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Scheduling with target start times

J.A. Hoogeveen a,*, S.L. van de Velde b

a Department of Computer Science, Utrecht University, P.O. Box 80089, 3508 TB Utrecht, Netherlandsb Rotterdam School of Management, Erasmus University, P.O. Box 1738, 3000 DR Rotterdam, Netherlands

Received 10 August 1999

Abstract

We address the single-machine problem of scheduling n independent jobs subject to target start times. Target start

times are essentially release times that may be violated at a certain cost. The objective is to minimize a bicriteria ob-

jective function that is composed of total completion time and maximum promptness, which measures the observance

of these target start times. We show that in case of a linear objective function the problem is solvable in O�n4� time if

preemption is allowed or if total completion time outweighs maximum promptness. Ó 2001 Elsevier Science B.V. All

rights reserved.

Keywords: Single-machine scheduling; Bicriteria scheduling; Target start times; Total completion time; Maximum

promptness

1. Introduction

We address a scheduling problem motivated bythe traditional con¯ict between internal and ex-ternal e�ciency faced by production companies.Internal e�ciency is the e�cient use of the scarceresources. Its aim is to reduce cost in order toquote more competitive prices or make morepro®t. External e�ciency is the extent by whichconditions set by external relations can be met.External e�ciency between the company and its

clients, that is, the extent by which the companysuccessfully copes with due dates, is called down-stream e�ciency; upstream external e�ciencymeasures the success of meeting the conditions atthe suppliers' side. Often, however, these condi-tions are subject to negotiations, which is espe-cially true in case of integrated planning within asupply chain. Then the company can increase itsinternal e�ciency by successfully negotiating withits customers or its suppliers.

There exist several single-machine schedulingmodels of the trade-o� between internal anddownstream external e�ciency. Van Wassenhoveand Gelders (1980), for instance, consider a modelfor making the trade-o� between work-in-processinventories and due date performance; see alsoHoogeveen and van de Velde (1995). Schutten et al.

European Journal of Operational Research 129 (2001) 87±94www.elsevier.com/locate/dsw

* Corresponding author. Tel.: +31-30-253-4089; fax: +31-30-

251-3791.

E-mail addresses: [email protected] (J.A. Hoogeveen), s.vel-

[email protected] (S.L. van de Velde).

0377-2217/01/$ - see front matter Ó 2001 Elsevier Science B.V. All rights reserved.

PII: S 0 3 7 7 - 2 2 1 7 ( 9 9 ) 0 0 4 2 6 - 9

(1996) consider a batching problem for balancingout the utilization of machine capacity against duedate performance. Single-machine problems seemto be oversimpli®ed models, but the study of thesemodels makes sense, if we think of a company as asingle-machine shop, or if there is a single bottle-neck. What is more, single-machine models serveas building-blocks for solving complex schedulingproblems.

In this paper, we look at the problem ofdealing simultaneously with internal and upstreamexternal e�ciency. We assume that the companycan negotiate on the prices and delivery times ofraw material, which may result in a higher in-ternal e�ciency at the expense of a higher pricefor getting the raw material delivered sooner. Wemodel this problem as a single-machine schedul-ing problem in the following way. A set of n in-dependent jobs has to be scheduled on a singlemachine that is continuously available from timezero onwards and that can process at most onejob at a time. Each job Jj �j � 1; . . . ; n� requiresprocessing during a positive time pj and has atarget start time sj. Without loss of generality, weassume that the processing times and target starttimes are integral. A schedule r speci®es for eachjob when it is executed while observing the ma-chine availability constraint; hence, a schedule rde®nes for each job Jj its start time Sj�r� and itscompletion time Cj�r�. The promptness Pj�r� ofjob Jj is de®ned as Pj�r� � sj ÿ Sj�r�, and themaximum promptness is de®ned as Pmax�r� �max16 j6 n Pj�r�. We note that the maximumpromptness Pmax�r� equals the maximum earlinessEmax�r��max16j6n�djÿCj�r�� if each Jj has a duedate dj for which sj�djÿpj and if Cj�r��Sj�r��pj, that is, if interruption of job processingis not allowed.

The problem we consider is to schedule the jobsso as to minimize total completion time

Pnj�1 Cj

and maximum promptness Pmax simultaneously.Total completion time

Pnj�1 Cj is a measure of the

work-in-process inventories as well as the averageleadtime. Hence, it is a performance measure forinternal e�ciency as well as downstream externale�ciency.

Maximum promptness measures the obser-vance of target start times. If it is positive, then it

signals an ine�ciency: at least one job is scheduledto start before its target start time. Generally, thisis possible only if we are willing to pay a penalty.In case the target start times are derived from thedelivery times of raw material, then this penalty isactually the price of a speedier delivery. In casethe target start times are derived from the com-pletion times of the parts in the preceding pro-duction stage, then this penalty may be anoverwork bonus to expedite the production. If themaximum promptness is negative, then it signalsa slack, which implies that we may increase thedeadlines that are used in the preceding produc-tion stage.

It is important to realize that the target starttimes are actually release times that may beviolated at a certain cost. In this sense, ourproblem comes close to the well-studied single-machine problem of minimizing total completiontime subject to release times; see, for instance,Lenstra et al. (1977) and Ahmadi and Bagchi(1990).

We now give a formal speci®cation of our ob-jective function. We associate with each schedule ra point �Pn

j�1 Cj�r�; Pmax�r�� in R2 and a valuea1

Pnj�1 Cj�r� � a2Pmax�r�. Since we want to mini-

mize both criteria, we assume that a1 and a2 arenon-negative. Extending the three-®eld notationscheme of Graham et al. (1979), we denote thisproblem by 1ka1

Pnj�1 Cj � a2Pmax.

In comparison to single-criterion problems,there are few papers on multicriteria schedulingproblems. We refer to Dileepan and Sen (1988)and Hoogeveen (1992) for an overview of prob-lems, polynomial algorithms, and complexity re-sults.

This paper is organized as follows. In Section 2,we make some general observations and outline ageneric strategy for solving the 1ka1

Pnj�1 Cj

�a2Pmax problem. In Section 3, we consider thevariant in which preemption is allowed, that is, ajob may be interrupted and resumed later; thisproblem is denoted as 1jpmtnja1

Pnj�1 Cj � a2Pmax,

where the acronym pmtn indicates that preemp-tion is allowed. Our main results are that1jpmtnja1

Pnj�1 Cj � a2Pmax and, in the case that

a1 P a2, also 1ka1

Pnj�1 Cj � a2Pmax are solvable in

O�n4� time.

88 J.A. Hoogeveen, S.L. van de Velde / European Journal of Operational Research 129 (2001) 87±94

2. General observations

One strategy that we could follow to solve ourproblem is to determine the trade-o� curve ofPn

j�1 Cj and Pmax, which is de®ned as the curve thatconnects all points �C; P �, where C is the outcomeof 1jPmax6 P jPn

j�1 Cj. If the trade-o� is known,then we can minimize our linear composite ob-jective function by computing the objective valuein each extreme point of the trade-o� curve, whichis de®ned as follows.

De®nition 1. A point of the trade-o� curve iscalled an extreme point if it corresponds to a ver-tex of the lower envelope of the trade-o� curve. Aschedule corresponding to an extreme point iscalled an extreme schedule.

Theorem 1. If the minimum is bounded and if a1 anda2 are non-negative, then there exists an extremeschedule that solves 1ka1


Proof. Suppose to the contrary that the minimumis attained in a point �C; P� on the trade-o� curvethat is not an extreme point. If this point is not onthe lower envelope, then there exists a point � �C; �P �with �C6C and �P 6 P on the lower envelope,which clearly dominates �C; P �, as a1 and a2 arenon-negative. On the other hand, each point on asegment of the lower envelope between two ex-treme points is dominated by one of these, sincethe objective function is linear. �

The question then becomes how di�cult it is to®nd the trade-o� curve for

Pnj�1 Cj and Pmax or,

su�ciently, the set of extreme points. This requiressolving a problem of the kind 1jPmax6 P jPn

j�1 Cj,for some value P. The constraint Pmax6 P isequivalent to the constraint that Sj P maxf0; sj

ÿPg, and hence the problem 1jPmax6 P jPnj�1 Cj is

equivalent to the problem 1jrjjPn

j�1 Cj, where rj

denotes a release date before which job Jj cannotbe started. Lenstra et al. (1977), however, haveshown that the problem 1jrjj

Pnj�1 Cj is NP-hard

in the strong sense.We therefore make the additional assumption

that preemption of jobs is allowed, that is, theexecution of any job may be interrupted and re-

sumed later on. This assumption implies a crucialrelaxation of the original problem; it has bothpositive and negative aspects. To start with thepositive part: we can solve the problem1jpmtn; Pmax6 P jPn

j�1 Cj, since the problem1jpmtn; rjj

Pnj�1 Cj is solvable in O�n log n� time

by Baker's algorithm (Baker, 1974): always keepthe machine assigned to the available job with min-imum remaining processing time. The disadvantageis that we lose the equivalence that existed betweenthe maximum promptness criterion and the maxi-mum earliness criterion in case sj � dj ÿ pj. This isso, since a given value Emax induces an earliestcompletion time for each job, not a release date.

In the next section we will work out our plan ofdetermining the set of extreme points of the trade-o� curve. We conclude this section with a discus-sion of the two single-criterion problems that areembedded within 1jpmtnja1

Pnj�1 Cj � a2Pmax, that

is, 1kPmax and 1kPnj�1 Cj (we do not include

`pmtn', since preemption is not advantageous forthese problems). The 1kPmax problem is clearlymeaningless, since we can improve upon each so-lution by inserting extra idle time at the beginningof the schedule. Hence, we impose the restrictionthat machine idle time before the processing of anyjob is prohibited, that is, all jobs are to be sched-uled in the interval �0;Pn

j�1 pj�. It is easily checkedthat in case of a given overall deadline D >

Pnj�1 pj

the optimal schedule is obtained by insertingDÿPn

j�1 pj units of idle time before the start ofthe ®rst job. In the three-®eld notation scheme, theno machine idle time constraint is denoted by theacronym nmit in the second ®eld. The 1jnmitjPmax

problem is solved by sequencing the jobs in orderof non-decreasing target start times sj. The1kPn

j�1 Cj problem is solved by sequencing thejobs in order of non-decreasing processing times pj

(Smith, 1956). Let now MTST be an optimalschedule for the 1jnmitjPmax problem in which tiesare settled to minimize total completion time;MTST is the abbreviation of minimum target starttime. In addition, let SPT be an optimal schedulefor the 1kPn

j�1 Cj problem, in which ties are set-tled to minimize maximum promptness; SPT is theabbreviation of shortest processing time. It thenfollows that P �max6 Pmax�r�6 Pmax�SPT� andPn

j�1 C�j 6Pn

j�1 Cj�r�6Pn

j�1 Cj�MTST� for any

J.A. Hoogeveen, S.L. van de Velde / European Journal of Operational Research 129 (2001) 87±94 89

extreme schedule r without idle time, where P �max

andPn

j�1 C�j denote the outcome of the respectivesingle-criterion problems. Note that Baker's algo-rithm always generates a schedule without ma-chine idle time if Pmax P P �max.

3. Determining extreme points of the trade-o� curve

If the extreme set can be found in polynomialtime and if its cardinality is polynomially bounded,then the 1ka1

Pnj�1 Cj � a2Pmax problem is solved in

polynomial time by computing the cost of eachextreme point and taking the minimum.

We start by analyzing the special case in whichmachine idle time before the processing of any jobis prohibited; we later show that any instance ofthe general problem can be dealt with by refor-mulating it as an instance of the problem with nomachine idle time allowed.

3.1. No machine idle time allowed

Recall that if machine idle time is not allowed,then all jobs are processed in the interval�0;Pn

j�1 pj�. Hence, we only have to consider Pmax

values in the interval �P �max; Pmax�SPT��, and foreach Pmax value P in this interval, Baker's algo-rithm provides an optimal schedule for the corre-sponding 1jpmtn; Pmax6 P jPn

j�1 Cj problem thatdoes not contain idle time; let r�P � denote thisschedule and let �P ;Pn

j�1 Cj�r�P �� denote thepoint in R2 corresponding to it.

We start with a three-job example for which wedetermine the trade-o� curve. The data are asfollows:

The trade-o� curve is depicted in Fig. 1.Breakpoints have been marked; the correspondingschedules are described below.

We denote the schedules by the order in whichthe jobs occur; if job Jj occurs more than once,

then it is preempted, and the successor of a pre-empted piece starts as soon as possible. Point (1)corresponds to the schedule �J1; J2; J3�. Thisschedule remains optimal until Pmax becomes morethan 1; at point (2), schedule �J1; J2; J1; J3� becomesoptimal. This order remains optimal untilPmax P 2; point (3) corresponds to �J1; J2; J1; J3; J1�.For 26 Pmax6 5, this order remains optimal; theoptimal schedule becomes �J2; J1; J3; J1� (point (4))at Pmax � 5, which remains optimal until Pmax P 7,at which point the optimal order becomes�J2; J3; J1� (point (5)). For Pmax 2 �7; 9� this order isoptimal; at Pmax � 9 the optimal order becomes�J2; J3; J2; J1�, from which ®nally at Pmax � 10 theSPT order �J3; J2; J1� is reached. The lower enve-lope is found by connecting the points (1), (4), (5),and (7), which are the extreme points.

The problem is of course to distinguish betweenan extreme schedule and an ordinary schedule thatcorresponds to a point on the trade-o� curve. Weknow that the lower envelope is continuous andpiecewise-linear, and most important, that in eachbreakpoint the gradient decreases. This impliesthat a necessary condition for a schedule r�P � tobe extreme is that increasing P by some � > 0yields a smaller decrease in

Pnj�1 Cj than a decrease

of P by the same amount � would cost. From theexample above, we conclude that a schedule canonly be extreme if a complete interchange has

Jj 1 2 3

pj 7 3 2sj 0 5 10

Fig. 1. Trade-o� curve.


occurred in r�P �, where an interchange is de®nedto be a complete interchange if there are two jobsJi and Jj such that Ji is started before Jj inr�P ÿ ��, whereas Jj is started before Ji in r�P �.

Lemma 1. If P > P �max, then the point �P ;Pnj�1

Cj�r�P �� can be extreme only if a complete inter-change has occurred in r�P�.

The next step is to determine the Pmax values Psuch that their corresponding points�P ;Pn

j�1Cj�r�P�� satisfy this necessary condition.Given a pair of jobs Ji and Jj with pi > pj and Ji

started before Jj in r�P �, we have to increase theupper bound on Pmax such that Jj can start at timeSi�r�P ��. This will lead to a complete interchangeof Ji and Jj in r�P 1�, unless Ji itself is started at anearlier time in the schedule r�P 1�, whereP 1 �sj ÿ Si�r�P�� is the value of the upper boundon Pmax that makes Jj available at time Si�r�P ��. Itis not possible to determine beforehand whether Ji

gets started earlier when the upper bound on Pmax

is increased from P to P 1, except for one situation:Ji is executed between the start and completiontime of a preemptive job Jk. In that case, increasingthe upper bound on Pmax will ®rst lead to a uniformshift forward of Ji and Jj at the expense of Jk; thecomplete interchange of Ji and Jj cannot take placebefore a complete interchange has taken placebetween Jk and both Ji and Jj.

Algorithm 1 exploits these observations togenerate each point �P ;Pn

j�1 Cj�r�Pmax�� forwhich a complete interchange in r�P� may takeplace. The variable aj (j � 1; . . . ; n� signi®es theincrease of the current Pmax value necessary to let acomplete interchange involving Jj and some suc-cessor take place. After having stated the algo-rithm, we apply it to our example.

Algorithm 1.

Step 0. Let P � P �max.Step 1. Let T 0 and aj 1 for j � 1; . . . ; n;determine r�P � through Baker's rule.Step 2. Let Jk be the job that starts at time T inr�P �. Consider the following two cases:

(a) Jk is a preempted job. Then ak is equal tothe length of this portion of Jk. SetT Ck�r�P ��.

(b) Jk is not a preempted job. Thenak minfsj ÿ P ÿ Sk�r�P �� jJj 2Vg, whereV denotes the set of jobs Jj for whichsj ÿ P > Sk�r�P�� and pj > pk. SetT Ck�r�P��.

Step 3. If T <Pn

j�1 pj, then go to Step 2.Step 4. Put P min16 j6 n aj � P .Step 5. If P � Pmax�SPT�, then stop; otherwisego to Step 1.

If we apply Algorithm 1 to our example, thenwe start with the schedule r�0� in which the jobsoccur in the order J1; J2; J3 and are not preempted.Step 2(b) puts a1 5 and a2 3, which impliesthat we put P 3. If we apply Baker's rule todetermine r�3�, then we see that the planned in-terchange of jobs J2 and J3 does not take placesince J2 is moved forward and preempts job J1, justlike job J3 (the corresponding order of the jobs inr�3� is �J1; J2; J1; J3; J1�). If we apply Algorithm 1to r�3�, then Step 2(a) puts a1 2, which removesthe preempted piece of job J1 to the end of theschedule. Hence, P 5, and r�5� is the schedulecorresponding to point (4). In the next iteration,we ®nd a2 5 and a1 2, and the preemptedpiece of J1 between J2 and J3 is removed (P 7),after which we ®nally ®nd P 10, which enablesthe full interchange of jobs J2 and J3, which bringsus in point (7). Algorithm 1 has found all extremeschedules and the unwanted schedule r�3�. We willnow give a formal proof of the correctness of Al-gorithm 1.

Theorem 2. Algorithm 1 generates all Pmax values Pfor which a complete interchange has taken place inthe corresponding schedule r�P �.

Proof. Suppose that a complete interchange of thejobs Ji and Jj with pi > pj took place in theschedule r�P �, where P was not detected by Al-gorithm 1. Hence, Si�r�Pmax�� must have been ig-nored in Step 2, which could have happened onlyin Step 2(a): Ji is started between the start andcompletion time of some preempted job Jk. This,however, con¯icts with the earlier observation thatthe interchange of Ji and Jj has to wait until Jk hasbeen interchanged with both Ji and Jj. �


As shown in the example, the algorithm maygenerate too many Pmax values P: in some of theschedules r�P� not a complete interchange hastaken place. This is due to Step 2(b). There weimplicitly assumed that the part of the schedulebefore Jk, which was de®ned as the job to be in-terchanged, would remain scheduled before Jk,that is, that Jk itself would not be started earlier.This is not necessarily the case, however, since anincrease of the upper bound on Pmax may cause Jk

to move forward at the expense of some job Jl withpl > pk, where the increase of the upper bound isnot large enough to allow a complete interchange;Jk will preempt Jl then. Nevertheless, we nowprove that the number of values Pmax generated byAlgorithm 1 is polynomially bounded, therebyestablishing that 1jpmtn; nmitja1

Pnj�1 Cj � a2Pmax

is polynomially solvable. We de®ne for a givenschedule r the indicator function dij�r� as

dij�r� � 2 if Ci�r�6 Sj�r� and pi > pj;0 otherwise:

�We further de®ne Dj�r� as

Pni�1 dij�r� plus the

number of preemptions of Jj, and we let D�r� �Pnj�1 Dj�r�.

Theorem 3. Let P 1 be the Pmax value that is foundby Algorithm 1 when applied to r�P �, where P is anyPmax value determined by Algorithm 1. We then havethat D�r�P 1�� < D�r�P ��.

Proof. As explained above, one of the followingthree things has happened in r�P 1� in comparisonto r�P�:

(i) a preemption has been removed (Step 2(a));(ii) two jobs not in SPT-order have been inter-changed (successful Step 2(b));(iii) a new preemption has been created (unsuc-cessful Step 2(b)).

All three cases have a negative e�ect on the valueof D, as is easily checked (in the third case we docreate an extra preemption (e�ect +1), but this pairof jobs is no longer in the wrong order (e�ect ÿ2)).Hence, we only have to show that there are nomoves possible that have an overall positive e�ecton the value of D. The candidates for such a moveare a switch of two jobs from SPT order to LPT

order and the addition of an extra preemption. We®rst investigate the e�ect of the `wrong' switch.

Suppose that there are two jobs Ji and Jj withpi > pj such that Ji succeeds Jj in r�P �, whereas theorder is reversed in r�P 1�. Since Baker's algorithmprefers Jj to Ji if both jobs are available, Ji startsearlier in r�P 1� than Jj in r�P �, which means thatthe execution of (a part of) some job Jk is post-poned until Ji is completed. See Fig. 2 for an il-lustration.

It is easily checked that we have D�r�P �� 4and D�r�P 1�� 3. All we have to do is to showthat the situation depicted in Fig. 2 is worst pos-sible for this con®guration. It is su�cient to provethat Jj is available at time Ci�r�P 1��, that is,sj ÿ P 16Ci�r�P 1�� si ÿ P 1 � pi; if so, Baker'salgorithm will prefer it to Jk, since the remainder ofJk has length at least equal to pi. Hence, we have toshow that sj6 si � pi. As Ji did not preempt Jk inr�P�, we must have si ÿ P � pi P Ck�r�P ��Psj ÿ P , where the last inequality follows from theavailability of Jj at time Ck�r�P��. Since thesmaller job is available as soon as the larger jobinvolved in the wrong switch is completed, theincrease of dij is compensated for by the decreaseof dki. Moreover, job Jk cannot trigger a set ofnested wrong switches, where we mean with a setof nested wrong switches that r�P � and r�P 1�contain the subschedules Jk; Jj; Ji; Jh andJh; Ji; Jj; Jk with pj < pi < ph < pk.

Now consider the situation that the number ofpreemptions of a job Jk increases. Hence, theremust be a job Ji with pi < pk that succeeds Jk inr�P� but not in r�P 1�, which move decreases the Dfunction by 1. �

Corollary 1. If preemption is allowed, then thenumber of extreme schedules with respect to�Pmax;

Pnj�1 Cj� is bounded by n�nÿ 1� � 1.

Fig. 2. `Wrong' switch.


Proof. We have that D�r�6 n�nÿ 1� for anyschedule r. Application of Theorem 3 yields thedesired result. �

It is easy to construct an instance for whichAlgorithm 1 determines O�n2� di�erent Pmax val-ues. We have not found an example with O�n2�extreme points yet.

Corollary 2. The 1jpmtn; nmitja1


problem is solvable in O�n4� time.

Theorem 4. If a1 � a2, then there exists a non-preemptive optimal schedule for the 1jpmtn; nmitja1Pn

j�1 Cj � a2Pmax problem. If a1 > a2, then any op-timal schedule for the 1jpmtn; nmitja1

Pnj�1 Cj

�a2Pmax problem is non-preemptive.

Proof. Suppose that the optimal schedule containsa preempted job. Start at time 0 and ®nd the ®rstpreempted job Ji immediately scheduled beforesome non-preempted job Jj. Consider the scheduleobtained by interchanging job Jj and this portionof job Ji. If the length of the portion of job Ji is �,then Pj is increased by �, while Cj is decreased by �.As a1 � a2, the interchange does not increase theobjective value. The argument can be repeateduntil a non-preemptive schedule remains. In casea1 > a2 such an interchange would decrease theobjective value, contradicting the optimality of theinitial schedule. �

3.2. The general case

We now drop the no machine idle time con-straint. Obviously, if total completion time out-weighs maximum promptness, then the insertionof machine idle time before the processing of anyjob makes no sense. Hence, we have the following.

Corollary 3. If a1 P a2, then 1ka1


is solvable in O�n4� time.

If a1 < a2, then the insertion of idle time maydecrease the value of the objective function. Wenow show that we can solve the 1jpmtnja1Pn

j�1 Cj � a2Pmax problem by using Algorithm 1,

which was initially designed for solving1jpmtn; nmitja1


Suppose that a1 and a2 are given. De®neq � a2=a1. If q > n, then it is always advantageousto decrease Pmax, which implies that the executionof the ®rst job will be delayed for ever and ever. Toprevent unbounded solutions, we therefore assumethat q6 n. A straightforward computation thenshows that in any optimal schedule at leastbnÿ q� 1c jobs are scheduled before the ®rst in-cidence of idle time. The smallest value Pmax�q� formaximum promptness that leads to such a sched-ule is readily obtained. Moreover, no optimalschedule with Pmax P P �max contains idle time.Therefore, we only have to describe how our ap-proach can be adjusted to deal with the casePmax�q�6 Pmax6 P �max.

Consider any instance I of 1jpmtnja1

Pnj�1

Cj � a2Pmax; let r�P� denote any optimal schedulefor I of 1jpmtn; rjj

PCj for any Pmax value P with

Pmax�q�6 P 6 P �max and rj � maxf0; sj ÿ Pg.We create a very large job J0 that is available

from time 0 onwards to saturate r�P� by ®lling inJ0 in the periods of idle time. In fact, J0 is so largethat Baker's rule prefers each job in I to it; thechoices s0 � Pmax�q� and p0 � P �max ÿ Pmax�q��max16 j6 n pj � 1 ensure such a saturation for anyPmax value P with Pmax�q�6 P 6 P �max. Let I0 denotethe instance I to which J0 is added. Due to thechoice of p0 and s0, we have that no optimalschedule for the instance I0 of 1jpmtn; nmitja1Pn

j�1�a2Pmax contains machine idle time, andmoreover, that by simply removing J0 and leavingthe rest of the schedule intact we obtain an optimalschedule for the original instance I of1jpmtnja1

Pnj�1 Cj � a2Pmax. After all, we have that

C0 �Pn

j�0 pj and that J0 does not assume Pmax.Hence, instead of solving 1jpmtnja1

Pnj�1

Cj � a2Pmax for I, we solve 1jpmtn; nmitja1

Pnj�0

Cj � a2Pmax for I0. This approach provides us withthe extreme points for �Pn

j�1 Cj; Pmax� withPmax�q�6 Pmax6 P �max. If q is unknown, then weobtain all bounded extreme points by running theabove procedure with q � n; this choice of q cor-responds to the smallest value Pmax�q� that maycorrespond to a bounded extreme point.

As the number of extreme points is at mostequal to n�n� 1� � 1 (we have n� 1 jobs now),


and as each Pmax value that corresponds to an ex-treme point is determined by Algorithm 1, the1jpmtnja1

Pnj�1 Cj � a2Pmax problem is solved in

O�n4� time.

4. Related problems

In this section we discuss a number of relatedproblems. The ®rst one deals with external boundson Pmax. Usually, Pmax cannot assume any value,but is bounded from below or above. Such addi-tional constraints are easy to handle: we simplyrestrict the interval over which we look for extremeschedules.

The second problem is to minimize Pmax in caseof an upper bound on

Pnj�1 Cj, that is, problem

1jPnj�1 Cj6CjPmax. It is hard to deal with an upper

bound on a sum-criterion explicitly, but in our casewe search the trade-o� curve using binary search.

The third problem has to do with a general,non-linear, composite objective functionF �Pn

j�1 Cj; Pmax�, which we may assume to be non-decreasing in both arguments. Instead of lookingfor the extreme points of the trade-o� curve, wethen must ®nd all non-dominated or Pareto opti-mal points. Unfortunately, this number is notbounded in case preemption or idle time is al-lowed, as can easily be seen from Fig. 1. By using aresult by Schrijver (see Hoogeveen, 1996), weconclude that 1jpmtnjF �Pn

j�1 Cj; Pmax� and1kF �Pn

j�1 Cj;Pmax� are NP-hard in the strongsense.

5. Conclusions

We have presented a simple single-machinescheduling model for negotiating delivery timeswith an external supplier or a preceding produc-

tion stage. A characteristic of our model is that alldelivery times are a�ected in the same way, whichcontrasts with the work in the so-called just-in-time models, where the delivery time of one job isindependent of the delivery time of another job.We refer to Hoogeveen and van de Velde (1996)for an overview of work done in this area.

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