HW 07 Solution

# HW 07 Solution - -5 4 x2 5-5 x5 x6 x 3 =λ=5 New Objective...

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1  HW 07 S o lutio n Minimize c*x s.t. A*x = b 1 1 1 = 10 -1 1 = -1 -1 -1 1 = -4 -1 -1 1 = -5 A1 A2 A3 A4 A5 A6 1 1 1 =A -1 1 -1 -1 1 -1 -1 1 1 2 4 5 3 6 =c

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HW 07 S o lutio n 2 BASIC = {x1, x2, x5, x6} B={A1,A2,A5,A6}= 1 1 -1 -1 1 -1 1 B -1= -1 1 1 1 1 1 1 1 1 1
HW 07 S o lutio n 3 x B =B -1 b 1 x1 9 x2 5 x5 0 x6 x N = 0 x3 0 x4 c1 c2 c5 c6 c B = 1 2 3 6 Objective Value =c B *x B = 34

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HW 07 S o lutio n 4 v=c B B -1 = 11 10 9 6 cbar3=c 3 -v*A 3 = -1 cbar4=c 4 -v*A 4 = 4 Since cbar3<0, x3 enters the basis! p=3 y=-B -1 A3= 0 x1 -1 x2 -1 x5 0 x6 For each yi < 0 λ ≤ x 2 /-y 2 = 9/1 =9 λ ≤ x 5 /-y 3 = 5/1 =5 λ =5 x5 leaves the basis!
HW 07 S o lutio n 5 x Bnew =x B +λy 1 + 0 = 1 x1 9

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Unformatted text preview: -5 4 x2 5-5 x5 x6 x 3 =λ=5 New Objective Value = Old Objective Value + λ*cbar3 = 34-5=29 BASIC = {x1, x2, x3, x6} HW 07 S o lutio n 6 B={A1,A2,A3,A6}= 1 1 1-1-1-1 1 B-1=-1-1 1 1 1 1 1 1 1 v=c B B-1 = 10 9 8 6 HW 07 S o lutio n 7 cbar4=c 4-v*A 4 = 4 cbar5=c 5-v*A 5 = 1 Since all cbar>0, solution is optimal Find cbar for variables not in basis x= 1 x1 4 x2 5 x3 x4 x5 x6...
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## This note was uploaded on 10/06/2008 for the course EMIS 3360 taught by Professor Dr.kennington during the Spring '08 term at SMU.

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HW 07 Solution - -5 4 x2 5-5 x5 x6 x 3 =λ=5 New Objective...

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