Polyhedral analysis for the two-item uncapacitated lot-sizing problem with one-way substitution

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Contents lists available atScienceDirect

Discrete Applied Mathematics

journal homepage:www.elsevier.com/locate/dam

Polyhedral analysis for the two-item uncapacitated lot-sizing problem

with one-way substitution

Hande Yaman

Bilkent University, Department of Industrial Engineering, Bilkent 06800 Ankara, Turkey

a r t i c l e i n f o

Article history:

Received 4 June 2008

Received in revised form 15 May 2009 Accepted 6 June 2009

Available online 26 June 2009

Keywords:

Lot-sizing Product substitution Polyhedral analysis Facet defining inequalities Convex hull

a b s t r a c t

We consider a production planning problem for two items where the high quality item can substitute the demand for the low quality item. Given the number of periods, the demands, the production, inventory holding, setup and substitution costs, the problem is to find a minimum cost production and substitution plan. This problem generalizes the well-known uncapacitated lot-sizing problem. We study the projection of the feasible set onto the space of production and setup variables and derive a family of facet defining inequalities for the associated convex hull. We prove that these inequalities together with the trivial facet defining inequalities describe the convex hull of the projection if the number of periods is two. We present the results of a computational study and discuss the quality of the bounds given by the linear programming relaxation of the model strengthened with these facet defining inequalities for larger number of periods.

1. Introduction

In this paper we consider the two-item uncapacitated lot-sizing problem with one-way substitution, that is the high qual-ity item can substitute the demand for the low qualqual-ity item. Given the planning horizon and the demands for the two items in each period, the aim of the problem is to propose a production and substitution plan which minimizes the production, inventory, setup and substitution costs over the planning horizon and meets the demand for the two items on time.

Most of the literature on product substitution is concerned with problems in a stochastic setting (see, e.g., [7,10,17,20,28, 30,31,36,37]). The literature in the deterministic setting is quite limited. Balakrishnan and Geunes [5] model the production planning problem with substitutable components and an arbitrary substitution structure. They derive properties of optimal solutions and propose a dynamic programming algorithm. The proposed algorithm finds a shortest path in a graph with

O

(

nm

₎

_{nodes and O}

₍

_nm+1

₎

_{arcs where n is the number of periods and m is the number of components. Hence the method’s}

worst case running time is exponential in the general case. When applied to the two-item problem, the algorithm runs in polynomial time and hence our problem is polynomially solvable. Geunes [18] models the same problem as an Uncapacitated

Facility Location (UFL) Problem, solves using the dual-ascent method of Erlenkotter [15] and presents computational results where the performance of the heuristic approach is tested in comparison to the exact shortest path approach.

Hsu, Li and Xiao [21] consider two versions of the production planning problem with one-way substitution. In the first version called Substitution with Conversion (SWC), a lower-index product requires a physical transformation to be able to substitute the demand of a higher-index product. Once the product undergoes this transformation it may be stocked as a higher-index product if it is not used immediately. In the second version of the problem called Substitution without Conversion

E-mail address:hyaman@bilkent.edu.tr.

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(SWO), there is no need for a physical transformation and the lower-index product can substitute the demand for a

higher-index product immediately but cannot be stocked as a higher-higher-index product. The authors prove that the SWO is a special case of SWC and is strongly NP-hard. They propose dynamic programming algorithms as well as a heuristic.

Li, Chen and Cai [25,26] consider product substitution together with remanufacturing. In [25], the authors propose a dynamic programming approach by extending the one of [5] to handle remanufacturing. They also propose a heuristic algorithm and present computational results. In [26], a genetic algorithm is proposed for a capacitated version of the problem with batch processing.

We have not encountered any study on the polyhedral analysis of the production planning problem with substitution. In this paper we focus on the simplest case of the problem, that is, we consider only two items and one-way substitution. We call this problem the Two-item Uncapacitated Lot-sizing Problem with One-way Substitution and abbreviate by 2ULS.

The problem we consider generalizes the well-known Uncapacitated Lot-sizing Problem (ULS). If the demand for the low quality item is zero for all periods in the planning horizon, then 2ULS reduces to ULS. There is a huge literature on the polyhedral properties of ULS and many of its variants (some examples are [2,4,3,6,12,13,19,23,24,27,29,32,34,38,39]). Even though ULS is a polynomially solvable problem, the knowledge on strong valid inequalities for the convex hull of its feasible solutions is useful in solving more complicated production planning problems for which ULS is a relaxation (see, e.g., [8,9,33,42]). Our motivation to study strong valid inequalities for 2ULS is similar. This simple problem may arise as a substructure in more complicated problems and the results derived in this paper can be used in devising solution algorithms for these problems.

Before concluding this section, we review the results on ULS and then summarize our results in this paper for 2ULS with emphasis on similarities and differences with the results for ULS.

There has been a lot of research on ULS and its variants since the seminal paper of Wagner and Whitin [41]. In [41], a dynamic programming algorithm that runs in O

(

n2

₎

_{time with n being the number of periods is given. Later, more efficient}

implementations that run in O

(

nlogn

)

time are proposed [1,16,40]. Barany et al. [6] give a description of the convex hull of feasible solutions using the so-called

(

l

,

S

)

-inequalities. Krarup and Bilde [22] give an extended formulation as a UFL and show that the linear programming (LP) relaxation of this formulation always has an optimal solution with integer setup variables. A shortest path extended formulation is given in Eppen and Martin [14]. For details, we refer the reader to Pochet and Wolsey [35].

In this paper, we first give a model of the 2ULS that we refer to as 2ULS model with substitution variables. Then we study the projection of the feasible set of the 2ULS model with substitution variables onto the space of production and setup variables. This projection gives a valid formulation for the case where the substitution costs are zero and the inventory holding costs are equal for the two items. We refer to this formulation as the 2ULS model without substitution variables and to the convex hull of its feasible solutions as the 2ULS polytope.

We also develop a UFL model and project it onto the space of production and setup variables. We characterize the nondominated projection inequalities; these inequalities are generalizations of the

(

l

,

S

)

-inequalities. We provide necessary and sufficient conditions for these inequalities to be facet defining for the 2ULS polytope. As in the case of ULS, the LP relaxation of the 2ULS model without substitution variables strengthened with these

(

l

,

S

)

-like-inequalities gives the same lower bound as the LP relaxation of the UFL model. Unlike in the case of ULS, if the planning horizon is longer than two periods, these LP relaxations may not have optimal solutions with integral setup variables. In other words, the projection of the UFL model onto the space of production and setup variables is not necessarily the same as the 2ULS polytope for three or more periods.

The rest of the paper is organized as follows. In Section2, we present the two models, the 2ULS model with substitution

variables and the 2ULS model without substitution variables and introduce a family of valid inequalities which generalize the

(

l

,

S

)

-inequalities. In Section3, we investigate the dimension and trivial facets of the 2ULS polytope, derive some properties of its facet defining inequalities, present a UFL model, project it onto the space of production and setup variables, and derive a class of facet defining inequalities. At the end of this section, we show that these inequalities together with trivial facet defining inequalities describe the 2ULS polytope if the number of periods is two. We report the results of a computational experiment to see the quality of the lower bounds obtained by solving the LP relaxation of the UFL model in Section4. We conclude with future research directions in Section5.

2. The models and the

(

l1

,

l2

,

S1

,

S2

)

-inequalities Let n be a positive integer and T

= {

1

, . . . ,

n

}

. Let pi_t

,

hi

t

,

qit, and ditdenote the unit production cost, the unit inventory

holding cost, the setup cost and the demand for item i

=

1

,

2 in period t

∈

T , respectively. The cost of substituting a unit

demand of item 2 with item 1 in period t

∈

T is denoted by ct. We assume that the costs are nonnegative and the starting

and ending inventories are zero for both items. For t1,t2

∈

T and i

=

1

,

2, we define Dit1t2

=

P

t2

t=t1d

i

tif t1

≤

t2and Dit1t2

=

0

if t1

>

t2.

To model the 2ULS, we define the following decision variables. Let xi

tbe the amount of production of item i

=

1

,

2 in

period t

∈

T , si_tbe the amount of item i

=

1

,

2 in inventory at the end of period t

∈

T

∪ {

0

}

, and yi_tbe 1 if production of item i

=

1

,

2 takes place in period t

∈

T and 0 otherwise. For t

∈

T , let a12

t and a22t be the amounts of items 1 and 2 used to

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Fig. 1. 5 period example as a fixed charge network model.

Now the 2ULS model with substitution variables is as follows:

z

=

min n

X

t=1 p1_tx1_t

+

p2_tx2_t

+

h1_ts1_t

+

h_t2s2_t

+

q1_ty1_t

+

q2_ty2_t

+

cta12t

(1) s.t. x1_t

+

s1_t₋₁

=

a12_t

+

d1_t

+

s1_t

∀

t

∈

T (2) x2_t

+

s2_t₋₁

=

a22_t

+

s2_t

∀

t

∈

T (3) a12_t

+

a22_t

=

d2_t

∀

t

∈

T (4) x1_t

≤

(

D1_tn

+

D2_tn

)

y1_t

∀

t

∈

T (5) x2_t

≤

D2_tny2_t

∀

t

∈

T (6) s1₀

=

s2₀

=

s1_n

=

s2_n

=

0 (7) s1_t

,

s2_t

,

a22_t

≥

0

∀

t

∈

T (8) a12_t

≥

0

∀

t

∈

T (9) x1_t

,

x2_t

≥

0

∀

t

∈

T (10) y1_t

,

y2_t

∈ {

0

,

1

}

∀

t

∈

T

.

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Constraints(2)and(3)are balance equations for items 1 and 2 respectively. Constraints(2)imply that the amount of item 1 on hand at period t is used to satisfy the demands of items 1 and 2 and the remaining amount will be the inventory at the end of this period. Constraints(3)are similar, but item 2 can only be used to satisfy its own demand. Constraints(4) ensure that the demand of item 2 is satisfied on time using items 1 and 2. Due to constraints(5)and(6), if there is no setup for a given item in a given period, then this item cannot be produced in that period. Constraints(7)fix the values of initial and ending inventories to zero. Constraints(8)–(11)are nonnegativity and binary requirements. The objective function(1) is the total production, inventory holding, setup and substitution costs for the two items over the planning horizon.

The multi-item problem with an arbitrary substitution structure is modeled in [5]. The above model is a simplified version of their model for 2 items and one-way substitution. Balakrishnan and Geunes [5] view their problem as a generalized network flow problem with concave costs defined on a specific directed network and derive some properties of optimal solutions. In our case, as we have only one-way substitution, the corresponding network also simplifies. An example for 5 periods is depicted inFig. 1. All production arcs originate at a given dummy node.

In this representation, for given setup vectors for the two items, the problem reduces to a minimum cost network flow problem. Thus any extreme point solution satisfies si

t−1xit

=

0 for all t

∈

T and i

=

1

,

2 and a12t a22t

=

0 for all t

∈

T . These

properties are given for the general case by Balakrishnan and Geunes [5] under the names Zero Inventory Production (ZIP)

Property and Homogeneous Product Lots (HPL) Property. As a result of these two properties, the authors conclude that the

problem has an optimal solution where the demand of each item is satisfied from the most recent setup of the same item or an item that can substitute it. This is called the Most Recent Usage (MRU) Property.

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Next, we present the 2ULS model without substitution variables. Let K

= −

P

n t=1(h 1 tD11t

+

h 2 tD21t

)

, p 1 t

=

p 1 t

+

P

n l=th 1 l, p2 t

=

p 2 t

+

P

n l=th2l and ct

=

P

n l=t

(

h2l

−

h1l

) +

ctfor t

∈

T .

Theorem 1. If ct

=

0 for all t

∈

T , then the 2ULS can be formulated as:

z

=

K

+

min n

X

t=1 p1_tx1_t

+

p2_tx2_t

+

q1_ty1_t

+

q2_ty2_t

(12) s.t. (5),(6),(10),(11) n

X

t=1 x1_t

+

n

X

t=1 x2_t

=

D1_1n

+

D2_1n (13) t

X

l=1 x1_l

≥

D1_1t

∀

t

∈

T (14) t1

X

l=1 x1_l

+

t2

X

l=1 x2_l

≥

D1_1t 1

+

D 2 1t2

∀

t1,t2

∈

T

:

t1

≥

t2. (15)

Proof. SeeAppendix Awhere we prove that the projection of the feasible set of the 2ULS model with substitution variables onto the space of production and setup variables is given by(5),(6),(10),(11)and(13)–(15).

To see the necessity of constraints(15)for t1

>

t2, consider a three period problem and assume that dit

=

1 for i

=

1

,

2

and t

=

1

,

2

,

3. The solution

(

x1

_,

_x2

_,

_y1

_,

_y2

₎

_{where x}1

1

=

x13

=

x22

=

2, x12

=

x21

=

x23

=

0, y11

=

y13

=

y22

=

1 and y1

2

=

y21

=

y23

=

0 satisfies constraints(5),(6),(10),(11),(13),(14), and constraints(15)for t1

=

t2. But this solution is

infeasible as there is no set up for item 1 in period 2 and no set up for item 2 in period 1, we need to have x1₁

≥

3. Now constraint(15)for t1

=

2 and t2

=

1 which reads x11

+

x12

+

x21

≥

3 eliminates this solution.

In this paper, we introduce a family of valid inequalities which generalize the

(

l

,

S

)

-inequalities for the ULS polytope. Let l1and l2in T be such that l1

≥

l2, l2

<

n, A1

= {

1

, . . . ,

l1

}

, A2

= {

1

, . . . ,

l2

}

, S1

⊆

A1, S2

⊆

A2with 1

∈

S1. The

(

l1,l2,S1

_,

_S2

₎

_-inequality

X

t∈A1_\_S1

(

D1_tl 1

+

D 2 tl2

)

y 1 t

+

X

t∈A2_\_S2 D2_tl 2y 2 t

+

X

t∈S1 x1_t

+

X

t∈S2 x2_t

≥

D1_1l 1

+

D 2 1l2

.

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is a valid inequality for the feasible sets of both the 2ULS model with substitution variables and the 2ULS model without

substitution variables.

We present an example of the

(

l1,l2,S1

_,

_S2

₎

_{-inequalities.} Example 1. Consider the following instance of 2ULS. Let n

=

3, d1

t

=

d2t

=

1, p1t

=

2, p2t

=

1

.

5, h1t

=

h2t

=

0

.

5, and

q1_t

=

q2_t

=

2 for t

=

1

,

2

,

3. When we solve the LP relaxation of the 2ULS model with substitution variables together with the constraint y1

1

=

1, we obtain the following optimal solution: x11

=

2, x12

=

1, x13

=

2, x22

=

1, a121

=

1, a123

=

1, a222

=

1, y1₁

=

1, y1₂

=

0

.

25, y1₃

=

1, y2₂

=

0

.

5 and other variables are zero.

Now if in a feasible solution, y1

2

=

y22

=

0, then we need x11

+

x21

≥

4 to be able to satisfy the demand for the first two

periods. Hence this inequality is valid when y1

2

=

y22

=

0. Next, we lift this inequality with variables y22and y12. Suppose that y2

2

=

1. As the production of item 2 in period 2 cannot be used to satisfy the demand of item 2 in period 1 and as it is not

possible to substitute item 2 for item 1, we still need x1₁

+

x2₁

≥

3. So inequality x₁1

+

x2₁

+

y2₂

≥

4 is a valid inequality when

y1

2

=

0. Finally, if y12

=

1, we still need x11

+

x21

+

y22

≥

2. Hence, inequality x11

+

x21

+

2y12

+

y22

≥

4 is a valid inequality.

This is the

(

l1,l2,S1

,

S2

)

-inequality for l1

=

l2

=

2, A1

= {

1

,

2

}

, A2

= {

1

,

2

}

, S1

= {

1

}

, and S2

= {

1

}

and is violated by the

optimal solution of the LP relaxation. 3. Polyhedral analysis

Let X be the set of vectors

(

x1

,

x2

,

y1

,

y2

) ∈

R2n+

× {

0

,

1

}

2nthat satisfy constraints(5),(6)and(13)–(15)and P

=

con

v(

X

)

be the 2ULS polytope.

In what follows, we assume that d1_t

>

0 and d2_t

>

0 for all t

∈

T . This implies that y1₁

=

1 for all

(

x1

,

x2

,

y1

,

y2

) ∈

X .

Let etbe the unit vector of size n with the tth entry equal to 1 and other entries equal to 0.

3.1. Dimension and trivial facets

All solutions

(

x1

_,

_x2

_,

_y1

_,

_y2

₎

_{in X satisfy}

P

t∈Tx1t

+

P

t∈Tx2t

=

D1n1

+

D21nand y11

=

1. In the theorem below, we show that

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Theorem 2. dim

(

P

) =

4n

−

2.

Proof. Suppose that all solutions

(

x1

_,

_x2

_,

_y1

_,

_y2

₎

_{in P satisfy}

_α

1_x1

₊

_α

2_x2

₊

_β

1_y1

₊

_β

2_y2

₌

_γ

_{. Let x}1

₌

₍

_D1 1n

+

D 2 1n)e1, x2

₌

_{0, y}1

₌

P

t∈Tet, y2

=

P

t∈Tet. The solution

(

x1

,

x2

,

y1

,

y2

)

is in P. For t

∈

T , both solutions

(

x1

+

et

−

e1,x2

,

y1

,

y2

)

and

(

x1

₋

_e_1,_x2

₊

_e

t

,

y1

,

y2

)

are in P for small enough

>

0. Hence

α

1t

=

α

t2

=

α

11for all t

∈

T . For t

∈

T

\ {

1

}

,

(

x1

,

x2

,

y1

−

et

,

y2

)

is in P. So

β

t1

=

0 for all t

∈

T

\ {

1

}

. For t

∈

T ,

(

x1

,

x2

,

y1

,

y2

−

et

)

is in P. Thus

β

t2

=

0 for all

t

∈

T . So all solutions

(

x1

_,

_x2

_,

_y1

_,

_y2

₎

_{in P satisfy}

P

j∈T

α

11x1j

+

P

j∈T

α

11x2j

+

β

11y11

=

γ

. As

(

x1

,

x2

,

y1

,

y2

)

is in P, we have

γ = β

1

+

α

11(D11n

+

D21n). Then

α

1x1

+

α

2x2

+

β

1y1

+

β

2y2

=

γ

is the sum of

α

11times

P

t∈Tx1t

+

P

t∈Tx2t

=

D11n

+

D21nand

β

1

1times y11

=

1. Hence the rank of the equality matrix of P is equal to 2. As P

⊆

R4n, it follows that dim

(

P

) =

4n

−

2.

Next, we prove that some of the constraints of the 2ULS model without substitution variables define facets of P. Theorem 3. The following inequalities define facets of P:

x1_t

≥

0 for t

∈

T

\ {

1

}

(17) x2_t

≥

0 for t

∈

T (18) y1_t

≤

1 for t

∈

T

\ {

1

}

(19) y2_t

≤

1 for t

∈

T (20) x1_t

≤

(

D1_tn

+

D_tn2

)

y1_t for t

∈

T

\ {

1

}

(21) x2_t

≤

D2_tny2_t for t

∈

T

.

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Proof. Let t

∈

T

\ {

1

}

and define F

= {

(

x1

,

x2

,

y1

,

y2

) ∈

P

:

x1_t

=

0

}

. Suppose that all solutions in F also satisfy

α

1_x1

₊

_α

2_x2

₊

_β

1_y1

₊

_β

2_y2

₌

_γ

_{. The solution}

₍

_x1

_,

_x2

_,

_y1

_,

_y2

₎

_{where x}1

₌

₍

_D1

1n

+

D21n)e1, x2

=

0, y1

=

P

j∈Tej, y2

=

P

j∈Tejis in F . For j

∈

T

\ {

t

}

, both solutions

(

x1

+

ej

−

e1,x2

,

y1

,

y2

)

and

(

x1

−

e1,x2

+

ej

,

y1

,

y2

)

are in F for small

enough

>

0. Also the solution

(

x1

−

e1,x2

+

et

,

y1

,

y2

)

is in F . Hence

α

1j

=

α

1

1 for all j

∈

T

\ {

t

}

and

α

2

j

=

α

1 1 for

all j

∈

T . For j

∈

T

\ {

1

}

, as

(

x1

_,

_x2

_,

_y1

₋

_e

j

,

y2

)

is in F , we have

β

j1

=

0. Similarly, for j

∈

T , as

(

x1

,

x2

,

y1

,

y2

−

ej

)

is

in F ,

β

2 j

=

0. As

(

x1

,

x2

,

y1

,

y2

)

is in F ,

γ = α

11

(

D11n

+

D21n) + β11. Hence

α

1x1

+

α

2x2

+

β

1y1

+

β

2y2

=

γ

is

P

j∈T

α

11x1j

+

P

j∈T

α

1 1x 2 j

+

(α

1t

−

α

1 1)x1 t

+

β

1 1y 1 1

=

α

1 1

(

D 1 1n

+

D 2 1n) + β1

1. This is a weighted sum of

P

j∈Tx 1 j

+

P

j∈Tx 2 j

=

D 1 1n

+

D 2 1n, y 1 1

=

1 and x1

t

=

0. The proof for inequalities(18)–(20)can be done in a similar way.

Let t

∈

T

\ {

1

}

, define F

= {

(

x1

_,

_x2

_,

_y1

_,

_y2

_{) ∈}

_P

_:

_x1

t

=

(

D1tn

+

D2tn

)

y1t

}

and suppose that all solutions in F also satisfy

α

1_x1

₊

_α

2_x2

₊

_β

1_y1

₊

_β

2_y2

₌

_γ

_{. Consider the solution}

₍

_x1

_,

_x2

_,

_y1

_,

_y2

₎

_{where x}1

₌

₍

_D1 1n

+

D 2 1n)e1, x2

=

0, y1

=

P

j∈T\{t}ej, y2

₌

P

j∈Tej. This solution is in F . Applying the same ideas as above, we can show that

α

j1

=

α

11and

β

j1

=

0 for all

j

∈

T

\ {

1

,

t

}

,

α

_j2

=

α

₁1and

β

_j2

=

0 for all j

∈

T . Moreover as

(

x1

,

x2

,

y1

,

y2

)

is in F , we also have

γ = α

11(D1_1n

+

D21n) + β₁1. Now consider the solution

(

x1

−

(

D1_tn

+

D2_tn

)

e1

+

(

D1tn

+

D

2

tn

)

et

,

x2

,

y1

+

et

,

y2

)

. This solution is also in F and hence

(α

1 t

−

α

11)(D1tn

+

D2tn

) + β

t1

=

0. So the equation

α

1x1

+

α

2x2

+

β

1y1

+

β

2y2

=

γ

is

P

j∈T

α

11x1j

+

P

j∈T

α

11x2j

+

(α

1t

−

α

1)1 x1t

+

β

1 1y 1 1

−

(α

1 t

−

α

11)(D1tn

+

D2tn

)

y1t

=

α

11(D11n

+

D 2 1n) + β1

1which is the sum of

α

1 1times

P

j∈Tx 1 j

+

P

j∈Tx 2 j

=

D11n

+

D 2 1n,

β

1 1 times y1

1

=

1 and

(α

t1

−

α

11

)

times x1t

−

(

D1tn

+

D2tn

)

y1t

=

0. The proof for inequalities(22)can be done in a similar way.

3.2. Properties of nontrivial facet defining inequalities of P

Now let inequality

P

j∈T

α

j1x1j

+

P

j∈T

α

j2x2j

+

P

j∈T

β

j1y1j

+

P

j∈T

β

j2y2j

≥

γ

be a facet defining inequality for P. Assume

that the facet defined by this inequality is different from those defined by inequalities(17)–(22). Without loss of generality, we can assume that

β

1

=

0,

α

1j

≥

0 and

α

j2

≥

0 for all j

∈

T , and

Q

j∈T

α

j1

α

2j

=

0 (add

−

β

11times y11

=

1 and

−

a times

P

j∈Tx 1 j

+

P

j∈Tx 2 j

=

D 1 1n

+

D 2

1nwhere a

=

min

{

minj∈T

{

α

j1

}

,

minj∈T

{

α

j2

}}

).

Next theorem presents some properties of the coefficients of variables in facet defining inequalities that are different from the trivial facet defining inequalities(17)–(22).

Theorem 4. If inequality

P

j∈T

α

1 jx 1 j

+

P

j∈T

α

2 jx 2 j

+

P

j∈T

β

1 jy 1 j

+

P

j∈T

β

2 jy 2

j

≥

γ

is facet defining for P, the facet defined by this

inequality is different from those defined by inequalities(17)–(22),

β

1

=

0,

α

1j

≥

0 and

α

j2

≥

0 for all j

∈

T and

Q

j∈T

α

j1

α

2j

=

0,

then

i.

β

1

j

≥

0 and

β

j2

≥

0 for all j

∈

T ,

ii.

γ >

0 and

α

1₁

>

0, iii. for j

∈

T , if

β

1

(6)

iv. for j

∈

T , if

β

_j2

=

0 then

α

_j2

≥

α

2_t for all t

∈

T with t

>

j,

v. for j

∈

T

\ {

1

}

and t

∈

T such that t

<

j and

β

1

t

=

0,

(α

1t

−

α

1j

)(

D1jn

+

D2jn

) ≥ β

j1,

vi. for j

∈

T

\ {

1

}

, if there exists t

∈

T such that t

<

j,

β

1

t

=

0 and

α

j1

=

α

t1, then

β

j1

=

0,

vii. for j

∈

T and t

∈

T such that t

≤

j and

β

1

t

=

0,

(α

t1

−

α

j2

)

D2jn

≥

β

j2,

viii. for j

∈

T and t

∈

T such that t

<

j and

β

_t2

=

0,

(α

_t2

−

α

2_j

)

D2_jn

≥

β

_j2,

ix. for j

∈

T , if there exists t

∈

T such that t

≤

j,

β

_t1

=

0 and

α

_j2

=

α

_t1, or if there exists t

∈

T such that t

<

j,

β

_t2

=

0 and

α

2

j

=

α

t2, then

β

2

j

=

0,

x. for j

∈

T

\ {

1

}

, if

β

_j1

=

0, then there exists t

∈

T with t

<

j such that

α

1_j

≥

α

_t1,

xi. for j

∈

T , if

β

_j2

=

0, then there exists t

∈

T with t

<

j such that

α

2_j

≥

min

{

α

1

t

, α

2t

}

or

α

j2

≥

α

1j, xii.

P

j∈T\{1}

β

j1

+

P

j∈T

β

j2

>

0, xiii. if

α

1

n

=

0 and

β

n1

=

0, then let t1be the smallest index in T with

α

t11

=

0 and

β

1

t1

=

0 and t2be the smallest index in T with

α

2 t2

=

0 and

β

2 t2

=

0. Then t2

≤

t1,

α

1 j

=

β

1

j

=

0 for all j

∈

T with j

≥

t1and

α

j2

=

β

2

j

=

0 for all j

∈

T with j

≥

t2,

xiv. if

α

1

n

+

β

n1

>

0,

α

2n

=

0, and

β

n2

=

0, let t2be the smallest index in T with

α

2t2

=

0 and

β

2 t2

=

0. Then

α

2 j

=

β

j2

=

0 for all j

∈

T with j

≥

t2, xv.

β

2 n

=

0. Proof. Let F

= {

(

x1

_,

_x2

_,

_y1

_,

_y2

_{) ∈}

_P

_:

P

j∈T

α

j1x1j

+

P

j∈T

α

j2x2j

+

P

j∈T

β

j1y1j

+

P

j∈T

β

j2y2j

=

γ }

. We prove the above items

one by one.

i. For j

∈

T

\ {

1

}

, since F is different from the facet defined by y1_j

≤

1, there exists a solution

(

x1

,

x2

,

y1

,

y2

)

in F such that

y1

j

<

1. The solution

(

x1

,

x2

,

y1

+

ej

,

y2

)

is in P for small enough

>

0. Hence,

β

j1

≥

0. Similarly, we can show that

β

2

j

≥

0 for all j

∈

T .

ii. In addition, as

α

1

j

≥

0 and

α

j2

≥

0 for all j

∈

T , x1, x2, y1and y2are nonnegative and F is different from the facets defined

by inequalities(17)–(22), we have

γ >

0. Assume

α

1

=

0. Then as

((

D11n

+

D21n)e1,0

,

e1,0

)

is in P, we have 0

≥

γ

. This contradicts

γ >

0. So

α

11

>

0.

iii. Let j

∈

T with

β

1

j

=

0. For t

∈

T with t

>

j, there exists a solution

(

x1

,

x2

,

y1

,

y2

)

in F such that x1t

>

0 since F is

different from the facet defined by x1

t

≥

0. The solution

(

x1

+

(

ej

−

et

),

x2

,

y1

−

(

y1j

−

1

)

ej

,

y2

)

>

0. Hence

α

1_j

≥

α

_t1. Similarly, we can prove that

α

1_j

≥

α

_t2for all j

∈

T with

β

_j1

=

0 and t

∈

T with t

≥

j.

iv. Similar to the proof of item iii.

v. Let j

∈

T

\ {

1

}

. There exists a solution

(

x1

_,

_x2

_,

_y1

_,

_y2

₎

_{in F such that x}1

j

< (

D1jn

+

D2jn

)

y1jsince F is different from the facet

defined by x1

j

≤

(

D1jn

+

D2jn

)

y1j. Let t be in T such that t

<

j. The solution

(

x1

−

x1j

(

ej

−

et

),

x2

,

y1

+

(

1

−

yt1

)

et

−

y1jej

,

y2

)

is in P. If

β

_t1

=

0, then

(α

_t1

−

α

_j1

)

x_j1

≥

β

_j1y1_j. As

α

1_t

≥

α

_j1and x1_j

< (

D1_jn

+

D2_jn

)

y1_j, we have

(α

_t1

−

α

_j1

)(

D_jn1

+

D2_jn

) ≥ β

_j1. vi. Let j

∈

T

\ {

1

}

. Suppose there exists t

∈

T such that t

<

j,

β

_t1

=

0 and

α

_j1

=

α

1

t. Then by (i) and (v), we have

β

j1

=

0.

vii. Let j

∈

T . There exists a solution

(

x1

_,

_x2

_,

_y1

_,

_y2

_{) ∈}

_{F such that x}2

j

<

D2jny2j since F is different from the facet defined by

x_j2

≤

D2_jny2_j. Let t be in T is such that t

≤

j and

β

_t1

=

0. As the solution

(

x1

+

x2_jet

,

x2

−

x2jej

,

y1

+

(

1

−

y1t

)

et

,

y2

−

y2jej

)

is in P, we have

(α

1

t

−

α

j2

)

D2jn

> β

j2.

viii. Similar to the proof of item (vii).

ix. Let j

∈

T . If there exists t

∈

T such that t

≤

j,

β

1

t

=

0 and

α

2j

=

α

1t, then by (i) and (vii), we have

β

j2

=

0. Similarly, if

there exists t

∈

T such that t

<

j,

β

_t2

=

0 and

α

_j2

=

α

_t2, then by (i) and (viii) we have

β

_j2

=

0. x. For j

∈

T

\ {

1

}

with

β

1

j

=

(

x1

,

x2

,

y1

,

y2

) ∈

F such that y1j

=

0. Then

P

j−1

i=1x1i

>

D11j−1and

P

j−1

i=1x1i

+

P

j−1

i=1x2i

>

D11j−1

+

D21j−1. Let t

∈

T be the largest index with t

<

j and x1t

>

0. Then

(

x1

+

(

ej

−

et

),

x2

,

y1

+

ej

,

y2

)

>

0. Hence

α

1j

≥

α

1

t.

xi. For j

∈

T with

β

2

j

=

(

x1

,

x2

,

y1

,

y2

) ∈

F such that y2j

=

0. Then

P

j

i=1x1i

+

P

j−1

i=1x2i

>

D11j

+

D21j−1.

Let t

∈

T be the largest index with t

≤

j and x1_t

>

0 or x2_t

>

0. Then either

(

x1

−

et

,

x2

+

ej

,

y1

,

y2

+

ej

)

or

(

x1

_,

_x2

₊

₍

_e

j

−

et

),

y1

,

y2

+

ej

)

>

0. Hence

α

j2

≥

min

{

α

t1

, α

t2

}

. If t

=

j, then

α

2j

≥

α

j1.

xii. If

β

_j1

=

β

_j2

=

0 for all j

∈

T , then by item (iii), we have

α

1₁

≥

α

₂1

≥

. . . ≥ α

1

n. Item (x) for j

=

2 implies that

α

1 2

≥

α

1 1.

Hence

α

1

2

=

α

11. Applying (x) for j

=

3 yields

α

31

=

α

12

=

α

11. Repeating this argument, we obtain

α

1j

=

a for all j

∈

T

and for some a

∈

_R.

Applying item (xi) for j

=

1, we get

α

2

1

≥

α

11. Together with

α

12

≤

α

11from item (iii), this gives

α

12

=

α

11. Now by item

(iv), we have

α

₁2

≥

α

₂2

≥

. . . ≥ α

_n2. Item (xi) for j

=

2 implies

α

2₂

≥

min

{

α

1, α1 2₁

}

or

α

₂2

≥

α

₂1. Both cases give

α

₂2

≥

α

₁1. As

α

2

≤

α

21from item (iii) and

α

21

=

α

11, we obtain

α

22

=

α

11. Repeating this argument iteratively, we can show that

α

2

j

=

a for all j

∈

T .

Hence we conclude that if

β

1

j

=

β

j2

=

0 for all j

∈

T , then

α

j1

=

α

2j

=

a for all j

∈

T and for some a

∈

R. But as

α

1 1

>

0 and

Q

j∈T

α

1 j

α

2

j

=

0 this is not possible. So

P

j∈T\{1}

β

1 j

+

P

j∈T

β

2 j

>

0.

(7)

xiii. Suppose that

α

1_n

=

0 and

β

_n1

=

0. Let t1be the smallest index in T with

α

t11

=

0 and

β

1

t1

=

0. Then (iii) implies that for j

∈

T with j

>

t1,

α

j1

≤

α

t11

=

0. As

α

1

j

≥

0, we have

α

1j

=

0. Now as

α

j1

=

α

t11, by (vi), we have

β

1

j

=

β

t11

=

0. Similarly, for

j

∈

T with j

≥

t1, (iii) implies that

α

j2

≤

α

t11

=

0. Again as

α

2

j

≥

0, we have

α

2j

=

0. Then by (ix), we have

β

j2

=

β

t11

=

0.

Let t2be the smallest index in T with

α

t22

=

0 and

β

2

t2

=

0. Then as

α

2

j

=

β

j2

=

0 for all j

∈

T with j

≥

t1, we have t2

≤

t1. Let j

∈

T with j

>

t2. By (iv), we have

α

2j

=

0. Then by (ix), we have

β

j2

=

0.

xiv. Similar to the proof of item (xiii).

xv. There exists a solution in

(

x1

_,

_x2

_,

_y1

_,

_y2

_{) ∈}

_F

_∩

_{X such that x}2

n

<

d2ny2nsince F is different from the facet defined by

x2

n

≤

d2ny2n. Now as

(

x1

,

x2

,

y1

,

y2

) ∈

X , we have yn2

=

1. If x2n

=

0, then as

(

x1

,

x2

,

y1

,

y2

−

en

)

is in P, we have

that

β

_n2

≤

0. As

β

_n2

≥

0 by (i), we have

β

_n2

=

0. Now suppose that x2_n

>

0. Then there exists t

∈

T such that

both

(

x1

₋

_e

t

,

x2

+

en

,

y1

,

y2

)

and

(

x1

+

et

,

x2

−

en

,

y1

,

y2

)

are in P or there exists t

∈

T

\ {

n

}

such that both

(

x1

,

x2

+

(

en

−

et

),

y1

,

y2

)

and

(

x1

,

x2

−

(

en

−

et

),

y1

,

y2

)

are in P. We will give the proof for the first case. So let t

∈

T be

such that both

(

x1

₋

_e

t

,

x2

+

en

,

y1

,

y2

)

and

(

x1

+

et

,

x2

−

en

,

y1

,

y2

)

are in P. Then

α

n2

=

α

1t. Now consider the solution

(

x1

+

x2_net

,

x2

−

xn2en

,

y1

,

y2

−

en

)

. This solution is also in P since

(

x1

,

x2

,

y1

,

y2

)

is in X and so y1t

=

1 and x1t

+

x2n

≤

D1tn

+

D2tn.

As

α

2

n

=

α

t1,

β

n2

≤

0. Together with (i), this implies

β

n2

=

0. The proof for the second case is similar.

By item (xii) ofTheorem 4, we know that

P

j∈T\{1}

β

j1

+

P

j∈T

β

j2

>

0 in any facet defining inequality that satisfies the

assumptions of the theorem. This leads to the following corollary. Corollary 1. If inequality

P

j∈T

α

1 jx 1 j

+

P

j∈T

α

2 jx 2

j

≥

α0

defines a face of P different from those defined by x

1

t

≥

0 for some

t

∈

T

\ {

1

}

and x2

t

≥

0 for some t

∈

T , then the inequality is not facet defining for P.

Item (ii) ofTheorem 4states that

α

₁1

>

0 in any facet defining inequality that satisfies the assumptions of the theorem. Since

Q

j∈T

α

j1

α

j2

=

0 is among the assumptions of the theorem, we have the following corollary.

Corollary 2. If inequality

P

j∈T

β

j1y1j

+

P

j∈T

β

j2y2j

≥

β0

defines a face of P different from those defined by y1t

≤

1 for some

t

∈

T

\ {

1

}

or y2

t

≤

1 for some t

∈

T , then the inequality is not facet defining for P.

These two corollaries imply that nontrivial facet defining inequalities of P involve both the production and setup variables. In the following section, we derive such a family of facet defining inequalities.

3.3. The UFL formulation and its projection onto the space of the production and setup variables

In this section, we derive the

(

l1,l2,S1

,

S2

)

-inequalities using the projection of the UFL formulation onto the space of production and setup variables.

Geunes [18] gives a UFL formulation for the multi-item problem with an arbitrary substitution structure. The proposed model in [18] is the so-called aggregate or weak model whereas here we focus on the strong UFL model.

For u and t in T such that u

≤

t, define

v

1

utand

v

ut2 to be the amount of production of items 1 and 2 in period u to satisfy

their own demands in period t, respectively. Define also

v

12

ut to be the amount of production of item 1 in period u to satisfy

the demand of item 2 in period t.

The UFL formulation for the 2ULS with ct

=

0, p1t

≥

0 and p2t

≥

0 for all t

∈

T is as follows.

z

=

K

+

min n

X

t=1 p1_tx1_t

+

p2_tx2_t

+

q1_ty1_t

+

q2_ty2_t

(23) s.t. (5),(6),(10),(11),(13) t

X

u=1

v

1 ut

=

d 1 t

∀

t

∈

T (24) t

X

u=1

v

12 ut

+

t

X

u=1

v

2 ut

=

d2t

∀

t

∈

T (25)

v

1 ut

≤

d1ty1u

∀

u

,

t

∈

T

:

u

≤

t (26)

v

12 ut

≤

d 2 ty 1 u

∀

u

,

t

∈

T

:

u

≤

t (27)

v

2 ut

≤

d2ty2u

∀

u

,

t

∈

T

:

u

≤

t (28) n

X

t=u

(v

1 ut

+

v

12 ut

) ≤

x 1 u

∀

u

∈

T (29)

(8)

n

X

t=u

v

2 ut

≤

x2u

∀

u

∈

T (30)

v

1 ut

, v

2 ut

, v

12 ut

≥

0

∀

u

,

t

∈

T

:

u

≤

t

.

(31)

Constraints(24)and(25)ensure that the demands of items 1 and 2 are satisfied on time. Constraints(26)–(28)imply that if a setup for an item does not take place in a given period, then the demand of later periods cannot be satisfied from production of that item in this period. Constraints(29)and(30)compute the amounts of production of items 1 and 2 in terms of

v

1_ut,

v

_ut2 and

v

_ut12variables. Here the use of inequality constraints rather than equations does not change the optimal value since p1

t

≥

0 and p2t

≥

0 for all t

∈

T . Finally, constraints(31)are nonnegativity constraints.

Now we project the feasible set of the LP relaxation of the above UFL formulation onto the space of production and setup variables.

Theorem 5. The projection of the feasible set of the LP relaxation of the UFL formulation onto the space of production and setup

variables is given by inequalities(5),(6),(10),(13), y1

t

≤

1 and y2t

≤

1 for all t

∈

T , and the

(

l1,l2,S1

,

S2

)

-inequalities(16)for

all l1and l2in T such that l1

≥

l2, l2

<

n, A1

= {

1

, . . . ,

l1

}

, A2

= {

1

, . . . ,

l2

}

, S1

⊆

A1, S2

⊆

A2with 1

∈

S1. Proof. For given values of x1

t

,

x2t

,

y1t, and y2tfor t

∈

T that satisfy(5),(6),(10),(13), and y1t

≤

1 and y2t

≤

1 for all t

∈

T , there

exists an assignment of values

v

1

ut

, v

12ut and

v

2utfor all u

,

t

∈

T such that u

≤

t satisfying(24)–(31)if and only if n

X

t=1 t

X

u=1

(

d1_ty1_u

β

_ut1

+

d2_ty1_u

β

_ut12

+

d2_ty2_u

β

_ut2

) +

n

X

t=1

(

x1_t

σ

_t1

+

x2_t

σ

_t2

) ≥

n

X

t=1

(

d1_t

α

_t1

+

d2_t

α

_t2

)

for all

(α, β, σ ) ≥

0 such that

β

1 ut

≥

α

1 t

−

σ

1 u

∀

u

,

t

∈

T

:

u

≤

t (32)

β

12 ut

≥

α

t2

−

σ

u1

∀

u

,

t

∈

T

:

u

≤

t (33)

β

2 ut

≥

α

2 t

−

σ

2 u

∀

u

,

t

∈

T

:

u

≤

t

.

(34)

Note here that we limited our attention to nonnegative

α

, since

σ ≥

0,

β ≥

0 and di

t

≥

0 for all i

=

1

,

2 and t

∈

T .

Let C

= {

(α, β, σ ) ≥

0

:

(32)–(34)

}

. Define B

= {

(

u

,

t

) :

u

,

t

∈

T

,

u

≤

t

}

. For a given

(α, β, σ ) ∈

C , define A1

_{= {}

_t

_∈

_T

_:

_α

1

t

>

0

}

, A2

= {

t

∈

T

:

α

2t

>

0

}

, S1

= {

t

∈

T

:

σ

t1

>

0

}

, S2

= {

t

∈

T

:

σ

t2

>

0

}

, B1

= {

(

u

,

t

) ∈

B

:

β

ut1

>

0

}

,

B2

= {

(

u

,

t

) ∈

B

:

β

_ut2

>

0

}

, and B12

= {

(

u

,

t

) ∈

B

:

β

_ut12

>

0

}

. Define also t_a1, t_a2, t_s1, and t_s2to be the largest indices in

A1

_,

_A2

_,

_S1_{, and S}2_{, respectively.}

Next, we investigate the extreme rays of the projection cone C . If

(α, β, σ )

is an extreme ray of C and

|

B1

∪

B12

∪

B2

∪

A1

_∪

_A2

_∪

_S1

_∪

_S2

_{| =}

_{1, then A}1

_∪

_A2

_{= ∅}

_{. These extreme rays give the redundant inequalities y}1

t

≥

0 and y2t

≥

0 for t

∈

T

and x1₁

≥

0 and the facet defining inequalities x1_t

≥

0 for t

∈

T

\ {

1

}

and x2_t

≥

0 for t

∈

T . Now, we study the remaining

extreme rays in the following lemma.

Lemma 1. If

(α, β, σ)

is an extreme ray of C ,

|

B1

_∪

_B12

_∪

_B2

_∪

_A1

_∪

_A2

_∪

_S1

_∪

_S2

_{| ≥}

_{2, then}

_α

1

t

=

ρ

for t

∈

A1,

α

t2

=

ρ

for

t

∈

A2,

σ

_t1

=

ρ

for t

∈

S1,

σ

_t2

=

ρ

for t

∈

S2,

β

_ut1

=

ρ

for

(

u

,

t

) ∈

B1,

β

_ut2

=

ρ

for

(

u

,

t

) ∈

B2, and

β

_ut12

=

ρ

for

(

u

,

t

) ∈

B12 for some

ρ >

0. Moreover

β

1

ut

=

(α

t1

−

σ

u1

)

+_,

_β

12 ut

=

(α

t2

−

σ

u1

)

+_{, and}

_β

2 ut

=

(α

2t

−

σ

u2

)

+_{for all}

₍

_u

_,

_t

_{) ∈}

_{B, t}1 s

≤

max

{

ta1

,

ta2

}

and t2 s

≤

ta2.

Proof. Let

(α, β, σ ) ∈

C be such that

|

B1

_∪

_B12

_∪

_B2

_∪

_A1

_∪

_A2

_∪

_S1

_∪

_S2

_{| ≥}

_{2. Then}

_{(α, β, σ ) =}

₁

_/

₂

_{(α, β, σ ) +}

₁

_/

₂

_{(α, β, σ)}

where

α

1_t

=

α

1 t

−

and

α

1t

=

α

1t

+

for t

∈

A1,

α

1 t

=

α

1t

=

α

t1for t

∈

T

\

A1,

α

2 t

=

α

2t

−

and

α

2t

=

α

2t

+

for t

∈

A2,

α

2 t

=

α

2t

=

α

t2for t

∈

T

\

A2,

σ

1 t

=

σ

t1

−

and

σ

1t

=

σ

t1

+

for t

∈

S1,

σ

1 t

=

σ

1t

=

σ

t1for t

∈

T

\

S1,

σ

2 t

=

σ

t2

−

and

σ

2 t

=

σ

t2

+

for t

∈

S2,

σ

2 t

=

σ

2t

=

σ

t2for t

∈

T

\

S2,

β

1 ut

=

β

ut1

−

and

β

1 ut

=

β

1 ut

+

if

(

u

,

t

) ∈

B1and

σ

u1

=

0,

β

1 ut

=

β

1 ut

=

β

1 utif

(

u

,

t

) ∈

B1and

σ

u1

>

0 or

(

u

,

t

) ∈

B

\

B1,

β

2 ut

=

β

ut2

−

and

β

2 ut

=

β

2 ut

+

if

(

u

,

t

) ∈

B2and

σ

u2

=

0,

β

2 ut

=

β

2_ut

=

β

ut2 if

(

u

,

t

) ∈

B2and

σ

u2

>

0 or

(

u

,

t

) ∈

B

\

B2,

β

12 ut

=

β

ut12

−

and

β

12_ut

=

β

ut12

+

if

(

u

,

t

) ∈

B12and

σ

u1

=

0,

β

12

ut

=

β

12_ut

=

β

ut12if

(

u

,

t

) ∈

B12and

σ

u1

>

0 or

(

u

,

t

) ∈

B

\

B12. Besides, both

(α, β, σ )

and

(α, β, σ)

are in C for some small

enough

>

0. Hence if

(α, β, σ)

is an extreme ray of C , then all its positive entries should be equal. The second part is easy to prove.