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HW9_sol - longest path with positive arc length so the...

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IE 335 Operations Research - Optimization Solutions to Homework 9 Spring 2011 Problem 38 9-2.(a) nodes : 1 , 2 , 3 , 4 , 5 arcs : (2 , 4) , (2 , 5) , 3 , 5) edges : (1 , 2) , (1 , 3) , (2 , 3) , (4 , 5) 9-2.(b) 3 - 2 - 1 - 3 - 5 : No. Node 3 repeats 3 - 2 - 4 : Yes 1 - 3 - 2 - 4 - 5 : Yes 1 - 3 - 5 - 2 : No. No arc go back to 2 from 5. 9-2.(c) Network: 9-4.(a) 2 : 1 - 2 3 : 1 - 2 - 3 4 : 1 - 2 - 5 - 4 5 : 1 - 2 - 5 9-4.(b) To 3 : 1 - 2 is the optimal subpath 2 - 3 is the optimal subpath 1
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So 1-2-3 is the optimal 1 to 3 path. Problem 39 9-23.(a) 9-23.(b) You can say that all the arcs connecting activities i and j have i < j . So the project network is acyclic. We can treat the start node in (a) as source node and the fin node as sink node. The time spent on each activity could be thought as the length of each arc. Complete this project, which means complete all the activities, is equivalent to finding out the longest critical path from source to sink. Then to turn the longest path problem into a shortest path problem, the only thing we need to do is to negate the sign of all the arc length. Because finding the shortest path with negative arc length is the same as finding the
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Unformatted text preview: longest path with positive arc length. so the shortest path going from source or start to fin or sink with negative arc length gives us the earliest time to complete the project. By inspection, the shortest path is start-1-2-3-5-6-7-8-finish . The earliest time to complete the project is 13. Problem 40 9-32.(a) 2 8 7 10 9 12 2 11 1 2 hours 3 hours 4 hours = $30 = $40 = $50 9-32.(b) Yes. There’s no arc going backward. The source node is node 7 , and the sink node is node 2 . The cost of each arc could be treated as the arc length, which means for each red arc the cost or length is 30, for the black arc the cost or length is 40, for the green arc the cost or length is 50. Therefore, finding out the minimum total cost is equivalent to finding out the shortest path going from node 7 to node 2. By inspection, the best schedule is 7-11-2. 3...
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