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IEOR162_hw11_sol

# IEOR162_hw11_sol - IEOR 162 Fall 2011 Suggested Solution to...

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IEOR 162, Fall 2011 Suggested Solution to Homework 11 Problem 1 The branch-and-bound tree is depicted below. The complete steps are: 1. We consider the four options on the last job. This creates the four nodes in the second level. For node “1”, we schedule job 1 as the last job. This implies that job 1 will be completed at time 27 with a 13-hour delay. This is a lower bound of the total delay of any schedule having job 1 at the last. We find the other three upper bounds using the same way. 2. We branch on node “3” because it is currently the best option and create three children “13”, “23”, and “43”. These are the three options on the third job given that the last job has been fixed to job 4. For node “13”, we know job 1, the third job to be processed, will be completed at time 23 with a 9-hour delay. 5 + 9 = 14 is thus the lower bound of the total delay of any schedule having jobs 1 and 3 at the last. We find the other two upper bounds using the same way. 3. We branch on node “43” because it is currently the best option in the third level. Eventually we find a schedule “2143” with a 7-hour total delay time. 4. Now we know nodes “1”, “2”, “13”, “23”, and “243” are all fathomed. We return to node “4” and branch it. Nodes “14” and “24” are fathomed immediately. 5. We branch on node “34” and eventually obtain a schedule “2134” that is better than “2143”. Node “234” is then fathomed. We then conclude that the optimal schedule is “2134” with a 6-hour total delay time. » » » » » » » » » » » » 1 13 ' ' ' ' ' ' 2 14 3 5 ' ' ' ' ' ' 13 14 23 15 H H H H H H 43 7 ' ' ' ' ' ' 143 7 2143 7 243 7 X X X X X X X X X X X X 4 6 14 8 H H H H H H 24 9 X X X X X X X X X X X X 34 6 ' ' ' ' ' ' 134 6 2134 6 234 6 1

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Problem 2 (Problem 9.Review.16) Let x i = production quantity in plant i , i = 1 , ..., 3, and z i = 1 if plant i is open 0 otherwise , i = 1 , ..., 3 be the decision variables. Also let F = (80000 , 40000 , 30000) be the vector of fixed costs, C = (20 , 25 , 30) be the vector of variable costs, K = (6000 , 7000 , 6000) be the vector of capacity levels. The complete formulation is min 3 i =1 F i z i + C i x i · s.t.
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IEOR162_hw11_sol - IEOR 162 Fall 2011 Suggested Solution to...

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