c 2007 Society for Industrial and Applied Mathematics

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1 SIAM J DISCRETE MATH Vol 21 No 3 pp c 2007 Soety for Industril nd Applied Mthemtics THE INTEGER KNAPSACK COVER POLYHEDRON HANDE YAMAN Astrct We study the integer knpsck cover polyhedron which is the convex hull of the set of vectors x Z n tht stisfy CT x with C Z n nd Z We present some generl results out the nontrivil fcet-defining inequlities Then we derive spefic fmilies of vlid inequlities nmely rounding residul cpty nd lifted rounding inequlities nd identify cses where they define fcets We lso study some known fmilies of vlid inequlities clled 2-prtition inequlities nd improve them using sequence-independent lifting Key words integer knpsck cover polyhedron vlid inequlities fcets sequence-independent lifting AMS suect clssifictions 90C10 90C57 DOI / Introduction The purpose of this pper is to study the integer knpsck cover polyhedron Let N {1 2n} Item i N hs cpty We would like to cover demnd of using integer mounts of items in N We ssume tht nd for i N re positive integers We re interested in the integer knpsck cover set { X x Z n : } (1) x i i N nd its convex hull PX conv(x) The constrint i N x i is clled the cover constrint Set X is relxtion of the fesile sets of mny optimiztion prolems involving demnds tht my e covered with different types of items Pochet nd Wolsey [15] study spel cse to derive vlid inequlities for network design prolem Mzur [11] uses the polyhedrl results on PX to generte strong vlid inequlities for the multiflity loction prolem Ymn [18] uses the sme relxtion to strengthen formultions for the heterogeneous vehicle routing prolem which generlizes the well-known cptted vehicle routing prolem y introdung the choice etween different vehicle types Ymn nd Sen [19] rrive t the sme relxtion in the context of the mnufcturer s mixed pllet design prolem where ech customer cn uy integer numers of pllets with different configurtions to stisfy its demnd Knowledge out polyhedrl properties of PX cn e used in deriving strong formultions for these prolems For recent work in understnding the structure of simple mixed integer nd integer sets see eg [ ] There hs een lot of work on the polytope of the 0/1 knpsck prolem (eg [ ]) The sitution is different for the integer knpsck cover polyhedron Despite the mny ppliction res where set X my pper s relxtion the literture on the polyhedrl properties of its convex hull is quite limited Pochet nd Wolsey [15] study the spel cse where 1 is n integer multiple of for ll i 1 2n 1 They derive the prtition inequlities nd show tht Received y the editors Septemer ; ccepted for puliction (in revised form) Mrch ; pulished electroniclly July Bilkent University Deprtment of Industril Engineering Bilkent Ankr Turkey (hymn@ilkentedutr) 551

2 552 HANDE YAMAN these inequlities define the convex hull together with the nonnegtivity constrints They derive conditions under which these inequlities re vlid in the generl cse Mzur [11] nd Mzur nd Hll [12] study the generl cse They show tht dim(px)n x i 0 defines fcet of PX for i N nd if i N α ix i α 0 is nontrivil fcet-defining inequlity of PX then α i > 0 for ll i N nd α 0 > 0 Let c 1c m e the distinct vlues tht re less thn An importnt result y Mzur [11] is tht if one knows the description of conv({x Z m : m i1 c i x i }) it is trivil to otin the description of PX The inequlity i N α ix i α 0 is nontrivil fcet-defining inequlity for PX if nd only if α i α for ll i N with α i α 0 for ll i N with nd m i1 α i x i α 0 is nontrivil fcet-defining inequlity for conv({x Z m : m i1 c i x i }) where α i α if c i for i 1 m nd N So interesting instnces stisfy c 1 <c 2 < <c n < Mzur nd Hll [12] lso study the integer cpty cover polyhedron defined s the convex hull of the set {(y x) {0 1} q Z n : i N x i q i1 y i} They use simultneous lifting to derive fcet-defining inequlities for this polyhedron using those of the integer knpsck cover polyhedron They remrk tht little is known out the polyhedrl properties of the ltter polyhedron nd it is difficult to identify its fcets Atmturk [1] presents fmily of fcet-defining inequlities nd lifting results for the polytope conv(x {x Z n : x u}) for u Z n In this pper we derive severl fmilies of vlid inequlities nd discuss when they define fcets of PX We investigte the domintion reltions etween these fmilies of vlid inequlities Most of our results on fcet-defining inequlities re for the spel cse where c 1 1 This work is motivted y the results of Mzur nd Hll [12] where vlid inequlities for the integer knpsck cover polyhedron re lifted to vlid inequlities for more complicted polyhedron the integer cpty cover polyhedron We re lso motivted y the positive results in [18 19] which demonstrte the use of simple vlid inequlities sed on the integer knpsck cover relxtion in closing the dulity gp for complicted mixed integer progrmming prolems studied in these ppers The pper is orgnized s follows In section 2 we give the generl properties of nontrivil fcet-defining inequlities of PX In sections 3 6 we introduce four fmilies of vlid inequlities nmely rounding residul cpty lifted rounding nd lifted 2-prtition inequlities We compre their reltive strengths nd give conditions under which they define fcets of PX In section 7 we investigte the use of lifted rounding nd lifted 2-prtition inequlities in solving the mnufcturer s mixed pllet design prolem introduced y Ymn nd Sen [19] We conclude in section 8 2 Generl results on fcet-defining inequlities In this section we derive generl properties of nontrivil fcet-defining inequlities of PX In the sequel we ssume tht c 1 c n nd re positive integers nd tht they stisfy c 1 <c 2 < <c n <(this ssumption is mde without loss of generlity due to the result of Mzur [11] mentioned ove) Let c e the gretest common divisor of s We replce with c for ech i N nd with c This does not chnge the set X ut strengthens the cover constrint Let e i denote the n-dimensionl unit vector with 1 t the ith plce nd 0 elsewhere Proposition 1 Let i N α ix i α 0 e nontrivil fcet-defining inequlity for PX Then 0 <α 1 α 2 α n α 0 min α i i N

3 THE INTEGER KNAPSACK COVER POLYHEDRON 553 Proof Suppose tht i N α ix i α 0 is nontrivil fcet-defining inequlity for PX The fct tht α i > 0 for i 0 1n is proved in [11 12] Let nd l e such tht <lnd x PX e such tht i N α ix i α 0 with x 1 Consider x x e e l As c l > x PX Then i N α ix i α 0 implying tht α l α Soα 1 α 2 α n Let x PX e such tht i N α ix i α 0 with x n 1 Then α n x n α 0 nd s x n 1 α n α 0 For i N x c ei i is in PX nd so α i α0 Thus α 0 min i N α i We hve necessry condition for nontrivil inequlity to e fcet-defining Theorem 1 Let i N α ix i α 0 e nontrivil fcet-defining inequlity for c PXLet rg mx i i N α i Then (α 0 α i ) c α for ll i N \{} Proof Assume tht there exists l N \{} such tht (α 0 α l ) c α c l < Let x X e such tht i N α ix i α 0 Then x α0 i N\{} αixi α The left-hnd side of the cover constrint evluted t x is i N x i i N\{} ( c α α i )x i c α α 0 This is less thn or equl to (c l c α α l )x l c α α 0 since c α α i 0 for ll i N \{} Now s (α 0 α l ) c α c l <nd c l c α α l 0 whenever x l 1 (c l c α α l )x l c α α 0 < This proves tht for ny x X such tht i N α ix i α 0 wehvex l 0 Next we give necessry nd suffient conditions for some inequlities to e fcet-defining Lter we use this result to identify spefic fmilies of fcet-defining inequlities Theorem 2 Let i N α ix i α 0 e vlid inequlity for PX with α i > 0 nd integer for ll i N {0} nd α 1 1 Let e the lrgest index with α 1 If α i for ll i 1n then the inequlity i N α ix i α 0 is fcet-defining for PX if nd only if (α 0 α i ) for i 1n nd (α 0 1) c 1 Proof If the conditions of the theorem re stisfied then α 0 e (α 0 1)e e i for i 1 1 nd (α 0 α i )e e i for i 1n re in PX; they stisfy i N α ix i α 0 nd re ffinely independent This proves tht the inequlity i N α ix i α 0 is fcet-defining for PX The necessity of the conditions re implied y Theorem 1 To conclude this section we investigte when the cover constrint is fcet-defining for PXIf divides for ll N then the nonnegtivity constrints nd the cover constrint descrie the polyhedron PX ie PX {x R n : N x } Using Theorem 2 we identify nother cse where the cover constrint is fcetdefining Corollry 1 If c 1 1 then the cover constrint is fcet-defining for PX The conclusion of Theorem 1 is trivilly stisfied for the cover constrint But the cover constrint is not necessrily fcet-defining for PX The following simple exmple proves this sttement Exmple 1 Let X 1 {x Z 2 :3x 1 4x 2 14} The polyhedron conv(x 1 ) {(x 1 x 2 ) R 2 : x 1 x 2 4 2x 1 3x 2 10} 3 Rounding inequlities In this section we derive fmily of vlid inequlities clled the rounding inequlities nd identify some cses where they re fcetdefining for PX For λ>0 the rounding inequlity (2) i N λ x i λ

4 554 HANDE YAMAN is vlid inequlity for PX It is otined using the well-known Chvtl Gomory procedure (see eg Nemhuser nd Wolsey [14]) These inequlities hve een used y Ymn [18] Here we investigte under which conditions these inequlities re fcetdefining for PX The inequlity for λ c n is i N x i c n Mzur [11] proves tht this inequlity is fcet-defining for PX if nd only if ( ) c n 1 cn c 1 Inequlity (2) for ny λ>c n is dominted y the corresponding inequlity for c n So we re interested in λ<c n The result elow is corollry to Theorem 2 Corollry 2 Let λ e such tht λ<1 for some {1n 1} If λ for ll i 1n then inequlity (2) is fcet-defining if nd only if ( ) λ 1 c c 1 nd ( λ ) λ c for ll i 1n Proof As λ for i N nd λ re positive integers c1λ 1 is the lrgest index with coeffient 1 in inequlity (2) nd λ for ll i 1n Theorem 2 pplies We hve necessry condition s corollry to Theorem 1 Corollry 3 Let λ>0 If there exists N such tht is divisile y λ nd if inequlity (2) is fcet-defining for PX then ( λ λ )λ for ll i N \{} Proof For i N rg mx i N λ λ λ nd we cn pply Theorem 1 So if N is such tht λ divides We consider the suset of inequlities (2) defined y λ equl to c 1 c n In the following corollry we generlize the result y Mzur [11] Corollry 4 For N the inequlity (3) i N x i is fcet-defining for PX if nd only if ( ) 1 c c 1 nd ( ) c c for ll i 1n Proof Tke λ As for ll i 1n we pply Corollry 2 to otin the result Atmturk [1] studies the polytope conv(x {x Z n : x u}) for u Z n nd proves tht inequlity (3) for N such tht u is fcet-defining if nd only if the conditions of Corollry 4 re stisfied We go ck to Exmple 1 nd see if rounding inequlities re fcet-defining Exmple 2 Consider set X 1 defined in Exmple 1 The rounding inequlity for λ c 1 is not fcet-defining since ( ) < 14 The inequlity is x 1 2x 2 5 nd is dominted y 2x 1 3x 2 10 We cn otin the ltter inequlity y lifting inequlity x 1 5 which is rounding inequlity when x 2 0 with vrile x 2 (see section 5) The rounding inequlity for λ c 2 is fcet-defining since ( ) This is the inequlity x 1 x 2 4 The convex hull of X 1 is descried y the nonnegtivity constrints rounding inequlity (x 1 x 2 4) nd lifted rounding inequlity (2x 1 3x 2 10) In the next exmple we see two sets tht re defined y prmeters which differ only in the right-hnd side of the cover constrint The rounding inequlities for λ c 2 c 3 c n re fcet-defining for the polyhedron when the right-hnd side is nd none re fcet-defining when the right-hnd side is 1

5 THE INTEGER KNAPSACK COVER POLYHEDRON 555 Exmple 3 Consider the set X 2 {x Z 4 : x 1 4x 2 5x 3 6x 4 61} The convex hull of X 2 is descried y the nonnegtivity constrints nd the following inequlities (these results re otined using PORTA [6]): (4) (5) (6) (7) (8) x 1 4x 2 5x 3 6x 4 61 x 1 2x 2 3x 3 3x 4 31 x 1 x 2 2x 3 2x 4 16 x 1 x 2 x 3 2x 4 13 x 1 x 2 x 3 x 4 11 Inequlity (4) is the cover constrint By Corollry 1 s c 1 1 we know tht the cover constrint is fcet-defining Inequlities (6) (8) re rounding inequlities It is esy to verify tht the conditions of Corollry 4 re stisfied Note tht inequlity (5) is the rounding inequlity for λ 2 nd the conditions of Corollry 3 re stisfied Now consider the set X 3 {x Z 4 : x 1 4x 2 5x 3 6x 4 62} The following inequlities together with the nonnegtivity constrints descrie the convex hull of X 3 : (9) (10) (11) (12) x 1 4x 2 5x 3 6x 4 62 x 1 2x 2 3x 3 4x 4 32 x 1 2x 2 2x 3 3x 4 26 x 1 2x 2 2x 3 2x 4 22 The cover constrint (9) is fcet-defining ut the rounding inequlities for λ c 2 c 3 c 4 do not define fcets Inequlity (10) domintes the rounding inequlity for λ c 2 which is x 1 x 2 2x 3 2x 4 16 (11) domintes inequlity x 1 x 2 x 3 2x 4 13 which is the rounding inequlity for λ c 3 nd (12) domintes x 1 x 2 x 3 x 4 11 which is the rounding inequlity for λ c 4 In the following section we will identify these inequlities (10) (12) 4 Residul cpty inequlities Residul cpty inequlities re introduced y Mgnnti Mirchndni nd Vchni [10] for the single rc design prolem Here we present inequlities tht re sed on similr ide Assume tht the demnd is covered using some item N Then t lest units of item need to e used If 1 units re used to full cpty then the cpty of the lst unit to e used is r ( 1) If only 1 units of item re used then the remining items should cover demnd equl to r This is expressed in the following vlid inequlity For N define N {1 2} nd N {i N : r } For N 0 N nd N 1 N \ N 0 let X h (N 1 ){x Z n : i N x i h x i 0 for ll i N 0 } Theorem 3 For N the inequlity (13) min{ r }x i i1 n i1 x i r is vlid for PX Proof If i N x i then the inequlity is stisfied If i N x i p for some p 1 then the fesiility of x implies i N \N x i n i1 x i

6 556 HANDE YAMAN i N x i i N x i r (p1) Asr (p1) r p inequlity (13) is stisfied For N ifr then is divisile y nd inequlity (13) is the sme s the cover constrint Theorem 4 If c 1 1for N the inequlity (14) min{ r }x i r i1 is fcet-defining for conv(x (N )) Proof Let F {x X (N ): i1 min{r }x i r } Assume tht ll x F stisfy i1 α ix i α 0 As e F we need α 0 α For i N ( 1)e e i F implying tht α i α Asc 1 1 we hve ( 1)e r e 1 F So α 1 α r Finlly for i N \ (N {1}) ( 1)e e i (r )e 1 F Hence α i α r r Then i1 α ix i α 0 is α r multiple of i1 min{r }x i For N ifr 1 then inequlity (14) is i1 x i nd is the sme s the rounding inequlity for λ for conv(x (N )) By Corollry 4 it is fcet-defining since ( 1 ) c c 1 r c 1 For n conv(x (N n )) PX nd the following result cn e deduced from Theorem 4 Corollry 5 If c 1 1 inequlity (13) for n is fcet-defining for PX Exmple 4 Consider the set X 3 given in Exmple 3 For item 2 r 2 2 nd c 2 16 Inequlity (13) for item 2 is x1 2x 2 5x 3 6x 4 32 nd is dominted y inequlity (10) For item 3 r 3 2 nd c 3 13 The corresponding inequlity (13) is x 1 2x 2 2x 3 6x 4 26 nd is dominted y inequlity (11) For item 4 r 4 2 nd c 4 11 Inequlity (13) is x1 2x 2 2x 3 2x 4 22 nd is the sme s inequlity (12) In the remining of this section we will try to identify inequlities (10) nd (11) We cn generlize inequlity (13) s follows Theorem 5 For N let μ 0 e such tht r (r μ) μ r nd r μ The inequlity (15) min{ r }x i i1 n i1 (r μ) x i r c is vlid for PX Proof If i N x i then the inequlity is stisfied If i N x i 1 then inequlity (15) simplifies to i N \N x i n (r μ) i1 c xi r By fesiility we need to hve i N \N x i n i1 x i r Using coeffient reduction we otin i N \N x i n i1 r x i r As (r μ) r for ll i 1n inequlity (15) is stisfied If i N x i p for some p 2 then inequlity (15) simplifies to i N \N x i n (r μ) i1 c xi r p The fesiility of x implies tht x i n i1 x i r (p1) We multiply this inequlity with rμ i N \N

7 THE INTEGER KNAPSACK COVER POLYHEDRON 557 nd otin r i N \N μ i x i n x i r(rμ) (p1)(r μ) Now s r μ nd so i N \N x i n (r μ) i1 c xi (r i N \N μ) i x i n i1 (rμ) x i we hve i N \N x i n (r μ) i1 c xi r(rμ) (p 1)(r μ) Since the left-hnd side is lwys n integer we round up the righthnd side nd get r (r μ) nd p 2 we otin i N \N i1 rμ (p1)μ (p1)r As r (r μ) μ r μ 0 x i n (r μ) i1 c xi r p So x stisfies inequlity (15) For μ r inequlity (15) is the sme s inequlity (13) As μ increses inequlity (15) gets weker So for given N we re interested in inequlity (15) defined y the smllest μ tht stisfies the condition r (r μ) r Let ɛ>0 e very smll We tke μ c(r1)r2 r μ ɛ if r 2 <r nd μ 0 otherwise Oserve tht nondominted residul cpty inequlities (15) re defined per item so there re O(n) of them Exmple 5 Consider gin the set X 3 of Exmple 3 For item 2 r 2 2 As r 2 2 c 2 1< 2r2 μ 2 4(21)4 24 ɛ ɛ The corresponding inequlity (15) is x 1 2x 2 3x 3 4x 4 32 nd is the sme s inequlity (10) For item 3 r 3 2 As r 2 3 c 3 1< 2r3 μ 3 5(21)4 25 ɛ 1 7 ɛ The corresponding inequlity (15) is x 1 2x 2 2x 3 3x 4 26 nd is the sme s inequlity (11) If r 1 then μ 0 nd inequlity (15) is the sme s the rounding inequlity (3) for λ If r then gin μ 0 This time inequlity (15) is the sme s the cover constrint We hve necessry condition for inequlity (15) to e fcet-defining Corollry 6 For N if inequlity (15) is fcet-defining for PX nd r < then c c r (r μ ) for ll i 1n Proof As c r min{ r } 0 for ll i 11 nd ( c r (r μ ) ) 0 for ll i 1n we pply Theorem 1 So if inequlity (15) is fcet-defining for PX then c min{ r } c r for i 11 nd c c r (r μ ) for ll i 1n For i N the condition is c The left-hnd side is equl to c c r Fori N \ N the condition is c r ( r ) (r) r c The left-hnd side is equl to r r i r since r nd r So the conditions of Theorem 1 re lwys stisfied for i N 5 Lifted rounding inequlities In this section we derive vlid inequlities using lifting For N 0 N nd N 1 N \N 0 let i N 1 α ix i α 0 e vlid inequlity for X (N 1 ) Suppose we lift inequlity i N 1 α ix i α 0 with x l with l N 0 The optiml lifting coeffient of x l is α l mx α 0 i N α ix 1 i x l st x l 1 x X (N 1 {l})

8 558 HANDE YAMAN Consider the cse where α i 1 for ll i N 1 rg mx i N 1 nd α 0 For l N 0 the nonliner lifting prolem simplifies to c α l mx (c l x l ) x l Z x l Clerly mximizing x l cnnot e lrger thn c l Hence we otin c α l mx (c l x l ) x l {12 c } x l l nd we cn compute α l y enumertion Exmple 6 Consider the set X 1 defined in Exmple 1 Inequlity x 1 5 is fcetdefining for conv(x 1 {x Z 2 : x 2 0}) We lift inequlity x 1 5 with vrile x 2 (144x2 ) 5 3 The optiml lifting coeffient α 2 mx x2 {1234} x 2 mx{ } 3 2 The corresponding inequlity is 2x 1 3x 2 10 nd is fcet-defining for conv(x 1 ) Computtion of the optiml lifting coeffients of vriles tht re lifted in lter in the sequence my ecome hrder So we re interested in sequence-independent lifting Atmturk [4] studies sequence-independent lifting for mixed integer progrmming The following cn e derived from his results Consider the lifting function Φ() α 0 min x X (N 1 ) i N α ix 1 i If this function is sudditive ie if Φ()Φ(d) Φ( d) for ll d R then the lifting is sequence-independent In this cse the inequlity i N α ix 1 i i N Φ()x 0 i α 0 is vlid inequlity for PX In the generl cse let Θ e sudditive function with Θ Φ Then the inequlity i N α ix 1 i i N Θ()x 0 i α 0 is vlid inequlity for PX If the inequlity i N α ix 1 i α 0 is fcet-defining for conv(x (N 1 )) nd Θ( )Φ( ) for ll i N 0 then inequlity i N α ix 1 i i N Θ()x 0 i α 0 is fcet-defining for PX Theorem 6 Let N 1 N nd i N α ix 1 i α 0 e vlid inequlity for X (N 1 ) If there exists N 1 such tht α i α for ll i N 1 \{} then the lifting function is Φ() α 0 α ( ) Proof Suppose there exists N 1 such tht α i α for ll i N 1 \ {} The lifting function is Φ() α 0 min x X (N 1 ) i N α ix 1 i Let x e n optiml solution to the minimiztion prolem Consider x x i N 1 \{} x ie i i N 1 \{} xi e Clerly x X (N 1 ) The oective function evluted t x is equl to α i x i α i x i i N 1 i N 1 i N 1 α i x i i N 1 \{} i N 1 \{} i N α i x i α 1 \{} c ix i α i x i α i N 1 \{} x i As α i α for ll i N 1 \{} i N 1 α ix i i N 1 α ix i nd so x is lso optiml Hence () e is lso optiml nd the optiml vlue is α ()

9 THE INTEGER KNAPSACK COVER POLYHEDRON 559 Θ( ) Φ( ) Fig 1 Lifting function Φ nd sudditive function Θ for 17nd 5 Suppose there exists N 1 such tht α i α for ll i N 1 \{} α 1 nd α 0 The lifting function for the inequlity i N α ix 1 i is Φ() () The function Φ is not sudditive An exmple where 17 nd 5 is depicted in Figure 1 Here for 11 nd d 6 we hve d 42433< d 404 For N nd R define () Lemm 1 For N if () > 0 the function Θ() min{ () () 1} (see Figure 1) is sudditive Proof Let d R Then Θ() Θ(d) min{ () 1} d () min{ (d) 1} There re two cses: (i) () () (d) ( d) nd (ii) () (d) ( d) In cse (i) since () (d) ( d) we hve d d If min{ () (d) () 1} 1 or min{ () 1} 1 then Θ()Θ(d) d 1 Θ( d) Otherwise min{ () () Then Θ()Θ(d) d 1} () () () (d) () d () s () (d) ( d) d d min{ (d) () nd min{ (d) () 1} 1 then Θ()Θ(d) d () (d) So d min{ (d) () () () 1} 1 then Θ() Θ(d) d 1} (d) () Θ()Θ(d) d () d () () (d) () () () nd min{ 1} () () Θ( d) In cse (ii) 1 If min{ () () 1 Θ( d) If min{ () () () 1} 1 nd () 1} () Since () (d) Θ(d) The cse where min{ () () is similr Finlly if min{ () () 1 () () (d) 1 (d) () c () d () d tht Θ is sudditive Now we will lift the inequlity i N 1 α ix i 1} 1 nd () () 1} () nd min{ 1} () 1 (d) () c Since () Θ( d) This proves (d) () using the function Θ Theorem 7 Let N 0 N N 1 N \ N 0 nd i N α ix 1 i α 0 e vlid inequlity for X (N 1 ) If there exists N 1 such tht α 1 α i for ll

10 560 HANDE YAMAN i N 1 \{} α 0 nd () > 0 then the inequlity ( ) (16) ()α i x i () min{ ( ) ()} x i () i N 1 i N 0 is vlid inequlity for PX Proof The inequlity i N α ix 1 i is vlid for X (N 1 ) Consider the sudditive function Θ() min{ () () 1} given in Lemm 1 We will show tht Θ Φ If <nd () < () then ( ) () () > 0 So Φ() () Θ() ()()c () If <nd () () then Θ() Φ() If then Φ() If then () () So Θ() Φ() If 1 then Θ() Φ() So the inequlity i N α ix 1 i ( i N 0 min{ () 1}) () x i is vlid inequlity for PX Multiplying oth sides with () we otin inequlity (16) Some of the inequlities (16) re dominted y others Indeed s given in the following proposition the numer of nondominted inequlities (16) is polynomil Proposition 2 For N with () > 0 the inequlity (17) min{ ()}x i i1 n i1 ( () ) min{ ( ) ()} x i () is vlid nd domintes inequlity (16) for N 0 N N 1 N \ N 0 such tht N 1 α 1 α i for ll i N 1 \{} nd α 0 Proof Inequlity (17) is vlid since it is the sme s inequlity (16) for N 1 {} Let N 0 N N 1 N \N 0 such tht N 1 α 1α i for ll i N 1 \{} nd α 0 Fori N 1 () min{ ( ) ()} () ()α i So the coeffient of x i in (17) is less thn or equl to its coeffient in (16) The coeffients of x i for i N 0 nd the right-hnd sides re the sme in oth inequlities Hence inequlity (17) domintes inequlity (16) We cll inequlities (17) lifted rounding inequlities The numer of lifted rounding inequlities tht re not dominted is O(n) It is interesting to note tht even though inequlities (16) re not inequlities (17) re spel cses of the multiflity cut-set inequlities derived y Atmturk [2] for the single commodity-multiflity network design prolem For N such tht () > 0 consider the inequlity x which is fcetdefining for conv(x ({})) If c 1 () then for i< () So Φ( ) Θ( )1 Fori>if ( )0or ( ) () then Φ( )Θ( ) By Theorem 5 in Atmturk [4] the resulting inequlity (18) x i i1 n i1 x i is fcet-defining for PX Notice tht this is the sme inequlity s the rounding inequlity (2) for λ The condition c 1 () implies tht ( ) 1 c c 1 For i<if ( ) 0 then ( ) c c c c If ( ) ()

11 THE INTEGER KNAPSACK COVER POLYHEDRON 561 then ( ) c ( () c() ) c () ( ) As result the conditions stted ove re the sme s the conditions of Corollry 4 However Corollry 4 is stronger result since it sttes tht these conditions re oth necessry nd suffient Now we compre inequlities (17) nd (3) The two following propositions re esy to prove Proposition 3 For N with () 1 inequlities (17) nd (3) re the sme Proposition 4 For N with () 2 inequlity (17) domintes inequlity (3) If for N () > 0 (or equivlently r < ) then () r So residul cpty inequlities (15) nd inequlities (17) look very similr Coeffients of vriles x i with i {1} re the sme in oth inequlities The right-hnd sides re lso the sme Only coeffients of vriles x i with i { 1n} mye different Proposition 5 For N ifr < nd r 2 r then inequlity (15) for μ 0nd inequlity (17) re the sme Proof If r 2 r then the coeffient of x i with i { 1n} is r in inequlity (15) with μ 0 This is equl to ( c c ( ))r r ( )r Since ( ) nd r ()r min{ ( )r } So the coeffient of x i in (15) is less thn or equl to its coeffient in (17) If ( ) r then ()r r 2 r Now ssume tht ( ) <r nd < ( ) Then ( )r ( ( ) 1) This is equivlent to ( ()r r ) ( ) Since r 2 r r 2 > (r 1) So > ( r )r > ( r ) ( ) This contrdicts ( r ) ( ) Hence if ( ) <r then ()r ( ) So the coeffients of vrile x i in inequlities (15) nd (17) re the sme Proposition 6 For N if r 2 c <r then inequlity (17) domintes inequlity (15) for μ μ Proof If r 2 c <r then the coeffient of x i with i> in (15) for μ μ is (r μ ) c r c μ ()(rμ) If ( ) r then c μ ()(r μ ) c μ r(rμ) Since c μ r(rμ) μ r (r μ ) r Assume tht ( ) <r nd c μ ()(rμ) c < ( ) Then ( )(r μ ) ( ( ) 1 c μ ) or equivlently ( )( r ) μ Since r (r μ ) μ >r 1 we hve tht >r ( r μ ) μ nd now since r > ( ) > ( )( r μ )μ Putting together with ( )( r )μ we otin ( )( r ) μ > ( )( r μ ) μ This is equivlent to ( ) > since μ > 0 But this is impossile So if ( ) <r then c μ ()(rμ) ( ) This proves tht the coeffient of x i in (15) is greter thn or equl to its coeffient in (17)

12 562 HANDE YAMAN These four propositions show tht for N with () > 0 the lifted rounding inequlity (17) domintes the rounding inequlity (2) for λ nd the residul cpty inequlity (15) for μ μ For spel cse these inequlities (17) re fcet-defining for PX Theorem 8 For N such tht () > 0 ifc 1 1 then inequlity (17) is fcet-defining for PX Proof Suppose tht () > 0 nd c 1 1 Assume tht ll points in X which stisfy inequlity (17) t equlity lso stisfy n i1 α ix i α 0 The point e is in X nd stisfies inequlity (17) t equlity So α 0 α Notice tht if we remove one item the remining demnd to e covered is () For i<if > () then consider the point e i ( 1)e It is esy to verify tht this point is lso in X nd tht inequlity (17) is tight t this point Then we hve α i α For i<if () then the point e i ( c 1)e ( () )e 1 is in X nd inequlity (17) is tight t this point So α i α ( () )α 1 Since c 1 1 () α we otin α 1 α Then α () i Hence for i< α () i min{ ()} α () For i>if ( ) 0 consider point e i ( )e The left-hnd side of inequlity (17) t this point is equl to () ( ) () c () So inequlity (17) is tight The left-hnd side of the cover constrint is equl to ( ) c Thus this point is in X Then we hve α i α Finlly for i> with ( ) > 0 consider e i ( c )e ( () ( )) e 1 The left-hnd side of inequlity (17) evluted t this point is equl to () min{ ( ) ()}( c ) ()( () ( )) () c ()( c ) () Since ( ) > 0 this is equl to () ( 1) () c () showing tht inequlity (17) is tight t this point The left-hnd side of the cover constrint is equl to ( ) (19) ( () ( )) If ( ) > () then (19) is equl to ( )c ( 1)c ( )( 1)c > ()( 1)c If ( ) () then (19) is equl to ( c )c () ( ) ( c )c () ( ) So this point is in X This proves tht α i α ( () ( )) α 1 () α α ( () ( )) If () () ( ) then α i α α ( 1) If () > ( ) then α i α ( () ( )) α α () ( 1)α ( ) α α ( () ) So for i < α () i α ( min{ () 1}) () α () ( () min{ ( ) ()}) Hence n i1 α ix i α 0 hs the form 1 min{ ()} α () x iα x i1 n 1 α () ( ()min{ ( ) ()} )x i α c This is α times () i1 min{ ()}x i n ( i1 () c min{ ( ) ()} ) x i ()

13 THE INTEGER KNAPSACK COVER POLYHEDRON 563 Exmple 7 Consider the set X 4 {x Z 7 : x 1 2x 2 3x 3 4x 4 5x 5 6x 6 7x 7 38} The convex hull of X 4 is descried y the nonnegtivity constrints nd the following inequlities (otined using PORTA [6]): (20) (21) (22) (23) (24) (25) (26) (27) (28) (29) x 1 2x 2 3x 3 4x 4 5x 5 6x 6 7x x 1 2x 2 4x 3 4x 4 5x 5 6x 6 6x 7 34 x 1 2x 2 3x 3 3x 4 4x 5 5x 6 5x 7 28 x 1 2x 2 2x 3 3x 4 4x 5 4x 6 5x 7 26 x 1 2x 2 3x 3 3x 4 3x 5 4x 6 5x 7 24 x 1 2x 2 2x 3 2x 4 3x 5 4x 6 4x 7 20 x 1 2x 2 3x 3 3x 4 3x 5 3x 6 3x 7 18 x 1 x 2 2x 3 2x 4 2x 5 3x 6 3x 7 16 x 1 2x 2 2x 3 2x 4 2x 5 2x 6 3x 7 14 x 1 x 2 2x 3 2x 4 2x 5 2x 6 2x 7 12 As c 1 1 the cover constrint (20) is fcet-defining for conv(x 4 ) None of the rounding inequlities for items λ c 2 c 7 is fcet-defining for conv(x 4 ) For item 2 2 (38) 0 For item 3 3 (38) 2 Inequlity (17) for 3 x 1 2x 2 2x 3 3x 4 4x 5 4x 6 5x 7 26 is vlid inequlity nd is fcet-defining since c 1 1 nd 3 (38) > 0 Indeed it is the sme s inequlity (23) For item 4 4 (38) 2 Inequlity (17) reds x 1 2x 2 2x 3 2x 4 3x 5 4x 6 4x 7 20 nd is vlid inequlity This is the sme s inequlity (25) nd is fcet-defining Note here tht μ 4 ɛ nd inequlity (15) for item 4 x 1 2x 2 2x 3 3x 4 3x 5 4x 6 4x 7 20 is dominted y inequlity (25) For item 5 5 (38) 3 Inequlity (17) x 1 2x 2 3x 3 3x 4 3x 5 4x 6 5x 7 24 is the sme s inequlity (24) For item 6 6 (38) 2 The corresponding inequlity (17) is x 1 2x 2 2x 3 2x 4 2x 5 2x 6 3x 7 14 nd is the sme s inequlity (28) For item 7 7 (38) 3 The inequlity x 1 2x 2 3x 3 3x 4 3x 5 3x 6 3x 7 18 is vlid nd fcet-defining for conv(x 4 ) This is the sme s inequlity (26) 6 Lifted 2-prtition inequlities Pochet nd Wolsey [15] derive prtition inequlities for PX where divides 1 for ll i 1n 1 Then they prove tht these inequlities re vlid for PX in generl under some conditions Let (i 1 1 )(i p p ) e prtition of N such tht i 1 1 p n nd i t t1 1 for ll t 2p Let β p For t p1 compute κ t β t t nd β t1 β t (κ t 1)t The inequlity (30) p t1 ( t1 ) t κ s s1 { c min c i it t κ t } x p κ s s1 is clled the prtition inequlity Pochet nd Wolsey [15] prove tht the prtition inequlity is vlid for PX if κ t1 t t1 for ll t 2p If divides 1 for ll i 1n 1 then the prtition inequlities re vlid without ny condition nd they descrie PX together with nonnegtivity constrints Consider the cse where i 1 1 nd 1 n Then inequlity (30) reduces to the inequlity n 1 min { c 1 κ1 } x κ 1 This is the sme s the rounding inequlity (2) for λ c 1 since κ 1 c 1 nd c <for ll N The next spel cse is when i i 2 nd 2 n Then κ 2 β1 ( 1)c Notice tht β 1 r Finlly κ 1 r c 1 Inequlity

14 564 HANDE YAMAN (30) ecomes (31) 1 min i1 { c 1 r c 1 } x i r c 1 n i x i r c 1 nd is vlid if r c c 1 c 1 We refer to these inequlities s 2-prtition inequlities Proposition 7 For N ifc 1 1 inequlity (31) is dominted y the cover constrint or inequlity (17) Proof Ifc 1 1 then the inequlity simplifies to (32) min{ r }x i r i1 n i1 x i r c nd is lwys vlid If moreover r then the inequlity ecomes i1 x i n i1 c c xi nd is dominted y the cover constrint If r < then r () nd () > 0 For i>if is divisile y then r () ( ) since ( )0 If is not divisile y then r () () So the coeffient of x i in (32) is greter thn or equl to its coeffient in inequlity (17) For i the vrile x i hs the sme coeffient in (32) nd (17) Also the right-hnd sides of (32) nd (17) re the sme Hence if c 1 1 nd r < inequlity (17) domintes inequlity (32) If r c 1 c 1 for ll i < then inequlity (31) simplifies to i1 x i n i1 c xi which is the rounding inequlity (2) for λ c Now we will improve the 2-prtition inequlities (31) using lifting Let N 0 N N 1 N \ N 0 min rg min i N 1 nd N 1 with min The 2-prtition inequlity for the prtition N {i N 1 : i<} nd N {i N 1 : i } is { } r r r (33) x i x i i N min min min min i N nd is vlid when x i 0 for ll i N 0 if r c min min The lifting function for inequlity (33) is r β() min ( { } r min x i x X (N 1 ) i N min min min r min min i N Lemm 2 If r 1 nd r c min min for R r () min min if <nd 0 < () <r r β() min if <nd () r or () 0 r min if Proof For d R let ( z(d) min x X d (N 1 ) { min c i N min r min } x i r min i N x i ) x i )

15 THE INTEGER KNAPSACK COVER POLYHEDRON 565 If d 0 then z(d) 0 Ifd>0 Pochet nd Wolsey [15] prove tht there exists n optiml solution where x i 0 for i min nd i nd x min r min 1 Consider such optiml solutions If d< then e or d c emin min is optiml Hence z(d) min{ r min d min } If d c then x d since otherwise xmin c min So d c e (d) c emin min or d c e is optiml nd z(d) min{ r d min (d) r min d min }Soif< then { } r r r ( ) β() min min min min min Consider <If ( ) () () nd () > 0 then ( ) r ( ) β() min min ( r () ) ( ) ( ) min min ( r () ) () () ( ) min min ( ) r () ( ) min min r ( ) min min If ( ) () () then ( ) r r β() min min ( r () ) ( ) 1 min ( r () ) () () 1 min r () min r min If () 0 then β() r min r min r min ( ( () ) r min ) () 1 Function β is not sudditive in generl Consider 18 5 nd min 2 Let 25 nd 55 Then β(25)1β(55) 2 nd β(8) 4 So β(25)

16 566 HANDE YAMAN β(55) <β(8) So to do lifting we need sudditive function which is greter thn or equl to β We first study the cse where min divides r Notice tht in this cse r c min min is lwys stisfied Theorem 9 Let N 0 N N 1 N \ N 0 min rg min i N 1 N 1 with min < r 1 nd min (r ) 0 N {i N 1 : i < } nd N {i N 1 : i } The inequlity } r x i r (34) { min c i N min ( r i N 0 min min min is vlid for PX Proof Consider the function σ() { ( ) min min i N r min x i }) x i r min r min min{ () min r min } Notice tht σ() r min Θ() for ll R Since Θ is sudditive (see Lemm 1) nd r min > 0 σ is sudditive So to prove the vlidity of (34) we need to show tht σ() β() for ll R If nd >nd 1 then σ() r min β() If r min β() If <nd 0 < () <r then () () So σ() then σ() r min () min nd β() r () r min min r min min () r min () min min σ() If <nd () r or () 0 then σ() β() Hence σ() β() for ll R These inequlities re not useful s they re dominted y the lifted rounding inequlities Proposition 8 For N with r 1 inequlity (17) domintes inequlity (34) for ll choices of N 0 N N 1 N \ N 0 with N 1 min rg min i N 1 min nd min (r )0 Proof Let N 0 N N 1 N \ N 0 with N 1 min rg min i N 1 min nd min (r ) 0 If we divide inequlity (17) y min we otin (35) i1 { min min r min } x i min n i1 ( r min { ( ) min min r min r min }) x i In inequlity (34) vrile x i hs the coeffient min { r min } { min min min if i N For i N r the vrile x i hs the coeffient r min min min { () r min } min The coeffient of xi for i N 0 nd the right-hnd sides re equl in inequlities (17) nd (34) Now we re interested in cses where min does not divide r Lemm 3 If r 1 r c min min nd min (r ) > 0 then the function { } r () γ() min min (r ) r for R is sudditive min r } min

17 THE INTEGER KNAPSACK COVER POLYHEDRON 567 Proof For d R if () (d) ( d) then d d If min { () min (r r } r { (d) ) min min or min min (r r } r ) min min then γ() γ(d) r d r r min min γ( d) Otherwise γ()γ(d) d min (d) min (r γ( d) If ) () (d) ( d) then d d 1 If min { () min (r r } r { (d) ) min min nd min min (r r } r ) min min then γ() γ(d) r d r { () min min γ( d) If min min (r r } ) min () { min (r ) nd min (d) min (r r } r r ) min min then γ()γ(d) d min () min (r r d ) min (d) min (r γ( d) The cse where min { () ) min (r ) r } r { (d) min min nd min min (r r } ) min (d) min (r ) is similr Finlly if we hve min { () min (r r } ) min () min (r nd min { (d) ) min (r r } ) min (d) min (r ) then γ() γ(d) r ( d ) min 1 (d) min (r Since r c ) min min nd min (r c ) min r min γ()γ(d) d min (d) min (r ) γ( d) So γ is sudditive Using function γ we will lift inequlity (33) Theorem 10 Let N 0 N N 1 N \ N 0 min rg min i N 1 N 1 with min < r 1 min (r ) > 0 nd r c min min N {i N 1 : i<} nd N {i N 1 : i } The lifted 2-prtition inequlity } r r x i (36) { min c i N min ( r i N 0 min min min min { ( ) min (r ) i N r min x i }) x i r min is vlid for PX Proof To prove the vlidity of (36) we need to show tht γ() β() for ll R For< with 0 < () <r ifmin { () min (r r } ) min () min (r then ) γ() β() r () min min (r ) r ( ) min min () min (r ) r ( ) min min () min (r ) r r () min min () min (r ) r min r min (r )min () min (r ) min min () min (r ) () min(r ) min min If () < min (r ) then () min (r )min min 0 nd γ() β() () min (r ) 0 If () min (r ) then γ() β() () min (r 1 () min (r ) ) min () min (r ) min () min (r ) min 0

18 568 HANDE YAMAN If min { () min (r r } r r ) min min γ() min β() For < with () 0γ() r min β() For < with () r () min (r ) r () min min (r ) r (r min )min min ()min min (r )(r min (r )min ) min (r )min r min min (r )(r min (r )min ) min (r )min r (min min (r )) min (r )( min (r )min ) min (r )min (r min (r ))(min min (r )) 0 min (r )min So γ() r min β() For if then () r nd γ() r min β() Otherwise 1 nd so γ() β() Hence γ() β() for ll R As in the cse of lifted rounding inequlities the lifted 2-prtition inequlities re lso dominted y suset of them which is polynomil in size Proposition 9 Let { min } N with min < r 1 min (r ) > 0 nd r c min min The inequlity min1 i1 (37) { min min (r ) n i r min ( r min } x i 1 { min c i min min min { ( ) min (r ) r min min (r ) }) x i r min r min } x i is vlid nd domintes inequlity (36) for N 0 N N 1 N \N 0 with { min } N 1 nd min rg min i N 1 Proof Let { min } N with min < r 1 min (r ) > 0 nd r c min min Consider N { min i<: min min (r } N ) {} N 1 N N nd N 0 N \ N 1 For this choice of susets inequlity (36) is the sme s inequlity (37) Let N 1 N with { min } N 1 nd min rg min i N 1 In inequlity (36) for i N 1 ifi< then x i hs the coeffient min { r } min min nd if i N 0 then it hs the coeffient min { min (r r } ) min In oth cses its coeffient in inequlity (36) is greter thn or equl to its coeffient in inequlity (37) If i> nd i N 1 then the coeffient of x i in inequlity (36) is r min nd is greter thn or equl to its coeffient in inequlity (37) Other vriles hve the sme coeffients in oth inequlities As the right-hnd sides re lso the sme we cn conclude tht inequlity (37) domintes inequlity (36) The numer of lifted 2-prtition inequlities tht re not dominted is O(n 2 ) 7 Preliminry computtionl results We mentioned in the introduction tht the inequlities presented in this pper could e used to solve some hrd mixed integer progrmming prolems such s the heterogeneous vehicle routing prolem (see

19 THE INTEGER KNAPSACK COVER POLYHEDRON 569 [18]) nd the mnufcturer s mixed pllet design prolem (MPD) (see [19]) Some preliminry results with the rounding inequlities nd the lifted rounding inequlities re presented in [18] nd [19] respectively In this section we investigte the effect of the lifted rounding inequlities nd the lifted 2-prtition inequlities in solving the MPD instnces The rounding inequlities for λ for some N nd the residul cpty inequlities re not included in this study s they re the sme s or dominted y the lifted rounding inequlities We first give rief definition of the MPD For detils we refer the reder to [19] Let C e the set of customers N e the set of products nd T {1 2τ} e the set of periods Ech customer k C hs demnd of d kit units for product i N in period t T Products re of identicl dimensions nd re sold in pllets Ech pllet hs Q 1 rows nd in ech row there re Q 2 units of product A pllet which contins more thn one product type is clled mixed pllet Let P denote the set of potentil mixed pllet designs nd q i denote the numer of rows of product i N in pllet design P The mnufcturer lso offers full pllets for ech product i N which consists of Q 1 Q 2 units of product i We denote y h kit nd π kit the unit inventory holding cost nd the unit cklogging cost respectively for product i N nd customer k C t the end of period t T No cklogging is permitted t the end of period τ The prolem is to select t most m mixed pllet designs from set P to minimize the sum of customers inventory holding nd cklogging costs in periods 1 2 τ Let p e 1 if mixed pllet design P is offered nd 0 otherwise Let P k denote the set of mixed pllets tht customer k C cn uy Define y kt to e the numer of pllets of type P k tht customer k C uys in period t T nd f kit to e the numer of full pllets of product type i N tht customer k C uys in period t T In ddition define I kit nd B kit to e the mount of product i N tht remins in inventory nd tht is cklogged t the end of period t T for customer k C respectively Let M e very lrge numer The MPD is formulted s follows in [19]: (38) (39) (40) (41) (42) (43) (44) (45) (46) min k C st (π kit B kit h kit I kit ) i N t T p m P I kit1 B kit1 Q 1 Q 2 f kit P k Q 2 q i y kt d kit I kit B kit y kt Mp k C P k t T I ki0 B ki0 B kiτ 0 k C i N I kit B kit 0 k C i Nt T f kit 0 nd integer k C i Nt T y kt 0 nd integer k C P k t T p {0 1} P k C i Nt T The oective function (38) is the sum of inventory holding nd cklogging costs over ll periods At most m mixed pllet designs cn e offered due to constrint (39) Constrints (40) re the lnce equtions Constrints (41) ensure tht customers do

20 570 HANDE YAMAN Tle 1 Results with nd without vlid inequlities Model1 Model2 Model3 Prolem Nodes CPU (17) % gp Nodes CPU (37) Nodes CPU not uy mixed pllets tht re not offered Constrints (42) re eginning nd ending conditions Constrints (43) (46) re nonnegtivity nd integrlity constrints Ymn nd Sen prove tht the optiml vlue of the liner progrmming relxtion of MPD is zero As result it is importnt to derive strong vlid inequlities for this prolem to e le to improve the liner progrmming-sed lower ounds For k C nd i N let D ki t T d kit/q 2 The inequlity ( min{q 1 D ki }f kit ) (47) min{q i D ki }y kt D ki t T P k is stisfied y ll fesile solutions of MPD Remrk tht the set of nonnegtive integer solutions stisfying inequlity (47) is n integer knpsck cover set Hence we cn generte vlid lifted rounding nd lifted 2-prtition inequlities for the MPD sed on inequlities (47) We test the use of these vlid inequlities on seven prolem instnces We strt with two se instnces In the first instnce the numer of products is two nd in the second instnce the numer of products is three In oth se instnces the numer of periods is three nd the mximum numer of mixed pllet designs to e offered is one Using the first se instnce we generted four prolems where the numer of customers tkes vlues nd 7 Using the second se instnce we generted three prolems with 5 6 nd 7 customers For ech prolem instnce we first solve the model without vlid inequlities We cll this Model1 We report the numer of nodes in the rnch nd ound tree (in column node) nd the CPU time in seconds (in column CPU) Then we form Model2y dding the nondominted lifted rounding inequlities (17) to Model1 For Model2 we report the numer of inequlities (17) dded (in column (17)) the percentge dulity gp (in column %gp where %gp optlp opt 100 opt is the optiml vlue nd lp is the lower ound otined from the liner progrmming relxtion) the numer of nodes in the rnch nd ound tree nd the CPU time in seconds Finlly we form Model3 y dding the nondominted lifted 2-prtition inequlities (37) to Model2 We report here the numer of inequlities (37) dded (in column (37)) the numer of nodes in the rnch nd ound tree nd the CPU time in seconds The percentge dulity gps remined the sme s the ones of Model2 nd so re not reported We solve the models using the mixed integer progrmming (MIP) solver of CPLEX 81 on n AMD Opteron 252 processor (26 GHz) with 2 GB of RAM The results re given in Tle 1 The results show tht oth fmilies of vlid inequlities hve een useful in decresing the numer of nodes in the rnch nd ound tree nd the solution times for these instnces The solution time for Model3 is lrger thn the one of Model2

21 THE INTEGER KNAPSACK COVER POLYHEDRON 571 for instnce five ut still it is out twenty times less thn the one of Model1 The verges of percentge improvements otined in the numer of nodes nd CPU time with the ddition of inequlities (17) re 9629% nd 9585% respectively The verges of percentge improvements otined in the numer of nodes nd CPU time compred to Model2 with the ddition of inequlities (37) re 3407% nd 2807% respectively 8 Conclusion We studied the polyhedrl properties of the convex hull of the integer knpsck cover set which ppers s relxtion of mny optimiztion prolems tht concern covering given demnd using integer numers of different types of items We derived four fmilies of vlid inequlities investigted when they dominte ech other nd gve some conditions under which some re fcet-defining We used sequence-independent lifting to derive tht lst two fmilies of vlid inequlities These inequlities cn e used to solve prolems such s those investigted in [ ] Except the rounding inequlities for ritrry λ vlues the vlid inequlities derived in this pper shre some common fetures There exists lwys n item N such tht the right-hnd side of the inequlity is equl to the coeffient of x times We know tht this is n upper ound on the vlue of the right-hnd side (see Proposition 1) Clerly there re fcet-defining inequlities which do not follow this rule For instnce the cover constrint is fcet-defining for conv({x Z 3 : 3x 1 4x 2 5x 3 13}) Agin excluding rounding inequlities nother common feture is tht the numer of inequlities tht re nondominted within fmily is polynomil even when the fmily hs n exponentil numer of inequlities These inequlities cn e further lifted or modified to define lrger fmilies of vlid inequlities for more complicted prolems in considertion For instnce n exponentil numer of vlid inequlities cn e derived for the integer cpty cover polyhedron using the inequlities of this pper nd the lifting results of Mzur nd Hll [12] REFERENCES [1] A Atmturk Cover nd pck inequlities for (mixed) integer progrmming Ann Oper Res 139 (2005) pp [2] A Atmturk On cptted network design cut-set polyhedr Mth Progrm 92 (2002) pp [3] A Atmturk On the fcets of the mixed-integer knpsck polyhedron Mth Progrm 98 (2003) pp [4] A Atmturk Sequence independent lifting for mixed-integer progrmming Oper Res 52 (2004) pp [5] E Bls Fcets of the knpsck polytope Mth Progrm 8 (1975) pp [6] T Christof PORTA - POlyhedron Representtion Trnsformtion Algorithm Version [7] S Dsh nd O Gunluk Vlid inequlities sed on simple mixed-integer sets Mth Progrm 105 (2006) pp [8] Z Gu G L Nemhuser nd M W P Svelsergh Cover inequlities for 0-1 liner progrms: Computtion INFORMS J Comput 10 (1998) pp [9] P L Hmmer E L Johnson nd U N Peled Fcets of regulr 0-1 polytopes Mth Progrm 8 (1975) pp [10] T L Mgnnti P Mirchndni nd R Vchni The convex hull of two core cptted network design prolems Mth Progrm 60 (1993) pp [11] D R Mzur Integer Progrmming Approches to Multi-Flity Loction Prolem PhD Thesis John Hopkins University Bltimore MD 1999

22 572 HANDE YAMAN [12] D R Mzur nd L A Hll Fcets of Polyhedron Closely Relted to the Integer Knpsck- Cover Prolem Technicl report [13] A J Miller nd L A Wolsey Tight formultions for some simple mixed integer progrms nd convex oective integer progrms Mth Progrm 98 (2003) pp [14] G L Nemhuser nd L A Wolsey Integer nd Comintoril Optimiztion Wiley New York 1988 [15] Y Pochet nd L A Wolsey Integer knpsck nd flow covers with divisile coeffients: Polyhedr optimiztion nd seprtion Discrete Appl Mth 59 (1995) pp [16] R Weismntel On the 0/1 knpsck polytope Mth Progrm 77 (1997) pp [17] L Wolsey Fces for liner inequlity in 0-1 vriles Mth Progrm 8 (1975) pp [18] H Ymn Formultions nd vlid inequlities for the heterogeneous vehicle routing prolem Mth Progrm 106 (2006) pp [19] H Ymn nd A Sen Mnufcturer s Mixed Pllet Design Prolem Europen Journl of Opertionl Reserch to pper [20] E Zemel Esily computle fcets of the knpsck polytope Mth Oper Res 14 (1989) pp