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Facets for Single Module and Multi-Module Capacitated Lot-Sizing Problems without Backlogging Manish Bansal Grado Department of Industrial and Systems Engineering Virginia Tech, Blacksburg, VA 24060 Email: [email protected] June 14, 2017 Abstract. In this paper, we consider the well-known constant-batch lot-sizing problem, which we refer to as the single module capacitated lot-sizing (SMLS) problem, and multi-module capacitated lot-sizing (MMLS) problem. We provide sufficient conditions under which the (k, l, S, I) inequalities of Pochet and Wolsey (Math of OR 18: 767-785, 1993), the mixed (k, l, S, I) inequalities, derived using mixing procedure of G¨ unl¨ uk and Pochet (Math. Prog. 90(3): 429-457, 2001), and the paired (k, l, S, I) inequalities, derived using sequential pairing procedure of Guan et al. (Discrete Optimization 4(1): 21-39, 2007), are facet-defining for the SMLS problem without backlogging. We also provide conditions under which the inequalities derived using the sequential pairing and the n-mixing procedure of Sanjeevi and Kianfar (Discrete Optimization 9:216-235, 2012) are facetdefining for the MMLS problem without backlogging.

1.

Introduction

Developing valid inequalities for mixed integer programming problems, in particular lot-sizing problems, by using facets (or valid inequalities) of simple mixed integer sets is a promising approach [3, 4, 5, 9, 10, 15, 16]. Using this approach, various families of valid inequalities have been developed for the classical capacitated lot-sizing problem and its generalizations [1, 3, 4, 9, 15, 16]. However, not much is known about the facet-defining properties of these inequalities [14]. In this paper, we investigate the facet-defining properties of the well-known families of valid inequalities for the constant-batch lot-sizing problem which we refer to as the single module capacitated lot-sizing (SMLS) problem and one of its generalizations, called multi-module capacitated lot-sizing (MMLS) problem. The MMLS without backlogging (MMLS-WB) is defined as follows: Let {α1 , . . . , αn } be the set of sizes of the n available capacity modules and the setup cost per module of size αt , t = 1, . . . , n in period p is denoted by fpt . Then the MMLS-WB is formulated as: ! n X X cp xp + hp sp + fpt zpt min t=1

p∈P

s.t. sp−1 + xp = dp + sp , n X xp ≤ αt zpt , z∈

t=1 |P |×n Z+ ,x

|P |

p∈P p∈P |P |+1

∈ R+ , s ∈ R+

where xp is the production in period p, sp is the inventory at the end of period p, and zpt is the number of capacity modules of size αt , t = 1, . . . , n, used in period p. In addition, the parameters dp , cp , and hp denote the demand, the production per unit cost, and the inventory per unit cost in period p, respectively. For n = 1, MMLS-WB reduces to SMLS without backlogging (SMLS-WB). 1

Pochet and Wolsey [12] consider the SMLS-WB problem where the capacity in each period can be some integer multiple of a single capacity module with a given size. They introduce the so-called (k, l, S, I) inequalities and show that these inequalities subsume facets of a certain form (described implicitly in [12]) for SMLS-WB. In this paper, we explicitly define a subclass of the (k, l, S, I) inequalities and show that under certain conditions, only the (k, l, S, I) inequalities belonging to this subclass can be facet-defining. More specifically, we show that for each (k, l, S, I) inequality which does not belong to this subclass, there exists a stronger valid inequality belonging to this subclass. Later, G¨ unl¨ uk and Pochet [9] generalized the mixed integer rounding (MIR) inequalities [11, 16] by introducing mixed MIR inequalities for a multi-constraint mixed integer set. They demonstrate how these inequalities can be utilized to develop valid inequalities for general mixed integer programs (MIPs) including SMLS-WB and referred to this cut-generation procedure as “mixing” (see Section 2.2 for details). We refer to a class of valid inequalities for SMLS-WB which can be derived using the mixing procedure as mixed (k, l, S, I) inequalities. In this paper, we provide sufficient conditions under which the mixed (k, l, S, I) inequalities are facet-defining for SMLS-WB. In another direction, Guan et al. [8] develop facet-defining inequalities for multi-stage stochastic uncapacitated lot-sizing problem, referred to as the (Q, SQ ) inequalities. Thereafter, Guan et al. [6] introduce a cut-generation procedure, referred to as “pairing”, for MIPs; see Section 2.4 for details. They use (sequential) pairing to derive valid inequalities for multi-stage stochastic mixed integer programs including multi-stage stochastic capacitated lot-sizing problem with variable capacities [7]. It is important to note that the (Q, SQ ) inequalities can be derived using the pairing procedure for multi-stage stochastic uncapacitated lot-sizing problem. Recently in [2], Bansal et al. present globally valid parametric inequalities and tight second stage formulations for four variants of the two-stage stochastic capacitated lot-sizing problems with uncertain demands and costs. In this paper, we present a class of valid inequalities for SMLS-WB derived using pairing procedure, investigate their facet-defining properties, and their relationship with the mixed (k, l, S, I) inequalities. We refer to this class of inequalities as the paired (k, l, S, I) inequalities. Sanjeevi and Kianfar [15] generalize the SMLS-WB to MMLS-WB problem, in which the total production capacity in each period can be the summation of some integer multiples of several capacity modules of different sizes. They show that the mixed n-step MIR inequalities (which generalize the mixed MIR inequalities [9] and n-step MIR inequalities [10]) can be used to generate valid inequalities for MMLS-WB. These inequalities are referred to as the multi-module mixed (k, l, S, I) inequalities as they also generalize the (k, l, S, I) inequalities [12]. Recently, Bansal and Kianfar [3, 4] further generalize the mixed n-step MIR inequalities [15] to n-step cycle inequalities and develop the so-called n-step (k, l, S, C) cycle inequalities for both MMLS-WB and MMLS with backlogging problems. The class of n-step (k, l, S, C) cycle inequalities subsumes the multi-module mixed (k, l, S, I) inequalities for MMLS-WB problem. Note that in [1], the n-step cycle inequalities are also used to generate new classes of valid inequalities for multi-module capacitated facility location (MMFL) and multi-module capacitated network design (MMND) problems. In this paper, we investigate the facet-defining properties of the multi-module mixed (k, l, S, I) inequalities for the MMLS-WB problem. In addition, we present a new class of valid inequalities for MMLS-WB using pairing procedure [6], investigate their facet-defining properties, and their relationship with the multi-module mixed (k, l, S, I) inequalities. Organization of this paper: In Section 2, we briefly review the cut-generation procedures: mixed integer rounding (MIR) [11, 16], mixing [9], n-mixing [15], and sequential pairing [6]. In Section 3, we provide sufficient conditions under which the (k, l, S, I) inequalities of Pochet and Wolsey [12], the mixed (k, l, S, I) inequalities, derived using mixing procedure of G¨ unl¨ uk and Pochet [9], and 2

paired (k, l, S, I) inequalities, derived using pairing procedure of Guan et al. [6], are facet-defining for SMLS-WB. In Section 4, we present conditions under which the inequalities derived using the pairing procedure [6] and the n-mixing cut-generation procedure of Sanjeevi and Kianfar [15] are facet-defining for the MMLS-WB. Finally in Section 5, we provide concluding remarks and discuss about potential extension of the results presented in this paper.

2.

Necessary Background

In this section, we briefly review the cut-generation procedures: mixed integer rounding (MIR) [11, 16], mixing [9], n-mixing [15], and sequential pairing [6] which we will use to develop valid inequalities for SMLS-WB and MMLS-WB problems in the subsequent sections.

2.1

Mixed Integer Rounding (MIR)

One fundamental procedure to develop cuts for general mixed integer programs is the MIR procedure [11, 16] which utilizes the facet of a single-constraint mixed integer set, Q := {(y, v) ∈ Z × R+ : α1 y + v ≥ β}, where α1 > 0 and β ∈ R (page 127 of [16]). This facet is referred to as the MIR facet and is given by         β β β v + β − α1 y ≥ β − α1 . (1) α1 α1 α1 In order to avoid trivial case, it is assumed that β/α1 ∈ / Z because then Inequality (1) will reduce to v ≥ 0.

2.2

Mixing

G¨ unl¨ uk and Pochet [9] studied a mixed integer set, defined by Q0 := {(y, v) ∈ Zm × R+ : α1 y i + v ≥ βi , i = 1, . . . , m}, where α1 ∈ R+ , β ∈ Rm , and βi /α1 ∈ / Z, i = 1, . . . , m. This set is referred to as the mixing set which is a multi-constraint generalization of the set Q that leads to the well-known MIR inequality (1). G¨ unl¨ uk and Pochet [9] derive the mixed MIR inequalities for the mixing set Q0 as follows: (1) Define β0 := 0, βi := βi − bβi /α1 c, i ∈ {0, . . . , m}, and without loss of generality assume that (1) (1) (1) β0 = 0 < βi−1 ≤ βi , i = 2, . . . , m. Let K := {i1 , . . . , i|K| }, where i1 < i2 < · · · < i|K| , be a non-empty subset of {1, . . . , m}. Then the inequalities, v≥

|K|  X

(1) βip

−

(1) βip−1

  β  ip

α1

p=1

v≥

|K|  X p=1

(1) βip

−

(1) βip−1

  β  ip

α1

−y

ip

3





+ α1 −

−y

ip

(1) βi|K|

 (2)

  β  ip

α1

−y

i1

 ,

(3)

are valid for Q0 and sufficient to describe the convex hull of Q0 . Inequalities (2) and (3) are referred to as the type I and type II mixed MIR inequalities, respectively. As mentioned in Section 1, G¨ unl¨ uk and Pochet used these inequalities to generate cuts for linear mixed integer programs and referred to this cut-generation procedure as mixing. In Section 3.1, we will show how to generate valid inequalities for SMLS-WB using mixing. In addition, G¨ unl¨ uk and Pochet mentioned that in a variant of the mixing set Q0 where y ∈ Zm , inequality (3) is redundant unless y i1 < bβi1 /α1 c for + some feasible y.

2.3

n-mixing

Sanjeevi and Kianfar [15] generalize the mixing procedure [9] to n-mixing cut-generation procedure by considering the n-mixing set, n X  1 n−1 m m Qm,n := (y , . . . , y , v) ∈ (Z × Z ) × R : αt yti + v ≥ βi , i = 1, . . . , m , + + 0 t=1

= Q0 . They develop the mixed nwhere αt > 0, t = 1, . . . , n, and β ∈ Rm . Note that Qm,1 0 (n) (n) m,n step MIR inequalities for Q0 as follows: Without loss of generality, we assume βi−1 ≤ βi , i = j k (t) (t−1) (t−1) (0) 2, . . . , m, where βi := βi − αt βi /αt , t = 1, . . . , n, and βi := βi . It is also assumed (t−1)

(t)

that β /αt ∈ / Z, i = 1, . . . , m. Note that 0 < βi < αt for t = 1, . . . , n. By definition, we set Qb Pb i a (.) = 1 if a > b. If the n-step MIR conditions, a (.) = 0 and l m (t−1) αt βi /αt ≤ αt−1 , t = 2, . . . , n, (4) hold for each constraint i ∈ K ⊆ {1, . . . , m}, then the inequalities v≥

|K|  X

 (n) (n) βip − βip−1 φnip (y ip ),

(5)

  
(n) (n) (n) βip − βip−1 φnip (y ip ) + αn − βi|K| φni1 (y i1 ) − 1 ,

(6)

p=1

v≥

|K|  X p=1

(n)

are valid for Qm,n 0 , where βi0 = 0 and & (l−1) ' & (l−1) ' n n Y n Y X β βi i φni (y i ) := − yti αl αl

(7)

t=1 l=t+1

l=1

for i ∈ K. Inequalities (5) and (6) are referred to as type I and type II mixed n-step MIR inequalities, respectively. Inequality (5) is shown to be facet-defining for Qm,n 0 . Inequality (6) also defines a m,n facet for Q0 if some additional conditions are satisfied; see [15] for details. Sanjeevi and Kianfar [15] used mixed n-step MIR inequalities to generate valid inequalities for MMLS-WB problem (see Section 4.1 for details) and referred to this cut-generation procedure as n-mixing. Note that the n-step MIR inequalities [10] are special cases of the type I mixed n-step MIR inequalities (5), and both type I and type II mixed n-step MIR inequalities, (5) and (6), are special cases of the n-step cycle inequalities of Bansal and Kianfar [3, 4, 5]. Remark 1. The n-step MIR conditions (4) are automatically satisfied when the coefficients, α1 , . . . , αn , are divisible, i.e. αt /αt−1 ∈ Z+ , t = 2, . . . , n. 4

2.4

Pairing of mixed integer inequalities

Given a mixed integer set Z := {(x, z) ∈ Zn+x × Rn+z : Ex x + Ez z ≥ eR } and two valid inequalities for Z: nx X

a1j xj +

nz X

j=1

j=1

nx X

nz X

a2j xj +

j=1

b1j zj ≥ c1R

(8)

b2j zj ≥ c2R .

(9)

j=1

Guan et al. [6] introduce a cut-generation procedure, referred to as pairing, to develop a new valid inequality (10) for Z. Theorem 1 ([6]). Assuming c2R ≥ c1R , the inequality, nx X

min {a1j

+

c2R

−

c1R , max {a1j , a2j }}xj

j=1

+

nz X

max {b1j , b2j }zj ≥ c2R ,

(10)

j=1

is valid for Z. For the sake of convenience, let Inequality (10) be written as nx nz X X 1 2 (aj ◦ aj )xj + (b1j ◦ b2j )zj ≥ c2R , j=1

j=1

and referred to as (1 ◦ 2) paired inequality. Assume that the inequalities nx X j=1

akj xj +

nz X

bkj zj ≥ ckR , for k = 3, . . . , m, ¯

(11)

j=1

¯ are also valid for Z where c2R ≤ c3R ≤ . . . ≤ cm R . Then pairing the (1 ◦ 2) paired inequality and inequality (11), for any k ∈ {3, . . . , m}, ¯ gives another valid inequality for Z, i.e. nx X

min {(a1j ◦ a2j ) + ckR − c2R , max {(a1j ◦ a2j ), akj }}xj

j=1

+

nz X

(12) max {(b1j ◦ b2j ), bkj }zj ≥ ckR ,

j=1

which is referred to as the ((1 ◦ 2) ◦ k) sequentially paired inequality. This cut-generation procedure is called sequential pairing. Guan et al. [6] provide the following result for these inequalities. Theorem 2 ([6]). Let K := {i1 , . . . , i|K| } be a subset of {1, . . . , m} ¯ such that i1 ≤ i2 ≤ . . . ≤ i|K| . For each of the following cases, there exists a subset K such that the ((. . . ((i1 ◦ i2 ) ◦ i3 ) ◦ . . .) ◦ i|K| ) sequentially paired inequality is at least as strong as any other sequentially paired inequality obtained by arbitrary sequence of pairing of the inequalities (11) for k = 1, . . . , m: ¯ 1. Nested case, i.e. akj ≤ ak+1 , for j = 1, . . . , nx and k = 1, . . . , m ¯ − 1, j 2. Disjoint case, i.e. akj akl = 0 for j 6= l, j, l ∈ {1, . . . , nx }, and k ∈ {1, . . . , m}. ¯ 5

3.

Single Module Lot-Sizing Problem without Backlogging

In this section, we first re-define the single module capacitated lot-sizing (SMLS) problem (referred to as the constant batch lot-sizing problem in [12]). Then we present valid inequalities for this problem and investigate their facet-defining properties. The SMLS without backlogging (SMLSWB) is defined as follows: Let P := {1, . . . , m} be the set of time periods and α1 be the size of the available capacity module (or batch). Given the setup cost per module, the demand, the production per unit cost, and the inventory per unit cost in period p, denoted by fp1 , dp , cp , and hp , respectively, P
1 1 SML where SMLS-WB can be formulated as: min p∈P cp xp + hp sp + fp zp : (z, x, s) ∈ X  m+1 m X SML := (z, x, s) ∈ Zm + × R+ × R+ sp−1 + xp = dp + sp , xp ≤

α1 zp1 ,

(13)

p∈P

(14)

p∈P ,

(15)

xp is the production in period p, sp is the inventory at the end of period p, and zp1 is the number of capacity modules of size α1 used in period p.

3.1

Valid Inequalities for SMLS-WB Problem

Similar to Pochet and Wolsey [12], we consider a subnetwork consisting of periods k, . . . , l, for any k, l ∈ P where k < l. Let S ⊆ {k, . . . , l} such that k ∈ S. For i ∈ S, let := S ∩ {k, . . . , i}, PSmii −1 mi = min{p : p ∈ S\Si } with mi = l + 1 if S\Si = ∅, and bi = p=k dp . We define (1)

(1)

(1)

(1)

I := {i1 , i2 , . . . , i|I| } ⊆ S such that bi0 = 0 < bi1 ≤ bi2 ≤ . . . ≤ bi|I| < α1 . Theorem 3 ([13]). The inequality,

sk−1 +

X

xp ≥

|I|  X

(1)

(1)

biq − biq−1

q=1

p∈{k,...,l}\S



    X b  iq − zp1  , α1

(16)

p∈Siq

referred to as the (k, l, S, I) inequality, is valid for X SML where s0 = sm = 0. These inequalities generate all facets for the convex hull of X SML ∩ {(s0 , sm ) : s0 = sm = 0} which are of the form X X sk−1 + xp + πp1 zp1 ≥ π0 p∈S

p∈{k,...,l}\S

for all 2 ≤ k ≤ l ≤ m and S ⊆ {k, . . . , l}. Example 1. Consider a SMLS-WB problem instance defined over six time periods, i.e. m = 6 and P = {1, . . . , 6}, where α1 = 5 and demand in each time period are as follows: d1 = 13, d2 = 2, d3 = 2, d4 = 23, d5 = 1, and d6 = 8. Figure 1 represents a flow diagram for this instance. Let the

6

set of feasible solutions, (13)-(15), for this instance be defined by X E := {(z, x, s) ∈ Z6+ × R6+ × R7+ : s0 + x1 = s1 + 13 s1 + x2 = s2 + 2 s2 + x3 = s3 + 2 s3 + x4 = s4 + 23 s4 + x5 = s5 + 1 s5 + x6 = s6 + 8 xp ≤ 5zp1 , p = 1, . . . , 6}.

Figure 1: Example of SMLS-WB Using the above defined notations, we derive valid inequalities for X E as follows. Let k = 2, l = 6, and S = {2, 4, 5}. Therefore, S2 = {2}, S4 = {2, 4}, S5 = {2, 4, 5}, and m2 = 4, m4 = 5, (1) m5 = 7. Also, b2 = d2 + d3 = 4 and b2 = 4 − 5 b4/5c = 4. Likewise, b4 = d2 + d3 + d4 = 27, P (1) (1) m5 −1 = 36 − 5 b36/5c = 1. Notice that b4 = 27 − 5 b27/5c = 2, and b5 = p=2 dp = 36, b5 (1)

(1)

(1)

(1)

b0 = 0 < b5 < b4 < b2 . Because of Theorem 3, the following (k, l, S, I) = (2, 6, {2, 4, 5}, {4, 2}) inequality is valid for X E where s0 = s6 = 0:        b  b4 (1) (1) (1) 2 1 1 1 s1 + x3 + x6 ≥ b4 − z2 − z4 + b2 − b4 − z2 α1 α1 = 2(6 − z21 − z41 ) + 2(1 − z21 ) = 14 − 4z21 − 2z41 . Similarly, the (k, l, S, I) = (2, 6, {2, 4, 5}, {5, 4}) inequality,        b  b5 (1) (1) (1) 4 1 1 1 1 1 s1 + x3 + x6 ≥ b5 − z2 − z4 − z5 + b4 − b5 − z2 − z 4 α1 α1 = (8 − z21 − z41 − z51 ) + (6 − z21 − z41 ) = 14 − 2z21 − 2z41 − z51 , the (k, l, S, I) = (2, 6, {2, 4, 5}, {5, 2}) inequality,        b  b5 (1) (1) (1) 2 1 1 1 1 s1 + x3 + x6 ≥ b5 − z2 − z4 − z5 + b2 − b5 − z2 α1 α1 = (8 − z21 − z41 − z51 ) + 3(1 − z21 ) = 11 − 4z21 − z41 − z51 ,

7

and the (k, l, S, I) = (2, 6, {2, 4, 5}, {5, 4, 2}) inequality,      b  b5 (1) (1) (1) 4 1 1 1 1 1 s1 + x3 + x6 ≥ b5 − z2 − z4 − z5 + b4 − b5 − z2 − z 4 α1 α1     b  (1) (1) 2 1 + b2 − b4 − z2 α1 =(8 − z21 − z41 − z51 ) + (6 − z21 − z41 ) + 2(1 − z21 ) = 16 − 4z21 − 2z41 − z51 , are also valid for X E where s0 = s6 = 0. In order to generate valid inequalities for X SML using the mixing procedure [9] and the pairing procedure [6], we aggregate equalities (14) from period k to period mi − 1, i ∈ I ⊆ S, and obtain

sk−1 +

m i −1 X

xp = bi + smi −1 .

(17)

p=k

P i −1 where bi = m p=k dp for i ∈ I. By definition Si ⊆ {k, . . . , mi − 1}. Therefore, relaxing xp , p ∈ Si , in (17) to its upper bound based on (15) and smi −1 in (17) to its lower bound based on (13) gives X

sk−1 +

X

xp + α1

zp1 ≥ bi ,

(18)

p∈Si

p∈{k,...,mi −1}\Si

Inequalities derived using the mixing procedure. Let iu = max {i : i ∈ I}. Then we get the following valid inequality for X SML : X

sk−1 +

xp + α1

X

zp1 ≥ bi ,

(19)

p∈Si

p∈{k,...,miu −1}\Si

P P for i ∈ I. Setting v := sk−1 + p∈{k,...,miu −1}\S xp and y i := p∈Si zp1 , inequality (19) becomes v + α1 y i ≥ bi , i ∈ I, which is of the same form as the defining inequalities of mixing set Q0 . Notice that v ∈ R+ , y i ∈ Z+ . Therefore, the type I mixed MIR inequality (2) written for X SML with K = I,

sk−1 +

X p∈{k,...,miu −1}\S

xp ≥

|I|  X

(1)

(1)

biq − biq−1

q=1



    X biq  − zp1  , α1 p∈Siq

is valid for X SML . We refer to inequality (20) as mixed (k, l, S, I) inequality. Example 1 (continued). For i ∈ S, inequalities (18) for X E are written as: For i = 2: s1 + x3 + 5z21 ≥ 4 For i = 4: s1 + x3 + 5(z21 + z41 ) ≥ 27 For i = 5: s1 + x3 + x6 + 5(z21 + z41 + z51 ) ≥ 38.

8

(20)

As discussed above, we apply the mixing procedure on these inequalities. For I = {4, 2}, we get the following valid mixed (k, l, S, I) = (2, 6, {2, 4, 5}, {4, 2}) inequality for X E where iu = 4:        b  b4 (1) (1) (1) 2 1 1 1 − z2 − z4 + b2 − b4 − z2 s1 + x3 ≥ b4 α1 α1 = 2(6 − z21 − z41 ) + 2(1 − z21 ) = 14 − 4z21 − 2z41 . Likewise, the mixed (k, l, S, I) = (2, 6, {2, 4, 5}, {5, 4}) inequality where iu = 5,        b  b5 (1) (1) (1) 5 1 1 1 1 1 s1 + x3 + x6 ≥ b5 − z2 − z4 − z5 + b4 − b5 − z2 − z 4 α1 α1 = (8 − z21 − z41 − z51 ) + (6 − z21 − z41 ) = 14 − 2z21 − 2z41 − z51 , the mixed (k, l, S, I) = (2, 6, {2, 4, 5}, {5, 2}) inequality where iu = 5,        b  b5 (1) (1) (1) 2 1 1 1 1 s1 + x3 + x6 ≥ b5 − z2 − z4 − z5 + b2 − b5 − z2 α1 α1 = (8 − z21 − z41 − z51 ) + 3(1 − z21 ) = 11 − 4z21 − z41 − z51 , and the mixed (k, l, S, I) = (2, 6, {2, 4, 5}, {5, 4, 2}) inequality where iu = 5,      b  b5 (1) (1) (1) 4 1 1 1 1 1 s1 + x3 + x6 ≥ b5 − z2 − z4 − z5 + b4 − b5 − z2 − z 4 α1 α1     b  (1) (1) 2 1 + b2 − b4 − z2 = 16 − 4z21 − 2z41 − z51 , α1 are also valid for X E . Inequalities derived using the pairing procedure. First,P for each i ∈ S, we applyP the MIR procedure on inequalities (18) as follows. By setting v := sk−1 + p∈{k,...,mi −1}\Si xp , y := p∈Si zp1 , and β = bi , Inequality (18) becomes v+α1 y ≥ β which is of the same form as the defining inequalities of set Q. Notice that v ∈ R+ and y ∈ Z+ . Therefore, the MIR inequality (1) written for inequality (18),     X X bi  (1) (1) sk−1 + xi + bi  zp1 − ≥ bi , (21) α1 p∈Si

p∈{k,...,mi −1}\Si

is valid for X SML . Next, we apply sequential pairing procedure P (discussed in Section 2.4) on inequalities (21). However, to do so, we have to assume that p∈Si zp1 ≥ bbi /α1 c for each i ∈ S. The sequentially paired inequalities obtained by arbitrary sequence of pairing of pairing of the (1) inequalities (21) for i ∈ I, in increasing order of bi , are valid for X SML and we refer to thus obtained inequalities as the paired (k, l, S, I) inequality for SMLS-WB.

3.2

Facets for SMLS-WB Problem

We provide sufficient conditions under which the (k, l, S, I) inequalities (16), the mixed (k, l, S, I) inequalities (20), and the paired (k, l, S, I) inequalities are facet-defining for X SML . Theorem 4. The (k, l, S, I) inequality (16) is either same as the mixed (k, l, S, I) inequality (20) or is dominated by the mixed (k, l, S, I) inequality (20). 9

Proof. Observe that the mixed (k, l, S, I) inequality (20) is actually the (k, miu − 1, S, I) inequality (16). This means that if l = miu −1, then the (k, l, S, I) inequality is the mixed (k, l, S, I) inequality. However, in case l 6= miu − 1, the left hand side of inequality (16) is greater than the left hand side of inequality (20), which implies that the (k, l, S, I) inequality is dominated by the mixed (k, l, S, I) inequality. Corollary 1. The (k, l, S, I) inequality (16) is valid for X SML for all 1 ≤ k ≤ l ≤ m and S ⊆ {k, . . . , l}. Corollary 2. The (k, l, S, I) inequality defines a facet for conv(X SML ) if and only if it is a facetdefining mixed (k, l, S, I) inequality. Example 1 (continued). Notice that the mixed {2, 6, {2, 4, 5}, {4, 2}} inequality dominates the {2, 6, {2, 4, 5}, {4, 2}} inequality because l = 6 6= miu − 1 = m4 − 1 = 4; whereas for (k, l, S) = (2, 6, {2, 4, 5}) and I ∈ {{5, 4}, {5, 2}, {5, 4, 2}}, the mixed (k, l, S, I) inequalities are same as the (k, l, S, I) inequalities as l = miu − 1 = m5 − 1 = 6. Lemma 1. Given a mixed (k, l, S, I) inequality (20), any feasible point (ˆ z, x ˆ, sˆ) ∈ X SML lies on the corresponding face if   X biq zˆp1 = 1, for q = 1, . . . , r (i) − α1 p∈Siq

 (ii)



X biq − zˆp1 = 0, for q = r + 1, . . . , |I| α1 p∈Siq

(iii) sˆk−1 +

P

(1)

p∈{k,...,miu −1}\S

x ˆp = bir (1)

for some r ∈ {0, 1, . . . , |I|} where bi0 = 0. Proof. Given (k, l, S, I), the hyperplane corresponding to (20) can be rewritten as   |I|   b  X X X i (1) (1) sk−1 + xp = biq − biq−1  q − zp1  α1 p∈{k,...,miu −1}\S

q=1

(22)

p∈Siq

Let (ˆ z, x ˆ, sˆ) ∈ X SML such that conditions (i) − (iii) hold and Γ = {(z, x, s) ∈ X SML : (22) holds} be the face of X SML defined by hyperplane (22). Now, it is easy to show that (ˆ z, x ˆ, sˆ) ∈ Γ as it satisfies (22). Theorem 5. The mixed (k, l, S, I) inequality (20) defines a facet for conv(X SML ) if the following condition holds for each i ∈ I: {j : j ∈ S, j < i, dbj /α1 e = dbi /α1 e} ⊂ I.

(23)

Proof. Let Γ = {(z, x, s) ∈ X SML : (22) holds} be the face of X SML defined by hyperplane (22) corresponding to the mixed (k, l, S, I) inequality (20). Assuming that for each i ∈ I, {j : j ∈ S, j < i, dbj /α1 e = dbi /α1 e} ⊂ I, we prove that a generic hyperplane passing through Γ, ν0 s0 +

m X

(νp sp + µp xp + λzp zp1 ) = η

p=1

10

(24)

where (λz1 , . . . , λzm , µ1 , . . . , µm , ν0 , ν1 , . . . , νm , η) ∈ R3m+2 , must be a scalar multiple of (22). First, we separately add equalities (14) from period i ∈ {1, . . . , k − 1} to period k − 1, and from period k to period i ∈ {k, . . . , m} to get si−1 +

k−1 X

xp =

p=i

sk−1 +

i X

k−1 X

dp + sk−1 ,

for i = 1, . . . , k − 1

(25)

dp + si ,

for i = k, . . . , m

(26)

p=i

xp =

p=k

i X p=k

respectively. Since each point belonging to Γ satisfies (25) and (26), we eliminate variables si , i = 0, . . . , k−2, k, . . . , m, from (24) by subtracting νi−1 times equality (25) for each i ∈ {1, . . . , k−1} from (24), and adding νi times equality (26) for each i ∈ {k, . . . , m} to (24). This gives λ0 sk−1 +

m X

λxp xp +

p=1

m X

λzp zp1 = θ

(27)

p=1

where θ=η+

k−1 X

 νi−1

k−1 X

i=1

λ0 = νk−1 +

dp  −

p=i k−1 X

k−1 X

m X

 νi

i=k

νi−1 −

i=1

λxp = µp −



m X

i X

 dp  ,

p=k

νi ,

i=k

νi−1 for p = 1, . . . , k − 1,

i=1

λxp = µp +

m X

νi

for p = k, . . . , m.

i=k 1 ,x ˆ1 , . . . , x ˆm , sˆ0 , . . . , sˆm ) ∈ Zm It is important to note that to have a point (ˆ z, x ˆ, sˆ) = (ˆ z11 , . . . , zˆm + × m+1 m SML 1 1 R+ ×R+ ∈ X , it is sufficient to know the value of zˆ1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , and sˆk−1 coordinates because the remaining ones can be obtained using equalities (25) and (26). Therefore, in the rest of 1 ,x the proof, we will define a point belonging to X SML by (ˆ z, x ˆ, sˆk−1 ) = (ˆ z11 , . . . , zˆm ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ m m Z+ × R+ × R+ (for the sake of convenience).

Next, assuming S := {w1 , . . . , w|S| } where w1 = k, consider the point A = (ˆ z, x ˆ, sˆk−1 ) = 1 ,x m × R such that s (ˆ z11 , . . . , zˆm ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zm × R ˆ = 0, + k−1 + + ( (ddp /α1 e , dp ) if p ∈ {1, . . . , m}\{k, . . . , l}, (ˆ zp1 , x ˆp ) := (0, 0) if p ∈ {k, . . . , l}\S, for p ∈ {1, . . . , m}\S and for i ∈ {1, . . . , |S|}, 1 (ˆ zw ,x ˆwi ) i

 :=

bwi α1





     bwi−1 bwi−1 bwi − , α1 − α1 , α1 α1 α1

where bw0 = 0. As mentioned before, coordinates sˆp , p ∈ {0, . . . , m}\{k − 1} can be obtained using (25)-(26), i.e. for p ∈ {0, . . . , m}\{k − 1}, 11

  0 if p ∈ {0, . . . , k − 2}     X  p   bwi  − dj if p ∈ {mwi−1 , . . . , mwi − 1} and i = 1, . . . , |S|, α1 sˆp := α1 j=k      bw|S|    − bw|S| if p ≥ mw|S| , α1 α1 Pmw −1 where mw0 = k. Recall that bwi = j=ki dj for wi ∈ S (by definition). It is easy to verify that A ∈ X SML and satisfies conditions (i) − (iii) of Lemma 1. Therefore A ∈ Γ and hence must satisfy (27). Substituting A into (27) gives X X λzp ddp /α1 e λxp dp + p∈{1,...,m}\{k,...,l}

p∈{1,...,m}\{k,...,l}

+

|S| X

(28) λxwi α1 + λzwi






dbwi /α1 e − bwi−1 /α1




= θ.

i=1

Using (28), hyperplane (27) reduces to X λ0 sk−1 + λxp (xp − dp ) + p∈{1,...,m}\{k,...,l}

X

λzp zp1 − ddp /α1 e




p∈{1,...,m}\{k,...,l} |S|

+

X


X x  
λxp xp + λzp zp1 + λwi (xwi − α1 dbwi /α1 e − bwi−1 /α1 )

=

|S| X

(29)

i=1

p∈{k,...,l}\S

 
1 λzwi dbwi /α1 e − bwi−1 /α1 − zw . i

i=1

Now, let S be a set of all disjoint subsets of S such that for each element (or time period) p of a subset, dbp /α1 e is same and the elements of each subset are arranged in the increasing order of the associated time period or index p. We denote a set containing only first element of the disjoint subsets of S (arranged in increasing order) by Ω := {ω1 , ω2 , . . . , ω|Ω| }. Note that dbω0 /α1 e = 0 < l m dbω1 /α1 e < dbω2 /α1 e < . . . < bω|Ω| /α1 and Ω ⊆ S. For each ωr ∈ Ω, we consider the points m 1 ,x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zm B ωr = (ˆ z, x ˆ, sˆk−1 ) = (ˆ z11 , . . . , zˆm + × R+ × R+ such that each coordinate of ω r B is same as the coordinates of the point A, except that          bωr−1 bωr−1 bωr bωr (1) 1 − , α1 − α1 + bω , (ˆ zωr , x ˆωr ) = α1 α1 α1 α1 (1)

(1)

where bω0 = 0 and bω = maxi=1,2,... {bωi }. It is easy to verify that B ωr ∈ X SML and satisfies conditions (i) − (iii) of Lemma 1. Therefore B ωr ∈ Γ and hence must satisfy (27). Substituting B ωr into (29) gives λxωr = 0

for ωr ∈ Ω.

(30)

Next, consider the points C1r for r ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, C2r for r ∈ {1, . . . , k − 2} ∪ {l + 1, . . . , m − 1}, and C3 such that each coordinate of C1r , C2r , and C3 are same as the coordinates of

12

the point A, except that in C1r , zˆr1 = ddp /α1 e + 1, in C2r , & P ' k−1  k−1  X d  i i=p   , di  if p ∈ {1, . . . , k − 1} ∩ {r},   α  1  i=p    ' & 1 Pm (ˆ zp , x ˆp ) := m X  i=p di    , di  if p ∈ {l + 1, . . . , m} ∩ {r},   α1   i=p   (0, 0) if p ∈ {1, . . . , k − 1, l + 1, . . . , m}, p > r, and in C3 , x ˆk−1 = 0 and sˆp = dk−1 for p = 0, . . . , k − 2. It is easy to verify that C1r for r ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, C2r for r ∈ {1, . . . , k − 2} ∪ {l + 1, . . . , m − 1}, and C3 belong to X SML and satisfy conditions (i) − (iii) of Lemma 1. Therefore C1r , C2r , C3 ∈ Γ and hence must satisfy (27). Substituting C1r into (29) gives λzr = 0 for all r ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}.

(31)

Likewise, one by one substituting the points C21 , . . . , C2k−2 , C2l+1 , . . . , C2m−1 , and C3 into (29) gives λxr = 0 for all r ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m − 1}. (1)

(32) (1)

(1)

By definition we know that I := {i1 , . . . , i|I| } ⊆ S such that 0 < bi1 ≤ bi2 ≤ . . . ≤ bi|I| ≤ α1 , iu = max {i : i ∈ I}, and Si = S ∩ {k, . . . , i}. We consider the points Dr = (ˆ z, x ˆ, sˆk−1 ) = 1 1 m m (ˆ z1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Z+ × R+ × R+ for r ∈ {k, . . . , l}\Siu such that sˆk−1 = 0,   (ddp /α1 e , dp ) 1 (ˆ zp , x ˆp ) := (0, 0)   (1, 0)

if p ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, if p ∈ {k, . . . , l}\S and p 6= r, if p ∈ {k, . . . , l}\S and p = r,

for p ∈ {1, . . . , m}\S and         bwi−1 bwi−1 bwi bwi   − , α1 − α1 if wi ∈ S\{r},   α1 α1 α1 α1  1      (ˆ zw ,x ˆwi ) :=  b   b i bwi−1 bwi wi−1 wi   − + 1, α − α if wi ∈ S ∩ {r}  1 1  α1 α1 α1 α1  ∩{miu , . . . , l}, for i = 1, . . . , |S|. It is easy to verify that Dr ∈ X SML and satisfies conditions (i)−(iii) of Lemma 1. Therefore Dr ∈ Γ and hence must satisfy (27). Substituting Dr into (29) gives λzr = 0 for all r ∈ {k, . . . , l}\Siu .

(33)

Using (30), (31), (33), and (32), (29) reduces to      bwi−1 bwi λ0 sk−1 + + xwi − α1 − α1 α1 p∈{k,...,l,m}\S      X bwi−1 bwi 1 = λzwi − − zw . i α1 α1 X

λxp xp

X

λxwi wi ∈S\Ω

wi ∈Siu

13

(34)

1 ,x m Now, consider the points E r = (ˆ z, x ˆ, sˆk−1 ) = (ˆ z11 , . . . , zˆm ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zm + × R+ × R+ for r ∈ {miu , . . . , l, m} where miu ≤ l ≤ m such that sˆk−1 = 0,   if p ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}\{r},  (ddp /α1 e , dp )  (dd /α e + 1, d + α ) if p ∈ {r} ∩ {m} and m 6= l, p 1 p 1 (ˆ zp1 , x ˆp ) :=  (0, 0) if p ∈ {k, . . . , l}\(S ∪ {r}),    (1, 1) if p ∈ {r} ∩ ({miu , . . . , l}\S),

for p ∈ {1, . . . , m}\S and      bwi−1 bwi  1  − , α1 zwi ,   α1 α1 1 (ˆ zwi , x ˆwi ) :=       bwi−1 bwi  1  − + 1, α1 zwi ,  α1 α1

if wi ∈ S\{r} if wi ∈ {r} ∩ {miu , . . . , l} ∩ S.

for i ∈ {1, . . . , |S|}. It is easy to verify that E r ∈ X SML and satisfies conditions (i) − (iii) of Lemma 1. Therefore E r ∈ Γ and hence must satisfy (27). Substituting E r into (29) gives λxr = 0 for all r ∈ {miu , . . . , l} ∪ {m}.

(35)

Using (35), (34) reduces to X

λ0 sk−1 +

λxp xp

+

λxwi wi ∈Siu \Ω



 1 − zw . i

p∈{k,...,miu −1}\S

=

X

λzwi



wi ∈Siu

bwi α1



 −

X

bwi−1 α1

     bwi−1 bwi xwi − α1 − α1 α1 (36)

Let Ia := {ia1 , . . . , ia|I| } be a set which has same elements as in the set I, except that ia1 < ia2 < . . . < ia|I| . In other words, set Ia is same as the set I, the only difference is that the elements of Ia are arranged in the increasing order of time periods. Now, for each iaζ ∈ I where ζ ∈ z, x ˆ, sˆk−1 ) = {1, . . . , |I|} and r ∈ Siaζ \(Siaζ−1 ∪ {iaζ }) where Sia0 = ∅, consider the points F r,aζ = (ˆ 1 1 m m (ˆ z1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Z+ × R+ × R+ such that sˆk−1 = 0, ( (ddp /α1 e , dp ) if p ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, (ˆ zp1 , x ˆp ) := (0, 0) if p ∈ {k, . . . , l}\S, for p ∈ {1, . . . , m}\S and      bwi−1 bwi 1   − , α1 zwi if wi ∈ S\{r, iaζ },     α1   α1     b bwi−1 wi 1 1 (ˆ zw , x ˆ ) := − + 1, α z if wi = r, wi 1 wi i  α1 α1           bwi−1 bwi  1  − − 1, α1 zwi if wi = iaζ , α1 α1 l m P for i = 1, . . . , |S|. Note that p∈Sia zˆp1 = biaζ /α1 for all iaζ ∈ I, and also in case dbwi /α1 e = ζ   bwi−1 /α1 for wi = iaζ ∈ I and wi−1 ∈ S then wi−1 = iaζ−1 where iaζ−1 ∈ I (because of Condition 14

(23)), and hence Siaζ \(Siaζ−1 ∪{iaζ }) = ∅. It is easy to verify that F r,aζ ∈ X SML under assumptions (23) and satisfies conditions (i)−(iii) of Lemma 1. Therefore F r,aζ ∈ Γ and  hence must satisfy (27). Next, for iaζ ∈ I where ζ ∈ {1, . . . , |I|} and r ∈ Siaζ \(Siaζ−1 ∪ {iaζ }) ∩ (Siu \Ω) where u = a|I| , r,a

1 ,x m z, x ˆ, sˆk−1 ) = (ˆ z11 , . . . , zˆm ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zm we define another point F1 ζ = (ˆ + × R+ × R+ whose coordinates are all exactly same as F r,aζ except that for wi = r,  
x ˆwi = α1 dbwi /α1 e − bwi−1 /α1 + b(1) w , (1)

r,a

(1)

where bw = maxwi ∈S {bwi }. Again, it is easy to verify that F1 ζ ∈ X SML and satisfies conditions r,a (i) − (iii) of Lemma 1. Therefore F1 ζ ∈ Γ and hence must satisfy (27). For each iaζ ∈ I and   r,a r ∈ Siaζ \(Siaζ−1 ∪ {iaζ }) ∩ (Siu \Ω), we substitute F r,aζ and F1 ζ into (36), and subtract one equality from the other. This gives λxr = 0 for r ∈ Siu \(I ∪ Ω).

(37)

  Again, by substituting the points F r,aζ , for iaζ ∈ I where ζ ∈ {1, . . . , |I|} and r ∈ Siaζ \(Siaζ−1 ∪ {iaζ }) , into (36) and using (37), we get λzr = λzia

ζ

for iaζ ∈ I, r ∈ Siaζ \(Siaζ−1 ∪ {iaq }).

(38)

Substituting (37) and (38) into (36) gives      bwi−1 bwi λ0 sk−1 + + xwi − α1 − α1 α1 p∈{k,...,miu −1}\S   & ' ' & |I| X X X b b iaζ−1  iaζ  λzia  = − zp1 − + zp1  , ζ α1 α1 X

λxp xp

X

λxwi wi ∈I\Ω

p∈Sia

ζ=1

p∈Sia

ζ

(39)

ζ−1

where bia0 = 0 and Sia0 = ∅. By rearranging the right-hand side of (39), we get      bwi−1 bwi λ0 sk−1 + + xwi − α1 − α1 α1 p∈{k,...,miu −1}\S     ' ' & & |I|−1   X  X X   bia|I|  biaζ = − zp1  + λzia  − zp1  . λzia − λzia  ζ ζ+1 |I| α1 α1 X

λxp xp

X

λxwi wi ∈I\Ω

p∈Sia

ζ=1

ζ

p∈Sia

|I|

Furthermore, the last equation can also be simplified to      X X bwi−1 bwi x x λ0 sk−1 + λ p xp + λwi xwi − α1 − α1 α1 p∈{k,...,miu −1}\S wi ∈I\Ω     |I| X X z  biq = γ iq − zp1  , α1 q=1

where for q = 1. . . . , |I|,

(40)

p∈Siq

γizq

=



λziq

−

λzia ζ+1



such that aζ = q for ζ ∈ {1, . . . , |I|} and λza|I|+1 = 0. 15

(1)

m Next, for iq ∈ I, consider the points G q = (ˆ z, x ˆ, sˆk−1 ) ∈ Zm ˆk−1 = biq , + × R+ × R+ , such that s

(ˆ zp1 , x ˆp )

( (ddp /α1 e , dp ) := (0, 0)

if p ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, if p ∈ {k, . . . , l}\S,

1 ,x 1 ) where χ for p ∈ {1, . . . , m}\S, and for i ∈ {1, . . . , |S|}, (ˆ zw ˆwi ) := (χwi − χwi−1 , α1 zˆw w0 = 0 and i i   dbwi /α1 e if wi ∈ S\I, χwi = bbwi /α1 c if wi ∈ {i1 , . . . , iq },   dbwi /α1 e if wi ∈ {iq+1 , . . . , i|I| }.

This implies that     bwi bwi   , α1  α1 α1          b  X X bwi wi 1   zˆp , x ˆp = , α1  α1 α1  p∈Swi p∈Swi         bwi bwi   , α1 α1 α1

if wi ∈ S\I, if wi ∈ {i1 , . . . , iq }, if wi ∈ {iq+1 , . . . , i|I| }.

  Since we assume that {j : j ∈ S, j < iq , dbj /α1 e = biq /α1 } ⊂ I for each iq ∈ I (Condition (23)), χwi ≥ χwi−1 for i = 1, . . . , |S|. Now, it is easy to verify that G q ∈ X SML and satisfies conditions (i) − (iii) of Lemma 1. Therefore G q ∈ Γ and hence must satisfy (27). Furthermore, because of our assumption (23), for wr ∈ I\Ω,     bwr−1 bwr (1) = , wr−1 ∈ I, and b(1) (41) wr−1 < bwr . α1 α1 Let Ξ := {ξ1 , . . . , ξ|Ξ| } = I\Ω := {ij1 , . . . , ij|Ξ| } such that ξ1 = ij1 < ξ2 = ij2 < . . . < ξ|Ξ| = ij|Ξ| . In other words, jr provides the positioning of ξr in the set I. For ijr ∈ I\Ω, we define another point (1) m G1r = (ˆ z, x ˆ, sˆk−1 ) ∈ Zm ˆk−1 = bij −1 where ijv = min{ij|Ξ| , ij|Ξ|−1 , . . . , ijr }, + × R+ × R+ , such that s v

( (ddp /α1 e , dp ) (ˆ zp1 , x ˆp ) := (0, 0)

if p ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, if p ∈ {k, . . . , l}\S,

for p ∈ {1, . . . , m}\S, and for i ∈ {1, . . . , |S|},   1  χ ¯0wi − χ ¯0wi−1 , α1 zw i 1  (ˆ zw ,x ˆwi ) :=  i (1)  χ ¯0wi − χ ¯0wi−1 , maxw∈S {bw }

if wi 6= ijr , if wi = ijr ,

where χ ¯0w0 = 0 and χ ¯0wi

  dbwi /α1 e = bbwi /α1 c   dbwi /α1 e

if wi ∈ S\I, if wi ∈ {i1 , . . . , ijv −1 }, if wi ∈ {ijv , . . . , i|I| }. 16

Notice that for i = 1, . . . , |S|,     bwi bwi   , α1 if wi ∈ S\I,   α1 α1        bwi bwi      , α1 if wi ∈ {i1 , . . . , ijv −1 },   X X α1 α1 1   x ˆp =  zˆp ,      bwi , α bwi p∈Swi p∈Swi  if wi ∈ {ijv , . . . , i|I| }\{ijr },  1  α1 α1            bwi bwi (1)   , α1 + max {bw } if wi = ijr . w∈S α1 α1 It is easy to verify that G1r ∈ X SML under conditions (23) and satisfies conditions (i) − (iii) of Lemma 1. Therefore G1r ∈ Γ and hence must satisfy (27). For each ijr ∈ I\Ω, we substitute G jv −1 , G1r into (40) and subtract one equality from the other. This gives λxijr = 0 for ijr ∈ I\Ω.

(42)

Using (42), we one by one substitute G 1 , G 2 , . . . , G |I| into (40) and get (1)

γiz1 = λ0 bi1 , γiz2 =

(1) λ0 bi2

(43)

− γiz1 = λ0



(1) bi2

−

(1) bi1



,

(44)

.. .     (1) (1) (1) γiz|I| = λ0 bi|I| − γiz1 + . . . + γiz|I|−1 = λ0 bi|I| − bi|I|−1 .

(45)

This implies that for q ∈ {1, . . . , |I|},   (1) (1) γizq = λ0 biq − biq−1

(46)

(1)

where bi0 = 0. For r ∈ {k, . . . , miu − 1}\S, consider the points Hr whose coordinates are same as (1)

the coordinates of G u except that sˆk−1 = 0 and (ˆ zr1 , x ˆr ) = (1, biu ). By substituting the points Hr into the (40) and subtracting from equality corresponding to G u , we get λxr = λ0 for r ∈ {k, . . . , miu − 1}\S.

(47)

Substituting (42), (46), and (47) into the (40) gives λ0 sk−1 +

X p∈{k,...,miu −1}\S

λ 0 xp =

|I| X



(1)

(1)

λ0 biq − biq−1

q=1



    X biq  − zp1  α1

(48)

p∈Siq

P SML ) if The identity (48) is λ0 = µk−1 + m i=k−1 ηi times (22). Hence Γ defines a facet for conv(X conditions (23) hold. This completes the proof. Example 1 (continued). According to Theorem 5, for (k, l, S) = (2, 6, {2, 4, 5}) and I ∈ {{5, 4}, {5, 2}, {4, 2}, {5, 4, 2}}, the mixed (k, l, S, I) inequalities are facet-defining for X E because in these inequalities for each i ∈ I, {j : j ∈ S, j < i, dbj /α1 e = dbi /α1 e} = ∅. 17

  P Theorem 6. Assuming that p∈Siq zp1 ≥ biq /α1 for each iq ∈ I, the paired (k, l, S, I) inequality is either dominated by the mixed (k, l, S, I) inequality (20) or is same as the mixed (k, l, S, I) inequality (20). Proof. First we repeatedly apply sequential pairing procedure on inequalities (21) for i ∈ I, in (1) increasing order of bi , to get ((. . . ((i1 ◦ i2 ) ◦ i3 ) ◦ . . .) ◦ i|I| ) sequentially paired inequality. The inequality thus obtained, referred to as the ((. . . ((i1 ◦i2 )◦i3 )◦. . .)◦i|I| ) sequentially paired (k, l, S, I) inequality, is same as the mixed (k, l, S, I) inequality (20). Moreover, the set of valid inequalities (21) for i ∈ I belong to the nested case. Therefore, according to Theorem 2, the mixed (k, l, S, I) inequality or the ((. . . ((i1 ◦ i2 ) ◦ i3 ) ◦ . . .) ◦ i|K| ) sequentially paired (k, l, S, I) inequality is at least as strong as any other sequentially paired inequality obtained by arbitrary sequence of pairing of the inequalities (21) for i ∈ I. Corollary 3. Assuming that

1 p∈Siq zp

P

  ≥ biq /α1 for each iq ∈ I, the paired (k, l, S, I) inequality

defines a facet for conv(X SML ) if and only if it is a facet-defining mixed (k, l, S, I) inequality.

4.

Multi-Module Lot-Sizing Problem without Backlogging

In this section, we redefine the multi-module capacitated lot-sizing (MMLS) problem (introduced in [15]). Then we present valid inequalities for this problem and investigate their facet-defining properties. The MMLS without backlogging (MMLS-WB) is defined as follows: Let {α1 , . . . , αn } be the set of sizes of the n available capacity modules and the setup cost per module of size αt , t = 1, . . . , n in period p is denoted by fpt . We formulate MMLS-WB as: ! X  n X t t MML min cp xp + hp sp + fp zp : (z, x, s) ∈ X t=1

p∈P

where  m+1 X MML := (z, x, s) ∈ Zm×n × Rm + × R+ + sp−1 + xp = dp + sp , n X xp ≤ αt zpt ,

(49)

p∈P

(50)

p ∈ P = {1, . . . , m} .

(51)

t=1

zpt

Here, is the number of capacity modules of size αt , t = 1, . . . , n, used in period p and parameters (dp , cp , hp ) and variables (xp , sp ) are same as defined for SMLS-WB. Notice that for n = 1, MMLS-WB reduces to SMLS-WB.

4.1

Valid Inequalities for MMLS-WB Problem

In order to generate valid inequalities for MMLS-WB problem using n-mixing procedure [15] and the pairing procedure [6], we use notations defined in Section 3.1 except that I := {i1 , i2 , . . . , i|I| } ⊆ S (n)

(n)

(n)

(n)

such that bi0 = 0 < bi1 ≤ bi2 ≤ . . . ≤ bi|I| < αn . Also, we assume that the n-step MIR conditions hold, i.e. for each i ∈ I, l m (t−1) αt bi /αt ≤ αt−1 , for t = 2, . . . , n, (52) 18

which are automatically satisfied when the sizes of the capacity modules are divisible, i.e. αt /αt−1 ∈ Z+ , t = 2, . . . , n. First, we aggregate equalities (50) from period k to period mi − 1, i ∈ S, and then relax xp , p ∈ Si , to its upper bound based on (51) and smi − 1 to its lower bound based on (49). This gives n X X X sk−1 + xi + αt zpt ≥ bp . (53) t=1

p∈{k,...,mi −1}\Si

p∈Si

Inequalities derived using the n-mixing procedure. Assuming that iu = max {i : i ∈ I}, we get the following valid inequality for X MML : sk−1 +

X

xi +

n X t=1

p∈{k,...,miu −1}\Si

αt

X

zpt ≥ bp .

(54)

p∈Si

P P Setting v := sk−1 + p∈{k,...,miu −1}\S xp and yti := p∈Si zpt , t = 1, . . . , n, inequality (54) becomes P v + nt=1 αt yti ≥ bi , i ∈ I, which is of the same form as the defining inequalities of n-mixing set i Qm,n 0 . Notice that v ∈ R+ , yt ∈ Z+ , t = 1, . . . , n. Therefore, the type I mixed n-step MIR inequality MML (5) written for X with K = I, (55)

    (ρ−1) (ρ−1) n n n Y X X Y b b  iq  iq   t φniq (z) :=  αρ  −  α ρ  zp ,  p∈Siq t=1 ρ=t+1   ρ=1 

(56)

X

xp ≥

q=1

p∈{k,...,miu −1}\S

where

|I|  X

 (n) (n) biq − biq−1 φniq (z)

sk−1 +

is valid for X MML if the n-step MIR conditions (52) hold. In the rest of the paper, we will refer to the inequality (55) as the n-mixed (k, l, S, I) inequality. Inequalities derived using the pairing procedure. Similar to SMLS-WB, first, we apply the n-step MIR procedure [10] on inequality (53) for all i ∈ S, which gives the following valid inequalities if the n-step MIR conditions hold: X (n) sk−1 + xp ≥ bi φni (z) (57) p∈{k,...,mi −1}\S

Assuming that φni (z) ≤ 1 for all i ∈ K, we repeatedly apply sequential pairing procedure on inequalities (57) for i ∈ K, in increasing order of i, to get ((. . . ((i1 ◦ i2 ) ◦ i3 ) ◦ . . .) ◦ i|K| ) sequentially paired inequality, which is same as inequality (55).

4.2

Facets for MMLS-WB Problem

We provide conditions under which the n-mixed (k, l, S, I) inequality (55) is facet-defining for the conv(X MML ). Lemma 2. Given the n-mixed (k, l, S, I) inequality (55), any feasible point (ˆ z, x ˆ, sˆ) ∈ X MML lies on the corresponding face if

19

(i) φniq (ˆ z ) = 1, for q = 1, . . . , r z ) = 0, for q = r + 1, . . . , |I| (ii) φniq (ˆ P (n) (iii) sˆk−1 + p∈{k,...,miu −1}\S x ˆp = bir (n)

for some r ∈ {0, 1, . . . , |I|} with bi0 = 0. Proof. Given (k, l, S, I), the hyperplane corresponding to (55) can be rewritten as X

sk−1 +

xp =

|I|  X

(n) biq

−

(n) biq−1



φniq (z)

(58)

q=1

p∈{k,...,miu −1}\S

Let (ˆ z, x ˆ, sˆ) ∈ X MML such that conditions (i) − (iii) hold and Γ = {(z, x, s) ∈ X MML : (58) holds} be the face of X MML defined by hyperplane (58). Now, it is easy to show that (ˆ z, x ˆ, sˆ) ∈ Γ as it satisfies (58). Theorem 7. Assuming that the n-step MIR conditions (52) hold, the n-mixed (k, l, S, I) inequality (55) defines a facet for the convex hull of X MML if the following conditions hold for each wr ∈ S: & ' & (t−1) ' (t−1) bwr−1 bwr > , t = 2, . . . , n. (59) αt αt

Proof. Let Γ = {(z, x, s) ∈ X MML : (58) holds} be the face of X MML defined by hyperplane (58) corresponding to the n-mixed (k, l, S, I) inequality (55). Assuming that Conditions (59) hold, we prove that a generic hyperplane passing through Γ, ! m n X X t t ν0 s0 + νp sp + µp xp + λ p zp = η (60) p=1

t=1

where (λ11 , . . . , λ1m , . . . , λn1 , . . . , λnm , µ1 , . . . , µm , ν0 , ν1 , . . . , νm , η) ∈ Rmn+2m+2 , must be a scalar multiple of (58). First, we separately add equalities (50) from period i ∈ {1, . . . , k − 1} to period k − 1, and from period k to period i ∈ {k, . . . , m} to get si−1 +

k−1 X

xp =

p=i

sk−1 +

i X p=k

k−1 X

dp + sk−1 ,

for i = 1, . . . , k − 1

(61)

dp + si ,

for i = k, . . . , m

(62)

p=i

xp =

i X p=k

respectively. Since each point belonging to Γ satisfies (61) and (62), we eliminate variables s0 , . . . , sk−2 , sk , . . ., sm−1 , and sm from (60) by subtracting νi−1 times equality (61) for each i ∈ {1, . . . , k − 1} from (60), and adding νi times equality (62) for each i ∈ {k, . . . , m} to (60). This gives λ0 sk−1 +

m X

λxp xp +

p=1

m X n X p=1 t=1

20

λtp zpt = θ

(63)

where θ=η+

k−1 X

 νi−1

i=1

k−1 X

k−1 X

k−1 X i=1 m X

m X

 νi

i=k

νi−1 −

i=1

λxp = µp +

dp  −

p=i

λ0 = νk−1 + λxp = µp −



m X

i X

 dp  ,

p=k

νi ,

i=k

νi−1 for p = 1, . . . , k − 1, νi

for p = k, . . . , m.

i=k

It is important to note that to have a point (ˆ z, x ˆ, sˆ) = (ˆ z1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , sˆ0 , . . . , sˆm ) ∈ Zmn + × m+1 m MML 1 n R+ × R+ ∈ X where zˆp = (ˆ zp , . . . , zˆp ) for p = 1, . . . , m, it is sufficient to know the value of zˆpt and x ˆp for t = 1, . . . , n and p = 1, . . . , m, and sˆk−1 coordinates because the remaining coordinates can be obtained using equalities (61) and (62). Therefore, in the rest of the proof, we will define a m point belonging to X MML by (ˆ z, x ˆ, sˆk−1 ) = (ˆ z1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zmn + × R+ × R+ . Next, assuming S := {w1 , . . . , w|S| } where w1 = k, consider the point J = (ˆ z, x ˆ, sˆk−1 ) = m × R such that s (ˆ z1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zmn × R ˆ = 0, + k−1 + +     dp , 0, . . . , 0, d if p ∈ {1, . . . , m}\{k, . . . , l}, p 1 2 n α1 (ˆ zp , zˆp , . . . , zˆp , x ˆp ) :=  (0, 0, . . . , 0, 0) if p ∈ {k, . . . , l}\S, for p ∈ {1, . . . , m}\S and for i ∈ {1, . . . , |S|}, 1 2 n (ˆ zw , zˆw , . . . , zˆw ,x ˆwi ) i i i

 :=

bwi α1





     bwi−1 bwi−1 bwi − , 0, . . . , 0, α1 − α1 , α1 α1 α1

Pmw −1 where bw0 = 0. Recall that bwi = p=ki dp for wi ∈ S (by definition). It is easy to verify that J ∈ X MML . Also, since for all wi ∈ I ⊆ S, & ' & ' n n n (ρ−1) (ρ−1) Y X X Y bwi bwi n − zˆpt φwi (ˆ z) = αρ αρ ρ=1 p∈Swi t=1 ρ=t+1 & & ' ' n n (ρ−1) (ρ−1) Y X Y bwi bwi = − zˆp1 = 0, αρ αρ ρ=1

ρ=2

p∈Swi

the point J satisfies conditions (i) − (iii) of Lemma 2. Therefore, J ∈ Γ and hence must satisfy (63). Substituting J into (63) gives    X x 1 dp λp dp + λp α1 p∈{1,...,m}\{k,...,l} (64)     |S| X
b b wi−1 wi + λxwi α1 + λ1wi − = θ. α1 α1 i=1

21

Using (64), hyperplane (63) reduces to X λxp (xp − dp ) + λ0 sk−1 +

X

λ1p



zp1

p∈{1,...,m}\{k,...,l}

p∈{1,...,m}\{k,...,l}



dp − α1



|S|

+

X p∈{k,...,l}\S

+

m X n X

    
X x bwi−1 bwi x 1 1 − λp xp + λp zp + λwi xwi − α1 α1 α1

(65)

i=1

λtp zpt =

p=1 t=2

|S| X i=1

λ1wi



bwi α1



 −

bwi−1 α1



 1 − zw . i

m For each wr ∈ S, we consider the points Lwr = (ˆ z, x ˆ, sˆk−1 ) ∈ Zmn + × R+ × R+ such that each coordinate of Lwr is same as the coordinates of the point J , except that     bwr−1 bwr x ˆwr = α1 − + b(1) w , α1 α1 (1)

(1)

where bw = maxi=1,2,... {bwi }. Under assumptions (59) for t = 1, it is easy to verify that Lwr ∈ X MML and satisfies conditions (i) − (iii) of Lemma 2. Therefore Lwr ∈ Γ and hence must satisfy (63). Substituting Lwr into (65) gives λxwr = 0

for wr ∈ S.

(66)

Next, consider the points Mr1 for r ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, Mr2 for r ∈ {1, . . . , k − 2} ∪ {l + 1, . . . , m − 1}, and M3 such that each coordinate of Mr1 , Mr2 , and M3 are same as the coordinates of the point J , except that in Mr1 , zˆr1 = ddp /α1 e + 1, in Mr2 , & P ' k−1  k−1  X  i=p di   , di    α  1  i=p    & ' 1 Pm (ˆ zp , x ˆp ) := m X  i=p di    , di    α1   i=p   (0, 0)

if p ∈ {1, . . . , k − 1} ∩ {r},

if p ∈ {l + 1, . . . , m} ∩ {r}, if p ∈ {1, . . . , k − 1, l + 1, . . . , m}, p > r,

and in M3 , x ˆk−1 = 0 and sˆp = dk−1 for p = 0, . . . , k − 2. It is easy to verify that Mr1 for r ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, Mr2 for r ∈ {1, . . . , k − 2} ∪ {l + 1, . . . , m − 1}, and M3 belong to X MML and satisfy conditions (i) − (iii) of Lemma 2. Therefore Mr1 , Mr2 , M3 ∈ Γ and hence must satisfy (63). Substituting Mr1 into (65) gives λ1r = 0 for all r ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}.

(67)

l+1 m−1 Likewise, one by one substituting the points M12 , . . . , Mk−2 , and M3 into (65) 2 , M2 , . . . , M2 gives

λxr = 0 for all r ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m − 1}.

(68)

For r ∈ {1, . . . , k − 2} ∪ {l + 1, . . . , m − 1} and τ ∈ {2, . . . , n}, we consider point Ntr such that coordinates of Nτr are same as the coordinates of the point J , except that $ % & ' !   (τ −2) (τ −1) dr dr dr 1 τ −1 τ τ +1 n ,..., , , 0, . . . , 0 . (ˆ zr , . . . , zˆr , zˆr , zˆr , . . . , zˆr ) := α1 ατ −1 ατ 22

Since

n X

αt zˆrt

=

t=1

τ X

$ αt

t=1

(t−1)

dr αt

% + ατ = dr + ατ − d(t) r ≥ dr

(t) dr ,

as αr > it is easy to verify that Ntr for r ∈ {1, . . . , k − 2} ∪ {l + 1, . . . , m − 1} and t ∈ {2, . . . , n} belongs to X MML and satisfy conditions (i) − (iii) of Lemma 2. Therefore Ntr ∈ Γ and hence must satisfy (63). Now, one by one substituting the points N2r , . . . , Nnr into (65) and using (67) gives λtr = 0 for all r ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m − 1}, t ∈ {2, . . . , n}.

(69)

By definition of the n-mixed (k, l, S, I) inequality, we know that I := {i1 , . . . , i|I| } ⊆ S such (n)

(n)

(n)

(n)

that bi0 = 0 < bi1 ≤ bi2 ≤ . . . ≤ bi|I| < α1 , iu = max {i : i ∈ I}, and Si = S ∩ {k, . . . , i}. z, x ˆ, sˆk−1 ) = (ˆ z1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ For r ∈ {k, . . . , l}\Siu , we consider the points Or = (ˆ m × R such that s t = ... = z t = 0 for t = 2, . . . , n, Zmn × R ˆ = 0, z ˆ ˆ + k−1 + + m 1   (ddp /α1 e , dp ) 1 (ˆ zp , x ˆp ) := (0, 0)   (1, 0)

if p ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, if p ∈ {k, . . . , l}\S and p 6= r, if p ∈ {k, . . . , l}\S and p = r,

for p ∈ {1, . . . , m}\S and         bwi−1 bwi−1 bwi bwi   − , α1 − α1 if wi ∈ S\{r},   α1 α1 α1 α1  1      (ˆ zw ,x ˆwi ) :=  b   b i bwi−1 bwi wi−1 wi   − + 1, α − α if wi ∈ S ∩ {r}  1 1  α1 α1 α1 α1  ∩{miu , . . . , l}, for i = 1, . . . , |S|. It is easy to verify that Or ∈ X MML and satisfies conditions (i)−(iii) of Lemma 2. Therefore Or ∈ Γ and hence must satisfy (63). Substituting Or into (65) gives λ1r = 0 for all r ∈ {k, . . . , l}\Siu .

(70)

z, x ˆ, sˆk−1 ) = (ˆ z1 , . . . , zˆm , For r ∈ {k, . . . , l}\Siu and t ∈ {2, . . . , n}, we also consider the point Otr = (ˆ m × R whose coordinates are same as the coordinates of the point J , x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zmn × R + + + except that (ˆ zr2 , . . . , zˆrt−1 , zˆrt , zˆrt+1 , . . . , zˆrn ) := (0, . . . , 0, 1, 0, . . . , 0). It is easy to verify that Otr ∈ X MML and satisfies conditions (i) − (iii) of Lemma 2. Therefore Otr ∈ Γ and hence must satisfy (63). Substituting Otr into (65) and using (70) gives λtr = 0 for all r ∈ {k, . . . , l}\Siu and t ∈ {2, . . . , n}.

(71)

Using (66), (67), (68), (69), (70), and (71), (65) reduces to X λ0 sk−1 + λxp xp p∈{k,...,l,m}\S

=

X wi ∈Siu

λ1wi



bwi α1





bwi−1 − α1



23

−

1 zw i

 −

n X X wi ∈Siu t=2

t λtwi zw . i

(72)

m Now, consider the points P r = (ˆ z, x ˆ, sˆk−1 ) = (ˆ z1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zmn + × R+ × R+ for t t r ∈ {miu , . . . , l, m} where miu ≤ l ≤ m such that sˆk−1 = 0, zˆ1 = . . . = zˆm = 0 for t = 2, . . . , n,   (ddp /α1 e , dp ) if p ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}\{r},    (dd /α e + 1, d + α ) if p ∈ {r} ∩ {m} and m 6= l, p 1 p 1 (ˆ zp1 , x ˆp ) :=  (0, 0) if p ∈ {k, . . . , l}\(S ∪ {r}),    (1, 1) if p ∈ {r} ∩ ({miu , . . . , l}\S),

for p ∈ {1, . . . , m}\S and      bwi−1 bwi  1  − , α1 zwi ,   α1 α1 1 (ˆ zwi , x ˆwi ) :=       bwi−1 bwi  1  − + 1, α1 zwi ,  α1 α1

if wi ∈ S\{r} if wi ∈ {r} ∩ {miu , . . . , l} ∩ S.

for i ∈ {1, . . . , |S|}. It is easy to verify that P r ∈ X MML and satisfies conditions (i) − (iii) of Lemma 2. Therefore P r ∈ Γ and hence must satisfy (63). Substituting P r into (65) gives λxr = 0 for all r ∈ {miu , . . . , l} ∪ {m}.

(73)

Using (73), (72) reduces to X

λ0 sk−1 +

λxp xp

p∈{k,...,miu −1}\S

X

=

wi ∈Siu

λ1wi



bwi α1





bwi−1 − α1

 −

1 zw i

 −

n X X

t λtwi zw . i

(74)

wi ∈Siu t=2

Similar to the proof of Theorem 5, let Ia := {ia1 , . . . , ia|I| } be a set which has same elements as in the set I, except that ia1 < ia2 < . . . < ia|I| . In other words, set Ia is same as the set I, the only difference is that the elements of Ia are arranged in the increasing order of time periods. Now, for each iaζ ∈ I where ζ ∈ {1, . . . , |I|} and r ∈ Siaζ \(Siaζ−1 ∪ {iaζ }) where Sia0 = ∅, we consider the m ˆk−1 = 0, points Qr,aζ = (ˆ z, x ˆ, sˆk−1 ) = (ˆ z1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zmn + × R+ × R+ such that s t zˆp = 0 for p = 1, . . . , m, and t = 2, . . . , n, (ˆ zp1 , x ˆp )

( (ddp /α1 e , dp ) := (0, 0)

if p ∈ {1, . . . , k − 1} ∪ {l + 1, . . . , m}, if p ∈ {k, . . . , l}\S,

for p ∈ {1, . . . , m}\S and      bwi−1 bwi 1   − , α1 zwi   α1 α1       b   b wi−1 wi 1 1 (ˆ zwi , x ˆwi ) := − + 1, α1 zwi  α1 α1           bwi−1 bwi  1  − − 1, α1 zwi α1 α1

24

if wi ∈ S\{r, iaζ }, if wi = r, if wi = iaζ ,

  for i = 1, . . . , |S|. Since dbwi /α1 e > bwi−1 /α1 for wi = iaζ ∈ I ⊆ S (Condition (59)), it is easy to verify that Qr,aζ ∈ X MML . Also, this point satisfies conditions (i) − (iii) of Lemma 2 because for all iaζ ∈ I,  (ρ−1)   (ρ−1)  n n n bia bi Y X X Y  ζ −  aζ  zˆpt φnia (ˆ z) =   αρ   ζ  p∈Sia t=1 ρ=t+1  αρ  ρ=1  ζ  (ρ−1)   (ρ−1)  n n bi bia Y X Y  aζ   ζ − = zˆp1  αρ   αρ   ρ=2   p∈Sia ρ=1  ζ  (ρ−1)  &  (ρ−1)  ' n n bia bia Y Y biaζ ζ ζ     =  αρ  α1 = 0.  αρ  −     ρ=2 ρ=1 Therefore Qr,aζ ∈ Γ and hence must satisfy (63). By substituting the points Qr,aζ , for iaζ ∈ I   where ζ ∈ {1, . . . , |I|} and r ∈ Siaζ \(Siaζ−1 ∪ {iaζ }) , into (74), we get λ1r = λ1ia

ζ

for iaζ ∈ I, r ∈ Siaζ \(Siaζ−1 ∪ {iaq }).

(75)

m Next, we consider a point Jτ = (ˆ z, x ˆ, sˆk−1 ) = (ˆ z1 , . . . , zˆm , x ˆ1 , . . . , x ˆm , sˆk−1 ) ∈ Zmn + × R+ × R+ for τ ∈ {2, . . . , n}, such that sˆk−1 = 0,     dp , 0, . . . , 0, d if p ∈ {1, . . . , m}\{k, . . . , l}, p α1 (ˆ zp1 , zˆp2 , . . . , zˆpn , x ˆp ) :=  (0, 0, . . . , 0, 0) if p ∈ {k, . . . , l}\S,

for p ∈ {1, . . . , m}\S and for i ∈ {1, . . . , |S|}, x ˆwi := $ % $ (t−1) % (t−1)  bwi−1 b  wi   −   αt αt   & ' & (t−1) ' (t−1) t zˆw := bwi−1 bwi i  −   αt αt      0

Pτ

t ˆw t=1 αt z i

where

if t ∈ {1, . . . , τ − 1}

if t = τ if t ∈ {τ + 1, . . . , n},

Pmw −1 for t = 1, . . . , n. Recall that bw0 = 0 and bwi = j=ki dj for wi ∈ S (by definition). Under assumptions (59), it is easy to verify that Jτ ∈ X MML because for each wi ∈ S, $ % τ τ (t−1) X X X X bwi ) t + ατ = bwi − b(τ x ˆp = αt zˆp = αt wi + ατ ≥ bwi . αt p∈Swi

t=1 p∈Swi

t=1

25

Also, since for all iq ∈ I and τ ∈ {2, . . . , n},     (ρ−1) (ρ−1) n τ n Y X Y X b b −   iq  iq φniq (ˆ z) = zˆpt  αρ   αρ   t=1 ρ=t+1   p∈Siq ρ=1          (τ −1) (ρ−1)  (t−1)  (ρ−1) (ρ−1) τ −1 n n n X Y Y Y b b b b b     iq  −   iq   iq  iq  iq =     αρ   αρ  −  α α α t ρ τ  t=1 ρ=t+1     ρ=1  ρ=τ +1           (ρ−1) (ρ−1) (ρ−1) (ρ−1) τ −1 Y n n n n X Y Y Y biq biq biq biq           =  αρ  −  αρ  −  αρ  −  αρ          ρ=τ ρ=t t=1 ρ=t+1 ρ=1 = 0, the point Jτ satisfies conditions (i) − (iii) of Lemma 2. Therefore, Jτ ∈ Γ and hence must satisfy (63). Again, for each iaζ ∈ I where ζ ∈ {1, . . . , |I|}, r ∈ Siaζ \(Siaζ−1 ∪ {iaζ }) where Sia0 = ∅, r,a

and τ ∈ {2, . . . , n}, we consider the point Qτ ζ = (ˆ z, x ˆ, sˆk−1 ) = (z1 , . . . , zm , x1 , . . . , xm , sk−1 ) ∈ m × R whose coordinates are same as the coordinates of point J , except that Zmn × R + τ + +  & ' & (τ −1) ' ! τ (τ −1) X  b b w  w i−1 i t  − + 1, α t zw for wi = r,   i  ατ ατ t=1 τ (ˆ zwi , x ˆwi ) := & ' & (τ −1) ' ! τ (τ −1)  X  b b w w  i−1 i t  − − 1, αt zwi for wi = iaζ .   ατ ατ t=1



Since

 (τ −1)

bwi ατ r,aζ ∈ Qτ

 >

 (τ −1)

bwi−1 ατ

for wi = iaζ ∈ I ⊆ S (Condition (59) for t = τ ), it is easy to verify r,a

X MML and satisfies conditions (i) − (iii) of Lemma 2. Therefore Qτ ζ ∈ Γ and that r,a hence must satisfy (63). By substituting the points Qτ ζ , for iaζ ∈ I where ζ ∈ {1, . . . , |I|},   r ∈ Siaζ \ Siaζ−1 ∪ {iaζ } , and τ ∈ {2, . . . , n}, into (74), we get λτr = λτia

ζ

for iaζ ∈ I, r ∈ Siaζ \(Siaζ−1 ∪ {iaq }), and τ = 2, . . . , n.

(76)

Using (75) and (76), Equation (74) reduces to

X

λ0 sk−1 +

λxp xp =

+

X

λ1ia

ζ

ζ=1

 λtia

ζ

ζ=1 t=2

p∈{k,...,miu −1}\S |I|

|I| n X X

 & ' ' & X biaζ−1  biaζ 1 − + zp −  α1 α1 p∈Sia

ζ

 X X   zpt − zpt   p∈Sia

 X p∈Sia

 zp1  ,

ζ−1

26

p∈Sia

ζ−1

ζ

(77)

where bia0 = 0 and Sia0 = ∅. By rearranging the right-hand side of (77), we get X

λ0 sk−1 +

λxp xp = −

+

X

λ1ia ζ+1

−



|I|

ζ+1

 X   zpt   p∈Sia

ζ



& ' X   biaζ − zp1   α1

(78)

p∈Sia

ζ=1

+ λ1ia

λtia − λtia ζ

 λ1ia ζ

 

ζ=1 t=2

p∈{k,...,miu −1}\S |I|−1 

|I|−1 n  XX

ζ

    & ' n b X X X  ia|I|    − zp1  − λtia  zpt  .  |I| α1 t=2

p∈Sia

|I|

p∈Sia

|I|

Furthermore, Equation (78) can also be simplified to X λxp xp λ0 sk−1 + p∈{k,...,miu −1}\S

=

|I| X q=1





    n X X X b γi1  iq − zp1  − γitq  zpt  q α1 

t=2

p∈Siq

where for q = 1, . . . , |I| and t = 1, . . . , n, γitq = λtiq − λtia

ζ+1

λta|I|+1

(79)

p∈Siq

such that aζ = q for ζ ∈ {1, . . . , |I|} and

= 0. We substitute the point Jn in (79) to get |I| X

 γi1 − q

q=1

n−1 X t=2

     (t−1)  (t−1) b b   i iq  γitq  q  − γinq   αt  = 0. αt  

(80)

Subtracting (80) from (79) gives miu −1

λ0 sk−1 +

X p=k p∈S /

λxp xp =

|I| X q=1

      (t−1)  n X X b   i  γitq  q  − zpt  + γinq  . αt t=1

(81)

p∈Siq

(n)

m Next, for ir ∈ I, we consider the point Rr = (ˆ z, x ˆ, sˆk−1 ) ∈ Zmn ˆk−1 = bir , + ×R+ ×R+ , such that s

(ˆ zp1 , zˆp2 , . . . , zˆpn , x ˆp )

( (ddp /α1 e , 0, . . . , 0, dp ) := (0, 0, . . . , 0, 0)

if p ∈ {1, . . . , m}\{k, . . . , l}, if p ∈ {k, . . . , l}\S,

for p ∈ {1, . . . , m}\S, and for i ∈ {1, . . . , |S|}, (ˆ zwi , x ˆwi ) :=

χwi − χwi−1 ,

n X t=1

27

! t αt zˆw i

where χw0 = (0, . . . , 0) ∈ Rn and $ % & '!    (n−2) (n−1) b b b  wi wi wi   ,..., ,   α α αn 1 n−1     $ % $ %!  (n−2) (n−1)  b  bwi bwi wi 1 n (χwi , . . . , χwi ) = ,..., ,  α1 αn−1 αn    $ % & '!     (n−2) (n−1)   b b b wi wi wi   ,..., ,  α1 αn−1 αn This implies that

P

p∈Swi

if wi ∈ S\I, if wi ∈ {i1 , . . . , ir }, if wi ∈ {ir+1 , . . . , i|I| }.

zˆp = χwi . Because of Conditions (59),

t zˆw = χtwi − χtwi−1 ≥ 0 for i = 1, . . . , |S| and t = 1, . . . , n. i

Now, it is easy to verify that Rr ∈ X MML and satisfies conditions (i) − (iii) of Lemma 1 because for wi ∈ {i1 , . . . , ir }, & ' ' & n n n (ρ−1) (ρ−1) Y X X Y b b w w i i φnwi (ˆ z) = − zˆpt αρ αρ t=1 ρ=t+1 ρ=1 p∈Swi ' & ' $ % $ % & n−1 n n (ρ−1) (t−1) (n−1) (ρ−1) X Y Y bwi bwi bwi bwi − − = αρ αρ αt αn t=1 ρ=t+1 ρ=1  & ' n−1 n & ' ' & n n (ρ−1) (ρ−1) (ρ−1) Y X Y Y bwi bwi bwi   = − − αρ α α ρ ρ ρ=t ρ=1 t=1 ρ=t+1 & ' (n−1) bwi − + 1 = 1, αn and for wi ∈ {ir+1 , . . . , i|I| }, ' & ' & n n n (ρ−1) (ρ−1) X Y X Y b b w w i i − zˆpt φnwi (ˆ z) = αρ αρ ρ=1 t=1 ρ=t+1 p∈Swi & ' & ' $ % & ' n n−1 n (ρ−1) (ρ−1) (t−1) (n−1) Y X Y bwi bwi bwi bwi = − − αρ αρ αt αn ρ=1 t=1 ρ=t+1  & ' n−1 n & ' & ' & ' n n (ρ−1) (ρ−1) (ρ−1) (n−1) Y X Y Y b bwi b b w w w i i i  − = − − αρ α α α ρ ρ n ρ=t ρ=1

t=1

ρ=t+1

= 0, Therefore Rr ∈ Γ and hence must satisfy (63). One by one we substitute R1 , R2 , . . . , R|I| into (81) and get

28

(n)

γin1 = λ0 bi1 ,   (n) (n) (n) γin2 = λ0 bi2 − γin1 = λ0 bi2 − bi1 , .. . |I|−1

γin|I|

=

(n) λ0 bi|I|

−

X

  (n) (n) γinq = λ0 bi|I| − bi|I|−1 .

q=1

This implies that for q ∈ {1, . . . , |I|},   (n) (n) γinq = λ0 biq − biq−1 ,

(82)

(n)

where bi0 = 0. For r ∈ {k, . . . , miu − 1}\S, consider the points S r whose coordinates are same as (n)

the coordinates of Ru , where iu = max{iq ∈ I}, except that sˆk−1 = 0 and (ˆ zr1 , x ˆr ) = (1, biu ). By r substituting the points S into the (81) and subtracting from equality corresponding to Ru , we get λxr = λ0 for r ∈ {k, . . . , miu − 1}\S.

(83)

For iaζ ∈ I where ζ ∈ {1, . . . , |I|} and τ ∈ {1, . . . , n − 1}, we consider the point T ζ,τ such that sk−1 = 0, ( (ddp /α1 e , 0, . . . , 0, dp ) if p ∈ {1, . . . , m}\{k, . . . , l}, 1 2 n (ˆ zp , zˆp , . . . , zˆp , x ˆp ) := (0, 0, . . . , 0, 0) if p ∈ {k, . . . , l}\S, for p ∈ {1, . . . , m}\S, and for i ∈ {1, . . . , |S|}, (ˆ zwi , x ˆwi ) :=

∆wi − ∆wi−1 ,

n X

! t αt zˆw i

t=1

where ∆wi ∈ Rn+ such that ∆w0 = (0, . . . , 0) and    bwi   , 0, . . . , 0   α1     $ % & ' !   (τ −2) (τ −1)   bwi bwi bwi ,..., , , 0, . . . , 0 ∆wi = α1 ατ −1 ατ    $ % & '!     (n−2) (n−1)   b b b w w wi  i i  ,..., ,  α1 αn−1 αn P

if wi ∈ {w1 , . . . , iaζ − 1}, if wi ∈ {iaζ }, if wi ∈ {iaζ + 1, . . . , w|S| }.

zˆp = ∆wi and because of assumptions (59), zˆwi = ∆wi − ∆wi−1 ≥ 0 for P all wi ∈ S. Also, since p∈Sw x ˆp ≥ bwi , it is easy to verify that T ζ,τ ∈ X MML and it satisfies

Observe that

p∈Swi

i

conditions (i) − (iii) of Lemma 1 because for all wi ∈ I, φnwi (ˆ z ) = 0. Therefore, T ζ,τ ∈ Γ and hence 1,n−1 1,1 must satisfy (63). One by one we substitute T , . . . , T , . . . , T |I|,n−1 , . . . , T |I|,1 into (81) and

29

get 

(n−1)

bia



1 , γin−1 = γina   aζ ζ  α n    (n−2)   b   n−2 n−1  iaζ  + γin γ ia = γ ia aζ ζ ζ αn−1



(n−1)

bia





n Y



(ρ−1)

bia



 ζ  γin ,  1 = aζ  αn   αρ     ρ=n−1 

.. .  (ρ−1)  n bi Y  aζ  γin , = aζ  αρ   ρ=2  

γi1a

ζ

for ζ = 1, . . . , |I|. This implies for τ = 1, . . . , n − 1,   (ρ−1)  n bi Y  aζ  γin , γiτa =  aζ  αρ  ζ  ρ=τ +1 

(84)

Substituting (82), (83), and (84) into Equation (81) gives miu −1

λ0 sk−1 +

X p=k p∈S /

where

λ 0 xp =

|I| X

  (n) (n) λ0 biq − biq−1 φniq (z)

(85)

q=1

    (ρ−1) (ρ−1) n n n X Y X Y b b  iq   t  iq φniq (z) :=  α ρ  zp .  αρ  −     ρ=1 p∈Siq t=1 ρ=t+1

The identity (85) is λ0 times (58). Hence Γ defines a facet for conv(X MML ) if conditions (59) hold. This completes the proof.

5.

Conclusion and Future Work

We provided sufficient conditions under which the (k, l, S, I) inequalities of Pochet and Wolsey [12], the mixed (k, l, S, I) inequalities, derived using mixing procedure of G¨ unl¨ uk and Pochet [9], and the paired (k, l, S, I) inequalities, derived using sequential pairing procedure of Guan et al. [6], are facet-defining for the single module (or constant batch) capacitated lot-sizing problem without backlogging (SMLS-WB). We also investigated the facet-defining properties of the inequalities derived using the sequential pairing and the n-mixing procedure of Sanjeevi and Kianfar [15] for the multi-module capacitated lot-sizing problem without backlogging. One potential extension would be to provide necessary conditions under which the mixed (k, l, S, I) inequalities for SMLS-WB problem are facet-defining. In order to proceed in this direction, the following example can be helpful as it showcases that in case the given k, l, S, and I do not satisfy the sufficient condition (23) then the associated mixed (k, l, S, I) inequality is dominated by another facet-defining inequality which satisfies the condition (23). 30

Example 1 (continued). Next, we consider k = 2, l = 6, and S = {2, 4, 5, 6}. Therefore, S2 = {2}, S4 = {2, 4}, S5 = {2, 4, 5}, S6 = {2, 4, 5, 6}, m2 = 4, m4 = 5, m5 = 5, m6 = 7, b2 = 4, b4 = 27, (1) (1) (1) (1) (1) (1) b5 = 28, b6 = 36, b2 = 4, b4 = 3, b5 = 4, and b6 = 1. Notice that b0 = 0 < b6 < (1) (1) (1) b4 < b2 = b5 and more importantly, db4 /α1 e = db5 /α1 e = 6. The mixed (2, 6, {2, 4, 5, 6}, {5, 2}) inequality where iu = 5,        b  b5 (1) (1) (1) 2 1 1 1 1 s1 + x3 ≥ b5 − z2 − z4 − z5 + b2 − b5 − z2 α1 α1 = 24 − 4z21 − 4z41 − 4z51 , does not satisfy the conditions (23) because for i = 5, there exist a j = 4 such that j ∈ S, j < i, dbj /α1 e = dbi /α1 e, and j ∈ / I. In addition, the facet-defining mixed (2, 6, {2, 4, 5, 6}, {4, 5, 2}) inequality where iu = 5,        b  b4 (1) (1) (1) 5 1 1 1 1 1 s1 + x3 ≥ b4 − z2 − z4 + b5 − b4 − z2 − z 4 − z5 α1 α1     b  (1) (1) 2 1 + b2 − b5 − z2 α1 = 3(6 − z21 − z41 ) + (6 − z21 − z41 − z51 ) = 24 − 4z21 − 4z41 − z51 , dominates the the mixed (2, 6, {2, 4, 5, 6}, {5, 2}) inequality because 24 − 4z21 − 4z41 − z51 > 24 − 4z21 − 4z41 − 4z51 as z51 ∈ Z+ .

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32