Cutting planes from two rows of simplex tableau

Transcription

1 Cutting planes from two rows of simplex tableau Based on talk by Andersen et al, IPCO-2007 Ashutosh Mahajan 1 1 Lehigh University Department of Industrial and Systems Engineering Cor@l Seminar Series - August 28, 2007

2 Split Cuts Subsume MIR, GMI, Lift and Project,... Effectively used in branch-and-cut algorithms Let (π, π 0 ) Z (n+1). Any valid inequality for P {x πx π 0 } and P {x πx π 0 + 1} is valid for P Split cuts alone are NOT sufficient to solve a general MIP problem Need of stronger classes of cuts How to split on multiple disjunctions? Also see: Mixing mixed integer inequalities by Günluk and Pochet (and Cor@l talk by Kumar Abhishek) Today s focus Cutting planes from two rows of a simplex tableau AndLouWeiWol-2may.pdf

3 Cook, Kannan and Schrijver s example x 3 min x 3 s.t. (0,0,0) (0.67,0.67,0.67) (0,2,0) x 2 x 3 x 1 x 3 x 2 x 1 + x 2 + x 3 2 (2,0,0) x 1 x 1, x 2 Z x 3 R + x 3 0 is a valid cut for the above polytope. Proof:...

4 Cook, Kannan and Schrijver s example x 3 min x 3 s.t. (0,0,0) (0.67,0.67,0.67) (0,2,0) x 2 x 3 x 1 x 3 x 2 x 1 + x 2 + x 3 2 (2,0,0) x 1 x 1, x 2 Z x 3 R + x 3 0 is a valid cut for the above polytope. Proof:... x 3 0 is not a split inequality. See Cook et al.

5 The simplex tableau I: set of integer variables, C: set of continuous variables B: set of basic variables, N: set of non-basic variables rows in simplex tableau: x i = f i + j N r j x j, i B If f i Z i B I, current solution is feasible if f i / Z then cuts may be derived from this row (MIR, GMI) Lets consider two rows (with change of notation) now: x 1 = f 1 + r j 1 s j j N x 2 = f 2 + r j 2 s j j N or x = f + j N rj s j x, f, r j are vectors in two dimensions

6 The simplex tableau I: set of integer variables, C: set of continuous variables B: set of basic variables, N: set of non-basic variables rows in simplex tableau: x i = f i + j N r j x j, i B If f i Z i B I, current solution is feasible if f i / Z then cuts may be derived from this row (MIR, GMI) Lets consider two rows (with change of notation) now: x 1 = f 1 + r j 1 s j j N x 2 = f 2 + r j 2 s j j N or x = f + j N rj s j x, f, r j are vectors in two dimensions

7 First steps P I = {(x, s) Z 2 R n + : x = f + r j s j } j N P LP = {(x, s) R 2 R n + : x = f + r j s j } j N r j : also called a ray, as in an LP P I may be empty. (Never so for 1-row case.) e.g. x 1 = s 1 + 4s 2 x 2 = s 1 + 4s 2 Lemma: P I is empty if and only if 1. All rays {r j } are parallel, and 2. The lines {f + r j s j : s j R} for j N do not contain any integer points

8 First steps P I = {(x, s) Z 2 R n + : x = f + r j s j } j N P LP = {(x, s) R 2 R n + : x = f + r j s j } j N r j : also called a ray, as in an LP P I may be empty. (Never so for 1-row case.) e.g. x 1 = s 1 + 4s 2 x 2 = s 1 + 4s 2 Lemma: P I is empty if and only if 1. All rays {r j } are parallel, and 2. The lines {f + r j s j : s j R} for j N do not contain any integer points

9 Structure of conv(p I ) (2,3) 5x 1 + 3x 2 1 x 1 5x 2 2 x 1, x 2 Z (0.5,0.5) P LP (3,1) 5x 1 + 3x 2 + s 1 = 1 x 1 5x 2 + s 2 = 2 x 1, x 2 Z s 1, s 2 R + (0,0) (2,3) conv(p ) I (3,1) (0.5,0.5) (0,0)

10 conv(p I ) (Not empty) Let P I = {(x, s) Z 2 R n + : x = f + j N r j s j } Lemma: 1. The extreme rays of conv(p I ) are (r j, e j ) for j N 2. The dimension of conv(p I ) is n(= N ) 3. The vertices (x I, s I ) of conv(p I ) take either of two forms: 3.1 (x I, s I ) = (x I, e j s I j), where x I = f + r j s I j Z 2 and j N. (integer point on ray {f + r j s j : s j 0) 3.2 (x I, s I ) = (x I, e j s I j + e k s I k ), where xi = f + r j s I j + r k s I k Z2 and j, k N. (integer point in the set f + cone({r j, r k }). Proof:... (Not all points satisfying above properties are vertices.) Corollary: Every non-trivial valid inequality for P I that is tight at a point ( x, s) P I can be written in the form α j s j 1, j N where α j 0 for all j N.

11 conv(p I ) (Not empty) Let P I = {(x, s) Z 2 R n + : x = f + j N r j s j } Lemma: 1. The extreme rays of conv(p I ) are (r j, e j ) for j N 2. The dimension of conv(p I ) is n(= N ) 3. The vertices (x I, s I ) of conv(p I ) take either of two forms: 3.1 (x I, s I ) = (x I, e j s I j), where x I = f + r j s I j Z 2 and j N. (integer point on ray {f + r j s j : s j 0) 3.2 (x I, s I ) = (x I, e j s I j + e k s I k ), where xi = f + r j s I j + r k s I k Z2 and j, k N. (integer point in the set f + cone({r j, r k }). Proof:... (Not all points satisfying above properties are vertices.) Corollary: Every non-trivial valid inequality for P I that is tight at a point ( x, s) P I can be written in the form α j s j 1, j N where α j 0 for all j N.

12 Valid inequality for conv(p I ) Let j N α js j 1 be a valid inequality for conv(p I ) that is tight for P I. Let L α = {x R 2 : s R n + s.t.(x, s) P LP and j n α j s j 1} Lemma: Let v j = f + 1 α j r j, j N\N 0 α, then 1. interior(l α ) P I = φ 2. if interior(l α ) φ, then f interior(l α ) 3. L α = conv({f } {v j } j N\N 0 α ) + cone({r j } j N 0 α )} Proof:... Let X α = {x Z 2 : s R n +s.t.(x, s) P LP and α j s j = 1} j N = L α Z 2 φ when α j s j = 1 is a facet j N not neccessarily true for faces or other valid inequalities

13 Split cuts Lemma: If, for a facet defining inequality j N α js j 1 for conv(p I ), N 0 α φ then (π, π 0 ) Z 2 Z s.t. L α {(x 1, x 2 ) : π 0 π 1 x 1 + π 2 x 2 π 0 + 1}. Proof: 1. Let k N 0 α. Then the line {f + µr k : µ R} does not pass through any integer points in R 2 2. All rays {r j } j N 0 α are parallel 3. Let π = ( r k 2, rk 1 ), π 0 = π f. Then, {f + µr k : µ R} = {x π x = π 0 } 4. Let π 1 0 = max{π 1x 1 π 2x 2 π 0, x Z 2 } π 2 0 = min{π 1x 1 π 2x 2 π 0, x Z 2 } S π = {x R 2 : π 1 0 π 1x 1 + π 2x 2 π 2 0} 5. L α S π 6. S π = {x R 2 : π 0 π 1 x 1 + π 2 x 2 π 0 + 1} for some (π, π 0 ) Z 2 Z.

14 When N 0 α = φ

15 When N 0 α = φ Stay tuned

16 Recap P I = {(x, s) Z 2 R n + : x = f + j N r j s j } P LP = {(x, s) R 2 R n + : x = f + j N r j s j } Basic solution: (f, 0) Objective: Find facet defining inequality(ies) for conv(p I ). Some of these inequalities may not be split cuts (of any rank). P LP is a cone. dim(conv(p I )) = n = N For any ( x, s) P I, either: 1. x = f + s j e j (ray point) or, 2. x = f + s j e j + s k e k

17 Recap Every non-trivial valid inequality for P I that is tight at a point ( x, s) P I can be written in form: α j s j 1, j N where α j 0, j N.

18 Recap Every non-trivial valid inequality for P I that is tight at a point ( x, s) P I can be written in form: α j s j 1, j N where α j 0, j N. s 2 = 0 P LP s 1 = 0

19 Example x 3 min x 3 s.t. (0,0,0) (0.67,0.67,0.67) (0,2,0) x 2 x 3 x 1 x 3 x 2 x 1 + x 2 + x 3 2 (2,0,0) x 1 x 1, x 2 Z x 3 R + P LP : x 1 = s s s 3 x 2 = s s s 3 facet definining inequality: x s s s 3 1

20 Split Cuts Let j N α js j 0 be a valid inequality for conv(p I ). Then: 1. Let N 0 α = {j : α j = 0}, 2. L α = {x R n s R n +s.t.(x, s) P LP and j N α js j 1} 3. X α = L α Z 2 Lemma: If, for a facet defining inequality j N α js j 1 for conv(p I ), N 0 α φ then (π, π 0 ) Z 2 Z s.t. L α {(x 1, x 2 ) : π 0 π 1 x 1 + π 2 x 2 π 0 + 1}. e.g. previous example. Converse is not true.

21 When N 0 α φ Main results of this paper: Every facet is derivable from at most four non-basic variables With every facet, one can associate three or four particular vertices of conv(p I ). These facets can be classified into: 1. Split Cuts 2. Dissection Cuts 3. Lifted two-variable cuts Dissection cuts are not split cuts Lifted two-variable cuts are not split cuts

22 conv(x α ) 1. Recall, X α = L α Z 2 2. conv(x α ) R 2 3. Extreme points of conv(x α ) are integers 4. How many such polygons exist? Which one is Cook s example?

23 conv(x α ) 1. Recall, X α = L α Z 2 2. conv(x α ) R 2 3. Extreme points of conv(x α ) are integers 4. How many such polygons exist? Which one is Cook s example?

24 conv(x α ) 1. Recall, X α = L α Z 2 2. conv(x α ) R 2 3. Extreme points of conv(x α ) are integers 4. How many such polygons exist? Which one is Cook s example?

25 What about L α The main theorem Let j N α js j 1 be a facet defining inequality that satisfies α j > 0 for all j N. Then L α is a polygon with at most four vertices. Proof: Follows from six lemmas. Also, there exists a set S N such that S 4 and j S α js j 1 is facet defining for conv(p I (S)) where, P I (S) = {(x, s) Z n R S + : x = f + s j r j } j S Find this inequality and do simultaneous lifting of coefficients for N\S to get the desire cut.

26 What about L α The main theorem Let j N α js j 1 be a facet defining inequality that satisfies α j > 0 for all j N. Then L α is a polygon with at most four vertices. Proof: Follows from six lemmas. Also, there exists a set S N such that S 4 and j S α js j 1 is facet defining for conv(p I (S)) where, P I (S) = {(x, s) Z n R S + : x = f + s j r j } j S Find this inequality and do simultaneous lifting of coefficients for N\S to get the desire cut.

27 What about L α The main theorem Let j N α js j 1 be a facet defining inequality that satisfies α j > 0 for all j N. Then L α is a polygon with at most four vertices. Proof: Follows from six lemmas. Also, there exists a set S N such that S 4 and j S α js j 1 is facet defining for conv(p I (S)) where, P I (S) = {(x, s) Z n R S + : x = f + s j r j } j S Find this inequality and do simultaneous lifting of coefficients for N\S to get the desire cut.

28 More notation Let k 4 denote the number of vertices of conv(x α ). K = {1,..., k}. Let the set {x ν } ν K denote vertices of conv(x α ). if x X α, is not a ray point, then x = f + s j1 r j1 + s j2 r j2, s j1, s j2 > 0, unique Such a pair (j 1, j 2 ) is said to give a representation of x. Additionally if α j1 s j1 + α j2 s j2, then (j 1, j 2 ) is said to give a tight representaion. If cone({r i1, r i2 }) cone({r j1, r j2 }), then the pair (i 1, i 2 ) is a sub-cone of (j 1, j 2 ). T α ( x) = {(j 1, j 2 ) : (j 1, j 2 ) gives a tight representation of x} Lemma: There exists a unique maximal representation of (j x 1, j x 2 ) T α( x) (One tight representation of x can be used).

29 Where does this lead to? Suppose j N α js j = 1 is a facet of conv(p I ). Then, There exist n affinely independent points in P I that satisfy this equality, Substituting values of s j and solving for α j should give this equality as the unique solution Project these n points on to plane of (x 1, x 2 ). These projections are either vertices of conv(x α ), or they lie on edges of conv(x α ). So...

30 Where does this lead to? After a lot of hand waving, we get: There is a set S, such that S 4 and j α α js j 1 is facet defining for P I (S) L α = conv({f } {ν j } j S )

31 Classification of cuts if each vertex of conv(x α ) belongs to a different L α : Dissection cut if exactly one facet of L α contains two vertices of conv(x α ): Lifted 2-variable cut two facets of L α contain 2 vertices of conv(x α ) each: split cuts What kind of cut is Cook s example?