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Further uses of random sampling and randomized rounding of linear programs in .NET Include gs1 datamatrix barcode in .NET Further uses of random sampling and randomized rounding of linear programs




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Further uses of random sampling and randomized rounding of linear programs generate, create 2d data matrix barcode none on .net projects Visal Basic .NET minimum te .net vs 2010 Data Matrix ECC200 rminal spanning tree at the start of the iteration, and let denote the optimal x fractional solution to the directed component relaxation, and let = C xC . Let C denote the directed full component selected in this iteration (which occurred with probability xC / ), and let T denote the resulting minimum terminal spanning tree.

Then, by taking into account the random selection of the component C, we see that E[c(T )] = c(T ) E[dropT (C)] = c(T ) (xC / )dropT (C) 1 c(T ) ( ) 1 1 2 OPT, 1 where OPT denotes the optimal value for the given Steiner tree input. In fact, we can strengthen this nal inequality by the following lemma, which relates the cost of the minimum terminal spanning tree to the optimal value of the directed component linear programming relaxation. Lemma 12.

11: For any input to the Steiner tree problem, the cost of the minimum terminal spanning tree T is at most twice the cost of the optimal fractional solution x for the directed component linear programming relaxation (12.3). Proof.

We rst transform x into a feasible fractional solution y to the bidirected cut relaxation for the input induced on the terminals R. The cost of y will be at most twice the cost of x, and by the integrality of the bidirected cut relaxation in this case (since then all nodes are terminals), we may conclude that the minimum terminal spanning tree costs no more than y. This completes the proof of the lemma.

To construct y, we initially set y = 0, then, in turn, consider each directed component C for which xC > 0. Consider the doubling of each edge in the Steiner tree for this component C (ignoring edge directions), to yield an Eulerian graph. This Eulerian tour can then be shortcut to yield a cycle on the terminals of C, R(C).

We delete one edge of the cycle (chosen arbitrarily), and then orient the edges of this terminal spanning tree on R(C) towards the root of the component, sink(C). For each of the edges in this directed spanning tree, we increment its current value ye by xC . If we view each component C as providing capacity for xC to ow from each terminal in R(C) to the node sink(C), then we see that we have provided exactly the same additional capacity from each terminal to sink(C) in the shortcut solution that uses only terminal nodes.

Since the feasibility of x ensures that for each node there is at least total capacity 1 from that node to the root, so must the modi ed construction for y. But this means that y is a feasible fractional solution for the bidirected cut relaxation, and this completes the proof of the lemma. Intuitively, if in each iteration we decrease the cost of a minimum terminal spanning tree by a factor of (1 1/ ), then if we apply the same technique for iterations, we decrease the cost of the minimum terminal spanning tree by a factor that is at most 1/e.

Therefore, if consider iterations, we decrease the resulting cost by a factor of (1/e) . By Lemma 12.11, we know in fact that, if we start with the minimum terminal spanning tree, the end result has expected cost at most 2(1/e) times the optimal value of the directed component relaxation.

However, in each iteration, this decrease is paid for by the connection cost c(C) of the selected component C. Due to the random selection rule, the expected cost incurred is equal to C (xC / )c(C), which Electronic web edition. Copyright 2010 by David P.

Williamson and David B. Shmoys. To be published by Cambridge University Press ( ).

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