hw5.sol - IE 335 Operations Research - Optimization...

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Unformatted text preview: IE 335 Operations Research - Optimization Solutions to Homework 5 Spring 2012 Problem 21 (a) 5-8.(a) The current basic solution: A = 13 2 3 1- 4 1- 1 b = 7- 1 T c = 8- 5 1 T We set the non basic variables x 1 , x 2 → 0 and using Ax ( ) = b we solve: x ( ) = 2 1 T (b) 5-8.(b) The simplex directions are: Δ x for x 1 = 1- 3- 4 T Δ x for x 2 = 1- 1 1 T (c) 5-8.(d) We compute the reduced cost for each direction by calculating ¯ c i = c · Δ x c · Δ x for x 1 = ¯ c 1 = 4 = ⇒ Does not improve minimization c · Δ x for x 2 = ¯ c 2 =- 4 = ⇒ Improves minimization Therefore, the simplex direction for x 2 improves the objective function. (d) 5-8.(e) We determine the maximum step for both directions and the corresponding new basis: Δ x for x 1 : λ = min 2- (- 3 ) , 1- (- 4 ) = 1 4 New basis = { x 1 , x 3 } Δ x for x 2 : λ = min 2- (- 1 ) = 2 New basis = { x 2 , x 4 } Problem 22 (a) 5-10.(a) 1 1 2 3 4 5 6 1 2 3 4 5 6 z 1 z 2 Graphical Solution to Problem 22 3z 1 + 2z 2 = 18 z 1 = 5 z 2 = 3 2z 1 + z 2 = 10 2z 1 + 5z 2 = 17 2z 1 + 5z 2 = 23 z (2) = (4,3) z (0) = (0,0) z (1) = (0,3) (b) 5-10.(b) The standard form model: max 2 z 1 + 5 z 2 s.t. 3 z 1 + 2 z 2 + z 3 = 18 z 1 + z 4...
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This note was uploaded on 02/22/2012 for the course IE 335 taught by Professor Jean-philippe,r during the Spring '08 term at Purdue University.

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hw5.sol - IE 335 Operations Research - Optimization...

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