Thus, lmted mache avalablty s commo practce. Kowledge about mache avalabltes mght be complete or complete. I a ole settg mache avalabltes are ot kow a

Transcription

1 Schedulg wth Uexpected Mache Breakdows Susae Albers Guter Schmdt y Abstract We vestgate a ole verso of a basc schedulg problem where a set of jobs has to be scheduled o a umber of detcal maches so as to mmze the makespa. The job processg tmes are kow advace ad preempto of jobs s allowed. Maches are o-cotuously avalable,.e., they ca break dow ad recover at arbtrary tme staces ot kow advace. New maches may be added as well. Thus mache avalabltes chage ole. We rst show that o ole algorthm ca costruct optmal schedules. We also show that o ole algorthm ca acheve a bouded compettve rato f there may be tme tervals where o mache s avalable. The we preset a ole algorthm that costructs schedules wth a optmal makespa of Cmax OP T f a lookahead of oe s gve,.e., the algorthm always kows the ext pot tme whe the set of avalable maches chages. Fally we gve a ole algorthm wthout lookahead that costructs schedules wth a early optmal makespa of Cmax OP T +, for ay > 0, f at ay tme at least oe mache s avalable. Our results demostrate that ot kowg mache avalabltes advace s of lttle harm. 1 Itroducto I schedulg theory the basc model assumes that a xed set of maches s cotuously avalable for processg throughout the plag horzo. Ths assumpto mght be justed some cases but t does ot apply f certa mateace requremets, breakdows or other costrats that cause the maches ot to be avalable for processg have to be cosdered. Mache avalablty costrats appear very ofte. Clearly, maches may be faulty ad break dow. Moreover, avalablty costrats arse o the operatoal level of producto schedulg. Here some jobs are xed terms of startg ad shg tmes ad resource assgmet. Whe ew jobs become avalable for processg, there are already jobs assged to tme tervals ad correspodg maches whle the ew oes have to be processed usg the remag free processg tervals. A smlar problem occurs operatg systems for sgle- ad mult-processors whe subprograms wth hgher prorty have to be scheduled before subprograms wth lower prorty. Max-Plack-Isttut fur Iformatk, Im Stadtwald, Saarbrucke, Germay. E-mal: albers@mp-sb.mpg.de. Part of ths work was doe whle vstg the Free Uverstat Berl. y Iformato ad Techology Maagemet, Uversty of Saarlad, Saarbrucke, Germay. Emal: gs@tm.u-sb.de. Part of ths work was doe whle vstg the ICSI Berkeley. Ths research was partally supported by INTAS (Project INTAS ). 1

2 Thus, lmted mache avalablty s commo practce. Kowledge about mache avalabltes mght be complete or complete. I a ole settg mache avalabltes are ot kow advace. Mache breakdows are a typcal example of evets that arse ole. Sometmes a scheduler has partal kowledge of the avalabltes,.e, he has some lookahead. He mght kow of the ext tme terval where a mache requres mateace or he mght kow whe a broke mache wll be avalable aga. I a oe settg all mache avalabltes are kow pror to schedule geerato. I ths paper we study a very basc schedulg problem wth respect to lmted mache avalablty: A set of jobs has to be scheduled o a set of detcal maches so as to mmze the makespa. More speccally, let J = fj j = 1; : : : ; g be a set of depedet jobs to be scheduled. Job J has a processg tme of p tme uts kow advace, 1. The jobs have to be scheduled o a set of maches that operate wth the same speed. At ay tme preempto of jobs s allowed at o pealty. Also, the mmum tme slce for preempto may be arbtrarly small. The curret state of a preempted job s saved for the mache system. If a job s preempted, the t may be resumed later o a ay mache. Each mache may work oly o oe job at a tme, ad each job may be processed by oly oe mache at a tme. We wsh to mmze the makespa,.e., the completo tme of the last job that shes. Maches may have deret tme tervals of avalablty. We emphasze here that we are terested the ole verso of the problem where the mache avalabltes are ot kow advace. We also call a terval where a mache s ot avalable a mache break dow. Maches may break dow or recover at arbtrary tme staces. New maches may be added as well. If a mache breaks dow, the the job curretly beg processed s smply preempted. We also cosder the ole problem wth lookahead oe,.e., a scheduler always kows the ext pot tme where the set of avalable maches chages. However, he does ot have to kow whch maches break dow or become avalable. I the prevous lterature [4, 6], ths settg s also referred to as early ole. Gve a schedulg algorthm A ad a problem stace, let Cmax A deote the makespa of the schedule produced by A. I partcular, Cmax OP T deotes the makespa of a optmal oe algorthm that kows the mache avalabltes advace. Followg [9] we call a ole schedulg algorthm A c-compettve f, for all problem staces, Cmax A c COP max T. Related work: Schmdt [7] was the rst who studed schedulg problems wth lmted mache avalablty. He cocetrated o the oe verso of the above problem whe all the mache breakdow tmes are kow advace. Note that f the dow tmes are detcal for all the maches, the a optmal schedule ca be costructed usg McNaughto's algorthm [3]. The algorthm rus O() tme ad uses o more tha S 1 preemptos, where S s the total umber of tervals where maches are avalable. Schmdt [7] studed the problem wth arbtrary mache avalabltes ad gave a algorthm that always costructs a optmal schedule. Hs algorthm has a rug tme of O( + m log m) ad uses at most S 1 preemptos f the tervals of avalablty are rearraged such that they form a starcase patter. Aga, S s the total umber of tervals where maches are avalable. I [8] the problem s geeralzed takg to accout deret job release tmes or deadles. 2

3 There are results for early ole problems,.e., the ext pot tme whe a mache breaks dow or recovers s kow. I [4], Salavlle presets a algorthm for the problem varat that jobs have release ad due dates ad the goal s to mmze maxmum lateess. At ay pot tme, the algorthm also has to kow the ext release date. The algorthm costructs optmal schedules for zgzag mache avalablty patters (oly m or m 1 maches are avalable at ay pot tme) but ot for arbtrary patters. The rug tme of the algorthm s O( 2 p max + T up ), where p max s the logest processg tme of the jobs ad T up s the total tme eeded to update the set of avalable maches. Salavlle [4] also reports that hs algorthm costructs optmal schedules for arbtrary avalablty patters f there are o release dates ad the objectve to mmze the makespa. However, ether the paper or a prvate commucato [5] cotas ay addtoal formato regardg the optmalty proof. As for the ole settg, schedulg wth uexpected mache breakdows was studed by Kalyaasudaram ad Pruhs [1, 2]. I [1] ole algorthms wth optmal compettve ratos are gve for varous umbers of faulty maches. The authors assume that f a mache breaks dow, the job curretly beg processed has to be restarted later from the begg. Also two specc types of breakdows are cosdered. I a permaet breakdow a mache does ot recover aga; a traset breakdow the mache s avalable aga rght after the breakdow. Ths s deret from the problem settg we cosder. I [2] Kalyaasudaram exame to whch extet redudacy ca help ole schedulg wth faulty maches. Our cotrbuto: I ths paper we study the schedulg problem deed above. As metoed before we are maly terested the ole verso of the problem. I Secto 2 we prove that o ole algorthm ca costruct optmal schedules f maches ca break dow ad recover at arbtrary tme staces. We also show that o ole algorthm ca acheve a bouded compettve rato f there may be tme tervals where o mache s avalable. I Secto 3 we preset a ole algorthm that costructs schedules wth a optmal makespa of Cmax OP T f a lookahead of oe s gve,.e., the algorthm always kows the ext pot tme whe the set of avalable maches chages. However, the algorthm does ot eed to kow whch maches break dow or become avalable. Our algorthm has a rug tme of O(a + T up ), where a s the umber of tme staces where the set of avalable maches chages ad T up s aga the tme to update the set of avalable maches. Note that our algorthm has a better rug tme tha Salavlle's f a < p max, whch wll be true practcal applcatos. If a p max, the the set of avalable maches chages after each tme ut. Fally, Secto 4 we gve a ole algorthm wthout lookahead that costructs schedules wth a early optmal makespa of Cmax OP T +, for ay > 0, f at ay tme at least oe mache s avalable. Ths mples that ot kowg mache avalabltes does ot really hurt the performace of a algorthm. 3

4 2 The performace of ole algorthms Frst ote that f at ay tme at most oe mache s avalable, a optmal ole schedule s trval to costruct. I the followg we cocetrate o problems wth a arbtrary set of maches. Theorem 1 No ole algorthm ca, geeral, costruct optmal schedules. If there may be tme tervals where o maches are avalable, the o ole algorthm ca acheve a bouded compettve rato. Proof: Let A be ay ole algorthm. Itally, at tme t = 0 oly oe of m maches s avalable. We cosder jobs J 1 ; : : : ; J, each of whch has a processg tme of 1 tme ut. We asume = m. At tme t = 0, algorthm A starts processg oe job J 0. Let t 0 be the rst tme stace such that A rst preempts J 0 or A shes processg J 0. At that tme t 0 all maches become avalable. A's makespa s at least t because oe of the jobs J, 6= 0, has bee processed so far. A optmal algorthm wll dvde the terval from 0 to t 0 evely amog the jobs so that ts makespa s Cmax OP T = t (t 0 =). Ths proves the rst part of the theorem. For the proof of the secod part we modfy the problem stace so that o mache s avalable durg the terval (Cmax OP T ; c COP max T ], for ay c > 1. The algorthm A caot sh before c Cmax OP T because t has jobs left at tme Cmax OP T. 2 3 Optmal schedules I ths secto we gve a algorthm that costructs optmal schedules wth a makespa of C OP T max. The algorthm s ole wth a lookahead of oe,.e., the algorthm always kows the ext pot tme whe the set of avalable maches chages. The algorthm does ot eed to kow, however, whch maches break dow or become avalable. Let J 1 ; : : : ; J be the gve jobs ad let p, 1, deote the processg tme of J. We assume that p s kow advace. Wthout loss of geeralty jobs are umbered such that p 1 p 2 : : : p. At ay tme durg the schedulg process, r deotes the remag processg tme of J, 1. We wll show later that the algorthm always matas the varat r 1 r 2 : : : r. Startg at tme t = 0, the algorthm repeatedly schedules tme tervals I = [t; t 0 ) whch the set of avalable maches remas the same. The avalablty chaged at t ad wll ext chage at tme t 0. I each terval, the algorthm schedules as much load as possble whle mmzg the legth of the largest remag processg tme. More speccally, suppose that the algorthm has already scheduled the terval [0; t) ad that the set of avalable maches chages at t. At tme t, usg lookahead formato, the algorthm determes the ext pot tme t 0 > t at whch the mache avalablty chages. Let = t 0 t ad m av be the umber of avalable maches I = [t; t 0 ). Itutvely, the algorthm ow tres to determe the largest possble, r 1 > 0, such that, for all jobs J k, 4

5 1 k, the remag processg tme excess to r 1 ca be scheduled I. Thus, at the ed of I, all jobs would have a remag processg tme of at most r 1. Fgure 1 shows a example. Pctorally, the algorthm determes a vertcal le such that the total shaded processg tme to the rght of ths le s equal to total processg capacty avalable I. Note that the total processg tme excess to r 1 s X k=1 maxf0; r k (r 1 )g ad that the total processg capacty avalable I s m av. Thus the algorthm computes a such that k=1 maxf0; r k (r 1 )g = m av. r 1 r 2 r 3 r Fgure 1: The choce of However, the algorthm has to satsfy P the costrat that at most tme uts of each job ca be scheduled I. Thus, f k=1 maxf0; r k (r 1 )g = m av for some >, the the algorthm caot schedule tme uts of J 1 I. Oly tme uts are permssble. For ths reaso, the algorthm rst determes a set of jobs that are scheduled for tme uts I, see les 6{8 of the code Fgure 4. Suppose that the algorthm has already scheduled tme uts of J 1 ; : : : ; J 1 the terval I. Let m av be the curret umber of avalable maches after these rst 1 jobs J 1 ; : : : ; J 1 have bee scheduled. Note that m av = m av ( 1). The total remag processg capacty I s equal to m av. The algorthm also schedules tme uts of J I f the total remag processg tme excess to P r s ot sucet to ll the processg capacty stll avalable ad r. (Formally, f k= maxf0; r k (r )g < m av ad r.) Suppose that the whle-loop les 6{8 termates ad >. The, the algorthm ca schedule o more jobs I. If, the there are two cases to cosder. (a) k= maxf0; r k (r )g m av I ths case, the algorthm determes the, 0 <, such that for all jobs J k, 5

6 r 1 r 1 r 1 r 1 r r r r Fgure 2: A example of case a) Fgure 3: A example of case b) k, the total remag processg tme excess to r s exactly equal to m av, see Fgure 2. Each of these jobs s scheduled I to a extet of maxf0; r k (r )g. (b) k= maxf0; r k (r )g < m av ad r < I ths case, the algorthm ca schedule the rest of J ; : : : ; J, f t exsts, I, see Fgure 3. I each case, the schedulg of the jobs s doe usg McNaughto's algorthm. Algorthm Lookahead (LA) 1. t := 0; 2. r := p, for 1 ; 3. whle there exst jobs wth postve remag processg tme do 4. t 0 := ext pot tme whe set of avalable maches chages; 5. := t 0 t; := 1; m av 1 := umber of maches avalable [t; t + ); 6. whle ad k= maxf0; r k (r )g < m av ad r do 7. Schedule tme uts of J [t; t + ); 8. r := r ; m av +1 := mav 1; := + 1; 9. f the 10. Compute the maxmum, mf; r g, such that k= maxf0; r k (r )g m av ; 11. For k = ; : : : ;, schedule maxf0; r k (r )g tme uts of J k [t; t + ) usg McNaughto's algorthm ad set r k = mfr k ; r g; 12. t := t 0 ; Fgure 4: The ole algorthm wth a lookahead of oe We aalyze the rug tme of the algorthm ad rst argue that wth a terato of the outer whle-loop, all executos of les 6{8 take O() tme. The crtcal part are the computatos of the sums S = k= maxf0; r k (r )g. Set S 0 = 0. We show that S +1 6

7 ca be easly derved from S. Whe computg S we determe the largest job dex l such that r l (r ) 0. We wll show below that r 1 r 2 : : : r, see Lemma 1. Gve l, we ca easly d l +1 by gog through the jobs startg wth J l +1 ad d P the largest dex l +1 l such that r l+1 (r +1 ) 0. The S +1 = S +(l )(r r +1 )+ +1 k=l +1 (r k (r +1 )). Thus all sums ca be computed O() tme. Smlarly, le 10, we ca compute the desred O() tme. Hece, the schedulg process each terval I = [t; t 0 ) ca be doe O() tme. Thus the total rug tme of our algorthm s O(a + T up ), where a s the umber of tmes staces where the set of avalable maches chages ad T up s the tme to update the set of avalable maches. If we represet the set of actve maches as a balaced tree, the each mache avalablty chage ca be mplemeted O(log m av max) tme, where m av max s the maxmum umber of maches ever avalable. Let B deote the total umber of mache breakdows. The T up = O(B log m av max). I the aalyss of the algorthm we cosder the sequece of tervals whch LA schedules jobs. Wth each terval, the set of avalable maches remas the same. Mache avalablty oly chages at the begg of a terval. We rst show that the algorthm works correctly. Whe the algorthm termates, all jobs have a remag processg tme of zero,.e. the schedulg process s complete. The codto le 6 of the algorthm esures that at most = t 0 t tme uts of each P job are scheduled a terval. The assgmet m av +1 := mav 1 le 8 ad the costrat k= maxf0; r k (r )g m av le 10 esure that the total amout of processg tme scheduled a terval s ot greater tha the avalable processg capacty. Next we prove two useful lemmas. Lemma 1 At the begg of each terval, r 1 r 2 : : : r. Proof: The varat holds at tme t = 0 because tally r k = p k, for 1 k, ad p 1 p 2 : : : p. Suppose that r 1 r 2 : : : r holds at the begg of some terval I. We show that the varat s also satsed at the ed of I. Let r1 0 ; : : : ; r0 deote the remag processg tmes at the ed of I. Suppose that whle executg the whle-loop les 6{8, the algorthm schedules tme uts of J 1 ; : : : ; J 1. The remag processg tme of each of these jobs decreases by ad thus r1 0 : : : r0 1. If >, we are doe. Otherwse we have to cosder two cases. (a) k= maxf0; r k (r )g m av If P > 1, the the last terato of the whle-loop, mplesp the codto le 6 was satsed,.e. k= 1 maxf0; r k (r 1 )g < m av 1 ; whch k= maxf0; r k (r 1 )g < m av : I le 10, the algorthm chooses a such that k= maxf0; r k (r )g = m av. Thus, f > 1, r 1 0 = r 1 > r = r 0. For ay 1, the varat ow follows because r 0 = : : : = r0, where l l s the largest job dex such that r l (r ) 0, ad rk 0 = r k for k > l. (b) k= maxf0; r k (r )g < m av ad r < I ths case, the rest of J ; : : : ; J s scheduled I,.e. r 0 = : : : = r0 = 0 ad the varat holds. 2 7

8 Now cosder ay other algorthm A for schedulg J 1 ; : : : ; J. I partcular, A may be a optmal algorthm that kows the mache breakdows advace. At ay tme cosder the sorted sequece q 1 q 2 : : : q of remag processg tmes mataed by A. That s, q s the -th value the sorted sequece, 1. Note that q s ot ecessarly the remag processg tme of J. Lemma 2 At the begg of each terval, r 1 q 1 ad k=1 r k k=1 q k. Proof: We show ductvely that at the begg of each terval jx k=1 r k jx k=1 q k for j = 1; : : : ; : (1) The lemma follows from the specal case j = 1 ad j =. The above equaltes hold at tme t = 0. Suppose that they hold at the begg of some terval I. We show that they are also satsed at the ed of I,.e. at the begg of the terval followg I. Let r1 0 ; : : : ; r0 ad q1 0 ; : : : ; q0 be the remag processg tmes at the ed of I. Recall that r 0 k s the remag processg tme of J k, 1 k. By Lemma 1, r1 0 : : : r0. We have q1 0 : : : q0 by the deto of the q-values. Note that q k ad q 0 k ca be the processg tmes of deret jobs. However, qk 0 q k for 1 k. Suppose that les 6{8, algorthm LA schedules tme uts of J 1 ; : : : ; J 1. The r 0 k = r k, for k = 1; : : : ; 1. We have q 0 k q k, for 1 k, because the processg tmes of jobs decrease by at most I. Thus, equalty (1) holds for j = 1; : : : ; 1. Aga, for, we cosder two cases. (a) k= maxf0; r k (r )g < m av ad r < The algorthm LA schedules the rest of J ; : : : ; J I so that r 0 = : : : = r 0 = 0. Iequalty (1) also holds for j = ; : : : ;. (b) k= maxf0; r k (r )g m av LA computes a, 0 <, such that k= maxf0; r k (r )g = m av. It reduces the remag processg tmes of J ; : : : ; J l to r, where l s the largest job dex such that r l (r ) 0. Let m av 1 be the umber of maches that P were tally P avalable I. Sce LA uses all j of the avalable processg capacty, k=1 r0 = j k k=1 r k m av 1 for j = l; : : : ;. Sce P P j k=1 q0 k j k=1 q k m av 1 for j = l; : : : ;, equalty (1) holds for j = l; : : : ;. It remas to show that the equalty s also satsed for j = ; : : : ; l 1. P P P P 1 Let R 1 = k=1 r0, k R l 2 = k= r0 ad smlarly k Q 1 1 = k=1 q0, k Q l 2 = k= q0 k. We have already show () R 1 Q 1 ad () R 1 + R 2 Q 1 + Q 2. Suppose that Q 1 = R 1 + x for some x 0. The () mples Q 2 + x R 2. Cosder the l + 1 values q 0 ; : : : ; q0. l Sce q 0 : : : q0 l, the sum of the rst values, for ay 1 l + 1, s at least Q 2 =(l + 1). Thus, for ay j wth j l, jx k=1 Q 2 q 0 k Q 1 + (j + 1) l + 1 = R 1 + x + (j + 1) l Q 2

9 = R 1 + (j + 1) Q 2 + x l + 1 R 1 + (j + 1) l + 1 jx k=1 r 0 k : The last equato follows because r = r 0 = r0 +1 = : : : = r0 l = R 2=(l + 1). 2 Theorem 2 For ay problem stace, C LA max = C OP T max. R 2 Proof: Gve a set of jobs J 1 ; : : : ; J, let I = [t; t 0 ) be the last terval whch LA has scheduled jobs,.e., t Cmax LA t0. Cosder the makespa Cmax OP T produced by a optmal oe algorthm. We dstgush two cases. (1) I the ole schedule, the terval from t to Cmax LA cotas o dle maches Thus, the ole schedule all maches sh at the same P tme. Lemma 2 mples that at the begg of I, the total remag processg tme k=1 r k of LA s ot greater tha the total remag processg tme k=1 q k of OPT. Thus, Cmax OP T CLA max. (2) I the ole schedule, the terval from t to Cmax LA cotas dle maches Sce LA schedules job portos usg McNaughto's algorthm, there must exst a job that spas the etre terval from t to Cmax. LA Thus, at the begg of I the largest remag processg tme r 1 equals Cmax LA t. By Lemma 2, the largest remag processg tme q 1 of OPT s ot smaller. Thus OPT caot sh earler tha LA. 2 4 Nearly optmal schedules I ths secto we study the problem that a ole algorthm has o formato about the future mache avalabltes. It does ot kow the ext pot tme whe the set of avalable maches chages. We preset a algorthm that always produces a makespa of Cmax OP T +, for ay > 0. It s assumed that at ay tme at least oe mache s avalable sce otherwse, by Theorem 1, o bouded performace guaratee ca be acheved. We umber the jobs to be scheduled such that p 1 p 2 : : : p. Gve a xed > 0, our ole algorthm, called ON(), computes = = 2. Startg at tme t = 0, the algorthm always schedules jobs wth the tme terval [t; t + ). Let m av be the umber of maches avalable at tme t. The algorthm determes the m av jobs wth the largest remag processg tmes (tes are broke arbtrarly) ad schedules them o the avalable maches. If a mache breaks dow or becomes avalable at some tme t + 0, 0 <, the the algorthm preempts the jobs curretly beg processed ad computes a ew schedule for the ext tme uts from t + 0 to t Otherwse, f the set of avalable maches remas the same throughout [t; t + ), the algorthm computes a ew partal schedule at tme t +. Let a be the umber of tme staces where the set of avalable maches chages. The total umber of tervals scheduled by the algorthm s at least a. A formal descrpto 9

10 Algorthm Ole() (ON()) 1. t := 0; = = 2 ; 2. r := p, for 1 ; 3. whle there exst jobs wth postve remag processg tme do 4. m av := umber of maches avalable at tme t; 5. av := umber of jobs wth postve remag processg tme; 6. S := set of the mfm av ; av g jobs wth the largest remag processg tme; 7. Process the jobs J, 2 S, o the avalable maches; 8. f maches break dow or become avalable at some tme t + 0, 0 < the 9. Set r := maxf0; r 0 g for 2 S; t := t + 0 ; 10. else 11. Set r := maxf0; r g for 2 S; t := t + ; Fgure 5: The ole algorthm ON() of the algorthm s gve Fgure 5. At ay tme r deotes the remag processg tme of J, 1. I the schedulg process, the algorthm repeatedly has to d jobs wth the largest remag processg tme. If we keep a prorty queue of the remag processg tmes, each such job ca be foud O(log ) tme. Let m av, 1 a, be the umber of maches that are avalable rght after P the -th chage; m av 0 s the umber of maches that are avalable tally. Let P = =1 p. Note P that the total umber of job portos scheduled by the a algorthm s o more tha P= + =0 mav. Ths s because at the ed of a scheduled job porto, tme uts have bee processed or the set P of avalable maches chages. Thus the total rug tme of the algorthm s O(( 2 a = + =0 mav ) log + T up ), where T up s the tme eeded to update the set of avalable maches. As the aalyss of the algorthm LA we ca show that T up = O(B log m av max), where B s the total umber of mache breakdows ad m av max = max 0a m av. Jobs are oly preempted at the ed of a terval of legth = = 2 or whe the set P of avalable maches chages. Thus the umber of preemptos s o more tha 2 a = + =0 mav. For the aalyss of the algorthm we partto the tme to tervals such that at the begg of a terval the ole algorthm computed a ew partal schedule,.e., t executed les 4{7. Note that tervals have a legth of at most ad that wth each terval the set of avalable maches remas the same. The algorthm ON() does ot mata the property that the remag processg tmes r 1 ; r 2 ; : : : ; r ecessarly form a o-creasg sequece (cf. Lemma 1). However, the ext lemma shows that f a job J j has a larger remag processg tme tha a job J ad < j, the the derece s bouded by. Lemma 3 At the begg of each terval, for ay two jobs J ad J j wth < j, r r j. Proof: The lemma holds at the begg of the rst terval because, tally, r 1 : : : r. 10

11 Suppose that the lemma holds at the begg of a terval I = [t; t + 0 ), for some 0. We show that the lemma s also satsed at the ed of I. Let ad j be the umber of tme uts for whch J ad J j are processed I. We have 0 ; j 0. If j, the there s othg to show. We study the case j <. Let r k ad rk 0, k 2 f; jg, deote the remag processg tmes at the begg ad at the ed of I. If j = 0,.e. oly J s processed I, the r r j ad thus r 0 = r r r j = rj 0. Fally, the case 0 < j < ca oly occur f the processg of J j shes durg I,.e., rj 0 = 0. The lemma holds because r I the followg aalyss, we have to boud the remag processg tmes mataed by ON() terms of the remag processg tmes mataed by a optmal oe algorthm. I the prevous secto, whe aalyzg the algorthm LA, we could show that the prex sum P j k=1 r k are bouded by the prex sum P j k=1 q k, for j = 1; : : : ;, see (1). Ufortuately, ths relato does ot hold the algorthm ON(). Problems arse f some terval there exst jobs J ad J j wth < j such that J s ot scheduled but J j s scheduled the terval. For ths reaso we mata a sequece of job sets S 1 ; : : : ; S l, for some 1 l, whch s a partto of the job sequece J 1 ; : : : ; J. Itutvely, a set S k, 1 k l, cossts of jobs that have \early the same" remag processg tme. Ths wll be made precse Lemma 5. If there are jobs J ad J j wth < j such that J s ot scheduled but J j s scheduled some terval, the we merge the sets cotag J ; : : : ; J j. Ths way we wll be able to boud the prex sums deed by the set S 1 ; : : : S l, see Lemma 4 below. Formally, the sets are mataed as follows. Itally, at tme 0, S cotas J, 1. At the ed of each terval I, the sets are updated as follows. Let be the smallest job dex such that J was ot processed I ad let j be the largest job dex such that J j was processed I. Suppose that J 2 S k ad J j 2 S kj. If k < k j, the replace S k ; S k +1; : : : ; S kj by the uo of these sets. Reumber the ew sequece of sets so that the k-th set the sequece has dex k. Fgure 6 shows a example of the update algorthm for sets. Suppose that some terval three maches avalable ad jobs J 1, J 3 ad J 4 are scheduled for tme uts (the shaded job portos). Job J 2 s the rst job ot scheduled ad J 5 s the last job scheduled that terval. Thus sets S 1 ad S 2 are merged. Note that, as metoed above, at ay tme the sequece of sets forms a parttog of the jobs J 1 ; : : : ; J. The update rule esures that every set cotas a sequece of cosecutve jobs wth respect to the job umberg. I the followg, let k deote the umber of jobs S k, ad let N k = 1 + : : : + k. At ay tme let l max deote the maxmum dex such that S 1 ; : : : ; S lmax cota oly jobs wth postve remag processg tmes. If there s o such set, the let l max = 0. Let A be ay other schedulg algorthm. I partcular, A may be a optmal oe algorthm. At ay tme cosder the sequece of remag processg tmes mataed by A, sorted o-creasg order. Let q be the -th value ths sorted sequece. 11

12 r 1 S 1 r 2 r 3 j r 4 S 2 r 5 r 6 r 7 S 3 Fgure 6: A example of the update rule for sets Lemma 4 At the begg of each terval, for k = 1; : : : ; l max, P N k =1 r P N k =1 q. Proof: The lemma holds tally because at tme t = 0, r = q = p for 1. Suppose that the lemma holds at the begg of a terval I = [t; t + 0 ), for some 0. Let S 1 ; : : : ; S l ad S1 0 ; : : : ; S0 0 l be the sequeces of job sets at the begg ad at the ed of I. Furthermore, let j be the largest dex such that all jobs S1 0 ; : : : ; S0 j were scheduled I ad stll have a postve remag processg tme. These sets were ot volved a merge operato at the ed of I ad, hece, each S 0 cotas the same jobs as k S k, 1 k j. Sce the jobs of these sets have a postve remag processg tme, all of them were scheduled for exactly 0 tme uts I. Let r ; r 0 ad q ; q 0 deote the remag processg tmes at the begg ad at the ed of I. Sce q 0 q 0, for 1, we obta =1 r 0 = =1 (r 0 ) for k = 1; : : : ; j. If j = l 0 max, the we are doe. =1 (q 0 ) Suppose that j < lmax. 0 By the deto of lmax, 0 the set Sj+1 0 does ot cota jobs wth zero remag processg tme. Also, by the deto of j, Sj+1 0 cotas jobs ot scheduled I. The update rule for job sets esures that Sj+1 0 cotas all jobs J, > N j, that were scheduled I. Let N be the umber of jobs Sj+1 0 scheduled I. All of these jobs were scheduled for 0 tme uts because they all have postve remag processg tme. The total umber of avalable maches I s N j + N sce, otherwse, the algorthm ON() would have scheduled more jobs of Sj+1 0 I. Thus ay other algorthm caot process more tha (N j + N) tme uts I. We coclude =1 r 0 = =1 r (N j + N) 0 12 =1 =1 q 0 ; q (N j + N) 0 =1 q 0 ;

13 for k = j + 1; : : : ; l 0 max. 2 Whle l max > 0, the above lemma esures that a optmal oe algorthm has a total ozero remag processg tme. Whe l max = 0, we have to be able to boud the total remag processg tme of ON(). For ths purpose we aalyze the derece remag tmes that ca occur a job set S k. Lemma 5 At the begg of each terval, for every set S k, 1 k l, ad jobs J ; J j 2 S k, jr r j j ( 1). The boud gve the above lemma s a overestmate, whch s sucet for the rest of the aalyss. However, there exst problem staces such that jr r j j (=2). Proof: We prove ductvely that at the begg of each terval, for every set S k ad jobs J ; J j 2 S k, jr r j j ( k 1): (2) Ths holds tally because at tme t = 0, every set cotas exactly oe job. Cosder a terval I = [t; t + 0 ), for some 0, ad suppose (2) holds at the begg of I. We rst show that (2) s mataed whle jobs are processed I ad before the update rule for the sets s appled. Gve a set S k, let J ; J j 2 S k be ay two jobs wth < j. Let r ; r 0 ad r j ; r 0 j be the remag processg tmes at the begg ad at the ed of I. If r0 r0 j, the by Lemma 3, jr 0 r0 j j = r0 j r0. If r 0 > r0 j, we have to cosder several cases. If oe of the two jobs was processed I or f both jobs were processed for 0 tme uts, the there s othg to show. Otherwse, let ad j be the umber of tme uts for whch J ad J j are processed I. If oly J j s processed I, the r j r ad thus jr 0 r0 j j = r0 r0 = j r (r j j ) j. The case that both J ad J j are scheduled I, but J j s processed for a loger perod, caot occur. Ths would mply that the processg of J s complete,.e. r 0 = 0, whch cotradcts r 0 > r0 j. Fally suppose that J s processed as least as log as J j I,.e. 0 j. The jr 0 r0 j j = r (r j j ) = r r j + j r r j ( k 1). Iequalty (2) s satsed. We ow study the eect whe the set update rule s appled at the ed of I. Suppose that a sequece of sets S k1 ; : : : ; S k2 s merged. Let J 2 S k1 be a job ot scheduled I ad let J j 2 S k2 be the job wth the largest dex scheduled I. Let J max be the job S k1 ; : : : ; S k2 wth the largest remag processg tme at tme t+ 0 ad let J m be the job S k1 ; : : : ; S k2 wth the smallest remag processg tme. We wll show jrmax 0 P rm 0 j ( k 1 + k2 1). k Ths completes the proof because the ewly merged set cotas 2 k=k 1 k k1 + k2 jobs. We have jr 0 max r 0 mj = r 0 max r 0 m = (r 0 max r 0 ) + (r 0 r 0 j) + (r 0 j r 0 m): If J max 2 S k1, the rmax 0 r 0 ( k 1 1). If J max =2 S k1, the rmax 0 r 0 by Lemma 3 because J max has a hgher dex tha J. I ay case rmax 0 r 0 ( k 1 1). Smlarly, f J m 2 S k2, the rj 0 r0 m ( k 2 1). If J m =2 S k2, the rj 0 r0 m by Lemma 3. I ay case rj 0 r0 m ( k 2 1). Sce J was ot scheduled I but J j was scheduled, r j r. 13

14 Job J j was scheduled for at most tme uts, whch mples r 0 = r r j rj 0 + ad hece r 0 r0 j. I summary we obta jr 0 max r0 m j = (r0 max r0 ) + (r 0 r0 j) + (r 0 j r0 m) ( k1 1) + + ( k2 1) = ( k1 + k2 1): 2 Theorem 3 For ay xed > 0 ad ay problem stace, C ON() max C OP T max +. Proof: Let I = [t; t + 0 ), 0, be the last terval such that l max > 0 at the begg of I. Cosder the total remag processg tme of the jobs S 1 ; : : : ; S lmax at tme t. By Lemma 4, the value of ON() s ot larger tha the value of a optmal oe algorthm. Thus Cmax OP T t. We aalyse ON()'s makespa. At tme t+0, S 1 cotas a job J wth zero remag processg tme. By Lemma 5, all jobs belogg to the rst set have a remag processg tme of at most ( 1). All jobs ot belogg to the rst set have a hgher dex tha J ad, by Lemma 3, they have a remag processg tme of at most. Thus at tme t + 0, we are left wth at most 1 jobs havg a remag processg tme of at most ( 1),.e., the total remag processg tme of ON() s at most ( 1) 2. Sce at ay tme at least oe mache s avalable Cmax ON() t+ 0 +( 1) 2 Cmax OP T + 2 Cmax OP T +: 2 Ackowledgmet We thak Olver Brau for may terestg dscussos. Moreover, we thak two aoymous referees for ther helpful commets mprovg the presetato of the paper. Refereces [1] B. Kalyaasudaram ad K.P. Pruhs. Fault-tolerat schedulg. I Proceedgs of the 26th Aual ACM Symposum o the Theory of Computg, pages 115{124, [2] B. Kalyaasudaram ad K.P. Pruhs. Fault-tolerat real-tme schedulg. I Proc. 5th Aual Europea Symposum o Algorthms (ESA), Sprger Lecture Notes Computer Scece, [3] R. McNaughto. Schedulg wth deadles ad loss fuctos. Maagemet Scece, 6:1-12, [4] E. Salavlle. Nearly o le schedulg of preemptve depedet tasks. Dscrete Appled Mathematcs, 57:229{241, [5] E. Salavlle. Prvate commucato, [6] E. Salavlle ad G. Schmdt. Mache schedulg wth avalablty costrats. Acta Iformatca, 35:795{811, [7] G. Schmdt. Schedulg o sem-detcal processors. Z. Oper. Res., 28:153{162,

15 [8] G. Schmdt. Schedulg depedet tasks wth deadles o sem-detcal processors. J. Oper. Res. Soc., 39:271{277, [9] D.D. Sleator ad R.E. Tarja. Amortzed ececy of lst update ad pagg rules. Commucatos of the ACM, 28:202{208,