Math 466 HW 9 Solutions - Math566 HW 9 Due on Thursday Oct...

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Math566 - HW 9 Due on Thursday, Oct 30 1 Network Optimization (Fall 2014) Brief Solutions to Homework 9 1. We first try to find a feasible flow in the given network. For this purpose, we can create an equivalent network with u 0 ij = u ij - l ij ( i, j ) A, and b ( i ) = X ( j,i ) A l ji - X ( i,j ) A l ij i N. Then we can solve a max flow problem to establish a feasible flow in this network, or prove that there is no feasible flow (see Application 6.1, pages 169-170 and 193-194 in AMO). Without loss of generality, let x be a feasible flow in the original network. We have l ij x ij u ij ( i, j ) A . To find the minimum flow, we would like to push back as much flow (from t to s ), still honoring the bounds. For this purpose, define a residual network with residual capacities defined as follows: r 0 ij = ( x ij - l ij ) + ( u ji - x ji ) . The first term is the amount of flow on ( i, j ) that we can push back still satisfying the lower bound l ij , while the second term is how much more flow we can send from j to i along the reverse arc. Recall that the default expression for the residual capacity of arc ( i, j ) is r ij = u ij - x ij + x ji . Hence, we re-write the expression for r 0 ij as follows: r 0 ij = ( u ij - l ij ) - ( u ij - x ij ) + ( u ji - x ji ) . Defining u 0 ij = u ij - l ij and x 0 ij = u ij - x ij , the above expression becomes r 0 ij = u 0 ij - x 0 ij + x 0 ji .
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