n-step cycle inequalities: facets for continuous n-mixing set and strong cuts for multi-module capacitated lot-sizing problem

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1 n-step cyce nequates: facets for contnuous n-mxng set and strong cuts for mut-modue capactated ot-szng probem Mansh Bansa and Kavash Kanfar Department of Industra and Systems Engneerng, Texas A&M Unversty, USA Emas: May 1, 2014 Abstract. In ths paper, we ntroduce a generazaton of the contnuous mxng set whch we refer to as the contnuous n-mxng set) Q m,n := {y, v, s) Z Z+ n 1 ) m R m+1 + : n t=1 α tyt + v + s β, = 1,..., m}. Ths set s cosey reated to the feasbe set of the mut-modue capactated ot-szng MML) probem wthout) backoggng. For each n {1,..., n}, we deveop a cass of vad nequates for ths set, referred to as n -step cyce nequates, and show that they are facetdefnng for convq m,n ) n many cases. The cyce nequates of Van Vyve Math of OR 30: , 2005), the n-step MIR nequates of Kanfar and Fath Math Progam 120: , 2009), and the mxed n-step MIR nequates of Saneev and Kanfar Dscrete Optmzaton 9: , 2012) form speca cases of the n-step cyce nequates for Q m,n. We aso present a compact extended formuaton for Q m,n and an exact separaton agorthm over the set of a n -step cyce nequates for any gven n {1,..., n}. We then use these nequates to generate vad nequates for the MML probem wthout) backoggng, referred to as the n -step k,, S, C) cyce nequates for n {1,..., n}. Our computatona resuts show that our cuts are very effectve n sovng the MML nstances wthout) backoggng, resutng n substanta reducton n the ntegraty gap, number of nodes, and tota souton tme. 1. Introducton Poyhedra study of the mxed nteger base sets whch consttute we-structured reaxatons of mportant mxed nteger programmng MIP) probems s a promsng approach n deveopng strong cuttng panes for these MIP probems. Ths s because oftentmes one can deveop procedures n whch the vad nequates or facets) deveoped for the base set are used to generate vad nequates or facets) for the orgna MIPs see [1, 6, 7, 8, 14, 18] for a few exampes among many others). One base set studed for ths purpose s the contnuous mxng set Q := {y, v, s) Z m R m+1 + : y + v + s β, = 1,..., m}, where β R, = 1,..., m [17]. Ths set s a generazaton of the we-studed mxng set {y, s) Z m R + : y + s β, = 1,..., m} [7], whch tsef s a mut-constrant generazaton of the base set {y, s) Z R + : y + s β} that eads to the we-known mxed nteger roundng MIR) nequaty page 127 of [18]). In a these base sets each constrant has ony one nteger varabe. Fg. 1 presents a summary of the generazaton reatonshp between these base sets and other base sets of nterest n ths paper. The set Q arses as a substructure n reaxatons of probems such as ot-szng producton pannng), capactated facty ocaton, and capactated network desgn. Mer and Wosey [10] presented an extended formuaton for convq) wth Om 2 ) varabes and Om 2 ) constrants. Later, Van Vyve [17] gave a compact and tght extended formuatons wth 1

2 Number of constrants n the base poyhedron ) Number of MIR Mxng MIR Contnuous mxng nteger varabes Wosey, 1998) Günük and Pochet, 2001) Van Vyve, 2005) n each constrant of Om) the base varabes and Om 2 ) constrants n-step for MIR convq) and ts reaxaton Mxed n-step to MIR the case where Contnuous s R. n-mxng He aso poyhedron ntroduced ) the so-caed cyce Kanfar nequates and Fath, 2009) caed 1-step Saneev cyce and nequates Kanfar, 2012) n ths paper) Ths Research) for these sets and showed that these nequates aong wth bound constrants are suffcent to descrbe the convex hus of these sets. The MIR nequates caed 1-step MIR nequates n ths paper) of Nemhauser and Wosey [11, 18] and the mxed 1-step) MIR nequates of Günük and Pochet [7] are speca cases of the 1-step cyce nequates for Q Fg. 1). Number of constrants n the base poyhedron ) Number of nteger varabes n each constrant of the base poyhedron ) MIR [11, 18] -step MIR [8] Mxed MIR [7] Mxed -step MIR [14] Contnuous mxng [17] Contnuous -mxng Ths Paper) Fgure 1: Generazatons of Mxed Integer Roundng MIR) In another drecton Fg. 1), Kanfar and Fath [8] generazed the 1-step MIR nequates [11] and deveoped the n-step MIR nequates for the mxed nteger knapsack set by studyng the base set { Q 1,n 0 = y, s) Z Z+ n 1 R + : n } α ty t + s β, t=1 where α t R + \{0}, t = 1,..., n and β R. Note that ths base set has a snge constrant and n nteger varabes n ths constrant. The n-step MIR nequates are vad and facet-defnng for the base set Q 1,n 0 f α t s and β satsfy the so-caed n-step MIR condtons see condtons 4) n Secton 2). However, n-step MIR nequates can aso be generated for a mxed nteger constrant wth no condtons mposed on the coeffcents. In that case, the externa parameters used n generatng the nequaty are pcked such that they satsfy the n-step MIR condtons see [8] for more detas). The n-step MIR nequates are facet-defnng for the mxed nteger knapsack set n many cases [2, 8]. The Gomory mxed nteger cut [13] and the 2-step MIR nequates [5] are the speca cases of n-step MIR nequates, correspondng to n = 1, 2, respectvey. Kanfar and Fath [8, 9] showed that the n-step MIR nequates defne new fames of facets for the fnte and nfnte group probems. Recenty, Saneev and Kanfar [14] showed that the procedure proposed by Günük and Pochet [7] to mx 1-step MIR nequates can be generazed and used to mx the n-step MIR nequates [8] Fg. 1). As a resut, they deveoped the mxed n-step MIR nequates for a generazaton of the mxng set caed the n-mxng set,.e. { Q m,n 0 = y, s) Z Z n 1 + )m R + : n } α ty t=1 t + s β, = 1,..., m, where α t R + \ {0}, t = 1,..., n, β R, = 1,..., m. Note that ths s a mut-constrant base set wth n nteger varabes n each constrant and a contnuous varabe whch s common among a constrants. The mxed n-step MIR nequates are vad f α t and β satsfy the n-step MIR condtons n the mxed rows. These nequates are aso facet-defnng for the convex hu of Q m,n 0 under certan condtons. Saneev and Kanfar [14] aso generazed the ot-szng probem wth constant batches [12] where the capacty n each perod can be some nteger mutpe of a snge 2

3 capacty modue wth a gven sze) and ntroduced the mut-modue capactated ot-szng MML) probem. In ths probem, the tota producton capacty n each perod can be the summaton of some nteger mutpes of severa capacty modues of dfferent szes. They showed that the mxed n-step MIR nequates can be used to generate vad nequates for the MML probem wthout backoggng whch we denote by MML-WB). They referred to these nequates as the mut-modue k,, S, I) nequates. These nequates generaze the k,, S, I) nequates whch were ntroduced for the ot-szng probem wth constant batches by Pochet and Wosey [12]. In ths paper, we generaze the concepts of contnuous mxng [17] and mxed n-step MIR [14] by ntroducng a more genera base set referred to as the contnuous n-mxng set whch we defne as Q m,n := { y, v, s) Z Z n 1 + )m R m+1 + : n } α ty t=1 t + v + s β, = 1,..., m, where α t > 0, t = 1,..., n and β R, = 1,..., m see Fg. 1). Note that ths set has mutpe m) constrants wth mutpe n) nteger varabes n each constrant; but t s more genera than the n-mxng set because n addton to the common contnuous varabe s, each constrant has a contnuous varabe v of ts own. The contnuous mxng set Q s the speca case of Q m,n, where n = 1 and α 1 = 1, and the n-mxng set of Saneev and Kanfar [14] s the proecton of Q m,n {v = 0} on y, s). The contnuous n-mxng set arses as a substructure n reaxatons of MML-WB, MML wth backoggng MML-B), MML wth stochastc demand, mut-modue facty ocaton probem, and mut-modue capactated network desgn probem. In Secton 3, for each n {1,..., n}, we deveop a cass of vad nequates for Q m,n whch we refer to as n -step cyce nequates, and dscuss how the n-step MIR nequates [8] and the mxed n-step MIR nequates [14] are speca cases of the n-step cyce nequates. We aso ntroduce a compact extended formuaton for Q m,n. In Secton 4, we obtan condtons under whch n -step cyce nequates are facet-defnng for convq m,n ). In Secton 5, we present an effcent exact separaton agorthm to separate over the set of a n -step cyce nequates, n {1,..., n}, for set Q m,n. Then, n Secton 6, we use these nequates to generate vad nequates for the MML-WB and MML-B. Our computatona resuts n Secton 7 on appyng 2-step cyce nequates usng our separaton agorthm show that our cuts are very effectve n sovng MML-WB and MML-B wth two capacty modues n = 2), resutng n consderabe reducton n the ntegraty gap on average 85.90% for MML-WB and 86.32% for MML-B) and the number of nodes on average 132 tmes for MML-WB and 31 tmes for MML-B). Aso, the tota tme taken to sove an nstance whch aso ncudes the cut generaton tme) s n average 58.3 tmes for MML-WB) and 9.9 tmes for MML-B) smaer than the tme taken by CPLEX wth defaut settngs except for very easy nstances). More nterestngy, n these nstances addng cuts by appyng 2-step cyce nequates over 1-step cyce nequates has mproved the cosed gap on average 19.47% for MML-WB and 15.96% for MML-B), the number of nodes on average 43 tmes for MML-WB and 14 tmes for MML-B), and the tota souton tme on average 18 tmes for MML-WB and 4 tmes for MML-B). 2. Necessary background In ths secton, we brefy revew the 1-step) cyce nequates for the contnuous mxng set [17], the n-step MIR nequates [8], and the mxed n-step MIR nequates [14] to the extent requred as background for the resuts n ths paper. 3

4 Van Vyve [17] generated the cyce nequates for the contnuous mxng set Q as foows: Defne β 0 := 0, f := β β, {0,..., m} and wthout oss of generaty assume that f 1 f, = 1,..., m. Let G := V, A) be a drected graph, where V := {0, 1,..., m} and A := {, ) :, V, f f }. Note that G s a compete graph except for the arcs, ) where f = f. An arc, ) A s caed a forward arc f < and a backward arc f >. To each arc, ) A, assocate a near functon ψ y, v, s) defned as { s + v + f f + 1)y Background Contnuous β ) n-mxng fset n-step Cyce Inequates Varant and speca cases ψ y, v, s) := v + f f )y β ) f, ) s a forward arc, f, ) s a backward arc, where v 0 = y 0 = 0. See Fg. 2. ψ 0 ψ m Fgure 2: Each cyce n graph G gves rse to a cyce nequaty. Theorem 1 [17]). Gven an eementary cyce C = V C, A C ) n the graph G, the nequaty Bansa and Kanfar - Texas A&M Unversty Contnuous n-mxng Set 17/27 referred to as the cyce nequaty, s vad for Q.,) A C ψ y, v, s) 0, 1) In [17], the vadty of the cyce nequaty 1) was proved ndrecty through the foowng extended formuaton for Q: Q δ = { y, v, s, δ) R m R m+1 + R m+1 : ψ y, v, s) δ δ for a, ) A, y + v + s β, = 1,..., m }. Note that the set of a orgna nequates, a cyce nequates, aong wth the bound constrants v, s 0, defne Pro y,v,s Q δ ). Van Vyve [17] showed that for every extreme pont or extreme ray) of Q, there exsts a pont or a ray) n ts extended formuaton Q δ. Ths mpes Q Pro y,v,s Q δ ), and hence, the cyce nequates are vad for Q. Furthermore, t was shown n [17] that convq) = Pro y,v,s Q δ ) and the separaton over convq) can be performed n Om 3 ) tme by fndng a negatve weght cyce n G. Smar resuts were presented for the reaxaton of Q to the case where s R. In another drecton, Kanfar and Fath [8] deveoped the n-step MIR nequates a generazaton of MIR nequates [11, 18]) for the set Q 1,n 0. Note that Q1,n 0 = Pro y,s Q 1,n {v = 0} ). The n-step MIR nequaty for ths set s 4

5 n s β n) =1 β 1) α β n) where the recursve remanders β t) are defned as n n t=1 =t+1 β 1) α y t ), 2) β t) := β t 1) α t β t 1) /α t, t = 1,..., n, 3) and β 0) := β note that 0 β t) < α t for t = 1,..., n). By defnton f a > b, then b a.) = 0 and b a.) = 1. For nequaty 2) to be non-trva, we assume that βt 1) /α t / Z, t = 1,..., n. Kanfar and Fath [8] showed that the n-step MIR nequaty 2) s vad and facet-defnng for the convex hu of Q 1,n 0 f the so-caed n-step MIR condtons,.e. α t β t 1) /α t α t 1, t = 2,..., n, 4) hod. As mentoned n Secton 1, they aso used nequates 2) to generate n-step MIR nequates for snge-constrant mxed nteger sets wth no condtons on the coeffcents. Later, Atamtürk and Kanfar [2] showed that those nequates aso have facet-defnng propertes n severa cases. Pease refer to [2, 8] for more detas. As mentoned n Secton 1, Saneev and Kanfar [14] generazed the MIR mxng procedure of Günük and Pochet [7] to the case of n-step MIR and deveoped the mxed n-step MIR nequates for the n-mxng set Q m,n 0. Note that Q m,n 0 = Pro y,s Q m,n {v = 0} ). These nequates are generated as foows: Wthout oss of generaty, we assume β n) 1 βn), = 2,..., m. Let K := { 1,..., K }, where 1 < 2 < < K, be a non-empty subset of {1,..., m}. If the n-step MIR condtons 4) hod for each constrant K,.e. α t β t 1) /α t α t 1, t = 2,..., n, then the nequates K ) s β n) p β n) p 1 φ n p y p ) 5) p=1 K ) ) φ s β n) p β n) p 1 φ n p y p ) + α n β n) n K 1 y 1 ) 1 ), 6) p=1 are vad for Q m,n 0, where β n) 0 = 0 and φ n y ) := n β 1) α =1 n n t=1 =t+1 β 1) yt 7) α for K. Inequates 5) and 6) are referred to as the type I and type II mxed n-step MIR nequates, respectvey. Inequaty 5) s shown to be facet-defnng for Q m,n 0. Inequaty 6) aso defnes a facet for Q m,n 0 f some addtona condtons are satsfed see [14] for detas). Note that the functon φ n y ) has the same form as the mutpe of β n) n the rght-hand sde of the n-step MIR nequaty 2). Ths functon can aternatvey be wrtten as foows see proof of Lemma 10 n [14]): 5

6 φ n y ) := 1 + n n t=1 =t+1 ) β 1) β t 1) yt. 8) α α t 3. n-step cyce nequates and an extended formuaton for Q m,n In ths secton, we generaze the concepts of contnuous mxng [17] and mxed n-step MIR [14] by studyng the contnuous n-mxng set Q m,n := { y, v, s) Z Z n 1 + )m R m+1 + : n t=1 } α t yt + v + s β, = 1,..., m ntroduced n Secton 1. Q m,n s a generazaton of both the contnuous mxng set Q and the n-mxng set Q m,n 0. We w show that for each n {1,..., n}, there exst a famy of vad nequates for Q m,n, whch we refer to as the n -step cyce nequates. In provng the vadty of these nequates, Theorem 1 w become necessary. As mentoned before, Van Vyve [17] proved Theorem 1 ndrecty by defnng the extended formuaton Q δ and showng that every extreme pont ray) of the set Q has a counterpart n Q δ see [17] for detas). We have deveoped a drect proof for Theorem 1, whch ony uses the orgna nequates and the cyce structure. We beeve ths proof can be nsghtfu n further pursut of research n ths area. Here, we present an aternatve form of Theorem 1 and provde our proof: Lemma 1. Let C = V C, A C ) be a drected Hamtonan cyce over q nodes, where V C = {1,..., q}, A C := {1, 2 ), 2, 3 ),..., q, 1)}, and 2,..., q {2,..., q} are dstnct. Let σ R, α R +, and to each node {1,..., q} assgn the vaues ω R +, κ Z, and γ R + such that γ < α, = 1,..., q, γ 1 < γ, = 2,..., q. If then the cyce nequaty σ + ω γ + γ γ + α) κ ) +,) F σ + ω + ακ γ = 1,..., q, 9),) B ω + γ γ ) κ ) 0, 10) s vad, where F and B are the sets of forward and backward arcs n A C, respectvey.e. F = {, ) A C : < } and B = {, ) A C : > }). Proof. For p {1,..., q}, et A p be the arcs n the path from 1 to p+1 n C,.e. A p := {1, 2 ), 2, 3 ),..., p, p+1 )} we defne q+1 := 1). Denote the set of forward and backward arcs n A p by F p and B p, respectvey note that f p < p, then A p A p, F p F p, and B p B p ). Aso, et T.) be an operator that, when apped on an arc set, returns the set of ta nodes of the arcs n that arc set. Defne the ndex g p { 1,..., p } recursvey as foows: g 1 := 1, and 6

7 g p 1 f p T F p ), g p 1 T F p 1 ), κ gp 1 κ p, p f p T F p ), g p 1 T F p 1 ), κ gp 1 < κ p, g p 1 f p T F p ), g p 1 T B p 1 ), κ gp 1 > κ p, p f p T F p ), g p 1 T B p 1 ), κ gp 1 κ p, g p := g p 1 f p T B p ), g p 1 T B p 1 ), κ gp 1 κ p, p f p T B p ), g p 1 T B p 1 ), κ gp 1 > κ p, g p 1 f p T B p ), g p 1 T F p 1 ), κ gp 1 < κ p, p f p T B p ), g p 1 T F p 1 ), κ gp 1 κ p, for p = 2,..., q and for p = 1,..., q, defne p = γ gp γ p+1, f g p T B p ), and 0 f g p T F p ). In order to prove the theorem, we frst show that the nequaty γ γ γ + α) κ ) + γ γ ) 1 κ ),) F p,) B p F p 1)σ + ω γ 1 γ p+1 + α ) κ gp + γ 1 + p, T A p)\{g p} 11) s vad for p = 1,..., q. We prove ths by nducton on p. For p = 1, the nequaty 11) s trva because A 1 = {1, 2 )}, F 1 = A 1, B 1 =, and 1 = 0, and therefore, both sdes of the nequaty reduce to γ 1 γ 1 γ 2 + α) κ 1. For smpcty, we denote the eft-hand and rght-hand sdes of nequaty 11) for p by L p and R p, respectvey. Now as the nducton hypothess we assume L p 1 R p 1. We then prove L p R p. Consder the foowng cases whch correspond to the cases n the g p defnton): I. p T F p ). Ths means γ p < γ p+1, F p = F p 1 { p, p+1 )}, and B p = B p 1. Therefore we can wrte L p = L p 1 + γ p γ p γ p+1 + α ) κ p F p 1 1)σ + T A p 1 )\{g p 1 } ω 12) γ 1 γ p + α ) κ gp 1 + γ 1 + p 1 + γ p γ p γ p+1 + α ) κ p where the ast nequaty s based on 11) for p 1. Now, consder the foowng subcases: I.1. g p 1 T F p 1 ), κ gp 1 κ p. Ths mpes g p = g p 1, and hence p = p 1 = 0. Now notce that 0 ) ) γ p γ p+1 κp κ gp 1, and by nequaty 9) for p, 0 σ +ω p +ακ p γ p. Addng these two nequates to nequaty 12), we get L p F p 1 σ + T A p 1 )\{g p 1 } ω + ω p γ 1 γ p+1 + α ) κ gp 1 + γ 1 + p 1 = F p 1)σ + T A p)\{g p} ω γ 1 γ p+1 + α ) κ gp + γ 1 + p = R p. 13) 7

8 The frst dentty s true because F p 1 = F p 1, T A p 1 ) { p } = T A p ), g p 1 = g p, and p 1 = p = 0). I.2. g p 1 T F p 1 ), κ gp 1 < κ p. Ths mpes g p = p, and hence g p T F p ). Therefore, p 1 = p = 0. Notce that 0 ) ) γ 1 γ p κgp κ p, 0 γgp 1 γ 1, and by nequaty 9) for g p 1, 0 σ + ω gp 1 + ακ gp 1 γ gp 1. By addng these three nequates to nequaty 12), we get L p F p 1 σ + ω + ω gp 1 T A p 1 )\{g p 1 } γ 1 γ p+1 + α ) κ p + γ 1 + p 1 = R p. The fna dentty s true because F p 1 = F p 1, T A p 1 ) = T A p )\{ p }, p = g p, and p 1 = p = 0). I.3. g p 1 T B p 1 ), κ gp 1 > κ p. Ths means g p = g p 1, p 1 = γ gp 1 γ p, and p = γ gp γ p+1 = γ gp 1 γ p+1. Addng vad nequates 0 ) ) γ p γ p+1 κp + 1 κ gp 1 and 0 σ + ω p + ακ p γ p to nequaty 12) gves L p F p 1 σ + T A p 1 )\{g p 1 } ω + ω p γ 1 γ p+1 + α ) κ gp 1 + γ 1 + p 1 + γ p γ p+1 = R p. The fna dentty hods because F p 1 = F p 1, T A p 1 ) { p } = T A p ), g p 1 = g p, and p 1 + γ p γ p+1 = γ gp 1 γ p+1 = p. I.4. g p 1 T B p 1 ), κ gp 1 κ p. Ths means g p = p, and hence g p T F p ). Therefore, p = 0. Aso, p 1 = γ gp 1 γ p. Now addng vad nequates 0 ) ) γ 1 γ p κgp 1 κ p and 0 σ + ω gp 1 + ακ gp 1 γ gp 1 to nequaty 12) gves L p F p 1 σ + T A p 1 )\{g p 1 } ω + ω gp 1 γ 1 γ p+1 + α ) κ p + γ 1 + p 1 + γ p γ gp 1 = R p. The fna dentty s true because F p 1 = F p 1, T A p 1 ) = T A p )\{ p }, p = g p, p 1 + γ p γ gp 1 = 0, and p = 0. II. p T B p ). Ths means γ p > γ p+1, F p := F p 1, and B p := B p 1 { p, p+1 )}. Therefore we can wrte L p = L p 1 + ) ) γ p γ p+1 1 κp F p 1 1)σ + ω 15) T A p 1 )\{g p 1 } γ 1 γ p + α ) κ gp 1 + γ 1 + p 1 + γ p γ p+1 ) 1 κp ) where the ast nequaty s based on 11) for p 1. Now, consder the foowng subcases: II.1. g p 1 T B p 1 ), κ gp 1 κ p. Ths means g p = g p 1, p 1 = γ gp 1 γ p, and p = γ gp γ p+1 = γ gp 1 γ p+1. Addng vad nequates 0 ) ) γ p γ p+1 κp κ gp 1 and 14) 8

9 0 ω p to nequaty 15), we get the same nequaty as 14) except for the coeffcent of σ whch w be F p 1 1. Ths nequaty s true for the same reasons stated n case I.3 and the fact that F p 1 = F p n ths case. II.2. g p 1 T B p 1 ), κ gp 1 > κ p. Ths means p 1 = γ gp 1 γ p. Aso, g p = p, and hence g p T B p ). Therefore, p = γ gp γ p+1 = γ p γ p+1. Addng vad nequates 0 γ 1 γ p + α ) κ gp 1 κ p 1 ), 0 γ 1 γ gp 1 + α, and 0 ω gp 1 to nequaty 15) gves L p F p 1 1) σ + ω + ω gp 1 T A p 1 )\{g p 1 } γ 1 γ p+1 + α ) κ p + γ 1 + p 1 + γ p γ gp 1 + γ p γ p+1 = R p. The fna dentty s true because F p 1 = F p, T A p 1 ) = T A p )\{ p }, p = g p, p 1 = γ gp 1 γ p, and γ p γ p+1 = p. II.3. g p 1 T F p 1 ), κ gp 1 < κ p. Ths mpes g p = g p 1, and hence p = p 1 = 0. Addng vad nequates 0 ) ) γ p γ p+1 κp 1 κ gp 1 and 0 ωp to nequaty 15), we get the same nequaty as 13) except for the coeffcent of σ whch w be F p 1 1. Ths nequaty s true for the same reasons stated n case I.1 and the fact that F p 1 = F p n ths case. II.4. g p 1 T F p 1 ), κ gp 1 κ p. Ths means p 1 = 0. Aso, g p = p, and hence g p T B p ). Therefore p = γ gp γ p+1 = γ p γ p+1. Addng vad nequates 0 γ 1 γ p + α ) ) κ gp 1 κ p and 0 ω gp 1 to nequaty 15) gves L p F p 1 σ + T A p 1 )\{g p 1 } ω + ω gp 1 γ 1 γ p+1 + α ) κ p + γ 1 + p 1 + γ p γ p+1 = R p. The fna dentty s true because F p 1 = F p, T A p 1 ) = T A p )\{ p }, p = g p, p 1 = 0, and γ p γ p+1 = p. A cases are exhausted, and therefore, nequaty 11) s vad for any p = 1,..., q. Now reca that q+1 = 1. Ths mpes A q = A C, and therefore, L q =,) F =,) F γ γ γ + α) κ ) +,) B γ γ γ + α) κ ),) B γ γ ) 1 κ ) γ γ ) κ 16) The second dentty s true because,) F γ +,) B γ γ ) =,) F γ γ + γ ) +,) B γ γ ) =,) F γ +,) A C γ γ ) =,) F γ. Note that,) A C γ γ ) = 0 because the arcs n A C form a cyce. Now based on nequaty 11) for p = q and nequaty 16), 9

10 we have,) F γ γ γ + α) κ ) F 1)σ + F σ + T A C ) T A C )\{g q},) B γ γ ) κ ω ακ gq + γ 1 + q, ω + q + γ 1 γ gq F σ + T A C ) where the second nequaty s true by addng the vad nequaty 0 σ + ω gq + ακ gq γ gq to the frst nequaty, and the thrd nequaty s true because we have ether q = 0 or q = γ gq γ q+1 = γ gq γ 1, and hence, q + γ 1 γ gq mn{γ 1 γ gq, 0} = γ 1 γ gq 0. By rearrangng the terms n nequaty 17), we get nequaty 10). Ths competes the proof. Now gven n {1,..., n}, we deveop the n -step cyce nequates for Q m,n as foows: Wthout oss of generaty, we assume β n ) 1 ) βn, = 2,..., m, where β n ) s defned as 3). Aso defne β 0 := 0. Now smar to the graph defned for the cyce nequates see Secton 2), here we defne a drected graph G n = V, A), where V := {0, 1,..., m} and A := {, ) :, V, β n ) β n ) }. G n s a compete graph except for the arcs, ) where β n ) = β n ). Here to each arc, ) A, we assocate the near functon ψ n y, v, s) defned as ψ n y, v, s) := s + v + v + n t=n n +1 t=n +1 α t y t + α t yt + β n ) β n ) ω, ) 1 φ n y ) f <, ) ) 1 φ n y ) f >, where β n ) := β n ) + α n for a, ) A, <, the functons φ n y ), = 1,..., m, are defned as 7) and by defnton, v 0 := 0, y 0 := 0, and φ n 0 y0 ) := 1. We show that each eementary cyce of graph G n corresponds to a vad nequaty for the set Q m,n, whch we refer to as the n -step cyce nequaty. To do ths n addton to Lemma 1, we need the foowng emma: Lemma 2. For {1,..., m} and n {1,..., n}, the nequaty s + v + n t=n +1 α t y t + α n 17) 18) ) 1 φ n y ) β n ) 19) s vad for Q m,n f the n -step MIR condtons 4) hod for constrant of Q m,n,.e. α t β t 1) /α t α t 1, t = 2,..., n. Proof. Kanfar and Fath [8] proved that the foowng nequaty s + v + + α n n t=n +1 n n α t y t t=1 =t+1 β 1) yt α n =1 β 1) + α ) β n 1) β n 1) α n 20) 10

11 s vad for the reaxaton of Q m,n defned by ts th constrant,.e. {y, v, s) Z Z+ n 1 ) R + R + : n t=1 α tyt + v + s β }, f the n -step MIR condtons for constrant hod. Therefore, t s aso vad for Q m,n. Subtractng α n β n 1) /α n from both sdes and rearrangng the terms n 20) gves 19). Theorem 2. Gven n {1,..., n} and an eementary cyce C = V C, A C ) of graph G n, the n -step cyce nequaty y, v, s) 0 21),) A C ψ n s vad for Q m,n f the n -step MIR condtons for V C,.e. α t β t 1) /α t α t 1, t = 2,..., n, V C. 22) Proof. Consder a pont ŷ, ˆv, ŝ) Q m,n. Based on Lemma 2, nequaty 19) s satsfed by the pont ŷ, ˆv, ŝ) for each V C \{0} because of 22). But notce that nequaty 19) for ths pont s the same as nequaty 9) f we defne σ := ŝ, α := α n, and ω := ˆv + n t=n +1 α tŷt, κ := 1 φ n ŷ ), γ := β n ), V C \{0}. Aso, n case 0 V C, f we defne ω 0, κ 0, and γ 0 n a smar way, nequaty 9) for = 0 reduces to the vad nequaty ŝ 0 because as we defned before y 0 := 0, v 0 := 0, φ n 0 y0 ) := 1, and β 0 := 0. Wth these defntons, we have ω 0, κ Z, V C and 0 = γ 0 γ 1 < γ 2 < < γ n < α n. Therefore, accordng to Lemma 1, nequaty 10) n whch σ, α and ω, κ, γ, V C are repaced wth the vaues defned here s vad. It s easy to see that ths nequaty s exacty the same as the n -step cyce nequaty 21) for the pont ŷ, ˆv, ŝ). Ths competes the proof. Speca Cases: The n-step MIR nequates [8] and the mxed n-step MIR nequates [14] are speca cases of the n-step cyce nequates. I. The n-step cyce nequaty 21) wrtten for cyce C = V C, A C ) such that A C = {0, ),, 0)} and v = 0 gves the n-step MIR nequaty 2) wrtten for constrant n Q m,n 0. II. The n-step cyce nequaty 21) wrtten for cyce C = V C, A C ) such that A C = { 1, 2 ),..., q, 1 )} wth ony one forward arc 1, 2 ), foowed by backward arcs 1, 2 ),..., q, 1 ) and v = 0 for a K, gves the foowng nequates for Q m,n 0 : the type I mxed n-step MIR nequaty 5) where K = { q,..., 2 }, f 1 = 0, and the type II mxed n-step MIR nequaty 6) where K = { q,..., 1 }, f 1 0. Remark: For the speca case where the parameters α 1,..., α n are dvsbe,.e. α t α t 1, t = 2,..., n, the n -step MIR condtons are automatcay satsfed no matter what the vaue of β s. 11

12 Exampe 1. Consder the foowng contnuous 2-mxng set wth 6 rows: Q 6,2 = {y, v, s) Z Z + ) 6 R 7 + : 50y y v 1 + s 87, 50y y v 2 + s 39, 50y y v 3 + s 141, 50y y v 4 + s 93, 50y y v 5 + s 45, 50y y v 6 + s 71}. So we have α = α 1, α 2 ) = 50, 12), β 1 = 87, β 2 = 39, β 3 = 141, β 4 = 93, β 5 = 45, β 6 = 71. Note that β 1) 6 = 21 < β 1) 1 = 37 < β 1) 2 = 39 < β 1) 3 = 41 < β 1) 4 = 43 < β 1) 5 = 45 and β 2) 1 = 1 < β 2) 2 = 3 < β 2) 3 = 5 < β 2) 4 = 7 < β 2) 5 = β 2) 6 = 9. Note that β 1) /α 2 = 4 for = 1,..., 5, β 1) 6 /α 2 = 3 and ceary the 2-step MIR condtons 22),.e. α 1 α 2 β 1) /α 2, are satsfed for = 1,..., 6. 2-step cyce nequates for Q 6,2 : Settng n = 2, the set of nodes and arcs of the graph G 2 w be V 2 = {0,..., 6} and A 2 = {, ) :, V 2 }\{5, 6), 6, 5)} because β 2) 5 = β 2) 6. The near functon ψ 2 y, v, s) assocated wth each arc, ) A 2 s defned by 1) where n = 2,.e. where φ 2 y ) = s + v + ψy, 2 v, s) := v + β 1) /α 2 β /α 1 β 2) ) 1 β 2) + α 2 φ 2 y ) ) β 2) f β 2) < β 2), ) 1 φ 2 y ) ) f β 2) > β 2), β 2) β 2) β 1) /α 2 y 1 y 2, for = 1,..., 6, and v 0 := 0, y 0 := 0, and φ 2 0 y0 ) := 1. Based on Theorem 2, the 2-step cyce nequates correspondng to the cyces n G 2 are vad for Q 6,2. For exampe, the 2-step cyce nequaty correspondng to a cyce C 2 1 = V C 2 1, A C 2 1 ) n G 2 where A C 2 1 = {1, 3), 3, 6), 6, 4), 4, 5), 5, 2), 2, 1)} s ψ ψ ψ ψ ψ ψ2 21 0,.e. s + v y y ) + s + v y y ) + v 6 + 6y y ) + s + v y y ) + v y y ) + v 2 + 8y y 2 2 6) 0. 23) Lkewse, for a cyce C2 2 n G 2 wth A C 2 2 ψ ψ ψ ψ2 52 0,.e. = {2, 4), 4, 3), 3, 5), 5, 2)}, the 2-step cyce nequaty s s + v y y ) + v 4 + 8y y ) + s + v y y ) + v y y ) 0, 24) and for a cyce C3 2 n G 2 wth A C 2 3 = {0, 6), 6, 4), 4, 1), 1, 0)}, the 2-step cyce nequaty s ψ ψ ψ ψ2 10 0,.e. s 9) + v 6 + 6y y ) + v y y ) + v 1 + 4y y 1 2 7) 0. 25) 12

13 1-step Cyce Inequates for Q 6,2 : Settng n = 1, the set of nodes and arcs of the graph G 1 w be V 1 = {0, 6, 1,..., 5} and A 1 = {, ) :, V 1 } because β 1) 6 < β 1) 1. The near functon ψ1 y, v, s) assocated wth each arc, ) A 1 s defned by 1) where n = 1,.e. ) 1 s + v + α 2 y ψy, β 1) β 1) + α 1 φ 1 y ) ) β 1) f β 1) < β 1), v, s) := ) 1 v + α 2 y2 + β 1) β 1) φ 1 y ) ) f β 1) > β 1), where φ 1 y ) = β /α 1 y1, for = 1,..., 5, and v 0 := 0, y 0 := 0, and φ 1 0 y0 ) := 1. Based on Theorem 2, the 1-step cyce nequates correspondng to the cyces n G 1 are vad for Q 6,2. For exampe, the 1-step cyce nequaty correspondng to a cyce C1 1 = V C1 1, A C1 1) n G 1 where = {6, 1), 1, 2), 2, 3), 3, 4), 4, 5), 5, 6)} s ψ ψ ψ ψ ψ ψ1 56 0,.e. A C 1 1 s + v y y ) + s + v y y ) + s + v y y ) + s + v y y ) + s + v y y ) + v y y 5 2) 0. 26) Lkewse, for a cyce C2 1 n G 1 wth A C 1 2 = {6, 2), 2, 5), 5, 6)}, the 1-step cyce nequaty s ψ ψ ψ1 56 0,.e. s + v y y ) + s + v y y ) + v y y 5 2) 0, 27) and for a cyce C3 1 n G 1 wth A C 1 3 = {0, 4), 4, 6), 6, 0)}, the 1-step cyce nequaty s ψ ψ ,.e. ψ 1 60 s 43) + v y y ) + v y y ) 0. 28) Theorem 3. The foowng near program s a compact extended formuaton for Q m,n, f condtons 22) hod. ψy, n v, s) δ n δ n for a, ) A, n {1,..., n} 29) n t=1 α tyt + v + s β, = 1,..., m 30) y R R n 1 + )m, v R m +, s R +, δ R nm+1). 31) Proof. Let Q m,n,δ := {y, v, s, δ) satsfyng 29)-31)}. Ceary P ro y,v,s Q m,n,δ ) s defned by the set of a n -step cyce nequates 21), for n = 1,..., n, and bound constrants s, v 0. Ths means a the nequates whch defne P ro y,v,s Q m,n,δ ) are vad for Q m,n f condtons 22) hod whch mpes Q m,n P ro y,v,s Q m,n,δ ) under the same condtons. Ths proves that Q m,n,δ s an extended formuaton for Q m,n. 4. Facet-defnng n-step cyce nequates In ths secton, we show that for any n {1,..., n}, the n -step cyce nequates defne facets for convq m,n ) under certan condtons. In order to prove ths, we frst defne some ponts and provde some propertes for them. 13

14 Defnton 1. For {1,..., m}, defne the ponts P,d, Q,d Z Z+ n 1, d = 1,..., n, as foows: β t 1) αt t = 1,..., d 1, P,d t := β t 1) αt t = d, 0 t = d + 1,..., n, β t 1) Q,d t := αt t = 1,..., d, 0 t = d + 1,..., n, and the pont R Z Z n 1 + assumng β n 1) /α n 1) as R = Q,n e n, where e n s the n th unt vector n R n. Aso, defne the ponts S,d Z Z+ n 1, d = 2,..., n, assumng β d 1) /α d 1, d = 2,..., n ) as foows: Q,n t t = 1,..., d 2, d + 1,..., n S,d β t 1) t := αt 1 t = d 1, β 2 t 1) αt + 1 t = d. Moreover, for, {1,..., m} such that β n ) > β n ), defne the ponts T,,d Z Z + n 1, d = n,..., n, as T,,d t := Q,n t for t = 1,..., n, d + 1,..., n and β n,t 1) α t β n,n ) := β n ) and β n,t) := β n,t 1) α t β n,t 1) /α t, t = n + 1,..., n. for t = n + 1,..., d, where Lemma 3. The pont ŷ, ˆv, ŝ) Z Z n 1 + )m R m+1 + satsfes constrant {1,..., m} of Q m,n f any of the foowng s true a). ŷ = P,d for some d {1,..., n} b). ŷ = Q,d for some d {1,..., n} and ˆv + ŝ β d), c). ŷ = R and ˆv + ŝ α n + β n ), d). ŷ = S,d for some d {2,..., n } and ˆv + ŝ β n ) + α d 1 α d β d 1) /α d, e). ŷ = T,,d for some d {n,..., n} and {1,..., m} and ˆv + ŝ β n ) + β n,d). Proof. Cases a) and b) can be easy proved smar to the proof of Lemma 5 n [14]. Cases c) and d) can aso be easy proved smar to the proof of Lemma 9 n [14]. For e), notce that by substtutng the pont ŷ, ˆv, ŝ) n constrant of Q m,n, we get n t=1 α t β t 1) /α t + d t=n +1 α t β n,t 1) /α t + ˆv + ŝ β, or ˆv + ŝ β n ) of case e). Lemma 4. For {1,..., m} and n {1,..., n}, a). φ n P,d ) = 0, d = 1,..., n, 14 + β n,d), whch s true by the assumpton

15 b). φ n Q,d ) = 1, d = 1,..., n, c). φ n R ) = 2, d). φ n S,d ) = 1, d = 2,..., n, e). φ n T,,d ) = 1, d = n,..., n, for {1,..., m} such that β n ) > β n ). Proof. Cases a), b) and e) can be proved smar to Lemma 6 of [14] and cases c) and d) can be proved smar to Lemma 10 of [14]. As before, gven a cyce C = V C, A C ) of G n, et F and B be the set of forward arcs and backward arcs of the cyce C, respectvey,.e. F := {, ) A C : < } and B := {, ) A C : < }. Theorem 4. For n {1,..., n}, the n -step cyce nequaty 21) for an eementary cyce C = V C, A C ) of graph G s facet-defnng for convq m,n ) f n addton to the n -step MIR condtons 22)) the foowng condtons hod a) β d 1) k /α d 1, d = 2,..., n, for a k, ) F, } b) β n ) β k max {α d 1 α d 1) k d αd, d = 2,..., n for a k, ) F, c) β n,d 1) k /α d 1, d = n + 1,..., n, for a k, ) B. Proof. Consder the supportng hyperpane of nequaty 21) for the cyce C. Note that ths hyperpane can be wrtten as n ) ) ) s + v + α t yt + β n ) + α n 1 φ n y ),) F =,) B t=n +1 ) β n ) φ n y ) n t=n +1 α t y t v ) because ),) F βn + ),) B β n ) = ),) F βn. Let Γ = {y, v, s) convq m,n ) : 32)} be the face of convq m,n ) defned by hyperpane 32). Frst, we prove that Γ s a facet of Q m,n under condtons a) note that under condtons a), 0 / V C because β 0 = 0 and does not satsfy condtons a)). Let m =1 t=1 n λ tyt + 32) m ρ v + ρ 0 s = θ 33) be a hyperpane passng through Γ. We prove that 33) must be a mutpe of 32). =1 Notce that for each k {1,..., m}\v C and d {1,..., n}, the unt vector E k,d 1 = y 1,..., y m, v 1,..., v m, s) Z Z n 1 + )m R m+1 +, n whch yk d = 1 and a other coordnates are zero, s a drecton for both the set Q m,n and the hyperpane defned by 32), and hence a drecton for the face Γ. Ths mpes that λ k d = 0 for a k {1,..., m}\v C and d = 1,..., n. By smar reasonng, for each k {1,..., m}\v C, the unt vector E2 k = y1,..., y m, v 1,..., v m, s) Z Z n 1 + )m R m+1 +, n 15

16 whch v k = 1 and a other coordnates are zero, s a drecton for the face Γ, mpyng that ρ k = 0, k {1,..., m}\v C. These reduce the hyperpane 33) to V C t=1 n λ tyt + V C ρ v + ρ 0 s = θ. 34) Next, consder the pont A = y, v, s) = y 1,..., y m, v 1,..., v m, 0) Z Z n 1 + )m R m+1 + such that, for = 1,..., m, y, v ) = Q,n, β n ) ) f T F ), and y, v ) = P,1, 0) f / T F ). Based on Lemma 3a,b), A Q m,n and usng Lemma 4a,b), t can be easy verfed that A satsfes 32). So, A Γ and hence must satsfy 34). Substtutng A nto 34) gves T F ) ρ β n ) + n t=1 λ t β t 1) ) /α t + T B) λ 1 β /α 1 = θ. 35) Usng 35), hyperpane 34) reduces to T F ) ) n ρ v + + ρ 0 s = T B) t=1 λ t ) yt β t 1) /α t + ) λ 1 β /α 1 y1 ) n λ tyt ρ v. t=2 n t=n +1 λ ty t ) 36) Now, consder the ponts B k,d = y, v, s) = y 1,..., y m, v 1,..., v m, 0) Z Z n 1 + )m R m+1 + for k T F ) and d = n + 1,..., n such that Q,n, β n ) y ) f T F )\{k},, v ) = Q,d, β d) ) f = k, P,1, 0) f / T F ), for = 1,..., m. By Lemma 3a,b), B k,d Q m,n, for a k T F ) and d = n + 1,..., n. Usng Lemma 4a,b), one can easy verfy that a these ponts aso satsfy 32). So for a k T F ) and d = n + 1,..., n, B k,d Γ, and hence must satsfy 36). Now f for each k T F ), we substtute the ponts B k,n +1,..., B k,n one after the other nto 36), snce condtons a) hods) we get λ k d = α dρ k, d = n + 1,..., n, k T F ) 37) Next, consder the ponts C k,d = y, v, s) = y 1,..., y m, v 1,..., v m, 0) Z Z n 1 + )m R m+1 + for k T B), d = 2,..., n such that Q,n, β n ) y ) f T F ),, v ) = P,d, 0) f = k, P,1, 0) f / T F ) {k}, for = 1,..., m. By Lemma 3a,b), C k,d Q m,n, for a k T B) and d = 2,..., n. Usng Lemma 4a,b), one can easy verfy that a these ponts aso satsfy 32). So for a k T B) and d = 2,..., n, C k,d Γ, and hence must satsfy 36). For each k T B), substtutng the ponts 16

17 C k,2,..., C k,n one after the other nto 36) gves λ k d 1 = λk d whch mpes λ k d = n λk n =d+1 hyperpane 36) to T F ) ρ + ρ 0 s = v + T B) β d 1) k /α d, d = 2,..., n, k T B), β 1) k /α, d = 1,..., n, k T B). Ths, aong wth 37), reduces n t=n +1 α t y t λ n φn y ) n t=n +1 ) n + t=1 λ t λ ty t ρ v ) ) ) yt β t 1) /α t. 38) Now, consder the pont D = y, v, s) = y 1,..., y m, v 1,..., v m, η) Z Z+ n 1 )m R m+1 +, where η = mn{β n) : T F )}, such that for = 1,..., m, y, v ) = Q,n, β n ) η) f T F ), and y, v ) = P,1, 0) f / T F ). By Lemma 3a,b), t s cear that D Q m,n and usng Lemma 4a,b), one can easy verfy that t aso satsfes 32). So D Γ, and hence must satsfy 38). Substtutng D nto 38) gves ρ 0 = T F ) ρ. 39) Now for V C, et N) be the node n V C such that, N)) A C. For each k, ) A C, snce condtons a) hods, consder the ponts F k, = y, v, s) = y 1,..., y m, v 1,..., v m, β n ) ) Z Z n 1 + )m R m+1 + such that R, β n ) + α n ) f T F ), N) < Q,n, 0) f T F ), < N) Q,n, β n ) y ) f T F ),, v ) = Q,n, 0) f T B), < Q,n, β n ) ) f T B), N) < P,1, 0) f T B), N) P,1, 0) f / V C, for = 1,..., m. By Lemma 3a,b,c), t s cear that F k, Q m,n for a k, ) A C. Usng Lemma 4a,b,c), f we substtute F k, nto 32), we get ) ) β n ) + β n ),) F ;,< + = +,) F ;<,) F ;<,) F ;< ) β n ) + β n ) β n ) +,) B;,<,) B;<,) B;<,) B;< β n ) β n ) = 0, ) β n ) 40) 17

18 whch s obvousy true. Therefore, the ponts F k,, for a k, ) A C, aso satsfy 32). Hence, they beong to Γ, and must satsfy 38). Now, note that n the pont F k,, k, ) F, by defnton we have y k, v k ) = Q k,n, 0). For each k, ) F, defne another pont F k, 1 = y, v, s) Z Z n 1 + )m R m+1 + whose coordnates are a exacty the same as F k, except that y k, v k ) = R k, β n ) k β n ) +α n ). For precsey the same reasons stated for F k,, the ponts F k, 1, k, ) F, must aso satsfy 38) note that substtutng F k, 1 n 32) gves dentty 40) agan). Now f for each k, ) F, we substtute F k, and F k, 1 nto 38) and subtract one equaty from the other, we get λ k n = ρ k β n ) k + α n ), for a k, ) F. 41) Next, for each k, ) F and d = 2,..., n, snce condtons a) hod, defne the pont F k,,d y, v, s) Z Z n 1 + )m R m+1 2 = + whose coordnates are a exacty the same as F k, except that y k, v k ) = S k,d, 0). By Lemma 3a,b,c,d) and because of condtons b), t s cear that F k,,d 2 Q m,n for a k, ) F and d = 2,..., n. Usng Lemma 4a,b,c,d), one can easy verfy that they aso satsfy 32) note that substtutng F k,,d 2 n 32) gves dentty 40) agan), and hence beong to Γ and must satsfy 38). Now f for each k, ) F and d = 2,..., n, we substtute the ponts F k, and F k,,d 2 nto 38) and subtract one equaty from the other, we get /α d, d {2,..., n }, k T F ). 42) Ths mpes λ k d 1 = λk d λ k d = n λk n p=d+1 β d 1) k β p 1) k /α p, d = 1,..., n, k T F ). 43) Next, note that n the pont F k,, k, ) B, by defnton we have y k, v k ) = P k,1, 0). For each k, ) B and d = n,..., n, defne the pont F k,,d 3 = y, v, s) Z Z n 1 + )m R m+1 + whose coordnates are a exacty the same as F k, except that y k, v k ) = T k,,d, β n,d) k ). By Lemma 3a,b,c,e), t s cear that F k,,d 3 Q m,n for a k, ) B and d = n,..., n. Usng Lemma 4a,b,c,e), we can easy verfy that they aso satsfy 32) note that substtutng F k,,d 3 n 32) gves dentty 40) agan), and hence beong to Γ and must satsfy 38). Now f for each k, ) B, we substtute F k, and F k,,n 3 nto 38) and subtract one equaty from the other, we get λ k n = ρ k ) β n ) k, for a k, ) B, 44) and f we contnue to do the same wth F k,,n +1 3,..., F k,,n 3 one after the other, n ght of condton c), we get λ k d = α dρ k, d = n + 1,..., n, for a k, ) B. 45) Based on 39), 41), 43), 44), 45), and usng 8), hyperpane 38) reduces to,) F =,) B ρ s + v + n t=n +1 ρ β n ) ) ) ) α t yt + β n ) + α n 1 φ n y ) ) φ n y ) n t=n +1 α t y t v ). 46) 18

19 Now, for V C, et P ) be the node n V C such that P ), ) A C, and defne a := mn{ V C : < } and b := max{ V C : < }. Aso et max = max{ : V C } and mn = mn{ : V C }. For V C \{ max }, f we substtute the pont F P ), and F P a),a nto 46) note that both ponts must satsfy 46) as argued for a ponts F k, ) and subtract the two equates, we get ) ρ β n ) a + ) ρ β n ) a = 0. Snce β n ) β n ), we get,) F <a,) B <a,) F ;< a ρ,) B;< a a ρ = 0. 47) Lkewse, for V C \{ mn }, f we substtute the pont F P b), b subtract the two equates, we get and F P ), nto equaty 46) and ρ ρ = 0 48),) F ;<,) B;< because β n ) b β n ). Notce that f = P max ), then a = max, and dentty 47) reduces to ρ P max) = ρ max 49) Aso f for each V C \{ mn, max }, we subtract 47) from 48), we get ρ P ) = ρ, V C \{ mn, max }. 50) Identtes 49) and 50) mpy that ρ P ) = ρ for a V C because P ) = mn for some V C \{ mn }). Therefore, ρ = ρ for a, V C 51) as C s a cyce. Ths reduces hyperpane 46) to a constant mutpe by 39) ths mutpe s ρ 0 / F ) of 32), whch competes the proof. Exampe 1 contnued). Notce that for n = 1, each cyce C = V C, A C ) n graph G 1 wth a set of backward arcs B = {, 6)}, for {1,..., 5}, satsfy the addtona condtons requred for Theorem 4,.e. a) β 1) k /α 2 = 3 1, for k {1,..., 5}, β 1) 6 /α 2 = 2 1, b) ths condton s ) automatcay satsfed for n = 1, and c) β 1,1) k,6 /α 2 = β 1) k β 1) 6 /α 2 1, for k = 1,..., 5. Therefore, the 1-step cyce nequaty 21) correspondng to a cyce C n G 1, where B = {, 6)} for {1,..., 5}, defnes facet for convq 6,2 ). In partcuar, the 1-step cyce nequates correspondng to the cyces C1 1 and C1 2 are facet-defnng for convq6,2 ). Now, for n = 2, the coeffcents of Q 6,2 aso satsfy the addtona condtons requred n Theorem 4,.e. a) β 1) k /α 2 = 3 1, for k {1,..., 5}, β 1) 6 /α 2 = 2 1, b) β 2) β 2) k 2 = α 1 α 2 β 1) k /α 2 for a k, ) A 2 such that 1 k < 6, and there s no condton c) for n = n = 2. Therefore, the 2-step cyce nequaty 21) correspondng to each cyce C = V C, A C ) n graph G 2, where V C {1,..., 6}, defnes a facet for convq 6,2 ). In partcuar, 2-step cyce nequates correspondng to the cyces C1 2 and C2 2 are facet-defnng for convq6,2 ). 19

20 Theorem 5. For n {1,..., n}, the n -step cyce nequaty 21) for an eementary cyce C = V C, A C ) of graph G s facet-defnng for convq m,n ) f n addton to the n -step MIR condtons 22)) the foowng condton hod a) T F ) = {0}, b) β n,d 1) k /α d 1, d = n + 1,..., n, for a k, ) B. Proof. As shown before, the supportng hyperpane of nequaty 21) can be wrtten as 32), whch for the C consdered n ths theorem reduces to s = ) ) n β n) β n) φ y ) α t yt v 52),) B t=n +1 because by condton a), the cyce C has ony one forward arc, whch goes out of node 0, and we have v 0 = 0, y 0 = 0 and φ n 0 y0 ) := 1 by defnton. Let Γ be the face of Q m,n defned by hyperpane 52). We prove that any generc hyperpane ρ 0 s + m ρ v + =1 m =1 t=1 n λ y = θ 53) that passes through Γ s a scaar mutpe of 52). By the same reasonng we reduced hyperpane 33) to 34) n Theorem 4, we can reduce hyperpane 53) to V C \{0} t=1 n λ tyt + V C \{0} ρ v + ρ 0 s = θ. 54) Now consder the foowng ponts correspondg to the ponts wth the same name n the proof of Theorem 4): The pont A = y 1,..., y m, v 1,..., v m, s) Z Z n 1 + )m R m+1 + such that y, v ) = P,1, 0), = 1,..., m, and s = 0; the ponts C k,d = y 1,..., y m, v 1,..., v m, s) Z Z n 1 + )m R m + R +, for k T B), d = 2,..., n, such that y k, v k ) = P k,d, 0) and y, v ) = P,1, 0) for {1,..., m}\t F ) {k}), and s = 0; the ponts F k, = y 1,..., y m, v 1,..., v m, s) Z Z n 1 + )m R m + R +, for k, ) B, such that Q,n, 0) f T B), y, v ) = P,1, 0) f T B), N) P,1, 0) f / V C, n for = 1,..., m, and s = β ) ; and the ponts F k,,d 3 Z Z n 1 + )m R m+1 +, for k, ) B, d = n,..., n, whose coordnates are a exacty the same as F k, except that y k, v k ) = T k,,d, β n,d) k ). By Lemma 3a,b,e), a the aforementoned ponts beong to Q m,n, and by Lemma 4a,b,e), t s easy to verfy that they aso satsfy 52). So, they beong to Γ, and hence must satsfy 54). Therefore, gven condtons b), a these ponts can be used n the same fashon the ponts wth smar names were used n the proof of Theorem 4 to reduce the hyperpane 54) to an equaty whch s ρ 0 tmes the hyperpane 52). Ths competes the proof. 20

21 Exampe 1 contnued). Notce that for n = 1, each cyce C = V C, A C ) n graph G 1 wth A C = {0, ),, 0)} for {1,..., 6} or A C = {0, ),, 6), 6, 0)} for {1,..., 5} satsfes) the condtons requred for Theorem 5,.e. a) T F ) = {0}, and b) β 1,1) k, /α 2 = β 1) k β 1) /α 2 1 for a k, ) {, ) A C : β 1) > β 1) }. Therefore, the 1-step cyce nequaty 21) correspondng to each cyce C defnes a facet for convq 6,2 ). In partcuar, 1-step cyce nequaty correspondng to the cyce C3 1 s facet-defnng for convq6,2 ). Moreover, the 2-step cyce nequaty 21) correspondng to each cyce C = V C, A C ) n G 2 = V 2, A 2 ), where T F ) = {0}, aso defnes facet for convq 6,2 ) because there s no condton b) for n = n = 2. In partcuar, 2-step cyce nequaty correspondng to the cyce C3 2 s facet-defnng for convq6,2 ). 5. Separaton agorthm Gven a pont ŷ, ˆv, ŝ) and n {1,..., n}, t s possbe to sove the exact separaton probem over a the n -step cyce nequates for the set Q m,n. The goa s to fnd an n -step cyce nequaty 21) that s voated by ŷ, ˆv, ŝ), f any. Ths can be done by detectng a negatve weght cyce f any) n the drected graph G n = V, A) wth weghts ψ n ŷ, ˆv, ŝ) for each arc, ) A. Ths means that the most negatve cyce n G n f t exsts) corresponds to the n -step cyce nequaty that s most voated by ŷ, ˆv, ŝ). However, the probem of fndng the most negatve cyce n a graph s strongy NP-hard [15]. A method proposed by Cherkassy and Godberg [3] whch s a combnaton of the cyce detecton strategy of Taran [16] and the Beman-Ford-Moore s abeng agorthm [4]), denoted by BFCT, s one of the fastest known agorthms to detect a negatve cyce. BFCT termnates when t fnds the frst negatve cyce; however, there may be cyces wth smaer weght n the graph whch woud ead to stronger nequates. Therefore, we devsed a modfed verson of BFCT, denoted by MBFCT, whch does not stop after fndng the frst negatve cyce and contnues the search for other negatve cyces f any) unt a certan termnaton condton s satsfed refer to Agorthm 1 for detas). Out of a the cyces found by MBFCT, the one wth the most negatve weght s used to generate the n -step cyce nequaty 21) that separates ŷ, ˆv, ŝ) wth the argest voaton among a generated cyces. 21

22 Agorthm 1 Separaton Agorthm for n -step Cyce Inequates 1: functon MBFCTG, ŷ, ˆv, ŝ), n ) 2: for V do 3: d) ; parent) N u; status) unreached ; 4: end for 5: NC, abe {0}; status0) abeed ; d0) 0; Count 0; 6: for abe and Count 3 V do FIFO seecton rue 7: for, ) A do 8: f d) + ψ n ŷ, ˆv, ŝ) < d) then 9: d) d) + ψ n ŷ, ˆv, ŝ); status) abeed ; parent) ; 10: A p {parent), ) : V, parent) Nu}; 11: Construct graph G p V, A p ) 12: f the subtree of G p rooted at contans then 13: NC NC { )} s the path from to n G p 14: ese 15: remove a the nodes of subtree except from G p 16: and change ther status to unreached 17: end f 18: end f 19: end for 20: abe abe\{}; status) scanned ; Count Count + 1; 21: end for 22: return the most negatve cyce n NC f exst) 23: end functon 6. Cuts for mut-modue capactated ot-szng probem In ths secton, we use n-step cyce nequates to deveop cuttng panes for MML-W)B probem. We defne MML-B as foows. Let P := {1,..., m} be the set of tme perods and {α 1,..., α n } be the set of szes of the n avaabe capacty modues. The setup cost per modue of sze α t, t = 1,..., n n perod p s denoted by fp. t Gven the demand, the producton per unt cost, the nventory per unt cost, and the per unt shortage backog) cost n perod p, denoted by d p, c p, h p, and b p, respectvey, the MML-B probem can be formuated as: mn c p x p + h p s p + b p r p + n fpz t p t 55) p P p P p P p P t=1 s p 1 r p 1 + x p = d p + s p r p, p P 56) n x p α t zp, t p P 57) t=1 z, x, r, s) Z m n + R m + R m+1 + R m ) where x p s the producton n perod p, s p and r p are the nventory and backog, respectvey, at the end of perod p, s 0 = r m = 0, and z t p s the number of capacty modues of sze α t, t = 1,..., n, used n perod p. Let X MML B denote the set of feasbe soutons to constrants 56)-58). Note that 22

23 every vad nequaty for X MML B aso gves a vad nequaty for the set of feasbe soutons to the MML-WB probem whch s the proecton of X MML B {r = 0} on z, x, s). In order to generate vad nequates for X MML B, we consder perods k,...,, for any k, P where k <. Let S {k,..., } such that k S. For S, et S := S {k,..., }, m = mn{p : p S\S } wth m = + 1 f S\S =, and b = m 1 p=k d p. Now, by addng equates 56) from perod k to perod m 1, we get m 1 s k 1 + r m 1 + x p = b + s m 1 + r k 1. 59) p=k Note that S {k,..., m 1} by defnton. If we reax x p, p S, n 59) to ts upper bound based on 57) and drop r k 1, s m 1 0), we get the foowng vad nequaty: s k 1 + r m 1 + p {k,...,m 1}\S x p + n α t t=1 p S z t p b. 60) Settng s := s k 1, v := r m 1 + x p, and yt := zp, t 61) p {k,...,m 1}\S p S nequaty 60) becomes s + v + n α t yt b, 62) t=1 whch s of the same form as the defnng nequates of contnuous n-mxng set notce that s, v R +, yt Z +, t = 1,..., n). Therefore we can form a set of base nequates consstng of nequates 60) for a S such that the n-step MIR condtons,.e. α t b t 1) /α t α t 1, t = 2,..., n, hod. We construct a drected graph for these base nequates n the same fashon as we dd for the contnuous n-mxng set Q m,n n Secton 3. The n-step cyce nequates correspondng to each eementary cyce C n ths graph s vad for X MML B. We refer to these nequates as the n-step k,, S, C) cyce nequates. The same procedure aso provdes a new cass of vad nequates for MML-WB whch subsume the vad nequates generated usng the mxed n-step MIR nequates [14] for MML-WB. Note that a procedure smar to what was presented above for n can aso be used to deveop n -step k,, S, C) cyce nequates for MML-W)B probem for any n {1,..., n} n genera. 7. Computatona Resuts In ths secton, we computatonay evauate the effectveness of the n -step cyce nequates, n {1,..., n}, for the MML-W)B probem usng our separaton agorthm dscussed n Secton 5). We chose n = 2 for our experments n ths paper and refer to the MML-WB and MML-B probem wth two capacty modues n = 2) as 2ML-WB and 2ML-B, respectvey. We created random 2ML-WB 23

24 and 2ML-B nstances wth 60 tme perods,.e. P = {1,..., 60}, and varyng cost and capacty characterstcs. The demand d p, producton cost c p, and hodng cost h p n each perod were drawn from nteger unform[10, 190 ], nteger unform[81, 119 ], and rea unform[1, 19 ], respectvey. For each nstance of 2ML-B, the backog cost b p n each perod equas h p pus a rea number drawn from unform[1, 10 ]. We used three sets of capacty modues α = α 1, α 2 ): 70, 34), 100, 35), and 180, 80), denoted by M a, M b, and M c respectvey, and four sets of setup costs f 1 p, f 2 p ), p P : 1000, 600), 5000, 2600), 10500, 6600), and 13000, 10600), denoted by F I, F II, F III, and F IV respectvey. Ths eads to 12 nstance categores where the frst set of setup costs.e. F I ) eads to easy nstances and the remanng three ead to hard nstances. Note that some of the nstance generaton deas we used here are nspred by the deas used n [14] for 2ML-WB. For each 2ML-W)B nstance, we frst soved the probem defned n Secton 6), for n = 2, wthout addng any of our own cuts usng CPLEX 11.0 wth ts defaut settngs 2ML-W)B-DEF). In a separate run, for each n {1, 2}, we used our cut generaton agorthm, denoted by CutGenn ), to add n -step k,, S, C) cyce nequates to the probem at the root node. The pseudocode of CutGen s presented n Agorthm 2. Ths agorthm cas our separaton agorthm n Lne 12 for severa choces of k,, S) see Lnes 3-9) to generate n -step k,, S, C) cyce nequates Lnes 10-12) that are voated by the LP reaxaton optma souton, whch s updated after addng each cut see Lnes 13-17). Note that each choce of k,, S) provdes one set of base nequates 60) where n = 2) and we sove an exact separaton probem over the set of a 2-step k,, S, C) cyce nequates correspondng to the base nequates whch satsfy the n-step MIR condtons dscussed n Secton 6). We then removed the nactve cuts and used CPLEX 11.0 wth ts defaut settngs to sove the probem 2ML-W)B-1CUTS for n = 1, and 2ML-W)B-2CUTS for n = 2). We mpemented our codes n Mcrosoft Vsua C and a the experments were run on a PC whch has two Inte Xeon E GHz processors and 12 GB of RAM. The resuts of our computatona experments are shown n Tabes 1 and 2. Each row of these tabes reports the average resuts for 10 nstances of the correspondng nstance category. Note that an nstance category correspondng to a set of setup costs say F I ) and a set of capacty modue say M a ) s denoted by I-a. We report the percentage of the ntegraty gap cosed by our cuts,.e. G% = 100 zcut zp)/zmp zp), where zp, zcut, and zmp are the optma obectve vaues of the LP reaxaton wthout our cuts, LP reaxaton wth our cuts, and MIP, respectvey. We aso report the number of branch-and-bound nodes Nodes), and the tme n seconds) to sove 2ML-W)B-DEF T Def ), 2ML-W)B-1CUTS TOpt 1 ), and 2ML-W)B-2CUTS T Opt 2 ) to optmaty. Note that TOpt 1 and TOpt 2 excude the cut generaton tme. For each n {1, 2}, the number of actve n -step k,, S, C) cyce cuts added at the root node Cuts), the tme n seconds) to generate n -step k,, S, C) cyce cuts denoted by TCut 1 for n = 1 and TCut 2 for n = 2), and the tota tme ncudng the cut generaton tme) to sove 2ML-W)B-1CUTS and 2ML-W)B-2CUTS, denoted by T 1 = TCut 1 + T Opt 1 and T 2 = TCut 2 + T Opt 2 respectvey, are aso reported. In Tabe 1, comparng the tme to optmze the 2ML-WB probem before and after addng the 2-step k,, S, C) cyce cuts.e. T 2 Opt vs. T Def ), we see sgnfcant mprovement obtaned by addng these cuts n both easy nstances on average 3 tmes) and hard nstances on average 112 tmes). There s aso a substanta reducton n the number of branch-and-bound nodes on average 6.5 tmes for easy nstances and 174 tmes for hard nstances). The percentage of ntegraty gap cosed by the 2-step k,, S, C) cyce cuts s between 80.32% and 91.15% the average s 85.90%). These resuts show the strength of 2-step k,, S, C) cyce nequates. Interestngy, n these nstances addng 2-step k,, S, C) cyce nequates over 1-step k,, S, C) cyce nequates has mproved the cosed ntegraty gap by 19.48% n average), the number of nodes by 43 tmes n average), 24