Computers & Industrial Engineering 61 (2011) 179–183
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Some unrelated parallel machine scheduling problems with past-sequence-dependent setup time and learning effects q Wen-Hung Kuo a,⇑, Chou-Jung Hsu b, Dar-Li Yang a a b
Department of Information Management, National Formosa University, Yun-Lin 632, Taiwan Department of Industrial Engineering and Management, Nan Kai University of Technology, Nan-Tou 542, Taiwan
a r t i c l e
i n f o
Article history: Received 4 February 2010 Received in revised form 2 March 2011 Accepted 14 March 2011 Available online 21 March 2011 Keywords: Past-sequence-dependent Learning effect Unrelated parallel machine Scheduling Total absolute deviation of job completion times Total load
a b s t r a c t In this paper, we study an unrelated parallel machine scheduling problem with setup time and learning effects simultaneously. The setup time is proportional to the length of the already processed jobs. That is, the setup time of each job is past-sequence-dependent. The objectives are to minimize the total absolute deviation of job completion times and the total load on all machines, respectively. We show that the proposed problem is polynomially solvable. We also discuss two special cases of the problem and show that they can be optimally solved by lower order algorithms. 2011 Elsevier Ltd. All rights reserved.
1. Introduction The setup time in classical scheduling problems is considered either sequence independent or sequence dependent (see Allahverdi, Gupta, & Aldowaisan, 1999). In the first case, the setup time is usually added to the job processing time while in the second case the setup time depends not only on the job currently being scheduled but also on the last scheduled job. Browne and Yechiali (1990) were among the first to introduce the latter type of setup time. Koulamas and Kyparisis (2008) further named it the past-sequence-dependent (p-s-d) setup time, which happens in high-tech testing environment, into a single-machine scheduling problem. They showed that the problem with the objectives of minimizing the makespan, the total completion time (TC), the total absolute deviation of job completion times (TADC) and a linear combination of TC and TADC can be optimally solved in polynomial time. Biskup and Herrmann (2008) considered single-machine scheduling problems with past-sequence-dependent setup times and due dates. They showed that the problems of minimizing total lateness, total tardiness (with agreeable due dates), maximum lateness (with agreeable due dates), maximum tardiness (with agreeable due dates) and the problem of Kanet (1981) remain polynomially solvable. For other problems, namely, minimizing the number of q
This manuscript was handled by area editor Hans Kellerer.
⇑ Corresponding author. Tel.: +886 5 631 5733; fax: +886 5 636 4127. E-mail address:
[email protected] (W.-H. Kuo). 0360-8352/$ - see front matter 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.cie.2011.03.008
tardy jobs, maximum lateness, and maximum tardiness, the wellknown solution procedures that find the optimal solution are no longer valid. On the other hand, the topic of learning effects is very popular in scheduling problems recently. The reader may refer to the following survey papers: Alidaee and Womer (1999), Cheng, Ding, and Lin (2004) and Biskup (2008). For more recent studies, Wu and Lee (2009) considered a general learning effect model in both single-machine and flowshop scheduling problems. Janiak, Janiak, Rudek, and Wielgus (2009) proposed some algorithms for the makespan minimization scheduling problem with a general learning model. Okolowski and Gawiejnowicz (2010) studied a parallel machine scheduling problem with DeJong’s learning effect. Cheng, Lee, and Wu (2010) investigated a two-agent single-machine scheduling problem with a truncated learning effect model. Wu, Yin, and Cheng (2011) introduced a truncation learning effect into some single-machine scheduling problems. Further, some scholars extended learning effect models to learning and deterioration models in scheduling problems. Some related studies includes Cheng, Wu, and Lee (2008), Wang and Liu (2009), Wang (2009), Sun (2009), Yang and Yang (2010), Yang and Kuo (2010), Cheng et al. (2010), Huang, Wang, Wang, Gao, and Wang (2010), Huang, Wang, and Wang (2011), etc. Aside from the learning and deterioration models, some scholars incorporated learning effect models into some special scheduling problems. For example, Yang and Kuo (2009) considered a single-machine scheduling problem with learning effects in a intermittent batch production system. Wu and
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Liu (2010) studied a single-machine scheduling problem with learning and unequal release time. Wang, Wang, and Wang (2010) studied a single-machine scheduling problem with learning effect and resource-dependent processing times. Zhu, Sun, Chu, and Liu (2011) combined learning effect and resource allocation in a single-machine group scheduling problem. In this note, we focus on scheduling problems with the p-s-d setup time and learning effects. Some previous studies are given as follows. Kuo and Yang (2007) incorproated the p-s-d setup time and learning effects into a single-machine scheduling problem. They considered the scheduling measures of the makespan, the total completion time, the total absolute deviation of job completion times and the sum of earliness, tardiness and common due-date penalty and showed that these problems all remain polynomially solvable. Wang (2008) studied a single-machine scheduling problem with p-s-d setup time and time-dependent learning effect and showed that the objective functions of the makespan, the total completion time, and the sum of the quadratic job completion time minimization can be solved in polynomial time, respectively. He also showed that the total weighted completion time minimization problem and the maximum lateness minimization problem can be solved in polynomial time under certain conditions. Wang et al. (2009) considered a single-machine scheduling problem with exponential time-dependent learning effect and p-s-d setup time. They showed that the makespan minimization problem, the total completion time minimization problem and the sum of the quadratic job completion time minimization problem can be solved by the shortest (normal) processing time first (SPT) rule. They also showed that the total weighted completion time minimization problem and the maximum lateness minimization problem can be solved in polynomial time under certain conditions. In this note, we will consider the total absolute deviation of job completion times and the total load on all machines as scheduling measures and study an unrelated parallel machine problem simultaneously with p-s-d setup time and learning effects. We show that the proposed problem remains polynomially solvable. In addition, two special cases are also discussed.
There are n jobs (J = {J1, J2, . . . , Jn}) to be processed on m unrelated parallel machines (Mj, j = 1, 2,. . ., m). Let Si denote the set of jobs assigned to Mi. Then Si \ Sj = £, "i – j, and Sm . . ; J n g. Let nj denote the number of jobs assigned to j¼1 Sj ¼ fJ 1 ; .P Mj and n ¼ m j¼1 nj . We assume, as in practical situations, that m < n. All jobs are non-resumable and available for processing at time zero. Each machine can handle at most one job at a time and cannot stand idle until the last job assigned to it has finished processing. Let pij denote the normal processing time of job i (i e {1, 2, . . . , n}) when assigned to machine Mj. Then the actual processing time of job i when scheduled in position r of Mj is given by:
J i 2 Sj # fJ 1 ; . . . ; J n g;
j ¼ 1; 2; . . . ; m;
ð1Þ
where the learning index aij 6 0 is dependent on different machines and jobs. Also, it is assumed that sjr the p-s-d setup time of any job when scheduled in position r of Mj is given by:
sj1 ¼ 0 and sjr ¼ bj
pijk ;
C 1 ¼ sj1 þ pAj½1 C 2 ¼ C 1 þ sj2 þ pAj½2 ¼ sj1 þ pAj½1 þ sj2 þ pAj½2 ¼
and r ¼ 2; 3; . . . ; nj
2 X
ðsjr þ pAj½r Þ
r¼1
C nj ¼
nj X ðsjr þ pAj½r Þ r¼1
where pAj½r is the actual processing time of a job when scheduled in position r of Mj. Then C 2 C 1 ¼ sj2 þ pAj½2 ¼
2 X ðsjr þ pAj½r Þ r¼2
3 3 X X C3 C1 ¼ ðsjr þ pAj½r Þ;C 3 C 2 ¼ ðsjr þ pAj½r Þ r¼2
r¼3
C nj C 1 ¼
nj nj nj X X X ðsjr þ pAj½r Þ;C nj C 2 ¼ ðsjr þ pAj½r Þ;...;C nj C nj 1 ¼ ðsjr þ pAj½r Þ: r¼2
r¼3
r¼nj
Therefore,
ðC r C 1 Þ ¼ ðnj 1Þ½sj2 þ pAj½2 þ ðnj 2Þ½sj3 þ pAj½3 þ þ ½sjnj þ pAj½nj
r¼1 nj
X
ðC r C 2 Þ ¼ ðnj 2Þ½sj3 þ pAj½3 þ ðnj 3Þ½sj4 þ pAj½4 þ þ ½sjnj þ pAj½nj
r¼2
nj X
ðC r C k Þ ¼ ðnj kÞ½sj;kþ1 þ pAj½kþ1 þ ðnj ðk þ 1ÞÞ½sj;kþ2 þ pAj½kþ2
r¼k
þ þ ½sjnj þ pAj½nj nj X
ðC r C nj 1 Þ ¼ ðnj ðnj 1ÞÞ½sj;nj þ pAj½nj
r¼nj 1
TADC ¼
nj nj X X l¼1
jC l C k j
k¼l
¼ jC 1 C 1 j þ jC 1 C 2 j þ jC 1 C 3 j þ þ jC 1 C nj j þ jC 2 C 2 j þ jC 2 C 3 j þ þ jC 2 C nj j þ jC nj 1 C nj j ¼ ðnj 1Þ½sj2 þ pAj½2 þ ð2Þðnj 2Þ½sj3 þ pAj½3 þ ð3Þðnj 3Þ½sj4
J i 2 Sj # fJ 1 ; . . . ; J n g; j
þ pAj½4 þ þ ðnj 1Þð1Þ½sj;nj þ pAj½nj :
k¼1
¼ 1; 2; . . . ; m;
In this section, we show that the problem Pm=pijr ¼ pij r aij ; spsd =TADC can be solved in polynomial time. First, if the number (nj) of jobs for machine j (j = 1, 2, . . . , m) is known in advance, similar to Kuo and Yang (2007), the total absolute deviation of job completion times in machine j can be calculated as follows:
and
and r
¼ 1; 2; . . . ; nj
r1 X
3. Minimization of the total deviation of job completion times
nj X
2. Problem formulation
pijr ¼ pij r aij ;
respectively. For convenience, we denote the p-s-d setup time given in Eq. (2) by spsd, TADC for the total absolute deviation of job compleP tion times and C jmax for the total load on all machines. Then, using the three-field notation introduced by Graham, Lawler, Lenstra, and Rinnooy Kan (1979), the corresponding scheduling problems are deP noted by Pm=pijr ¼ pij r aij ; spsd =TADC and Pm=pijr ¼ pij raij ; spsd = C jmax , respectively.
ð2Þ
where bj P 0. The objectives are to minimize the total absolute deviation of job completion times and the total load on all machines,
Let Ji e Sj. That is, considering those jobs assigned to machine j, the total absolute deviation of job completion times in machine j can be rewritten as follows.
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TADC ¼
nj nj X X l¼1
¼
jC l C k j ¼
r¼1 J i 2Sj
k¼l
nj X
nj X ðr 1Þðnj r þ 1Þðsjr þ pij raij Þ
"
ðr 1Þðnj r þ 1Þ þ bj
r¼1 J i 2Sj
nj X
known, similar to Kuo and Yang (2007), the total load on all machines can be calculated as follows:
# ðl 1Þðnj l þ 1Þ pij r
m X aij
nj n X m X X i¼1
Subject to
n X
j¼1 r¼1
xijr ¼ 1;
ðr 1Þðnj r þ 1Þ þ bj
nj X
j¼1
xijr ¼ 1;
ðsjr þ pij raij Þ ¼
r¼1
i2Sj
nj m X X X ½bj ðnj rÞ þ 1pij raij j¼1
i2Sj
r¼1
Hence, taking advantage of the analysis in the previous section, if the number (nj) of jobs for machine j (j = 1, 2, . . . , m) is known, the corresponding problem can also be formulated as the following assignment problem.
Minimize
nj n X m X X ½bj ðnj rÞ þ 1pij r aij xijr i¼1
Subject to
n X
j¼1 r¼1
xijr ¼ 1;
j ¼ 1; 2; . . . ; m;
r ¼ 1; 2; . . . ; nj
i¼1 nj m X X j¼1
xijr ¼ 1;
i ¼ 1; 2; . . . ; n
r¼1
xijr ¼ 0 or 1;
i ¼ 1; 2; . . . ; n; j ¼ 1; 2; . . . ; m;
r ¼ 1; 2; . . . ; nj
l¼rþ1
ðnj l þ 1Þ pij raij xijr j ¼ 1; 2; . . . ; m;
In addition, by Lemma 1, the following theorem follows immediately.
r ¼ 1; 2; . . . ; nj
nj
X
nj m X X X
ðl 1Þ
i¼1 m X
j¼1
l¼rþ1
Therefore, if the number (nj) of jobs for machine j is known, the problem of Pm=pijr ¼ pij r aij ; spsd =TADC minimization on unrelated parallel machines can be formulated as an assignment problem. Let P(n, m) = (n1, n2, . . . , nm) denote the allocation vectors. For a given P(n, m) vector, the problem can be described by a standard bi-partite graph with n job nodes and n position nodes. We denote a job node by i (i = 1, 2, . . . , n), and a position node by (j, r) (position r on machine Mj, r = 1, 2, . . . , nj, j = 1, 2, . . . , m). Then the cost of the arc that connects i and (j, r) is given by ðr 1Þðnj r þ 1Þ þ bj Pnj aij l¼rþ1 ðl 1Þðnj l þ 1Þpij r . Similar to Biskup (1999), let xijr be a 0/1 variable such that xijr = 1 if Ji is the rth job to be processed on machine Mj and xijr = 0 otherwise. Then the scheduling problem can be formulated as the " following assignment problem.
Minimize
C jmax ¼
i ¼ 1; 2; . . . ; n
P
C jmax can be solved
5. Special cases
j¼1 r¼1
xijr ¼ 0 or 1; i ¼ 1; 2; . . . ; n;
Theorem 2. The problem Pm=pijr ¼ pij r aij ; spsd = in O(nm+3) time.
j ¼ 1; 2; . . . ; m;
r ¼ 1; 2; . . . ; nj :
However, it is necessary to obtain the total number of P(n, m) vectors to completely solve the problem Pm=pijr ¼ pij r aij ; spsd =TADC. Though there is no simple expression for P(n, m), a simple polynomial upper bound of the number of P(n, m) vectors is given in Lemma 1. Lemma 1 Mosheiov (2001). The number of P(n, m) vectors is bounded by (2n)m/m!. Theorem 1. The problem Pm=pijr ¼ pij r aij ; spsd =TADC can be solved in O(nm+3) time. Proof. According to Lemma 1, to solve the problem Pm=pijr ¼ pij raij ; spsd =TADC, a polynomial number of assignment problems need to be solved. Each assignment problem is solved in O(n3) time. Hence, the time complexity of the problem is O(nm+3) time. h Corollary 1. The problem P2=pijr ¼ pij r aij ; spsd =TADC is solved in O(n4) time. Proof. In a two-machine problem, if P(n, 2) = (n1, n2) vector is given, the problem is solved as an assignment problem in O(n3) time. We have to solve such a problem for each (n1, n2) vector. The relevant pairs are (n, 0), (n 1, 1), . . . , (1, n 1), (0, n). Therefore, we have to solve such an assignment problem n times. Thus, the problem P2=pijr ¼ pij raij ; spsd =TADC is solved in O(n4) time. h 4. Minimization of the total load on all machines Let C jmax denote the largest completion time of a job on machine j, j = 1, 2, . . . , m. The total load is the sum of the largest completion times on all the machines or the total time that all machines work. Therefore, if the number (nj) of jobs for machine j (j = 1, 2, . . . , m) is
In this section, we will discuss some special cases of the unrelated parallel machine problem. First, a technique to obtain the minimum sum of the product of two sequences of numbers is introduced in the Lemma 2. Lemma 2 Hardy, Littlewood, and Polya (1967). Let there be two P sequences of numbers xi and yi. The sum ixiyi of products of the corresponding elements is the least (largest) if the sequences are monotonic in the opposite (same) sense. P 5.1. Pm/pijr = pira, spsd, bj = b/TADC and Pm=pijr ¼pi ra ;spsd ;bj ¼b= C jmax In this subsection, the special case of pij = pi, aij = a and bj = b for the problem is discussed. That is, the unrelated parallel machine problem becomes an identical parallel machine problem. The normal processing of job i is pi no matter what machine the job is assigned to. Also, the setup time is b for all machines and the learning effect is job-independent. We consider the two objectives of TADC P j and C max in this case. Theorem 3. The problem Pm/pijr = pira, spsd, bj = b/TADC can be optimally solved in O(nm+1 log n). Proof. Suppose that P(n, m) = (n1, . . . , nm) vector is known. That is, the number of jobs assigned to each machine is given. Let Jj[r] denote the job assigned to the rth position on machine Mj. ThereS fore, Sj ¼ fJ j½1 ; J j½2 ; . . . ; J j½nj g and m j¼1 Sj ¼ fJ 1 ; . . . ; J n g. Let pj[r] denote the normal processing time of job Jj[r]. Then the total completion time of all jobs in these machines is calculated as follows:
TADC ¼
nj nj m X X X ½ðr 1Þðnj r þ 1Þ þ bj ðl 1Þðnj l j¼1 r¼1
þ 1Þpj½r r a
l¼rþ1
ð3Þ
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Pnj
Let wjr ¼ ½ðr 1Þðnj r þ 1Þ þ bj l¼rþ1 ðl 1Þðnj l þ 1Þr a for j = 1, 2, . . . , m and r = 1, 2, . . . , nj. Then Eq. (3) can be viewed as the scalar product of the wjr and pj[r] vectors (j = 1, 2, . . . , m and r = 1, 2, . . . , nj). Therefore, by Lemma 2, all jobs are sorted in nondecreasing order of their normal processing times first. Then the job with the longest normal processing time is assigned to the position with the smallest value of wjr, the job with the second longer normal processing time to the position with the second smaller value of wjr, . . . , and so on. The time complexity of a sorting algorithm is O(n log n) and the number of P(n, m) vectors is bounded by (2n)m/ m!. Therefore, the problem Pm/pijr = pira, spsd, bj = b/TADC can be solved in O(nm+1 log n). h P Theorem 4. The problem Pm=pijr ¼ pi ra ; spsd ; bj ¼ b= C jmax can be m+1 optimally solved in O(n log n). Proof. In this problem, the total load on all machines is calculated as follows: m X
C jmax ¼
nj m X X
ðsjr þ pj½r r a Þ ¼
j¼1 r¼1
j¼1
nj m X X ½bðnj rÞ þ 1pj½r r a
ð4Þ
j¼1 r¼1
Let wjr = [b(nj - r) + 1]ra for j = 1, 2, . . . , m and r = 1, 2, . . . , nj. Then, Eq. (4) can be viewed as the product of the wjr and pj[r] vectors (j = 1, 2, . . . , m and r = 1, 2, . . . , nj). Therefore, by Lemma 2, the theorem follows easily. h P 5.2. Pm=pijr ¼ pi ; spsd ; bj ¼ b= C jmax In this subsection, the special case of aij = 0 and bj = b for the problem is discussed. That is, there are no learning effects of jobs in the identical parallel machine problem. Here, we consider the P j objective of C max in this case. 5.2.1. Group balance principle Suppose there are n jobs to be assigned to m groups, the number of jobs in each group is either l 1 or l where l ¼ dmn e denotes the smallest integer larger than or equal to mn . P Theorem 5. For the problem Pm=pijr ¼ pi ; spsd ; bj ¼ b= C jmax , there exists an optimal solution such that the number of jobs in each group satisfies the group balance principle. Proof. Consider an optimal schedule p = {p1, . . . , pk, . . . , pj, . . . , pm}, where pj denotes the sequence of the jobs scheduled on machine Mj. Suppose the number of jobs in each machine does not satisfy the group balance principle, then there must be at least two machines, say Mk and Mi, such that the difference of the numbers of the two machines is greater than one, that is, nk ni > 1. Let p0 = {p1, . . . , p0i , . . . , p0k , . . . , pm}, where p0k and p0i denote the same sequences of pk and pi except for moving the first job from pk to the first position of pi, respectively. Then, n0k ¼ nk 1, n0i = ni + 1, pi0 [1] = pk[1], pi0 [r] = pi[r1] (r = 2, 3, . . . , ni + 1) and pk0 [r] = pk[r+1] (r = 1, 2, . . . , nk 1). P j Let C max ðpk þ pi Þ denote the total load of Mk and Mi and P j C max ðp0k þ p0i Þ denote the total load of M 0k and M 0i . Then
X
C jmax ðpk þ pi Þ ¼
nk ni X X ½bðnk rÞ þ 1pk½r þ ½bðni rÞ þ 1pi½r r¼1
r¼1
and
X
0
C jmax ð
0 k
0 iÞ
p þp
0
nk ni X X ¼ ½bðn0k rÞ þ 1pk0 ½r þ ½bðn0i rÞ þ 1pi0 ½r r¼1
r¼1
where pj[r] denotes the normal processing time of a job assigned to the rth position in sequence pj (j = 1, 2, . . . , m). Therefore, the differP P ence of C jmax ðpk þ pi Þ and C jmax ðp0k þ p0i Þ is calculated as follows:
X
C jmax ðp0 Þ
X
C jmax ðpÞ ¼
X
C jmax ðp0k þ p0i Þ
X
C jmax ðpk þ pi Þ
¼ ½bðn0i 1Þ þ 1pi0 ½1 ½bðnk 1Þ þ 1pk½1 6 0 since nk ni > 1; ðn0i 1Þ ¼ ni < ðnk 1Þ; b P 0, and pi0 [1] = pk[1] > 0. This contradicts the optimality of p and then Theorem 5 follows. h P Theorem 6. For the problem Pm=pijr ¼ pi ; spsd ; bj ¼ b= C jmax , there exists an optimal schedule in which jobs are sorted in non-decreasing order of their normal processing times (pi) and then one by one the jobs in the sequence are assigned to each machine in turn. That is, P the problem Pm=pijr ¼ pi ; spsd ; bj ¼ b= C jmax can be solved in O(n log n). Proof. We use the same notations in the proof of Theorem 2. Then the total completion time of all jobs in these machines is calculated as follows: m X j¼1
C jmax ¼
nj nj m X m X X X ðsjr þ pj½r Þ ¼ ½bðnj rÞ þ 1pj½r j¼1
r¼1
ð5Þ
j¼1 r¼1
Let wjr = [b(nj r) + 1] for j = 1, 2, . . . , m and r = 1, 2, . . . , nj. Similarly, Eq. (5) can also be viewed as the scalar product of the wjr and pj[r] vectors (j = 1, 2, . . . , m and r = 1, 2, . . . , nj). Then, by Theorem 5, an optimal schedule of this case has to satisfy the group balance principle. Thus, we have jni nj j 6 1; 8i–j. Without loss of generality, let n1 P n2 P P nm. Then
w11 P w21 P P wm1 P w12 P w22 P P wm2 P P w1n1 P w2n2 P P wmnm : Therefore, by Lemma 2, for the problem P Pm=pijr ¼ pi ; spsd ; bj ¼ b= C jmax , we can find an optimal schedule by the following steps.
Step 1. All jobs are sorted in non-decreasing order of their normal processing times (pi). Step 2. One by one the jobs of the sequence are assigned to a schedule from the first position of the first machine to the first position of the last machine, and then from the second position of the first machine to the second position of the last machine . . . and so on. The time complexity of Step 1 is O(n log n) and that for Step 2 is P O(1). Therefore, the problem Pm=pijr ¼ pi ; spsd ; bj ¼ b= C jmax can be solved in O(n log n). This completes the proof. h 6. Conclusion In this paper, an unrelated parallel machine scheduling problem with the p-s-d setup time and learning effects is studied. The two scheduling measures of the total absolute deviation of job completion times and the total load on all machines are considered. We show that the problem with the two scheduling measures is polynomially solvable. Besides, two special cases of the proposed problem is discussed and can be solved by lower order algorithms. Future research may focus on other scheduling performance measures or a similar problem with other learning effects and setup time models. Acknowledgements We would like to thank the referee for his valuable comments and suggestions on an earlier version of this paper. This research
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Further reading Cheng, T. C. E., Cheng, S. R., Wu, W. H., Hsu, P. H., & Wu, C. C. (2011). A two-agent single-machine scheduling problem with truncated sum-of-processing-timesbased learning considerations. Computers and Industrial Engineering, 60(4), 534–541.