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Homework 3

# Homework 3 - 7 Pp 68#7 The optimal solution to a bounded LP...

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x 2 6 5 6 x 1 4 3 5 4 3 2 1 2 1 x 1 x 2 1200 1000 1200 1000 800 600 400 200 1000 800 600 400 200 x 2 50 40 30 10 20 60 50 40 30 20 10 x 1 x 1 x 2 x 1 x 2 18 9 12 6 x 2 x 1 x 2 x 2 x 1 x 1 DOR# 182 Homework #3 1. Pp. 63 #1 max 30x 1 + 100x 2 s.t. x 1 +x 2 7 4x 1 +10x 2 40 x 1 30 x 2 0 Solution x 1 = 3 x 2 = 2.8 30(3) + 100(2.8) = 370 2. Pp 63 #2 max 300x 1 + 500x 2 s.t. x 1 +(8/7)x 2 800

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x 1 +(5/4)x 2 1500 x 1 ,x 2 0 Solution x 1 =0 x 2 =750 300(0)+500(750) = 375,000
Pp 63 #6 max 200x 1 +300x 2 s.t. x 1 +x 2 45 3x 1 +2x 2 100 2x 1 +4x 2 120 x 1 ,x 2 0 Solution x 1 =20 x 2 =20 200(20)+300(20)=10000 3. Pp 68 #1-4 1. Case 3: Infeasible LP 2. Case 2: Multiple Optimal Solutions 3. Case 4: Unbounded LP 4. Case 1: Unique Optimal Solution 5. Pp 68 #5 True. If an LP has a bounded feasible region, then it is bounded. Otherwise, it is unbounded. 6. Pp 68 #6 False. An LP may have an optimal solution even if it has an unbounded feasible region. For example,

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if the feasible region expands infinitely in the positive direction, but the optimization function requires a minimization, the LP may have an optimal solution.

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Unformatted text preview: 7. Pp 68 #7 The optimal solution to a bounded LP always falls at an extreme point of the feasible region. The method of comparing z-values at extreme points may fail if the LP’s feasible region is unbounded because the z-value of a point in infinity may be more than the z-value at an extreme point. 8. Pp 68 #8 min z = x 1 – x 2 s.t. x 1 +x 2 6 ≤ x 1-x 2 ≥ x 2-x 1 3 ≥ x 1 ,x 2 ≥ Solution There are no feasible solutions to the LP, and therefore no optimal solutions. 9. Pp 68 #9 x 1 = 0 x 1 = 12 x 2 = 18 x 2 = 0 10. Pp 68 #10 max x 1 – 0.25x 2 x 1 ,x 2 ≥ There are an infinite number of solutions to the LP because there are no constraints on the number of initial dollars or francs....
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