GT_HW6.pdf - MATH 4022 HOMEWORK 6(DUE NOVEMBER 10 AT...

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MATH 4022: HOMEWORK 6 (DUE NOVEMBER 10 AT 11:59PM) You are strongly encouraged to typeset your homework solutions using L A T E X. Warm-up and bonus questions are not for credit. Problem 5 is an (optional) extra credit problem. Problem 1. Ford–Fulkerson algorithm, part 1 . Let N = ( V, s, t, c ) be a network. The Ford–Fulkerson algorithm is the following process that attempts to find a maximum flow in N : Start by letting f 0 be the zero flow in N . Once we have constructed a flow f n , we consider two cases. Case 1 : If there are no f n -augmenting paths, then f n is a maximum flow, so the algorithm stops and outputs f n . Case 2 : Otherwise, we build the next flow f n + 1 as follows. Pick an arbitrary f n -augmenting path s = x 0 , x 1 , ..., x k = t . Let ε n := min { c ( x i , x i + 1 ) - f n ( x i , x i + 1 ) : 0 6 i 6 k - 1 } and define f n + 1 ( x, y ) := f n ( x, y ) + ε n if x = x i and y = x i + 1 for some 0 6 i 6 k - 1 ; f n ( x, y ) - ε n if x = x i + 1 and y = x i for some 0 6 i 6 k - 1 ; f n ( x, y ) otherwise . As we discussed in class, f n + 1 is then a feasible flow with val ( f n + 1 ) = val ( f n ) + ε n . In other words, if at some point we get a flow with no augmenting paths, then we are done, and otherwise we use some augmenting path to increase the value of the flow. An important point to observe is that if there are several augmenting paths available, then we choose an arbitrary one (i.e., we don’t try to decide which one of them is “best”). ( a ) Show that if the capacity function c is integral (i.e., if c ( x, y ) is an integer for all x , y V ), then the Ford–Fulkerson algorithm stops after a finite number of steps and outputs an integral maximum flow (regardless of what augmenting paths are chosen at each step). ( b ) Say that a capacity function c (resp. a flow f ) is rational if c ( x, y ) (resp. f ( x, y ) ) is a rational number for all x , y V . Show that if the capacity function c is rational, then the Ford–Fulkerson

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