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Extra resources for Studies on Graphs and Discrete Programming
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Corresponds to row i k of the original matrix A. An optimal solution to fi and I1 is given by k-1 y * --Xi=, eil. Remark. These algorithms at first sight may seem to involve the solution of many linear programs, and hence quite complicated. But really not much work is involved. The reason is outlined in the following results. They also show why these algorithms work. Discussion on Algorithm 4 Lemma 1. Let an optimal solution at Step l(a) in the kth cycle be x k . ) I f dk= O k- 1, then x k is also optimal to the linear program in Step l(b).
C) Find i k 3 Afk * Bf. = 1, j E S k , s 1, j $ S k . Polynomial algorithms for totally dual integral systems 41 Such an index exists, and corresponds to row :i of the original matrix A. )+; [ ( x + = ( x : , . . ,x : ) where x + =max(O, xi)]. If any component of W k + ' = 0, delete that component of W k f l , the corresponding columns of A k and B k to get w k + ' ,A k + 'and B k f ' respectively. Go to Step l(a). Step 2: Stop; an optimal solution to i and I is y* =ei:. If:: Remarks. (1) It is assumed that to begin with wo>O; for otherwise we simply drop redundant constraints and get a reduced system that does satisfy this condition.
11: Packing Problems: max b'y: y 2 0, A'y w, y int. In both these problems, we assume A, b, and w are integral. If in addition A is 0/1 and w 3 0 we have set covering and set packing problems respectively. The cases in which A is 0/1, w 3 0 , b = e have received a great deal of attention, most notably in the work of Fulkerson et al. e. whether or not there exists an integral solution that is optimal to the linear program, whenever the linear program has an optimal solution. Even when this is the case, it is not easy to get an integer optimal solution; for example the simplex method might not produce it.