# sol6 - AMS/MBA 546 (Fall, 2010) Estie Arkin Network Flows:...

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Unformatted text preview: AMS/MBA 546 (Fall, 2010) Estie Arkin Network Flows: Solution sketch to homework set # 6 1). (a). We have a node s and a node t . In addition there are 3 sets of nodes N 1 , N 2 and N 3 . N 1 has a node for each man, an arc from s to this node of capacity 1. N 3 has a node for each woman with an arc of capacity 1 from these nodes to t . N 2 has a node for each broker, with an arc of capacity 1 from man j to broker i if man j is on the list of broker i . Similarly, there is an arc from broker i to each woman l on his list, also with capacity 1. The max flow in this graph corresponds to the max number of marriages that can be arranged. (b). Here we add a capacity a i on the flow through the node corresponding to broker i . This is equivalent to replacing the node by 2 nodes with an arc between them of capacity a i . 2). [6.32] An airline has p flight legs that it wishes to service by the fewest possible planes. To do so, it must determine the most efficient way to combine these legs into flight schedules. The starting time for flightdetermine the most efficient way to combine these legs into flight schedules....
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## This note was uploaded on 01/31/2011 for the course AMS 546 taught by Professor Arkin,e during the Fall '08 term at SUNY Stony Brook.

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