Minimizing Maximum Flow-time on Related Machines

Transcription

1 Mnmzng Maxmum Flow-tme on Related Machnes Nkhl Bansal and Bouke Cloostermans Endhoven Unversty of Technology P.O. Box 513, 5600 Endhoven, The Netherlands Abstract We consder the onlne problem of mnmzng the mum flow-tme on related machnes. Ths s a natural generalzaton of the extensvely studed makespan mnmzaton problem to the settng where jobs arrve over tme. Interestngly, natural algorthms such as Greedy or Slowft that work for the smpler dentcal machnes case or for makespan mnmzaton on related machnes, are not O(1)-compettve. Our man result s a new O(1)-compettve algorthm for the problem. Prevously, O(1)-compettve algorthms were known only wth resource augmentaton, and n fact no O(1) approxmaton was known even n the offlne case ACM Subject Classfcaton F.2.2 Nonnumercal Algorthms and Problems Keywords and phrases Related machnes schedulng, Maxmum flow-tme mnmzaton, On-lne algorthm, Approxmaton algorthm Dgtal Object Identfer /LIPIcs.APPROX-RANDOM Introducton Schedulng a set of jobs on machnes to optmze some qualty of servce measure s one of the most well studed problems n computer scence. A very natural measure of servce receved by a job s the flow-tme, defned as the amount of tme the job spends n the system. In partcular, f a job j arrvng at tme r j completes ts processng at tme C j, then ts flow-tme F j s defned as C j r j ;.e., ts completon tme mnus ts arrval tme. Over the last few years, several varants of flow-tme related problems have receved a lot of attenton: on sngle and multple machnes, n onlne or offlne settng, for dfferent objectves such as total flow-tme, l p norms of flow-tme, stretch etc., wth or wthout resource augmentaton, n weghted or unweghted settng and so on. We refer the reader to [15, 13, 12, 4] for a survey of some of these results. In ths paper we focus on the objectve of mnmzng the mum flow-tme. Ths s desrable when we want to guarantee that each job has a small delay. Maxmum flow-tme s also a very natural generalzaton of the mnmum makespan or the load-balancng problem, that has been studed extensvely (see e.g. [5, 9, 1] for a survey). In partcular, f all jobs have dentcal release tmes, then the mum flow-tme value s precsely equal to the makespan. Mnmzng the mum flow-tme s also related to deadlne schedulng problems. In partcular, the mum flow-tme s at most D f and only f each job j completes by r j + D. Moreover, note that arbtrary deadlnes d j can be modeled by consderng the weghted verson of mum flow-tme 1. Supported by NWO grant and ERC consoldator grant In deadlne schedulng however, the deadlnes are typcally consdered fxed and the focus s on mzng the throughput. Nkhl Bansal and Bouke Cloostermans; lcensed under Creatve Commons Lcense CC-BY 18th Int l Workshop on Approxmaton Algorthms for Combnatoral Optmzaton Problems (APPROX 15) / 19th Int l Workshop on Randomzaton and Computaton (RANDOM 15). Edtors: Naveen Garg, Klaus Jansen, Anup Rao, and José D. P. Rolm; pp Lebnz Internatonal Proceedngs n Informatcs Schloss Dagstuhl Lebnz-Zentrum für Informatk, Dagstuhl Publshng, Germany

2 86 Mnmzng Maxmum Flow-tme on Related Machnes Known results for mum flow-tme For a sngle machne, t s easy to see that Frst In Frst Out (FIFO) s an optmal (onlne) algorthm for mnmzng the mum flow-tme. For dentcal multple machnes, Bender et al. [8] showed that the Greedy algorthm, that schedules the ncomng job on the least loaded machne s 3 2/m compettve, where m s the number of machnes. They also showed that ths bound s tght for the Greedy algorthm. If jobs can be preempted and mgrated (moved from one machne to another), [2] gave a 2-compettve algorthm. A systematc nvestgaton of the problem for varous machne models was ntated recently by Anand et al. [3]. Recall that n the related machnes model each machne has speed s, and processng job j on machne takes p j = p j /s unts of tme. In the more general unrelated machnes model, p j can be completely arbtrary. Among other results, Anand et al. [3] gave a (1 + ɛ)-speed O(1/ɛ)-compettve algorthm for the unrelated machne case, for any gven ɛ > 0. Here the onlne algorthm can process 1 + ɛ unts of work per tme step, but s compared to an offlne optmum that does not have ths extra resource augmentaton [14, 15]. They also showed that n the unrelated settng any algorthm wthout resource augmentaton must be Ω(m) compettve. For the weghted mum flow-tme objectve, they gave a (1 + ɛ)-speed, O(1/ɛ 3 )-compettve algorthm for the related machnes settng, and showed that no O(1)-speed, O(1)-compettve algorthm exsts n the unrelated settng. A natural queston that remans s the complexty of the problem for related machnes: Is there an O(1)-compettve algorthm for the related machnes settng, wthout usng resource augmentaton? Ths queston s partcularly ntrgung as t s not at all clear what the rght algorthm should be [2]. In fact, no O(1)-approxmaton s known even n the offlne case. One ssue s that the natural Slow-ft algorthm, that s O(1)-compettve for makespan mnmzaton (even when the jobs are temporary and have unknown duratons [6]), s not O(1)-compettve for mum flow-tme (Lemma 2 below). The algorthm of [3] for weghted mum flow-tme wth resource augmentaton s also a varant of Slow-ft. Recently, [7] obtaned an O(log n) approxmaton for mnmzng mum flow-tme on unrelated machnes, where n s the number of machnes. However ther technques do not seem to gve anythng better for the related machnes settng ether. Our man result s the followng. Theorem 1. There s a 13.5 compettve algorthm, Double-ft, for mnmzng mum flow-tme on related machnes. Ths also gves the frst O(1) approxmaton for the offlne problem. We also show that no such result s possble n the weghted case (wthout resource augmentaton), and gve an Ω(W ) lower bound on the compettve rato where W s the mum to mnmum weght rato. Hgh-level approach There are two competng trade-offs whle schedulng on related machnes. On one hand the algorthm should keep as many machnes busy as possble, otherwse load mght accumulate and delay future jobs. Ths accumulated load could be mpossble to get rd of f there s no resource augmentaton. On the other hand, the algorthm should keep fast machnes empty for processng large jobs that mght arrve later. In partcular, fast machnes are a scarce resource that should not be wasted on processng small jobs unnecessarly. It s nstructve

3 N. Bansal and B. Cloostermans 87 to consder the lower bounds n Secton 2, where both Slow-ft and Greedy are shown to perform badly due to these opposte reasons. To get around ths, we desgn an algorthm that combnes the good propertes of both Slow-ft and Greedy. In partcular, the algorthm uses a two phase strategy whle assgnng jobs to machnes at each step. Frst, the jobs are spread out to ensure that machnes are busy as much as possble. Once machnes are saturated, the algorthm shfts nto a Slow-ft mode, whch ensures that small jobs do not unnecessarly go on fast machnes. The key dffculty n the analyss s to control how the two phases nteract wth each other. To do ths, we mantan two nvarants that capture the dynamcs of the algorthm, and control how much the onlne algorthm s load on a subset of machnes devates from the offlne algorthm s load on those machnes. The man part of the argument s to show nductvely that these nvarants are mantaned over tme. Notaton and formal problem descrpton There are m machnes ndexed by non-decreasng order of speeds s 1 s 2... s m. The processng requrement of job j s p j, and t requres tme p j /s on machne. We wll call p j the work of j, and p j /s ts load on machne. Jobs arrve onlne over tme and p j s known mmedately upon ts release tme r j. The goal s to fnd a schedule that mnmzes the mum flow-tme, and we assume that a job cannot be mgrated from one machne to another. We use Opt to denote some fxed optmum offlne schedule, and also to denote the value of ths soluton. 2 Lower bounds on Slow-ft and Greedy Slow-ft Algorthm Slow-ft takes as nput a threshold F opt (the current guess on optmum), and schedules every ncomng job on the slowest possble machne whle keepng the load below F opt. If the jobs cannot be feasbly scheduled on any machne, the algorthm fals and the threshold s doubled. Lemma 2. Slow-ft has a compettve rato of Ω(m). Proof. We descrbe an nstance where the threshold F opt keeps doublng untl t reaches m even though Opt = 2. There are m dentcal machnes (but we arbtrarly order them from slow to fast). Next, we assume that F opt 2, whch can be acheved by gvng 2m unt-sze jobs ntally at t = 0. At each tme step t 2, m unt-length jobs arrve. As Slow-ft wll not use all m machnes ntally, there wll be some tme t 0 at whch all the machnes 1,..., m 1 have load F opt. At tme t 0 + 1, when these ntal m 1 machnes have F opt 1 pendng jobs, we release 2m unt-sze jobs. As there s at most m 1 + F opt total capacty avalable, these jobs cannot be scheduled feasbly f F opt m. On the other hand, at each tme step Opt dstrbutes the ncomng jobs over all machnes and acheves value 2. Intutvely, Slow-ft unnecessarly bulds up load on slow machnes whle keepng the fast machnes empty, and cannot recover f there s small burst of jobs. APPROX/RANDOM 15

4 88 Mnmzng Maxmum Flow-tme on Related Machnes Greedy When a job j arrves, Greedy schedules j on the machne that mnmzes the flow-tme of j (assumng FIFO order). Tes are broken arbtrarly. The followng bound s well-known [11], but we sketch t here for completeness. The dea s that Greedy puts too many slow jobs on fast machnes, whch causes problems when large jobs arrve. Lemma 3. Greedy has a compettve rato of Ω(log m). Proof. Consder an nstance where we have k groups of machnes where group G contans 2 2k 2 machnes of speed 2. Note that the total processng power n group G s equal to S = 2 2k. The processng power of groups,..., k combned s thus equal to P = k 22k 2S. We receve k sets of jobs, all at tme 0, but n order. For all = 1,..., k, set J contans 2 2k 2 jobs of sze 2. Agan, note that the total sze of jobs n set J s equal to 2 2k. Greedy wll spread jobs from set over groups,..., k. Group k (contanng only a sngle machne of speed 2 k ) wll receve a S k /P 1 2 S k/s = 2 k+ 1 fracton of these jobs. Ths means group k receves k =1 22k 2 k+ 1 = k2 k 1 work. Snce group k has a sngle machne of speed 2 k, fnshng these jobs takes Ω(k) tme. However, optmum can schedule the -th batch of jobs on group machnes, ncurrng a mum load of 1 (.e., t does Slow-ft wth threshold 1). 3 The Algorthm Double-ft We descrbe our algorthm, denoted by Double-ft hereafter. Double-ft takes an nput a parameter F opt, whch s supposed to be our estmate of Opt. By a slght varaton on the doublng trck that loses an addtonal factor of 1.5 (see Secton 3.4), we wll assume henceforth that F opt [Opt, 1.5Opt). We dvde tme nto ntervals I k of sze as I k = [3(k 1)F opt, 3kF opt ). We refer to tme 3kF opt as the k-th epoch. For each k = 1, 2,..., Double-ft batches the jobs that arrve durng I k and schedules them at epoch k usng the algorthm n Fgure 1. We use [ : m] to denote the machnes,..., m. If the total remanng work on jobs on machne s w() at tme t, we say that t has load w()/s. 1. Let J denote the set of jobs arrvng durng I k. 2. Partton jobs n J nto classes J 1,..., J m, where each job j s n class J wth the smallest ndex such that p j s F opt. 3. For = m, m 1,..., 1 4. Consder the jobs j n J n arbtrary order and assgn them as follows: 5. (Saturaton Phase:) If some machne n [ : m] s loaded below 6. schedule j on the slowest such machne. 7. (Slow-ft Phase:) Else schedule j on the slowest machne n [ : m] 8. such that ts load stays below 6F opt. 9. If no such machne exsts return FAIL. Fgure 1 Algorthm Double-ft for the epoch k.

5 N. Bansal and B. Cloostermans 89 Descrpton Frst, Double-ft classfes the jobs arrvng durng I k dependng on the smallest machne on whch they have sze no larger than F opt. Note that as F opt Opt, f job j s put n class J, then Opt cannot schedule job j onto a machne smaller than ether. Double-ft consders jobs from classes J m down to J 1 (ths orderng wll be used crucally). Each class s scheduled n two phases. In the saturaton phase, when schedulng a job j, t checks f there s some machne n [ : m] wth load less than. If so, j s scheduled on the slowest such machne. If no such machne exsts, the algorthm enters the Slow-ft phase (for class J ), and performs Slow-ft for class J on machnes [ : m] wth threshold 6F opt. 3.1 Analyss Our goal n ths secton s to show the followng result. Theorem 4. If F opt Opt, then the algorthm never fals. Ths drectly mples Theorem 1 as follows. Each job spends at most tme watng to be assgned, and at most 6F opt on ts desgnated machne, thus the flow-tme of any job s at most 9F opt. As F opt 1.5Opt by the doublng trck, ths mples a compettve rato of 13.5 For the purpose of analyss, t wll be convenent to consder a restrcted Opt that also batches jobs and schedules the jobs arrvng n I k at epoch k. Note that such a restrcted algorthm has objectve at most + Opt 4F opt (as we can take the orgnal schedule and delay every job by ). To prove theorem 4, we wll n fact prove the followng stronger result: Double-ft never fals for any nstance where the restrcted Opt has value at most 4F opt. The Invarants Fx an epoch k. Let A (k) and B (k) denote the total work on machnes [ : m] n Doubleft s schedule just before and just after all the jobs from nterval I k are scheduled respectvely. Smlarly, let A opt (k) and B opt (k) be the total work remanng on machnes [ : m] n Opt s schedule. We wll show that the followng two nvarants hold at each epoch k. A (k) A opt B (k) s. (1) s, B opt (k) m + F opt s. (2) Roughly speakng, nvarants (1) and (2) show that the load on any suffx of Double-ft s machnes stays close to Opt s load on those machnes, both before and after the jobs are scheduled n epoch k. We wll prove that (1) and (2) hold by a careful nducton over and k. Before we prove these nvarants, let us frst see why they mply Theorem 4. Proof of Theorem 4. Consder a fxed epoch k. As the (restrcted) Opt has mum flow-tme at most 4F opt, for each t must hold that B opt m (k) 4F opt s. Thus by (2) t follows that B opt m (k) 5F opt s for each. Choosng = m, ths mples that APPROX/RANDOM 15

6 90 Mnmzng Maxmum Flow-tme on Related Machnes Double-ft never loads machne m above 5F opt and thus never fals (as machne m always has room for an addtonal job). Provng the Invarants The strategy for provng that (1) and (2) hold at all epochs k wll be to show the followng two lemmas. Lemma 5. If at epoch k, (1) holds for all machnes, then (2) also holds for all machnes. The next step wll be to relate the condtons at epochs k and k + 1. Lemma 6. If at any epoch k, (2) holds for all machnes, then (1) also holds for all machnes at epoch k + 1. As (1) trvally holds for k = 0 (as A (0) = A opt (0) = 0 for all ), applyng Lemma 5 and Lemma 6 alternately mples that (1) and (2) hold for all k. 3.2 Proof of Lemma 5 We frst show that Double-ft s conservatve n schedulng small jobs on fast machnes. Lemma 7. Let 1 < 2. If some job j of class 1 s scheduled by Double-ft onto machne 2 durng the saturaton phase (.e. usng threshold ), then all jobs of class for 1 < 2 are also scheduled durng the saturaton phase. Proof. Consder the state of Double-ft s machnes just before j was scheduled. As j s scheduled on machne 2 durng the saturaton phase, the load on 2 must be below at that pont. As jobs of class for 1 < 2 were consdered before class 1 -jobs, the load on 2 was also below after schedulng class jobs, and thus Double-ft must have never swtched to the Slow-ft phase whle consderng class. Next we defne the noton of separated machnes, whch wll play a crucal role n the analyss. Defnton 8. Machnes 1 and 2 ( 1 < 2 ) are separated at epoch k f Double-ft scheduled no jobs from classes [1 : 1 ] onto machnes [ 2 : m] at epoch k. The followng lemma shows that f two consecutve machnes are separated, t s easy to relate epochs k and k + 1. Lemma 9. If machnes 1 and are separated at epoch k, then (1) mples (2) for machne. Moreover ths trvally holds for machne = 1. Proof. As machnes 1 and are separated at epoch k, no jobs from class [1 : 1] were scheduled onto machnes [ : m] at epoch k. Thus B (k) = A (k) + J, (3) where J represents the total work of all jobs n J. As jobs from J cannot be scheduled onto machnes [1 : 1] n an optmal schedule, we also obtan B opt (k) A opt (k) + J. (4)

7 N. Bansal and B. Cloostermans 91 Ths mples that B (k) = A (k) + J A (k) + B opt (k) A opt (k) B opt s, (5) where the last step follows by our assumpton that (1) holds for (, k). Fnally for = 1, we observe that both (3) and (4) hold wth equalty, and hence the result holds trvally. We now have all the tools we need to prove Lemma 5. Proof of Lemma 5. We use nducton over n the order of larger to smaller. In partcular, to prove that (2) holds for some par (, k), we assume that (1) holds for all (, k) and that (2) holds for all (, k) wth >. As the base case note that ths s vacuously true for = m + 1 (as all relevant quanttes are 0). We consder three cases dependng on how Double-ft assgns jobs from classes [1 : 1] to machnes [ : m]. 1. No jobs from class [1 : 1] were scheduled onto machnes [ : m]: In ths case, machnes 1 and are separated and (2) follows from Lemma Jobs from classes [1 : 1] are only scheduled onto machnes [ : m] durng the saturaton phase: Let denote the smallest ndex such that machnes 1 and + 1 are separated (f no such machne exsts, set = m). By the nductve hypothess, we can assume that (2) holds for + 1. In the case where = m, ths holds vacuously. As jobs from classes [1 : 1] are assgned to [ : m] (and hence to ) durng the saturaton phase, Lemma 7 mples that all jobs n classes [ : ] were also scheduled durng the saturaton phase, whch mples that all machnes [ : ] are loaded below 4F opt. Ths gves us the followng: B (k) 4F opt s + B +1(k) 4F opt s + = s + s, B opt (k) s, B opt +1 (k) + F opt s, B opt +1 (k) m + F opt s, + F opt +1 where the second nequalty follows from the nductve hypothess for machne Some job j from class [1 : 1] was scheduled onto machnes [ : m] durng Slow-ft phase (usng threshold 6F opt ): We assume that > 1, otherwse the result follows from case 1. Let mn < denote the largest ndex such that machnes [ mn : 1] have load more than 5F opt and machne mn 1 has load at most 5F opt. If no such machne exsts, set mn = 1. mn s well-defned as > 1 and machne 1 must have load more than 5F opt as job j from class [1 : 1] was assgned to a machne n [ : m] durng the Slow-ft phase. Clam 1. Machnes mn 1 and mn are separated or mn = 1. s s APPROX/RANDOM 15

8 92 Mnmzng Maxmum Flow-tme on Related Machnes Proof. Ths s trvally true f mn = 1. If mn > 1, suppose that some job j from class [1 : mn 1] was scheduled onto machnes [ mn : m]. Now j cannot be scheduled durng the Slow-ft phase as ths would mply that the load on mn 1 was more than 5F opt, whch contradcts the choce of mn. So all jobs n [1 : mn 1] that were assgned to [ mn : m] must have been assgned durng the saturaton phase. Let mn denote the largest ndex where such a job s assgned. By Lemma 7, t must be that all machnes [ mn : ] were assgned load durng the saturaton phase and must have load at most 4F opt. Ths contradcts that mn has load more than 5F opt. By Lemma 9 appled to mn, we get that (2) holds for machne mn and thus m m B mn (k) s, B opt mn s. (6) mn mn Furthermore, by choce of mn all the machnes n [ mn : 1] are loaded above 5F opt. Ths mples that B (k) B mn (k) 5F opt 1 mn s. (7) As every machne s loaded below 4F opt n an optmal schedule, we also have B opt 1 mn (k) B opt (k) + 4F opt s. (8) mn Addng (6) and (7) we obtan that B (k) = 4F opt 4F opt mn s, B opt mn (k) m whch mples that (2) holds for. 3.3 Proof of Lemma 6 s, B opt mn s, B opt s, B opt (k) We now prove Lemma 6, whch s relatvely easer. + F opt s 5F opt mn 1 s 4F opt s m + F opt s, mn s By (8) 1 mn s Proof of Lemma 6. We wll apply nducton over (n decreasng order of machnes). Consder epoch k. We assume that (2) holds for all at epoch k, and that (1) holds for all > at epoch k + 1. For the base case of = m + 1 the lemma follows trvally snce all the relevant quanttes are 0. Consder some machne. We consder two cases dependng on the load of machne after the jobs were scheduled at epoch k.

9 N. Bansal and B. Cloostermans Machne has load at most 4F opt after epoch k,.e., B (k) B +1 (k) 4F opt s : At epoch k+1 before the jobs arrvng durng nterval I k+1 are scheduled, the load of machne wll be at most F opt. Thus we have that A (k + 1) A +1 (k + 1) + F opt s A opt +1 (k + 1) + F opt A opt (k + 1) + F opt +1 s. s + F opt s Here the second nequalty follows by the nductve hypothess for machne + 1, and the thrd nequalty follows as A opt (k + 1) s non-decreasng as decreases. 2. Machne s loaded above 4F opt after epoch k,.e., B (k) B +1 (k) > 4F opt s : In ths case, some job j must have been scheduled onto machne durng the Slow-ft phase. Ths only happens f j could not be scheduled durng the saturaton phase. In partcular, ths mples that all the machnes [ : m] (whch s surely a subset of machnes where j could have been scheduled) were loaded above. So the total work on all machnes m [ : m] decreases by exactly s durng nterval I k+1. Thus we have that A (k + 1) = B (k) m Smlarly, as Opt can complete at most s nterval, we have A opt (k + 1) B opt (k) s. (9) on machnes [ : m] durng ths s. (10) As (2) holds for each at epoch k, we obtan that m A (k + 1) B (k) s B opt A opt s (k + 1) + F opt s, m s By (2) and hence (1) holds for at epoch k + 1, whch completes the proof. 3.4 Removng the assumpton of knowledge of Opt We descrbe a varant of the standard doublng trck where we ncrease the onlne estmate of Opt by only 1.5 tmes at each step. Consder some epoch k where the algorthm frst fals wth the current guess of F opt. It must be that (2) does not hold. In partcular, (1) holds at epoch k as (2) holds at k 1. Now, Lemma 5 mples that F opt < Opt. We then abort epoch k, and do not schedule any jobs. Instead, we set F opt = 1.5F opt and redefne the new epoch to be the tme (k 1)F opt + 3F opt. Note that between these epochs 4.5F opt tme passes, so at the next epoch the load on all machnes n the schedule of Double-ft wll be at most 6F opt 3F opt = 1.5F opt = F opt. APPROX/RANDOM 15

10 94 Mnmzng Maxmum Flow-tme on Related Machnes Ths mples that for all A (k) F opt m s A opt s. The crucal pont s that (1) holds for all machnes at ths new epoch rrespectve of the workload of the new restrcted Opt (wth parameter F opt). Thus, (2) holds f F opt Opt and Double-ft proceeds as normal. 4 Other Lower bounds We also show smple (but strong) lower bounds for weghted mum flow-tme and mum stretch. Lemma 10. Any algorthm for mnmzng mum weghted flow-tme on dentcal machnes must have a compettve rato of Ω(W ), where W s the rato between the largest and smallest weght. Proof. Consder the followng nstance on 2 machnes. At tme t = 0 we receve 2 jobs of sze w wth weght 1. Now, any algorthm has three optons: () t schedules both jobs on the same machne, () t schedules both jobs on dfferent machnes, or () t wats before assgnng jobs. In all three cases, we show that t wll end up tralng by at least w work behnd an optmal schedule. In opton (), w work remans at tme t = w, whle optmum s empty. In opton () we nstantly receve another 2w-szed job wth weght 1, so that one of our machnes has load 3w. At tme t = 2w we have w load remanng whle an optmal schedule s empty. In opton () we receve no jobs untl algorthm decde to choose () or (). If we have not chosen by tme t = w we are tralng 2w work behnd an optmal schedule. Once we tral w work behnd optmum, at every unt tme step we receve 2 unt-sze jobs of weght w. If the tralng jobs are ever to be fnshed, at least w/2 delay s ncurred on the weght w jobs, mplyng an objectve value of Ω(w 2 ). Opt on other hand has value O(w). A lower bound of Ω(W 0.4 ) for mum weghted flow-tme follows from [10], usng the analogy between delay factor and weghted mum flow-tme descrbed n [3]. By replacng the unt sze jobs by unt weght jobs n the above lower bound nstance, ths also drectly mples an Ω(S) lower bound on the compettve rato for mum stretch [3] where S s the rato between the sze of the largest and the smallest job. 5 Concludng Remarks Note that our algorthm Double-ft s not mmedate dspatch,.e., t does not dspatch a job to a machne mmedately upon arrval. We are unable to extend the deas here to obtan an O(1)-compettve mmedate dspatch algorthm, and t s not clear to us whether such an algorthm exsts. Gven that n the unrelated settng, there can be no O(1)-speed, O(1)-compettve mmedate dspatch algorthm [3] (whle there s a (1 + ɛ)-speed, O(1/ɛ)-compettve algorthm), t would be qute nterestng to resolve ths queston.

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