brahimi n.-mathemtical models and lagrangian heuristics

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8th International Conference of Modeling and Simulation - MOSIM10 - May 10-12, 2010 - Hammamet - Tunisia

Evaluation and optimization of innovative production systems of goods and services

MATHEMATICAL MODELS AND LAGRANGIAN HEURISTICS

FOR A TWO-LEVEL LOT-SIZING PROBLEM WITH BOUNDED

INVENTORY

N. BRAHIMI N. ABSI S. DAUZERE-PERES S. KEDAD-SIDHOUM

University of Sharjah Ecole des Mines de Saint-Etienne Universite Pierre

Department of Industrial Centre Microelectronique de Provence et Marie Curie LIP6

Engineering and Management 880 route de Mimet 4 Place Jussieu

P.O. Box 27272 Sharjah F-13541 Gardanne, France F-75252 Paris Cedex, France

United Arab Emirates {absi, dauzere-peres}@emse.fr [email protected]

[email protected]

ABSTRACT: We consider a two-level lot-sizing problem where the first level consists of N end productscompeting for a single type of raw material (second level), which is supposed to be critical. In particular, thestorage capacity of raw materials is limited and must be carefully managed. The goal is to simultaneouslydetermine an optimal replenishment plan for the raw material and optimal production plans for the endproducts on a horizon of T periods. The problem is modeled as an integer linear program and solved using botha Lagrangian relaxation-based heuristic and a commercial optimization software. The results obtained usingthe Lagrangian heuristic are promising and new ideas are generated to further improve the quality of the solution.

KEYWORDS: Production Planning, Lot-Sizing, Lagrangian, Heuristic, Multi-level.

1 INTRODUCTION

In this paper, we address the two-level multi-item lot-sizing problem with bounded inventory (2LLSP-BI).It is a production planning problem where a timevarying demand of N end products has to be pro-duced over a planning horizon of length T. The endproducts require a critical raw material. Indeed, thestorage capacity for the raw material can be limitedor the raw material is shipped over long distances.Both situations imply that the consumption by theend products of the raw material must be carefullymanaged. In this work, we focus on both determin-ing an optimal replenishment plan for the raw mate-

rial and optimal production plans for the end prod-ucts over the planning horizon. The problem is closeto the disassembly planning problem with the objec-tive of minimizing the amount of waste by meansof recycling and remanufacturing (see Lambert andGupta (2005)).

Drexl and Kimms (1997) give a detailed review of thedefinition of different lot-sizing problems from simplesingle-level uncapacitated to complex multi-level ca-pacitated problems. Eftekharzadeh (1993) as well asRizk and Martel (2001) provide a review paper onmulti-stage lot-sizing problems. We also quote the

survey papers of Brahimi et al. (2003) and Karimi

et al. (2003) for single-level problems. Crowston and

Wagner (1997) and Kim et al. (2007) present a liter-ature review for the planning problem in disassemblysystems.

For the two-level disassembly problem, Kim etal. (2008) present a polynomial algorithm where aproduct (usually mechanical or electronic equipment)is to be disassembled while satisfying the demand ofleaf items over a given planning horizon with the ob-jective of minimizing the sum of setup and inventoryholding costs. The differences between our problemand the one suggested in Kim et al. (2008) are pre-sented in the following assumptions of their paper.

There is no shortage for the used product (to be dis-assembled). Whenever the disassembly process is setup, there are always enough used product to process.Moreover, no partial extraction of components is al-lowed, when a certain amount of the used product isprocessed all its components are extracted in a pro-portional way (based on the gozinto factor).

In the problem addressed in this paper, the equivalentof the used product to disassemble is a Raw Material(RM) which can be involved in the production of dif-ferent types of derived products. The whole quantity

of processed RM can be used to make one single typeof derived product. We note that the RM can be pur-


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chased and kept in stock. This is different from thetwo-level disassembly problem presented by Kim etal. (2008) where they assume that the raw material(the product to be processed) is always available. Inthis paper, we must decide on when to order (pro-duce) the RM. Additionally, they do not consider thecost related to purchasing and/or holding the prod-

uct, the only costs considered in their model are thesetup related to the disassembly process and the hold-ing cost of the components. We consider (at leastinitially) production (purchasing), holding and setupcost of the RM and all derived products. Moreover,the production variables in their model must be in-teger (they usually represent equipment to be disas-sembled).

The structure of the paper is as follows. Mathemati-cal formulations of the problem are presented in Sec-tion 2 as well as a complexity analysis. Section 3proposes a Lagrangian-based heuristic to solve the

problem. The efficiency of the solving method is il-lustrated by computational experiments in Section 4.Finally, some concluding remarks and research issuesare discussed in Section 5

2 MATHEMATICAL PROGRAMMINGMODELS

In this section, we present two mathematical formula-tions, namely a straightforward aggregate model anda more efficient disaggregate formulation.

2.1 Aggregate formulationThe elements of the mathematical programmingmodel are given in Table 1.

Symbols Description

Ranges

i = 0, 1,..,N Index of products. i = 0

corresponds to the raw material

t = 1,..,T Index of time periods

Parameters

dit Demand for product i in period t

pit Production or purchasing cost of

product i in period t

hit

Inventory holding cost of product iin period t

sit Setup cost of product i in period t

gi Number of units of raw material

required to make one unit of item i

Imax0t

Warehouse capacity for raw material

Variables

xit Production level of item i in period t

Iit Inventory level of item i in period t

yit Binary setup variable equal to 1

if xit > 0 and zero otherwise

Table 1: Elements describing the aggregate model

The aggregate formulation of the 2LLSP-BI problemis as follows:

minNi=0

Tt=1

(pitxit + sityit + hitIit) (1)

s.t.

Ii,t1 + xit = Iit + dit, i = 1, . . . , N , t = 1, . . . , T (2)

I0,t1 + x0t =Ni=1

gixit + I0t, t = 1, . . . , T (3)

I0t Imax0t , t = 1, . . . , T (4)

x0t y0t

Ts=t

Ni=1

gidis, t = 1, . . . , T (5)

xit yit

Ts=t

dis, i = 1, . . . , N , t = 1, . . . , T (6)

yit {0, 1} , i = 0, . . . , N , t = 1, . . . , T (7)

xit, Iit 0, i = 0, . . . , N, t = 1, . . . , T (8)

The objective (1) is to minimize the total production,purchasing, setup and inventory holding costs. Con-straints (2) and (3) are the inventory balance equa-tions for the end products and raw material, respec-tively. Constraint (4) restricts the inventory level tobe less than the maximum warehouse capacity re-served for the raw material. Constraints (5) and (6)link the continuous production variables xit to thebinary setup variables yit. Both of them translatethe condition yit = 1 if xit > 0 and 0 otherwise fori = 0, , N and t = 1, , T. Constraint (5) is writ-

ten for the raw material separately because the bestvalue of the big-M associated with this constraint isdifferent from that of Constraint (6). Finally, theintegrity and non-negativity Constraints are (7) and(8).

2.2 Disaggregate formulation

The aggregate formulation presented in the previoussection is very intuitive and easy to understand. How-ever, we expect an integer linear programming solverto take too much time to find an optimal solution ifthis model is used. This is why we derived a suppos-

edly more efficient formulation where the productionvariables xit are disaggregated into variables z0ist asfollows for i = 1, . . . , N , t = 1, . . . , T and s = 1, . . . , t:

xit =1

gi

Tt=s

z0ist

Here, the variable z0ist represents the quantity of rawmaterial processed in period s to satisfy (part of) thedemand of product i in period t. This is why we haveadded the dummy index 0 to z0ist.

For a more elegant presentation of the model, we in-troduce another continuous variable 0ks which corre-


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sponds to the quantity of raw material produced (orordered) in period s to satisfy the demand of periodt. Figure (1) illustrates the relationship between 0st,z0ist and the demand dit. The raw material receivedin period k will be used in part of the production inperiod s. This production will be used to partiallysatisfy the demand in period t. A new parameter in

the disaggregate model is the cost aist for i = 0,..,N,t = 1,..,T and s = 1, . . ,t where aist = pis +

t1l=s

hil.The disaggregate formulation of the 2LLSP-BI prob-lem is as follows:

k

0ks

iz0ist

dit

ts

Figure 1: Relationship between continuous variables

and demands

min

Ni=1

Tt=1

sityit +Tt=1

ts=1

a0st0st

+

Ni=1

Tt=1

ts=1

aistgiz0ist (9)

s.t.t

s=1

z0ist = gidit, i = 1, , N , t = 1, , T (10)

ts=1

0st =Tl=t

Ni=1

z0itl, t = 1, , T (11)

Tj=t

0tj Ni=1

Tj=t

z0itj Imax0t , t = 1, , T (12)

0st y0s

Tl=t

Ni=1

gidil, s = 1, ,t, t = 1, , T (13)

z0ist yisgidit, i = 1, ,N, s,t = 1, , T , s t (14)

yit {0, 1} , i = 0, , N , t = 1, , T (15)

0st 0, t = 1, , T , s = 1, , t (16)

z0ist 0, i = 1, , N , t = 1, , T , s = 1, , t (17)

To our surprise, the standard solver solves the aggre-gate formulation in Section 2.1 faster than the dis-aggregate formulation presented here. A comparisonbetween the results obtained using the two models isshown in Section 4.2.

Although it is not detailed in this paper, it can beshown that this problem is as hard as the classical Ca-

pacited Lot-Sizing Problem (CLSP), which is knownto be strongly NP-hard (see Chen and Thizy (1990)).

2.3 Problem complexity

The Uncapacitated Dynamic Demand Joint Replen-ishment Problem (UDJRP) consists in determiningthe replenishment/production schedule of productsbelonging to the same family. Whenever there is pro-duction of a subset of items in a given period, a joint

(family) setup is incurred in addition to the setups ofthe individual items. One of the earliest publicationson the the subject is Zangwill (1966). For a recent lit-erature review of the problem, the reader is referredto Robinson et al. (2009). The UDJRP is an NP-hardproblem (Arkin et al. (1989)) that has been treatedextensively in the literature. We will show that theUDJRP is a special case of the 2LLSP-BI and thusthat the latter is NP-hard.

The classical MILP formulation of the UDJRP isshown below. St and Zt represent the family setupcost and binary setup variable respectively.

min

Tt=1

(StZt +Ni=1

(pitxit + sityit + hitIit)) (18)

s.t.

Ii,t1 + xit = Iit + dit, i = 1, . . . , N , t = 1, . . . , T (19)

xit yit

Ts=t

dis, i = 1, . . . , N , t = 1, . . . , T (20)

N

i=1

yit N Zt, t = 1, . . . , T (21)

Zt {0, 1} , t = 1, . . . , T (22)

yit {0, 1} , i = 1, . . . , N , t = 1, . . . , T (23)

xit, Iit 0, i = 1, . . . , N , t = 1, . . . , T (24)

This model can be obtained from the aggregate for-mulation of the 2LLSP-BI through the followingsteps:

1. Set the raw material inventory capacity to zero.

2. Set the production and inventory holding costsof the raw material to zero.

3. Let Zt = y0t, St = s0t, t.

4. Considering Constraints (3), (4) and I0t 0(now =0); and knowing that the received rawmaterial in a period t must be totally used in theproduction in the same period, we obtain the fol-lowing constraint: x0t =

Ni=1

xit. Though thisconstraint becomes redundant in the model (itjust calculates x0t), it states that If raw ma-terial is received in period t, then there must

be production of at least one product. Thisis translated by

Ni=1

yit N Zt, t = 1, . . . , T


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and constraint (5) (x0t y0t

Ts=t

Ni=1

gidis or

x0t ZtT

s=t

Ni=1

gidis) can be removed fromthe model.

Thus, we conclude that the UDJRP is a special caseof the 2LLSP-BI and consequently the latter is NP-hard.

Theorem. The 2LLSP-BI is NP-hard.

3 LAGRANGIAN RELAXATION BASEDHEURISTIC

A Lagrangian relaxation approach is often used tosolve multi-item lot-sizing problems with couplingconstraints. The main idea of this method is todecompose a multi-item lot-sizing problem into sev-eral easy to solve single item lot-sizing problems byrelaxing the complicating constraints. When con-sidering the two-level lot-sizing problem defined byConstraints (1)-(8), the coupling constraints can beidentified as Constraints (3) and (4). If we considerthe relaxation of Constraints (3) and (4) simultane-

ously, we obtain a problem that decomposes into Nclassical single-item Uncapacitated Lot-Sizing prob-lem (ULS) and a particular single item problem thatcan be solved analytically. Each ULS sub-problem issolved using an O(T log T) dynamic programming al-gorithm (see Wagelmans et al., 1992). In the remain-der of this section, we present the general skeleton ofthe Lagrangian heuristic, the models resulting fromthe Lagrangian relaxation based on the aggregate for-mulation of the 2LLSP-BI problem, and some validinequalities that can improve the Lagrangian basedlower bound.

3.1 Lagrangian relaxation approach

The following algorithm summarizes the generalscheme of our Lagrangian relaxation.

3.2 Restructuring and decomposing theproblem

We relax in the following model Constraints (3) and(4). We can combine Constraints (3) and the non-negativity constraints of the inventory variables forraw material to derive an aggregate formulation with-

out inventory variables for the raw material. The La-grangian relaxation will be based on this model.

Algorithm 1 Lagrangian relaxation algorithm

Step 1: Initialization.a. Initialize all multipliers to 0.b. Set iteration number k = 1.c. Initialize step length, 1 (set to 2 in our case).d. Initialize the lower bound LB = and the up-per bound U B = .

Step 2: Solving the relaxed problem. Solve La-grangian problem Lk (See Section 3.3) and calculatecurrent lower bound LBk.Step 3:Incumbent saving. If L B < L Bk, thenLB := LBk.Step 4: Smoothing heuristic. Use the values of X,I and Y obtained in Step 2 and heuristic procedureto determine a feasible solution and an upper boundU Bk. The procedure described in Section 3.5 is usedin our case. If UB > UBk, then U B := U Bk

Step 5: Updating multipliers. Lagrangian multi-pliers are updated using the subgradient optimization

method (See Parker and Rardin, 1988)Step 6: Stopping conditions. If any stopping con-dition is met, then stop. The stopping conditions areeither an optimal solution is found, i.e. LB = U B,or the number of iterations k reaches 150.Step 7: Update step length. k+1 = k/2.0 if thebest lower bound is not improved during the last 10iterationsStep 8:Increment k and go to Step 2.

min

Ni=1

Tt=1

(pitxit + sityit + hitIit)

+Tt=1

(p0tx0t + s0ty0t)

+ h0t(t

k=1

x0k t

k=1

Ni=1

gixik) (25)

s.t.

Ii,t1 + xit = Iit + dit, i = 1, , N , t = 1, , T (26)

t

k=1N

i=1 gixik t

k=1x0k, t = 1, , T, (27)t

k=1

x0k t

k=1

Ni=1

gixik Imax0t , t = 1, , T, (28)

x0t y0t

Tk=t

Ni=1

gidik, t = 1, , T (29)

xit yit

Tk=t

dik, i = 1, , N , t = 1, , T (30)

yit {0, 1} , i = 1, , N , t = 1, , T (31)

xit, Iit 0, i = 1, , N , t = 1, , T (32)


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The non-satisfaction of Constraints (27) and (28) ispenalized in the objective function using Lagrangianmultipliers. We associate the positive Lagrangianmultipliers t and t with Constraints (27) and (28),respectively. When using the Lagrangian relaxationof Constraints (27) and (28) to derive a lower boundfor the above problem, we obtain the following prob-

lem denoted 2LLSP-BI-LAG that must be solved op-timally:

min T CL =Ni=1

Tt=1

(sityit + hitIit)

+

Ni=1

Tt=1

xit

pit + gi

Tk=t

(k h0k k)

+Tt=1

x0t

p0t +

Tk=t

(h0k k + k)

+

T

t=1

(s0ty0t)

+Tt=1

t (Imax0t ) (33)

s.t. (26), (29), (30), (31), (32).

Note thatTt=1

t (Imax0t ) is a constant.

3.3 Solving the sub-problems

The 2LLSP-BI-LAG problem defined by (33) andConstraints (26), (29), (30), (31), and (32) can bedecomposed into two types of problems:

The first problem is defined by the followingmodel:

min T CL1 =Ni=1

Tt=1

(sityit + hitIit)

+N

i=1

T

t=1

pit + giT

k=t

(k h0k k) xit(34)

s.t. (26), (30), (31), (32). This problem de-composes into N (i = 1, . . . , N ) independentclassical single-item uncapacitated lot-sizing sub-problems with T periods. Each sub-problem canbe solved using an O(T log T) dynamic program-ming algorithm (see Wagelmans et al. (1992)).

The second problem denoted by 1LLSP-RM is aparticular single item lot-sizing problem for i = 0

defined by the following mathematical formula-tion:

min

Tt=1

s0ty0t +Tt=1

x0tpL0t (35)

s.t.

x0t y0t

T

k=t

N

i=1

gidik t = 1, , T (36)

y0t {0, 1} t = 1, , T (37)

x0t 0 t = 1, , T (38)

where pL0t = p0t +T

k=t (h0k k + k) is the La-grangian production cost.

This problem can be further decomposed into T singleperiod problems where the objective is to minimizes0ty0t + x0tp

L0t subject to Constraints (36), (37), and

(38) for every period t.

Note that x0t will have a positive value (hence y0t =1) ifs0t+x0tp

L0t < 0, which can only happen ifp

L0t < 0.

This implies that x0t > s0t

pL0t.

Combining this result with Constraint (36), we con-clude that the solution of the single period problemis for each period t = 1, , T:

x0t =

0, ifTk=t

Ni=1

gidik s0t

pL0tT

k=t

N

i=1

gidik, otherwise

3.3.1 Improving the Lagrangian lower bound

Proposition. The following constraints are valid in-equalities for the problem defined by (25)-(32).

tk=1

x0k Ni=1

gi

tk=1

dik, t = 1, , T (39)

Proof. We first rewrite Constraint (27):

tk=1

x0k Ni=1

gi

tk=1

xik, t = 1, , T

We know, from the inventory balance equations andthe non-negativity constraints of the inventory vari-ables that, for every item i,

tk=1

xik t

k=1

dik, t = 1, , T

Hence:

tk=1

x0k

Ni=1

gi

tk=1

dik, t = 1, , T


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Which can also be written in a more convenient wayas:

tk=1

x0k t

k=1

Ni=1

gidik, t = 1, , T

This constraint will be added to the subproblems rep-resented by (35) to (38) which will improve the La-grangian lower bounds.

3.3.2 Solving the resulting raw material

problem

After adding valid inequalities (39) to the 1LLSP-RMmodel, the derived model denoted by 1MLLSP-RMbecomes a single item lot-sizing problem with zeroinventory cost. The O(T log T) algorithm of Wagel-mans et al. (1992) is well adapted to solve problemswith this structure. In this case, the input of the algo-rithm of Wagelmans et al. (1992) are for t = 1, , T:

Demand: d0t =Ni=1

gidit

Setup cost: s0t

Marginal cost: = pL0t

3.4 Obtaining an initial feasible solution

In this section, we present three construction heuris-tics in the chronological order of their development.The three heuristics build an initial solution thatwill eventually be improved in the iterations of theLagrangian heuristic. At the end of the tests, weadopted the third heuristic. The aim of presentingthe two first heuristics here is to present the stepsthat we followed before building the most efficientheuristic WHK-BI.

3.4.1 L4L-L4L heuristic

This heuristic implements first applies the Lot-for-

Lot (L4L) algorithm to derive the production plansof end products. The replenishment levels of the rawmaterial are equal to the aggregation of the produc-tion quantities of the end products in each period.This can be summarized as: This is equivalent to saythat the solution of the problem using the L4L-L4Lis calculated using the formulas:

xit = dit, i = 0, . . . , N , t = 1, . . . , T

and

x0t =

Ni=1

gidit, t = 1, . . . , T

3.4.2 WHK-L4L heuristic

Before calculating the raw material production usingthe L4L algorithm, the production plan of the endproducts is calculated using the algorithm of Wagel-mans et al. (1992) (WHK) which runs in O(T log T).

Note that in the WHK-L4L and L4L-L4L heuristicswe are sure that the resulting solution is feasible; sinceall demands are respected and the inventory level ofthe raw material is always equal to zero. Thus the in-ventory capacity constraint is always respected. Themain drawback of these two heuristics is that the L4Lsolution for the raw material generates too many se-tups and thus considerably increases the total costs.

3.4.3 WHK-BI heuristic

In this heuristic the plans of the end products arealso calculated using the WHK algorithm. The raw

material demands correspond to the aggregation ofthe resulting production levels of end products. Theraw material production plan is determined by solvinga single item problem with bounded inventory (ex-plicit consideration of the inventory capacity for rawmaterial). The single item problem with boundedinventory was first solved by Love (1973) using anO(T3) algorithm. Recently, Liu (2008) designed anO(T2) algorithm based on the geometric techniquesof Wagelmans et al. (1992).

On the other hand, it was shown by Wolsey (2006)that the single item problem with bounded inven-tory is equivalent to the single item problem withnon-customer specific time windows (introduced byDauzere-Peres et al. (2002)), which was first solvedin O(T4) by Dauzere-Peres et al. (2002), and thenin O(T2) by Wolsey (2006). It is very easy to con-vert the data of the bounded inventory problem intoa non-customer specific time window problem usinga simple heuristic with linear time complexity.

Our heuristic is based on the latter approach. The al-gorithm that converts the data of the bounded inven-tory problem into a non-customer specific time win-

dow problem and is summarized below. This pro-cedure first calculates the total available and duedemand for all periods (Wolsey (2006)), then usesthese aggregate demands to calculate the individualtime window demands (Dauzere-Peres et al. (2002)).Note that the algorithm mostly adopts the notationused in Wolsey (2006). Once the time window de-mands are built, the O(T2) algorithm proposed byWolsey (2006) is used to solve the problem.


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Algorithm 2 Procedure to convert demands d0t tonon-customer specific time windows problems.

Declarations:d0t: Aggregate demand of raw materialD0t =

tk=1 d0t: Cumulative demand of RM from

period 1 to period tv0t: Aggregate demand available for processing in

period tV0t =

tk=1 v0t: Cumulative available demand from

period 1 to period tdw0k: demand of time window (interval) k withavailability date bk and delivery date ek

Compute d0t =N

i=1gixit , t

Compute D0tCompute the available demand based on the inven-tory upper bound:V0,t = D0,t + I

max0,t , for t = 1 to T

v0,t = D0t + Imax0t V0,t1 for t = 1 to T (we suppose

that V0,0 = 0)

Compute the time window demandsLet Lt = Vt, for t = 1 to TLet Rt = Dt, for t = 1 to TL0 = R0 = 0 and k = 1While L, R = 0

Set = mint {Lt > 0}, = mint {Rt > 0}Set dw

0k = min {L, R}, bk = , ek =

L L dw0k, R R dw0k

k k 1End While

3.5 Improvement (smoothing) heuristic

The smoothing heuristic takes the solution of the La-grangian problem and modifies it to build a good fea-sible solution. This heuristic is based on the WHK-BIheuristic presented in Section 3.4.3. The productionquantities of the end products (xit, i = 1,...,N,t =1, ...T) obtained from the solution of the Lagrangianproblem are used to calculate a new demand for theraw material as d0t =

N

i=1 gixit for every period t.

The production plan of the raw material is built usingthe same approach in Section 3.4.3. That is, we firstconvert the data of the problem into a non-customerspecific time window problem before using the O(T2)algorithm of Wolsey (2006) to solve the new problem.

4 EXPERIMENTAL TESTS

4.1 Generated data sets

The tested data sets are built based on Trigeiro etal. (1989) data sets (TTM). The TTM data sets were

generated for the capacitated single level problemwith setup times.

In our problem, we consider the set of 540 problemsin TTM. The generated production capacity valuesare replaced with generated inventory limits Imax0t .At each period, Imax0t is either equal to zero (zeroinventory capacity), to the total average demand ofall products, or to twice the total average demand.

Imax0t =

0

1 1T

Ni=1

Tt=1

dit

2 1T

Ni=1

Tt=1 dit

This approach allows us to keep the same set of prob-lems. However, since the setup time is varied (twodifferent levels) in TTM, we just need 270 problems(the remaining problems become duplicate for us).

The two other parameters that we varied in our data

sets were the holding cost of raw material with respectto that of end items and its Time-Between-Orders(TBO).

The holding cost of the raw material is usually smallerthan the smallest holding cost of all end products.The values that we have chosen are:

h0 =

0.2 mini {hi}

0.6 mini {hi}

0.9 mini {hi}

Note here that hit = hi, for t in TTM data. Wesuppose that this is also the case for raw material.

The TBO of raw material is usually larger than that ofend products, but we have tried to cover most possiblecases by setting it equal to:

TBOz =

1 TBOi

2 TBOi

3 TBOi

Where TBOi is the time between orders of end itemsin TTM. Given TBOz and h0, we can calculate thesetup cost s0 as follows:

s0 =1

2Th0 (TBOz)

2

Ni=1

Tt=1

dit

We suppose that the utilization coefficient of raw ma-

terial gi is equal to 1 for all products. That is one unitof each product i needs one unit of raw material.


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By combining all possible values ofh0 and TBOz withthe already existing problems, the total number oftested problems is: 54033 = 4860 problems wherethe parameters are fixed for every 10 problems.

4.2 Numerical results

We have implemented our algorithms in the C++ lan-guage using Microsoft Visual C++ 6.0. The mathe-matical models were implemented on Xpress-MP withoptimizer version 18.1. Both Xpress-MP and theC++ code were run on a personal computer withCPU Intel Core2 Duo T7300 and 1GB of RAM.

The average CPU time of the Lagrangian heuristicis less than 0.05 sec. and at most 0.08 seconds. Wedecided to fix the maximum execution time of Xpress-MP to 10 seconds.

The gaps were calculated using the formula:

Gap = 100 U B LB

LB(40)

It is worth noting here that, by default, the gaps inXpress-MP are equal to 100 UBLB

UB. Instead of

using this formula, we extracted the upper boundsand lower bounds in Xpress-MP and used formula(40).

In the next paragraphs, we analyze the results in twosteps. We first compare the gaps of the Lagrangianheuristics with those obtained by running Xpress-MPon both aggregate and disaggregate formulations. Inthe second step, we compare the upper bounds andlower bounds of the lagrangian heuristic with the bestlower bounds and upper bounds available.

Table 2 summarizes the results of our experimen-tal tests and compares the different (upper bound)-(lower bound) gaps. The columns of the table repre-sent from left to right the changing parameters andtheir values and the gaps for the disaggregate model(see Section 2.2) MIP-FAL, the aggregate model (see

Section 2.1) MIP-AGG and the Lagrangian relaxationLR. The CV parameter represents the coefficient ofvariation of demand. 0.35 represents a medium varia-tion when 0.59 represents a high variation. The Imax

parameter represents the coefficient used to calculateinventory capacity.

It is clear from these numerical results that the La-grangian heuristic gives better (smaller) gaps thanones obtained using Xpress-MP for the aggregateformulation with an average gap of 9.29% for theLagrangian relaxation and 11.96% for Xpress-MP.The worst gaps were obtained using the solver on

the disaggregate formulation (with an average gap of50.89%).

GapParameter Value MIP-FAL MIP-AGG LR

h0 0.2 22.28 9.68 9.100.6 56.90 11.73 9.920.9 73.47 14.48 8.85

TBOz 1 13.02 12.37 7.52 50.29 13.87 8.64

3 89.35 9.65 11.72N 10 35.18 0.10 8.44

20 56.86 5.83 9.5130 60.61 29.96 9.91

Imax 0 4.85 6.54 14.301 77.46 16.65 7.642 70.35 12.70 5.92

TBO 1 12.38 1.91 7.542 55.03 16.78 11.744 85.25 17.20 8.58

CV 0.35 50.94 11.48 7.280.59 50.83 12.45 11.29

Global Average 50.89 11.96 9.29

Table 2: Average gaps for the two solution ap-proaches.

If we compare aggregate formulation gaps with thoseof the Lagrangian relaxation for the different param-eters, we notice that the solver gives better gaps forlarge TBOs of the raw material, for small size prob-lems (for N = 10 and N = 20), for problems with zeroinventory capacity, and when the TBOs of individualitems is small. We have to insist here on the factthat Xpress-MP was run for 10 seconds which gaveit time to solve a lot of small size problems and rel-atively easy problems. An example of the easy prob-lems is when the TBO of individual items is small.This means that the setup costs are small and, evenif the MIP solution is suboptimal, the consequencesof extra setup costs are less harmful to the solution.

Compared to Xpress-MP results, the gaps of the La-grangian relaxation are relatively stable even if wevary the different parameters. However, we can noticethat increasing the TBO of raw material increases the

gap. This is related to the principle of the smoothingheuristic. This heuristic shifts extra amounts of rawmaterial to other periods which generates extra setupcosts. They are directly proportional to the value ofTBOz. As the inventory capacity increases, the gapsdecrease. This is due to the fact that, for larger inven-tory capacities, fewer constraints are violated duringthe solution of the sub-problems (relaxed solution).And the smoothing heuristic has to make very fewchanges to this solution to build a feasible plan.

In Table 3, we report the gap between Lagrangianheuristic lower bounds (LBLR) and upper bounds

(UBLR) and the best known lower bounds (LBMIP)and upper bounds (UBMIP), respectively. The latter


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are obtained by running Xpress-MP for 60 secondson the aggregate formulation. These results clearlyshow that the Lagrangian relaxation lower boundsare of good quality. This is further supported by thefact that out of the 4860 problems solved, the LBLRwas better than LBMIP on 363 problems (7.47%).This is not the case for the UBLR. This means that

more effort has to be made to improve the smoothingheuristic in the Lagrangian relaxation.

Gap between Gap betweenLBLR and UBLR and

Parameter Value LBMIP UBMIPh0 0.2 2.30 5.65

0.6 1.85 7.080.9 1.76 6.22

TBOz 1 2.14 3.812 1.64 6.063 2.12 9.09

N 10 2.53 5.7420 2.60 6.6230 0.78 6.60

Imax 0 2.90 10.691 1.75 4.602 1.26 3.68

TBO 1 1.84 5.562 2.56 8.374 1.50 5.02

CV 0.35 1.14 5.210.59 2.79 7.43

Global Average 1.97 6.32

Table 3: Gaps of lower and upper bounds of La-grangian heuristic to the best available bounds.

5 CONCLUSION

We have successfully analyzed and solved a two-level lot-sizing problem with bounded storage capac-ity for raw material. Different models and solutionapproaches were tested and compared. The best ap-proach was the Lagrangian relaxation heuristic which

showed a good stability for varying parameters suchas TBO and the inventory capacity.

Though the Lagrangian relaxation heuristic is goodcompared to the standard solver, we believe that it ispossible to obtain better results using this approach.We noticed that the lower bounds of the Lagrangianrelaxation are very good. Thus we intend to focus onimproved smoothing heuristics.

ACKNOWLEDGMENTS

The authors are grateful to an anonymous referee for

his helpful comments that improved the presentationof the theoretical and experimental results.

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