Tight Bounds for Online Scheduling in the One-Fast-Many-Slow Machines Setting

Tigh t Bounds for Online Sc heduling in the One-F ast-Man y-Slo w Mac hines Setting John Jeang and Vladimir P o dolskii T ufts Univ ersity F ebruary 26, 2026 Abstract In the One-F ast-Many-Slo w decision problem, in tro duced b y Sheffield and W esto v er [17], a sc heduler, with access to one fast mac hine and infinitely many slow machines, receiv es a series of tasks and m ust decide how to allocate the w ork among its mac hines. The goal here is to achiev e the minimal o verhead of an online algorithm o ver the optimal offline algorithm that kno ws all tasks and their arriv al times in adv ance. Three v ersions of this setting were considered: Instan tly-committing sc hedulers that m ust assign a task to a mac hine immediately and irrev o cably , Even tually-committing sc hedulers whose assignments are irrevocable but can o ccur an ytime after a task arriv es, and Never-committing sc hedulers that can in terrupt and restart a task on a differen t mac hine. Sheffield and W estov er constructed online algorithms in all three settings. In the Instantly-committing mo del the optimal comp etitive ratio is equal to 2. In the Even tually-committing mo del, they give an algorithm achieving a 1.678 competitive ratio, while a low er b ound of 1.618 w as established b y Kuszmaul and W esto ver in [12]. In the Nev er-committing mo del, Sheffield and W estov er sho w that the optimal comp etitive ratio lies in the in terv al [1.366, 1.5]. In the latter tw o settings, the exact v alues of the optimal comp etitiv e ratios were left as an open problem, moreov er Kuszmaul and W esto ver [12] conjectured that the lo wer b ound in the Even tually-committing model is tigh t. In this pap er we resolve this problem by providing tight b ounds for the competitive ratios in the Ev entually-committing and Never-committing settings. F or the Even tually-committing mo del we prov e Kuszmaul and W estov er’s conjecture b y pro viding an algorithm that ac hieves a comp etitiv e ratio equal to the previous low er b ound of 1+ √ 5 2 ≈ 1 . 618. Notably , our algorithm o ccasionally leav es the fast mac hine free even in the presence of unassigned tasks, as was shown to b e necessary for any impro vemen t by Sheffield and W estov er. F or the Nev er-committing mo del we pro vide an explicit T ask Arriv al Process (T AP) low er b ounding the competitive ratio to the previous upper bound of 1.5. Unlike the T AP used to pro ve the previous low er bound, whic h uses only tw o tasks, we construct a series of T APs with a growing n umber of tasks that approac h the upp er bound of 1.5. 1 1 In tro duction Online algorithms deal with streaming data in whic h decisions are made without complete knowl- edge of the entire problem instance. An online algorithm’s p erformance is measured relative to that of the optimal offline algorithm that knows the entire problem instance prior to all decisions. Optimizing this metric, kno wn as the c omp etitive r atio , for online algorithms has a wide range of applications to sc heduling, net working, and learning theory [21]. Sc heduling algorithms aim to efficiently distribute incoming tasks to optimize for some goal (e.g. mak espan, tardiness, fairness). Efficient sc heduling algorithms hav e b een well-studied in b oth the offline and online regime for their p oten tial application to a v ariety of settings such as cloud computing, service, and industrial en vironments [10, 16, 18]. Sometimes in online scheduling, tasks can tak e adv an tage of sp ecialized hardware to exp edite pro cessing. How ever, when resources are limited, it is unclear how to prioritize tasks. One of the approac hes to study this problem was suggested b y Sheffield and W estov er [17] when they in tro duced the One-F ast-Many-Slo w Decision Problem (OFMS). In OFMS, an algorithm, with access to a single instance of accelerated hardware and infinitely man y weak pro cessors, receiv es tasks, whose sizes and arriv al times are unkno wn in adv ance, in an online stream and must allo cate resources to minimize the completion of the set of tasks as a whole. The OFMS problem is motiv ated by its relev ance to the Serial to P arallel Decision Problem (SPDP) [12], which addresses multiprocessor scheduling with parallelizable tasks. The SPDP ques- tion concerns itself with the scheduling of p erfectly parallelizable tasks on p identical pro cessors. Indeed, Sheffield and W estov er reduce SPDP to OFMS in the limit where the num b er of pro cessors greatly outnum b ers the n umber of tasks. The reduction w orks b y isolating a small n umber of pro- cessors to b e used only for the serial implementation of tasks and treating the rest of the pro cessors as a single high p erformance collection for tasks implemented in parallel; for more information w e refer the reader to the pap er [17]. In b oth settings the tasks are unknown ahead of time and arrive in an online stream and the authors analyze b ounds on the comp etitiv e ratios of online algorithms. In the case of OFMS the follo wing bounds are known: • Instan tly-committing schedulers: T asks must b e assigned to a sp ecific machine immedi- ately upon arriv al and cannot be mov ed. The matc hing lo w er [12] and upper [17] b ounds of 2 on the comp etitiv e ratio are kno wn. • Ev en tually-committing sc hedulers: T asks can b e assigned any time after arriv al, but cannot b e mov ed. The b est known low er b ound is 1+ √ 5 2 ≈ 1 . 618 [12] and the best kno wn upp er b ound is ≈ 1 . 678 [17]. • Nev er-committing sc hedulers: T asks can b e mov ed and restarted on a different machine. The b est known low er and upper b ounds are 1+ √ 3 2 ≈ 1 . 366 and 1 . 5 resp ectiv ely [17]. Th us, although in the Instan tly-committing model the optimal comp etitiv e ratio is known, its v alue in the other t w o settings w ere left in [17] as an op en problem. Moreo v er, it is conjectured in [12] that the low er b ound in the Even tually-committing mo del is tigh t. 2 Results. In this pap er we resolve the op en question p ertaining to the optimal comp etitiv e ratio of an online algorithm in the OFMS problem in the Ev entually-committing and Nev er-committing settings. Sp ecifically w e pro ve the conjecture made in [12] by providing an optimal Ev entually- committing scheduler achieving a comp etitive ratio of 1+ √ 5 2 = φ meeting the previous lo wer b ound from [12]. In the Never-committing model, w e give an explicit task arriv al pro cess that low er b ounds the comp etitiv e ratio of any online algorithm to 1.5, matching the previous upp er b ound from [17]. T ec hniques. Our main technical contribution is providing an optimal φ -comp etitiv e online al- gorithm, in the Even tually-committing setting. In [17], the authors pro vide the optimal non- pr o cr astinating algorithm for the OFMS problem. That is, their algorithm is optimal among algo- rithms that do not dela y task assignments when the fast machine is free. Their analysis consist of b ounding the amount of time spent b y their algorithm’s fast mac hine on tasks assigned differ ently from the optimal algorithm. At a high lev el, their algorithm do es excellen t for man y T ask Arriv al Pro cesses (T AP for short), but struggles in the case where it misallo cates a small task that should ha ve b een held on to in wait of more information from the T AP . Our algorithm mo difies the previous algorithm by o ccasionally holding onto tasks ev en when the fast mac hine is free. The main tec hnical work is showing that this modification do es not incur to o m uch delay as to hinder the strengths of the previous algorithm, but provides significant enough delay such that tasks are assigned more accurately . The key insight is making this delay indep enden t of a tasks arriv al time and solely dep enden t on the size of the task. This allo ws us to b ound the time at whic h tasks will receive additional delay thereb y allo wing it to p erform similarly to the previous algorithm in [17], late in the T AP , but significantly differently at the b eginning of the T AP . F rom here we can use similar techniques to bound the amount of time sp en t, by our algorithm’s fast mac hine, on misallo cated tasks. In the Nev er-committing setting, we come up with an explicit sequence of T APs lo wer b ounding the comp etitiv e ratio of online algorithms in this mo del to the previous upp er b ound of 1.5. These T APs, consist of a growing n umber of tasks, differing significantly from the t wo-task T AP giving the previous 1+ √ 3 2 lo wer b ound. Other related works. Muc h work has b een done in the field of scheduling in b oth the offline and online setting. In the offline setting, in whic h an algorithm knows the entire problem instance, optimal s c heduling has b een shown to b e NP -hard in a v ariet y of settings (see, e.g., [4, 3, 11]), and progress has b een made to wards approxi mation algorithms [15, 14, 20]. In [19], they examine heterogeneous hardw are in the offline setting, and [2] deals with robust single-mac hine sc heduling where jobs are known in adv ance but the release date of a job is given as an in terv al, rather than a p oin t, in which the job arriv es. The online setting has t ypically b een explored in the con text of unkno wn jobs and arriv al times, dealing with either a single pro cessor [6] or m ultiple iden tical processors [1]. In [9], they consider the problem of sc heduling jobs that ma y require multiple machines to pro cess. The case of sto chastic sc heduling concerns itself with tasks coming from a known distribution of sizes and arriv al times 3 [7]. Finally the case of flexible constrain ts has b een considered in b oth fuzzy sc heduling as w ell as in case studies such as the n urse sc heduling problem [5, 8, 13]. Outline. The rest of the pap er is organized as follows. In Section 2, we state the problem and pro vide the notation needed for the analysis. In Section 3, w e giv e a φ -comp etitiv e online algorithm for the Ev en tually-committing setting of OFMS. The analysis in this section is the bulk of our technical w ork, so w e first b egin b y proving some general prop erties of our algorithm and the OFMS problem, and then pro ceed to ev aluating our algorithm’s p erformance. Finally in Section 4, w e give a pro cedure for generating a family of T APs that result in a tight lo wer bound of 1.5 for online algorithms in the Nev er-committing setting. 2 Preliminaries 2.1 The Problem In this section w e formally state the, previously in tro duced, One-F ast-Man y-Slow Decision Problem (OFMS), where an online algorithm with access to infinitely many slo w machines and a single fast mac hines m ust assign tasks to mac hines with the goal of minimizing the total completion time (makespan) of the set of tasks as a whole. W e first define a task τ i = ( f i , s i , t i ), representing the runtime on the fast machine, runtime on a slo w machine, and arriv al time resp ectiv ely . W e enforce that t i ≥ 0 and s i ≥ f i > 0. An instance of OFMS is a T ask Arriv al Pro cess (T AP) T = { τ 1 , . . . , τ n } , with t j ≥ t i for j > i . An online sc heduling algorithm M incremen tally receives p ortions of T upon the arriv al of tasks. That is, M learns τ i at t i , whereas and offline sc heduling algorithm N receives the en tirety of T at the start. A scheduling algorithm’s goal is to assign tasks to b e run on mac hines to minimize the comple- tion time of T . The completion time is defined to be the minim um time t at which all tasks in T ha ve finished running. In the OFMS problem, each mac hine can only run a single task at a time. A mac hine is said to b e free in the close d interv als in which it is not running an y tasks (and thus there is a notion of “the last time a machine is free”). W e are interested in t wo previously defined mo dels for this problem: 1. Ev en tually-committing sc hedulers : A scheduler may hold on to a task without imme- diately assigning/starting it, but once the task b egins running, the assignmen t cannot be c hanged. 2. Nev er-committing sc hedulers : A scheduler ma y hold on to a task without immediately assigning/starting it, and even after the task b egins running, the task can b e mo ved to start o ver on a differen t mac hine. In [17], W estov er and Sheffield also define Instantly-c ommitting schedulers in whic h a scheduler m ust commit and queue a task to a particular machine up on arriv al. In this case the comp etitiv e ratio of the optimal online algorithm is kno wn to b e 2. 4 W e note that there are several algorithms, with p ossibly different assignmen ts, that solve the offline v ersion of OFMS. W e describ e one such algorithm, OPT , that we’ll use for the remainder of the pap er. Giv en an OFMS problem instance T , OPT first calculates the completion time of the assignment, q 0 , that allo cates all tasks in T to the fast machine. Then for every task τ i ∈ T , OPT calculates the completion time of an assignment q i defined as follo ws: for all τ j ∈ T , τ j ’s assignment =    a slow machine if s j + t j ≤ s i + t i the fast mac hine otherwise, (1) Let K ( q i ) denote the completion time of assignmen t q i . OPT then simply uses the assignmen ts asso ciated with the minimum completion time c = min i ( K ( q i )). Intuitiv ely , OPT assigns a task to the slo w mac hine whenev er it can do so without increasing the completion time of T . T o see that this algorithm pro duces an optimal assignment, consider an optimal assignmen t G . If G assigns all tasks in T to the fast machine then this is exactly the q 0 assignmen t, otherwise consider G ’s last- ending slow mac hine, w orking on τ k , that ends at s k + t k . First note that all of the slow machines in the q k assignmen t will finish b efore s k + t k . Second, notice that G ’s fast machine will contain ev ery task that is on q k ’s fast mac hine b ecause it contains every task τ i suc h that s i + t i > s k + t k . Th us q k ’s fast machine finishes at least as quickly as G ’s fast machine and so q k m ust b e an optimal assignmen t. Imp ortan tly , in this algorithm w e ha ve that for all tasks τ j suc h that s j + t j ≤ c , OPT assigns τ j to a slo w mac hine, and we make frequen t use of this prop ert y in our pro ofs. 2.2 Notation A t every given p oin t in time t , there is an optimal offline allo cation of just the subset of tasks in T that ha ve arriv al time less than or equal to t . W e use OPT ( t ) to denote the optimal offline assignmen t of this subset of tasks and w e use OPT ( ∞ ) for the optimal offline assignment of the whole T . W e refer to the completion time of OPT ( t ) as C t , and to the total completion of OPT ( ∞ ) as C ∞ . W e sometimes drop the argumen t and use OPT to denote the final assignment when it is clear from context. F urthermore, we use e C t to refer to the runtime of OPT ( t )’s fast mac hine. Finally , when talking ab out the optimal offline algorithm without reference to a sp ecific time we just use OPT . F or an online algorithm M , w e use M f to denote the run time of M ’s fast machine after the en tire T AP has arriv ed. R ( P ) will denote the earliest time at which an online algorithm M , kno ws that the optimal offline algorithm m ust tak e at least P time (e.g. by calculating C t ). F or ease of communication, we define a notion of lar geness , where we sa y that a task τ x is larger than τ y if and only if s x + t x > s y + t y . Note that this notion do es not care ab out the time it takes the tasks to run on the fast mac hine. 5 3 Ev en tually-committing sc heduler 3.1 A φ -Comp etitiv e Even tually-committing Scheduler W e define φ = 1+ √ 5 2 , the golden ratio. Throughout the pap er we use the fact that φ is the p ositiv e solution to the equation φ = 1 + 1 φ . In this section w e provide a φ -comp etitiv e algorithm for the ev entually-committing setting of OFMS. In [17], Sheffield and W estov er define non-pr o cr astinating algorithms as those that only dela y the machine assignment of tasks if the fast mac hine is o ccupied. They sho w that a competitive ratio of ≈ 1 . 678 is optimal for non-pro crastinating algorithms, and thus to ac hieve a comp etitiv e ratio of φ ≈ 1 . 618 our algorithm needs to occasionally dela y assigning a task to a machine even if the fast mac hine is free. T o do so we first define the notion of an eligible task. A task τ i is eligible at time t if t ≥ f i φ . (2) Our algorithm will only assign eligible tasks to the fast mac hine and therefore ma y sometimes pro crastinate. Finally we note that an online algorithm M can (efficiently) calculate OPT ( t ) and C t at time t , by running the algorithm for OPT on the tasks that hav e arrived at or before time t . This giv es M a low er b ound for C ∞ at time t . W e say that M safely assigns a task to a slo w machine if the finishing time of that mac hine is guaranteed to b e within the φ -comp etitiv e ratio with the completion time of the optimal offline algorithm. W e no w giv e the formal description of our Ev entually-committing sc heduler H . Algorithm H : A t t 0 = 0, H initializes an empty standb y set S . Whenever a task τ i arriv es, it is immediately added to S . Additionally H contin uously ev aluates all tasks in S at all times t as follo ws: 1. If s i + t ≤ φ C t , then start τ i on an un used slo w mac hine and remov e it from S . 2. If al l of the following conditions hold: (a) the fast mac hine is free or just finished a task, (b) τ i has the largest s i + t i v alue among all tasks in S , (c) t ≥ f i φ ( τ i is eligible), then start τ i on the fast machine and remov e it from S . W e refer to S as the standby set, and denote S t to b e the set of task on standb y at time t . A t a high level, H first do es a “slow-c heck” to see if a task can safely b e assigned to a slow mac hine. If this c heck fails, H then do es a “fast-chec k” to see if the task can b e assigned to the fast machine. W e will frequen tly use these c heck conditions to characterize the p erformance of H . Although H makes these c hecks con tinuously , we observ e the chec ks need only to b e made during sp ecific even ts. Notice that the slow-c hec ks only need to o ccur up on the arriv al of a task, 6 b ecause s i + t only increases in t while φ C t do es not change in b et ween the arriv al of tasks. The fast-c hecks only need to occur upon the arriv al of a task, when a task on the fast machine finishes, or when the largest task in S b ecomes eligible. Therefore, the algorithm can actually b e made discrete without c hanging its b eha vior b ecause there are only finitely man y even ts warran ting chec ks. This observ ation allo ws us to pick δ small enough suc h that for any ev ent happ ening at time t , there are no even ts b etw een t − δ and t . F rom this we can use the notation t − = t − δ (3) for what can b e thought of as “the time just b efore t ”. The eligibility condition, (c) in the fast-chec k, is what causes H to sometimes procrastinate. Ho wev er notice that eligibility is concerned with the absolute time t but not the arriv al time t i of a particular task τ i . In tuitively , this is ho w H manages to pro crastinate mostly at the b eginning of the T AP , but not to o m uch near the end of the T AP . F urther notice that the eligibility condition is not present in the slo w-chec k. This is b ecause there are infinitely many slow machines, so we are not worried ab out k eeping any of them free and th us H simply runs tasks on slow mac hines whenev er it can safely do so. Finally , we emphasize that the conditions in the fast chec k must b e met simultaneously . T o run a task τ i on the fast machine at time t , it is not sufficient for τ i to merely b e the largest among all eligible tasks in S t ; it m ust be the largest among all tasks in S t . If there is a largeness tie, and at least one task participating in the tie is eligible, then H assigns one of the eligible tasks to the fast mac hine, breaking ties arbitrarily among the eligible tasks participating in the tie. A t this p oin t it is also useful to in tro duce the follo wing sets of tasks that will help us with our analysis. F or a given T AP T w e define the following: • A (“Actual”): The set of tasks that both H and OPT ( ∞ ) assign to the fast mac hine; • F (“F ak e”): The set of tasks that H assigns to the fast machine but OPT ( ∞ ) assigns to a slo w mac hine; • F big ⊂ F : The set of tasks in F such that s i + t i > C ∞ φ ; • F small ⊂ F : The set of tasks in F such that s i + t i ≤ C ∞ φ . T o av oid confusion, it is also imp ortan t to note some subtleties. Firstly , we note that A ∪ F do es not capture an y of the tasks that H assigns to its slow mac hine; most of the time we simply do not care ab out these tasks since they finish within a φ comp etitive ratio of OPT ( ∞ )’s completion time, by the design of the algorithm. Ho w ever, there is no explicit mechanism that prohibits the p ossibilit y of a task τ from b eing assigned to OPT ( ∞ )’s fast machine, while being assigned to the slo w mac hine by H . While these tasks are not relev ant for most of our analysis, it is imp ortan t to remem b er that they exist since they can sho w up in standby set S and affect OPT ’s fast machine. Finally , w e note that since H only safely assigns tasks to slo w mac hines, then to pro ve that H is φ -comp etitiv e, it suffices to consider just the p erformance of its fast mac hine. 7 3.2 Initial Observ ations and Prop erties Before analyzing the p erformance of our algorithm, it is useful to familiarize ourselves with some of the basic prop erties of H , OPT , F big , F small and A that will greatly simplify our analysis. W e b egin b y pro ving some prop erties of OPT . Note that for t 1 < t 2 it is p ossible for OPT ( t 1 ) to assign a task τ i to the fast machine while OPT ( t 2 ) assigns τ i to a slow machine (note that OPT is an offline algorithm, and so this sw app ed assignmen t is retroactive and th us effective at t i and not t 2 ). This is b ecause OPT ( t 2 ) ma y b enefit from freeing up space on the fast machine for other tasks arriving at t 1 < t ≤ t 2 . In this case we sa y that OPT has swapp e d τ i ’s assignment. This next lemma tells us that tasks only swap from the fast mac hine to a slow machine. Lemma 1. If τ i is assigne d to a slow machine by OPT ( t 1 ) , then it wil l b e assigne d to a slow machine by OPT ( t 2 ) for al l t 2 > t 1 . Pr o of. Consider a task τ i assigned to a slo w mac hine by OPT ( t 1 ). Let t 2 b e any time greater than t 1 . W e ha v e s i + t i ≤ C t 1 ≤ C t 2 , (4) where the first inequality comes from the fact that OPT ( t 1 )’s completion time must b e at least the completion time of τ i . The second inequality comes from the fact that all task parameters are non-negativ e and thus C t is monotone in t . Since OPT ( t ) assigns a task τ i to a slow mac hine whenev er s i + t i ≤ C t , w e conclude that τ i m ust be assigned to a slow mac hine b y OPT ( t 2 ). Lemma 1 helps us further characterize some of OPT ’s b eha vior. Lemma 2. Each task τ i ∈ A ∪ F is assigne d to the fast machine by OPT ( t i ) . Pr o of. First consider a task τ a ∈ A . W e know that OPT ( ∞ ) assigns τ a to the fast machine, and from Lemma 1 we know that OPT do es not swap task assignments from slo w to fast. Therefore τ a m ust be assigned to the fast mac hine the entire time since its arriv al at t a . No w consider a task τ z ∈ F . Suppose for the sake of contradiction that OPT ( t z ) assigns τ z to the slow machine. Then during H ’s slo w-chec k we hav e s z + t z C t z ≤ s z + t z s z + t z | {z } OPT ’s completion of τ z = 1 < φ. (5) This means that H f w ould also assign τ z to a slo w mac hine at t z . How ev er this is a con tradiction since H ev entually assigns all tasks in F to the fast mac hine, and H is an Ev en tually-committing sc heduler that cannot c hange the assignment of tasks. Therefore OPT ( t z ) assigns τ z to the fast mac hine. Finally , in our next lemma, we build a classification of the tasks assigned to H ’s fast mac hine. 8 Lemma 3. F or any τ a ∈ A , τ b ∈ F big , and τ m ∈ F smal l we have the fol lowing: 1. s m + t m ≤ C ∞ φ < s b + t b ≤ C ∞ < s a + t a , 2. f b ≤ C ∞ φ , 3. f m ≤ C ∞ φ 2 . Pr o of. The first t wo inequalities from item 1 come from the definition of F small and F big . The third inequality from item 1 comes from the fact that τ b is ultimately put on one of OPT ’s slow mac hines and the completion times of all of OPT ’s slo w machines are b ounded b y OPT ’s total completion time. Finally the last inequality from item 1 comes from the fact that τ a ultimately go es on OPT ( ∞ )’s fast mac hine, and therefore m ust be unable to go on OPT ’s slow mac hine without increasing the completion time since OPT puts tasks on a slow mac hine whenev er possible. T o see items 2 and 3, it suffices to sho w that for any τ i ∈ F big ∪ F small w e hav e that f i < s i φ , and then we can just apply item 1. Consider τ i ∈ F big ∪ F small . T o show f i < s i φ , we recall that b y Lemma 2 we kno w OPT ( t i ) assigns τ i to the fast mac hine, but H do esn’t assign τ i to a slow mac hine at this time so we hav e φ < s i + t i C t i ≤ s i + t i e C t i ≤ τ i on H ’s slow mac hine z }| { s i + t i f i + t i | {z } minimum time on OPT ’s fast machine = ⇒ f i ≤ s i φ . (6) With these initial observ ations in hand we can pro ceed to the main analysis. 3.3 Analysis Our goal is to pro v e the following theorem. Theorem 1. H is a φ -c omp etitive Eventual ly-c ommitting sche duler. Recall that w e only hav e to concern ourselves with the performance of H ’s fast machine. Our first strategy for analyzing H f will b e to characterize the tasks from F . These can b e thought of as misassigned “extra work” that contribute to H f since the optimal offline algorithm assigns them to slo w mac hines once the entire T AP has arrived. Our second strategy will b e to bound the amount of time that H ’s fast machine sp ends pro crastinating. W e will need sev eral k ey lemmas to com bine these t wo ideas and b ound H f . W e start with pro ving that misassigned tasks that H runs on the fast mac hine, cannot arriv e too late. Lemma 4. Al l tasks τ i ∈ F arrive (strictly) b efor e R ( C ∞ /φ ) . Pr o of. All tasks τ i ∈ F are run on the fast machine by H , but ultimately on a slow machine by OPT once the entire T AP has arriv ed. This implies 9 φ C t i < s i + t i ≤ C ∞ ≤ φ C R ( C ∞ /φ ) . (7) The first inequalit y comes from the fact that H do esn’t immediately assign τ i to a slo w mac hine. The second inequality comes from the fact that the run time of τ i cannot exceed the total runtime of the entire T AP since OPT ultimately runs τ i on a slow machine. The last inequalit y comes from the fact that C R ( C ∞ /φ ) ≥ C ∞ /φ by definition of R . Thus t i < R ( C ∞ /φ )). F or further analysis w e need to define a few key sets of tasks. Let τ l b e the last task to arriv e strictly b efore R ( C ∞ /φ ): • α = S t l ∩ A , • β = S t l ∩ F big , F urthermore, for t ≥ R ( C ∞ /φ ) we define β t = S t ∩ F big . (8) Imp ortan tly , w e note that for t 2 ≥ t 1 ≥ R ( C ∞ /φ ), β t 2 ⊆ β t 1 ⊆ β , (9) b ecause as time progresses for t > R ( C ∞ /φ ), no additional tasks from F arriv e (by Lemma 4) but some may b e remov ed from standb y . F or an y set X we define Σ X = P i ∈ X f i , the total load X can add to the fast machine if all tasks in X are run successively without pause. Using this notation, we see that Σ β t 2 ≤ Σ β t 1 ≤ Σ β . In our next lemma, w e show that Σ α and Σ β cannot b e to o large. Lemma 5. Σ α + Σ β < C ∞ φ . Pr o of. Observ e that any task τ i ∈ α ∪ β must be assigned to the fast mac hine by OPT ( t l ) or else C t l w ould exceed C ∞ /φ , by Lemma 3, whic h is not possible since t l < R ( C ∞ /φ ). Thus w e ha v e Σ α + Σ β < e C t l ≤ C t l < C ∞ φ . (10) F rom this w e define 0 < ε = C ∞ φ − (Σ α + Σ β ) . (11) Our next lemma b ounds how long tasks from A can stay ineligible. Lemma 6. L et τ a b e a task in A . If either τ a ∈ α or b oth t a ≥ R ( C ∞ /φ ) and β t a  = ∅ , then τ a is eligible, for H ’s fast machine, for al l t ≥ max( t a , Σ α , ε + Σ β − Σ β t a ) . 10 Pr o of. Consider a task τ a ∈ A arriving at t a ≥ R ( C ∞ /φ ) with β t a  = ∅ . If the task is eligible up on arriv al then w e’re done. Otherwise we hav e that t a < f a φ . F rom Lemma 2, w e kno w that τ a is assigned to the fast mac hine b y OPT ( t a ). F urthermore we kno w that for an y τ i ∈ β t a , H do es not start τ i on a slow machine at t a b ecause all tasks in β t a are ultimately run on H ’s fast machine. Th us similarly to (6) w e ha v e φ < τ i on H ’s slow mac hine at t a z }| { s i + t a C t a ≤ s i + t a e C t a ≤ s i + t a f a + t a = ⇒ s i φ > f a . (12) This tells us that t a < f a φ < s i φ 2 ≤ s i + t i φ 2 . F rom this we can see that if any τ i ∈ β t a w ere assigned to a slow mac hine b y OPT ( t a ) then H would assign τ i to a slow mac hine b ecause then we would ha ve the following slow-c hec k for τ i : s i + t a C t a ≤ ( s i + t i ) + s i + t i φ 2 s i + t i | {z } τ i on OPT ( t a )’s slow mac hine = 1 + 1 φ 2 ≤ φ. (13) This gives us Σ α + Σ β t a + f a ≤ e C t a (14) b ecause all of α ∪ β t a ∪ { τ a } are on OPT ( t a )’s f ast mac hine. F urthermore, recall that for an y τ i ∈ β t a , w e hav e s i ≤ C ∞ from Lemma 3 because β t a ⊆ F big . T aken together w e can get a stronger version of (12). That is, we no w ha v e that for an y τ i ∈ β t a φ < s i + t a e C t a ≤ C ∞ + f a φ Σ α + Σ β t a + f a = C ∞ + f a φ ( C ∞ φ − ε ) + (Σ β t a − Σ β ) + f a . (15) Rearranging the inequalit y from the first and last expression giv es us φ ( ε + Σ β − Σ β t a ) > ( φ − 1 φ ) f a = f a and thus τ a will b e eligible for t ≥ ε + Σ β − Σ β t a . No w consider a task τ a ∈ α . It is clear that it will b e eligible for all t ≥ Σ α φ since f a ≤ Σ α . Therefore τ a will certainly be eligible for t ≥ Σ α . Com bining these tw o results gives us the lemma. Next we hav e a useful lemma that simplifies our analysis. F or this we remind the reader of the notion of t − , defined in (3), as the timestep “just b efore t ”. Lemma 7. Assume ther e exist some time b efor e H f when H ’s fast machine is fr e e, and let t b e the last of such times. If al l of the fol lowing c onditions hold 1. t ≥ R ( C ∞ /φ ) , 2. F smal l ∩ S t − = ∅ , 3. A ∩ S t −  = ∅ , 4. A l l tasks in A ∩ S t − ar e eligible (at t − ), 11 then H f ≤ φ C ∞ . Pr o of. Supp ose that the four conditions of the lemma hold. Observ e that conditions 1 and 2 imply that the remaining contribution to H f from F -tasks after time t is b ounded by Σ β b ecause w e can assume by Lemma 4 that no additional tasks from F arriv e after R ( C ∞ /φ ). Conditions 3 and 4 imply that there is at least one large task on standb y prev enting the tasks from A from running on H ’s fast mac hine at t − . This is b ecause H ’s fast mac hine is free at t − if t is the last time it is free. Th us there exists a non-empt y subset of ineligible tasks V ⊂ S t − suc h that V ∩ A = ∅ and that for ev ery task τ v ∈ V and τ a ∈ A ∩ S t − w e ha v e t v + s v > t a + s a > C ∞ , (16) where the last inequality comes from Lemma 3. This implies that all tasks in V are assigned to the fast machine b y C ∞ , otherwise OPT ’s final completion time would b e larger than C ∞ . Let τ w ∈ V b e suc h that f w = max τ i ∈ V ( f i ) . (17) In other words, τ w is the task with the longest fast run time in V . Since τ w / ∈ A , it do es not con tribute to H f . W e can now c haracterize the contribution to H f from A as follows: C ∞ ≥ e C ∞ ≥ f w + X τ i ∈A f i = ⇒ C ∞ − f w ≥ X τ i ∈A f i . (18) Since τ w is ineligible, then t < f w φ . Since t is the last time H ’s fast mac hine is free, we hav e H f ≤ t + work from F z}|{ Σ β + X τ i ∈A f i | {z } work from A < f w φ + C ∞ φ + ( C ∞ − f w ) ≤  1 + 1 φ  C ∞ = φ C ∞ , (19) where in the second inequalit y we use the fact that Σ β < C ∞ φ from Lemma 5. Next we prov e a lemma that narrows do wn our analysis. Lemma 8. If H starts a task τ ∈ F smal l at any time t ≥ R ( C ∞ /φ ) then H f ≤ φ C ∞ . Pr o of. First observ e that if any task τ m ∈ F small is w aiting on standb y at t = C ∞ , then H will b e a ware that OPT tak es at least C ∞ time and will run τ m on a slo w mac hine b ecause w e ha ve s m + t | {z } τ m on H ’s slow mac hine = s m + C ∞ < s m + t m + C ∞ ≤ C ∞ φ + C ∞ ≤ φ C ∞ , (20) whic h is imp ossible since tasks from F small are all run on H ’s fast mac hine. Since no tasks can arriv e after C ∞ , this implies that no task from F small will start after C ∞ . 12 No w let t ∗ ≥ R ( C ∞ /φ ) b e the last time that H starts a task τ m ∈ F small . Since H prioritizes tasks with larger s i + t i , then by Lemma 3 there m ust b e no tasks τ ∈ F big ∪ A in S t ∗ . By Lemma 4, no tasks from F will arriv e from this p oin t forward, and thus all remaining work contributing to H f is from A . F urther w e note that from this point forw ard at most C ∞ − t ∗ amoun t of work can arriv e from A b ecause OPT m ust complete these tasks on the fast mac hine; notice that b ecause t ∗ ≤ C ∞ this amount of w ork is v alid. Since for τ a ∈ A w e hav e f a < C ∞ , we know that all tasks from A will b e eligible by C ∞ φ . Th us if H runs contin uously from max( t ∗ + f m , C ∞ φ ) then w e ha ve H f ≤ max( t ∗ + f m , C ∞ φ ) + ( C ∞ − t ∗ ) (21) = max( C ∞ + f m , (1 + 1 φ ) C ∞ − t ∗ ) (22) ≤ max((1 + 1 φ 2 ) C ∞ , (1 + 1 φ ) C ∞ ) (23) = φ C ∞ , (24) where in the last inequalit y we used the fact that f m ≤ C ∞ φ 2 from Lemma 3. Otherwise H ’s fast mac hine is free at some p oin t after max( t ∗ + f m , C ∞ φ ). Let t > max( t ∗ + f m , C ∞ φ ) b e the last time H ’s fast mac hine is free. Supp ose we hav e A ∩ S t − = ∅ . Since F ∩ S t = ∅ , this implies that from t onw ard H ’s fast mac hine only w orks on tasks from A , with arriv al time greater than or equal to t . Since H ’s fast mac hine do es this w ork contin uously it must finish processing the set A at least as quic kly as OPT ’s fast machine. This gives us H f ≤ C ∞ . Otherwise w e hav e that A ∩ S t −  = ∅ . W e note that since all tasks in A are eligible at max( t ∗ + f m , C ∞ φ ) < t , then they are all also eligible at t − . Thus we hav e all the necessary conditions to in vok e Lemma 7. Lemma 8 basically tells us that we can assume that all tasks in F small start b efore R ( C ∞ /φ ). Another wa y to think ab out it is that no tasks from F small are in S t for any t ≥ R ( C ∞ /φ ). The next lemma strengthens Lemma 7 and tells us that H p erforms w ell, unconditionally , if its fast machine is ev er free at a late time. Lemma 9. If H ’s fast machine is fr e e at any time R ( C ∞ /φ ) ≤ t < H f then H f ≤ φ C ∞ . Pr o of. If t ≥ C ∞ then all tasks hav e arrived and all tasks are eligible. T o see the latter observe that for an y task τ i , we hav e f i ≤ C ∞ < φ C ∞ ≤ φt, (25) where the first inequality comes from the fact that OPT finishes processing τ i on some machine prior to C ∞ . In this case if H is free, then there m ust not b e an y tasks on standb y and th us H ’s fast mac hine has already completed and th us t ≥ H f and the lemma is v acuously true. So we only fo cus on the case where t < C ∞ . Consider the last time t ≥ R ( C ∞ /φ ) such that H ’s fast mac hine is free. 13 Due to Lemma 8, w e can assume that F small ∩ S t = ∅ . Thus w e ha v e that Σ ( F small ∪F big ) ∩ S t = Σ F big ∩ S t = Σ β t . (26) In other w ords, the total contribution of F tasks on standby , to H f , is bounded b y Σ β t . Case 1 ( S t − ∩ A = ∅ ): In this case, from time t , the total con tribution from A -tasks to H f is at most C ∞ − t and the total con tribution from F -tasks is Σ β t ≤ Σ β . Therefore w e hav e H f ≤ t + ( C ∞ − t ) + Σ β t ≤ t + ( C ∞ − t ) + Σ β ≤ (1 + 1 φ ) C ∞ = φ C ∞ , (27) where the last inequality comes from Lemma 5. Case 2 ( S t − ∩ A  = ∅ and all tasks from S t − ∩ A are eligible): This case giv es us all the conditions necessary to inv ok e Lemma 7. Case 3 ( S t − ∩ A  = ∅ but some tasks in S t − ∩ A are ineligible): Let us define τ a to b e the last arriving task in S t − ∩ A suc h that R ( C ∞ /φ ) ≤ t a < t . If τ a do esn’t exist then the con tribution from S t to H f is b ounded by Σ α + Σ β < C ∞ φ . Since the maxim um load for all future tasks from A is C ∞ − t w e get H f ≤ t + ( C ∞ − t ) + (Σ α + Σ β ) < C ∞ + C ∞ φ = φ C ∞ . Th us for the remainder of this case we’ll assume that τ a exists. Now supp ose that β t a = ∅ . This means that from t on ward, H f do es not get any contribution from F . F urthermore, since some task in S t ∩ A is not eligible w e kno w that t < C ∞ φ . Thus w e ha v e H f ≤ t + C ∞ |{z} All work comes from A ≤ C ∞ φ + C ∞ = φ C ∞ . (28) Th us for the remainder of this case w e’ll assume that β t a  = ∅ . No w consider some ineligible task τ i ∈ S t − ∩ A . W e hav e that t i ≤ t a and that β t a ⊆ β t i  = ∅ . W e can no w in vok e Lemma 6 to b ound t . Concretely we hav e t ≤ max( t i , Σ α , ε + Σ β − Σ β t i ). By assumption the maximum cannot resolv e to t i since t i < t . If the maxim um resolv es to Σ α w e ha v e H f ≤ t + ( C ∞ + Σ β t ) | {z } maximum work from A and F ≤ Σ α + ( C ∞ + Σ β ) < C ∞ + C ∞ φ = φ C ∞ , (29) where the last inequality comes from Lemma 5. Otherwise the maxim um resolv es to ε + Σ β − Σ β t i and we hav e H f ≤ t + ( C ∞ + Σ β t ) ≤ ( ε + Σ β − Σ β t i ) + ( C ∞ + Σ β t ) ≤ ( ε + Σ β ) + C ∞ ≤ C ∞ φ + C ∞ = φ C ∞ , (30) 14 where in the p en ultimate inequalit y we used the fact that Σ β t i ≥ Σ β t b ecause t ≥ t i . W e are now ready to pro ve the main theorem, restated here, which requires some case analysis. Theorem (Restatement of Theorem 1) . H is a φ -c omp etitive in the Eventual ly-c ommitting task mo del. Pr o of. As previously mentioned, it suffices to just consider the p erformance of H ’s fast machine. F urthermore b y Lemma 9 w e can assume for the remainder of the proof that H ’s fast mac hine is not free for an y t ≥ R ( C ∞ /φ ) (and is thus op erating contin uously for t ≥ R ( C ∞ /φ )). Imp ortan tly this implies that some task is running at t = R ( C ∞ /φ ), we call this task the “stuck task” τ u . W e clarify that τ u can b e a task that starts at t = R ( C ∞ /φ ), but not a task that finishes at t = R ( C ∞ /φ ). There are t wo cases to consider. Case 1 ( s u + t u > C ∞ /φ ): Define A early = { τ i ∈ A : t i < R ( C ∞ /φ ) } and A late = A \ A early . In this case, b y largeness, τ u m ust b e a member of one of A early , A late , or F big . Our strategy will be to characterize the con tribution of these sets to H f . Let t < R ( C ∞ /φ ) b e a time when all tasks in F big ∪ A early ha ve already arriv ed; this must exist due to Lemma 4. Since C t < C ∞ /φ all of these tasks m ust b e on OPT ( t )’s fast machine at this time or else OPT ( t )’s runtime would b e greater than C ∞ φ . Therefore w e ha v e Σ F big ∪A early ≤ C ∞ /φ . Notice that A late can con tribute at most C ∞ − R ( C ∞ /φ ) to H f since OPT m ust complete the w ork on its fast mac hine. F urthermore, b y Lemma 8, we kno w that no tasks from F small can con tribute to H f after R ( C ∞ /φ ). Thus w e ha v e H f ≤ R ( C ∞ /φ ) + Σ F big ∪A early + Σ A late (31) ≤ R ( C ∞ /φ ) + C ∞ φ + ( C ∞ − R ( C ∞ /φ )) (32) ≤  1 + 1 φ  C ∞ (33) ≤ φ C ∞ . (34) Case 2 ( s u + t u ≤ C ∞ /φ ): In this case w e ha ve τ u ∈ F small b y Lemma 3. Let t ∗ b e the time in whic h H starts τ u on the fast mac hine. Since τ u ∈ F small , by Lemma 8, w e hav e the strict inequalit y t ∗ < R ( C ∞ /φ ) Now we can define some additional task sets (see Figure 1): • F ′ big : The set of tasks τ i ∈ F big suc h that t i ≥ t ∗ . Note that b y Lemma 4 we also know t i < R ( C ∞ /φ ). • A 1 : The set of tasks τ i ∈ A such that R ( C ∞ /φ ) > t i ≥ t ∗ . • A 2 : The set of tasks τ i ∈ A such that t ∗ + f u > t i ≥ R ( C ∞ /φ ). • A 3 : The set of tasks τ i ∈ A such that t i > t ∗ + f u . 15 0 t ∗ R ( C ∞ φ ) C ∞ F ′ big , A 1 A 2 A 3 H ’s fast mac hine f u Figure 1: Main theorem, Case 2, arriv al times of tasks relativ e to the stuck task τ u . Our strategy will b e to characterize the contribution of all of the tasks assigned to H f ’s fast mac hine. Since H prioritizes tasks with large s j + t j there are no tasks from F big ∪ A on standby at t ∗ . F urthermore by Lemma 8 we can assume there aren’t any task from F small in S t ∗ . Thus we can conclude that there is no contribution from S t ∗ to H f . Next, we recall that t ∗ ≥ f u φ b ecause of the eligibilit y condition. No w w e show that all tasks in F ′ big are assigned to the fast mac hine by OPT ( t ) for all t ∗ ≤ t ≤ f u + t ∗ . T o see this we note that if any task τ i ∈ F ′ big w ere assigned to one of OPT ( t )’s slow mac hine at time t then H would assign τ i to a slo w mac hine at time t b ecause τ i on H ’s slow mac hine z }| { s i + t C t ≤ s i + t ∗ + f u s i + t i ≤ s i + t i + f u s i + t i ≤ C ∞ /φ + C ∞ /φ 2 C ∞ /φ = φ, (35) where in the second inequalit y we ha ve t ∗ ≤ t i b y the definition of F ′ big and in the last inequalit y w e use the facts that f u ≤ C ∞ φ 2 and s i + t i > C ∞ φ from Lemma 3. Note that all tasks in A are assigned to OPT ’s fast mac hine at all times after arriv al. This is b ecause they are on OPT ’s fast mac hine at the end of the T AP , and Lemma 1 tells us OPT nev er sw aps tasks from the slo w mac hine to the fast mac hine. Let t y < R ( C ∞ /φ ) b e the arriv al time of the last task in F ′ big ∪ A 1 . W e therefore hav e, t ∗ + Σ F ′ big + Σ A 1 | {z } on OPT ( t y )’s fast machine ≤ C t y < C ∞ φ . (36) F rom here w e define ∆ = C ∞ φ − ( t ∗ + Σ F ′ big + Σ A 1 ) . (37) No w w e hav e to b ound Σ A 2 . F or this we use the follo wing claim. Claim 1. If F ′ big = ∅ , then H f ≤ φ C ∞ . Pr o of. Supp ose that F ′ big = ∅ . Recall from Lemma 4, that no tasks from F will arriv e after R ( C ∞ /φ ). F urthermore, as we previously argued, there is no contribution to H f from S t ∗ . Th us w e actually hav e that there is no further con tribution to H f from F . F rom t ∗ on ward no more than a total w orkload of C ∞ − t ∗ can arriv e from tasks from A since OPT ’s fast mac hine needs to do this work. 16 Th us since H ’s fast machine runs contin uously after t ≥ t ∗ + f u > R ( C ∞ /φ ), we hav e H f = ( t ∗ + f u ) + ( C ∞ − t ∗ ) ≤ C ∞ φ 2 + C ∞ ≤ φ C ∞ . (38) No w w e may assume that F ′ big  = ∅ . Let t z ≤ t ∗ + f u b e a time by whic h all tasks in A 2 ha ve arriv ed. As noted in (35), all of the tasks in F ′ big remain on OPT ( t )’s fast mac hine for t ∗ ≤ t ≤ f u + t ∗ and are therefore assigned to the fast machine by OPT ( t z ). Th us at t z , all of A 1 , F ′ big , and A 2 are assigned to the fast machine b y OPT ( t z ). Since H assigns none of the tasks τ i ∈ F ′ big to a slo w mac hine, w e ha v e φ ≤ s i + t z C t z ≤ s i + ( t ∗ + f u ) e C t z ≤ s i + ( t ∗ + f u ) t ∗ + Σ F ′ big + Σ A 1 + Σ A 2 . (39) Rearranging the first and last expression and using that fact that t ∗ + f F ′ big + f A 1 = C ∞ φ − ∆ giv es Σ A 2 ≤ ( t i + s i ) − C ∞ φ + ( t ∗ − t i ) φ + f u φ + ∆ ≤ f u φ + ∆ , (40) where in the second inequalit y we used the fact that t i ≥ t ∗ and that C ∞ ≥ s i + t i from Lemma 3. Lastly , w e ha ve Σ A 3 ≤ C ∞ − ( f u + t ∗ ) (41) b ecause OPT has to finish all of these tasks on the fast mac hine. T aken together, we can bound H f as follows: H f = t ∗ + Σ F ′ big + Σ A 1 + Σ A 2 + f u + Σ A 3 (42) ≤ C ∞ φ − ∆ + f u φ + ∆ + f u + C ∞ − ( f u + t ∗ ) (43) ≤  1 + 1 φ  C ∞ (44) = φ C ∞ . (45) 4 Nev er-committing sc heduler No w w e mov e on to the next considered mo del in whic h task assignmen t can b e altered. In this section we sho w that for any Nev er-committing sc heduler M and ε > 0, there exists a T AP T for which M cannot achiev e a comp etitiv e of 1 . 5 − ε on T . W e note that this matches the 17 upp er b ound of [17]. As a reminder, in this setting a scheduler M has the ability to restart a task on a different mac hine, so our proof will hav e to consider this p ossibilit y . Theorem 2. No online algorithm M c an achieve a c omp etitive r atio of 1 . 5 − ε for any ε > 0 , in the Never-c ommitting task mo del. Pr o of. Assume for the sake of contradiction that there exists an algorithm M ac hieving a comp et- itiv e ratio of 1 . 5 − ε for all T APs T . W e give a strategy for an adversary by generating a T AP for M . First, let k b e any ev en in teger such that k ≥ 1 ε . Consider a T AP T consisting of • τ 0 = (1 , 1 . 5 , 0), • τ i = ( 1 k , ∞ , 1 . 5 i k ) , ∀ i ∈ [1 , k 2 − 1], • τ L = (0 . 75 , ∞ , 0 . 75), where the use of s i , s L = ∞ just forces M to put τ i on the fast machine; ∞ can be substituted by a large enough constant. T o clarify , this is a T AP that an adv ersary has in mind, but the adv ersary can c ho ose to truncate the T AP (not send any remaining tasks from T ) whenev er they w ant to. F rom this T AP description, w e can easily compute the optimal offline completion time at each arriv al time. Clearly C 0 = 1 by assigning τ 0 to the fast mac hine. F or an y i ∈ [1 , k 2 − 1] we ha ve that C t i = max( 1 + i k | {z } work on fast mac hine , 1 . 5 i k + 1 k | {z } start τ i at t i ) = 1 + i k for i < k 2 , (46) whic h is achiev ed by assigning all task that ha ve arriv ed up to t i to the fast machine. This is apparen t b ecause if an y of the tasks are assigned to a slow machine, then the completion time w ould be at least 1.5 which is greater than 1 + i k for i < k 2 . Finally , upon the arriv al of τ L , we hav e C t l = C 0 . 75 = 1 . 5 , (47) whic h is ac hieved by assigning τ 0 to a slo w mac hine and all other tasks to the fast mac hine. No w w e use these completion times to complete the argumen t. Our basic strategy will b e to sho w that M can nev er put τ 0 on a slo w mac hine. Notice that up on the arriv al of τ 0 , M cannot put this task on the slow machine, or else the adv ersary will simply not send any additional tasks from T and the competitive ratio will b e 1.5. Up on the arriv al of an y task τ i , M will hav e to assign τ i to the fast machine b ecause its slow run time is infinite. F urthermore, if M mov es τ 0 to a slow machine at this time the comp etitiv e ratio would b e given b y at least 18 s 0 + t i C t i = 1 . 5 + 1 . 5 i k 1 + i k = 1 . 5 . (48) Th us if M mov es τ 0 to a slow mac hine, then the adv ersary can simply stop sending tasks and M will ha ve failed to meet a competitive ratio of 1 . 5 − ε . Finally , consider the arriv al of τ L . If M mov es τ 0 to the slow machine at this time, the comp etitiv e ratio will b e at least t l + s 0 C t l = 0 . 75 + 1 . 5 C 0 . 75 = 2 . 25 1 . 5 = 1 . 5 , (49) whic h does not achiev e the desired competitive ratio of 1 . 5 − ε . This means that τ 0 m ust b e on M ’s fast machine. Moreov er, all other tasks m ust also b e on M ’s fast machine because their slo w run times are infinite. Th us all the tasks m ust b e on the fast mac hine for M , yielding a total completion time of 1 +  k 2 − 1   1 k  + 0 . 75 = 2 . 25 −  1 k  , (50) and thereby giving a comp etitiv e ratio of 2 . 25 − 1 k 1 . 5 ≥ 1 . 5 − ε 1 . 5 > 1 . 5 − ε. (51) This is a con tradiction since M is a (1 . 5 − ε )-comp etitiv e online algorithm. 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