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Dynamic Programming -Solutions. 9-26. (a) Dynamic programming requires that future decisions depend only on optimal ways to reach prior states. This will.

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Presentation on theme: "Dynamic Programming -Solutions. 9-26. (a) Dynamic programming requires that future decisions depend only on optimal ways to reach prior states. This will."— Presentation transcript:

1 Dynamic Programming -Solutions

2 9-26. (a) Dynamic programming requires that future decisions depend only on optimal ways to reach prior states. This will be the case in this problem if a period is reached with no backlog and no inventory. (b) Arcs correspond to setting up and producing for a series of periods, holding as necessary for periods after the one where production occurs.

3 (c) Every period's demand must be covered. With each arc representing a sequence of periods, a path from 1 to 5 covers all. (d) v[1) = 0, v[2] = 190, d{2] = 1, v[3] = 365, d[3] = 2, v[4] = 445, d[4] = 2, v[5] = 650, d{5] = 4. The implied plan produces in periods 1, 2 and 4 in quantities 30, 35, and 35, respectively. (e) This is the optimal path for node 3, where we will arrive with no inventory. It produces 30 in the first period and 25 in the second.

4 9-27. (a) Dynamic programming requires that future decisions depend only on optimal ways to reach prior states. This will be the case in this problem if a period is reached with no backlog and no inventory. (b) Arcs correspond to setting up and producing for a series of periods holding as necessary for periods after the one where production occurs, states for just w = 10, 20, 30, 40, 50 and 60.

5 (c) Every period's demand must be covered. With each arc representing a sequence of periods, a path from I to 5 covers all. (d) v[l] = 0, v[2] = 190, d[2] = 1, v[3] = 315, d[3] = I, v[4] = 385, d{4] = 1, v[5] = 590, d{5] = 4. The implied plan produces in periods I and 4 in quantities 65 and 35, respectively. (e) This is the optimal path to node 3, where we will arrive with no inventory. It produces only in period 1 for the 55 units needed in the first two periods.

6 9-28. (a) Decisions must be made about each of the 4 pieces of equipment, and those decisions may depend on how much of the weight limit has already been expended. With all weights multiples of 10 pounds, it is sufficient to have states for just w = 10. 20. 30. 40. 50 and 60. (b) Horizontal arcs corresponding to not selecting the equipment for that stage, and downright arcs correspond to selecting the equipment. When the equipment is selected, we gain its probability of use. (c) Starting with stage 1. weight 60, we must find a highest value path through all stages, (d) We take nodes in order. state within stage, setting v[] and d[] labels. An optimal path takes items I and 3 at value 80%.


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