co227_f19_a9_sln.pdf - CO 227 Fall 2019 Assignment 9 Solutions 1(a min s.t 5y1 y1 2y1 3y1 \u2212 4y1 \u2212 y1 \u2265 0 2y2 3y3 2y2 \u2265 3 3y3 \u2265 \u22121 y2 = 4 3y2

co227_f19_a9_sln.pdf - CO 227 Fall 2019 Assignment 9...

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CO 227 Fall 2019: Assignment 9 Solutions 1. (a) min 5 y 1 + 2 y 2 + 3 y 3 s.t. y 1 + 2 y 2 3 2 y 1 + 3 y 3 - 1 3 y 1 - y 2 = 4 4 y 1 - 3 y 2 + y 3 1 y 1 0 , y 2 free , y 3 0 (b) max 5 y 1 + y 2 + 9 y 3 s.t. 2 y 1 + y 2 - 3 y 3 1 y 1 - y 2 + 5 y 3 = - 3 y 1 0 , y 2 0 , y 3 0 2. (a) min 2 y 1 + 5 y 2 s.t. y 1 + 2 y 2 4 4 y 1 - 3 y 2 - 6 y 1 - 3 y 2 - 15 y 2 1 - y 1 - y 2 - 13 y 1 , y 2 0 (b) We add slack variables x 6 , x 7 to the primal LP and obtain the following initial dictio- nary/tableau: z = 0 + 4 x 1 - 6 x 2 - 15 x 3 + x 4 - 13 x 5 x 6 = 2 - x 1 - 4 x 2 - x 3 + x 5 x 7 = 5 - 2 x 1 + 3 x 2 + 3 x 3 - x 4 + x 5 - 4 6 15 - 1 13 0 0 0 1 4 1 0 - 1 1 0 2 2 - 3 - 3 1 - 1 0 1 5 Solving this using simplex gives us the following final dictionary/tableau: z = 9 - 11 x 2 - 14 x 3 - 10 x 5 - 2 x 6 - x 7 x 1 = 2 - 4 x 2 - x 3 + x 5 - x 6 x 4 = 1 + 11 x 2 + 5 x 3 - x 5 + 2 x 6 - x 7 0 11 14 0 10 2 1 9 1 4 1 0 - 1 1 0 2 0 - 11 - 5 1 1 - 2 1 1 So the primal optimal solution with slack variables is (2 , 0 , 0 , 1 , 0 , 0 , 0) T , and without slack variables is ¯ x = (2 , 0 , 0 , 1 , 0) T . The dual optimal solution is the coefficient of the primal slack variables in the objective row of the tableau, which is ¯ y = (2 , 1) T . (c) We first check that ¯ x = (2 , 0 , 0 , 1 , 0) T is feasible for the primal. We have ¯ x 0, and 1 + 0 + 0 + 0 + 0 = 1 2 2 · 2 + 0 + 0 + 1 + 0 = 5 5
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We now check that ¯ y = (2 , 1) T is feasible for the dual. We have ¯
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