Algorithms, Probability, and Computing Spa2 Solutions Hs14 Solution 1: Minimum Cuts Revisited

Clearly, the optimum objective value of (P’) is at most μ̃s(G) because any relaxed s-t-cut c : V → [0, 1] induces a feasible solution with objective value size(c) by defining xv := c(v) for all variables xv. Conversely, every feasible solution x for (P’) induces a relaxed s-t-cut c by defining c(v) := xv for all v ∈ V. We conclude that the optimum objective value of (P’) is at least (and therefore equal to) μ̃s(G). In a second step, we rewrite (P’) in order to obtain a linear program. We use the strategy from the lecture in order to get rid of the absolute values in the objective. That is, besides the old variables xv we introduce m new real variables ae (one variable per edge e ∈ E) which can be seen to represent the absolute differences |xu − xv| (where e = {u, v}) in an optimum solution (obviously, |xu − xv| 6 ae, and if the inequality was strict, then we could decrease the objective function by decreasing ae).