Single-machine scheduling with piece-rate maintenance and interval ...
Applied Mathematics and Computation 226 (2014) 415–422
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Single-machine scheduling with piece-rate maintenance and interval constrained position-dependent processing times Pengfei Xue a, Yulin Zhang a,⇑, Xianyu Yu a,b a b
School of Economics and Management, Southeast University, Nanjing 210096, China School of Science, East China Institute of Technology, Fuzhou, Jiangxi 344000, China
a r t i c l e
i n f o
Keywords: Single-machine Scheduling Position-dependent Maintenance Total cost
a b s t r a c t This paper investigates single-machine scheduling problems with piece-rate machine maintenance and interval constrained actual processing time. The actual processing time of a job is a general function of the normal job processing time and the position in job sequence, and it is required to restrict in given interval otherwise earliness or tardiness penalty should be paid. The maintenance duration studied in the paper is a time-dependent linear function. The objective is to find jointly the optimal frequency to perform maintenances and the optimal job sequence to minimize the total cost, which is a linear function of the makespan, total earliness and total tardiness. There is shown that the problem can be optimally solved in O(n4) time. There is also shown that two special cases of the problem can be optimally solved by lower order algorithms. Ó 2013 Elsevier Inc. All rights reserved.
1. Introduction Recent developments in the field of job scheduling have triggered into a growing interest towards learning or aging effect. In case of the learning effect, the actual processing time of a job will be shorter when it is scheduled later in a sequence. While in case of the aging effect, the actual processing time of a job will be longer when it is scheduled later in a sequence. Some 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 scheduling problems with position-dependent effect, the position-dependent function as specific linear or exponential function was considered by Cheng and Wang [6], Bachman and Janiak [7], Wang and Xia [8], Biskup [9], Kuo and Yang [10], Yang, Yang and Cheng [11], Yang and Yang [12] and Zhao and Tang [13]. Some relevant references for the combination of the exponential function and linear function are: Lee [14], Cheng, Wu and Lee [15], Wang [16], Wang and Cheng [17], Cheng and Lee [18], Yin et al. [19] and Lai and Lee [20]. However, all these papers consider learning or aging model with specific processing time function. Only a few papers study general position-dependent processing times. Mosheiov and Sidney [21] and Wang and Wang [22] considered a scheduling problem with a general position-dependent learning function. Mosheiov [23] studied a scheduling problem with position-dependent job processing times which is not restricted to a monotone function. In addition, it is well-known that the production efficiency can be improved by performing maintenance in manufacturing processes. Some scheduling researchers considered that the machine is maintained exactly once during the planning period, and the duration of maintenance is not fixed. Some recent relevant references are: Kubzin and Strusevich [24], Mosheiov and Sidney [25], Cheng et al. [26], Mor and Mosheiov [27] and Cheng et al. [28]. Some scheduling researchers assumed that the duration of each maintenance is given. Liao and Chen [29] and Ji et al. [30] considered single-machine scheduling ⇑ Corresponding author. E-mail address: [email protected] (Y.L. Zhang). 0096-3003/$ - see front matter Ó 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2013.10.034
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problems with periodic maintenance activities, where each maintenance activity is required after a periodic time interval. Ji and Cheng [31] and Yang et al. [32] assumed that the machine may have multiple rate-modifying activities over the scheduling horizon. Another popular topic in recent years is scheduling problem with job completion time due window. To cope with global competition and to improve customer demand, jobs should be finished as close as possible to their due-dates. If a job is finished earlier than its due-date, it would result in storage costs; on the other hand, if a job is finished later than its due-date, it would violate contractual obligation with the customer. Cheng [33] initiated research on scheduling with due windows. In modern manufacturing management, scheduling problems with due window assignment are important issues. Many papers have been conducted on these issues in different scheduling environments. Yin et al. [34] studied a single-machine scheduling problem with batch delivery cost and an assignable common due window simultaneously. They provided polynomialtime solutions for considered problems. Yeung et al. [35] studied a non-preemptive two-stage flowshop scheduling problem under the environment of a common due window. The objective is to minimize the earliness and tardiness. They showed that the problem is NP-complete in the strong sense, and developed a branch and bound algorithm and a heuristic to solve the problem. For more recent works on scheduling with a common due window, we refer the reader to the comprehensive surveys by Cheng and Gupta [36], Gordon et al. [37], and the recent papers by Yeung et al. [38], Cheng [39], Baker and Scudder [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 is required to restrict in given interval otherwise earliness or tardiness penalty should be paid. For example, in porcelain manufacturing processes, the actual processing time can not exceed a given interval otherwise the porcelain may have quality flaws. To the best of our knowledge, however, scheduling with simultaneous consideration of interval constrained actual processing time and machine maintenance has not been explored. Motivated by these points, in this paper we consider the single machine scheduling with interval constrained actual processing time and piece-rate maintenance, where the piece-rate maintenance is proposed by Yu et al. [49]. We explore to find the optimal polynomial algorithm to minimize the total cost, which is assumed to conclude production fee and total earliness and tardiness cost. Moreover, we investigate two special cases where the actual processing time of each job is assumed to be a product function, and explore to find more effective solving algorithm to minimize the total cost for the cases. This paper is organized as follows. We introduce the notation and terminology of this paper in the next section. In Section 3, we propose the main results of this paper. In Section 4, we conclude this paper, and suggest some topics for future. 2. Notations and problem formulation A set J ¼ ½J 1 ; J 2 ; . . . ; J n of n independent jobs are partitioned into m þ 1 groups and are all available for processing at time zero. The machine can handle one job at a time. In the manufacturing process, the jobs are non-preemptive. Each job j 2 J is associated with the normal processing time pj . We extend the general processing time function of [23] and construct a more general model of the actual job processing time
prj ¼ fj ðpj ; rÞ;
ð1 6 j 6 n; 1 6 r 6 nj Þ:
ð1Þ
Observing from Eq. (1), different jobs may have different actual processing time functions which are not restricted to be specific and monotone. During the maintenance, the machine is stopped. After the maintenance activity, the machine will revert to its initial condition. We assume that the machine need to be maintained one time when every k jobs are completed and maintenance is just finished at time zero. The jobs will be processed from a group continuously. Thus, we denote the schedule as r=[G1 ; M1 ; . . . ; Gi ; Mi ; . . . ; Mm ; Gmþ1 ], where Gi denotes the ith group and Mi denotes the ith maintenance. Let ni denote the number of jobs in the ith group, and m be the maintenance frequency, which is equal to the integer part of nk, i.e., m ¼ bnkc. P We can obtain that mþ1 i¼1 ni ¼ n. Let C ½l;r be the completion time of the job scheduled in rth position of the lth group. t Let Gi denote the time length of ith group, and ti be the duration of the maintenance, i.e., ti ¼ aGti þ b, where a and b are two constants, a P 0. As in Zhao and Tang [13], we denote the group Gi as Gi =[J ½i;1 ; J ½i;2 ,. . .,J ½i;ni 1 ; J ½i;ni , where J ½i;r is the rth job in the group Gi . Let p½l;r denote the normal processing time of job J ½l;r and pr½l;r be actual processing time of job J ½l;r , where pr½l;r =f½l;r (p½l;r ,r). Let aj and bj denote the lower limit and upper limit of the interval constraint of every job, respectively. Then, the earliness of job j is denoted as Ej , i.e., Ej =maxf0; aj prj g. The tardiness of job j is denoted as T j , i.e., T j =max{0; prj bj g. Let C max denote the makespan, i.e., C max =max{C j jj ¼ 1; 2; . . . ; n}, where C j denotes the completion time of the job j. The total earliness of all jobs is P P denoted by nj¼1 Ej , and the total tardiness of all jobs is denoted by nj¼1 T j . In the manufacturing process, the production fee is determined by the length of the working time. Moreover, the earliness cost and tardiness cost are assumed to be linear relationship with the total earliness and tardiness of all jobs, respectively. Thus, We define the total cost as follows: n n X X TC ¼ aC max þ b Ej þ c T j ; j¼1
j¼1
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where a; b and c are the unit production fee, the unit earliness cost and the unit tardiness cost, respectively. a; b and c should be positive numbers, i.e., a; b; c > 0. 3. The main results The completion time of all jobs can be obtained as follows: n
k mþ1 C ½mþ1;nmþ1 ¼ ða þ 1ÞRm l¼1 Rr¼1 f½l;r ðp½l;r ; rÞ þ Rr¼1 f½mþ1;r ðp½mþ1;r ; rÞ þ mb:
ð2Þ
From the Eq. (2), it can be obtained that n
k mþ1 C max ¼ C ½mþ1;nmþ1 ¼ ða þ 1ÞRm l¼1 Rr¼1 f½l;r ðp½l;r ; rÞ þ Rr¼1 f½mþ1;r ðp½mþ1;r ; rÞ þ mb:
ð3Þ
The total earliness is given by nmþ1 ni m k Rmþ1 l¼1 Rr¼1 E½l;r ¼ Rl¼1 Rr¼1 maxf0; a½l;r f½l;r ðp½l;r ; rÞg þ Rr¼1 maxf0; a½mþ1;r f½mþ1;r ðp½mþ1;r ; rÞg:
ð4Þ
Similarly, the total tardiness is given by nmþ1 ni m k Rmþ1 l¼1 Rr¼1 T ½l;r ¼ Rl¼1 Rr¼1 maxf0; f½l;r ðp½l;r ; rÞ b½l;r g þ Rr¼1 maxf0; f½mþ1;r ðp½mþ1;r ; rÞ b½mþ1;r g:
ð5Þ
Then, the total cost can be obtained as follows: n
k mþ1 TC ¼ aC max þ bRnj¼1 Ej þ cRnj¼1 T j ¼ aðRm l¼1 Rr¼1 ða þ 1Þf½l;r ðp½l;r ;rÞ þ mb þ Rr¼1 f½mþ1;r ðp½mþ1;r ;rÞÞ n
k mþ1 þ bðRm l¼1 Rr¼1 maxf0;a½l;r f½l;r ðp½l;r ;rÞg þ Rr¼1 maxf0;a½mþ1;r f½mþ1;r ðp½mþ1;r ;rÞgÞ n
k mþ1 þ cðRm l¼1 Rr¼1 maxf0;f½l;r ðp½l;r ;rÞ b½l;r g þ Rr¼1 maxf0;f½mþ1;r ðp½mþ1;r ;rÞ b½mþ1;r gÞ k ¼ Rm l¼1 Rr¼1 ðaða þ 1Þf½l;r ðp½l;r ;rÞ þ bmaxf0;a½l;r f½l;r ðp½l;r ;rÞg þ cmaxf0;f½l;r ðp½l;r ;rÞ b½l;r gÞ n
mþ1 þ Rr¼1 ðaf½mþ1;r ðp½mþ1;r ;rÞ þ bmaxf0;a½mþ1;r f½mþ1;r ðp½mþ1;r ;rÞg þ cmaxf0;f½mþ1;r ðp½mþ1;r ;rÞ b½mþ1;r gÞ þ amb:
ð6Þ
For simplicity, let b maxf0; a½l;r f½l;r ðp½l;r ; rÞg þ c maxf0; f½l;r ðp½l;r ; rÞ b½l;r g be denoted by Z. Then, we introduce a new ð1Þ function f½l;r ðp½l;r ; rÞ as follows:
(
ð1Þ f½l;r ðp½l;r ; rÞ
¼
aða þ 1Þf½l;r ðp½l;r ; rÞ þ Z; l ¼ 1; 2; . . . ; m; af½l;r ðp½l;r ; rÞ þ Z; l ¼ m þ 1:
ð7Þ
Combining (6) and (7), we can obtain ni mþ1X X ð1Þ f½l;r ðp½l;r ; rÞ þ amb:
TC ½mþ1;nmþ1 ¼
l¼1 r¼1
The problem we studied in this section can be denoted by 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; t i ¼ aGti þ bjTC. For a given the number of jobs k, we can get the number of groups by equation m ¼ bnkc. Then, we can obtain that mb is a constant. We explore to find a polynomial to solve the problem. The problem 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; ti ¼ aGti þ bjTC can be transformed as the following standard assignment problem:
min
ni n m þ1X X X wjlr xjlr þ amb;
st:
n X xjlr ¼ 1;
j¼1 l¼1 r¼1
l ¼ 1; 2; . . . ; m þ 1; r ¼ 1; 2; . . . ; ni ;
j¼1 ni mþ1 XX
xjlr ¼ 1;
ð8Þ
j ¼ 1; 2; . . . ; n;
l¼1 r¼1
xjlr ¼ 0 or 1;
j ¼ 1; 2; . . . ; n; l ¼ 1; 2; . . . ; m þ 1; r ¼ 1; 2; . . . ni ;
ð1Þ f½l;r ðp½l;r ; rÞ.
where wjlr ¼ We define xjlr ¼ 0 or 1 such that xjlr ¼ 1 if job J j is scheduled in the rth position in the group Gl and xjlr ¼ 0 otherwise. In case of k ¼ n, all jobs are completed in one group, and the machine does not require maintenance. Thus, P Pmþ1 Pni Pn Pmþ1 Pni the objective of the assignment problem is not nj¼1 l¼1 j¼1 l¼1 r¼1 wjlr xjlr þ amb, but r¼1 wjlr xjlr . In order to minimize the total cost, we propose a polynomial time algorithm to determine jointly the optimal k, and the optimal job sequence. Algorithm 1. Step 1. For each k (k=1,2,. . .,n-1,n), solve the assignment problem (8) and calculate the objective value TCðkÞ. Step 2. Let ðTCðkÞÞ =min ðTCðkÞ, (k=1,2,. . .,n)).
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Theorem 1. The 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; ti ¼ aGti þ bjTC problem can be optimal solved by Algorithm 1 in Oðn4 Þ time. Proof. For a fixed the number of jobs k, the problem 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; t i ¼ aGti þ bjTC can be optimally solved via the assignment problem (8), and the computational complexity of obtaining the optimal job sequence is Oðn3 Þ. Since k has n possible values, the computational complexity of solving the problem 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; t i ¼ aGti þ bjTC is Oðn4 Þ. h In what follows, we investigate two special cases of the 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; t i ¼ aGti þ bjTC problem, and explore to find a more efficient algorithm for each case. We assume that pr½l;r ¼ p½l;r r a0 ði:e:; prj ¼ pj ra0 Þ; t i ¼ bðb > 0Þ and a P b. The first case concerns the position-dependent learning effect, a0 < 0. In this case, we assume that every job has its own constrained interval and has no tardiness. Let pj b0 ð0 < b0 < 1Þ and pj denote the lower limit and upper limit of the constrained interval of the job j, respectively. The second case concerns the position-dependent aging effect, a0 > 0. In this case, we assume that every job has its own constrained interval and has no earliness. Let pj and pj b0 ðb0 > 1Þ denote the lower limit and upper limit of the constrained interval of the job j, respectively. We first consider the case of the position-dependent learning effect. It is obviously that machine does not need maintenance in the optimal scheduling. Then, we can denote the scheduling problem with learning effect as 1jprj ¼ pj r a0 jTC. Theorem 2. If a0 < 0, the problem 1jprj ¼ pj r a0 jTC can be optimally solved by the shortest processing time first (SPT) rule in Oðn log nÞ time. Proof. Assume that an optimal schedule r1 is not the SPT sequence. Then, there must exist a pair of adjacent jobs i and j so that pi P pj , with job j following job i. We assume that job i is scheduled in the rth position. Now let a new job schedule r2 be formed from r1 by interchanging jobs i and j and keeping the other jobs in the same position 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 j in both schedules, and B is the set of jobs following jobs i and j in both schedules. Let TCðr1 Þ and TCðr2 Þ denote the total cost of r1 and r2 , respectively. Since the positions of the jobs in A and B remain unchanged, the cost of processing and the earliness of the jobs in A and B remains unchanged. Since not all jobs in the sequence have earliness in case of learning effect, we aim to find the tipping position r , whose pffiffiffiffiffi illustration is proposed in Fig. 2. Let pi b0 ¼ pi ra0 , we can obtain r ¼ b a0 b0 c. In case of r 6 r , since pi b0 6 pi r a0 , it can be seen that the job before the position (r þ 1) has no earliness. While in case of r > r , since pi b0 > pi ra0 , it can be seen that the job behind the position r has earliness. In order to prove Theorem 2, we consider the following three cases: (1) r < r ; (2) r ¼ r ; (3) r > r . Case (1). r < r In sequence r1 and
r2 , the jobs i and j have no earliness. Then, it can be obtained
TCðr1 Þ TCðr2 Þ ¼ aðpi r a0 þ pj ðr þ 1Þa0 Þ aðpj r a0 þ pi ðr þ 1Þa0 Þ ¼ aðpi pj Þðra0 ðr þ 1Þa0 Þ P 0: Then, the schedule r1 is dominated by the schedule r2 in this case. Case (2). r ¼ r In 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 can be obtained
TCðr1 Þ TCðr2 Þ ¼ ðaðpi r a0 þ pj ðr þ 1Þa0 Þ þ bðpj b0 pj ðr þ 1Þa0 ÞÞ ðaðpj r a0 þ pi ðr þ 1Þa0 Þ þ bðpi b0 pi ðr þ 1Þa0 ÞÞ ¼ ðpi pj Þðaðr a0 ðr þ 1Þa0 Þ bðb0 ðr þ 1Þa0 ÞÞ P 0: Then, the schedule
r1 is dominated by the schedule r2 in this case.
Fig. 1. The illustration of the exchange of jobs Ji and J j .
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Fig. 2. The illustration of the tipping position r .
Case (3). r > r In sequence r1 and
r2 , the jobs i and j have earliness. Then, it can be obtained
TCðr1 Þ TCðr2 Þ ¼ ðaðpi r a0 þ pj ðr þ 1Þa0 Þ þ bðpi b0 pi r a0 Þ þ bðpj b0 pj ðr þ 1Þa0 ÞÞ ðaðpj r a0 þ pi ðr þ 1Þa0 Þ þ bðpj b0 pj r a0 Þ þ bðpi b0 pi ðr þ 1Þa0 ÞÞ ¼ ðra0 ðr þ 1Þa0 Þðpi pj Þða bÞ P 0: Then, the schedule r1 is dominated by the schedule r2 in this case. Summing the above argument, in any case, we obtain that the interchange of jobs i and j will result in a decrease in total cost. By the repeated application of this argument, any sequence that is not a SPT sequence can be improved with respect to the total cost. Then, the problem 1jprj ¼ pj ra0 jTC can be optimally solved by the SPT rule in Oðn log nÞ time. h In what follows, we discuss the case of the position-dependent aging effect. It is necessary to perform maintenance for the machine in the optimal schedule of the 1jprj ¼ pj r a0 ; PRM ¼ k; ti ¼ bjTC problem. We present a modified Longest Processing Time first (LPT) rule for the 1jprj ¼ pj r a0 ; PRM ¼ k; ti ¼ bjTC problem. The modified LPT rule. At first, we sequence the jobs in non-increasing order of their normal processing times. Then, schedule the job in the first position of each group one by one. If the first position of each group is filled, then schedule the remaining job in the second position of each group one by one. If all the second positions are filled, fill the third position, and so on, until all jobs are scheduled. Theorem 3. If a0 > 0, the problem 1jprj ¼ pj ra0 ; PRM ¼ k; ti ¼ bjTC can be optimally solved by the modified LPT rule in Oðn2 lognÞ time. Proof. Assume that an optimal schedule r3 is not sequenced by the modified LPT rule. We consider jobs J ½i;r and J ½j;rþ1 . Job J ½i;r is sequenced in rth position of the group Gi . Job J ½j;rþ1 is sequenced in ðr þ 1Þth position of the group Gj . p½i;r and p½j;rþ1 denote the normal processing times of jobs J ½i;r and J ½j;rþ1 , respectively. Moreover, assume that p½i;r 6 p½j;rþ1 . Now let a new job schedule r4 be formed from r3 by interchanging jobs J ½i;r and J ½j;rþ1 and keeping the other jobs in the same position as in r3 . The exchange of the jobs J ½i;r and J ½j;rþ1 is illustrated by Fig. 3, where p1 ; p2 and p3 are partial schedule of the schedule r3 . Let TCðr3 Þ and TCðr4 Þ denote the total cost of r3 and r4 , respectively. For the reason that the positions of the jobs in p1 ; p2 and p3 unchanged, the cost of processing and the tardiness of every job in p1 ; p2 and p3 remains unchanged. Since not all jobs in the sequence have tardiness in case of aging effect, so we aim to find the tipping position r , whose pffiffiffiffiffi a illustration is proposed in Fig. 4. Let pi b0 ¼ pi r a0 , we can obtain r ¼ b 0 b0 c. In case of r 6 r , since pi b0 P pi r a0 , the job before the position (r þ 1) has no tardiness. While in case of r > r , since pi b0 < pi ra0 , the job behind the position r has tardiness. In order to prove Theorem 3, we consider the following three cases: (1) r < r ; (2) r ¼ r ; (3) r > r .
Fig. 3. The illustration of the exchange of jobs J½i;r and J½j;rþ1 .
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Fig. 4. The illustration of the tipping position r .
case (1). r < r In sequence
r3 and r4 , the jobs J½i;r and J½j;rþ1 have no tardiness. Then, it can be obtained
TCðr3 Þ TCðr4 Þ ¼ aðp½i;r r a0 þ p½j;rþ1 ðr þ 1Þa0 Þ aðp½j;rþ1 r a0 þ p½i;r ðr þ 1Þa0 Þ ¼ aðp½i;r p½j;rþ1 Þðr a0 ðr þ 1Þa0 Þ P 0: Then, the schedule case (2). r ¼ r
r3 is dominated by the schedule r4 .
In sequence r3 , the job J ½i;r has no tardiness and the job J ½j;rþ1 has tardiness, but in the sequence r4 the case is converse. Then, it can be obtained
TCðr3 Þ TCðr4 Þ ¼ ðaðp½i;r ra0 þ p½j;rþ1 ðr þ 1Þa0 Þ þ cðp½j;rþ1 ðr þ 1Þa0 p½j;rþ1 b0 ÞÞ ðaðp½j;rþ1 ra0 þ p½i;r ðr þ 1Þa0 Þ þ cðp½i;r ðr þ 1Þa0 p½i;r b0 ÞÞ ¼ ðp½i;r p½j;rþ1 Þðaðr a0 ðr þ 1Þa0 Þ þ rðb0 ðr þ 1Þa0 ÞÞ P 0: Then, the schedule case (3). r > r In sequence
r4 dominates the schedule r3 in this case.
r3 and r4 , the jobs J½i;r and J½j;rþ1 have tardiness. Then, it can be obtained
TCðr3 Þ TCðr4 Þ ¼ ðaðp½i;r ra0 þ p½j;rþ1 ðr þ 1Þa0 Þ þ cðp½i;r r a0 p½i;r b0 Þ þ cðp½j;rþ1 ðr þ 1Þa0 p½j;rþ1 b0 ÞÞ ðaðp½j;rþ1 r a0 þ p½i;r ðr þ 1Þa0 Þ þ cðp½j;rþ1 ra0 p½j;rþ1 b0 Þ þ cðp½i;r ðr þ 1Þa0 p½i;r b0 ÞÞ ¼ ðra0 ðr þ 1Þa0 Þðp½i;r p½j;rþ1 Þða þ cÞ P 0: Then, the schedule r3 is dominated by the schedule r4 in this case. Based on the above argument, we obtain that the interchange of jobs J ½i;r and J ½j;rþ1 will result in a decrease in total cost. By the repeated application of this argument, any sequence that is not a modified LPT sequence can be improved with respect to the total cost. For a fixed k, the 1jprj ¼ pj r a0 ; PRM ¼ k; ti ¼ bjTC problem can be optimally solved by the modified LPT rule, and the computational complexity is Oðn log nÞ. Since kðk ¼ 1; 2; . . . ; nÞ has n possible values, the total computational complexity of solving the problem 1jprj ¼ pj ra0 ; PRM ¼ k; t i ¼ bjTC is Oðn2 lognÞ. h 4. Conclusions The paper considered the scheduling problem with piece-rate machine maintenance and interval constrained actual processing time. The job actual processing time is required to restrict in given interval otherwise earliness or tardiness penalty should be paid. The objective is to minimize the total cost that is a linear function of the makespan, the earliness and tardiness penalties of all jobs. For scheduling problem with general position-dependent processing times, we showed that the total cost minimization problem can be optimally solved in Oðn4 Þ time. Moreover, for the two special cases, we showed that the total cost minimization problem with learning effect can be solved by SPT rule in Oðn log nÞ time, and the total cost minimization problem with aging effect can be solved by the modified LPT rule in Oðn2 lognÞ time. For the future research, it is suggested to investigate the scheduling problem with piece-rate machine maintenance and interval constrained actual processing time in the context of parallel machines scheduling problems or job-shop scheduling problems. Acknowledgements The authors wish to thank editors and anonymous referees for their helpful comments. This work is supported by National Nature Science Foundation Project of China (71171046) and the Scientific Research Foundation of Graduate School of Southeast University (YBJJ1239). References [1] A. Bachman, T.C.E. Cheng, J. Janiak, C.T. Ng, Scheduling start time dependent jobs to minimize the total weighted completion time, Journal of the Operational Research Society 53 (2002) 688–693.
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Single-machine scheduling with piece-rate maintenance and interval constrained position-dependent processing times Pengfei Xue a, Yulin Zhang a,⇑, Xianyu Yu a,b a b
School of Economics and Management, Southeast University, Nanjing 210096, China School of Science, East China Institute of Technology, Fuzhou, Jiangxi 344000, China
a r t i c l e
i n f o
Keywords: Single-machine Scheduling Position-dependent Maintenance Total cost
a b s t r a c t This paper investigates single-machine scheduling problems with piece-rate machine maintenance and interval constrained actual processing time. The actual processing time of a job is a general function of the normal job processing time and the position in job sequence, and it is required to restrict in given interval otherwise earliness or tardiness penalty should be paid. The maintenance duration studied in the paper is a time-dependent linear function. The objective is to find jointly the optimal frequency to perform maintenances and the optimal job sequence to minimize the total cost, which is a linear function of the makespan, total earliness and total tardiness. There is shown that the problem can be optimally solved in O(n4) time. There is also shown that two special cases of the problem can be optimally solved by lower order algorithms. Ó 2013 Elsevier Inc. All rights reserved.
1. Introduction Recent developments in the field of job scheduling have triggered into a growing interest towards learning or aging effect. In case of the learning effect, the actual processing time of a job will be shorter when it is scheduled later in a sequence. While in case of the aging effect, the actual processing time of a job will be longer when it is scheduled later in a sequence. Some 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 scheduling problems with position-dependent effect, the position-dependent function as specific linear or exponential function was considered by Cheng and Wang [6], Bachman and Janiak [7], Wang and Xia [8], Biskup [9], Kuo and Yang [10], Yang, Yang and Cheng [11], Yang and Yang [12] and Zhao and Tang [13]. Some relevant references for the combination of the exponential function and linear function are: Lee [14], Cheng, Wu and Lee [15], Wang [16], Wang and Cheng [17], Cheng and Lee [18], Yin et al. [19] and Lai and Lee [20]. However, all these papers consider learning or aging model with specific processing time function. Only a few papers study general position-dependent processing times. Mosheiov and Sidney [21] and Wang and Wang [22] considered a scheduling problem with a general position-dependent learning function. Mosheiov [23] studied a scheduling problem with position-dependent job processing times which is not restricted to a monotone function. In addition, it is well-known that the production efficiency can be improved by performing maintenance in manufacturing processes. Some scheduling researchers considered that the machine is maintained exactly once during the planning period, and the duration of maintenance is not fixed. Some recent relevant references are: Kubzin and Strusevich [24], Mosheiov and Sidney [25], Cheng et al. [26], Mor and Mosheiov [27] and Cheng et al. [28]. Some scheduling researchers assumed that the duration of each maintenance is given. Liao and Chen [29] and Ji et al. [30] considered single-machine scheduling ⇑ Corresponding author. E-mail address: [email protected] (Y.L. Zhang). 0096-3003/$ - see front matter Ó 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2013.10.034
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problems with periodic maintenance activities, where each maintenance activity is required after a periodic time interval. Ji and Cheng [31] and Yang et al. [32] assumed that the machine may have multiple rate-modifying activities over the scheduling horizon. Another popular topic in recent years is scheduling problem with job completion time due window. To cope with global competition and to improve customer demand, jobs should be finished as close as possible to their due-dates. If a job is finished earlier than its due-date, it would result in storage costs; on the other hand, if a job is finished later than its due-date, it would violate contractual obligation with the customer. Cheng [33] initiated research on scheduling with due windows. In modern manufacturing management, scheduling problems with due window assignment are important issues. Many papers have been conducted on these issues in different scheduling environments. Yin et al. [34] studied a single-machine scheduling problem with batch delivery cost and an assignable common due window simultaneously. They provided polynomialtime solutions for considered problems. Yeung et al. [35] studied a non-preemptive two-stage flowshop scheduling problem under the environment of a common due window. The objective is to minimize the earliness and tardiness. They showed that the problem is NP-complete in the strong sense, and developed a branch and bound algorithm and a heuristic to solve the problem. For more recent works on scheduling with a common due window, we refer the reader to the comprehensive surveys by Cheng and Gupta [36], Gordon et al. [37], and the recent papers by Yeung et al. [38], Cheng [39], Baker and Scudder [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 is required to restrict in given interval otherwise earliness or tardiness penalty should be paid. For example, in porcelain manufacturing processes, the actual processing time can not exceed a given interval otherwise the porcelain may have quality flaws. To the best of our knowledge, however, scheduling with simultaneous consideration of interval constrained actual processing time and machine maintenance has not been explored. Motivated by these points, in this paper we consider the single machine scheduling with interval constrained actual processing time and piece-rate maintenance, where the piece-rate maintenance is proposed by Yu et al. [49]. We explore to find the optimal polynomial algorithm to minimize the total cost, which is assumed to conclude production fee and total earliness and tardiness cost. Moreover, we investigate two special cases where the actual processing time of each job is assumed to be a product function, and explore to find more effective solving algorithm to minimize the total cost for the cases. This paper is organized as follows. We introduce the notation and terminology of this paper in the next section. In Section 3, we propose the main results of this paper. In Section 4, we conclude this paper, and suggest some topics for future. 2. Notations and problem formulation A set J ¼ ½J 1 ; J 2 ; . . . ; J n of n independent jobs are partitioned into m þ 1 groups and are all available for processing at time zero. The machine can handle one job at a time. In the manufacturing process, the jobs are non-preemptive. Each job j 2 J is associated with the normal processing time pj . We extend the general processing time function of [23] and construct a more general model of the actual job processing time
prj ¼ fj ðpj ; rÞ;
ð1 6 j 6 n; 1 6 r 6 nj Þ:
ð1Þ
Observing from Eq. (1), different jobs may have different actual processing time functions which are not restricted to be specific and monotone. During the maintenance, the machine is stopped. After the maintenance activity, the machine will revert to its initial condition. We assume that the machine need to be maintained one time when every k jobs are completed and maintenance is just finished at time zero. The jobs will be processed from a group continuously. Thus, we denote the schedule as r=[G1 ; M1 ; . . . ; Gi ; Mi ; . . . ; Mm ; Gmþ1 ], where Gi denotes the ith group and Mi denotes the ith maintenance. Let ni denote the number of jobs in the ith group, and m be the maintenance frequency, which is equal to the integer part of nk, i.e., m ¼ bnkc. P We can obtain that mþ1 i¼1 ni ¼ n. Let C ½l;r be the completion time of the job scheduled in rth position of the lth group. t Let Gi denote the time length of ith group, and ti be the duration of the maintenance, i.e., ti ¼ aGti þ b, where a and b are two constants, a P 0. As in Zhao and Tang [13], we denote the group Gi as Gi =[J ½i;1 ; J ½i;2 ,. . .,J ½i;ni 1 ; J ½i;ni , where J ½i;r is the rth job in the group Gi . Let p½l;r denote the normal processing time of job J ½l;r and pr½l;r be actual processing time of job J ½l;r , where pr½l;r =f½l;r (p½l;r ,r). Let aj and bj denote the lower limit and upper limit of the interval constraint of every job, respectively. Then, the earliness of job j is denoted as Ej , i.e., Ej =maxf0; aj prj g. The tardiness of job j is denoted as T j , i.e., T j =max{0; prj bj g. Let C max denote the makespan, i.e., C max =max{C j jj ¼ 1; 2; . . . ; n}, where C j denotes the completion time of the job j. The total earliness of all jobs is P P denoted by nj¼1 Ej , and the total tardiness of all jobs is denoted by nj¼1 T j . In the manufacturing process, the production fee is determined by the length of the working time. Moreover, the earliness cost and tardiness cost are assumed to be linear relationship with the total earliness and tardiness of all jobs, respectively. Thus, We define the total cost as follows: n n X X TC ¼ aC max þ b Ej þ c T j ; j¼1
j¼1
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where a; b and c are the unit production fee, the unit earliness cost and the unit tardiness cost, respectively. a; b and c should be positive numbers, i.e., a; b; c > 0. 3. The main results The completion time of all jobs can be obtained as follows: n
k mþ1 C ½mþ1;nmþ1 ¼ ða þ 1ÞRm l¼1 Rr¼1 f½l;r ðp½l;r ; rÞ þ Rr¼1 f½mþ1;r ðp½mþ1;r ; rÞ þ mb:
ð2Þ
From the Eq. (2), it can be obtained that n
k mþ1 C max ¼ C ½mþ1;nmþ1 ¼ ða þ 1ÞRm l¼1 Rr¼1 f½l;r ðp½l;r ; rÞ þ Rr¼1 f½mþ1;r ðp½mþ1;r ; rÞ þ mb:
ð3Þ
The total earliness is given by nmþ1 ni m k Rmþ1 l¼1 Rr¼1 E½l;r ¼ Rl¼1 Rr¼1 maxf0; a½l;r f½l;r ðp½l;r ; rÞg þ Rr¼1 maxf0; a½mþ1;r f½mþ1;r ðp½mþ1;r ; rÞg:
ð4Þ
Similarly, the total tardiness is given by nmþ1 ni m k Rmþ1 l¼1 Rr¼1 T ½l;r ¼ Rl¼1 Rr¼1 maxf0; f½l;r ðp½l;r ; rÞ b½l;r g þ Rr¼1 maxf0; f½mþ1;r ðp½mþ1;r ; rÞ b½mþ1;r g:
ð5Þ
Then, the total cost can be obtained as follows: n
k mþ1 TC ¼ aC max þ bRnj¼1 Ej þ cRnj¼1 T j ¼ aðRm l¼1 Rr¼1 ða þ 1Þf½l;r ðp½l;r ;rÞ þ mb þ Rr¼1 f½mþ1;r ðp½mþ1;r ;rÞÞ n
k mþ1 þ bðRm l¼1 Rr¼1 maxf0;a½l;r f½l;r ðp½l;r ;rÞg þ Rr¼1 maxf0;a½mþ1;r f½mþ1;r ðp½mþ1;r ;rÞgÞ n
k mþ1 þ cðRm l¼1 Rr¼1 maxf0;f½l;r ðp½l;r ;rÞ b½l;r g þ Rr¼1 maxf0;f½mþ1;r ðp½mþ1;r ;rÞ b½mþ1;r gÞ k ¼ Rm l¼1 Rr¼1 ðaða þ 1Þf½l;r ðp½l;r ;rÞ þ bmaxf0;a½l;r f½l;r ðp½l;r ;rÞg þ cmaxf0;f½l;r ðp½l;r ;rÞ b½l;r gÞ n
mþ1 þ Rr¼1 ðaf½mþ1;r ðp½mþ1;r ;rÞ þ bmaxf0;a½mþ1;r f½mþ1;r ðp½mþ1;r ;rÞg þ cmaxf0;f½mþ1;r ðp½mþ1;r ;rÞ b½mþ1;r gÞ þ amb:
ð6Þ
For simplicity, let b maxf0; a½l;r f½l;r ðp½l;r ; rÞg þ c maxf0; f½l;r ðp½l;r ; rÞ b½l;r g be denoted by Z. Then, we introduce a new ð1Þ function f½l;r ðp½l;r ; rÞ as follows:
(
ð1Þ f½l;r ðp½l;r ; rÞ
¼
aða þ 1Þf½l;r ðp½l;r ; rÞ þ Z; l ¼ 1; 2; . . . ; m; af½l;r ðp½l;r ; rÞ þ Z; l ¼ m þ 1:
ð7Þ
Combining (6) and (7), we can obtain ni mþ1X X ð1Þ f½l;r ðp½l;r ; rÞ þ amb:
TC ½mþ1;nmþ1 ¼
l¼1 r¼1
The problem we studied in this section can be denoted by 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; t i ¼ aGti þ bjTC. For a given the number of jobs k, we can get the number of groups by equation m ¼ bnkc. Then, we can obtain that mb is a constant. We explore to find a polynomial to solve the problem. The problem 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; ti ¼ aGti þ bjTC can be transformed as the following standard assignment problem:
min
ni n m þ1X X X wjlr xjlr þ amb;
st:
n X xjlr ¼ 1;
j¼1 l¼1 r¼1
l ¼ 1; 2; . . . ; m þ 1; r ¼ 1; 2; . . . ; ni ;
j¼1 ni mþ1 XX
xjlr ¼ 1;
ð8Þ
j ¼ 1; 2; . . . ; n;
l¼1 r¼1
xjlr ¼ 0 or 1;
j ¼ 1; 2; . . . ; n; l ¼ 1; 2; . . . ; m þ 1; r ¼ 1; 2; . . . ni ;
ð1Þ f½l;r ðp½l;r ; rÞ.
where wjlr ¼ We define xjlr ¼ 0 or 1 such that xjlr ¼ 1 if job J j is scheduled in the rth position in the group Gl and xjlr ¼ 0 otherwise. In case of k ¼ n, all jobs are completed in one group, and the machine does not require maintenance. Thus, P Pmþ1 Pni Pn Pmþ1 Pni the objective of the assignment problem is not nj¼1 l¼1 j¼1 l¼1 r¼1 wjlr xjlr þ amb, but r¼1 wjlr xjlr . In order to minimize the total cost, we propose a polynomial time algorithm to determine jointly the optimal k, and the optimal job sequence. Algorithm 1. Step 1. For each k (k=1,2,. . .,n-1,n), solve the assignment problem (8) and calculate the objective value TCðkÞ. Step 2. Let ðTCðkÞÞ =min ðTCðkÞ, (k=1,2,. . .,n)).
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Theorem 1. The 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; ti ¼ aGti þ bjTC problem can be optimal solved by Algorithm 1 in Oðn4 Þ time. Proof. For a fixed the number of jobs k, the problem 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; t i ¼ aGti þ bjTC can be optimally solved via the assignment problem (8), and the computational complexity of obtaining the optimal job sequence is Oðn3 Þ. Since k has n possible values, the computational complexity of solving the problem 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; t i ¼ aGti þ bjTC is Oðn4 Þ. h In what follows, we investigate two special cases of the 1jprj ¼ fj ðpj ; rÞ; PRM ¼ k; t i ¼ aGti þ bjTC problem, and explore to find a more efficient algorithm for each case. We assume that pr½l;r ¼ p½l;r r a0 ði:e:; prj ¼ pj ra0 Þ; t i ¼ bðb > 0Þ and a P b. The first case concerns the position-dependent learning effect, a0 < 0. In this case, we assume that every job has its own constrained interval and has no tardiness. Let pj b0 ð0 < b0 < 1Þ and pj denote the lower limit and upper limit of the constrained interval of the job j, respectively. The second case concerns the position-dependent aging effect, a0 > 0. In this case, we assume that every job has its own constrained interval and has no earliness. Let pj and pj b0 ðb0 > 1Þ denote the lower limit and upper limit of the constrained interval of the job j, respectively. We first consider the case of the position-dependent learning effect. It is obviously that machine does not need maintenance in the optimal scheduling. Then, we can denote the scheduling problem with learning effect as 1jprj ¼ pj r a0 jTC. Theorem 2. If a0 < 0, the problem 1jprj ¼ pj r a0 jTC can be optimally solved by the shortest processing time first (SPT) rule in Oðn log nÞ time. Proof. Assume that an optimal schedule r1 is not the SPT sequence. Then, there must exist a pair of adjacent jobs i and j so that pi P pj , with job j following job i. We assume that job i is scheduled in the rth position. Now let a new job schedule r2 be formed from r1 by interchanging jobs i and j and keeping the other jobs in the same position 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 j in both schedules, and B is the set of jobs following jobs i and j in both schedules. Let TCðr1 Þ and TCðr2 Þ denote the total cost of r1 and r2 , respectively. Since the positions of the jobs in A and B remain unchanged, the cost of processing and the earliness of the jobs in A and B remains unchanged. Since not all jobs in the sequence have earliness in case of learning effect, we aim to find the tipping position r , whose pffiffiffiffiffi illustration is proposed in Fig. 2. Let pi b0 ¼ pi ra0 , we can obtain r ¼ b a0 b0 c. In case of r 6 r , since pi b0 6 pi r a0 , it can be seen that the job before the position (r þ 1) has no earliness. While in case of r > r , since pi b0 > pi ra0 , it can be seen that the job behind the position r has earliness. In order to prove Theorem 2, we consider the following three cases: (1) r < r ; (2) r ¼ r ; (3) r > r . Case (1). r < r In sequence r1 and
r2 , the jobs i and j have no earliness. Then, it can be obtained
TCðr1 Þ TCðr2 Þ ¼ aðpi r a0 þ pj ðr þ 1Þa0 Þ aðpj r a0 þ pi ðr þ 1Þa0 Þ ¼ aðpi pj Þðra0 ðr þ 1Þa0 Þ P 0: Then, the schedule r1 is dominated by the schedule r2 in this case. Case (2). r ¼ r In 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 can be obtained
TCðr1 Þ TCðr2 Þ ¼ ðaðpi r a0 þ pj ðr þ 1Þa0 Þ þ bðpj b0 pj ðr þ 1Þa0 ÞÞ ðaðpj r a0 þ pi ðr þ 1Þa0 Þ þ bðpi b0 pi ðr þ 1Þa0 ÞÞ ¼ ðpi pj Þðaðr a0 ðr þ 1Þa0 Þ bðb0 ðr þ 1Þa0 ÞÞ P 0: Then, the schedule
r1 is dominated by the schedule r2 in this case.
Fig. 1. The illustration of the exchange of jobs Ji and J j .
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Fig. 2. The illustration of the tipping position r .
Case (3). r > r In sequence r1 and
r2 , the jobs i and j have earliness. Then, it can be obtained
TCðr1 Þ TCðr2 Þ ¼ ðaðpi r a0 þ pj ðr þ 1Þa0 Þ þ bðpi b0 pi r a0 Þ þ bðpj b0 pj ðr þ 1Þa0 ÞÞ ðaðpj r a0 þ pi ðr þ 1Þa0 Þ þ bðpj b0 pj r a0 Þ þ bðpi b0 pi ðr þ 1Þa0 ÞÞ ¼ ðra0 ðr þ 1Þa0 Þðpi pj Þða bÞ P 0: Then, the schedule r1 is dominated by the schedule r2 in this case. Summing the above argument, in any case, we obtain that the interchange of jobs i and j will result in a decrease in total cost. By the repeated application of this argument, any sequence that is not a SPT sequence can be improved with respect to the total cost. Then, the problem 1jprj ¼ pj ra0 jTC can be optimally solved by the SPT rule in Oðn log nÞ time. h In what follows, we discuss the case of the position-dependent aging effect. It is necessary to perform maintenance for the machine in the optimal schedule of the 1jprj ¼ pj r a0 ; PRM ¼ k; ti ¼ bjTC problem. We present a modified Longest Processing Time first (LPT) rule for the 1jprj ¼ pj r a0 ; PRM ¼ k; ti ¼ bjTC problem. The modified LPT rule. At first, we sequence the jobs in non-increasing order of their normal processing times. Then, schedule the job in the first position of each group one by one. If the first position of each group is filled, then schedule the remaining job in the second position of each group one by one. If all the second positions are filled, fill the third position, and so on, until all jobs are scheduled. Theorem 3. If a0 > 0, the problem 1jprj ¼ pj ra0 ; PRM ¼ k; ti ¼ bjTC can be optimally solved by the modified LPT rule in Oðn2 lognÞ time. Proof. Assume that an optimal schedule r3 is not sequenced by the modified LPT rule. We consider jobs J ½i;r and J ½j;rþ1 . Job J ½i;r is sequenced in rth position of the group Gi . Job J ½j;rþ1 is sequenced in ðr þ 1Þth position of the group Gj . p½i;r and p½j;rþ1 denote the normal processing times of jobs J ½i;r and J ½j;rþ1 , respectively. Moreover, assume that p½i;r 6 p½j;rþ1 . Now let a new job schedule r4 be formed from r3 by interchanging jobs J ½i;r and J ½j;rþ1 and keeping the other jobs in the same position as in r3 . The exchange of the jobs J ½i;r and J ½j;rþ1 is illustrated by Fig. 3, where p1 ; p2 and p3 are partial schedule of the schedule r3 . Let TCðr3 Þ and TCðr4 Þ denote the total cost of r3 and r4 , respectively. For the reason that the positions of the jobs in p1 ; p2 and p3 unchanged, the cost of processing and the tardiness of every job in p1 ; p2 and p3 remains unchanged. Since not all jobs in the sequence have tardiness in case of aging effect, so we aim to find the tipping position r , whose pffiffiffiffiffi a illustration is proposed in Fig. 4. Let pi b0 ¼ pi r a0 , we can obtain r ¼ b 0 b0 c. In case of r 6 r , since pi b0 P pi r a0 , the job before the position (r þ 1) has no tardiness. While in case of r > r , since pi b0 < pi ra0 , the job behind the position r has tardiness. In order to prove Theorem 3, we consider the following three cases: (1) r < r ; (2) r ¼ r ; (3) r > r .
Fig. 3. The illustration of the exchange of jobs J½i;r and J½j;rþ1 .
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Fig. 4. The illustration of the tipping position r .
case (1). r < r In sequence
r3 and r4 , the jobs J½i;r and J½j;rþ1 have no tardiness. Then, it can be obtained
TCðr3 Þ TCðr4 Þ ¼ aðp½i;r r a0 þ p½j;rþ1 ðr þ 1Þa0 Þ aðp½j;rþ1 r a0 þ p½i;r ðr þ 1Þa0 Þ ¼ aðp½i;r p½j;rþ1 Þðr a0 ðr þ 1Þa0 Þ P 0: Then, the schedule case (2). r ¼ r
r3 is dominated by the schedule r4 .
In sequence r3 , the job J ½i;r has no tardiness and the job J ½j;rþ1 has tardiness, but in the sequence r4 the case is converse. Then, it can be obtained
TCðr3 Þ TCðr4 Þ ¼ ðaðp½i;r ra0 þ p½j;rþ1 ðr þ 1Þa0 Þ þ cðp½j;rþ1 ðr þ 1Þa0 p½j;rþ1 b0 ÞÞ ðaðp½j;rþ1 ra0 þ p½i;r ðr þ 1Þa0 Þ þ cðp½i;r ðr þ 1Þa0 p½i;r b0 ÞÞ ¼ ðp½i;r p½j;rþ1 Þðaðr a0 ðr þ 1Þa0 Þ þ rðb0 ðr þ 1Þa0 ÞÞ P 0: Then, the schedule case (3). r > r In sequence
r4 dominates the schedule r3 in this case.
r3 and r4 , the jobs J½i;r and J½j;rþ1 have tardiness. Then, it can be obtained
TCðr3 Þ TCðr4 Þ ¼ ðaðp½i;r ra0 þ p½j;rþ1 ðr þ 1Þa0 Þ þ cðp½i;r r a0 p½i;r b0 Þ þ cðp½j;rþ1 ðr þ 1Þa0 p½j;rþ1 b0 ÞÞ ðaðp½j;rþ1 r a0 þ p½i;r ðr þ 1Þa0 Þ þ cðp½j;rþ1 ra0 p½j;rþ1 b0 Þ þ cðp½i;r ðr þ 1Þa0 p½i;r b0 ÞÞ ¼ ðra0 ðr þ 1Þa0 Þðp½i;r p½j;rþ1 Þða þ cÞ P 0: Then, the schedule r3 is dominated by the schedule r4 in this case. Based on the above argument, we obtain that the interchange of jobs J ½i;r and J ½j;rþ1 will result in a decrease in total cost. By the repeated application of this argument, any sequence that is not a modified LPT sequence can be improved with respect to the total cost. For a fixed k, the 1jprj ¼ pj r a0 ; PRM ¼ k; ti ¼ bjTC problem can be optimally solved by the modified LPT rule, and the computational complexity is Oðn log nÞ. Since kðk ¼ 1; 2; . . . ; nÞ has n possible values, the total computational complexity of solving the problem 1jprj ¼ pj ra0 ; PRM ¼ k; t i ¼ bjTC is Oðn2 lognÞ. h 4. Conclusions The paper considered the scheduling problem with piece-rate machine maintenance and interval constrained actual processing time. The job actual processing time is required to restrict in given interval otherwise earliness or tardiness penalty should be paid. The objective is to minimize the total cost that is a linear function of the makespan, the earliness and tardiness penalties of all jobs. For scheduling problem with general position-dependent processing times, we showed that the total cost minimization problem can be optimally solved in Oðn4 Þ time. Moreover, for the two special cases, we showed that the total cost minimization problem with learning effect can be solved by SPT rule in Oðn log nÞ time, and the total cost minimization problem with aging effect can be solved by the modified LPT rule in Oðn2 lognÞ time. For the future research, it is suggested to investigate the scheduling problem with piece-rate machine maintenance and interval constrained actual processing time in the context of parallel machines scheduling problems or job-shop scheduling problems. Acknowledgements The authors wish to thank editors and anonymous referees for their helpful comments. This work is supported by National Nature Science Foundation Project of China (71171046) and the Scientific Research Foundation of Graduate School of Southeast University (YBJJ1239). References [1] A. Bachman, T.C.E. Cheng, J. Janiak, C.T. Ng, Scheduling start time dependent jobs to minimize the total weighted completion time, Journal of the Operational Research Society 53 (2002) 688–693.
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