hw6.sol[1]

# hw6.sol[1] - IE 335 Operations Research - Optimization...

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Unformatted text preview: IE 335 Operations Research - Optimization Solutions to Homework 6 Fall 2011 Problem 25 (a) 0.5 1 1.5 2 2.5 3 3.5 4 0.5 1 1.5 2 2.5 3 3.5 4 y 1 y 2 Graphical Solution to Problem 25 y 1 + y 2 = 2 y 1 = 3 (b) Add slacks y 3 and y 4 to place the model in standard form max 2 y 1 + 8 y 2 s.t. y 1 + y 2 + y 3 = 2 y 1- y 4 = 3 y 1 ,y 2 ,y 3 ,y 4 (c) min y 5 + y 6 s.t. y 1 + y 2 + y 3 + y 5 = 2 y 1- y 4 + y 6 = 3 y 1 ,y 2 ,y 3 ,y 4 ,y 5 ,y 6 A = 1 1 1 1 0 1 0 0- 1 0 1 b = 2 3 T 1 c = 0 0 0 0 1 1 T (d) The simplex computation is presented in the following table: y 1 y 2 y 3 y 4 y 5 y 6 minimize d 1 1 b 1 1 1 1 2 1-1 1 3 N N N N B B y (0) 2 3 d y (0) = 5 y for y 1 1-1-1 c 1 =-2 y for y 2 1-1 c 2 =- 1 y for y 3 1-1 c 3 =- 1 y for y 4 1 1 c 4 = 1 ---- 2 1 3 1 B N N N N B y (1) 2 1 d y (1) = 1 y for y 2-1 1 1 c 2 = 1 y for y 3-1 1 1 c 3 = 1 y for y 4 1 1 c 4 = 1 y for y 5-1 1 1 c 4 = 2 Infeasible The original model is infeasible because the Phase I optimum has an artificial total 1. Problem 26 5-21.(b) No, because all basic variables are positive. Please refer to definition 5.35 for the definition of a degenerate solution and the following sample exercise.degenerate solution and the following sample exercise....
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## This note was uploaded on 12/14/2011 for the course IE 335 taught by Professor Jean-philippe,r during the Fall '08 term at Purdue.

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hw6.sol[1] - IE 335 Operations Research - Optimization...

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