Single-machine scheduling with piece-rate maintenance and interval constrained position-dependent processing times

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<ul><li><p>dent linear function. The objective is to nd jointly the optimal frequency to performmain-tenances and the optimal job sequence to minimize the total cost, which is a linear function</p><p>of jobtual pactu</p><p>et al. [19] and Lai and Lee [20]. However, all these papers consider learning or aging model with specic processing time1] and Wang andeiov [23]e function</p><p>In addition, it is well-known that the production efciency can be improved by performing maintenance in manuing processes. Some scheduling researchers considered that the machine is maintained exactly once during the planniiod, and the duration of maintenance is not xed. Some recent relevant references are: Kubzin and Strusevich [24], Moand Sidney [25], Cheng et al. [26], Mor and Mosheiov [27] and Cheng et al. [28]. Some scheduling researchers assumed thatthe duration of each maintenance is given. Liao and Chen [29] and Ji et al. [30] considered single-machine scheduling</p><p>0096-3003/$ - see front matter 2013 Elsevier Inc. All rights reserved.</p><p> Corresponding author.E-mail address: zhangyl@seu.edu.cn (Y.L. Zhang).</p><p>Applied Mathematics and Computation 226 (2014) 415422</p><p>Contents lists available at ScienceDirect</p><p>Applied Mathematics and Computationhttp://dx.doi.org/10.1016/j.amc.2013.10.034function. Only a few papers study general position-dependent processing times. Mosheiov and Sidney [2Wang [22] considered a scheduling problem with a general position-dependent learning function. Mosha scheduling problem with position-dependent job processing times which is not restricted to a monotonstudied.factur-ng per-sheiovSome papers discussed time-dependent scheduling problems. The relevant papers include Bachman et al. [1], Cheng et al. [2]and Ji and Cheng [3]. Mosheiov and Oron [4] and Hsu et al. [5] discussed position-dependent scheduling problems. For sched-uling problems with position-dependent effect, the position-dependent function as specic linear or exponential functionwas considered by Cheng andWang [6], Bachman and Janiak [7], Wang and Xia [8], Biskup [9], Kuo and Yang [10], Yang, Yangand Cheng [11], Yang and Yang [12] and Zhao and Tang [13]. Some relevant references for the combination of the exponentialfunction and linear function are: Lee [14], Cheng, Wu and Lee [15], Wang [16], Wang and Cheng [17], Cheng and Lee [18], Yin1. Introduction</p><p>Recent developments in the eldIn case of the learning effect, the acWhile in case of the aging effect, theof the makespan, total earliness and total tardiness. There is shown that the problem can beoptimally solved in O(n4) time. There is also shown that two special cases of the problemcan be optimally solved by lower order algorithms.</p><p> 2013 Elsevier Inc. All rights reserved.</p><p>scheduling have triggered into a growing interest towards learning or aging effect.rocessing time of a job will be shorter when it is scheduled later in a sequence.al processing time of a job will be longer when it is scheduled later in a sequence.Single-machine scheduling with piece-rate maintenanceand interval constrained position-dependent processing times</p><p>Pengfei Xue a, Yulin Zhang a,, Xianyu Yu a,ba School of Economics and Management, Southeast University, Nanjing 210096, Chinab School of Science, East China Institute of Technology, Fuzhou, Jiangxi 344000, China</p><p>a r t i c l e i n f o</p><p>Keywords:Single-machineSchedulingPosition-dependentMaintenanceTotal cost</p><p>a b s t r a c t</p><p>This paper investigates single-machine scheduling problems with piece-rate machinemaintenance and interval constrained actual processing time. The actual processing timeof a job is a general function of the normal job processing time and the position in jobsequence, and it is required to restrict in given interval otherwise earliness or tardinesspenalty should be paid. The maintenance duration studied in the paper is a time-depen-</p><p>journal homepage: www.elsevier .com/ locate/amc</p></li><li><p>probleJi andsched</p><p>maint</p><p>416 P. Xue et al. / Applied Mathematics and Computation 226 (2014) 415422which is assumed to conclude production fee and total earliness and tardiness cost. Moreover, we investigate two specialcases where the actual processing time of each job is assumed to be a product function, and explore to nd more effectivesolving algorithm to minimize the total cost for the cases.</p><p>This paper is organized as follows. We introduce the notation and terminology of this paper in the next section. In Sec-tion 3, we propose the main results of this paper. In Section 4, we conclude this paper, and suggest some topics for future.</p><p>2. Notations and problem formulation</p><p>A set J J1; J2; . . . ; Jn of n independent jobs are partitioned into m 1 groups and are all available for processing at timezero. The machine can handle one job at a time. In the manufacturing process, the jobs are non-preemptive. Each job j 2 J isassociated with the normal processing time pj. We extend the general processing time function of [23] and construct a moregeneral model of the actual job processing time</p><p>prj fjpj; r; 1 6 j 6 n; 1 6 r 6 nj: 1Observing from Eq. (1), different jobs may have different actual processing time functions which are not restricted to be spe-cic and monotone.</p><p>During the maintenance, the machine is stopped. After the maintenance activity, the machine will revert to its initial con-dition. We assume that the machine need to be maintained one time when every k jobs are completed and maintenance isjust nished at time zero. The jobs will be processed from a group continuously. Thus, we denote the schedule asr=[G1;M1; . . . ;Gi;Mi; . . . ;Mm;Gm1], where Gi denotes the ith group and Mi denotes the ith maintenance. Let ni denote thenumber of jobs in the ith group, and m be the maintenance frequency, which is equal to the integer part of nk, i.e., m bnkc.We can obtain that</p><p>Pm1i1 ni n. Let Cl;r be the completion time of the job scheduled in rth position of the lth group.</p><p>Let Gti denote the time length of ith group, and ti be the duration of the maintenance, i.e., ti aGti b, where a and b aretwo constants, aP 0. As in Zhao and Tang [13], we denote the group Gi as Gi=[Ji;1; Ji;2,. . .,Ji;ni1; Ji;ni , where Ji;r is the rth jobin the group Gi.</p><p>Let pl;r denote the normal processing time of job Jl;r and prl;r be actual processing time of job Jl;r, where prl;r=fl;r(pl;r,r).</p><p>Let aj and bj denote the lower limit and upper limit of the interval constraint of every job, respectively. Then, the earliness ofjob j is denoted as Ej, i.e., Ej=maxf0; aj prj g. The tardiness of job j is denoted as Tj, i.e., Tj=max{0; prj bjg. Let Cmax denote themakespan, i.e., Cmax=max{Cjjj 1;2; . . . ;n}, where Cj denotes the completion time of the job j. The total earliness of all jobs isdenoted by</p><p>Pnj1Ej, and the total tardiness of all jobs is denoted by</p><p>Pnj1Tj.</p><p>In the manufacturing process, the production fee is determined by the length of the working time. Moreover, the earlinesscost and tardiness cost are assumed to be linear relationship with the total earliness and tardiness of all jobs, respectively.Thus, We dene the total cost as follows:</p><p>TC aCmax bXnj1</p><p>Ej cXnj1</p><p>Tj;enance is proposed by Yu et al. [49]. We explore to nd the optimal polynomial algorithm to minimize the total cost,Another popular topic in recent years is scheduling problem with job completion time due window. To cope with globalcompetition and to improve customer demand, jobs should be nished as close as possible to their due-dates. If a job is n-ished earlier than its due-date, it would result in storage costs; on the other hand, if a job is nished later than its due-date, itwould violate contractual obligation with the customer. Cheng [33] initiated research on scheduling with due windows. Inmodern manufacturing management, scheduling problems with due window assignment are important issues. Many papershave been conducted on these issues in different scheduling environments. Yin et al. [34] studied a single-machine sched-uling problem with batch delivery cost and an assignable common due window simultaneously. They provided polynomial-time solutions for considered problems. Yeung et al. [35] studied a non-preemptive two-stage owshop scheduling problemunder the environment of a common due window. The objective is to minimize the earliness and tardiness. They showedthat the problem is NP-complete in the strong sense, and developed a branch and bound algorithm and a heuristic to solvethe problem. For more recent works on scheduling with a common due window, we refer the reader to the comprehensivesurveys by Cheng and Gupta [36], Gordon et al. [37], and the recent papers by Yeung et al. [38], Cheng [39], Baker and Scud-der [40], Liman et al. [41], Yeung et al. [42], Yeung [43], Mosheiov and Sarig [44], Yeung[45], Yang et al. [46], Yin et al. [47]and Yin et al. [48]. In this paper, we assume that every job has its own interval constrained actual processing time, which isrequired to restrict in given interval otherwise earliness or tardiness penalty should be paid. For example, in porcelain man-ufacturing processes, the actual processing time can not exceed a given interval otherwise the porcelain may have qualityaws.</p><p>To the best of our knowledge, however, scheduling with simultaneous consideration of interval constrained actual pro-cessing time and machine maintenance has not been explored. Motivated by these points, in this paper we consider the sin-gle machine scheduling with interval constrained actual processing time and piece-rate maintenance, where the piece-ratems with periodic maintenance activities, where each maintenance activity is required after a periodic time interval.Cheng [31] and Yang et al. [32] assumed that the machine may have multiple rate-modifying activities over theuling horizon.</p></li><li><p>The to</p><p>Simila</p><p>Then,</p><p>functi</p><p>Thof jobsa polystanda</p><p>xjlr 0</p><p>P. Xue et al. / Applied Mathematics and Computation 226 (2014) 415422 417the objective of the assignment problem is not j1 l1 r1wjlrxjlr amb, but j1 l1 r1wjlrxjlr .In order to minimize the total cost, we propose a polynomial time algorithm to determine jointly the optimal k, and the</p><p>optimal job sequence.</p><p>Algorithm 1.Step 1. For each k (k=1,2,. . .,n-1,n), solve the assignment problem (8) and calculate the objective value TCk.Step 2. Let TCk=min TCk, (k=1,2,. . .,n)).otherwise. In case of k n, all jobs are completed in one group, and the machine does not require maintenance. Thus,Pn Pm1Pni Pn Pm1Pni</p><p>where wjlr f 1l;r pl;r; r. We dene xjlr 0 or 1 such that xjlr 1 if job Jj is scheduled in the rth position in the group Gl andl1 r1xjlr 0 or 1; j 1;2; . . . ;n; l 1;2; . . . ;m 1; r 1;2; . . .ni;Xm1Xnixjlr 1; j 1;2; . . . ; n;j1jlr i 8st:</p><p>Xnx 1; l 1;2; . . . ;m 1; r 1;2; . . . ;n ;minj1 l1 r1</p><p>wjlrxjlr amb;</p><p>Xn Xm1XniTCm1;nm1 Xm1l1</p><p>Xnir1</p><p>f 1l;r pl;r; r amb:</p><p>e problem we studied in this section can be denoted by 1jprj fjpj; r; PRM k; ti aGti bjTC. For a given the numberk, we can get the number of groups by equationm bnkc. Then, we can obtain thatmb is a constant. We explore to ndnomial to solve the problem. The problem 1jprj fjpj; r; PRM k; ti aGti bjTC can be transformed as the followingrd assignment problem:fl;r pl;r; r afl;rpl;r; r Z; l m 1:7</p><p>Combining (6) and (7), we can obtainon f 1l;r pl;r; r as follows:</p><p>1 aa 1fl;rpl;r; r Z; l 1;2; . . . ;m;(For simplicity, let bmaxf0; al;r fl;rpl;r; rg cmaxf0; fl;rpl;r; r bl;rg be denoted by Z. Then, we introduce a newRml1Rkr1aa1fl;rpl;r;rbmaxf0;al;r fl;rpl;r;rgcmaxf0;fl;rpl;r;rbl;rgRnm1r1 afm1;rpm1;r;rbmaxf0;am1;r fm1;rpm1;r;rgcmaxf0;fm1;rpm1;r;rbm1;rgamb: 6TCaCmaxbRnj1EjcRnj1TjaRml1Rkr1a1fl;rpl;r;rmbRnm1r1 fm1;rpm1;r;rbRml1Rkr1maxf0;al;r fl;rpl;r;rgRnm1r1 maxf0;am1;r fm1;rpm1;r;rgcRml1Rkr1maxf0;fl;rpl;r;rbl;rgRnm1r1 maxf0;fm1;rpm1;r;rbm1;rgCmax Cm1;nm1 a 1Rml1Rkr1fl;rpl;r; r Rnm1r1 fm1;rpm1;r; r mb: 3tal earliness is given by</p><p>Rm1l1 Rnir1El;r Rml1Rkr1maxf0; al;r fl;rpl;r; rg Rnm1r1 maxf0; am1;r fm1;rpm1;r; rg: 4</p><p>rly, the total tardiness is given by</p><p>Rm1l1 Rnir1T l;r Rml1Rkr1maxf0; fl;rpl;r; r bl;rg Rnm1r1 maxf0; fm1;rpm1;r; r bm1;rg: 5</p><p>the total cost can be obtained as follows:where a; b and c are the unit production fee, the unit earliness cost and the unit tardiness cost, respectively. a; b and c shouldbe positive numbers, i.e., a; b; c &gt; 0.</p><p>3. The main results</p><p>The completion time of all jobs can be obtained as follows:</p><p>C m1;nm1 a 1Rml1Rkr1fl;rpl;r; r Rnm1r1 fm1;rpm1;r; r mb: 2From the Eq. (2), it can be obtained that</p></li><li><p>Theorem 1. The 1jprj fjpj; r; PRM k; ti aGti bjTC problem can be optimal solved by Algorithm 1 in On4 time.</p><p>Proof. For a xed the number of jobs k, the problem 1jprj fjpj; r; PRM k; ti aGti bjTC can be optimally solved via theassignment problem (8), and the computational complexity of obtaining the optimal job sequence is On3. Since k has n pos-sible values, the computational complexity of solving the problem 1jprj fjpj; r; PRM k; ti aGti bjTC is On4. h</p><p>In what follows, we investigate two special cases of the 1jprj fjpj; r; PRM k; ti aGti bjTC problem, and explore tond a more efcient algorithm for each case. We assume that pr p ra0 i:e:; pr p ra0 ; ti bb &gt; 0 and aP b.</p><p>constrstrain</p><p>Th</p><p>TheorOn lo</p><p>1 2</p><p>418 P. Xue et al. / Applied Mathematics and Computation 226 (2014) 415422Fig. 1. The illustration of the exchange of jobs Ji and Jj.TCr1 TCr2 apira0 pjr 1a0 apjra0 pir 1a0 api pjra0 r 1a0P 0:</p><p>Then, the schedule r1 is dominated by the schedule r2 in this case.Case (2). r rIn sequence r1, the job i has no earliness and the job j has earliness, but in the sequence r2 the case is converse. Then, it</p><p>can be obtained</p><p>TCr1 TCr2 apira0 pjr 1a0 bpjb0 pjr 1a0 apjra0 pir 1a0 bpib0 pir 1a0 pi pjara0 r 1a0 bb0 r 1a0P 0:</p><p>Then, the schedule r1 is dominated by the schedule r2 in this case.Proof. Assume that an optimal schedule r1 is not the SPT sequence. Then, there must exist a pair of adjacent jobs i and j sothat pi P pj, with job j following job i. We assume that job i is scheduled in the rth position.</p><p>Now let a new job schedule r2 be formed from r1 by interchanging jobs i and j and keeping the other jobs in the sameposition as in r1. The exchange of the jobs i and j is illustrated by Fig. 1, where A denotes the set of jobs preceding jobs i and jin both schedules, and B is the set of jobs following jobs i and j in both schedules. Let TCr1 and TCr2 denote the total costof r1 and r2, respectively. Since the positions of the jobs in A and B remain unchanged, the cost of processing and theearliness of the jobs in A and B remains unchanged.</p><p>Since not all jobs in the sequence have earliness in case of learning effect, we aim to nd the tipping position r, whoseillustration is proposed in Fig. 2. Let pib0 pira0 , we can obtain r b</p><p>b0</p><p>a0p</p><p>c. In case of r 6 r, since pib0 6 pira0 , it can be seenthat the job before the position (r 1) has no earliness. While in case of r &gt; r, since pib0 &gt; pira0 , it can be seen that the jobbehind the position r has earliness. In order to prove Theorem 2, we consider...</p></li></ul>

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