By Raffaele Cerulli, Satoru Fujishige, A. Ridha Mahjoub
This publication constitutes the completely refereed post-conference complaints of the 4th foreign Symposium on Combinatorial Optimization, ISCO 2016, held in Vietri sul Mare, Italy, in may well 2016. The 38 revised complete papers offered during this publication have been conscientiously reviewed and chosen from ninety eight submissions. They current unique examine on all features of combinatorial optimization, akin to algorithms and complexity; mathematical programming; operations examine; stochastic optimization; and graphs and combinatorics.
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Additional info for Combinatorial Optimization: 4th International Symposium, ISCO 2016, Vietri sul Mare, Italy, May 16-18, 2016, Revised Selected Papers
Manag. Sci. 24(3), 312–319 (1977) 17. : A Benders decomposition approach for the robust spanning tree problem with interval data. Eur. J. Oper. Res. 174(3), 1479–1490 (2006) 18. : The robust traveling salesman problem with interval data. Transp. Sci. 41(3), 366–381 (2007) 19. : The robust shortest path problem with interval data via Benders decomposition. 4OR 3(4), 315–328 (2005) 20. : Exact and heuristic algorithms for the interval data robust assignment problem. Comput. Oper. Res. 38(8), 1153–1163 (2011) 21.
4g) The newly introduced variable η accounts for the worst case storage cost over all d ∈ D (a so-called, partial, epigraph reformulation). Assuming, as it is the case for a discrete scenario approach, a ﬁnite D, LS-ROB2 calls for a vector u(d) ≥ 0 for each d ∈ D satisfying Constraints (4b)–(4d). Clearly, LS-ROB2 can be solved directly by employing a mixed-integer linear programming solver, although at the cost of expressing Constraints (4b)–(4d) and (4f) |D| times. 3 Solving LS-ROB2 as an Instance of LS-DET We now present a more eﬃcient way to solve LS-ROB2 as an instance of LS-DET with suitably chosen demand and storage upper bound vectors d and U .
4) l=1 i=1 Let us show that it is a valid inequality for P , that is to say, that every L−1 n i extreme point (y, h) of P veriﬁes it. If h ≥ L, then l=1 i=1 (l − L)yl = 0 ˜ i and we are done. If h ≤ L − 1, then there exists ˜i ∈ 1, n such that yh = 1. Combining this with the validity of (3) implies L−1 n (l−L)yli +L = l=1 i=1 L−1 n (l−γ)yli +γ+ l=1 i=1 L−1 n (γ−L)yli +L−γ ≤ h+γ−L+L−γ = h. l=1 i=1 Moreover, if (y, h) is an extreme point of P saturating (3), then there exists ˜i ∈ ˜ L−1 n i 1, n such that yhi = 1 and h ≤ hmax = L−1 which yields l=1 i=1 (l −γ)yl + L−1 n L−1 n i i γ = h = l=1 i=1 (l − γ)yl + h − γ + γ.