By Ascheuer N., Junger M., Reinelt G.
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Additional resources for A Branch & Cut Algorithm for the Asymmetric Traveling Salesman Problem with Precedence Constraints
Since M ≥ 22k , this will leave space for exactly one item of each size 2−j M + 1, j = i + 1, . . , . Thus, the bin gets ﬁlled to (1 − 2− )M + ε, where ε = (2i − 1) + ( − i) is the number of items packed in the bin. This number is increasing with i, and starting with the second smallest items, the bin gets ﬁlled −1 to (1 − 2− )M + 2 −1 . Thus, since 1 − 2− + 2M < α, FFDα will start with the smallest items, using i=1 2in−1 bins. The (nonpolynomial) algorithm, Knapsack, that simply ﬁlls each bin as much as possible (considering all possible combinations of items) will behave as FFDα on the sequences of the proof of Theorem 7.
The purpose of this partition into types is that if the load caused by the small intervals is very low, then opening a color of capacity 1 right away might be an overkill for the small intervals. Speciﬁcally, we want to show an absolute competitive ratio of 4, which would be impossible if a unit capacity color was opened immediately. Lemma 6. The total cost of the colors used for small requests is at most 4·OPT. Proof. If there is no type 2 small request, then the claim holds since the doubling procedure is a 4-competitive algorithm.
Obviously, all items y, such that bo (y) = b , ﬁt in the bin b . For any two chains, x0 , . . , x and y0 , . . , ym , x = ym , since no two chains intersect. In addition, all items on a chain from some x are at least as large as x. Hence, for any b ∈ L, |b| ≥ x | b (x)=b |x| and thus, e(L) + s(L) ≥ s(F < ). (2) Most bins in L are ﬁlled to at least α: Consider two bins, b, b ∈ L, where b occurs before b and FFDα ﬁlls both with at least two items, but to less than α full. Let x and y be two items in b .
A Branch & Cut Algorithm for the Asymmetric Traveling Salesman Problem with Precedence Constraints by Ascheuer N., Junger M., Reinelt G.