Discrete Mathematics with Graph Theory (3rd Edition) 396

Discrete Mathematics with Graph Theory (3rd Edition) 396 -...

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394 Solutions to Review Exercises (d) No. This path contains the saturated arc bf. (e) No. The flow can be increased along the path sbdt (or along seft). 2. The network is now as shown on the left. (a) sadt and saet both contain the saturated arc sa; sbdt contains the saturated arc bd; sbft contains the saturated arc bf; sbet and seet both contain the saturated arc et, and seft contains the saturated arc ef. (b) The slack of sbeadt is 1, since Csb - fsb = 7 - 6. (c) Using the flow-augmenting chain in (b), we increase the flow as shown on the right. a s t c 2,2 Note that we still have another flow-augmenting chain, namely seeadt. The slack in this chain is now
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Unformatted text preview: 1: Cad -fad = 4 -3. We increase the flow as shown and claim that this flow is now maximal, of value 15. To prove this, we construct an (s, t)-cut using the method of Theorem 14.2.2. Thus S = {s, e, e, a, b} and T = {d, f, t}. We have cap(S, T) = Cad +Cbd + CbJ + eeJ + Cet = 4+ 1 +4+ 2 +4 = 15, which is the same as the value of our flow. a a--+-----j;! c 2,2 3. Pushing unit flows through until we have a saturated arc on each path from s to t gives the network on the left. a d lC--+---2i) c 5,5 f c 5,5 f This is not a maximal flow; for example, sbdaet is a flow-augmenting chain with slack 4. We increase the flow to that shown on the right above....
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