Split Rank of Triangle and Quadrilateral Inequalities Quentin - PowerPoint PPT Presentation

Split Rank of Triangle and Quadrilateral Inequalities Quentin Louveaux Universit´ e de Li` ege - Montefiore Institute January 2009 Joint work with Santanu Dey (CORE) Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 1 / 1

Outline Cuts from two rows of the simplex tableau The different cases to consider Split cuts and split ranks Finiteness proofs for the triangles The ideas for the quadrilaterals Conclusion Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 2 / 1

Cuts from two rows of the simplex tableau Consider a mixed-integer program min c T x s.t. Ax = b x ∈ Z n 1 + × R n 2 + . We consider the problem of finding valid inequalities cutting off the linear relaxation optimum. We consider the simplex tableau a 1 n s n = ¯ x 1 − ¯ a 11 s 1 −· · ·− ¯ b 1 . ... . . a mn s n = ¯ x m − ¯ a m 1 s 1 −· · ·− ¯ b m . - Select two rows - Relax the integrality requirements of the non-basic variables - Relax the nonnegativity requirements of the basic variables but keeping integrality Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 3 / 1

Cuts from two rows of the simplex tableau Consider a mixed-integer program min c T x s.t. Ax = b x ∈ Z n 1 + × R n 2 + . We consider the problem of finding valid inequalities cutting off the linear relaxation optimum. We consider the simplex tableau a 1 n s n = ¯ x 1 − ¯ a 11 s 1 −· · ·− ¯ b 1 . ... . . a mn s n = ¯ x m − ¯ a m 1 s 1 −· · ·− ¯ b m . - Select two rows - Relax the integrality requirements of the non-basic variables - Relax the nonnegativity requirements of the basic variables but keeping integrality Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 3 / 1

Cuts from two rows of the simplex tableau Consider a mixed-integer program min c T x s.t. Ax = b x ∈ Z n 1 + × R n 2 + . We consider the problem of finding valid inequalities cutting off the linear relaxation optimum. We consider the simplex tableau a 1 n s n = ¯ x 1 − ¯ a 11 s 1 −· · ·− ¯ b 1 . ... . . a mn s n = ¯ x m − ¯ a m 1 s 1 −· · ·− ¯ b m . - Select two rows - Relax the integrality requirements of the non-basic variables - Relax the nonnegativity requirements of the basic variables but keeping integrality Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 3 / 1

The 2 row-model The model „ x 1 „ f 1 „ r j n « « « X 1 = + s j , x 1 , x 2 ∈ Z , s j ∈ R + r j x 2 f 2 2 j =1 Model studied in [Andersen, Louveaux, Weismantel, Wolsey, IPCO2007] (for the finite case) and [Cornu´ ejols, Margot, 2009] (for the infinite case). Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 4 / 1

The 2 row-model The model „ x 1 „ f 1 „ r j n « « « X 1 = + s j , x 1 , x 2 ∈ Z , s j ∈ R + r j x 2 f 2 2 j =1 The geometry „ 1 / 4 „ 2 „ 1 „ x 1 „ − 3 « « « « « „ « „ « 0 1 = + s 1 + s 2 + s 3 + s 4 + s 5 1 / 2 1 1 2 − 1 − 2 x 2 3 r x 2 2 r 1 r f x 1 4 r 5 r Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 4 / 1

The geometry The projection picture 2 s 1 + 2 s 2 + 4 s 3 + s 4 + 12 7 s 5 ≥ 1 We project the n + 2-dim space onto 3 the x -space r x 2 The facet is represented by a polygon 2 r 1 r L α There is no integer point in the interior of L α f The coefficients are a ratio of x 1 distances on the figure 4 r α 1 α 3 5 r Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 5 / 1

The geometry The projection picture 2 s 1 + 2 s 2 + 4 s 3 + s 4 + 12 7 s 5 ≥ 1 We project the n + 2-dim space onto 3 the x -space r x 2 The facet is represented by a polygon 2 r 1 r L α There is no integer point in the interior of L α f The coefficients are a ratio of x 1 distances on the figure 4 r α 1 α 3 5 r Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 5 / 1

The geometry The projection picture 2 s 1 + 2 s 2 + 4 s 3 + s 4 + 12 7 s 5 ≥ 1 We project the n + 2-dim space onto 3 the x -space r x 2 The facet is represented by a polygon 2 r 1 r L α ������� ������� ������� ������� ������� ������� ������� ������� ������� ������� ������� ������� ������� ������� There is no integer point in the interior of L α f The coefficients are a ratio of x 1 distances on the figure 4 r α 1 α 3 5 r Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 5 / 1

The geometry The projection picture 2 s 1 + 2 s 2 + 4 s 3 + s 4 + 12 7 s 5 ≥ 1 We project the n + 2-dim space onto 3 the x -space r x 2 The facet is represented by a polygon 2 r 1 r L α There is no integer point in the interior of L α f The coefficients are a ratio of x 1 distances on the figure 4 r α 1 α 3 5 r Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 5 / 1

The geometry The projection picture 2 s 1 + 2 s 2 + 4 s 3 + s 4 + 12 7 s 5 ≥ 1 We project the n + 2-dim space onto 3 the x -space r x 2 The facet is represented by a polygon 2 r 1 r L α There is no integer point in the interior of L α f The coefficients are a ratio of x 1 distances on the figure 4 r α 1 α 3 5 r Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 5 / 1

The geometry The projection picture 2 s 1 + 2 s 2 + 4 s 3 + s 4 + 12 7 s 5 ≥ 1 We project the n + 2-dim space onto 3 the x -space r x 2 The facet is represented by a polygon 2 1 r r L α There is no integer point in the interior of L α f The coefficients are a ratio of x 1 distances on the figure 4 r α 1 α 3 5 r Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 5 / 1

Classification of all possible facet-defining inequalities Theorem : All facets are projected to triangles and quadrilaterals [Andersen et al 2007]. Quadrilateral Cut Triangle Cut Split Cut Disection Triangle Cook−Kannan−Schrijver Disection Quadrilateral Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 6 / 1

The split rank question Split cut : applying a disjunction π T x ≤ π 0 ∨ π T x ≥ π 0 + 1 to a polyhedron P x = f + RS s 1 ≥ 0 . . . s n ≥ 0 π T x ≤ π 0 The first split closure P 1 of P is what you obtain after having applied all possible split disjunctions π . The split rank of a valid inequality is the minimum i such that the inequality is valid for P i Most inequalities used in commercial softwares are split cuts Question : what is the split rank of the 2 row-inequalities ? In how many rounds of split cuts only can we generate the inequalities ? The Cook-Kannan-Schrijver has infinite rank and we prove that the other triangles have finite rank. Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 7 / 1

The split rank question Split cut : applying a disjunction π T x ≤ π 0 ∨ π T x ≥ π 0 + 1 to a polyhedron P x = f + RS s 1 ≥ 0 . . . s n ≥ 0 π T x ≤ π 0 The first split closure P 1 of P is what you obtain after having applied all possible Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 7 / 1

The split rank question Split cut : applying a disjunction π T x ≤ π 0 ∨ π T x ≥ π 0 + 1 to a polyhedron P x = f + RS s 1 ≥ 0 . . . s n ≥ 0 π T x ≤ π 0 The first split closure P 1 of P is what you obtain after having applied all possible Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 7 / 1

The split rank question Split cut : applying a disjunction π T x ≤ π 0 ∨ π T x ≥ π 0 + 1 to a polyhedron P x = f + RS s 1 ≥ 0 . . . s n ≥ 0 π T x ≤ π 0 The first split closure P 1 of P is what you obtain after having applied all possible Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 7 / 1

The split rank question Split cut : applying a disjunction π T x ≤ π 0 ∨ π T x ≥ π 0 + 1 to a polyhedron P x = f + RS s 1 ≥ 0 . . . s n ≥ 0 π T x ≤ π 0 The first split closure P 1 of P is what you obtain after having applied all possible Quentin Louveaux (Universit´ e de Li` ege - Montefiore Institute) Split Rank of Triangles and Quadrilaterals January 2009 7 / 1