CORE DISCUSSION PAPER2000/52

On the Polyhedral Structure of a Multi–Item

Production Planning Model with Setup Times

Andrew J. Miller 1, George L. Nemhauser 2, and Martin W.P. Savelsbergh 2

November 2000

Abstract

We present and study a mixed integer programming model that arises as a sub-structure in many industrial applications. This model provides a relaxation of vari-ous capacitated production planning problems, more general fixed charge networkflow problems, and other structured mixed integer programs. After distinguishingthe general case, which is NP–hard, from a polynomially solvable case, we analyzethe polyhedral structure of the convex hull of this model, as well as of a strengthenedLP relaxation. Among other results, we present valid inequalities that induce facetsof the convex hull in the general case. These inequalities suffice to solve this modelby linear programming in the polynomially solvable case mentioned above, and theyhave proven computationally effective in solving capacitated lot–sizing problems (seealso Miller et al. [2000a, 2000b]).

Keywords: Mixed integer programming, production planning, polyhedral combin-atorics, capacitated lot–sizing, fixed charge network flow

1CORE, Universite Catholique de Louvain, Belgium2Georgia Institute of Technology, School of Industrial and Systems Engineering, Atlanta, GA

30332-0205, USA

This paper presents research results of the Belgian Program on Interuniversity Poles of Attrac-tion initiated by the Belgian State, Prime Minister’s Office, Science Policy Programming. The scientificresponsibility is assumed by the authors.This research was also supported by NSF Grant No. DMI-9700285 and by Philips Electronics NorthAmerica.

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1 Introduction

One of the most successful techniques that has been used to solve integer programming(IP) and mixed integer programming (MIP) problems in recent years is the applicationof polyhedral results for structured relaxations of these problems. Often facet–definingfamilies of valid inequalities can be defined for the relaxed models. Since these inequal-ities are also valid for the original problem, they can be applied in a branch–and–cutalgorithm.

In this paper we introduce a model that occurs as a relaxation of many structuredMIP problems. We derived this relaxation as a single–period relaxation of the multi–itemcapacitated lot–sizing problem with setup times (MCL), which is described in Section2. In this relaxation we consider, among other decision variables, those that representthe amount of inventory carried over from the preceding period; therefore we call thisrelaxation PI, for preceding inventory. PI can be formulated as follows:

minP∑i=1

pixi +P∑i=1

qiyi +P∑i=1

hisi (1)

subject toxi + si ≥ di, i = 1, ..., P, (2)P∑i=1

xi +P∑i=1

tiyi ≤ c, (3)

xi ≤ (c− ti)yi, i = 1, ..., P, (4)xi, si ≥ 0, i = 1, ..., P, (5)yi ∈ {0, 1}, i = 1, ..., P. (6)

Perhaps the simplest way to view PI is as a single–node, multi–item fixed charge flowmodel with two inbound arcs (see Figure 1.) The first of these arcs has capacity c, whichis jointly imposed on P items. Moreover, a fixed capacity usage ti ≥ 0 must be incurredif any amount of item i is shipped along this arc. (If capacity is measured in time units,then ti can be considered a setup time; we therefore refer to this quantity as the setuptime for item i throughout this paper.) Thus xi is the amount of flow shipped along thisarc with unit cost pi, and the binary variable yi indicates whether the fixed capacityusage for item i is incurred; the cost for setting yi = 1 is qi. The second arc can alsobe used by each of these commodities. This arc is uncapacitated, and there are no fixedcosts associated with flow on it. The variables si represent the amount of flow on thissecond arc for item i; the unit cost for shipping i along this arc is hi. The total amountof flow shipped along these two arcs must be at least di, i = 1, ..., P .

We let P = {1, ..., P}, and we denote the set of points defined by (2)–(6) as XPI .

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Figure 1: A Single–Node Multi–Commodity Model with Two Arcs

If hi < 0 for any i ∈ P, PI has unbounded optimum. We will therefore assume thathi ≥ 0, i = 1, ..., P . Also, we assume that qi > 0, i = 1, ..., P , which ensures that in everyoptimal solution, yi = 1 if and only if xi > 0.

During the last two decades researchers have often used strong valid inequalitiesfor structured MIP relaxations to solve more complicated models. Among those whopioneered such methods were Crowder, Johnson, and Padberg [1983], who demonstratedthat the use of inequalities for knapsack relaxations defined by single constraints of 0–1IP problems can be effective in solving IP’s. Today the use of valid inequalities for theknapsack polytope is a common feature in IP solvers. Padberg, Van Roy, and Wolsey[1985] derived flow cover inequalities for a relaxation of MIP’s defined by a single–nodefixed charge (SNFC) network flow model, and the use of inequalities based on their workto solve MIP’s has also become common.

When ti = 0, i = 1, ..., P , PI resembles SNFC, but even in this case PI has a different,more complicated structure than SNFC. Both PI and SNFC have flow variables xi, setupvariables yi, and a joint capacity constraint (3). Moreover, both have variable upperbounds similar to (4), although in SNFC these are generally tighter than in PI. PI, inaddition, has constraints (2), which impose a demand on each of the flow variables, andvariables si, which serve as slack variables for (2) and, for example, allow demand inexcess of capacity to be satisfied.

All the results that we present for PI are valid for the case ti = 0, i = 1, ..., P . Thefact that our results also hold when setup times for some or all of the items are strictlypositive only serves to make these results more applicable to industrial problems such asMCL (Miller et al. [2000b]).

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An outline of the rest of the paper follows. In Section 2, we motivate our discussionof PI further by discussing applications for which it serves as a relaxation. These includenot only MCL, but other lot–sizing problems and more general MIP’s as well.

In Section 3, we analyze the polyhedral structure of PI. We first discuss briefly thecomplexity of PI. Although PI is NP–hard, under certain conditions it is polynomiallysolvable. Thus, unless NP = co–NP, it is in general not possible to derive a goodcharacterization of conv(XPI) by families of linear inequalities (Karp and Papadimitrou[1982]). However, for the polynomially solvable case, we can hope to do so.

After identifying trivial facets, we analyze the extreme points and rays of conv(XPI),and we show which of these extreme points are possible optimal solutions of PI, giventhe conditions on the objective function coefficients. This analysis not only allows us tocharacterize conv(XPI) by extreme points and rays, it also allows us to describe whichpart of the polyhedron is the most interesting for our purposes. We perform a similaranalysis for an LP relaxation of XPI that is strengthened with a family of facet–inducingvalid inequalities analogous to the path inequalities (Barany, Van Roy, and Wolsey [1984],Van Roy and Wolsey [1987]).

In Section 4, we derive new valid inequalities for PI. We give general conditionsunder which each of these classes of inequalities define facets of conv(XPI), and we alsoshow that these new inequalities cut off all possible fractional optimal solutions of thestrengthened LP relaxation mentioned above. In defining the second of these classes, weuse the concepts of superadditivity and lifting to expand the class into a more completeset of inequalities. After thus expanding this class, we can state that, under conditionsthat ensure that PI is polynomially solvable, the inequalities that we present allow it tobe solved by linear programming (Miller et al. [2000a]).

2 Motivation

Our original motivation in studying PI was to derive strong valid inequalities for MCL byconsidering simplified models obtained from a single time period. A standard formulationfor MCL is

minP∑i=1

J∑j=1

pijxij +

P∑i=1

J∑j=1

qijyij +

P∑i=1

J∑j=1

hijsij (7)

subject toxij + sij−1 − sij = dij , i = 1, ..., P, j = 1, ..., J, (8)P∑i=1

xij +P∑i=1

tijyij ≤ cj , j = 1, ..., J, (9)

xij ≤ (cj − tij)yij , i = 1, ..., P, j = 1, ..., J, (10)

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xij , sij ≥ 0, i = 1, ..., P, j = 1, ..., J, (11)

yij ∈ {0, 1}, i = 1, ..., P, j = 1, ..., J. (12)

The number of time periods in the problem is J . The production capacity, setup times,and demands in period j are given, respectively, by cj , tij , i = 1, ..., P , and dij , i = 1, ..., P ;pij , q

ij , and hij represent production, setup, and holding costs, respectively. The variables

xij represent production, sij represent ending inventory, and yij are the binary setupvariables, for item i in period j.

It is clear that an instance of PI can be obtained from each period j of MCL byconsidering xij , y

ij , s

ij−1, and the constraints that link them. (Note also that constraints

(8) are relaxed by not considering variables sij .) It is also possible to consider tighterrelaxations of MCL that are also instances of PI but that consider demand for multipleperiods. This is discussed in detail in Miller et al. [2000b], where successful computa-tional experience in applying and extending results for PI to solve instances of MCL isalso given.

There are a number of other production planning problems for which PI serves as arelaxation. For example, Pochet and Wolsey [1991] showed how to model the multi–levelcapacitated lot–sizing problem using echelon stocks. The resulting model has a structurethat is similar to (8)–(12), with additional constraints linking the echelon inventoryvariables. Using these results, it is possible to show that for any time period j, PIprovides a single–period relaxation for this model as well.

In addition, PI provides a relaxation of more general MIP problems, such as fixedcharge network flow problems. For example, PI can be used to model an instance ofSNFC with side constraints. In such a relaxation constraints (3) and (4) correspond tothe basic model SNFC, and (2) provide a potential relaxation for additional constraintsinvolving the flow variables. For more on SNFC and related models, see, among others,Padberg, Van Roy, and Wolsey [1985], Gu, Nemhauser, and Savelsbergh [1999], andMarchand and Wolsey [1999].

3 Basic Polyhedral Results

In this section we present basic results concerning the polyhedral structure of PI. Whereproofs are omitted, they can be found in Miller [1999], unless otherwise stated. The firstbasic result we state for PI concerns its complexity.

Proposition 1 (Miller et al. [2000a]) If both setup times and demand are constant forall items (i.e., ti = t ≥ 0, di = d ≥ 0, i = 1, ..., P ), then PI is polynomially solvable.Otherwise, PI is NP–hard.

In Miller et al. [2000a], we present an O(P 2) algorithm for the case of constant setuptimes and demand. Because of Proposition 1, it is doubtful that we can find a complete

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description of conv(XPI) given by facet–inducing inequalities in the general case.

Proposition 2 If c > ti, i = 1, ..., P , conv(XPI) is full–dimensional. Moreover, fori = 1, ..., P , (2), (4), and the bound xi ≥ 0 induce facets of conv(XPI).

Proposition 3 If, for every (i, j) ∈ P × P, i 6= j, c > ti + tj, then (3) and the boundsyi ≤ 1, i = 1, ..., P induce facets of conv(XPI).

Proposition 4 If c > ti + di, the valid inequalities

si + diyi ≥ di (13)

induce facets of conv(XPI), i = 1, ..., P .

Note that (13) implies the bound si ≥ 0, i = 1, ..., P ; therefore this bound never inducesa facet. Also note that these inequalities are similar to the (l, S) inequalities for theuncapacitated lot–sizing problem, introduced by Barany, Van Roy, and Wolsey [1984].Van Roy and Wolsey [1987] later generalized these to the path inequalities for moregeneral fixed charge network flow problems.

We will now discuss the extreme points and rays of conv(XPI) and of an LP relaxationof XPI . Given an extreme point (x, y, s) of conv(XPI), let Q = {i ∈ P : yi = 1}. Also,let Qu = {i ∈ Q : xi = di}, Ql = {i ∈ Q : xi = 0}, and let Qr = {i ∈ Q : xi > 0, xi 6= di}.Moreover, for a given i, we define types of (xi, yi, si) triples:

Type 1: xi = di, yi = 1, si = 0,Type 2: xi = 0, yi = 1, si = di,

Type 3: xi = 0, yi = 0, si = di.

Proposition 5 (Characterization of the extreme points of conv(XPI)) In every extremepoint (x, y, s) of conv(XPI), (xi, yi, si) is of Type 1, i ∈ Qu; (xi, yi, si) is of Type 2,i ∈ Ql; and (xi, yi, si) is of Type 3, i ∈ P \Q. Moreover, either Qr = ∅, or |Qr| = 1 and

xr = c−∑i∈Qu

(ti + di)−∑i∈Ql

ti − tr, yr = 1, sr = (dr − xr)+, (14)

where Qr = {r}.

The fact that |Qr| ≤ 1 in every extreme point of conv(XPI) can be shown by contra-diction. In effect, this says that there in only one i ∈ P for which the constraint (3)

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plays a part in determining the value xi. For the other i ∈ P, xi is determined by someinteraction among (2), the variable upper bound (4), and the bound xi ≥ 0. If |Qr| > 1does not hold for some (x, y, s) ∈ XPI , then it is not difficult to construct two new pointsin XPI such that (x, y, s) is a convex combination of the two.

Now define XPILP to be the set of feasible points to the LP relaxation of (2)–(6), (13);

we can characterize its extreme points as well. Given an extreme point (x, y, s) of XPILP ,

let Q, Qu, and Ql be defined as for XPI , and let R = {i ∈ P : 0 < yi < 1}.

Proposition 6 (Characterization of the extreme points of XPILP ) Let (x, y, s) be an ex-

treme point of XPILP . Then (xi, yi, si) is of Type 1, i ∈ Qu; (xi, yi, si) is of Type 2, i ∈ Ql;

and (xi, yi, si) is of Type 3, i ∈ P \ {Q∪R}. Moreover, if R is non–empty, it consists ofa singleton {r}, and exactly one of the following statements is true:

1. (xr, yr, sr) is defined as in (14), and thus (x, y, s) is an extreme point of conv(XPI);

2.

yr =c−

∑i∈Qu(ti + di)−

∑i∈Ql t

i

tr, xr = 0, sr = dr; (15)

3.

yr =c−

∑i∈Qu(ti + di)−

∑i∈Ql t

i

tr + dr, xr = dryr, sr = dr − xr; (16)

4.

yr =c−

∑i∈Qu(ti + di)−

∑i∈Ql t

i

c, xr = (c− tr)yr, sr = dr(1− yr). (17)

The reasoning that |R| ≤ 1 is similar to the reasoning that |Qr| ≤ 1 in Proposition 5.For fractional extreme points defined by (15), inequality (2) and the bound xr ≥ 0 areboth tight for r, and inequalities (4) and (13) are not. For fractional extreme pointsdefined by (16), inequalities (2) and (13) are both tight for r, inequality (4) is not, andxr > 0. For fractional extreme points defined by (17), inequalities (4) and (13) are tight,inequality (2) is not, and xr > 0.

Proposition 7 All extreme rays of both conv(XPI) and XPILP have the form si = 1, xi =

yi = 0, for some i ∈ P, and sj = xj = yj = 0, j 6= i.

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Thus PI has a bounded optimum if and only if hi ≥ 0, i = 1, ..., P , and PI has anunbounded optimal solution if and only if hi ≤ 0, for some i ∈ P. If hi ≥ 0, i = 1, ..., P ,and there is an element i for which hi = 0, then PI has an infinite number of optimalsolutions. So, given our assumption that hi ≥ 0, i = 1, ..., P , every optimal solution ofPI is either bounded, or is the sum of another optimal solution that is bounded and ofa linear combination of extreme rays.

These results provide a characterization of both conv(XPI) and conv(XPILP ) by their

extreme points and rays. Next, we discuss which extreme points are possible optimalsolutions of XPI and XPI

LP . In discussing a set of points X ⊂ {(x, y, s) : xi, si ≥ 0, 0 ≤yi ≤ 1, i = 1, ..., P}, we will use

Definition 1 The point (x, y, s) is a dominant solution of X if there exists some costvector (p, q, h), such that qi > 0, hi ≥ 0, i = 1, ..., P , for which optimizing (1) over Xyields (x, y, s) as the unique optimal solution.

This definition allows us to obtain

Proposition 8 The point (x, y, s) is a dominant solution of XPI for PI if and only if(1) (x, y, s) is an extreme point of conv(XPI); and (2) yi = 1 if and only if xi > 0, i =1, ..., P .

Thus, the condition qi > 0, i = 1, ..., P , ensures that a setup for an item does nottake place unless production of that item also occurs. This is a reasonable conditionto enforce; for example, in both lot–sizing and fixed charge problems, a fixed chargeto produce/transport an item is not paid unless some positive quantity of that item isproduced/transported. A similar result holds for XPI

LP .

Proposition 9 Given an instance of PI, no dominant solution of XPILP has the form

(15). Moreover, every dominant solution of XPILP is either a dominant solution of XPI ,

or else it is of one of the forms (16) or (17) with Ql = ∅.

4 New Valid Inequalities for PI

Now we will derive two new families of valid inequalities for XPI and present resultsconcerning their strength. In order to define these inequalities, we first define a coverof PI to be a set S such that λ =

∑i∈S(ti + di) − c ≥ 0. Similarly, define a reverse

cover of PI to be a set S 6= ∅ such that µ = c−∑

i∈S(ti + di) > 0. Note that a reversecover S does not have to be maximal in any sense, merely non-empty. Because thedevelopment needed to define reverse cover inequalities is much simpler than that forcover inequalities, we will discuss the former class first.

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4.1 Reverse Cover Inequalities

Before formally defining this class, we provide a brief motivation. Given (x, y, s) ∈ XPI ,let S be a reverse cover, and let i′ be some element not in S. If xi

′= (c− ti′)yi′ , either

1. yi′

= 0; or

2. yi′

= 1 and no capacity is left for i ∈ S; therefore yi = 0, i ∈ S, and∑i∈S s

i ≥∑

i∈S di.

Such reasoning yields

∑i∈S

si ≥ (∑i∈S

(ti + di))yi′ −∑i∈S

ti(1− yi)− ((c− ti′)yi′ − xi′),

which is generalized in the following proposition.

Proposition 10 (Reverse Cover Inequalities) Let S be a reverse cover of PI, let T =P \ S, and let (T ′, T ′′) be any partition of T . Then

∑i∈S

si ≥ (∑i∈S

(ti + di))∑i∈T ′

yi −∑i∈S

ti(1− yi)−∑i∈T ′

((c− ti)yi − xi) (18)

is valid for XPI .

Proof: Let (x, y, s) be any point in XPI . If yi = 0 (and thus xi = 0), i ∈ T ′, then (18)is obviously valid. Let T ′1 = {j ∈ T ′ : yj = 1}; consider first the case in which T ′1 = {i′}is a singleton. Then (18) becomes

∑i∈S

si ≥∑i∈S

di +∑i∈S

tiyi − (c− ti′) + xi′.

This inequality holds because

(c− ti′) ≥∑

i∈S(xi + tiyi) + xi′

(by (3))≥

∑i∈S(di − si + tiyi) + xi

′. (by (2))

When |T ′1| > 1, the argument is similar, except that in this case the inequality isweaker and cannot ever hold at equality. 2

Note that T ′ can be seen as a candidate set for the choice of the element i′ 6∈ Smentioned in the paragraph motivating Proposition 10.

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Example 1

Consider an instance of PI with P = 4, with c = 16, and with demand and setup timesgiven by

i 1 2 3 4ti 2 2 1 3di 5 4 5 10

Choose S = {2, 3} as a reverse cover; thus µ = 16− (2 + 4)− (1 + 5) = 4. Let T ′ = {1};then (18) yields

s2 + s3 ≥ 12y1 − 2(1− y2)− (1− y3)− (14y1 − x1)= −3− 2y1 + x1 + 2y2 + y3, (19)

a valid inequality for XPI . 2

One property of reverse cover inequalities concerns the fractional extreme points ofXPILP that they cut off.

Proposition 11 For PI, every fractional dominant solution of XPILP is cut off by an

inequality of the form (18).

We can prove this proposition by considering any fractional dominant solution (as char-acterized by Propositions 6 and 9) and taking S = Qu and T ′ = {r} to define (18).It is interesting to recall here that XPI

LP is defined by a formulation that has alreadybeen strengthened with a family of strong inequalities (13). We now state very generalconditions under which reverse cover inequalities define facets of conv(XPI).

Proposition 12 If S 6= ∅, T ′ 6= ∅, and ti < µ, i ∈ T , (18) induces a facet of conv(XPI).

Proof: The conditions of the Proposition imply that c > ti, i = 1, ..., P , and so fromProposition 2 conv(XPI) is full–dimensional. It therefore suffices to show that the in-tersection of the hyperplane defined by (18) and of conv(XPI) has dimension 3P − 1.Note that the |T | rays si = 1, i ∈ T , are extreme rays of conv(XPI) and also lie in thehyperplane defined by (18). To obtain the results, therefore, it suffices to show 3P − |T |linearly independent points in XPI that lie in this hyperplane, and that are also linearlyindependent of each of these extreme rays. Let i∗ be some element in T ′; consider thepoints

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• for each i′ ∈ S,

xi′

= 0, yi′

= 1, si′

= di′

xi∗

= c− ti∗ − ti′ , yi∗ = 1, si∗

= (di∗ − xi∗)+,

xi = yi = 0, si = di, i 6= i′, i 6= i∗

(|S| points)

• for each i′ ∈ S,

xi′

= di′, yi′

= 1, si′

= 0xi∗

= c− ti∗ − ti′ − di′ , yi∗ = 1, si∗

= (di∗ − xi∗)+

xi = yi = 0, si = di, i 6= i′, i 6= i∗

(|S| points)

• for each i′ ∈ S,

xi′

= c− ti′ − ti∗ , yi′ = 1, si′

= 0xi∗

= 0, yi∗

= 1, si∗

= di∗

xi = yi = 0, si = di, i 6= i′, i 6= i∗

(|S| points)

• for each i′ ∈ T ′ \ i∗,

xi′

= µ− ti′ , yi′ = 1, si′

= (di′ − xi′)+

xi = di, yi = 1, si = 0, i ∈ Sxi = yi = 0, si = di, i ∈ T \ {i′}

(|T ′| − 1 points)

• for each i′ ∈ T ′,

xi′

= c− ti′ , yi′ = 1, si′

= (di′ − xi′)+

xi = yi = 0, si = di, i 6= i′

(|T ′| points)

• for each i′ ∈ T ′′,

xi′

= 0, yi′

= 1, si′

= di′

xi = di, yi = 1, si = 0, i ∈ Sxi = yi = 0, si = di, i ∈ T \ {i′}

(|T ′′| points)

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• for each i′ ∈ T ′′,

xi′

= µ− ti′ , yi′ = 1, si′

= (di′ − xi′)+

xi = di, yi = 1, si = 0, i ∈ Sxi = yi = 0, si = di, i ∈ T \ {i′}

(|T ′′| points)

• the point

xi = di, yi = 1, si = 0, i ∈ Sxi = yi = 0, si = di, i ∈ T.

It can be checked that the points listed are linearly independent of each other and of the|T | rays si = 1, i ∈ T . 2

Example 1 (continued)The inequality (19) satisfies the conditions of Proposition 12 and is therefore a facet ofconv(XPI). 2

4.2 Cover Inequalities

The form of the full set of cover inequalities is rather complex. Therefore, in presentingthese inequalities, we first present the most basic form of the inequality. Next, we presenta second form that, though more complex than the first, still does not encompass theentire class. We then lift additional variables into this second form, thus expanding thisclass into the full set of cover inequalities.

4.2.1 Basic Cover Inequalities

To motivate this class, consider a cover S of PI. Recall that

λ =∑i∈S

(ti + di)− c ≥ 0.

So if yi = 1, i ∈ S,

∑i∈S

si ≥∑i∈S

di −∑i∈S

xi ≥∑i∈S

di − (c−∑i∈S

ti) =∑i∈S

(ti + di)− c ≥ λ.

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Moreover, if yi′

= 0 for exactly one i′ ∈ S, then both

∑i∈S

si ≥ λ− ti

and

∑i∈S

si ≥ si′ ≥ di′

must hold. From such reasoning we can derive that

∑i∈S

si ≥ λ+∑i∈S

max{−ti, di − λ}(1− yi) (20)

is valid for PI. A more formal statement and proof follows.

Proposition 13 (Cover Inequalities, basic form) Given a cover S of PI, (20) is validfor XPI .

Proof: Let (x, y, s) be any point in XPI . Let S1 = {i ∈ S : yi = 1}, and let S0 = {i ∈S : yi = 0}. If {i ∈ S0 : ti + di > λ} = ∅, then

∑i∈S

si =∑i∈S0

si +∑i∈S1

si

≥∑i∈S0

di + (∑i∈S1

(ti + di)− c)+

=∑i∈S0

(ti + di − ti(1− yi)) + (∑i∈S1

(ti + di)− c)+

≥∑i∈S

(ti + di)− c−∑i∈S0

(ti(1− yi))

= λ+∑i∈S

max{−ti, di − λ}(1− yi). (21)

If {i ∈ S0 : ti + di > λ} 6= ∅, then

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∑i∈S

si ≥∑i∈S0

si (22)

≥∑i∈S0

di

≥ λ+∑

i∈S0:ti+di>λ

(di − λ) +∑

i∈S0:ti+di≤λ

di

≥ λ+∑i∈S

max{−ti, di − λ}(1− yi), (23)

and so the proposition holds. 2

Example 1 (continued)Choose as a cover S = {1, 2, 3}; then λ = 7 + 6 + 6− 16 = 3, and (20) yields

s1 + s2 + s3 ≥ 3 + 2(1− y1) + (1− y2) + 2(1− y3) = 8− 2y1 − y2 − 2y3, (24)

a valid inequality for XPI . 2

The family of inequalities defined by (20) cuts off many fractional extreme points.Specifically, we have

Proposition 14 For PI, every dominant fractional solution of XPILP of the form (16) is

cut off by an inequality of the form (20).

The proof is obtained by observing that Ql = ∅ (from Proposition 9), and taking S =Qu ∪ {r} in order to define (20).

Under general conditions, inequalities of the form (20) define facets of conv(XPI).

Proposition 15 Given an inequality of the form (20), order the i ∈ S such that t[1] +d[1] ≥ ... ≥ t[|S|] + d[|S|], let µ1 = t[1] + d[1] − λ, and define T = P \ S. If λ > 0,t[2] + d[2] > λ, and ti < µ1, i ∈ T , then (20) induces a facet of conv(XPI).

Proof: Similarly to the proof of Proposition 12, it suffices to show 3P − |T | points and|T | rays in XPI that are linearly independent and that lie in the hyperplane defined by(25). To simplify notation, we assume first that di + d[1] ≥ λ, i ∈ S \ [1]. Consider the3|P | − |T | points

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• for each i′ ∈ S \ {[1]},

xi′

= (di′ − λ)+, yi

′= 1, si

′= di

′ − xi′

x[1] = d[1] − (λ− di′)+, y[1] = 1, s[1] = d[1] − x[1], ( if i′ 6= [1])xi = di, yi = 1, si = 0, i ∈ S \ {i′ ∪ [1]}

xi = yi = 0, si = di, i 6∈ S

(|S| − 1 points)

• the point

x[1] = (d[1] − λ)+, y[1] = 1, s[1] = d[1] − x[1]

x[2] = d[2] − (λ− d[1])+, y[2] = 1, s[1] = d[1] − x[1],

xi = di, yi = 1, si = 0, i ∈ S \ {i′ ∪ [1]}xi = yi = 0, si = di, i 6∈ S

• for each i′ ∈ S \ {[1]},

xi′

= yi′

= 0, si′

= di′

x[1] = d[1] − (λ− ti′ − di′)+, y[1] = 1, s[1] = d[1] − x[1]

xi = di, yi = 1, si = 0, i ∈ S \ {i′ ∪ [1]}xi = yi = 0, si = di, i 6∈ S

(|S| − 1 points)

• the point

x[1] = 0, y[1] = 0, s[1] = d[1]

xi = di, yi = 1, si = 0, i ∈ S \ {[1]}xi = yi = 0, si = di, i 6∈ S

• for each i′ ∈ S \ [1],

xi′

= di′+ µ1, yi

′= 1, si

′= 0

x[1] = y[1] = 0, s[1] = d[1]

xi = di, yi = 1, si = 0, i ∈ S \ {i′ ∪ [1]}xi = yi = 0, si = di, i 6∈ S

(|S| − 1 points)

15

• the point

x[1] = d[1] + t[2] + d[2] − λ, y[1] = 1, s[1] = 0x[2] = y[2] = 0, s[2] = d[2]

xi = di, yi = 1, si = 0, i ∈ S \ {i′ ∪ [2]}xi = yi = 0, si = di, i 6∈ S

• for each i′ ∈ T ,

xi′

= 0, yi′

= 1, si′

= di′

x[1] = y[1] = 0, s[1] = d[1]

xi = di, yi = 1, si = 0, i ∈ S \ [1]xi = yi = 0, si = di, i 6∈ S ∪ i′

(|T | points)

• for each i′ ∈ T ,

xi′

= µ1 − ti′ , yi′ = 1, si′

= (di′ − xi′)+

x[1] = y[1] = 0, s[1] = d[1]

xi = di, yi = 1, si = 0, i ∈ S \ [1]xi = yi = 0, si = di, i 6∈ S ∪ i′

(|T | points).

It can be checked that these points and the |T | rays si = 1, i ∈ T , are linearly indepen-dent. When the assumption di + d[1] ≥ λ, i ∈ S \ [1], does not hold, 3P − |T | linearlyindependent points can be chosen similarly, but either a larger number of subsets ormore complex notation is required to list them. 2

Example 1 (continued)

The inequality (24) satisfies the conditions of Proposition 15, and is therefore a facet ofconv(XPI). 2

16

4.2.2 Second Form of Cover Inequalities

We now expand the class of cover inequalities by incorporating another set into theirdefinition. This allows us to cut off more (in many cases, all) fractional dominant solu-tions and to define more facets of conv(XPI), and it is essential to the sufficiency resultthat is stated at the end of this section. This new set plays a role in cover inequalitiessimilar to that played by the set T ′ in the reverse cover inequalities (18).

Proposition 16 (Cover Inequalities, second form) Given a cover S of PI, order thei ∈ S such that t[1] + d[1] ≥ ... ≥ t[|S|] + d[|S|]. Let T = P \ S, and let (T ′, T ′′) be anypartition of T . Define µ1 = t[1] + d[1] − λ; if µ1 ≥ 0 and |S| ≥ 2, then

∑i∈S

si ≥ λ+∑i∈S

max{−ti, di − λ}(1− yi) +∑

i∈S min{ti + di, λ} − λ(|S| − 1)(t[2] + d[2])

∑i∈T ′

(xi − (µ1 − ti)yi)

(25)

is valid for XPI .

Proof: Let (x, y, s) be any point in XPI . Let S1 = {i ∈ S : yi = 1}, let S0 = {i ∈S : yi = 0}, and let T ′1 = {i ∈ T ′ : yi = 1}. If T ′1 = ∅, then the validity argument isexactly the case as for Proposition 13. So let T ′1 6= ∅; we first assume that T ′1 = {i′}, asingleton. To show that (x, y, s) satisfies (25), it suffices to show that

∑i∈S

si +∑i∈S

max{−ti, di − λ}yi ≥ λ+∑i∈S

max{−ti, di − λ}

+∑

i∈S min{ti + di, λ} − λ(|S| − 1)(t[2] + d[2])

∑i∈T ′

(xi − (µ1 − ti)yi).

(26)

Now

∑i∈S

si +∑i∈S

max{−ti, di − λ}yi ≥ min(x,y,s)∈XPI :xi′=xi′ ,yi′=1

{∑i∈S

si +∑i∈S

max{−ti, di − λ}yi}.

(27)

This is true because the right hand side is the minimum value that the left hand sidecan achieve, provided that yi

′= 1 and xi

′= xi

′are both still true. Therefore we will

consider the minimization problem

17

min(x,y,s)∈XPI :xi′=xi′ ,yi′=1

{∑i∈S

si +∑i∈S

max{−ti, di − λ}yi}, (28)

and show that the optimal solution to this problem has value greater than or equal tothat of the right hand side of (26).

To characterize the optimal value to the above minimization problem, we can considerthe problem item by item. Note that if in a solution of (28) we set yi

∗= 1, xi

∗=

di∗, si∗

= 0, where ti∗

+ di∗ ≤ λ, then we obtain a savings in the objective function of

ti∗

+ di∗

over the same solution in which yi∗

= xi∗

= 0, si∗

= di∗. Moreover, setting

yi∗

= 1, xi∗

= di∗, si∗

= 0, requires the usage of ti∗

+ di∗

units of capacity. However, ifti∗

+ di∗> λ, then setting yi

∗= 1, xi

∗= di

∗, si∗

= 0, results in a savings in the objectivefunction of only λ over the same solution in which yi

∗= xi

∗= 0, si

∗= di

∗, although

doing so still uses ti∗

+ di∗

units of capacity. Thus, as ti∗

+ di∗

increases above λ, itbecomes less economical, given the objective function of (28), to satisfy demand fromproduction.

Therefore, it is possible to solve (28) greedily, by satisfying demand for i ∈ S fromproduction in nondecreasing order of ti + di. Given this ordering, the demand for itemsin S is satisfied by production as long as there is sufficient capacity left over after theproduction of i′ to do so. After capacity is used up, the remaining demand for items inS is satisfied from inventory. That is, for some integer j with 1 ≤ j ≤ P , this solutionhas objective function value

j−1∑i=1

d[i] +|S|∑i=j

max{−t[i], d[i] − λ}+ (|S|∑i=j

(t[i] + d[i]) + ti′+ xi

′ − c)+. (29)

The value of j is determined by the number of items S for which it is possible to satisfydemand by production, given the value of xi

′and the fact that we produce in nonde-

creasing order of ti + di. So let j′ be defined by

j′ = min{i′′ :|S|∑i=i′′

(t[i] + d[i]) < c− ti′ − xi′}.

Then the choice of j for which (29) attains the optimal value to (28) is

j =

{j′ − 1 if

∑|S|i=j′−1(t[i] + d[i])− (c− ti′ − xi′) ≤ λ

j′ otherwise.

18

Depending on the value of xi′, the capacity constraint (3) may or may not be tight at the

optimal solution obtained in this way. If (3) is tight, then demand for item [j] cannotbe fully met. In this case j = j′ − 1 and

|S|∑i=j

(t[i] + d[i]) + ti′+ xi

′ − c (30)

will be positive. The sum (30) is the inventory necessary to satisfy the demand for [j] thatcannot be satisfied from production, given that all the demand for items [j + 1], ..., [S],is satisfied from production. If (3) is not tight, then demand for all of [j], ..., |S|, can besatisfied entirely from production. In this case j = j′, and (30) is nonpositive.

So we have that

min(x,y,s)∈XPI :xi′=xi′ ,yi′=1

{∑i∈S s

i +∑

i∈S max{−ti, di − λ}yi}

=∑j−1i=1 d

[i] +∑|S|

i=j max{−t[i], d[i] − λ}+ (∑|S|

i=j(t[i] + d[i]) + ti

′+ xi

′ − c)+ =

λ+∑

i∈S max{−ti, di − λ}+∑j−1

i=1 min{t[i] + d[i], λ} − λ+ (∑|S|

i=j(t[i] + d[i]) + ti

′+ xi

′ − c)+,

where the last equality holds because

d[i] = max{−t[i], d[i] − λ}+ min{t[i] + d[i], λ}, i = 1, ..., j − 1.

So now we need to show that

λ+∑

i∈S max{−ti, di − λ}+∑j−1

i=1 min{t[i] + d[i], λ} − λ+ (∑|S|

i=j(t[i] + d[i]) + ti

′+ xi

′ − c)+ ≥

λ+∑

i∈S max{−ti, di − λ}+∑|S|i=1 min{t[i]+d[i],λ}−λ

(|S|−1)(t[2]+d[2])(xi′ − (µ1 − ti′)),

or equivalently, that

∑j−1i=1 min{t[i] + d[i], λ} − λ+ (

∑|S|i=j(t

[i] + d[i]) + ti′+ xi

′ − c)+ ≥∑|S|i=1 min{t[i]+d[i],λ}−λ

(|S|−1)(t[2]+d[2])(xi′ − (µ1 − ti′)). (31)

From here the proof proceeds by cases. First consider the case in which

xi′

= µ1 +j−1∑i=2

(t[i] + d[i])− ti′ .

19

In this case, by definition of µ1,∑|S|

i=j(t[i] + d[i]) + ti

′+ xi

′ − c = 0. Also,

j−1∑i=1

min{t[i] + d[i], λ} − λ =j−1∑i=2

min{t[i] + d[i], λ}

≥|S|∑i=2

min{t[i] + d[i], λ}∑j−1

i=2 (t[i] + d[i])(|S| − 1)(t[2] + d[2])

=∑|S|

i=1 min{t[i] + d[i], λ} − λ(|S| − 1)(t[2] + d[2])

(xi′ − (µ1 − ti′)).

The above inequality follows from the fact that t[1] + d[1] ≥ λ, and from the inequality

(|S| − 1)(t[2] + d[2])j−1∑i=2

min{t[i] + d[i], λ} ≥|S|∑i=2

min{t[i] + d[i], λ}j−1∑i=2

(t[i] + d[i]).

To see that this inequality holds, note that

(t[2] + d[2]) min{t[j] + d[j], λ} ≥ min{t[k] + d[k], λ}(t[2] + d[2])

when 2 ≤ j < k; therefore

(|S| − 1)(t[2] + d[2])j−1∑i=2

min{t[i] + d[i], λ} =j−1∑i=2

(t[2] + d[2])j−1∑i=2

min{t[i] + d[i], λ}

+|S|∑i=j

(t[2] + d[2])j−1∑i=2

min{t[i] + d[i], λ}

≥j−1∑i=2

min{t[i] + d[i], λ}j−1∑i=2

(t[2] + d[2])

+|S|∑i=j

min{t[i] + d[i], λ}j−1∑i=2

(t[2] + d[2])

=|S|∑i=2

min{t[i] + d[i], λ}j−1∑i=2

(t[2] + d[2])

≥|S|∑i=2

min{t[i] + d[i], λ}j−1∑i=2

(t[i] + d[i]).

20

So (31) holds, and (x, y, s) satisfies (25), when xi′

= µ1 +∑j−1

i=2 (t[i] + d[i]) − ti′ . Giventhis, it is not difficult to see that (x, y, s) still satisfies (25) when

xi′> µ1 +

j−1∑i=2

(t[i] + d[i])− ti′ ,

because the left hand side of (31) increases one unit for every unit of xi′

above µ1 +∑j−1i=2 (t[i] + d[i]) − ti

′, but the right hand side increases less than one unit per unit

increase of xi′. Moreover, when

xi′< µ1 +

j−1∑i=2

(t[i] + d[i])− ti′ ,

one can see that (x, y, s) still satisfies (25), because the left hand side of (31) remainsconstant as xi

′decreases below µ1 +

∑j−1i=2 (t[i] + d[i]) − ti

′, while the right hand side

decreases.When there is more than one element in T ′1, the proof proceeds analogously. Having

more than one element in T ′1 weakens (25), however, and there are no tight points inwhich |T ′1| > 1. 2

Example 1 (continued)

Again choose S = {1, 2, 3}, and recall that λ = 3. Then t[1] + d[1] = t1 + d1 = 7,and µ1 = 4. Also note that t[2] + d[2] = t2 + d2 = t3 + d3 = 6. Take T ′ = {4}, then

∑i∈S min{ti + di, λ} − λ(|S| − 1)(t[2] + d[2])

=3 + 3 + 3− 3

2 ∗ 6=

12,

and (25) yields

s1 + s2 + s3 ≥ 3 + 2(1− y1) + (1− y2) + 2(1− y3) +12

(x4 − y4)

= 8− 2y1 − y2 − 2y3 +12

(x4 − y4), (32)

a valid inequality for XPI . 2

21

In addition to the fractional dominant solutions mentioned in Proposition 14, undercertain conditions, inequalities of the form (25) cut off all fractional dominant solutionsof XPI

LP . For example, we have

Proposition 17 For PI, if setup times and demand are constant for all items, thenevery dominant fractional solution of XPI

LP of the form (17) is cut off by an inequality ofthe form (25).

To prove this proposition, first observe again that Ql = ∅. Then to define (25) letT ′ = {r}, and let S be any set such that |S| = d c

t+de and S ⊃ Qu. Thus, givenProposition 9, in this case cover inequalities cut off all fractional dominant solutions ofXPILP . It is again interesting to recall at this point that XPI

LP is an LP relaxation of XPI

that has already been strengthened with the family of facet–inducing inequalities (13).It is also interesting to note that the family of reverse cover inequalities is stronger thanthe family of cover inequalities, in the sense that the former always cuts off all possiblefractional optimal solutions to XPI

LP .There are facets induced by inequalities of the form (25) with T ′ 6= ∅. We can extend

Proposition 15 to establish conditions under which these inequalities induce facets. Forexample, we have

Proposition 18 If λ > 0, t[2]+d[2] > λ, ti < µ1, i ∈ T , and ti+di = t[2]+d[2],∀i ∈ S\[1],then (25) induces a facet of conv(XPI).

A proof of this proposition can be found in Miller [1999].

Example 1 (continued)

The inequality (32) satisfies the conditions of Proposition 18, and is therefore a facet ofconv(XPI). 2

4.2.3 Lifted Cover Inequalities

In order to expand further this set of inequalities, we need to extend (25) by lifting intoit variables associated with some subset U ⊂ P \ {S ∪ T ′}. Expanding (25) to includecoefficients on variables associated with items not in S∪T ′ is important, because even inthe simplest cases (for example, the polynomially solvable case of Proposition 1), thereare facets defined by such inequalities.

To lift si and yi, i ∈ U ⊂ P \ {S ∪ T ′}, into (25), we use a superadditive function toensure that the lifting coefficients are sequence independent. However, unlike the usualapplication of lifting, the function we use does not immediately yield the coefficients of

22

the variables lifted into (25). A second, related difference is that the constant term inthe inequality changes as variables are lifted in. That is, the constant term in (25) is

λ+∑i∈S

max{−ti, di − λ},

but for each item i ∈ U for which we lift si and yi into (25), this term will be increasedby di.

Let

k′ = |{i ∈ S : ti + di > λ}|, (33)

and define

φS(u) =

0 if u ≤ µ1,

(j − 1)λ+ (u− [µ1 +∑j

i=2(t[i] + d[i])]) if µ1 +∑j

i=2(t[i] + d[i])≤ u ≤

µ1 + λ+∑j

i=2(t[i] + d[i]),1 ≤ j ≤ k′ − 1,

jλ if µ1 + λ+∑j

i=2(t[i] + d[i])≤ u ≤

µ1 +∑j+1

i=2 (t[i] + d[i]),1 ≤ j ≤ k′ − 1,

(k′ − 1)λ+ (u− [µ1 +∑k′

i=2(t[i] + d[i])]) if u ≥ µ1 +∑k′

i=2(t[i] + d[i]).(34)

It can be shown that the function φS(·) is superadditive on IR+. A graph of an exampleof the function φS(·) appears in Figure 2; in this example k′ = 3.

Intuitively, for each i′ ∈ P \S, φS(ti′+ di

′) is defined to be the maximum amount by

which the right hand side of (25) can be increased if si′

is added to the left hand side,provided that yi

′= 1 and T ′ = ∅. That is, we have

Lemma 19 Given an instance of PI and a cover S, then

φS(ti′+ di

′) = min

(x,y,s)∈XPI ,yi′=1

{∑S∪i′

si − (λ+∑i∈S

max{−ti, di − λ}(1− yi))},

where φS(·) is defined as in (34).

23

Figure 2: Graph of the function φS(·)

The proof of this Lemma is long and tedious. Moreover, it resembles the developmentof a similar lifting function in Marchand and Wolsey [1999]; therefore it is omitted here.

For each i′ ∈ P \ S, if yi′

= 0 and T ′ = ∅, it is clear that the maximum amount thatthe right hand side can be increased is di

′, if si

′is added to the left hand side; that is,

min(x,y,s)∈XPI ,yi

′=0

{∑S∪i′

si − (λ+∑i∈S

max{−ti, di − λ}(1− yi))}

= di′.

Therefore we define a function f(·, ·) : IR2+ → IR to represent the difference between these

two quantities:

f(t0, d0) = φS(t0 + d0)− d0.

It can be verified that

24

f(t0, d0) =

−d0 if t0 + d0 ≤ µ1,

t0 + (j − 1)λ− [µ1 +∑j

i=2(t[i] + d[i])] if µ1 +∑j

i=2(t[i] + d[i]) ≤ t0 + d0 ≤µ1 + λ+

∑ji=2(t[i] + d[i]), j ≤ k′ − 1

jλ− d0 if µ1 + λ+∑j

i=2(t[i] + d[i]) ≤ t0 + d0 ≤µ1 +

∑j+1i=2 (t[i] + d[i]), j ≤ k′ − 1

t0 + (k′ − 1)λ− [µ1 +∑k′

i=2(t[i] + d[i])], if t0 + d0 ≥ µ1 +∑k′

i=2(t[i] + d[i]).(35)

Now we can state the complete class of cover inequalities:

Proposition 20 (Lifted Cover Inequalities) Given a cover S of PI, order the i ∈ S asin Proposition 16. Also, define µ1 as in Proposition 16. Let (T,U) be any partition ofP \ S, let (T ′, T ′′) be any partition of T , and define

D′ = max{t[2] + d[2],maxi∈U{ti + di}}

If µ1 ≥ 0 and |S| ≥ 2, then

∑i∈S∪U s

i ≥ λ+∑

i∈S max{−ti, di − λ}(1− yi) +∑i∈U f(t, d)yi +

∑i∈U d

i +∑i∈S min{ti+di,λ}+

∑i∈U φ(ti+di)−λ

(|S|+|U |−1)D′∑

i∈T ′(xi − (µ1 − ti)yi) (36)

is valid for XPI .

Proof: Consider any point (x, y, s) ∈ XPI . Let U1 = {i ∈ U : yi = 1}, and letU0 = {i ∈ U : yi = 0}. First, consider the case when T ′1 = {i ∈ T ′ : yi = 1} = ∅.Let tU

1=∑

i∈U1 ti, and let dU1

=∑

i∈U1 di. Then Lemma 19 and the fact that φS(·) issuperadditive yield

∑i∈U1

φS(ti + di) ≤ φS(tUi

+ dUi)

= min(x,y,s)∈XPI ;yi=1,i∈U1

{∑S∪U1

si − (λ+∑i∈S

max{−ti, di − λ}(1− yi))}

≤∑S∪U1

si − (λ+∑i∈S

max{−ti, di − λ}(1− yi)).

25

This, the fact that si ≥ di, i ∈ U0, and the fact that φS(ti + di) = f(ti, di)yi + di, i ∈ U ,imply that (36) holds for (x, y, s).

If T ′1 6= ∅, then validity can be proven by considering a minimization problem similarto that described in the proof of Proposition 16. 2

Example 1 (continued)Again take S = {1, 2, 3} as our cover, so that λ = 3, t[1] + d[1] = t1 + d1 = 7, and µ1 = 4.Then k′ = 3, and

φS(u) =

0 if u ≤ 4,u− 4 if 4 ≤ u ≤ 7

3 if 7 ≤ u ≤ 103 + (u− 10) if 10 ≤ u ≤ 13

6 if 13 ≤ u ≤ 166 + (u− 16) if u ≥ 16,

(37)

Let U = {4}. Then φS(t4 + d4) = φS(13) = 6, and so f(t4, d4) = φS(t4 + d4) − d4 =6− 10 = −4. Then (25) yields

s1 + s2 + s3 + s4 ≥ 3 + 2(1− y1) + (1− y2) + 2(1− y3)− 4y4 + 10= 18− 2y1 − y2 − 2y3 − 4y4, (38)

a valid inequality for XPI . 2

Under certain conditions, cover inequalities define facets of conv(XPI). For example,we have

Proposition 21 Given an inequality of the form (36), let λ > 0, t[2] + d[2] > λ, ti <µ1, i ∈ T , and T ′ = ∅, Also, let k′ be defined as in (33); for each i ∈ U , assume thatti + di =

∑kj=1(t[j] + d[j]), for some k : 1 ≤ k ≤ k′ − 1. Then (36) induces a facet of

conv(XPI).

Proof: Again it suffices to exhibit 3P − T points that are in the intersection of thehyperplane defined by (36) and of XPI , and that are linearly independent of each otherand of the |T | rays si = 1, i ∈ T ,

To simplify notation we first assume that that di+d[1] ≥ λ, i ∈ {S\[1]}∪U . Note thatif we add the qualification xi = yi = 0, si = di, i ∈ U , to each of the points presented inthe proof of Proposition 15, then we have 3|S|+2|T | points that are linearly independentof each other and of the extreme directions, and they all satisfy (36) at equality. (This istrue because of the condition, implied in Proposition 15 and included in the statement of

26

Proposition 21, that T ′ = ∅.) Therefore, it suffices to show 3|U | points in the hyperplanedefined by (36) that are linearly independent of each other, of the 3|S| + 2|T | pointsjust mentioned, and of the |T | extreme rays. That is, for each item i′ ∈ U , it sufficesto show three points in the hyperplane defined by (36) that are linearly independent ofeach other and the points and rays already mentioned.

Let k′ = |{i ∈ S : ti + di > λ}|. Recall that, for each i′ ∈ U ,

ti′+ di

′=

k∑i=1

(t[i] + d[i]) = µ1 + λ+k∑i=2

(t[i] + d[i]),

for some k such that 1 ≤ k ≤ k′ − 1. So for each i′ ∈ U , consider the points

1.

xi′

= (di′ − λ)+, yi

′= 1, si

′= min{ti′ + di

′+∑|S|

i=k+1(t[i] + d[i])− c, di′} = min{λ, di′}x[i] = y[i] = 0, s[i] = d[i], i = 1, ..., k

x[k+1] = d[k+1] − (λ− di′)+, y[i] = 1, s[i] = (λ− di′)+,

x[i] = d[i], y[i] = 1, s[i] = 0, i = k + 2, ..., |S|xi = yi = 0, si = di, i 6∈ S ∪ i′;

2.

xi′

= di′, yi′

= 1, si′

= 0x[i] = y[i] = 0, s[i] = d[i], i = 1, ..., k + 1

x[i] = d[i], y[i] = 1, s[i] = 0, i = k + 2, ..., |S|xi = yi = 0, si = di, i 6∈ S ∪ i′;

3.

xi′

= di′+ t[k+1] + d[k+1] − λ, yi′ = 1, si

′= 0

x[i] = y[i] = 0, s[i] = d[i], i = 1, ..., k + 1x[i] = d[i], y[i] = 1, s[i] = 0, i = k + 2, ..., |S|

xi = yi = 0, si = di, i 6∈ S ∪ i′.

It can be checked that the 3|U | points defined in this manner are linearly independentof each other and of the previously given points and rays.

As before, when the assumption di + d[1] ≥ λ, i ∈ {S \ [1]} ∪ U , does not hold, theproof is similar. 2

27

There are facets induced by inequalities of the form (36) in which T ′ 6= ∅; conditionscan be stated that are similar to those in Proposition 18.

Although the conditions in Proposition 21 on the elements in U (and T ′) are restric-tive, it is not difficult to show that (36) induces faces of dimension at least 3P−|U |−|T ′|,if the restrictions on items in U and T ′ do not hold but the other conditions of the propo-sition do. The proof of this proceeds along the same lines as that given for Proposition 21,and uses many of the same linearly independent points. Moreover, when setup times anddemand are constant for all items, under mild conditions all cover inequalities (includingthose with T ′ 6= ∅) induce facets, since ti + di = t[1] + d[1], i ∈ P.

Example 1 (continued)

The inequality (38) satisfies the conditions of Proposition 21; in particular, note thatt4 + d4 = t1 + d1 + t2 + d2. Therefore (38) defines a facet of conv(XPI). 2

4.3 Completeness Issues

An interesting question is what other classes of inequalities define facets of conv(XPI). Incertain special cases, it may be possible to find a complete description of conv(XPI). Thequestion of defining a set of inequalities that characterizes conv(XPI) in the polynomiallysolvable case mentioned in Proposition 1 remains open. We can, however, state a similarresult that, given the cost conditions qi > 0, hi ≥ 0, i = 1, ..., P , has the same algorithmicimplications as a complete description would have.

Proposition 22 Given that qi > 0, hi ≥ 0, i = 1, ..., P , and given an instance of PIin which setup times and demand are constant for all items, optimizing (1) over thepolyhedron defined by (2)–(5), (13), (18), (36), and the bounds yi ≤ 1, i = 1, ..., P ,solves PI.

The extensive development necessary to prove this result can be found in Miller et al.[2000a]. Thus, the set of inequalities that we have defined for PI suffices to solve itby linear programming in this polynomially solvable case. We remark here that theextensions of the class of cover inequalities accomplished in Propositions 16 and 20 areessential for obtaining this result.

5 Conclusions and Future Directions

In this paper we have introduced a model PI that provides a relaxation of a number oflot–sizing and more general mixed integer models. We have stated results concerningthe structure and complexity of this model; among these, we have given three nontrivial

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classes of facet–inducing valid inequalities for PI. The first of these is analogous to thewell–known path inequalities; the other two are new. For one of the new classes, we usethe concepts of superadditivity and lifting in order to obtain the complete class. Wehave shown conditions under which both of these classes of inequalities define facets ofthe convex hull. Moreover, these classes cut off all fractional extreme optimal solutionsof XPI

LP , and they are sufficient to allow XPI to be solved by LP in a polynomial solvablecase.

In Miller [1999], a number of extensions to PI are considered. For example, whenvariable lower bounds are added to the formulation of MCL, it is possible to derive anextension of PI that is analogous to a single period of this new lot–sizing model. Theseextensions provide potentially useful relaxations for even more general models than thosefor which PI is a relaxation. Analysis of these other generalizations of PI is one area offuture work.

Computational experience using the inequalities presented for PI in solving MCLconfirms the expectation that they should be effective in solving MIP problems thathave substructures of the form PI (Miller et al. [2000b]). We expect that additionalcomputational experience will show that valid inequalities for PI and similar modelsprove useful in solving instances of the extended lot–sizing models mentioned above. Wealso expect that they will be useful in solving more general problems such as the fixedcharge network flow model mentioned in Section 2.

6 Acknowledgements

The authors would like to thank Laurence Wolsey for many helpful comments thatimproved the presentation of the results in this paper.

References

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Z. Gu, G.L. Nemhauser, and M.W.P. Savelsbergh. Lifted flow cover inequalities formixed 0–1 integer programs. Mathematical Programming, 85:439–467, 1999.

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