MIE376 Lec 12 DP.page13

MIE376 Lec 12 DP.page13 - Æ x* 2 (0) = 5 Æ Produce 5...

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MIE 376 Mathematical Programming 13 Stage 1 (Demand = 1) ± f 1 (0) = Min [3 + 1*1 + f 2 (0), 3 + 2*1 + f 2 (1), 3 + 3*1 + f 2 (2), 3 + 4*1 + f 2 (3), 3 + 5*1 + f 2 (4)] = Min [20, 20.5, 21, 20.5, 20.5] = 20 and x* 1 (0) = 1 ± Thus the minimum cost is 20M$. ± The optimal solution is: ² s1 = 0 Æ x* 1 (0) = 1 Æ Produce 1 unit in month 1 Æ ² s2 = 0 + 1 – 1 = 0
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Unformatted text preview: Æ x* 2 (0) = 5 Æ Produce 5 units in month 2 Æ ² s3 = 0 + 5 – 3 = 2 Æ x* 3 (2) = 0 Æ Produce 0 in month 3 Æ ² s4 = 2 + 0 – 2 = 0 Æ x* 4 (0) = 4 Æ Produce 4 units in month 4 ± Check the total cost: ² Setup Cost = 3*3 ² Production Cost = 10*1 ² Holding Cost = 2*0.5 ² Total Cost = 20 √...
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This note was uploaded on 07/10/2011 for the course MIE 376 taught by Professor Daniel during the Spring '11 term at University of Toronto.

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