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
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 On4 time ifpreemption is allowed or if total completion time outweighs maximum promptness. 2001 Elsevier Science B.V. Allrights reserved.
Keywords: Single-machine scheduling; Bicriteria scheduling; Target start times; Total completion time; Maximum
We address a scheduling problem motivated bythe traditional conflict between internal and ex-ternal eciency faced by production companies.Internal eciency is the ecient use of the scarceresources. Its aim is to reduce cost in order toquote more competitive prices or make moreprofit. External eciency is the extent by whichconditions set by external relations can be met.External eciency between the company and its
clients, that is, the extent by which the companysuccessfully copes with due dates, is called down-stream eciency; upstream external eciencymeasures 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 eciency by successfully negotiating withits customers or its suppliers.
There exist several single-machine schedulingmodels of the trade-o between internal anddownstream external eciency. 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) 8794www.elsevier.com/locate/dsw
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(1996) consider a batching problem for balancingout the utilization of machine capacity against duedate performance. Single-machine problems seemto be oversimplified 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 eciency. We assume that the companycan negotiate on the prices and delivery times ofraw material, which may result in a higher in-ternal eciency 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 specifies for eachjob when it is executed while observing the ma-chine availability constraint; hence, a schedule rdefines for each job Jj its start time Sjr and itscompletion time Cjr. The promptness Pjr ofjob Jj is defined as Pjr sj Sjr, and themaximum promptness is defined as Pmaxr max16 j6 n Pjr. We note that the maximumpromptness Pmaxr equals the maximum earlinessEmaxrmax16j6ndjCjr if each Jj has a duedate dj for which sjdjpj and if CjrSjrpj, that is, if interruption of job processingis not allowed.
The problem we consider is to schedule the jobsso as to minimize total completion time
and maximum promptness Pmax simultaneously.Total completion time
Pnj1 Cj is a measure of the
work-in-process inventories as well as the averageleadtime. Hence, it is a performance measure forinternal eciency as well as downstream externaleciency.
Maximum promptness measures the obser-vance of target start times. If it is positive, then it
signals an ineciency: 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 specification of our ob-jective function. We associate with each schedule ra point Pnj1 Cjr; Pmaxr in R2 and a valuea1Pn
j1 Cjr a2Pmaxr. Since we want to mini-mize both criteria, we assume that a1 and a2 arenon-negative. Extending the three-field notationscheme of Graham et al. (1979), we denote thisproblem by 1ka1
Pnj1 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
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
Pnj1 Cj a2Pmax,
where the acronym pmtn indicates that preemp-tion is allowed. Our main results are that1jpmtnja1
Pnj1 Cj a2Pmax and, in the case that
a1 P a2, also 1ka1Pn
j1 Cj a2Pmax are solvable inOn4 time.
88 J.A. Hoogeveen, S.L. van de Velde / European Journal of Operational Research 129 (2001) 8794
2. General observations
One strategy that we could follow to solve ourproblem is to determine the trade-o curve ofPn
j1 Cj and Pmax, which is defined as the curve thatconnects all points C; P , where C is the outcomeof 1jPmax6 P j
Pnj1 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 defined as follows.
Definition 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
Pnj1 Cj a2Pmax.
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 dicult it is tofind the trade-o curve for
Pnj1 Cj and Pmax or,
suciently, the set of extreme points. This requiressolving a problem of the kind 1jPmax6 P j
for some value P. The constraint Pmax6 P isequivalent to the constraint that Sj P maxf0; sjPg, and hence the problem 1jPmax6 P j
Pnj1 Cj is
equivalent to the problem 1jrjjPn
j1 Cj, where rjdenotes a release date before which job Jj cannotbe started. Lenstra et al. (1977), however, haveshown that the problem 1jrjj
Pnj1 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 j
Pnj1 Cj, since the problem
j1 Cj is solvable in On log n timeby Bakers 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 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
Pnj1 Cj a2Pmax, that
is, 1kPmax and 1kPn
j1 Cj (we do not includepmtn, 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;Pnj1 pj. It is easily checkedthat in case of a given overall deadline D >
the optimal schedule is obtained by insertingDPnj1 pj units of idle time before the start ofthe first job. In the three-field notation scheme, theno machine idle time constraint is denoted by theacronym nmit in the second field. The 1jnmitjPmaxproblem is solved by sequencing the jobs in orderof non-decreasing target start times sj. The1kPnj1 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 1kPnj1 Cj problem, in which ties are set-tled to minimize maximum promptness; SPT is theabbreviation of shortest processing time. It thenfollows that P max6 Pmaxr6 PmaxSPT andPn
j1 Cj 6
Pnj1 CjMTST for any
J.A. Hoogeveen, S.L. van de Velde / European Journal of Operational Research 129 (2001) 8794 89
extreme schedule r without idle time, where P maxand
j denote the outcome of the respective
single-criterion problems. Note that Bakers 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
Pnj1 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 interval0;Pnj1 pj. Hence, we only have to consider Pmaxvalues in the interval P max; PmaxSPT, and foreach Pmax value P in this interval, Bakers algo-rithm provides an optimal schedule for the corre-sponding 1jpmtn; Pmax6 P j
Pnj1 Cj problem that
does not contain idle time; let rP denote thisschedule and let P ;Pnj1 CjrP 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 becomesJ2; J3; J1 (point (5)). For Pmax 2 7; 9 this order isoptimal; at Pmax 9 the optimal order becomesJ2; J3; J2; J1, from which finally 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 rP tobe extreme is that increasing P by some > 0yields a smaller decrease in
Pnj1 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.
90 J.A. Hoogeveen, S.L. van de Velde / European Journal of Operational Research 129 (2001) 8794
occurred in rP , where an interchange is definedto be a complete interchange if there are two jobsJi and Jj such that Ji is started before Jj inrP , whereas Jj is started before Ji in rP .
Lemma 1. If P > P max, then the point P ;Pn
j1CjrP can be extreme only if a complete inter-change has occurred in rP.
The next step is to determine the Pmax values Psuch that their corresponding pointsP ;Pnj1CjrP satisfy this necessary condition.Given a pair of jobs Ji and Jj with pi > pj and Jistarted before Jj in rP , we have to increase theupper bound on Pmax such that Jj can start at timeSirP . This will...