7687: ISyE Advanced Integer Programming
Fall 2008
Lecture 3 -- Aug/25/2008
Lecturer: Dr. Nemhauser Scribes: J. Antonio Carbajal and Chien-Hung Chen
Abstract In this lecture we describe the polyhedral characterization of superadditive valid inequalities for IP in the case of packing or independence systems. We also talk about the idea of master independence systems and examples are provided. Lastly, we prove the necessary and sufficient conditions for superadditive inequalities to be maximal.
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Polyhedral Characterization of Superadditive Valid Inequalities for IP
Independence Systems
1.1
Definition 1. A set S Zn is called an independence system if + 0 S and if x1 is feasible and 0 x2 x1 then x2 is also feasible. Consider the polyhedron S={x Zn , Ax b}. We assume that aij 0 and bi > 0 , and they are + all integral data. This is an independence system. We also assume that aij bi j so that the unit vectors ej , j = 1, . . . , n, are feasible.
1.2
Form of Valid Inequalities
xj 0 j are facet-defining since affinely independent points (0, e1 , . . . , ej-1 , ej+1 , . . . , en ) are on xj = 0, j. Consider the form x 0 : 0 0 since x = 0 is feasible. If 0 = 0, x 0 immediately implies j 0 j since unit vectors ej j are feasible. Therefore, maximal inequalities of the form x 0 are simply nonnegative constraints(-xj 0). If 0 = 0, we can normalize so that 0 = 1. Hence, all the remaining facet-defining inequalities are of the form x 1, 0, = 0.
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1.3
Superadditive Inequalities
Given function F: D Rm R+ , Nonnegativity and superadditivity of F implies that F is nondecreasing. F(d1 ) F(d1 ) + F(d2 ) F(d1 + d2 ) These equations for every feasible d1 , d2 , d1 + d2 , together with F (b) = 1 (due to the normalization of 0 = 1) characterize the convex hull of independence systems.
1.4
Master Independence System
aj xj b, we have
"Master" means all possible columns. For the case of a single constraint (knapsack problem)
jN
1x1 + 2x2 + . . . + kxk b = k An interpretation of all possible columns in the case of the knapsack problem corresponds to having available items of every possible size (1, . . . , k). Consider the concrete example for k = 3: 1x1 + 2x2 + 3x3 3, xj 0, j. The feasibe solutions of this knapsack are: 0 1 2 3 0 0 1 x1 x= x2 = 0 , 0 , 0 , 0 , 1 , 0 , 1 x3 0 0 0 0 0 1 0 The convex hull of this knapsack is defined by the following inequalities:
1 2 3 x1 + 3 x2 x2 + x3
x1 , x2 , x3
+ x3 1 1 0
Let us show how these equations correspond to supperaditive valid inequalities. Note that we have: F (1) 0, F (2) 0 and F (3) = 1. Taking all possible values of d1 , d2 , d1 + d2 which are feasible For yields: d1 = d2 = 1: F (1) + F (1) F (2) 2F (1) F (2) For d1 = 1, d2 = 2: F (1) + F (2) F (3) F (1) + F (2) 1 Figure 1 shows the feasible solutions that satisfy this conditions in the (F (1), F (2)) space. The extreme points in the (F (1), F (2), F (3)) space are (0, 0, 1), (0, 1, 1) and ( 1 , 2 , 1) but note that 3 3 the first one is not maximal since it is dominated by the second one. Also, note that the inequalities defining the convex hull correspond exactly to the maximal extreme points. This is always the case as will be stated by the following proposition. 2
Figure 1: Feasible region in the (F (1), F (2)) space Proposition 2. Supperaditive inequalities are maximal iff F (d) + F (b - d) = 1, d Proof. () If F (d) + F (b - d) = 1, d, we cannot increase any component of F and still maintain feasibility. This implies that F is maximal. () We will prove the contrapositive, namely that if d0 s.t. F (d0 ) + F (b - d0 ) < 1, then the inequality of the form F (d)xd 1 will not be maximal. Basically, we will show that in such case there exists another valid inequality (d)xd 1 s.t. (d) = F (d), d = d0 and (d0 ) > F (d0 ). We will break this into two cases: 1. d0 = bi , bi even, i = 1, . . . , m i 2 b b b (d)xd 1, with (d) = We have F ( 2 ) + F ( 2 ) < 1, thus F ( 2 ) < 1 and let us show that 0 , ( b ) = 1 is valid. Note that the feasible values for x are {0, 1, 2}. F (d), d = d d 2 2 If x(d0 ) = 0, (d0 )x(d0 ) = 0 and therefore If x(d0 ) = 2, x(d) = 0, d = d0 . Hence, (d)xd = F (d)xd 1. (d)xd = 2(d0 ) = 1.
If x(d0 ) = 1: b 0 d=d0 dxd b - d xd0 = 2 b 1 d=d0 F (d)xd F ( d=d0 dxd ) F ( 2 ) < 2 [The first inequality follows because F is super additive and the second one because F is nondecreasing]. (d)xd = d=d0 F (d)xd + (d0 ) < 1 + 1 < 1. 2 2 Therefore, (d)xd 1 is a valid inequality that dominates F (d)xd 1 .
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2. d0 = bi , bi even, for some i {1, . . . , m} i 2 We have F (d0 ) + F (b - d0 ) < 1 and WLOG we can assume that d0 > bi (if d0 < bi , let i i 2 2 d0 := b - d0 > bi ). Note that the feasible values for xd are {0, 1} and let us show that i i 2 (d)xd 1, with (d) = F (d), d = d0 , (d0 ) = 1 - F (b - d0 ) is valid. If x(d0 ) = 0, (d0 )x(d0 ) = 0 and therefore If x(d0 ) (d)xd = F (d)xd 1.
= 1: dxd b - d0 xd0 = b - d0 . d=d0 0 d=d0 F (d)xd F ( d=d0 dxd ) F (b - d ) [The first inequality follows because F is super additive and the second one because F is nondecreasing]. (d)xd = d=d0 F (d)xd + (d0 ) F (b - d0 ) + (1 - F (b - d0 )) = 1. (d)xd 1 is a valid inequality that dominates F (d)xd 1 .
Therefore,
References
[1] G.L. Nemhauser and L.A. Wolsey. Integer and Combinatorial Optimization. Wiley, 1999.
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