parallel machine scheduling of machine-dependent jobs with unit-length
TRANSCRIPT
European Journal of Operational Research 156 (2004) 261–266
www.elsevier.com/locate/dsw
Short Communication
Parallel machine scheduling of machine-dependentjobs with unit-length q
Yixun Lin *, Wenhua Li
Department of Mathematics, Zhengzhou University, Zhengzhou, Henan 450052, China
Received 24 April 2002; accepted 5 December 2002
Abstract
In this paper we study the parallel machine scheduling problem with unit-length jobs in which each job j is only
allowed to be processed on a specified subset Mj of machines. For the problem P jpj ¼ 1;MjjCmax where the subset
family fMjg is nested, Pinedo [Scheduling: Theory, Algorithms, and Systems, Prentice-Hall, Englewood Cliffs, NJ,
1995] established an least flexible job first algorithm. We further present polynomial algorithms for the problem with
general subset family fMjg. As a generalization of the problem for nested family, we consider the problem for convex
subset family fMjg and present an efficient algorithm for solving it.
� 2003 Elsevier B.V. All rights reserved.
Keywords: Scheduling; Parallel machine; Unit-length job; Machine eligibility constraint
1. Introduction
The parallel machine scheduling problems have
been extensively studied in the literature [3,4,7]. In
a basic model, there is a set M ¼ fM1;M2; . . . ;Mmgof identical machines, which are used to process njobs J1; J2; . . . ; Jn. Each job Jj has one operation
which can be processed on any machine with
processing time pj. The objective is to minimize themakespan Cmax. By the well-known three-field
representation [3], this problem is denoted by
PkCmax.
In some practical applications, the machines
may have different capabilities (or eligibility re-
qProject supported by NSFC (grant no. 10371112).* Corresponding author.
E-mail address: [email protected] (Y. Lin).
0377-2217/$ - see front matter � 2003 Elsevier B.V. All rights reserv
doi:10.1016/S0377-2217(02)00914-1
strictions). This gives rise to a generalization thateach job Jj is only allowed to be processed on a
specified subset Mj of machines (Mj � M). This
subset Mj is called the admissible set of job Jj(16 j6 n). The generalized model can be denoted
by P jMjjCmax. In fact, it can be regarded as a
special case of unrelated parallel machine sched-
uling with processing times
pij ¼pj; if Mi 2 Mj;1; if Mi 62 Mj:
�
Since PkCmax is NP-hard, it is clear that the gen-
eralization P jMjjCmax is also NP-hard.
Pinedo [9] proposed the problem P jpj ¼ 1;MjjCmax, in which each job has a processing timeof unit. When the admissible subset family fMjg is
nested, i.e.,
ed.
Fig. 1. Network N .
262 Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266
Mi \Mj 6¼ ; ) Mi � Mj or
Mj � Mi for all i and j;
he established a least flexible job first (LFJ) algo-
rithm. The computational complexity of
P jpj ¼ 1;MjjCmax for general subset family fMjgwas left open. Later, Centeno and Armacost [2]
studied a special case of P jrj;MjjLmax, namely, the
problem of minimizing the maximum lateness Lmax
with release times rj and due dates dj ¼ rj þ c.Bhatia et al. [1] also considered the machine ca-
pability constraint, and they focused attention on
a more complicated model with precedence and
loading (setup) time constraints, which was shownto be NP-hard.
In this paper we mainly investigate the problem
P jpj ¼ 1;MjjCmax with general subset family
fMjg. First we present an Oðn3 log nÞ-time algo-
rithm, which can be naturally generalized to the
case where the machines have different speeds.
Then we give a more efficient algorithm for the
problem in which the family fMjg is convex,which is a generalization of nested family. Here, a
subset family fMjg is said to be convex if we can
order m machines in a sequence, say (M1;M2; . . . ;Mm), such that each subset Mj consists of consec-
utive terms of the sequence. This is similar to
the concept of ‘‘convex bipartite graph’’ [6]. In
the case of fMjg being convex, we obtain an
Oðmðmþ nÞÞ algorithm.The organization of the paper is as follows. In
Section 2 we establish the basic algorithm for
P jpj ¼ 1;MjjCmax with general constraint. Then,
we apply the algorithm to Qjpj ¼ 1;MjjCmax. Sec-
tion 3 is devoted to the case of convex subset
family fMjg and to develop a revised LFJ algo-
rithm. We provide some concluding remarks in
Section 4.
2. Binary search algorithm
Let us consider the problem P jpj ¼ 1;MjjCmax
for which M ¼ fM1;M2; . . . ;Mmg is the set of midentical machines; J1; J2; . . . ; Jn are the jobs with
unit-length (pj ¼ 1); the machine capability con-straint is represented by a family of admissible
subsets Mj ¼ fMi 2 MjJj can be processed on Mig(j ¼ 1; 2; . . . ; n); and the objective function is the
makespan Cmax (the final completion time of all jobs).
We will use a network flow based method on
the problem. The network flow theory can beconsulted in [6,8]. For any given subset family
fMjg we can construct a network N as follows.
Let X ¼ fx1; x2; . . . ; xmg correspond to the set of mmachines, Y ¼ fy1; y2; . . . ; yng to the set of n jobs.
The vertex set of N is V ¼ fs; tg [ X [ Y , where s isa source and t is a sink. The arc set A of N consists
of the following arcs:
ðs; xiÞ for all xi 2 X ði ¼ 1; . . . ;mÞ;ðxi; yjÞ for Mi 2 Mj ð16 i6m; 16 j6 nÞ;ðyj; tÞ for all yj 2 Y ðj ¼ 1; . . . ; nÞ:
All arcs ðxi; yjÞ and ðyj; tÞ have unit capacity; and
all arcs ðs; xiÞ have capacity c, a variable parame-
ter. Without loss of generality, we may assume
that c is a positive integer. We denote this network
by N ¼ ðV ;A; cÞ, as shown in Fig. 1.
Theorem 2.1. The problem P jpj ¼ 1;MjjCmax has afeasible schedule of makespan Cmax 6 c if and only ifthe value of max-flow in network N ¼ ðV ;A; cÞ is n.
Proof. If the scheduling problem has a feasible
schedule of Cmax 6 c, then at most c jobs are as-
signed on each machine. We may define a flow f in
network N as follows:
f ðxi; yjÞ ¼1; if Jj is assigned to Mi;
0; otherwise;
�
f ðyj; tÞ ¼Xmi¼1
f ðxi; yjÞ ¼ 1 for j ¼ 1; . . . ; n;
Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266 263
f ðs; xiÞ ¼Xnj¼1
f ðxi; yjÞ6 c for i ¼ 1; . . . ;m:
It is easy to see that this is a max-flow in N withvalue n. Conversely, if the value of a max-flow in Nis n, then there exists an integral max-flow f such
that all arcs ðyj; tÞ are saturated. This is because
that all these arcs have unit capacity and they form
a cut of capacity n. Thus the n unit flow f must
saturate this cut. So we can get a feasible schedule
by assigning job Jj to machine Mi if f ðxi; yjÞ ¼ 1.
Since each arc ðs; xiÞ has capacity c, it follows thatat most c jobs are assigned on each machine.
Therefore the makespan Cmax is not greater than c.This completes the proof. �
Based on this theorem, the problem P jpj ¼1;MjjCmax can be solved by searching for the
minimal parameter c. Thus we obtain the follow-
ing binary search algorithm.
BS Algorithm
Step 0: Construct the network N ¼ ðV ;A; cÞ fromthe subset family fMjg (where c is an un-
known parameter for the time being). Let
l ¼ 1 and u ¼ n.Step 1: If l ¼ u, then the optimal value is reached:
C�max ¼ l; return the schedule of makespan
l, stop.Step 2: Let c ¼ b1
2ðlþ uÞc. Find the max-flow in
network N ¼ ðV ;A; cÞ. If the value of
max-flow is exactly n, namely C�max 6 c,
then set u :¼ c and write down the sched-
ule of makespan c by means of the max-
flow. Otherwise, the value of max-flow is
less than n, namely, C�max > c, we set
l :¼ cþ 1. Go back to Step 1.
Fig. 2. Network N .
Theorem 2.2. The BS algorithm correctly solves theproblem P jpj ¼ 1;MjjCmax for any subset familyfMjg in Oðn3 log nÞ time.
Proof. By Theorem 2.1, we can assert that
l6C�max 6 u holds during the process of the algo-
rithm. As soon as the algorithm terminates,
l ¼ C�max ¼ u, an optimal solution is obtained. As
to the running time of the algorithm, it is clear that
the binary search has log n stages each of which is
a procedure of finding max-flow in network N . It is
well-known that there are Oðn3Þ algorithms for
the max-flow problem [6,8]. Therefore the time
bound of our algorithm is Oðn3 log nÞ. The theo-
rem follows. �
In fact, the network N considered in our algo-
rithm has special structure (it can be changed into
a �simple network� if each vertex xi is replaced by cvertices). So, the max-flow algorithm on it would
have complexity less than Oðn3Þ (see [8] for de-
tails). Therefore the complexity result in Theorem
2.2 can be improved slightly.
We can further consider the problemQjpj ¼ 1;MjjCmax, i.e., the case of uniform parallel
machines. In this case, m machines have different
speeds; the speed of machine Mi is denoted by vi.Accordingly, the actual processing time of job Jjon machine Mi is pj=vi (at the moment pj ¼ 1).
Without loss of generality we may assume that all
speeds v1; v2; . . . ; vm are positive integers (the ma-
chines with vi ¼ 1 are regarded as standard ma-chines). Let v be the least common multiple of
v1; v2; . . . ; vm. Then the completion time of every
job will be an integer if we take 1=v as a time unit.
We may construct the network N ¼ ðV ;A; cÞ asbefore, except that the capacity of arc ðs; xiÞ is
defined to be vic, as shown in Fig. 2.
Similar to Theorem 2.1, we have
Theorem 2.3. The problem Qjpj ¼ 1;MjjCmax has afeasible schedule of makespan Cmax 6 c if and only ifthe value of max-flow in network N ¼ ðV ;A; cÞ is n.
264 Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266
So we can use the binary search algorithm todetermine the minimal makespan C�
max and the
optimal schedule as before. However, the value of
parameter c now may not be integral. The range
½1=v; n� of c must be divided into vn� 1 sections by
the time unit 1=v and the binary search is executed
in these sections. Hence, there will be Oðlog nvÞstages in the BS algorithm. We then come to the
following conclusion.
Theorem 2.4. The BS algorithm correctly solves theproblem Qjpj ¼ 1;MjjCmax for any subset familyfMjg in Oðn3 log nvÞ time.
3. Convex subset family
We discuss a special case––the problem P jpj ¼1;MjjCmax where fMjg is a convex subset family.
We first address the problem of determining the
convexity of a subset family fMjg. In fact, a
family fMjg can be represented by a ð0; 1Þ-matrix
in which a row corresponds an element Mi and a
column corresponds a subset Mj, and the entry
ði; jÞ is 1 if and only if Mi 2 Mj. Then fMjg isconvex if and only if the matrix has �the consecu-
tive 1�s property for columns�, namely, the rows of
the matrix can be permuted in such a way that the
1�s in each column occur consecutively. This is also
equivalent to the interval hypergraph recognition
problem. It is well-known in the literature [5] that
the recognition of the consecutive 1�s property (as
well as that of interval graphs or interval hyper-graphs) can be solved in linear time. Hence, we
may assume here that the convexity of family
fMjg has been shown.
For the sake of simplicity, suppose that
M ¼ f1; 2; . . . ;mg. Moreover, by an integer inter-
val ½a; b�, we mean the set of integers
fi 2 Z j a6 i6 bg ¼ fa; aþ 1; . . . ; b� 1; bg (a6 b).For the case of convex family, we may assumethat m machines have been ordered as (1; 2; . . . ;m)such that each admissible set Mj is an integer in-
terval:
Mj ¼ ½aj; bj� ðaj; bj 2 M; aj 6 bjÞ:For any a; b 2 Mða6 bÞ, we define
J ½a; b� ¼ fJjja6 aj 6 bj 6 bgas a set of jobs whose admissible sets are included
in ½a; b�.
Theorem 3.1. The minimal makespan of the prob-lem is
C�max ¼ max
16 a6 b6m
jJ ½a; b�jb � a þ 1
� �:
Proof. Denote
k ¼ max16 a6b6m
jJ ½a; b�jb � a þ 1
� �:
We are going to prove C�max ¼ k. For any given
a and b (16 a6b6m), the jobs in J ½a; b� are onlyallowed to be processed on the b � a þ 1 machinesin ½a; b�. So, the final completion time c½a;b� of
these machines satisfies
c½a; b�P jJ ½a; b�jb � a þ 1
:
By the arbitrariness of a and b, we have C�max P k.
In order to show C�max 6 k, it suffices to con-
struct a feasible schedule of makespan k. We give
an algorithm as follows.
Revised LFJ Algorithm
Step 0: Let i ¼ 1; S ¼ ; (where S denotes the set of
jobs which have been assigned).
Step 1: Let Ti ¼ fJjjJj 62 S; aj 6 i6 bjg be the set
of jobs which can be assigned to machine
i. If Ti ¼ ;, go to the next step. If
0 < jTij6 k, then assign all jobs in Ti tomachine i; If jTij > k, then, according tothe non-decreasing order of fbjg (the
�LFJ� rule), assign the first k jobs in Ti tomachines i. Put all assigned jobs into S.
Step 2: Stop if i ¼ m; otherwise let i :¼ iþ 1, go
back to Step 1.
We proceed to show that the algorithm can
produce a feasible schedule of makespan k. If not,there must be a job Jj which cannot be assigned in
any machine. Suppose that this Jj is the first job
not being assigned (i.e., the one with the least bj
Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266 265
among those failed jobs). For brevity, the ma-chines which have received k jobs in the algorithm
are called �full machines�. Then, for aj 6 i6 bj, themachines i must be full, so that there is no room to
hold Jj. Besides, there may be more full machines.
Let b ¼ bj and let a be the least number such that
all machines in ½a; b� are full. Then 16 a6 aj. Note
that the machine ða � 1Þ is not full if a > 1. It can
be seen that Jj and all jobs assigned to the ma-chines in ½a; b� belong to J ½a; b�. In fact, for a job Jiassigned in ½a; b�, the rule in Step 1 implies that
bi 6 bj ¼ b. On the other hand, we have ai P a.Otherwise, ai 6 a � 1, Ji would be assigned to ma-
chine a � 1, as it is not full. So Ji 2 J ½a; b�. Hence
jJ ½a; b�jP ðb � a þ 1Þk þ 1;
thus
jJ ½a; b�jb � a þ 1
> k;
which contradicts the definition of k. Therefore
C�max ¼ k. The theorem follows. �
Theorem 3.2. The revised LFJ algorithm correctlysolves the problem P jpj ¼ 1;MjjCmax with convexsubset family fMjg in Oðmðmþ nÞÞ time.
Proof. The correctness has been proved in Theo-
rem 3.1. Let us see the time bound of the algo-
rithm. Note that we have to compute k in advance.For doing this, there are two steps as follows.
First, we should calculate an m� mmatrix kab
� �where
kab ¼ jJ ½a; b�j; if a6 b;0; otherwise:
�
This can be done by constructing the sets J ½a;�� ¼fJjja6 ajg and sets J ½�; b� ¼ fJjjbj 6 bg in OðmnÞtime (for each a or b, scan all n jobs); and com-
puting all jJ ½a; b�j ¼ jJ ½a;�� \ J ½�; b�j in Oðm2Þtime.
Second, compute
k ¼ max16 a6 b6m
kab
b � a þ 1
� �
in Oðm2Þ time.
After k has been computed, the revised LFJ
algorithm can be carried out in OðmnÞ time. In
fact, the algorithm has m stages and in each stagewe assign the jobs in Ti in OðnÞ time.
To summarize, the overall complexity of the
algorithm is Oðmðmþ nÞÞ. This completes the
proof. �
Pinedo [9] studied the case of nested subset
family fMjg (i.e., Mi \Mj 6¼ ; ) Mi � Mj or
Mj � Mi). It is easy to show by induction on nthat if a family fMjg is nested, then it is convex.
But the converse is not true. Theorem 3.2 implies
the following.
Corollary 3.3. For nested subset family fMjg, theproblem P jpj ¼ 1;MjjCmax has an Oðmðmþ nÞÞ al-gorithm.
The following example in [9] illustrates that the
LFJ algorithm may not work if fMjg is not nes-
ted:
M1 ¼ f1; 2g; M2 ¼ M3 ¼ f1; 3; 4g;M4 ¼ f2g; M5 ¼ M6 ¼ M7 ¼ M8 ¼ f3; 4g:
However, this fMjg is convex in viewing the order
(2,1,3,4) of machines. Our algorithm provides the
following solution:
4. Concluding remarks
In the foregoing sections we have investigated
the parallel machine scheduling problems withadmissible subset constraint––each job j is only
allowed to be processed on the machines in subset
Mj. These models come up in connection with
many application backgrounds. Meanwhile, they
have theoretic interest in the viewpoint of
266 Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266
combinatorics. In fact, they can be regarded asgeneralizations of the system of distinct represen-
tatives (SDR) problem: the existence of SDR is
equivalent to determining whether the problem
P jpj ¼ 1; MjjCmax (mP n) has a schedule of
Cmax ¼ 1.
There are many aspects worthwhile to further
study. For example, we have considered the ob-
jective Cmax. How about the others, e.g.,P
Cj,Lmax,
PUj, etc? In the case of ordinary processing
times (not necessarily unit-length), the problem
P jMjjCmax is NP-hard. It is significant to study the
approximation algorithms and other polynomially
solvable cases. As a generalization of nested subset
family, we have taken the convex subset family
fMjg into account. More results on other special
cases would be expected.
Acknowledgements
The authors would like to thank the referees for
their helpful comments on improving the repre-
sentation of the paper.
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