- Email: [email protected]
Outsourcing and scheduling for two-machine ordered flow shop scheduling problems
European Journal of Operational Research 226 (2013) 46–52
Contents lists available at SciVerse ScienceDirect
European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor
Discrete Optimization
Outsourcing and scheduling for two-machine ordered flow shop scheduling problems q Dae-Young Chung a, Byung-Cheon Choi b,⇑ a b
SAP Korea Ltd., 27-28F, MMAA Building, 467-12 Dogok-dong, Gangnam-gu, Seoul 135-700, Republic of Korea Department of Business Administration, Chungnam National University, 79 Daehangno, Yuseong-gu, Daejeon 305-704, Republic of Korea
a r t i c l e
i n f o
Article history: Received 21 July 2011 Accepted 29 October 2012 Available online 10 November 2012 Keywords: Scheduling Outsourcing Ordered flow shop Computational complexity
a b s t r a c t This paper considers a two-machine ordered flow shop problem, where each job is processed through the in-house system or outsourced to a subcontractor. For in-house jobs, a schedule is constructed and its performance is measured by the makespan. Jobs processed by subcontractors require paying an outsourcing cost. The objective is to minimize the sum of the makespan and the total outsourcing cost. Since this problem is NP-hard, we present an approximation algorithm. Furthermore, we consider three special cases in which job j has a processing time requirement pj, and machine i a characteristic qi. The first case assumes the time job j occupies machine i is equal to the processing requirement divided by a characteristic value of machine i, that is, pj/qi. The second (third) case assumes that the time job j occupies machine i is equal to the maximum (minimum) of its processing requirement and a characteristic value of the machine, that is, max{pj, qi} (min{pj, qi}). We show that the first and the second cases are NP-hard and the third case is polynomially solvable. Ó 2012 Elsevier B.V. All rights reserved.
1. Introduction In today business environment, many companies are looking to outsource, that is, entrust some jobs to an external provider. Through proper outsourcing, they can deliver goods to consumers in time, reduce operating costs and make their organizations more flexible. Thus, proficient use of outsourcing can make a company more competitive (Cachon and Harker, 2002; Webster et al., 1997), which provides the authors with motivation to consider a scheduling problem with an outsourcing option. This paper considers the two-machine, n-job ordered flow shop problem, where the processing time of job j on machine i, pi,j, follows two special characteristics (Smith et al., 1975, 1976): If, for some 1 6 i0 6 2, pi0 ;j0 6 pi0 ;j , then pi;j0 6 pi;j for i = 1, 2, and if pi0 ;j0 6 pi;j0 , then pi0 ;j 6 pi;j for j = 1, . . . , n. Furthermore, there are two options for processing jobs: The first is to process the job by utilizing in-house machines, while the second is to outsource the job to a subcontractor. For in-house jobs, a schedule r is constructed and its performance is measured by the makespan, that is, the latest completion time of jobs processed in-house, denoted Cmax(r). If job j is processed by subcontractors, an outsourcing cost q This study was financially supported by the research fund of Chungnam National University in 2011. ⇑ Corresponding author. E-mail addresses: [email protected] (D.-Y. Chung), [email protected] (B.-C. Choi).
0377-2217/$ - see front matter Ó 2012 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.ejor.2012.10.048
oj should be paid. Note that the total outsourcing costs can be P described as j2O oj , where O is the set of jobs outsourced. The objective is to minimize the sum of the makespan and total outsourcing costs. The problem considered in this paper is defined below. Problem P. In a two-machine, n-job ordered flow shop, determine a schedule (r, O) that minimizes T(r, O), defined as
Tðr; OÞ ¼ C max ðrÞ þ
X oj : j2O
Since Problem P is known to be NP-hard (Choi and Chung, 2011), we present an approximation algorithm and consider three special cases such that job j has a processing time pj that affects the time job j spends on each of two machines, and machine i has a characteristic value qi that affects the time each job spends on machine i. First, we consider a setting in which the processing time of job j on machine i can be expressed as pi,j = pj/qi. In this case, qi can be interpreted as the speed of machine i, which implies that job processing times are inversely proportional to machine speeds. Let this case be referred to as Problem I. The second case considers a setting in which job j must spend a minimum time qi on machine i. The processing time of job j on machine i can then be expressed
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
as pi,j = max{pj, qi}. This minimum amount of time can also be due to setting or fine-tuning the machine; however, in this case the job can already start while the machine is being set up. Let this case be referred to as Problem II. Third, if, on the other hand, job j can spend at most time qi on machine i, then the processing time of job j on machine i can be expressed as pi,j = min{pj, qi}. This last case can occur when the time of machine i is expensive and must be rationed; no job is allowed more than a maximum time qi on the machine. Let this case be referred to as Problem III. It is easy to see that Problems I–III are ordered flow shops. The rest of the paper is organized as follows. Section 2 discusses the related literature. Section 3 presents an approximation algorithm for Problem P. Section 4 proves that Problems I and II are NP-hard. Section 5 shows that Problem III can be solved in polynomial time. Finally, Section 6 draws our conclusions.
2. Literature review Many researchers have focused on specially structured flow shops, such as the ordered flow shop (Panwalkar and Khan, 1976; Smith et al., 1975, 1976) and the proportionate flow shop (Allahverdi, 1996; Hou and Hoogeveen, 2003; Ow, 1985), where the processing time of job j on all the machines is the same and equal to pj, that is, pij = pj (Pinedo, 2002). Smith et al. (1975, 1976) considered an m-machine ordered flow shop with the makespan objective. Smith et al. (1975) considered a special case in which the job processing times on the first (last) machine are the longest and showed that the problem can be solved in polynomial time. The authors showed that if the first (last) machine is the slowest, the LPT (SPT) sequence is an optimal permutation schedule. Smith et al. (1976) showed that there exists an optimal schedule that has an inverted V-shape. Hou and Hoogeveen (2003) and Choi et al. (2007) considered a three-machine flow shop with pi,j = pj/qi with the objective of minimizing the makespan. They showed that the problem is NP-hard in the ordinary sense. Furthermore, Choi et al. (2007) have proposed a heuristic of which tight worst-case bound has been proven to be 3/2 in Koulamas and Kyparisis (2009). Choi et al. (in press) considered an m-machine flow shop with pi,j = pj + qi, with the objective of minimizing the makespan. The authors showed that the problem is solvable in polynomial time when the number of machines is fixed. Choi et al. (2010) considered two m-machine flow shops with pi,j = max{pj, qi} and then pi,j = min{pj, qi}, with the objective of minimizing the makespan. They showed that the first problem is NP-hard, and the second is solvable in polynomial time. Under a proportionate flow shop environment, the makespan is constant independently of the sequence if the permutation schedules are considered (Pinedo, 2002). The literature on scheduling with an outsourcing option has studied the problem of simultaneously deciding which jobs to be outsourced and how to schedule in-house jobs. Qi (2008) considered a single-machine scheduling problem that allows some jobs to be outsourced to a single outsourcing machine, with a transportation delay and transportation cost. Lee and Sung (2008a,b) considered a single-machine scheduling problem to minimize the weighted sum of the outsourcing cost and the associated scheduling cost with an outsourcing budget constraint, where the scheduling cost is the one of the following: the maximum lateness, denoted by Lmax = max16j6n{Cj dj}; the total P P tardiness, denoted by T j ¼ 16j6n maxf0; C j dj g; and the total P completion time, denoted by C j , where Cj and dj are the completion time and due date of job j, respectively. Chen and Li (2008) considered a parallel machine scheduling problem with an outsourcing option to minimize the total production cost subject to a constraint on the makespan. Qi (2009) considered a
47
two-machine flow shop scheduling problem that allows only operations on the first machine to be outsourced. For practical reasons, the author’s model considered a single-batch system with a transportation delay, with the objective of minimizing the makespan subject to a constraint on the maximum outsourcing cost. Qi (2011) extended this model to include other forms of outsourcing. Lee and Choi (2011) considered a two-machine flow shop problem that allows any operations to be outsourced in which the objective is to minimize the sum of the makespan and the total outsourcing costs. Choi and Chung (2011) considered two-machine flow shop problem such that two operations comprising the job must be oursourced when a job is determined to be outsourced. The objective is to minimize the sum of the makespan and total oursourcing costs. Chung et al. (2005) considered a job shop scheduling problem with an outsourcing option. The problem is the same as a typical job shop scheduling problem, except that the capacity constraints can be violated with costs, which implies each operation can be outsourced, with operation-dependent costs. The author presented a heuristic algorithm that decomposes the problem into smaller subproblems and solves each subproblem optimally, and demonstrated its performance experimentally.
3. Approximation algorithm for Problem P This section presents an approximation algorithm, referred to as Heuristic H, for Problem P, which is defined as follows. Heuristic H: Outsource a job if its outsourcing cost is not greater than the larger of its processing times, that is, In case of p1,j < p2,j, outsource job j if p2,j P oj. In case of p1,j P p2,j, outsource job j if p1,j P oj. Henceforth, we consider only the case where p1,j < p2,j, j = 1, 2, . . . , n. From the reversibility of flow shops, it follows that the results of the case under consideration also apply to that case where p1,j P p2,j, j = 1, 2, . . . , n. Let (r⁄, O⁄) be an optimal schedule and (rH, OH) be the schedule obtained by Heuristic H. Let I⁄ = {1, 2, . . . , n}nO⁄ and IH = {1, 2, . . . , n}nOH. It is assumed that jobs in I⁄ and IH are sequenced according to Johnson’s rule. Lemma 1. If OH = {1, 2, . . . , n}, then (rH, OH) becomes an optimal schedule for the two-machine ordered flow shop problem with an outsourcing option. Proof. Since OH = {1, 2, . . . , n}, it is observed that p2,j P oj, j = 1, 2, . . . , n. Suppose that there exists an optimal schedule (r⁄, O⁄) such that I⁄ – ;. Without loss of generality, let job k be positioned last on an in-house machine. We can construct a new schedule ; OÞ by letting O ¼ O [ fkg. Then, it is observed that the makeðr span is decreased by at least p2,j and the outsourcing cost is increased by ok. From these observations, one obtains the inequality.
; OÞ 6 Tðr ; O Þ p2;k þ ok : Tðr Note, however, that p2,k P ok. Thus, by applying this argument repeatedly, Lemma 1 can be obtained or a contradiction occurs. h By Lemma 1, henceforth, we only consider instances where for some k, p2,k < ok. Lemma 2.
Tðr ; O Þ P
X j2OH
oj þ
X p2;j : j2IH
48
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
Proof. Consider Problem P0 equivalent to Problem P except that the processing times of all jobs are zero on machine 1, that is, p1,j = 0, j = 1, 2, . . . , n. Clearly, it is observed that The optimal value of Problem P0 is less than that of Problem P, and (rH, OH) is an optimal schedule for Problem P0 , and the optimal P P value can be expressed as j2IH p2;j . j2OH oj þ
Lemma 3.
X
oj þ p1;max þ
j2OH
X p2;j ; j2IH
Proof. Consider the schedule (r0 , OH) equivalent to the schedule (rH, OH) except that the jobs in IH are sequenced by decreasing order of p1,j, that is,
X
C max ðr Þ ¼ p1;max þ
p2;j :
Theorem 3. The decision version of Problem II is NP-complete. Proof. The proof of this theorem is given in Appendix B. h
5. Polynomiality of Problem III
Since rH is an optimal sequence for the two-machine flow shop with minimizing the makespan as the objective,
C max ðrH Þ 6 C max ðr0 Þ:
ð2Þ
From inequalities (1) and (2), Lemma 3 holds. h The worst-case performance ratio of Heuristic H, q, can be defined as
(
h
ð1Þ
j2IH
q ¼ inf r :
Proof. The proof of this theorem is given in Appendix A.
and jSj ¼ 2g ? Without loss of generality, assume that g P 10 and aj P 2, j = 1, 2, . . . , g.
where p1,max = max{p1,jjj 2 IH}.
0
Theorem 2. The decision version of Problem I is NP-complete.
The decision version of Problem II is stated as follows: Given a threshold k2, is there a schedule (r, O) such that T(r, O) 6 k2? To prove the NP-hardness of Problem II, we introduce the equal cardinality partition problem, which is stated as follows: Given g integers a1, a2, . . . , ag such that g is an even integer and P Ph j¼1 aj ¼ 2A, is there a subset S # {1, 2, . . . , g} such that j2S aj ¼ A
From the above observations, Lemma 2 holds. h
TðrH ; OH Þ 6
To prove the NP-hardness of Problem I, we introduce the partition problem, which is stated as follows: Given g integers a1, a2, Pg . . . , ag such that j¼1 aj ¼ 2A, is there a subset S # {1, 2, . . . , g} P such that j2S aj ¼ A?
) TðrH ; OH Þ 6 r : Tðr ; O Þ
This section shows that Problem III is polynomially solvable. We consider only the case where q1 < q2. From the reversibility of flow shops, it follows that the results of the case under consideration also apply to that case where q1 P q2. Let (r⁄, O⁄) be an optimal schedule. Let J1 = {jjpj < q1}, J2 = {jjq1 6 pj < q2}, and J3 = {jjq2 6 pj}. Then,
8 > < ðpj ; pj Þ for j 2 J 1 ; ðp1;j ; p2;j Þ ¼ ðq1 ; pj Þ for j 2 J 2 ; > : ðq1 ; q2 Þ for j 2 J 3 : Let job j be referred to as a critical job if it satisfies the following conditions:
p
P Theorem 1. We have a tight bound q ¼ 1 þ P o 1;max , þ p j2OH j j2IH 2;j where p1,max = max{p1,jjj 2 IH}.
Proof. From Lemmas 2 and 3,
TðrH ; OH Þ 6 Tðr ; O Þ
P
P þ p1;max þ j2IH p2;j p1;max P P 61þP : H oj þ H p2;j H oj þ j2I j2IH p2;j j2O j2O
j2OH oj
P
To complete the proof, we show, by example, that this bound is tight in the limit: Consider a two-job, two-machine ordered flow shop in which p1,1 = , p1,2 = 4, p2,1 = 100, p2,2 = 100 + , o1 = 100 , and o2 = 200, where > 0 is sufficiently small. Since OH = {1}, IH = {2}, O⁄ = ;, and I⁄ = {1, 2},
q¼1þ
4 ; 200
Thus, as
? 0,
q
Tðr ; O Þ ¼ 200 þ 2 and TðrH ; OH Þ ¼ 200 þ 4:
TðrH ; OH Þ 4 2 ! 0: ¼ 1 þ 1 þ Tðr ; O Þ 200 200 þ 2
4. Computational complexity of Problems I and II This section shows that Problems I and II are NP-hard. The decision version of Problem I is stated as follows: Given a threshold k1, is there a schedule (r, O) such that T(r, O) 6 k1?
job j can be processed on machine 2 immediately after it is completed on machine 1, and the in-house jobs positioned after job j can be processed continuously on machine 2 immediately after job j is completed. Note that since p1,j 6 p2,j, j = 1, 2, . . . , n, an increasing sequence of pj minimizes the makespan according to Johnson’s rule. Thus, henceforth, we consider only the schedule such that in-house jobs are processed in increasing order of pj. For simplicity, we assume that job k is the critical job in an optimal schedule (r⁄, O⁄) throughout Lemmas 4–6. Lemma 4. If k 2 J1, then O⁄ = O1 [ O2, where O1 = {jjpj > pk for j = 1, 2, . . . , n} and O2 = {jjpj > oj and pj 6 pk}. Proof. Without loss of generality, assume that if pk0 ¼ pk and k0 2 I⁄, then job k is processed after job k0 in (r⁄, O⁄). Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l1 in O1 is processed in-house immediately after job k. Since pk < q1 and pl1 > pk , C 1;l1 ðr Þ > C 2;k ðr Þ, which implies that job k is not critical job. This is a contradiction. Henceforth, we consider only schedules with O1 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that ; OÞ be the schedule such job l2 in O2 is processed in-house. Let ðr that O ¼ O [ fl2 g. Since pl2 6 pk , job k remains a critical job in ; OÞ. Thus, ðr
49
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
; OÞ ¼ Tðr ; O Þ pl2 þ ol2 : Tðr
; OÞ < Tðr ; O Þ. This is a contradiction. HenceSince pl2 > ol2 , Tðr forth, we only consider schedules with O1 [ O2 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l3 R O1 [ O2 is outsourced. Since job l3 R O1 [ O2, it is observed that
l3 2 J 1 ;
pl3 6 pk
and pl3 6 ol3 :
b be the schedule such that O b ¼ O n fl3 g. Since p 6 p , job ^ ; OÞ Let ðr l3 k b Thus, ^ ; OÞ. k remains a critical job in ðr
b ¼ Tðr ; O Þ þ p ol : ^ ; OÞ Tðr l3 3
; OÞ 6 Tðr ; O Þ. By applying this argument Since pl3 6 ol3 , Tðr repeatedly, Lemma 4 is obtained or a contradiction occurs. h Note that since the frameworks of the proofs of Lemmas 5 and 6 below are similar to the proof of Lemma 4, these proofs are given in Appendices C and D, respectively.
Lemma 5. If k 2 J2, then O⁄ = O1 [ O2 [ O3, where O1 = {jjpj < pk for j 2 J2}, O2 = {jjpj > oj for j 2 J1}, and O3 = {jjp2,j > oj and pj P pk for j 2 J2 [ J3}.
follows. Consider the set of (g + 1) jobs {1, 2, . . . , g + 1} such that q1 = A2, q2 = 1,
( pj ¼
aj A
3
for j ¼ 1; . . . ; g; oj ¼ for j ¼ g þ 1;
8 < aj aAj
for j ¼ 1; . . . ; g;
:
for j ¼ g þ 1;
A
4
and k1 ¼ A3 þ 2A 1 þ A1. This reduction can be carried out in polynomial time. Suppose that there is a set S to the partition problem such that P ; OÞ such that a ¼ A. Then we can construct schedule ðr j2S j jobs in f1; 2; . . . ; gg n S are outsourced, that is, O ¼ f1; 2; . . . ; gg nS, jobs in S [ fg þ 1g are processed in-house, and ¼ ðS; g þ 1Þ, jobs in S are processed in an arbitrary order. given r Then, since
P
a j2O j
¼ A 1,
X X ; OÞ ¼ Tðr p1;j þ p1;gþ1 þ p2;gþ1 þ oj j2S
j2O
X 1 1 1 ¼ A3 þ 2A 1 þ ¼ k1 : ¼ þ A þ A3 þ aj 1 A A A j2O
Proof. The proof of this lemma is given in Appendix C. h
b for Problem I such that ^ ; OÞ Suppose that there exists a schedule ðr b 6 k1 . Note that since p1,j 6 p2,j, j = 1, 2, . . . , g + 1, an increasing ^ ; OÞ Tðr
Lemma 6. If k 2 J3, then O⁄ = O1 [ O2, where O1 = J2 and O2 = {jjp2,j > oj for j 2 J1 [ J3}.
sequence of pj minimizes the makespan according to Johnson’s rule. In addition, since A4 > k1, job g + 1 should be processed in-house b Thus, we only consider schedules such that r ^ is an ^ ; OÞ. under ðr increasing sequence of pj and job n + 1 is processed in-house. Let b Then, r bI ¼ f1; 2; . . . ; gg n O. ^ ¼ ðbI; g þ 1Þ, where jobs in bI are
Proof. The proof of this lemma is given in Appendix D. h Based on Lemmas 4–6, we can construct the following algorithm. Algorithm for Problem III Step 1: Set j = 1, O⁄ = ;, and (r⁄, O⁄) = (;, {1, 2, . . . , n}). Step 2: If j = n + 1, then (r⁄, O⁄) is an optimal schedule and STOP. Otherwise, construct schedule (r0 , O0 ) with j as a critical job, based on Lemmas 4–6. Step 3: If T(r0 , O0 ) < T(r⁄, O⁄), then set (r⁄, O⁄) = (r0 , O0 ). Step 4: Let j = j + 1 and go to Step 2. Note that the algorithm for Problem III can be carried out in O(n2) time. 6. Conclusions We consider a two-machine ordered flow shop problem with an outsourcing option, which is known to be NP-hard. Thus, we present an approximation algorithm and consider three special cases, referred to as Problems I–III. We show that Problems I and II are NP-hard and present a polynomial-time algorithm for Problem III. For future research, it would be interesting to develop an FPTAS, if exists, for our problem. Note that since the pseudo-polynomial time algorithm for the general version of our problem in Choi and Chung (2011) does not satisfy some conditions for the existence of an FPTAS in Woeginger (2000), it is necessary to develop a dynamic program different from Choi and Chung’s algorithm. Furthermore, it is another interesting problem to analyze the computational complexity of Problem III with more than two machines. Appendix A. Proof of Theorem 2 It is clear that the decision version is in NP. Given an instance of the partition problem, we can construct an instance of Problem I as
sequenced by increasing order of pj, and
8 9 > X b ¼ max ^ ; OÞ Tðr p1;j þ p1;gþ1 ; p1;r^ ð1Þ þ p2;j þ p2;gþ1 þ oj > > : b ; j2 I j2b I j2b O 8 9 > > <1 X X pr^ ð1Þ X = ¼ max pj þ A; 2 þ pj þ A3 þ oj ; 2 > > A A : ; j2b I j2b I j2b O ^ . Consider now where pr^ ð1Þ is the processing time of the first job in r the following three cases. Case 1:
P
pj > A
j2b I
Without loss of generality, let
P
pj ¼ A þ k and
j2b I
1 A
where k is a positive integer. Since A þ þ P aj ¼ A k, j2b O
k A2
P
j2b O
pj ¼ A k,
pr^ ð1Þ A2
and
X X 1 3 p þ A þ a 1 j j A A2 j2b I j2b O pr^ ð1Þ 1 ¼ 2 þ A þ k þ A3 þ ðA kÞ 1 A A k pr^ ð1Þ 3 ¼ A þ 2A 1 þ þ 2 > k1 : A A
b ¼ ^ ; OÞ Tðr
pr^ ð1Þ
þ
This is a contradiction. P Case 2: bpj < A j2 I
Without loss of generality, let
P
pj ¼ A k and
j2b I
1 A
where k is a positive integer. Since A þ P b aj ¼ A þ k, j2 O
k A2
P
j2b O
pj ¼ A þ k,
>Akþ
pr^ ð1Þ A2
and
50
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
X X 1 3 b ¼ 1 ^ ; OÞ Tðr p þ A þ A þ a 1 j j A A2 b b j2 I
j2 O
1 k 1 þ A þ A3 þ ðA þ kÞð1 Þ A A2 A 1 1 1 ¼ A3 þ 2A 1 þ þ kð1 2 Þ > k1 : A A A
¼
This is a contradiction. P Case 3: bpj ¼ A
b jbIj ¼ j Oj b ¼ g. ^ ; OÞ, Claim. In schedule ðr 2
j2 I
Since
P
j2b O
pj ¼
P
p ¼ A,
j j2b I X 1 1 1 b ¼ þ A þ A3 þ ^ ; OÞ ¼ A3 þ 2A 1 þ ¼ k1 : Tðr pj 1 A A A j2b O
By Cases 1–3,
P
X X X b ¼p ^ ; OÞ Tðr p1;j þp1;nþ2 ;p2;gþ1 þ p2;j gþp2;gþ2 þ oj 1;gþ1 þmaxf b b b j2 I j2 I j2 O 8 9 > > > 2 2 : b ; j2 I j2b O
Proof. Suppose that jbIj ¼ 2g þ k, where k is a positive integer. Note b ¼ g k. Then, that j Oj 2
X 2 ðA A þ aj Þ j2b O g g X g þ k A2 þ A2 þ k ðA2 AÞ þ aj ¼ 1 þ A2 þ 2 2 2 j2b O g X 3g þ 1 A2 k Aþ1þ ¼ aj > k2 : 2 2 j2b O
b P 1 þ A2 þ jbIjA2 þ g A2 þ ^ ; OÞ Tðr 2
b Thus, the set bI is a solution to the ^ ; OÞ. aj ¼ A in ðr
j2b I
partition problem. The proof is complete. Appendix B. Proof of Theorem 3 It is clear that the decision version is in NP. Given an instance of the equal cardinality partition problem, we can construct an instance of Problem II as follows. Consider the set of (g + 2) jobs {1,2, . . . , g + 2} such that q1 = 0, q2 = A2,
8 ( 2 > < aj A for j ¼ 1; . . . ; g; A A þ aj for j ¼ 1; . . . ; g; for j ¼ g þ 1; pj ¼ 1 oj ¼ > :n 2 for j ¼ g þ 1; g þ 2; A4 A for j ¼ g þ 2; 2 And k2 ¼ 32 g þ 1 A2 2g 1 A þ 1. This reduction can be carried out in polynomial time. Suppose that there is a set S in the equal cardinality partition P a ¼ A and jSj ¼ 2g . Then we can construct problem such that j2S j ; OÞ such that a schedule ðr jobs in f1; 2; . . . ; gg n S are outsourced, that is, O ¼ f1; 2; . . . ; gg nS, jobs in S [ fg þ 1; g þ 2g are processed in-house, and ¼ ðg þ 1; S; g þ 2Þ, jobs in S are processed in an arbitrary Given r order. Then, since
P
a j2S j
¼
P
a j2O j
¼ A and jOj ¼ 2g ,
X X ; OÞ ¼ p1;gþ1 þ Tðr p1;j þ p1;gþ2 þ p2;gþ2 þ oj j2S
¼1þ
X j2S
j2O
X 2 g g aj A þ A2 þ A2 þ ðA A þ aj Þ 2 2 j2O
g 3g 2 g ¼ 1 þ A þ gA þ ðA2 AÞ þ A ¼ A 1 Aþ1 2 2 2 ¼ k2 : 2
This is a contradiction. Suppose that jbIj ¼ 2g k, where k is a positive integer. Note that b ¼ g þ k. Thus, O 2
X X 2 g g aj A þ A2 þ A2 þ ðA A þ aj Þ 2 2 j2b I j2b O g X X þ k ðA2 AÞ þ aj A þ nA2 þ aj ¼1þ 2 j2b I j2b O 0 1 X X 3n g B C þ k A2 @ þ k ¼ aj AA þ 1 þ aj 2 2 j2b I j2b O 0 1 X BX C 2 ¼ k2 þ ðk 1ÞA þ @ aj k 1AA þ aj : b b j2 I j2 O
b P1þ ^ ; OÞ Tðr
Now, we only show that
! X X ðk 1ÞA þ aj k 1 A þ aj > 0: j2I2 j2b O 2
If k = 1, then inequality (3) can be rewritten as
0 1 X X B C aj > 0: @ aj 2AA þ j2b I j2b O
2
b for Problem II such ^ ; OÞ Suppose that there exists a schedule ðr b 6 k2 . Note that since p1,j 6 p2,j, j = 1, 2, . . . , g + 1, an ^ ; OÞ that Tðr increasing sequence of pj minimizes the makespan according to Johnson’s rule. In addition, since A4 > k2, jobs g + 1 and g + 2 should b Thus, we only consider ^ ; OÞ. be processed in-house under ðr ^ schedules such that r is an increasing sequence of pj and jobs b g + 1 and g + 2 are processed in-house. Let bI ¼ f1; 2; . . . ; gg n O. ^ ¼ ðg þ 1; bI; g þ 2Þ, where jobs in bI are sequenced in Then, r increasing order of pj. Since p2,j = max{pj, q2} = A2, j = 1, 2, . . . , g,
ð3Þ
Note that bI – ;. Since n P 10 and aj P 2, (3) holds. If k P 2, then
ðk 1ÞA2 þ ð
P
aj > 2. Thus, inequality
j2b I
X X aj k 1ÞA þ aj > A2 ðk þ 1ÞA > 0: j2b I j2b O
Note that A > k + 1. Thus, inequality (3) holds. Thus, since inequality b > k2 . This is a contradiction. Thus, the proof of the ^ ; OÞ (3) holds, Tðr claim is complete. h By the claim, henceforth, we consider only schedules with b ¼ h. Then, jbIj ¼ j Oj 2
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
8 9 > > > 2 2 2 2 : b ; j2 I j2b O 8 9 > > > 2 : b ; 2 j2 I j2b O Consider now the following three cases. Case 1:
P
aj > A
j2b I
P P Without loss of generality, let aj ¼ A þ k and aj ¼ A k, j2b I j2b O where k is a positive integer. Since A > 1,
b ¼ 1 þ AðA þ kÞ þ 3g A2 g A þ A k ^ ; OÞ Tðr 2 2 g 3g 2 þ1 A 1 A þ 1 þ ðA 1Þk > k2 : ¼ 2 2
j2 I
P
aj ¼ A k and Without loss of generality, let j2b I where k is a positive integer. Since k > 0,
P
j2b O
aj ¼ A þ k,
b ¼ 1 þ A2 þ 3g A2 g A þ A þ k ^ ; OÞ Tðr 2 2 g 3g 3 þ1 A 1 A þ 1 þ k > k2 : ¼ 2 2 This is a contradiction. P Case 3: baj ¼ A j2 I
Since
j2b O
aj ¼
P
aj ¼ A,
j2b I
b ¼ 1 þ A2 þ 3g A2 g A þ A ^ ; OÞ Tðr 2 2 g 3g 3 þ1 A 1 A þ 1 ¼ k2 : ¼ 2 2 By Cases 1–3,
P
b Since p1,j = q1 6 p2,k = pk and p1,j < p2.j for ^ ; OÞ. tioned after job k in ðr b ^ ; OÞ. Thus, j 2 I k , job k remains a critical job in ðr
b ¼ Tðr ; O Þ p þ ol : ^ ; OÞ Tðr l3 3 b < Tðr ; O Þ. This is a contradiction. Thus, ^ ; OÞ Since pl3 > ol3 , Tðr henceforth, we only consider schedules with O1 [ O2 [ O3 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l4 R O1 [ O2 [ O3 is outsourced, which implies that pl4 6 ol4 . Let e be the schedule such that O e ¼ O n fl4 g. If l4 2 J1, job k re~ ; OÞ ðr e because p 6 q , whereas if l4 2 J2 [ J3, ~ ; OÞ mains a critical job in ðr l4 1 e because p ¼ q 6 p . Thus, ~ ; OÞ job k remains a critical job in ðr 1;l4 1 2;k e and ~ ; OÞ it is observed that job k remains a critical job in ðr
e ¼ Tðr ; O Þ þ p ol : ~ ; OÞ Tðr l4 4 e 6 Tðr ; O Þ. By applying this argument ~ ; OÞ Since pl4 6 ol4 , Tðr repeatedly, Lemma 5 is obtained or a contradiction occurs. The proof is complete. Appendix D. Proof of Lemma 6
This is a contradiction. P Case 2: baj < A
P
51
b Thus, the set bI is a solution to the ^ ; OÞ. aj ¼ A in ðr
j2b I
equal cardinality partition problem. The proof is complete. Appendix C. Proof of Lemma 5 Without loss of generality, assume that if pk0 = pk, then job k is processed before job k0 in (r⁄, O⁄). Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l1 in O1 is processed in-house immediately before job k. Since p1;k ¼ q1 < p2;l1 , however, C 1;k ðr Þ < C 2;l1 ðr Þ, implying job k is not a critical job. This is a contradiction. Thus, henceforth, we only consider schedules with O 1 # O ⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l2 in O2 is processed in-house. Note that job k is positioned first among in-house jobs in J2 in schedule (r⁄, O⁄). Thus, job l2 is posi ; OÞ be the schedule such tioned immediately before job k. Let ðr that O ¼ O [ fl2 g. Since p1;l2 ¼ p2;l2 ¼ pl2 < q1 , job k remains a crit ; OÞ. Thus, ical job in ðr
; OÞ ¼ Tðr ; O Þ pl2 þ ol2 : Tðr ; OÞ < Tðr ; O Þ. This is a contradiction. Thus, Since pl2 > ol2 , Tðr henceforth, we only consider schedules with O1 [ O2 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that b be the sche^ ; OÞ job l3 in O3 is processed in-house after job k. Let ðr b ¼ O [ fl3 g, and bI k be the set of in-house jobs posidule such that O
Without loss of generality, assume that job k is positioned first among in-house jobs in J3. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l1 in O1 is processed immediately before job k on an in-house machine. Since p1;k ¼ q1 < p2;l1 ¼ pl1 , however, C1,k(r⁄) < C2,l(r⁄). This is a contradiction. Thus, henceforth, we only consider schedules with O1 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that ; OÞ be the schedule such job l2 in O2 is processed in-house. Let ðr ; OÞ bethat O ¼ O [ fl2 g. If l2 2 J1, job k remains a critical job in ðr cause p1,j = p2,j = pj and pj < q1 for j 2 J1, whereas if l2 2 J3, job k re ; OÞ because (p1,j, p2,j) = (q1, q2) for j 2 J3 mains a critical job in ðr and q1 < q2. Thus, it is observed that job k remains a critical job ; OÞ and in ðr
; OÞ ¼ Tðr ; O Þ p2;l2 þ ol2 : Tðr ; OÞ < Tðr ; O Þ. This is a contradiction. Thus, Since p2;l2 > ol2 , Tðr henceforth, we only consider schedules with O1 [ O2 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l3 R O1 [ O2 is outsourced. Since job l3 R O1 [ O2, it is observed that
l3 2 J 1 [ J 3 and p2;l3 6 ol3 : b be the schedule such that O b ¼ O n fl3 g. If l3 2 J1, job k re^ ; OÞ Let ðr b ^ ; OÞ because p1;l3 ¼ p2;l3 ¼ pl3 < pk , whereas mains a critical job in ðr b because p ¼ p ¼ q ^ ; OÞ if l3 2 J3, job k remains a critical job in ðr 1;l3 1;k 1 and p2;l3 ¼ p2;k ¼ q2 . Thus,
b ¼ Tðr ; O Þ þ p ol : ^ ; OÞ Tðr l3 3 ; OÞ 6 Tðr ; O Þ. By applying this argument Since pl3 6 ol3 ; Tðr repeatedly, Lemma 6 is obtained or a contradiction occurs. The proof is complete. References Allahverdi, A., 1996. Two-machine proportionate flowshop scheduling with breakdown to minimize maximum lateness. Computers and Operations Research 23, 909–916. Cachon, G.P., Harker, P.T., 2002. Competition and outsourcing with scale economies. Management Science 48, 1314–1333. Chen, Z.L., Li, C.L., 2008. Scheduling with subcontracting options. IIE Transactions 40, 1171–1184. Choi, B.C., Chung, J.B., 2011. Two-machine flow shop scheduling problem with an outsourcing option. European Journal of Operational Research 213, 66–72. Choi, B.C., Leung, J.Y.T., Pinedo, M.L., 2011. Minimizing makespan in an ordered flow shop with machine-dependent processing times. Journal of Combinatorial Optimization. 22, 797–818. Choi, B.C., Leung, J.Y.T., Pinedo, M.L., 2010. A note on makespan minimization in proportionate flow shops. Information Processing Letters 111, 77–81.
52
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
Choi, B.C., Yoon, S.H., Chung, S.J., 2007. Minimizing maximum completion time in a proportionate flow shop with one machine of different speed. European Journal of Operational Research 176, 964–974. Chung, D., Lee, K., Shin, K., Park, J., 2005. A new approach to job shop scheduling problems with due date constraints considering operation subcontracts. International Journal of Production Economics 98, 238–250. Hou, S., Hoogeveen, H., 2003. The three-machine proportionate flow shop with unequal machine speeds. Operations Research Letters 31, 225–231. Koulamas, C., Kyparisis, G.J., 2009. A note on the proportional flow shop with a bottleneck machine. European Journal of Operational Research 193, 644–645. Lee, I.S., Sung, C.S., 2008a. Minimizing due date related measures for a single machine scheduling problem with outsourcing allowed. European Journal of Operational Research 186, 931–952. Lee, I.S., Sung, C.S., 2008b. Single machine scheduling with outsourcing allowed. International Journal of Production Economics 101, 623–634. Lee, K.B., Choi, B.C., 2011. Two-stage production scheduling with an outsourcing option. European Journal of Operational Research 213, 489–497. Ow, P.S., 1985. Focused scheduling in proportionate flow shops. Management Science 31, 852–869. Panwalkar, S.S., Khan, A.W., 1976. An ordered flow shop sequencing problem with mean completion time criterion. International Journal of Production Research 14, 631–635.
Pinedo, M., 2002. Scheduling: Theory, Algorithms and Systems, second ed. Prentice Hall, New York. Smith, M.L., Panwalkar, S.S., Dudek, R.A., 1975. Flow shop sequencing problem with ordered processing time matrices. Management Science 21, 544–549. Smith, M.L., Panwalkar, S.S., Dudek, R.A., 1976. Flow shop sequencing problem with ordered processing time matrices: A general case. Naval Research Logistics Quarterly 23, 481–486. Qi, X., 2008. Coordinated logistics scheduling for in-house production and outsourcing. IEEE Transactions on Automation Science and Engineering 5, 188–192. Qi, X., 2009. Two-stage production scheduling with an option of outsourcing from a remote supplier. Journal of System Science and System Engineering 18, 1–15. Qi, X., 2011. Outsourcing and production scheduling for a two-stage flow shop. International Journal of Production Economics 129, 43–50. Webster, M., Alder, C., Muhlemann, A.P., 1997. Subcontracting within the supply chain for electronics assembly manufacture. International Journal of Operations and Production Management 17, 827–841. Woeginger, G.J., 2000. When does a dynamic programming formulation guarantee the existence of a fully polynomial time approximation scheme? INFORMS Journal on Computing 12, 57–74.
Contents lists available at SciVerse ScienceDirect
European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor
Discrete Optimization
Outsourcing and scheduling for two-machine ordered flow shop scheduling problems q Dae-Young Chung a, Byung-Cheon Choi b,⇑ a b
SAP Korea Ltd., 27-28F, MMAA Building, 467-12 Dogok-dong, Gangnam-gu, Seoul 135-700, Republic of Korea Department of Business Administration, Chungnam National University, 79 Daehangno, Yuseong-gu, Daejeon 305-704, Republic of Korea
a r t i c l e
i n f o
Article history: Received 21 July 2011 Accepted 29 October 2012 Available online 10 November 2012 Keywords: Scheduling Outsourcing Ordered flow shop Computational complexity
a b s t r a c t This paper considers a two-machine ordered flow shop problem, where each job is processed through the in-house system or outsourced to a subcontractor. For in-house jobs, a schedule is constructed and its performance is measured by the makespan. Jobs processed by subcontractors require paying an outsourcing cost. The objective is to minimize the sum of the makespan and the total outsourcing cost. Since this problem is NP-hard, we present an approximation algorithm. Furthermore, we consider three special cases in which job j has a processing time requirement pj, and machine i a characteristic qi. The first case assumes the time job j occupies machine i is equal to the processing requirement divided by a characteristic value of machine i, that is, pj/qi. The second (third) case assumes that the time job j occupies machine i is equal to the maximum (minimum) of its processing requirement and a characteristic value of the machine, that is, max{pj, qi} (min{pj, qi}). We show that the first and the second cases are NP-hard and the third case is polynomially solvable. Ó 2012 Elsevier B.V. All rights reserved.
1. Introduction In today business environment, many companies are looking to outsource, that is, entrust some jobs to an external provider. Through proper outsourcing, they can deliver goods to consumers in time, reduce operating costs and make their organizations more flexible. Thus, proficient use of outsourcing can make a company more competitive (Cachon and Harker, 2002; Webster et al., 1997), which provides the authors with motivation to consider a scheduling problem with an outsourcing option. This paper considers the two-machine, n-job ordered flow shop problem, where the processing time of job j on machine i, pi,j, follows two special characteristics (Smith et al., 1975, 1976): If, for some 1 6 i0 6 2, pi0 ;j0 6 pi0 ;j , then pi;j0 6 pi;j for i = 1, 2, and if pi0 ;j0 6 pi;j0 , then pi0 ;j 6 pi;j for j = 1, . . . , n. Furthermore, there are two options for processing jobs: The first is to process the job by utilizing in-house machines, while the second is to outsource the job to a subcontractor. For in-house jobs, a schedule r is constructed and its performance is measured by the makespan, that is, the latest completion time of jobs processed in-house, denoted Cmax(r). If job j is processed by subcontractors, an outsourcing cost q This study was financially supported by the research fund of Chungnam National University in 2011. ⇑ Corresponding author. E-mail addresses: [email protected] (D.-Y. Chung), [email protected] (B.-C. Choi).
0377-2217/$ - see front matter Ó 2012 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.ejor.2012.10.048
oj should be paid. Note that the total outsourcing costs can be P described as j2O oj , where O is the set of jobs outsourced. The objective is to minimize the sum of the makespan and total outsourcing costs. The problem considered in this paper is defined below. Problem P. In a two-machine, n-job ordered flow shop, determine a schedule (r, O) that minimizes T(r, O), defined as
Tðr; OÞ ¼ C max ðrÞ þ
X oj : j2O
Since Problem P is known to be NP-hard (Choi and Chung, 2011), we present an approximation algorithm and consider three special cases such that job j has a processing time pj that affects the time job j spends on each of two machines, and machine i has a characteristic value qi that affects the time each job spends on machine i. First, we consider a setting in which the processing time of job j on machine i can be expressed as pi,j = pj/qi. In this case, qi can be interpreted as the speed of machine i, which implies that job processing times are inversely proportional to machine speeds. Let this case be referred to as Problem I. The second case considers a setting in which job j must spend a minimum time qi on machine i. The processing time of job j on machine i can then be expressed
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
as pi,j = max{pj, qi}. This minimum amount of time can also be due to setting or fine-tuning the machine; however, in this case the job can already start while the machine is being set up. Let this case be referred to as Problem II. Third, if, on the other hand, job j can spend at most time qi on machine i, then the processing time of job j on machine i can be expressed as pi,j = min{pj, qi}. This last case can occur when the time of machine i is expensive and must be rationed; no job is allowed more than a maximum time qi on the machine. Let this case be referred to as Problem III. It is easy to see that Problems I–III are ordered flow shops. The rest of the paper is organized as follows. Section 2 discusses the related literature. Section 3 presents an approximation algorithm for Problem P. Section 4 proves that Problems I and II are NP-hard. Section 5 shows that Problem III can be solved in polynomial time. Finally, Section 6 draws our conclusions.
2. Literature review Many researchers have focused on specially structured flow shops, such as the ordered flow shop (Panwalkar and Khan, 1976; Smith et al., 1975, 1976) and the proportionate flow shop (Allahverdi, 1996; Hou and Hoogeveen, 2003; Ow, 1985), where the processing time of job j on all the machines is the same and equal to pj, that is, pij = pj (Pinedo, 2002). Smith et al. (1975, 1976) considered an m-machine ordered flow shop with the makespan objective. Smith et al. (1975) considered a special case in which the job processing times on the first (last) machine are the longest and showed that the problem can be solved in polynomial time. The authors showed that if the first (last) machine is the slowest, the LPT (SPT) sequence is an optimal permutation schedule. Smith et al. (1976) showed that there exists an optimal schedule that has an inverted V-shape. Hou and Hoogeveen (2003) and Choi et al. (2007) considered a three-machine flow shop with pi,j = pj/qi with the objective of minimizing the makespan. They showed that the problem is NP-hard in the ordinary sense. Furthermore, Choi et al. (2007) have proposed a heuristic of which tight worst-case bound has been proven to be 3/2 in Koulamas and Kyparisis (2009). Choi et al. (in press) considered an m-machine flow shop with pi,j = pj + qi, with the objective of minimizing the makespan. The authors showed that the problem is solvable in polynomial time when the number of machines is fixed. Choi et al. (2010) considered two m-machine flow shops with pi,j = max{pj, qi} and then pi,j = min{pj, qi}, with the objective of minimizing the makespan. They showed that the first problem is NP-hard, and the second is solvable in polynomial time. Under a proportionate flow shop environment, the makespan is constant independently of the sequence if the permutation schedules are considered (Pinedo, 2002). The literature on scheduling with an outsourcing option has studied the problem of simultaneously deciding which jobs to be outsourced and how to schedule in-house jobs. Qi (2008) considered a single-machine scheduling problem that allows some jobs to be outsourced to a single outsourcing machine, with a transportation delay and transportation cost. Lee and Sung (2008a,b) considered a single-machine scheduling problem to minimize the weighted sum of the outsourcing cost and the associated scheduling cost with an outsourcing budget constraint, where the scheduling cost is the one of the following: the maximum lateness, denoted by Lmax = max16j6n{Cj dj}; the total P P tardiness, denoted by T j ¼ 16j6n maxf0; C j dj g; and the total P completion time, denoted by C j , where Cj and dj are the completion time and due date of job j, respectively. Chen and Li (2008) considered a parallel machine scheduling problem with an outsourcing option to minimize the total production cost subject to a constraint on the makespan. Qi (2009) considered a
47
two-machine flow shop scheduling problem that allows only operations on the first machine to be outsourced. For practical reasons, the author’s model considered a single-batch system with a transportation delay, with the objective of minimizing the makespan subject to a constraint on the maximum outsourcing cost. Qi (2011) extended this model to include other forms of outsourcing. Lee and Choi (2011) considered a two-machine flow shop problem that allows any operations to be outsourced in which the objective is to minimize the sum of the makespan and the total outsourcing costs. Choi and Chung (2011) considered two-machine flow shop problem such that two operations comprising the job must be oursourced when a job is determined to be outsourced. The objective is to minimize the sum of the makespan and total oursourcing costs. Chung et al. (2005) considered a job shop scheduling problem with an outsourcing option. The problem is the same as a typical job shop scheduling problem, except that the capacity constraints can be violated with costs, which implies each operation can be outsourced, with operation-dependent costs. The author presented a heuristic algorithm that decomposes the problem into smaller subproblems and solves each subproblem optimally, and demonstrated its performance experimentally.
3. Approximation algorithm for Problem P This section presents an approximation algorithm, referred to as Heuristic H, for Problem P, which is defined as follows. Heuristic H: Outsource a job if its outsourcing cost is not greater than the larger of its processing times, that is, In case of p1,j < p2,j, outsource job j if p2,j P oj. In case of p1,j P p2,j, outsource job j if p1,j P oj. Henceforth, we consider only the case where p1,j < p2,j, j = 1, 2, . . . , n. From the reversibility of flow shops, it follows that the results of the case under consideration also apply to that case where p1,j P p2,j, j = 1, 2, . . . , n. Let (r⁄, O⁄) be an optimal schedule and (rH, OH) be the schedule obtained by Heuristic H. Let I⁄ = {1, 2, . . . , n}nO⁄ and IH = {1, 2, . . . , n}nOH. It is assumed that jobs in I⁄ and IH are sequenced according to Johnson’s rule. Lemma 1. If OH = {1, 2, . . . , n}, then (rH, OH) becomes an optimal schedule for the two-machine ordered flow shop problem with an outsourcing option. Proof. Since OH = {1, 2, . . . , n}, it is observed that p2,j P oj, j = 1, 2, . . . , n. Suppose that there exists an optimal schedule (r⁄, O⁄) such that I⁄ – ;. Without loss of generality, let job k be positioned last on an in-house machine. We can construct a new schedule ; OÞ by letting O ¼ O [ fkg. Then, it is observed that the makeðr span is decreased by at least p2,j and the outsourcing cost is increased by ok. From these observations, one obtains the inequality.
; OÞ 6 Tðr ; O Þ p2;k þ ok : Tðr Note, however, that p2,k P ok. Thus, by applying this argument repeatedly, Lemma 1 can be obtained or a contradiction occurs. h By Lemma 1, henceforth, we only consider instances where for some k, p2,k < ok. Lemma 2.
Tðr ; O Þ P
X j2OH
oj þ
X p2;j : j2IH
48
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
Proof. Consider Problem P0 equivalent to Problem P except that the processing times of all jobs are zero on machine 1, that is, p1,j = 0, j = 1, 2, . . . , n. Clearly, it is observed that The optimal value of Problem P0 is less than that of Problem P, and (rH, OH) is an optimal schedule for Problem P0 , and the optimal P P value can be expressed as j2IH p2;j . j2OH oj þ
Lemma 3.
X
oj þ p1;max þ
j2OH
X p2;j ; j2IH
Proof. Consider the schedule (r0 , OH) equivalent to the schedule (rH, OH) except that the jobs in IH are sequenced by decreasing order of p1,j, that is,
X
C max ðr Þ ¼ p1;max þ
p2;j :
Theorem 3. The decision version of Problem II is NP-complete. Proof. The proof of this theorem is given in Appendix B. h
5. Polynomiality of Problem III
Since rH is an optimal sequence for the two-machine flow shop with minimizing the makespan as the objective,
C max ðrH Þ 6 C max ðr0 Þ:
ð2Þ
From inequalities (1) and (2), Lemma 3 holds. h The worst-case performance ratio of Heuristic H, q, can be defined as
(
h
ð1Þ
j2IH
q ¼ inf r :
Proof. The proof of this theorem is given in Appendix A.
and jSj ¼ 2g ? Without loss of generality, assume that g P 10 and aj P 2, j = 1, 2, . . . , g.
where p1,max = max{p1,jjj 2 IH}.
0
Theorem 2. The decision version of Problem I is NP-complete.
The decision version of Problem II is stated as follows: Given a threshold k2, is there a schedule (r, O) such that T(r, O) 6 k2? To prove the NP-hardness of Problem II, we introduce the equal cardinality partition problem, which is stated as follows: Given g integers a1, a2, . . . , ag such that g is an even integer and P Ph j¼1 aj ¼ 2A, is there a subset S # {1, 2, . . . , g} such that j2S aj ¼ A
From the above observations, Lemma 2 holds. h
TðrH ; OH Þ 6
To prove the NP-hardness of Problem I, we introduce the partition problem, which is stated as follows: Given g integers a1, a2, Pg . . . , ag such that j¼1 aj ¼ 2A, is there a subset S # {1, 2, . . . , g} P such that j2S aj ¼ A?
) TðrH ; OH Þ 6 r : Tðr ; O Þ
This section shows that Problem III is polynomially solvable. We consider only the case where q1 < q2. From the reversibility of flow shops, it follows that the results of the case under consideration also apply to that case where q1 P q2. Let (r⁄, O⁄) be an optimal schedule. Let J1 = {jjpj < q1}, J2 = {jjq1 6 pj < q2}, and J3 = {jjq2 6 pj}. Then,
8 > < ðpj ; pj Þ for j 2 J 1 ; ðp1;j ; p2;j Þ ¼ ðq1 ; pj Þ for j 2 J 2 ; > : ðq1 ; q2 Þ for j 2 J 3 : Let job j be referred to as a critical job if it satisfies the following conditions:
p
P Theorem 1. We have a tight bound q ¼ 1 þ P o 1;max , þ p j2OH j j2IH 2;j where p1,max = max{p1,jjj 2 IH}.
Proof. From Lemmas 2 and 3,
TðrH ; OH Þ 6 Tðr ; O Þ
P
P þ p1;max þ j2IH p2;j p1;max P P 61þP : H oj þ H p2;j H oj þ j2I j2IH p2;j j2O j2O
j2OH oj
P
To complete the proof, we show, by example, that this bound is tight in the limit: Consider a two-job, two-machine ordered flow shop in which p1,1 = , p1,2 = 4, p2,1 = 100, p2,2 = 100 + , o1 = 100 , and o2 = 200, where > 0 is sufficiently small. Since OH = {1}, IH = {2}, O⁄ = ;, and I⁄ = {1, 2},
q¼1þ
4 ; 200
Thus, as
? 0,
q
Tðr ; O Þ ¼ 200 þ 2 and TðrH ; OH Þ ¼ 200 þ 4:
TðrH ; OH Þ 4 2 ! 0: ¼ 1 þ 1 þ Tðr ; O Þ 200 200 þ 2
4. Computational complexity of Problems I and II This section shows that Problems I and II are NP-hard. The decision version of Problem I is stated as follows: Given a threshold k1, is there a schedule (r, O) such that T(r, O) 6 k1?
job j can be processed on machine 2 immediately after it is completed on machine 1, and the in-house jobs positioned after job j can be processed continuously on machine 2 immediately after job j is completed. Note that since p1,j 6 p2,j, j = 1, 2, . . . , n, an increasing sequence of pj minimizes the makespan according to Johnson’s rule. Thus, henceforth, we consider only the schedule such that in-house jobs are processed in increasing order of pj. For simplicity, we assume that job k is the critical job in an optimal schedule (r⁄, O⁄) throughout Lemmas 4–6. Lemma 4. If k 2 J1, then O⁄ = O1 [ O2, where O1 = {jjpj > pk for j = 1, 2, . . . , n} and O2 = {jjpj > oj and pj 6 pk}. Proof. Without loss of generality, assume that if pk0 ¼ pk and k0 2 I⁄, then job k is processed after job k0 in (r⁄, O⁄). Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l1 in O1 is processed in-house immediately after job k. Since pk < q1 and pl1 > pk , C 1;l1 ðr Þ > C 2;k ðr Þ, which implies that job k is not critical job. This is a contradiction. Henceforth, we consider only schedules with O1 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that ; OÞ be the schedule such job l2 in O2 is processed in-house. Let ðr that O ¼ O [ fl2 g. Since pl2 6 pk , job k remains a critical job in ; OÞ. Thus, ðr
49
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
; OÞ ¼ Tðr ; O Þ pl2 þ ol2 : Tðr
; OÞ < Tðr ; O Þ. This is a contradiction. HenceSince pl2 > ol2 , Tðr forth, we only consider schedules with O1 [ O2 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l3 R O1 [ O2 is outsourced. Since job l3 R O1 [ O2, it is observed that
l3 2 J 1 ;
pl3 6 pk
and pl3 6 ol3 :
b be the schedule such that O b ¼ O n fl3 g. Since p 6 p , job ^ ; OÞ Let ðr l3 k b Thus, ^ ; OÞ. k remains a critical job in ðr
b ¼ Tðr ; O Þ þ p ol : ^ ; OÞ Tðr l3 3
; OÞ 6 Tðr ; O Þ. By applying this argument Since pl3 6 ol3 , Tðr repeatedly, Lemma 4 is obtained or a contradiction occurs. h Note that since the frameworks of the proofs of Lemmas 5 and 6 below are similar to the proof of Lemma 4, these proofs are given in Appendices C and D, respectively.
Lemma 5. If k 2 J2, then O⁄ = O1 [ O2 [ O3, where O1 = {jjpj < pk for j 2 J2}, O2 = {jjpj > oj for j 2 J1}, and O3 = {jjp2,j > oj and pj P pk for j 2 J2 [ J3}.
follows. Consider the set of (g + 1) jobs {1, 2, . . . , g + 1} such that q1 = A2, q2 = 1,
( pj ¼
aj A
3
for j ¼ 1; . . . ; g; oj ¼ for j ¼ g þ 1;
8 < aj aAj
for j ¼ 1; . . . ; g;
:
for j ¼ g þ 1;
A
4
and k1 ¼ A3 þ 2A 1 þ A1. This reduction can be carried out in polynomial time. Suppose that there is a set S to the partition problem such that P ; OÞ such that a ¼ A. Then we can construct schedule ðr j2S j jobs in f1; 2; . . . ; gg n S are outsourced, that is, O ¼ f1; 2; . . . ; gg nS, jobs in S [ fg þ 1g are processed in-house, and ¼ ðS; g þ 1Þ, jobs in S are processed in an arbitrary order. given r Then, since
P
a j2O j
¼ A 1,
X X ; OÞ ¼ Tðr p1;j þ p1;gþ1 þ p2;gþ1 þ oj j2S
j2O
X 1 1 1 ¼ A3 þ 2A 1 þ ¼ k1 : ¼ þ A þ A3 þ aj 1 A A A j2O
Proof. The proof of this lemma is given in Appendix C. h
b for Problem I such that ^ ; OÞ Suppose that there exists a schedule ðr b 6 k1 . Note that since p1,j 6 p2,j, j = 1, 2, . . . , g + 1, an increasing ^ ; OÞ Tðr
Lemma 6. If k 2 J3, then O⁄ = O1 [ O2, where O1 = J2 and O2 = {jjp2,j > oj for j 2 J1 [ J3}.
sequence of pj minimizes the makespan according to Johnson’s rule. In addition, since A4 > k1, job g + 1 should be processed in-house b Thus, we only consider schedules such that r ^ is an ^ ; OÞ. under ðr increasing sequence of pj and job n + 1 is processed in-house. Let b Then, r bI ¼ f1; 2; . . . ; gg n O. ^ ¼ ðbI; g þ 1Þ, where jobs in bI are
Proof. The proof of this lemma is given in Appendix D. h Based on Lemmas 4–6, we can construct the following algorithm. Algorithm for Problem III Step 1: Set j = 1, O⁄ = ;, and (r⁄, O⁄) = (;, {1, 2, . . . , n}). Step 2: If j = n + 1, then (r⁄, O⁄) is an optimal schedule and STOP. Otherwise, construct schedule (r0 , O0 ) with j as a critical job, based on Lemmas 4–6. Step 3: If T(r0 , O0 ) < T(r⁄, O⁄), then set (r⁄, O⁄) = (r0 , O0 ). Step 4: Let j = j + 1 and go to Step 2. Note that the algorithm for Problem III can be carried out in O(n2) time. 6. Conclusions We consider a two-machine ordered flow shop problem with an outsourcing option, which is known to be NP-hard. Thus, we present an approximation algorithm and consider three special cases, referred to as Problems I–III. We show that Problems I and II are NP-hard and present a polynomial-time algorithm for Problem III. For future research, it would be interesting to develop an FPTAS, if exists, for our problem. Note that since the pseudo-polynomial time algorithm for the general version of our problem in Choi and Chung (2011) does not satisfy some conditions for the existence of an FPTAS in Woeginger (2000), it is necessary to develop a dynamic program different from Choi and Chung’s algorithm. Furthermore, it is another interesting problem to analyze the computational complexity of Problem III with more than two machines. Appendix A. Proof of Theorem 2 It is clear that the decision version is in NP. Given an instance of the partition problem, we can construct an instance of Problem I as
sequenced by increasing order of pj, and
8 9 >
P
pj > A
j2b I
Without loss of generality, let
P
pj ¼ A þ k and
j2b I
1 A
where k is a positive integer. Since A þ þ P aj ¼ A k, j2b O
k A2
P
j2b O
pj ¼ A k,
pr^ ð1Þ A2
and
X X 1 3 p þ A þ a 1 j j A A2 j2b I j2b O pr^ ð1Þ 1 ¼ 2 þ A þ k þ A3 þ ðA kÞ 1 A A k pr^ ð1Þ 3 ¼ A þ 2A 1 þ þ 2 > k1 : A A
b ¼ ^ ; OÞ Tðr
pr^ ð1Þ
þ
This is a contradiction. P Case 2: bpj < A j2 I
Without loss of generality, let
P
pj ¼ A k and
j2b I
1 A
where k is a positive integer. Since A þ P b aj ¼ A þ k, j2 O
k A2
P
j2b O
pj ¼ A þ k,
>Akþ
pr^ ð1Þ A2
and
50
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
X X 1 3 b ¼ 1 ^ ; OÞ Tðr p þ A þ A þ a 1 j j A A2 b b j2 I
j2 O
1 k 1 þ A þ A3 þ ðA þ kÞð1 Þ A A2 A 1 1 1 ¼ A3 þ 2A 1 þ þ kð1 2 Þ > k1 : A A A
¼
This is a contradiction. P Case 3: bpj ¼ A
b jbIj ¼ j Oj b ¼ g. ^ ; OÞ, Claim. In schedule ðr 2
j2 I
Since
P
j2b O
pj ¼
P
p ¼ A,
j j2b I X 1 1 1 b ¼ þ A þ A3 þ ^ ; OÞ ¼ A3 þ 2A 1 þ ¼ k1 : Tðr pj 1 A A A j2b O
By Cases 1–3,
P
X X X b ¼p ^ ; OÞ Tðr p1;j þp1;nþ2 ;p2;gþ1 þ p2;j gþp2;gþ2 þ oj 1;gþ1 þmaxf b b b j2 I j2 I j2 O 8 9 > >
Proof. Suppose that jbIj ¼ 2g þ k, where k is a positive integer. Note b ¼ g k. Then, that j Oj 2
X 2 ðA A þ aj Þ j2b O g g X g þ k A2 þ A2 þ k ðA2 AÞ þ aj ¼ 1 þ A2 þ 2 2 2 j2b O g X 3g þ 1 A2 k Aþ1þ ¼ aj > k2 : 2 2 j2b O
b P 1 þ A2 þ jbIjA2 þ g A2 þ ^ ; OÞ Tðr 2
b Thus, the set bI is a solution to the ^ ; OÞ. aj ¼ A in ðr
j2b I
partition problem. The proof is complete. Appendix B. Proof of Theorem 3 It is clear that the decision version is in NP. Given an instance of the equal cardinality partition problem, we can construct an instance of Problem II as follows. Consider the set of (g + 2) jobs {1,2, . . . , g + 2} such that q1 = 0, q2 = A2,
8 ( 2 > < aj A for j ¼ 1; . . . ; g; A A þ aj for j ¼ 1; . . . ; g; for j ¼ g þ 1; pj ¼ 1 oj ¼ > :n 2 for j ¼ g þ 1; g þ 2; A4 A for j ¼ g þ 2; 2 And k2 ¼ 32 g þ 1 A2 2g 1 A þ 1. This reduction can be carried out in polynomial time. Suppose that there is a set S in the equal cardinality partition P a ¼ A and jSj ¼ 2g . Then we can construct problem such that j2S j ; OÞ such that a schedule ðr jobs in f1; 2; . . . ; gg n S are outsourced, that is, O ¼ f1; 2; . . . ; gg nS, jobs in S [ fg þ 1; g þ 2g are processed in-house, and ¼ ðg þ 1; S; g þ 2Þ, jobs in S are processed in an arbitrary Given r order. Then, since
P
a j2S j
¼
P
a j2O j
¼ A and jOj ¼ 2g ,
X X ; OÞ ¼ p1;gþ1 þ Tðr p1;j þ p1;gþ2 þ p2;gþ2 þ oj j2S
¼1þ
X j2S
j2O
X 2 g g aj A þ A2 þ A2 þ ðA A þ aj Þ 2 2 j2O
g 3g 2 g ¼ 1 þ A þ gA þ ðA2 AÞ þ A ¼ A 1 Aþ1 2 2 2 ¼ k2 : 2
This is a contradiction. Suppose that jbIj ¼ 2g k, where k is a positive integer. Note that b ¼ g þ k. Thus, O 2
X X 2 g g aj A þ A2 þ A2 þ ðA A þ aj Þ 2 2 j2b I j2b O g X X þ k ðA2 AÞ þ aj A þ nA2 þ aj ¼1þ 2 j2b I j2b O 0 1 X X 3n g B C þ k A2 @ þ k ¼ aj AA þ 1 þ aj 2 2 j2b I j2b O 0 1 X BX C 2 ¼ k2 þ ðk 1ÞA þ @ aj k 1AA þ aj : b b j2 I j2 O
b P1þ ^ ; OÞ Tðr
Now, we only show that
! X X ðk 1ÞA þ aj k 1 A þ aj > 0: j2I2 j2b O 2
If k = 1, then inequality (3) can be rewritten as
0 1 X X B C aj > 0: @ aj 2AA þ j2b I j2b O
2
b for Problem II such ^ ; OÞ Suppose that there exists a schedule ðr b 6 k2 . Note that since p1,j 6 p2,j, j = 1, 2, . . . , g + 1, an ^ ; OÞ that Tðr increasing sequence of pj minimizes the makespan according to Johnson’s rule. In addition, since A4 > k2, jobs g + 1 and g + 2 should b Thus, we only consider ^ ; OÞ. be processed in-house under ðr ^ schedules such that r is an increasing sequence of pj and jobs b g + 1 and g + 2 are processed in-house. Let bI ¼ f1; 2; . . . ; gg n O. ^ ¼ ðg þ 1; bI; g þ 2Þ, where jobs in bI are sequenced in Then, r increasing order of pj. Since p2,j = max{pj, q2} = A2, j = 1, 2, . . . , g,
ð3Þ
Note that bI – ;. Since n P 10 and aj P 2, (3) holds. If k P 2, then
ðk 1ÞA2 þ ð
P
aj > 2. Thus, inequality
j2b I
X X aj k 1ÞA þ aj > A2 ðk þ 1ÞA > 0: j2b I j2b O
Note that A > k + 1. Thus, inequality (3) holds. Thus, since inequality b > k2 . This is a contradiction. Thus, the proof of the ^ ; OÞ (3) holds, Tðr claim is complete. h By the claim, henceforth, we consider only schedules with b ¼ h. Then, jbIj ¼ j Oj 2
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
8 9 > >
P
aj > A
j2b I
P P Without loss of generality, let aj ¼ A þ k and aj ¼ A k, j2b I j2b O where k is a positive integer. Since A > 1,
b ¼ 1 þ AðA þ kÞ þ 3g A2 g A þ A k ^ ; OÞ Tðr 2 2 g 3g 2 þ1 A 1 A þ 1 þ ðA 1Þk > k2 : ¼ 2 2
j2 I
P
aj ¼ A k and Without loss of generality, let j2b I where k is a positive integer. Since k > 0,
P
j2b O
aj ¼ A þ k,
b ¼ 1 þ A2 þ 3g A2 g A þ A þ k ^ ; OÞ Tðr 2 2 g 3g 3 þ1 A 1 A þ 1 þ k > k2 : ¼ 2 2 This is a contradiction. P Case 3: baj ¼ A j2 I
Since
j2b O
aj ¼
P
aj ¼ A,
j2b I
b ¼ 1 þ A2 þ 3g A2 g A þ A ^ ; OÞ Tðr 2 2 g 3g 3 þ1 A 1 A þ 1 ¼ k2 : ¼ 2 2 By Cases 1–3,
P
b Since p1,j = q1 6 p2,k = pk and p1,j < p2.j for ^ ; OÞ. tioned after job k in ðr b ^ ; OÞ. Thus, j 2 I k , job k remains a critical job in ðr
b ¼ Tðr ; O Þ p þ ol : ^ ; OÞ Tðr l3 3 b < Tðr ; O Þ. This is a contradiction. Thus, ^ ; OÞ Since pl3 > ol3 , Tðr henceforth, we only consider schedules with O1 [ O2 [ O3 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l4 R O1 [ O2 [ O3 is outsourced, which implies that pl4 6 ol4 . Let e be the schedule such that O e ¼ O n fl4 g. If l4 2 J1, job k re~ ; OÞ ðr e because p 6 q , whereas if l4 2 J2 [ J3, ~ ; OÞ mains a critical job in ðr l4 1 e because p ¼ q 6 p . Thus, ~ ; OÞ job k remains a critical job in ðr 1;l4 1 2;k e and ~ ; OÞ it is observed that job k remains a critical job in ðr
e ¼ Tðr ; O Þ þ p ol : ~ ; OÞ Tðr l4 4 e 6 Tðr ; O Þ. By applying this argument ~ ; OÞ Since pl4 6 ol4 , Tðr repeatedly, Lemma 5 is obtained or a contradiction occurs. The proof is complete. Appendix D. Proof of Lemma 6
This is a contradiction. P Case 2: baj < A
P
51
b Thus, the set bI is a solution to the ^ ; OÞ. aj ¼ A in ðr
j2b I
equal cardinality partition problem. The proof is complete. Appendix C. Proof of Lemma 5 Without loss of generality, assume that if pk0 = pk, then job k is processed before job k0 in (r⁄, O⁄). Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l1 in O1 is processed in-house immediately before job k. Since p1;k ¼ q1 < p2;l1 , however, C 1;k ðr Þ < C 2;l1 ðr Þ, implying job k is not a critical job. This is a contradiction. Thus, henceforth, we only consider schedules with O 1 # O ⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l2 in O2 is processed in-house. Note that job k is positioned first among in-house jobs in J2 in schedule (r⁄, O⁄). Thus, job l2 is posi ; OÞ be the schedule such tioned immediately before job k. Let ðr that O ¼ O [ fl2 g. Since p1;l2 ¼ p2;l2 ¼ pl2 < q1 , job k remains a crit ; OÞ. Thus, ical job in ðr
; OÞ ¼ Tðr ; O Þ pl2 þ ol2 : Tðr ; OÞ < Tðr ; O Þ. This is a contradiction. Thus, Since pl2 > ol2 , Tðr henceforth, we only consider schedules with O1 [ O2 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that b be the sche^ ; OÞ job l3 in O3 is processed in-house after job k. Let ðr b ¼ O [ fl3 g, and bI k be the set of in-house jobs posidule such that O
Without loss of generality, assume that job k is positioned first among in-house jobs in J3. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l1 in O1 is processed immediately before job k on an in-house machine. Since p1;k ¼ q1 < p2;l1 ¼ pl1 , however, C1,k(r⁄) < C2,l(r⁄). This is a contradiction. Thus, henceforth, we only consider schedules with O1 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that ; OÞ be the schedule such job l2 in O2 is processed in-house. Let ðr ; OÞ bethat O ¼ O [ fl2 g. If l2 2 J1, job k remains a critical job in ðr cause p1,j = p2,j = pj and pj < q1 for j 2 J1, whereas if l2 2 J3, job k re ; OÞ because (p1,j, p2,j) = (q1, q2) for j 2 J3 mains a critical job in ðr and q1 < q2. Thus, it is observed that job k remains a critical job ; OÞ and in ðr
; OÞ ¼ Tðr ; O Þ p2;l2 þ ol2 : Tðr ; OÞ < Tðr ; O Þ. This is a contradiction. Thus, Since p2;l2 > ol2 , Tðr henceforth, we only consider schedules with O1 [ O2 # O⁄. Suppose that there exists an optimal schedule (r⁄, O⁄) such that job l3 R O1 [ O2 is outsourced. Since job l3 R O1 [ O2, it is observed that
l3 2 J 1 [ J 3 and p2;l3 6 ol3 : b be the schedule such that O b ¼ O n fl3 g. If l3 2 J1, job k re^ ; OÞ Let ðr b ^ ; OÞ because p1;l3 ¼ p2;l3 ¼ pl3 < pk , whereas mains a critical job in ðr b because p ¼ p ¼ q ^ ; OÞ if l3 2 J3, job k remains a critical job in ðr 1;l3 1;k 1 and p2;l3 ¼ p2;k ¼ q2 . Thus,
b ¼ Tðr ; O Þ þ p ol : ^ ; OÞ Tðr l3 3 ; OÞ 6 Tðr ; O Þ. By applying this argument Since pl3 6 ol3 ; Tðr repeatedly, Lemma 6 is obtained or a contradiction occurs. The proof is complete. References Allahverdi, A., 1996. Two-machine proportionate flowshop scheduling with breakdown to minimize maximum lateness. Computers and Operations Research 23, 909–916. Cachon, G.P., Harker, P.T., 2002. Competition and outsourcing with scale economies. Management Science 48, 1314–1333. Chen, Z.L., Li, C.L., 2008. Scheduling with subcontracting options. IIE Transactions 40, 1171–1184. Choi, B.C., Chung, J.B., 2011. Two-machine flow shop scheduling problem with an outsourcing option. European Journal of Operational Research 213, 66–72. Choi, B.C., Leung, J.Y.T., Pinedo, M.L., 2011. Minimizing makespan in an ordered flow shop with machine-dependent processing times. Journal of Combinatorial Optimization. 22, 797–818. Choi, B.C., Leung, J.Y.T., Pinedo, M.L., 2010. A note on makespan minimization in proportionate flow shops. Information Processing Letters 111, 77–81.
52
D.-Y. Chung, B.-C. Choi / European Journal of Operational Research 226 (2013) 46–52
Choi, B.C., Yoon, S.H., Chung, S.J., 2007. Minimizing maximum completion time in a proportionate flow shop with one machine of different speed. European Journal of Operational Research 176, 964–974. Chung, D., Lee, K., Shin, K., Park, J., 2005. A new approach to job shop scheduling problems with due date constraints considering operation subcontracts. International Journal of Production Economics 98, 238–250. Hou, S., Hoogeveen, H., 2003. The three-machine proportionate flow shop with unequal machine speeds. Operations Research Letters 31, 225–231. Koulamas, C., Kyparisis, G.J., 2009. A note on the proportional flow shop with a bottleneck machine. European Journal of Operational Research 193, 644–645. Lee, I.S., Sung, C.S., 2008a. Minimizing due date related measures for a single machine scheduling problem with outsourcing allowed. European Journal of Operational Research 186, 931–952. Lee, I.S., Sung, C.S., 2008b. Single machine scheduling with outsourcing allowed. International Journal of Production Economics 101, 623–634. Lee, K.B., Choi, B.C., 2011. Two-stage production scheduling with an outsourcing option. European Journal of Operational Research 213, 489–497. Ow, P.S., 1985. Focused scheduling in proportionate flow shops. Management Science 31, 852–869. Panwalkar, S.S., Khan, A.W., 1976. An ordered flow shop sequencing problem with mean completion time criterion. International Journal of Production Research 14, 631–635.
Pinedo, M., 2002. Scheduling: Theory, Algorithms and Systems, second ed. Prentice Hall, New York. Smith, M.L., Panwalkar, S.S., Dudek, R.A., 1975. Flow shop sequencing problem with ordered processing time matrices. Management Science 21, 544–549. Smith, M.L., Panwalkar, S.S., Dudek, R.A., 1976. Flow shop sequencing problem with ordered processing time matrices: A general case. Naval Research Logistics Quarterly 23, 481–486. Qi, X., 2008. Coordinated logistics scheduling for in-house production and outsourcing. IEEE Transactions on Automation Science and Engineering 5, 188–192. Qi, X., 2009. Two-stage production scheduling with an option of outsourcing from a remote supplier. Journal of System Science and System Engineering 18, 1–15. Qi, X., 2011. Outsourcing and production scheduling for a two-stage flow shop. International Journal of Production Economics 129, 43–50. Webster, M., Alder, C., Muhlemann, A.P., 1997. Subcontracting within the supply chain for electronics assembly manufacture. International Journal of Operations and Production Management 17, 827–841. Woeginger, G.J., 2000. When does a dynamic programming formulation guarantee the existence of a fully polynomial time approximation scheme? INFORMS Journal on Computing 12, 57–74.