Embedded Systems Development

Transcription

1 Embedded Systems Development Leture eal-tme Shedulng Dr. Danel Kästner AbsInt Angewandte Informat GmbH

2 Leture Evaluaton Please do leture evaluaton Deadlne tomorrow!

3 3 Deadlne Monoton Shedulng Let eah proess have a unque prorty P based on ts relatve deadlne d. We assume that the shorter the deadlne, the hgher the prorty, e d < d P > P. Same as rate monoton, f eah tas s relatve deadlne equals ts perod. Example shedule: T wth π d 3 and 0.5, T wth π, d and and T 3 wth π 3 d 3 6 and 3. T T T 3

4 Shedulablty Analyss The rate monoton shedulablty test an be appled also to deadlne monoton shedulng, by redung perods to relatve deadlnes: N N N( ) d However, ths test sgnfantly overestmates the worload on the proessor. Observatons: The worst-ase proessor demand ours when all tass are released at ther rtal nstants. For eah tas T the sum of ts proessng tme and the nterferene (preempton) mposed by hgher prorty tass must be less than or equal to ts deadlne d.

5 Shedulablty Analyss 5 Assume that tass are ordered by nreasng relatve deadlnes: < d <d P >P. Then a tas set Γ { T }..N s shedulable f the followng ondton s satsfed: where I s a measure of the nterferene of T, whh an be omputed as the sum of the proessng tmes of all hgher-prorty tass released before d : I N : d π I d T T d

6 Shedulablty Analyss 6 T T d Note that ths test s suffent but not neessary. I s alulated by assumng that eah hgher-prorty tas exatly d nterferes tmes durng the exeuton tme of T. However, π sne T may termnate earler, the atual nterferene may be smaller. A suffent and neessary shedulablty test for DM must tae the exat nterleavng of hgher-prorty tass nto aount for eah proess.

7 esponse Tme Analyss 7 The longest response tme of a perod tas T s omputed as the sum of ts omputaton tme and the nterferene due to preempton by hgher-prorty tass at the rtal nstant. I where I π suh that (*) π The worst-ase response tme s the smallest value of that satsfes Eq.(*).

8 8 esponse Tme Analyss Soluton: Fxed pont teraton. Let () be the -th value of and let I () be the nterferene on tas T n the nterval [0, () ]: Let (0) be the frst pont n tme that T ould possbly omplete: For > 0 repeatedly ompute () untl () (). I ) ( ) ( π (0) I ) ( ) ( ) ( ) ( π

9 esponse Tme Analyss 9 The tas set s shedulable f d holds for the fxed pont. TA s neessary and suffent. Let N be the number of tass and m the number of teratons of the fxed pont algorthm. Then the omplexty of the TA algorthm s O(Nm).

10 0 esponse Tme Analyss Example Consder the tas set on the rght sde. Assume T-T3 have been shown to be shedulable. Is also the tas set wth T shedulable? Proess Perod π WCET Deadlne d T 3 T 5 T3 6 5 T ) ( ) ( ) ( ) ( ) ( 5 () (5) (3) () () (3) () () 3 3 (0) (0) (0) (0) () (0) π π π π π π π π

11 Earlest Deadlne Frst Earlest Deadlne Frst (EDF) s a dynam shedulng sheme that selets tass aordng to ther absolute deadlne. Tass wth earler deadlnes wll be exeuted at hgher prortes. So far: d relatve deadlne, e the tme between T beomng avalable and the tme untl whh T has to fnsh exeuton. Let T, denote the -th nstane of tas T. Let r, be the release tme of the -th nstane of tas T. Let Φ denote the phase of tas T, e the release tme of ts frst nstane (Φ r, ). d, denotes the absolute deadlne of the -th nstane of tas T whh s gven by d, Φ ( ) π d EDF assumes tass are preemptve; tass an be perod or aperod.

12 Earlest Deadlne Frst Theorem: A set of perod tass s shedulable wth EDF f and only f Proof: Same as before. dle T U N π T T ov Tme overflow T m t t Assume that U and the tas set s not shedulable. Let t be the nstant where the deadlne volaton ours. Let [t,t ] be the longest nterval of ontnuous utlzaton before the overflow, suh that only nstanes wth deadlne t are exeuted n [t,t ].

13 3 Earlest Deadlne Frst Let C p (t,t ) be the total omputaton tme demanded by perod tass n [t,t ]. Then Sne a deadlne s mssed at t, C p (t,t ) must be greater than the avalable proessor tme t t. Thus T T T ov T m dle t t Tme overflow U t t t t t t t t C N N t d t r p ) ( ), (, π π ) ( ), ( > < U U t t t t C t t p

14 EDF vs. M Fxed-prorty shedulng s easer to mplement sne prortes are stat. Dynam shemes requre a more omplex run-tme system whh wll have hgher overhead. It s easer to norporate proesses wthout deadlnes nto M; gvng a proess an arbtrary deadlne s more artfal.

15 5 EDF vs. M Durng overload stuatons M s more predtable. Low prorty proesses mss ther deadlnes frst. EDF s unpredtable; a domno effet an our n whh a large number of proesses mss deadlnes. To ounter ths detrmental domno effet, many on-lne shemes have two mehansms: an admssons ontrol module that lmts the number of proesses that are allowed to ompete for the proessors, and an EDF dspathng routne for those proesses that are admtted An deal admssons algorthm prevents the proessors gettng overloaded so that the EDF routne wors effetvely

16 6 esoure Aess Protools esoures: data strutures, fles, deves,... prvate resoure: dedated to partular tas shared resoure: avalable to more than one tas To ensure onssteny of shared resoures, tass must be granted exlusve aess mutually exlusve resoures. Program setons durng whh exlusve aess to a resoure s requred are alled rtal setons. A tas watng for a mutually exlusve resoure s alled bloed on that resoure. Any tas whh needs to enter a rtal seton must wat untl no other tas s holdng the resoure. Otherwse the tas enters the rtal seton and hold the resoure. When the tas leaves the rtal seton, the resoure beomes free agan.

17 7 esoure Aess Protools Classal approah: Eah mutually exlusve resoure s proteted by a semaphore S. Eah rtal seton on must begn wth wat(s ) and end wth sgnal(s ) the only operatons supported on semaphores. Tas state graph: dspathng, aordng to prorty atvaton eady unnng termnaton preempton sgnal Watng wat Problem: Prorty Inverson.

18 Prorty Inverson 8 Let two tass T and T wth prortes P >P be gven that share a mutually exlusve resoure. T bloed T T Crtal seton Normal exeutng t t T s atvated frst, enters the rtal seton and los the semaphore. When T s released, t preempts T sne ts prorty s hgher. However, when attemptng to enter ts rtal seton at t, T s bloed on the semaphore, so T resumes although ts prorty s lower. In [t,t ] a prorty nverson ours.

19 9 Prorty Inverson Nave soluton: Dsallow preempton durng exeuton of rtal setons. May ause unneessary blong for a long perod of tme. Example: Assume P >P >P 3. T s bloed for a long tme although t does not use any resoure. T bloed T T T 3 t t Better solutons requred.

20 Prorty Inhertane Protool (PIP) 0 Idea: modfy the prorty of the blong tass. Let J denotes a ob, e a gener nstane of tas T. When a ob J blos one or more hgher-prorty tass, t temporarly nherts the hghest prorty of the bloed tass. Ths prevents medum-prorty tass from preemptng J. Let Γ {T } be a set of perod tass ooperatng through M shared resoures,..., M. Eah resoure s guarded by a dstnt semaphore S. Assume d π for all tass T. The protool an modfy the prorty of tass. Thus: nomnal prorty P atve prorty p (p P ), whh s dynam and ntally set to P.

21 Prorty Inhertane Protool (PIP) Only one ob at a tme an be wthn the rtal seton orrespondng to a partular semaphore S. Let z, denote the th rtal seton of ob J. The S, s the semaphore guardng z, and, s the resoure assoated wth z,. Let u, denote the duraton of z,, e the tme needed by J to exeute z, wthout nterrupton. We assume prorty orderng for obs J, J,..., J n wrt nomnal prortes suh that P P... P n. Crtal setons are perfetly nested, e ether z, z, or z, z,, or z, z,.

22 Defnton of Prorty Inhertane Protool Jobs are sheduled based on atve prortes. Jobs wth the same prorty are exeuted frst ome frst served. When a ob J tres to enter a rtal seton z, and resoure, s already held by a lower-prorty ob, J wll be bloed. Otherwse J enters z,. When a ob J s bloed on a semaphore, t transmts ts atve prorty to the ob J that holds that semaphore. Then J resumes and exeutes the rest of ts rtal seton wth the nherted prorty p p. When J exts a rtal seton, t unlos the semaphore and the hghestprorty ob bloed on that semaphore s awaened. The atve prorty of J s updated as follows: f no other obs are bloed by J, p s set to ts nomnal prorty P, otherwse t s set to the hghest prorty of the obs bloed by J. Prorty nhertane s transtve; f J 3 blos J and J blos J then J 3 nherts the prorty of J va J.

23 3 Prorty Inhertane Protool Example dret blong push-through blong J J Crtal seton Normal exeutng J 3 P P 3 P 3 Dret blong: a hgh-prorty ob tres to aqure a resoure held by a lower-prorty ob. Neessary to ensure onssteny of shared resoures. Push-through blong: a medum-prorty ob s bloed by a lowerprorty ob that has nherted a hgher prorty from a ob t dretly blos. Neessary to avod unbounded prorty nverson.

24 Prorty Inhertane Protool Propertes Lemma: If there are n lower-prorty obs that an blo a ob J, then J an be bloed for at most the duraton of n rtal setons (one for eah of the n lower-prorty obs), regardless of the number of semaphores used by J. Proof: A ob J an be bloed by a lower-prorty ob J only f J has been preempted wthn a rtal seton z, and s stll suspended n the moment when J s ntated. One J exts z,, t an be preempted by J ; thus J annot be bloed by J agan. The same stuaton may happen for eah of the n lower-prorty obs; therefore J an be bloed at most n tmes.

25 Prorty Inhertane Protool Propertes 5 Lemma: If there are m dstnt semaphores that an blo a ob J, then J an be bloed for at most the duraton of m rtal setons, one for eah of the m semaphores. Proof: Sne semaphores are bnary, only one of the lower-prorty obs J an be wthn a blong rtal seton orrespondng to a partular semaphore S. One S s unloed, J an be preempted and an no longer blo J. If all m semaphores that an blo J are loed by m lower-prorty obs, then J an be bloed at most m tmes.

26 Prorty Inhertane Protool Propertes 6 Theorem (Sha-aumar-Lehozy): Under the Prorty Inhertane Protool, a ob J an be bloed for at most the duraton of mn(n,m) rtal setons, where n s the number of low-prorty obs that ould blo J and m s the number of dstnt semaphores that an be used to blo J. Proof: mmedately follows from the two prevous lemmas.

27 PIP Shedulablty Analyss Lu/Layland: N π N( N ) Let B be the maxmum blong tme, due to lower-prorty obs, that a ob J may experene. (*) 7 Theorem: A set of n perod tass usng the Prorty Inhertane Protool an be sheduled by the ate-monoton algorthm f n : ( ) π π Proof: If the rteron holds then a ob J has enough tme even f t lasted for B, tang nto aount the preempton /π from hgher prorty obs. B

28 8 PIP esponse Tme Analyss To tae resoures nto aount, the blong fator B must be added to the omputaton tme of eah tas. Ths gves the followng response tme equaton The orrespondng reurrene equaton s: B I B π ) ( ) ( ) ( B π

29 PIP Computng the Blong Tme 9 Let the elng C(S ) of a semaphore S be defned as C ( S ) max{ P ob J uses S } Let D, denote the duraton of the longest rtal seton of tas T among those guarded by semaphore S. Let a set of N perod tass that use M bnary semaphores be gven. Then the maxmum blong tme B for eah tas T an be determned as follows: B l N max{ D, C(S ) P} B s M max{ D >, C(S ) P} B mn( B l, B s )

30 PIP Example 30 Let a set of four tass wth three semaphores be gven. The table shows the values D, for a ob J and a semaphore S. The semaphore elngs are gven n parentheses. D, S (P ) S (P ) S 3 (P ) J 0 J J J 6 5 Then the blong fators for ob J are omputed as follows: l B max{ D, C(S ) P } B s B M max{ D 7 >, C(S ) P }

31 PIP Chaned Blong 3 J a b J J 3 a b a b Crtal seton Normal exeutng In the worst ase, f J aesses m dstnt semaphores that have been loed by m lower-prorty obs, then J wll be bloed for the duraton of m rtal setons.

32 3 PIP Deadlos bloed on S b bloed on S a J a Crtal seton J b b Normal exeutng t t t 3 t t 5 t : J los S b. t : J s preempted by the hgher-prorty ob J. t 3 : J los S a. t : J s bloed on S b. J resumes and ontnues exeuton at the prorty of J. t 5 : J attempts to lo S a. > Deadlo! Note: deadlo s aused by erroneous use of semaphores.

33 33 The Prorty Celng Protool Eah semaphore S s assgned a prorty elng C(S ), equal to prorty of the hghest-prorty tas that an lo t. Note that C(S ) s a stat value that an be omputed offlne. When a tas T wants to lo a semaphore S, let H be the set of semaphores held by tass dfferent from T and P* max{c(s ) S H }. Tas T gets the lo S only f P >P*. Note that P* s ndependent from the semaphore S. When a ob J s bloed on a semaphore t transmts ts prorty to the ob J that holds the semaphore. Hene, J resumes and exeutes the rest of ts rtal seton wth the prorty of J. J s sad to nhert the prorty of J. When J exts a rtal seton, t unlos the semaphore and the hghestprorty ob, f any, bloed on that semaphore s awaened. The atve prorty of J s set to the normal prorty J f no other obs are bloed by J, otherwse t s set to the hghest prorty of the obs bloed by J.

34 3 The Prorty Celng Protool Example Let three obs J, J, and J 3 havng dereasng prortes be gven. J sequentally aesses two rtal setons guarded by semaphores S and S. J only aesses a rtal seton guarded by S 3. J 3 uses semaphore S 3 and then maes a nested aess to S. Ths gves the followng prorty elngs: C(S )P, C(S ) P, C(S 3 ) P. elng blong J S S J S 3 Crtal seton J 3 S 3 S 3 S S S 3 Normal exeutng P 3 P P 3

35 The Prorty Celng Protool - Propertes 35 A hgh-prorty proess an be bloed at most one durng ts exeuton by any lower-prorty proess. Deadlos are prevented. Transtve blong s prevented.

36 Model-based Software Development 36 Generator Lustre programs Esterel programs Compler C Code Compler Esterel SCADE - SCADE language - SynCharts Bnary Code at WCET Analyzer - Tmng Valdaton SymTA/S - System-level Shedulng & Shedulablty Analyss