Lagrangian Primal Dual Algorithms in Online Scheduling

Transcription

1 Lagrangan Prmal Dual Algorthms n Onlne Schedulng Nguyen Km Thang IBISC, Unversty of Evry Val d Essonne, France Abstract We present a prmal-dual approach to desgn algorthms n onlne schedulng. Our approach makes use of the Lagrangan weak dualty and convexty to derve dual programs for problems whch could be formulated as convex assgnment problems. The constrants of the duals explctly ndcate the onlne decsons and naturally lead to compettve algorthms. We llustrate the advantages and the flexblty of the approach through problems n dfferent settng: from sngle machne to unrelated machne envronments, from typcal compettve analyss to the one wth resource augmentaton, from convex relaxatons to non-convex relaxatons. 1 Introducton In the onlne settng, tems arrve over tme and one must determne how to serve tems n order to optmze a qualty of servce wthout the knowledge about future. A popular measure for studyng the performance of onlne algorthms s compettve rato n the model of the worst-case analyss. An algorthm s sad to be c-compettve f for any nstance ts obectve value s wthn factor c of the optmal offlne algorthm s one. Moreover, to remedy the lmtaton of pathologcal nstances n the worst-case analyss, there s other model called resource augmentaton [11]. In the latter, onlne algorthms are gven an extra power and are compared to the optmal offlne algorthm wthout that addtonal resource. Ths model has successfully provded theoretcal evdence for heurstcs wth good performance n practce, especally n onlne schedulng where obs arrve onlne and need to be processed on machnes. We say a schedulng algorthm s s-speed c-compettve f for any nput nstance the obectve value of the algorthm wth machnes of speed s s at most c tmes the obectve value of the optmal offlne scheduler wth unt speed machnes. The most successful tool untl now to analyze onlne algorthms s the potental functon method. Potental functons have been desgned to show that the correspondng algorthms behave well n an amortzed sense. However, desgnng such potental functons s far from trval and often yelds lttle nsght about how to desgn such potental functons and algorthms for related problems. Recently, nterestng approaches [1, 8, 14] based on mathematcal programmng have been presented n the search for a prncpled tool to desgn and analyze onlne schedulng algorthms. The approaches gve nsght about the nature of many schedulng problems, hence lead to algorthms whch are usually smple and compettve. thang@bsc.fr 1

2 1.1 Contrbuton and technques In ths paper, we present a prmal-dual approach to desgn algorthms for an onlne convex assgnment and applcatons to onlne schedulng. The onlne convex assgnment conssts of a set of agents and a set of tems whch arrve over tme. At the arrval of tem, the tem needs to be fractonally assgned to some agents. Let x s the amount of assgned to. The problem s to mnmze f a x under the constrants g b x and h c x for every, where functons f, g, h s are convex. In offlne settng, the optmal solutons are completely characterzed by the KKT condtons see [3] for example. However, for onlne settng, the condtons could not be satsfed due to the lack of knowledge on the nputs. Our approach s the followng. We frst consder the problem as a prmal convex mathematcal program. Then we derve a Lagrangan dual program by the standard Lagrangan dualty. Instead of analyzng drectly the correspondng Lagrangan functons where n general one can not dsentangle the obectve and the constrants as well as the prmal and dual varables, we explot the convexty property of gven functons and construct a dual program. In the latter, dual varables are separated from the prmal ones. The constructon s shown n Secton 2. As the prce of the separaton procedure, the strong dualty property s not guaranteed. However, the weak dualty always holds and that s crucal and enough to deduce a lower bound for the gven problem. An advantage of the approach les n the dual program n whch the constrants could be mantaned onlne. Moreover, the dual constrants explctly ndcate the onlne decsons and naturally lead to a compettve algorthm n the prmal-dual sense. We llustrate the advantages and the flexblty of the approach through problems n onlne schedulng. The problems that we study vary n dfferent settng: from sngle machne to unrelated machne envronments, from typcal compettve analyss to the one wth resource augmentaton. Moreover, we show how to combne the prmal-dual approach n ths paper and the dual-fttng approach presented n [14] to desgn and analyze algorthms for problem wth non-convex relaxaton. 1. We consder the problem n whch there s a machne and obs arrve over tme. Each ob s released at tme r, has deadlne d, processng volume p and a value a. Jobs could be executed preemptvely and at a tme t, the scheduler has to choose a set of obs and the machne speed st n order to process such obs. The energy cost of a schedule s P stdt where P s a gven convex energy power. Typcally, P z = z α for some constant α 1. The obectve of the problem s to mnmze energy cost plus the lost value whch s the total value of uncompleted obs. Usng the prmal-dual approach, we derve an algorthm where the compettve rato s characterzed by a system of dfferental equatons. For typcal power functon P z = z α, the compettve rato turns out to be α α and redscover the result n [12]. Wth the prmal-dual framework, the result s more general and the analyss s smpler. 2. We consder other tradeoff between value and energy. Smlar to the prevous problem, obs arrve onlne, each one has a deadlne, a value and could be processed preemptvely. However, there are a set of unrelated machnes and a ob may have dfferent processng volumes on dfferent machnes. At a tme t, the scheduler has to choose a set of obs to be executed on each machne, and the speeds s t s for every machne to process such obs. Job mgraton s not allowed,.e., no ob could be executed on more than one machne. The obectve now s to maxmze the total value of completed obs mnus the energy cost whch s P s tdt where P s a gven convex power functon. It has been shown that wthout resource augmentaton no algorthm has bounded compettve rato even for a sngle machne 2

3 [13]. We study the problem n the resource augmentaton model. We gve a prmal-dual algorthm whch s 1 + ɛ-speed and 1/ɛ-compettve for every ɛ ɛp > where ɛp depends on functon P. For typcal functon P z = z α, ɛp = 1 α 1/α whch s closed to for α large. 3. We study the problem of speed scalng wth power down schedulng on a sngle machne. Agan, obs are released onlne, each ob has a deadlne, a processng volume and could be processed preemptvely. In the problem, all obs have to be executed. The machne has could be transtoned nto a sleep state or n the actve state. In the sleep state, the energy consumpton s. In the actve state, the power energy consumpton at tme t s P st = st α + g where α 1 and g are constant. A transton cost from the sleep state to the actve state s A, whch represents the wake-up cost. The obectve s to mnmze the total energy the consumed energy n actve state plus the wake-up energy whle executng all obs. For the problem, t s unclear how to formalze a relaxaton as a convex program. Therefore, we consder a natural relaxaton whch s not convex. We frst study a specal case wth no wake-up cost. In ths case, the relaxaton becomes convex and our framework could be appled to show a α α -compettve algorthm the algorthm s n fact algorthm Optmal Avalable [15]. Next we study the problem n general wth wakeup cost. The specal case effectvely gves dea to determne the machne speed n actve state. Thus we consder an algorthm whch uses that procedure to mantan the machne speed n actve state as a subroutne. The algorthm turns out to be algorthm Sleep-aware Optmal Avalable SOA [9] wth dfferent descrpton due to the prmal-dual vew. Han et al. [9] proved that SOA has compettve rato max{4, α α + 2}. We prove that SOA s ndeed max{4, α α }-compettve by the dual-fttng technque presented n [14]. Although the mprovement s slght, the analyss s tght 1 and t suggests that the dualty-based approach s seemngly a rght tool for onlne schedulng. Through the problem, we llustrate the desgn and analyss of an algorthm for non-convex relaxaton. The prmal-dual framework presented n ths paper gves deas for the constructon of an algorthm whle the analyss s done usng dual-fttng technque. 1.2 Related work Anand et al. [1] was the frst who proposed studyng onlne schedulng by lnear convex programmng and dual fttng. By ths approach, they gave smple algorthms and smple analyses wth mproved performance for problems where the analyses based on potental functons are complex or t s unclear how to desgn such functons. Subsequently, Nguyen [14] generalzed the approach n [1] and proposed to study onlne schedulng by non-convex programmng and the weak Lagrangan dualty. Usng that technque, [14] desgned and analyzed non-water-fllng algorthms for problems related to weghted flow-tme. Buchbnder and Naor [4] presented the prmal-dual method for onlne packng and coverng problems. Ther method unfes several prevous potental functon based analyss and s a powerful tool to desgn and analyze algorthms for problems wth lnear relaxatons. Gupta et al. [8] gave a prmal-dual algorthm for a general class of schedulng problems wth cost functon fz = z α. Devanur and Jan [7] also used the prmal-dual approach to derve optmal compettve ratos for onlne matchng wth concave return. Later on, Devanur and Huang [6] consdered the problem of 1 The algorthm has compettve rato exactly α α even wthout wake-up cost [2]. 3

4 onlne schedulng to mnmze the sum of energy and weghted flow-tme on unrelated machnes. They gave an algorthm wth almost optmal compettve rato for arbtrary convex power functon by prmal-dual approach. The constructon of dual programs n [6, 7] s based on convex conugates and Fenchel dualty for prmal convex programs n whch the obectve s convex and the constrants are lnear. An nterestng qualty of servce n onlne schedulng s the tradeoff between energy and throughput. The onlne problem to mnmze the consumed energy plus lost values wth the energy power P z = z α s frst studed by [5] where a α α + 2eα-compettve algorthm s gven for a sngle machne. Subsequently, Klng and Petrzyk [12] derved an mproved α α -compettve for dentcal machnes wth mgraton usng the technque n [8]. The onlne problem to maxmze the total value of completed obs mnus the consumed energy for a sngle machne has been consdered n [13]. Pruhs and Sten [13] proved that the compettve rato wthout resource augmentaton s unbounded and gave an 1 + ɛ-speed, O1/ɛ 3 -compettve algorthm. The problem of speed scalng wth power down schedulng on a sngle machne has been frst studed n [1]. Iran et al. [1] derved an algorthm wth compettve rato 2 2α 2 α α + 2 α Subsequently, Han et al. [9] presented algorthm SOA whch s max{4, α α + 2}-compettve. 2 Framework for Onlne Convex Assgnment Consder the assgnment problem where tems arrve onlne and need to be fractonally assgned to some agents wth the followng obectve and constrants. mn P x := f a x P subect to g b x h c x x, where x ndcates the amount of tem assgned to agent and functons f, g, h are convex, dfferental for every, and a, b. Denote k f tem k s released before tem. Let F be the set of feasble solutons of P. The Lagrangan dual s max λ,γ mn x F Lx, λ, γ 4

5 where L s the followng Lagrangan functon Lx, λ, γ = f a x + λ g [ x x a f a k x k, k + f a x + λ g [ x x a f a k x k, + f a x k + b x + +λ b g b x k +λ b g λ g b x γ h c x ] b k x k +γ c h c x + k γ h c x ] b k x k +γ c h c x + γ h c x where the nequaltes holds for any x due the convexty of functons f, g, h s. In the frst nequalty, we use f z f z + z z f z smlarly for functons g, h s and n the second nequalty, we use the monotoncty of f and smlarly for g. Denote [ ] Mx, x, λ, γ :=, Nx, λ, γ := x x f a x a f + k a k x k λ g +λ b g b x k + b k x k γ h +γ c h c x c x We have k Lx, λ, γ Mx, x, λ, γ + Nx, λ, γ 1 Intutvely, one could magne that x s the soluton of an algorthm or a functon on the soluton of an algorthm. We emphasze that x s not a soluton of an optmal assgnment. The goal s to desgn an algorthm, whch produces x and derves dual varables λ, γ, n such a way that the prmal obectve s bounded by a desred factor from the dual one. Inequalty 1 naturally leads to the followng dea of an algorthm. For any tem, we mantan the followng nvarants a f a k x k +λ b g b k x k +γ c h c x k k a f a k x k +λ b g b k x k +γ c h c x = f x > k Whenever the nvarants hold for every, Mx, x, λ, γ snce x for every,. Therefore, Lx, λ, γ Nx, λ, γ and so the dual s lower-bounded by Nx, λ, γ, whch does not depend anymore on x. The procedure of mantanng the nvarants dctate the decson x of an algorthm and ndcates the choce of dual varables. 5

6 Consder the followng dual subect to max Nx, λ, γ D a f a k x k +λ b g b k x k +γ c h c x, k k k a f a k x k +λ b g b k x k +γ c h c x = f x >, k x, λ, γ, Lemma 1 Weak Dualty Let OP T P and OP T D be optmal values of prmal program P and dual program D, respectvely. Then OP T P OP T D. Proof It holds that OP T P max λ,γ mn x F Lx, λ, γ Nx, λ, γ where the nequaltes follow the weak Lagrangan dualty and the constrants of D for every feasble soluton x, λ, γ. Therefore, the lemma follows. Hence, our framework conssts of mantanng the nvarants for every onlne tem and among feasble set of dual varables constraned by the nvarants choose the ones whch optmze the rato between the prmal and dual values. If an algorthm wth output x satsfes P x rnx, λ, γ for some factor r then the algorthm s r-compettve. 3 Mnmzng Total Energy plus Lost Values The problem. We are gven a machne wth a convex energy power P and obs arrve over tme. Each ob s released at tme r, has deadlne d, processng volume p and a value a. Jobs could be executed preemptvely and at a tme t, the scheduler has to choose a set of pendng obs.e., r t < d and a machne speed st n order to process such obs. The energy cost of a schedule s P stdt. Typcally, P z = z α for some constant α 1. The obectve of the problem s to mnmze energy cost plus the lost value whch s the total value of uncompleted obs. Formulaton. Let x and y be varables ndcatng whether ob s completed or t s not. We denote varable s t as the speed that the machne processes ob at tme t. The problem could be relaxed as the followng convex program. mn P stdt + a y subect to st = s t t d x + y 1 s tdt p x r x, y, s t, t 6

7 In the relaxaton, the second constrant ndcates that ether ob s completed or t s not. The thrd constrant guarantees the necessary amount of work done n order to complete ob. Applyng the framework, we have the followng dual. max P v t dt d λ v tdt + γ r subect to 1. For any ob, γ p λ. Moreover, f x > then γ = p λ. 2. For any ob, γ a and f y > then γ = a. 3. For any ob and any t [r, d ], t holds that λ P k v k t. Partcularly, f v t > then λ = P k v k t. Note that v t s not equal to s t the machne speed on ob accordng to our algorthm but t s a functon dependng on s t. That s the reason we use v t nstead of s t. We wll choose v t s n order to optmze the compettve rato. To smplfy the notaton, we drop out the star symbol n the superscrpt of every varable f one has that. Algorthm. The dual constrants naturally leads to the followng algorthm. We frst descrbe nformally the algorthm. In the algorthm, we mantan a varable u t representng the vrtual machne speed on ob. The vrtual speed on ob means that ob wll be processed wth that speed f t s accepted; otherwse, the real speed on wll be set to. Consder the arrval of ob. Observe that by the thrd constrant, we should always ncrease the machne speed on ob at arg mn P vt n order to ncrease λ. Hence, at the arrval of a ob, ncrease contnuously the vrtual speed u t of ob at arg mn P vt for r t d. Moreover, functon vt s also smultaneously updated as a functon of ut = k u kt accordng to a system of dfferental equatons 2 n order to optmze the compettve rato. The teraton on ob termnates whether one of the frst two constrants becomes tght. If the frst one holds, then accept the ob and set the real speed equal to the vrtual one. Otherwse, reect the ob. Defne Qz := P z zp z. Consder the followng system of dfferental equatons wth boundary condtons: Qv = f u =. Q v dv du + P v P u, r r 1P u + rq v dv, 2 du dv du >, where r s some constant. Let r 1 be a smallest constant such that the system has a soluton. The formal algorthm s gven n Algorthm 1. In the algorthm, machne processes accepted ob wth speed s t at tme t. As the algorthm completes all accepted obs, t s equvalent to state that the machne processes accepted obs n the earlest deadlne frst fashon wth speed st at tme t. By the algorthm, the dual varables are feasble. In the followng we bound the values of the prmal and dual obectves. 7

8 Algorthm 1 Mnmzng the consumed energy plus lost values. 1: Intally, set st, s t and u t s and vt, v t s equal to. 2: Let r 1 be the smallest constant such that 2 has a soluton. Durng the algorthm, keep vt as a soluton of 2 wth constant r and ut = u t for every tme t. 3: for a ob arrves do 4: Intally, u t. 5: whle d r u tdt < p and λ p < a do 6: Contnuously ncrease u t at arg mn P vt for r t d and update ut k u kt + u t and vt as a functon of ut and λ mn r t d P vt smultaneously. 7: end whle 8: Set v t vt k v kt. 9: f λ p = a and d r u tdt < p then 1: Reect ob. 11: Set γ a. 12: else 13: Accept ob. 14: Set s t u t, st st + s t and γ λ p. 15: end f 16: end for Lemma 2 It holds that P stdt + a y r P v t dt d λ v tdt + r γ Proof By the algorthm, λ = P vt at every tme t such that v t > for every ob. Hence, t s suffcent to show that P stdt + a y r Q v t dt + γ 3 where recall Qz = P z zp z. We wll prove the nequalty 3 by nducton on the number of obs n the nstance. For the base case where there s no ob, the nequalty holds trvally. Suppose that the nequalty holds before the arrval of a ob. In the followng, we consder dfferent cases. Job s accepted. Consder any moment τ n the whle loop related to ob. We emphasze that τ s a moment n the executon of the algorthm, not the one n the tme axs t. Suppose that at moment τ, an amount du t s ncreased allocated at t. Note that du t = dut as ut = u t. As s accepted, y = and the ncrease at τ n the left hand-sde of 3 s P utdut 8

9 Let vt 1, τ 1 be the value of vt 1 at moment τ 1 n the whle loop. By the algorthm, the dual varable γ satsfes γ = λ p = d τ 1 d τ 1 u t 2 dt 2 mn P vt 1, τ 1 dτ 1 r r t 1 d r u t 2 mn r t 1 d P vt 1, τ 1 dt 2 dτ 1 where the nequalty s due to the fact that at the end of the whle loop, d r u tdt = p by the loop condton and P s ncreasng. Therefore, at moment τ, dγ mn r t 1 d P vt 1, τdu t = P vtdut where the equalty follows snce t arg mn r t 1 d P vt 1, τ. Hence, the ncrease n the rght hand-sde of 3 s at least r [Q vtdvt + P vtdut]. Due to the system of nequatons 2 and the choce of r, at any moment n the executon of the algorthm, the ncrease n the left hand-sde of 3 s at most that n the rght hand-sde. Thus, the nducton step follows. Job s reected. If s reected then y = 1 and so the ncrease n the left hand-sde of 3 s a. Moreover, by the algorthm γ = a. So we need to prove that after the teraton of the for loop on ob, t holds that r 1a + r Qvtdt. As s reected, a = λ p > Therefore, t s suffcent to prove that d r mn P vt 1 u tdt r t 1 d d r 1 mn P vt 1 u tdt + r r r t 1 d Qvtdt 4 Before the teraton of the whle loop, the left-hand sde of 4 s. Smlar as the analyss of the prevous case, durng the executon of the algorthm the ncrease rate of the left-hand sde s r 1P vtdut + r Q vtdvt, whch s non-negatve by equaton 2. Thus, nequalty 4 holds. By both cases, the lemma follows. Theorem 1 The algorthm s r -compettve. Partcularly, f the energy power functon P z = z α then the algorthm s α α -compettve Proof The theorem follows by the framework and Lemma 2. If the power energy functon P z = z α then r = α α and ut = vt/α satsfy the system 2. Thus, the algorthm s α α -compettve. 4 Maxmzng the Total Value mnus Energy The problem. We are gven unrelated machnes and obs arrve over tme. Each ob s released at tme r, has deadlne d, a value a and processng volume p f t s executed on machne. 9

10 Jobs could be executed preemptvely but mgraton s not allowed,.e., no ob could be executed on more than one machne. At a tme t, the scheduler has to choose a set of pendng obs.e., r t < d to be processed on each machne, and the speeds s t s for every machne to execute such obs. The obectve now s to maxmze the total value of completed obs mnus the energy P s tdt where P s a gven convex power functon. cost whch s It s shown that wthout resource augmentaton, the compettve rato s unbounded [13]. In ths secton, we consder the problem wth resource augmentaton, meanng that wth the same speed z the energy power for the algorthm s P 1 ɛz, whereas the one for the adversary s P z. Let ɛp > be the smallest constant such that zp 1 ɛp z P z for all z >. For the typcal energy power P z = z α, ɛp = 1 α 1/α whch s closed to for α large. Formulaton. Let x be varable ndcatng whether s completed n machne. Moreover, varable y equals 1 f s not completed and equals otherwse. Let s t be the varable representng the speed that machne processes ob at tme t. The problem could be formulated as the followng convex program. max subect to a x, d P 1 ɛs tdt s t = s t, t x 1 s tdt p x, r x, s t,, t Note that n the obectve, by resource augmentaton the consumed energy s P 1 ɛs tdt. Applyng the framework, we have the followng dual. subect to mn γ P u t dt +, 1. For any machne and any ob, γ + p λ a. d λ u tdt 2. For any machne, any ob and any t [r, d ], λ P 1 ɛ k u k t where the sum s taken over all obs k released before,.e., k. Partcularly, f u t > then λ = P 1 ɛ k u k t. Smlar as n the prevous secton, the constrants naturally lead to Algorthm 2. In the algorthm and the analyss, to smplfy the notaton we drop out the star symbol n the superscrpt of every varable f one has that. Lemma 3 Dual varables constructed by Algorthm 2 are feasble. r 1

11 Algorthm 2 Mnmzng the throughput mnus consumed energy. 1: Intally, set st and ut equal to. 2: The algorthm always runs accepted obs wth speed st n the earlest deadlne frst fashon. 3: for a ob arrves do 4: Intally, u t for every t and let I be the set of all machnes, I. 5: whle I do 6: For every I, ncrease u t at arg mn P u t n the contnuous manner for r t d and update u t = k u kt + u t and λ = mn r t d P 1 ɛu t smultaneously. 7: f λ p = a and d r u tdt < p for some machne then 8: I I \ {}. 9: end f 1: f λ p < a and d r u tdt = p for some machne then 11: I I {} and I I \ {}. 12: end f 13: end whle 14: f I = then 15: Reect ob and set γ note that p λ = a. 16: else 17: Let = arg mn I p λ. 18: Accept and assgn ob to machne,.e., x = 1. 19: Set s t u t, s t s t + s t and γ a λ p. 2: end f 21: end for Proof By the algorthm lne 6, λ P 1 ɛ k u kt where the sum s taken over all obs k released before k and f u t > then λ = P 1 ɛ k u kt. Consder the frst constrant. If s reected then γ = a. Otherwse, by the assgnment lne 17, t always holds that γ + p λ a for every,. In the followng we bound the values of the prmal and dual obectves n the resource augmentaton model. Lemma 4 For every ɛ ɛp, t holds that 1 a x ɛ, P 1 ɛs tdt γ P u t dt +, d λ u tdt r 11

12 Proof We have d λ u tdt P u tdt, r max λ u tdt :r t d P 1 ɛu tu tdt P u tdt P u tdt where the second nequalty s due to the algorthm, and the last nequalty follows by the defnton of ɛp. Therefore, t s suffcent to prove that 1 a x P 1 ɛs tdt γ 5 ɛ, We prove nequalty 5 by nducton on the number of released obs n the nstance. For the base case where there s no ob, the nequalty holds trvally. Suppose that the nequalty holds before the arrval of a ob. If s reected then x =, s t = for every, t and γ =. Therefore, the ncreases n both sde of nequalty 5 are. Hence, the nducton step follows. In the rest, assume that s accepted and for smplcty let be the machne to whch s assgned. We have [ P 1 ɛu t P 1 ɛ ] 1 ɛu t u t dt 1 ɛ P 1 ɛu t u tdt P u tu tdt = 1 ɛλ u tdt = 1 ɛλ p 1 ɛa The frst nequalty s due to the convexty of P P 1 ɛu t u t P 1 ɛu t 1 ɛu tp 1 ɛu t. The second nequalty holds because P s ncreasng. The frst equalty follows snce u t only at t such that P u t = λ by the algorthm. The last nequalty s due to the loop condton n the algorthm. Thus, at the end of the teraton related to ob n the for loop, the ncrease n the left hand sde of nequalty 5 s 1 ɛ a x [ P 1 ɛu t P ] 1 ɛu t u t dt 1 a 1 ɛa = a ɛ Besdes, the ncrease n the rght hand-sde of nequalty 5 s γ a. Hence, the nducton step follows; so does the lemma. Theorem 2 The algorthm s 1 + ɛ-augmentaton, 1/ɛ-compettve for ɛ ɛp. Proof By resource augmentaton, wth the same speed z the energy power for the algorthm s P 1 ɛz, whereas the one for the adversary s P z. So n the nequalty of Lemma 4, the left hand-sde represents exactly the obectve value of the algorthm n the model of resource augmentaton. Hence, the theorem follows. 12

13 5 Speed Scalng wth Power Down Schedulng The problem. We are gven a sngle machne that could be transtoned nto a sleep state or an actve state. Each transton from the sleep state to the actve state costs A >, whch s called the wake-up cost. Jobs arrve onlne, each ob has a released tme r, a deadlne d, a processng volume p and could be processed preemptvely. In the problem, all obs have to be completed. In the sleep state, the energy consumpton of the machne s. In the actve state, the power energy consumpton at tme t s P st = st α + g where α 1 and g are constant. Thus, the consumed energy of the machne n actve state s P stdt, that can be decomposed nto dynamc energy st α dt and statc energy gdt where the ntegral s taken over t at whch the machne s n actve state. At any tme t, the scheduler has to decde the state of the machne and the speed f the machne s n actve state n order to execute and complete all obs. The obectve s to mnmze the total energy the consumed energy n actve state plus the wake-up energy. Formulaton. Let Ht be the Heavsde step functon,.e., Ht = f t < and Ht = 1 f t. Then Ht s the ntegral of the Drac delta functon H = δ and t holds that + δtdt = 1. We use ths fact to formulate a relaxaton for the problem. Let F t be a functon ndcatng whether the machne s n the actve state,.e., F t = 1 f the machne s n the actve state and equals otherwse. Assume that ntally the machne s n the sleep state. Then A + F t dt equals twce the wake-up energy. Let s t be the varable representng the speed that machne speed on ob at tme t. The problem could be formulated as the followng non-convex program. mn subect to P s t F tdt + A 2 d + F t dt r s tf tdt p s t, F t {, 1}, t 5.1 Speed Scalng wthout Wake-Up Cost The problem wthout wake-up cost A = has been extensvely studed. We reconsder the problem throughout our prmal-dual approach. In case A =, the machne s put n actve state whenever there s some pendng ob thus the functon F t s useless and could be removed from the formulaton. In ths case, the relaxaton above becomes a convex program. Applyng the framework and by the same observaton as n prevous sectons, we derve the followng algorthm. At the arrval of ob, ncrease contnuously s t at arg mn P st for r t d and update smultaneously st st + s t untl d r s t dt = p. It turns out that the machne speed st of the algorthm equals max t >t V t, t /t t where V t, t s the remanng processng volume of obs arrvng at or before t wth deadlne n t, t ]. So the algorthm s ndeed algorthm Optmal Avalable [15] that s α α -compettve [2]. However, the prmal-dual vew of the algorthm gves more nsght and that s useful for the general model wth wake-up cost see Lemma 5. 13

14 5.2 Speed Scalng wth Wake-Up Cost The Algorthm. Defne the crtcal speed s c = arg mn s> P s/s. In the algorthm, the machne speed s always at least s c f t executes some ob. Intally, set st and s t equal for every tme t and obs. If a ob s released then t s marked as actve. Intutvely, a ob s actve f ts speed s t has not been settled yet. Let τ be the current moment. Consder currently actve obs n the earlest deadlne frst EDF order. Increase contnuously s t at arg mn P st for r t d and update smultaneously st st + s τ untl d r s t dt = p. Now consder dfferent states of the machne at the current tme τ. We dstngush three machne states: 1 n workng state the machne s actve and s processng some obs; 2 n dle state the machne s actve but ts speed equals ; and 3 n sleep state the machne s nactve. In workng state. If sτ > then keep process obs wth the earlest deadlne by speed max{sτ, s c }. Mark all currently pendng obs as nactve. If sτ =, swtch to dle state. In dle state. If sτ s c then swtch to workng state. If s c > st >. Mark all currently pendng obs as actve. Notce that f there s no new ob released then the actve obs wll be processed at speed s c durng some perod n the future. Otherwse, f the total duraton of dle state from the last wake-up equals A/g then swtch to sleep state. In sleep state. If st s c then swtch to the workng state. In the rest, we denote s t as the machne speed at tme t by the algorthm. Analyss. The Lagrangan dual s max λ mn s,f Ls, F, λ where the mnmum s taken over s, F feasble solutons of the prmal and L s the followng Lagrangan functon Ls, F, λ = P s t F tdt + A 2 λ p d r + F t dt + s tf t λ P st st d λ p s tf tdt r 1 {st>} 1 {F t=1} dt + A 2 + F t dt where st = s t. By weak dualty, the optmal value of the prmal s always larger than the one of the correspondng Lagrangan dual. In the followng, we bound the Lagrangan dual value n functon of the prmal obectve value due to the algorthm and derve the compettve rato. That s done by the dual-fttng approach presented n [14]. Dual varables Let < β be some constant to be chosen later. For obs such that s t > for every t [r, d ] then defne λ such that λ p /β equals the margnal ncrease n the dynamc energy due to the arrval of ob. For obs such that s t = for some moment t [r, d ] then defne λ such that λ p equals the margnal ncrease n the energy due to the arrval of ob. Lemma 5 Let be an arbtrary ob. 14

15 1. If s t > for every t [r, d ] then λ βp s t for every t [r, d ]. 2. Moreover, f s t = for some t [r, d ] then λ = P s c /s c. Proof We prove the frst clam. For any tme t, speed s t s non-decreasng as long as new obs arrve. Hence, t s suffcent to prove the clam assumng that no other ob s released after. So s t s the machne speed after the arrval of. The margnal ncrease n the dynamc energy due to the arrval of could be wrtten as 1 β λ p = d r d P s t P s t s t dt = mn P s t d r r s tdt = mn P s tp P s ts tdt where mn P s t s taken over t [r, d ] such that s t >. The nequalty s due to the convexty of P and the frst equalty follows by the algorthm. Moreover, mn P s t P s t for t [r, d ]; so the lemma follows. We are now showng the second clam. By the algorthm, the fact that s t = for some t [r, d ] means that ob wll be processed at speed s c n some nterval [a, b] [r, d ]. The margnal ncrease n the energy s P s c b a whle p could be expressed as s c b a. Therefore, λ = P s c /s c. Theorem 3 The algorthm has compettve rato at most max{4, α α }. Proof Let E1 be the dynamc energy of the algorthm schedule,.e., E 1 = [P s t P ]dt λ p /β due to the defnton of λ s and < β 1. Moreover, let E2 be the statc energy plus the wake-up energy of the algorthm,.e., E2 = P F tdt+ A 2 F t dt. We wll bound the Lagrangan dual obectve. By Lemma 5, for every ob such that s t = for some t [r, d ], λ = P sc s. By the c defnton of the crtcal speed, λ P z z for any z >. Therefore, d s tf t λ P st dt 6 st r 15

16 where n the sum s taken over obs such that s t = for some t [r, d ]. Therefore, d s tf t λ P st st 1 {st>} 1 {F t=1} dt L 1 s, λ := λ p r d [ βe1 max s tf t βp s t P st s,f r st [ βe1 max st βp s t P st ] 1 s st {st>} 1 {F t=1} 1 {s t>}dt [ ] βe1 βp s t st P st 1 {st>} 1 {F t=1} 1 {s t>}dt βe1 1 [ ] βp s t st P st 1 2 {s t>}dt 1 [ ] βp s t st P st 1 2 {F t=1} dt ] 1 {st>} 1 {F t=1} 1 {s t>}dt where n the second lne, the sum s taken over obs such that s t > for all t [r, d ]. The frst nequalty follows 6. The second nequalty holds snce F t 1 and st = s t. The thrd nequalty s due to the frst order dervatve and st s the soluton of equaton P zt = βp s t. In fact st maxmzes functon βp s t P st/st. As the energy power functon P z = z α + g where α 1 and g, st α 1 = βs t α 1. Therefore, L 1 s, λ βe1 1 βαs t α 1 st st α g 1 2 {s t>}dt 1 βαs t α 1 st st α g 1 2 {F t=1} dt = βe1 α 1β α/α 1 s t α dt + 1 g1 2 {s t>}dt + 1 g1 2 {F t=1} dt ] = [β α 1β α/α 1 E1 + 1 g1 2 {s t>}dt + 1 g1 2 {F t=1} dt Choose β = 1/α α 1, we have that Ls, F, λ 1 α α E g1 {s t>}dt g1 {F t=1} dt + A 2 F t dt In the followng, we clam that L 2 F := 1 2 g1 {s t>}dt+ 1 2 g1 {F t=1} dt+ A 2 F t dt s at least E2 /4. Consder the algorthm schedule. An end-tme u s a moment n the schedule such that the machne swtches from the dle state to the sleep state. Conventonally, the frst end-tme n the schedule s. Partton the tme lne nto phases. A phase [u, v s a tme nterval such that u, v are consecutve end-tmes. Observe that n a phase, the schedule has transton cost A and there s always a new ob released n a phase otherwse the machnes would not swtch to non-sleep state. We wll prove the clam on every phase. In the followng, we are nterested n phase [u, v and whenever we menton L 2 F, t refers to 1 v 2 u g1 {s t>}dt + 1 v 2 u g1 {F t=1}dt + A v 2 u F t dt. 16

17 By the algorthm, the statc energy of the schedule durng the dle tme s A,.e., v u g1 {s t=}dt = A. Let s, F be a feasble of soluton of the relaxaton. If durng [u, v, the machne followng soluton s, F makes a transton from non-sleep state to sleep state or nversely then L 2 F 1 v 2 u g1 {s t>}dt + A 2. Hence L 2 F 1 v v g1 4 {s t>}dt + g1 {s t=}dt + A = 1 u u 4 E 2 [u,v. If durng [u, v, the machne followng soluton s, F makes no transton from non-sleep statc to sleep state or nversely then F t = 1 durng [u, v n order to process obs released n the phase. Therefore, L 2 F 1 2 v u v g1 {s t>}dt g1 2 {s t>}dt + 1 u 4 1 v g1 4 {s t>}dt + u v u v u v u g1 {F t=1} dt = 1 2 v g1 {s t=}dt + A 4 g1 {s t=}dt + A = 1 4 E 2 u g1 {s t>}dt where the second nequalty s due to the fact that the machne swtches to sleep state at tme v, meanng that the total dle duraton n [u, v ncurs a cost A. In concluson, the dual Ls, F, λ E1 /αα + E2 /4 whereas the prmal s E 1 + E 2. Thus, the compettve rato s at most max{4, α α }. [u,v v u gdt References [1] S. Anand, Naveen Garg, and Amt Kumar. Resource augmentaton for weghted flow-tme explaned by dual fttng. In Proc. 23rd ACM-SIAM Symposum on Dscrete Algorthms, pages , 212. [2] Nkhl Bansal, Tracy Kmbrel, and Krk Pruhs. Speed scalng to manage energy and temperature. J. ACM, 541, 27. [3] Stephen Boyd and Leven Vandenberghe. Convex Optmzaton. Cambrdge Unversty Press, 24. [4] Nv Buchbnder and Joseph Naor. The desgn of compettve onlne algorthms va a prmaldual approach. Foundatons and Trends n Theoretcal Computer Scence, 32-3:93 263, 29. [5] Ho-Leung Chan, Tak Wah Lam, and Rongbn L. Tradeoff between energy and throughput for onlne deadlne schedulng. In Proc. 8th Workshop on Approxmaton and Onlne Algorthms, pages 59 7, 21. [6] Nkhl R. Devanur and Zhy Huang. Prmal dual gves almost optmal energy effcent onlne algorthms. In Proc. 25th ACM-SIAM Symposum on Dscrete Algorthms, 214. [7] Nkhl R. Devanur and Kamal Jan. Onlne matchng wth concave returns. In Proc. 44th ACM Symposum on Theory of Computng, pages ,

18 [8] Anupam Gupta, Ravshankar Krshnaswamy, and Krk Pruhs. Onlne prmal-dual for nonlnear optmzaton wth applcatons to speed scalng. In Proc. 1th Workshop on Approxmaton and Onlne Algorthms, pages , 212. [9] Xn Han, Tak Wah Lam, Lap-Ke Lee, Isaac Kar-Keung To, and Prudence W. H. Wong. Deadlne schedulng and power management for speed bounded processors. Theor. Comput. Sc., : , 21. [1] Sandy Iran, Sandeep K. Shukla, and Raesh Gupta. Algorthms for power savngs. ACM Transactons on Algorthms, 34, 27. [11] Bala Kalyanasundaram and Krk Pruhs. Speed s as powerful as clarvoyance. J. ACM, 474: , 2. [12] Peter Klng and Peter Petrzyk. Proftable schedulng on multple speed-scalable processors. In Proc. 25th Symposum on Parallelsm n Algorthms and Archtectures, 213. [13] Krk Pruhs and Clfford Sten. How to schedule when you have to buy your energy. In APPROX-RANDOM, pages , 21. [14] Nguyen Km Thang. Lagrangan dualty n onlne schedulng wth resource augmentaton and speed scalng. In Proc. 21st European Symposum on Algorthms, 213. [15] F. Frances Yao, Alan J. Demers, and Scott Shenker. A schedulng model for reduced cpu energy. In FOCS, pages ,