Operations Research Letters 40 (2012) 349–352

Contents lists available at SciVerse ScienceDirect

Operations Research Letters

journal homepage: www.elsevier.com/locate/orl

Algorithms better than LPT for semi-online scheduling with decreasingprocessing timesT.C.E. Cheng a, Hans Kellerer b,∗, Vladimir Kotov c

a Department of Logistics and Maritime Studies, The Hong Kong Polytechnic University, Kowloon, Hong Kongb Institut für Statistik und Operations Research, Universität Graz, Universitätsstraße 15, A–8010 Graz, Austriac Belarussian State University, Faculty of Applied Mathematics and Computer Science, Prospekt Nezavisimosty 4, 220030 Minsk, Belarus

a r t i c l e i n f o

Article history:Received 20 January 2012Received in revised form8 May 2012Accepted 16 May 2012Available online 4 June 2012

Keywords:Online algorithmsSemi-online algorithmsCompetitive ratioMultiprocessor scheduling

a b s t r a c t

We consider the semi-online multiprocessor scheduling problem with m identical, parallel machinesto minimize the makespan, where the jobs arrive in decreasing order of processing times. The famousLongest Processing Time (LPT) algorithm by Graham (1966) [4] for the classical offline multiprocessorscheduling problemschedules the jobs in decreasing order of processing times andhas aworst-case boundof 4/3−1/(3m). So far, no algorithmwith a better competitive ratio than the LPT algorithmhas been givenfor the semi-online scheduling problemwith decreasing processing times. In this note, we present a 5/4-competitive algorithm for m ≥ 3 and an algorithm that is the best possible for m = 3, i.e. an algorithmwith competitive ratio (1 +

√37)/6.

© 2012 Elsevier B.V. All rights reserved.

1. Introduction

In the classical multiprocessor scheduling problem, thereare given n independent jobs with nonnegative processingtimes p1, . . . , pn that have to be scheduled nonpreemptively onm identical, parallel machines M1, . . . ,Mm with the objectiveof minimizing the makespan. In the online version of themultiprocessor scheduling problem, each jobmust be immediatelyand irrevocably assigned to one of the machines without anyknowledge of future jobs. The performance of an online algorithmis measured by its competitive ratio, i.e., the worst-case ratio withrespect to the optimum solution of the corresponding offlineproblem. In the sequel wewill denote by C∗ the value of an optimalsolution of the offline problem. The total sum of the job processingtimes assigned to a machine M is called the load of machine M ,and is denoted by ℓ(M). Moreover, we will identify jobs with theircorresponding processing times, i.e., a job greater than x means ajob with processing time greater than x.

The online multiprocessor scheduling problem was firstinvestigated by Graham, who showed that the list schedulingalgorithm has a competitive ratio of exactly 2 − 1/m [5,4] and it isthe best possible for m ≤ 3 [2]. A long list of improved algorithms

∗ Corresponding author.E-mail addresses: lgtcheng@polyu.edu.hk (T.C.E. Cheng),

hans.kellerer@uni-graz.at (H. Kellerer), kotovVM@yandex.ru (V. Kotov).

0167-6377/$ – see front matter© 2012 Elsevier B.V. All rights reserved.doi:10.1016/j.orl.2012.05.009

has since been published. The best heuristic is due to Fleischerand Wahl [3]. They designed an algorithm with a competitiveratio smaller than 1.9201, when the number of machines tends toinfinity.

In practice, problems are often not really online or offlinebut somewhere in between. This means that, with respect tothe online problem, some further information about the jobs isavailable. This additional information allows the improvement ofthe performance of the best solution algorithm with respect tothe online problem. These problems are often vaguely called semi-online problems. We will investigate the semi-online schedulingproblemwith decreasing processing times (SOSDP), i.e., we know thatpj ≥ pj+1 for all j ≥ 1.

It is well-known that the offline multiprocessor schedulingproblem is strongly NP-hard by reduction from the 3-partition.Themost popular heuristic is the Longest Processing Time algorithm(LPT for short) proposed by Graham [4]. The LPT algorithm sortsjobs in decreasing order of processing times and assigns themone at a time in this order to the machine with the smallestcurrent load. The LPT algorithmhas aworst-case performance ratioof 4/3 − 1/(3m). Obviously, LPT can be used as an algorithmfor SOSDP. Many other approximation algorithms for offlinemultiprocessor scheduling have been developed and analyzed. Letus especially mention the paper by Hochbaum and Shmoys [6],which provides a PTAS for this problem by using a family of dualapproximation algorithms.

The problem SOSDP was first studied by Seiden et al. [7]. Theyshow that LPT with a competitive ratio of 7/6 for m = 2 is

350 T.C.E. Cheng et al. / Operations Research Letters 40 (2012) 349–352

best possible for two machines, and give a lower bound of (1 +√37)/6 ≈ 1.18046 for m ≥ 3. So far, no algorithm with a better

competitive ratio than LPT has been found for SOSDP.Problem SOSDP on two uniformly related machines is investi-

gated by Epstein and Favrholdt [1]. They show that LPT has optimalcompetitive ratio 1

4 (1 +√17) overall, but does not have optimal

competitive ratio for every value of the speed ratio between thetwo machines. They also construct different algorithms of optimalcompetitive ratio for the intervals where LPT fails.

In this note, we present a 5/4-competitive algorithm form ≥ 3and a best possible algorithm for m = 3, i.e. an algorithm witha competitive ratio (1 +

√37)/6. Both algorithms are elementary

and have linear running time. Essentially, they assign, if possible,a job to a machine for which the load will not exceed some lowerbound multiplied by the competitive ratio; otherwise they assignthe job to the machine with the smallest load.

2. A 5/4-competitive algorithm for an arbitrary number ofmachines

In this section we will present an algorithm, denoted by A, forSOSDP with a competitive ratio 5/4. Since the LPT algorithm has aperformance ratio 4/3 − 1/(3m), it outperforms algorithm A onlyfor m ≤ 3. The algorithm can be described as follows:

Algorithm A. 1. Assign job pj to machine Mj, j = 1, . . . ,m.2. Set L := pm + pm+1.3. Assign the current job pj, j ≥ m+ 1, to the machineM with the

largest load ℓ(M) such that ℓ(M) + pj ≤54 L. Otherwise, assign

it to the machine with the smallest load.

Theorem 1. Algorithm A is a deterministic algorithm for SOSDP withcompetitive ratio 5/4.

Proof. W.l.o.g. assume that C∗= 1. The proof is established by

contradiction. Then there is a counterexample with a failure job zthat shall be the first job being assigned by A to a machineM suchthat ℓ(M) > 5/4. Moreover, we assume that the counterexampleisminimal, i.e., there is no counterexample with a smaller numberof jobs. Consequently, z is the last job, i.e. z = pn.

The jobs are divided into several classes according to theirprocessing times: Let X1 := {pj | pj > 3/4}, X2 := {pj | 3/4 ≥

pj > 1/2}, X3 := {pj | 1/2 ≥ pj > 1/4} and X4 := {pj | pj < 1/4}.Denote by xi the cardinality of set Xi, i = 1, . . . , 4. Since the failurejob z is assigned to the machine M with the smallest load, z hasprocessing time greater than 1/4 and x4 = 0. Moreover, at most 3jobs are assigned to each machine in the optimal solution.

Let M∗

1 , M∗

2 and M∗

3 denote the machines in the optimalsolution with one, two and three jobs, respectively. The numbersm∗

1 , m∗

2 , m∗

3 shall denote the corresponding cardinalities.In the case that at most two jobs are assigned to each machine

in the optimal solution, we have C∗≥ maxj=1,...,m{pj + p2m−j+1}.

It is obvious that in this case Algorithm A is 5/4-competitive. Thus,m∗

3 > 0 and z ≤ 1/3 hold.Let nv denote the number of jobs greater than v. Then,

n1/2 ≤ m∗

1 + m∗

2. (1)

Because each job is greater than 1/4, the second job on each of themachines in M∗

3 is smaller than 3/8. Then,

n3/8 ≤ m∗

1 + 2m∗

2 + m∗

3 = m + m∗

2. (2)

We will distinguish two casesCase 1. p1 + pm+1 ≤ 5/4 L:SetM1 := {M1, . . . ,Mm∗

1+m∗2} andM2 := {Mm∗

1+m∗2+1, . . . ,Mm}.

Due to p1 + pm+1 ≤ 5/4 L Algorithm A assigns the m∗

2 jobspm+1, . . . , pm+m∗

2to machines in M1.

By (1) all jobs of M2 have processing time less than or equalto 1/2 and by (2) all second jobs of M2 have processing time lessthan or equal to 3/8. Since at least 2(m∗

1 + m∗

2) jobs (except z) areassigned to themachines ofM1, there is at least onemachine ofM2,which has (at most) two jobs immediately before z is processed.But then z can be assigned as a third job to that machine, sincethe load would not exceed 1/2 + 3/8 + 1/3 = 29/24 < 5/4,contradicting that z is a failure job.

Case 2. p1 + pm+1 > 5/4 LIf there is a machine in the optimal solution, which contains

only one job, denoted by x, we can exchange x with p1, so that p1is alone on a machine without that the makespan of the optimalsolution is increasing. Create a new instance by removing onemachine and all jobs except z, which are together with p1 onM1 inthe heuristic solution. The makespan of the optimal solution of thenew instancewill not increase. Because p1+pm+1 > 5/4 L, job pm+1is not assigned to M1. Thus, L will not change in the new instanceand all jobs except z are assigned to the same machines as in theoriginal instance. Only z is assigned to a differentmachine (if it wasprocessed onM1 before). In any case, themakespanwill remain thesame or it will possibly increase in the case that z was assigned toM1. Since our counterexample is minimal, we get m∗

1 = 0. Fromz > 1/4, we conclude x1 = 0.

We distinguish two subcases:(2a) 5/4 L ≥ 1LetM∗

1 denote themachinewhere p1 is processed in the optimalsolution. From p1 + pm+1 > 5/4 L ≥ 1 follows that p1 > 1/2, andwe haveM∗

1 ∈ M∗

2 . Thus, there is a largest job pℓ with p1 + pℓ ≤ 1.W.l.o.g. pℓ ∈ M∗

1 andM∗

1 consists of the two jobs p1 and pℓ.From p1 + pm+1 > 1 we get ℓ > m + 1 and with 5/4 L ≥ 1

follows that there is a job pℓ′ , ℓ ≥ ℓ′ > m + 1, which is assignedby Algorithm A to machine M1 such that p1 + pℓ′ ≤ 5/4 L. Get anew instance by removing jobs of M1, i.e., p1, pℓ′ plus possibly athird job, and one machine. Similarly as above, L will not changeand Algorithm A will not improve. Consider the optimal solutionfor m machines. Remove machine M∗

1 and replace pℓ′ by pℓ in theoriginal optimal assignment. Since pℓ′ ≥ pℓ, the makespan of thissolution for the new instance is not larger than for the makespanof the optimal solution for the original instance. Again this yields acontradiction to the minimality of our counterexample.

(2b) 5/4 L < 1Denote byA the assignment of jobs tomachines in the heuristic

solution immediately before the failure job z is assigned. Letm1,m2,m3 denote the corresponding number of machines with one, twoor at least three jobs in A.

If there is a machineM in A with only one job, we have ℓ(M)+

z ≤ 3/4+1/3 < 5/4. Thus, every machine contains at least 2 jobsbefore z arrives, i.e.m1 = 0.

Assume there is a machine in A with exactly 2 jobs y1, y2 suchthat y1 ≥ y2 and y1 ∈ X3. We get with y2 ≤ pm+1 ≤ L/2 thaty1 +y2 + z ≤ 1/2+L/2+1/3 ≤ 5/4. Hence, y1 ∈ X2 andm2 ≤ x2,which implies withm = m2 + m3 that

n = 2m2 + 3m3 + 1 = 3m − m2 + 1 ≥ 3m − x2 + 1.

On the other side, every job of X2 is assigned in the optimal solutionto a unique machine, together with at most one other job. Hence,n ≤ 2x2 + 3(m − x2), which is equivalent to n ≤ 3m − x2. We geta contradiction to the above inequality. �

3. A best possible algorithm for three machines

In this section we will present an algorithm for SOSDP for threemachines with a competitive ratio c := (1 +

√37)/6. According

to the lower bound given in [7], this algorithm is best possible

T.C.E. Cheng et al. / Operations Research Letters 40 (2012) 349–352 351

in terms of the competitive ratio. Formally, Algorithm A3 can bedescribed as follows:

Algorithm A3. 1. Assign jobs p1, p2, p3 to machines M1, M2,M3.2. Set L0 := max{p1, p3 +p4}. If p1 +p4 ≤ cL0, assign p4 toM1 and

p5 to M2 and the remaining jobs to the machine with smallestload.

3. If p1 + p4 > cL0, assign p4 to M3. Set L := max{L0, min{p2 +

p5, p3 +p4 +p5}}. IfM1 contains only job p1 and if p1 +pj ≤ cL,assign the current job pj, j ≥ 5, to machine M1. Otherwise,assign pj to the machine with smallest load.

Theorem 2. There exists a deterministic algorithm for SOSDP for m =

3 with competitive ratio c = (1 +√37)/6 ≈ 1.18046.

Proof. To distinguish between assignments of Algorithm A3 andoptimal assignments we denote the machines of the optimalassignment by M∗

1 , M∗

2 , M∗

3 . We may identify a machine with thejobs assigned to that machine. W.l.o.g. p1 is assigned to the firstmachine in an optimal assignment, i.e. p1 ∈ M∗

1 . W.l.o.g. C∗= 1

holds. Therefore, ℓ(M∗

i ) ≤ 1 for i = 1, 2, 3 and the total processingtime of the jobs is at most 3.

As in Theorem 1, the proof is established by contradiction. Thenthere is a counterexample with a failure job z that shall be the firstjob being assigned by A3 to a machine M such that ℓ(M) > c.W.l.o.g. z is the last job. Note that z is always assigned to amachinewith the smallest load.

It can be easily seen that the bounds L0, L calculated by A3 arelower bounds on the optimal makespan, i.e. L0 ≤ L ≤ 1, and thatthat A3 yields the claimed bound for an instance with at most 5jobs. Note that the competitive ratio c is the root of a quadraticequation. More precisely, we have

Observation O1. (c − 1/3)c = 1.Because the total load of the machines is at most 3, there is a

machinewith load less than or equal to 1−z/3 before z is assigned.Since z is assigned to the machine with smallest load, we have(1 − z/3) + z > c , which results in

Observation O2. z > 32 (c − 1) ≈ 0.27069.

Consequently, we get

Observation O3. All jobs have a processing time greater than32 (c − 1), each machine in the optimal solution has at most 3 jobsand the counterexample contains n ≤ 9 jobs.

Assume that n = 9. After assigning 8 jobs, there is a machinewith at most two jobs and load not exceeding p1 + p4. Thus, themakespan obtained by A3 is at most p1 + p4 + p9. Each machineof the optimal solution contains exactly 3 jobs. Thus, two jobs of{p1, p2, p3, p4} are on one machine and we get p3 + p4 + p9 ≤

1 and hence p4 + p9 ≤ 2/3. Obviously, p1 + p8 + p9 ≤ 1,which gives with (O2) that p1 ≤ 1 − 2z ≈ 0.45862. Therefore,p1+p4+p9 ≤ 1−2z+2/3 < c , which contradicts the assumptionof a counterexample. Recall that A3 yields the claimed bound for atmost 5 jobs. Thus, we have shown

Observation O4. The counterexample contains 6, 7 or 8 jobs.Elementary calculations using (O3) and the fact that the load of

each machine in the optimal solution is at most one, deliver thefollowing bounds summarized in

Observation O5. If n ≥ 7, then p3 ≤ 1/2, p7 ≤ 1/3 and p6 + p7≤ 2/3.

If n = 8, then p2 ≤ 1/2 and p5 + p8 ≤ 2/3.

Assume that p1 + p4 ≤ cL0. In this case p4 ∈ M1. Moreover, p5can be assigned toM2 and p6 toM3, respectively. We get from (O5)that p3 + p6 + p7 ≤ 7/6 < c and p7 fits on machine M3. If n = 8,

p2 + p5 + p8 ≤ c and p8 fits on machine M2. Since by (O4) acounterexample contains at most 8 jobs, there is no failure job andwe have proven

Observation O6. p1 + p4 > cL0, i.e., job p4 is assigned to machineM3.

Let A = {a1, a2, . . . , ar}, B = {b1, b2, . . . , bs} be two sets ofnonnegative integers both sorted in decreasing order with r ≥ s.We say that A dominates B if a1 ≥ b1, a2 ≥ b2, . . . , as ≥ bs.

Assume that M∗

1 consists only of job p1. Then, all other jobsare processed on machines M∗

2 , M∗

3 in the optimal solution. Sincep1 ∈ M1, the jobs processed on machines M2, M3, denoted by J,are dominated by the jobs on machines M∗

2 , M∗

3 , denoted by J∗.Thus, the optimal solution for an instance with two machines andjobs of J is not larger than for an instance with two machinesand jobs J∗, which means it is at most one. For two machines LPThas a worst-case performance of 7/6. Consequently, no subset ofJ has a load greater than 7/6 when assigned to two machines byLPT. But by (O6) no job will be assigned in Step 2 of Algorithm A3and the assignment of jobs to M2 and M3 by A3 is identical to theLPT algorithm. Thus, immediately before z is assigned, there is amachine M ∈ {M2,M3} with ℓ(M) + z ≤ 7/6. Since z is assignedto the machine with the smallest load, we get a contradiction tothe assumption that z is the failure job. Applying (O2) we obtain

Observation O7. M∗

1 contains at least two jobs and p1 ≤ 1 − z ≤

0.73.Now assume that L0 ≥ c − 1/3. It follows with (O1) and (O6)

that p1 +p4 > cL0 ≥ 1. Hence, p1 > 1/2 andmachineM∗

1 containsexactly 2 jobs. Since cL ≥ cL0 > 1, the second job on M1 is at leastas large as the second job onM∗

1 . Consequently, the jobs processedon machines M2, M3 are dominated by the jobs on machines M∗

2 ,M∗

3 . Equivalently to the proof of (O7) we get a contradiction to theexistence of a counterexample.

Observation O8. L0 < c − 1/3.Assume that p1 + p6 > c. Thus, M∗

1 contains no job of theset {p2, p3, . . . , p6} and there is a machine in the optimal solution,which contains at least three jobs of {p2, p3, . . . , p6}. This impliesp6 ≤ 1/3. Since by (O7) p1 ≤ 0.73, we conclude that

Observation O9. p1 + p6 ≤ c.

For the main part of the proof we distinguish two cases:Case 1. p2 ≥ p3 + p4Since p1 ≥ p2 ≥ p3 + p4 > 0.54 by (O2), machines M∗

1 and M∗

2contain at most 2 jobs and we have by (O4) that n ≤ 7.

By the definition of L, p3 + p4 + p5 ≤ L ≤ 1. Thus, if one thejobs p5, p6, p7 is assigned by A3 tomachineM3, the load ofM3 is nolarger than c. From (O9) we know that p2 +p6 ≤ c and if p5 or p6 isassigned by A3 to machineM2 the load ofM2 is also no larger thanc. Finally, by (O7) machine M∗

1 contains at least two jobs, whichmeans p1 + p7 ≤ 1. Thus, p7 can be assigned to machineM1 by A3without exceeding c. Since A3 always assigns jobs to the machinewith smallest load or to a machine for which the total load doesnot exceed c , none of the jobs p1, . . . , p7 can be a failure job.

Case 2. p2 < p3 + p4Since by (O6) p1 + p4 > cL0, job p4 is assigned to machine M3

and we get that L = max{p1, p2 + p5, p3 + p4} ≤ 1.We distinguish several subcases depending on the locations of

p5 and p6.Assume p5 is assigned to M1. Thus, because of p2 < p3 + p4,

job p6 must be processed on M2. By (O9) p6 is not the failure job.By (O5) p7 ≤ 1/3 and p7 can be assigned to machine M3 sincep3 + p4 + p7 ≤ L0 + 1/3 < c by (O8). Job p8 can be assigned tomachineM2 since due to (O5) p2+p5+p8 ≤ 1/2+2/3 = 7/6 < c.

Since p2 < p3 + p4, it remains to consider that p5 is assigned toM2. If p6 is assigned toM1 we proceed as above with p7 and p8.

352 T.C.E. Cheng et al. / Operations Research Letters 40 (2012) 349–352

If p6 is assigned toM3, thenM3 is themachinewith smallest loadafter five jobs andwe get from (O9) that p3+p4+p6 ≤ p1+p6 ≤ c .This implies p1 + p7 ≤ c and p7 can be assigned to M1. Finally, p8can be assigned to M2 since p2 + p5 + p8 ≤ 1/2 + 2/3 = 7/6by (O5).

If p6 is assigned to M2, as machine with smallest load, we getanalogously p2 + p5 + p6 ≤ p1 + p6 ≤ c and p1 + p7 ≤ c . Recallthat p3+p4 ≤ L0 and p8 ≤ 1/3 by (O5). Thus, job p8 can be assignedtoM3 since due to (O8) p3+p4+p8 ≤ L0+1/3 ≤ c . This completesthe proof of our theorem. �

4. Conclusions

In this note we have presented a simple algorithm for problemSOSDP with a competitive ratio better than LPT and a best possiblealgorithm for SOSDP with three machines. For general m thereis still a significant gap between 5/4 and the lower bound (1 +√37)/6. It is an interesting open questionwhether there are better

lower bounds than (1 +√37)/6, or whether for more than three

machines, algorithms with competitive ratio (1 +√37)/6 exist.

Acknowledgments

This research was partially supported by The Hong KongPolytechnic University under grant number G-T397. The authorsare thankful to an anonymous referee for his comments, whichgreatly improved the presentation of the paper.

References

[1] L. Epstein, L.M. Favrholdt, Optimal non-preemptive semi-online scheduling ontwo related machines, Journal of Algorithms 57 (2005) 49–73.

[2] U. Faigle, W. Kern, G. Turan, On the performance of on-line algorithms forpartition problems, Acta Cybernetica 9 (1989) 107–119.

[3] R. Fleischer, M. Wahl, On-line scheduling revisited, Journal of Scheduling 3(2000) 343–353.

[4] R.L. Graham, Bounds on multiprocessing timing anomalies, SIAM Journal onApplied Mathematics 17 (1969) 263–269.

[5] R.L. Graham, Bounds for certain multiprocessor anomalies, Bell SystemTechnical Journal 45 (1966) 1563–1581.

[6] D.S. Hochbaum, D. Shmoys, Using dual approximation algorithms for schedul-ing problems: theoretical and practical results, Journal of the Association forComputing Machinery 34 (1987) 144–162.

[7] S. Seiden, J. Sgall, G. Woeginger, Semi-online scheduling with decreasing jobsizes, Operations Research Letters 27 (2000) 215–222.