HW6_solution

HW6_solution - solution of homework 6 Problem 1 a. It only...

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Unformatted text preview: solution of homework 6 Problem 1 a. It only fits the convex case. b. Let D denote the demand vector D = ( d 1 ,d 2 ,d 3 ,d 4 ,d 5 ), x denote the optimal production vector x = ( x 1 ,x 2 ,x 3 ,x 4 ,x 5 ). M i denote the marginal contribution to cost of x i . M ? denote the minimum of M i when 1 i 5. It is obvious that when D = (0 , , , , 0), x = (0 , , , , 0) is the optimal solution. 1. D=(0,0,0,0,0), x=(0,0,0,0,0) 2. D=(1,0,0,0,0), since no backlog is allowed, x=(1,0,0,0,0) 3. D=(2,0,0,0,0), x=(2,0,0,0,0) 4. D=(2,1,0,0,0), M ? = min ( M 1 ,M 2 ,M 3 ,M 4 ,M 5 ) = min (16+1 , 16 , , , ) = 16. So x=(2,1,0,0,0) 5. D=(2,2,0,0,0), M ? = min ( M 1 ,M 2 ,M 3 ,M 4 ,M 5 ) = min (16+1 , 16 , , , ) = 16. So x=(2,2,0,0,0) 6. D=(2,3,0,0,0), M ? = min ( M 1 ,M 2 ,M 3 ,M 4 ,M 5 ) = min (16+1 , 16 , , , ) = 16. So x=(2,3,0,0,0) 7. D=(2,3,1,0,0), M ? = min ( M 1 ,M 2 ,M 3 ,M 4 ,M 5 ) = min (16+1+1 , 16+1 , 17 , , ) = 17. So x=(2,3,1,0,0) or x=(2,4,0,0,0). It does not affect the final result to choose either one of them. Here I will choose x=(2,3,1,0,0)either one of them....
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HW6_solution - solution of homework 6 Problem 1 a. It only...

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