Homework_5_Solutions

# Homework_5_Solutions - because y 1 6 = 0 and y 2 6 = 0. 2...

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HOMEWORK 5 SOLUTIONS 1. Problem 1 (a) y 1 = . 5 ,y 2 = 1. See the dual linear program and graphical solution below. 4 pts min z D = 6 y 1 + 4 y 2 2 y 1 1 3 y 1 + y 2 1 y 2 1 y 1 ,y 2 0 (b) x 2 = 0, because the second constraint of the dual linear program is not tight (3( . 5) + 1 = 2 . 5 > 1). 2 pts (c) x 1 = 3 ,x 3 = 4. Use the fact that the primal constraints must be tight
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Unformatted text preview: because y 1 6 = 0 and y 2 6 = 0. 2 pts (d) Yes, the primal objective here is 7 (3 + 0 + 4), which is the same as the dual objective from part (a) (6( . 5) + 4(1)), so this primal solution is optimal by strong duality. 2 pts 2. Problem 3 (a) y 1 = 11 10 ,y 2 = 9 20 ,y 3 = 1 4 10 pts 1...
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## This note was uploaded on 02/08/2012 for the course IEOR 4004 taught by Professor Sethuraman during the Fall '10 term at Columbia.

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