, the flow on arc (A1 , B2 ) represents the number of ambulances moved from Station A to Station B. The minimum cost flow of the above model represents a strategy for the emergency services department to minimize the total cost of repositioning the ambulances. Note that we only need to consider movements from today to tomorrow; we do not need to consider movements within the same day.

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