Critical sections. Using semaphores. Using semaphores. Using semaphores. How long is blocking time? 17/10/2016. Problems caused by mutual exclusion

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1 rtcal sectons Problems caused by mutual excluson crtcal secton wat(s) x = ; y = 5; sgnal(s) wrte global memory buffer nt x; nt y; read wat(s) a = x+; b = y+; c = x+y; crtcal secton sgnal(s) Usng semaphores Mae crtcal sectons as short as possble. nt x, y; // these are global shared varables mutex s; // ths s the semaphore to protect them tas reader() { nt ; // these are local varables float d, v[im];... an we shorten ths crtcal secton? wat(s); d = sqrt(x*x + y*y); for (=; ++; <IM) { v[] = *(x + y); f (v[] < x*y) v[] = x + y; sgnal(s);... crtcal secton length Usng semaphores possblty s to copy global varables nto local varables: tas reader() { nt ; // these are local varables float d, v[im]; float a, b; // two new local varables... wat(s); // copy global vars a = x; b = y; // to local vars sgnal(s); d = sqrt(a*a + b*b); // mae computaton for (=; ++; <IM) { // usng local vars v[] = *(a + b); f (v[] < a*b) v[\] = a + b; wat(s); x = a; y = b; sgnal(s);... // copy local vars // to global vars crtcal secton length crtcal secton length 4 Usng semaphores How long s blocng tme? Mae crtcal sectons as short as possble. Try to avod nested crtcal sectons. vod mang crtcal sectons across loops or condtonal statements. Ths code s very UNSFE, snce the... sgnal could never be executed, and wat(s); could be bloced forever! results = x + y; whle (result > ) { v[] = *(x + y); f (v[] < x*y) results = results - y; else sgnal(s); 5 S S P > P It seems that the maxmum blocng tme for s equal to the length of the crtcal secton (S )of, but 6

2 Schedule wth no conflcts onflct on a crtcal secton prorty prorty 7 8 onflct on a crtcal secton Prorty Inverson prorty hgh prorty tas s bloced by a lower prorty tas a for an unbounded nterval of tme Soluton Introduce a concurrency control protocol for accessng crtcal sectons. 9 Protocol ey aspects Rules for classcal semaphores ccess Rule: ecdes whether to bloc and when. The followng rules are normally used for classcal semaphores: Progress Rule: ecdes how to execute nsde a crtcal secton. Release Rule: ecdes how to order the pendng requests of the bloced tass. Other aspects nalyss: Implementaton: estmates the worst-case blocng tmes. fnds the smplest way to encode the protocol rules. ccess Rule (ecdes whether to bloc and when): Enter a crtcal secton f the resource s free, bloc f the resource s loced. Progress Rule (ecdes how to execute n a crtcal secton): Execute the crtcal secton wth the nomnal prorty. Release Rule (ecdes how to order pendng requests): Wae up the bloced tas n FIFO order. Wae up the bloced tas wth the hghest prorty.

3 Resource ccess Protocols lasscal semaphores (No protocol) ssumpton rtcal sectons are correctly accessed by tass: Non Preemptve Protocol (NPP) Hghest Locer Prorty (HLP) Prorty Inhertance Protocol (PIP) Prorty elng Protocol (PP) Stac Resource Polcy (SRP) wat(s ) wat(s ) sgnal(s ) sgnal(s ) wat(s ) wat(s ) sgnal(s ) sgnal(s ) 4 Non Preemptve Protocol onflct on a crtcal secton ccess Rule: tas never blocs at the entrance of a crtcal secton, but at ts actvaton tme. prorty (usng classcal semaphores) Progress Rule: sable preempton when executng nsde a crtcal secton. Release Rule: t ext, enable preempton so that the resource s assgned to the pendng tas wth the hghest prorty. 5 6 NPP: example NPP: mplementaton notes prorty Each tas must be assgned two prortes: a nomnal prorty P (fxed) assgned by the applcaton developer; a dynamc prorty p (ntalzed to P ) used to schedule the tas and affected by the protocol. 7 Then, the protocol can be mplemented by changng the behavor of the wat and sgnal prmtves: wat(s): p = max(p,, P n ) sgnal(s): p = P 8

4 NPP: pro & cons VNTGES: smplcty and effcency. Semaphores queues are not needed, because tass never bloc on a wat(s). Each tas can bloc at most on a sngle crtcal secton. It prevents deadlocs and allows stac sharng. It s transparent to the programmer. NPP: problem Long crtcal sectons delay all hgh prorty tass: prorty s useless: cannot preempt, although t could! PROLEMS:. Tass may bloc even f they do not use resources.. Snce tass are bloced at actvaton, blocng could be unnecessary (pessmstc assumpton). 9 Prorty assgned to nsde crtcal sectons: p = P max = max(p,, P n ) NPP: problem Hghest Locer Prorty tas could bloc even f not accessng a crtcal secton: blocs just n case... ccess Rule: tas never blocs at the entrance of a crtcal secton, but at ts actvaton tme. test Progress Rule: Insde resource R, a tas executes at the hghest prorty of the tass that use R. S S p Release Rule: t ext, the dynamc prorty of the tas s reset to ts nomnal prorty P. P max P HLP: example HLP: mplementaton notes prorty s bloced, but can preempt Each tas s assgned a nomnal prorty P and a dynamc prorty p. Each semaphore S s assgned a resource celng (S): (S) = max { P uses S p Then, the protocol can be mplemented by changng the behavor of the wat and sgnal prmtves: P P wat(s): sgnal(s): p = (S) p = P Prorty assgned to nsde a resource R: p (R) = max { P j j uses R Note: HLP s also nown as Immedate Prorty elng (IP). 4 4

5 HLP: pro & cons Prorty Inhertance Protocol VNTGES: smplcty and effcency. Semaphores queues are not needed, because tass never bloc on a wat(s). Each tas can bloc at most on a sngle crtcal secton. It prevents deadlocs. It allows stac sharng. PROLEMS: Snce tass are bloced at actvaton, blocng could be unnecessary (same pessmsm as for NPP). It s not transparent to programmers (due to celngs). 5 ccess Rule: Progress Rule: Release Rule: tas blocs at the entrance of a crtcal secton f the resource s loced. Insde resource R, a tas executes wth the hghest prorty of the tass bloced on R. t ext, the dynamc prorty of the tas s reset to ts nomnal prorty P. 6 PIP: example PIP: types of blocng prorty drect blocng push-through blocng rect blocng tas blocs on a loced semaphore Indrect blocng (Push-through blocng) tas blocs because a lower prorty tas nherted a hgher prorty. nherts the prorty of P LOKING: a delay caused by lower prorty tass P 7 8 PIP: mplementaton notes Insde a resource R the dynamc prorty p s set as p (R) = max { P h h bloced on R Identfyng blocng resources Under PIP, a tas can be bloced on a semaphore S only f: wat(s): f (s == ) { <suspend the callng tas exe n the semaphore queue> <fnd the tas that s locng the semaphore s> p = P exe // prorty nhertance <call the scheduler> else s = ; sgnal(s): f (there are bloced tass) { <awae the hghest prorty tas n the semaphore queue> p exe = P exe <call the scheduler>. S s drectly shared between and lower prorty tass (drect blocng) else s = ; 9 OR. S s shared between tass wth prorty lower than and tass havng prorty hgher than (push-through blocng). 5

6 Identfyng blocng resources Lemma : tas can be bloced at most once by a lower prorty tas. Identfyng blocng resources Lemma : tas can be bloced at most once on a semaphore S. If there are n tass wth prorty lower than, then can be bloced at most at most n tmes, ndependently of the number of crtcal sectons that can bloc. If there are m dstnct semaphores that can bloc atas, then can be bloced at most m tmes, ndependently of the number of crtcal sectons that can bloc. oundng blocng tmes Example theorem follows from the prevous lemmas: Theorem: can be bloced at most for the duraton of = mn(n,m ) crtcal sectons. prorty n = number of tass wth prorty less than m = number of semaphores that can bloc (ether drectly or ndrectly). can be bloced once by (on or ) and once by (on or ) can be bloced once by (on or ) cannot be bloced 4 prorty Example Example n whch s bloced on by push-through Identfyng blocng tmes To derve a general analyss, we defne the followng notaton: Z longest (external) crtcal secton used by protected by semaphore S. P worst-case duraton of Z set of the longest crtcal sectons used by for each semaphore S : ={Z S used by j set of crtcal sectons used by j that can bloc set of crtcal sectons that can bloc maxmum number of crtcal sectons that can bloc 5 worst-case blocng tme for 6 6

7 Identfyng blocng tmes For the other protocols.s. of j that can bloc for drect blocng: dr j j NPP j { Z j ( Z j j ) N ( Pj P ).S. of j that can bloc for push-through blocng:.s. of j that can bloc j dr j.s. that can bloc pt j pt j h { { h n h j j j h j 7 HLP j PIP j { Z ( Z ) N ( P P ) N { h j h j.s. that can bloc j j j n j j ( ( S ) P ) 8 Identfyng blocng tme Example. Identfy the set j for all lower prorty tass. Identfy the set. ompute 4. ompute as the hghest sum of the duratons of Z NOTE: The crtcal sectons selected from must belong to dfferent tass (for Lemma ); must refer to dfferent semaphores (for Lemma ); 9 onsder the followng applcaton: IN prorty E E (R ) E 4 4 OUT T Example Identfcaton of E E 4 4 T E From the tas code we can derve the followng table: 4 4 T E = {,,, 4 can only experence drect blocng because t s the hghest prorty tas. 4 7

8 Identfcaton of Identfcaton of 4 T E T E = {,,, 4 = {,,, 4 = {, 4, 4 = {, 4, 4 can be bloced drectly by 4 and 4, and ndrectly by and 4. 4 can be bloced = { 4, 4, E 4 drectly by E 4 and ndrectly by 4 and 4 44 Identfcaton of 4 Identfcaton of 4 4 T E T E n m {,,, {, 4, { 4, 4, E { = {,,, 4 = {, 4, 4 = { 4, 4, E 4 4 = { = mn(n, m ) number of tass wth prorty less than number of semaphores that can bloc (ether drectly or ndrectly) Identfcaton of PIP: pro & cons 4 T E {,,, {, 4, { 4, 4, E { VNTGES: It removes the pessmsms of NPP and HLP (a tas s bloced only when really needed). It s transparent to the programmer. NOTES For,fweselect, we cannot select 4, because each semaphore can bloc only once (Lemma ). For, we cannot select 4 and 4, because each tas can bloc only once (Lemma ). 47 PROLEMS: More complex to mplement (especally to support nested crtcal sectons). It s prone to chaned blocng. It does not avod deadlocs. 48 8

9 PIP: haned blocng Prorty elng Protocol,, prorty 4 ccess Rule: tas can access a resource only f t passes the PP access test. 4 Progress Rule: Release Rule: Insde resource R, a tas executes wth the hghest prorty of the tass bloced on R. t ext, the dynamc prorty of the tas s reset to ts nomnal prorty P. NOTE: can be bloced at most once for each lower prorty tas. NOTE: PP can be vewed as PIP + access test 49 5 vodng chaned blocng Resource elngs prorty 4 To eep trac of resource usage by hgh-prorty tass, each resource s assgned a resource celng: (s ) = max { P uses s 4 To avod multple blocng of we must prevent and to enter ther crtcal sectons (even f they are free), because a low prorty tas ( 4 ) s holdng a resource used by. 5 Then a tas can enter a crtcal secton only f ts prorty s hgher than the maxmum celng of the loced semaphores: PP access test P > max { (s ) : s loced by tass 5 PP: example PP: propertes s (s ) = P Theorem prorty s (s ) = P Under PP, a tas can bloc at most on a sngle crtcal secton. celng blocng Theorem PP prevents chaned blocng. t t : s bloced by the PP, snce P < (s ) Theorem PP prevents deadlocs

10 Typcal deadloc It can only occur wth nested crtcal sectons: PP: deadloc avodance It can only occur wth nested crtcal sectons: (S ) = P P > P P > P (S ) = P bloced bloced by PP bloced PP: pro & cons Questons VNTGES: It lmts blocng to the length of a sngle crtcal secton. It avods deadlocs when usng nested crtcal sectons. PROLEMS: It s complex to mplement (le PIP). It can create unnecessary blocng (t s pessmstc le HLP). It s not transparent to the programmer: resource celngs must be specfed n the source code.. If a tas uses several crtcal sectons, can t be bloced on the second one?. Each tas has a nomnal prorty (P ) and a dynamc prorty (p P ) changed by the protocol to prevent prorty nverson. If blocs because P max {(s ) : s loced by tass can t be that p > max {(s ) : s loced by tass? nalyss under shared resources nalyss under RM. Select a schedulng algorthm to manage tass and a protocol for accessng shared resources. preempton by HP tass. ompute the maxmum blocng tme for each tas. blocng by LP tass. Perform the guarantee test ncludng the blocng terms. T T / 59 6

11 Hyperbolc ound Response Tme nalyss preempton by HP tass R R T blocng by LP tass Iteratve soluton: T T R R ( s) terate whle R T ( s) s R R ( s) 6 6 Resource Sharng under EF The protocols analyzed so far have been orgnally developed for fxed prorty schedulng schemes. However: NPP can also be used under EF PIP has been extended under EF by Spur (997). PP has been extended under EF by hen-ln (99) In 99, aer proposed a new access protocol, called Stac Resource Polcy (SRP) that wors both under fxed prortes and EF. Stac Resource Polcy Ths protocol satsfes the same PP propertes: t avods unbounded prorty nverson; t prevents deadlocs and chaned blocng; each tas can be bloced at most once. In addton: It allows usng mult-unt resources; t can be used under fxed and dynamc prortes; t allows tass to share the same stac space Stac Resource Polcy Each resource R s characterzed by: a maxmum number of unts N a current number of avalable unts n Preempton level / Note that, under EF, a tas can preempt only f > (that s, f < ) Each tas s characterzed by: prortyp (statc or dynamc), e.g.: preempton level: / set of resource requrements: R (R ) specfes how many RM: p /T M: p / EF: p /d No preempton can occur f (that s, f ): unts of R are used by 65 66

12 Resource celng Each resource R s assgned a dynamc celng equal to the hghest preempton level of the tass that may bloc on R : R (n ) = max {, : n(r ) < (R ) elng property R (n ) = max {, : n(r ) < (R ) Lemma If > R (n ) then there exst enough unts of R NOTE: R (n ) ncreases only when a resource s loced R (n ) decreases only when a resource s released. to satsfy the requrements of. to satsfy the requrements of all tass that can mae preempton on Stac Resource Polcy omputng Resource elngs Fnally, a system celng s defned as: s = max { R (n ) R R N R () () () SRP preempton rule () ready tas can preempt the executng tas exe f and only f 5 p > p exe and > s 69 7 omputng Resource elngs SRP: example R R N R 5 () () () () N R R () R () R () R () R R - R () R () R () R () R s R - 7 tas blocs when attemptng to preempt 7

13 PP: example SRP: propertes P P P p () () () P () s (s ) = P s (s ) = P tas s bloced when accessng a resource 7 Theorem Under SRP, each tas can be bloced at most on a sngle crtcal secton. onsder the followng scenaro where blocs on two: t* Ths s not possble, because could not preempt because, at tme t*, < s 74 Theorem SRP: propertes Theorem SRP: propertes If > s then wll never bloc once started. SRP prevents deadlocs. Proof Snce s = max{ R (n ), then there are enough resources to satsfy the requrements of and those of all tass that can preempt. Proof From Theorem, f a tas can never bloc once started, then no deadloc can occur. Queston If a tas can never bloc once started, can we get rd of the wat / sgnal prmtves? SRP: deadloc avodance Schedulablty analyss under EF > When = T bloced by SRP tas set s schedulable f T T s can be computed as under PP and refers to the length of longest crtcal secton that can bloc

14 Schedulablty analyss under EF When T tas set s schedulable f U < and L n ( L) L T T L Stac sharng Each tas normally uses a prvate stac for savng context (regster values) managng functons storng local varables where: (L) = max { jh ( j > L) and ( h L) stac ponter PUSH = { d d max ( max, mn(h, L * )) POP max = max (,, n ) H = lcm(t,, T n ) * L n ( T ) U U 79 stac 8 Stac sharng Why stac cannot be normally shared? Stac sharng Why stac can be shared under SRP? Suppose tass share a resource: bloced bg problems SP SP stac SP SP SP stac 8 8 Savng stac memory To really save stac sze, we should use a small number of preempton levels. tass Kb stac per tas preempton levels tass per group stac sze = Mb stac sze = Kb stac savng = 9 % omparson Prot pro. pessm. bloc. nstant NPP any hgh HLP fxed medum PIP fxed low mn (n, m ) PP fxed medum SRP any medum on arrval on arrval on access on access on arrval transparent dloc avod. stac sharng mpl. YES YES YES easy NO YES YES easy YES NO NO hard NO YES NO hard NO YES YES easy