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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 flowaugmenting chain with slack 4. We increase the flow to that shown on the right above....
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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