Jon Lee's A First Course in Combinatorial Optimization PDF

By Jon Lee

ISBN-10: 0521010128

ISBN-13: 9780521010122

ISBN-10: 0521811511

ISBN-13: 9780521811514

Jon Lee specializes in key mathematical rules resulting in helpful versions and algorithms, instead of on info buildings and implementation information, during this introductory graduate-level textual content for college kids of operations study, arithmetic, and desktop technology. the perspective is polyhedral, and Lee additionally makes use of matroids as a unifying proposal. issues contain linear and integer programming, polytopes, matroids and matroid optimization, shortest paths, and community flows. difficulties and routines are incorporated all through in addition to references for extra examine.

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Extra resources for A First Course in Combinatorial Optimization

Example text

Aβi−1 , b, Aβi+1 , . . , Aβm . Because Aiβ is integer valued, we have that det(Aiβ ) is an integer. Also, because A is totally unimodular and Aβ is nonsingular, we have det(Aβ ) = ±1. Therefore, xβ∗ is integer valued. A “near-converse” result is also easy to establish. Theorem (Integrality implies unimodularity). Let A be an integer matrix. If the extreme points of {x ∈ Rn : Ax ≤ b, x ≥ 0} are integer valued, for all integer vectors b, then A is totally unimodular. Proof. The hypothesis implies that the extreme points of P := {x ∈ Rn+m : Ax = b, x ≥ 0} are integer valued for all integer vectors b, where A := [A, I ].

Such a decomposition of c ∈ Rn is called a weight splitting of c. For k = 1, 2, . . , p, consider the linear programs n (Pk ) ckj x j : x ∈ Pk . max j=1 Proposition (Sufficiency of weight splitting). Given a weight splitting of c, if x ∈ Rn is optimal for all Pk (k = 1, 2, . . , p), then x is optimal for P. Proof. Suppose that x is optimal for all Pk (k = 1, 2, . . , p). Let y k be optimal to the dual of Pk : m(k) yik bik min i=1 (Dk ) subject to: m(k) yik aikj ≥ ckj , for j = 1, 2, . . , n; i=1 yik ≥ 0, for i = 1, 2, .

If, at termination, we have n + 1 in the basis, then P is infeasible. 3 Basic Solutions and the Primal Simplex Method 25 Issue 2: Because Aβ = [Aβ1 , . . , Aβi−1 , Aη j , Aβi−1 , . . , Aβi ] = Aβ [e1 , . . , ei−1 , Aη j , ei+1 , . . , em ], we have 1 i−1 A−1 , Aη j , ei+1 , . . , em ]−1 A−1 β , β = [e , . . , e and xβ∗ = A−1 β b = [e1 , . . , ei−1 , Aη j , ei+1 , . . , em ]−1 A−1 β b = [e1 , . . , ei−1 , Aη j , ei+1 , . . , em ]−1 xβ∗ . Now, the inverse of [e1 , . . , ei−1 , Aη j , ei+1 , .

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A First Course in Combinatorial Optimization by Jon Lee


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