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Scheduling with target start times
European Journal of Operational Research 129 (2001) 87±94
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Scheduling with target start times J.A. Hoogeveen b
a,*
, S.L. van de Velde
b
a Department of Computer Science, Utrecht University, P.O. Box 80089, 3508 TB Utrecht, Netherlands 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 objective 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 by the traditional con¯ict between internal and external eciency faced by production companies. Internal eciency is the ecient use of the scarce resources. Its aim is to reduce cost in order to quote more competitive prices or make more pro®t. External eciency is the extent by which conditions set by external relations can be met. External eciency between the company and its
* Corresponding author. Tel.: +31-30-253-4089; fax: +31-30251-3791. E-mail addresses: [email protected] (J.A. Hoogeveen), [email protected] (S.L. van de Velde).
clients, that is, the extent by which the company successfully copes with due dates, is called downstream eciency; upstream external eciency measures the success of meeting the conditions at the suppliers' side. Often, however, these conditions are subject to negotiations, which is especially true in case of integrated planning within a supply chain. Then the company can increase its internal eciency by successfully negotiating with its customers or its suppliers. There exist several single-machine scheduling models of the trade-o between internal and downstream external eciency. Van Wassenhove and Gelders (1980), for instance, consider a model for making the trade-o between work-in-process inventories and due date performance; see also Hoogeveen and van de Velde (1995). Schutten et al.
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
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(1996) consider a batching problem for balancing out the utilization of machine capacity against due date performance. Single-machine problems seem to be oversimpli®ed models, but the study of these models makes sense, if we think of a company as a single-machine shop, or if there is a single bottleneck. What is more, single-machine models serve as building-blocks for solving complex scheduling problems. In this paper, we look at the problem of dealing simultaneously with internal and upstream external eciency. We assume that the company can negotiate on the prices and delivery times of raw material, which may result in a higher internal eciency at the expense of a higher price for getting the raw material delivered sooner. We model this problem as a single-machine scheduling problem in the following way. A set of n independent jobs has to be scheduled on a single machine that is continuously available from time zero onwards and that can process at most one job at a time. Each job Jj j 1; . . . ; n requires processing during a positive time pj and has a target start time sj . Without loss of generality, we assume that the processing times and target start times are integral. A schedule r speci®es for each job when it is executed while observing the machine availability constraint; hence, a schedule r de®nes for each job Jj its start time Sj r and its completion time Cj r. The promptness Pj r of job Jj is de®ned as Pj r sj ÿ Sj r, and the maximum promptness is de®ned as Pmax r max1 6 j 6 n Pj r. We note that the maximum promptness Pmax r equals the maximum earliness Emax rmax16j6n dj ÿCj r if each Jj has a due date dj for which sj dj ÿpj and if Cj r Sj rpj , that is, if interruption of job processing is not allowed. The problem we consider is to schedule the Pn jobs so as to minimize total completion time j1 Cj and maximum promptness Pn Pmax simultaneously. Total completion time j1 Cj is a measure of the work-in-process inventories as well as the average leadtime. Hence, it is a performance measure for internal eciency as well as downstream external eciency. Maximum promptness measures the observance of target start times. If it is positive, then it
signals an ineciency: at least one job is scheduled to start before its target start time. Generally, this is possible only if we are willing to pay a penalty. In case the target start times are derived from the delivery times of raw material, then this penalty is actually the price of a speedier delivery. In case the target start times are derived from the completion times of the parts in the preceding production stage, then this penalty may be an overwork bonus to expedite the production. If the maximum promptness is negative, then it signals a slack, which implies that we may increase the deadlines that are used in the preceding production stage. It is important to realize that the target start times are actually release times that may be violated at a certain cost. In this sense, our problem comes close to the well-studied singlemachine problem of minimizing total completion time 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 objective function. We associate with each schedule r Pn 2 a point C j1 j r; Pmax r in R and a value Pn a1 j1 Cj r a2 Pmax r. Since we want to minimize both criteria, we assume that a1 and a2 are non-negative. Extending the three-®eld notation scheme of Graham P et al. (1979), we denote this problem by 1ka1 nj1 Cj a2 Pmax . In comparison to single-criterion problems, there are few papers on multicriteria scheduling problems. We refer to Dileepan and Sen (1988) and Hoogeveen (1992) for an overview of problems, polynomial algorithms, and complexity results. This paper is organized as follows. In Section 2, we make some general observations and outline Pn a generic strategy for solving the 1ka1 j1 Cj a2 Pmax problem. In Section 3, we consider the variant in which preemption is allowed, that is, a job may be interrupted and resumed later; this Pn problem is denoted as 1jpmtnja1 j1 Cj a2 Pmax , where the acronym pmtn indicates that preemption is allowed. Our main results are that P a2 Pmax and, in the case that 1jpmtnja1 nj1 Cj P n a1 P a2 , also 1ka1 j1 Cj a2 Pmax are solvable in 4 O n time.
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 our problem is to determine the trade-o curve of Pn C and Pmax , which is de®ned as the curve that j1 j connects all points Pn C; P , where C is the outcome of 1jPmax 6 P j j1 Cj . If the trade-o is known, then we can minimize our linear composite objective function by computing the objective value in each extreme point of the trade-o curve, which is de®ned as follows. De®nition 1. A point of the trade-o curve is called an extreme point if it corresponds to a vertex of the lower envelope of the trade-o curve. A schedule corresponding to an extreme point is called an extreme schedule. Theorem 1. If the minimum is bounded and if a1 and a2 are non-negative, thenPthere exists an extreme schedule that solves 1ka1 nj1 Cj a2 Pmax . Proof. Suppose to the contrary that the minimum is attained in a point C; P on the trade-o curve that is not an extreme point. If this point is not on P the lower envelope, then there exists a point C; with C 6 C and P 6 P on the lower envelope, which clearly dominates C; P , as a1 and a2 are non-negative. On the other hand, each point on a segment of the lower envelope between two extreme points is dominated by one of these, since the objective function is linear. The question then becomes dicult it is to Phow n ®nd the trade-o curve for j1 Cj and Pmax or, suciently, the set of extreme points. This P requires n solving a problem of the kind 1jPmax 6 P j j1 Cj , for some value P. The constraint Pmax 6 P is maxf0; sj equivalent to the constraint that Sj PP n ÿP g, and hence the problem 1jPP max 6 P j j1 Cj is n equivalent to the problem 1jrj j j1 Cj , where rj denotes a release date before which job Jj cannot be started. Lenstra et al. (1977), Pn however, have shown that the problem 1jrj j j1 Cj is NP-hard in the strong sense. We therefore make the additional assumption that preemption of jobs is allowed, that is, the execution of any job may be interrupted and re-
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sumed later on. This assumption implies a crucial relaxation of the original problem; it has both positive and negative aspects. To start with the positive part: we Pn can solve the problem 1jpmtn; Pmax 6 P j since the problem j1 Cj , P 1jpmtn; rj j nj1 Cj is solvable in O n log n time by Baker's algorithm (Baker, 1974): always keep the machine assigned to the available job with minimum remaining processing time. The disadvantage is that we lose the equivalence that existed between the maximum promptness criterion and the maximum earliness criterion in case sj dj ÿ pj . This is so, since a given value Emax induces an earliest completion time for each job, not a release date. In the next section we will work out our plan of determining the set of extreme points of the tradeo curve. We conclude this section with a discussion of the two single-criterion Pn problems that are embedded within 1jpmtnja 1 j1 Cj a2 Pmax , that P is, 1kPmax and 1k nj1 Cj (we do not include `pmtn', since preemption is not advantageous for these problems). The 1kPmax problem is clearly meaningless, since we can improve upon each solution by inserting extra idle time at the beginning of the schedule. Hence, we impose the restriction that machine idle time before the processing of any job is prohibited, that P is, all jobs are to be scheduled in the interval 0; nj1 pj . It is easily checked P that in case of a given overall deadline D > nj1 pj the P optimal schedule is obtained by inserting n D ÿ j1 pj units of idle time before the start of the ®rst job. In the three-®eld notation scheme, the no machine idle time constraint is denoted by the acronym nmit in the second ®eld. The 1jnmitjPmax problem is solved by sequencing the jobs in order of Pnon-decreasing target start times sj . The 1k nj1 Cj problem is solved by sequencing the jobs in order of non-decreasing processing times pj (Smith, 1956). Let now MTST be an optimal schedule for the 1jnmitjPmax problem in which ties are settled to minimize total completion time; MTST is the abbreviation of minimum target start time. In addition, let SPT be an optimal schedule P for the 1k nj1 Cj problem, in which ties are settled to minimize maximum promptness; SPT is the abbreviation of shortest processing time. It then and follows that max r 6 Pmax SPT Pn Pmax 6 PP P n n C 6 C r 6 C MTST for any j j j1 j j1 j1
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extreme r without idle time, where Pmax Pn schedule and j1 Cj denote the outcome of the respective single-criterion problems. Note that Baker's algorithm always generates a schedule without ma . chine idle time if Pmax P Pmax
3. Determining extreme points of the trade-o curve If the extreme set can be found in polynomial time and if itsP cardinality is polynomially bounded, n then the 1ka1 j1 Cj a2 Pmax problem is solved in polynomial time by computing the cost of each extreme point and taking the minimum. We start by analyzing the special case in which machine idle time before the processing of any job is prohibited; we later show that any instance of the general problem can be dealt with by reformulating it as an instance of the problem with no machine idle time allowed.
3.1. No machine idle time allowed Recall that if machine idle time is not allowed, then P all jobs are processed in the interval 0; nj1 pj . Hence, we only have to consider Pmax values in the interval Pmax ; Pmax SPT, and for each Pmax value P in this interval, Baker's algorithm provides an optimal schedule for the correPn sponding 1jpmtn; Pmax 6 P j j1 Cj problem that does not contain idle P time; let r P denote this schedule and let P ; nj1 Cj r P denote the point in R2 corresponding to it. We start with a three-job example for which we determine the trade-o curve. The data are as follows: Jj
1
2
3
pj sj
7 0
3 5
2 10
The trade-o curve is depicted in Fig. 1. Breakpoints have been marked; the corresponding schedules are described below. We denote the schedules by the order in which the jobs occur; if job Jj occurs more than once,
Fig. 1. Trade-o curve.
then it is preempted, and the successor of a preempted piece starts as soon as possible. Point (1) corresponds to the schedule J1 ; J2 ; J3 . This schedule remains optimal until Pmax becomes more than 1; at point (2), schedule J1 ; J2 ; J1 ; J3 becomes optimal. This order remains optimal until Pmax P 2; point (3) corresponds to J1 ; J2 ; J1 ; J3 ; J1 . For 2 6 Pmax 6 5, this order remains optimal; the optimal 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 is optimal; at Pmax 9 the optimal order becomes J2 ; J3 ; J2 ; J1 , from which ®nally at Pmax 10 the SPT order J3 ; J2 ; J1 is reached. The lower envelope is found by connecting the points (1), (4), (5), and (7), which are the extreme points. The problem is of course to distinguish between an extreme schedule and an ordinary schedule that corresponds to a point on the trade-o curve. We know that the lower envelope is continuous and piecewise-linear, and most important, that in each breakpoint the gradient decreases. This implies that a necessary condition for a schedule r P to be extreme is that increasing P P by some > 0 yields a smaller decrease in nj1 Cj than a decrease of P by the same amount would cost. From the example above, we conclude that a schedule can only be extreme if a complete interchange has
J.A. Hoogeveen, S.L. van de Velde / European Journal of Operational Research 129 (2001) 87±94
occurred in r P , where an interchange is de®ned to be a complete interchange if there are two jobs Ji and Jj such that Ji is started before Jj in r P ÿ , whereas Jj is started before Ji in r P . P , then the point P ; nj1 Lemma 1. If P > Pmax Cj r P can be extreme only if a complete interchange has occurred in r P . The next step is to determine the Pmax values P suchP that their corresponding points P ; nj1 Cj 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 the upper bound on Pmax such that Jj can start at time Si r P . This will lead to a complete interchange of Ji and Jj in r P 1 , unless Ji itself is started at an earlier time in the schedule r P 1 , where P 1 sj ÿ Si r P is the value of the upper bound on Pmax that makes Jj available at time Si r P . It is 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 completion time of a preemptive job Jk . In that case, increasing the upper bound on Pmax will ®rst lead to a uniform shift forward of Ji and Jj at the expense of Jk ; the complete interchange of Ji and Jj cannot take place before a complete interchange has taken place between Jk and both Ji and Jj . Algorithm 1 exploits these Pn observations to generate each point P ; j1 Cj r Pmax for which a complete interchange in r P may take place. The variable aj (j 1; . . . ; n signi®es the increase of the current Pmax value necessary to let a complete interchange involving Jj and some successor take place. After having stated the algorithm, we apply it to our example. Algorithm 1. . Step 0. Let P Pmax 1 for j 1; . . . ; n; Step 1. Let T 0 and aj determine r P through Baker's rule. Step 2. Let Jk be the job that starts at time T in r P . Consider the following two cases: (a) Jk is a preempted job. Then ak is equal to the length of this portion of Jk . Set T Ck r P .
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(b) Jk is not a preempted job. Then ak minfsj ÿ P ÿ Sk r P j Jj 2 Vg, where V denotes the set of jobs Jj for which and pj > pk . Set sj ÿ P > Sk r P T Ck r P . P Step 3. If T < nj1 pj , then go to Step 2. Step 4. Put P min1 6 j 6 n aj P . Step 5. If P Pmax SPT, then stop; otherwise go to Step 1. If we apply Algorithm 1 to our example, then we start with the schedule r 0 in which the jobs occur in the order J1 ; J2 ; J3 and are not preempted. 5 and a2 3, which implies Step 2(b) puts a1 that we put P 3. If we apply Baker's rule to determine r 3, then we see that the planned interchange of jobs J2 and J3 does not take place since J2 is moved forward and preempts job J1 , just like job J3 (the corresponding order of the jobs in r 3 is J1 ; J2 ; J1 ; J3 ; J1 ). If we apply Algorithm 1 to r 3, then Step 2(a) puts a1 2, which removes the preempted piece of job J1 to the end of the schedule. Hence, P 5, and r 5 is the schedule corresponding to point (4). In the next iteration, 5 and a1 2, and the preempted we ®nd a2 7), piece of J1 between J2 and J3 is removed (P after which we ®nally ®nd P 10, which enables the full interchange of jobs J2 and J3 , which brings us in point (7). Algorithm 1 has found all extreme schedules and the unwanted schedule r 3. We will now give a formal proof of the correctness of Algorithm 1. Theorem 2. Algorithm 1 generates all Pmax values P for which a complete interchange has taken place in the corresponding schedule r P . Proof. Suppose that a complete interchange of the jobs Ji and Jj with pi > pj took place in the schedule r P , where P was not detected by Algorithm 1. Hence, Si r Pmax must have been ignored in Step 2, which could have happened only in Step 2(a): Ji is started between the start and completion time of some preempted job Jk . This, however, con¯icts with the earlier observation that the interchange of Ji and Jj has to wait until Jk has been interchanged with both Ji and Jj .
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As shown in the example, the algorithm may generate too many Pmax values P: in some of the schedules r P not a complete interchange has taken place. This is due to Step 2(b). There we implicitly assumed that the part of the schedule before Jk , which was de®ned as the job to be interchanged, would remain scheduled before Jk , that is, that Jk itself would not be started earlier. This is not necessarily the case, however, since an increase of the upper bound on Pmax may cause Jk to move forward at the expense of some job Jl with pl > pk , where the increase of the upper bound is not large enough to allow a complete interchange; Jk will preempt Jl then. Nevertheless, we now prove that the number of values Pmax generated by Algorithm 1 is polynomially bounded, thereby Pn establishing that 1jpmtn; nmitja1 j1 Cj a2 Pmax is polynomially solvable. We de®ne for a given schedule r the indicator function dij r as 2 if Ci r 6 Sj r and pi > pj ; dij r 0 otherwise: Pn We further de®ne Dj r as i1 dij r plus the number of preemptions of J , and we let D r j Pn D r. j j1 Theorem 3. Let P 1 be the Pmax value that is found by Algorithm 1 when applied to r P , where P is any Pmax value determined by Algorithm 1. We then have that D r P 1 < D r P . Proof. As explained above, one of the following three things has happened in r P 1 in comparison to r P : (i) a preemption has been removed (Step 2(a)); (ii) two jobs not in SPT-order have been interchanged (successful Step 2(b)); (iii) a new preemption has been created (unsuccessful Step 2(b)). All three cases have a negative eect on the value of D, as is easily checked (in the third case we do create an extra preemption (eect +1), but this pair of jobs is no longer in the wrong order (eect ÿ2)). Hence, we only have to show that there are no moves possible that have an overall positive eect on the value of D. The candidates for such a move are a switch of two jobs from SPT order to LPT
order and the addition of an extra preemption. We ®rst investigate the eect of the `wrong' switch. Suppose that there are two jobs Ji and Jj with pi > pj such that Ji succeeds Jj in r P , whereas the order is reversed in r P 1 . Since Baker's algorithm prefers Jj to Ji if both jobs are available, Ji starts earlier in r P 1 than Jj in r P , which means that the execution of (a part of) some job Jk is postponed until Ji is completed. See Fig. 2 for an illustration. It is easily checked that we have D r P 4 and D r P 1 3. All we have to do is to show that the situation depicted in Fig. 2 is worst possible for this con®guration. It is sucient to prove that Jj is available at time Ci r P 1 , that is, sj ÿ P 1 6 Ci r P 1 si ÿ P 1 pi ; if so, Baker's algorithm will prefer it to Jk , since the remainder of Jk has length at least equal to pi . Hence, we have to show that sj 6 si pi . As Ji did not preempt Jk in r P , we must have si ÿ P pi P Ck r P P sj ÿ P , where the last inequality follows from the availability of Jj at time Ck r P . Since the smaller job is available as soon as the larger job involved in the wrong switch is completed, the increase of dij is compensated for by the decrease of dki . Moreover, job Jk cannot trigger a set of nested wrong switches, where we mean with a set of nested wrong switches that r P and r P 1 contain the subschedules Jk ; Jj ; Ji ; Jh and Jh ; Ji ; Jj ; Jk with pj < pi < ph < pk . Now consider the situation that the number of preemptions of a job Jk increases. Hence, there must be a job Ji with pi < pk that succeeds Jk in r P but not in r P 1 , which move decreases the D function by 1. Corollary 1. If preemption is allowed, then the number Pnof extreme schedules with respect to Pmax ; j1 Cj is bounded by n n ÿ 1 1.
Fig. 2. `Wrong' switch.
J.A. Hoogeveen, S.L. van de Velde / European Journal of Operational Research 129 (2001) 87±94
Proof. We have that D r 6 n n ÿ 1 for any schedule r. Application of Theorem 3 yields the desired result. It is easy to construct an instance for which Algorithm 1 determines O n2 dierent Pmax values. We have not found an example with O n2 extreme points yet. Corollary 2. The 1jpmtn; nmitja1 problem is solvable in O n4 time.
Pn
j1
Cj a2 Pmax
Theorem 4. If a1 a2 , then there exists a nonpreemptive optimal schedule for the 1jpmtn; nmitja1 Pn C a any op2 Pmax problem. If a1 > a2 , then P j1 j n timal schedule for the 1jpmtn; nmitja1 j1 Cj a2 Pmax problem is non-preemptive. Proof. Suppose that the optimal schedule contains a preempted job. Start at time 0 and ®nd the ®rst preempted job Ji immediately scheduled before some non-preempted job Jj . Consider the schedule obtained by interchanging job Jj and this portion of 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 the objective value. The argument can be repeated until a non-preemptive schedule remains. In case a1 > a2 such an interchange would decrease the objective value, contradicting the optimality of the initial schedule. 3.2. The general case We now drop the no machine idle time constraint. Obviously, if total completion time outweighs maximum promptness, then the insertion of machine idle time before the processing of any job makes no sense. Hence, we have the following. Corollary 3. If a1 P a2 , then 1ka1 is solvable in O n4 time.
Pn
j1
Cj a2 Pmax
If a1 < a2 , then the insertion of idle time may decrease the value of the objective function. We now Pn show that we can solve the 1jpmtnja1 j1 Cj a2 Pmax problem by using Algorithm 1,
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which was initially designed for solving P 1jpmtn; nmitja1 nj1 Cj a2 Pmax . Suppose that a1 and a2 are given. De®ne q a2 =a1 . If q > n, then it is always advantageous to decrease Pmax , which implies that the execution of the ®rst job will be delayed for ever and ever. To prevent unbounded solutions, we therefore assume that q 6 n. A straightforward computation then shows that in any optimal schedule at least bn ÿ q 1c jobs are scheduled before the ®rst incidence of idle time. The smallest value Pmax q for maximum promptness that leads to such a schedule is readily obtained. Moreover, no optimal contains idle time. schedule with Pmax P Pmax Therefore, we only have to describe how our approach can be adjusted to deal with the case . Pmax q 6 Pmax 6 Pmax Pn Consider any instance I of 1jpmtnja1 j1 Cj a2 Pmax ; let r P P denote any optimal schedule for I of 1jpmtn; rj j Cj for any Pmax value P with and rj maxf0; sj ÿ P g. Pmax q 6 P 6 Pmax We create a very large job J0 that is available from time 0 onwards to saturate r P by ®lling in J0 in the periods of idle time. In fact, J0 is so large that Baker's rule prefers each job in I to it; the ÿ Pmax q choices s0 Pmax q and p0 Pmax max1 6 j 6 n pj 1 ensure such a saturation for any Pmax value P with Pmax q 6 P 6 Pmax . Let I0 denote the instance I to which J0 is added. Due to the choice of p0 and s0 , we have that no optimal schedule for the instance I0 of 1jpmtn; nmitja1 Pn j1 a2 Pmax contains machine idle time, and moreover, that by simply removing J0 and leaving the rest of the schedule intact we obtain an optimal schedule P for the original instance I of n 1jpmtnja 1 j1 Cj a2 Pmax . After all, we have that Pn Pmax . C0 j0 pj and that J0 does not assume P Hence, instead of solving 1jpmtnja1 Pnj1 n Cj a2 Pmax for I, we solve 1jpmtn; nmitja1 j0 0 Cj a2 Pmax for I . This approach Pn provides us with the extreme points for j1 Cj ; Pmax with . If q is unknown, then we Pmax q 6 Pmax 6 Pmax obtain all bounded extreme points by running the above procedure with q n; this choice of q corresponds to the smallest value Pmax q that may correspond to a bounded extreme point. As the number of extreme points is at most equal to n n 1 1 (we have n 1 jobs now),
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and as each Pmax value that corresponds to an extreme point Pn is determined by Algorithm 1, the 1jpmtnja1 j1 Cj a2 Pmax problem is solved in O n4 time. 4. Related problems In this section we discuss a number of related problems. The ®rst one deals with external bounds on Pmax . Usually, Pmax cannot assume any value, but is bounded from below or above. Such additional constraints are easy to handle: we simply restrict the interval over which we look for extreme schedules. The second problem is minimize Pmax in case Pto n of P an upper bound on j1 Cj , that is, problem n 1j j1 Cj 6 CjPmax . It is hard to deal with an upper bound on a sum-criterion explicitly, but in our case we search the trade-o curve using binary search. The third problem has to do with a general, non-linear, composite objective function Pn F j1 Cj ; Pmax , which we may assume to be nondecreasing in both arguments. Instead of looking for the extreme points of the trade-o curve, we then must ®nd all non-dominated or Pareto optimal points. Unfortunately, this number is not bounded in case preemption or idle time is allowed, as can easily be seen from Fig. 1. By using a result by Schrijver (see Hoogeveen, 1996), we Pn C ; P and conclude that 1jpmtnjF j max j1 Pn 1kF j1 Cj ;Pmax are NP-hard in the strong sense. 5. Conclusions We have presented a simple single-machine scheduling model for negotiating delivery times with an external supplier or a preceding produc-
tion stage. A characteristic of our model is that all delivery times are aected in the same way, which contrasts with the work in the so-called just-intime models, where the delivery time of one job is independent 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.
References Ahmadi, R., Bagchi, U., 1990. Lower bounds for singlemachine scheduling problems. Naval Research Logistics Quarterly 37, 967±979. Baker, K.R., 1974. Introduction to Sequencing and Scheduling. Wiley, New York. Dileepan, P., Sen, T., 1988. Bicriterion static scheduling research for a single machine. Omega 16, 53±59. Graham, R.L., Lawler, E.L., Lenstra, J.K., Rinnooy Kan, A.H.G., 1979. Optimization and approximation in deterministic sequencing and scheduling: a survey. Annals of Discrete Mathematics 5, 287±326. Hoogeveen, J.A., 1992. Single-machine bicriteria scheduling. Ph.D. Thesis, CWI, Amsterdam. Hoogeveen, J.A., 1996. Minimizing maximum promptness and maximum lateness on a single machine. Mathematics of Operations Research 21, 100±114. Hoogeveen, J.A., van de Velde, S.L., 1995. Minimizing total completion and maximum cost simultaneously is solvable in polynomial time. Operations Research Letters 17, 205±208. Hoogeveen, J.A., van de Velde, S.L., 1996. A branch-andbound algorithm for single-machine earliness-tardiness scheduling with idle time. INFORMS Journal on Computing 8, 402±412. Lenstra, J.K., Rinnooy Kan, A.H.G., Brucker, P., 1977. Complexity of machine scheduling problems. Annals of Discrete Mathematics 1, 343±362. Schutten, M., van de Velde, S.L., Zijm, W.H.M., 1996. Singlemachine scheduling with release dates, due date, and family setup times. Management Science 42, 1165±1174. Smith, W.E., 1956. Various optimizers for single-stage production. Naval Research Logistics Quarterly 1, 59±66. Van Wassenhove, L.N., Gelders, F., 1980. Solving a bicriterion scheduling problem. European Journal of Operational Research 4, 42±48.
www.elsevier.com/locate/dsw
Scheduling with target start times J.A. Hoogeveen b
a,*
, S.L. van de Velde
b
a Department of Computer Science, Utrecht University, P.O. Box 80089, 3508 TB Utrecht, Netherlands 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 objective 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 by the traditional con¯ict between internal and external eciency faced by production companies. Internal eciency is the ecient use of the scarce resources. Its aim is to reduce cost in order to quote more competitive prices or make more pro®t. External eciency is the extent by which conditions set by external relations can be met. External eciency between the company and its
* Corresponding author. Tel.: +31-30-253-4089; fax: +31-30251-3791. E-mail addresses: [email protected] (J.A. Hoogeveen), [email protected] (S.L. van de Velde).
clients, that is, the extent by which the company successfully copes with due dates, is called downstream eciency; upstream external eciency measures the success of meeting the conditions at the suppliers' side. Often, however, these conditions are subject to negotiations, which is especially true in case of integrated planning within a supply chain. Then the company can increase its internal eciency by successfully negotiating with its customers or its suppliers. There exist several single-machine scheduling models of the trade-o between internal and downstream external eciency. Van Wassenhove and Gelders (1980), for instance, consider a model for making the trade-o between work-in-process inventories and due date performance; see also Hoogeveen and van de Velde (1995). Schutten et al.
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
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(1996) consider a batching problem for balancing out the utilization of machine capacity against due date performance. Single-machine problems seem to be oversimpli®ed models, but the study of these models makes sense, if we think of a company as a single-machine shop, or if there is a single bottleneck. What is more, single-machine models serve as building-blocks for solving complex scheduling problems. In this paper, we look at the problem of dealing simultaneously with internal and upstream external eciency. We assume that the company can negotiate on the prices and delivery times of raw material, which may result in a higher internal eciency at the expense of a higher price for getting the raw material delivered sooner. We model this problem as a single-machine scheduling problem in the following way. A set of n independent jobs has to be scheduled on a single machine that is continuously available from time zero onwards and that can process at most one job at a time. Each job Jj j 1; . . . ; n requires processing during a positive time pj and has a target start time sj . Without loss of generality, we assume that the processing times and target start times are integral. A schedule r speci®es for each job when it is executed while observing the machine availability constraint; hence, a schedule r de®nes for each job Jj its start time Sj r and its completion time Cj r. The promptness Pj r of job Jj is de®ned as Pj r sj ÿ Sj r, and the maximum promptness is de®ned as Pmax r max1 6 j 6 n Pj r. We note that the maximum promptness Pmax r equals the maximum earliness Emax rmax16j6n dj ÿCj r if each Jj has a due date dj for which sj dj ÿpj and if Cj r Sj rpj , that is, if interruption of job processing is not allowed. The problem we consider is to schedule the Pn jobs so as to minimize total completion time j1 Cj and maximum promptness Pn Pmax simultaneously. Total completion time j1 Cj is a measure of the work-in-process inventories as well as the average leadtime. Hence, it is a performance measure for internal eciency as well as downstream external eciency. Maximum promptness measures the observance of target start times. If it is positive, then it
signals an ineciency: at least one job is scheduled to start before its target start time. Generally, this is possible only if we are willing to pay a penalty. In case the target start times are derived from the delivery times of raw material, then this penalty is actually the price of a speedier delivery. In case the target start times are derived from the completion times of the parts in the preceding production stage, then this penalty may be an overwork bonus to expedite the production. If the maximum promptness is negative, then it signals a slack, which implies that we may increase the deadlines that are used in the preceding production stage. It is important to realize that the target start times are actually release times that may be violated at a certain cost. In this sense, our problem comes close to the well-studied singlemachine problem of minimizing total completion time 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 objective function. We associate with each schedule r Pn 2 a point C j1 j r; Pmax r in R and a value Pn a1 j1 Cj r a2 Pmax r. Since we want to minimize both criteria, we assume that a1 and a2 are non-negative. Extending the three-®eld notation scheme of Graham P et al. (1979), we denote this problem by 1ka1 nj1 Cj a2 Pmax . In comparison to single-criterion problems, there are few papers on multicriteria scheduling problems. We refer to Dileepan and Sen (1988) and Hoogeveen (1992) for an overview of problems, polynomial algorithms, and complexity results. This paper is organized as follows. In Section 2, we make some general observations and outline Pn a generic strategy for solving the 1ka1 j1 Cj a2 Pmax problem. In Section 3, we consider the variant in which preemption is allowed, that is, a job may be interrupted and resumed later; this Pn problem is denoted as 1jpmtnja1 j1 Cj a2 Pmax , where the acronym pmtn indicates that preemption is allowed. Our main results are that P a2 Pmax and, in the case that 1jpmtnja1 nj1 Cj P n a1 P a2 , also 1ka1 j1 Cj a2 Pmax are solvable in 4 O n time.
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2. General observations One strategy that we could follow to solve our problem is to determine the trade-o curve of Pn C and Pmax , which is de®ned as the curve that j1 j connects all points Pn C; P , where C is the outcome of 1jPmax 6 P j j1 Cj . If the trade-o is known, then we can minimize our linear composite objective function by computing the objective value in each extreme point of the trade-o curve, which is de®ned as follows. De®nition 1. A point of the trade-o curve is called an extreme point if it corresponds to a vertex of the lower envelope of the trade-o curve. A schedule corresponding to an extreme point is called an extreme schedule. Theorem 1. If the minimum is bounded and if a1 and a2 are non-negative, thenPthere exists an extreme schedule that solves 1ka1 nj1 Cj a2 Pmax . Proof. Suppose to the contrary that the minimum is attained in a point C; P on the trade-o curve that is not an extreme point. If this point is not on P the lower envelope, then there exists a point C; with C 6 C and P 6 P on the lower envelope, which clearly dominates C; P , as a1 and a2 are non-negative. On the other hand, each point on a segment of the lower envelope between two extreme points is dominated by one of these, since the objective function is linear. The question then becomes dicult it is to Phow n ®nd the trade-o curve for j1 Cj and Pmax or, suciently, the set of extreme points. This P requires n solving a problem of the kind 1jPmax 6 P j j1 Cj , for some value P. The constraint Pmax 6 P is maxf0; sj equivalent to the constraint that Sj PP n ÿP g, and hence the problem 1jPP max 6 P j j1 Cj is n equivalent to the problem 1jrj j j1 Cj , where rj denotes a release date before which job Jj cannot be started. Lenstra et al. (1977), Pn however, have shown that the problem 1jrj j j1 Cj is NP-hard in the strong sense. We therefore make the additional assumption that preemption of jobs is allowed, that is, the execution of any job may be interrupted and re-
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sumed later on. This assumption implies a crucial relaxation of the original problem; it has both positive and negative aspects. To start with the positive part: we Pn can solve the problem 1jpmtn; Pmax 6 P j since the problem j1 Cj , P 1jpmtn; rj j nj1 Cj is solvable in O n log n time by Baker's algorithm (Baker, 1974): always keep the machine assigned to the available job with minimum remaining processing time. The disadvantage is that we lose the equivalence that existed between the maximum promptness criterion and the maximum earliness criterion in case sj dj ÿ pj . This is so, since a given value Emax induces an earliest completion time for each job, not a release date. In the next section we will work out our plan of determining the set of extreme points of the tradeo curve. We conclude this section with a discussion of the two single-criterion Pn problems that are embedded within 1jpmtnja 1 j1 Cj a2 Pmax , that P is, 1kPmax and 1k nj1 Cj (we do not include `pmtn', since preemption is not advantageous for these problems). The 1kPmax problem is clearly meaningless, since we can improve upon each solution by inserting extra idle time at the beginning of the schedule. Hence, we impose the restriction that machine idle time before the processing of any job is prohibited, that P is, all jobs are to be scheduled in the interval 0; nj1 pj . It is easily checked P that in case of a given overall deadline D > nj1 pj the P optimal schedule is obtained by inserting n D ÿ j1 pj units of idle time before the start of the ®rst job. In the three-®eld notation scheme, the no machine idle time constraint is denoted by the acronym nmit in the second ®eld. The 1jnmitjPmax problem is solved by sequencing the jobs in order of Pnon-decreasing target start times sj . The 1k nj1 Cj problem is solved by sequencing the jobs in order of non-decreasing processing times pj (Smith, 1956). Let now MTST be an optimal schedule for the 1jnmitjPmax problem in which ties are settled to minimize total completion time; MTST is the abbreviation of minimum target start time. In addition, let SPT be an optimal schedule P for the 1k nj1 Cj problem, in which ties are settled to minimize maximum promptness; SPT is the abbreviation of shortest processing time. It then and follows that max r 6 Pmax SPT Pn Pmax 6 PP P n n C 6 C r 6 C MTST for any j j j1 j j1 j1
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extreme r without idle time, where Pmax Pn schedule and j1 Cj denote the outcome of the respective single-criterion problems. Note that Baker's algorithm always generates a schedule without ma . chine idle time if Pmax P Pmax
3. Determining extreme points of the trade-o curve If the extreme set can be found in polynomial time and if itsP cardinality is polynomially bounded, n then the 1ka1 j1 Cj a2 Pmax problem is solved in polynomial time by computing the cost of each extreme point and taking the minimum. We start by analyzing the special case in which machine idle time before the processing of any job is prohibited; we later show that any instance of the general problem can be dealt with by reformulating it as an instance of the problem with no machine idle time allowed.
3.1. No machine idle time allowed Recall that if machine idle time is not allowed, then P all jobs are processed in the interval 0; nj1 pj . Hence, we only have to consider Pmax values in the interval Pmax ; Pmax SPT, and for each Pmax value P in this interval, Baker's algorithm provides an optimal schedule for the correPn sponding 1jpmtn; Pmax 6 P j j1 Cj problem that does not contain idle P time; let r P denote this schedule and let P ; nj1 Cj r P denote the point in R2 corresponding to it. We start with a three-job example for which we determine the trade-o curve. The data are as follows: Jj
1
2
3
pj sj
7 0
3 5
2 10
The trade-o curve is depicted in Fig. 1. Breakpoints have been marked; the corresponding schedules are described below. We denote the schedules by the order in which the jobs occur; if job Jj occurs more than once,
Fig. 1. Trade-o curve.
then it is preempted, and the successor of a preempted piece starts as soon as possible. Point (1) corresponds to the schedule J1 ; J2 ; J3 . This schedule remains optimal until Pmax becomes more than 1; at point (2), schedule J1 ; J2 ; J1 ; J3 becomes optimal. This order remains optimal until Pmax P 2; point (3) corresponds to J1 ; J2 ; J1 ; J3 ; J1 . For 2 6 Pmax 6 5, this order remains optimal; the optimal 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 is optimal; at Pmax 9 the optimal order becomes J2 ; J3 ; J2 ; J1 , from which ®nally at Pmax 10 the SPT order J3 ; J2 ; J1 is reached. The lower envelope is found by connecting the points (1), (4), (5), and (7), which are the extreme points. The problem is of course to distinguish between an extreme schedule and an ordinary schedule that corresponds to a point on the trade-o curve. We know that the lower envelope is continuous and piecewise-linear, and most important, that in each breakpoint the gradient decreases. This implies that a necessary condition for a schedule r P to be extreme is that increasing P P by some > 0 yields a smaller decrease in nj1 Cj than a decrease of P by the same amount would cost. From the example above, we conclude that a schedule can only be extreme if a complete interchange has
J.A. Hoogeveen, S.L. van de Velde / European Journal of Operational Research 129 (2001) 87±94
occurred in r P , where an interchange is de®ned to be a complete interchange if there are two jobs Ji and Jj such that Ji is started before Jj in r P ÿ , whereas Jj is started before Ji in r P . P , then the point P ; nj1 Lemma 1. If P > Pmax Cj r P can be extreme only if a complete interchange has occurred in r P . The next step is to determine the Pmax values P suchP that their corresponding points P ; nj1 Cj 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 the upper bound on Pmax such that Jj can start at time Si r P . This will lead to a complete interchange of Ji and Jj in r P 1 , unless Ji itself is started at an earlier time in the schedule r P 1 , where P 1 sj ÿ Si r P is the value of the upper bound on Pmax that makes Jj available at time Si r P . It is 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 completion time of a preemptive job Jk . In that case, increasing the upper bound on Pmax will ®rst lead to a uniform shift forward of Ji and Jj at the expense of Jk ; the complete interchange of Ji and Jj cannot take place before a complete interchange has taken place between Jk and both Ji and Jj . Algorithm 1 exploits these Pn observations to generate each point P ; j1 Cj r Pmax for which a complete interchange in r P may take place. The variable aj (j 1; . . . ; n signi®es the increase of the current Pmax value necessary to let a complete interchange involving Jj and some successor take place. After having stated the algorithm, we apply it to our example. Algorithm 1. . Step 0. Let P Pmax 1 for j 1; . . . ; n; Step 1. Let T 0 and aj determine r P through Baker's rule. Step 2. Let Jk be the job that starts at time T in r P . Consider the following two cases: (a) Jk is a preempted job. Then ak is equal to the length of this portion of Jk . Set T Ck r P .
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(b) Jk is not a preempted job. Then ak minfsj ÿ P ÿ Sk r P j Jj 2 Vg, where V denotes the set of jobs Jj for which and pj > pk . Set sj ÿ P > Sk r P T Ck r P . P Step 3. If T < nj1 pj , then go to Step 2. Step 4. Put P min1 6 j 6 n aj P . Step 5. If P Pmax SPT, then stop; otherwise go to Step 1. If we apply Algorithm 1 to our example, then we start with the schedule r 0 in which the jobs occur in the order J1 ; J2 ; J3 and are not preempted. 5 and a2 3, which implies Step 2(b) puts a1 that we put P 3. If we apply Baker's rule to determine r 3, then we see that the planned interchange of jobs J2 and J3 does not take place since J2 is moved forward and preempts job J1 , just like job J3 (the corresponding order of the jobs in r 3 is J1 ; J2 ; J1 ; J3 ; J1 ). If we apply Algorithm 1 to r 3, then Step 2(a) puts a1 2, which removes the preempted piece of job J1 to the end of the schedule. Hence, P 5, and r 5 is the schedule corresponding to point (4). In the next iteration, 5 and a1 2, and the preempted we ®nd a2 7), piece of J1 between J2 and J3 is removed (P after which we ®nally ®nd P 10, which enables the full interchange of jobs J2 and J3 , which brings us in point (7). Algorithm 1 has found all extreme schedules and the unwanted schedule r 3. We will now give a formal proof of the correctness of Algorithm 1. Theorem 2. Algorithm 1 generates all Pmax values P for which a complete interchange has taken place in the corresponding schedule r P . Proof. Suppose that a complete interchange of the jobs Ji and Jj with pi > pj took place in the schedule r P , where P was not detected by Algorithm 1. Hence, Si r Pmax must have been ignored in Step 2, which could have happened only in Step 2(a): Ji is started between the start and completion time of some preempted job Jk . This, however, con¯icts with the earlier observation that the interchange of Ji and Jj has to wait until Jk has been interchanged with both Ji and Jj .
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As shown in the example, the algorithm may generate too many Pmax values P: in some of the schedules r P not a complete interchange has taken place. This is due to Step 2(b). There we implicitly assumed that the part of the schedule before Jk , which was de®ned as the job to be interchanged, would remain scheduled before Jk , that is, that Jk itself would not be started earlier. This is not necessarily the case, however, since an increase of the upper bound on Pmax may cause Jk to move forward at the expense of some job Jl with pl > pk , where the increase of the upper bound is not large enough to allow a complete interchange; Jk will preempt Jl then. Nevertheless, we now prove that the number of values Pmax generated by Algorithm 1 is polynomially bounded, thereby Pn establishing that 1jpmtn; nmitja1 j1 Cj a2 Pmax is polynomially solvable. We de®ne for a given schedule r the indicator function dij r as 2 if Ci r 6 Sj r and pi > pj ; dij r 0 otherwise: Pn We further de®ne Dj r as i1 dij r plus the number of preemptions of J , and we let D r j Pn D r. j j1 Theorem 3. Let P 1 be the Pmax value that is found by Algorithm 1 when applied to r P , where P is any Pmax value determined by Algorithm 1. We then have that D r P 1 < D r P . Proof. As explained above, one of the following three things has happened in r P 1 in comparison to r P : (i) a preemption has been removed (Step 2(a)); (ii) two jobs not in SPT-order have been interchanged (successful Step 2(b)); (iii) a new preemption has been created (unsuccessful Step 2(b)). All three cases have a negative eect on the value of D, as is easily checked (in the third case we do create an extra preemption (eect +1), but this pair of jobs is no longer in the wrong order (eect ÿ2)). Hence, we only have to show that there are no moves possible that have an overall positive eect on the value of D. The candidates for such a move are a switch of two jobs from SPT order to LPT
order and the addition of an extra preemption. We ®rst investigate the eect of the `wrong' switch. Suppose that there are two jobs Ji and Jj with pi > pj such that Ji succeeds Jj in r P , whereas the order is reversed in r P 1 . Since Baker's algorithm prefers Jj to Ji if both jobs are available, Ji starts earlier in r P 1 than Jj in r P , which means that the execution of (a part of) some job Jk is postponed until Ji is completed. See Fig. 2 for an illustration. It is easily checked that we have D r P 4 and D r P 1 3. All we have to do is to show that the situation depicted in Fig. 2 is worst possible for this con®guration. It is sucient to prove that Jj is available at time Ci r P 1 , that is, sj ÿ P 1 6 Ci r P 1 si ÿ P 1 pi ; if so, Baker's algorithm will prefer it to Jk , since the remainder of Jk has length at least equal to pi . Hence, we have to show that sj 6 si pi . As Ji did not preempt Jk in r P , we must have si ÿ P pi P Ck r P P sj ÿ P , where the last inequality follows from the availability of Jj at time Ck r P . Since the smaller job is available as soon as the larger job involved in the wrong switch is completed, the increase of dij is compensated for by the decrease of dki . Moreover, job Jk cannot trigger a set of nested wrong switches, where we mean with a set of nested wrong switches that r P and r P 1 contain the subschedules Jk ; Jj ; Ji ; Jh and Jh ; Ji ; Jj ; Jk with pj < pi < ph < pk . Now consider the situation that the number of preemptions of a job Jk increases. Hence, there must be a job Ji with pi < pk that succeeds Jk in r P but not in r P 1 , which move decreases the D function by 1. Corollary 1. If preemption is allowed, then the number Pnof extreme schedules with respect to Pmax ; j1 Cj is bounded by n n ÿ 1 1.
Fig. 2. `Wrong' switch.
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Proof. We have that D r 6 n n ÿ 1 for any schedule r. Application of Theorem 3 yields the desired result. It is easy to construct an instance for which Algorithm 1 determines O n2 dierent Pmax values. We have not found an example with O n2 extreme points yet. Corollary 2. The 1jpmtn; nmitja1 problem is solvable in O n4 time.
Pn
j1
Cj a2 Pmax
Theorem 4. If a1 a2 , then there exists a nonpreemptive optimal schedule for the 1jpmtn; nmitja1 Pn C a any op2 Pmax problem. If a1 > a2 , then P j1 j n timal schedule for the 1jpmtn; nmitja1 j1 Cj a2 Pmax problem is non-preemptive. Proof. Suppose that the optimal schedule contains a preempted job. Start at time 0 and ®nd the ®rst preempted job Ji immediately scheduled before some non-preempted job Jj . Consider the schedule obtained by interchanging job Jj and this portion of 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 the objective value. The argument can be repeated until a non-preemptive schedule remains. In case a1 > a2 such an interchange would decrease the objective value, contradicting the optimality of the initial schedule. 3.2. The general case We now drop the no machine idle time constraint. Obviously, if total completion time outweighs maximum promptness, then the insertion of machine idle time before the processing of any job makes no sense. Hence, we have the following. Corollary 3. If a1 P a2 , then 1ka1 is solvable in O n4 time.
Pn
j1
Cj a2 Pmax
If a1 < a2 , then the insertion of idle time may decrease the value of the objective function. We now Pn show that we can solve the 1jpmtnja1 j1 Cj a2 Pmax problem by using Algorithm 1,
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which was initially designed for solving P 1jpmtn; nmitja1 nj1 Cj a2 Pmax . Suppose that a1 and a2 are given. De®ne q a2 =a1 . If q > n, then it is always advantageous to decrease Pmax , which implies that the execution of the ®rst job will be delayed for ever and ever. To prevent unbounded solutions, we therefore assume that q 6 n. A straightforward computation then shows that in any optimal schedule at least bn ÿ q 1c jobs are scheduled before the ®rst incidence of idle time. The smallest value Pmax q for maximum promptness that leads to such a schedule is readily obtained. Moreover, no optimal contains idle time. schedule with Pmax P Pmax Therefore, we only have to describe how our approach can be adjusted to deal with the case . Pmax q 6 Pmax 6 Pmax Pn Consider any instance I of 1jpmtnja1 j1 Cj a2 Pmax ; let r P P denote any optimal schedule for I of 1jpmtn; rj j Cj for any Pmax value P with and rj maxf0; sj ÿ P g. Pmax q 6 P 6 Pmax We create a very large job J0 that is available from time 0 onwards to saturate r P by ®lling in J0 in the periods of idle time. In fact, J0 is so large that Baker's rule prefers each job in I to it; the ÿ Pmax q choices s0 Pmax q and p0 Pmax max1 6 j 6 n pj 1 ensure such a saturation for any Pmax value P with Pmax q 6 P 6 Pmax . Let I0 denote the instance I to which J0 is added. Due to the choice of p0 and s0 , we have that no optimal schedule for the instance I0 of 1jpmtn; nmitja1 Pn j1 a2 Pmax contains machine idle time, and moreover, that by simply removing J0 and leaving the rest of the schedule intact we obtain an optimal schedule P for the original instance I of n 1jpmtnja 1 j1 Cj a2 Pmax . After all, we have that Pn Pmax . C0 j0 pj and that J0 does not assume P Hence, instead of solving 1jpmtnja1 Pnj1 n Cj a2 Pmax for I, we solve 1jpmtn; nmitja1 j0 0 Cj a2 Pmax for I . This approach Pn provides us with the extreme points for j1 Cj ; Pmax with . If q is unknown, then we Pmax q 6 Pmax 6 Pmax obtain all bounded extreme points by running the above procedure with q n; this choice of q corresponds to the smallest value Pmax q that may correspond to a bounded extreme point. As the number of extreme points is at most equal to n n 1 1 (we have n 1 jobs now),
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and as each Pmax value that corresponds to an extreme point Pn is determined by Algorithm 1, the 1jpmtnja1 j1 Cj a2 Pmax problem is solved in O n4 time. 4. Related problems In this section we discuss a number of related problems. The ®rst one deals with external bounds on Pmax . Usually, Pmax cannot assume any value, but is bounded from below or above. Such additional constraints are easy to handle: we simply restrict the interval over which we look for extreme schedules. The second problem is minimize Pmax in case Pto n of P an upper bound on j1 Cj , that is, problem n 1j j1 Cj 6 CjPmax . It is hard to deal with an upper bound on a sum-criterion explicitly, but in our case we search the trade-o curve using binary search. The third problem has to do with a general, non-linear, composite objective function Pn F j1 Cj ; Pmax , which we may assume to be nondecreasing in both arguments. Instead of looking for the extreme points of the trade-o curve, we then must ®nd all non-dominated or Pareto optimal points. Unfortunately, this number is not bounded in case preemption or idle time is allowed, as can easily be seen from Fig. 1. By using a result by Schrijver (see Hoogeveen, 1996), we Pn C ; P and conclude that 1jpmtnjF j max j1 Pn 1kF j1 Cj ;Pmax are NP-hard in the strong sense. 5. Conclusions We have presented a simple single-machine scheduling model for negotiating delivery times with an external supplier or a preceding produc-
tion stage. A characteristic of our model is that all delivery times are aected in the same way, which contrasts with the work in the so-called just-intime models, where the delivery time of one job is independent 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.
References Ahmadi, R., Bagchi, U., 1990. Lower bounds for singlemachine scheduling problems. Naval Research Logistics Quarterly 37, 967±979. Baker, K.R., 1974. Introduction to Sequencing and Scheduling. Wiley, New York. Dileepan, P., Sen, T., 1988. Bicriterion static scheduling research for a single machine. Omega 16, 53±59. Graham, R.L., Lawler, E.L., Lenstra, J.K., Rinnooy Kan, A.H.G., 1979. Optimization and approximation in deterministic sequencing and scheduling: a survey. Annals of Discrete Mathematics 5, 287±326. Hoogeveen, J.A., 1992. Single-machine bicriteria scheduling. Ph.D. Thesis, CWI, Amsterdam. Hoogeveen, J.A., 1996. Minimizing maximum promptness and maximum lateness on a single machine. Mathematics of Operations Research 21, 100±114. Hoogeveen, J.A., van de Velde, S.L., 1995. Minimizing total completion and maximum cost simultaneously is solvable in polynomial time. Operations Research Letters 17, 205±208. Hoogeveen, J.A., van de Velde, S.L., 1996. A branch-andbound algorithm for single-machine earliness-tardiness scheduling with idle time. INFORMS Journal on Computing 8, 402±412. Lenstra, J.K., Rinnooy Kan, A.H.G., Brucker, P., 1977. Complexity of machine scheduling problems. Annals of Discrete Mathematics 1, 343±362. Schutten, M., van de Velde, S.L., Zijm, W.H.M., 1996. Singlemachine scheduling with release dates, due date, and family setup times. Management Science 42, 1165±1174. Smith, W.E., 1956. Various optimizers for single-stage production. Naval Research Logistics Quarterly 1, 59±66. Van Wassenhove, L.N., Gelders, F., 1980. Solving a bicriterion scheduling problem. European Journal of Operational Research 4, 42±48.