Lifting Gomory Cuts With Bounded Variables

Transcription

1 Liting Gomoy Cuts With Bounded Vaiables Géad Conuéjols Teppe School o Business, Canegie Mellon Univesity, PA gc0v@andew.cmu.edu Tamás Kis Compute and Automation Reseach Institute, Hungaian Academy o Sciences, Hungay tamas.kis@sztaki.hu Maco Molinao Teppe School o Business, Canegie Mellon Univesity, PA molinao@cmu.edu Decembe 13, 011 Abstact Recently, Balas and Qualizza intoduced a new cut o mixed 0,1 pogams, called lopsided cut. Hee we pesent a amily o cuts that compises the Gomoy mixed intege cut at one exteme and the lopsided cut at the othe. We show that evey inequality in this amily is exteme o the appopiate ininite elaxation. We also show that these inequalities ae split cuts. Finally we povide computational esults. 1 Intoduction Recently, Balas and Qualizza [] intoduced a new cut o mixed 0,1 pogams, called lopsided cut. Thei deivation is based on the Balas-Jeoslow modulaization technique [1]. Hee we povide a geometic deivation o the lopsided cut and we genealize it to an ininite amily. Ou appoach is to stat om the classical Gomoy unction o continuous vaiables and to lit the coeicient o the intege vaiables using one o the bounds (say the uppe bound o 1) ollowing the technique intoduced in [5]. It is convenient to pesent ou geometic deivation using the ininite model ist intoduced by Gomoy and Johnson [6]. Setup We conside a linea equation whee a bounded intege vaiable x is expessed in tems on nonnegative vaiables. By a change o vaiable, we may assume that x 1. Conside the uppe Suppoted by NSF gant CMMI and ONR gant N Suppoted by the Hungaian Reseach Fund OTKA K Suppoted by NSF gant CMMI and ONR gant N

2 bounded system x = + R s + R y x {,..., 0, 1} s 0 (IP()) y Z + (s, y) has inite suppot. Fo simplicity o exposition, we ocus on the case 0 < < 1. Note howeve that the appoach that we pesent below can also be used when is uthe om the bound. We will ix (0, 1) om now own. We use IP to denote the set o (x, s, y) easible solutions o the above. Let ψ GMI denote the classical Gomoy unction o the coeicients o the continuous vaiables ψ GMI () = { < Ou goal is to lit ψ GMI o the integal vaiables, namely ind π such that ψ GMI ()s + π()y 1 is satisied by all (x, s, y) IP. Since integality constaints on basic vaiables (i.e. on the let-hand side o the equation in IP()) ae moe easily handled, the idea is to tanse the integality o y to a basic vaiable. Fo that, we ist conside the extended system ( x z ) = ( 0 ) + R ( 0 ) s + R x {,..., 0, 1} z Z s 0 y Z + (s, y) has inite suppot. ( ) y l() (IP(l)) We use IP (l) to denote the set o easible solutions (x, z, s, y) o the above and obseve that, when l() is integal o all R, this extended system is equivalent to the oiginal one. Poposition 1. IP = Poj x,y,s IP (l) o all l : R Z. We assume l : R Z in the emainde o the pape. Now we elax the integality o the non-basics to obtain the system that we wok with: ( ) ( ) x = + ( ) s + ( ) y z 0 0 l() R R x {,..., 0, 1} (Z(l)) z Z s 0 y 0 (s, y) has inite suppot.

3 Notice that this system patially captues the oiginal integality o y and hence (its pojection onto the (x, y, s)-space) is tighte than the elaxation obtained by completely dopping the integality o y om IP. 3 Wedge inequalities We now constuct the amily o cuts which ae the object o study in this pape. Let S = {,..., 0, 1} Z and notice that evey easible solution (x, z, s, y) o Z(l) satisies (x, z) S. Fo α (0, 1], we conside the S-ee convex set K α = {(x, z) : a 1 [(x, z) (, 0)] 1, a [(x, z) (, 0)] 1} with a 1 = ( 1, 1 ) a = ( ) 1 α(1 ), 1. 1 α(1 ) z K α x Figue 1: S-ee convex set K α. The slope o the lowe acet o K α is α 1 α(1 ). The S-ee convex set K α contains (, 0) in its inteio and theeoe it can be used to deive an intesection cut o Z(l). Moe speciically, om the theoy o S-ee cuts [3], we obtain the valid inequality ψ()s + π α()y l 1 (1) with ψ() = max{a 1 (, 0), a (, 0)} and π α() l = max{a 1 (, l()), a (, l())}. We have the ollowing explicit omula o the coeicients (see Figue ): { < 0 ψ() = 1 0 π l α() = { +l() 1 l()(1 α(1 )) α(1 ) l() > α l() α. Inequality (1) will be called wedge inequality in the emainde. Since ψ = ψ GMI, the unction ( ψ, π l α) is a liting o the Gomoy unction. Geometically this is clea: the Gomoy unction ψ GMI is obtained by consideing the lattice-ee set [0, 1] in IP (). Fo all α the unction ψ is obtained om the set K α along ays (, 0), and theeoe the elevant pat o K α is K α {z = 0} which is the segment [0, 1] {0}. We emak that i x is a 0,1 vaiable, anothe cut, exploiting the lowe bound o 0, can be obtained by substituting x by 1 x. 3

4 lopsided cut π 0.5 GMI Figue : Gaph o π 0.5 and o unctions coesponding to lopsided and GMI cuts. 3.1 Optimizing l Notice that the cut above is valid o evey l : R Z; we can choose the one which gives the best coeicients, i.e. the smallest value o π l α(). Thankully, each ay is associated to a dieent component o l, so we can actually get the best coeicient o all the ays simultaneously. Poposition. Fo given and α, the value o l that minimizes π α() l is l α () = α(+ 1). Futhemoe π α lα () = min{ + α, 1 α (1 α(1 )) α(1 ) }. Poo. Fix R. Note that π α() l as a unction o l() is a piece-wise linea unction which is deceasing in the inteval (, α] and inceasing in the inteval [α, ), hence with minimum at l() = α. Theeoe, the minimum ove all intege values o l() is attained at eithe l() = α o α. This shows the second pat o the poposition. To pove the ist pat, let l be the (unique) value l α < l + 1 such that π l α () = π l+1 α (). A simple calculation gives l = α( + 1). Since thee is only one intege in the ange [ l, l+1), the optimum choice o the intege l() is l α () = α( + 1). This has the ollowing geometic intepetation: The optimum choice o l is such that the ay (, l()) belongs to the stip (, 0) + R (see Figue 3). This is elated to the egion o best possible litings o wedges as intoduced in [5] Section 3.1. To simpliy the notation, let π α = π l α α. 3. Limit cases Now we conside the exteme cases α = 1 and α 0. Poposition 3. Fo α = 1, π 1 () = min{ 4, 1 }, i.e. we get the GMI cut.

5 z (,0)+R x Figue 3: Region (, 0) + R o best possible litings. Now we ocus on the case α 0. Let π 0 () = < 0 1 [0, 1 ] +1 > 1. Lemma 1. Fix R. Then thee exists α 0 > 0 such that o any 0 < α α 0, we have l α () = 0 i 1 and l α () = 1 i > 1. Hence, o all 0 < α α 0 we have π α () = π 0 (). The above lemma diectly gives the behavio o π α with α 0. Poposition 4. The pointwise limit lim α 0 π α equals π 0, i.e. we get the lopsided cut o Balas and Qualizza []. 4 Stength o ( ψ, π α ) The main goal o this section is to show that the unction ( ψ, π α ) is exteme o (IP()). In act we will pove an even stonge esult, namely that this unction is exteme in the 0, 1 case. That is, we conside the moe esticted system x = + R s + R y x {0, 1} s 0 (B()) y Z + (s, y) has inite suppot. Since IP() is a elaxation o B(), ( ψ, π α ) is valid o the latte o all α [0, 1]. We say that a valid unction (ψ, π) is exteme o (B()) i thee ae no distinct valid unctions (ψ 1, π 1 ) and (ψ, π ) such that ψ = ψ 1 + ψ and π = π 1 + π. The ollowing theoem omally states the main esult o this section. Theoem 1. Fo all α [0, 1], the unction ( ψ, π α ) is exteme o (B()). In paticula, this implies that we cannot impove the coeicients o the valid inequality o (IP()) ψ()s + π α ()y 1 5

6 even i we use the additional inomation that x {0, 1}. weake, namely that this inequality is minimal. We stat by showing something 4.1 Minimality We say that a valid unction (ψ, π) o B() is minimal i thee is no othe valid unction (ψ, π ) such that ψ ψ and π π. The ollowing lowe bound on valid unctions is the main obsevation to pove that ( ψ, π α ) is minimal. Lemma. Let (ψ, π) be a valid unction o B(). Then: 1. ψ ψ. π() + π(1 ) 1 o all R. Poo. Conside the ist popety. Notice that o > 0, setting s = 1 and all othe s s equal to 0 gives a easible solution o (B()); the validity o (ψ, π) implies that ψ() s 1, o equivalently that ψ() ψ(). Fo < 0, we employ the same easoning to the solution given by s = and s = 0 o to obtain ψ() ψ(). Finally, we use the solution s 0 = M > 0, s 1 = 1 and s = 0 o / {0, 1} to obtain that ψ(0) 1 ψ(1)(1 ) M ; taking M gives ψ(0) 0 = ψ(0). The second popety is poved similaly by consideing the easible solution obtained by setting ȳ = 1, ȳ 1 = 1 and evey othe component o ȳ to 0. Lemma 3. Fo all α [0, 1], the unction ( ψ, π α ) is minimal o (B()). Poo. Conside (ψ, π) such that ψ ψ and π π α. The ist pat o Lemma shows that ψ = ψ. Moeove, elementay calculations show that o all, π α () + π α (1 ) = 1. This shows that π = π α. Hence the pai ( ψ, π α ) is minimal. 4. Extemality Fist we ocus on poving Theoem 1 o the case α (0, 1]. It is easy to check that the unction π α is piecewise linea; uthemoe, o α > 0 its beakpoints occu at k α and k α + 1 o k Z; notice that the ist and the second set o beakpoints ae espectively the local minima and maxima o π α. Let... < x < x 1 < x 0 = 0 < x 1 < x <... be this set o beakpoints; accoding to this deinition, x i is a local minimum o i even. It is easy to check that π α is quasipeiodic, namely o all i Z and [x i 1, x i+1 ] we have π α () = π α (x i ) + π α ( x i ). In ode to pove Theoem 1, conside valid unctions (ψ 1, π 1 ) and (ψ, π ) satisying ψ = ψ 1 + ψ and π α = π 1 + π ; we show that ψ = ψ 1 = ψ and π α = π 1 = π. Fist notice that Lemma implies ψ 1 ψ and ψ ψ; but then since ψ 1 + ψ = ψ, it is clea that we must have the equality ψ 1 = ψ = ψ. Theeoe, we only need to pove π 1 = π = π α. It is easy to see that ( ψ, π 1 ) and ( ψ, π ) ae minimal: i thee wee, say, a valid π 1 π 1 such that π 1 π 1, then ( ψ, π 1 + π ) would be a valid unction contadicting the minimality o ( ψ, π α ). Now we evoke the ollowing popety about minimal valid unctions. Lemma 4. Conside a minimal valid unction (ψ, π) o (B()). Then π is subadditive, namely o all 1, R, π( 1 + ) π( 1 ) + π( ). Ate poving that π(0) 0 as in the ist pat o Lemma, the poo o Lemma 5. given in [4] goes though to pove the above lemma; details ae omitted. Claim 1. Fo [x 1, x 1 ], π 1 () = π () = ψ() = π α (). 6

7 Poo. Fix j {1, }. Using standad techniques (see [5]), one can show that since ( ψ, π j ) is valid then ( ψ, min{ ψ, π j }) is also valid. The minimality o ( ψ, π j ) then implies that π j min{ ψ, π j } ψ. Howeve, notice that o [x 1, x 1 ], ψ() = πα (). As beoe, o [x 1, x 1 ] the act that ψ() = π 1 () + π () togethe with the pevious paagaph implies π 1 () = π () = ψ(). The esult then ollows. Claim. Fo [x i 1, x i+1 ] and j = 1,, we have π j () π j (x i ) = π α () π α (x i ). Poo. Since π j is minimal, Lemma 4 gives that π j () π j (x i ) + π j ( x i ). But since x i [x 1, x 1 ], it ollows om Claim 1 and the quasipeiodicity o π α that π j ( x i ) = π α ( x i ) = π α () π α (x i ); this gives that π j () π j (x i ) π α () π α (x i ) o j = 1,. Adding these inequalities o j = 1,, dividing by and using the act that π α = π 1 + π, we again obtain that we must have the equality π j () π j (x i ) = π α () π α (x i ) o j = 1,. This concludes the poo. Now take i N, and [x i 1, x i+1 ]; we will show that π 1 () = π () = π α (). Fo that, simply wite π j () = (π j () π j (x i )) + i k=1 (π j(x k ) π j (x k 1 )) + π j (x 0 ). Applying Claim to each paenthesized expession and Claim 1 to the last tem, we obtain that π j () = π α (), giving the desied esult. The case i N can be handled analogously, which then poves Theoem 1 when α (0, 1]. The case α = 0 needs to be handled sepaately. As beoe, we still have ψ 1 = ψ = ψ. Moeove, as in Claim 1, the act that π 0 () = ψ() o all 1 implies that π 1 () = π () = π 0 () o all 1. Fo > 1, we claim that we have an equality analogous to Claim, namely that π j () π j (1 ) = π 0 () π 0 (1 ) o j = 1,. To see this, ist notice that o > 1 we have π 0 () = π 0 (1 ) π 0 (1 ). Then using the subadditivity o π j and the act that 1 1, we get π j (1 ) π j (1 ) + π j () = π 0 (1 ) π 0 () + π j (), and the claim ollows as in Claim. Since π j (1 ) = π 0 (1 ), this equation gives that π j () = π 0 () o all > 1 and j = 1,. This concludes the poo o Theoem Split cuts Let LP denote the linea elaxation o omulation IP(), i.e. it is obtained om IP() by eplacing the integality conditions x {,..., 0, 1}, y Z + by x 1, y 0. We say that an inequality ψ()s + π()y 1 is a split inequality i it is satisied by all (x, s, y) in LP ({ax + b y c} {ax + b y c + 1}), o some a, c Z, b Z o all R. Note that split disjunctions {ax + b y c} {ax + b y c + 1} ae valid o IP() even when we elax the uppe bound on x, namely when we simply take x to be an intege. In othe wods, split inequalities do not use the uppe bound inomation on x. This is in contast with the wedge inequalities (1), which use the uppe bound on x, being deived om the disjunction {(x, z, s, y) : (x, z) / intk α } (ecall Figue 1). It is theeoe somewhat supising that evey wedge inequality is dominated by a split inequality, as we show next. Theoem. Fo all α (0, 1] and all l : R Z, the wedge inequality (1) is a split inequality. Poo. To pove the theoem, we show that the wedge inequality (1) is satisied by all (x, s, y) in LP ({x l()y } {x 1 l()y }). Let β =. Notice that 0 < β 1 since 0 < α 1. α 1 α(1 ) Claim 1 I ( x, s, ȳ) LP ({x l()y } {x 1 l()y }), then ( x, s, ȳ) LP ({x l()y } {β(x 1) l()y }). 7

8 Poo. I ( x, s, ȳ) LP {x l()y }, then the claim holds. Now suppose ( x, s, ȳ) LP {x 1 l()y }. Since 0 < β 1, and x 1 is a constaint o LP, we have (1 β)( x 1) 0. Hence, l()ȳ ( x 1) (1 β)( x 1) = β( x 1) and we ae done. To inish the poo o the theoem, obseve that (1) is valid o LP ({x l()y } {β(x 1) l()y }). To see this, conside the unction ψ : R R deined as ψ(, l) = max{a 1 (, l), a (, l)}. It is easy to veiy that ψ is convex, and positively homogeneous, and thus it is subadditive, i.e., ψ(, l) + ψ(, l ) ψ( +, l + l ). Moeove, ψ(x, z) = 1 on the bounday o the set K α. By convexity o ψ and K α, we get that ψ(x, z) 1 i (x, z) intk α, i.e. x z o β(x 1) z. Using the equations o (x, z) in the deinition o LP (l), we deive the ollowing valid inequality o all (s, y) such that (x, z) satisies x z o β(x 1) z: 1 ψ(x, z) = ψ( (, 0)s + (, l())y ) ψ(, 0)s + ψ(, l())y. Substituting l()y o z into the inequalities x z o β(x 1) z gives the desied esult. We obtain as a coollay a esult o the cuts ( ψ, π α ), including the case α = 0. Coollay 1. Fo all α [0, 1], the inequality geneated by ( ψ, π α ) is a split inequality. Poo. It suices to pove o α = 0. Conside any easible solution (x, s, y) in LP ({(x, s, y) : x >1 y } {(x, s, y) : x 1 + >1 y }). Since s and y have inite suppot, Lemma 1 implies that thee exists α 0 > 0 such that: (i) l α0 () equals 0 i 1 and equals 1 i > 1 ; (ii) π α0 () = π 0 () o all such that s 0 o y 0. Popety (i) and the poo o Theoem give that (x, s, y) satisies the inequality deined by ( ψ, π α0 ). Popety (ii) then shows that (x, s, y) also satisies the inequality deined by ( ψ, π 0 ). This concludes the poo. 5 Computations We peomed computational tests to assess the pactical consequences o Theoem o Section 4.3. We have selected 63 instances om MIPLIB 010 with binay and continuous vaiables only. We have geneated wedge cuts ( ψ, π α ) with α {0, 0.5, 0.8, 0.85, 0.9, 0.95, 1} o the most actional binay vaiables in the optimal basis o the LP elaxation. Fo the expeiments we have used the LP solve o FICO Xpess with no pepocessing at all. Fo each instance, and each ixed α < 1, we geneated one ound o wedge cuts o stengthening the LP elaxation (50 o each one o the two oientations o the cone K α, c. Section 3). Fo α = 1, when wedge cuts ae equivalent to GMI cuts by Poposition, we have geneated one ound o 50 cuts. Hence, o each instance we have a total o 7 uns o ou cutting plane algoithm, and in each un we have added cuts with a single ixed α paamete only. Let LB(I) and LB α + (I) denote the optimum value o the LP elaxation, and that ate adding one ound o cuts o paamete α, espectively, on instance I. Fo each instance I, and paamete α, we have computed the quantity A α (I) = (LB α + (I) LB(I))/ LB(I) + 1, whee the denominato is petubed by 1 to handle those cases with LB(I) = 0. Let Āα denote the aveage o the A α (I) values ove the 63 instances o each α. Table 1 depicts the aveages. Obseve that as α tends to 1 (i.e., the cuts appoach GMI cuts), the impovement ove the LP optimum stictly inceases. Ou indings complement those o Balas and Qualizza [] which showed that in pactice lopsided cuts do not impove much ove GMI cuts om the tableau, although they ae occasionally stonge. On aveage, they tend to be weake. What these esults indicate is that it does not pay to use bounds on the basic vaiables when geneating 8

9 α (GMI) Ā α Table 1: Aveage impovement ove the LP optimum o inceasing α values. GMI cuts om tableau ows. Instead it is peeable to geneate the standad GMI cut and to simply use the bounds diectly in the omulation. This seems counteintuitive but it is in ageement with the poo o Theoem. Reeences [1] Egon Balas and Robet B. Jeoslow, Stengthening Cuts o Mixed Intege Pogams, Euopean Jounal o Opeations Reseach 4 (1980) [] Egon Balas and Andea Qualizza, Stonge Cuts om Weake Disjunctions, woking pape, Teppe School o Business, August 010. [3] Amitabh Basu, Michele Conoti, Gead Conuejols and Giacomo Zambelli, Minimal Inequalities o an Ininite Relaxation o Intege Pogams, SIAM Jounal on Discete Mathematics 4 (010) [4] Michele Conoti, Gead Conuejols and Giacomo Zambelli, Cone Polyhedon and Intesection Cuts, Suveys in Opeations Reseach and Management Science 16 (011) [5] Michele Conoti, Gead Conuejols and Giacomo Zambelli, A Geometic Pespective on Liting, Opeations Reseach 59 (011) [6] Ralph E. Gomoy and Ellis L. Johnson, Some Continuous Functions Related to Cone Polyheda I, Mathematical Pogamming 3 (197)