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Computers & Operations Research 30 (2003) 13351347www.elsevier.com/locate/dsw

Single-machine scheduling with periodic maintenance andnonresumable jobs

C.J. Liao , W.J. ChenDepartment of Industrial Management, National Taiwan University of Science and Technology, 43 Keelung Road,

Section 4, Taipei, Taiwan

Received 1 July 2001; received in revised form 1 January 2002

Abstract

We study a single-machine scheduling problem where periodic maintenance is required in a completeschedule. Although the scheduling problem with maintenance has attracted some researchers attention, most ofthem consider only one maintenance period. In this paper, several maintenance periods are under considerationwhere each maintenance is required after a periodic time interval. Speci2cally, the problem is to minimizethe maximum tardiness with periodic maintenance and nonresumable jobs. A branch-and-bound algorithmthat utilizes several inherent theorems is developed to derive the optimal schedule for the problem. To solvelarge-sized problems, a heuristic is also developed. Computational results are provided to demonstrate thee6ectiveness of the heuristic.

Scope and purpose

A survey of scheduling literature reveals that few researches consider the scheduling problem with main-tenance. However, machines subject to maintenance are found prevalently in process industries and manu-facturing systems. The purpose of this paper is to solve a single-machine scheduling problem with periodicmaintenance. The criterion considered is to minimize the maximum tardiness. We propose a heuristic for2nding the near-optimal solution for large-sized problems. A branch-and-bound algorithm is also provided toderive the optimal solution. Computational results demonstrate the e6ectiveness and e8ciency of the heuristic.? 2002 Published by Elsevier Science Ltd.

Keywords: Scheduling; Periodic maintenance; Maximum tardiness; Nonresumable job

Corresponding author. Tel.: +886-2-2737-6437; fax: +886-2737-6360.E-mail address: [email protected] (C.J. Liao).

0305-0548/03/$ - see front matter ? 2002 Published by Elsevier Science Ltd.PII: S0305-0548(02)00074-6

1336 C.J. Liao, W.J. Chen / Computers & Operations Research 30 (2003) 13351347

1. Introduction

Most scheduling models assume that machines are available at all times. However, this assumptionmay not be true in real industrial settings. For example, a preventive maintenance activity is requiredor a machine breakdown occurs. According to practical experience, we sometimes 2nd that some ofthe machines wait for maintenance while there are jobs waiting to be processed by these machines.This is due to the lack of coordination between maintenance planning and production scheduling.Therefore, this paper aims to improve the situation by deriving a satisfactory schedule that considersboth jobs and maintenance simultaneously.An uncertain breakdown will make the shop behavior hard to predict, and thereby reducing the

e8ciency of the production system. Maintenance can reduce the breakdown rate with minor sacri-2ces in production time. The importance of maintenance has gradually recognized by the decisionmaker. Therefore, it has been scheduled periodically in many manufacturing systems, especially ina just-in-time environment. The work of periodic maintenance includes periodic inspection, periodicrepair, and preventive maintenance. With proper planning of periodic maintenance, the shop canconcentrate on production e8ciency and safety, resulting in increasing productivity and achievinghigh safety awareness [1].Scheduling with maintenance is usually treated as availability constraint in the scheduling literature.

Although such an availability problem is important, it is relatively unexplored. For the single-machineproblem, Adiri et al. [2] assume that the unavailable time is unknown but with a probabilisticdistribution. They distinguish two cases of a breakdown, i.e., the resumable and nonresumable (orrestarting) cases. By applying the earliest due date (EDD) and modi2ed shortest processing time(MSPT) rules for the two cases, respectively, the criterion of number of tardy jobs can be minimized.For the two-machine problem, Mosheiov [3] solves the two-parallel-machine-scheduling problem forminimizing total completion time by assuming each machine is available in a speci2ed interval. Leeand Liman [4] consider a similar problem where one machine is not available for a speci2ed periodof time. They provide a pseudo-polynomial dynamic programming algorithm to solve the problem.Lee [58] studies in this area for other machine con2gurations including single and parallel machines,and di6erent performance measures such as makespan, total weight completion time, tardiness, andnumber of tardy jobs.In all the above papers, there is only one unavailability or availability period for each machine. As

stated earlier, however, maintenance has been scheduled regularly, or periodically, in many manufac-turing systems. Therefore, there is a need to develop scheduling method to deal with systems with pe-riodic maintenance, which usually has more than one maintenance periods. In our problem, there areseveral maintenance periods where each maintenance is required after a 2xed time interval. The objec-tive is to minimize the maximum tardiness subject to periodic maintenance and nonresumable jobs.

2. Notation and problem setting

The following notation will be used throughout this paper:

n = number of jobs for processing at time zeroJj = job number j

C.J. Liao, W.J. Chen / Computers & Operations Research 30 (2003) 13351347 1337

Fig. 1. A schedule with periodic maintenance for nonresumable case.

Mi =maintenance period ipj = processing time of job jdj = due date of job jCj = completion time of job jCMi = completion time of maintenance period iLj = lateness of job j, where Lj = Cj djTj = tardiness of job j, where Tj =max{0; Lj}Tmax = maxj{Tj}t = amount of time to perform one maintenanceT = time interval between two maintenance periods

In addition, J[j] denotes the job placed at the jth position, and p[j]; d[j]; C[j]; L[j], and T[j] arede2ned accordingly.We consider a single-machine problem where jobs are nonresumable, i.e., once a job is started

it must be completed without interruption. The setup time of job is sequence independent andincluded in the processing time. Also, it is assumed that all jobs are ready for processing at timezero and that, for simplicity, processing times and due dates can take only integral values. For eachmaintenance, it requires an amount of time t for the performance. There is a time interval T betweentwo consecutive maintenance periods, in which jobs can be scheduled for processing (see Fig. 1).The problem under consideration is to 2nd a schedule that minimizes the maximum tardiness subjectto periodic maintenance and nonresumable jobs. It is clear that the problem is NP-hard since theproblem that minimizes the maximum lateness subject to one unavailability period and nonresumablejobs is NP-hard [6].

3. The proposed heuristic

De2ne a batch as a set of jobs in each interval T . Then we have the following theorems that willbe used in the heuristic.

Theorem 1. Suppose Tp is the maximum tardiness of a schedule; where Jp is in batch B. Denoteby Ji a job preceding Jp in B. If di is greater than or equal to dp; then a smaller Tmax value canbe obtained by interchanging Jp with Ji.

Proof. Let partial schedule S=(PS;Ml; 1; Ji; 2; Jp); where PS; 1 and 2 are partial schedules. LetS be obtained from S by interchanging Ji with Jp. Denote by P1 and P2 the total processingtimes of 1 and 2; respectively. Then the tardiness of Jp in S is

Tp = CMl + P1 + pi + P2 + pp dp:


The tardiness of Ji in S is

Ti = CMl + P1 + pp + P2 + pi di:If di is greater than or equal to dp; then Ti6Tp = Tmax. Thus; S has a Tmax value smaller than orequal to that of S. This completes the proof.

Theorem 2. Suppose Tp is the maximum tardiness of a schedule; where Jp is in batch B. Denoteby Jj a job in the batch preceding B. If djpj is greater than or equal to dppp; then a smallerTmax value can be obtained by interchanging Jp with Jj.

Proof. Let partial schedule S = (PS;Mk; 1; Jj; ;Ml; 2; Jp); where PS and are partial schedules;and 1 and 2 are partial schedules of the same batches as Jj and Jp; respectively. Let S be obtainedfrom S by interchanging Jj with Jp. Denote by P1 and P2 the total processing times of 1 and2; respectively. Then the tardiness of Jp in S is

Tp = CMl + P2 + pp dp:The tardiness of Jj in S is

Tj = CMl + P2 + pj dj:If djpj is greater than or equal to dppp; then Tj6Tp=Tmax. Thus; S has a Tmax value smallerthan or equal to that of S. This completes the proof.

For explanatory convenience, we de2ne two terms that are needed in the heuristic.

Denition 1. The slack time of a batch is de2ned as the amount of time unscheduled in a batch.

Denition 2. The potential position is de2ned as the position immediately after the last position ofthe 2rst batch where its slack time is greater than or equal to the processing time of the consideredjob.

The steps of the proposed heuristic are outlined as follows:Step 1: Obtain a schedule with the smallest number of maintenance periods by the following

procedure:

(i) Generate a batch by grouping a set of jobs such that the total processing time of jobs in thebatch is equal to T (i.e., the bin-packing technique [9]). Repeat by grouping the remaining jobsto form other batches.

(ii) Repeat (i) except that the total processing time is changed to T 1; T 2; : : : ; until all jobs aregrouped.

Step 2: Sequence the jobs in each batch in EDD order. If two jobs have the same due date, placethe job with smaller processing time in the front.Step 3: Sequence the batches in EDD order by letting the smallest due date of jobs in each batch

be the due date of the batch.


Step 4: Denote by the resulting complete schedule, which is composed of the schedules in allbatches and the necessary maintenance periods. Calculate T[j] for all j in .Step 5: If T[j] = 0 for all j, stop; otherwise, 2nd J[p] with Tmax.Step 6: Let be the schedule obtained by interchanging J[p] with each of the jobs preceding J[p],

say J[q]. If J[p] and J[q] belong to the same batch, we apply Theorem 1 to determine whether anyinterchange of J[p] and J[q] is advantageous; otherwise, Theorem 2 is applied to determine whetherany interchange of J[p] and J[q] is advantageous. If it is, go to Step 7. Otherwise, the interchange isrepeated until no J[q] can be found. Go to Step 8.Step 7: Replace with if both the following conditions are satis2ed:

(i) The total processing times in the two batches are not larger than T (if J[p] and J[q] belong todi6erent batches).

(ii) The Tmax value is reduced.

If the replacement is achieved, calculate T[j] for all j. Return to Step 5.Step 8: Perform the movement if any of the following cases is satis2ed:

(i) Moving J[p] forward to the potential position can satisfy both the conditions in Step 7.(ii) Moving any job preceding J[p] in the same batch as J[p] to the potential position can satisfy

both the conditions in Step 7.(iii) Moving any job preceding J[p], say J[r], to the potential position and moving J[p] to J[r] can

satisfy both the conditions in Step 7.

Step 9: Reset the jobs in each batch in EDD order. Stop.

We now elaborate the steps in detail. It is clear that the smallest number of maintenance periodsis obtained in Step 1. In Steps 24, we construct an initial schedule, where both jobs and batchesare sequenced in EDD and the maintenance periods are incorporated. In Step 6, Theorems 1 and 2are used to determine whether any interchange is advantageous. In Step 7, the two conditions mustbe satis2ed for any improvement because the batch of jobs cannot be completed in T if condition (i)cannot be met. In Step 8, we check whether it is advantageous to move the job with Tmax forward((i) and (iii)) or to reduce the completion time of the job ((ii)).To calculate the time complexity of the proposed heuristic, note that the bin-packing technique in

Step 1 can be carried out in O(n log n). The complexities of Steps 2, 3, and 9 are O(n log n). Step5 needs O(n). Steps 6 and 7 require O(n2

nk=1 p[k]). Step 8 needs O(n

2p[k]). Thus, we havethe overall time complexity of the proposed heuristic as O(n2

ni=1 pi).

4. A numerical example

Example 1. As an illustration of the heuristic; consider a single-machine scheduling problem withnine jobs; as given in Table 1; where T = 8 and t = 2.

Applying Steps 15, we obtain an initial schedule as given in Table 2, where there are threemaintenance periods with Tmax=24. Since J9 is associated with Tmax, it is selected to be interchanged.


Table 1The data for Example 1 (in h)

Ji J1 J2 J3 J4 J5 J6 J7 J8 J9

pi 1 05 3 05 02 02 03 04 04di 1 13 2 30 10 13 20 12 14

Table 2The initial schedule for Example 1 (in h)

Ji J1 J2 M1 J3 J4 M2 J5 J6 J7 M3 J8 J9

pi 1 05 03 05 02 02 03 04 04Ci 1 06 13 18 22 24 27 34 38di 1 13 02 30 10 13 20 12 14Ti 0 00 11 00 12 11 07 22 24

Table 3The second schedule for Example 1 (in h)

Ji J1 J2 M1 J3 J4 M2 J5 J6 J9 M3 J8 J7

pi 1 05 03 05 02 02 04 04 03Ci 1 06 13 18 22 24 28 34 37di 1 13 02 30 10 13 14 12 20Ti 0 00 11 00 12 11 14 22 17

Table 4The third schedule for Example 1 (in h)

Ji J1 J2 M1 J3 J8 M2 J5 J6 J9 M3 J4 J7

pi 1 5 3 4 2 2 4 5 3Ci 1 6 13 17 22 24 28 35 38di 1 13 2 12 10 13 14 30 20Ti 0 0 11 5 12 11 14 5 18

In Steps 6 and 7, we interchange J9 with J7 and obtain Tmax = 22 (see Table 3). Now J8 = Tmaxbecomes the candidate to be interchanged. There are three jobs preceding J8 (i.e., J2; J4, and J9)being considered and d4p4 is the largest, so J4 is selected (by Theorem 2). After the interchange,we obtain a schedule with Tmax = 18 (see Table 4). Following the same procedure, we furtherinterchange J7 with J4 to reduce Tmax to 14 (see Table 5).According to Theorems 1 and 2, any interchange cannot improve the schedule any further. Thus,

we perform Step 8 to check other possible moves. We 2nd that J5, which precedes J9, can be movedto the potential position. After the jobs in each batch are reset in EDD (Step 9), we obtain the 2nalschedule with Tmax = 13 (see Table 6).


Table 5The fourth schedule for Example 1 (in h)

Ji J1 J2 M1 J3 J8 M2 J5 J6 J9 M3 J7 J4

pi 1 05 03 04 02 02 04 03 05Ci 1 06 13 17 22 24 28 33 38di 1 13 02 12 10 13 14 20 30Ti 0 00 11 05 12 11 14 13 08

Table 6The 2nal schedule for Example 1 (in h)

Ji J1 J5 J2 M1 J3 J8 M2 J6 J9 M3 J7 J4

pi 1 02 05 03 04 02 04 03 05Ci 1 03 08 13 17 22 26 33 38di 1 10 13 02 12 13 14 20 30Ti 0 00 00 11 05 09 12 13 08

5. The branch-and-bound algorithm

In this section, a branch-and-bound (B&B) algorithm is presented to provide the optimal schedule,which can be used to evaluate the proposed heuristic. There are several dominance rules stated inthe theorems that will be used in the B&B algorithm. Let PS; S, and S be partial schedules.

Theorem 3. For two adjacent jobs in the same batch; there exists an optimal schedule for whichthe job with smaller due date is placed before the other job.

Proof. The result follows immediately from the EDD order.

Theorem 4. Let S be created by appending (Jp;Mi; Jq) to PS. Let S be obtained from S byinterchanging Jq with Jp. If dp ppdq pq; then S can be dominated by S .

Proof. Let ST be the minimum of the starting times of Jp and Jq. Let L be the maximum lateness ofPS under S and L be the corresponding quantity under S . Let Lp; Lq be the lateness of Jp; Jq underS and Lp; Lq be the corresponding quantity under S . We; therefore; have the maximum latenessunder S

Lmax = (L; Lp; Lq);

and the maximum lateness under S

Lmax = (L; Lp; L

q)

= (L; Lp; Lq):

Now under S

Lp = ST + pp dp; Lq = CMi + pq dq;


and under S

Lq = ST + pq dq; Lp = CMi + pp dp:Therefore; Lq Lq (since CMi ST ); and LqLp (since dpppdqpq). Hence Lqmax(Lp; Lq).Since LqLp; we have

max(L; Lp; Lq) = max(L; Lq)max(L; Lp; L

q)

= max(L; Lp; Lq):

So LmaxLmax Tmax(S) = max{0; Lmax}max{0; Lmax} = Tmax(S ). Hence; S can be dominatedby S .

Theorem 5. Let Jp and Jq be any two jobs in S; where Jp is placed before Jq. Let S be obtainedfrom S by interchanging Jq with Jp. Suppose that the total processing time of each batch in S isno larger than T. If Tmax = Tq for Jq in both S and S ; then S can be dominated by S .

Proof. Suppose that both Jp and Jq are in the same batch in both S and S . Then the completiontime of Jq in S is smaller than that of Jq in S. Since Tmax = Tq for Jq in both S and S ; the Tmaxvalue of S is smaller than that of S. Hence; S can be dominated by S .Suppose that Jp and Jq are in di6erent batches in both S and S . The proof is the same as in the

same batch.

Theorem 6. Consider two partial schedules composed of the same set of jobs with the same com-pletion time. Then the partial schedule with larger Tmax value can be dominated by the other.

Proof. Denote by S and S the two considered partial schedules; where S has larger Tmax value.Let R be the partial schedule composed of the remaining jobs. It is clear that appending R to Swill result in a complete schedule with Tmax value no smaller than that of the complete schedule byappending R to S . Thus; S can be dominated by S .

For a partial schedule, the lower bound of Tmax, Tmax, is calculated based on a completion of thepartial schedule. The complete schedule is constructed by sequencing the unscheduled jobs in EDDorder. Then, the steps of the B&B algorithm can be stated as follows:Step 1: Initialization. Use the heuristic to obtain an initial schedule.Step 2: Branching. Select a partial schedule with the smallest Tmax value among all unfathomed

partial schedules. Let J[i] be the last job of the partial schedule. Create new partial schedules byplacing each of the unscheduled jobs at position i + 1. Note that if the job requires a processingtime greater than the slack time, it is scheduled in the next time interval T . If the number ofmaintenance periods of the new partial schedule is greater than that of the incumbent schedule, thenit is eliminated. Use the dominance rules given in Theorems 36 to eliminate any possible newlycreated partial schedules.Step 3: Bounding. For each newly created partial schedule, calculate Tmax by using the proposed

scheme stated earlier.


Table 7The computational results (Data Set I)

n T t B&B algorithm Heuristic PEDa

Comp. time (s) Comp. time (s) Min. Mean Max.

05 10 2 0.08 0.10 0.00 0.00 0.004 0.08 0.11 0.00 0.01 0.05

14 2 0.07 0.08 0.00 0.00 0.004 0.05 0.08 0.00 0.00 0.00

18 2 0.05 0.06 0.00 0.00 0.004 0.05 0.05 0.00 0.00 0.00

10 10 2 5.21 0.61 0.00 0.62 4.514 5.46 0.66 0.00 0.77 6.22

14 2 4.47 0.55 0.00 0.66 3.634 4.83 0.59 0.00 0.58 5.81

18 2 3.71 0.51 0.00 0.44 3.294 4.24 0.57 0.00 0.53 3.65

20 10 2 65.57 3.66 0.00 0.95 7.714 71.03 3.85 0.00 1.02 9.93

14 2 59.89 2.99 0.00 0.83 3.964 59.37 3.07 0.00 0.89 6.52

18 2 52.64 2.45 0.00 0.62 3.794 59.14 2.52 0.00 0.75 4.08

30 10 2 411.87 9.13 0.01 1.05 9.194 413.59 9.88 0.00 1.08 9.67

14 2 388.12 9.23 0.00 0.92 5.214 376.81 9.37 0.00 0.98 6.35

18 2 361.86 9.01 0.00 0.75 6.844 363.36 9.07 0.00 0.78 4.16

aPED: percentage error deviation.

Step 4: Fathoming. Fathom those newly created partial schedules whose Tmax values are no betterthan the incumbent schedule. Update the incumbent schedule if a new partial schedule has a smallerTmax value.Step 5: Stop. If there are nodes remaining unfathomed, go to Step 2; otherwise, stop.

To determine the time complexity of the branch-and-bound algorithm, we see that a typical branch-ing tree has 2n nodes for an n-job schedule. Thus, the overall time complexity of the algorithm isO(2n

ni=1 pi).

6. Computational results

To compare the heuristic with the B&B algorithm, an experimental work was conducted. Boththe heuristic and the B&B algorithm were coded in Visual BASIC and run on a PC-686. The test


Table 8The computational results (Data Set II)



5 10 2 0.08 0.09 0.00 0.01 0.034 0.09 0.12 0.00 0.01 0.08

14 2 0.06 0.08 0.00 0.00 0.004 0.05 0.09 0.00 0.00 0.00

18 2 0.04 0.06 0.00 0.00 0.004 0.05 0.06 0.00 0.00 0.00

10 10 2 5.17 0.67 0.00 0.58 4.364 5.51 0.74 0.00 0.76 7.54

14 2 4.58 0.59 0.00 0.59 3.714 4.83 0.63 0.00 0.63 5.25

18 2 3.68 0.55 0.00 0.48 4.114 4.11 0.62 0.00 0.51 4.33

20 10 2 66.92 3.63 0.00 0.99 6.284 68.14 3.81 0.01 0.98 9.17

14 2 61.35 2.92 0.00 0.82 4.684 63.71 3.16 0.00 0.91 5.15

18 2 52.63 2.56 0.00 0.65 3.744 57.35 2.71 0.00 0.77 4.19

30 10 2 421.17 9.19 0.01 1.06 6.724 426.36 9.42 0.01 1.13 8.36

14 2 391.72 9.04 0.00 0.92 4.524 378.85 9.16 0.00 0.96 8.24

18 2 358.95 8.93 0.00 0.77 5.884 375.13 9.11 0.00 0.81 4.31


problems were randomly generated based on the following scheme: processing times were selectedfrom a discrete uniform distribution (DU) over [1,10]. The due dates were selected from anotherDU over [(1C Q=2)nk=1 p[k], (1C +Q=2)

nk=1 p[k], with restriction d[i] 0, where Q and

C denote the due date range and tardiness factor, respectively. The experimental procedure consistsof a full factorial design with two settings of Q (Q=0:2; 0:6) and two settings of C (C =0:2; 0:6).For convenience, the four combinations C = 0:2, Q = 0:2; C = 0:2, Q = 0:6; C = 0:6, Q = 0:2; andC=0:6, Q=0:6 are referred to as Data Sets I, II, III, and IV, respectively. The experiments consistof three levels of T (10; 14; 18) and two levels of t(2; 4).The computational results are summarized in Tables 710 for Data Sets I, II, III, and IV, re-

spectively. In each complete trial, we randomly drew n (n = 5; 10; 20; 30) jobs. The tables providethe information on the average percentage error deviation and computation time for each combi-nation of n, T and t. The number of problems solved in each combination is 24. The percentageerror deviation (PED) used to validate the performance of the heuristic is computed as follows:


Table 9The computational results (Data Set III)



5 10 2 0.08 0.11 0.00 0.01 0.024 0.10 0.11 0.00 0.01 0.04

14 2 0.06 0.08 0.00 0.00 0.004 0.07 0.09 0.00 0.00 0.00

18 2 0.06 0.07 0.00 0.00 0.004 0.06 0.07 0.00 0.00 0.00

10 10 2 5.93 1.21 0.00 0.88 8.174 5.75 1.18 0.00 1.05 5.51

14 2 5.36 0.91 0.00 0.71 5.034 5.40 0.87 0.00 0.76 6.94

18 2 4.35 0.72 0.00 0.61 4.634 4.41 0.63 0.00 0.63 4.85

20 10 2 72.11 4.15 0.00 1.27 6.454 72.69 4.36 0.00 1.44 9.29

14 2 68.96 3.33 0.00 1.29 6.284 69.71 3.52 0.00 1.31 7.05

18 2 60.25 3.05 0.00 1.22 5.114 62.44 2.77 0.00 1.23 5.23

30 10 2 426.33 10.75 0.02 1.57 7.144 438.17 10.55 0.00 1.69 8.51

14 2 409.58 10.56 0.00 1.31 6.964 401.92 10.35 0.00 1.45 7.13

18 2 399.06 9.27 0.00 1.23 6.284 396.68 9.96 0.00 1.26 8.31


[(Tmax by heuristic Tmax by B&B algorithm)=Tmax by B&B algorithm] 100. It is observed fromthe tables that problems with smaller C value produce smaller PED and spend less computationtime. Also, it is found that problems with larger T value and smaller t value produce smallerPED. This is because in this case a smaller number of maintenance periods is generated, andhence a smaller number of potential positions is produced. This results in a Tmax value of theheuristic closer to the optimal value. Moreover, the PED increases slightly as the number of jobsincreases. The average PED is only 0.725 with a maximum of 1.69. It is seen that the heuristicspends much less computation time than the B&B algorithm. Therefore, the heuristic is applicablefor large-sized problems, while the B&B algorithm can only be used for small-to-medium-sizedproblems.


Table 10The computational results (Data Set IV)



5 10 2 0.07 0.10 0.00 0.01 0.064 0.09 0.11 0.00 0.01 0.09

14 2 0.06 0.07 0.00 0.00 0.004 0.06 0.08 0.00 0.00 0.00

18 2 0.05 0.06 0.00 0.00 0.004 0.06 0.06 0.00 0.00 0.00

10 10 2 5.89 1.06 0.00 0.83 7.114 5.74 1.14 0.00 1.01 6.80

14 2 5.24 0.82 0.00 0.79 5.564 5.41 0.88 0.00 0.77 7.79

18 2 4.23 0.65 0.00 0.56 5.644 4.47 0.61 0.00 0.61 5.07

20 10 2 70.09 3.98 0.00 1.19 5.874 72.13 4.03 0.00 1.30 9.76

14 2 69.40 3.17 0.00 1.21 6.134 70.65 3.64 0.00 1.26 7.82

18 2 58.47 2.82 0.00 1.07 5.814 63.16 2.77 0.00 1.26 6.39

30 10 2 441.15 11.25 0.00 1.42 7.384 456.07 10.79 0.02 1.56 9.11

14 2 426.39 10.43 0.00 1.31 6.724 437.90 10.22 0.00 1.35 9.61

18 2 402.55 9.58 0.00 1.24 6.584 394.26 10.26 0.00 1.30 8.19


7. Conclusions

The importance of maintenance has been gradually recognized by the decision maker. Therefore,it becomes a common practice to schedule maintenance periodically in many manufacturing systems.Unfortunately, most papers discussing maintenance assume there is only one maintenance period. Inthis paper, periodic maintenance that consists of several maintenance periods has been addressed.In particular, we have addressed a single-machine scheduling problem with periodic maintenanceand nonresumable jobs. The objective is to minimize the maximum tardiness subject to periodicmaintenance and nonresumable jobs.An e8cient heuristic has been proposed to provide the near-optimal solution for the problem. The

performance of the heuristic has been evaluated by comparing its solution with the optimal solutionderived by the developed branch-and-bound algorithm. Several properties associated with the problemhave also been investigated and implemented in the algorithm. Computational results have shown


that both the heuristic and branch-and-bound algorithm perform satisfactorily. Direct application ofthe results of this study to those companies where maintenance is performed periodically is easyand worthwhile.

References

[1] Art RHPM, Knapp GM, Lawrence MJ. Some aspects of measuring maintenance performance in the process industry.Journal of Quality in Maintenance Engineering 1998;4:611.

[2] Adiri I, Frostig E, Rinnooy Kan AHG. Scheduling on a single machine with a single breakdown to minimizestochastically the number of tardy jobs. Naval Research Logistics 1991;38:26171.

[3] Mosheiov G. Minimizing the sum of job completion times on capacitated parallel machines. Mathematical andComputer Modeling 1994;20:919.

[4] Lee CY, Liman SD. Capacitated two-parallel machine scheduling to minimize sum of job completion time. DiscreteApplied Mathematics 1993;41:21122.

[5] Lee CY. Parallel machines scheduling with non-simultaneous machine available time. Discrete Applied Mathematics1991;30:5361.

[6] Lee CY. Machine scheduling with an availability constraint. Journal of Global Optimization 1996;9:395416.[7] Lee CY. Minimizing the makespan in the two-machine Qowshop scheduling problem with an availability constraint.

Operations Research Letters 1997;20:12939.[8] Lee CY. Two-machine Qowshop scheduling with availability constraints. European Journal of Operational Research

1999;114:4209.[9] Knodel W. A bin packing algorithm with complexity O(n log n) in the stochastic limit. Mathematical Foundations of

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Ching-Jong Liao is a Professor of Industrial Management at National Taiwan University of Science and Technology.He holds the Ph.D. in Industrial Engineering from the Pennsylvania State University. His current research interests includeproduction scheduling and inventory control, on which he has published several papers.

Wen-Jinn Chen is a Ph.D. student at the Department of Industrial Management, National Taiwan University of Scienceand Technology. He currently is an instructor of Industrial Management at Oriental Institute of Technology. His currentresearch interests are scheduling and production management.

Single-machine scheduling with periodic maintenance and nonresumable jobsIntroductionNotation and problem settingThe proposed heuristicA numerical exampleThe branch-and-bound algorithmComputational resultsConclusionsReferences

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