Lifted Flow Cover Inequalities forMixed 0-1 Integer Programs 1Zonghao GuGeorge L. NemhauserMartin W.P. SavelsberghGeorgia Institute of TechnologySchool of Industrial and Systems EngineeringAtlanta, GA 30332-0205AbstractWe investigate strong inequalities for mixed 0-1 integer programs derived from ow cover inequalities. Flow cover inequalities are usually not facet de�ning and needto be lifted to obtain stronger inequalities. However, because of the sequential natureof the standard lifting techniques and the complexity of the optimization problemsthat have to be solved to obtain lifting coe�cients, lifting of ow cover inequalitiesis computationally very demanding. We present a computationally e�cient wayto lift ow cover inequalities based on sequence independent lifting techniques andcomputational results that justify the e�ectiveness of our lifting procedures.1 IntroductionA mixed integer program (MIP) with binary integer variables (BMIP) is the appropriatemathematical model for many practical optimization problems. This model is used, forexample, for facility location problems, distribution problems, network design problemsand more generally when �xed or concave costs are required in the objective function ofan otherwise linear system.Traditionally, BMIPs are solved by linear programming based branch-and-bound al-gorithms. The more modern technology of branch-and-cut, which is widely used forbinary pure integer programs (BIP), has seen only limited use in solving BMIPs. Forexample, both CPLEX and OSL use branch-and-cut with lifted cover inequalities forsolving BIPs but only branch-and-bound for solving BMIPs. This successful commer-cialization of branch-and-cut for BIPs has occurred because there are e�cient algorithmsfor �nding strong inequalities (lifted cover inequalities) that cut o� fractional solutionsto LP relaxations.1This research was supported by US Army Research O�ce DAAH04-94-G-0017, NSF Grant No.DDM-9115768, and NSF Grant No. DMI-9410102 1

The purpose of this paper is to present strong inequalities for BMIPs that can beused successfully in branch-and-cut. Our inequalities are derived from the ow coverinequalities that were introduced by Padberg, Van Roy, and Wolsey [1985] and VanRoy and Wolsey [1986] and implemented by Van Roy and Wolsey [1987]. Balas et al.[1993] use disjunctive cuts and Balas et al. [1996] use Gomory cuts to solve BMIPs bya branch-and-cut algorithm.Usually, ow cover inequalities are not facet-de�ning and need to be lifted to obtainstronger inequalities. However, because of the sequential nature of the standard liftingtechniques and the complexity of the optimization problems that have to be solvedto obtain lifting coe�cients, lifting of ow cover inequalities is computationally verydemanding. Our main contribution is a computationally e�cient way to lift ow coverinequalities. This is accomplished by a technique called sequence independent lifting,which we describe in detail in Gu, Nemhauser and Savelsbergh [1995].To demonstrate the e�ectiveness of these inequalites, we have incorporated them intothe MINTO branch-and-cut system, see Nemhauser, Savelsbergh and Sigismondi [1994],and have tested the system on numerous benchmark problems from MIPLIB and othersources. The results are striking, especially on di�cult �xed-charge problems. We areable to solve fast, without producing large search trees, problems that cannot be solvedwithin reasonable time limits by widely used commercial codes. The reason is simply thatthe LP relaxations without the lifted ow cover inequalities are just too weak to prunethe nodes. The time spent on �nding the lifted cover inequalities pays o� handsomelyby tightening the LP relaxations to the point where the size of the search tree is keptwithin reasonable limits.The paper is organized as follows. In Section 2, we review some results on ow coverinequalities. In Section 3, we brie y review sequential and sequence independent lifting.In Section 4, we derive characteristics of the lifting problems associated with single-node ow models that are independent of the inequality under consideration. In Sections 5and 6, we study lifted ow cover inequalities and lifted simple generalized ow coverinequalities. In Section 7, we present computational experiments.2 Flow Cover InequalitiesFlow cover inequalities are valid inequalities for the systemX = f(x; y) 2 Rn+ �Bn : Xj2N+ xj � Xj2N� xj � d; xj � mjyj ; j 2 Ng; (1)where N = N+ [ N� and n = jN j. The system (1) can be viewed as a single nodecapacitated network design model, see Figure 1. The x variables are arc ows that areconstrained by the conservation inequality with external demand of d and the variableupper bound constraints where mj is the capacity of arc j and yj is a 0-1 variable that2

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Figure 1: Single node ow modelindicates whether arc j is open. The usefulness of this system for general BMIPs arisesfrom the fact that a general inequality in a BMIP that contains both real and binaryvariables can be relaxed to (1), see Nemhauser and Wolsey [1988].Families of valid inequalites for (1) have been derived by Padberg, Van Roy andWolsey [1985] and Van Roy and Wolsey [1986]. Also see Nemhauser and Wolsey [1988]and Wolsey [1989] for an exposition and survey of these results.First, we consider a single-node ow model with only in ow arcs, i.e., N� = ;. A setC+ � N+ is called a ow cover if Pj2C+ mj > d. The inequality0 � d� Xj2C+ xj � Xj2C++(mj � �)(1� yj); (2)where � =Pj2C+ mj�d and C++ = fj 2 C+ :mj > �g, is called a ow cover inequalityand is valid for (1) when N� = ;.Next, we consider a general single-node ow model. A set C = C+ [ C� is called a ow cover if C+ � N+, C� � N� and Pj2C+ mj �Pj2C�mj = d+ � with � > 0. Theinequality0 � d+ Xj2C�mj � Xj2C+ xj � Xj2C++(mj � �)(1� yj) + Xj2L� �yj + Xj2L�� xj ; (3)where L� = fj 2 N� n C� : mj > �g and L�� = N� n (L� [ C�), is called a simplegeneralized ow cover inequality (SGFCI) and is valid for (1). The inequality0 � d+ Xj2C�mj � Xj2C+[L+ xj � Xj2C++(mj � �)(1� yj) +Xj2L+(maxfm1; mjg � �)yj � Xj2C�minf�; [mj � (m1 � �)]+g(1� yj) + (4)3

Xj2L�maxf�;mj � (m1 � �)gyj + Xj2L�� xj ;where L+ � N+ nC+, L� � N� nC�, and L�� = N� n (C� [L�) is called an extendedgeneralized ow cover inequality (EGFCI) and is also valid for (1).Both the simple and the extended generalized ow cover inequalities have been in-coporated in the mixed integer optimizers MPSARX [Van Roy and Wolsey 1987] andMINTO [Nemhauser, Savelsbergh, and Sigismondi 1994].After reviewing the lifting theory of Gu, Nemhauser and Savelsbergh [1995], we willpresent two new families of ow cover inequalities, one of which dominates EGFCIs andthe other dominates SGFCIs.3 Sequential and sequence independent liftingIn this section, we summarize some of the results of Gu, Nemhauser, and Savelsbergh[1995]. Consider the set of feasible points for a BMIP given byX = fx 2 RjN j : Xj2N ajxj � d0;Xj2Ck wjxj � rk; k = 0; : : : ; t;xj 2 f0; 1g; j 2 I � Ng;where fCk : k = 0; :::; tg is a partition of N , and aj ; d0; wj; r and rk are vectors ofappropriate dimension. Also suppose that each variable xj has at least one �nite boundgiven by bj , i.e., xj � bj or xj � bj. Initially, we consider the subset of X with xj = bjfor j 2 N n C0 given byX0 = fx 2 RjC0j : Xj2C0 ajxj � d;Xj2C0 wjxj � r0;xj 2 f0; 1g; j 2 I \ C0g;where d = d0 �Pj2NnC0 ajbj.Let 0 � �0 � Xj2C0 �jxj (5)be an arbitrary valid inequality for X0. To construct a valid inequality for X of the form0 � �0 � Xj2C0 �jxj � X1�k�t Xj2Ck �j(xj � bj); (6)4

we start with (5) and lift the variables in N nC0. Without loss of generality, we assumethat the variables with indices in C1; : : : ; Ct are lifted sequentially in that order and thatin a given set Ck they are lifted simultaneously. Note that this contains as special casessimultaneous lifting of all variables (t = 1) and sequential lifting of all variables (jCij = 1for i = 1; :::; t).The intermediate sets of feasible points X i for i = 1; :::; t are de�ned byX i = fx 2 RP0�k�i jCkj : Xj2C0 ajxj + X1�k�i Xj2Ck aj(xj � bj) � d;Xj2Ck wjxj � rk; k = 0; : : : ; i;xj 2 f0; 1g; j 2 I \ [[ik=0Ck]g:The lifting problem Li associated with Ci is, given a valid inequality0 � �0 � Xj2C0 �jxj � X1�k<i Xj2Ck �j(xj � bj) (7)for X i�1, �nd �j for j 2 Ci such thatXj2Ci �j(xj � bj) � �0 � Xj2C0 �jxj � X1�k<i Xj2Ck �j(xj � bj) (8)is a valid inequality for X i.LetZi = fz : 9x 2 X i : Xj2Ci aj(xj � bj) = z and Xj2C0 ajxj + X1�k<i Xj2Ck aj(xj � bj) � d� zg:Furthermore, let hi(z) = max Xj2Ci �j(xj � bj)s.t. Xj2Ci aj(xj � bj) = zXj2Ci wjxj � rixj 2 f0; 1g; j 2 Ci \ I;and let fi(z) = min �0� Xj2C0 �jxj � X1�k<i Xj2Ck �j(xj � bj)5

s.t. Xj2C0 ajxj + X1�k<i Xj2Ck aj(xj � bj) � d� zXj2Ck wjxj � rk; k = 0; : : : ; i� 1xj 2 f0; 1g; j 2 I \ C0 \ : : :\ Ci�1:Theorem 1 If (7) is valid for X i�1 and hi(z) � fi(z) for z 2 Zi, then (8) is valid forX i, i = 1; :::; t.When �j for j 2 Ci are such that hi(z) = fi(z) has jCij solutions x1; x2; : : : ; xjCijsuch that the components in Ci of x1 � b; x2 � b; : : : ; xjCij � b are linearly independent,we say that the lifting is maximal.Theorem 2 If (7) de�nes a facet of conv(X i�1), conv(X i�1) and conv(X i) are both fulldimensional, and hi(z) and fi(z) are such that the lifting is maximal, then (8) de�nes afacet of conv(X i), i = 1; :::; t.Lfting coe�cients are, in general, dependent on the lifting sequence C1; C2; : : : ; Ct.Superadditive functions are important to the development of sequence independent liftingtechniques.De�nition. A function f is superadditive on Z if f is bounded for all z 2 Z andf(z1) + f(z2) � f(z1 + z2) for all z1, z2 and z1 + z2 in Z.Let Z be a bounded convex set such that Zi � Z for i = 1; :::; t.De�nition. The lifting function f with respect to valid inequality (5) for X0 is de�nedto be f(z) = f1(z) for all z 2 Z.The following result is fundamental to the development of sequence independent liftingtechniques.Theorem 3 If f is superadditive on Z, then f(z) = fi(z) for z 2 Z and for i = 2; : : : ; t.Obviously, a superadditive lifting function greatly reduces the computational burdenof the lifting process. Instead of having to compute lifting functions fi for all i, weonly have to compute f . Unfortunately f is generally not superadditive. To be ableto pro�t from the computational advantages of sequence independent lifting, we need asuperadditive functions that yields a valid inequality.Corollary 1 If g is superadditive, g(z) � f(z) for all z 2 Z, and if �j for j 2 Ci aresuch that hi(z) � g(z) for z 2 Zi and for i = 1; : : : ; t, then the lifted inequality (6) isvalid for X. 6

A function g that satis�es the conditions of Corollary (1) is called a superadditive validlifting function for f .Next, we address the problem of choosing a `good' superadditive valid lifting function.A desirable property is that g should not be dominated by another superadditive validlifting function g0, i.e., there is no superadditive g0 with g(z) � g0(z) for all z 2 Z andg(z0) < g0(z0) for some z0 2 Z.A more interesting property is maximality. Let E = fz 2 Z : fi(z) = f(z) fori = 1; :::; t independent of the coe�cients aj for j 2 Ck and for k = 1; :::; tg. Note that iff is superadditive, then E = Z. We say that g is a maximal superadditive valid liftingfunction if g(z) = f(z) for all z 2 E.4 Lifting for single-node ow modelsAgain, consider the set X given by (1). In this section, we analyze the simultaneouslifting of a single variable pair (xp; yp). Let�X = f(x; y) 2 X : (xp; yp) = (bx; by)g;where (bx; by) is equal to (0; 0) or (mp; 1), i.e., X is the subset of feasible solutions inwhich xp and yp are �xed to either their lower or upper bounds.Suppose that both X and X are full dimensional, in Rn+ � Bn and Rn�1+ � Bn�1respectively, and that 0 � �0 � Xj2N+[N�nfpg(�jxj + �jyj) (9)de�nes a facet of conv( �X).Lifting the variable pair (xp; yp) means �nding coe�cients (�p; �p) such that�p(xp � bx) + �p(yp � by) � �0 � Xj2N+[N�nfpg(�jxj + �jyj) (10)is a valid inequality for X .Let h(z) = max �p(xp � bx) + �p(yp � by)s.t. �p(xp � bx) = z0 � xp � mpypyp 2 f0; 1g;7

where �p = 1 if p 2 N+ and �p = �1 if p 2 N�, and letf(z) = min �0 � Xj2N+[N�nfpg(�jxj + �jyj)s.t Xj2N+ xj � Xj2N� xj � d� z0 � xj � mjyj ; yj 2 f0; 1g; j 2 N+ [N�xp = bx; yp = by:To ensure that (10) is a valid inequality for X , we require that h(z) � f(z) forall z. Furthermore, (10) will be facet inducing if and only if h(z) = f(z) has twosolutions (x1; y1) and (x2; y2) such that (x1p � bx; y1p � by) and (x2p � bx; y2p � by) arelinearly independent. Note that h is independent of the speci�c facet inducing inequality(9) we are trying to lift. We will use the above observations to analyze the form of h forthe lifted inequality (10) to be facet inducing. We distinguish four cases depending onwhether p 2 N+ or p 2 N� and on whether (bx; by) = (0; 0) or (bx; by) = (mp; 1).Theorem 4 1. If p 2 N+, (bx; by) = (0; 0) and (10) de�nes a facet of conv(X), thenh(z) = ( 0 z = 0�pz + �p z > 0with �p � 0 and �p � 0.2. If p 2 N+, (bx; by) = (mp; 1) and (10) de�nes a facet of conv(X), thenh(z) = ( �pz �mp < z � 0��pmp � �p z = �mpwith �p � 0 and �p � 0.3. If p 2 N�, (bx; by) = (0; 0) and (10) de�nes a facet of conv(X), thenh(z) = ( 0 z = 0��pz + �p �mp � z < 0with �p � 0 and �p � 0.4. If p 2 N�, (bx; by) = (mp; 1) and (10) de�nes a facet of conv(X), thenh(z) = ( ��pz z � mp��pmp � �p z = mpwith �p � 0 and �p � 0. 8

Proof.Case 1 Since �p = 1, bx = 0 and xp � 0, we have z � 0. We prove �p � 0. Suppose not.Since (9) de�nes a facet, f(0) = 0 < �p = h(0), which is a contradiction. Next we show�p � 0. If �p < 0, then h(z) < 0 for z > 0. Since f(z) � 0 for z � 0, h(z) < f(z) forz > 0, and h(z) = f(z) cannot have two solutions. Hence �p � 0.Case 2 Since �p = 1, bx = mp and xp � mp, we have z � 0. If �p > 0, then any optimalsolution (xp; yp) must have yp = 1. Then f(z) = h(z) cannot have two solutions withlinearly independent (x1p �mp; y1p � 1) and (x2p �mp; y2p � 1). Hence �p � 0. If �p < 0,then h(z) = �pz > 0 for �mp < z < 0. Obviously f(z) � 0 for z � 0. So h(z) > f(z)for �mp < z < 0, which gives a contradiction.Case 3 Similar to the proof of Case 1.Case 4 Similar to the proof of Case 2. 2In Figures 2 and 3, we illustrate the �rst two cases discussed in Theorem 4. Observethat f is a monotone nondecreasing function with f(0) = 0 (since we have assumed that(9) is facet inducing). Furthermore, observe that h(0) = 0 in all four cases and that h iscontinuous at 0 in Cases 2 and 4, but not continuous at 0 in Cases 1 and 3.First, we consider Case 1. Figure 2 shows a typical realization of the function f onthe interval [0; mp]. In this situation, there are �ve possible functions hi i = 1; :::; 5 thatsatisfy the conditions of Case 1 and give rise to facets. To be facet inducing f(z) = hi(z)needs to have two solutions (x1; y1) and (x2; y2) such that (x1p; y1p) and (x2p; y2p) are linearlyindependent. Note that each hi de�nes a di�erent pair of lifting coe�cients (�p; �p).Next, we consider Case 2. Figure 3 shows a typical realization of the function f onthe interval [�mp; 0]. In this situation, there is only one function h that satis�es theconditions of Case 2 and gives rise to a facet. This is caused by the fact that h(0) = 0and h has to be continuous at 0. Since h is unique, the pair of lifting coe�cients (�p; �p)is unique as well.Observe that if we know f , then we can obtain all facets. Unfortunately, f is usuallyvery hard to compute. However, the LP relaxation fLP of the lifting function f is easyto compute and can be used to do approximate lifting. LetfLP (z) = �0� max Xj2N+[N�nfpg(�jxj + �jyj)s.t Xj2N+ xj � Xj2N� xj � d� z0 � xj � mjyj ; 0 � yp � 1; j 2 N+ [N�xp = bx; yp = by :If h � fLP , then h � f , since fLP � f . Therefore, by Theorem 1, the inequality(10) is valid but may no longer de�ne a facet of conv(X). However, in some situations,9

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Figure 3: f(z) and h(z) for Case 210

we may still get facets by using fLP . In Figure 4, we show fLP for Case 1. If we usefLP , we still get facets with h3 and h4. Note that the use of fLP has the possibility ofyielding facets only for Cases 1 and 3. We have to know f(�mp) for Case 2 and f(mp)for Case 4 to get a facet...................................................................................................................................................................................................................................................................pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pp...........................................................................................................................

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f(z)h(z)fLP (z)h1(z)h2(z)mph4(z)h3(z)

Figure 4: f(z), h(z) and fLP (z) for Case 1In the next two sections, we show how sequence independent lifting can be used toderive classes of lifted ow cover inequalities.5 Lifted ow cover inequalitiesConsider the set X of feasible solutions of the single-node ow model de�ned by (1).Given a ow cover C, if we �x variable pairs (xj ; yj) to (0; 0) for j 2 (N+ [N�) nC andto (mj ; 1) for j 2 C�, then we get a single-node ow model with only in ow arcs and anassociated subset X0 of feasible solutionsX0 = f(x; y) : Xj2C+ xj � d+ Xj2C�mj0 � xj � mjyjyj 2 f0; 1g; j 2 C+g:In this case, the ow cover inequality (2) de�nes a facet of conv(X0), see Nemhauser andWolsey [1988]. 11

Let d0 = d+Pj2C�mj . The lifting function associated with (2) isf(z) = min d0 � Xj2C+ xj � Xj2C++(mj � �)(1� yj)s.t. Xj2C+ xj � d0 � z (11)xj � mjyj ; for j 2 C+(x; y) 2 RjC+j+ � BjC+ j:Let C++ = fj1; : : : ; jrg with mji � mji+1 for i = 1; : : : ; r � 1. Let M0 = 0 andMi =Pik=1mjk for i = 1; : : : ; r.Theorem 5 The lifting function f is given byf(z) = 8>>>>><>>>>>: �� z � ��z �� � z � 0i� Mi � z �Mi+1 � � i = 0; :::; r� 1z �Mi + i� Mi � � � z �Mi i = 1:::; r� 1z �Mr + r� Mr � � � z � d0:Proof. Observe that in any optimal solution either xj � mj � � and yj = 1 or xj = 0and yj = 0 for j 2 C++, either xj > 0 and yj = 1 or xj = 0 and yj = 0 for j 2 C+ nC++,and that if we cannot set all xj 's to their upper bounds, we should decrease the variablesin C++ in the order j1; j2; :::; jr since mji > mji+1 for i = 1; :::; r � 1. This gives thefunction f . 2Theorem 6 The function f is superadditive on [0; d0], but not on (�1; d0].Proof. To show that f is superadditive on [0; d0], we use the superadditive function g2de�ned in the Appendix. Let l = �, vi = � for all i, and ui = mi � � for all i. Theng2(z) = ( h� Mh � z �Mh+1 � �z �Mh + h� Mh � � < z < Mh;and f(z) = g2(z) for 0 � z � d0.To show that f is not superadditive on (�1; d0], observe that f(�M1��)+f(M1) =��+ � = 0 > f(��). 2To lift (2), we de�ne the following superadditive valid lifting function gg(z) = ( i� im1 � z � (i+ 1)m1 � �; i = 0;�1;�2; : : :z � im1 + i� im1 � � � z � im1; i = 0;�1;�2; : : :Figure 5 shows functions f and g. 12

................................................................................................ ................................................................................................................................................................................................................................................................................................ .................... .......... .......... .......................... .......... .................... .......... .......... ................................................................................... ........................................f(z).......... .......... .......... g(z) �2���Figure 5: Functions f and g.Theorem 7 The function g is a maximal and nondominated superadditive valid liftingfunction for f .Proof. First, we show that g(z) � f(z). Comparing g(z) and f(z), we see thatg(z) � f(z) for z � 0 and for z > d0. Since im1 � Mi, if Mi � z � Mi+1 � �, thenz � imi � � and thus g(z) � i� = f(z). If Mi � � � z � Mi, then if im1 � � > z,g(z) � i� � f(z), and if im1� � � z, z � im1 � z �Mi and thus g(z) � f(z). Similarlywe can show g(z) � f(z) for Mr � r � z � d0.Second, we show that g(z) is superadditive. If we let l = v1 = �, and w1 = m1, theng(z) is the function g1(z) de�ned in the Appendix. Hence g(z) is superadditive.Third, we show that g(z) is nondominated. Suppose not. Then there is anothersupperadditive lifting function g0(z) such that g0(z) � g(z) for all z and there exists a z0with g0(z0) > g(z0).Case z0 < 0: Let z� be the largest z � 0 such that g0(z�) > g(z�). If z� 2 [i�m1 ��; i�m1], then i� < �1, since g(z) = f(z) for �m1 � z � m1, and i�m1 � z� � �. Sog0[m1� �+ (i�m1� z�)] = g[m1��+ (i�m1� z�)] = i�m1� z� and g0[(i�+1)m1� �] =g[(i�+ 1)m1 � �] = i��. Thus, by superadditivity of g0(z), we haveg0(z�) � g0(m1 � �+ i�m1)� g0[m1 � �+ (i�m1 � z�)]= i��� (i�m1 � z�)= g(z�);which is a contradiction. Since g0(z) must be nondecreasing in z, z� cannot be in [im1; (i+1)m1 � �]. 13

Case z0 � 0: If z� 2 [im1; (i+1)m1��], then (�i+1)m1�� � �z0+m1�� � (�i+1)m1.Thus, by superadditivity of g0(z), we haveg0(z0) � g0(m1 � �)� g0(�z0 +m1 � �)� f(m1 � �)� g(�z0 +m1 � �)= 0� [�z +m1 � �� (�i+ 1)m1 + (�i+ 1)�]= z � im1 + i�= g(z);which is a contradiction. Since g0(z) must be nondecreasing in z, z0 cannot be in[im1; (i+ 1)m1 � �]. Hence there does not exist a z0 such that g(z0) > g(z) and g(z) isnondominated.Finally, we show that g(z) is maximal. For any z > m1, if we �rst lift a variable pairj 2 N� nC� with mj = �z �m1, we get lifting function f2(z), see Section 2. It is easyto see that f2(z) = 0 < f(z). So z =2 E(Z). Similarly we can prove that if z < �m1,then z =2 E(Z). Thus E(Z) � [�m1; m1]. Since g(z) = f(z) for z 2 [�m1; m1], g(z) ismaximal. 2We can use g(z) to do sequence independent lifting for variables in N+ n C+, C� andN� n C�. Lifting of variable pairs in N+ n C+ is covered by Theorem 4, Case 1, liftingof variable pairs in C� is covered by Theorem 4, Case 4, and lifting of variable pairs inN� n C� is covered by Theorem 4, Case 3.Theorem 8 The lifted ow cover inequality (LFCI)Xj2C+[L+ xj + Xj2C++(mj � �)(1� yj)� Xj2L+ �jyj� d0 � Xj2C� g(mj)(1� yj) + (12)Xj2L��g(�mj)yj + Xj2L�� xj ;where �j = mj�i� if im1 � mj � (i+1)m1�� and �j = i(m1��) if im1�� < z < im1,is valid for X.Proof. To prove validity, based on Theorem 1, we have to show that for each of thelifted variable pairs we have h(z) � g(z) for all z. Below we list the lifting coe�cients(�j ; �j) that have been used to obtain (12). Note that they satisfy the conditions ofTheorem 4 and that for these choices h(z) � g(z).� For j 2 L+, (�j ; �j) = (1;�mj + i�) if im1 � z � (i + 1)m1 � � and (�j ; �j) =(1;�i(m1� �)) if im1 � � � z � im1.14

� For j 2 N+ n (C+ [ L+), (�j ; �j) = (0; 0).� For j 2 C�, (�j ; �j) = (0;�g(mj)).� For j 2 L�, (�j ; �j) = (0; g(�mj)).� For j 2 L��, (�j ; �j) = (�1; 0). 2An LFCI is a strengthening of an EGFCI as shown in the following theorem.Theorem 9 The inequality (12) is at least as strong as (4).Proof. It is easy to check that �j � maxfm1; mjg�� for j 2 L+, g(mj) � minf�; [mj�(m1 � �)]+g and �g(�mj) � maxf�;mj � (m1 � �)g. Hence (12) is at least as strongas (4). 26 Lifted simple generalized ow cover inequalitiesAgain, consider the set X of feasible solutions of a single-node ow model de�ned by(1). Given a ow cover C, if we �x variable pairs (xj ; yj) to (0; 0) for j 2 N+ n C+ andto (mj ; 1) for j 2 C�, then we get a reduced single-node ow model and an associatedsubset X0 of feasible solutionsX0 = f(x; y) : Xj2C+ xj � Xj2N�nC� xj � d+ Xj2C�mj0 � xj � mjyj ; for j 2 C+ [ (N� n C�)yj 2 f0; 1g; for j 2 C+ [ (N� n C�)g:In this case, the simple generalized ow cover inequality (3) de�nes a facet of conv(X0),see Nemhauser and Wolsey [1988].Let d0 = d+Pj2C�mj . The lifting function f(z) associated with (3) isf(z) = min d0 � Xj2C+ xj � Xj2C++(mj � �)(1� yj) + Xj2L� �yj + Xj2L�� xjXj2C+ xj � Xj2N�nC� xj � d0 � zxj � mjyj ; for j 2 C+ [ (N� n C�)(x; y) 2 RjC+[(N�nC�)j+ �BjC+[(N�nC�)j:Let r = jC++ [L�j, C++ [ L� = fj1; j2; : : :jrg with mji � mji+1 for i = 1; : : :r� 1,M0 = 0, and Mi =Pik=1mjk for i = 1; : : : ; r. Furthermore, let mp = minj2C++ mj , m =15

Pj2C+nC++ mj +Pj2L��mj , ml = min(m;�), t be the largest index in C++ [L� suchthat mjt = mp. Without loss of generality, we assume t 2 C++. Let �i = max[0; mi+1�(mp� �)�ml] for i = t; : : : ; r� 1.Theorem 10 The lifting function f is given byf(z) = 8>>>>>>>>>>>>>><>>>>>>>>>>>>>>: i� Mi � z �Mi+1 � �; i = 0; : : : ; t� 1z �Mi + i� Mi � � � z �Mi; i = 1; : : : ; t� 1z �Mi + i� Mi � � � z �Mi � �+ml;i = t; : : : ; r� 1z �Mi + (i+ 1)��ml � �i Mi � �+ml < z �Mi � �+ml+ �i;i = t; : : : ; r� 1i� Mi � �+ml + �i � z �Mi+1 � �;i = t; : : : ; r� 1z �Mr + r� Mr � � � z � d00:Proof. First, we rewrite f asf(z) = min d0 + Xj2N�nC�mj � Xj2C+ xj � Xj2C++(mj � �)(1� yj)�Xj2L�(mj � �yj)� Xj2L��(mj � xj)Xj2C+ xj + Xj2N�nC�(mj � xj) � d0 + Xj2N�nC�mj � zxj � mjyj ; for j 2 C+ [ (N� n C�)(x; y) 2 RjC+[(N�nC�)j+ �BjC+[(N�nC�)j:Next, we observe that variables xj ; j 2 C+ nC++ and x0j = mj � xj ; j 2 L�� behavesimilarly in the optimization. Note that for these variables we can assume,without lossof generality, that yj = 1. For convenience, we aggregate these variables into a singlevariables x =Pj2C+nC++ xj +Pj2L��(mj � xj). Let d00 = d0 +Pj2N�nC�mj , then weget f(z) = min d00 � x� Xj2C++(mj � �)(1� yj)� Xj2L�(mj � �yj)x+ Xj2C++ xj + Xj2L�(mj � xj) � d00 � zx � mxj � mjyj ; for j 2 C++ [ L�(x; y) 2 RjC+[(N�nC�)j+ �BjC+[(N�nC�)j:16

Now observe that in any optimal solution either xj > mj � � and yj = 1 or xj = 0and yj = 0 for j 2 C++, xj = 0 and yj = 0 or xj = mj and yj = 1 for j 2 L�, x � 0,and that if we cannot set all xj 's to their upper bounds we should �rst decrease variablesin C++ or L�. More speci�cally, we should decrease the variables in C++ [ L� in theorder j1; j2; :::; jr since mji > mji+1 for i = 1; :::; r� 1. Also observe that 0 � x � m.We proceed by giving a feasible solution for each of the six cases distinguished in thede�nition of f . From the above discussion on optimality conditions, it is easy to see thateach of the given solutions is also optimal.Since � =Pj2C+ mj �Pj2C� mj � d,d00 = d+ Xj2N�mj= Xj2C+mj � �� Xj2C�mj + Xj2N�mj= Xj2C++mj + Xj2C+nC++mj � �+ Xj2L�mj + Xj2L��mj= m+ Xj2C++ mj + Xj2L�mj � �= m+Mr � �:Since showing feasibility proceeds along the same lines in all cases, we only show itexplicitly for Case 1.Case 1. Mi � z �Mi+1 � �, i = 0; : : : ; t� 1.If jk 2 C++, xjk = 0, for k � i+ 1, xjk = mjk for k > i+ 1. If jk 2 L�, xjk = mjk ,for k � i+ 1, xjk = 0 for k > i+ 1. x =m.Then it is easy to seex+ Xj2C++ xj + Xj2L�(mj � xj) = m+Mr �Mi+1:Since d00� z � m+Mr ��� (Mi+1��) = m+Mr �Mi+1, the solution is feasible. Theobjective value isd00 � x� Xj2C++[xj + (mj � �)(1� yj)]� Xj2L�(mj � �yj)= m+Mr � ��m� Xj2C++ mj � Xj2L�mj + (i+ 1)�= i�:Case 2. Mi � � � z �Mi, i = 1; : : : ; t� 1.17

If jk 2 C++ n fjtg, xjk = 0, for k � i, xjk = mj for k > i. xjt = mp� z �Mi � �. Ifjk 2 L�, xjk = mj for k � i, xjk = 0 for k > i. x = m.Case 3. Mi � � � z �Mi � �+ml, i = t; : : : ; r� 1.If j 2 C++, xj = 0. If jk 2 L�, xjk = mj for k � i, xjk = 0 for k > i. x =m+Mi � �� z.Case 4. Mi � �+ml < z �Mi � �+ml + �i, i = t; : : : ; r� 1.If j 2 C++ n fjtg, xj = 0. xjt = Mi+1 � � � z. If jk 2 L�, xjk = mj for k � i+ 1,xjk = 0 for k > i+ 1. x =m.Case 5. Mi � �+ml+ �i � z �Mi+1 � �, i = t; : : : ; r� 1.If j 2 C++, xj = 0. If jk 2 L�, xjk = mj for k � i+1, xjk = 0 for k > i+ 1. x = m.Case 6. Mr � � � z � d00.If j 2 C++, xj = 0. If j 2 L�, xj = mj . x = m+Mr � �� z. 2Let g(z) = 8>>>>>>>>>><>>>>>>>>>>: i� Mi � z �Mi+1 � �; i = 0; : : : ; t� 1z �Mi + i� Mi � � � z �Mi; i = 1; : : : ; t� 1z �Mi + i� Mi � � � z �Mi � �+ml + �i;i = t; : : : ; r� 1i� Mi � �+ml+ �i � z �Mi+1 � �;i = t; : : : ; r� 1z �Mr + r� Mr � � � z � d00:Theorem 11 The function g is a superadditive valid lifting function for f .Proof. If �i = 0, then � � ml � �i = � � ml � 0. If �i > 0, then � � ml � �i =��ml � [mi+1 � (mp� �)�ml] = mp�mi+1 > 0. Thus g(z) � f(z).The function g is superadditive because we can choose the parameters ui and vi ofthe function g3 de�ned in the Appendix in such a way that g3 = g on 0 � z � d00. Tosee this, observe that the form of both functions is identical, that the break points of gare 0;M1��;M1;M2��; : : : ;Mt��;Mt��+ml+�t;Mt+1��; : : : ; d00, that the breakpoints of g3 are 0; u1;W1;W1 + u2;W2; : : :, and that Wi = Wi�1 + ui + vi. Now, it is asimple calculation to �nd values for ui and vi such that g3 = g. 2Corollary 2 The lifting function f(z) is superadditive if any one of the following con-ditions is satis�ed.(1) t = r, i.e. minj2C++ mj � minj2L�mj.(2) m � �, i.e. Pj2C+nC++ mj �Pj2L��mj � �.(3) mt+1 � (mp� �)�m > 0 if t 6= r and m < �.18

Proof. It is easy to check that if any one of the above conditions is satis�ed, thenf(z) = g(z). 2We can use g to perform sequence independent lifting on variables in N+ nC+ and C�.Lifting of variable pairs in N+ n C+ is covered by Proposition 4, Case 1 and lifting ofvariable pairs in C� is covered by Proposition 4, Case 4.Theorem 12 The lifted simple generalized ow cover inequality (LSGFCI)Xj2C+ xj + Xj2C++(mj � �)(1� yj) + Xj2N+nC+ �jxj � Xj2N+nC+ �jyj� d0 � Xj2C� g(mj)(1� yj) + Xj2L� �yj + Xj2L�� xjwhere (�j ; �j) = (0; 0) if Mi � mj � Mi+1 � � and (�j ; �j) = (1;Mi � i�) if Mi � � <mj < Mi, is valid for X.Proof. The proof is similar to that of Theorem 8. 2Note that f has a very good chance of being superadditive, since it is very likely that oneof the conditions of Corollary 2 is satis�ed. Recall that even if f is not superadditive,the use of the superadditive valid lifting function g may still give a facet.ExampleConsider the single-node ow model given belowX = f(x; y) 2 R4+ �B4 : x1 + x2 � x3 � x4 � 8; x1 � 14; x2 � 10; x3 � 12; x4 � 2g:For the ow cover given by C+ = f1g and C� = ;, we get the following parameters:� = 6; C++ = f1g; L� = f3g; L�� = f4g; r = 2; M0 = 0; M1 = 14; M2 = 26; m =2; mp = 14; t = 1; ml = 2; �1 = 2. Figure 6 gives f(z) and g(z). The resulting LSGFCIis x1 � 8(1� y1) + x2 � 8y2 � 8 + 8y3 + x4:7 A computational studyIn this section, we will compare the performance of the di�erent variants of lifted owcover inequalities discussed above, i.e., SGFCI, EGFCI, LFCI, and LSGFCI. However,before doing so, we address some important implementation issues. First, we discuss19

................................................................................................................................................................................................................................................. ...................................................................................................................................................................................................................................................................................................................................................................... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .................... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .................... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ........... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .........................................................qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq...................................... g(z)

q q q q q q qq q q q q

8 10 12 20 22 z2468f(z) = g(z)f(z)

Figure 6: f(z) and g(z)how the initial cover inequalities are derived. Second, we indicate how we select thelifting coe�cients for a pair of variables if there are many choices, as is the case whenwe apply Case 1 and Case 3 of Theorem 4.7.1 Initial ow coverThe initial ow cover C = C+ [ C� is obtained by solving the problemmax f Xj2N+(y�j � 1)�j + Xj2N� y�j�jgs.t. Xj2N+mj�j � Xj2N�mj�j > d� 2 BjN+j; � 2 BjN�j;where (x�; y�) is the current LP solution and where � and � represent incidence vectorsof C� and C+ respectively.This problem is a relaxation of the separation problem for simple generalized owcover inequalities (3), in which it is assumesd that L� = N� n C� and C++ = C+, andin which x� is replaced by mjy�j . For a more elaborate treatment we refer the reader toNemhauser and Wolsey [1988]. Since the separation problem is a 0-1 knapsack problem,which is NP-Hard, we use its LP relaxation to solve it approximately.20

A ow cover C is all that is required to de�ne a ow cover inequality and a simplegeneralized ow cover inequality. However, for an extended generalized ow cover, wealso need to determine sets L+ and L�. We useL+ = fj 2 N+ nC+ : x�j > �jy�j gand L� = fj 2 N� n C� : �g(�mj)y�j < x�jg;where (x�; y�) is the current LP solution.Whenever, there are many choices for the lifting coe�cients of a variable pair, we pickthose lifting coe�cients that result in the largest violation for the current LP solution.7.2 MINTOWe have used MINTO, a Mixed INTeger Optimizer [1994], to implement the di�erentvariants of lifted ow cover inequalities. MINTO is a software package for the solutionof mixed integer programming problems. It has many built-in features, such as prepro-cessing and probing, branching, and cut generation. In addition, MINTO allows the userto customize the code to take advantage of problem speci�c characteristics. Among theclasses of cuts generated by MINTO are LCIs and LGCIs. MINTO was used for all ourcomputational experiments and was invoked with run time options �p1� cikg � e, i.e.,simple preprocessing, cut generation for cover inequalities only, best-bound search of thetree, branching on a fractional variable closest to 0.5, and calling a primal heuristic basedon recursive rounding every 25 nodes. All computational experiments were performedon an IBM RS/6000 model 550 using MINTO 2.0 on top of CPLEX 3.0.7.3 Test setOur test set consists of 26 mixed 0-1 integer programs, 11 are taken from MIPLIB [Bixbyet al. 1992, Bixby et al. 1996], 5 are real-world problems that we obtained from di�erentsources, and 10 are randomly generated capacitated facility location problems accordingto the scheme proposed by Cornuejols, Sridharen and Thizy [1991].MIPLIB has 28 mixed integer programming problems. Of the 28 problems 9 are veryeasy, i.e., a plain LP based branch-and-bound algorithm solves them very fast, so wedon't use them. Of the remaining 19 problems 8 are very di�cult. Although MINTOcan solve most of them, some take quite a bit of time and some require other types of cutsas well, such as 0-1 knapsack cover inequalities. Furthermore, some have general integervariables and ow cover inequalities are most useful for mixed 0-1 integer programs.Therefore, we have decided not to use these either. That leaves us with 11 problemsfrom MIPLIB. 21

Cornuejols et al. [1991] developed a set of instances of capacitated facility loca-tion problems to be used to compare the performance of heuristics and relaxationsfor the capacitated facility location problem. Each instance is represented by a string`c pvxxyy:z', where v is the capacity level, xx the number of clients, yy the number ofdepots, and z the number of the instance in the series of test problems having the samevalues of v, xx, and yy. The computational results of Cornuejols et al. [1991] and Aardal[1994] show that instances with 50 clients, 33 depots, and small capacity levels are themost di�cult ones. Therefore, we included 10 such instances with capacity levels 1 and2. Table 1 gives a summary of the test problems. Note that several of these instances(�xnet3, �xnet4, �xnet6, modglob, set1al, set1cl, and �ber) cannot be solved in a rea-sonable amount of time by state-of-the-art commercial packages.7.4 Computational results for LFCI, SGFCI, LSGFCI, EGFCIThere are many alternative ways of using the ow cover inequalities developed in theprevious sections. We may choose to generate only one speci�c type or we may chooseto generate several types and select the most violated one among them. Since the goalof our experiments is to study the e�ectiveness of each type of ow cover inequality, weapply them separately with one exception. We also include the combination of SGFCIand EGFCI which is was incorporated in MINTO 2.0. We did experiment with usingother combinations, but combining di�erent types of ow cover inequalities does notyield better results.Table 2 shows the computational results for LSGFCI. This choice yields the bestperformance. For all other choices, we present only the di�erences in the performancewith respect to the best choice.Table 3 shows the computational results for SGFCI, Table 4 for EGFCI, Table 5 forLFCI, Table 6 for combined SGFCI and EGFCI. We see that with EGFCI we cannot solvethe problems \khb05250", \modglob" and \�ber" within 10000 nodes. The performanceof SGFCI is much better than that of EGFCI. The reason, we believe, is that the chancethat SGFCI de�nes a facet is high, while the chance that EGFCI de�nes a facet is small.Combined SGFCI and EGFCI does not produce better results than SGFCI. However,using a superadditive valid lifting function to lift SGFCI to LGSFCI reduces total CPU-time by about 40% and always reduces the number of evaluated nodes of the branchingtree. 22

Table 1: Test problem summaryName VARS 0-1 VARS ROWS Zlp Zipegout 141 55 98 149.589 568.101�xnet3 878 378 478 40717.018 51973�xnet4 878 378 478 4257.97 8936�xnet6 878 378 478 1200.88 3983khb05250 1350 24 101 95919464.0 106940226mod013 96 48 62 256.02 280.95modglob 422 98 291 20430947.0 20740508rentacar 9557 55 6803 28806137.644 30356761rgn 180 100 24 48.7999 82.1999set1al 712 240 492 11145.63 15869.75set1cl 712 240 492 1671.96 6484.25soda1 675 75 1182 13818.58 22606.82soda2 675 75 1182 13818.54 22603.16�ber 1298 44 363 156082.52 405935.18utrans.2 240 120 150 169.786 239.217utrans.3 284 142 182 313.798 432.285c p15033.1 1716 33 1767 26786.65 27189c p15033.2 1716 33 1767 27564.55 27891c p15033.3 1716 33 1767 28142.19 28194c p15033.4 1716 33 1767 28777.61 29080c p15033.5 1716 33 1767 28494.22 28963c p25033.1 1716 33 1767 22230.69 22570c p25033.2 1716 33 1767 24079.72 24362c p25033.3 1716 33 1767 23230.48 23591c p25033.4 1716 33 1767 24551.45 24718c p25033.5 1716 33 1767 23640.82 2396223

Table 2: Computational results for LSGFCIProblem Time Nodes Zroot LPs Cutsegout 0.82 25 556.4 27 14�xnet3 3.36 5 51880.8 18 83�xnet4 34.64 111 8307.8 170 190�xnet6 122.98 465 3507.4 545 169khb05250 6.16 13 106608880.0 35 122mod013 2.46 57 267.3 80 54modglob 132.22 651 20662084.0 704 366rentacar 253.57 35 29219168.0 57 60rgn 91.66 2881 64.6 2922 74set1al 5.55 21 15867.2 23 400set1cl 3.57 1 6484.2 2 400soda1 31.82 41 17644.7 57 147soda2 41.99 83 17644.5 106 164�ber 71.47 137 381837.8 310 360utrans.2 5.66 53 233.6 83 88utrans.3 36.97 449 416.4 506 144c p15033.1 55.92 39 26835.1 92 78c p15033.2 75.11 63 27786.5 116 78c p15033.3 7.87 1 28194.0 2 2c p15033.4 109.43 87 28784.3 175 130c p15033.5 30.85 29 28877.5 57 45c p25033.1 25.01 11 22245.2 26 23c p25033.2 93.12 47 24081.6 108 82c p25033.3 55.78 27 23438.7 62 48c p25033.4 27.37 21 24678.8 34 18c p25033.5 62.64 61 23656.5 109 69total 1388.00 5414 157253423.0 6426 340824

Table 3: Computational results for SGFCIProblem �Time �Nodes �Zroot �LPs �Cutsegout -0.07 0 0.0 0 0�xnet3 -0.14 0 0.0 0 0�xnet4 42.99 208 15.0 228 14�xnet6 89.88 338 0.0 378 46khb05250 -0.16 0 0.0 0 -1mod013 0.26 12 -0.1 13 6modglob -29.39 0 15684.0 -5 -65rentacar -1.73 0 0.0 0 0rgn 77.64 328 0.0 517 140set1al 0.89 0 0.0 0 0set1cl -0.05 0 0.0 0 0soda1 4.22 18 -68.7 20 -10soda2 -4.62 8 -56.6 1 -18�ber 10.07 24 -504.5 25 33utrans.2 -0.48 0 -0.3 -1 -5utrans.3 8.60 102 -0.4 115 5c p15033.1 146.04 184 -36.9 269 110c p15033.2 97.55 112 -189.2 197 119c p15033.3 0.00 0 0.0 0 0c p15033.4 149.53 208 0.0 317 133c p15033.5 142.11 198 -288.7 323 165c p25033.1 77.42 76 0.0 116 59c p25033.2 133.64 144 0.0 222 120c p25033.3 -18.29 6 -183.1 -10 -22c p25033.4 52.41 46 -107.6 77 40c p25033.5 83.50 88 -0.0 147 81total 1061.82 2100 14262.9 2949 95025

Table 4: Computational results for EGFCIProblem �Time �Nodes �Zroot �LPs �Cutsegout 0.08 6 -17.6 10 1�xnet3 9.46 82 -1465.1 91 -60�xnet4 38.85 326 -602.7 308 -149�xnet6 195.84 1014 -314.2 993 -109khb05250 1932.90 9987 -10269456.0 10668 258mod013 2.75 50 -2.5 77 19modglob 1443.24 9349 -230954.0 9538 -204rentacar 54.36 12 -290788.0 -10 -60rgn 68.31 1850 -15.8 1986 30set1al -1.45 0 0.0 0 -200set1cl -1.17 0 0.0 0 -200soda1 7.91 22 0.9 26 5soda2 -4.87 -12 0.9 -16 -12�ber 3498.17 9863 -213965.7 11767 437utrans.2 126.03 1116 -17.1 1234 105utrans.3 157.91 1538 -14.5 1632 38c p15033.1 301.18 444 -47.7 686 218c p15033.2 70.26 156 -209.3 240 59c p15033.3 0.30 0 0.0 0 -1c p15033.4 165.67 250 0.0 382 90c p15033.5 275.34 512 -380.7 792 261c p25033.1 101.11 90 94.2 142 44c p25033.2 300.32 196 0.4 405 188c p25033.3 28.76 70 -186.8 86 3c p25033.4 53.06 46 -107.6 78 27c p25033.5 165.27 160 4.4 270 89total 8989.59 37127 -11008444.4 41385 87726

Table 5: Computational results for LFCIProblem �Time �Nodes �Zroot �LPs �Cutsegout 1.80 106 -36.3 117 6�xnet3 9.47 82 -1465.1 91 -60�xnet4 38.42 326 -602.7 308 -149�xnet6 195.35 1014 -314.2 993 -109khb05250 1922.65 9987 -10269456.0 10668 258mod013 2.17 74 -3.0 92 6modglob 1436.00 9349 -230954.0 9538 -204rentacar 53.41 12 -290788.0 -10 -60rgn 66.73 1850 -15.8 1986 30set1al -1.46 0 0.0 0 -200set1cl -1.21 0 0.0 0 -200soda1 4.43 18 -68.7 20 -10soda2 -4.46 8 -56.6 1 -18�ber 4272.93 9863 -193520.2 12992 1068utrans.2 150.47 1484 -17.6 1624 92utrans.3 234.81 2064 -15.9 2206 53c p15033.1 243.30 390 -47.7 569 154c p15033.2 69.22 156 -209.3 238 58c p15033.3 0.02 0 0.0 0 -1c p15033.4 161.59 250 0.0 381 89c p15033.5 271.87 512 -380.7 792 261c p25033.1 128.71 138 -7.3 203 57c p25033.2 232.51 188 0.0 387 178c p25033.3 23.60 66 -186.8 78 -1c p25033.4 46.81 44 -107.6 73 24c p25033.5 125.02 132 -0.0 218 65total 9684.16 38113 -10988253.3 43565 138727

Table 6: Computational results for combining SGFCI and EGFCIProblem �Time �Nodes �Zroot �LPs �Cutsegout -0.07 0 0.0 0 0�xnet3 -0.14 0 0.0 0 3�xnet4 61.44 282 13.2 302 21�xnet6 87.23 362 -0.2 390 40khb05250 -0.06 0 0.0 0 2mod013 0.37 12 1.9 12 10modglob -14.13 2 15684.0 0 -60rentacar 2.13 0 0.0 0 0rgn 79.91 328 0.0 517 140set1al -0.11 0 0.0 0 0set1cl 0.01 0 0.0 0 0soda1 8.59 22 0.9 26 5soda2 -4.65 -12 0.9 -16 -12�ber 10.15 34 85.0 48 76utrans.2 2.43 24 0.0 39 18utrans.3 13.24 100 -0.4 120 27c p15033.1 187.43 192 -36.9 295 134c p15033.2 99.64 112 -189.2 198 120c p15033.3 0.22 0 0.0 0 0c p15033.4 149.43 208 0.0 317 133c p15033.5 141.78 198 -288.7 323 165c p25033.1 53.87 42 94.2 83 63c p25033.2 159.00 146 0.4 247 140c p25033.3 -18.80 6 -183.1 -10 -22c p25033.4 53.19 42 -107.6 73 42c p25033.5 127.47 136 4.4 215 104total 1199.57 2236 15078.8 3179 114928

ReferencesK. Aardal (1994). Capacitated facility location: separation algorithms and computa-tional experience. Report CCOP 261, Cornell University, Ithaca.E. Balas, S. Ceria, and G. Cornuejols (1993). A lift-and-project cutting planealgorithm for mixed 0-1 programs, Mathematical Programming 58, 295-324.E. Balas, S. Ceria, G. Cornuejols, and N. Natraj (1996). Gomory cuts revisited,Operations Research Letters, to appear.R.E. Bixby, E.A. Boyd, S.S. Dadmehr, and R.R. Indovina (1992). The MIPLIBMixed Integer Programming Library, Technical Report R92-36, Rice University.R.E. Bixby, S. Ceria, C.M. McZeal, and M.W.P. Savelsbergh (1996). AnUpdate Mixed Integer Programming Library: MIPLIB 3.0, Report LEC-96-03, GeorgiaInstitute of Technology, Atlanta.G. Cornuejols, R. Sridharan, and J.-M. Thizy (1991). A comparison of heuristicsand relaxations for the capacitated plant location problem. European Journal of Opera-tions Research 50, 280-297.H. Crowder, E.L. Johnson, and M.W. Padberg (1983). Solving Large Scale Zero-One Linear Programming Problems. Operations Research 31, 803-834.Z. Gu, G.L. Nemhauser, and M.W.P. Savelsbergh (1994). Lifted Cover Inequal-ities for 0-1 Integer Programs: Computation. Report LEC-94-09, Georgia Institute ofTechnology, Atlanta.Z. Gu, G.L. Nemhauser, and M.W.P. Savelsbergh (1995). Sequence IndependentLifting. Report LEC-95-08, Georgia Institute of Technology, Atlanta.G.L. Nemhauser, M.W.P. Savelsbergh, and G.C. Sigismondi (1994). MINTO, aMixed INTeger Optimizer, Operations Research Letters 15, 47-58.G.L. Nemhauser and L.A. Wolsey (1988). Integer and Combinatorial Optimization.Wiley, New York.M.W. Padberg, T.J. Van Roy, and L.A. Wolsey (1985). Valid Linear Inequalitiesfor Fixed Charge Problems. Operations Research 33, 842-861.T.J. Van Roy and L.A. Wolsey (1986). Valid Inequalities for Mixed 0-1 Programs.Discrete Applied Mathematics 14, 199-213.T.J. Van Roy and L.A. Wolsey (1987). Solving mixed integer programming prob-lems using automatic reformulation. Operations Research 35, 45-57.L.A. Wolsey (1989). Strong formulations for mixed integer programming: a survey.Mathematical Programming 45, 173-191 29

AppendixGiven l > 0;ui � 0; ui � ui+1; for i = 1; 2; : : : ;1;vi � 0; vi � vi+1; for i = 1; 2; : : : ;1:ui + vi > 0; for i = 1; 2; : : : ;1;let wi = ui + vi; for i = 1; 2; : : : ;1;Wh = hXi=1wi; for h = 1; 2; : : : ;1:We de�neg1(z) = ( hl hw1 � z � hw1 + u1; h is an integerhl + l[z � (hw1 + u1)]=v1 hw1 + u1 < z < (h+ 1)w1; h is an integerg2(z) = ( hl Wh � z � Wh + uh+1; h = 0; 1; : : : ;1l[h+ 1� (Wh+1 � z)=v1] Wh + uh+1 < z < Wh+1; h = 0; 1; : : : ;1g3(z) = ( hl Wh � z � Wh + uh+1; h = 0; 1; : : : ;1l[h+ (z �Wh � uh+1)=v1] Wh + uh+1 < z < Wh+1; h = 0; 1; : : : ;1The functions gi(z) for i = 1; 2; 3 are shown in Figures 7 and 8. Let Zi be the regionwhere gi(z) is de�ned, then Z1 = (�1;+1) and Zi = [0;+1) for i = 2; 3. Obviously,the functions gi(z); i = 1; 2; 3 are nondecreasing in z. LetDi = maxfgi(z1) + gi(z2)� gi(z1 + z2) : z1; z2; z1 + z2 2 Zig;then gi(z) is superadditive if and only if Di � 0.Theorem 13 The functions gi(z) for i = 1; 2; 3 are superadditive.Proof for g1(z): First, we show that to compute D1, we only need to consider z1 = hw1.Suppose hw1 < z1 � hw1 + u1. Then, since g1(z) is nondecreasing, g1(z1 + z2) �g1(hw1 + z2). Consequentlyg1(z1) + g1(z2)� g1(z1 + z2) = hl + g1(z2)� g1(z1 + z2)� g1(hw1) + g1(z2)� g1(hw1 + z2):30

pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp.................... ....... ....... .............. .............. ................................................................. .........................................-2l... zu1 w1 w1+u1 2w1-v1-w1-w1-v1 g1(z)0l2l-lFigure 7: Function g1(z)

......................................ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp ppppppppppppppppppppp ppppppppppppppppppppp ppppppppppppppppppppp pppppppppppppppp.............. .............. ........................... ................................................................... ................................................................................................................................................... ............................................................................................... ...ppppppppppppppppppppp pppppl2l3l0 u1 W1 W1+u2 W2 zW3g2(z)=g3(z)g2(z)g3(z)

Figure 8: Functions g2(z) and g3(z)31

Suppose hw1+u1 < z1 < (h+1)w1. Then g1(z+�z) � g1(z)+ (l=v1)�z. Consequentlyg1(z1) + g1(z2)� g1(z1 + z2)= hl + l(z1 � hw1 � u1)=v1 + g1(z2)� g1(z1 + z2)= (h+ 1)l+ l(z1 � hw1 � u1 � v1)=v1 + g1(z2)� g1(z1 + z2)= (h+ 1)l+ g1(z2)� g1(z1 + z2)� l[(h+ 1)w1 � z1]=v1� (h+ 1)l+ g1(z2)� g1[(h+ 1)w1 + z2]= g1[(h+ 1)w1] + g1(z2)� g1[(h+ 1)w1 + z2]:Secondly, let z1 = h1w1 and z2 = h2w2, theng1(z1) + g2(z2)� g(z1 + z2) = lh1 + lh2 � g[(h1 + h2)w1]= lh1 + lh2 � l(h1 + h2)= 0:Hence D1 = 0 and g1(z) is superadditive.Proof for g2(z): Suppose Wh < z1 � Wh + uh+1. Since g2(z1) is nondecreasing,g2(z1 + z2) � g2(Wh + z2). Consequentlyg2(z1) + g2(z2)� g2(z1 + z2) = hl + g2(z2)� g2(z1 + z2)� g2(Wh) + g2(z2)� g2(Wh + z2):Therefore, to compute D2, we only need to consider the case where Wh1 + uh1+1 < z1 �Wh1+1 and Wh2 + uh2+1 < z2 � Wh2+1. Sincez1 + z2 > Wh1 + uh1+1 +Wh2 + uh2+1= h1Xi=1wi + h2Xi=1wi + uh1+1 + uh2+1� h1Xi=1wi + h1+h2Xi=h1+1wi + uh1+1 + uh2+1= Wh1+h2 + uh1+1 + uh2+1� Wh1+h2 + uh1+h2+1;we have the following cases.Case 1: z1 + z2 � Wh1+h2+2.Then g2(z1 + z2) � (h1 + h2 + 2)l;32

g2(z1) � (h1 + 1)l;g2(z2) � (h2 + 1)l:So g2(z1) + g2(z2)� g2(z1 + z2) � 0:Case 2: Wh1+h2+1 + uh1+h2+2 � z1 + z2 < Wh1+h2+2.Then g2(z1) + g2(z2)� g2(z1 + z2)� l[h1 + 1� (Wh1+1 � z1)=v1] + l[h1 + 1� (Wh2+1 � z1)=v1]�l[(h1 + h2 + 2)� (Wh1+h2+2 � z1 � z2)=v1]= �(l=v1)(Wh1+1 +Wh2+1 �Wh1+h2+2)� 0:Case 3: Wh1+h2+1 � z1 + z2 < Wh1+h2+1 + uh1+h2+2.Because g2(z) is nondecreasing and the result for Case 2, we haveg2(z1) + g2(z2)� g2(z1 + z2)= g2(z1) + g2(z2)� g2(Wh1+h2+1 + uh1+h2+2)� g2(z1 +Wh1+h2+1 + uh1+h2+2 � z1 � z2) + g2(z2)� g2(Wh1+h2+1 + uh1+h2+2)� 0:Case 4: Wh1+h2 + uh1+h2+1 � z1 + z2 < Wh1+h2+1.Then g2(z1) + g2(z2)� g2(z1 + z2)� l[h1 + 1� (Wh1+1 � z1)=v1] + l[h1 + 1� (Wh2+1 � z1)=v1]�l[(h1 + h2 + 1)� (Wh1+h2+1 � z1 � z2)=v1]= l � (l=v1)(Wh1+1 +Wh2+1 �Wh1+h2+1)= l � (l=v1)(u1 + v1)� (l=v1)(h1+1Xi=2 wi + h2+1Xi=1 wi � h1+h2+1Xi=1 wi)� �(l=v1) h1+1Xi=2 (wi � wi+h1)� 0:Hence D2 � 0 and g2(z) is superadditive.Proof for g3(z): Similar to the proof for g2(z), we �nd that to computeD3, we only needto consider the case where Wh1 + uh1+1 < z1 � Wh1+1 and Wh2 + uh2+1 < z2 � Wh2+1.33

Case 1: z1 + z2 2 [Wh;Wh + uh+1].Then g3(z1 + z2) = hl = g2(z1 + z2). Since g3(z) � g2(z); 8z 2 [0;+1),g3(z1) + g3(z2)� g3(z1 + z2) � g2(z1) + g2(z2)� g3(z1 + z2)= g2(z1) + g2(z2)� g2(z1 + z2)= 0:Case 2: z1 + z2 2 (Wh + uh+1;Wh+1) and z1 � Wh1 � uh1+1 + z2 � Wh2 � uh2+1 �z1 + z2 �Wh � uh+1.It is easy to see that if we �x z2 and slightly decrease z1, then g3(z1) and g3(z1+ z2)will decrease by the same amount. It is also easy to see that if we decrease z1 and z2appropriately, we can decrease z1+z2 toWh without changing g3(z1)+g3(z2)�g3(z1+z2).By the result for Case 1, we haveg3(z1) + g3(z2)� g3(z1 + z2) � 0:Case 3: z1 + z2 2 (Wh + uh+1;Wh+1) and z1 � Wh1 � uh1+1 + z2 � Wh2 � uh2+1 <z1 + z2 �Wh � uh+1.Using the same idea that was used in Case 2, we can decrease z1 and z2 toWh1+uh1+1and Wh2 + uh2+1 respectively, without changing g3(z1) + g3(z2) � g3(z1 + z2). Thisimplies that we only need to consider the case z1 = Wh1 + uh1+1, z2 = Wh2 + uh2+1 andz1 + z2 2 (Wh + uh+1;Wh+1). Since g2(z1) + g2(z2)� g2(z1 + z2) � 0,0 � d(g2(z1) + g2(z2)� g2(z1 + z2))=lel= g2(z1) + g2(z2)� bg2(z1 + z2)=lcl= g3(z1) + g3(z2)� bg3(z1 + z2)=lcl� g3(z1) + g3(z2)� g3(z1 + z2):Hence D3 � 0 and g3(z) is superadditive. 234