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solutions2 - CO350/CM340/CO352 Linear Optimization Fall...

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CO350/CM340/CO352 Linear Optimization - Fall 2006 Midterm Solutions Problem 1: Basic Solutions (a) (i) No. (ii) Yes. (iii) No. (b) We have bracketleftbigg 0 2 2 0 bracketrightbiggbracketleftbigg x 3 x 4 bracketrightbigg = bracketleftbigg 8 4 bracketrightbigg . Now 2 x 4 = 8 implies x 4 = 4, and 2 x 3 = 4 implies x 3 = 2. So the corresponding basic solution is [0 , 0 , 4 , 2 , 0 , 0] T . (c) Yes, x is a basic solution with basis B = { 1 , 6 } . (d) No, x is not feasible for Ax = b . (e) No, since any basic solution of Ax = b can have at most 2 nonzero entries. Problem 2: Complementary Slackness (a) The dual is min 4 y 1 + 2 y 2 + y 3 s.t. y 1 + 2 y 2 + 2 y 3 1 y 1 y 3 2 2 y 1 y 2 + y 3 1 y 1 , y 2 , y 3 0 (b) The complementary slackness conditions are (1) x 1 = 0 or y 1 + 2 y 2 + 2 y 3 = 1 (2) x 2 = 0 or y 1 y 3 = 2 (3) x 3 = 0 or 2 y 1 y 2 + y 3 = 1 (4) y 1 = 0 or x 1 + x 2 + 2 x 3 = 4 (5) y 2 = 0 or 2 x 1 x 3 = 2 (6) y 3 = 0 or 2 x 1 x 2 + x 3 = 1 (c) First note that x = [0 , 4 , 0] T is feasible for the primal problem. If x is an optimal solution, there must exist a y that is feasible for the dual such that x and y satisfy the complementary slackness conditions. We now attempt to find such a
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