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By Bauer P.

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Example text

Referring to Fig. 2, we know that lau 1 lau 1 + lab + lbv1 = (3k + 1)/2 + 2i, for some i ∈ + lau2 + lab + lbv2 = (3k + 1)/2 + 2( j − 1), for some j ∈ + lau 2 + lbv1 + lbv2 = k − 2. 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 ).

1325. 1 344 P. Bauer et al. 8 2883. 4 1447. 4 1033. 3773. 6 1506. 5 2381. 2. 7 153 64 1400 237 142 3 265 145 45 45 45 7 5 1 1 1 7 5 17 345 1 39 45 11 133 17 29 9 17 4 55 17 1 23 16 7 551 181 273 171 144 117 279 231 115 116 69 31 95 37 23 57 7 4082 653 268 1681 1291 625 18 947 591 236 148 83 23 16 7 7 7 26 26 39 1143 8 100 150 27 392 57 50 31 50 22 167 37 5 66 38 15 2410 1150 1301 808 406 168 1188 970 491 390 200 56 293 60 38 173 16 4614 3501. 2505. 8 5360. 5448. 5 1651. 5657. 5649. 6052. 2092. 6 4056.

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. e∈(C\ f )∩(E\T ) Therefore, |C| ≥ |P Tf ∩ C| + m ≥ k + 3 + |PeT \ P Tf | ≥ k + 3, e∈E\T \ f which gives the contradiction |C| > k. We have shown that in all possible cases, if the coefficient of an edge e ∈ E \ T in the original tree inequality cannot be improved to wlTe , then there is a contradiction.

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


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