Conens Inroducion 2 Preliminary deniions and noaion 3 3 Schedulabliy and a bound on uilizaion 6 4 Rae Monoonic Scheduling 7 4. Deniion

Transcription

1 A Foray ino Uniprocessor Real-Time Scheduling Algorihms and Inracibiliy Ed Overon Drs. T. Brylawski and J. Anderson 2, advisors December 3, 997 UNC-CH Deparmen ofmahemaics 2 UNC-CH Deparmen of Compuer Science

2 Conens Inroducion 2 Preliminary deniions and noaion 3 3 Schedulabliy and a bound on uilizaion 6 4 Rae Monoonic Scheduling 7 4. Deniion Examples Example Example RM scheduling as an opimal scheduler Necessary Condiions Preliminary Lemmas Proof of opimaliy Uilizaion Resuls Uilizaion leas upper bound Complexiy of feasibiliy ess Deadline Monoonic Scheduling and Asynchronous Task Ses 56 i

3 5. Deniion Example DM as an opimal scheduler Asynchronous ask ses and a feasibiliy es Complexiy of feasibiliy ess Earlies Deadline Firs Deniion Example EDF as an opimal scheduler A feasibiliy es Complexiy of feasibiliy ess Modied Leas Laxiy Firs Deniion Example MLLF as an opimal scheduler Necessary condiions for opimaliy Proof of opimaliy MLLF as an opimal scheduler Complexiy of feasibiliy ess ii

4 8 Conclusions 9 9 Glossary 93 iii

5 I would like o hank Drs. Brylawski and Anderson for heir coninued eors o help me wih his paper. In paricular, I hank Dr. Brylawski for his exibiliy in working wih me, and I hank Dr. Anderson for his ime and energy working wih a suden ouside of his deparmen. iv

6 Inroducion Scheduling heory may be hough of as he sudy of how o accomplish cerain asks by cerain deadlines. As humans, we handle scheduling issues every day. For example, a suden mus accomplish homework by he appropriae due dae, a professor mus complee he rough draf of a paper by he submission dae, ec. Were we o have only one ask o accomplish, meeing ha deadline probably would be very simple. Our lives, however, conain many asks ha have deadlines { ax forms, car inspecions, meeings, classes, ec. Thus, we mus use some sor of scheduling echnique o \juggle" our various asks, so ha hey all are compleed by heir appropriae deadlines. Clearly, here are some asks whose deadlines are no sric { one may reques an exension o submi ones axes, one may balance he chance of geing ickeed agains missing ones car inspecion dae, ec. However, here are hose asks in life whose deadlines are much more sric { cour daes, gran proposals, ec. In such siuaions, if he deadline is missed, here are dire consequences (e.g., being sen o jail, no receiving funding). These examples show hawelive life in real ime: a siuaion where homework and gran proposals mus be properly submied, bu also one where hey mus be submied on ime. In his paper, we will consider hard-real-ime sysems. A real-ime sysem is one where compuaions no only mus produce correc oupu, bu also mus produce ha oupu in a imely fashion (namely, bygiven deadlines). Hard-real-ime sysems are real-ime sysems where he cos associaed wih unimely oupu (i.e., missing a deadline) is very high. An example of a hard-real-ime sysem is a compuer ha conrols he landing of an airplane: he rudders and aps mus respond o he sensors inpus wihin a given imeframe. If he responses are incorrec or are oo lae, he plane may crash. Due o he naure of hardreal-ime sysems, consideraion is focused primarily (and in his paper, focused solely) on wors-case behavior. If a scheduling sysem produces wonderful oupu in he average case, buisknown o fail in some siuaions, one would no wish o rus such a sysem o landing an airplane. Clearly, one would desire a guaranee of a correc landing. Hard-real-ime sysems are usually considered as a se of asks ha are repeaedly requesed. Tasks may be periodic, where here is a consan amoun of ime such ha he ask is requesed every ime unis (e.g., a digial wach changing is display every second), or sporadic, where each ask reques mus arrive a leas a consan ime afer he previous reques (e.g., reseing he ime on a digial wach). In his paper, we will focus on periodic asks, and discuss he eecs of considering sporadic asks where appropriae. Eiher way, each ask reques has an associaed deadline, by which he ask mus complee execuion. Addiionally, he asks may have shared resources { objecs which some asks require (exclusively) a some poin during execuion. An example of a shared objec would be a compuers

7 hard drive { one program may wish o wrie is daa while he oher wishes o read daa from anoher locaion on he disk. Since a disk canno wrie from one locaion and read from anoher a he same ime, he read and wrie asks require exclusive access o he disk during he appropriae read and wrie porions of heir execuion. A scheduling algorihm is one ha uses he informaion from he se of asks, and discerns when o schedule wha ask. All scheduling algorihms may be considered as prioriy driven { he ask wih he highes prioriy ha has execuion remaining should be scheduled. In ha regard, here are wo subses of scheduling algorihms: ones where prioriies are xed (saic prioriies), and ones where prioriies may change over ime (dynamic prioriies). We will consider four scheduling algorihms for ask ses of periodic asks wihou shared resources: rae monoonic [LL 73, LSD 89, La 74], and deadline monoonic [LW 82] use saic prioriies earlies deadline rs [LL 73, LM 80, BRH 90, BHR 93], and modied leas laxiy rs use dynamic prioriies. The modied leas laxiy rs scheduling algorihm was developed by he auhor o generalize wo dynamic prioriy scheduling algorihms { earlies deadline rs, and leas laxiy rs [Mo 83]. We will dene each scheduling algorihm, and deermine in which siuaions he algorihm is opimal. We will also derive feasibiliy ess in order o deermine if a given ask se will have avalid schedule under a given scheduling algorihm. We will also deermine he complexiyof he feasibiliy es. The complexiy is a signican facor in using he scheduling algorihms, since i gives a rough idea of he ime ha is involved wih deermining a priori wheher a given ask se has a valid schedule under he algorihm. In a hard-real-ime seing, one would no wish o simply sar up he scheduler and hope for he bes { one would wan o know wih cerainy ha all deadlines will be me. However, since we are considering hard-real-ime sysems, ime may be criical, and he ime i akes o deermine feasibiliy migh defer he sar of he schedule while feasibiliy is deermined. In such a case, one would clearly wan a \quick" feasibiliy es. Following he work of [LL 73, LSD 89, La 74], we will deermine ha for he condiions where rae-monoonic scheduling is opimal, here is a linear ime feasibiliy es for rae mononic scheduling ha deermines necessiy, bu no suciency. We will show ha here exiss a pseudo-polynomial ime necessary and sucien es ha indicaes he feasibiliy quesion for rae monononic scheduling is in NP. We will hen follow [LW 82] o show i is a generalizaion of rae-monoonic scheduling, and o deermine he condiions under which deadline-monoonic scheduling is opimal. We will hen begin o consider ask ses where he asks are no released simulaneously, and see ha here is a necessary and sucien es for feasibiliy in his case. The es, however, is 2

8 hen shown o be co-np-complee in he srong sense. Earlies-deadline-rs scheduling is a very powerful scheduling algorihm, as we shall see. I is opimal among dynamic scheduling algorihms and oers he rs polynomial ime necessary and sucien feasibiliy es for feasibiliy. However, under cerain circumsances ha es is no valid, and we will show ha he feasibiliy es for hose cases is co-np-complee in he srong sense. We will lasly develop a new scheduling algorihm, modied leas laxiy rs. The algorihm will be shown as a generalizaion of earlies-deadline-rs scheduling, and is herefore opimal. Addiionally, i hen inheris he feasibiliy ess from earlies-deadline-rs { under some condiions, we have a polynomial ime es, and under ohers he es is co-np-complee in he srong sense. 2 Preliminary deniions and noaion We dene a periodic ask wihou resources, i, o be he 4-uple (e i d i r i ) where e i, d i, and are posiive real numbers, and r i is a non-negaive real number. The ask i is said o have execuion ime e i, a deadline span of d i, a period of, and an iniial release ime of r i. i is said o have release imes a r i k, k 2 Z +, where r i k+ r i + k. r i k is said o be he k h release of i. Each release r i k has an associaed deadline, r i k + d i, he k h deadline of i. We dene a ask se of n asks wihou shared resources, T, o be f i g n i, where i (e i d i r i )asabove. As a convenion in his paper, T will represen a ask se, subscriped s will represen asks in T,andn will represen he number of asks in T. A schedule of T is a funcion g : R + 7! T [ f g. We say ha i is scheduled a ime or on he processor a ime if g() i, and ha he processor is idle a ime if g(). We say a ask se is synchronous if here exiss some r such ha for all i r i r. Wihou loss of generaliy for synchronous ask ses, we will also assume ha r 0: Given a ask se T f i g n i of n asks ha are synchronous, we dene T 0 o be he se fig 0 n i such ha 0 i (e i d i 0). I is clear ha if g() isheschedule of T produced by agiven scheduling algorihm, hen g 0 (), he schedule of T 0 produced by ha scheduling algorihm, is exacly g( + r). Addiionally, here will be no ask scheduled on [0 r)ing since no ask is released unil ime r. Hence, g is valid if and only if g 0 is valid. 3

9 Given a funcion f : A 7! B, and some elemen b 2 B, we dene ( 0 : f(a) 6 b f b (a) : f(a) b i is said o be acive a ime if here exiss k 2 Z + such ha r i k < r i k + d i and R r i k g i (x) dx < e. Informally, he ask has been released, bu has no compleed is execuion corresponding o ha release. i is said o overow or miss is deadline a if here exiss k 2 Z + such ha r i k + d i (i.e., is he k h deadline of i )and R r i k g i (x) dx < e. If R r i k g i (x) dx e, we say i mees is deadline a. These deniions correspond o he inuiive noion of a deadline { if he ask hasn execued \enough", hen he deadline is missed. The response ime of he k h release of a ask is he dierence beween he ime he ask nishes execuing ha invocaion and he ime i was released, which can be seen as he ime i akes he ask o complee is execuion. A criical insan of a ask (under a given scheduling algorihm) is a release ha yields he longes possible response ime of ha ask for he given ask se. A schedule is said o be valid if all deadlines of all asks are me. We say ha, under a given scheduling algorihm, he processor is fully uilized for a given ask se if he algorihm produces a valid schedule for he given ask se, bu an increase in he execuion ime of any process in he ask se would yield an overow. We call a scheduling algorihm opimal if, when here exiss a valid schedule for some ask se T,hen he scheduling algorihm also produces a valid schedule for T. A prioriy-based scheduling algorihm is one where each ask i is assigned a corresponding prioriy, P i. These prioriies may be eiher saic or dynamic. Lower prioriy numbers correspond o higher prioriies. Tha is o say, if P andp 2 2 hen ask has higher prioriy han ask 2. All prioriy based scheduling algorihms use he following deniion for heir schedule: g P () ( i : i is acive a ime and 8j 6 i P j <P i ) P j is no acive. : here is no acive ask a ime Noe ha if wo (or more) acive process have he same prioriy, ies may be broken arbirarily. Thus, in his paper, for saic prioriy scheduling algorihms, we will assume ha no wo asks have he same prioriy (since one mus be chosen over anoher, and ha choice mus be saic). By convenion, we will also assume ha for a saic prioriy scheduling algorihm, he ask ses under consideraion are ordered by prioriy: P <P 2 <:::<P n. The uilizaion funcion corresponds o he noion of \how busy he processor is". Formally, 4

10 U : T 7! R + is dened by U(T ) e i p i i i is clear ha U(T ) 0 for all T, since for all i e i > 0. In Secion 3 we will show ha U(T ) is a necessary condiion o produce a valid schedule of T for any uniprocessor scheduling algorihm. To faciliae how he scheduling algorihms work, here are several graphs of example ask ses. The key o hose graphs is as follows: Task release Sample graph Task deadline Task deadline concurren wih nex release Task 3 2 (Lower line) Task acive bu no scheduled (Upper line) Task scheduled Acive ask begins o execue Time Preemped ask is scheduled Scheduled ask is preemped Task complees execuion (On ime axis) Processor idle For example, in he sample graph we have 3 asks. has an execuion ime of 2, a deadline span of 5, a period of 7, and a release ime of 0. 2 has an execuion ime of 2, a deadline span of, a period of, and a release ime of. 3 has an execuion ime of 4, a deadline span of 6, a period of 6, and a release ime of 2. Thus, a ime 0, is released and execues. A ime, 2 is released, bu is no scheduled since is of higher prioriy. A ime 2, complees execuion and 3 is released. Since 2 is he higher prioriy ask, i execues o compleion a ime 4, when 3 begins execuion. A ime5, has a deadline (ha is me, since nished execuion a ime 2). A ime 7, is released and preemps 3 unil ime 9, when 3 is again scheduled. 3 complees a ime 0, when here is no acive ask. Thus, he processor is idle unil a ask is released, namely a ime 2 ( 2 is released). The res of he graph should be clear. 5

11 3 Schedulabliy and a bound on uilizaion We rs show one of he fundamenal heorems in scheduling heory, which saes ha no ask se wih a uilizaion greaer han one is schedulable. Inuiively, his heorem should agree wih he readers noions abou uilizaion { uilizaion represens he fracion of he ime he processor mus be acive for a valid schedule of he ask se. If ha fracion is greaer han one, hen here is more \work" han ime available, and he ask se has no valid schedule. We follow he work of [ARJ 97]. Theorem 3. ([ARJ 97]) If a ask se is schedulable, is uilizaion mus be a mos. Proof: Assume ha here is a valid schedule for some ask se T. Then every deadline of he ask se will be me. Thus, we know ha each j kask i is scheduled for e i ime unis every ;r ime unis afer r i. Thus, for r i, i has i saised deadlines over [ri ). Noe ha on [r i ), i is hen scheduled for j ;r i k e i ime unis. So, le 2 R + such ha max i 2T fr i g. Since any valid schedule mus mee every deadline, he ime available for execuion (namely, ) mus be greaer han he amoun of execuion corresponding o deadlines a or prior o : since bxc >x; for all x 2 R +,we have rearranging erms, we have > i i i i i $ % ; ri e i ; r i ; e i ; r i e i! e i ; e i!! e i ; r i e i + e i! > 6 i i r i e i + e i!!! e i ;

12 Noe ha his equaion holds for all 2 R + such ha max i 2T fr i g. Since he lef-hand side of he equaion is a bounded non-negaive consan, and he righ-hand side is linear in, we mus have i!! e i ; i 0, i.e., e i Thus, if here is a valid schedule for he given ask se, hen he ask ses uilizaion is a mos. 2 4 Rae Monoonic Scheduling Rae monoonic scheduling (RM) was he focus of one of he seminal papers in hard-realime scheduling heory, [LL 73]. The paper laid mos of he ground work for much of he developmen of saic-prioriy scheduling. RM is easy o undersand and simple o implemen, ye i yields several signican resuls. Addiionally, RM is an opimal scheduling algorihm for saic-prioriy scheduling algorihms under cerain circumsances. Due o is signicance in he eld, we devoe a reasonable amoun of aenion o is developmen and resuls. 4. Deniion Rae monoonic scheduling is a saic-prioriy scheduling algorihm for periodic asks. In RM, prioriies are equal o he periods of he associaed asks. Hence, he ask wih he shores period has he highes prioriy, and he ask wih he longes period has he lowes prioriy. Inuiively, his prioriizaion makes sense, since he ask ha has he shores period will be he rs one o be re-released. Hence, i should be he rs one o complee (so ha i will be ready for is nex release). In [LL 73], RM is considered in he case where each asks deadline is concurren wih he asks nex release (hus, d i ). [LW 82] laer showed ha RM is no opimal when his case does no hold, and developed deadline monoonic scheduling (DM). We will consider DM in Secion 5. Formally, RM is a prioriybased algorihm, such ha P i for each ask i in he given ask se. 7

13 4.2 Examples The wo following examples display wo exremes relaed o RM scheduling. The rs example considers a ask se ha fully uilizes he processor, ye whose schedule conains idle ime. The second example shows ha here are cases under which RM produces a valid schedule for ask ses whose uilizaion is equal o one Example Le T be he ask se f i g 3 i such ha ( 3 3 0), 2 ( 4 4 0), and 3 ( 5 5 0). Noe ha by he deniion of RM, has a higher prioriy han 2, which has a higher prioriy han 3. Hence, a ime 0, all hree asks are acive, and hus execues. Since is he highes-prioriy ask, i will no be preemped by any oher ask. Hence, execues o compleion a ime. A ime, he highes prioriy acive ask is 2, and i execues unil ime 2. A ime 2, 3 is he only acive ask, so i execues (unil ime 3). A ime 3, is released and is he highes prioriy ask (and he only acive ask). Hence, execues unil ime 4, when 2 is he only acive ask { 2 execues unilime5,when 3 is released. This process coninues, andhermschedule of T is displayed in he below graph from ime 0 o ime 30. I is worh noing ha if 3 s execuion ime were increased, he ask se could no be scheduled wih RM, because here is only one ime uni (from 0 unil 5) for 3 o execue. Thus, T fully uilizes he processor. Task Time I is ineresing o noe ha he uilizaion of T is exacly There are 47 of he 60 ime unis where a ask is scheduled, and 3 where he processor is idle. 8

14 4.2.2 Example 2 Le T be he ask se f i g 4 i such ha ( 4 4 0), 2 ( ), 3 ( ), and 4 ( ). Noe ha he uilizaion of T is exacly { laer we will show ha RM canno guaranee schedulabiliy of a ask se wih n asks if is uilizaion is greaer han n(2 n ; ), bu ha proof does no preclude he possibiliy ha RM can schedule some ask ses of uilizaions higher han he bound. The graph below displays he RM schedule of T from ime 0 o ime 32. Task Time Again, we see ha he uilizaion of he ask se is represened by he number of ime unis where a ask is scheduled. Namely, he uilizaion is (we leave unsimplied o show he relaion of he graph o he leas common muliple of he ask periods). There are 32 ime unis ou of 32 where a ask is scheduled, and here is no idle ime. 4.3 RM scheduling as an opimal scheduler We will now show ha under cerain condiions, RM is opimal among saic-prioriy scheduling algorihms. These condiions are no oo demanding, and since RM is easy o undersand and simple o implemen, i is clear why RM is commonly used in hard real-ime scheduling Necessary Condiions For RM, we assume in wha follows ha deadlines of a given ask are concurren wih is releases: for all i d i. As well, we assume he sysem o be synchronous. Since RM is 9

15 a saic-prioriy scheduling algorihm, (by convenion) we assume ha asks are ordered by heir prioriies, and hus + for all i<n Preliminary Lemmas We now proceed o prove wo lemmas, designed o provide a necessary and sucien condiion for schedule validiy under a given saic-prioriy scheduling algorihm. We will hen apply ha es o RM in order o show ha RM is opimal under cerain circumsances. Lemma 4. ([LL 73]) Given a synchronous periodic ask se and a xed-prioriy scheduling algorihm, a criical insan in he resulan schedule of a given ask occurs when ha ask is requesed simulaneously wih all higher prioriy asks. Proof: Firs, we noe ha here exiss a ime when such all asks are released simulaneously { a ime 0 (since he ask se is synchronous). Le 2 2 R + be such ha ask i has a criical insan a, and complees execuion for ha release a ime 2. Thus, 2 ; is he longes possible response ime for ask i. l m We claim ha each ask j wih higher-prioriyhan i has exacly 2 ; releases on [ 2 ). l Assume, oherwise, ha some higher-prioriyask j has l< 2 m ; releases on [ 2 ). Thus, on [ 2 ), j execues l form a oal of l e j ime unis. However, were j released a ime,i would l be released 2 ; imes on [ 2 ). Thus, i s compleion a 2 would be delayed a m leas 2 ; ; l e j addiional ime unis, and he saisfacion of i s release a would be laer han ime 2. Again, his implies is no a criical insan. Addiionally, l i implies m ha if every higher prioriy ask j were released a, hen each j would have 2 ; releases on [ 2 ). Hence, a criical insan for a given ask occurs when ha ask is requesed simulaneously wih all higher prioriy asks. 2 I is ineresing o noe ha (in he erms of Lemma 4.) eiher 0,orhere exiss some >0suchhaon[ ; ), here is no acive ask wih higher prioriyhan i. Assume ha for > 0 and for each >0 here is some j acive on [ ; ) wih higher prioriy han i. Then eiher j or anoher ask wih higher prioriy isscheduled for he inerval [ ; ). Consider ha if i were released a ime ;, i would be preemped by higher-prioriy 0

16 asks on [ ; ). Hence, a release of ask i a ime ; would be saised a ime 2, since i would execue on exacly he same inervals as if i were released a. Thus, he release a ; would be saised a 2, and i would have a response ime of 2 ; +. Hence, would no be a criical insan. Now ha we have a handle on a single ask (by way of is criical insan), we procede o a resul regarding schedulabiliy of an enire ask se. Lemma 4.2 ([LL 73]) A saic-prioriy scheduling algorihm produces a valid schedule for a synchronous ask se if and only if he rs deadline of each ask is me. Proof: Clearly, if he rs deadline of any ask is missed, hen he ask se is no schedulable. Le us hen assume ha he scheduling algorihm has produced a (possibly valid) schedule for he given ask se, and under ha schedule, he rs deadline of each ask is me. From Lemma 4., we know ha a criical insan occurs when a ask is requesed simulaneously wih all higher prioriy asks. Since all asks are rs requesed simulaneously, all asks have a criical insan a ha rs release. Because all asks mee heir rs deadline, he longes response ime for any ask is exacly he response ime for he rs release of ha ask -andsince each ask mees ha rs deadline, here is no release of any ask ha will no be me Proof of opimaliy We now proceed o show RM o be an opimal saic-prioriy scheduling algorihm under he condiion ha for each ask in he ask se, he asks deadline span is idenical o is period. Noe ha we will acually prove a more general resul ha will be used in Secion 5. Theorem 4. ([LL 73]) Any saic-prioriy scheduling algorihm where prioriies are ordered idenically wih he asks deadline spans is an opimal scheduling policy among saicprioriy scheduling algorihms for synchronous ask ses. Proof: This opimaliy is shown in [LL 73] via a prioriy swapping argumen. Le T be aaskse of n asks, and since we are considering a saic-prioriy scheduling algorihm, we

17 know ha P <P 2 <:::< P n. Thus, by he heorem assumpion, d d 2 ::: d n. Now le us assume ha here is some valid saic-prioriy schedule g of T. Le he prioriies used in g be P g i for each ask i. If P g i P g i+ for all i 2 f 2 :::n; g, hen he prioriies are ordered idenically wih he ask deadline spans. Hence, he saic-prioriy scheduling algorihm will produce a valid schedule (as i produces exacly he schedule g). So, assume here exiss i<j2f 2 :::ng such ha d i <d j and P g i >P g j. Wihou loss of generaliy, we may assume ha here is no k such ha P g i >P g k >P g j. Consider he schedule h produced by swapping he prioriies of i and j. Formally, P h i P g j P h j P g i P h k P g k for all k 6 i j We will prove ha h is also a valid schedule. By swapping he prioriies of all such i s and j s, we produce a schedule of he given saic-prioriy scheduling algorihm. Hence, if h is shown o be valid, hen he saic-prioriy scheduling algorihm will produce a valid schedule as well. By Lemma 4.2, if we show ha all rs deadlines are me by h, hen h is valid. Le k be some ask of T. Since g is a valid schedule, he rs deadline of k is me in g. We can express his resul as follows: X l:p g l P g k p l e l for some d k which saes ha here is a ime d k such ha k and all higher prioriy askssaisfy all of heir releases by ime. Addiionally, ifwe subsiue h for g and can nd such a, hen k will mee is deadline in h. We now show ha all asks mee heir deadlines in h by considering hree cases. Case : k 6 i and k 6 j. See hen ha fl : P g l P g k g fl : P h l P h k g. Since k mees is deadline in g, here exiss a d k such ha X e l p l by subsiuion, we have l:p g l P g k X l:p h l P h k p l e l d k 2

18 Thus, k mees is deadline in h. Case 2: k j. Since P g i P h j, fl : P g l P g i g fl : P h l P h j g. Since i mees is deadline in g, here is some i d i such ha by subsiuion, we have Thus, j mees is deadline in h. X l:p h l P h j X l:p g l P g i i i e l i p l e l i d i d j p l Prior o examining he oher case, we rs noe (as in Case 2) ha here exiss i d i such ha X i e l i p l l:p g l P g i Recalling ha here is no k wih P g j <P g k <P g i, we separae he sum as follows: X 0 i e l A X i e l A i p l p l l:p g l P g j l:p g l P g i By assumpion for saic-prioriy algorihms, here are no idenical prioriies: X i p l l:p g l P g j e l A + i e i i () Case 3: k i. Le i be as described in Case 2. Then we have 0 X X i l:p h l P h i e l p l l:p g l P g j p l e l A ; i e j + i e i by equaion (), X l:p h l P h i i e l i ; p l i i d i e j 3

19 Hence, i mees is deadline in h. Thus, for any k 2 T, if k mees is rs deadline in g, hen k mees is rs deadline in h. Therefore, if g is a valid schedule of T, hen so is h. 2 Corollary RM is an opimal scheduling algorihm among saic-prioriy scheduling algorihms for synchronous ask ses where each asks deadline span and period are idenical. This corollary follows direcly from Theorem 4. since under RM, prioriies are ordered by period lenghs. Since we are assuming period lenghs are equal o deadline spans, hen he heorem holds for RM under he given resricions. 4.4 Uilizaion Resuls One of he more powerful resuls of [LL 73] is he derivaion of a sucien es of schedulabiliy under RM relaed o he uilizaion of he given ask se. This developmen hinges on he deerminaion of a\wors case" ask se { one ha minimizes uilizaion while fully uilizing he processor. In his secion, we will re-develop some of [LL 73]s resuls based upon several lemmas ha we creae from he works of [LL 73] and [LSD 89]. We now proceed o develop mahemaical ess for schedulabiliy and full uilizaion. Wih hese lemmas in hand, we will be able o answer he \wors case" ask se quesion, and dene he ask ses ha minimize uilizaion while fully uilizing he processor. For schedulabiliy, by Lemma 4.2, we only need o concern ourselves wih a asks rs deadline. The nex Lemma provides an equaion o deermine if a given asks rs deadline is me. Lemma 4.3 Le T be a synchronous ask se and g be a saic prioriy schedule of T such ha P < P 2 < ::: < P n. Given i 2 T, here exiss a P l i such ha m e j if and only if i saises is rs release a or before ime. Proof: We rs assume ha here exiss a such ha P l i m e j. We will show ha by ime, ask i saises is rs release a or before ime. Assume oherwise, ha i is sill acive aime. Then R 0 g i (x)dx<e j. Le w R 0 g i (x)dx. The oal amoun of 4

20 work requesed by asks wih higher prioriy han i on [0 )is P l i; m e j. The maximum amoun of work due o i and higher prioriy asks on [0 ) is hen w + P i; P i l l m e j < m e j. Therefore, here is some ime 0 2 [0 )such ha a ime 0, he processor is idle, or alower prioriy (han ha of i ) ask is scheduled. Noe, hough, ha a 0, i is acive since 0 <and i is acive aime. By he deniion of a saic prioriy scheduling algorihm, i (or a higher prioriy ask) mus be scheduled a ime 0. By conradicion, i is no acive a ime, and herefore has saised is release a ime 0. We now assume ha here exiss a 0 such ha i saises is rs release a or before ime 0. We mus show ha here exiss a such ha P i l m e j. Since i has saised is rs relase by ime 0, here is some ime 0 such ha i has saised is rs release by ime, and for any >0, i is sill acive a ime ;. Therefore, we know ha here exiss a such ha i is scheduled in g on he inerval [ ). Thus, by deniion of a saic prioriy scheduling algorihm, i is he highes prioriy acive ask a any ime on [ ). Thus, a ime, we know ha asks 2 ::: i; have saised all heir releases prior o ime. Therefore, every release of asks 2 ::: i on [0 ) is saised a or before ime. Hence, P i l m e j, and. 2 Having shown how o deermine if a paricular ask mees is rs deadline, we now apply ha knowledge o he whole ask se o creae a necessary and sucienesofschedulabiliy. Lemma 4.4 ([LSD 89]) Le T be a synchronous ask se and g be a saic prioriy schedule of T such ha P <P 2 <:::< P n. g is a valid schedule of T if and only if >< < max min i n 2T >: 2 k jji k2f ::: j pi o : ix > e 9 p kg j j > (2) Proof: here exiss a such ha P i By Lemma 4.3, we know ha a given ask i mees is rs deadline if and only if l m e j. By Lemma 4.2, we know ha every ask mees is rs deadline if and only if he schedule is valid. Therefore we have he following chain of equivalen saemens: The schedule is valid if and only if for each i 2 T, here exiss a such ha P i l m e j. The schedule is valid if and only if for each i 2 T, here exiss a such ha P i l m e j. 5

21 The schedule is valid if and only if for all i 2 T, min 2[0 ] 8 < : ix e j 9 : The schedule is valid if and only if max i 2T 8 < : min 2[0 ] 8 < : ix 9 e 9 j : (3) Noe ha we may resric our consideraion for he value of from he se [0 ] o he se n j S k jj pi k o i k 2f ::: g. The se S represens every deadline on [0 pi ] of all P asks j wih P j P i. We claim ha i l m e j achieves is minimums on he se S. P Tha is o say, we claim ha for all 2 S, here exiss a 2 2 S such ha i l m > P i l 2 m 2. Le 2 [0 ) such ha 2 S. Since < and 2 S, here is a 2 2 S such ha l < m 2. l Le 2 be he minimal elemen in S ha is greaer han. Thus, for j i, 2 m l. To see his claim, assume oherwise, ha here exiss a j 2f ::: ig m l such ha 6 2 m l m l. Since < 2, < 2 m l. Le k m. We know 6 k since 2 S. Also, k 2 S by he deniion of S. Therefore, <k < 2. However, we dened 2 o be he minimal elemen of S ha is greaer han. By conradicion, we have shown ha for j i, l m l 2 m. Thus, ix e j ix > 2 ix Therefore, for any 2 S here is a 2 2 S such ha P i l m e j will acheive is minimum on S P i e j e j l 2 m P e j < i l m e j. Hence, Therefore, we have min 2[0 ] 8 < : ix e j 9 n 2 min k jji k2f ::: j pi kg 8 < o : ix e j9 (4) 6

22 Combining equaions (3) and (4), we have he desired resul: g is a valid schedule of T if and only if >< < max min i n o 2T : ix > e 9 j >: 2 k jji k2f ::: j pi kg > 2 Thus, our rs goal of his secion has been me: We have a compuable es for schedulabiliy. We now build upon ha knowledge for a es of full uilizaion. The rs sen ha process is o show ha full uilizaion is based upon idle ime prior o he las rs deadline. Lemma 4.5 ([LL 73]) A synchronous ask se fully uilizes he processor under a saicprioriy scheduling algorihm if and only if here is no idle ime prior o ime p n and all deadlines a or prior o p n are saised. Proof: By Lemma 4.2, all deadlines prior o p n mus be saised for he ask se o be schedulable. By deniion, a ask se ha fully uilizes he processor mus be schedulable. Thus, we focus our consideraion on idle ime prior o p n. If here is idle ime prior o p n, hen he execuion ime of he ask wih he longes period may be increased by he amoun of idle ime. Since all rs deadlines are sill me, he saic-prioriy scheduling algorihm (hereafer denoed SPSA) does yield a valid schedule for he modied ask se. Since a asks execuion could be increased and sill he ask se would be schedulable, he original ask se did no fully uilize he processor. Thus, if a ask se fully uilizes he processor under SPSA, here is no idle ime prior o maxf g. If here is no idle ime prior o p n and here is some j 2 T and >0such ha replacing e j by e j + yields a valid schedule under SPSA, consider ask n. In he original schedule, le w be he amoun of ime he processor is devoed o asks 2 ::: n; on [0 p n ). Thus, since here is no idle ime prior o p n, w + e n p n. However, in he modied ask se, n will miss is deadline a p n : Consider ha in he SPSA schedule of he modied ask se, if j <n,henheamoun of ime necessary for he asks 2 ::: n; on he inerval [0 p n ) will be a leas w+, since j has a leas one deadline prior o p n. Since n may only execue when oher asks are inacive (because n is he lowes prioriy ask), in he modied ask se schedule, n has p n ; w ; ime unis o execue. Since p n ; w ; < e n, n will miss is deadline a p n. If j n, hen consider ha asks 2 ::: n; occupy w ime unis in (0 p n ) in he modied ask se schedule, leaving e n ime unis for n o complee e n + 7

23 ime unis of compuaion. Thus, here is no execuion ime ha may be increased in he original ask se wihou yielding an invalid schedule under SPSA, and he original ask se fully uilizes he processor under SPSA. 2 Since we have seen full uilizaion is based on idle ime prior o he las rs deadline, we now consider idle ime prior o a given asks rs deadline. We use Lemma 4.4 o deermine if a saic prioriyschedule of a synchronous ask se conains idle ime prior o a given asks rs deadline. Lemma 4.6 Le T be a synchronous ask se and g be a valid saic prioriy schedule of T such ha P <P 2 <:::< P n. For any ask i 2 T, here isno idle ime in g on he inerval [0 ) if and only if 8 n min 2 k jji k2f ::: j pi p kg j < o : ix 9 e j : (5) Proof: n Le i 2 T. For ease of noaion, we dene he se S k jj i k 2f ::: j pi k o g. Assume here is no idle ime on [0 ). Since he schedule is valid, hen by Lemma 4.4, min 2S n P i l m o e j. By conradicion, we will show ha equaion (5) holds. As- l m e j <. Then here exiss some 0 < sume ha here exiss a 2 S such ha P i l such ha P m i e j 0. Since here is no idle ime on [0 0 ), all ask requess on [0 ) mus be saised by ime 0. Hence, here is no acive ask on [ 0 ) { he processor is hen idle, conradicing our assumpion ha here is no idle ime on [0 ). Thus, min 2S n P i l m e j o. n P Assume ha equaion (5) holds: min i l m o 2S e j. We mus show ha here is no idle ime on [0 ). Again, we do his by conradicion. Assume here is idle ime [ 0 ) on [0 ). Since here is an acive ask when a release occurs, we may assume ha a release occurs a ime, denoing he end of he idle period. Therefore, is in he se S. Since equaion (5) holds, P l i m e j. However, if he processor is idle on [ 0 ), hen we know all ask requess a or before are saised by ime 0. Thus, P l i 0 m e j 0. Since 0 <, we have a conradicion, and here can be no such idle ime [ 0 ). We have shown ha given a valid schedule, here is no idle ime on [0 ) if and only if 8

24 min 2S np i l m e j o. 2 Having a means of esing full uilizaion in erms of idle ime and rs deadlines, and a compuaional means o deermine idle ime, we have he ools we need o proceed. We combine he previous wo resuls o provide a compuaional es o deermine if a ask se fully uilizes he processor under a given schedule. Lemma 4.7 Le T be a synchronous ask se and g be a saic prioriy schedule of T such ha P < P 2 < ::: < P n. T fully uilizes he processor under g if and only if for all i 2 T i<n ( $ % ) pi ix here exiss a 2 k jj i k 2f ::: g such ha e j (6) and n min 2 k jjn k2f ::: j pi p kg j 8 < o : 9 e j : (7) Proof: By Lemma 4.5, we know ha a ask se fully uilizes a processor if and only if he he schedule is valid, and here is no idle ime prior o ime p n. We have seenin Lemma 4.4 ha a ask ses schedule is valid if and only if equaion (2) holds. By Lemma 4.6, we know ha given a valid schedule, here is no idle ime prior o p n if and only if equaion (7) holds. We rs assume ha equaions (6) and (7) hold. Clearly, hese wo equaions imply equaion (2). As well, equaion (7) implies here is no idle ime prior o ime p n. Thus, by Lemma 4.5, we know ha equaions (6) and (7) imply he ask se fully uilizes he processor under he given schedule. Assuming he ask se fully uilizes he processor, hen he schedule is valid and here is no idle ime prior o ime p n. Thus, equaion (2) holds, and here is no idle ime prior o p n. Equaion (2) implies equaion (6), and he lack of idle ime implies equaion (7). Thus, if a ask se fully uilizes he processor, equaions (6) and (7) hold. 2 We will work exensively wih Lemma 4.7 in he following proof, and before we begin he proof proper, we will make use of some assumpions on period lengh o considerably simplify he se over which is considered in equaions (6) and (7). 9

25 Lemma 4.8 For f g n wih i p p 2 ::: p n and pn p 2, ( ( $ %)) pi k jj i k 2 2 ::: f g i Proof: Since p p 2 ::: p n and pn p 2, we know ha for any i 2 T j <i, we have and 2. Thus, j k 2 wih equaliy if and only if 2pj. The remainder of he proof should be clear. 2 [LL 73] saes he following resul, bu heir proof is fauly (as will be shown below). The heorem lays he groundwork for providing a necessary condiion for schedulabiliy under RM. The signicance of he condiion is ha i may be esed in linear ime, whereas a necessary and sucien es of schedulabiliy requires psuedo-polynomial ime (as will be seen in Secion 4.6). Theorem 4.2 ([LL 73]) Over he se of synchronous ask ses wih n asks ha fully uilize he processor under RM such ha p p 2 p n and pn p 2, he execuion imes e i + ; for i<nand e n 2p ; p n minimize uilizaion. We prove his Theorem by subdividing ino hree cases based on he execuion imes of asks hrough n;. In cases and 2, we will modify he ask se in such a way ha he modied ask se fully uilizes he processor and whose uilizaion is less han or equal o ha of he original ask se. Repeaed modicaions will conver he ask se ino one where he execuion imes for asks hrough n; are idenical o he imes lised in he saemen of he Theorem. Case 3willshow ha given such execuion imes for hrough n;,ask n mus have he execuion ime specied above. Throughou his heorem, we will make use of Lemma 4.8. Since he ask se saises he condiions of ha lemma, he se over which we mus consider in equaions (6) and (7) is merely f g i. Proof: Since T fully uilizes he processor, we know ha i saises he wo condiions of Lemma 4.7, namely equaion (6): for all k 2 T, kx here exiss a 2f g k such ha e j 20

26 and equaion (7): min 2f g n 8 < : e j 9 : We aim o prove he same for T 0, dened below. Case : There exiss an i<nsuch ha e i >+ ;. Le i<nbe he lowes indexed such e i. Then le e i ;(+ ; ). Hence, e i + ; +. Consider he ask se T 0, idenical o T excep for execuion imes: e 0 i + ; e 0 i+ e i+ + e 0 j e j for all j 6 i i + Le g be he schedule produced by RM for T, and g 0 be he schedule produced by RM for T 0. Here is an example of how hose schedules appear. Noe he change ha occurs immediaely afer ime +. i+ e i+ i Task i < p 2 - p - < p 3 - p 2 - < < p 2 - p - < p 3 - p - < - p p p 2 p 3 - p i+ Time Case sample graph of T wih period lenghs indicaed 2

27 i+ e i+ + Task i i < p 2 - p - < p 3 - p 2 - < p i 0 - < p 2 - p - < p 3 - p - < - p - < + - p p p 2 p Time Case corresponding graph of T wih period lenghs indicaed Case : Subproof ha T 0 fully uilizes he processor: We mus saisfy he wo condiions of Lemma 4.7, equaions (6) and (7) above. Firs, we handle equaion (6). Le k 2 T. Since T fully uilizes he processor, by equaion (6) we have Since for all j n, here exiss a 2f g k such ha here exiss a 2fg k such ha kx kx e j e j (8) Now we divide our consideraion ino hree subcases based on he value of k in relaion o i. Our goal in each case is o show ha here exiss a 2fg k such ha kx e 0 j hereby proving T 0 saises equaion (6). Subcase.A: k<i. In his case, e j e 0 j for all j k. Therefore, equaion (8) becomes here exiss a 2fg k such ha kx p 0 j e 0 j 22

28 Thus, equaion (6) holds for subcase.a. Subcase.B: k i. We now break up he sum from equaion (8) o produce he desired resul: ix Xi; e j e i + p 0 i p 0 i p 0 i p 0 i ix e j Xi; (+ ; +)+ p 0 e 0 j j Xi; (+ ; )+ p 0 + i X e 0 i; i ++ p 0 e 0 j + j p 0 i e 0 j + l Since 0 < p 0 i for all 2fg i, pim and we have 0 ix ix e j e 0 j + Therefore, by equaion (8), we have p 0 i here exiss a 2fg i such ha ix here exiss a 2fg i such ha Thus, equaion (6) holds for subcase.b. Subcase.C: k>i. ix e 0 j e 0 j + e 0 j < As in subcase.b, we will break down equaion (8)s sum for analysis. Again, we use he fac ha for all j 2f 2 ::: ng j 2 fi i +g, we have e 0 j e j. kx Xi; e j e 0 kx j + e i + e i+ + p 0 i p 0 i+ 23 ji+2 e 0 j

29 Therefore, if l p 0 i Xi; p 0 j Xi; Xi; + p 0 j kx m p 0 i+ e 0 kx j + p 0 (+ ; +)+ i p 0 (e 0 i+ ; ) + i+ i+2 j e 0 j + p 0 (+ ; )+ i p 0 + i p 0 e 0 i+ ; i+ p 0 i+ kx p 0 e 0 j j e 0 j + p 0 e 0 kx i + i p 0 ; i p 0 + i+ p 0 e 0 j ji+ j! e 0 + j p 0 ; i p 0 i+ ji+2, kx e j which would yield our desired resul for his subcase. So we now focus our aenion on he se of values of, namely f g k. For (herefore j i), we know +, so l m and l kx m l +. For p 0 j >+,we know + <<2 2+,so m 6 and l + m 2. Therefore, he only value of 2 fpj g k e 0 j where l p 0 i p 0 i+ e 0 j m 2 is + (when 6 + ). Thus, my goal is o show ha when 6 +, equaion (8) holds rue for some value of oher han +. To do so, we mus show ha kx pi+ e j >+ l Firs, we noe ha since < +, for all j pi+ m i 2 and for all i + j k l pi+ m. Now, we analyze he sum kx pi+ e j ix Xi; Xi; 2e j + kx ji+ 2e j +e i + 2e j + 24 kx ji e j kx ji+ e j +e i e j + e i (9)

30 Now, by he case assumpion, i is he lowes indexed ask such ha e i > + ;. Therefore, e i; ; ;. Thus, since e i; > 0, we know ; <. We have all j <i, for all j i. Combining ha knowledge wih equaion (9), we ge kx pi+ e j Xi; pi p 0 j kx pi e j + kx ji pi e j + e i 2for e j + e i (0) Now, since T fully uilizes he processor, hen here is no idle ime prior o ime p n. More imporanly here, here is no idle ime prior o ime. Therefore, by equaion (5), we know Since k>i, ix kx pi pi Combined wih equaion (0), we hen have kx pi+ e j e j > e j > + e i +(+ ; +) + + Therefore, we know ha when 6 +, for +, P k holds rue, hen i holds rue for some 6 +. We hen have Thus, is such ha l p 0 i here exiss a 2fg k 6 + such ha m p 0 i+,and kx e j kx kx e j >. Since equaion (8) e j e 0 j () 25

31 yielding Therefore, here exiss a 2fg k 6 + such ha kx here exiss a 2fg k such ha and equaion (6) holds for subcase.c. kx e 0 j e 0 j Having considered all subcases, wehaveshown ha in case, he ask se T 0 saises equaion (6). We now prove ha for case, T 0 saises equaion (7). Firs noe ha k n falls ino subcase.c above since k>ifor all i<n, and case assumes i<n. Wih ha knowledge, for k n we have he following by equaion (): e j Since T fully uilizes he processor, equaion (7) holds: min 2f g n 8 < : e 0 j e j 9 : Combining he previous wo equaions wih he knowledge ha for all j n, we have 9 e 0 j : min 2f gn 8 < : Thus, in case, T 0 saises equaion (7). Since T 0 saises boh requiremens of Lemma 4.7, hen we have shown ha in case, T 0 fully uilizes he processor. Case : Subproof ha he uilizaion of T 0 is a mos ha of T : The uilizaion of T is U e + e e i; + + ; + + e i+ + + e n p p 2 ; + p n 26

32 The uilizaion of T 0 is U 0 e + e e i; + + ; + e i e n p p 2 ; + p n Hence, he dierence in uilizaion is U ; U 0 ; + + ; + + ; + since,, and + are posiive, and +,we have U ; U 0 + ; + 0 wih equaliy if and only if +. Thus, he uilizaion of T 0 is less han or equal o he uilizaion of T. Thus, for case, we have provided a ask se wih a uilizaion a mos ha of T ha fully uilizes he processor. Addiionally, we know ha for T 0, here is one less ask (han in T ) 0 i such ha e 0 i >p0 i+ ; p 0 i. Since here are a nie number of asks in he ask se, we may apply he case ransformaion repeaedly, unil we know ha for all i<n, e i + ;. Specically, repeaed ransformaions will evenually yield a ask se whose uilizaion is a mos ha of he original ask se, ha fully uilizes he processor, and which falls ino case 2or3below. Case 2: For all i<n, e i + ; and here is some e i such ha e i <+ ;. Le i<nbe he lowes indexed such e i. Then le (+ ; );e i. Hence, e i + ; ;. Consider he ask se T 0, idenical o T excep for execuion imes: e 0 i + ; e 0 n e n ; 2 e 0 j e j for all j 6 i i + Le g be he schedule produced by RM for he T, and g 0 be he schedule produced by RM for T 0. 27

33 Here is an example of how hose schedules appear. Noe he change ha occurs immediaely before imes + ; p and +. Task n i i- 3 2 (n) (n) (i+) 0 p 2 - p p 3 - p - p + - p p p 2 p 3 - p i+ Time Case 2 sample graph of T Task n i i- 3 2 (n) (i+) 0 p 2 - p p 3 - p - p + - p p p 2 p 3 - p i+ Time Case 2 corresponding graph of T Case 2: Subproof ha T 0 fully uilizes he processor: We mus saisfy he wo condiions of Lemma 4.7, equaion (6): for all 0 k 2 T 0, kx here exiss a 2fp 0 jg k such ha e 0 j and equaion (7): min 2f gn 8 < : e 0 j 9 : 28

34 Firs, we handle equaion (6). Le k 2 T. Since T fully uilizes he processor, here exiss a 2f g k such ha kx e j Since for all j n, here exiss a 2fg k such ha kx p 0 j e j (2) Now we divide our consideraion ino wo subcases based on he value of k in relaion o i. Our goal in each case is o show ha here exiss a 2f gk such ha kx e 0 j hereby proving T 0 saises equaion (6). Subcase 2.A: k < n. By he case 2 assumpion, we know ha for all j wih j k, e j + ;. Addiionally, noe ha he conversions for T 0 preserve his saemen. Namely, for all 0 j wih j k, e 0 j + ;. Le p 0. Noe ha since p 0 for all 0 j wih j k, hen. We hen have kx pk e 0 j kx kx pk (+ ; ) (+ ; ) Now, by he assumpions on period lengh for his heorem, p k+ ; p (3) Combining equaions (3) and (4), we have p k+ 2p p k+ ; p p (4) kx pk p 0 j e 0 j p 29

35 Therefore, for 0 k 2 T 0 such ha k<n, here exiss a 2fg k such ha kx p 0 j e 0 j Thus, for subcase 2.A, equaion (6) holds. Subcase 2.B: k n. By assumpion, T fully uilizes he processor. Therefore, here exiss a 2f g n such ha e j We will break our consideraion down ino wo subsubcases based on he value of. Subsubcase 2.B.i: p k. In his subsubcase, we will show ha we canno saisfy equaion (6) for T. The goal is o eliminae his subsubcase from consideraion, so ha we know equaion (6) is saised for T from a value of considered in subsubcase 2.B.ii. We dene p l such ha p l is he highes indexed period such hap l <p k. Therefore, we know ha for j l, l p k m 2, and ha for l<j n, l pk m. Knowing hese equaliies, we have pk e j lx lx lx lx pk e j + 2e j + e j + e j + jl+ p jl+ e j e j pk e j e j (5) Since T saises equaion (7), hen we know ha for p, p e j p p j p e j p (6) 30

36 Combining equaions (5) and (6), we have pk e j lx e j + p And by he assumpions of Case 2, we know ha for j l<i, we have e j + ;. Therefore, pk lx e j (+ ; )+p p l+ ; p + p p l+ Since p l is he highes indexed period ha is sricly less han p k, p l+ mus be equal o p k. Therefore, we have pk e j p k (7) wih equaliy if and only if Recall ha we are rying o show ha p e j p pk e j p (8) e j >p k and herefore we wish o show ha equaion (8) is false. Noe ha by he deniions of Case 2, we have e i < + ; l. Since e j > 0 for all j n, hen clearly + >. p Therefore, for all j i, i+ m l 2, and for all i < j n, pi+ m. Knowing hese equaliies, we have pi+ e j ix ix Xi; pi+ 2e j + e j + ji+ e j + e i + 3 ji+ e j e j pi+ e j

37 By he deniions of Case 2, recall ha for j <i, e j + ; and e i + ; ;. Therefore, pi+ e j Xi; (+ ; )+e i + e j ; p +(+ ; ; ) + + ; p ; + Consider ha since T fully uilizes he processor, hen by equaion (7), Applying equaion (9), pi+ + ; p ; + e j + e j + Which holds for all p k. Equaion (7) hen becomes pk e j >p k Therefore we know ha for all 2f g n wih e j e j (9) e j p + (20) e j > (2) Therefore, for n in T, equaion (6) mus be saised for some p k >. Tha is o say, mus fall ino in subsubcase 2.B.ii, since we know equaion (6) is rue for T. Subsubcase 2.B.ii: p k >. In his subsubcase, we know ha l p k m 2. As well, since l m p p k p n, k p n. Wih hose consideraions in mind, we see ha pk pk e j e 0 pk pk pk j + e i + e n ; p n p 0 e 0 i ; i pk e 0 j +2e i + e n ; 2e 0 i ; e 0 n 32 pk p 0 n e 0 n

38 By he deniions of e i and e 0 n, we hen have pk pk e j e 0 j +2(e 0 i ; ) + e n ; 2e 0 i ; (e n ; 2) p j p j pk e 0 j +2e 0 i ; 2 + e n ; 2e 0 i ; e n +2 pk e 0 j p 0 k p 0 e 0 j (22) j Now, since T fully uilizes he processor, by equaion (6), we know ha here exiss a 2f g n such ha e j However, considering equaion (2), he above becomes here exiss a 2f g n > such ha And since for all j n, here exiss a 2fg n >p0 i such ha e j e j Then by equaion (22), we have here exiss a 2fg n >p0 i such ha Therefore, we know ha for k n, here exiss a 2fg k such ha kx e 0 j e 0 j Thus, since subsubcase 2.B.i canno hold, and since equaion (6) holds for subsubcase 2.B.ii, hen equaion (6) holds for subcase 2.B. Addiionally, since equaion (6) holds for subcase 2.A, hen i holds for case 2asa whole. So, we have shown ha for all 0 k in T 0, kx here exiss a 2fp 0 jg k such ha e 0 j 33

39 And herefore, in case 2, ask se T 0 saises equaion (6). We now prove ha, in case 2, T 0 saises equaion (7). Since T fully uilizes he processor, hen by equaion (7) min 2f g n 8 < : e j 9 Since were considering a value where k n, hen we may use he resuls from subsubcase 2.B. Then by equaion (2), Since for all j n, min 2f g n > min 2f gn >p0 i 8 < : 8 < : e j 9 which, when combined wih equaion (22), yields min 2f gn >p0 i 8 < : e j 9 e 0 j 9 (23) Therefore, for equaion (7) o hold for T 0, i remains o prove ha 9 e 0 j min 2f gn p0 i 8 < : Firs, we will prove a preliminary resul... e 0 j e j + e 0 i + e 0 n ; e i ; e n e j +(e 0 i ; e i )+(e 0 n ; e n ) By he case 2 deniions of e i e 0 i e n and e n,we hen have e 0 j 34 e j +; 2 e j ;