sol7 - AMS/MBA 546 (Fall, 2010) Estie Arkin Network Flows:...

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AMS/MBA 546 (Fall, 2010) Estie Arkin Network Flows: Solution sketch to homework set # 7 1). [6.47] Let B be any subset of blue arcs with the property that no two arcs in B can be in any directed path. Note that the number of paths needed to cover all blue arcs must be at least | B | . Thus it su±ces to show that there is some set B and some collection of paths P covering all blue arcs, such that | B | = | P | . Formulate the problem of covering paths as minimum ²ow: Add a sources node s and a sink node t with arcs ( s, i ) and ( i, t ) for each node i . For each blue arc, let the lower bound be 1. All other lower bounds are zero, and all upper bounds are in³nity. The minimum value of ²ow satisfying these bounds is equal to the minimum number of paths needed to cover all blue arcs. Simply think of each path as an augmenting path with s added at the beginning and t at the end of the directed path. The min ²ow is equal to max( i S,j / S l ij - i/ S,j S u ij ). Since all upper bounds are in³nity, we can consider only cuts with no arcs directed from ¯ S to S . Since such arcs do not exists, any two arcs in the cut (which must be from S to ¯ S ) cannot be in the same directed path. Thus the min ²ow is equal to the number of blue arcs in the cut (since other arcs have lower bound zero). If the graph contains directed cycles, this result is not valid. Lower bounds can be satis³ed, while having zero ²ow from source to sink, by using cycles in the original graph. Thus, the proof we used above, does not
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This note was uploaded on 01/31/2011 for the course AMS 546 taught by Professor Arkin,e during the Fall '08 term at SUNY Stony Brook.

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sol7 - AMS/MBA 546 (Fall, 2010) Estie Arkin Network Flows:...

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