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Computers & Operations Research 36 (2009) 2740 -- 2747
Contents lists available at ScienceDirect
Computers & Operations Research
journal homepage: www.elsevier.com/locate/cor
The two-stage assembly scheduling problem to minimize total completion time
with setup times
Ali Allahverdia,∗, Fawaz S. Al-Anzib
aDepartment of Industrial and Management Systems Engineering, Kuwait University, P.O. Box 5969, Safat, Kuwait
bDepartment of Computer Engineering, Kuwait University, P.O. Box 5969, Safat, Kuwait
ARTICLE INFO ABSTRACT
Available online 9 December 2008
Keywords:
Scheduling
Assembly flowshop
Total completion time
Heuristic
Dominance relation
We address the two-stage assembly scheduling problem where there are mmachines at the first stage
and an assembly machine at the second stage. The objective is to schedule the available njobs so that to-
tal completion time of all njobs is minimized. Setup times are treated as separate from processing times.
This problem is NP-hard, and therefore we present a dominance relation and propose three heuristics.
The heuristics are evaluated based on randomly generated data. One of the proposed heuristics is known
to be the best heuristic for the case of zero setup times while another heuristic is known to perform
well for such problems. A new version of the latter heuristic, which utilizes the dominance relation, is
proposed and shown to perform much better than the other two heuristics.
© 2008 Elsevier Ltd. All rights reserved.
1. Introduction
The two-stage assembly scheduling problem has many applica-
tions in industry, and hence, has received an increasing attention of
researchers recently. Lee et al. [1] described an application in a fire
engine assembly plant while Potts et al. [2] described an applica-
tion in personal computer manufacturing. Another application of the
problem is in the area of queries scheduling on distributed database
systems [3]. In short, many real life problems can be modeled as
a two-stage assembly flowshop scheduling problem. In particular,
manufacturing of almost all items may be modeled as a two-stage
assembly scheduling problem.
The two-stage assembly flowshop problem consists of two stages
where there are mmachines at the first stage while there is only a
single assembly machine at the second stage. There are njobs to be
scheduled and each job has m+1 operations. For each job, the first m
operations are conducted at the first stage by mmachines in parallel
and a final operation in the second stage by the assembly machine.
The last operation at the second stage may start only after all m
operations at the first stage are completed.
The two-stage assembly flowshop scheduling problem has been
addressed with respect to different criteria. For example, Lee et al.
[1], Potts et al. [2], Hariri and Potts [4], Haouari and Daouas [5],
∗Corresponding author. Tel.: +965 2498 7874; fax: +965 2481 6137.
E-mail addresses: allahverdi@kuniv.edu.kw (A. Allahverdi),
alanzif@eng.kuniv.edu.kw (F.S. Al-Anzi).
0305-0548/$ - see front matter ©2008 Elsevier Ltd. All rights reserved.
doi:10.1016/j.cor.2008.12.001
Sun et al. [6], and Allahverdi and Al-Anzi [3] and Al-Anzi and
Allahverdi [8] addressed the problem with respect to makespan
criterion. On the other hand, Allahverdi and Al-Anzi [3,8] addressed
the problem with respect to maximum lateness criterion.
There are real life situations in which each completed job is
needed as soon as it is processed. In such situations, one is interested
in minimizing total completion time (TCT) of all jobs. This objec-
tive is particularly important in real life situations where reducing
inventory or holding cost is of primary concern. The literature sur-
vey reveals that the only researchers addressing TCT criterion in a
two-stage flowshop problem are Tozkapan et al. [9] and Al-Anzi and
Allahverdi [10]. Tozkapan et al. [9] developed a lower bound and
a dominance relation, and utilized the lower bound and the domi-
nance relation in a branch and bound algorithm. They also proposed
two heuristics to find an upper bound for their branch and bound
algorithm. On the other hand, Al-Anzi and Allahverdi [10] proposed
two algorithms and showed that one algorithm is optimal under
certain conditions. They also proposed a tabu search and a simu-
lated annealing heuristic for the problem. Moreover, they proposed
a hybrid tabu search heuristic and showed by computational analy-
sis that their proposed hybrid tabu search heuristic is more efficient
and can easily be used for large sized problems.
Both Tozkapan et al. [9] and Al-Anzi and Allahverdi [10] assumed
that setup times are zero or included in processing times. While
this assumption simplifies the analysis and/or reflects certain appli-
cations, it harmfully affects the solution quality for many applica-
tions which require separate and non-zero treatment of setup times
[11,12]. The significance of considering setup times as separate is
addressed by Allahverdi and Soroush [13].
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In this paper, we consider the two-stage flowshop scheduling
problem with respect to TCT criterion where setup times are treated
as separate from processing times. This problem is NP-hard since
its special case when m=1 (which is a regular two-machine flow-
shop problem) is NP-hard, Garey et al. [14]. Therefore, we propose
a dominance relation and present heuristics to solve the problem.
A dominance relation is presented in the next section. In Section 3,
three heuristics are presented while the evaluation of these heuris-
tics is conducted in Section 4. Finally, a summary of the work and
direction for the future research are given in Section 5.
2. A dominance relation
In this section we present a dominance relation for the problem.
Dominance relations are very useful for eliminating certain solutions
while searching for the optimal solution, and are usually used in
implicit enumeration techniques such as a branch-and-bound algo-
rithm. The objective of this paper is to present heuristics (in the next
section) to solve the problem rather than to present an enumeration
technique. Therefore, we use the dominance relation in heuristics.
Before developing the dominance relation, we first need to de-
velop the objective function. We assume that njobs are simultane-
ously available at time zero and that preemption is not allowed, i.e.,
any started operation has to be completed without interruptions.
Each job consists of a set of m+1 operations. The first moperations
are completed at stage one in parallel while the last operation is
performed at stage two. Let
ti,joperation time of job ion machine j(at stage one),
i=1,...,n, j =1,...,m
t[i,j]operation time of the job in position ion machine j
(at stage one)
si,jsetup time of job ion machine j(at stage one), i=1, . .. ,n,
j=1,...,m
s[i,j]setup time of the job in position ion machine j(at stage
one)
pioperation time of job ion assembly machine
(at stage two)
p[i]operation time of the job in position ion assembly machine
(at stage two)
sisetup time of job ion assembly machine (at stage two)
s[i]setup time of the job in position ion assembly machine
(at stage two)
C[i]completion time of the job in position i
TCT total completion time
Note that job kis complete once all of its operations sk,j,tk,j
(j=1,...,m)andskand pkare completed where the operation pk
may start only after all operations sk,jand tk,j(j=1,...,m) have been
completed. Tozkapan et al. [9] showed that permutation schedules
are dominant with respect to total flowtime (completion time) cri-
terion. Similarly, it can easily be shown that permutation schedules
are dominant for the problem with setup times. Therefore, we re-
strict our search for the optimal solution to permutation schedules.
In other words, the sequence of jobs on all of the machines, includ-
ing the assembly machine, is the same.
The completion time of the job in position jof a given sequence
can be computed as
C[j]=max ⎧
⎨
⎩
max
k=1,...,m
⎧
⎨
⎩
j
i=1
(s[i,k]+t[i,k])⎫
⎬
⎭
,C[j−1] +s[j]⎫
⎬
⎭
+p[j]where C[0] =0(1)
Let
j=max
k=1,...,m
⎧
⎨
⎩
j
i=1
(s[i,k]+t[i,k])⎫
⎬
⎭
−
j−1
i=1
(s[i]+p[i])−s[j](2)
and
j=max{0, 1,2,...,j}(3)
Then it can be shown that
C[j]=
j
i=1
(s[i]+p[i])+j(4)
and therefore, the TCT is computed as
TCT =
n
i=1
C[i](5)
Lemma 1. Assume that F(k)is any given value for k =1,...,m where
F(k)might be a different value for each k. Then,max
k=1,...,m{F(k)+
sw,k+tw,k}maxk=1,...,m{F(k)+sq,k+tq,k}if sq,k+tq,ksw,k+tw,kfor
k=1,...,m.
Proof. For each k and any given F(k) value, F(k)+sq,k+tq,kF(k)+
sw,k+tw,ksince for each k,sq,k+tq,ksw,k+tw,k. Since this relationship
is true for each term, it must be also true for the maximum of the
term.
Lemma 2. Assume that F(k)is any given value for k =1,...,m where
F(k)might be a different value for each k. Then,max
k=1,...,m{F(k)+sv,k+
tv,k}−sw−pwmaxk=1,...,m{F(k)}if maxk=1,...,m{sv,k+tv,k}sw+pw.
Proof. For any given kand any given F(k) value, F(k)+sv,k+tv,k−sw−
pwF(k) since maxk=1,...,m{sw,k+tw,k}sw+pw. Since the left-hand
side is less than or equal to the right-hand side for the maximum
value of sw,k+tw,k, it must be less than or equal to for any smaller
value of sw,k+tw,k.
Consider two sequences 1and 2such that 1has job iin an
arbitrary position and job jin position +1. The sequence 2is
exactly the same as 1except that job jis in position and job iin
position +1.
Lemma 3. C[r](2)=C[r](1)for i =1,...,−1.
Proof. Since both sequences 1and 2have the same job in positions
1,...,−1, C[r](2)=C[r](1) for r=1,...,−1.
Lemma 4. C[r](2)C[r](1)for r =+2,...,nifmax{(2),
+1(2)} max{(1), +1 (1)}.
Proof. Both sequences 1and 2have the same job in all positions
except positions and +1. Therefore, it follows from Eq. (2) that
r(2)=r(1) for r=1, . .. ,−1,+2, . .. ,n. Then, by Eqs. (3) and (4),
C[r](2)C[r](1)ifmax{(2), +1(2)}max{(1), +1(1)}
for r=+2,...,n.
Theorem 1. Consider a two-stage assembly flowshop scheduling prob-
lem and assume that two adjacent jobs i and j satisfy the following
three conditions:(i)sj+pjsi+pi, (ii) sj,k+tj,k−sjsi,k+ti,k−si
for k =1,...,m,and either (iiia) maxk=1,...,m{si,k+ti,k}si+pjor (iiib)
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maxk=1,...,m{sj,k+tj,k}sj+pj.Then,there exists an optimal solution
that minimizes TCT in which job j precedes job i.
Proof. Consider the two sequences 1and 2specified earlier. From
Eq. (2),
(1)=max
k=1,...,m
⎧
⎨
⎩
−1
r=1
(s[r,k]+t[r,k])+si,k+ti,k⎫
⎬
⎭
−
−1
r=1
(s[r]+p[r])−si(6)
(2)=max
k=1,...,m
⎧
⎨
⎩
−1
r=1
(s[r,k]+t[r,k])+sj,k+tj,k⎫
⎬
⎭
−
−1
r=1
(s[r]+p[r])−sj(7)
+1(1)=max
k=1,...,m
⎧
⎨
⎩
−1
r=1
(s[r,k]+t[r,k])+si,k+ti,k+sj,k+tj,k⎫
⎬
⎭
−
−1
r=1
(s[r]+p[r])−si−pi−sj(8)
+1(2)=max
k=1,...,m
⎧
⎨
⎩
−1
r=1
(s[r,k]+t[r,k])+sj,k+tj,k+si,k+ti,k⎫
⎬
⎭
−
−1
r=1
(s[r]+p[r])−sj−pj−si(9)
It follows from Eqs. (6) and (7) and Lemma 1 that
(2)(1) (10)
since by hypothesis sj,k+tj,k−sjsi,k+ti,k−sifor k=1,...,m.And
by Eqs. (7) and (9) and Lemma 2,
+1(2)(2) (11)
if maxk=1,...,msi,k+ti,ksi+pj. Furthermore, by Eqs. (6) and (9)
and Lemma 2
+1(2)(1) (12)
if maxk=1,...,m{sj,k+tj,k}sj+pj. Hence, if Eq (10) and either (11)
or (12) hold, then
max{(2), +1(2)}max{(1), +1(1)}(13)
Now by Eq (4), we have the following for the two sequences:
C[](1)=
−1
r=1
(s[r]+p[r])+si+pi+max{−1(1), (1)}(14)
C[](2)=
−1
r=1
(s[r]+p[r])+sj+pj+max{−1(2), (2)}(15)
C[+1](1)=
−1
r=1
(s[r]+p[r])+si+pi+sj+pj
+max{−1(1), (1), +1(1)}(16)
C[+1](2)=
−1
r=1
(s[r]+p[r])+sj+pj+si+pi
+max{−1(2), (2), +1(2)}(17)
It should be noted that from the above four equations,
(r)=max{−1(r), (r)}and
+1(r)=max{−1(r), (r), +1(r)}
for r=1 and 2. This follows by the definition of rfrom Eq. (3).
From Eqs. (14) to (17),
[C[](2)+C[+1](2)] −[C[](1)+C[+1] (1)]
=sj+pj−si−pi+max{−1(2), (2)}
−max{−1(1), (1)}
+max{−1(2), (2), +1(2)}
−max{−1(1), (1), +1(1)}(18)
Now it follows from Eqs. (10), (13), and (18) and the hypothesis of
sj+pjsi+pithat
[C[](2)+C[+1](2)] [C[](1)+C[+1] (1)] (19)
Then, it follows from Eqs. (13) and (19), Lemma 3, and Lemma 4 that
TCT(2)TCT (1)
Corollary 1. Consider a two-stage assembly flowshop scheduling
problem where setup times are zero (or included in processing times).
Assume that two adjacent jobs i and j satisfy the following three
conditions:(i)pjpi, (ii) tj,kti,kfor k=1, .. . ,m,and either (iiia)
maxk=1,...,m{ti,k}pjor (iiib) maxk=1,...,m{tj,k}pj.Then,there exists
an optimal solution that minimizes TCT in which job j precedes job i.
Proof. The proof of this corollary directly follows from that of
Theorem 1.
3. Proposed heuristics
In this section, we propose three heuristics; a hybrid tabu search,
a self-adaptive differential evolution (SDE), and a new self-adaptive
differential evolution (NSDE). These three heuristics are described in
the following three subsections.
3.1. Hybrid tabu search (Ntabu)
Al-Anzi and Allahverdi [10] proposed three heuristics, namely, a
simulated annealing, a tabu search, and a hybrid tabu search (Ntabu)
for the problem addressed in this paper by ignoring setup times. They
showed that Ntabu outperforms both of the simulated annealing and
tabu search heuristics by a large margin. They also showed that Ntabu
significantly outperforms the two heuristics proposed by Tozkapan
et al. [9], who also considered the same problem by ignoring setup
times. Since Ntabu is known to be the best heuristic for the problem
without setup times and there exist no heuristics for the problem
with setup times, we consider Ntabu as one of the heuristics. In the
evaluation of the heuristics, in order to have a fair comparison, we
also consider the case when setup times are zero. However, we have
fine tuned the parameters to the current problem, see Table 1.
3.2. A self-adaptive differential evolution (SDE)
Differential evolution (DE) heuristics have been applied to solve a
wide range of optimization problems in different areas, and recently
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A. Allahverdi, F.S. Al-Anzi / Computers & Operations Research 36 (2009) 2740 -- 2747 2743
Table 1
Parameter fine tuning for the proposed heuristics.
Heuristic Parameter Range Best value
Ntabu Imax (maximum number of iterations) 20,000–50,000 with an increment of 5000 35,000
h(tabu list size) 2–7 with an increment of 1 3
SDE VApop n–5nwith an increment of n2n
Vgn–5nwith an increment of n4n
Vcp n–5nwith an increment of n2n
Y1/6–5/6 with an increment of 1/6 3/6
Pri0.1−0.2 with an increment of 0.005 0.135
NSDE VApop n–5nwith an increment of n2n
Vgn–5nwith an increment of n4n
Vcp n–5nwith an increment of n2n
Y1/6–5/6 with an increment of 1/6 3/6
Cnew 0.01–0.5 with an increment of 0.01 0.2
Pri0.1–0.2 with an increment of 0.005 0.135
in scheduling, e.g., Onwubolu and Davendra [15]. In DE, the user has
to find the best values for the problem-dependent control param-
eters. Finding the best values for the control parameters is a time
consuming task. Therefore, a new version of DE has been proposed
by Omran et al. [16] where the control parameters are self-adaptive.
This new version is called SDE. Al-Anzi and Allahverdi [8] adapted
this SDE to the two-stage assembly flowshop scheduling problem
to minimize maximum lateness with separate setup times. They
showed that SDE performs much better than a tabu search heuristic
and a particle swarm optimization heuristic. Since the problem ad-
dressed in this paper is also a two-stage assembly flowshop problem
treating setup times as separate from processing time, we use this
SDE as one of the heuristics for the current problem, of course, by
adapting it to the new performance measure of TCT. Similar to Ntabu,
we have fine tuned the parameters of SDE to the current problem,
see Table 1.
3.3. A new self-adaptive differential evolution (NSDE)
In this paper, we propose a modification to the SDE that was
proposed by Al-Anzi and Allahverdi [8], and call it NSDE. We will not
present NSDE in detail since the detail of NSDE is given by Al-Anzi and
Allahverdi [8], and hence, we will only describe the difference next.
However, a complete pseudo code will be presented. It should be
noted that the developed dominance relation (Theorem 1) is utilized
in the mutation process of the proposed NSDE.
The difference between the SDE that was proposed by Al-Anzi
and Allahverdi [8] and the NSDE proposed in this paper is the in-
troduction of a new step (Step 5-iii) in the algorithm. In this step, a
random pair wise exchange is conducted which results in children of
crossover operator with a probability Pnew. This probability is com-
puted as follows:
Pnew =e−Cnew∗dwhere d=1−
Vg−xi
Vg
where xiis the current generation number, Vgis the maximum num-
ber of generations, and Cnew is an adjusting factor that needs to be
fine tuned for this specific problem. In this computation of probabil-
ity, Pnew is highly probable when the NSDE algorithm is in its earlier
stages and it gets less probable when the algorithm reaches its final
stages.
The steps of the NSDE heuristic are as follows.
Step 1: Initialize a population, VApop, of random sequences.
Step 2: Randomly initialize mutation probabilities for each se-
quence iin VApop to Pri.
Step 3: Compute the TCT of each sequence in VApop.
Step 4: Order the sequences in VApop according to TCT from the
best to the worst.
Step 5: Repeat Steps (i)–(vii) for Vgtimes.
(i) Set the neighborhood size sto be 1/Vgof total population size.
(ii) Repeat steps (a) to (d) for Vcp times:
(a) Randomly choose two different compatible parents to
crossover;
(b) select compatible segments in the two parents;
(c) swap the segments;
(d) save the new sequences in VAchild and compute TCT of each.
(iii) With probability Pnew do a pair wise exchange in VAchild and
compute its TCT. If the move improves the objective function,
TCT, of the child then keep it. Otherwise, reverse this pair wise
exchange.
(iv) Order VAchild with respect to TCT.
(v) Replace the worst ysequences of VApop with the best yse-
quences in VAchild maintaining order with respect to TCT.
(vi) Mutate each sequence iin VApop as follows:
(a) select a random position kbetween 1, . .. ,n;
(b) for each job jin position [j] in the sequence ido the fol-
lowing:
•if j=kor with probability Priselect three random se-
quences i1,i2,i3in the neighborhood sof the sequence i,
where i1i2i3I;
•let jobs j1,j2,j3be in position [j] of sequences i1,i2,i3,
respectively, then compute j4=(j1+j2+j3)/3;
•replace job at position [j] in sequence iwith j4(and fixing
inconsistency in sequence iby replacing the other (original)
job of value j4in sequence iwith job j);
•exchange every two adjacent jobs. If the dominance relation
developed in Theorem 1 applies, then keep the exchange,
otherwise, return the two jobs to their original positions;
•if sequence iafter mutation has a better objective function
then;
•update the probability Pri=Pri1+(Pri2−Pri3)∗Random(0, 1)
else reject mutation.
(vii) Compute TCT and order VApop.
Step 6: Store the best solution from VApop as the final solution.
Step 7. Improve the final solution by applying a pairwise exchange
procedure.
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4. Computational experiments
In this section, we first describe how the parameters of the pro-
posed heuristics are set in Section 4.1, and then the evaluation of
heuristics is described in Section 4.2.
4.1. Setting heuristic parameters
To optimize the performance of the proposed heuristics, fine tun-
ing of their parameters has been performed. An initial estimate for
the best value of a given parameter of a heuristic is obtained by
changing the values of that parameter while keeping all other pa-
rameters of the heuristic as constant. After some experimentations
and after no major changes in the performance have been noticed,
the parameters were set as given in Table 1.
4.2. Proposed heuristic evaluation
The proposed heuristics of Ntabu, SDE, and NSDE were imple-
mented in C under GCC-3.4.2 compiler using the built-in math li-
brary. The machine used was a Sun Fire V880 with 4 CPU processors
of 900 MHz running under Solaris Version 9.0 operating system with
8 GB RAM. To measure the effectiveness of the heuristics, we com-
pared the performance of the three heuristics against each other and
against a random solution.
The processing times were randomly generated from a uniform
distribution [1, 100] on all mmachines at the first stage as well as the
assembly machine at the second stage. In the scheduling literature,
most researchers have used this distribution in their experimenta-
tion. The reason for using a uniform distribution with a wide range
is that the variance of this distribution is large and if a heuristic per-
forms well with such a distribution, it is likely to perform well with
other distributions.
Setup times on both stages are generated from a uniform distri-
bution [1, 100k]. The parameter kis the expected ratio of setup time
to processing time (si,j/ti,j). The kvalue for each data set was set to
0, 0.3, 0.6, 0.9 and 1.2. Note that when k=0, the problem is reduced
to the problem addressed by Al-Anzi and Allahverdi [10]. Therefore,
we can compare the newly proposed heuristic of NSDE with the best
heuristic Ntabu of Al-Anzi and Allahverdi [10].
Problem data were generated for different number of jobs: 20, 30,
40, 50, 60, and 70. The experimentation was conducted for different
number of machines at the first stage as 3, 6, or 9. We compared the
performance of the heuristics using two measures: average percent-
age error (Error) and standard deviation (Std) out of thirty replicates.
The percentage error is defined as 100∗(TCT of the heuristic−TCT of
the best heuristic)/(TCT of the best heuristic).
There are 90 combinations for different values of n(20, 30, 40,
50, 60, 70), m(3,6,9),andk(0, 0.3, 0.6, 0.9, 1.2). Thirty replicates
were generated for each combination, and therefore, a total of 2700
instances were generated and evaluated. For the sake of brevity,
the results will not be tabulated. The summary of the results are
presented in Figs. 1–7. A random solution was also considered for
comparison purposes. However, the average error for the random
solution was very large (on average, more than 25 times the error of
the worst heuristic) compared with the other heuristics, and there-
fore, is not reported in the figures.
The overall average errors and standard deviation of the errors of
Ntabu, SDE, and NSDE are summarized in Figs. 1 and 2, respectively.
Fig. 1(2) illustrates the overall average errors (standard deviation of
the errors) with respect to the number of jobs (n). As can be seen
from the figures, both Ntabu and NSDE perform much better than
SDE. The figures also show that NSDE performs better than Ntabu.
The comparison of NSDE and Ntabu is more explicitly shown in
Figs. 3 and 4. It is clear from Figs. 3 and 4 that the performance
0
1
2
3
4
5
6
7
20
n
Avg. Error
NSDE
Ntabu
SDE
30 40 50 60 70
Fig. 1. The average error versus the number of jobs.
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.018
0.02
20
n
Std
NSDE
Ntabu
SDE
30 40 50 60 70
Fig. 2. The standard deviation of the error versus the number of jobs.
0
0.2
0.4
0.6
0.8
1
1.2
20
n
Avg. Error
NSDE
Ntabu
30 40 50 60 70
Fig. 3. The average error comparisons of Ntabu and NSDE.
of NSDE gets better as n increases. A statistical comparison of the
heuristics will also be conducted.
The heuristic performances were also investigated for different
values of the number of machines at the first stage (m). The results
are given in Fig. 5. As can be seen, the performance of SDE gets better
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0
0.0005
0.001
0.0015
0.002
0.0025
0.003
0.0035
0.004
20
n
Std
NSDE
Ntabu
30 40 50 60 70
Fig. 4. The Std comparisons of Ntabu and NSDE.
0
1
2
3
4
5
6
3
m
Avg. Error
NSDE
Ntabu
SDE
69
Fig. 5. The average error versus the number of machines at the first stage.
0
1
2
3
4
5
6
7
0
k
Avg. Error
NSDE
Ntabu
SDE
0.3 0.6 0.9 1.2
Fig. 6. The average error versus setup to processing time ratio.
as m increases while there does not seem to be any difference for
the performance of the other two heuristics of Ntabu and NSDE.
Fig. 6 indicates the performance of heuristics for different values
of k, setup time to processing time ratio. As can be seen from the
0
50
100
150
200
250
300
20
n
CPU time (in seconds)
NSDE
Ntabu
SDE
30 40 50 60 70
Fig. 7. The average CPU time (in seconds) versus number of jobs.
figure, the performance of SDE gets better as kincreases. The per-
formance of the other two heuristics does not seem to be affected
by the value of k.
We have also conducted a test of hypothesis for comparison of
the heuristics by using a paired ttest for all 90 combinations of n,
m,andk. The following hypothesis testing was conducted:
H0(1).The average error of Ntabu =the average error of SDE.
H1(1).The average error of Ntabu <the average error of SDE.
The null hypotheses were rejected for all 90 combinations at 99%
significance level. This implies that the average error of Ntabu is
statistically smaller than that of SDE.
Similarly, the following hypothesis testing was conducted:
H0(2).The average error of NSDE =the average error of SDE.
H1(2).The average error of NSDE <the average error of SDE.
The null hypotheses were rejected for all 90 combinations at 99%
significance level. This implies that the average error of NSDE is
statistically smaller than that of SDE.
Finally, the following hypothesis testing was conducted:
H0(3).The average error of NSDE =the average error of Ntabu.
H1(3).The average error of NSDE <the average error of Ntabu.
The null hypotheses were also rejected for all 90 combinations at
99% significance level. This implies that the average error of NSDE is
statistically smaller than that of Ntabu. It should be noted that the
above results for all the three tests were also valid for the significance
level of 99.9%.
Before concluding that NSDE outperforms Ntabu, one has to also
consider CPU time in addition to the average error. The CPU times of
all the heuristics are summarized in Fig. 7. As can be seen from the
figure, the CPU times of NSDE are larger than that of Ntabu. However,
a fair comparison would be to consider both the error and CPU time
at the same time. The overall average error of Ntabu is about 70
times (69.6 to be exact) that of NSDE while the overall average CPU
time of NSDE is only about 3 times (3.12 to be exact) that of Ntabu.
Therefore, it can now be stated that NSDE outperforms Ntabu.
It should be noted that Ntabu is known to be the best heuristic
for the problem without setup times. Computational analysis shows
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0.02
0.025
0.03
0.035
0.04
0.045
0.05
0.055
0.06
0.065
0.07
0
k
Avg. Error
NSDE (without DR)
NSDE
0.3 0.6 0.9 1.2
Fig. 8. Evaluation of the developed dominance relation.
that NSDE performs better than Ntabu even for the case of zero setup
times. Hence, it can be stated that the new proposed heuristic of
NSDE is the best heuristic for the problem with or without setup
times.
Finally, in order to evaluate the effectiveness of the developed
dominance relation (Theorem 1), we compared the performance of
the proposed heuristic NSDE as described in Section 3.3, which uses
the dominance relation in the mutation process in Step 5, with that
of the proposed heuristic without using the dominance relation in
Step 5, which is represented by NSDE(without DR). An overall im-
provement of 7.5% has been observed. This improvement seems to
be affected neither by nnor by m. However, it seems that the im-
provement is affected by the setup to processing time ratio k,see
Fig. 8. As can be seen from the figure, the improvement gets better
as the ratio of setup to processing time gets smaller.
5. Summary and future research
The scheduling problem of a two-stage assembly flowshop is
considered with the objective of minimizing TCT of all navailable
jobs, where job setup times are treated as separate from processing
times. The problem is NP-hard since it is known that the problem
is NP-hard when m=1, a special case of the considered problem.
Hence, for such problems, one solution is to come up with efficient
heuristics.
Three heuristics are proposed and evaluated on randomly gen-
erated data. The three heuristics are a hybrid tabu search (Ntabu),
SDE, and a new version of self-adaptive differential evolution (NSDE)
which is introduced in this paper. The Ntabu is known to be the best
for the case when setup times are zero. It is shown that the newly
proposed NSDE performs much better than SDE and Ntabu (even for
the case when setup times are zero). Therefore, NSDE is also the best
heuristic for the problem where setup times are ignored.
Researches use two different approaches to solve scheduling
problems. One approach is to use an implicit enumeration technique
such as a branch-and-bound algorithm while the other approach
is to use some heuristic to solve the problem. Each approach has
advantages and disadvantages. We have opted using the second
approach and hence proposed different heuristics. A possible exten-
sion to this research is to come up with some implicit enumeration
techniques. In such a case, the proposed dominance relation in this
paper can be used to eliminate some feasible solutions in order
to reach the optimal solution faster. Therefore, a possible research
area is to construct a branch-and-bound algorithm for this problem
by utilizing the dominance relation established in this paper or to
develop a Lagrangian relaxation approach similar to the one used
by Augusto et al. [17] or Gourgand et al. [18].
Another possible extension is to consider the problem with re-
spect to other objective functions such as job waiting time variance
(e.g., [19]) or an objective function taking into account early and
tardy penalties (e.g., [20]).
In this paper, it is assumed that setup times are sequence inde-
pendent. This assumption is valid for some scheduling environments.
However, the assumption may not be valid for some other schedul-
ing environments, e.g., Yu et al. [21], Hendizadeh et al. [22], Pessan
et al. [23], and Chandrasekaran et al. [24]. Therefore, another pos-
sible extension is to consider the problem addressed in this paper
with sequence dependent setup times. Yet another possible exten-
sion to the problem addressed in this paper is to consider a hybrid
assembly flowshop where at each stage there might be more than
a single machine available, similar to a hybrid flowshop e.g., Ben
Hmida et al. [25]. Finally, it is assumed that there is an infinite buffer
space between the two stages. This assumption may not necessar-
ily be realistic for some scheduling problems, e.g., see Fahmy et al.
[26]. Therefore, one more possible research area is to address the
problem with a limited buffer space between the two stages.
Acknowledgments
This research was supported by Kuwait University Research Ad-
ministration Grant no. EO 06/06. The authors would like to thank
anonymous referees for their helpful comments that improved the
quality of the paper significantly.
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