10/04/2018. Definitions. The scheduling problem. Feasibility vs. schedulability. Complexity. Why do we care about complexity? P R

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1 //8 Defos schedule s sad o be feasble f sasfes a se of cosras. as se s sad o be feasble, f here exss a algorhm ha geeraes a feasble schedule for. asse s sad o be schedulable wh a algorhm, f geeraes a feasble schedule. Examples of cosras mg cosras: acvao, perod, deadle, jer. Precedece: order of execuo bewee ass. Resources: sychrozao for muual excluso. Feasbly vs. schedulably he schedulg problem ufeasble as se Feasble as ses Gvease of ass, a se P of p processors, ad a se R of r resources, fd a assgme of P ad R o ha produces a feasble schedule uder a se of cosras. Space of all as ses as ses schedulable wh alg. as ses schedulable wh alg. P R Schedulg algorhm feasble cosras omplexy I 97, Garey ad Johso showed ha he geeral schedulg problem s NP hard. I pracce, meas ha he me for fdg a feasble schedule grows expoeally wh he umber of ass. Foruaely, polyomal me algorhms ca be foud uder parcular codos. Why do we care abou complexy? Le s cosder a applcao wh =ass o a processor whch he elemeary sep aes s osder algorhms wh he followg complexy: : O() : O( 8 ) : O(8 ) s 8 hours. bllo years

2 //8 Smplfyg assumpos as se assumpos Sgle processor Homogeeous as ses Fully preempve ass Smulaeous acvaos No precedece cosras No resource cosras We cosder algorhms for dffere ypes of ass: Sgle-job ass (oe sho) ass wh a sgle acvao (o recurre) Perodc ass recurre ass regularly acvaed by a mer (each as poeally geeraes fe jobs) perodc/sporadc ass recurre ass rregularly acvaed by eves (each as poeally geeraes fe jobs) Mxed as ses lasscal schedulg polces Frs ome Frs Served Shores Job Frs Prory Schedulg Roud Rob Frs ome Frs Served I assgs he PU o ass based o her arrval mes (rscally o preempve): Ready queue PU No sued for real-me sysems a a a a s s s f Very upredcable Frs ome Frs Served respose mes srogly deped o as arrvals: a a a R = R = R = 8 8 a a a R = R = 8 R = 8 8 Shores Job Frs (SJF) I selecs he ready as wh he shores compuao me. Sac ( s a cosa parameer) I ca be used o le or off le a be preempve or o preempve I mmzes he average respose me

3 //8 SJF - Opmaly SJF - Opmaly SJF r S L L S f S < f L f L = f S... * R ( ) R( ') R( '')... R(*) R( ') f S + f L ( f ' r ) f S + f L ( f r ) R( ) R( SJF ) * = SJF s he mmum respose me achevable by ay algorhm Is SJF sued for Real-me? Prory Schedulg I s o opmal he sese of feasbly SJF feasble d d d 8 8 Each as has a prory P,ypcallyP [, ] he as wh he hghes prory s seleced for execuo. asswh hesameproryareserved FFS d d d SJF o feasble 8 8 NOE: p / SJF p /a FFS Prory Schedulg Problem: sarvao low prory ass may experece log delays due o he preempo of hgh prory ass. possble soluo: agg prory creases wh wag me Roud Rob he ready queue s served wh FFS, bu... Each as cao execue for more ha Q me us (Q = me quaum). Whe Q expres, s pu bac he queue. REDY queue PU Q expred 7 8

4 //8 Roud Rob = umber of as he sysem Q Q Roud Rob f Q>max( ) he RR FFS f Q coex swch me () he Q me sharg R ( Q) Q Each as rus as was execug aloe o a vrual processor mes slower ha he real oe. Q + (Q + ) ) (Q + ) ) R ( Q ) Q Q Q 9 Mul-Level Schedulg Mul-Level Schedulg Hgh prory sysem ass PRIORIY prory Medum prory eracve ass RR PU PU Low prory bach ass FFS How o schedule R ass? How o schedule R ass o maxmze feasbly? D d

5 //8 Earles Due Dae [Jacso ] Gve a se of real-me ass arrved smulaeously, execug hem by creasg deadle wll mmze he maxmum laeess (L max ). - - maxmum Laeess EDD EDD - Opmaly r f a < f b f b = f a d a d b -8 L max = L a = f a d a - L a = f a d a < f a d a NOE: No oher scheduler ca decrease L max Preempo s o requred L b = f b d b < f a d a L max < L max EDD - Opmaly EDD guaraee es (off le)... * L ) L ( ') L ( '')... L max ( *) max ( max max * = EDD L ( ) s he mmum value max EDD achevable by ay algorhm f f f f as se s feasble ff f f d D 7 8 Earles Deadle Frs If ass arrve dyamcally, he maxmum laeess ca be mmzed execug hem by creasg absolue deadle, bu preempo mus be eabled. EDF Guaraee es (o le) c () c () c () c () 9 c ( ) d

6 //8 omplexy EDF opmaly EDD Scheduler (queue orderg): Feasbly es (guaraee es): O( log ) O() EDF Scheduler (sero he queue): O() Feasbly es (guaraee sgle as): O() I he sese of feasbly [Derouzos 97] algorhm s opmal he sese of feasbly f geeraes a feasble schedule, f here exss oe. Demosrao mehod I s suffce o prove ha, gve a arbrary feasble schedule, he schedule geeraed by EDF s also feasble. propery of opmal algorhms If a as se s o schedulable by a opmal algorhm, he cao be scheduled by ay oher algorhm. If a algorhm mmzes L max he s also opmal he sese of feasbly. he oppose s o rue. Problem formulao We cosder a compug sysem ha has o execue a se of perodc real-me ass: = {,, } Each as s characerzed by: wors-case compuao me acvao perod D relave deadle al arrval me (phase) Problem formulao (,,, D ) job a d For each perodc as we mus guaraee ha: each job s acvaed a a = + (-) each job complees wh d = a + D here are several wrog ways o acheve hs goal.

7 //8 farm schedulg problem Frs ry Feed cow for m / m lerae pg wh cow Pg ow 8 Feed pg for m / m Evaluao: Pg ges hugry ow ges fa 7 8 Secod ry hrd ry Feed pg ad cow m each Feed pg ad cow m each Pg 8 Pg 8 ow ow Evaluao: Pg s OK Evaluao: Pg s OK, ow s OK ow s o happy bu he farmer s red 9 Opmal algorhm Wha do we lear? Pg ow Feed he mos sarvg amal ( EDF) 8 Evaluao: Everybody s happy Reducg he execuo me wdow, we ge closer o a feasble soluo. he me s spl proporoally bewee he amals. I he example, each amal requred food for % of he me, bu how ca we geeralze he soluo f he amals requre dffere fraco of me? 7

8 //8 ew schedulg problem Feed cow for m / m Proporoal share algorhm asc dea Dvde he mele o slos of equal legh. Wh each slo serve each as for a me proporoal o s ulzao: Pg ulzao facor = / = / ow ulzao facor = / = / Feed pg for m / m Pg / ow / 8 Proporoal share algorhm I geeral Le: U = requred feedg fraco Δ = GD (, ) = 8 execue each as for = U Δ each slo Δ Pg / ow / 8 NOE: U Δ esures, fac: ( /Δ) = Feasbly es: Δ.e. U Proporoal share algorhm hs mehod approxmaes a flud sysem, where execuo progresses proporoally o U he major problem s ha f perods are o harmoc, Δ = GD (,, ) s small ad a as s fragmeed o may chus: /Δ of small durao = U Δ. oo much overhead Wor ad Sleep Wor ad Sleep ccordg o hs mehod, a as execues for us ad he suspeds for us: as Sleep me 8 7 fuco(); sleep(); fuco(); sleep(8); fuco(); sleep(7); Example : (/) (/) (/) as 7 8 Sleep 8 7 I wors well for small compuao mes

9 //8 Wor ad Sleep Loop Schedulg Example : as 8 Sleep Problem Low prory ass experece log delays I s a smple rc o schedule perodc acves a dffere raes usg a sgle loop (ofe used rduo): cou = ; // relave me = ; // perod ms = ; // perod ms = 8; // perod ms (/) (/) 7 7 whle () { f (cou% == ) fuco(); f (cou% == ) fuco(); f (cou% == ) fuco(); (/) cou++; f (cou == **) cou = ; } delay(); // wa for ms Loop Schedulg Noe ha he couer mus be rese a he leas commo mulple of he perods, called he hyperperod (H): cou++; f (cou == **) cou = ; Q: How may bs are eeded o represe he hyperperod? = = = = = = H= bs = log = 9.8 = I does o f o a log eger. We are rouble! Loop Schedulg beer way s o rely o a sysem call ha reurs he sysem me: = ge_me(); a+ = ge_me(); a+ a = a = fuco(); fuco(); Ialzao = ge_me(); a = ; a = ; Loop Schedulg Loop Schedulg Implemeao: #defe N // umber of ass me ; // curre me me a[n], [N]; // ac. mes, perods alze_perods(); // e.g., read from fle Example : as = ge_me(); for (=; <N; ++) a[] = []; whle () { for (=; <N; ++) { = ge_me(); f ( >= a[] + []) { a[] = ; fuco(); } } } (/) (/) (/)

10 //8 Loop Schedulg Loop Schedulg Example : (/) (/) (/) as 9 9 Problem ass experece delays from he oher ass If he scheduler s o he oly hread, a sleep mus be sered: #defe N // umber of ass #defe DEL // mllsecods me ; // curre me me a[n], [N]; // ac. mes, perods alze_perods(); // e.g., read from fle = ge_me(); me(); for (=; <N; ++) a[] = []; whle () { for (=; <N; ++) { = ge_me(); f (-a[] >= []) { a[] = ; fuco(); } } sleep(del); // susped for ms } Loop Schedulg mele Schedulg Example : DEL = as Problem he expereced delay s gve by oher ass + suspeso lso ow as cyclc schedulg, has bee used for years mlary sysems, avgao, ad moorg sysems. Examples (/) (/) r raffc corol sysems Space Shule oeg 777 (/) rbus avgao sysem 9 NOE: Suspeso me ca be hgher due o oher ass 7 8 mele Schedulg mele Schedulg Mehod he me axs s dvded ervals of equal legh (me slos). Example: as ms ms ms ms ms ms = GD (mor cycle) = lcm (major cycle) Each as s sacally allocaed a slo order o mee he desred reques rae. he execuo each slo s acvaed by a mer Guaraee: + +

11 //8 mele Schedulg yclg Schedulg Implemeao: odg: mer mer mer mer mor cycle major cycle #defe MINOR // mor cycle = ms alze_mer(minor); // errup every ms whle () { syc(); // bloc ul errup fuco_(); fuco_(); syc(); // bloc ul errup fuco_(); fuco_(); syc(); // bloc ul errup fuco_(); fuco_(); syc(); // bloc ul errup fuco_(); } mele schedulg dvaages Smple mplemeao (o ROS s requred). Low ru-me overhead. ll ass ru wh very low jer. Dsadvaages I s o robus durg overloads. I s dffcul o expad he schedule. I s o easy o hadle aperodc acves. Problems durg overloads Wha do we do durg as overrus? Le he as coue we ca have a domo effec o all he oher ass (mele brea) bor he as he sysem ca rema cosse saes. Expadbly Expadbly If oe or more ass eed o be upgraded, we may have o re-desg he whole schedule aga. Example: s updaed so ha = ms ow + > We have o spl as wo subass (, ) ad re-buld he schedule: 7 Guaraee: + + +

12 //8 Expadbly Example If he frequecy of some as s chaged, he mpac ca be eve more sgfca: as old ew ms ms ms ms ms ms mor cycle: major cycle: = = = = syc. per cycle! 7 8 Prory Schedulg Mehod. ssg prores o each as based o s mg cosras.. Verfy he feasbly of he schedule usg aalycal echques.. Execue ass o a prory-based erel. 7 How o assg prores? ypcally, as prores are assged based o he her relave mporace. Prory vs. mporace If s more mpora ha ad s assged hgher prory, he schedule may o be feasble: However, dffere prory assgmes ca lead o dffere processor ulzao bouds. P > P deadle mss P > P 7 7

13 //8 Prory vs. mporace If prory are o properly assged, he ulzao boud ca be arbrarly small: applcao ca be ufeasble eve whe he processor s almos empy! P > P U = deadle mss + Opmal prory assgmes Rae Moooc (RM): P / (sac) Deadle Moooc (DM): P /D (sac) Earles Deadle Frs (EDF): P /d d, = r, + D (dyamc) opmal amog FP alg s for = D opmal amog FP alg s for D opmal amog all alg s 7 Rae Moooc s opmal RM s opmal amog all fxed prory algorhms (f D = ): Deadle Moooc s opmal If D he he opmal prory assgme s gve by Deadle Moooc (DM): If here exss a fxed prory assgme whch leads o a feasble schedule, he he RM schedule s feasble. DM P > P If a as se s o schedulable by RM, he cao be scheduled by ay fxed prory assgme. RM P > P 7 7 EDF Opmaly Opmaly EDF s opmal amog all algorhms: EDF DM If here exss a feasble schedule for a as se, he EDF wll geerae a feasble schedule. If a as se s o schedulable by EDF, he cao be scheduled by ay algorhm. dyamc prory (D ) fxed prory (D ) RM fxed prory (D = ) 77 78

14 //8 Rae Moooc (RM) ufeasble RM schedule Each as s assged a fxed prory proporoal o s rae [Lu & Laylad 7]. 7 U p Noe ha small parameer varaos are auomacally hadled by he scheduler whou ay erveo. 9 8 deadle mss 79 EDF Schedule How ca we verfy feasbly? D = U p Each as uses he processor for a fraco of me: U Hece he oal processor ulzao s: U p U p s a measure of he processor load. 8 Idefyg he wors case rcal Isa Feasbly may deped o he al acvaos (phases): U p 9.9 For ay as, he loges respose me occurs whe arrves ogeher wh all hgher prory ass deadle mss R R 8

15 //8 rcal Isa ecessary codo For depede preempve ass uder fxed prores, he crcal sa of, occurs whe arrves ogeher wh all hgher prory ass. ecessary codo for havg a feasble schedule s ha U p. / /8 / I fac, f U p >he processor s overloaded hece he as se cao be schedulable. Idle me / However, here are cases whch U p bu he as se s o schedulable by RM. ufeasble RM schedule Ulzao upper boud U p 9.9 U p deadle mss Gve hs as se (perod cofgurao), wha s he hgher ulzao ha guaraees feasbly? NOE: If or s creased, wll mss s deadle! dffere upper boud dffere upper boud U ub.9 U p NOE: he upper boud U ub depeds o he specfc as se. NOE: he upper boud U ub depeds o he specfc as se. 89 9

16 //8 he leas upper boud suffce codo U ub If U p U lub he as se s ceraly schedulable wh he RM algorhm. U lub... NOE If U lub < U p we cao say ayhg abou he feasbly of ha as se. 9 9 U lub for RM RM Leas Upper oud I 97, Lu ad Laylad proved ha for a se of perodc ass: PU% U lub RM lb / 9% for U lub l 9 specal case If ass have harmoc perods U lub =. U p 8 8 RM Guaraee es We compue he processor ulzao as: U p Guaraee es (oly suffce): U p / 8 9

17 //8 asc ssumpos he Hyperbolc oud. s cosa for every job of. s cosa for every job of I, e al. proved ha a se of perodc ass s schedulable by RM f:. For each as, D =. ass are depede: o precedece relaos o resource cosras o blocg o I/O operaos ( U ) H vs. LL Exeso o ass wh D < U.8 LL U ( U / ) H ( U ) D r, d, r,+ Schedulg algorhms Deadle Moooc: p /D (sac) Earles Deadle Frs: p /d (dyamc).8 U 99 Deadle Moooc Respose me alyss [udsley '9] 8 8 Problem wh he Ulzao oud U p. D bu he as se s schedulable. For each as compue he erferece due o hgher prory ass: I D D compue s respose me as R = +I verfy ha R D 7

18 //8 ompug Ierferece ompug Respose mes R R R Ierferece of o he erval [, R ]: Ierferece o by hgh-prory ass: I I R R Ierave soluo: R R s R ( s) erae whle s R R ( s) Earles Deadle Frs (EDF) Each job receves a absolue deadle: d, =r, +D ay me, he processor s assged o he job wh he earles absolue deadle. Uder EDF, ay as se ca ulze he processor up o %. EDF Example Ufeasble uder RM U p 9.9 U p 9.9 D = deadle mss 7 8 8

19 //8 EDF Opmaly EDF schedulably EDF s opmal amog all algorhms: I 97, Lu ad Laylad proved ha for a se of perodc ass: If here exss a feasble schedule for a as se, he EDF wll geerae a feasble schedule. U EDF lub If s o schedulable by EDF, he cao be scheduled by ay algorhm. hs meas ha a as se s schedulable by EDF f ad oly f U p 9 EDF wh D Processor Demad Schedulably alyss Processor Demad rero [aruah 9] I ay erval of legh L, he compuaoal demad g(,l) of he as se mus be o greaer ha he avalable me ha erval. he demad [, ] s he compuao me of hose ass sared a or afer wh deadle less ha or equal o : L, g(, L) L g(, ) d r Demad of a perodc as Example L L D g (, L ) g(, L) L D g(, L) 8 8 L L 9

20 //8 oudg complexy oudg complexy Sce g(,l) s a sep fuco, we ca chec feasbly oly a deadle pos. If ass are sychroous ad U p <,weca chec feasbly up o he hyperperod p H: H = lcm(,, ) Moreover we oe ha: g(, L) G(, L) G(, L) L D L ( D ) LU ( D ) U Lmg L Processor Demad es G(, L) LU * L ( U D ) U ( D ) U L * L G(, L) g(, L) for L > L * g(,l) G(,L) < L L L D, g(, L) L D = {d * d m (H, L )} H = lcm(,, ) L * ( D ) U U 7 8 Summary hree schedulg approaches: Off-le cosruco (mele) Fxed prory (RM, DM) Dyamc prory (EDF) hree aalyss echques: Processor Ulzao oud U U lub Respose me alyss R D Processor Demad rerol g(,l) L RM EDF Schedulably alyss D = D Suff.: polyomal O() LL: U ( / ) H: U +) Exac pseudo-polyomal R polyomal: O() pseudo-polyomal Respose me alyss R D R U L, g(, L) L R pseudo-polyomal Processor Demad alyss 9