Common due date assignment and scheduling with ready times

Computers & Operations Research 29 (2002) 1957–1967www.elsevier.com/locate/dsw

Common due date assignment andscheduling with ready times

T.C.E. Chenga ; ∗, Z.-L. Chenb, N.V. Shakhlevichc

aDepartment of Management, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong KongbDepartment of Systems Engineering, University of Pennsylvania, Philadelphia, PA 19104, USA

cInstitute of Engineering Cybernetics, Surganov Street 6, 220012 Minsk, Belarus

Received 1 May 2000; received in revised form 1 May 2001

Abstract

We consider the problem of scheduling a set of nonsimultaneously available jobs on one machine.Each job has a ready time only at or after which the job can be processed. All the jobs have a commondue date, which needs to be determined. The problem is to determine a due date and a schedule soas to minimize a total penalty depending on the earliness, tardiness and due date. We show that thisproblem is strongly NP-hard and give an e7cient algorithm that 8nds an optimal due date and schedulewhen either the job sequence is predetermined or all jobs have the same processing time. We alsopropose three approximation algorithms for the general and special cases together with their experimentalanalysis.

Scope and purpose

We consider the single machine due date assignment problem for scheduling jobs which are readyfor processing at di;erent times. The problem under consideration arises in production planning andscheduling concerning the setting of appropriate due dates for a number of customer orders arrivingover time. Most of the earlier publications on this subject assumed that the jobs are ready for process-ing simultaneously. This assumption is too restrictive for real-life production systems where jobs arriveat di;erent times. We show that the problem with unequal ready times is NP-hard and develop fastheuristic algorithms for it, and exact algorithms for two special cases. ? 2002 Elsevier Science Ltd. Allrights reserved.

Keywords: Single-machine scheduling; Earliness; Tardiness; Due date assignment

∗ Corresponding author. Tel.: +852-2766-5215; fax: +852-2364-5245.E-mail address: [email protected] (T.C.E. Cheng).

0305-0548/02/$ - see front matter ? 2002 Elsevier Science Ltd. All rights reserved.PII: S 0305-0548(01)00067-3

1958 T.C.E. Cheng et al. / Computers & Operations Research 29 (2002) 1957–1967

1. Introduction

There are n nonsimultaneously available jobs, N = {1; 2; : : : ; n}, to be processed on a machine.Each job i has a processing time ti ¿ 0, and a ready time ri¿0 only at or after which job ican be processed. All the jobs have a common due date d, which needs to be determined. Themachine is available at time 0 and job preemption is not allowed. For any given schedule s, let

Si(s) = start time of job i,

Ci(s) = Si(s) + ti, completion time of job i,

Ei(s) = max{0; d− Ci(s)}, earliness of job i,

Ti(s) = max{0; Ci(s) − d}, tardiness of job i,

F(d; s) =∑

i∈N (p1d+p2Ei(s)+p3Ti(s)), total penalty function, where p1¿0, p2¿0, and p3¿0are the penalty weights for the due date, earliness and tardiness, respectively.

In this paper, we consider the problem of determining an optimal combination of duedate d∗ and schedule s∗ so that F(d; s) is minimized.

By way of background, when r1 = r2 = · · ·= rn, Panwalkar et al. [1] solved the probleme7ciently by constructing an O(n log n) algorithm. The more general case with unequal readytimes arises when each job i has to be preprocessed which takes ri time units or the jobsthat are released by the suppliers at the times ri speci8ed by the suppliers. To the best of ourknowledge, there are no published results for unequal ready times. In what follows, we willdenote this problem by 1|ri|F(d; s) adopting the three-8eld classi8cation scheme from [2], itsoptimal schedule by s∗, the relaxed problem with zero ready times by 1||F(d; s), and its optimalschedule, which may be constructed by the algorithm from [1], by sPSS.

In Section 2, we derive some basic results about the problem under study. In Section 3, wedevelop an e7cient algorithm that 8nds an optimal solution when either the job sequence is pre-determined or all jobs have the same processing time. Finally, in Section 4, we propose severalapproximation algorithms for the general case and for some special cases. Computational resultsshow that the approximation algorithms are able to produce near-optimal solutions rapidly.

2. Preliminary analysis

In this section, we present some elementary results. For brevity, the proofs are omitted. Recallthat d∗ denotes an optimal due date for the problem.

Proposition 2.1. If p1 = 0; then there exists an optimal schedule with the job sequence de:nedby schedule sPSS; which is optimal for the relaxed problem with zero ready times.

Indeed, any increase in d can reduce the total penalty of earliness and tardiness as the jobscan be resequenced without violating their ready time constraints. The optimal job sequencethus corresponds to that of schedule sPSS. Starting the schedule from the maximum ready timeR= maxi∈N {ri} ensures its feasibility.

T.C.E. Cheng et al. / Computers & Operations Research 29 (2002) 1957–1967 1959

Proposition 2.2. If p2 = 0; then the problem is strongly NP-hard.

Observe that while the performance measure for the general case is nonregular, for the specialcase with p2 = 0, it is nondecreasing in job completion times. Optimal sequencing of late jobsis equivalent to 1|ri|

∑Ci, which is strongly NP-hard (see [3]).

Proposition 2.3. If p3 = 0; then d∗ = 0 and any feasible schedule is optimal.

Proposition 2.4. If p1¿p3; then d∗ = 0 and the problem is strongly NP-hard.

In this case, it is again equivalent to the problem 1|ri|∑

Ci.Throughout the remainder of this paper, we suppose p1 ¿ 0, p2¿0, p3 ¿ 0 and p1 ¡p3.

We eliminate the case p1¿p3 because the problem 1|ri|∑

Ci is very well studied.

Proposition 2.5. The problem 1|ri|F(d; s) is strongly NP-hard.

This result follows immediately from Proposition 2.2. Actually, even the restricted subproblemwith a given partition of jobs into two subsets of early and tardy jobs and a given commondue date d is strongly NP-hard, because the optimal sequencing of the late jobs is equivalentto the strongly NP-hard problem 1|ri|

∑Ci.

Proposition 2.6. For any speci:ed job sequence; there exists an optimal due dateequal to the completion time of the job which is processed in position k = �n(p3 − p1)=(p2 + p3)�.

Here, �x� denotes the smallest integer greater than or equal to x.When r1 = r2 = · · ·= rn, this result was proved in [1]. In fact, whether the ready times are

equal or not does not a;ect the correctness of the result. So, Proposition 2.6 can be similarlyestablished.

3. Solvable cases

In this section, we consider two special cases of the problem. For the 8rst one, the jobs areto be processed according to a predetermined sequence. For the second one, all jobs have thesame processing time. We give an e7cient algorithm that 8nds an optimal solution for each ofthese two cases.

3.1. Predetermined job sequence

It is assumed that the job sequence is predetermined. What we need to do is toselect a common due date and a start time for each job so as to minimize the totalpenalty.


Proposition 3.1. Given a predetermined sequence � = (1; 2; : : : ; n); there is no idle time betweenthe jobs (1; 2; : : : ; u) in the optimal schedule s∗(�); where u= �n(p3 − p1)=p3�.

Proof. Let s0 be a schedule with the start times of the jobs de8ned recursively:

S1(s0) = r1; Si(s0) = max{ri; Ci−1(s0)}; i = 2; : : : ; n (1)

and let k be the job occupying the position de8ned in Proposition 2.6.It is clear that Si(s0) is the minimum value of the start time of job i among all feasible

schedules. To obtain an optimal schedule, some jobs have to be postponed to ensure the min-imum performance measure value. Evidently, in the optimal schedule, early jobs 1; : : : ; k areprocessed as late as possible, i.e., they are processed without idle time among them. Let usshift jobs 1; : : : ; k − 1 to job k, eliminating all idle times between them, and denote the newschedule by s1.

Consider the 8rst idle time after job k is completed. Let it occur between jobs j and j + 1,j¿k, and let its duration be �. We construct schedule s2 by shifting jobs 1; : : : ; k; : : : ; j to jobj + 1. The change in performance measure value is

F(d; s2) − F(d; s1) =p1n�− p3(n− j)�:

The latter expression is negative for any j ¡n(p3 − p1)=p3 and is nonnegative, otherwise.It means that the 8rst job which should not be shifted to the right is job �n(p3 − p1)=p3�,i.e., job u.

From the proof of Proposition 3.1 it immediately follows that the following algorithm con-structs an optimal schedule for a predetermined job sequence.

Algorithm 3.1.

Phase 1. Construct an initial schedule with start times according to (1).Phase 2. Determine u= �n(p3 − p1)=p3�.Phase 3. Shift jobs 1; : : : ; u− 1 to job u to be processed without idle times among them.Phase 4. Determine k = �n(p3 −p1)=(p2 + p3)� and set the due date equal to the completion

time of the job processed in position k.

3.2. Equal processing times

We now consider the case when all jobs have the same processing time: t1 = · · ·= tn. First,we renumber the jobs such that r16r26· · ·6rn. Next, we obtain the following result.

Proposition 3.2. There exists an optimal schedule in which the job sequence is (1; 2; : : : ; n).

The result can be easily proved by the adjacent job interchanging argument.Thus, to construct an optimal schedule, we 8rst 8x the job sequence as (1; 2; : : : ; n) and then

use Algorithm 3.1 to solve the problem.


4. Approximation algorithms

4.1. General case

One viable approach to solve 1|ri|F(d; s) is to use the optimal schedule for the relaxedproblem 1||�(d; s) with zero ready times and the same objective function �(d; s) =F(d; s). Theformal description can be given as follows:

Algorithm 4.1.

(1) Construct an optimal job sequence for the relaxed problem 1||�(d; s).(2) Use the job sequence thus determined to construct a heuristic schedule for the original

problem 1|ri|F(d; s) by Algorithm 3.1.

It may happen that the job with the largest ready time R= maxi∈N {ri} is processed 8rst.Such an undesirable schedule can be obtained without using Algorithm 3.1. It corresponds tothe optimal schedule for 1||�(d; s), but starts R units later.

Let sPSS be an optimal schedule for 1||�(d; s) constructed by the algorithm from [1]. Recallthat its due date dPSS is equal to the completion time of the job which is processed in positionk. This position number is the same as that for the problem with unequal ready times, asde8ned in Proposition 2.6. To estimate a schedule constructed by Algorithm 4.1, we constructa heuristic schedule sh with unequal ready times by starting schedule sPSS R units later. If s∗is an optimal schedule for the original problem and d∗ its optimal due date, then

F(dh; sh) =�(dPSS; sPSS) + p1nR6F(d∗; s∗) + p1nR;

and hence,

F(dh; sh)=F(d∗; s∗)61 + R;

where dh is the due date for sh.Next, we propose a 2-approximation algorithm for the special cases when the penalty for

early jobs is zero, the set of early jobs is predetermined, or it can be e7ciently constructed.Our algorithm is based on the 2-approximation algorithm for 1|ri|

∑Ci as suggested in [4].

4.2. The case of zero penalty for early jobs

If p2 = 0, then the performance measure is nondecreasing in job completion times and thereis no need to optimize the sequence of early jobs or to postpone them. Due to Proposition2.6, for any 8xed job sequence, the optimal d value corresponds to the completion time ofthe job processed in position k = �n(p3 − p1)=p3�. It follows that k = u, where u is de8ned inProposition 3.1.

Consider the preemptive counterpart of our problem, i.e., 1|ri; pmtn|�C[k] + p3∑n

i=k+1 C[i],where � =p1n − p3(n − k) and C[i] is the completion time of the job in positioni; i = k; k+1; : : : ; n. The shortest remaining processing time rule developed in [5] for the problem1|ri; pmtn|∑Ci solves the problem under consideration optimally, since for the resulting


schedule spmtn and for an arbitrary preemptive schedule s feasible with respect to ready times,the following property holds:

C[i](spmtn)6C[i](s):

To construct a nonpreemptive heuristic schedule sh based on the job sequence in spmtn, weuse the following algorithm.

Algorithm 4.2.

(1) Construct an optimal schedule spmtn for the preemptive problem 1|ri; pmtn|�C[k] +p3

∑ni=k+1 C[i] using the shortest remaining processing time rule from [5].

(2) Determine a set of early jobs as those k = �n(p3 − p1)=p3� jobs in schedule spmtn, whichare completed 8rst. Sequence these jobs in nondecreasing order of their ready times.

(3) Sequence the remaining jobs in the order of their completion times in spmtn.(4) Set the due date equal to the completion time of the job processed in position k.

As in the previous section, we denote an optimal schedule for the original problem by s∗and its optimal due date by d∗. If sh is a heuristic schedule constructed by Algorithm 4:2 anddh is its due date, then

dh =C[k](sh) =C[k](spmtn)6d∗:

Besides, as it has been proved in Philips et al. [4] for the transformation used in Step 3,

C[i](sh)62C[i](spmtn); i = k + 1; : : : ; n:

Using the relation

C[i](spmtn)6C[i](s∗); i = k + 1; : : : ; n;

we obtain

F(dh; sh) = �dh + p3

n∑i=k+1

C[i](sh)6�d∗ + 2p3

n∑i=k+1

C[i](s∗)62F(d∗; s∗);

i.e., the approximation algorithm has an error ratio guarantee of 2.

4.3. Predetermined set of early jobs

Consider the case when there are a given set of early jobs NE and a given set of tardy jobsNT, N =NE ∪NT, and |NE|= k, where k is de8ned in Proposition 2.6. We describe an algorithmwhich is based on the enumeration of nondominated sequences for the early jobs and on a2-approximation algorithm for the late jobs. If the subset of early jobs is not predetermined,then it can be created by choosing the k jobs with the smallest ready times.

We split the problem 1|ri|F(d; s) into two subproblems with respect to the subsets NE andNT. In Stage 1, we determine the job sequences for the subsets NE and NT, separately. InStage 2, we concatenate these two sequences, apply Algorithm 3:1 and reconstruct the scheduleto ensure the error ratio guarantee of 2.


Stage 1: Constructing partial schedules for NE and NT. For subset NE, we consider theproblem 1|ri|p1nCmax +p2

∑Ei with earliness calculated with respect to d=C[k] =Cmax, which

is the completion time of the last job processed in position k, and solve it optimally under therestriction

CE6Cmax6 LCE: (2)

Here, CE is the minimum makespan value for processing jobs from NE and this value can bedetermined by scheduling jobs in nondecreasing order of their ready times; LCE is the makespanvalue of the schedule which starts from the maximum ready time maxi∈NE{ri} and processes jobsfrom NE without idle time. Observe that the maximum

∑Ei-value corresponds to the schedule

with Cmax =CE. The minimum∑

Ei-value corresponds to the schedule with Cmax = LCE with thejobs sequenced in nonincreasing order of the processing times. We do not consider scheduleswith Cmax ¿ LCE; because the job sequence minimizing

∑Ei is the same for all such schedules

and the∑

Ei-value is the same as well.To solve the problem with the jobs NE under restriction (2), we enumerate the Pareto optimal

points for the bicriterion problem 1|ri; Cmax6C|∑Ei, where the constant C satis8es the restric-tion CE6C6 LCE. The latter problem is equivalent to the problem 1|di|

∑Ci of minimizing the

total completion time under the restriction on the deadlines di =C − ri, which can be solvede7ciently by the O(n log n) Smith’s rule [6]. Modi8ed for the problem under consideration, therule can be formulated in the following way.Modi8ed Smith’s Rule

� :=C −∑i∈NE

ti; (the starting time of the schedule)l := 1; (the current position)WHILE NE = ∅ DO

A := {j | rj6�} (a set of jobs available at time �);Choose job j∈A with the maximum processing time tj;Schedule it in the interval [�; � + tj];l := l + 1; NE :=NE \ {j}; � := � + tj;

END WHILE.

Constructing the Pareto optimal schedules by satisfying (2) for the jobs from NE and se-quencing the jobs from NT can be done as follows.

Algorithm 4.3 (Stage 1).

Pareto optimal schedules for the jobs from NE:

(1) Construct the 8rst Pareto optimal point �i; i = 1; by the modi8ed Smith’s rule with C= LCE.Shift (if possible) the jobs to the left until at least one of the jobs starts exactly at itsready time.

(2) WHILE Cmax(�i)¿CE(3) Construct the next Pareto optimal point �i+1 by applying Smith’s rule with C=Cmax(�i)−1.

i := i + 1. Shift (if possible) the jobs to the left until at least one of the jobs starts exactlyat its ready time. i := i + 1.


(4) END WHILESequencing jobs from NT:

(5) Construct a 2-approximate schedule for 1|ri|∑

Ci by the algorithm from [4].

The complexity of the modi8ed Smith’s rule is O(n log n), and the number of Pareto optimalpoints O(n2) (the proof for the similar problem 1||F(

∑Ci; fmax) is presented in [7]). Hence, the

overall complexity of constructing Pareto optimal points for the jobs in NE is O(n3 log n). Ob-serve that the algorithm from [4] for sequencing jobs from NT is in fact described in Section 4.2and its time complexity is O(n log n) time. Thus, the overall complexity of Stage 1 of Algorithm4:3 is O(n3 log n).Stage 2: Constructing heuristic schedule for N =NE∪NT. We consider di;erent job sequences

corresponding to di;erent Pareto optimal points for NE and concatenate each of these sequenceswith the job sequence obtained for NT. The following algorithm constructs a heuristic schedulebased on a particular Pareto optimal schedule for the jobs from NE and on the 2-approximateschedule obtained for NT.

Algorithm 4.3 (Stage 2).

(1) Concatenate job sequences for NE and NT. Apply Algorithm 3:1.Let sh

1 be a schedule obtained in Phase 1 of that algorithm and sh2 be a schedule

obtained in Phase 3 of that algorithm after shifting the 8rst u − 1 jobs to the job inposition u.

(2) IF the starting time � for jobs NT increases for sh1 → sh

2, THEN resequence the jobs fromNT by applying the algorithm from [4] once more with larger ready times ri := max{ri; �};i∈NT, producing the resulting schedule sh

3.

Thus, for each Pareto optimal point we perform transformations sh1 → sh

2 → sh3, consisting

of shifting the 8rst u − 1 jobs and resequencing the last n − u jobs, which takes O(n log n)time. Enumerating the resulting schedules for di;erent Pareto optimal schedules, we choosethe schedule with the minimum performance measure value. Thus, the overall complexity ofAlgorithm 4:3 is O(n3 log n).

4.3.1. Analysis of the algorithmThe Pareto optimal points for the 8rst subproblem correspond to di;erent nondominant sched-

ules for the early jobs. Hence, the optimal schedule for the original problem has the job sequencefor the early jobs which coincides with one of the Pareto optimal points. In order to estimate theaccuracy of our algorithm, one may consider O(n2) classes of schedules with a 8xed sequenceof the early jobs, with each class corresponding to a particular Pareto optimal point. For each8xed class of schedules, we compare the heuristic schedule sh

3 and the optimal schedule s∗,which has the same order of the early jobs as sh

3 and may di;er from sh3 by a sequence of the

tardy jobs.We start with the 8rst auxiliary schedule sh

1 and compare it with the auxiliary schedule s∗1,which has the same job sequence as s∗, but the 8rst u − 1 jobs are processed at their earliesttimes (we may assume that schedule s∗ is obtained from s∗1 by shifting the 8rst u − 1 jobs to


the job in position u). For schedules sh1 and s∗1, we have

C[i](sh1)

{= C[i](s∗1) for i∈NE;

6 2C[i](s∗1) for i∈NT;

F(dh1; s

h1)62F(d∗

1; s∗1): (3)

Observe that within schedules sh1; s

h2; s

h3, job earliness is calculated with respect to dh

16dh2 =dh

3,and within schedules s∗1 ; s

∗, job earliness is calculated with respect to d∗16d∗.

To proceed with the next schedules sh2 and s∗, we introduce �[i; i+1](sh

1) and �[i; i+1](s∗1) as thetime span between the completion of job [i] and starting of job [i + 1] within schedules sh

1 ands∗1, respectively. Then

u−1∑i=k

�[i; i+1](sh1)6 2

u−1∑i=k

�[i; i+1](s∗1):

Due to the latter relation and due to inequality (3),

dh2 =dh

1 +u−1∑i=k

�[i; i+1](sh1)6d∗

1 + 2u−1∑i=k

�[i; i+1](s∗1)6 2d∗: (4)

Finally, to show that T[i](dh3; s

h3)6 2T[i](d∗; s∗), we consider two auxiliary problems with

smaller ready times: ri = max{ri; dh3} − dh

3 and ri = max{ri; d∗} − d∗. For the 8rst auxiliaryproblem, we consider a heuristic schedule sh with the same job sequence as sh

3:

C[i](sh) =C[i](sh3) − dh

3:

For the second auxiliary problem, we consider a schedule s∗ with the same job sequence as s∗:

C[i](s∗) =C[i](s∗) − d∗:

Since ri6 ri, the optimal schedule s∗ for the 8rst auxiliary problem dominates the schedules∗:

C[i](s∗)6C[i](s∗):

Besides, due to the properties of the algorithm from [4],

C[i](sh)6 2C[i](s∗):

Hence, C[i](sh)6 2C[i](s∗); i.e.,

T[i](dh3; s

h3)6 2T[i](d∗; s∗): (5)

Due to the transformations sh1 → sh

2 → sh3 and s∗1 → s∗, the total earliness

∑i∈NE

(C[k] −C[i]) doesnot change and, due to relations (4) and (5), we obtain:

F(dh3; s

h3) = p1ndh

3 + p2

∑Ei(dh

3; sh3) + p3

∑Ti(dh

3; sh3)

6 2p1nd∗ + p2

∑Ei(d∗; s∗) + 2p3

∑Ti(d∗; s∗)6 2F(d∗; s∗):

Hence, the algorithm developed is a 2-approximation algorithm with running timeO(n3 log n).


Table 1Computational results for three approximation algorithms (each entry represents 10 randomly generated examples)

Error Ratio= Heuristic SolutionOptimal Solution Error Ratio= Heuristic Solution

Lower Bound

Job number n 5 5 5 100 100 500 500 500Maximum ri 100 500 1000 500 1000 1000 2500 5000

Algorithm 4.1Minimum error 1.000 1.000 1.020 1.336 1.884 1.159 1.493 1.996Average error 1.175 1.071 1.064 1.447 1.947 1.187 1.525 2.053Maximum error 1.380 1.216 1.151 1.578 2.003 1.221 1.569 2.155CPU time (s) 0.00 0.00 0.00 0.00 0.00 0.24 0.26 0.23



4.4. Computational experience

The analysis of the approximation algorithms suggested in Sections 4.1–4.3 does not providetight error bounds. Moreover, the theoretical estimate of Algorithm 4:1 is overstated. The ex-perimental results show that the heuristic solutions are often better than the theoretical estimatesand in most cases, they are very close to the optimal solutions.

For our experiments, we randomly generated 80 problems: 30 examples with 5 jobs, 20examples with 100 jobs, and 30 examples with 500 jobs. The processing times were ran-domly drawn from a uniform distribution over [1; 20], while the ready times were randomlydrawn from a uniform distribution over [0; 100]; [0; 500], [0; 1000], [0; 2500], [0; 5000]. Thepenalty weights for the small-size problems were taken from the sample example from [1]:p1 = 5; p2 = 11; p3 = 18. For the large-size problems with n¿100, smaller penalty weightswere used to avoid memory overQow: p1 = 1; p2 = 2; p3 = 3. The error estimate was calculatedas the ratio of the heuristic solution value to the optimal value for the problems with n= 5 or tothe lower bound of the optimal solution determined by relaxing the restriction on ready times.

The algorithms were coded in TurboPascal and run on an IBM PC with a clock speed of133 MHz. The results of the computer experiments were generalized for each group of 10similar problems and summarized in Table 1. For Algorithm 4:1; the solutions obtained exceedthe optimal solutions or their lower bounds by no more than 2:16 times and this is essentiallyless than the theoretical estimate 1 + R. The worst examples correspond to the problems withscattered ready times. The algorithm is very simple and its running time is the smallest amongthe approximation algorithms, while the accuracy of the solution is the worst.

Algorithm 4:2 developed for the special case with zero penalty for early jobs constructs theoptimal schedules almost every time for n= 5 and has very small errors for large-size problems.


As is expected, the better solutions for the general problem were obtained by Algorithm 4:3.However, its running time is longer due to its complexity bound. The error of the solutionsobtained does not exceed 34%. In fact, the real error ratios for large-size problems should besmaller than those given in Table 1 if the optimal solutions were used instead of their lowerbounds, which are rather rough. The lower bounds of the optimal solutions were calculated bysetting all ready times equal to zero, which should be far from the optimal solutions, espe-cially for large-size problems with ri ∈ [0; 5000]. In our examples, the set of early jobs was notpredetermined and we chose the k jobs with the smallest ready times as early.

For all approximation algorithms, the accuracy of their solutions was much higher than thattheoretically estimated. In general, however, more accurate solutions were generated for prob-lems with smaller maximum ready times.

Acknowledgements

This research was partially supported by The Hong Kong Polytechnic University underGrant number G-S818. The research of the third author was partially supported by INTAS(Project INTAS-96-0820) and by the Byelorussian Foundation of Fundamental Research, ProjectsM96-052 and U97-147.

References

[1] Panwalkar S, Smith M, Seidmann A. Common due date assignments to minimize total penalty for the onemachine scheduling problem. Operations Research 1982;30:391–9.

[2] Graham RL, Lawler EL, Lenstra JK, Rinnooy Kan AHG. Optimization and approximation in deterministicsequencing and scheduling: a survey. Annals of Discrete Mathematics 1979;5:287–326.

[3] Lenstra JK, Rinnooy Kan AHG, Brucker P. Complexity of machine scheduling problems. Annals of DiscreteMathematics 1977;1:343–62.

[4] Phillips C, Stein C, Wein J. Minimizing average completion time in the presence of release dates. MathematicalProgramming 1998;82:199–223.

[5] Baker KR. Introduction to sequencing and scheduling. New York: Wiley, 1974.[6] Smith WE. Various optimizers for single-stage production. Naval Research Logistics Quarterly 1956;3:56–66.[7] Hoogeveen JA, Van de Velde SL. Minimizing total completion time and maximum cost simultaneously is

solvable in polynomial time. Operations Research Letters 1995;17:205–8.

T.C. Edwin Cheng is Chair Professor of Management in The Hong Kong Polytechnic University, Hong KongSpecial Administrative Region, China. He received his Ph.D. in Operations Research from the University of Cam-bridge in 1984. His research interests include Operations Research and Operations Management. Professor Chengis a member of the editorial board of this journal and has published extensively in a variety of academic andprofessional journals and two books.

Z.-L. Chen is an Assistant Professor in the University of Pennsylvania. He has published a number of papersin di;erent mathematical and operations research journals. His research interests are in Combinatorial Optimizationand Operations Research.

Natalia V. Shakhlevich is a Senior Researcher in the Institute of Engineering Cybernetics, National Academy ofSciences of Belarus, Minsk, Belarus. She received the Candidate of Science degree in Physics and Mathematics in1993. Her main research area is Deterministic Scheduling.

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Common due date assignment and scheduling with ready times