midtermSoln09

midtermSoln09 - CO 350 Linear Optimization Winter 2009...

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Unformatted text preview: CO 350 Linear Optimization Winter 2009 Midterm Examination February 11 7:00 – 9:00 p.m. SOLUTIONS. 1. [10 marks = 4+6] (a) Convert the following LP problem to standard equality form. [4] maximize 2 x 1 − 3 x 2 subject to 2 x 1 − x 2 + 4 x 3 ≤ 4 x 2 − 5 x 3 + x 4 ≥ 7 x 1 , x 2 , x 3 ≥ Solution: We let x 4 = u 4 − v 4 where u 4 , v 4 ≥ 0, and introduce two slack variables x 5 , x 6 ≥ 0. maximize 2 x 1 − 3 x 2 subject to 2 x 1 − x 2 + 4 x 3 + x 5 = 4 − x 2 + 5 x 3 + u 4 − v 4 + x 6 = − 7 x 1 , x 2 , x 3 , u 4 , v 4 , x 5 , x 6 ≥ (b) Convert the following LP problem to standard inequality form. [6] minimize 2 x 2 − x 4 + 9 subject to 2 x 1 − x 2 + 4 x 3 = 4 x 2 − x 3 + x 4 = 3 x 1 + 3 x 3 + 4 x 4 ≥ 1 x 1 ≤ 5 x 2 , x 4 ≥ Solution: Let x ′ 1 = − x 1 + 5, that is x 1 = − x ′ 1 + 5, then x 1 ≤ 5 iff x ′ 1 ≥ 0. Solution 1: We eliminate the free variable x 3 using the second equality constraint: x 3 = x 2 + x 4 − 3 Then the first constraint becomes 2 x 1 − x 2 + 4 x 2 + 4 x 4 − 12 = 4, that is, 2 x 1 + 3 x 2 + 4 x 4 = 16 With x 1 = − x ′ 1 + 5, this becomes − 2 x ′ 1 + 3 x 2 + 4 x 4 = 6. The third constraint becomes x 1 +3 x 2 +3 x 4 − 9+4 x 4 ≥ 1, that is, x 1 +3 x 2 +7 x 4 ≥ 10. With x 1 = − x ′ 1 + 5, this becomes − x ′ 1 + 3 x 2 + 7 x 4 ≥ 5. Altogether: maximize − 2 x 2 + x 4 subject to − 2 x ′ 1 + 3 x 2 + 4 x 4 ≤ 6 2 x ′ 1 − 3 x 2 − 4 x 4 ≤ − 6 x ′ 1 − 3 x 2 − 7 x 4 ≤ − 5 x ′ 1 , x 2 , x 4 ≥ Solution 2: We write x 3 = u 3 − v 3 where u 3 , v 3 ≥ 0. In all constraints, we substitute x 1 = − x ′ 1 + 5, and x 3 = u 3 − v 3 . Altogether: maximize − 2 x 2 + x 4 subject to − 2 x ′ 1 − x 2 + 4 u 3 − 4 v 3 ≤ − 6 2 x ′ 1 + x 2 − 4 u 3 + 4 v 3 ≤ 6 x 2 − u 3 + v 3 + x 4 ≤ 3 − x 2 + u 3 − v 3 − x 4 ≤ − 3 x ′ 1 − 3 u 3 + 3 v 3 − 4 x 4 ≤ 4 x ′ 1 , x 2 , u 3 , v 3 , x 4 ≥ 2. [11 marks = 4+2+2+3] (a) Write down the dual problem of the following LP problem: [4] maximize z = 6 x 1 + 4 x 2 − x 3 ( P ) subject to x 1 + x 2 + 2 x 3 ≤ 5 x 1 + 3 x 3 ≤ 3 2 x 1 + x 2 + 7 x 3 ≤ 10 x 1 , x 2 , x 3 ≥ Solution: minimize 5 y 1 + 3 y 2 + 10 y 3 subject to y 1 + y 2 + 2 y 3 ≥ 6 y 1 + y 3 ≥ 4 2 y 1 + 3 y 2 + 7 y 3 ≥ − 1 y 1 , y 2 , y 3 ≥ (b) Is ˆ x = [3 , 2 , 0] T a feasible solution of ( P )? Justify your answer. [2] Solution: Yes. 3 + 2 + 0 = 5 ≤ 5 3 + 0 = 3 ≤ 3 2 · 3 + 2 + 0 = 8 ≤ 10 (c) Is ˆ y = [4 , 2 , 0] T a feasible solution of the dual problem? Justify your answer. [2] Solution: Yes. 4 + 2 + 0 = 6 ≥ 6 4 + 0 = 4 ≥ 4 2 · 4 + 3 · 2 + 0 = 14 ≥ − 1 (d) Find an optimal solution x ∗ of ( P ), if there exists one. Briefly explain why x ∗ [3] is optimal. If (P) has no optimal solution, then explain why it has none....
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This note was uploaded on 02/05/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Spring '07 term at Waterloo.

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midtermSoln09 - CO 350 Linear Optimization Winter 2009...

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