Competitive Analysis of Task Scheduling Algorithms on a Fault-Prone Machine and the Impact of Resource Augmentation

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1 Competitive Analyi of Tak Scheduling Algorithm on a Fault-Prone Machine and the Impact of Reource Augmentation Antonio Fernández Anta a, Chryi Georgiou b, Dariuz R. Kowalki c, Elli Zavou a,d,1 a Intitute IMDEA Network b Univerity of Cypru c Univerity of Liverpool d Univeridad Carlo III de Madrid Abtract Reliable tak execution in machine that are prone to unpredictable crahe and retart i challenging and of high importance. However, not much work exit on the wort cae analyi of uch ytem. In thi paper, we analyze the fault-tolerant propertie of four popular cheduling algorithm: Longet In Sytem (LIS), Shortet In Sytem (SIS), Larget Proceing Time (LPT) and Shortet Proceing Time (SPT), under wort cae cenario on a fault-prone machine. We ue three metric for the evaluation and comparion of their competitive performance, namely, completed time, pending time and latency. We alo invetigate the effect of reource augmentation in their performance, by increaing the peed of the machine. To do o, we compare the behavior of the algorithm for different peed interval and how that between LIS, SIS and SPT there i no clear winner with repect to all the three conidered metric, while LPT i not better than SPT. Keyword: Scheduling, Online Algorithm, Different Tak Proceing Time, Failure, Competitive Analyi, Reource Augmentation 1. Introduction Motivation. The demand for proceing computationally-intenive job ha been increaing dramatically during the lat decade, and o ha the reearch to face the many challenge it preent. In addition, with the preence of machine failure (and retart) thing get even wore, epecially when a maliciou entity i cauing them. In thi work, we apply peed augmentation [9, 1] in order to overcome uch failure, that i, we increae the computational power of the ytem machine. Thi i an alternative to increaing the number of proceing entitie, a done in multiproceor ytem. Hence, we conider a peedup 1, under which the machine perform a job time fater than the baeline execution time. More preciely, we conider a etting with a ingle machine prone to crahe and retart that are being controlled by an adverary, and a cheduler that aign injected job (or tak) to be executed by the machine. Thee tak arrive continuouly and have different computational demand and hence proceing time. Specifically we aume that each tak ha proceing time π [, π max ], where and π max are the mallet and larget value, repectively. Since the cheduling deciion mut be made continuouly and without knowledge of the future, neither the tak injection nor the machine crahe and retart, we look at the problem a an online cheduling problem [10, 4, 3, 1, 14]. The importance of uing peedup lie on thi online nature of the problem; the future failure, and the 1 Partially upported by FPU Grant from MECD

2 intant of arrival of future new tak and their proceing time are unpredictable. Thu, the need to overcome thi lack of information. We evaluate the performance of the different cheduling policie (online algorithm) under wortcae cenario, which guarantee efficient cheduling even in the wort of cae. For that, we perform competitive analyi [13]. The four cheduling policie we conider are Longet In Sytem (LIS), Shortet In Sytem (SIS), Larget Proceing Time (LPT) and Shortet Proceing Time (SPT). Achieving reliable and table computation in uch an environment withhold everal challenge. One of our main goal i therefore to confront thee challenge conidering the ue of the mallet poible peedup. However, our primary intention i to dicover the dependence of the efficiency meaure for each cheduling policy with repect to the peedup ued. Contribution. In thi paper we explore the behavior of ome of the mot widely ued algorithm in cheduling, analyzing their fault-tolerant propertie under wort-cae combination of tak injection and crah/retart pattern, a decribed above. A already mentioned, the four algorithm we conider are: (1) LIS: the tak that ha been waiting the longet i cheduled; i.e., it follow the FIFO (Firt In Firt Out) policy, () SIS: the tak that ha been injected the latet i cheduled; i.e., it follow the LIFO (Lat In Firt Out) policy, (3) LPT: the tak with the longet proceing time i cheduled, and (4) SPT: the tak with the hortet proceing time i cheduled. We focu on three evaluation metric, which we regard to embody the mot important quality of ervice parameter: the completed time, which i the aggregate proceing time of all the tak that have completed their execution uccefully, the pending time, which i the aggregate proceing time of all the tak that are in the queue waiting to be completed, and the latency, which i the larget time a tak pend in the ytem, from the time of it arrival until it i fully executed. They repreent the machine throughput, queue ize and delay repectively, all of which we conider eential. They how how efficient the cheduling algorithm are in a fault-prone etting from different angle: machine utilization (completed time), buffering (pending time) and fairne (latency). The performance of an algorithm ALG i evaluated under thee three metric by mean of competitive analyi, in which the value of the metric achieved by ALG when the machine ue peedup 1 i compared with the bet value achieved by any algorithm X running without peedup ( = 1) under the ame pattern of tak arrival and machine failure, at all time intant of an execution. Table 1 ummarize the reult we have obtained for the four algorithm. Although not clearly tated in the table, the firt reult we how in our work apply to any work conerving cheduling algorithm one that do not idle a long a there are pending tak and they do not break the execution of a tak unle the machine crahe. Thee reult are: (a) When ρ = πmax, the completed time competitive ratio i lower bounded by 1/ρ and the pending time competitive ratio i upper bounded by ρ. (b) When 1 + ρ, the completed time competitive ratio i lower bounded by 1 and the pending time competitive ratio i upper bounded by 1 (i.e., they are 1-competitive). Then, for pecific cae of peedup le than 1 + ρ we obtain better lower and upper bound for the different algorithm. However, it i clear that none of the algorithm i better than the ret. With the exception of SPT, no algorithm i competitive in any of the three metric conidered when < ρ. In particular, algorithm SPT i competitive in term of completed time, except in the cae when tak may have any arbitrary proceing time in the range [, π max ] and the machine ha no peedup ( = 1). In term of latency, only algorithm LIS i competitive, when ρ, which might not be very urpriing ince algorithm LIS give priority to the tak that have been waiting the longet in the ytem. Another intereting obervation i that algorithm LPT and SPT become 1-competitive a oon a ρ, both in term of completed and pending time, wherea LIS and SIS require greater peedup to achieve thi. What i more, thi i the firt thorough and rigorou online analyi of popular cheduling algorithm in a faultprone etting. In ome ene, collectively our reult demontrate in a clear way the difference between two clae of policie: the one that give priority baed on the arrival time of the tak in the ytem (LIS and SIS) and the one

3 LIS SIS LPT SPT Condition Completed Time, C Pending Time, P Latency, L < ρ 0 [ρ, 1 + 1/ρ) [ 1 ρ, ρ ] [ 1+ρ, ρ] (0, 1] [max{ρ, 1 + 1/ρ}, ) [ 1 ρ, 1 + πmin π ] [ 1 + π, ρ] (0, 1] max{ρ, } [1, ] 1 (0, 1] < ρ 0 [ρ, 1 + ρ) [ 1 ρ, πmin π ] [ π /+ρ 1+ρ, ρ] 1 + ρ [1, ] 1 < ρ 0 ρ [1, ] 1 < ρ [ 1 +ρ, ˆρ ˆρ+ρ ] (*) ρ [1, ] 1 Table 1: Metric comparion of the four cheduling algorithm. Recall that repreent the peedup of the ytem machine, π max and the larget and mallet tak proceing time repectively. Note that ρ = πmax, ˆρ = ρ 1 and π, π (, π max) are tak proceing time uch that π < and 1 π < +π max. Note alo that by definition, 0-completed-time competitivene ratio equal to non-competitivene, a oppoed to the other two metric where non-competitivene correpond to an competitivene ratio. Finally note that the lower bound of reult (*) i for injection pattern with tak of only two proceing time and π max. that give priority baed on the required proceing time of the tak (LPT and SPT). Oberve that different algorithm cale differently with repect to the peedup, in the ene that with the increae of the machine peed the competitive performance of each algorithm change in a different way. Related Work. We relate our work with the online verion of the bin packing problem [15], where the object to be packed are the tak and the bin are the time period between two conecutive failure of the machine (i.e., alive interval). Wide reearch ha taken place over the year around thi problem, ome of which we conider related to our. For example, Johnon et al. [8] analyzed the wort cae performance of two imple algorithm (Bet Fit and Next Fit) for the bin packing problem, giving upper bound on the number of bin needed (correponding to the completed time in our work). Eptein et al. [6] (ee alo [15]) conidered online bin packing with reource augmentation in the ize of the bin (correponding to the length of alive interval in our work). Oberve that the eential difference of the online bin packing problem with the one that we are looking at in thi work, i that in our ytem the bin and their ize (correponding to the machine alive interval) are unknown. On a different tone, Boyar and Ellen [5] have looked into a problem imilar to both online bin packing problem and our, conidering job cheduling in the grid. The main difference with our etting i that they conider everal machine (or proceor), but mainly the fact that the arriving item are proceor with limited memory capacitie and there i a fixed amount of job in the ytem that mut be completed. They alo ue fixed job ize and achieve lower and upper bound that only depend on the fraction of uch job in the ytem. Another related problem i packet cheduling in a link. Andrew and Zhang [] conider online packet cheduling over a wirele channel whoe rate varie dynamically, and perform wort cae analyi regarding both the channel condition and the packet arrival. A imilar work i [3], where packet of two different ize were cheduled through an unreliable link. In that work, the goodne metric i the long-term competitive ratio, which i called relative throughput and online algorithm a well a bound for any online cheduling protocol are given. We can alo directly relate our work with reearch done on machine cheduling with availability contraint (e.g., [11, 7]). One of the mot important reult in that area i the neceity of online algorithm in cae of unex- 3

4 pected machine breakdown. However, in mot related work preemptive cheduling i conidered and optimality i hown only for nearly online algorithm (need to know the time of the next job or machine availability). In a previou work [4], we looked at a ytem of multiple machine and at leat two different tak cot (i.e., proceing time in our work). We applied ditributed cheduling and performed wort-cae competitive analyi, conidering the pending cot competitivene (correponding pending time competitivene in the preent work) a our main evaluation metric. We howed the NP-hardne of the offline verion of the problem and uggeted the ue of peedup in order to achieve competitivene. We defined ρ = πmax and proved that if both condition (a) < ρ and (b) < 1 + γ/ρ hold for the ytem machine (γ i ome contant that depend on and π max ), then no determinitic algorithm i competitive with repect to the queue ize (pending cot). Additionally, we propoed online algorithm to how that relaxing any of the two condition i ufficient to achieve competitivene. In fact, [4] motivated thi paper, ince it made evident the need of a thorough tudy of imple algorithm even under the implet baic model of one machine and cheduler.. Model and Definition Computing Setting. We conider a ytem of one machine prone to crahe and retart with a Scheduler reponible for the tak aignment to the machine following ome algorithm. The client ubmit job (or tak) of different ize (proceing time) to the cheduler, which in it turn aign them to be executed by the machine. Tak. Tak are injected to the cheduler by the client of the ytem, an operation which i controlled by an arrival pattern A (a equence of tak injection). Each tak τ ha an arrival time a(τ) and a proceing time π(τ), being the time it require to be completed by a machine running with = 1. We ue the term π-tak to refer to a tak of proceing time π [, π max ] throughout the paper. We alo aume tak to be atomic with repect to their completion; in other word preemption i not allowed (tak mut be fully executed without interruption). Machine failure. The crahe and retart of the machine are controlled by an error pattern E, which we aume to coordinate with the arrival pattern in order to give the wort-cae cenario. We conider that the tak being executed at the time of the machine failure i not completed, and it i therefore till pending in the cheduler until it i eventually re-cheduled (it i not dicarded). The machine i active in the time interval [t, t ], if it i executing ome tak at time t and ha not crahed by time t. Hence, an error pattern E can be een a a equence of active interval of the machine. Reource augmentation / Speedup. We alo conider a form of reource augmentation by peeding up the machine and the goal i to keep it a low a poible. A mentioned earlier, we denote the peedup with 1. Notation. Let u denote here ome notation that will be extenively ued throughout the paper. Becaue it i eential to clarify the tak being accounted at each timepoint in an execution, we introduce et I t (A), N t (X, A, E) and Q t(x, A, E), where X i an algorithm, A and E the arrival and error pattern repectively, t the time intant we are looking at and the peedup of the machine. I t (A) repreent the et of injected tak within the interval [0, t], N t (X, A, E) the et of completed tak within [0, t] and Q t (X, A, E) the et of pending tak at time intant t. Q t(x, A, E) contain the tak that were injected by time t incluively, but not the one completed before and up to time t. Oberve that I t (A) = N t (X, A, E) Q t (X, A, E) and note that et I depend only on the arrival pattern A, while et N and Q alo depend on the error pattern E, the algorithm run by the cheduler, X, and the peedup of the machine,. Note that the upercipt i omitted in further ection of the paper for implicity. However, the appropriate peedup in each cae i clearly tated. Efficiency Meaure. Conidering an algorithm ALG running with peedup under arrival and error pattern A and E, we look at the current time t and focu on three meaure; the Completed Time, which i the um of proceing time of the completed tak 4

5 C t (ALG, A, E) = τ N t (ALG,A,E) π(τ), the Pending Time, which i the um of proceing time of the pending tak Pt (ALG, A, E) = τ Q t (ALG,A,E) π(τ), and the Latency, which i the maximum amount of time a tak ha pent in the ytem { } L f(τ) a(τ), τ Nt (ALG, A, E) t(alg, A, E) = max t a(τ), τ Q, t(alg, A, E) where f(τ) i the time of completion of tak τ. Computing the chedule (and hence finding the algorithm) that minimize or maximize correpondingly the meaure Ct (X, A, E), Pt (X, A, E), and L t(x, A, E) offline (having the knowledge of the pattern A and E), i an NP-hard problem [4]. Due to the dynamicity of the tak arrival and machine failure, we view the cheduling of tak a an online problem and purue competitive analyi uing the three metric. Note that for each metric, we conider any time t of an execution, combination of arrival and error pattern A and E, and any algorithm X deigned to olve the cheduling problem: An algorithm ALG running with peedup, i conidered α-completed-time-competitive if Ct (ALG, A, E) α Ct 1 (X, A, E) + C, for ome parameter C that doe not depend on t, X, A or E; α i the completed-time competitive ratio of ALG, which we denote by C(ALG). Similarly, it i conidered α-pending-time-competitive if Pt (ALG, A, E) α Pt 1 (X, A, E) + P, for parameter P which doe not depend on t, X, A or E. In thi cae, α i the pending-time competitive ratio of ALG, which we denote by P(ALG). It i alo conidered α-latency-competitive if L t(alg, A, E) α L 1 t (X, A, E) + L, where L i a parameter independent of t, X, A and E. In thi cae, α i the latency competitive ratio of ALG, which we denote by L(ALG). Finally, note that α, i independent of t, X, A and E, for the three metric accordingly. Both completed and pending time meaure are important. Oberve that they are not complementary of one another. An algorithm may be completed-time-competitive but not pending-time-competitive, even though the um of proceing time of the uccefully completed tak complement the um of proceing time of the pending one. For example, think of an online algorithm that manage to complete uccefully half of the total injected tak proceing time up to any point in any execution. Thi give a completed time competitivene ratio C(ALG) = 1/. However, it i not necearily pending-time-competitive ince in an execution with infinite tak arrival it total pending time increae unboundedly and there might exit an algorithm X that manage to keep it total pending time contant under the ame arrival and error pattern. Thi i further demontrated by our reult ummarized in Table Propertie of Work Conerving Algorithm In thi ection we preent ome general propertie for all online work-conerving algorithm running with peedup. They alo apply to the four policie we focu on in the ret of the paper. Obervation 1. Any work-conerving algorithm ALG running with peedup, ha a completed-time competitive ratio C(ALG) 1 when we allow Q t (ALG) = infinitely many time, or C(ALG) when the queue never become empty after a point in time. Parameter C, P, L a well a α may depend on ytem parameter like, π max or, which are not conidered a input of the problem. 5

6 Proof: Let u firt conider the cae at which the adverary allow the queue of pending tak of ALG to become empty infinitely many time in an execution. In particular, let u conider the arrival and error pattern A and E, uch that there are time intant t k, k = 0, 1,... where t k = t k 1 + π and t 0 = 0. At each t k there i a machine failure (crah and retart) and exactly one π-tak (π [, π max ]) injected. We name time interval T i = [t i, t i+1 ]. Oberve that an algorithm X (running with = 1) complete π-tak injected at t i in interval T i, while any work-conerving algorithm ALG running with peedup will complete the ame tak at time t i + π/ < t i+1 reulting in an empty queue. Hence, C(ALG) = 1 a claimed. Let u now conider the cae at which the adverary never let the queue of ALG to become empty after ome point in time. More preciely we conider the arrival and error pattern A and E to be uch that ALG alway ha at leat one pending tak of any proceing time π [, π max ] available to chedule. Therefore, in a period of time π an algorithm X X (running with = 1) i able to complete a π-tak while ALG may complete up to π proceing time. Thi will reult in a completed time competitive ratio C(ALG) a claimed. Obervation. Any work-conerving algorithm ALG running with peedup, ha a pending-time competitive ratio P(ALG) 1 when it queue of pending tak never become empty after a point in time. Proof: Let u conider arrival and error pattern A and E uch that algorithm ALG alway ha at leat one pending tak of any proceing time π [, π max ] available to chedule. We conider phae of arbitrarily choen length π, defined a interval T i = [t k, t k+1 ] where t k+1 = t k + π, t 0 = 0 and k = 0, 1,... being intant of machine failure. A a reult, in a phae of length π an algorithm X X will be able to complete a π-tak, while ALG will complete up to π proceing time. Auming that there are no phae of length le than, the complementing pending time at a time t k will therefore be P tk (X) I tk (A) t k and P tk (ALG) I tk (A) t k. The pending time competitive ratio become P(ALG) I(A) t I(A) t, which yield to P(ALG) 1, ince we can make I(A) infinitely big. Lemma 1. No algorithm X (running without peedup) complete more tak than a work conerving algorithm ALG running with peedup ρ. Formally, t, A A and E E, N t (ALG, A, E) N t (X, A, E), and hence Q t (ALG, A, E) Q t (X, A, E). Proof: We will prove that t, A A and E E, Q t (ALG, A, E) Q t (X, A, E), which implie that N t (ALG, A, E) N t (X, A, E). Oberve that the claim trivially hold for t = 0. We now ue induction on t to prove the general cae. Conider any time t > 0 and the cloet time t < t uch that t i either a failure/retart time point or a point where ALG pending queue i empty. Let u aume by induction hypothei that Q t (ALG) Q t (X). Let i T be the number of tak injected in the interval T = (t, t]. Since ALG i work conerving, it i continuouly executing tak in the interval T. Alo, ALG need at mot π max / time to execute any tak uing peedup ρ, regardle of the tak being executed. Then it hold that t t t t Q t (ALG) Q t (ALG) + i T Q t (ALG) + i T. π max / On the other hand, X can complete at mot one tak every time. Hence, t t Q t (X) Q t (X) + i T. A a reult, we have that 6

7 t t t t Q t (X) Q t (ALG) Q t (X) + i T Q t (ALG) i T + 0. Since thi hold for all time t, the claim follow. Theorem 1. Any work-conerving algorithm ALG running with peedup ρ ha completed-time competitive ratio C(ALG) 1/ρ and pending-time competitive ratio P(ALG) ρ. Proof: It follow directly from Lemma 1, ince for any algorithm X the proceing time of every tak completed by ALG i at leat πmin π max time the proceing time of every tak completed by X. From Lemma 1, ALG alway ha at leat a many completed tak a X, and hence it completed-time competitive ratio C(ALG) i at leat πmin π max = 1/ρ. Complementary, ince the proceing time of any pending tak in the queue of ALG i at mot πmax time bigger than the proceing time of any pending tak in X, it pending-time competitive ratio P(ALG) i at mot πmax = ρ. Theorem. Any work-conerving algorithm ALG running with peedup 1 + ρ, ha completed-time competitive ratio C(ALG) 1 and pending-time competitive ratio P(ALG) 1. Proof: Conider an execution of any work-conerving algorithm ALG running with peedup 1 + ρ under any arrival and error pattern A and E, a well a an algorithm X. Then, looking at any time t of an execution, we define time intant t < t to be the latet time before t at which one of the following event happen: (1) an active period tart (after a machine crah/retart), () algorithm X ha uccefully completed a tak, or (3) the queue of pending tak of ALG i empty, Q t (ALG) =. It i trivial that P 0 (ALG, A, E) P 0 (X, A, E) hold at the beginning of the execution. Now auming that P t (ALG, A, E) P t (X, A, E) hold at time t, we prove by induction that P t (ALG, A, E) P t (X, A, E) till hold at time t. Thi alo mean that the tak uccefully completed by ALG by time t have at leat the ame total proceing time a the one completed by X. Conidering the interval T = (t, t], there are two cae: Cae 1: X i not able to complete any tak in the interval T. Then, it hold that P t (X, A, E) = P t (X, A, E) + i T, where i T denote the proceing time of the tak injected during the interval T. Similarly, it hold that P t (ALG, A, E) P t (ALG, A, E) + i T even if ALG i not able to complete uccefully any tak in T. Cae : X complete uccefully a tak in the interval T. Note that by definition of time t, during interval T there can only be one tak completed by X, and it mut be completed at time t. (If that were not the cae, t would not be well defined.) There are two ubcae. (a) Firt, t i from cae (3) of it definition. Hence, Q t (ALG) = and P t (ALG, A, E) i T. At time t algorithm X wa executing the tak that wa completed at time t. Hence, the tak wa injected before t, and X ha not completed any of the tak injected in T. Then, P t (X, A, E) i T P t (ALG, A, E). (b) Second, t i from cae (1) or () of it definition. Then, the interval T ha length π, which i the proceing time of the tak completed by X. In that interval ALG i continuouly executing tak. Hence, in the interval (t, t] if complete tak whoe aggregate proceing time i at leat π π max. Then, the pending time at time intant t of both algorithm atify P t (X, A, E) = P t (X, A, E) + i T π while P t (ALG, A, E) < P t (ALG, A, E) + i T (π π max ). Oberve that 1 + ρ implie that π π π max. Hence, from the induction hypothei, P t (ALG, A, E) P t (X, A, E). Thi implie a completed-time competitive ratio C(ALG) 1 and a pending-time competitive ratio P(ALG) 1, a claimed. 7

8 4. Completed and Pending Time Competitivene In thi ection we preent a detailed analyi of the four algorithm with repect to the completed and pending time metric, firt for peedup < ρ and then for peedup ρ Speedup < ρ Let u tart with ome negative reult, whoe proof involve pecifying the combination of arrival and error pattern that force the claimed bad performance of the algorithm. We alo give ome poitive reult for SPT, the only algorithm that can achieve a non-zero completed-time competitivene under ome cae. Lemma. When algorithm LIS and LPT run with peedup < ρ, they both have a completed-time competitive ratio C(LIS) = C(LPT) = 0 and a pending-time competitive ratio P(LIS) = P(LPT) =. Proof: Let u ue the ame combination of algorithm X, arrival and error pattern A and E for the proof of noncompetitivene for the two algorithm. We conider an infinite arrival pattern which inject one π max -tak at the beginning of the execution, t = 0, and after that it keep injecting one -tak every time. Conider alo an infinite error pattern that et the machine failure point (crah immediately followed by a retart) at time intant t i = i, where i = 1,,.... It can be eaily een, that an algorithm X running with no peedup ( = 1), will be able to complete the -tak injected, while neither LIS nor LPT will manage to complete any tak, ince they will both init on cheduling the π max -tak injected at the beginning. In an interval of length, algorithm X i able to complete a -tak but neither LIS nor LPT can complete the π max -tak ince it need time πmax >. Thi mean that the number of pending tak in the queue of both LIS and LPT will be continuouly increaing with time, while X i able to keep them bounded, with no more than one π max and one tak, and o will their pending proceing time. A for the total proceing time of completed tak, C(LIS, A, E) = C(LPT, A, E) = 0 at all time of the execution, while the one of X grow to infinite a t goe to infinity. Hence, for peedup < ρ, algorithm LIS and LPT have completed-time competitive ratio C(LIS) = C(LPT) = 0 and pending-time competitive ratio P(LIS) = P(LPT) = a claimed, which complete the proof. Lemma 3. When algorithm SIS run with peedup < ρ, it ha a completed-time competitive ratio C(SIS) = 0 and a pending-time competitive ratio P(SIS) =. Proof: Let u divide the proof in two part giving different combination of arrival and error pattern for the completed time and the pending time repectively. We firt conider a combination of arrival and error pattern A and E that behave a follow. We define time intant t k where k = 1,,... and t i = t i 1 + with time t 0 = 0 being the beginning of the execution. At every uch time intant there i a crah and retart at the machine and then an immediate injection of a -tak followed by a π max -tak. Thi create active interval [t i, t i+1 ) of length. It i eay to oberve that the pattern decribed caue algorithm SIS to aign the lat π max -tak injected, every time it ha to make a cheduling deciion, ince it i the lat tak injected. Since the alive interval are of length and SIS need πmax > time to complete the π max -tak, it i not able to complete any of the tak it tart executing, giving C t (SIS) = 0 at all time t (and in particular at t k time intant). On the ame time, an algorithm X i able to chedule and complete all the -tak injected, one in every alive interval, giving a completed time of C tk (X) = k at every t k time intant. 8

9 Now let u conider another combination of arrival and error pattern, A and E repectively a well a algorithm X. We define time intant t where = 1,,... and t = t k 1 + ( + π max ) with time t 0 = 0 being the beginning of the execution. At every uch time intant there i a -tak injected followed by a π max -tak. The crahe of the machine are et every time t and t +, creating active interval of length or π max alternatively. Thi behavior caue algorithm SIS to chedule the lat π max -tak injected, after every t time intant, ince it i the lat tak injected. The firt alive interval after a t time intant i of length and thu the π max -tak cheduled by SIS cannot be completed. On the ame time, an algorithm X i able to complete the lat -tak injected. On the next alive period though, SIS will be able to complete uccefully the π max -tak ince it i of length π max, while X will do the ame. A a reult, looking at time intant t, before the injection take place, P t (SIS) = while P t (X) = 0, ince algorithm SIS will never complete any -tak injected but X will be able to complete all tak injected up to that time. Therefore, for peedup < ρ algorithm SIS ha completed-time competitive ratio C(SIS) = 0 and pending-time competitive ratio P(SIS) = a claimed. Combining now Lemma and 3 we have the following theorem. Theorem 3. NONE of the three algorithm LIS, LPT and SIS i competitive when peedup < ρ, with repect to completed or pending time, even in the cae of only two tak proceing time (i.e., and π max ). Theorem 4. If tak can be of only two proceing time ( and π max ), algorithm SPT can achieve a completedtime competitive ratio C(SPT) 1 +ρ, for any peedup 1. In particular, C t(spt) 1 +ρ C t(x) π max, for any time t. Proof: Let u aume fixed arrival and error pattern A and E repectively, a well a an algorithm X, and let u look at any time t in the execution of SPT. Let τ be a tak completed by X by time t (i.e., τ N t (X)), where t τ i the time τ wa cheduled and f(τ) t the time it completed it execution. We aociate τ with the following tak in N t (SPT): The ame tak τ. The tak w being executed by SPT at time t τ, if it wa not later interrupted by a crah. Not every tak in N t (X) i aociated to ome tak in N t (SPT), but we how now that mot tak are. In fact, we how that the aggregate proceing time of the tak in N t (X) that are not aociated with any tak in N t (SPT) i at mot π max. More pecifically, there i only one tak execution of a π max -tak, namely w, by SPT uch that the -tak cheduled and completed by X concurrently with the execution of w fall in thi cla. Conidering the generic tak τ N t (X) from above, we conider the following cae: If τ N t (SPT) then tak τ i aociated at leat with itelf in the execution of SPT, regardle of τ proceing time. If τ / N t (SPT), τ i in the queue of SPT at time t τ. By it greedy nature, SPT i executing ome tak w at time t τ. If π(τ) π(w), then tak w will complete by time f(τ) and hence it i aociated with τ. If π(τ) < π(w) (i.e., π(τ) = and π(w) = π max ), then τ wa injected after w wa cheduled by SPT. If thi execution of tak w i completed by time t, then tak w i aociated with τ. Otherwie, tak τ i not aociated to any tak in N t (SPT) if a crah occur or the time t i reached before w i completed. 9

10 Let t the time either event occur. Since τ / N t (SPT), τ i not completed by SPT in the interval [t, t]. Then, SPT never chedule a π max -tak in the interval [t, t], and the cae that a tak from N t (X) i not aociated to any tak in N t (SPT) cannot occur again in that interval. Hence, all the tak τ N t (X) that are not aociated to tak in N t (SPT) are -tak and have been cheduled and completed during the execution of the ame π max -tak by SPT. Hence, their aggregate proceing time i at mot π max. Now let u evaluate the proceing time of the tak in N t (X) aociated to a tak in w N t (SPT). Let u conider any tak w uccefully completed by SPT at a time f(w) t. Tak w can be aociated at mot with itelf and all the tak that X cheduled within the interval T w = [f(w) π(w), f(w)]. The latter et can include tak whoe aggregate proceing time i at mot π(w) + π max, ince the firt uch tak tart it execution no earlier than f(w) π(w) and in the extreme cae a π max -tak could have been cheduled af the end of T w and completed at t w +π max. Hence, if tak w i a -tak, it will be aociated with tak completed by X that have a total proceing time of at mot + π max, and if w i a π max -tak, it will be aociated with tak completed by X that have a π total proceing time of at mot 3π max. Oberve that min +π max < πmax 3π max. A a reult, we can conclude that C t (SPT) C t (X) π max = 1 + π max + ρ C t(x) π max. Conjecture 1. The above lower bound on completed time, till hold in the cae of any contant number of tak proceing time in the range [, π max ]. Theorem 5. For peedup < ρ, algorithm SPT cannot have a completed-time competitive ratio more than x for x < ˆρ ˆρ+ρ. In other word, C(SPT) ˆρ ˆρ+ρ. Additionally, it i NOT competitive with repect to the pending time, i.e., P(SPT) =. Proof: Auming peedup < ρ we conider the following combination of arrival and error pattern A and E repectively: We define time point t k, where k = 0, 1,..., uch that t 0 i the beginning of the execution and t k = t k 1 + π max + ˆρ (recall that ˆρ = ρ 1 < ρ). At every t k time intant there are ˆρ tak of proceing time injected along with one π max -tak. What i more, the crah and retart of the ytem machine are et at time t k + π max and then after every time until t k+1 i reached. By the arrival and error pattern decribed, every epoch; time interval [t k, t k+1 ], reult in the ame behavior. Algorithm SPT i able to complete only the ˆρ tak of proceing time, while X i able to complete all tak that have been injected at the beginning of the epoch. From the nature of SPT, it chedule firt the mallet tak, and therefore the π max one never have the time to be executed; a π max -tak i cheduled at the lat phae of each epoch which i of ize (recall < ρ < ˆρ). Hence, at time t k, C tk (SPT, A, E) = kˆρ and C tk (X, A, E) = kˆρ + kπ max. Looking at the pending time at uch point, we can eaily ee that SPT i contantly increaing, while X i able to have pending time zero; P tk (SPT, A, E) = kπ max but P tk (X, A, E) = 0. A a reult, we have a maximum completed-time competitive ratio C(SPT) pending time P(SPT) =. k ˆρ k ˆρ+kπ max = ˆρ ˆρ+ρ and a Theorem 6. If tak can have any proceing time in the range [, π max ], and there i no peedup ( = 1), Algorithm SPT i NOT competitive with repect to completed time, i.e., C(SPT) = 0. 10

11 Proof: Auming = 1, we conider the following cenario a a reult of adverarial arrival and error pattern A and E repectively. Let u fix ome ɛ (0, 1) and ue the notation (k) = (π max )ɛ k. Then, let w k be a tak with proceing time π(w k ) = + (k), for all k = 0, 1,, 3,.... Oberve that k, π(w k ) (, π max ] and π(w k+1 ) < π(w k ). Let u alo define time point t k, uch that t 0 = 0 (the beginning of the execution) and t k+1 = t k ɛ (k). The arrival pattern A i uch (k). Let u alo define time point t k = t k ɛ that tak w 0 = π max i injected in the ytem at time intant t 0. Then, for k = 1,,... tak w k i injected at time t k. The error pattern E i uch that at every time intant t k there i a crah and retart. We compare SPT with an algorithm X of our choice. In the execution of SPT, tak w 0 i cheduled a oon a it arrive, at time t 0. On the other hand, X wait until time t 1 for the arrival of w 1 and chedule it immediately. When the proceor crahe at time t 1 the tak w 0 executed by SPT i interrupted, ince t 1 t 0 = + 1+ɛ (0) < π(w 0 ) = + (0). However, X i able to complete tak w 1 becaue t 1 + π(w 1 ) = 1 ɛ (1) + + (1) = + 1+ɛ (0) = t 1. After the retart at t 1 SPT chedule tak w 1, while X wait until t to chedule w. The general proce i a follow. At time intant t k, SPT chedule tak w k while X wait until tak w k+1 i injected at time t k+1 and chedule it. When the proceor crahe at time t k+1 the tak w k of SPT i interrupted, ince t k+1 t k = + 1+ɛ (k) < π(w k) = + (k). However, X i able to complete tak w k+1 becaue t k+1 + π(w k+1) = 1 ɛ (k + 1) + + (k + 1) = + 1+ɛ (k) = t k+1. Letting thi adverarial behavior run to infinity we ee that at any point in time t, C t (SPT) = 0, while X will keep completing the injected tak. Thi, reult to a completed-time competitive ratio C(SPT) = Speedup ρ Firt, recall that in Theorem 1 we have hown that any work conerving algorithm running with peedup ρ ha pending-time competitive ratio at mot ρ and completed-time competitive ratio at leat 1/ρ. So do the four algorithm LIS, LPT, SIS and SPT. A natural quetion that rie i whether we can improve thee ratio. Let u tart from ome negative reult again, focuing at firt on the two policie that chedule tak according to their arrival time, algorithm LIS and SIS. Lemma 4. When algorithm LIS run with peedup [ρ, 1+1/ρ), it ha a completed-time competitive ratio C(LIS) ρ ρ and a pending-time competitive ratio P(LIS). Proof: Let peedup [ρ, 1 + 1/ρ). We define a combination of arrival and error pattern A and E, and algorithm X. Pattern A and E behave a follow: Initially, there i a -tak injected, followed by a π max -tak. After every period of π max time the ame injection equence i repeated, when alo the machine i crahed and retarted. Thi behavior reult to the following execution. There are only active phae of ize π max, during which an algorithm X can uccefully execute the π max tak injected, while LIS i forced to chedule the tak in the order they arrive. Oberve that, ince < 1 + 1/ρ = ( + π max )/π max, LIS i able to complete only one tak in each phae. Oberve alo, that after k phae, where k i an multiple of, there will be exactly k tak of ize pending in the queue of X, while LIS will have pending half of the tak injected, half of which are of proceing time and the other half π max. Hence, the pending-time competitive ratio of the algorithm become P(LIS) = πmin+πmax = 1+ρ and the completed-time competitive ratio C(LIS) = πmin+πmax π max = ρ, which complete the proof. Lemma 5. When algorithm LIS run with peedup [1+1/ρ, ), it ha a completed-time competitive ratio C(LIS) 1 + πmin π and a pending-time competitive ratio P(LIS) 1 + π for any proceing time π (, π max ) uch that π < πmin 1. 11

12 Proof: Let peedup [1 + 1/ρ, ). We define a combination of arrival and error pattern A and E, and algorithm X to behave a follow: We define time intant t k, where k = 0, 1,,... and t 0 = 0 being the beginning of the execution and t k = t k 1 + π, where π (, π max ) and i uch that πmin+π > π π < πmin 1. At each t k time intant there i a machine crah and retart followed by an injection of a -tak and then a tak of proceing time π. Thi behavior reult to the following execution. All phae are of ize π, during which algorithm X complete uccefully the π-tak injected at the beginning of the phae, while LIS i able to complete either a -tak or a π-tak. Algorithm LIS follow the order of tak arrival to chedule the tak. However, by the definition of proceing time π, in a period of length π LIS cannot complete both a and a π-tak. Oberve that at every time intant t k where k i a multiple of, LIS will be able to complete k/ tak of proceing time and k/ tak of proceing time π while X will complete k tak of proceing time π. Repectively, at uch time intant P tk (LIS) = k(πmin+π) while P tk (X) = k. Thi lead to a completed-time competitive ratio C(LIS) = 1 + πmin π while time goe to infinity, and a pending-time competitive ratio P(LIS) = 1 + π a claimed. Lemma 6. When algorithm LIS run with peedup [, 1+ρ), it ha a completed-time competitive ratio C(LIS) 1 and a pending-time competitive ratio P(LIS) 1. Proof: Let peedup [, 1 + ρ) and let u analyze firt the completed time metric. Let t be the firt time in an execution, at which by mean of contradiction, C t (LIS) < C t (X) 3πmax hold. Alo, let time t < t be the latet time intance uch that for every t [t, t ], C t (LIS) < C t (X) hold. Note that thi implie that the queue of pending tak of LIS i never empty within the interval [t, t ]. What i more, both intant t and t are time at which algorithm X complete a tak. By definition of t, it alo hold that C t (LIS) C t (X) π max. We then break the interval [t, t ] into conecutive period [t, t 1 ] and (t i 1, t i ] for i =, 3..., k, called period i. Time intance t k = t, and the ret of t i are the proceor crahing point within the interval. Let u denote by C i (X) and C i (LIS) the proceing time completed in period i by X and LIS repectively. We dicard the period in which C i (LIS) = 0 ince C i (X) = 0 will hold a well. After dicarding thee period we renumber the ret in equence from 1 to. In order to prove the theorem, we need to how that the total completed time by X within the interval [t, t ] i larger than the total completed time by LIS within the ame interval by at leat an additive term of 3πmax π max. If in a period j, algorithm LIS complete more tak proceing time than X, it mut be the cae that j 1 C i(x) j 1 C i(lis) > C j (LIS) C j (X), otherwie time t i not well defined. Ele if in a period j < algorithm X complete more than LIS, i.e., C j (X) > C j (LIS), (1) then the following hold, C j (LIS) + π(τ j+1 ) > C j (X) C j (X) π(τ j+1 ) < C j (LIS), () where τ j+1 i the lat tak intended for execution by LIS in period j but i not completed, it end at the head of the queue of LIS at the end of period j. Hence it will be the firt one to be completed in the next period. Therefore j [, ], C j (LIS) π(τ j ). (3) From equation 1 and, we have that C j (X) > C j (LIS) > C j (X) π(τ j+1 ). Since, it implie ( 1) C j (X) < π(τ j+1 ) C j (X) < π(τ j+1 ). 1

13 What i more, from equation 1 and 3 we have that C j (X) > π(τ j ) and hence the following order of relationhip hold π(τ j ) C j (LIS) < C j (X) < π(τ j+1 ) C j+1 (LIS). Combining thi with equation : C j (X) C j (LIS) < π(τ j+1 ) C i (X) C i (LIS) < π(τ i+1 ) = k +1 i= π(τ i ) C i (X) C i (LIS) < C i (LIS) + π(τ +1) i= C i (X) < C i (LIS) C 1 (LIS) + π(τ k +1) C i (X) < C i (LIS) + π(τ +1) C 1 (LIS) C i (X) < C i (LIS) + π max. Combining thi with the fact that C t (LIS) C t (X) π max, we have that C t (X) = C t (X) + C i (X) < C t (LIS) + π max + C i (LIS) + π max = C t (LIS) + π max + π max C t (LIS) + 3π max, which contradict the initial claim and the definition of time t. Note that again, the lat inequality follow from the fact that peedup. Hence, even if algorithm X manage to complete more tak proceing time in ome period, LIS will eventually urpa it performance. Since the pending time i complementary to the completed time we can claim the following: C t (LIS) C t (X) 3π max I t C t (LIS) I t C t (X) + 3π max P t (LIS) P t (X) + 3π max. which complete the proof for both completed-time and pending-time competitive ratio being optimal for algorithm LIS when peedup [, 1 + ρ). Combining Lemma 4, 5, 6 and Theorem we have the following theorem. 13

14 Theorem { 7. Algorithm LIS ha a completed-time competitive ratio 1 C(LIS) + 1 ρ [ρ, 1 + 1/ρ) 1 +, and C(LIS) 1 when max{ρ, }. πmin π [1 + 1/ρ, ) It alo ha{ a pending-time competitive ratio 1+ρ P(LIS) [ρ, 1 + 1/ρ) 1 + π, and P(LIS) 1 when max{ρ, }. [1 + 1/ρ, ) Proceing time π (, π max ) i uch that π < πmin 1. Alo, combining Lemma 6 and Theorem, yield a completed-time competitive ratio C(LIS) 1 and a pendingtime competitive ratio P(LIS) 1, when peedup max{ρ, }. Recall that ρ 1 which mean that 1 + ρ. Theorem 8. When algorithm SIS run with peedup [ρ, 1+ρ), it ha a completed-time competitive ratio C(SIS) π and a pending-time competitive ratio P(SIS) π+πmax +π max, where π < πmin+πmax and π (, π max ). Proof: Let peedup [ρ, 1 + ρ). We define a combination of arrival and error pattern A and E, algorithm X and conider tak of proceing time π max, and π, where π (, π max ), uch that π < πmin+πmax. Note that, uch a value π alway exit ince < 1 + ρ. Pattern A and E behave a follow: We define time intant t k, where k i an increaing poitive integer (k = 0, 1,,... ), with t = 0 being the beginning of the execution and t k = t k 1 + π. At time t k there i exactly one π-tak injected, followed by one π max -tak, followed by one -tak. Crahe and retart are alo et at time t k, cauing active interval of π duration. Thi behavior reult to execution where an algorithm X i able to complete the lat π-tak injected, while SIS i forced to chedule the latet -tak followed by the latet π max -tak, being able to complete only the -tak. Therefore, at the end of each alive interval, C tk (SIS) = k, C tk (X) = kπ, P tk (SIS) = k(π + π max ) and P tk (X) = k( + π max ). Hence, the completed-time competitive ratio of algorithm SIS become C(SIS) = πmin π and it pending-time competitive ratio P(SIS) = π+πmax +π max. The nature of algorithm LPT and SPT however (cheduling according to the proceing time of tak rather than their arrival time), give better reult for both the completed and pending time meaure. Lemma 7. When algorithm LPT run under peedup ρ, it ha a completed-time competitive ratio C(LPT) 1. Proof: A proven in Lemma 1, the number of completed tak of any work conerving algorithm under any combination of arrival and error pattern A and E, and peedup ρ, i never maller than the number of completed tak of X. The ame hold for algorithm LPT, N t (LPT) N t (X). Since the policy of LPT i to chedule firt the tak with the bigget proceing time, the one completed will be of the maximum ize available at all time, which trivially reult to a total completed proceing time at leat a much a the one of X, C t (LPT) C t (X) at any time t. Thi give a completed-time competitive ratio of at leat 1 a claimed. Theorem 9. When algorithm LPT run with peedup ρ, it ha a completed-time competitive ratio C(LPT) 1 and a pending-time competitive ratio P(LPT) 1. Proof: The completed-time competitivene of algorithm LPT follow from Lemma 7 above. Here we how it pending-time complexity; the proof i much more challenging. We conider a time t and define the ordered et of pending tak at the cheduler a Q t (LPT) = v 1, v,..., v z in an execution of algorithm LPT, and a Q t (X) = u 1, u,..., u z in an execution of an algorithm X. We ort the 14

15 tak in both et in decending lexicographic order according to their proceing time and ID, o v 1 v v z and u 1 u u z. By Lemma 1, we know that Q t (LPT) Q t (X) at any time t, which mean that z z and alo N t (LPT) N t (X). We conider any execution and focu on the time intance where algorithm LPT ha to make a cheduling deciion, denoting them a t k where k = 0, 1,,.... We then define the following property: (*) At time t, for any 1 < i z and j = N t (LPT) N t (X), the proceing time of the i th tak in the et Q t (LPT) i not bigger than the proceing time of (i + j 1) t tak in the et Q t (X); in other word v i u i+j 1. We now make three claim regarding property (*). Firt, Claim 1 argue that the property hold at all time intant t k. In Claim we conider the property at time of injection and how that it i till true. Finally, Claim 3 argue that the property hold at all time between time intant t k. Combining the reult of the three claim yield that property (*) hold at all time in any execution. Hence, the total pending proceing time of algorithm LPT i never bigger than that of X plu π max which might be the difference in their et of pending tak. Thi reult to a pending-time competitive ratio P(LPT) = 1, a claimed. So the proof of the theorem conclude by tating and proving the three claim. Claim 1: For all time t k where k = 0, 1,,... property (*) hold. We now prove Claim 1. Taking the beginning of any execution, t 0, it i clear that the property (*) hold, a none of the two algorithm ha computed any tak yet and the pending et are exactly the ame, i, v i = u i. To be more precie, ince N 0 (LPT) = N 0 (X) = 0, j = 0 and for any i where 1 < i z, v i u i 1 alo hold. Now conider by contradiction time t k, the mallet time in an execution, where (*) doe not hold. It i clear that it mut be a point at which the machine complete a tak in the execution of LPT. We know that right before uch point N t k (LPT) N t (X). So there are two cae: k (1) If the two algorithm had the ame amount of completed tak before t k (and hence j = 0), then for 1 < i z, v i u i +j 1 = u i 1, where i and j are the correponding parameter right before time t k. At time t k, the firt tak in Q tk (LPT), v 1, ha been completed and hence removed from the et, cauing the ret of the tak to move up in the ordered et by one poition. Then index i = i 1 and thu v i = v i 1 u i 1 till hold. Alo, ince N tk (LPT) = N tk (X) + 1, index j = j + 1 = 1 and u i+j 1 = u i 1+j +1 1 = u i 1. Therefore, v i u i+j 1 and the property (*) till hold. () If N t k (LPT) > N t (X), then v i u i +j 1 right before t k. At time t k all the tak in the et of LPT will k move up the lit by one poition; i = i 1 and thu v i = v i 1. Alo, the difference between the completed tak of LPT and X increae by 1; j = j + 1 and thu u i +j 1 = u i+j = u i+j 1 (the tak remain at the ame poition for the et of X). Since v i u i +j 1 = u i+j 1 v i = v i 1 u i+j 1 hold and o doe the property (*). In both cae we have come to contradiction, o at time t k the property (*) till hold. Thi complete the proof of Claim 1. Claim : When a tak τ i injected, property (*) till hold. We now prove Claim. Aume a time t when a tak τ arrive in the ytem, o it i added to both Q t (LPT) and Q t (X). For the purpoe of analyi we will look at time t k where k = 0, 1,,..., ince during a time interval (t k 1, t k ) LPT complete exactly one tak and even if X ha alo executed a tak and cheduled the next one (even one that ha jut been injected), it won t be able to compute it uccefully within the interval. Thu, ince it wa not in any of the et at time t k 1 but i in both of them at time t k, we conider the time t k to how that an injection of a tak doe not affect the property (*). 15

16 Hence, conider by contradiction a time t k, uch that it i the mallet time in an execution where there i an injection of tak τ and the property (*) doe not hold. We know that N t (LPT) N k t (X), o before time t k, k v i u i +j 1. After the injection, j = j (remain the ame), a τ i added in both et of pending tak. Aume that in Q tk (LPT) tak τ i placed between v i and v i +1. Thi mean that v i τ v i +1. By property (*) it i trivial that v i u i +j 1, thu τ u i +j 1 a well. So, we know that τ will be placed after u i +j 1 in Q tk (X). It i alo clear that v i and u i +j 1 do not change poition in the et and thu at time t k we can ay that v i become v i, till having proceing time le than or equal to u i +j 1 that become u i+j 1. From property (*) we alo know that v i +1 u i +j. If τ u i +j, then it will alo be placed right between u i +j 1 and u i +j in the ordered et of X, Q t k (X). They will be renamed to u i+j 1 and u i+j repectively and u i+j 1 τ u i+j will hold a well a the property (*), ince the mapping between tak did not change and the new tak in Q tk (LPT) i mapped to itelf in Q tk (X), τ τ. If τ u i +j, then it will be added omewhere after u i +j in the et Q t k (X). Tak u i +j doe not change poition in the et and thu at time t k we can ay it become u i+j and i mapped with tak τ from the et of LPT, τ u i+j. Now let u aume that τ i placed after tak u i+j+m in Q tk (X). So for all the tak v i+c in Q tk (LPT) and all the tak u i+j+c in Q tk (X) where 1 c m, it i true that, ince τ v i+c and τ u i+j+c, v i+c u i+j+c. For the next tak in the et of LPT, v i+m+1 we can ay for ure that v i+m+1 τ ince τ i inerted in a higher place in Q tk (LPT). For the ret of tak in both ordered et we can afely ay that, ince τ ha been added to both in higher poition, they have the mapping they had before the injection of τ. A a reult, property (*) till hold after a tak injection. Thi complete the proof of Claim. Claim 3: For every time t in the interval (t i 1, t i ), property (*) till hold. Finally, we prove Claim 3. From Claim 1 we have that at time t i 1, property (*) hold, o for any 1 < m z and j = N ti (LPT) N ti (X), it hold v m u m+j 1. We want to prove that thi i true for all time t in the interval (t i 1, t i ). Aume by contradiction, that there i a time t in that interval, the firt intance where the property doe not hold. Note that for thi to occur, X mut complete a tak at time t. We know by definition, that during the interval (t i 1, t i ) LPT can not fully execute any tak, a the point where tak are completed are time t k where k = 1,, 3,.... There are therefore, two cae to conider: (1) During the interval neither X complete any tak. Thi mean that there i no change in the et of pending tak of the two algorithm and thu at any point in the interval (t i 1, t i ), for all 1 < m z and j = N ti (LPT) N ti (X), v m u m+j 1. () During the interval, X uccefully complete a tak. Note that it can complete only one tak (ince ρ), and that it finihe at time t. Alo note that before time t, N t (LPT) N t (X) = j > 0 mut hold becaue if the amount of completed tak were equal, after the removal of the jut fully computed tak of X, LPT would have more pending tak than X, which by Lemma 1 i not poible. So, we know that the et Q t (LPT) remain the ame, wherea in the Q t (X) the tak that wa jut completed i removed. Recall that, before t, v i u i +j 1. Uing imilar argument a in cae () of the proof of Claim 1, we may conclude the proof: By X uccefully completing a tak, ay the k th tak u k, where 1 < k N t (X), all the tak with i > k will move up in the lit of X by one poition, to form the new equence u 1, u,.... Hence N t (LPT) N t (X) remain true and N t (LPT) N t (X) = j = j 1. We want to prove that v i u i+j 1. For all the tak v i uch that i + j 1 < k, the tak u i +j 1 did not change it poition in Q(X). Hence, v i u i +j 1 = u i+j u i+j 1 (i = i ). However, for all the tak v i uch that i + j 1 k, the tak u i +j 1 ha moved up one poition in Q(X). Hence, v i u i +j 1 = u i+j = u i+j 1 (i =). Thi complete the proof of Claim 3 and of the theorem. Theorem 10. When algorithm SPT run with peedup ρ, it ha a completed-time competitive ratio C(SPT) 1 16