A Note on Online Scheduling for Jobs with Arbitrary Release Times

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1 A Note on Online Scheduling for Jobs with Arbitrary Release Ties Jihuan Ding, and Guochuan Zhang College of Operations Research and Manageent Science, Qufu Noral University, Rizhao 7686, China College of Coputer Science, Zhejiang University Hangzhou 3007, China Abstract. We investigate the proble of on-line scheduling for jobs with arbitrary release ties on identical parallel achines. The goal is to iniize the akespan. We derive a best possible online algorith with copetitive ratio of for =. For a special case that all the jobs have unit processing ties, we prove that Algorith LS has a tight bound of 3/ for general achines. Keywords: On-line scheduling, Makespan, Copetitive ratio, Identical parallel achines. Introduction Online scheduling has received great attention in decades. The ost basic odel is the classical on-line scheduling proble on identical parallel achines which was proposed by Graha []. In the classical on-line scheduling proble, jobs arrive one by one (or over a job list), we do not know any inforation about the job list in advance, whenever a job arrives, it ust be scheduled iediately on one of the achines without knowledge of any future jobs. Only after the current job is scheduled can the next job appear. The objective is to iniize the akespan. It is well known that the first online algorith in scheduling theory, naely List Scheduling (LS) algorith, for the proble was proposed by Graha []. Note that in the above odel all jobs in the job list have release ties zero. A ore general version of the classical on-line scheduling proble on identical parallel achines was given by Li and Huang []. In the general version, all jobs appear in for of orders at tie zero. When a job appears, the scheduler is infored of the release tie and processing tie of the job. The proble can be forally defined as follows. A list of jobs is to be scheduled on identical parallel achines, M,M,..., M. We assue that all jobs appear on-line in a job list. Whenever a job J j with release tie r j and processing tie p j appears, the This work has been supported by NSFC ( ). D.-Z. Du, X. Hu, and P.M. Pardalos (Eds.): COCOA 009, LNCS 5573, pp , 009. c Springer-Verlag Berlin Heidelberg 009

2 A Note on Online Scheduling for Jobs with Arbitrary Release Ties 355 scheduler has to assign a achine and a processing slot for J j irrevocably without knowledge of any future jobs. The goal is to iniize the akespan. In this general on-line situation, the jobs release ties are assued arbitrary, whereas in the existing literature the jobs release ties are norally non-decreasing. That is to say, although a job ay appear first in the job sequence, but its release tie ay be greater than the release tie of the job which appears later in the job sequence. It is obvious that the above odel is just the odel of classical on-line scheduling proble on identical parallel achines if all the jobs have release tie zero. The proble is called on-line scheduling proble for jobs with arbitrary release ties [3]. The quality of an online algorith A is usually easured by its copetitive ratio C R(, A) =sup ax(l) A L Cax(L) where Cax(L) A andcax(l) denote the akespans of the schedule produced by algorith A and an optial off-line algorith, respectively. Li and Huang considered the on-line scheduling proble for jobs with arbitrary release ties first. For the general odel of the proble where job processing ties are arbitrary, Li and Huang [] gave a lower bound of, and then showed an algorith LS with tight bound of 3 /. A odified algorith MLS is also proposed which is better than LS for any, and the copetitive ratio of MLS is bounded by.939 for any. For the special odel where the job length is in [p, rp](r ), the perforance of algorith LS is analyzed in [3]. They first gave an upper bound of R(, LS) = 3 r, r ; + r +r + ( )r (+r), r<, for general and showed that the tight bound for =is+ +r.when =, they presented a tight bound of the copetitive ratio + 5r+4 (r+) for r 4; For r<4, they gave a lower bound and showed that provides an upper bound for the copetitive ratio. Our results. In this paper, we consider the on-line scheduling proble for jobs with arbitrary release ties. For the general odel where job processing ties are arbitrary, we derive a best possible online algorith with copetitive ratio of for =. Then we show that the idea of the algorith for = cannot be extended to the case with 3 achines by a counterexaple. Finally, for a special odel that all the jobs have unit processing ties, we prove that algorith LS has a tight bound of 3/ for general achines. A Best Possible Online Algorith for = Before we give the best possible online algorith for =, we first describe the following definition which is defined in Li and Huang []. r

3 356 J. Ding and G. Zhang Definition. Suppose that J j is the current job with release tie r j and processing tie p j. We say that achine M i has an idle tie interval for job J j, if there exists a tie interval [T,T ] satisfying the following conditions:. Machine M i is currently idle in interval [T,T ] and a job has been assigned on M i to start processing at T.. T ax{t,r j } p j. For the job set J(i) ={J,J,...,J i } that has been scheduled by the algorith, let M i =ax ji(r j + p j ), M i = i j= p j and M i =ax{m i,mi }.Itis obvious that M i is a lower bound of the optial akespan for job set J(i) = {J,J,...,J i }. Algorith:. Let L i and L i be the copletion tie of the last job on achine M and M iediately after J i has been scheduled by the algorith, respectively. Let J i+ be a new job given to the algorith.. If there exist soe achines which have idle intervals for job J i+, select a achine M i which has an idle interval [T,T ]forjobj i+ with inial T.ThenwestartjobJ i+ on achine M i at tie ax{t,r i+ } in the idle interval. Otherwise, go to step If r i+ > ax{l i,l i }, then we assign job J i+ to achine M to start at tie r i+. Otherwise, go to step If r i+ ax{l i,li } and Li >Li and Li + p i+ M i+, then we assign job J i+ to achine M tostartattiel i. Otherwise, go to step Start job J i+ on achine M at tie ax{l i,r i+ }. When it is our turn to schedule the next job J i+,ifr i+ > ax{l i,li }, then we always assign job J i+ to achine M tostartattier i+.forthe case that r i+ ax{l i,li }, unless there is idle interval for job J i+, theain idea of the algorith is that we always assign job J i+ to achine with larger load if the copletion tie of job J i+ is no ore than M i+.otherwisewe will assign job J i+ to another achine to start as soon as possible. Theore. The copetitive ratio of the algorith is. Proof. Let J(n) = {J,J,...,J n } be an arbitrary job list. The optiu akespan for job set J(i) = {J,J,...,J i } is denoted by OPT i. For each i, i n, we will prove that ax{l i,l i } OPT i holds iediately after job J i has been scheduled by the algorith. If i =, then it is obvious that the conclusion holds. Now we assue ax{l i,li }OPT i holds iediately after job J i has been scheduled by the algorith. Next we will show ax{l i+ }OPT i+ holds iediately after job J i+ has been scheduled by the algorith. It is obvious that OPT i+ M i+.sowehave ax{l i+ }OPT i+ if job J i+ is scheduled by Step or Step 3 or

4 A Note on Online Scheduling for Jobs with Arbitrary Release Ties 357 Step 4 of the algorith. In the following we will show ax{l i+ } OPT i+ if job J i+ is scheduled by Step 5 of the algorith. Case. L i >Li and Li + p i+ > M i+. Case.. If r i+ L i, then the algorith will assign job J i+ to achine M to start at tie r i+. By the inductive assuption and the fact that OPT i+ M i+,wehaveax{l i+ }OPT i+. Case.. If r i+ <L i, then the algorith will assign job J i+ to achine M to start at tie L i.lett be the least tie point fro which achine M is busy to L i + p i+,naelyl i+.letl = L i T.Wecandescribe the copletion tie of job J i+ as T + l + p i+.ift = 0, then by the fact that OPT i+ (l + p i+ )= Li+ and the inductive assuption we have ax{l i+ }OPT i+.ift > 0, then let J k (k<i+ ) be the first job scheduled on achine M in tie interval [T,L i ]. According to the algorith, just before the algorith will schedule job J k,weusthavel k >L k and L k + p k > M k.otherwise,jobj k will be scheduled on achine M.Sothe start tie of job J k is r k. Let T be the least tie point fro which achine M is busy to L k.let l = L k T,thenwehaveL k + p k = T + l + p k > (r k + p k ). Furtherore we have M k T,soT +l +p k > M k T. Cobining the above inequalities we can get l >r k. Therefore we have L i+ = T + l + p i+ r k + l + p i+ l + l + p i+ OPT i+. So ax{l i+ }OPT i+. Case. L i Li. By the algorith we have r i+ L i,sothecopletion tie of J n is L i + p i+. Siilar to the Case., We can find the sae T and J k on achine M,andT on achine M. Copletely repeat the above process, we can get L i+ = T + l + p i+ r k + l + p i+ l + l + p n OPT i+. Hence we have ax{l i+ }OPT i+. Although the above algorith is very siple, but the idea of it is ipossible to extended to ore achines case, even for 3 achines. In the following, a counterexaple I for 3 achines is given to show the fact. For the sake of siplicity, let j =(r j,p j ) denote job j which has release tie r j and processing tie p j. I = {( ɛ, ɛ), (0, ), ( ɛ, ɛ), (/3 ɛ, ɛ), (/3 ɛ, ɛ), (/3 ɛ, ɛ), (/3 ɛ, ɛ), (0, /3), (0, /3), (0, /3), ( ɛ, ɛ), (0, ), (0, )}. It is easy to verify the optial akespan for I is ɛ.LetH be the algorith designed copletely by the idea of the algorith for two-achine case. Then the akespan of Algorith H for I is 3, see Fig.. 3 A Special Case for Jobs with Unit Processing Tie In this part, we give a tight bound of Algorith LS for a special case that all the jobs have unit processing tie. In the following we will describe Algorith

5 358 J. Ding and G. Zhang M 3 M M Fig.. A counterexaple shows that the ratio is not saller than label on each job denotes not only the order but also the job. 3 +/3 = 9 4.The LS which is defined in Li and Huang []. Essentially, the algorith assigns a job to be processed as early as possible when its order arrives. Algorith LS. Assue that L i is the scheduled copletion tie of achine M i (i =,,). Reorder achines so that L L L and let J n be a new job given to the algorith with release tie r n and running tie p n =. Let t =ax{r n,l }.. If there exist soe achines which have idle intervals for job J n, select a achine M i which has an idle interval [T,T ]forjobj n with inial T. Then start job J n on achine M i at tie ax{t,r n } in the idle interval. Otherwise, assign job J n to achine M to start the processing at tie t. In order to give the proof of the result. We first give soe explanation of sybols. [s]: the integral part of s; (s): the fractional part of s; s : the sallest integer that is not saller than s. The following Lea gives a estiate of the nuber of jobs copleted in tie interval [0,T] on one achine in schedule of Algorith LS on the assuption that there is no idle tie interval with length equal to or greater than. Lea. Let N denote the nuber of jobs copleted on one achine in tie interval [0,T]. If the length of all the idle intervals in schedule of Algorith LS is saller than, then we have () N T/ if T is even; () N (T )/ if T is odd and there is one job with start tie saller than T which is copleted after tie T.(3)N (T +)/ if T is odd and there is no job with start tie saller than T which is copleted after tie T. Proof. At ost N + idle intervals are generated by the process of N jobs in [0,T] on one achine. Then the total length of these idle intervals is saller than N +.Sowehave(N +)+N>T. Fro the inequality we can get N>(T )/. Therefore N T/ifT is even;

6 A Note on Online Scheduling for Jobs with Arbitrary Release Ties 359 If T is odd and there is one job with start tie saller than T which is copleted after tie T.Thenwealsohave(N +)+N>T. So we can get k (T )/. If T is odd and there is no job with start tie saller than T which is copleted after tie T.Thenwehave(N +)+N>T,naelyN (T +)/. The proof is then finished. Theore. The copetitive ratio of LS is R LS =3/. Proof. We assue L = {J,J,...,J n } is an arbitrary job list. Let Cax(L) LS and Cax (L) denote the akespans of the schedule produced by Algorith LS and an optial schedule, respectively. Without loss of generality, we assue J n is the last job copleted in schedule of Algorith LS and J n is the only job with copletion tie Cax LS (L). Let s n be the start tie of job J n in schedule of Algorith LS. If s n = r n, then it is obvious that R LS 3/. If s n >r n, then job J n ust be assigned on achine M to start at tie s n = L by the algorith. In schedule of Algorith LS, let s be the least tie point fro which all the achines are busy to L,andletk be the nuber of jobs(including job J n ) copleted after tie s on achine M, then it is obvious that at least (k ) jobs are copleted after tie s on each of the other achines. So at least (k ) + jobs in total are copleted after tie s in schedule of Algorith LS. We have Cax LS (L) s + k and C ax (L) s +, because at least one job has release tie s by the algorith. We first consider the case that all the idle intervals in schedule of Algorith LS have length saller than. Case. [s] is even. By Lea, it is easy to know the optial schedule copletes at ost [s]/ ore jobs before tie [s] than the schedule of Algorith LS does. Case.. If k s +3,thenwehave: Cax(L) LS Cax(L) s + k s + s +3 = 3 s + (s +). Case.. If k s +3=[s] +++(s), naely k s +4.Then atleast (k ) + [s]/ jobs are processed after tie [s] in optial schedule. So we have [ Cax (L) [s]+ (k ) + [s] ] Therefore: = [s]+k. ax (L) C ax (L) s + k s +[s]+4 [s]+k ([s]+)

7 360 J. Ding and G. Zhang = 3([s]+)+(s) ([s]+) 3. Case. [s] is odd. Case.. If k s + 3, then: Cax(L) LS Cax(L) s + k s + s +3 = 3 s + (s +). Case.. If k >s+3=[s]+++(s), naely, k s + 5. By Lea, one achine in optial schedule copletes at ost [s]+ ore jobs before [s] than it does in schedule of Algorith LS, and at least one achine in optial schedule(the achine is idle iediately before s) copletes at ost [s] ore jobs [ before [s] than it ] does in schedule of Algorith LS. So at least [(k )+] ( ) [s]+ + [s] jobs ust be processed after tie [s] in optial schedule. Hence we can derive [ ] [(k ) + ] ( ) [s]+ Cax + [s] (L) [s]+ =[s]+(k ) [s]+ + = [s]+k. Case... If (s), then: ax(l) C ax(l) s + k [s]+k s +[s]+5 [s]+[s]+5 3([s]+)+(s [s]) = ([s]+) 3. Case... (s) >. In this case we need to estiate the nuber of jobs copleted after tie s in schedule of Algorith LS ore carefully. Case...a. Thetiepoints appears exactly on achine M,naelyM is idle iediately before tie s, andp jobs with start tie saller than [s] are copleted after tie s in schedule of Algorith LS. For the p achines that process such p jobs, k jobs are copleted after tie s on each of the in schedule of Algorith LS. By Lea, each of the p achines in optial schedule

8 A Note on Online Scheduling for Jobs with Arbitrary Release Ties 36 copletes at ost [s]+ ore jobs before tie [s] than it does in schedule of Algorith LS and each of the the other p achines in optial schedule copletesatost([s] )/ ore jobs before tie [s] than it does in schedule of Algorith LS. So we have: C ax (L) [s]+ { [(k ) + p +] =[s]+(k ) [s] = [s]+k +. + [ p [s]+ +( p) [s] ]} Therefore: ax(l) C ax (L) = s + k [s]+k + s +[s]+5 ([s]+3) 3([s]+3)+(s [s]) 4 ([s]+3) 3. Case...b. The tie point s does not appear on achine M and the start tie of the first copleted job after tie s on achine M is not greater than [s]. Then we have ax(l) k +[s]. Hence ax(l) C ax(l) [s]+k [s]+k [s]+[s]+5 [s]+[s]+5 = 3([s]+) ([s]+) 3. Case...c. The tie point s does not appear on achine M,andpjobs with start tie saller than [s] are copleted after tie s in schedule of Algorith LS, and the start tie of the first copleted job after tie s on achine M is greater than [s]. Siilar to the analysis in Case...a, we have C ax(l) [s]+ { [(k ) + p +] =[s]+(k ) [s] = [s]+k +. + [ p [s]+ +( p) [s] ]}

9 36 J. Ding and G. Zhang So ax(l) C ax(l) = s + k [s]+k + 3. s +[s]+5 ([s]+3) 3([s]+3)+(s [s]) 4 ([s]+3) Secondly, if there exist idle tie intervals with length not saller than in schedule of Algorith LS. We select a such idle tie interval with latest over tie, say b, andchooseatiepointa such that b a =. By the Algorith LS, all the jobs with start tie greater than a ust have release tie greater than a also. So siilar to the analysis in the first part of the proof, we can get R LS 3/. Finally, the following instance shows that the bound of Algorith LS is tight. The instance consists jobs in total, the first jobs have release tie ɛ and the last jobs have release tie zero. It is easy to know the akespan of Algorith LS for the instance is 3 ɛ, but the optial akespan is. So R LS (3 ɛ)/, let ɛ tend to zero, we get the bound is tight. 4 Final Rearks We consider the proble of scheduling for jobs with arbitrary release ties on identical parallel achines. For =, we gave a best possible online algorith. It ust be a challenging work to design best possible online algorith for 3. For a special odel where all the jobs have processing tie, we derived that Algorith LS has a tight bound of 3/. Furtherore, soe other special odels of the proble are also worth to considering in future. References. Graha, R.L.: Bounds on Multiprocessing Tiing Anoalies. SIAM Journal on Applied Matheatics 7, (969). Li, R., Huang, H.C.: On-line Scheduling for Jobs with Arbitrary Release Ties. Coputing 73, (004) 3. Li, R., Huang, H.C.: List Scheduling for Jobs with Arbitrary Release Ties and Siilar Lengths. Journal of Scheduling 0, (007)

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