By Jon Lee

ISBN-10: 0521010128

ISBN-13: 9780521010122

ISBN-10: 0521811511

ISBN-13: 9780521811514

Jon Lee specializes in key mathematical principles resulting in precious types and algorithms, instead of on info constructions and implementation information, during this introductory graduate-level textual content for college kids of operations learn, arithmetic, and machine technology. the perspective is polyhedral, and Lee additionally makes use of matroids as a unifying thought. themes contain linear and integer programming, polytopes, matroids and matroid optimization, shortest paths, and community flows. difficulties and workouts are integrated all through in addition to references for extra examine.

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**Additional info for A First Course in Combinatorial Optimization **

**Sample text**

N; v j ≥ 0, for j = 1, 2, . . , n; si ≥ 0, for i = 1, 2, . . , m; τ ≥ 0; m n bi u i + i=1 n cjv j − j=1 cjv j <0 j=1 has no solution. Making the substitution, h j := v j − v j , for j = 1, 2, . . cls 16 T1: IML December 11, 2003 16:30 Char Count= 0 0 Polytopes and Linear Programming we get the equivalent system: τ ≥ 0; m u i ai j = c j τ , for j = 1, 2, . . , n; i=1 u i ≥ 0, (I I ) for i = 1, 2, . . , m; n ai j h j ≤ bi τ , for i = 1, 2, . . , m; j=1 h j ≥ 0, m bi u i < i=1 for j = 1, 2, .

N; i=1 u i ≥ 0, (I I ) for i = 1, 2, . . , m; n ai j h j ≤ bi τ , for i = 1, 2, . . , m; j=1 h j ≥ 0, m bi u i < i=1 for j = 1, 2, . . , n; n cjh j j=1 First, we suppose that P and D are feasible. The conclusion that we seek is that I is feasible. If not, then I I has a feasible solution. We investigate two cases: Case 1: τ > 0 in the solution of I I . Then we consider the points x ∈ Rn and y ∈ Rm deﬁned by x j := τ1 h j , for j = 1, 2, . . , n, and yi := τ1 u i , for i = 1, 2, . . , m.

If x ∈ Rn is optimal for P, then there exists a weight splitting of c so that x is optimal for all Pk (k = 1, 2, . . , p). Proof. Suppose that x is optimal for P. Obviously x is feasible for Pk (k = 1, 2, . . , p). Let (y 1 , y 2 , . . , y p ) be an optimal solution of D. Let m(k) k k y i ai j , so y k is feasible for Dk . Note that it is not claimed that ckj := i=1 this is a weight splitting of c. However, because (y 1 , y 2 , . . , y p ) is feasible for D, we do have p p ckj = y ik aikj ≥ c j .

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