Linear Programming by Robert J. Vanderbei

Author:Robert J. Vanderbei
Language: eng
Format: epub
Publisher: Springer US, Boston, MA

Fig. 14.9The tree solution at the end of the first iteration.

The Second Iteration. The tree solution shown in Figure 14.9 has only one remaining infeasibility: z ba = −10. Arc (b,a) must therefore enter the spanning tree. Adding it, we create a cycle consisting of nodes “a”, “b”, and “c”. The leaving arc must be pointing in the opposite direction from the entering arc. Here, there is only one such arc, (b,c). It must be the leaving arc. The leaving arc’s flow decreases from 3 to 0. The flow on the other two cycle arcs must increase by 3 to preserve flow balance.

The two subtrees formed by removing the leaving arc are shown in Figure 14.10 The dual variables on the non-root-containing subtree get incremented by the dual slack on the entering arc z ba = −10. The dual slacks for the spanning arcs also change by 10 either up or down depending on which way they bridge the two subtrees. The resulting tree solution is shown in Figure 14.11.

Fig. 14.10The two disjoint subtrees arising in the second iteration.

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