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Scheduling on uniform parallel machines to minimize maximum
lateness
Christos Koulamas
∗, George J. Kyparisis
Department of Decision Sciences and Information Systems, College of Business Administration, Florida International University, University Park Campus, Miami, FL 33199, USA
Received 1 January 1999; received in revised form 1 October 1999
Abstract
We consider the uniform parallel machine scheduling problem with the objective of minimizing maximum lateness. We show that an extension of the EDD rule to a uniform parallel machine setting yields a maximum lateness value which does not exceed the optimal value by more thanpmax, wherepmax is the maximum job processing time. c2000 Elsevier Science B.V. All rights reserved.
Keywords:Scheduling; Parallel machines; Approximation/heuristic; Worst-case analysis
1. Introduction
One of the earliest results in scheduling theory is that the earliest due date (EDD) rule minimizes max-imum lateness on a single machine [6]. The objective of this paper is to extend the EDD rule to a uniform parallel machine setting where the maximum lateness problem is known to be NP-hard. We show that the implementation of the EDD rule to the uniform paral-lel machine problem yields a maximum lateness value that does not exceed the optimal value by more than
pmax, where pmax is the maximum job processing
time.
We formally dene the uniform parallel machines scheduling problem with the maximum lateness
ob-∗Corresponding author. Fax: +1-305-348-4126.
E-mail address:[email protected] (C. Koulamas)
jective,Qm==Lmax, as follows. There arenjobs to be
processed without preemption on m continuously available uniform parallel machines. Each machine can process only one job at a time, and each job can be processed on only one machine. JobJj; j= 1; : : : ; n,
becomes available at time zero, requires pj units
of processing and has a due date dj. Machine
Mi; i= 1; : : : ; m, has a speed si; si¿1. The impact
of speedsi is that machineMi can carry outsi units
of processing in one time unit. Without loss of gen-erality, we may assume that s1¿s2¿· · ·¿sm= 1.
If job Jj is assigned to machine Mi then it requires
pj=si time units to be completed. The objective is to
determine a schedule so that the maximum lateness
Lmax= max16j6nLjis minimized, whereLj=Cj−dj
is the lateness andCjis the completion time of jobJj.
The Qm==Lmax problem is known to be NP-hard
even form= 2 [3]. Let SH be a schedule generated
by a heuristic H and let S∗ be an optimal
sched-ule for Qm==Lmax. Heuristic H is said to provide
the worst-case ratio bound if for any problem in-stance Lmax(SH)=Lmax(S∗)6. Since it is possible
thatLmax(S∗) = 0, Lenstra [8] suggested an equivalent
formulation of thePm==Lmax problem (with identical
parallel machines) which avoids this diculty. Let
qj =dmax −dj; j = 1; : : : ; n, be the delivery time
(or tail) of job Jj, where dmax = max16j6ndj is
the maximum due date. Consider the related prob-lem Pm=qj=Cmax, whereCmax= max16j6n(Cj+qj).
Observe that Cmax = max16j6n(Cj+dmax −dj) =
max16j6n(Cj − dj) + dmax = Lmax + dmax. Thus
Pm==Lmax andPm=qj=Cmax are equivalent. The above
relationships are applicable to theQm==Lmax case as
well. Until now, no worst-case ratio bounds have been obtained in the literature for either Qm==Lmax
or Qm=qj=Cmax. Guseld [4] implemented the EDD
heuristic for the Pm=rj=Lmax problem (with
un-equal job release times rj) and obtained the bound
Lmax(SEDD)−Lmax(S∗)6[(2m−1)=m]pmax. Using
the same heuristic as Guseld [4], Masuda et al. [9] obtained a modied worst-case ratio bound for the Pm==Lmax problem of the form (Lmax(SEDD)−
Lmax(S∗))=(Lmax(S∗) +dmax)61 −1=m. Carlier [1]
also considered the same heuristic as Guseld [4] applied to the Pm=rj; qj=Cmax problem and obtained
the bound Cmax(SEDD) − Cmax(S∗)62(pmax −1).
Finally, Hall and Shmoys [5] considered the problem
Pm=rj; qj;prec=Cmax (with precedence constraints)
and proved that for general list scheduling heuristic (LS) Cmax(SLS)=Cmax(S∗)¡2. They also developed
a polynomial approximation scheme for this problem. For more on the related literature, see Lawler et al. [7] and Tanaev et al. [10].
In this paper, we obtain ratio bounds for the
Qm==Lmax and Qm=qj=Cmax problems. Our bounds
yield the result of Masuda et al. [9] when si =
1; i= 1; : : : ; m, that is whenQm==Lmaxreduces to the
Pm==Lmax problem.
2. Absolute performance bounds forQmLmax
Let= (j1; : : : ; jn) be an arbitrary permutation of
the n jobs for the Qm==Lmax problem. Given , the
Modied List Scheduling (LS′) rule creates a
sched-ule forQm==Lmaxby assigning the job to be scheduled
next (in the order) to the uniform machine on which it will nish earliest (see [2]). In the next lemma we prove some properties of the schedule constructed us-ing the LS′ rule.
Lemma 1. Let=(1; : : : ; n)be an arbitrary permuta-tion; apply the LS′ rule to in order to create a
schedule forQm==Lmax.Then;for any partial schedule
of jobsJj; j= 1;2; : : : ; k;the following inequality is
rule. This implies that if jobJk were assigned to any
other machineMi; i= 1; : : : ; m; i6=ik, it would have
nished not earlier on that machine. Consequently, its completion time Ck =Pkj=1p
ik
j=sik on machine Mik
should not exceed the length of the partial sched-ule of jobs from the subset{Jj; j= 1; : : : ; k}on
ma-chineMi; i= 1; : : : ; m; i 6=ik (given by the quantity
Pk
j=1pij=si) plus the quantitypk=si(which represents
the processing time of jobJk if it were scheduled on
machineMi; i= 1; : : : ; m; i6=ik), that is
Inequality (2) can be written as
Together, (3) and (4) imply that
The division of both sides of inequality (5) byPm
i=1si
yields inequality (1).
Lemma 1 is needed to provide upper bounds on the job completion times inQm==Lmax. The next lemma
will be used to provide lower bounds on the job com-pletion times inQm==Lmax.
Lemma 2. Let S be an arbitrary schedule for the
Qm==Lmaxproblem;let= (1; : : : ; n)be the
permuta-tion of the job complepermuta-tion times inS(with ties broken arbitrarily).Then;for any partial schedule comprised of the rstkjobs in permutation; Jj; j= 1;2; : : : ; k; the following inequality is true:
k
Proof. Consider any partial schedule of the rst k
jobs in permutation; Jj; j= 1;2; : : : ; k. JobJkis the
job with the latest completion time among all jobs
Jj; j= 1;2; : : : ; k, therefore
can be written as
si By dividing both sides of inequality (9) byPm
i=1si,
we obtain inequality (6).
For any instance of theQm==Lmaxproblem, dene an
associated single-machine maximum lateness problem 1=p1
ule for Qm==Lmax and SEDD be the schedule for the
Qm==Lmax problem obtained by applying the LS′ rule
to the EDD sequence. The following result provides a bound onLmax(SEDD)−Lmax(S∗) forQm==Lmax in
terms of the maximum job processing time pmax =
max16j6n{pj}.
Proposition 1. For theQm==Lmax problem;
Lmax(SEDD)−Lmax(S∗)
and the bounds in inequality(10)are tight.
Proof. LetC1
[k]andC[k] denote the completion times
of the job in position k of the EDD sequence ( job
J[k]) in theSEDD1 andSEDDschedules, respectively. Let
L1
[k] =C[1k] −d[k] and L[k] =C[k] −d[k] denote the
corresponding lateness values. Using the notation of Lemma 1 and assuming that the permutationused in Lemma 1 is the EDD sequence, the completion times
C[k]andC[1k] are given by By combining (1) and (11), we obtain
C[k]6C[1k]+
m−1
Pm
i=1si
By subtractingd[k]from both sides in (12) and taking
the maxima overk, we obtain
max
16k6n{C[k]−d[k]}
6 max
16k6n{C
1
[k]−d[k]}+
m−1
Pm
i=1si
pmax: (13)
Observe that Lmax(SEDD) = max16k6n{L[k]} and
Lmax(SEDD1 ) = max16k6n{L1[k]}. Therefore, (13) can
be written as
Lmax(SEDD)6Lmax(SEDD1 ) +
m−1
Pm
i=1si
pmax: (14)
Now, consider an optimal scheduleS∗; let∗be the
permutation of the job completion times inS∗ (with
ties broken arbitrarily) and letS1∗be the 1=p1
j; d1j=Lmax
schedule corresponding to the permutation∗. LetC∗
[k]
andC1∗
[k] denote the completion times (in theS∗ and
S1∗ schedules, respectively) of the job in position k
( jobJ[k]) in the∗ sequence. By applying inequality
(6) of Lemma 2 to scheduleS∗(and permutation∗),
we obtain
C∗
[k]=
k
X
j=1
pik
[j]=sik¿
k
X
j=1
p[j]
, m
X
i=1
si
!
=C1∗
[k]:
(15) Inequality (15) implies thatL∗
[k]=C[∗k]−d[k]¿C[1k∗]−
d[k] =L1[k∗]. Since Lmax(S∗) = max16k6n{L∗[k]} and
Lmax(S1∗) = max16k6n{L1[k∗]}, this in turn implies
that
Lmax(S∗)¿Lmax(S1∗)¿Lmax(SEDD1 ) (16)
where we use the fact thatSEDD1 is an optimal schedule for 1=p1j; d1j=Lmax. Inequalities (14) and (16) yield the
rst inequality in (10); then, the second inequality in (10) follows from the observation that (recall that
s1¿s2¿· · ·¿sm= 1)
pmax=s16 max 16k6n{C
∗
[k]}
= max
16k6n{C
∗
[k]−d[k]+d[k]}
6 max
16k6n{C
∗
[k]−d[k]}+ max 16k6n{d[k]}
=Lmax(S∗) +dmax:
To prove that the bounds in (10) are tight, consider the following example. We assume for simplicity that
mis even (a similar example can be constructed for the case wheremis odd by augmenting the problem data below with one additional machine and one additional job of lengthm). Suppose that there aremmachines with speedssi=1+(m−i); i=1; : : : ; m, andn=2m−1
jobs withp2j−1=p2j=m−j; j= 1; : : : ; m−1; p2m−1
=m, and dj =m+j; j= 1; : : : ;2m−1,
respec-tively, where 1≫¿0. TheSEDDschedule assigns jobs
Jj; j=1; : : : ;2m−1, in the order (1; : : : ;2m−1) which
by applying the LS′ rule results in jobsJ
1andJ2m−1
assigned to machineM1, jobJ2 assigned toM2, jobs
Jk andJ2m−k assigned toMk; k= 3;5; : : : ; m−1, and
jobsJk andJ2m−k+2 assigned to Mk; k= 4;6; : : : ; m
(all sequenced in the stated order). It is not dicult to show that, as→0; Lmax(SEDD)→m−1.
In an optimal scheduleS∗, jobsJ
kandJ2m−k−1are
assigned to machine Mk; k = 1; : : : ; m−1, (in that
order) and jobJ2m−1 is assigned toMm. It is easy to
see that as → 0; Lmax(S∗) → 0. Sincepmax=m;
dmax → m; s1 → 1, and Pim=1si → m as → 0,
inequality (10) converges to the double equality (m−1)−0 = [(m−1)=m]m= [(m−1)=m](0 +m) as
→0.
Corollary 1. For thePm==Lmaxproblem;we have
Lmax(SEDD)−Lmax(S∗)
6m−1
m pmax6 m−1
m (Lmax(S ∗) +d
max): (17)
Proof. By substitutingsi= 1; i= 1; : : : ; m, in (10) we
obtain (17).
Inequality L=Lmax(SEDD)−Lmax(S∗)6[(m−
1)=m]pmax in (17) provides a reduction of
Gus-eld’s (1984) result (L6[(2m−1)=m]pmax) for the
Pm=rj=Lmaxproblem to thePm==Lmax case. Inequality
L6[(m−1)=m]pmax6[(m−1)=m](Lmax(S∗)+dmax)
in (17) rearms the result of Masuda et al. (1983) for thePm==Lmax problem.
Corollary 2. For theQm=qj=Cmaxproblem;we have
Cmax(SEDD)=Cmax(S∗)6
(m−1)s1
Pm
i=1si
Proof. We observed in the Introduction that Cmax=
Lmax+dmax. Thus,
Cmax(SEDD)=Cmax(S∗) =
Cmax(SEDD)−Cmax(S∗)
Cmax(S∗)
+ 1
=(Lmax(SEDD) +dmax)−(Lmax(S
∗) +d
max)
Lmax(S∗) +dmax
+ 1
=Lmax(SEDD)−Lmax(S
∗) Lmax(S∗) +dmax
+ 1:
In view of (10), this implies (18).
In thePm=qj=Cmax case, inequality (18) reduces to
Cmax(SEDD)=Cmax(S∗)62 −1=m and becomes
simi-lar to the resultCmax(SLS)=Cmax(S∗)¡2 of Hall and
Shmoys [5] for general list scheduling heuristic for a much more general problem.
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