Refer again to the network for problem 1. Assume now that the numbers on each edge are the maximum c

Refer again to the network for problem 1. Assume now that the numbers on each edge are the maximum capacity that can be carried along the arc. Below I have reproduced the diagram, but the numbers now refer to a flow from Node 0 to Node 8. 1 8 10 MM MM MM MM MM MM M 5 – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 4 15 . . . . . . . . . . . . . . 5 20 > > > > > > > 0 23 20 62 . . . . . . . . . . . . . . 2 10 MM MM MM MM MM MM M 10 < < < < < < < < < < < < < < < < < 8 6 50 3 7 10 40 qq qq qq qq qq qq q 5 7 20 (a) Check that the flow is feasible. (Explicitly state everything you must check to show that the flow is feasible.) You need to check that the flow into each node is equal to the flow out; that all flows are non-negative; and that the flow along each edge is no greater than the edge’s capacity. All of these things are true. (b) Compute the value of the flow. It is 105 (sum the flow out of 0 for example). (c) Demonstrate that the flow is not maximal by constructing another feasible flow with a strictly greater value. I illustrate a better flow (obtained by using the algorithm). 4 1 8 10 MM MM MM MM MM MM M 5 – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 4 20 . . . . . . . . . . . . . . 5 20 > > > > > > > 0 23 20 67 . . . . . . . . . . . . . . 2 10 MM MM MM MM MM MM M 10

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