Problem Set 10 Solutions
نویسنده
چکیده
(b) In the relaxation we can get many fractional paths between each demand pair of total unit value. These paths can be read out of the solution in polynomial time in the following way. We find an edge e of the smallest positive flow value c, and subtract flow of this value from a cycle or a path going through e over edges of positive flow value. Repeating this procedure we will decompose the flow into at most m cycles and paths, where m is the number of edges. Furthermore, we can assume that all paths are simple paths, and no vertex appears twice on a path.
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