Permutation Flow Shop Problem with Shortening Job ... - World Scientific

2nd Reading July 14, 2016 16:4 WSPC/S0217-5959

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Asia-Pacific Journal of Operational Research Vol. 33, No. 4 (2016) 1650032 (14 pages) c World Scientific Publishing Co. & Operational Research Society of Singapore DOI: 10.1142/S0217595916500329

Permutation Flow Shop Problem with Shortening Job Processing Times

Zhenyou Wang Faculty of Applied Mathematics Guangdong University of Technology Guangzhou 510520, P. R. China [email protected]

Cai-Min Wei∗ Department of Mathematics, Shantou University Shantou 515063, P. R. China Guangdong Provincial Key Lab of Digital Signals and Image Processing Shantou University, Shantou, Guangdong, 515063, P. R. China [email protected]

Yuan-Yuan Lu College of Mathematics, Jilin Normal University Siping, Jilin 136000, P. R. China luyuanyuan [email protected] Received 4 February 2015 Revised 15 February 2016 Accepted 19 April 2016 Published 18 July 2016 In this paper, we consider a three-machine makespan minimization permutation flow shop scheduling problem with shortening job processing times. Shortening job processing times means that its processing time is a nonincreasing function of its execution start time. Optimal solutions are obtained for some special cases. For the general case, several dominance properties and two lower bounds are developed to construct a branch-andbound (B&B) algorithm. Furthermore, we propose a heuristic algorithm to overcome the inefficiency of the branch-and-bound algorithm. Keywords: Scheduling; permutation flow shop; shortening job processing times; makespan; branch-and-bound algorithm; heuristic algorithm.

1. Introduction Machine scheduling problems with start time dependent processing times (i.e. deteriorating jobs) have received increasing attentions in recent years (Kung and Shu, ∗ Corresponding

author. 1650032-1


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Z. Wang, C.-M. Wei & Y.-Y. Lu

2015; Li et al., 2015, Wang et al., 2015; Zhao and Tang, 2015). Researchers have formulated this phenomenon into different models and solved different versions of the problems for various criteria. An extensive survey of different scheduling models and problems involving start time dependent processing times (i.e., deteriorating jobs) can be found in Gawiejnowicz (2008). There is a significant number of papers that focus on the flow shop scheduling with deteriorating jobs (Mosheiov (2002); Wang et al. (2006); Wang and Xia (2006a, 2006b); Shiau et al. (2007); Lee et al. (2008)). More recently, Tang and Liu (2009), Ng et al. (2010), Yang and Wang (2011), Bank et al. (2012) and Zhao and Tang (2012) considered two-machine flow shop scheduling with deteriorating jobs. Wang et al. (2010), and Wang and Wang (2013a, 2013b) considered three-machine flow shop scheduling with deteriorating jobs. Lee et al. (2009), Wang (2010), Mosheiov et al. (2010), Ng et al. (2011), Th¨ ornblad and Patriksson (2011), Sun et al. (2012), Lee et al. (2014) and Yin and Kang (2015) considered m-machine flow shop scheduling with deteriorating jobs, where m is the number of machines. However, the flow shop scheduling with shortening job processing times are relatively unexplored. Wang and Liu (2009) considered two-machine flow shop scheduling with shortening job processing times. Wang (2007) and Wang et al. (2011) considered m-machine flow shop scheduling with shortening job processing times. This paper studies the three-machine makespan minimization flow shop scheduling with shortening job processing times (Ho et al., 1993; Wang and Xia, 2005). Since the classical three-machine flow shop scheduling to minimize the makespan is NP-hard, the problem of three-machine flow shop scheduling to minimize makespan with shortening job processing times is NP-hard. Therefore, this paper considers some special cases in which the three-machine problem can be solved optimally, and develops a branch-and-bound algorithm and a heuristic algorithm for general problem. The paper is organized as follows. In the next section, we give the flow shop scheduling problem description. In Sec. 3, we consider some polynomially solvable special cases. In Sec. 4, we propose several elimination rules and two lower bounds. In Sec. 5, we develop a heuristic algorithm and a branch-and-bound algorithm. In Sec. 6 we present computational experiments of the proposed algorithms. The last section concludes the paper. 2. Problem Description There are n jobs J1 , J2 , . . . , Jn to be processed successively on three machines M = {M1 , M2 , M3 } in that order. Each job can be processed on no more than one machine at any time, while each machine can handle only one job at a time and the processing of a job may not be interrupted. All the jobs are available for processing at time t0 ≥ 0. The operation of job Jj on machine Mi is denoted by Oij . Following Ho et al. (1993) and Wang and Xia (2005), we assume that the actual processing time 1650032-2

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Permutation Flow Shop Problem

of operation Oij is given by pij (t) = aij (1 − ut) (i = 1, 2, 3; j = 1, 2, . . . , n),

(1)

where aij > 0 denotes the normal processing time of operation Oij , t is its starting 3 n time. It is assumed that u satisfies the following condition: u(t0 + i=1 j=1 aij − amin ) < 1, where amin = mini=1,2,3;j=1,2,...,n {aij }. The condition ensures that all the job processing times are positive in a feasible schedule (Ho et al., 1993; Wang and Xia, 2005). Our goal is to find a schedule that minimizes the maximum completion time (i.e., the makespan). We assume unlimited intermediate storage between successive machines for the general flow shop scheduling problem. Let Cij (S) denote the completion time of job Jj on machine Mi under some schedule S, and Ci[j] (S) denote the completion time of the jth job on machine Mi under schedule S, where [j] denote the job in the jth position of a schedule. Thus, the completion time of job Jj is Cj = C3j . Using the three-field natation for problem classification (Gawiejnowicz, 2008), the problem can be represented as F 3 | pij = aij (1 − ut)|Cmax . For u = 0, the problem F 3 | pij = aij (1 − ut)|Cmax is the classical problem F 3||Cmax . For ease of exposition, we denote a1j by αj , a2j by βj and a3j by γj , j = 1, 2, . . . , n. Since unlimited intermediate storage is assumed, we have (Wang and Xia, 2005) j 1 1 (2) C1[j] = t0 − (1 − uα[i] ) + , for u > 0, j = 1, 2, . . . , n. u i=1 u 3. Solvable Cases Lemma 1. There exists an optimal schedule in which the job sequence is identical on three machines. Proof. Similar to the proof of Kononov and Gawiejnowicz (2001). The conclusion of Lemma 1 is that only permutation schedules need be considered for this problem. For two machines M1 , M2 , Wang and Xia (2005) proved that the problem F 2|pij = aij (1 − ut)|Cmax can be solved by the following O(n log n) rule. Modified Johnson’s Rule (SPT(M 1 )-LPT(M 2 )) (Wang and Xia (2005)): Step 1. Partition the set of jobs N = {J1 , J2 , . . . , Jn } into two subsets N1 = {Jj | αj ≤ βj } and N2 = {Jj | αj > βj }. Step 2. The jobs in N1 go first in the order of nondecreasing αj , the jobs in N2 follow in the order of nonincreasing βj . In what follows, we will show that the problem can be solved in polynomial time for some special cases. Now we consider a special case: flow shop scheduling with dominant machines. Following Wang and Xia (2007), machine M1 (M2 ) is dominated by M2 (M1 ), or M2 1650032-3


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(M1 ) dominates M1 (M2 ) iff max{αj | j = 1, 2, . . . , n} ≤ min{βj | j = 1, 2, . . . , n} (max{βj | j = 1, 2, . . . , n} ≤ min{αj | j = 1, 2, . . . , n}) (denoted as M1 < M2 (M1 > M2 )). Similarly, machine M2 (M3 ) is dominated by M3 (M2 ), or M3 (M2 ) dominates M2 (M3 ) iff max{βj | j = 1, 2, . . . , n} ≤ min{γj | j = 1, 2, . . . , n} (max{γj | j = 1, 2, . . . , n} ≤ min{βj | j = 1, 2, . . . , n}) (denoted as M2 < M3 (M2 > M3 )). Theorem 1. For the problem F 3 | pij = aij (1 − ut), M1 > M2 | Cmax , an optimal schedule can be obtained by Modified Johnson’s Rule, where αj = (1 − uαj )(1 − uβj ), βj = (1 − uβj )(1 − uγj ). Proof. Let S1 = (π, Jj , Jk , π ) denote an optimal schedule which is not obtained by Modified Johnson’s Rule, where π and π are partial sequences, such that one of the following three conditions is satisfied: Condition (a): αj > βj and αk ≤ βk , i.e., αj < γj and αk ≥ γk (job Jj is in Set N2 and job Jk is in Set N1 ). Condition (b): αj ≤ βj , αk ≤ βk and αj > αk , i.e., αj ≥ γj , αk ≥ γk and (1 − uαj )(1 − uβj ) > (1 − uαk )(1 − uβk ) (jobs Jj and Jk are in Set N1 and arranged in a nonincreasing order of αj ). Condition (c): αj > βj , αk > βk and βj < βk , i.e., αj < γj , αk < γk and (1 − uβj )(1 − uγj ) < (1 − uβk )(1 − uγk ) (jobs Jj and Jk are in Set N2 and arranged in a nondecreasing order of βj ). Denote by S2 the schedule obtained after a pairwise interchange of jobs Jj and Jk . It suffices to show that under any of these conditions the makespan can only be reduced, i.e. Cj (S2 ) ≤ Ck (S1 ). Let A = the last completion time prior to jobs Jj and Jk on machine M1 in S1 , C = the last completion time prior to jobs Jj and Jk on machine M3 in S1 . Clearly, A and C remain unchanged in schedule S2 . Under the condition M1 > M2 , the completion time of job Jk on machine M3 under schedule S1 is 1 1 Ck (S1 ) = max max C − , A − (1 − uαj )(1 − uβj ) (1 − uγj ), u u 1 1 A− (1 − uαj )(1 − uαk )(1 − uβk ) (1 − uγk ) + u u = max

1 1 C− (1 − uγj )(1 − uγk ), A − (1 − uαj )(1 − uβj ), u u

1 (1 − uγj )(1 − uγk ) A − (1 − uαj )(1 − uαk )(1 − uβk )(1 − uγk ) u 1 + . u

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Under the condition M1 > M2 , the completion time of job Jj on machine M3 under schedule S2 is 1 1 Cj (S2 ) = max C − (1 − uγk )(1 − uγj ), A − (1 − uαk )(1 − uβk ) u u 1 × (1 − uγk )(1 − uγj ), A − (1 − uαk )(1 − uαj )(1 − uβj )(1 − uγj ) u 1 + . u

(4)

For the Condition (a), we have first term in (4) = first term in (3), second term in (4) ≤ third term in (3), third term in (4) ≤ second term in (3). Under the Condition (b), we have first term in (4) = first term in (3), second term in (4) ≤ second term in (3), third term in (4) ≤ second term in (3). Under the Condition (c), we have first term in (4) = first term in (3), second term in (4) ≤ third term in (3), third term in (4) ≤ third term in (3). We conclude that in all Conditions (a), (b) and (c), Cj (S2 ) ≤ Ck (S1 ). By repeating this procedure we obtain an optimal schedule without any of these conditions, i.e. a schedule produced by Modified Johnson’s Rule, where αj = (1 − uαj )(1 − uβj ), βj = (1 − uβj )(1 − uγj ). Similarly, we have the following theorems: Theorem 2. For the problem F 3 | pij = aij (1 − ut), M3 > M2 |Cmax , an optimal schedule can be obtained by Modified Johnson’s Rule, where αj = (1 − uαj )(1 − uβj ), βj = (1 − uβj )(1 − uγj ). Theorem 3. For the problem F 3 | pij = aij (1 − ut), M2 > M1 |Cmax , an optimal schedule can be obtained by the following algorithm: Solve the problem by Modified Johnson’s Rule, where αj = βj , βj = γj . Let Job Jk denote the first job in this schedule. Generate additional schedules by inserting in first position jobs with αj < αk . Among these schedules, the one with the smallest makespan in the three machines problem is optimal Theorem 4. For the problem F 3 | pij = aij (1 − ut), M2 > M3 |Cmax , an optimal schedule can be obtained by the following algorithm: Solve the problem by Modified Johnson’s Rule, where αj = αj , βj = βj . Let Job Jk denote the last job in this schedule. Generate additional schedules by inserting in first position jobs with γj < γk . Among these schedules, the one with the smallest makespan in the three machines problem is optimal. Theorem 5. If βj ≤ min{αj , γj } for all j, then an optimal schedule can be obtained by Modified Johnson’s Rule, where αj = (1 − uαj )(1 − uβj ), βj = (1 − uβj )(1 − uγj ). 1650032-5


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4. Dominance Properties and Lower Bound The dominance test is an important issue in branch-and-bound algorithms for scheduling problems. In the following some dominance relations are given. Let S1 = (π, Jj , Jk , π ), and S2 = (π, Jk , Jj , π ) be obtained from S1 by only interchanging jobs Jj and Jk , where π and π are partial sequences. To show that S2 dominates S1 , it is sufficient to show that Cj (S2 ) ≤ Ck (S1 ). Let A = the last completion time prior to jobs Jj and Jk on machine M1 in S1 , B = the last completion time prior to jobs Jj and Jk on machine M2 in S1 , C = the last completion time prior to jobs Jj and Jk on machine M3 in S1 . Clearly, A, B and C remain unchanged in schedule S2 . Proposition 1. Suppose that two jobs Jk and Jj satisfy the following conditions: (i) Either βk ≤ βj or (B − u1 )(1 − uγj ) ≤ (A − u1 )(1 − uαj )(1 − uαk ) or γj ≤ βj or (B − u1 )(1 − uβk ) ≤ C − u1 ; (ii) Either (A− u1 )(1−uαk )(1−uαj ) ≤ (B − u1 )(1−uγk ) or (1−uβj )(1−uγj ) ≤ (1− uβk )(1−uγk ) or (A− u1 )(1−uαk )(1−uαj )(1−uγj ) ≤ (B − u1 )(1−uβk )(1−uγk ) or (A − u1 )(1 − uαk )(1 − uαj )(1 − uβj ) ≤ (C − u1 )(1 − uγk ); (iii) Either βk ≤ γk or (B− u1 )(1−uβj )(1−uγj ) ≤ (A− u1 )(1−uαj )(1−uαk )(1−uγk ) or γj ≤ γk or (B − u1 )(1 − uβk )(1 − uβj ) ≤ (C − u1 )(1 − uγk ). Then S2 dominates S1 . Proof. Now we only consider the case: βk ≤ βj , A(1 − uαk )(1 − uαj ) ≤ B(1 − uγk ) and βk ≤ γk . The other cases are similar to this case. From Theorem 1, we have 1 1 Ck (S1 ) = max max C − , B − (1 − uβj ) (1 − uγj ), u u 1 max A − (1 − uαj )(1 − uαk ), u 1 1 B− (1 − uβj ) (1 − uβk ) (1 − uγk ) + u u 1 1 = max C − (1 − uγj )(1 − uγk ), B − u u × (1 − uβj )(1 − uγj )(1 − uγk ), 1 (1 − uαj )(1 − uαk )(1 − uβk )(1 − uγk ), A− u 1 1 B− (1 − uβj )(1 − uβk )(1 − uγk ) + u u 1650032-6

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and

1 Cj (S2 ) = max C − (1 − uγk )(1 − uγj ), u 1 B− (1 − uβk )(1 − uγk )(1 − uγj ), u 1 A− (1 − uαk )(1 − uαj )(1 − uβj )(1 − uγj ), u 1 1 B− (1 − uβk )(1 − uβj )(1 − uγj ) + . u u

(6)

Owing to the fact that βk ≤ min{βj , γk } and A(1 − uαk )(1 − uαj ) ≤ B(1 − uγk ), we have: first term in (6) = first term in (5), second term in (6) ≤ second term in (5), third term in (6) ≤ second term in (5), fourth term in (6) ≤ second term in (5). Hence, Cj (S2 ) ≤ Ck (S1 ), this completes the proof. Similarly, we have Proposition 2. For S1 = (π, Jj , π , Jk , π ) and S2 = (π, Jk , π , Jj , π ), if jobs Jj and Jk satisfy αk ≤ αj , βk = βj and γk ≥ γj , then S2 dominates S1 , where π, π and π are partial sequences. Let S = (P S, U S) be a schedule of jobs in which P S is the scheduled part (suppose there are k jobs in P S), and U S is a unscheduled part. For the problem F 3 | pij = aij (1 − ut)|Cmax , a lower bound that is adapted from Ignall and Schrage (1965) is:   1     C (P S) − (1 − uα )   j 1[k] US   u       1       × maxUS [(1 − uβj )(1 − uγj )] + ,     u . LBa = max 1 1     , (P S) − (1 − uβ ) × max (1 − uγ ) + C   j US j 2[k] US  u u           

1   1     C3[k] (P S) − (1 − uγ ) + j US u u For the problem F 2|pij = aij (1 − ut)|Cmax can be solved by Modified Johnson’s Rule, hence, we have   1 1     C1[k] (P S) + M1,2 − maxUS (1 − uγj ) + ,     u u       1 1 , LBb = max max C1[k] (P S) − maxUS (1 − uαj ) + , C2[k] (P S) + M2,3 ,    u u           C1[k] (P S) + M1,3 1650032-7


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where Ml,k is the solution of solving unscheduled jobs U S by Modified Johnson’s (Ml , Mk ) Rule. In order to make the lower bound tighter, we choose the maximum value of LBa and LBb as a lower bound for S = (P S, U S), i.e. LB = max{LBa , LBb }. 5. Algorithms In this section, we propose an efficient heuristic algorithm and an exact branchand-bound algorithm. 5.1. The heuristic algorithm In this subsection a heuristic algorithm is proposed. From Theorems 1–5, we used the different normal processing times for the artificial two-machine flow shop scheduling problem as a heuristic algorithm. Furthermore, a local improvement procedure is added to the heuristic algorithm. Now, we give the steps of the heuristic algorithm. The Algorithm 1 Phase I Step 1. Run Modified Johnson’s Rule to obtain an optimal schedule S1 for the artificial two-machine flow shop scheduling problem with normal processing times αj = (1 − uαj )(1 − uβj ), βj = (1 − uβj )(1 − uγj ), (j = 1, 2, . . . , n). Step 2. Run Modified Johnson’s Rule to obtain an optimal schedule S2 for the artificial two-machine flow shop scheduling problem with normal processing times αj = αj , βj = βj , (j = 1, 2, . . . , n). Step 3. Run Modified Johnson’s Rule to obtain an optimal schedule S3 for the artificial two-machine flow shop scheduling problem with normal processing times αj = βj , βj = γj , (j = 1, 2, . . . , n). Step 4. Run Modified Johnson’s Rule to obtain an optimal schedule S4 for the artificial two-machine flow shop scheduling problem with normal processing times αj = αj , βj = γj , (j = 1, 2, . . . , n). Step 5. Choose a best solution from {S1 , S2 , S3 , S4 }. Phase II Step 1. Let S0 be the initial schedule obtained from Phase I. Step 2. Set k = 1 and i = k + 1. Step 3. Create a new sequence S1 by moving J[i] forward to position k in S0 . Replace S0 by S1 if the value of the total completion time of S1 is smaller than that of S0 . Step 4. If i < n, then set i = i + 1, go to Step 3. Step 5. If k < n − 1, then set k = k + 1, go to Step 2. Otherwise, stop. 1650032-8

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Since Phase I takes O(n log n) time and Phase II takes O(n2 ) time, the overall time complexity of Algorithm 1 is O(n2 ). 5.2. The branch-and-bound (B&B) algorithm The B&B algorithm uses the depth first search strategy. It requires very little storage space and can be used for large-sized problems. This algorithm assigns jobs in a forward manner starting from the first position. In the searching tree, we choose a branch and systematically work down the tree until we either eliminate it on grounds of a lower bound or reach its final node, which is either used as a substitute for the initial solution or eliminated. Similar to the B&B algorithm of Lee et al. (2009), our B&B algorithm is given as follows. The B&B algorithm Step 1. Algorithm 1 is applied to obtain an initial solution. Step 2. Start the assignment of jobs at the beginning of a schedule and move forward one step at a time. Step 3. In the kth level node, the first k positions are occupied by k specific jobs. Select one of the remaining n − k jobs for the node at level k + 1. Step 4. First apply Proposition 2, then Proposition 1, to eliminate the dominated partial sequences. Step 5. Calculate the lower bound for the makespan (Cmax ) of the unfathomed partial schedules or the makespan (Cmax ) of the completed schedules. During the search, any unfathomed branch that has a lower bound which is larger than or equal to the initial solution is fathomed, i.e., eliminated from further consideration. If the value of the completed schedule is less than the initial solution, use it as the new initial solution. Otherwise, eliminate it. Step 6. Continue to search all the nodes, and the remaining initial solution is optimal. 6. Computational Experiments Computational experiments were conducted to evaluate the effectiveness of the B&B algorithm and the heuristic algorithm. The Algorithm 1 and the B&B algorithm were coded in VC++ 6.0 and ran the computational experiments on a Pentium 4-2.4G personal computer with a RAM size of 1G. The normal job processing times on M1 , M2 and M3 were randomly generated according to the uniform distribution from the following interval: aij ∈ (0, 10). For all the tests, t0 = 0, the values u = 0.1G, u = 0.3G, u = 0.5G u = 0.7G and u = 0.9G were used, where G = 1/[ ni=1 (αi + βi + γi ) − mini=1,2,...,n {αi , βi , γi }]. In order to test the B&B algorithm, 8 different job sizes, n = 10, 11, 12, 13, 14, 15, 16 and 17 were used. As a consequence, 8 experimental conditions were examined 1650032-9


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and 50 replications were randomly generated for each of them. A total of 2000 problems were tested. For the B&B algorithm, the average number of nodes, the maximum number of nodes, the average time, and the maximum time (in milliseconds) are reported. The contribution of the dominance properties and the lower bound is shown by the algorithm efficiency, which is calculated in terms of the number of nodes explored compared to the total number of nodes. For the heuristic algorithm, the average and maximum error percentage are recorded. The error percentage of the solution produced by the heuristic algorithm is calculated as (Heur-V ∗ )/V ∗ , where Heur is the solution value of Algorithm 1 and V ∗ is the optimal value of the objective function obtained by the B&B algorithm. The results are summarized in Table 1. As shown in Table 1, it can be observed that the B&B algorithm can solve most of the problems in a reasonable amount of time when the job size is less than or equal to 17. The running times of the B&B algorithm strongly depend on the value of u, i.e. the instances become easily to solve when the value of u is getting larger. The reason is that the lower bound and the elimination properties are more efficient if the value of u is getting larger. As to the performance of Algorithm 1, all mean error percentages are less than 0.4%. In addition, their performances also are not affected as the values of u. All of the maximum values of the worst cases of Algorithm 1 were less 6%. Thus, Algorithm 1 is recommended due to its accuracy.

Table 1. Results of B&B algorithm and heuristic algorithm. u

n

B&B algorithm CPU time (ms) mean

Heuristic algorithm (Heur-V ∗ )/V ∗

Node number

max

mean

max

mean

max

u = 0.1G

10 242.950 11 1382.800 12 4734.350 13 32053.100 14 92628.900 15 177621.850 16 1176281.860 17 4372385.828

4500 27640 94656 394250 1650187 3545406 15845064 43864048

12551 46707 143619 794483 1560624 4257042 42567446 512586787

123013 933867 2872135 9995995 19157477 84972624 259766224 2553767291

0.000 0.003 0.002 0.002 0.001 0.005 0.004 0.003

0.001 0.035 0.021 0.018 0.016 0.057 0.051 0.048

u = 0.3G

10 110.550 11 530.300 12 1164.850 13 4360.200 14 22896.050 15 125060.950 16 959165.830 17 2391516.753

2166 18966 13297 66172 452406 2501187 15181728 858312798

6397 48313 150792 198901 530786 1361303 8661593 68695136

71705 858579 5015606 9493433 10494262 87225792 307627719 2307921781

0.000 0.004 0.001 0.004 0.001 0.001 0.002 0.002

0.001 0.065 0.012 0.047 0.021 0.012 0.022 0.022

u = 0.5G

10 11 12

956 1178 3406

1903 1912 2478

44127 92621 90683

0.001 0.002 0.001

0.046 0.036 0.026

30.120 42.500 98.120

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Permutation Flow Shop Problem Table 1. (Continued) u

n

B&B algorithm CPU time (ms)

Heuristic algorithm

Node number mean

(Heur-V ∗ )/V ∗

mean

max

max

mean

max

13 14 15 16 17

2414.680 9338.440 83825.000 251868.625 971286.653

114922 915969 1676484 8012852 28618254

60079 197273 2025723 9189673 89819767

2869425 13345988 40514168 151548512 2155848152

0.001 0.003 0.001 0.002 0.003

0.014 0.054 0.022 0.021 0.051

u = 0.7G

10 11 12 13 14 15 16 17

10.750 20.010 39.940 570.940 1454.060 4650.000 23882.151 93582.155

215 389 969 26922 90078 230859 1082475 5085457

674 2531 5335 12922 19555 100740 789639 7586933

9031 18187 94586 611019 926100 5003632 25361513 275365316

0.003 0.002 0.001 0.001 0.002 0.004 0.003 0.002

0.042 0.044 0.025 0.022 0.048 0.049 0.043 0.049

u = 0.9G

10 11 12 13 14 15 16 17

8.550 12.350 23.400 68.000 92.500 578.100 2821.125 10281.153

116 147 453 1344 8219 11515 108541 618451

376 660 710 1718 2596 11059 59631 195613

3315 12972 33964 34069 31636 220882 1145123 6152138

0.001 0.001 0.002 0.002 0.001 0.001 0.001 0.002

0.018 0.010 0.042 0.031 0.013 0.010 0.011 0.031

7. Conclusion We considered in this paper a three-machine flow shop scheduling problem with makespan criterion. We also assumed that the jobs follow a decreasing timedependent job processing times function. Some polynomially solvable special cases were given. Several dominance conditions and two lower bounds were implemented in the proposed B&B algorithm to search for the optimal solution. A heuristic algorithm was also provided to overcome the inefficiency of the B&B algorithm for large-sized problems. Computational results showed that the heuristic algorithm is shown to perform well in obtaining near-optimal solutions, and hence for larger problems it can be used as an efficient heuristic algorithm.

Acknowledgments We are grateful to the editor and two anonymous referees for their helpful comments on an earlier version of this paper. This research was supported by the National Natural Science Foundation of China (Grant Nos. 11401115, 11471012 and 71501082). 1650032-11


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Biography Zhenyou Wang is an Associate Professor of the Department of Mathematics, Guangdong University of Technology, China. His research interests include operations research and medical image translation. He has published about 20 papers, including BMC Medical Image, Advances in Difference Equations, Computational and Applied Mathematics Journal, Clinical & Medical Biochemistry, and so on. 1650032-13


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Cai-Min Wei is a Professor at the Department of Mathematics, Shantou University, China. His research interests the theory and method of stochastic model including Queue Theory, Scheduling Theory and Financial Mathematics. He has published about 30 papers, including Filomat, Applied Mathematical Modelling, Applied Mathematics and Computation, International Journal of Production Economics, International Journal of Advanced Manufacturing Technology, Solitons and Fractals, Journal of Applied Mathematics and Computing, and so on. Yuan-Yuan Lu is an Associate Professor at the College of Mathematics, Jilin Normal University, China. Her research interests include operations research and planning scheduling. She has published about 10 papers, including Asia-Pacific Journal of Operational Research, Applied Mathematical Modelling, Optimization Letters, Applied Mathematics and Computation, and so on.

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