Directory UMM :Data Elmu:jurnal:O:Operations Research Letters:Vol26.Issue4.2000:
Operations Research Letters 26 (2000) 175–179
www.elsevier.com/locate/dsw
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
c 2000 Elsevier Science
not exceed the optimal value by more than pmax , where pmax is the maximum job processing time.
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 maximum 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 parallel 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 de ne the uniform parallel machines
scheduling problem with the maximum lateness obCorresponding author. Fax: +1-305-348-4126.
E-mail address: koulamas@ u.edu (C. Koulamas)
∗
jective, Qm==Lmax , as follows. There are n jobs 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. Job Jj ; 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 speed si is that machine Mi can carry out si units
of processing in one time unit. Without loss of generality, 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 = max16j6n Lj is minimized, where Lj = Cj − dj
is the lateness and Cj is the completion time of job Jj .
The Qm==Lmax problem is known to be NP-hard
even for m = 2 [3]. Let SH be a schedule generated
c 2000 Elsevier Science B.V. All rights reserved.
0167-6377/00/$ - see front matter
PII: S 0 1 6 7 - 6 3 7 7 ( 0 0 ) 0 0 0 1 8 - 3
176
C. Koulamas, G.J. Kyparisis / Operations Research Letters 26 (2000) 175–179
by a heuristic H and let S ∗ be an optimal schedule for Qm==Lmax . Heuristic H is said to provide
the worst-case ratio bound if for any problem instance Lmax (SH )=Lmax (S ∗ )6. Since it is possible
that Lmax (S ∗ ) = 0, Lenstra [8] suggested an equivalent
formulation of the Pm==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 = max16j6n dj is
the maximum due date. Consider the related problem Pm=qj =Cmax , where Cmax = max16j6n (Cj + qj ).
Observe that Cmax = max16j6n (Cj + dmax − dj ) =
max16j6n (Cj − dj ) + dmax = Lmax + dmax . Thus
Pm==Lmax and Pm=qj =Cmax are equivalent. The above
relationships are applicable to the Qm==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 . Gus eld [4] implemented the EDD
heuristic for the Pm=rj =Lmax problem (with unequal job release times rj ) and obtained the bound
Lmax (SEDD ) − Lmax (S ∗ )6[(2m − 1)=m]pmax . Using
the same heuristic as Gus eld [4], Masuda et al.
[9] obtained a modi ed 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 Gus eld [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 when Qm==Lmax reduces to the
Pm==Lmax problem.
2. Absolute performance bounds for Qm Lmax
Let = (j1 ; : : : ; jn ) be an arbitrary permutation of
the n jobs for the Qm==Lmax problem. Given , the
Modi ed List Scheduling (LS′ ) rule creates a schedule for Qm==Lmax by 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 using the LS′ rule.
Lemma 1. Let =(1; : : : ; n) be an arbitrary permutation; apply the LS′ rule to in order to create a
schedule for Qm==Lmax . Then; for any partial schedule
of jobs Jj ; j = 1; 2; : : : ; k; the following inequality is
true:
,
!
k
k
m
X
X
X
pjik =sik 6
pj
si
j=1
j=1
i=1
+(m − 1)pk
,
m
X
i=1
si
!
;
(1)
where Mik ; 16ik 6m; denotes the machine to which
job Jk has been assigned; also pji = pj if job Jj is
assigned to Mi and pji = 0 otherwise; i = 1; : : : ; m.
Proof. Job Jk is assigned to machine Mik by the LS′
rule. This implies that if job Jk were assigned to any
other machine Mi ; i = 1; : : : ; m; i 6= ik , it would have
nished not earlier on that machine. Consequently, its
Pk
completion time Ck = j=1 pjik =sik on machine Mik
should not exceed the length of the partial schedule of jobs from the subset {Jj ; j = 1; : : : ; k} on machine Mi ; i = 1; : : : ; m; i 6= ik (given by the quantity
Pk
i
j=1 pj =si ) plus the quantity pk =si (which represents
the processing time of job Jk if it were scheduled on
machine Mi ; i = 1; : : : ; m; i 6= ik ), that is
k
X
pjik =sik 6
j=1
k
X
pji =si + pk =si
(2)
j=1
for i = 1; : : : ; m; i 6= ik .
Inequality (2) can be written as
k
k
X
X
pjik =sik 6
pji + pk
si
j=1
(3)
j=1
for i = 1; : : : ; m; i 6= ik . We also have, trivially, that
k
k
X
X
pjik :
(4)
pjik =sik 6
s ik
j=1
j=1
177
C. Koulamas, G.J. Kyparisis / Operations Research Letters 26 (2000) 175–179
yields
Together, (3) and (4) imply that
! k
m
X
X i
si
pjk =sik
i=1
m
X
j=1
i=1
m
k
X
X
pji + (m − 1)pk
6
i=1
=
k
X
pj + (m − 1)pk :
(5)
j=1
Pm
i=1 si
Lemma 1 is needed to provide upper bounds on the
job completion times in Qm==Lmax . The next lemma
will be used to provide lower bounds on the job completion times in Qm==Lmax .
Lemma 2. Let S be an arbitrary schedule for the
Qm==Lmax problem; let = (1; : : : ; n) be the permutation of the job completion times in S (with ties broken
arbitrarily). Then; for any partial schedule comprised
of the rst k jobs in permutation ; Jj ; j = 1; 2; : : : ; k;
the following inequality is true:
,
!
m
k
k
X
X
X
ik
si ;
pj =sik ¿
pj
(6)
j=1
i=1
j=1
j=1
i=1
j=1
j=1
(9)
By dividing both sides of inequality (9) by
we obtain inequality (6).
j=1
The division of both sides of inequality (5) by
yields inequality (1).
! k
k
k
m
X
X
X
X i
pj
pji =
si
pjk =sik ¿
Pm
i=1 si ,
For any instance of the Qm==Lmax problem, de ne an
associated single-machine maximum lateness P
problem
m
1=pj1 ; d1j =Lmax with processing times pj1 =pj =( i=1 si )
1
be the optimal (EDD)
and due dates d1j = dj . Let SEDD
1 1
schedule for 1=pj ; dj =Lmax ; S ∗ be an optimal schedule 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 on Lmax (SEDD ) − Lmax (S ∗ ) for Qm==Lmax in
terms of the maximum job processing time pmax =
max16j6n {pj }.
Proposition 1. For the Qm==Lmax problem;
Lmax (SEDD ) − Lmax (S ∗ )
m−1
pmax
6 Pm
i=1 si
(m − 1)s1
(Lmax (S ∗ ) + dmax )
6 Pm
i=1 si
(10)
where ik ; Mik and pji are de ned as in Lemma 1.
and the bounds in inequality (10) are tight.
Proof. Consider any partial schedule of the rst k
jobs in permutation ; Jj ; j = 1; 2; : : : ; k. Job Jk is the
job with the latest completion time among all jobs
Jj ; j = 1; 2; : : : ; k, therefore
1
and C[k] denote the completion times
Proof. Let C[k]
of the job in position k of the EDD sequence (job
1
and SEDD schedules, respectively. Let
J[k] ) in the SEDD
1
1
L[k] = C[k] − d[k] and L[k] = C[k] − d[k] denote the
corresponding lateness values. Using the notation of
Lemma 1 and assuming that the permutation used in
Lemma 1 is the EDD sequence, the completion times
1
are given by
C[k] and C[k]
,
!
k
k
m
X
X
X
ik
1
p[ j] =sik ; C[k] =
p[ j]
si :
C[k] =
Ck =
k
X
j=1
pjik =sik ¿
k
X
pji =si
(7)
j=1
Pk
i
for i = 1; : : : ; m, where
j=1 pj =si is the length of
the partial schedule of jobs from the subset {Jj ; j =
1; 2; : : : ; k} on machine Mi ; i=1; : : : ; m. Inequality (7)
can be written as
k
k
X
X
pjik =sik ¿
pji
(8)
si
j=1
j=1
for i = 1; : : : ; m. The summation of inequalities (8)
j=1
j=1
i=1
(11)
By combining (1) and (11), we obtain
m−1
1
p[k] :
+ Pm
C[k] 6C[k]
i=1 si
(12)
178
C. Koulamas, G.J. Kyparisis / Operations Research Letters 26 (2000) 175–179
By subtracting d[k] from both sides in (12) and taking
the maxima over k, we obtain
max {C[k] − d[k] }
16k6n
m−1
1
− d[k] } + Pm
pmax :
6 max {C[k]
16k6n
i=1 si
(13)
Observe that Lmax (SEDD ) = max16k6n {L[k] } and
1
) = max16k6n {L1[k] }. Therefore, (13) can
Lmax (SEDD
be written as
m−1
1
pmax :
) + Pm
Lmax (SEDD )6Lmax (SEDD
i=1 si
(14)
Now, consider an optimal schedule S ∗ ; let ∗ be the
permutation of the job completion times in S ∗ (with
ties broken arbitrarily) and let S 1∗ be the 1=pj1 ; d1j =Lmax
∗
schedule corresponding to the permutation ∗ . Let C[k]
1∗
and C[k]
denote the completion times (in the S ∗ and
1∗
S schedules, respectively) of the job in position k
( job J[k] ) in the ∗ sequence. By applying inequality
(6) of Lemma 2 to schedule S ∗ (and permutation ∗ ),
we obtain
,
!
m
k
k
X
X
X
ik
1∗
∗
:
si = C[k]
p[ j] =sik ¿
p[ j]
C[k] =
j=1
i=1
j=1
(15)
∗
1∗
− d[k] ¿C[k]
−
Inequality (15) implies that L∗[k] = C[k]
∗
∗
}
and
.
Since
L
(S
)
=
max
{L
d[k] = L1∗
max
16k6n
[k]
[k]
},
this
in
turn
implies
Lmax (S 1∗ ) = max16k6n {L1∗
[k]
that
1
)
Lmax (S ∗ )¿Lmax (S 1∗ )¿Lmax (SEDD
(16)
1
is an optimal schedule
where we use the fact that SEDD
1 1
for 1=pj ; dj =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 =s1 6 max {C[k]
16k6n
∗
− d[k] + d[k] }
= max {C[k]
16k6n
To prove that the bounds in (10) are tight, consider
the following example. We assume for simplicity that
m is even (a similar example can be constructed for the
case where m is odd by augmenting the problem data
below with one additional machine and one additional
job of length m). Suppose that there are m machines
with speeds si =1+(m−i); i=1; : : : ; m, and n=2m−1
jobs with p2j−1 =p2j =m−j; j =1; : : : ; m−1; p2m−1
= m, and dj = m + j; j = 1; : : : ; 2m − 1, respectively, where 1≫¿0. The SEDD schedule assigns jobs
Jj ; j=1; : : : ; 2m−1, in the order (1; : : : ; 2m−1) which
by applying the LS′ rule results in jobs J1 and J2m−1
assigned to machine M1 , job J2 assigned to M2 , jobs
Jk and J2m−k assigned to Mk ; k = 3; 5; : : : ; m − 1, and
jobs Jk and J2m−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 schedule S ∗ , jobs Jk and J2m−k−1 are
assigned to machine Mk ; k = 1; : : : ; m − 1, (in that
order) and job J2m−1 is assigned to Mm . It is easy to
) → 0. Since pmax = m;
see that as → 0; Lmax (S ∗P
m
dmax → m; s1 → 1, and i=1 si → 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 the Pm==Lmax problem; we have
Lmax (SEDD ) − Lmax (S ∗ )
6
m−1
m−1
pmax 6
(Lmax (S ∗ ) + dmax ):
m
m
(17)
Proof. By substituting si = 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 =Lmax problem to the Pm==Lmax case. Inequality
L6[(m−1)=m]pmax 6[(m−1)=m](Lmax (S ∗ )+dmax )
in (17) rearms the result of Masuda et al. (1983)
for the Pm==Lmax problem.
Corollary 2. For the Qm=qj =Cmax problem; we have
∗
− d[k] } + max {d[k] }
6 max {C[k]
16k6n
16k6n
∗
= Lmax (S ) + dmax :
(m − 1)s1
+ 1:
Cmax (SEDD )=Cmax (S ∗ )6 Pm
i=1 si
(18)
C. Koulamas, G.J. Kyparisis / Operations Research Letters 26 (2000) 175–179
Proof. We observed in the Introduction that Cmax =
Lmax + dmax . Thus,
Cmax (SEDD )=Cmax (S ∗ ) =
Cmax (SEDD ) − Cmax (S ∗ )
+1
Cmax (S ∗ )
=
(Lmax (SEDD ) + dmax ) − (Lmax (S ∗ ) + dmax )
+1
Lmax (S ∗ ) + dmax
=
Lmax (SEDD ) − Lmax (S ∗ )
+ 1:
Lmax (S ∗ ) + dmax
In view of (10), this implies (18).
In the Pm=qj =Cmax case, inequality (18) reduces to
Cmax (SEDD )=Cmax (S ∗ )62 − 1=m and becomes similar to the result Cmax (SLS )=Cmax (S ∗ ) ¡ 2 of Hall and
Shmoys [5] for general list scheduling heuristic for a
much more general problem.
References
[1] J. Carlier, Scheduling jobs with release dates and tails on
identical machines to minimize the makespan, European J.
Oper. Res. 29 (1987) 298–306.
[2] Y. Cho, S. Sahni, Bounds for list schedules on uniform
processors, SIAM J. Comput. 9 (1980) 91–103.
179
[3] M.R. Garey, D.S. Johnson, Computers and Intractability: A
Guide to the Theory of NP-completeness, Freeman, New
York, 1979.
[4] D. Gus eld, Bounds for naive multiple machine scheduling
with release times and deadlines, J. Algorithms 5 (1984) 1–6.
[5] L.A. Hall, D.B. Shmoys, Approximation schemes for
constrained scheduling problems, Proceedings of the 30th
IEEE Symposium Foundations of Computer Science, 1989,
pp. 134 –139.
[6] J.R. Jackson, Scheduling a production line to minimize
maximum tardiness, Research Report 43, Management
Science Research Project, University of California, Los
Angeles, 1955.
[7] E.L. Lawler, J.K. Lenstra, A.H.G. Rinnooy Kan, D.B.
Shmoys, Sequencing and scheduling: algorithms and
complexity, in: S.C. Graves, A.H.G. Rinnooy Kan, P.H.
Zipkin (Eds.), Handbooks in Operations Research and
Management Science, Vol. 4: Logistics of Production and
Inventory, North-Holland, Amsterdam, 1993, pp. 445–522.
[8] J.K. Lenstra, Sequencing by Enumerative Methods,
Mathematical Centre Tracts, Vol. 69, Centre for Mathematics
and Computer Science, Amsterdam, 1977.
[9] T. Masuda, H. Ishii, T. Nishida, Some bounds on
approximation algorithms for n=m=1=Lmax and n=2=F=Lmax
scheduling problems, J. Oper. Res. Soc. Japan 26 (1983)
212–224.
[10] V.S. Tanaev, V.S. Gordon, Y.M. Shafransky, Scheduling
Theory, Single-Stage Systems, Kluwer, Dordrecht, 1994.
www.elsevier.com/locate/dsw
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
c 2000 Elsevier Science
not exceed the optimal value by more than pmax , where pmax is the maximum job processing time.
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 maximum 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 parallel 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 de ne the uniform parallel machines
scheduling problem with the maximum lateness obCorresponding author. Fax: +1-305-348-4126.
E-mail address: koulamas@ u.edu (C. Koulamas)
∗
jective, Qm==Lmax , as follows. There are n jobs 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. Job Jj ; 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 speed si is that machine Mi can carry out si units
of processing in one time unit. Without loss of generality, 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 = max16j6n Lj is minimized, where Lj = Cj − dj
is the lateness and Cj is the completion time of job Jj .
The Qm==Lmax problem is known to be NP-hard
even for m = 2 [3]. Let SH be a schedule generated
c 2000 Elsevier Science B.V. All rights reserved.
0167-6377/00/$ - see front matter
PII: S 0 1 6 7 - 6 3 7 7 ( 0 0 ) 0 0 0 1 8 - 3
176
C. Koulamas, G.J. Kyparisis / Operations Research Letters 26 (2000) 175–179
by a heuristic H and let S ∗ be an optimal schedule for Qm==Lmax . Heuristic H is said to provide
the worst-case ratio bound if for any problem instance Lmax (SH )=Lmax (S ∗ )6. Since it is possible
that Lmax (S ∗ ) = 0, Lenstra [8] suggested an equivalent
formulation of the Pm==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 = max16j6n dj is
the maximum due date. Consider the related problem Pm=qj =Cmax , where Cmax = max16j6n (Cj + qj ).
Observe that Cmax = max16j6n (Cj + dmax − dj ) =
max16j6n (Cj − dj ) + dmax = Lmax + dmax . Thus
Pm==Lmax and Pm=qj =Cmax are equivalent. The above
relationships are applicable to the Qm==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 . Gus eld [4] implemented the EDD
heuristic for the Pm=rj =Lmax problem (with unequal job release times rj ) and obtained the bound
Lmax (SEDD ) − Lmax (S ∗ )6[(2m − 1)=m]pmax . Using
the same heuristic as Gus eld [4], Masuda et al.
[9] obtained a modi ed 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 Gus eld [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 when Qm==Lmax reduces to the
Pm==Lmax problem.
2. Absolute performance bounds for Qm Lmax
Let = (j1 ; : : : ; jn ) be an arbitrary permutation of
the n jobs for the Qm==Lmax problem. Given , the
Modi ed List Scheduling (LS′ ) rule creates a schedule for Qm==Lmax by 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 using the LS′ rule.
Lemma 1. Let =(1; : : : ; n) be an arbitrary permutation; apply the LS′ rule to in order to create a
schedule for Qm==Lmax . Then; for any partial schedule
of jobs Jj ; j = 1; 2; : : : ; k; the following inequality is
true:
,
!
k
k
m
X
X
X
pjik =sik 6
pj
si
j=1
j=1
i=1
+(m − 1)pk
,
m
X
i=1
si
!
;
(1)
where Mik ; 16ik 6m; denotes the machine to which
job Jk has been assigned; also pji = pj if job Jj is
assigned to Mi and pji = 0 otherwise; i = 1; : : : ; m.
Proof. Job Jk is assigned to machine Mik by the LS′
rule. This implies that if job Jk were assigned to any
other machine Mi ; i = 1; : : : ; m; i 6= ik , it would have
nished not earlier on that machine. Consequently, its
Pk
completion time Ck = j=1 pjik =sik on machine Mik
should not exceed the length of the partial schedule of jobs from the subset {Jj ; j = 1; : : : ; k} on machine Mi ; i = 1; : : : ; m; i 6= ik (given by the quantity
Pk
i
j=1 pj =si ) plus the quantity pk =si (which represents
the processing time of job Jk if it were scheduled on
machine Mi ; i = 1; : : : ; m; i 6= ik ), that is
k
X
pjik =sik 6
j=1
k
X
pji =si + pk =si
(2)
j=1
for i = 1; : : : ; m; i 6= ik .
Inequality (2) can be written as
k
k
X
X
pjik =sik 6
pji + pk
si
j=1
(3)
j=1
for i = 1; : : : ; m; i 6= ik . We also have, trivially, that
k
k
X
X
pjik :
(4)
pjik =sik 6
s ik
j=1
j=1
177
C. Koulamas, G.J. Kyparisis / Operations Research Letters 26 (2000) 175–179
yields
Together, (3) and (4) imply that
! k
m
X
X i
si
pjk =sik
i=1
m
X
j=1
i=1
m
k
X
X
pji + (m − 1)pk
6
i=1
=
k
X
pj + (m − 1)pk :
(5)
j=1
Pm
i=1 si
Lemma 1 is needed to provide upper bounds on the
job completion times in Qm==Lmax . The next lemma
will be used to provide lower bounds on the job completion times in Qm==Lmax .
Lemma 2. Let S be an arbitrary schedule for the
Qm==Lmax problem; let = (1; : : : ; n) be the permutation of the job completion times in S (with ties broken
arbitrarily). Then; for any partial schedule comprised
of the rst k jobs in permutation ; Jj ; j = 1; 2; : : : ; k;
the following inequality is true:
,
!
m
k
k
X
X
X
ik
si ;
pj =sik ¿
pj
(6)
j=1
i=1
j=1
j=1
i=1
j=1
j=1
(9)
By dividing both sides of inequality (9) by
we obtain inequality (6).
j=1
The division of both sides of inequality (5) by
yields inequality (1).
! k
k
k
m
X
X
X
X i
pj
pji =
si
pjk =sik ¿
Pm
i=1 si ,
For any instance of the Qm==Lmax problem, de ne an
associated single-machine maximum lateness P
problem
m
1=pj1 ; d1j =Lmax with processing times pj1 =pj =( i=1 si )
1
be the optimal (EDD)
and due dates d1j = dj . Let SEDD
1 1
schedule for 1=pj ; dj =Lmax ; S ∗ be an optimal schedule 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 on Lmax (SEDD ) − Lmax (S ∗ ) for Qm==Lmax in
terms of the maximum job processing time pmax =
max16j6n {pj }.
Proposition 1. For the Qm==Lmax problem;
Lmax (SEDD ) − Lmax (S ∗ )
m−1
pmax
6 Pm
i=1 si
(m − 1)s1
(Lmax (S ∗ ) + dmax )
6 Pm
i=1 si
(10)
where ik ; Mik and pji are de ned as in Lemma 1.
and the bounds in inequality (10) are tight.
Proof. Consider any partial schedule of the rst k
jobs in permutation ; Jj ; j = 1; 2; : : : ; k. Job Jk is the
job with the latest completion time among all jobs
Jj ; j = 1; 2; : : : ; k, therefore
1
and C[k] denote the completion times
Proof. Let C[k]
of the job in position k of the EDD sequence (job
1
and SEDD schedules, respectively. Let
J[k] ) in the SEDD
1
1
L[k] = C[k] − d[k] and L[k] = C[k] − d[k] denote the
corresponding lateness values. Using the notation of
Lemma 1 and assuming that the permutation used in
Lemma 1 is the EDD sequence, the completion times
1
are given by
C[k] and C[k]
,
!
k
k
m
X
X
X
ik
1
p[ j] =sik ; C[k] =
p[ j]
si :
C[k] =
Ck =
k
X
j=1
pjik =sik ¿
k
X
pji =si
(7)
j=1
Pk
i
for i = 1; : : : ; m, where
j=1 pj =si is the length of
the partial schedule of jobs from the subset {Jj ; j =
1; 2; : : : ; k} on machine Mi ; i=1; : : : ; m. Inequality (7)
can be written as
k
k
X
X
pjik =sik ¿
pji
(8)
si
j=1
j=1
for i = 1; : : : ; m. The summation of inequalities (8)
j=1
j=1
i=1
(11)
By combining (1) and (11), we obtain
m−1
1
p[k] :
+ Pm
C[k] 6C[k]
i=1 si
(12)
178
C. Koulamas, G.J. Kyparisis / Operations Research Letters 26 (2000) 175–179
By subtracting d[k] from both sides in (12) and taking
the maxima over k, we obtain
max {C[k] − d[k] }
16k6n
m−1
1
− d[k] } + Pm
pmax :
6 max {C[k]
16k6n
i=1 si
(13)
Observe that Lmax (SEDD ) = max16k6n {L[k] } and
1
) = max16k6n {L1[k] }. Therefore, (13) can
Lmax (SEDD
be written as
m−1
1
pmax :
) + Pm
Lmax (SEDD )6Lmax (SEDD
i=1 si
(14)
Now, consider an optimal schedule S ∗ ; let ∗ be the
permutation of the job completion times in S ∗ (with
ties broken arbitrarily) and let S 1∗ be the 1=pj1 ; d1j =Lmax
∗
schedule corresponding to the permutation ∗ . Let C[k]
1∗
and C[k]
denote the completion times (in the S ∗ and
1∗
S schedules, respectively) of the job in position k
( job J[k] ) in the ∗ sequence. By applying inequality
(6) of Lemma 2 to schedule S ∗ (and permutation ∗ ),
we obtain
,
!
m
k
k
X
X
X
ik
1∗
∗
:
si = C[k]
p[ j] =sik ¿
p[ j]
C[k] =
j=1
i=1
j=1
(15)
∗
1∗
− d[k] ¿C[k]
−
Inequality (15) implies that L∗[k] = C[k]
∗
∗
}
and
.
Since
L
(S
)
=
max
{L
d[k] = L1∗
max
16k6n
[k]
[k]
},
this
in
turn
implies
Lmax (S 1∗ ) = max16k6n {L1∗
[k]
that
1
)
Lmax (S ∗ )¿Lmax (S 1∗ )¿Lmax (SEDD
(16)
1
is an optimal schedule
where we use the fact that SEDD
1 1
for 1=pj ; dj =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 =s1 6 max {C[k]
16k6n
∗
− d[k] + d[k] }
= max {C[k]
16k6n
To prove that the bounds in (10) are tight, consider
the following example. We assume for simplicity that
m is even (a similar example can be constructed for the
case where m is odd by augmenting the problem data
below with one additional machine and one additional
job of length m). Suppose that there are m machines
with speeds si =1+(m−i); i=1; : : : ; m, and n=2m−1
jobs with p2j−1 =p2j =m−j; j =1; : : : ; m−1; p2m−1
= m, and dj = m + j; j = 1; : : : ; 2m − 1, respectively, where 1≫¿0. The SEDD schedule assigns jobs
Jj ; j=1; : : : ; 2m−1, in the order (1; : : : ; 2m−1) which
by applying the LS′ rule results in jobs J1 and J2m−1
assigned to machine M1 , job J2 assigned to M2 , jobs
Jk and J2m−k assigned to Mk ; k = 3; 5; : : : ; m − 1, and
jobs Jk and J2m−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 schedule S ∗ , jobs Jk and J2m−k−1 are
assigned to machine Mk ; k = 1; : : : ; m − 1, (in that
order) and job J2m−1 is assigned to Mm . It is easy to
) → 0. Since pmax = m;
see that as → 0; Lmax (S ∗P
m
dmax → m; s1 → 1, and i=1 si → 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 the Pm==Lmax problem; we have
Lmax (SEDD ) − Lmax (S ∗ )
6
m−1
m−1
pmax 6
(Lmax (S ∗ ) + dmax ):
m
m
(17)
Proof. By substituting si = 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 =Lmax problem to the Pm==Lmax case. Inequality
L6[(m−1)=m]pmax 6[(m−1)=m](Lmax (S ∗ )+dmax )
in (17) rearms the result of Masuda et al. (1983)
for the Pm==Lmax problem.
Corollary 2. For the Qm=qj =Cmax problem; we have
∗
− d[k] } + max {d[k] }
6 max {C[k]
16k6n
16k6n
∗
= Lmax (S ) + dmax :
(m − 1)s1
+ 1:
Cmax (SEDD )=Cmax (S ∗ )6 Pm
i=1 si
(18)
C. Koulamas, G.J. Kyparisis / Operations Research Letters 26 (2000) 175–179
Proof. We observed in the Introduction that Cmax =
Lmax + dmax . Thus,
Cmax (SEDD )=Cmax (S ∗ ) =
Cmax (SEDD ) − Cmax (S ∗ )
+1
Cmax (S ∗ )
=
(Lmax (SEDD ) + dmax ) − (Lmax (S ∗ ) + dmax )
+1
Lmax (S ∗ ) + dmax
=
Lmax (SEDD ) − Lmax (S ∗ )
+ 1:
Lmax (S ∗ ) + dmax
In view of (10), this implies (18).
In the Pm=qj =Cmax case, inequality (18) reduces to
Cmax (SEDD )=Cmax (S ∗ )62 − 1=m and becomes similar to the result Cmax (SLS )=Cmax (S ∗ ) ¡ 2 of Hall and
Shmoys [5] for general list scheduling heuristic for a
much more general problem.
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