a new lp-based lower bound for the cumulative scheduling problem

A new LP-based lower bound for the cumulative schedulingproblem

Jacques Carlier *, Emmanuel N�eron

U.M.R. C.N.R.S. 6599 HEUDIASYC, Centre de Recherches de Royallieu, Universit�e de Technologie de Compi�egne,

60205 Compi�egne Cedex, France

Abstract

This paper deals with the computation of lower bounds for Cumulative Scheduling Problems. Based on a new linear

programming formulation, these lower bounds take into account how resource requirements can be satis®ed simul-

taneously for a given resource capacity. One of the main interests of this paper is that the solutions of the LP can be

tabulated, for a given value of resource capacity. Thus, even if it is based on a linear programming formulation, the

computation of the bounds is low time consuming as con®rmed by our computational results on the Resource Con-

strained Project Scheduling Problem. Ó 2000 Elsevier Science B.V. All rights reserved.

Keywords: Cumulative scheduling problem; Linear programming; Resource constraints

1. Introduction

The aim of this paper is to present new lowerbounds for Cumulative Scheduling Problems. Allthese bounds are deduced from a linear program-ming formulation, which takes into account theway resource requirements of activities are simul-taneously satis®ed. These bounds can also be ap-plied, in turn, to each resource of a multi-resourcescheduling problem such as the Resource Con-strained Project Scheduling Problem (RCPSP) [8].

Recently, Baptiste et al. [1] have formalised thewell-known Cumulative Scheduling Problem

(CuSP), which consists of a set of n non-preemp-tive activities I � f1; . . . ; ng to be scheduled on aresource of capacity B. Each activity i has a releasedate ri, a processing time pi and a tail qi. More-over, activity i requires a constant amount bi ofresource throughout its processing. Baptiste et al.were interested in determining necessary condi-tions for the existence of a schedule whose make-span is smaller than or equal to UB. Thus for eachactivity, a due date di � UBÿ qi is computed. Thenon-preemptive CuSP is the extension of the par-allel machine problem [4], where resource re-quirements are not necessarily equal to 1. Thus it isNP-hard in the strong sense. It is a known fact thatthis problem is useful from a theoretical and apractical point of view for solving more complexproblems, like the RCPSP.

European Journal of Operational Research 127 (2000) 363±382www.elsevier.com/locate/dsw

* Corresponding author. Fax: +33-2-44234477.

E-mail addresses: [email protected] (J. Carlier), em-

[email protected] (E. NeÂron).

0377-2217/00/$ - see front matter Ó 2000 Elsevier Science B.V. All rights reserved.

PII: S 0 3 7 7 - 2 2 1 7 ( 9 9 ) 0 0 4 9 4 - 4

A large amount of literature has been devotedto lower bounds for the RCPSP. Some of them arebased on B-machine problems generated from theinitial RCPSP [6]. A basic lower bound formakespan is the quantity G0�I� de®ned as [4]

G0�I� � �1=B� � ri1

� ri2 � � � � � riB �

Xi�1;...;n

pi

� qj1� qj2

� � � � � qjB

!;

where ri1 ; . . . ; riB are the B minimal release dates,qj1; . . . ; qjB are the B minimal tails and

Pi2I pi is the

sum of the processing times.In the B-machine problem, each activity is

processed on one machine at a time, which can beseen as a resource requirement of one unit of theresource. The capacity of the resource is B.

Some lower bounds may improve the boundG0�I� as maxI 0�I �G0�I 0��, the so-called Sub-SetBound. An O�n2� algorithm was proposed forcomputing the Sub-Set Bound by Perregaard [13],and Vandevelde [16]. Another improvement of theG0�I� bound is the Adjusted Sub-Set Bound whichtakes into account the fact that an operation iseither processed alone on a machine, or its releasedate and its tail could not be both useful forcomputing G0�I�. Perregaard [13] has proposed anO�2mm2� for computing the ASSB bound �m � B�.

Many other bounds have been developed forthe m-machine problems. Perregaard [13] hastested a lower bound based on network ¯ow for-mulation. This bound is a graph formulation ofthe preemptive m-machine problem. But the worstcase complexity associated with this method isO�n4�. Jackson's Pseudo Preemptive Schedule,proposed by Carlier and Pinson [7], is also an ef-®cient lower bound for the m-machine problem,based on the construction of Jackson's preemptiveschedule adapted to the cumulative case. It can becomputed in O�mn� n logn�. Baptiste et al. [1]have presented feasibility tests that can be appliedto the m-machine problems. These tests are basedon energetic reasoning: they compute the workthat must be processed during some relevant timeintervals �t1; t2�. If this work is greater thanB� �t2 ÿ t1�, there is no feasible solution. A qua-

dratic algorithm is presented to compute thesetests.

Other works based on linear programmingformulations for the RCPSP have been proposedby Mingozzi et al. [11] and Brucker et al. [3]. Au-thors relax non-preemption and associate a vari-able with each subset of activities that can beprocessed simultaneously without violating eitherthe resource constraints or the precedence con-straints. Consequently the linear program may bevery large, but it can be tackled by column gen-eration.

Our aim is to present a new formulation forcomputing the work that must be processed duringa time interval. This work is related to the way aresource can be ®lled with some resource require-ments. Thus we do not take care of precedenceconstraints between activities: only their resourcerequirements and the resource capacity are in-volved in the LP-based bound that we present. Soour linear programs are small and their solutionscan be tabulated. Then we present several methodsto take into account release dates and tails of ac-tivities in order to get e�cient lower bounds.

The paper is divided into seven sections. InSection 2, we explain our relaxation. Section 3 isdedicated to linear programming formulations. InSection 4, we solve the linear program for speci®edvalues of B. In Section 5 we take into accountrelease dates and tails. Finally, Section 6 is devotedto computational results, and Section 7 to theconclusion.

2. A new relaxation

We can use the preemptive relaxation of theCuSP for computing a lower bound. But we areinterested in a stronger relaxation than the classi-cal preemption, which is NP-hard [1]. This relax-ation is more e�cient than the Fully Elasticrelaxation de®ned by Baptiste et al. [1]:

De®nition 1. In the Fully Elastic CuSP, each ac-tivity i 2 f1; 2; . . . ; ng, requires a variable amountof resource in �0::B� such that for any time t be-tween its starting and its completion time, any

364 J. Carlier, E. N�eron / European Journal of Operational Research 127 (2000) 363±382

amount bi(t)2 [0..B], bi(t) integer can be allocatedprovided thatZ 1

0

bi�t�dt � pi � bi:

In the relaxation that we propose, at every timet the amount of resource allocated to an activity isa multiple of its resource requirement. Thus weobtain what we call a Multiple-Elastic Preemptiverelaxation of the CuSP.

De®nition 2. A Multiple-Elastic Preemptive sched-ule of the CuSP is feasible i� any amountbi(t)2 [0..B] can be allocated to activityi 2 f1; 2; . . . ; ng provided that

8t 2 R�; 9mi�t� 2 N=bi�t� � mi�t� � bi

andZ 1

0

bi�t�dt � pi � bi:

mi(t) is called the multiplying factor at time t foractivity i. Moreover an activity in process can bepreempted (see Fig. 1).

3. Linear programming formulation

In this section we are interested in the Multiple-Elastic Preemptive relaxation of the CuSP, inwhich the time bounds on activities are loose(ri � 0; qi � 0 8i 2 I). Then in this section D de-notes an instance of a CuSP in which the release

dates and tails are equal to 0. Some methods fortaking into account the release dates and the tailsof the activities will be presented in Section 5, andD will denote a global instance of the CuSP (timebounds included).

Before introducing the new linear programmingformulation, let us present some additional de®-nitions.

De®nition 3. For a resource capacity B, we say thatCB � fc1; c2; . . . ; cK j8k 2 �1::K�; ck 2 N�g is afeasible configuration if c1 � c2 � � � � � cK 6B andc1 P c2 P � � � P cK . As B is ®xed, it will be omittedin the notation.

C�b� denotes the number of occurrences of b inthe set fc1; . . . ; cKg. If there exists a feasibleschedule, we can compute at any time t the feasiblecon®guration that occurs. For example, if weconsider a resource capacity of 6, C � f3; 3; 1g isnot a feasible con®guration. On a resource of ca-pacity 6, activities having resource requirementsequal to 3, 3 and 1 units cannot be processed at thesame time. But C� {2, 2, 1} is a feasible con®gu-ration for B � 6. In this case C�1� � 1 andC�2� � 2.

Without loss of generality, we assume that if acon®guration is feasible with any C�b�P 1; C�b�corresponds either to a multiplying factor of anoriginal activity with a resource requirement equalto b, according to the Multiple-Elastic relaxation,or to C�b� activities with a resource requirementequal to b, in process at the same time.

A lexicographic decreasing order can be de®nedon con®gurations, thus Ch denotes a con®gura-

Fig. 1. On the left Gantt chart, an activity i (pi � 4; bi � 3) is processed between its release date (ri � 3) and its due date (di � 9). The

resource capacity is 8, the resource requirement of the activity is 3. The right Gantt chart displays a feasible solution for processing the

same activity according to the Multiple-Elastic Preemptive relaxation.

J. Carlier, E. N�eron / European Journal of Operational Research 127 (2000) 363±382 365

tion, and fCh=h 2 �1::H �g denotes the ordered setof feasible con®gurations.

De®nition 4. Let· S be a feasible schedule of I on a resource of ca-

pacity B;· C�S; t� be the con®guration that occurs at time t

in S, 8t9h 2 �1::H �=C�S; t� � Ch;· d�C�S; t�; Ch� � 1 if at time t C�S; t� � Ch and 0

otherwise.We de®ne, the con®guration-time, xh(S) as the

sum of the time intervals during which the con-®guration Ch occurs in the schedule (see Fig. 2):

xh�S� �Z 1

0

d�C�S; t�; Ch�dt:

We call a time/machine unit, the resource oc-cupation due to an activity whose processing timeand resource requirement are equal to 1. We alsode®ne PI�b� �

Pfi2I=bi�bg pi, as the sum of time

intervals during which a resource requirement of bunits must be in process. Without loss of general-ity, in order to simplify the notations, Pb denotesPI�b�.

3.1. A new linear programming formulation

From these de®nitions and notation, we intro-duce the new linear programming formulation. At®rst we present an example illustrating our work.Then a global formulation is given.

A simple example. Let us consider a set of ac-tivities, which must be scheduled on a resource of

capacity 5. Each activity is supposed to have aresource requirement equal to 2 or 3 units. Thenthere exist four di�erent feasible con®gurations forprocessing the activities (see Fig. 3).

The problem can be written as follows:

Min x1 � x2 � x3 � x4 �III�x1 � x2 � P3 �I�2x4 � x3 � x1 � P2 �II�x1 P 0; x2 P 0; x3 P 0; x4 P 0:

8><>:The constraint (I) states that the time/machineunits corresponding to activities which have a re-source requirement equal to 3 are processed eitherin con®guration C1 or in con®guration C2.

The constraint (II) states that all the activitiesverifying bi � 2 are processed in one of the threecon®gurations in which a resource requirement of2 units appears. Moreover if f2; 2g occurs, theamount of time units corresponding to a resourcerequirement of 2, is C4�2� � x4, with C4�2� � 2 thenumber of 2 in f2; 2g.

The cost function (III) corresponds to themakespan minimization.

To be convinced of the interest of such an ap-proach, let us consider a numerical example, whichcorresponds to the LP formulation below. It con-sists of four activities (see Fig. 4):

Fig. 2. This Gantt chart presents a 3-activity schedule (S).

From this schedule we deduce: C1 � f2g; x1�S� � 1; C3 �f3;2g; x3�S� � 2:5; C2 � f3g; x2�S� � 1:5; C4 � f8g; x4�S� � 1.

Fig. 3. For each time unit spent in con®guration f3; 2g, one

time unit of resource requirement 2, and one time unit of re-

source requirement 3 are processed.

Fig. 4. A solution for this example is: x1 � 5; x4 � 2:5. The

corresponding cost is equal to 7.5 time units.


· Two activities with a resource requirement of 3and with processing times equal to 3 and 2,P3 � 5.

· Two activities with a resource requirement of 2and with processing times equal to 6 and 4,P2 � 10.On this example the value of the lower bound is

strictly greater than the naive evaluation:�Pi pibi�=B � 35=5.

The general model. We now present the generalformulation of the LP-based bound. Recall that inthis section, D � �I ;B� denotes an instance ofCuSP such that f8i 2 I ; ri � 0 and qi � 0g.

De®nition 5. The Multiple-Elastic PreemptiveBound is de®ned as the solution of

MEPB�D� �Minxh P 0

Xh�1;...;H

xh

!P

h�1;...;HCh�1� � xh � P1; �1�P

h�1;...;HCh�2� � xh � P2; �2�

..

.Ph�1;...;H

Ch�b� � xh � Pb; �b�

..

.Ph�1;...;H

Ch�B� � xh � PB: �B�

8>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>:Theorem 1. MEPB(D) matches an optimal solutionof the Multiple-Elastic Preemptive relaxation of D.

Proof. In a feasible solution all equality constraintsare veri®ed. Thus, due to the ®rst equality con-straint (1), the total amount of time/machine unitscoming from operations which have a resourcerequirement equal to 1 are processed. We can

generalise it to all the time/machine units for theoperations which have a resource requirementequal to b (b 2 �1::B�; Eq. (b)). Moreover onlyfeasible con®gurations appear, thus the resourceconstraint is never violated. Finally, among thesesolutions, an optimal one minimizes the total timeduring which the resource is idle; it corresponds tominimizing the makespan. �

3.2. Property of this new formulation

The linear program we present is based on theenumeration of the set of feasible con®gurations.An algorithm for generating this set can easily bedesigned. Moreover such a generation is done onlyonce before the beginning of any treatment be-cause of the parametric formulation that we pre-sent in Section 3.3. Thus even if this algorithm isfar from an optimal one, its impact on time con-sumption may be ignored.

For some special instances, some resource re-quirements never occur (for example bi�B). Inthese cases the number of con®gurations to takeinto account decreases drastically.

Size of the model. Of course the size of themodel we present, determines if it is useful inpractice. Its number of constraints is equal to B.So, generally the number of constraints of the LPis tractable. But it is harder to evaluate the numberof variables involved. To each feasible con®gura-tion corresponds a variable in the LP, which is thetime the resource spends in this con®guration.Thus we have to evaluate the number of feasiblecon®gurations that exist, for the di�erent values ofB. Indeed we see in Table 1 that it is not so largefor B6 10. For higher resource capacity, if Part(B)is the number of exact partitions of B, then thenumber of variables is:

Pb�1;...;B Part(b). Exact and

Table 1

The number of variables for some values of the resource capacity with and without reductions due to both theorems

Value of B 2 3 4 5 6 7 8 9 10

bi 2 �1::B� 3 6 11 18 29 44 66 96 138

bi 2 �2::B� (Theorem 2) 1 2 4 6 10 14 21 29 41

bi 2 �2::B� No dominated

(Theorems 2 and 3)

1 2 3 5 7 10 14 19 25


asymptotic values of Part(B) can be computed byusing the formulation proposed by Euler, and laterby Hardy and Ramanujan [14].

How to reduce the model. We present a result inorder to leave the activities with unit resource re-quirements, out of the feasible con®gurations.Dbi>1 denotes the CuSP in which the set of activi-ties Ibi>1 is deduced from I, by ignoring activitieshaving a unitary resource requirement.

Theorem 2. To compute the Multiple-Elastic Pre-emptive Bound, the activities which have a resourcerequirement equal to 1 either fill the resource and:EL�D�� 1=B��P1�2�P2�3�P3��B�PB�reaches the optimum, or these activities have noeffect on the computation of MEPB(D), i.e., theoptimum does not change if the activities with aunitary resource requirement are taken into con-sideration or not.

MEPB�D� � max�EL�D�;MEPB�Dbi>1��:

Proof. Let· S be an optimal solution of the Multiple-Elastic

preemptive relaxation of I.· Cmax(S) be the makespan of this optimal solu-

tion.· N�S� be the number of idle time/machine units

in this solution between t � 0 and t � Cmax�S�.· Sbi>1 denote an optimal solution of the relaxed

problem without the unitary resource require-ments.It is obvious that EL(D) is a lower bound of the

completion time: it is the weighted sum of theprocessing times, divided by the resource capacity.If the resource is ®lled (no idle times on the re-source from 0 to Cmax(S)), this value is achieved.

Let N�Sbi>1� be the number of idle time/machineunits in Sbi>1. Cmax�Sbi>1� is the cost associated withthis solution.

Let us consider:· N�Sbi>1�P P1: the number of idle time/machine

units in this solution is greater than the sum ofthe processing times of the activities, which havea unitary resource requirement. Thus these ac-tivities, due to the fact that preemption is al-lowed, could be processed during the idle timesof the resource. Then we obtain a solution, that

takes into account the unitary resource require-ments and which has the same cost as the opti-mal solution without unitary resourcerequirements: Cmax�S� � Cmax�Sbi>1�.

· N�Sbi>1� < P1: the number of idle time/machineunits in this solution is smaller than the sumof the processing times of the activities, whichhave a unitary resource requirement. Thus theresource in the initial problem is ®lledN�S� � 0. EL(D) is reached. �

Thanks to Theorem 2, we have not to take intoaccount the activities, which have a resource re-quirement equal to 1 unit. Consequently, thenumber of feasible con®gurations is drasticallyreduced (see Table 1). Nevertheless the boundcomputed from this new problem must be com-pared to the elementary bound if N�S�P P1 (seeFig. 5).

To build the model, we have to consider all thefeasible con®gurations. But some of them neveroccur in an optimal solution. Before presenting adominance criterion on con®gurations, let us in-troduce some notions for describing a con®gura-tion (see Fig. 6):· the consumption of a con®guration:

Pb�1::B

Ch�b� � b;· the slack of a con®guration: BÿPb�1::B

Ch�b� � b;· the minimal resource requirement of a con®gu-

ration: minb�1::Bfb=Ch�b� 6� 0g.

Theorem 3. Configurations that have a slack largerthan or equal to their minimal resource requirementare dominated, i.e., they can always be replaced by aset of configurations, all of them having a slackstrictly smaller than their minimal resource re-quirement, without increasing the cost of the solu-tion.

Proof. Let S be a solution in which a dominatedcon®guration Cdomin occurs. Then we can replaceCdomin in S by Cdomin1 and Cdomin2. Cdomin1 is de-duced from Cdomin by deleting the minimal re-source requirement. Cdomin2 is deduced byreplacing one minimal resource requirement by m(m P 2) minimal resource requirements. This op-eration is feasible according to the multiple elastic


relaxation. Notice that Cdomin2 is not dominatedand the slack of Cdomin1 is strictly larger than theslack of Cdomin.

As the preemption is allowed, we can buildfrom S, a solution by replacing Cdomin by Cdomin2

on (xdomin/m) time units, Cdomin1 on the otherxdomin ÿ �xdomin=m� time units. This new solutionhas the same cost as S, which is strictly smaller ifCdomin1 does not exist.

If Cdomin1 exists but is dominated, then we it-erate by replacing it by two other con®gurations.We are sure to get in a ®nite number of iterations aset of non-dominated con®gurations since theslack strictly increases at each iteration (see Fig. 7).�

Thanks to Theorems 2 and 3, the number ofvariables in the LP formulation is drastically re-duced as it is shown in Table 1.

Moreover in some speci®c instances of a CuSP,if a given resource requirement never occurs, thenumber of con®gurations can be drastically re-duced. For example if the capacity is equal to 10,the number of non-dominated con®gurations,without unitary resource requirements is equal to25. If we consider an instance in which there is noactivity requirement of 3 units of the resource,then the number of con®gurations to take intoaccount decreases from 25 down to 16.

3.3. Primal and dual approach

Parametric approach of the LP. We now restrictour study to the case where B is given and con-stant. Our problem can be written as

Min fOT � X ; CT � X � P ; X P 0g;where

Ob � 1 8b 2 �1::B�;Pb �

Xi2I=bi�b

pi 8b 2 �2::B�;

Cb;h � Ch�b� 8b 2 �2::B� 8h 2 �1::H �;Xh � xh 8h 2 �1::H �:

Notice that if we consider two di�erent instancesof a Multiple-Elastic Preemptive CuSP, only vector

Fig. 5. Illustration of Theorem 2 with B � 7.

Fig. 6. B � 9; C � f3; 2; 2g; Consumption of this con®gura-

tion: 7; Slack: 2; Minimal resource requirement: 2.


P has changed, i.e., for a given capacity, the set ofcon®gurations is the same, thus the matrix C isconstant, even if the processing times and the re-source requirements are di�erent. So the basis so-lutions of the dual linear program are notdependent on the parameters (see Section 3.2) andthe value of a solution can be written as a linearexpression of the Pb.

De®nition 6. If we now consider the Pb as param-eters, we denote Sopt�Pb; b 2 �2::B�) as an optimalsolution of the primal LP for P and MEPB thecorresponding objective function which is a linearform of the Pb.

Then, thanks to the duality theory,Cmax�Sopt�P�� is a lower bound for the initialCuSP. Moreover the number of Sopt�P � is ®nite.Our purpose in Section 4 is to build such para-

metric solutions for any vector P. The cost asso-ciated with these solutions is a linear expressionof the Pb values, whose coe�cients depend on theparameters. To illustrate this parametric point ofview, let us present an example. Without provingthe optimality of the solution we give (see Section4.3), this example presents the parametric ap-proach for computing the linear expressions (seeFig. 8).

In this case corresponding to B � 5, two situa-tions may occur. If P3 > P2: resource require-ments of 2 units can be matched with resourcerequirements of 3 units. The cost associatedwith vector P is P3 � P4 � P5. If P36 P2: resourcerequirements 2 and 3 are matched, as muchas possible, but time/machine units coming fromactivities which have requirements of 2 units re-main. They have to be processed in the 2; 2 con-®guration.

Fig. 8. Parametric linear expressions of the lower bounds (B � 5).

Fig. 7. The initial con®guration (on the left) is dominated, then we can replace it by two other con®gurations in order to get only

dominant con®gurations. Here Cdomin1 is also dominated, thus it should be replaced by two other con®gurations by applying the same

mechanism.


Thus the cost associated with vector P is

1=2 � �P2 ÿ P3� � P3 � P4 � P5

� 1=2� P2 � 1=2� P3 � P4 � P5:

We obtain two linear expressions whose coe�-cients depend on some conditions on the values ofPb. We have to determine which one of these ex-pressions is the good one.

How to use the parametric expressions. If we con-sider that Ub are the dual variables, the dual linearformulation is:

MaxX

b

Pb

� Ub

!P

b�2::BC1�b� � Ub6 1;

..

.

Pb�2::B

Ch�b� � Ub6 1;

..

.

Pb�2::B

CH �b� � Ub6 1:

8>>>>>>>>>>>>>><>>>>>>>>>>>>>>:If we consider a linear form on the Pb, associatedwith a dual feasible solution, it is a feasible cost ofthe dual program. Indeed the cost associated witheach dual feasible solution is smaller (equal at theoptimum) than one of any primal feasible solution.

Then all the linear expressions determined arelower bounds of our problem.

Theorem 4. fSgopt�P�; . . . ; SG

opt�P �g are feasible solu-tions for the dual formulation, that implies

MEPB�D� � maxg�1;...;G

�Cmax�Sgopt�P��;

where G is finite.

Proof. Even if we have not yet de®ned how tobuild Sg

opt�P � according to the con®gurations con-straints, and the parametric approach, we haveassumed that these solutions have two properties:· they are built without violating the resource

constraints, i.e., according to the con®gurationconstraints, which are the dual constraints;

· they are optimal for some values of the Pb.Due to the linear programming theory the

maximum computed among the Sgopt�P � matches

the dual optimum, which is equal to the primaloptimum. Moreover the number of Sg

opt�P � is ®nitebecause they correspond to basis solutions of thedual polyhedron which is independent of the pa-rameters. �

After presenting the new formulation and in-troducing the basic concept of a parametric reso-lution, we propose methods for computing thelower bound for a resource capacity from 2 to 10units.

4. Exact formulation of the optimum for somevalues of B

In this section our goal is to present how tocompute the MEPB(D) bounds, for some values ofthe resource capacity, and for all the Pb, withoutsolving directly the linear program by using aniterative method. So we are searching for optimalsolutions whose costs are expressed as linear ex-pressions of the Pb, and obtained without violatingany constraint of the linear program. These linearexpressions are denoted by MEPBg(D).

We explain how to build the Sgopt�P �, according

to the principle of the non-violation of the con-®guration constraints and to optimality. Thus wecan deduce the MEPBg(D) bounds. MEPBg(D) isthe cost of Sg

opt�P � and the ®nal MEPB(D) ex-pression, thanks to Theorem 4. Here index l de-notes a case we have to study according to somevalues of Pb. The rank of the linear program isBÿ 1 because the ®rst equation is not used. Sothere are at most Bÿ 1 con®gurations in an opti-mal solution. The objective function is equal toP

i xi. It is computed every time to get the corre-sponding linear expression.

For B � 2, the formulation we present has nointerest, because it is the same as the naive boundpresented previously.

4.1. B � 3

In any solution of a Multiple Elastic Preemp-tive CuSP having a resource capacity equal to 3,the bound verifying the con®guration constraints


and the time/machine unit constraints is:MEPB�D� � P2 � P3 (see Fig. 9).

4.2. B � 4

This case is also easy to solve. In any solution,the activities having a resource requirement equalto 4 ®ll the resource. Only one con®guration ispossible to process them. No activity can be pro-cessed at the same time as an activity which needs 3resource units (thanks to Theorem 2, the unitaryresource requirements are ignored). Only onecon®guration is possible to process them. The partsof activities having a resource requirement equal to2, can either be processed without being `multi-plied' (in the sense de®ned in the Multiple Elasticrelaxation), or with a multiplying factor equal to 2.But if they are processed in the f2g con®gurationon a given time interval, due to the multiple re-laxation, they can be processed in a f2; 2g con®g-uration on the half of this interval. Thus thesolution obtained is better than the initial one.

We conclude that MEPB�D� � P3 � P4 � P2=2(see Fig. 10).

4.3. B � 5

This case is the ®rst one for which some con-ditions on the values of Pb occur, for determiningthe lower bound. As the ideas of the proofs are thesame in all the cases we present, we do not describethem in detail.

In any optimal solution for a given vector P, theparts of operations which require 5 or 4 resourceunits can only be processed in one con®guration.There exist two feasible con®gurations to process

the activities that need 3 units of the resource: f3gor f3; 2g. If a part of an activity requiring 3 units isprocessed alone on the resource during a giventime interval, and during another time interval ofthe solution, the con®guration f2g (or f2; 2g) oc-curs during the same interval, the solution whichconsists of replacing f3g and f2g by f3; 2g is bet-ter. We can make the same remark if f2g occurs,we can replace it by f2; 2g during half of the initialtime interval. Combining both remarks we noticethat in any solution f3g, f2; 2g, f2g can be re-placed by f3; 2g and if necessary f2; 2g.

Thus we have to take into account two situa-tions (see Fig. 11):· If P3 > P2: f3gf2g and f2; 2g in an initial solu-

tion are optimally replaced by f3; 2g and f3g.Thus: MEPB1�D� � P3 � P4 � P5 (case 1).

· If P36 P2: f3g; f2g and f2; 2g in an initial solu-tion are optimally replaced with f3; 2g; f2; 2gand the time con®guration of f2; 2g is equal to�P2 ÿ P3�=2. MEPB2�D� � P2=2� P3=2� P4 � P5

(case 2).

4.4. B � 6

First of all, notice that all activities or parts ofactivities, which need 6 or 5 units of the resource,cannot be processed simultaneously with anotheractivity. They must be processed respectively in thecon®guration {6} and {5}. {3}, {2} and f2; 2g aredominated and can be replaced by a non-domi-nated con®guration. If in any solution the con®g-uration f3; 2g occurs during a time interval T,f3; 3g can replace it during T/2 and f2; 2; 2g duringT/3. Finally, the idle times on the resource de-crease, and the time spent to process the sameoperations is smaller than the previous one. Thus

Fig. 9. Optimal solution for B � 3.Fig. 10. Optimal solution for B � 4.


in an optimal solution, f3; 2g cannot be present.There exists only one optimal con®guration forprocessing the activities, which need 3 units: {3,3}.If in any solution, {4} and f2; 2; 2g occur, respec-tively, during T1 and T2, they can be replaced byf4;2g during T1 and f2;2;2g during (T2 ´ 3ÿT1) /3.The initial con®gurations are not optimal, sincethe time spent in these new con®gurations issmaller than the previous one. Thus in an optimalsolution, {4} and f2; 2; 2g could not both occur.

According to these remarks, two cases have tobe considered (see Fig. 12):· P4 > P2: f4; 2g, {4} are the two con®gurations

used and MEPB1�D� � P6 � P5 � P4 � P3=2.· P46 P2: f4; 2g and f2; 2; 2g are the two con®gu-

rations used to process the operations that need4 and 2 units of the resource, and MEPB2�D� �P6 � P5� P3=2� P4 � �P2 ÿ P4�=3.

4.5. B � 7

The number of cases to consider grows with theresource capacity. Thus it is more complex than inthe previous section, even if the ideas to reach an

optimal solution are the same. Activities that need6 and 7 units of the resource must be processedalone on the resource. If {5} and any con®gura-tion, which has a resource requirement of 2 units(f5; 2g excepted) both occur, we can replace themby f5; 2g and the previous con®guration withoutthe requirement of 2 units. The solution is notmodi®ed. Thus we have to consider two cases:P56 P2 and P5 > P2. In the ®rst one we know thatthere exists a solution in which all the 5 are mat-ched with 2 in f5; 2g, and we de®ne P 02 � P2 ÿ P5 asthe remaining time corresponding to a resourcerequirement of 2 units. The same discussion can bedone with P4 and P3. We can de®ne P 03 � P3 ÿ P4 ifP3 > P4 and P 04 � P4 ÿ P3 if P4 P P3. Finally if P 02 isde®ned, we have to consider how to process opti-mally, either resource requirements equal to 4 and2, respectively, with processing times equal to P 04and P 02, or resource requirements equal to 3 and 2,respectively, with processing times equal to P 03 andP 02.

The results are summed up in the formulasbelow. The redundant formulas are not reported.· If P56P2�P 02 � P2 ÿ P5� and P46P3 �P 03 � P3 ÿ P4�

and P 036 P 02=2:

Fig. 12. Optimal solutions for B � 6.

Fig. 11. Optimal solutions for B � 5.


MEPB1�D� � P7 � P6 � 2� P5=3� 2� P4=3

� P3=3� P2=3;

· if P56P2 �P 02 � P2 ÿ P5� and P46P3 �P 03 � P3 ÿ P4�and P 03 > P 02=2:

MEPB2�D� � P7 � P6 � 3� P5=4� P4=2� P3=2

� P2=4;

· if P56P2 �P 02 � P2ÿ P5� and P4 > P3 �P 04 � P4ÿ P3�and P 04 > P 02:

MEPB3�D� � P7 � P6 � P5 � P4;

· if P5 > P2 �P 05 � P5 ÿ P2� and P4 6 P3 �P 03 � P3

ÿP4�MEPB4�D� � P7 � P6 � P5 � P4=2� P3=2:

4.6. B � 8

We explain how to build optimal solutions inthe case B � 8. Before enumerating the di�erentcases, let us notice that activities, which require 7or 8 units, ®ll the resource and cannot be processedat the same time than any other part of activity. If{6} and any con®guration which has a resourcerequirement of 2 units of the resource (f6; 2g ex-cepted) occurs in the same solution, we can replace{6}, by f6; 2g without modifying the makespan ofthe solution. Two cases may occur, either P6 > P2

and all resource requirements of 2 can be matchedwith 6, or P66 P2 and thus we de®ne P 02 � P6 ÿ P2.The same remark can be done with P5 and P3: if{5} exists, it can be matched with a 3 withoutmodifying the makespan. If P5 > P3 we de®neP 05 � P5 ÿ P3, otherwise P 03 � P3 ÿ P5. If any con-®guration f4; rg occurs with r � 2 or 3 on a timeinterval T1, we can modify it in f4; 4g on T1/2 andfr; rg on T1/2 without modifying the makespan.Thus we can build an optimal solution in which allthe 4 are processed by two. Finally we have toconsider the other cases that may occur: how tooptimally schedule 2 and 5 resource requirements,and 2 and 3 resource requirements. Notice that forprocessing P 05 and P 02, f5; 2g is not dominated, but{5} and {2} are dominated. Thus we have toconsider two cases P 05 > P 02 and its opposite case.

For processing P 03 and P 02, notice that f3; 3gf3; 3; 2gand f2; 2; 2; 2g are non-dominated con®gurations.Moreover if f3; 3g and f2; 2; 2; 2g occur, then wecan replace them by f3; 3; 2g and f2; 2; 2; 2g ifnecessary, without increasing the makespan. Thusin an optimal solution f3; 3g and f2; 2; 2; 2g bothcannot be present.

To illustrate the resolution of the dual LP, thedi�erent cases and the corresponding optimal so-lutions are presented in Appendix A. In this sec-tion, we only report the results that are useful tocompute the MEPB bound.

MEPB1�D� � P8 � P7 � P6 � P5 � P4=2;

MEPB2�D� � P8 � P7 � P6 � P5=2� P4=2� P3=2;

MEPB4�D� � P8 � P7 � 3� P6=4� 3� P5=4

� P4=2� P3=4� P2=4;

MEPB6�D� � P8 � P7 � 3� P6=4� 5� P5=8

� P4=2� 3� P3=8� P2=4:

4.7. B � 9

The more the resource capacity increases, themore the computation of optimal solutions be-comes complex. Based on the same ideas, we pre-sent the di�erent cases that may occur. To helpunderstand, let us notice that activities that haverequirements of 9 and 8 cannot be processed si-multaneously with any other activity. The more wecan match requirements of 7 units with require-ments of 2 units, the better is the solution. Thisremark can be applied to requirements of 6 and 3,and 5 and 4. Thus we have to consider for each ofthese couples two di�erent cases. f5; 2; 2g ®lls theresource, so {5} and any con®guration, which hasa 2 and does not ®ll the resource cannot occur. Forscheduling resource requirements of 4, 3, 2 units,one optimal con®guration is f4; 3; 2g. Thus weintroduce a new test which consists in taking into


account the minimum among three values and notonly the minimum among two values.

The results are summed up in the formulasbelow. The redundant formulas are not reported.

· If P7 > P2 and P6 > P3 and P5 > P4,

MEPB1�D� � P9 � P8 � P7 � P6 � P5;

· if P7 > P2 and P6 > P3 and P56 P4,

MEPB2�D� � P9 � P8 � P7 � P6 � P5=2� P4=2;

· if P7 > P2 and P66 P3 �P 03 � P3 ÿ P6� and P5 > P4

�P 05 � P5 ÿ P4� and P 03 > P 05,

MEPB3�D� � P9 � P8 � P7 � 2� P6=3� 2

� P5=3� P4=3� P3=3;

· if P7 > P2 and P6 6 P3 �P 03 � P3 ÿ P6� and P5 6P4 �P 04 � P4 ÿ P5� and P 03 > P 05,

MEPB4�D� � P9 � P8 � P7 � 2� P6=3� P5=2

� P4=2� P3=3;

· if P76P2�P 02 � P2 ÿ P7� and P6 > P3 �P 06 � P6 ÿ P3�and P5 > P4 �P 05 � P5 ÿ P4� and P 02 > P 06� 2� P 05,

MEPB5�D� � P9 � P8 � 3� P7=4� 3� P6=4

� P5=2� P4=2� P3=4� P2=2;

· if P76P2 �P 02 � P2ÿ P7� and P66P3 �P 03 � P3ÿ P6�and P5 > P4 �P 05 � P5ÿ P4� and P 05 > P 02=2 �P 005 � P 05ÿP 02=2� and P 03 PP 005 ,

MEPB7�D� � P9 � P8 � 5� P7=6� 2� P6=3� 2

� P5=3� P4=3� P3=3� P2=6;

· if P76P2�P 02 � P2 ÿ P7� and P66P3�P 03 � P3 ÿ P6�and P5 6 P4�P 04 � P4 ÿ P5� and min�P 04; P 03; P 02�� P 04,

MEPB9�D� � P9 � P8 � 3� P7=4� 2� P6=3� 7

� P5=12� 5� P4=12� P3=3

� P2=4;

· if P76 P2�P 02 � P2 ÿ P7� and P66 P3�P 03 � P3 ÿ P6�and P5 6 P4�P 04 � P4 ÿ P5� and min�P 04; P 03; P 02�� P 02,

MEPB11�D� � P9 � P8 � 5� P7=6� 2� P6=3

� P5=2� P4=2� P3=3� P2=6:

4.8. B � 10

This case is the last one we present. For largervalues of the resource capacity the computation ofthe MEPB bound becomes too complex and is avery challenging problem. For B � 10, only resultsare given. The proof of the optimality of the so-lutions we present is the same as previously. Theresults are summed up in the formulas below. Theredundant formulas are not reported.· If P8 > P2 and P7 > P3 and P6 > P4,

MEPB1:1�D� � P10 � P9 � P8 � P7 � P6 � P5=2;

· if P8 > P2 and P7 > P3 and P66 P4,

MEPB1:2�D� � P10 � P9 � P8 � P7 � P6=2� P5=2

� P4=2;

· if P8 > P2 and P7 6 P3 �P �1�3 � P3 ÿ P7� and P6 >P4 �P �1�6 � P6 ÿ P �1�3 � and P �1�3 > P �1�6 ,

MEPB2�D� � P10 � P9 � P8 � 2� P7=3� 2

� P6=3� P5=2� P4=3� P3=3;

· if P8 > P2 and P76 P3 �P �1�3 � P3 ÿ P7� and P66P4 �P �1�4 � P4 ÿ P6� and P �1�4 > P �1�3 =2,

MEPB3�D� � P10 � P9 � P8 � 3� P7=4� P6=2

� P5=2� P4=2� P3=4;

· if P86 P2�P �1�2 � P2 ÿ P8� and P7 > P3�P �1�7 � P7ÿP3� and P6 > P4�P �1�6 � P6 ÿ P4� and P �1�6 < P �2�2 =2,

MEPB5�D� � P10 � P9 � 4� P8=5� 4� P7=5

� 3� P6=5� P5=2� 2� P4=5

� P3=5� P2=5;

· if P86 P2 �P �1�2 � P2 ÿ P8� and P7 > P3 �P �1�7 �P7 ÿ P3� and P6 6 P4 �P �1�4 � P4 ÿ P6� and P �1�4 =2

< P �2�2 =2,

MEPB6�D� � P10 � P9 � 4� P8=5� 4� P7=5

� 7� P6=10� P5=2� 3� P4=10

� P3=5� P2=5;

· if P86 P2 �P �1�2 � P2 ÿ P8� and P76 P3 �P �1�3 �P3ÿP7� and P6 > P4 �P �1�6 � P6ÿP4� and P �1�6 > P �1�2

=2 �P �2�6 � P �1�6 ÿ P �1�2 =2� and P �2�6 6 P �1�3 ,


MEPB8�D� � P10 � P9 � 5� P8=6� 2� P7=3

� 2� P6=3� P5=2� P4=3� P3=3

� P2=6;

· if P8 6 P2 �P �1�2 � P2 ÿ P8� and P7 6 P3 �P �1�3 � P3

ÿP7� and P6 > P4�P �1�6 � P6 ÿ P4� and P �1�6 6P �1�2 =2�P �5�2 � P �1�2 ÿ 2� P �1�6 � and P �1�3 6 P �5�2 ,

MEPB10�D� � P10 � P9 � 4� P8=5� 7� P7=10

� 3� P6=5� P5=2� 2� P4=5

� 3� P3=10� P2=5:

5. Taking into account �ri ; qi�

The lower bounds we have presented are notdesigned to take into account the release dates andthe tails of the activities. Nevertheless release datesand tails must be included into the computation ofthe lower bounds to be e�cient. After presentingthe simplest way to deal with release dates andtails, we give two methods for carrying them. The®rst one is based on an energetic reasoning [1,10],the second one uses Jackson's Preemptive Sched-ule [5].

5.1. A basic way to use MEPB

We have determined that the MEPB bound is abetter approximation of the time during which theresource is busier than �1=B��Pi bi � pi), because ittakes into account both the work to process andthe idle times that may occur. Thus the simplestway to associate release dates and tails with theMEPB bound is:

Theorem 5. Let D be an instance of CuSP,GMEPB�D� � ri1 �MEPB�D� � qj1

is a lower boundfor the makespan of D where ri1 is the minimal re-lease date among the activities of I, qj1

is the mini-mal tail among the activities of I.

Proof. Let S be an optimal solution with a make-span equal to Cmax(S). No activity can be pro-cessed before ri1, and during the time interval

�Cmax�S� ÿ qj1; Cmax�S�]. By adding the MEPB(D)

bound, which is a lower bound of the time duringwhich the resource is busy, and the idle times, weget the lower bound GMEPB(D). �

But this MEPB bound is not as e�cient as thebounds we have presented in Section 1 because ittakes into account only the minimal release dateand the minimal tail. To get more e�cient lowerbounds for the CuSP, we have to use more in-formation on release dates and tails. At ®rst ri

and UB ± qi are used to determine the extremitiesof relevant time intervals in the energetic ap-proach, then they are introduced in the JPS al-gorithm.

5.2. An energetic approach

In this section, we determine necessary condi-tions for the existence of a solution having amakespan smaller than or equal to an overalldeadline UB. Thus, we associate with activity i, adue date di � UBÿ qi.

For example, these conditions are useful in abranch and bound method based on the resolutionof several decision variants of the cumulativeproblem, i.e., given an instance of the problem andan overall deadline UB, it determines whetherthere exists a schedule with a makespan smallerthan or equal to UB.

The relaxation we have introduced is deducedfrom the Fully Elastic relaxation proposed byBaptiste et al. [1]. Naturally, we adapt the resultsthey get for the elastic relaxation, to the MultipleElastic relaxation. Especially they have proposedsatis®ability tests for determining if a solution witha makespan smaller than or equal to UB may exist.These results are based on the computation of thework that must be processed in a time interval�t1; t2�: if this work is greater than the availableenergy, which is equal to �t2 ÿ t1� � B, no solutionhaving a makespan smaller than or equal to UB,can exist.

We propose to improve the estimation of thework that must be processed in the interval byusing the MEPB bound.


De®nition 7. Let D be an instance of CuSP, wede®ne the following parameters:· MinProcess �i; t1; t2� � min�t2 ÿ t1; pi; max�0; ri

� pi ÿ t1�; max�0; t2 ÿ di � pi��. (MinProcess�i; t1; t2� is the minimal temporal part of activityi that must be processed between t1 and t2).

· MinProcess �D; b; t1; t2� �P

i=bi�b MinProcess�i; t1; t2�. (MinProcess �D; b; t1; t2� is the sum ofthe temporal parts of the activities having a re-source requirement equal to b, that must be pro-cessed between t1 and t2) (see Fig. 13).

De®nition 8. We de®ne an extension of the MEPBbound applied to the minimum resource require-ments over �t1; t2� as

MEPB�D; t1; t2� �Minxh P 0

Xh�1::H

xh

!P

h�1::HCh�1� � xh �MinProcess�D; 1; t1; t2�P

h�1::HCh�2� � xh �MinProcess�D; 2; t1; t2�

..

.Ph�1::H

Ch�b� � xh �MinProcess�D; b; t1; t2�

..

.Ph�1::H

Ch�B� � xh �MinProcess�D;B; t1; t2�

8>>>>>>>>>>>>><>>>>>>>>>>>>>:Theorem 6. There does not exist for the instance Dof a CuSP a solution with a makespan smaller thanor equal to UB if there exist t1 and t2, t1 > t2 suchthat MEPB �D; t1; t2� > t2 ÿ t1.

Proof. If there exist t1 and t2 such thatMEPB�t1; t2� > t2 ÿ t1, then on the time interval�t1; t2� the work that must be processed plus the idletime units that may occur due to the optimalcon®gurations, is larger than the resource capaci-ty. �

Our aim in this section was not to generalizeall the results that Baptiste et al. have presentedfor the Elastic relaxation. Indeed it does not seemeasy to determine the relevant time intervals inthe case of the Multiple Elastic relaxation, sincethe evaluation of the work is no more a linearfunction of the processing times. Another direc-tion of research is to adapt adjustments of headsand tails.

5.3. Using JPS for computing a lower bound forCuSP

Jackson's Preemptive Schedule (JPS) is the listschedule for the one-machine problem associatedwith the earliest due date priority rule [5].

Let:· I be a set of activities f1; 2; . . . ; ng,· ri, pi, qi the release date, the processing time and

the tail associated with activity i,· Cmax(JPS(I)) the makespan of JPS(I).

The following property was proved [5]:

Cmax� JPS�I�� maxJ�I

h�J� with

h�J� � minj2J

rj �Xj2J

pj �minj2J

qj:

Fig. 13. This Gantt chart illustrates the minimal temporal parts of three activities that must be processed between t1 and t2. Activity 1

cannot start before r1 then MinProcess �1; t1; t2� � r1 � p1 ÿ t1 � 3. Activity 2 cannot be completed after d2 then MinProcess

�2; t1; t2� � t2 ÿ d2 � p2 � 3. Activity 3 must start after t1 and end before t2 then MinProcess �3; t1; t2� � p3 � 5. If we consider only

these three activities: MinProcess �D; 3; t1; t2� � 3 and MinProcess �D; 2; t1; t2� � 8.


We use these results for taking into account therelease dates and the tails. Let us introduce a one-machine problem:

De®nition 9. We associate with an instance D ofCuSP and a linear expression: a1 � P1 � a2 � P2

� � � � � aB � PB , the one-machine problem de®nedby I 0 a set of n activities and r0i; p

0i; q0i the release

date, the processing time and the tail of activity iverifying (see Table 2)

ri � ri; p0i � abi � pi; q0i � qi:

Proposition 3.

Cmax�JPS�I 0�� maxJ�I

minj2J

rj

"�Xj2J

abi � pj �minj2J

qj

#:

Proof. It is clear that Cmax�JPS�I 0�� maxJ�I

�minj2J r0j�P

j2J p0j � minj2J q0j� because we havede®ned r0i � ri; p0i � abi � pi; q0i � qi. �

We have already proved that MEPB(D)�maxg MEPBg(D) (see Theorem 4), and theMEPBg(D) are linear expressions of the Pb (thanksto the dual formulation).

Theorem 7. Let D be an instance of CuSP, Ig theone machine problem associated with the CuSP and

MEPBg�D� � ag1 � P1 � ag

2 � P2 � � � � � agB � PB:

maxg�Cmax�JPS�Ig�� is a lower bound of the make-

span of the CuSP :

Proof. Thanks to Proposition 3, we can deducethat

maxg�Cmax� JPS�Ig��

� maxg

maxJ�I

minj2J

rj

""�Xj2J

agbi� pj �min

j2Jqj

##;

maxg�Cmax� JPS�Ig��

� maxJ�I

minj2J

rj

"�max

g

Xj2J

agbi

� pj

!�min

j2Jqj

#:

Let us introduce the Pb which have been de®nedas: Pb �

Pi2J=bi�b pi, thus

maxg

Xj2J

agbi

� pj

!� max

g�ag

1 � P1 � ag2 � P2 � � � � � ag

B � PB�:

This expression is the MEPB bound as it has beenpreviously de®ned for D in which all the releasedates and tails are equal to 0.

Thus thanks to Theorem 7:

max1�Cmax� JPS�Ig�� max

J�I�GMEPB�DJ ��;

where GMEPB(DJ ) is the lower bound of DJ . DJ isan instance of CuSP which consists of a set J,J � I , and a resource capacity B. �

This lower bound takes into account both re-lease dates and tails and the MEPB bound tocompute a lower bound for a CuSP in which ac-tivities have non-zero release dates and tails.Moreover this new bound is extremely useful froma practical point of view. Since JPS(I) can becomputed in O�n � log2n� [5], by using heapstructures, the new bound can also be computed inO�n � log2n�, once linear forms are computed.

Table 2

Illustration of how to build the one machine problem from an initial CuSP and a linear forma

B � 6; n � 5 1 2 3 4 5

bi 3 2 3 4 5

pi 4 6 4 7 8

p0i 4� �1=2� � 2 6� �1=3� � 2 4� �1=2� � 2 7� �2=3� � 14=3 8� 1 � 8

a Here the linear form is: P6 � P5 � �2=3�P4 � 1=2� P3 � �1=3�P2. Notice that the processing times of the one machine problem deduced

are not necessarily integer.


6. Experimental results

To prove the e�ciency of the lower boundspresented, we have tested them into a branch andbound procedure solving RCPSP instances, andthe bounds de®ned for the CuSP are used in turnon each resource of the problem.

As we precise in Section 5.1, several decisionvariants of the RCPSP are solved one after an-other in order to determine whether there exist afeasible schedule of activities with a makespansmaller than or equal to UB. The branchingscheme is a chronological one. It consists in de-ciding at each node of the search tree if an activityis scheduled or if it has to be delayed. In this caseits release date is adjusted. The energetic adjust-ments, which are e�cient for highly cumulativeproblems are also performed at each node of thesearch tree [2]. Moreover the left-shifting domi-nance rule (e.g. [8]) is used to reduce the searchtree: if an activity can be shifted without violatingeither the precedence constraints or the resourceconstraints, then the schedule is dominated. Thusthe generated schedules are semi-active [15].

We have tested the method on a large set ofproblems. It has been proved [2] that most of theclassical benchmarks (Patterson [12], and KSD [9])are disjunctive ones, i.e., there exist many pairs ofactivities �i; j� such that i and j cannot be executedsimultaneously because either of their resourcerequirements or the precedence constraints. Forthese kinds of instances the MEPB bound haspractically no e�ect. Thus we have tested ourmethod on two other sets of instances.

The ®rst one was proposed by Baptiste and LePape [2]: instances were especially designed to becumulative ones. The results obtained were notsigni®cantly better if the MEPB bound is used. Butthere exist numerical reasons to explain these re-sults. In all these instances there are three re-sources, and the minimal resource capacity is equalto 3 and the maximal one is equal to 6. For theresource having a capacity equal to 3 and 4, allresource requirements are equal to one or two.Then the MEPB bound is the same as the ELbound, which is equal to

Ppi � bi. In the in-

stances, which have a resource of capacity 6, thereare only resource requirements equal to 1, 2 and 3.

Because 6 is a multiple of 2 and 3, the MEPBbound matches the EL bound. On problems inwhich one of the resource capacity is 5 (only 16among 40 instances), it appears that the resourceswhich have a capacity of 5 are less critical, i.e.,their ratios �1=B� � �P pi � bi� are always smallerthan the ratio of other resources. There exist only®ve problems, which have a resource of capacity 5as critical as the other ones, but all these ®ve in-stances are easy to solve, and the e�ciency of theMEPB bound cannot be proved on these instanc-es.

The second benchmark tested was generated toprove that MEPB could be e�cient for a givenrange of instances: resource capacities are ran-domly chosen among 5, 6, 7, 8, and 9. 26 problemswere generated. The smallest consists of 17 activ-ities and the largest of 35 activities. We have al-ways three resources. Processing times wererandomly generated. Only few precedence con-straints are added, to allow a large amount ofactivities to be processed simultaneously. We haveno resource with a capacity greater than 10 units;so a parametric version of the bound can be used.We have tested three versions of the MEPB bound:(1) simple, (2) designed according to JPS, (3) in-cluded in an energetic computation procedure.Moreover we have added to the procedure that wehave already described (energetic computation,bound,. . .), a procedure to detect if there existsome one-machine problems. In this case adjust-ments proposed by Carlier and Pinson [5] deducedfrom JPS, are applied (see Table 3).

To sum up our experimental results let us noticethat:· The column `without MEPB' gives the results

obtained, if the original energetic approach isused, only including necessary conditions of ex-istence and time bound adjustments deducedfrom the non-interruptible relaxation [1]. Noticethat the authors have shown that the non-inter-ruptible approach leads to better results thanthe fully elastic relaxation, on highly cumulativeproblems. The three other methods used areadded to this initial one, in order to prove thatthe MEPB bound can deduce more informationthan the original non-interruptible energetic ap-proach.


· The simple version of the bound GMEPB�D� �ri1 � MEPB�D� � qj1

is e�cient since it allowsto solve 50% of problems more than the initialmethod.

· There are few di�erences in terms of search timeand number of explored nodes between the sim-ple version of GMEPB and the MEPB boundbased on JPS.

· The energetic approach that uses MEPB boundseems to be the most e�cient way to use MEPB.As it has been explained previously, it seems dif-®cult to determine the exact formulations of rel-evant time intervals for the MEPB formulation.Thus we decide to compute the MEPB bound ontime intervals �t1; t2� such that t1; t2 2 fri; ri � pi;di ÿ pi; di=i 2 Ig�. Notice that this set of timepoints is the set used to compute the original en-ergetic bound, because they are very useful inpractice. Future research may determine theset of relevant time points. We want to pointout that most of our problems consist of 25±30 activities. If we apply this approach to largerproblems the search time could increase toomuch, due to the high time complexity of thisapproach.

· Even if these experiments are encouraging, wemust test our method on many other kinds of in-stances to con®rm the real e�ciency of thebounds we have proposed.

7. Conclusion

In this paper we have presented a new linearformulation that leads to e�cient lower boundsfor Cumulative Scheduling Problem, as theRCPSP. These bounds do not only take into ac-count the time/units during which the resource isbusy, but also the idle times that may occur on the

resource. The experimental results have provedthat for some classes of instances, the MEPBbound is e�cient. So it is very interesting from atheoretical and a practical point of view.

We show how to use it directly, and in aparametric approach. The parametric formula-tion, solved for some values of the resource ca-pacity, seems to be extremely e�cient since itallows us to compute the bound without using alinear solver.

Moreover we present di�erent ways for takinginto account release dates and tails of opera-tions. Even if the energetic approach leads tohigh time consuming algorithms, it seems inter-esting from a practical point of view. For thiskind of approach, we are able to compute ad-justments of release dates and tails which seemto be powerful. Their theoretical presentationand the optimisation of the algorithm related tothe energetic approach will be the subject offurther works.

The bound has been described for capacity upto 10 units. Such a parametric approach can beextended to resource capacities of 11, but it seemshard to generalize this approach to higher values.So a big challenge is to compute the MEPB boundfor B � 12; 13; . . . Then the primal linear programcan be solved with a linear solver. In this casespecial constraints can be added to the linearprogram in order to consider for example a mini-mum time for a con®guration. Of course if thelinear program is solved with a linear solver, theenergetic approach cannot be used, due to the factthat the bound must be computed for each timeinterval.

Moreover, we can intuitively see that the largeris the resource capacity, the larger is the number ofoptimal con®gurations. Thus the MEPB boundcould lead to the same result as the sum of the

Table 3

Experimental results which prove that MEPB bound is useful on this set of randomly generated instances, on a Ultra Sparc 2000

running Solaris. The search time is limited to 15 minutes

Time search limited to 500

seconds

Without MEPB

bound

MEPB bound

version GMEPB(I)

MEPB bound JPS MEPB bound

energetic

% of problems solved 35% 85% 85% 96%

Av. search time 51.1 s 17.9 s 16.7 s 16.4 s

Number of explored nodes 27 441 5662 5156 5347


weighted processing times divided by the resourcecapacity.

Finally, even if we have exhibited some classesof instances for which the bound is not useful,further work should identify more precisely thespeci®c classes of instances, for which the MEPBbound is e�cient.

Acknowledgements

The authors would like to thank PhilippeBaptiste for many enlightening discussions on theenergetic approach. Finally we are grateful to thereferees for their constructive comments.

Appendix A

A.1. Details for computing the MEPB bounds forB=8

Here we present the solutions that correspondto the di�erent cases described in the diagram.· Case 1: MEPB1�D� � P8 � P7 � P6 � P5 � P4=2

· Case 2: MEPB2�D� � P8 � P7 � P6 � P5 � P4=2� P 03=2 with P 03 � P3 ÿ P5

· Case 3: MEPB3�D� � P8 � P7 � P6 � P5 � P4=2

· Case 4: MEPB4�D� � P8 � P7 � P6 � P5� P4=2��P 02 ÿ P 05�=4 with P 02 � P2 ÿ P6 and P 05 � P5 ÿ P3

· Case 5: MEPB5�D� � P8 � P7 � P6 � P5 � P4=2�P 03=2 with P 03 � P3 ÿ P5

· Case 6: MEPB6�D� � P8 � P7 � P6 � P5 � P4=2�P 03=2� �P 02 ÿ P 03=2�=4 with P 03 � P3 ÿ P5; P 02 � P2

ÿ P6

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