CH10 - Chapter 10 MP Solutions SECTION 10.1 SOLUTIONS 1....

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Unformatted text preview: Chapter 10 MP Solutions SECTION 10.1 SOLUTIONS 1. max z = 3x 1 + x 2 + x 3 s.t. x 1 + x 2 + x 3 + s 1 = 6 2x 1- x 3 + s 2 = 4 x 2 + x 3 + s 3 = 2 x 1 ,x 2 ,x 3 > BV(0) = {s 1 , s 2 , s 3 } NBV(0) = {x 1 , x 2 , x 3 } = =- 1 1 1 1 B B c BV = [0 0 0] c BV B-1 = [0 0 0] c 1 = [0 0 0] 2 1-3 = -3 c 2 = [0 0 0] 1 1-1 = -1 c 3 = [0 0 0] - 1 1 1-1 = -1 Column for x 1 = 1 1 1 2 1 = 2 1 1 Right Hand Side = 1 1 1 2 4 6 = 2 4 6 BV(1) = {s 1 , x 1 , s 3 } NBV(1) = {x 2 , x 3 , s 2 } B 1-1 = - 1 2 / 1 2 / 1 1 c BV B 1-1 =[0 3 0] - 1 2 / 1 2 / 1 = [0 3/2 0] c 2 = [0 3/2 0] 1 1-1 = -1 c for s 2 = 3/2 _ c 3 = [0 3/2 0] - 1 1 1-1 = -5/2 Column for x 3 = - 1 2 / 1 2 / 1 1 - 1 1 1 = - 1 2 / 1 2 / 3 Right Hand side tableau 1 = - 1 2 / 1 2 / 1 1 2 4 6 = 2 2 4 BV(2) = {s 1 , x 1 , x 3 } NBV(2) = {x 2 , s 2 , s 3 } B 2-1 = -- 1 2 / 1 2 / 1 2 / 3 2 / 1 1 c BV B 2-1 = [0 3 1] -- 1 2 / 1 2 / 1 2 / 3 2 / 1 1 2 =[0 3/2 5/2] c 2 = [0 3/2 5/2] 1 1- 1 = 3/2, c for s 2 >0, c for s 3 >0 so this an optimal tableau. Right Hand Side = -- 1 2 / 1 2 / 1 2 / 3 2 / 1 1 2 4 6 = 2 3 1 X X S 3 1 1 = 2 3 1 z = c BV B 2-1 b = [0 3/2 5/2] 2 4 6 = 11 2. max z = 4x 1 + x 2- Ma 2- Ma 3 st x 1 + x 2 + s 1 = 4 2x 1 + x 2- e 2 + a 2 = 6 3x 2- e 3 + a 3 = 6 BV(0) = {s 1 , a 2 , a 3 } B-1 = 1 1 1 c BV B-1 = [0 -M -M] _ _ _ _ c 1 = -2M -4 c 2 = -4M - 1 c for s 1 = 0 c for e 2 = M _ c for e 3 = M Enter x 2 into basis. x 2 column = 3 1 1 RHS = 6 6 4 Thus x 2 enters basis in row 3 3 BV(1) = {s 1 , a 2 , x 2 } B 1-1 = -- 3 / 1 3 / 1 1 3 / 1 1 c BV B 1-1 = [0 -M 1] -- 3 / 1 3 / 1 1 3 / 1 1 = [0 -M (M+1)/3] Pricing out the non-basic variables we obtain _ _ _ c 1 = -2M - 4, c for e 2 = -M, c for e...
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CH10 - Chapter 10 MP Solutions SECTION 10.1 SOLUTIONS 1....

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