Discrete Optimization
Fast approximation algorithms for bi-criteria scheduling with machine
assignment costs
Kangbok Lee
a, Joseph Y-T. Leung
b,1, Zhao-hong Jia
c,2, Wenhua Li
d,3, Michael L. Pinedo
e,⇑,4,
Bertrand M.T. Lin
f,5a
Department of Business & Economics, York College, The City University of New York, 94-20 Guy R. Brewer Blvd, Jamaica, NY 11451, USA b
Department of Computer Science, New Jersey Institute of Technology, Newark, NJ 07102, USA c
Key Lab of Intelligent Computing & Signal Processing of Ministry of Education, Anhui University, Hefei, Anhui 230039, PR China dSchool of Mathematics and Statistics, Zhengzhou University, Zhengzhou, Henan 450001, PR China
e
Department of Information, Operations & Management Sciences, Stern School of Business, New York University, 44 West 4th Street, New York, NY 10012-1126, USA f
Department of Information & Finance Management, National Chiao Tung University, Taiwan
a r t i c l e
i n f o
Article history: Received 28 May 2013 Accepted 17 March 2014 Available online 29 March 2014 Keywords:
Bi-criteria scheduling Maximum machine cost Total machine cost Makespan
Total completion time Heuristics
a b s t r a c t
We consider parallel machine scheduling problems where the processing of the jobs on the machines involves two types of objectives. The first type is one of two classical objective functions in scheduling theory: either the total completion time or the makespan. The second type involves an actual cost asso-ciated with the processing of a specific job on a given machine; each job-machine combination may have a different cost. Two bi-criteria scheduling problems are considered: (1) minimize the maximum machine cost subject to the total completion time being at its minimum, and (2) minimize the total machine cost subject to the makespan being at its minimum. Since both problems are strongly NP-hard, we propose fast heuristics and establish their worst-case performance bounds.
Ó 2014 Elsevier B.V. All rights reserved.
1. Introduction
In traditional scheduling theory, most problems are concerned with the minimization of certain functions of the completion times of the jobs. This type of objectives relate in certain ways to cus-tomer satisfaction since they tend to result in schedules that have either early completion times or on-time completions of the jobs. In reality, there are other important aspects in the evaluation of schedules, namely, aspects that are related to the machines them-selves. When a job is assigned to a machine, the assignment results in a cost (or profit) that depends on the job as well as on the
ma-chine. The machines’ objective may be the minimization of the to-tal cost incurred by all the machines or the maximum cost incurred by any machine.
This type of multi-criteria scheduling problem occurs nowadays quite often in manufacturing as well as in services industries. Con-sider, for example, a manufacturing company with multiple plants in different locations. The production cost of a customer order at one plant may be completely different from the production cost at another plant. The company has to worry about the timing of the production (in order to provide good customer service) as well as the production cost (in order for the company to remain profitable). In the services industries also, such multi-criteria scheduling problems have become more and more important in recent years. Consider, for example, an organization that provides professional services and has, say, m service providers (e.g., medical doctors, teams of consultants, lawyers, etc.) and n tasks (e.g., patients, pro-jects, legal cases, etc.). Each task has to be handled by one of the service providers. A service provider could be regarded as a ‘‘ma-chine’’ and a task may be regarded as a ‘‘job’’. The service providers in such an environment are typically not identical, i.e., different providers have different skill sets and different experience levels and charge therefore differently. The tasks may also be different in such a way that each may require a provider with a specific skill
http://dx.doi.org/10.1016/j.ejor.2014.03.026
0377-2217/Ó 2014 Elsevier B.V. All rights reserved.
⇑Corresponding author. Tel.: +1 212 998 0287; fax: +1 212 995 4227. E-mail addresses: [email protected] (K. Lee), [email protected]
(J.Y-T. Leung), [email protected] (Z.-h. Jia), [email protected] (W. Li),
[email protected](M.L. Pinedo),[email protected](B.M.T. Lin). 1
Work supported in part by the NSF Grant CMMI-0969830. 2
Work supported in part by the Science Foundation of Anhui University Grant 33050044, NSFC Grant 71171184, and China Scholarship Council Grant 201206505002.
3
Work supported in part by the NSFC Grant 11171313. 4 Work supported in part by the NSF Grant CMMI-0969755. 5
Work supported in part by the National Science Council of Taiwan Grant NCS-101-2918-I-009-008.
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set. From the organization’s point of view (e.g., a clinic, a consulting company, a legal firm, etc.), the objective is to minimize the total completion time of the tasks or to minimize the latest completion time so as to increase the clients’ satisfaction levels, while at the same time minimize the total service providers’ cost in order to maximize the organization’s profit or minimize the maximum profit of any service provider in order to balance the profits over the service providers. Balancing the profits of the service providers is an important goal in maintaining the morale of the employees in the company.
Clearly, the framework considered in this paper applies to any organization that has to assign a resource to a specific task or pro-ject while taking into account obpro-jectives related to customer satis-faction levels as well as objectives related to the costs of utilization of resources.
In this paper we consider such bi-criteria scheduling problems with the second objective (either maximum machine cost or total machine cost) to be minimized subject to the constraint that the first objective (either makespan or total completion time) is at its minimum.
Many papers have dealt with bi- or multi-objective scheduling
problems and there are several survey papers in this area.T’kindt
and Billaut (2001) and T’kindt and Billaut (2006)presented a
re-view on scheduling problems with various different machine envi-ronments including single machine, parallel machines, and
flowshop environments. Hoogeveen (2005) paid more attention
to due-date related objectives and scheduling with controllable
processing times.Lei (2009)provided a more recent review.
None-theless, most research initiatives focused on two or more schedul-ing objectives that are all related to the customers’ perspectives. Thus, all the objectives are typically functions of the completion times.
Typical examples are papers that focus on minimizing two
tra-ditional scheduling objectives.Smith (1956) considered a single
machine scheduling problem to minimize the total completion time subject to the constraint that all jobs should be completed
at their due dates or before.Leung and Young (1989)studied a
par-allel machine scheduling problem to minimize the makespan sub-ject to the constraint that the total completion time is at its
minimum.Leung and Pinedo (2003)considered a parallel machine
scheduling problem to minimize the total completion time subject to the constraint that the makespan is at its minimum.
Multi-objective scheduling can be found in many different set-tings. There are studies on agent scheduling problems that focus on multiple objectives which agents selfishly optimize; see, for exam-ple,Agnetis, Mirchandani, Pacciarelli, and Pacifici (2004), Agnetis, Pacciarelli, and Pacifici (2007), Lee, Choi, Leung, and Pinedo
(2009) and Leung, Pinedo, and Wan (2010). Distributed scheduling
can also lead to multi-objective scheduling problems in which each job’s objective as well as the overall objective function are opti-mized at the same time. Several other papers focus on coordination
mechanisms; see, for example,Lee, Leung, and Pinedo (2011) and
Lee, Leung, and Pinedo (2012).
In the scheduling literature, there are two streams of research that deal with service providers’ objectives. The first and most pop-ular stream involves a machine activation cost. The number of ma-chines used is a variable and the overall objective is to minimize the sum of a traditional scheduling objective (e.g., makespan)
and the cost of activating machines; see, for example,Imreh and
Noga (1999) and Dòsa and He (2004).
Another approach assumes a machine assignment cost. When a job is scheduled on a machine, a machine assignment cost is in-curred. In most cases, the objective is either the sum of a scheduling objective and the total machine assignment cost or a prioritization
of the two objectives; see, for example, Shmoys and Tardos
(1993), Vignier, Sonntag, and Portmann (1999), T’kindt, Billaut,
and Proust (2001), Khuller, Li, and Saha (2010), and Leung, Lee, and Pinedo (2012).Khuller et al. (2010)considered a problem with
machine activation cost as well as machine assignment cost.
Shm-oys and Tardos (1993)formulated a general bi-objective scheduling
problem with machine assignment costs as a generalized assign-ment problem. They considered a prioritization of the two objec-tives and provided an approximation approach by combining a
linear programming algorithm with a rounding technique.Leung
et al. (2012) described a bi-objective scheduling problem to
minimize the scheduling objective and the total machine assign-ment cost at the same time. The makespan and the total completion time were considered as scheduling objectives and both non-pre-emptive and prenon-pre-emptive versions were dealt with. They considered weighted combinations of the two objectives as well as prioritiza-tions of the two objectives. They restricted themselves to an analy-sis of the problems in terms of their time complexity under different assumptions; they proved NP-hardness for some cases and presented polynomial time algorithms for other cases.
In this paper, we will follow in our problem formulations the
framework presented inLeung et al. (2012). However, we consider
several aspects that are different. UnlikeLeung et al. (2012), we
only focus on a prioritization of objectives, which implies that we try to optimize one objective first and then try to optimize the second one with the first one being at its minimum. Also, we
develop approximation algorithms with their worst-case
performance analyses.
In Section2, we formally describe the problem and introduce
notations. We provide in Section3approximation algorithms for
the problem in which the total completion time and maximum
machine cost have to be minimized. We present in Section 4
approximation algorithms for minimizing the makespan and the total machine cost. We conclude the paper in the last section with a discussion on future research directions.
2. Problem description
We consider the problem of scheduling n jobs on m identical machines to minimize at the same time a customers’ objective and a service providers’ objective. The processing time of job j is
pj. Let Cjdenote the completion time of job j. The customers’
objec-tive is to minimize either the total completion time (PCj) or the
last completion time, commonly referred to as the makespan
(Cmax). In addition, a cost cijis incurred when job j; 1 6 j 6 n, is
pro-cessed on machine i; 1 6 i 6 m. Let xij¼ 1 if job j is processed on
machine i and xij¼ 0 otherwise. Thus, the total machine cost, for
short TMC, isPmi¼1Pnj¼1cijxij, and the maximum machine cost, for
short MMC, is maxm
i¼1f
Pn
j¼1cijxijg. The service providers’ objective
is to minimize either TMC or MMC. The goal is to find a schedule that minimizes either the total completion time or the makespan in addition to the total machine cost or the maximum machine cost. We consider in what follows only nonpreemptive scheduling models. Let M ¼ f1; 2; . . . ; mg denote the set of m machines and J ¼ fJ1;J2; . . . ;Jng denote the set of n jobs.
We consider three types of cost functions: (i) cij¼ cj,
(ii) cij¼ aiþ cj, and
(iii) cij¼ ai cj,
where aiis a non-negative integer parameter belonging to machine
i and cjis a non-negative integer parameter belonging to job j. The
first cost function (i) is the case of identical costs; i.e., the process-ing cost of a job is independent of the machine the job is assigned to. The second cost function (ii) plays a role when the assignment
machine dependent. The third cost function (iii) plays a role when one machine operates in a more cost efficient manner than another machine; i.e., the efficiency factor of machine i is captured by the
factor ai. For cases (ii) and (iii), we can assume, without loss of
gen-erality, that a16a26. . . 6am. Obviously, (i) is a special case of (ii)
when ai¼ 0 for all i 2 M and (i) is a special case of (iii) when ai¼ 1
for all i 2 M. Throughout the paper, we also assume that, for case
(ii) a1¼ 0 and for case (iii) a1¼ 1.
Basically, the bi-criteria problems we study can be referred to as
LEX (
c
1;c
2). In LEX (c
1;c
2), thec
1represents the primary objectiveand the
c
2represents the secondary objective; i.e., the problem isto minimize objective
c
2 subject to the constraint that objectivec
1is at its minimum.Since we consider eitherPCj or Cmaxas the customers’
objec-tive, and consider either TMC or MMC as the service providers’
(ma-chines’) objective, we have four possible combinations:
(PCj;MMC), (PCj;TMC), (Cmax, MMC) and (Cmax;TMC). Leung
et al. (2012)already showed that LEX (PCj;TMC) can be solved
optimally in polynomial time. The problem LEX (Cmax;MMC) is
not completely investigated, but a special case of this problem
was considered by Lee, Leung, and Pinedo (2013). In this paper
we will focus on the following two problems:
1. LEX (PCj;MMC); i.e., minimize the maximum machine
cost subject to the constraint thatPCjis minimized.
2. LEX (Cmax;TMC); i.e., minimize the total machine cost
sub-ject to the constraint that Cmaxis minimized.
The two scheduling problems above can be shown to be unary
NP-hard via a simple reduction from the 3-Partition problem (
Gar-ey & Johnson, 1979). For LEX (Cmax;TMC), the primary objective can
be reduced from the 3-Partition problem. For LEX (PCj;MMC), the
secondary objective can be reduced from 3-Partition.
Because of the computational complexity of the problems, we are interested in fast heuristics. We say that heuristic A is an
ð
a
;bÞ-approximation for problem LEX (c
1;c
2) if A produces asche-dule where
c
1ðAÞ 6a
c
1and
c
2ðAÞ 6 bc
2. Here,c
1is theopti-mal value for the first objective, and
c
2is the optimal value for the
second objective provided that the schedule attains the optimal va-lue (
c
1) for the first objective. For example, if A is a ð32;
2Þ-approx-imation for LEX (Cmax;TMC), then CmaxðAÞ 632 C
max and
TMCðAÞ 6 2 TMC, where TMCis the optimal value for the total
machine cost provided that the makespan of the schedule is C
max.
Shmoys and Tardos (1993)consider a bi-objective scheduling
problem that is more general than those we consider; they assume
processing times pij(i.e., the processing time depends on the job as
well as on the machine), and assignment costs cij(i.e., the
assign-ment costs depend also on the job as well as on the machine). They do develop a polynomial time approximation algorithm for this framework; however, their algorithm has a higher time complexity than ours. Moreover their algorithm seems to be hard to implement, since it depends on the solution of a Linear Program, followed by the solution of a matching problem. We will compare our results with their results in the conclusion section.
The following notation will be used throughout this paper. Let
pmax¼ maxðp1; . . . ;pnÞ
and
cmax¼ maxðc1; . . . ;cnÞ:
For a given set of jobs J0, let PðJ0Þ ¼Pj2J0pj and VðJ0Þ ¼
P j2J0cj. Let C max and P C
j be the optimal objective values of the makespan
and the total completion time, respectively. Let TMC and MMC
be the optimal objective values of TMC and MMC, respectively, when the primary objective is at its minimum.
3. Total completion time and maximum machine cost
In this section we consider the problem LEX (PCj;MMC). It is
well-known that the Shortest-Processing-Time first (SPT) rule
yields an optimal schedule forPCj; seePinedo (2012). The SPT
rule, in its most general form, operates as follows: Let integer k be such that n ¼ ðk 1Þm þ ‘ for some integer ‘; 1 6 ‘ 6 m. Sort the jobs in ascending order of their processing times, i.e.,
p16p26. . . 6pn. We refer to the last m jobs (i.e., jobs
n m þ 1; n m þ 2; . . . ; n,) as the rank-k jobs, the second last m jobs (i.e., jobs n 2m þ 1; . . . ; n m) as the rank-(k 1) jobs, and so on; the first ‘ jobs (i.e., jobs 1; . . . ; ‘) are the rank-1 jobs. The rank-1 jobs are assigned first, one job per machine, followed by the rank-2 jobs, again one job per machine. This process is repeated until jobs in the last rank (rank-k) are assigned. Clearly, there may be many different schedules that can be generated according to this SPT rule and all these
sched-ules have a minimum PCj. A schedule generated by the SPT
rule is referred to as an SPT schedule. While limiting ourselves to SPT schedules, we try to find for the problem LEX
(PCj;MMC) a ð1; bÞ-approximation algorithm with the smallest
possible b.
In order to guarantee a minimumPCj, we will only consider
SPT schedules. The following procedure to setup ranks will be used as a subroutine in the algorithms proposed.
Setup Ranks
(1) Sort the jobs in ascending order of their processing time; i.e.,
p16p26 6 pn. Let n ¼ ðk 1Þm þ ‘, where 1 6 ‘ 6 m.
(2) Assign the last m jobs to set Rk, the second last m jobs to set
Rk1, and so on. Assign the first ‘ jobs to set R1.
(3) Return hR1;R2; . . . ;Rki.
Note that the Setup Ranks procedure returns a possible rank composition and there may be a different rank composi-tion for SPT schedules when there are jobs with equal pro-cessing times.
3.1. The special case cij¼ cj
We first consider the cost function cij¼ cjfor all 1 6 i 6 m. We
consider a heuristic, H1, that schedules jobs according to SPT, but jobs in each rank are judiciously assigned to the machine so as to minimize the maximum machine cost. For each 1 6 i 6 m, we let
Bidenote the set of jobs assigned to machine i. Below is a
descrip-tion of H1.
Heuristic H1ðJ; MÞ (1) Call Setup Ranks
(2) Bi:¼ ; for each 1 6 i 6 m.
(3) For r from 1 to k do
(a) Sort the machines in ascending order of their costs; i.e., VðBl1Þ 6 VðBl2Þ 6 6 VðBlmÞ.
(b) Let fJi1;Ji2; . . . ;Jiyg be all the jobs in Rr(i.e., Rrhas y jobs)
where the jobs are sorted in descending order of their costs; i.e., ci1Pci2P P ciy.
(c) For each j from 1 to y do
(i) Blj:¼ Blj[ fJijg.
The analysis of the performance bound for heuristic H1 is sim-ilar to the analysis of a list scheduling algorithm for the classical
PkCmax problem. Before we prove the bound for heuristic H1, we
Lemma 1. Let Vði; rÞ denote the total cost of jobs assigned to machine i after assigning the jobs in rank r by heuristic H1. Then, we have
0 6 jVði; rÞ Vðh; rÞj 6 cmaxfor 1 6 i; h 6 m:
Proof. We will prove the lemma by induction on r. The base case,
r ¼ 1, is obvious. This is because each of Biand Bhhas at most one
job and thus 0 6 Vði; 1Þ 6 cmaxand 0 6 Vðh; 1Þ 6 cmax. Thus, we have
0 6 jVði; 1Þ Vðh; 1Þj 6 cmax:
Assume the lemma is true for all ranks up to r 1. We want to show that the lemma is true for rank-r. If Vði; r 1Þ ¼ Vðh; r 1Þ, then it
immediately holds that jVði; rÞ Vðh; rÞj 6 cmax. Otherwise, by
sym-metry, we may assume that Vði; r 1Þ > Vðh; r 1Þ. Then, from the induction hypothesis, we have
0 < Vði; r 1Þ Vðh; r 1Þ 6 cmax:
By heuristic H1, the job assigned to machine h has a cost that is at least as large as the cost of the job assigned to machine i. Since the cost of the job assigned to machine h is simply Vðh; rÞ Vðh; r 1Þ and the cost of the job assigned to machine i is simply Vði; rÞ Vði; r 1Þ, we have
Vðh; rÞ Vðh; r 1Þ P Vði; rÞ Vði; r 1Þ;
and hence
Vði; rÞ Vðh; rÞ 6 Vði; r 1Þ Vðh; r 1Þ 6 cmax:
Moreover,
Vði;rÞ Vðh;rÞ P Vði;r 1Þ Vðh;rÞ P Vði;r 1Þ ðVðh;r 1Þ þ cmaxÞ P ðVði; r 1Þ Vðh;r 1ÞÞ cmaxPcmax;
where the last inequality is due to the fact that
Vði; r 1Þ > Vðh; r 1Þ. Therefore, we have
jVði; rÞ Vðh; rÞj 6 cmax:
By induction, the lemma holds for all r. h
Theorem 1. Heuristic H1 is a (1; 2 1=m)-approximation for the
problem LEX (PCj;MMC) when cij¼ cj for all 1 6 i 6 m. Moreover,
the bound is tight.
Proof. Since heuristic H1 schedules jobs according to the SPT rule,
it generates an optimal solution for thePCj objective. Consider
now the MMC objective. ByLemma 1, we have
max
i2MfVði; kÞg mini2MfVði; kÞg 6 cmax
6MMC:
There are two cases to consider.
Case 1: mini2MfVði; kÞg 6m1m MMC
.
In this case, we have
max
i2MfVði; kÞg 6 mini2MfVði; kÞg þ cmax
6m 1 m MMC þ MMC ¼ 2 1 m MMC :
Case 2: mini2MfVði; kÞg >m1m MMC
.
A lower bound for the total cost on all the machines is
maxi2MfVði; kÞg þ ðm 1Þ mini2MfVði; kÞg; i.e., one machine has
the maximum cost and m 1 machines have the minimum cost. Thus, we have
max
i2MfVði; kÞg þ ðm 1Þ mini2MfVði; kÞg 6
Xm i¼1 Vði; kÞ 6 m MMC; and hence, max i2MfVði; kÞg 6 m MMC ðm 1Þ min i2MfVði; kÞg < m ðm 1Þ 2 m ( ) MMC¼ 2 1 m MMC:
To show that the bound is tight, consider 3m jobs, where the rank-1 jobs all have a processing time of 1 unit, the rank-2 jobs all have a processing time of 2 units, and the rank-3 jobs all have a processing time of 3 units. The costs of the rank-1 and rank-2 jobs are 0; 1; 2; . . . ; m 1, while the cost of the rank-3 jobs are zero except one job that has a cost of m. Heuristic H1 will produce a schedule
with MMCðH1Þ ¼ 2m 1, while the optimal schedule has
MMC¼ m. Therefore, the ratio is 2 1=m. h
3.2. The special case cij¼ aiþ cj
We now consider the cost function cij¼ aiþ cj. We propose a
different heuristic, H2, which is slightly more complicated. First, the Setup Ranks routine is called to obtain a possible rank composition for the minimum total completion time objective. In a SPT schedule with k ranks such that n ¼ kðm 1Þ þ ‘; ‘ machines have k jobs each and ðm ‘Þ machines have ðk 1Þ jobs each. Thus,
the cost at machine i is jBijaiþ VðBiÞ where jBij ¼ k or ðk 1Þ. The
basic idea of heuristic H2 is to construct two schedules – the first one for jobs in rank 1 and the second one for jobs in ranks 2; . . . ; k – and combine these to create a final schedule.
For the first schedule, the jobs in R1are sorted in descending
or-der of their costs; i.e., cj1Pcj2P P cj‘. Then, for each 1 6 i 6 ‘,
job Jji is assigned to machine i and Ai is set to be k aiþ cji. For
each ‘ þ 1 6 i 6 m; Aiis set to be ðk 1Þ ai. Then, the machines
are sorted in descending order of Ai; i.e., the machines l1;l2; . . . ;lm
are such that
Al1PAl2P P Alm:
Now, heuristic H1 is then called to generate the second schedule
with the set of jobs R2[ R3[ [ Rk. When heuristic H1 returns,
let Bi be the set of jobs assigned to machine i. Then, the Bi’s are
sorted in an ascending order of the total cost of the jobs in Bi; i.e.,
VðBi1Þ 6 VðBi2Þ 6 6 VðBimÞ:
Finally, for the minimum maximum cost, for each 1 6 h 6 m, the
jobs in Bih are assigned to machine lh. The cost of machine lh is
Alhþ VðBihÞ. Shown below is a description of heuristic H2.
Heuristic H2 ðJ; MÞ (1) Call Setup Ranks
(2) Let fJj1;Jj2; . . . ;Jj‘g be the jobs in R1, sorted in descending
order of their costs; i.e., cj1Pcj2P P cj‘.
(3) Sort the machines in ascending order of ai; i.e.,
a16a26 6 am.
(4) For i from 1 to ‘, assign job Jji to machine i and let
Ai¼ k aiþ cji.
(5) For i from ‘ þ 1 to m, let Ai¼ ðk 1Þ ai.
(6) Sort Aivalues such that Al1PAl2P P Alm.
(7) Call heuristic H1(R2[ R3[ [ Rk;M) to schedule jobs in
R2[ R3[ [ Rk on m identical machines. When heuristic
H1 returns, let Bi be the set of jobs assigned to machine i
in the schedule produced by heuristic H1.
(8) Sort the Bi in ascending order of VðBiÞ; such that
VðBi1Þ 6 VðBi2Þ 6 6 VðBimÞ.
(9) For each 1 6 h 6 m, assign the jobs in Bih to machine lh and
Theorem 2. Heuristic H2 is a (1; 2 1=m)-approximation for the
problem LEX (PCj;MMC) when cij¼ aiþ cj for all 1 6 i 6 m.
More-over, the bound is tight.
Proof. For each 1 6 i 6 m, let Vibe VðBiÞ for simplicity. Recall that
Biis the set of jobs scheduled on machine i from R2[ R3[ [ Rk.
Thus, the cost of machine i is Aiþ Vi.
We claim that maxi2MfAig 6 MMCby the following reason. We
consider another set of jobs J0where all costs of jobs in rank 1 is the
same as the original problem instance and all costs of jobs in ranks
2; . . . ; k become zero. Then, maxi2MfAig is the optimal MMC value
for J0. Obviously, the optimal MMC value for J is greater than or
equal to the optimal MMC value for J0. Therefore,
maxi2MfAig 6 MMC.
Thus, jAi Ahj 6 MMCfor 1 6 i; h 6 m. ByLemma 1, we have
jVi Vhj 6 MMCfor 1 6 i; h 6 m.
Without loss of generality, we may assume that AiPAh. Thus,
0 6 Ai Ah6MMC. Since AiPAh, we have Vi6Vh, by the nature
of heuristic H2. Thus, we have
ðAiþ ViÞ ðAhþ VhÞ 6 Ai Ah6MMC;
and
ðAiþ ViÞ ðAhþ VhÞ P Vi VhPMMC:
Therefore, we have jðAiþ ViÞ ðAhþ VhÞj 6 MMC. We now
consider two cases. Let Aiþ Vi¼ maxl2MfAlþ Vlg and Ahþ Vh¼
minl2MfAlþ Vlg.
Case 1: Ahþ Vh6m1m MMC
.
In this case, we have
Aiþ Vi6Ahþ Vhþ MMC6 m 1 m MMC þ MMC¼ 2 1 m MMC : Case 2: Ahþ Vh>m1m MMC .
In this case, we have
Aiþ Viþ ðm 1ÞðAhþ VhÞ 6 X l2M ðAlþ VlÞ 6 m MMC: Therefore, we have Aiþ Vi6m MMC ðm 1ÞðAhþ VhÞ < m ðm 1Þ 2 m ( ) MMC ¼ 2 1 m MMC:
The set of jobs achieving a tight bound inTheorem 1can be used to
show the tight bound for this theorem as well. Simply let ai¼ 0 for
all i. h
3.3. The special case cij¼ ai cj
We now consider the cost function cij¼ ai cj. Recall that we
assume that a1¼ 1 and ai6aiþ1 for i ¼ 1; . . . ; m 1. We can
assume that n ¼ km; otherwise without changing the problem structure we can add dn=mem n dummy jobs with zero process-ing times and zero costs.
We consider the third heuristic, H3, which is similar to heuristic H1. Again, jobs are ordered according to the SPT rule, rank by rank. Jobs within the same rank are assigned to machines as follows:
Jobs are considered in descending order of their cost cj. When job
j is being assigned, we consider all the eligible machines and assign job j to that machine i that results in the smallest total cost on ma-chine i. After job j is assigned to mama-chine i, mama-chine i becomes inel-igible until we deal with the jobs in the next rank. The heuristic is described as follows.
Heuristic H3(J; M) (1) Call Setup Ranks
(2) Bi:¼ ; for each 1 6 i 6 m.
(3) For r from 1 to k do
(a) Let M0
¼ f1; 2; . . . ; mg be the eligible machines.
(b) Let Ri¼ fJi1;Ji2; . . . ;Jimg, sorted in descending order of
their cost; i.e., ci1Pci2P P cim.
(c) For each j from 1 to m do (i) l :¼ arg minf 2M0fVðBfÞ þ af ci
jg, with ties broken by
choosing the machine with a larger af.
(ii) Bl:¼ Bl[ fJijg.
(iii) M0:¼ M0n flg.
Even though heuristic H3 looks reasonable, its theoretical bound may be poor. We can derive a lower bound of the ratio from
the literature.Cho and Sahni (1980)considered a list scheduling
algorithm for a uniform parallel machine scheduling problem to minimize the makespan. They showed that the worst case perfor-mance ratio of the list scheduling algorithm is at least Oðlog mÞ by presenting a problem instance. We can construct a problem
in-stance from the prohlem inin-stance inCho and Sahni (1980) and
show that the approximation ratio of heuristic H3, for problem
LEX (PCj;MMC) when cij¼ ai cj for all 1 6 i 6 m, is at least
(1, Oðlog mÞ). However, we will prove an unbounded approxima-tion ratio of heuristic H3 in the following theorem.
Theorem 3. The approximation ratio of heuristic H3, for problem LEX
(PCj;MMC) when cij¼ ai cjfor all 1 6 i 6 m, is at least (1; OðamÞ).
Proof. Since heuristic H3 schedules jobs according to SPT, it
gener-ates an optimal solution for thePCjobjective. Thus, we will prove
the lower bound of MMCðH3Þ=MMCby providing an example.
Consider two machines and four jobs. The four jobs have
identical processing times, say 1 unit. Their costs are:
c1¼ c2¼ 1; c3¼ c4¼ a, where a 1. For machine 1, a1¼ 1 and
for machine 2, a2¼ a.
Suppose the algorithm puts jobs 1 and 2 in the first rank, and
jobs 3 and 4 in the second rank, then VðB1Þ ¼ VðB2Þ ¼ a þ 1. In this
case, we have MMCðH3Þ ¼ aða þ 1Þ.
On the other hand, the optimal solution puts jobs 1 and 3 in the first rank, and jobs 2 and 4 in the second rank. Then we assign the job with the smallest cost in each rank to machine 2. In this case, we have jobs 1 and 2 in machine 2, and jobs 3 and 4 in machine 1.
We have MMC
¼ 2a.
The approximation ratio is aða þ 1Þ=ð2aÞ = ða þ 1Þ=2, which approaches infinity when a gets large. This completes the proof. h
Since heuristic H3 does not have a constant approximation
ra-tio, we consider a special case of LEX (PCj;MMC) where
cij¼ ai cj;a1¼ ¼ am1¼ 1 and amP1. This setting implies
that there is only one inefficient machine. For this special case, we present an approximation algorithm with a constant ratio.
The basic idea is that we consider two cases for the am value
separately. When amis not so large, we just use heuristic H1 and
modify the schedule. When amis large, we need to assign jobs with
small costs to machine m within the structure of SPT schedules. However, if we fix ranks by Setup Ranks like heuristics H1, H2 and H3, it might be impossible to assign jobs with small costs to machine m. Therefore, we first consider a procedure to assign a job from each rank to machine m to minimize the total cost of jobs assigned to machine m among all possible rank compositions. The algorithm can be described as follows:
Minimum Cost for Machine m (1) Call Setup Ranks
(2) Construct a weighted bipartite graph G ¼ ðU1[ U2;E; wÞ
where
(a) U1¼ fjjJj2 Jg; U2¼ f1; . . . ; k þ 1g
(b) E1¼ fð
l
;m
Þjminj2Rmfpjg 6 pl6maxj2Rmfpjgforl
2 U1;m
2 U2n fk þ 1gg,E2¼ fð
l
;k þ 1Þjl
2 U1g andE ¼ E1[ E2.
(c) wl;m¼ cl for
l
2 U1;m
2 U2n fk þ 1g.(3) Solve the following transportation problem on G and obtain
an optimal solution y l;m. minimize X ðl;mÞ2E1 wl;myl;m subject to X m:ðl;mÞ2E yl;m¼ 1 for
l
2 U1 X l:ðl;mÞ2E1 yl;m¼ 1 form
2 U2n fk þ 1g X l:ðl;kþ1Þ2E2 yl;kþ1¼ n k yl;m2 f0; 1g for ðl
;m
Þ 2 E(4) Job Jl is scheduled on rank
m
of machine m if and only ify
l;m¼ 1 for ð
l
;m
Þ 2 E1.Instead of solving the mathematical program described above through a transportation algorithm, we will solve it by an alternative procedure that will significantly reduce the running time. This alternative procedure is described in the Appendix. Its running time is Oðn log nÞ.
Note that VðBiÞ is the sum of cj values of the jobs assigned to
machine i. The cost of machine i is VðBiÞ for i ¼ 1; . . . ; m 1 and
is amVðBmÞ for i ¼ m and thus
MMC ¼ maxfmaxfVðBiÞji 2 f1; . . . ; m 1gg; amVðBmÞg:
Now, we are ready to present an algorithm with a constant ratio for
LEX (PCj;MMC) where cij¼ ai cj;a1¼ ¼ am1¼ 1 and amP1.
Heuristic H4
(1) If amP2,
(a) Call Minimum Cost for Machine m. Let J0be the set of
jobs scheduled on machine m.
(b) Apply H1ðJ n J0;M n fmgÞ for the schedule of the
remain-ing jobs on machines 1; . . . ; m 1. (2) If 1 6 am<2,
(a) Apply H1(J; M) and obtain Bifor i 2 M.
(b) Let h :¼ arg minfVðBiÞji 2 Mg. Swap Bmand Bh.
Theorem 4. Heuristic H4 is a ð1; 2 1
2ðm1ÞÞ-approximation for
problem LEX (PCj;MMC) when cij¼ ai cj;a1¼ ¼ am1¼ 1 and
amP1.
Proof. Case 1. Suppose that amP2. We consider two sub-cases: (i)
MMC is determined at machine m, and (ii) MMC is determined at machine i, for i – m.
(i) If MMC is determined by machine m, by the property of the schedule by Minimum Cost for Machine m, the current schedule is optimal.
(ii) Suppose that MMC is determined by a machine that is not
machine m. By Lemma 1, jVðBiÞ VðBhÞj 6 cmax for
i; h 2 M n fmg. If min i2MnfmgfVðBiÞg 6 1 1 2ðm 1Þ MMC; then max
i2MnfmgfVðBiÞg 6 mini2MnfmgfVðBiÞg þ cmax
6 2 1 2ðm 1Þ MMC: If min i2MnfmgfVðBiÞg > 1 1 2ðm 1Þ MMC ;
then by the lower bound for MMC, we have MMCPVðJÞ=
ðm 1 þ 1=amÞ. We consider the following relationship:
max
i2MnfmgfVðBiÞg þ ðm 2Þ mini2MnfmgfVðBiÞg 6 VðJÞ
6ðm 1 þ 1=amÞMMC: Thus, we have max i2MnfmgfVðBiÞg < ðm 1 þ 1=amÞMMC ðm 2Þ 1 1 2ðm 1Þ MMC ¼ 3 2þ 1 am 1 2ðm 1Þ MMC 6 2 1 2ðm 1Þ MMC :
Case 2. Suppose that 1 6 am<2. We again consider two subcases:
(i) MMC is determined by machine m, and (ii) MMC is determined by machine i, for i – m.
(i) Suppose that MMC is determined at machine m. If VðBmÞ >m1ðm 1 þa1mÞMMC , thenPi2MVðBiÞ > ðm 1 þa1mÞ MMC , which is a contradiction. If VðBmÞ 6m1 ðm 1 þa1mÞMMC , then we have amVðBmÞ 6 am m m 1 þ 1 am MMC¼ amð1 1 mÞ þ 1 m MMC 6 2 1 m MMC:
(ii) Suppose that MMC is determined at a machine that is not
machine m. ByLemma 1, jVðBiÞ VðBhÞj 6 cmaxfor i; h 2 M.
If VðBmÞ 6 1 m1MMC, then max i2MnfmgfVðBiÞg 6 VðBmÞ þ cmax 6 2 1 m MMC: If VðBmÞ > 1 m1
MMC, then consider the relationship
max i2MfVðBiÞg þ ðm 1ÞVðBmÞ 6 VðJÞ 6 ðm 1 þ 1=amÞMMC : Thus, we have max i2MfVðBiÞg 6 ðm 1 þ 1=amÞMMC ðm 1Þ 1 1 m MMC ¼ 1 amþ m 1 m MMC 6 2 1 m MMC: Note that 2 1
m62 2ðm1Þ1 for m P 2. The proof is complete. h
4. Makespan and total machine cost
In this section we consider the problem LEX (Cmax;TMC). For the
cost function cij¼ cj, this problem is equivalent to the problem
PjjCmax, for which many heuristics have been proposed and
On both cost functions cij¼ aiþ cj and cij¼ ai cj, we will
ob-tain an upper bound of the optimal makespan by the following steps. We first order the n jobs on the m machines according to the Largest-Processing-Time (LPT) rule, ignoring the cost of the jobs. The LPT rule schedules the jobs in descending order of their processing times. The next job will be assigned to the machine that finishes the earliest. The makespan of the LPT schedule is denoted by L and is used as an upper bound for the optimal makespan. It is
well known that L 6 ð4
33m1ÞC
max, where C
maxis the optimal
make-span; seeGraham (1969).
Since the primary objective, the makespan, is hard to optimize, we may consider using LPT to optimize the primary objective. Unfortunately, such an algorithm can perform very poorly with
re-gard to the secondary objective for both cost functions cij¼ aiþ cj
and cij¼ ai cj.
For the cost function cij¼ aiþ cj, we consider an example with
two machines having a1¼ 0 and a2¼ a > 0 and n jobs, where n
is even and n P 6. The job information is as follows:
p1¼ p2¼ 1=2 and p3¼ ¼ pn¼ 1=ðn 2Þ and c1¼ ¼ cn¼ 0.
Obviously, Cmax¼ 1. In the LPT schedule, jobs J1and J2are
sched-uled on different machines and the remaining jobs are evenly scheduled on both machines. Thus, TMCðLPTÞ ¼ n=2 a. However,
in the optimal schedule, jobs J1and J2are scheduled on machine 2
and the other jobs are scheduled on machine 1. Thus, TMC
¼ 2 a.
Therefore, TMCðLPTÞ=TMC
¼ OðnÞ.
For the cost function cij¼ ai cj, we consider an example with
two machines having a1¼ 1 and a2¼ a > 1 and n jobs, where n
is even and n P 6. The job information is as follows:
p1¼ p2¼ 1=2 and p3¼ ¼ pn¼ 1=ðn 2Þ and c1¼ c2¼ c=2 and
c3¼ ¼ cn¼ 0. Obviously, Cmax¼ 1. In the LPT schedule, jobs J1
and J2are scheduled on different machines and the remaining jobs
are evenly scheduled on both machines. Thus,
TMCðLPTÞ ¼c
2ð1 þ aÞ. However, in the optimal schedule, jobs J1
and J2are scheduled on machine 1 and the other jobs are
sched-uled on machine 2. Thus, TMC
¼ c. Since
TMCðLPTÞ TMC ¼
1 þ a 2
and a can be arbitrarily large, TMCðLPTÞ=TMCis unbounded.
There-fore, we need to design an algorithm that considers both objectives simultaneously.
4.1. The special case cij¼ aiþ cj
For the cost function cij¼ aiþ cj, we propose a fast heuristic,
heuristic H5, which works as follows. Let the makespan of the LPT schedule be L. Then the jobs are sorted in ascending order of
their processing times; i.e., p16p26 6 pn. The jobs are
sched-uled on the machines, starting with the first machine. Specifically, the first k jobs are scheduled on machine 1, where k is the smallest index such that the total processing time of the first k jobs is larger than L. We then delete the k jobs and machine 1 from consider-ation, and repeat the process on machine 2. This process is re-peated until all jobs have been scheduled. Shown below is a description of heuristic H5.
Heuristic H5(J; M)
(1) Disregarding the cost of the jobs, schedule the n jobs on the m machines according to LPT. Let L denote the makespan of the LPT schedule.
(2) Sort the jobs in ascending order of their processing times. Let J ¼ ðJ1;J2; . . . ;JnÞ be a list of the n jobs such that p16p26
6 pn.
(3) Set i ¼ 1.
(4) If the total processing time of the jobs in J is less than or equal to L, then schedule all jobs in J on machine i and stop. (5) Let k be the smallest integer such that the total processing
time of the first k jobs is larger than or equal to L. (6) Assign the first k jobs on machine i.
(7) Delete the first k jobs from J. (8) i :¼ i þ 1.
(9) Goto Step 4.
Let Bi and Bi be the sets of jobs assigned to machine i in the
schedule generated by heuristic H5 and in the optimal schedule, respectively. Let B ¼ hB1;B2; . . . ;Bmi and B¼ hB1;B
2; . . . ;B
mi. With
this notation, we will prove a lemma that is instrumental in
provingTheorem 5.
Lemma 2. Let Bi be the set of jobs assigned to machine i in the
schedule generated by heuristic H5. Then, we have
Pi
h¼1PðBhÞ P minfi L;Pnj¼1pjg for all i 2 M.
Proof. If PðBhÞ P L for all h; h ¼ 1; . . . ; i, then we have
Pi
h¼1PðBhÞ P i L. Otherwise, there is a machine f such that
PðBfÞ < L for 1 6 f 6 i. This means that all n jobs have been
sched-uled on machines 1; 2; . . . ; f . Thus, we havePi
h¼1PðBhÞ ¼Pnj¼1pj.
h
We are now ready to proveTheorem 5.
Theorem 5. Heuristic H5 is a (ð7
3 1
3mÞ,1)-approximation for problem
LEX (Cmax;TMC) when cij¼ aiþ cj.
Proof. First, we want to show that heuristic H5 would be able to schedule all the jobs on m machines so that each machine finishes
by time L þ pmax. Suppose not. Then there is a job that cannot be
scheduled on any machine such that it completes by time
L þ pmax. Since pj6p
max for all j, no machine becomes available
before time L when this job is being considered for scheduling. This means that the total processing time of all the jobs that have been
scheduled so far is more than mL. This is impossible since L P Cmax.
By the result ofGraham (1969), we have
Cmax6L 6 4 3 1 3m Cmax:
Furthermore, we have pmax6Cmax. Therefore, we have
CmaxðH5Þ 6 7 3 1 3m C max:
We now consider the TMC objective. We will show that
TMCðH5Þ 6 TMC, where TMC is the optimal total machine cost
among all schedules with makespan equal to C
max.
Let ni¼ jBij and ni ¼ jB
ij. Note that
P
i2Mni¼Pi2Mni ¼ n. Let S
and Sbe the schedule produced by heuristic H5 and the optimal
schedule, respectively. In schedule S, the smallestPih¼1nh jobs,
with respect to processing times, are scheduled on machines
1; . . . ; i. Suppose that, in schedule S;Pi
h¼1nh> Pi h¼1nh. Then, obviously, we havePih¼1PðB hÞ >P i h¼1PðBhÞ. ByLemma 2, we have Pi h¼1PðBhÞ P i L orPih¼1PðBhÞ ¼Pnj¼1pjfor all i 2 M. IfPih¼1PðBhÞ P i L, then we have Xi h¼1 PðB hÞ > Xi h¼1 PðBhÞ P i L P i Cmax;
which contradicts the fact that Cmaxis the optimal makespan.
Xi h¼1 PðB hÞ > Xi h¼1 PðBhÞ ¼ Xn j¼1 pj;
which again leads to a contradiction. Thus, we have
Xi h¼1 nhP Xi h¼1 n h
for all i 2 M. Now, TMCðH5Þ ¼Pi2Mni aiþPj¼1n cj. SincePnj¼1cjis a
constant independent of the assignment, we will show that P
i2Mni ai6Pi2Mni ai, and hence TMCðH5Þ 6 TMC.
Let ai¼ ai ai1for all i 2 M, where a0¼ 0. Since aiPai1, we
have aiP0 for all i 2 M.
X i2M niai¼ n1a1þ n2ða1þ a2Þ þ þ nm Xm h¼1 ah ! ¼ ðn1þ n2þ þ nmÞa1 þ ðn2þ þ nmÞa2þ þ nmam ¼ na1þ ðn n1Þa2 þ ðn n1 n2Þa3þ ðn n1 n2 nm1Þam ¼ nða1þ a2þ þ amÞ X m1 i¼1 Xi h¼1 nh ! aiþ1 ( ) :
Since nða1þ a2þ þ amÞ is a constant, Pih¼1nhPPih¼1nh for all
i 2 M and aiP0 for all i 2 M, we have TMCðH5Þ 6 TMC. h
4.2. The special case cij¼ ai cj
We now consider the cost function cij¼ ai cj. We propose a
heuristic, heuristic H6, which works as follows. Like in heuristic H5, let the makespan of the LPT schedule be L. We then sort the
jobs in descending order of the ratiocj
pj; i.e., c1 p1P c2 p2P P cn pn. Jobs
are now assigned to the machines, starting with the first machine. The first k jobs will be assigned to machine 1, where k is the small-est index such that the total processing time of the first k jobs is larger than L. These k jobs and machine 1 will be deleted from consideration, and we proceed to schedule jobs on machine 2. This process is iterated until all jobs are assigned. Below is a description of heuristic H6.
Heuristic H6(J; M)
(1) Disregarding the cost of the jobs, schedule the n jobs on the m machines by the LPT rule. Let L denote the makespan of this schedule.
(2) Sort the jobs in descending order of the ratios cj
pj. Let
J ¼ ðJ1;J2; . . . ;JnÞ be a list of the n jobs such that
c1 p1P c2 p2P P cn pn. (3) Set i ¼ 1.
(4) If the total processing time of the jobs in J is less than or equal to L then assign all jobs in J to machine i and stop. (5) Let k be the smallest integer such that the total processing
time of the first k jobs is greater than or equal to L. (6) Assign the first k jobs to machine i.
(7) Delete the first k jobs from J. (8) i :¼ i þ 1.
(9) Goto Step 4.
In the proof ofTheorem 6, we will assume that the jobs have
been sorted in descending order ofcj
pj(i.e., c1 p1P c2 p2P P cn pn). Let
Biand Bi be the sets of jobs assigned to machine i in the schedule
generated by heuristic H6 and in the optimal schedule,
respec-tively. Thus, TMCðH6Þ ¼Pm
i¼1ai VðBiÞ and TMC¼Pmi¼1ai VðBiÞ.
Theorem 6. Heuristic H6 is a (ð7
33m1Þ,1)-approximation for the
problem LEX (Cmax;TMC) when cij¼ ai cjfor all 1 6 i 6 m.
Proof. Similar to the proof ofTheorem 5, we can show that
heuris-tic H6 must be able to schedule the n jobs on the m machines so
that each machine finishes by time L þ pmax. Since
L 6 ð4 33m1ÞC
maxand pmax6Cmax, we have CmaxðH6Þ 6 ð733m1ÞC
max.
We now show that TMCðH6Þ 6 TMC, where TMC is the optimal
total machine cost among all schedules with a makespan of C
max.
The ratio cj=pjis the cost per unit processing time. Let kj¼ cj=pj
for each 1 6 j 6 n; hence cj¼ kj pj. We have k1P k2P P kn.
Since heuristic H6 schedules jobs on each machine with total
processing time larger than L and since L P Cmax, we have for each
1 6 i 6 m, Xi h¼1 PðBhÞ P Xi h¼1 PðB hÞ: Now, Xi h¼1 VðBhÞ ¼ X j2ðB1[[BhÞ cj¼ X j2ðB1[[BhÞ kj pj:
Since heuristic H6 schedules the jobs in ascending order of the
in-dexes (and hence larger kj values) and since Pih¼1PðBhÞ P
Pi h¼1PðB hÞ, we have Xi h¼1 VðBhÞ P Xi h¼1 VðB hÞ:
Let ai¼ ai ai1 for all i 2 M, where a0¼ 0 and a1¼ 1. Since
aiPai1, we have aiP0 for all i 2 M.
X i2M ai VðBiÞ ¼ VðB1Þa1þ VðB2Þða1þ a2Þ þ þ VðBmÞ Xm h¼1 ah ! ¼ ðVðJÞÞa1þ ðVðJÞ VðB1ÞÞa2þ þ VðBmÞam ¼ VðJÞða1þ a2þ þ amÞ X m1 i¼1 Xi h¼1 VðBhÞ ! aiþ1 ( ) : Since VðJÞða1þ a2þ þ amÞ is a constant,Pih¼1VðBhÞ PPih¼1VðB hÞ
for all i 2 M, and aiP0 for all i 2 M, we have TMCðH6Þ 6 TMC. h
In heuristics H5 and H6, we calculate an upper bound for the optimal makespan by using LPT and thus have a competitive ratio
of ð1 þ4
33m1;1Þ for our problems. However, if we use a better
approximation algorithm for the makespan with a worst-case
per-formance ratio of
q
, we can have a ð1 þq
;1Þ-approximationalgo-rithm. For example, if we compute an upper bound by MULTIFIT, we have a ð1 þ 13=11; 1Þ-approximation algorithm because the
worst-case performance ratio of MULTIFIT is 13/11 byYue (1990).
The MULTIFIT algorithm is an approximation algorithm for
PjjCmax, which operates as follows. First, it computes a lower bound
(LB) and an upper bound (UB) for C
max. An obvious lower bound is
LB ¼ max pmax;
Pn
j¼1pj m
, and an obvious upper bound is
UB ¼ 2 LB. It then conducts a binary search in the interval [LB; UB]. For each value C obtained in the binary search (i.e., C ¼LBþUB
2 ), it tries to schedulle the jobs by the First-Fit-Decreasing
(FFD) rule so that no job completes after time C. If it is successful in scheduling all n jobs, then it sets UB to be C; otherwise, it sets LB to be C. This process is iterated until UB ¼ LB þ 1. Now, if all jobs can be scheduled by the FFD rule so that no job completes after time LB, then it returns LB; otherwise, it returns UB.
Shmoys and Tardos (1993)considered the following scheduling
incurs a cost cij. Assume T is given as an upper bound of the
make-span, what is the machine-job assignment that minimizes the total assignment cost? They modeled this as the following linear pro-gram. Given T and C, for any t P T, integer solutions to the linear program, LP (T; C : t), have a one-to-one correspondence with schedules that have a cost of at most C and a makespan of at most T. The LP (T; C : t) problem can be formulated as follows:
Xm i¼1 Xn j¼1 cijxij6C; Xm i¼1 xij¼ 1 for j ¼ 1; . . . ; n; Xn j¼1 pijxij6T for i ¼ 1; . . . ; m; xijP0 for i ¼ 1; . . . ; m ; j ¼ 1; . . . ; n; xij¼ 0 if pij>t; i ¼ 1; . . . ; m; j ¼ 1; . . . ; n;
They proved that if LP (T; C : t) has a feasible solution, then there ex-ists a schedule that has a makespan of at most T þ t and a cost of at most C. They also provided an algorithm that converts a feasible solution to LP (T; C : t) into the required schedule.
We can apply their result to our problem LEX (Cmax;TMC). Note
that in our problem pij¼ pj and cij¼ aiþ cj or cij¼ ai cj, which
implies that the model inShmoys and Tardos (1993)is more
gen-eral than ours. According to the above result, we can get an
ð1 þ
q
;1Þ approximation algorithm by the following procedurewhere
q
is an approximation ratio of any approximation algorithmfor minimizing the makespan in an environment with identical machines in parallel.
Step 1. Apply an approximation algorithm for minimizing the makespan with identical machines in parallel and obtain the
makespan T0. (For example, when LPT is applied, we have T0such
that Cmax6T 06 ð4 3 1 3mÞC max).
Step 2. Regard C as a variable, solve LP (T0;C : p
max) with the
objective C to be minimized and get an optimal value C0.
Step 3. Convert an optimal solution to LP (T0;C0
:pmax) into a
fea-sible solution.
Since T0PC
maxand we allow split jobs to multiple machines in
a linear program solution, we have C06TMC. Furthermore,
pmax6Cmaxand T 06
q
Cmax. The optimal solution from Step 2 is
fea-sible to LP (T0;C0:p
max), and thus, we can get a schedule with a
makespan of at most T0þ p
maxand cost at most C
0
, which implies
we have a ð1 þ
q
;1Þ -approximation algorithm for LEX (Cmax;TMC).However, to find such an approximation, it is necessary to apply two procedures (which determine the overall time complexity), namely, a linear programming algorithm in order to obtain a frac-tional matching solution (in Step 2) and a weighted bipartite
matching algorithm for the rounding (in Step 3). This LP is consid-ered a fractional packing problem and thus can be solved in
Oðmn2log nÞ; the weighted bipartite graph has OðnÞ nodes and
OðnÞ edges and thus the weighted bipartitie matching can be
solved in Oðn2log nÞ. Therefore, the overall time complexity is
Oðmn2log nÞ.
The proposed algorithms for our problem LEX (Cmax;TMC) are in
both cases (i.e., cij¼ aiþ cjand cij¼ ai cj) much simpler and
eas-ier to implement; their time complexity is Oðn log nÞ. 5. Conclusions
We consider bi-criteria scheduling problems where customers’ objectives as well as service providers’ objectives have to be opti-mized at the same time. As a customers’ objective, we consider either the total completion time or the makespan and as a service providers’ objective, we consider either the total assignment cost or the maximum assignment cost. The primary objective is the cus-tomers’ objective and the secondary objective is the service provid-ers’ objective. For three machine assignment cost functions (cij¼ cj;cij¼ aiþ cj, and cij¼ ai cj), we provide ð
a
;bÞ-approxima-tion algorithms.
Our models are all special cases of the more general model
con-sidered byShmoys and Tardos (1993), who did provide a
polyno-mial time approximation algorithm that is applicable to all models within their framework. However, their approximation scheme is clearly computationally slower than those we propose and it does not provide any intuitive insights into the respective problems. Our schemes are based on prioritization and are
there-fore more intuitive and easier to implement (seeTable 1). All the
heuristics proposed in this paper run in Oðn log nÞ, whereas the
algorithm developed by Shmoys and Tardos runs in Oðmn2log nÞ.
Our worst case approximation ratios are in three cases better than the ratio in Shmoys and Tardos; in two other cases they are the same and in one case we were not able to determine a bounded worst case approximation ratio.
We have considered only traditional scheduling objectives that are functions of completion times as the primary objective. How-ever, considering MMC or TMC as the primary objective may lead to different approximation ratios. We have only considered cases
where either
a
¼ 1 or b ¼ 1. Finding an ða
;bÞ-approximationalgo-rithm for
a
>1 and b > 1 may be a very promising research topic.Appendix A. Procedure for solving minimum cost for machine m problem
For SPT schedules, we sort the jobs in non-decreasing order of their processing times and in case of a tie we sort the jobs in non-decreasing order of their costs. When there are jobs with iden-tical processing times, those jobs may be eligible to multiple
con-secutive ranks. Thus, we define Fr0;r00to be the set of jobs that can be
Table 1
Comparisons of heuristics and approximation algorithms.
Problem cij Our paper Shmoys and Tardos (1993)
Alg. Ratio Ratio
LEXðPCj;MMCÞ cj H1 ð1; 2 m1Þ ð1; 2Þ aiþ cj H2 ð1; 2 m1Þ ai cj H3 (1, unbounded) ai cj; ay H4 ð1; 2 2ðm1Þ1 Þ LEXðCmax;TMCÞ aiþ cj H5 ð1 þqz;1Þ ð1 þqz;1Þ ai cj H6 ð1 þqz;1Þ ay:a 1¼ ¼ am1¼ 1; am>1.
scheduled in rank r for r06r 6 r00. Since the number of jobs in Fr0;r00
that can be scheduled on machine m in ranks ½r0;r00 is at most
r00 r0þ 1, we can define F
r0;r00to be the set of (r00 r0þ 1) jobs with
minimum costs from Fr0;r00. Now, the problem is to select and assign
jobs from Fr0;r00 to rank r for r06r 6 r00such that each rank has
ex-actly one job and the total cost of assigned jobs is minimum.
If r00Pr0þ 2, then ranks r0þ 1; r0þ 2; . . . ; r00 1 must be filled
by jobs in Fr0;r00. So, we can assign ðr00 r0 1Þ jobs to those ranks
first and redefine the problem by taking off those ranks along with the scheduled jobs. Thus, without loss of generality, we can say
that r00¼ r0 or r00¼ r0þ 1. Moreover, F
r;r has at most one job and
Fr;rþ1has at most two jobs. The number of possible ðr0;r00Þ pairs
for Fr0;r00 is bounded by k þ ðk 1Þ ¼ 2k 1 and the number of jobs
in Fr0;r00 is bounded by k þ 2ð2k 1Þ ¼ 5k 2.
In order to describe the subsequent steps, we use following notation.
Let c1ðF
r;rÞ be the cost of the job in Fr;r if jFr;rj ¼ 1 and 1
otherwise. Let c1ðF
r;rþ1Þ be the cost of the job with the lower cost in Fr;rþ1if
jFr;rþ1j P 1 and 1 otherwise.
Let c2ðF
r;rþ1Þ be the cost of the job with the higher cost in Fr;rþ1if
jFr;rþ1j ¼ 2 and 1 otherwise.
We define the following shortest path problem from a source node ð0; 0Þ to the destination node ðk þ 1; k þ 1Þ in a layered graph (SeeFig. 1).
Nodes are defined as follows: Node (0, 0) is a dummy source node.
Node (k þ 1; k þ 1) is a dummy destination node.
Node ðr; r þ 1Þ denotes that rank r is assigned from a job in Fr;rþ1
for r ¼ 1; . . . ; k 1.
Node ðr; rÞ denotes that rank r is assigned from a job in Fr;rfor
r ¼ 1; . . . ; k.
Node ðr; r 1Þ denotes that rank r is assigned from a job in Fr1;r
for r ¼ 2; . . . ; k.
The arcs and their costs are defined as follows:
ðk; kÞ and ðk þ 1; k þ 1Þ are connected and the arc cost is 0. ðk; k 1Þ and ðk þ 1; k þ 1Þ are connected and the arc cost is 0.
ðr 1; rÞ and ðr; r þ 1Þ are connected and the arc cost is c1ðF
r;rþ1Þ.
ðr 1; r 1Þ and ðr; r þ 1Þ are connected and the arc cost is c1ðF
r;rþ1Þ.
ðr 1; r 2Þ and ðr; r þ 1Þ are connected and the arc cost is c1ðF
r;rþ1Þ.
ðr 1; rÞ and ðr; rÞ are connected and the arc cost is c1ðF
r;rÞ.
ðr 1; r 1Þ and ðr; rÞ are connected and the arc cost is c1ðF
r;rÞ.
ðr 1; r 2Þ and ðr; rÞ are connected and the arc cost is c1ðF
r;rÞ.
ðr 1; rÞ and ðr; r 1Þ are connected and the arc cost is c2ðF
r1;rÞ.
ðr 1; r 1Þ and ðr; r 1Þ are connected and the arc cost is c1ðF
r1;rÞ.
ðr 1; r 2Þ and ðr; r 1Þ are connected and the arc cost is c1ðF
r1;rÞ.
From the arcs of a shortest path, we can construct an optimal solution.
Now, consider the running time of the proposed algorithm. It takes Oðn log nÞ time to sort the jobs. It takes OðnÞ time to construct
Fr0;r00 and Fr0;r00. As for the shortest path problem, the number of arcs
is less than 9k ¼ OðkÞ 6 OðnÞ and a shortest path can be obtained in OðnÞ time. Therefore, the proposed algorithm can be implemented in Oðn log nÞ time.
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