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7), every element in the basis satisfies it as an equality. 12. Since we assume a x s a is not a (0-1)-constraint, therefore, every basis of M U T contains at least one tour. ) So we must have T f 0. We will show that for any tour t E T, either E ( t )C E+ or else E ( t ) induces a hamilton path in G+ . Then there is j E E ( t )with ai= 0 and E ( t )- { j } consists of the edges of an even length path of G since G has an odd number of nodes. This path can be expressed as the average of two complementary np-matchings x1and x2 of G, and at = 0 .

Let the graph G’ be obtained from G [ S ]by ‘splitting’ i into two G. Pulleyblank 34 nodes i’ and i” such that all the edges j of 6 ( i )n y ( S ) for which ai takes on the minimum value are incident with i’ and all the others are adjacent with i”. Since G [ S ] was nonseparable, G’ is connected and in addition must have a perfect 1-matching. 8)), there would exist a nonempty subset Y of the node of G’ such that deleting Y creates at least I Yl + 2 odd cardinality components. Let Y’be the subset of S consisting of‘all nodes of Y n S together with i if at least one of i’, i” belongs to Y.

Therefore T U M is affinely independent of cardinality IE,( so F’ is a facet of P,, and F‘f l P,, = F. 0 Perhaps surprisingly, there are presently only three classes of facets for Q T appearing in the literature. 21. 9) for P,,. 11. For any such S, the subtour elimination constraints corresponding to S and V,, - S induce the same facet of Q+. ) In particular, the edge ‘capacity’ constraints xj =s1 for j E E induce the same facets as the subtour elimination constraints for the cardinality n - 2 subsets of V,.

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