7
Arab J Sci Eng (2014) 39:1489–1495 DOI 10.1007/s13369-013-0716-9 RESEARCH ARTICLE - SYSTEMS ENGINEERING Single Machine Scheduling with Aging Effect and Upper-Bounded Actual Processing Times Xianyu Yu · Yulin Zhang Received: 23 February 2012 / Accepted: 15 July 2012 / Published online: 2 October 2013 © King Fahd University of Petroleum and Minerals 2013 Abstract Four single machine-scheduling problems with general position-dependent aging effect and upper-bounded actual processing times are considered. For the first two prob- lems without machine maintenance consideration, it is shown that they are polynomially solvable. For the last two prob- lems with machine maintenance consideration, two polyno- mial algorithms are proposed. Keywords Scheduling · Single machine · Aging effect · Maintenance activities · Bounded job processing times 1 Introduction In traditional scheduling problems, the processing times of the jobs under consideration are assumed to be constant val- ues. However, in many real situations, the actual processing time of a job may be subject to change due to aging effect X. Yu · Y. Zhang (B ) School of Economics and Management, Southeast University, Nanjing 211189, China e-mail: [email protected] X. Yu School of Science, East China Institute of Technology, Jiangxi 344000, China e-mail: [email protected] (also known as deteriorating effect), where the later a given job is scheduled in the sequence, the longer is its processing time. In the last decade, scheduling problems with aging effect have received increasing attention. There are mainly two cat- egories of aging effect models, i.e., time-dependent aging effect models and position-dependent aging effect models. For the former category, see, e.g., Gawiejnowicz [1], Chang et al. [2], and Cheng et al. [3]; for the latter one, see, e.g., Bachman and Janiak [4], Biskup [5], Li et al. [6], and Zhao and Tang [7]. In a recent paper, Inderfurth et al. [8] studied the problem of planning the production of new and recovering defective items of the same product manufactured in the same facility. They introduced a given deterioration time limit. A defec- tive item that is decided not to be reworked or cannot be reworked because it will exceed the deterioration time limit is disposed immediately after its work operation is complete. Gawiejnowicz [9] considered a single machine-scheduling problem with independent, non-preemptive and proportion- ally deteriorating jobs. Instead of proposing a deterioration time limit, Gawiejnowicz assumed that each job has a ready time and a deadline. In this paper, we consider four single machine-scheduling problems, of which the first two problems are subject to aging effect and bounded job actual processing times, where the actual processing time of a job is assumed to be a gen- eral strictly increasing function of its normal processing time and its position in the sequence. The assumption regarding bounded processing times is motivated by some practical manufacturing processes, e.g., the porcelain-forming process where the actual processing time of producing porcelain is not permitted to exceed the upper bound; otherwise, the out- put may have some flaws in color or pattern. The objective of the first problem is to minimize makespan, while that of the 123

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Arab J Sci Eng (2014) 39:1489–1495DOI 10.1007/s13369-013-0716-9

RESEARCH ARTICLE - SYSTEMS ENGINEERING

Single Machine Scheduling with Aging Effect and Upper-BoundedActual Processing Times

Xianyu Yu · Yulin Zhang

Received: 23 February 2012 / Accepted: 15 July 2012 / Published online: 2 October 2013© King Fahd University of Petroleum and Minerals 2013

Abstract Four single machine-scheduling problems withgeneral position-dependent aging effect and upper-boundedactual processing times are considered. For the first two prob-lems without machine maintenance consideration, it is shownthat they are polynomially solvable. For the last two prob-lems with machine maintenance consideration, two polyno-mial algorithms are proposed.

Keywords Scheduling · Single machine · Aging effect ·Maintenance activities · Bounded job processing times

1 Introduction

In traditional scheduling problems, the processing times ofthe jobs under consideration are assumed to be constant val-ues. However, in many real situations, the actual processingtime of a job may be subject to change due to aging effect

X. Yu · Y. Zhang (B)School of Economics and Management, Southeast University,Nanjing 211189, Chinae-mail: [email protected]

X. YuSchool of Science, East China Institute of Technology,Jiangxi 344000, Chinae-mail: [email protected]

(also known as deteriorating effect), where the later a givenjob is scheduled in the sequence, the longer is its processingtime.

In the last decade, scheduling problems with aging effecthave received increasing attention. There are mainly two cat-egories of aging effect models, i.e., time-dependent agingeffect models and position-dependent aging effect models.For the former category, see, e.g., Gawiejnowicz [1], Changet al. [2], and Cheng et al. [3]; for the latter one, see, e.g.,Bachman and Janiak [4], Biskup [5], Li et al. [6], and Zhaoand Tang [7].

In a recent paper, Inderfurth et al. [8] studied the problemof planning the production of new and recovering defectiveitems of the same product manufactured in the same facility.They introduced a given deterioration time limit. A defec-tive item that is decided not to be reworked or cannot bereworked because it will exceed the deterioration time limitis disposed immediately after its work operation is complete.Gawiejnowicz [9] considered a single machine-schedulingproblem with independent, non-preemptive and proportion-ally deteriorating jobs. Instead of proposing a deteriorationtime limit, Gawiejnowicz assumed that each job has a readytime and a deadline.

In this paper, we consider four single machine-schedulingproblems, of which the first two problems are subject toaging effect and bounded job actual processing times, wherethe actual processing time of a job is assumed to be a gen-eral strictly increasing function of its normal processing timeand its position in the sequence. The assumption regardingbounded processing times is motivated by some practicalmanufacturing processes, e.g., the porcelain-forming processwhere the actual processing time of producing porcelain isnot permitted to exceed the upper bound; otherwise, the out-put may have some flaws in color or pattern. The objective ofthe first problem is to minimize makespan, while that of the

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1490 Arab J Sci Eng (2014) 39:1489–1495

second is to minimize total completion time. We show thatthese two problems are polynomially solvable.

On the other hand, note that in some real production situ-ations, machines may not be always available due to variousreasons such as machine maintenance or sudden break-down. As is well known, preventive maintenance can reducemachine breakdown rate with minor sacrifices in many real-istic situations. As a result, it has received many researchers’attention. Recently, some researchers have started conduct-ing research on scheduling problems with aging effect andmachine maintenance. For the details on this stream ofresearch, the reader can refer to recent papers by Ji et al.[10], Kuo and Yang [11], Gawiejnowicz and Kononov [12],Yang and Yang [13], Zhao and Tang [14], and Yang and Yang[15].

Motivated by the importance of preventive maintenance,in this paper, we consider two other single machine-scheduling problems, where aging effect, machine main-tenance, and bounded job actual processing times areconsidered simultaneously. Both the objectives are to min-imize makespan. The main difference between the twoproblem is that one assumes that the upper bound of the main-tenance frequency equals n −1, while the other assumes thatthe upper bound of the maintenance frequency is prefixedand strictly less than n − 1, where n is the total number ofjobs under consideration. We provide polynomial algorithmsfor these two problems.

The remainder of this paper is organized as follows. InSect. 2, we introduce the formulations and notations. In Sect.3, we show that the two scheduling problems without theconsideration of maintenance are polynomially solvable. InSect. 4, we propose a polynomial algorithm for each of thescheduling problems with maintenance. In the last section,we present some concluding remarks.

2 Formulation and Notation

There are n jobs J1, J2, . . . , Jn to be scheduled on a sin-gle machine. All jobs are non-preemptive and available forprocessing at time zero. Let p j and C j be the normal process-ing time and the completion time of job J j , respectively. Theduration of each maintenance activity (if any) is t . Duringeach maintenance activity, the machine is turned off and theproduction is stopped. Let M represent the maintenance fre-quency. The inequality M ≤ μ means that each feasibleschedule contains at most μ maintenance activities on themachine.

Let k be the total number of maintenance activitiesin a feasible schedule. It is easy to see that the jobsprocessed continuously form a group, and thus the sched-ule can be denoted as [G1, M1, G2, M2, . . . , Mk, Gk+1],where Gi (1 ≤ i ≤ k + 1) denotes the i th group of jobs

and Mi (1 ≤ i ≤ k) represents the i th maintenance activityin the schedule. Let ni be the number of jobs in group Gi .Clearly,

∑k+1i=1 ni = n. If there is no maintenance activity in

the schedule, there is clearly only one group in the schedule.In this paper, the actual processing time of job J j when it

is assigned to the r th position in group Gi is defined by

pri j = f j (p j , i, r), 1 ≤ i ≤ (k + 1), 1 ≤ j ≤

n and 1 ≤ r ≤ ni , (1)

where f j (p j , i, r) is a general function and is assumed to bestrictly increasing in p j and r .

After each maintenance activity, the machine will berestored to its initial state. Therefore, if job J j is assignedto the first position in group Gi , the aging effect of job J j

does not take place. Hence, we have

p1i j = f j (p j , i, 1) = p j , 1 ≤ i ≤ (k + 1), 1 ≤ j ≤ n.

(2)

For the same reason, the actual processing time of job J j

depends on its scheduled position r in a group only, no mat-ter which group it belongs to. Thus, Eq. (1) can be reformu-lated as

pri j = pr

j = f j (p j , r), 1 ≤ i ≤ (k + 1), 1 ≤ j ≤ n (3)

and 1 ≤ r ≤ ni ,

where f j (p j , r) is a general function, and its definitionmay include elementary arithmetic operations or functions.Assume that f j (p j , r) is strictly increasing in p j and r ,and can be calculated in a constant time. It is obvious thatf j (p j , r) ≥ p j , 1 ≤ r ≤ ni .

Let the upper-bounded actual processing time of job J j

be U j . It is reasonable to assume that

p j ≤ U j , 1 ≤ j ≤ n, (4)

as otherwise there are clearly no feasible schedules due to theaging effect. Clearly, pr

j ≤ U j must hold if job J j is sched-uled to the r th position of a group in any feasible schedule.

Following the three-field notation proposed by Grahamet al. [16] and extended by Agnetis et al. [17] and Kuo andYang [11], the makespan minimization problem with agingeffect and bounded processing times is denoted as

1∣∣∣pr

j = f j (p j , r)

∣∣∣ Cmax : pr

j ≤ U j . (5)

Similarly, the total completion time minimization problemwith aging effect and bounded processing times is denotedas

1∣∣∣pr

j = f j (p j , r)

∣∣∣∑

C j : prj ≤ U j . (6)

The makespan minimization problem with aging effect,maintenance frequency upper bound n − 1, and boundedactual processing times is denoted as

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Arab J Sci Eng (2014) 39:1489–1495 1491

1, M ≤ n − 1∣∣∣pr

j = f j (p j , r)

∣∣∣ Cmax : pr

j ≤ U j . (7)

Similarly, the makespan minimization problem with agingeffect, maintenance frequency upper bound k0 that is strictlyless than n − 1, and bounded actual processing times isdenoted as

1, M ≤ k0 < n − 1∣∣∣pr

j = f j (p j , r)

∣∣∣ Cmax : pr

j ≤ U j .

(8)

3 Scheduling Without Maintenance

3.1 Problem 1|prj = f j (p j , r)|Cmax : pr

j ≤ U j

In this subsection, we will show that the scheduling problem1|pr

j = f j (p j , r)|Cmax : prj ≤ U j is polynomially solv-

able. Firstly, we consider the problem 1|prj = f j (p j , r)|

Cmax, which is useful for solving the problem under study.

Lemma 3.1 The problem 1|prj = f j (p j , r)|Cmax can be

optimally solved in O(n3) time.

Proof The makespan can be given by

Cmax =n∑

j=1

n∑

r=1

Prj x jr =

n∑

j=1

n∑

r=1

f j (Pj , r)x jr , (9)

where x jr = 1 if job J j is assigned in the r th positionin the schedule; otherwise, x jr = 0. Then, the problem1|pr

j = f j (p j , r)|Cmax can be transformed into the follow-ing standard assignment problem.

Minimizen∑

j=1

n∑

r=1

(p j , r)x jr (10)

s.t.n∑

j=1

x jr = 1, 1 ≤ r ≤ n, (11)

n∑

r=1

x jr = 1 = 1, 1 ≤ j ≤ n, (12)

x jr = 0 or 1, 1 ≤ j ≤ n, 1 ≤ r ≤ n. (13)

It is well known that the standard assignment problem canbe optimally solved in O(n3) time by the classic Hungarianalgorithm. Hence, the problem 1|pr

j = f j (p j , r)|Cmax can

also be optimally solved in O(n3) time. ��We let

pr(1)j = f (1)

j (p j , r)

={

A1 + f j (p j , r) if f j (p j , r) > U j ,

f j (p j , r) if f j (p j , r) ≤ U j ,(14)

where A1 = ∑nj=1 f j (p j , n). Note that A1 is the sum of the

actual processing time of jobs which are all assumed to be

scheduled in the nth position in the sequence. In any feasibleschedule σ , only one job can be scheduled in the nth positionand f j (p j , r) strictly increases in r . Hence, the makespan ofany feasible schedule must be smaller than A1.

We let

T L1 =n∑

j=1

n∑

r=1

pr(1)j x jr =

n∑

j=1

n∑

r=1

f (1)j (p j , r)x jr . (15)

Note that Eq. (15) is quite similar with Eq. (9). It is easy toget the following lemma by using a similar method in theproof of Lemma 3.1.

Lemma 3.2 The problem 1|pr(1)j = f (1)

j (p j , r)|T L1 can be

optimally solved in O(n3) time.

For a scheduling problem under consideration, if thereexists at least one feasible schedule, we regard it as a feasi-ble problem, otherwise the problem is infeasible. Let T L∗

1

be the optimal objective value of the problem 1|pr(1)j =

f (1)j (p j , r)|T L1. Let C∗

max be the optimal objective valueof the problem 1|pr

j = f j (p j , r)|Cmax : prj ≤ U j if the

scheduling problem is feasible.

Lemma 3.3 For the problem 1|pr(1)j = f (1)

j (p j , r)|T L1

and the problem 1|prj = f j (p j , r)|Cmax : pr

j ≤ U j , ifT L∗

1 < A1, then the scheduling problem 1|prj = f j (p j , r)|

Cmax : prj ≤ U j is feasible and its optimal objective value

is T L∗1, or else the scheduling problem is infeasible.

Proof From Eq. (14), we can see that the makespan of anyfeasible schedule for the problem 1|pr

j = f j (p j , r)|Cmax :pr

j ≤ U j must be smaller than A1.If T L∗

1 < A1, then no actual job processing time exceeds

the upper bound. From Eq. (14), we can get f (1)j (p j , r) =

f j (p j , r), 1 ≤ j ≤ n, 1 ≤ r ≤ n. In this case, the prob-lem 1|pr

j = f j (p j , r)|Cmax : prj ≤ U j is equivalent to the

problem 1|pr(1)j = f (1)

j (p j , r)|T L1. It can be seen that theproblem 1|pr

j = f j (p j , r)|Cmax : prj ≤ U j is feasible and

its optimal objective value C∗max = T L∗

1.If T L1 ≥ A1, then there is at least one job whose actual

processing time exceeds the upper bound. Hence, the prob-lem 1|pr

j = f j (p j , r)|Cmax : prj ≤ U j is infeasible. ��

Theorem 3.1 The problem 1|prj = f j (p j , r)|Cmax : pr

j ≤U j can be solved in O(n3) time.

Proof In the first step, from Lemma 3.2, we can optimallysolve the problem 1|pr(1)

j = f (1)j (p j , r)|T L1 and obtain the

optimal result T L∗1 and its corresponding schedule in O(n3)

time.In the second step, from Lemma 3.3, we can get the opti-

mal result of the problem 1|prj = f j (p j , r)|Cmax : pr

j ≤ U j .If T L∗

1 < A1, then the problem 1|prj = f j (p j , r)|Cmax :

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1492 Arab J Sci Eng (2014) 39:1489–1495

prj ≤ U j is feasible, and C∗

max = T L∗1. The correspond-

ing schedule of T L∗1 is the optimal schedule of the problem

1|prj = f j (p j , r)|Cmax : pr

j ≤ U j . In the other case thatT L∗

1 ≥ A1, the problem 1|prj = f j (p j , r)|Cmax : pr

j ≤ U j

is infeasible. The computational complexity of the secondstep is O(1).

Hence, to solve the problem 1|prj = f j (p j , r)|Cmax :

prj ≤ U j , the total computational complexity is O(n3). ��

3.2 Problem 1|prj = f j (p j , r)| ∑ C j : pr

j ≤ U j

Similar to the previous subsection, we firstly consider theproblem 1|pr

j = f j (p j , r)| ∑ C j , which is useful for solvingthe problem 1|pr

j = f j (p j , r)| ∑ C j : prj ≤ U j .

Lemma 3.4 The problem 1|prj = f j (p j , r)| ∑ C j can be

optimally solved in O(n3) time.

Proof If job J j is assigned in the r th position in the sequence,its contribution to the objective value is (n −r +1) f j (p j , r).Then, we get

∑C j =

n∑

j=1

n∑

r=1

(n − r + 1) f j (p j , r)x jr , (16)

where x jr = 1 if job J j is assigned in the r th position inthe schedule, otherwise x jr = 0. Similar to the problem1|pr

j = f j (p j , r)|Cmax, the problem 1|prj = f j (p j , r)|∑

C j can be transformed into the∑

C j minimizationassignment problem, which can be solved in O(n3) time bythe classic Hungarian algorithm. This completes the proof.

��We let

pr(2)j = f (2)

j (p j , r)

={

A2 + f j (p j , r) if f j (p j , r) > U j ,

f j (p j , r) if f j (p j , r) ≤ U j ,

(17)

where A2 = ∑nj=1 n fi (p j , n). If job J j is assigned in the

r th position in the sequence, its contribution to the objectiveis (n − r + 1) f j (p j , r). It is obvious that

(n − r + 1) f j (p j , r) < n f j (p j , n), 1 ≤ j ≤ n, 2 ≤ r ≤ n.

Hence,∑

C j of any feasible schedules must be smaller thanA2. Let

T C =n∑

j=1

n∑

r=1

f (2)j (p j , r)(n − r + 1)x jr . (18)

Similar to Lemma 3.2, it is easy to obtain the followinglemma.

Lemma 3.5 The problem 1|pr(1)j = f (2)

j (p j , r)|T C can be

optimally solved in O(n3) time.

Let T C∗ be the optimal result of the problem 1|pr(1)j = f (1)

j

(p j , r)|T C. Let∑

C∗j be the optimal objective value of the

problem 1|prj = f j (p j , r)| ∑ C j : pr

j ≤ U j if the problemis feasible. With the similar proof method in Lemma 3.3, wecan get the following lemma.

Lemma 3.6 For the problem 1|prj = f j (p j , r)| ∑ C j :

prj ≤ U j and the problem 1|pr(1)

j = f j (p j , r)(2)|T C, ifT C∗ < A2, the problem 1|pr

j = f j (p j , r)| ∑ C j : prj ≤ U j

is feasible and∑

C∗j = T C∗; else the problem 1|pr

j =f j (p j , r)| ∑ C j : pr

j ≤ U j is infeasible.

Combining Lemma 3.5 and 3.6, it is easy to obtain the fol-lowing theorem by the similar proof method in Theorem 3.1.

Theorem 3.2 The problem 1|prj = f j (p j , r)| ∑ C j : pr

j ≤U j can be solved in O(n3) time.

4 Scheduling with Maintenances

4.1 Problem1, M ≤ n − 1|pr

j = f j (p j , r)|Cmax : prj ≤ U j

In this subsection, we will propose a polynomial algorithmto solve the problem 1, M ≤ n − 1|pr

j = f j (p j , r)|Cmax :pr

j ≤ U j .

Lemma 4.1 There exists at least one feasible schedule forthe problem 1, M ≤ n−1|pr

j = f j (p j , r)|Cmax : prj ≤ U j .

Proof To prove the existence of the feasible schedule, it isonly necessary to find a feasible schedule for the problem1, M ≤ n − 1|pr

j = f j (p j , r)|Cmax : prj ≤ U j . If the

maintenance frequency is (n − 1) in a schedule σ , there aren groups in the schedule, and there is only one job in eachgroup. From Eq. (2), it can be seen that the actual processingtime of each job is equal to its normal processing time. Basedon condition (4), we obtain that pr

j ≤ U j (1 ≤ j ≤ n) in theschedule σ . Hence, σ is a feasible schedule for the problem1, M ≤ n − 1|pr

j = f j (p j , r)|Cmax : prj ≤ U j . ��

Group Balance Principle (Kuo and Yang [11]) If themachine is maintained k times in a schedule, the jobs aredivided into k +1 groups: G1, . . . , Gi , . . . , Gk+1, where Gi

has ni jobs (1 ≤ i ≤ k + 1), then [n/(k + 1)] ≤ ni ≤[n/(k + 1)] + 1.

Similar to Eq. (14), we let

pr(3)j = f (3)

j (p j , r)

={

A3 + f j (p j , r) if f j (p j , r) > U j ,

f j (p j , r) if f j (p j , r) ≤ U j ,

(19)

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Arab J Sci Eng (2014) 39:1489–1495 1493

where A3 = ∑nj=1 f j (p j , n) + (n − 1)t. Since f j (p j , r)

is a strictly increasing function of p j and r , it is easy to

prove that f (3)j (p j , r) is also a strictly increasing function

of p j and r . Note that A3 is the sum of (n − 1) mainte-nance durations and the actual processing time of jobs whichare all assumed to be scheduled in the nth position in thesequence. In any feasible schedule for the problem 1|pr

j =f j (p j , r), Mleqn − 1|Cmax : pr

j ≤ U j , only one job can bescheduled in the nth position and the maintenance frequencyis no more than (n −1). Hence, the makespan of any feasibleschedules for the problem 1, M ≤ n−1|pr

j = f j (p j , r)|Cx :pr

j ≤ U j must be smaller than A3. The makespan of the prob-lem 1, M ≤ n − 1|pr

j = f j (p j , r)|Cmax : prj ≤ U j can be

given by

Cmax =k+1∑

i=1

n∑

j=1

n∑

r=1

f j (Pj , r)xi jr + kt, (20)

where xi jr = 1 if job J j is assigned in the r th position of thei th group in the schedule, otherwise xi jr = 0. We let

T L2 =k+1∑

i=1

n∑

j=1

n∑

r=1

f (3)j (Pj , r)xi jr + kt. (21)

Similar to the proof in Zhao and Tang [14], we will prove thatthe Group Balance Principle remains valid for the problem1, M ≤ n − 1|pr(3)

j = f (3)j (p j , r)|T L2.

Lemma 4.2 For the problem 1, M ≤ n − 1|pr(3)j = f (3)

j(p j , r)|T L2, there exists an optimal solution such that thenumbers of jobs in groups satisfy the Group Balance Princi-ple in the case that k is fixed.

Proof Consider an optimal schedule σ and assume that themachine is maintained k times, then there are (k + 1) groupsof jobs in σ . Let G1, . . . , Gi , . . . , Gk+1 denote the groupsin the schedule ma. Suppose that the numbers of jobs in thegroups do not satisfy the Group Balance Principle, then theremust be at least two groups, say Gi and G j , where Gi hasni jobs and G j has n j jobs, such that ni − n j > 1. Letσ = [σ1, Gi , σ2, G j , σ3], where σ1, σ2 and σ3 are thepartial schedules of σ . Let Gi = [Ji1, Ji2, . . . , Ji,ni −1, Ji,ni ]and G j = [J j1, J j2, . . . , J j,n j ]. By moving job Ji,ni fromgroup Gi to the last position of group G j , we get a newschedule σ ′ = [σ1, Gi , σ2, G j , σ3]. Without loss ofgenerality, assume that job Js is the job that is scheduledin the last position in the group Gi in the schedule σ (i.e.,Ji,ni is Js). Then, we denote Gi = [Ji1, Ji2, . . . , Ji,ni −1],G j = [J j1, J j2, . . . , J j,n j , Ji,n j +1(Js)]. In the schedule σ ′,

the actual processing time of Js is pn j +1s = fs(ps, n j + 1).

From Eq. (19), we obtain

pn j +1(3)s = f (3)

s (p js, n j + 1)

={

A3+ fs(ps, n j +1) if fs(ps, n j + 1) > Us,

fs(ps, n j + 1) if fs(ps, n j + 1) ≤ Us .

(22)

Similar to the above analysis, we obtain the actual processingtime of Js as follows:

pni (3)s = f (3)

s (ps, ni )

={

A3 + fs(ps, ni ) if fs(ps, ni ) > Us,

fs(ps, ni ) if fs(ps, ni ) ≤ Us .

(23)

Note that f j (p j , r) is a strictly increasing function in r , andni −n j > 1; it can be seen that fs(ps, ni ) > fs(ps, n j +1).Combining (22) and (23), we get

f (3)s (ps, ni ) > f (3)

s (ps, n j + 1). (24)

Since we only move job Js from group Gi to G j , the otherjobs and the maintenance activities remain unchanged, theactual processing times of the jobs except job Js also remainunchanged. Hence, based on (24), the value of the objectivefunction of the schedule σ ′ is smaller than that of σ . Thiscontradicts the optimality of σ and proves the lemma. ��Lemma 4.3 For a prefixed k, the problem 1, M ≤ n −1|pr(3)

j = f (3)j (p j , r)|T L2 can be optimally solved in O(n3)

time.

Proof Based on Lemma 4.2 and the analysis in Zhao andTang [14], we can minimize T L2 via solving the followingassignment problem.

Minimizek+1∑

j=1

n∑

j=1

n∑

r=1

f (3)j (p j , r)xi jr + kt (25)

s.t.n∑

j=1

xi jr = 1, 1 ≤ i ≤ k + 1 and 1 ≤ r ≤ ni , (26)

k+1∑

j=1

ni∑

j=1

xi jr = 1, 1 ≤ j ≤ n, (27)

xi jr ∈ {0, 1}, 1 ≤ i ≤ k + 1, 1 ≤ j ≤ n and 1 ≤ r ≤ ni .

(28)

Hence, the problem 1, M ≤ n − 1|pr(3)j = f (3)

j (p j , r)|T L2 can be solved in O(n3) time.

Let T L2(k) denote the optimal objective value of the prob-lem with k (k = 0, 1, . . . , n − 1) maintenance activities,and T L∗

2 denote the optimal objective value of the problem

1, M ≤ n − 1|pr(3)j = f (3)

j (p j , r)|T L2. Combining Lem-mas 4.2 and 4.3, we will propose the solving algorithm of theproblem 1, M ≤ n − 1|pr(3)

j = f (3)j (p j , r)|T L2 as follows.

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1494 Arab J Sci Eng (2014) 39:1489–1495

Algorithm 4.1Input: n t, , (3) ( ) 1 2j jf p r i r n, , , = , ,...,

Output: 2TL∗ , σ

2 0TL =for 0k = to 1n − dosolve the corresponding assignment problem

(3)2

1

( ) ( )n

j jj

TL k f p r kt=

= , +∑endfor 0k = to 1n − do

2 2min ( )TL TL k∗ =end

Lemma 4.4 The problem 1, M ≤ n −1|pr(3)j = f (3)

j (p j , r)|T L2 can be optimally solved in O(n4) time.

Proof We solve the problem 1, M ≤ n − 1|pr(3)j = f (3)

j(p j , r)|T L2 by Algorithm 4.1. For a fixed k, we can get theoptimal objective value T L2(k) via the assignment problem(25), (26), (27) and (28) in O(n3) time. Because there are npossible values of k, to obtain all the values T L2(k), (k =0, 1, . . . , n − 1), the computational complexity is O(n4).The step to obtain the minimum T L2(k) will take O(nlogn)

time. Therefore, the total computational complexity of Algo-rithm 4.1 is O(n4). This completes the proof. ��

Let C∗max be the optimal objective value of the problem

1, M ≤ n − 1|prj = f j (p j , r)|Cmax : pr

j ≤ U j . Based onthe Lemma 4.1, we can obtain the following lemma by thesimilar proof method in Lemma 3.3.

Lemma 4.5 The problem 1, M ≤ n − 1|prj = f j (p j , r)|

Cmax : prj ≤ U j has the same optimal objective value and

schedule as those of the problem 1, M ≤ n − 1|pr(3)j =

f (3)j (p j , r)|T L2, respectively.

Theorem 4.1 The problem 1, M ≤ n − 1|prj = f j (p j , r)|

Cmax : prj ≤ U j can be optimally solved in O(n4) time.

Proof First, we can optimally solve the problem 1, M ≤ n−1|pr(3)

j = f (3)j (p j , r)|T L2 in O(n4) time by Algorithm 4.1.

Second, from Lemma 4.5, we can use the result of the prob-lem 1, M ≤ n−1|pr(3)

j = f (3)j (p j , r)|T L2 to solve the prob-

lem 1, M ≤ n − 1|prj = f j (p j , r)|Cmax : pr

j ≤ U j . Hence,it is easy to get that the total computational complexity ofsolving the problem 1, M ≤ n − 1|pr

j = f j (p j , r)|Cmax :pr

j ≤ U j is O(n4). ��

4.2 Problem 1, M ≤ k0 < n − 1|prj = f j (p j , r)|

Cmax : prj ≤ U j

In the subsection, we explore solving the problem 1, M ≤k0 < n − 1|pr

j = f j (p j , r)|Cmax : prj ≤ U j , which is very

similar to the problem 1, M ≤ n − 1|prj = f j (p j , r)|Cmax :

prj ≤ U j . Similar to the analysis in the above subsection, we

provide Algorithm 4.2 to solve the problem 1, M ≤ k0 <

n − 1|pr(3)j = f (3)

j (p j , r)|T L2.

Algorithm 4.2Input: n , t , 0k , (3) ( ) 1 2j jf p r j r n, , , = , ,...,

Output: 2TL∗ , σ

2 0TL =for 0k = to 0k dosolve the corresponding assignment problem

(3)2

1

( ) ( )n

j jj

TL k f p r kt=

= , +∑endfor 0k = to 0k do

2 2min ( )TL TL k∗ =end

Lemma 4.6 The problem 1, M ≤ k0 < n − 1|pr(3)j = f (3)

j(p j , r)|T L2 can be optimally solved by Algorithm 4.2 inO(k0n3) time.

Proof For a fixed k, based on Lemma 4.3, we solve the prob-lem 1, M ≤ n − 1|pr(3)

j = f (3)j (p j , r)|T L2 in O(n3) time.

Because there are k0 + 1 possible values of k, to obtainT L2(k), (k = 0, 1, . . . , k0), the computational complexityis O((k0 + 1)n3). The step to obtain the minimum T L2(k)

will take O((k0+1)log(k0+1)) time. Since O((k0+1)n3) =O((k0)n3), the problem 1, M ≤ k0 < n − 1|pr(3)

j =f (3)

j (p j , r)|T L2 can be solved in O((k0)n3) time.Let the optimal objective function value of the problem

1, M ≤ k0 < n−1|prj = f j (p j , r)|Cmax : pr

j ≤ U j be C∗max.

With the similar proof method in Lemma 3.3, we show therelationship between the problem 1, M ≤ k0 < n −1|pr

j =f j (p j , r)|Cmax : pr

j ≤ U j and the problem 1, M ≤ k0 <

n−1|pr(3)j = f (3)

j (p j , r)|T L2 in the following lemma. ��

Lemma 4.7 For the problem 1, M ≤ k0 < n − 1|prj = f j

(p j , r)|Cmax : prj ≤ U j and the problem 1, M ≤ k0 <

n − 1|pr(3)j = f (3)

j (p j , r)|T L2, if T L∗2 < A3, the problem

1, M ≤ k0 < n − 1|prj = f j (p j , r)|Cmax : pr

j ≤ U j isfeasible and C∗

max = T L∗2; else the problem 1, M ≤ k0 <

n − 1|prj = f j (p j , r)|Cmax : pr

j ≤ U j is infeasible.

Based on Lemmas 4.6 and 4.7, we can obtain the followingtheorem by the similar proof method in Theorem 3.1.

Theorem 4.2 The problem 1, M ≤ k0 < n − 1|prj = f j

(p j , r)|Cmax : prj ≤ U j can be solved in O(k0 n3) time.

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Arab J Sci Eng (2014) 39:1489–1495 1495

5 Conclusions

In this paper, we considered the single machine-schedulingproblems with general position-dependent aging effect,maintenance activities, and upper-bounded actual process-ing times. For scheduling without maintenance, we aimedto minimize the makespan and the total completion time, asfor scheduling with maintenance activities, the objective isto determine the maintenance frequency and find a feasibleschedule such that the makespan is minimized. We proposedpolynomial time solutions for all the problems under study.Our future research will be directed to investigate other mod-els of learning or aging effect and variable machine mainte-nance, or other performance measures.

Acknowledgments The authors would like to express their warmestthanks to Professor Bassam El Ali (Managing Editor) for communi-cating the paper. The authors also would like to express their warmestthanks to the referees for their valuable comments, which improved thepaper substantially. This work is supported by the National Nature Sci-ence Foundation of China (71171046), the College Graduate Researchand Innovation Foundation of Jiangsu Province (CXLX 0162) and theScientific Research Foundation of Graduate School of Southeast Uni-versity (YBJJ1239).

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