Journal of Applied Mathematics & Bioinformatics, vol.2, no.3, 2012, 33-38
ISSN: 1792-6602 (print), 1792-6939 (online)
Scienpress Ltd, 2012
Minimizing Total Completion Time in a Flow-shop
Scheduling Problems with a Single Server
Shi ling1 and Cheng xue-guang2
Abstract
We consider the problem of two-machine flow-shop scheduling with a single
server and equal processing times, we show that this problem is NP -hard in the
strong sense and present a busy schedule for it with worst-case bound 7 / 6 .
Mathematics Subject Classification: 90B35
Keywords: flow-shop scheduling problem, total completion time, worst-case
1 Introduction
We consider the two-machine flow-shop scheduling problem with
minimizing
1
2
total
completion
time
and
equal
processing
times,
that
Department of Mathematics, Hubei University for Nationalities, shiling59@126.com,
School of Mathematics and Statistics, Wuhan University,
e-mail: chengxueguang6011@msn.com.
Article Info: Received : September 22, 2012. Revised : October 29, 2012
Published online : December 30, 2012
34
Minimizing Total Completion Time in a Flow-shop ...
is F 2 p1, j  p  C j . Complexity results for F 2
C
j
problem obtained by
Garey, et al [1], J.A. Hoogereen [2] studied some special cases for two-machine
flow-shop problems with minimizing total completion times, and proved that the
problem with equal processing times on first machine, that is F 2 p1, j  p  C j ,is
NP -hard in the strong sense, and present an O(n log n) approximation algorithm
for it with worst-case bound 4 / 3 .Complexity results for flow-shop problems with
a single server was obtained by Brucher, et al [3]. In this paper, we derive some
new complexity results for two-machine problem with a single server, introduce
an improved algorithm, and prove that its worst case is 7 / 6 , the bound is tight.
2 Complexity of the F 2, S1 pi, j  p  C j problem
Let C i , j denote the completion times of job J j on machine M i . If there are
no idle times on M 1 and M 2 , we have C1,1  s1,1  p1,1 , C 2,1  s1,1  p1,1  s 2,1  p 2,1 ,
C1, j  C1, j 1  s1, j  p1, j , C 2, j  max{ C 2, j 1 , C1, j }  s 2, j  p 2, j , for j  2,...,n .
Theorem 1 The problem of deciding whether for a given instance of the
F 2, S1 pi , j  p  C j problem there exists a schedule with cost no more than a
given threshold value y is NP -hard in the strong sense.
Proof.
Our proof is based upon a reduction from the problem Numerical
Matching with Target Sums or, in short, T S, which is known to be NP -hard in
the strong sense.
T S Given two multisets X  {x1 ,..., x n } and Y  { y1 ,..., y n } of positive integers
and an target vector {z1 ,..., z n } ,where  j 1 ( x j  y j )   j 1 z j ,is there a position
n
n
of the set X  Y into n disjoint set Z 1 ,..., Z n ,each containing exactly one element
Shi ling and Cheng xue-guang
35
from each of X and Y , such that the sum of the numbers in Z j equal z j , for
1,...,n ?
(1) P -jobs: s1,i  b, p1,i  b; s2,i  b  xi , p2,i  b(i  1,..., n) ,
(2) Q -jobs: s1,i  0, p1,i  b; s2,i  b  yi , p2,i  b(i  1,..., n) ,
(3) R -jobs: s1,i  0, p1,i  b; s2,i  b  zi , p2,i  b(i  1,..., n) ,
(4) U -jobs: s1,i  0, p1,i  b; s2,i  0, p2,i  b(i  1,..., n) ,
(5) V -jobs: s1,i  0, p1,i  b; s2,i  0, p2,i  b(i  1,..., n) ,
(6) W -jobs: s1,i  0, p1,i  b; s2,i  0, p2,i  b(i  1,..., n) ,
(7) L -jobs: s1,i  4b, p1,i  b; s2,i  b, p2,i  b(i  1,..., n) .
Observe that all processing times are equal to b .To prove the theorem we show
that in this constructed if the F 2, S1 pi , j  p  C j problem a schedule S 0
satisfying
C
(S 0 )  y  i 1 xi  i 1 ( xi  yi )  (77n 2  13n  4)b / 2 exists if
n
j
n
and only if T S has a solution. Suppose that T S has a solution. The desired
schedule S 0 exists and can be described as follows. No machine has intermediate
idle time. M 1 process the jobs in order of the sequence  , i.e., in the sequence
  { P , Q , R , U , V , W , L . . .,P
1 , 1
1 , 1
1 , 1
1 , 1
1 , 1
1 , 1n
1 , n1
,Q
,R
,
n1 ,
,V
n 1 U
,
n
U,
V,
1 ,
,W
n
1 ,
n
,L
1 ,
}
1 ,
While M 2 process the jobs in the sequence
  { P , Q , R , U , V , W , L , . . .P,
2 , 1
2 , 1
2 , 1
2 , 1
2 , 1
2 , 1n
2 , 1n
Q,
,
n 2 R
,
n
2 ,
n
2 ,
n
W,
2 ,n
L,
as indicated in Figure 1.
Then we define the sequence  and  shown in Figure 1. Obviously,
these sequence  and  fulfills C (S )  C ( , )  y . Conversely, assume that the
flow-shop scheduling problem has a solution  and  with C ( S )  y .
Considering
the
path
composed
of
{P1,1 , Q1,1 , R1,1 , U1,1 , V1,1 , W1,1} , M 2 operations of jobs
M1
operations
of
jobs
2 ,
}2 ,
2 ,
36
Minimizing Total Completion Time in a Flow-shop ...
{R2,1 , U 2,1V2,1 , W2,1 , L2,1 ,... , R2,n , U 2,n , V2,n , W2,n , L2,n } ,
we obtain that
C ( S )  3b  x1  5b  x1  y1  7b  x1  y1  z1  8b  9b  10b  ... 
(3  (n  1)11)b  x  (5  (n  1)11)b  xn  yn  (7  (n  1)11)b
 ...  (11n  1)b
  i 1 xi   i 1 ( xi  yi )  (77n 2  13n  4)b / 2  y.
n
n
So we have C (S )  y .
1.
P1,1
Q1,1
R1,1
U1,1
P2,1
P1,n
Q1,n
R1,n
V1,1
Q2,1
U1,n
P2,n
V1,n
W1,1
W
L1,1
R2,1
U2,1
V2,1
W1,n
Q2,n
W2,1
L2,1
L1,n
R2,n
U2,n
V2,n
W2,n
L2,n
Figure 1: Gant chart for the F 2, S1 pi , j  p  C j problem
(a) If S has a partition  , then there is a schedule with finish times y . One such
schedule is shown in Figure 1.
(b) If S has no partition, then all schedule must have a finish times  y . Since
S has no partition, then xi  y i  z i (i  1,..., n) . Let  i  xi  y i  z i (i  1,..., n) , so
 C (S )  
j
n
x   i 1 ( xi  yi )  (77n 2  13n  4)b / 2  5 i 1 i
n
n
i 1 i
 10 i 1 i  ...  5n1  y
n 1
3 Worst-case for the F 2, S1 pi, j  p  C j problem
In examining “worst” schedule, we restrict ourselves to busy schedule. A
busy schedule is a schedule in which at all times from start to finish at least one
server is processing a task.
Shi ling and Cheng xue-guang
37
Theorem 2 The F 2, S1 pi , j  p  C j problem, let S 0 be a busy schedule for this
problem, S * be the optimal solution for the F 2, S1 pi , j  p  C j problem, then
C (S ) / C (S
j
0
*
j
)  7 / 6 , the bound is tight.
Proof. For a schedule S , let I i , j ( S )( i  1,2; j  1,..., n) denote the total idle times of
job J j on M i .Considering the path composed of M 1 operations of jobs
1,..., j , M 2 operation of job j , we obtain that
C j  i 1 ( s1,i  p1,i )  I1, j  s 2, j  p2, j
j
(1)
Considering the path composed of M 1 operations of jobs 1 , M 2 operation of
job 1, 2,..., j , we obtain that
C j  s1,1  p1,1  i 1 ( s 2,i  p2,i )  I 2, j
j
(2)
Considering the path composed of M 1 operations of jobs 1,...,l , M 2 operation of
job l ,..., j , we obtain that
C j  i 1 (s1,i  p1,i )  I1,i  i l (s 2,i  p2,i )  I 2, j
l
j
(3)
So we have
6 C j ( S0 )  (2( i 1 ( s1,i  p1,i )  I1, j ))  (2( i 1 ( s2,i  p2,i )  I 2, j )) 
j
j
(2( i 1 ( s1,i  p1,i )  I1, j   i 1 ( s2,i  p2,i  (2( s1,1  p1,1 )  2( s2, j  p2, j ))
l
j
 7 C j ( S * )
C (S ) / C (S
j
0
j
*
)  7/6.
To prove the bound is tight, introduce the following example as follows and show
in Figure 2 and Figure 3.
(1) P -jobs: s1,i  2b, p1,i  b, s 2,i  2b, p 2,i  b(i  1,2) ;
(2) Q -jobs: s1,i  0, p1,i  b, s 2,i  0, p 2,i  b(i  3,4) .
38
Minimizing Total Completion Time in a Flow-shop ...
Figure 3:
Figure 2:
C (S )
 C (S )
 30b
*
j
j
0
 35b
So we have C j (S0 ) / C j (S * )  35b / 30b  7 / 6 , the bound is tight.
References
[1] M.R.Garey, D.S.Johson, and R.Sethi. The complexity of flowshop and
jobshop scheduling, Math. Oper.Res.m, 12, (1976), 117-129.
[2] J.A. Hoogereen, T.Kaweguchi. Minimizing total completion time in a
two-machine flowshop: analysis of special cases, Processing of the 5th IPCO
conference, (1996), 374-388.
[3] P. Brucker, S. Knust,G.Q. Wang,et al. Complexity of results for flow-shop
problems with a single server, European J. Oper. Res., 165(2), (2005),
398-407.