Bauer P.'s A branch and cut approach to the cardinality constrained PDF

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14) yields: 2(lau 2 + lbv2 ) = k − 2 + 2(i − j + 1). The left hand side of this equation is even and the right hand side is odd, a contradiction. ii. k is odd, l Tf = 3k/2 +2i for some i ∈ such that l gT ≥ k. + , and ∃g ∈ C ∩(E \ T \ f ) 342 P. Bauer et al. In this subcase, we know that wlTf = (5−k)/2 and we = 2−leT ∀e ∈ (C\ f )∩(E\T ). 3) implies that (2 − leT ) ≥ 3 − (5 − k)/2 + e∈(C\ f )∩(E\T ) Using the fact that e∈T xe. 16) and applying Lemma 3, we get that |P Tf ∩ C| − |PeT \ P Tf | ≥ k − m + 3.

Bauer, P. (1997): The circuit polytope: facets. Mathematics of Operations Research 22, 110–145 7. P. (1998): A branch and cut approach to the cardinality constrained circuit problem. Technical Report TLI-98-04, The Logistics Institute, Georgia Institute of Technology, 1998. pdf 8. , Williamson, D. (1993): A note on the prize collecting traveling salesman problem. Mathematical Programming 59, 413–420 9. , Simchi-Levi, D. (1996): The capacitated prize-collecting traveling salesman problem. Technical Report TR 96-10, Northwestern University 10.

Case II. l Tf ≥ 3k/2 − 1. 1. k is even, and ∃g ∈ (C \ f ) ∩ (E \ T ) such that l gT ≥ k. In this subcase we know by definition of the weights wlTe that wlTf = 2 − k/2, and wg ≤ 2 − k/2. 1. 2. k is even, and ∃g ∈ (C \ f ) ∩ (E \ T ) such that l gT ≥ k. In this subcase, we know that wlTf = 2−k/2 and we = 2−leT ∀e ∈ (C \ f )∩(E \ T ). 3), we can get the following inequalities: 2 − k/2 + (2 − leT ) ≥ 3 − e∈(C\ f )∩(E\T ) 2 − k/2 + 2(m − 1) − xe e∈T leT ≥ 3 + m − k e∈(C\ f )∩(E\T ) − leT e∈(C\ f )∩(E\T ) ≥ 3 − m − k/2.

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A branch and cut approach to the cardinality constrained circuit problem by Bauer P.


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