Buildlt Construction uses steel beams with a uniform cross section but of different lengths in its...
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Buildlt Construction uses steel beams with a uniform cross section but of different lengths in its various projects. For its projects over the next month, it will require beams of lengths 4, 7, 9, 10, 14, 18, 23, and 30; the number of beams needed are 40, 30, 60, 110, 100, 75, 90, and 125, respectively. While it can hold all required beams in its storage center, BuildIt typically will store only some of these beams lengths and use them to satisfy the demand for smaller lengths, with the scrap pieces being discarded. For example, an 18 ft beam can be trimmed to satisfy all required lengths except for the 23 and 30 ft beams. However, for example, if it is used to satisfy the demand for 7 ft beam, then each time it is used there will be a scrap of 11 ft which cannot be used to satisfy the demand of any other beams. This extra 11 ft should be discarded. It costs BuildIt $100 to prepare the storage center for holding beams of any particular length, and a per-beam amount of which is given below for each beam length. 4 ft 7 ft 9 ft Holding cost/beam (S) 8 9 11 10 ft 14 ft 13 20 18 ft 25 23 ft 32 30 ft 38 Thus, if it were told to hold 200 beams of length 18 ft, the cost would be $100+200($25) $5100. BuildIt wants to minimize the total cost of storing the required beams. Formulate this problem as a shortest path problem. Drawing the network/graph, putting down the costs/lengths of each arc/edge and explaining the objective (such as we are trying to find the shortest path from node x to node y) on this graph is sufficient. You do not have to write down the formulation explicitly. (Hint: If we were to store beams of lengths 10 and 30 ft, would we use a beam of length 30 ft to satisfy the demand for a beam of length 7 ft in an optimal solution?) Buildlt Construction uses steel beams with a uniform cross section but of different lengths in its various projects. For its projects over the next month, it will require beams of lengths 4, 7, 9, 10, 14, 18, 23, and 30; the number of beams needed are 40, 30, 60, 110, 100, 75, 90, and 125, respectively. While it can hold all required beams in its storage center, BuildIt typically will store only some of these beams lengths and use them to satisfy the demand for smaller lengths, with the scrap pieces being discarded. For example, an 18 ft beam can be trimmed to satisfy all required lengths except for the 23 and 30 ft beams. However, for example, if it is used to satisfy the demand for 7 ft beam, then each time it is used there will be a scrap of 11 ft which cannot be used to satisfy the demand of any other beams. This extra 11 ft should be discarded. It costs BuildIt $100 to prepare the storage center for holding beams of any particular length, and a per-beam amount of which is given below for each beam length. 4 ft 7 ft 9 ft Holding cost/beam (S) 8 9 11 10 ft 14 ft 13 20 18 ft 25 23 ft 32 30 ft 38 Thus, if it were told to hold 200 beams of length 18 ft, the cost would be $100+200($25) $5100. BuildIt wants to minimize the total cost of storing the required beams. Formulate this problem as a shortest path problem. Drawing the network/graph, putting down the costs/lengths of each arc/edge and explaining the objective (such as we are trying to find the shortest path from node x to node y) on this graph is sufficient. You do not have to write down the formulation explicitly. (Hint: If we were to store beams of lengths 10 and 30 ft, would we use a beam of length 30 ft to satisfy the demand for a beam of length 7 ft in an optimal solution?)
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To formulate the problem as a shortest path problem we can represent it as a directed graph Each node in the graph represents a beam length and the edges between nodes represent the possibility of usi... View the full answer
Related Book For
Shigleys Mechanical Engineering Design
ISBN: 978-1121345317
9th edition
Authors: Richard G. Budynas, J. Keith Nisbett
Posted Date:
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