handout05solutions

# handout05solutions - max 3 x 1 4 x 2-x 3 s.t 4 x 1-2 x 2 3...

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IEOR 162 Linear Programming Spring 2007 Discussion Section Handout 5 Solutions 1. Problem written in standard form: max 5 x 1 + 3 x 2 s.t. x 1 + 2 x 2 + x 3 = 3 2 x 1 + x 2 + x 4 = 3 x 1 , x 2 , x 3 , x 4 0 Iteration 1: B = { 3 , 4 } , N = { 1 , 2 } , z = 0 , x B = A - 1 B b = ¯ b = ± 3 3 ² compute reduced costs: ¯ c N = c N - c B A - 1 B A N = ³ 5 3 ´ choose x 1 to enter the basis d = A - 1 B A 1 = ± 1 2 ² α * = min { ¯ b i d i : d i > 0 } = min { 3 1 , 3 2 } = 3 2 x 1 = 3 2 x 3 = 3 - 3 2 = 3 2 , x 4 = 3 - 2 3 2 = 0 x 4 leaves the basis z = 0 + ¯ c 1 x 1 = 5 3 2 = 15 2 Iteration 2: B = { 1 , 3 } , N = { 2 , 4 } , z = 15 2 , x B = A - 1 B b = ¯ b = ± 3 2 3 2 ² compute reduced costs: ¯ c N = c N - c B A - 1 B A N = ³ 1 2 - 5 2 ´ choose x 2 to enter the basis d = A - 1 B A 2 = ± 1 2 3 2 ² α * = min { ¯ b i d i : d i > 0 } = min { 3 , 1 } = 1 x 2 = 1 x 1 = 3 2 - 1 2 = 1 , x 3 = 3 2 - 3 2 = 0 x 3 leaves the basis z = 15 2 + ¯ c 2 x 2 = 8 Iteration 3: B = { 1 , 2 } , N = { 3 , 4 } , z = 8 , x B = A - 1 B b = ¯ b = ± 1 1 ² compute reduced costs: ¯ c N = c N - c B A - 1 B A N = ³ - 1 3 - 7 3 ´ x B is optimal . Optimal solution is x = (1 , 1 , 0 , 0) , z = 8. 2. Problem written in standard form:

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Unformatted text preview: max 3 x 1 + 4 x 2-x 3 s.t. 4 x 1-2 x 2 + 3 x 3 + s 1 = 10 x 1 + 4 x 2-2 x 3 + s 2 = 8 x 1 , x 2 , x 3 , s 1 , s 2 ≥ Solve using tableaus: ↓ z x 1 x 2 x 3 s 1 s 2 RHS ratio 1-3-4 1 4-2 3 1 10-1 4-2 1 8 2 1 ↓ z x 1 x 2 x 3 s 1 s 2 RHS ratio 1-2-1 1 8 9 2 2 1 1 2 14 7 1 4 1-1 2 1 4 2-z x 1 x 2 x 3 s 1 s 2 RHS 1 1 4 1 2 5 4 15 9 4 1 1 2 1 4 7 11 8 1 1 4 3 8 11 2 Optimal solution is (0, 11 2 ,7,0,0) with objective value 15. 2...
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handout05solutions - max 3 x 1 4 x 2-x 3 s.t 4 x 1-2 x 2 3...

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