Worst Case Interrupt Response Time Draft, Fall 2007

Transcription

1 Worst Case Interrupt esponse Te Draft, Fall 7 Phlp Koopan Carnege Mellon Unversty Copyrght 7, Phlp Koopan eproducton and dssenaton beyond students of CMU ECE s prohbted.. Overvew: Interrupt Servce outnes (ISs) are coonly used to provde fast response tes to external events or ted events. Because the pont of provdng fast response s to eet deadlnes, t s portant to know the worst case executon te of ultple concurrent nterrupts copetng for processor resources. The usual schedulng theory ath doesn t work that well for ths case because ost schedulng theory assues preeptve task swtchng, whle ISs are usually wrtten to be non-preeptve (.e., nterrupts rean asked whle the IS s runnng). Ths s an nstance of the ore general proble of deternng the axu response te for a prortzed, non-preeptve taskng envronent.. Iportance: If only a sngle nterrupt s used n a syste, deternng nterrupt servce latency s relatvely easy. However, f ultple prortzed nterrupts can occur, then soe wll be servced quckly, and others wll be servced ore slowly. There wll be soe worst-case stuaton n whch lower prorty nterrupts wll have to wat for one (or ore) executons of all hgher prorty nterrupts. Ensurng that the worst case latency of lower prorty nterrupts s fast enough to eet real te requreents s an portant analyss ssue. Unfortunately, t s often dffcult or possble to create worst-case stuatons n a testng stuaton, so analytc approaches ust be used to ake sure that testng doesn t ss a partcularly bad tng proble. The IS response scenaro s an nstance of the ore general stuaton of a prortzed cooperatve schedulng taskng syste. In such a syste each task has a prorty, but tasks run to copleton (.e., tasks are non-preeptve). 3. Graphcal Approach For ths dscusson, we assue that there s a collecton of prortzed tasks that needs to be executed perodcally. Those tasks could be ISs, threads, processes, or any xture of the above so long as there s a statc total orderng of prorty across all tasks (.e., fxed task prorty). Ths would be the case, for exaple, for prortzed ISs whch keep nterrupts asked whle executng, deferrng any hgher prorty nterrupt servcng untl the currently executng IS has 7, Phlp Koopan Interrupt esponse Te

2 copleted. Ths s usually how prortzed nterrupts are executed. (The excepton s when a software developer explctly re-enables nterrupts durng an IS, but that s usually bad practce.) Frst, let s work out an exaple graphcally to understand what s nvolved. Consder the below exaple task set ncludng prortzed nterrupts, executon tes, and perods: takes 5 sec to execute and occurs at ost once every 5 sec takes 6 sec and occurs at ost once every sec takes 7 sec and occurs at ost once every sec takes 9 sec and occurs at ost once every 5 sec IS4 takes 3 sec and occurs at ost once every 6 sec where s the hghest prorty and IS4 s the lowest prorty. No IS can preept any other IS. When an IS copletes executon, the hghest prorty IS that s ready to execute wll execute next. We assue that any underlyng tasks don t dsable nterrupts. There are a nuber of other assuptons we are akng to splfy ths analyss, but those wll be dscussed later n the analytc approach secton. The frst queston we want to ask s, what s the worst case latency for? For exaple f ust coplete wthn 5 sec of the te the nterrupt s frst requested, s there a case where that won t happen? The proble s that s not the only task runnng other ISs are copetng for processor resources. A bad case s when another lower prorty task than has just started to run when s trggered for executon. In partcular, the worst case s when the task wth the longest runnng te havng lower prorty than has just started to run. For ths task set that s ( and IS4 are both lower prorty than, but has a uch longer run te of 9 sec copared to the 3 sec run te of IS4). Why dd we pck a lower prorty nstead of a hgher prorty nterrupt to start wth? The reason s that n the worst case, the CPU wll be unavalable for a whle when an nterrupt arrves, causng other nterrupts to ple up before the one we are nterested even has a chance to copete for CPU te. Because no nterrupt wth lower prorty wll execute after becoes ready to run, selectng one wth a lower prorty adds ore work to the tasks that ust be copleted before can be started. We ll take all the hgher prorty nterrupts nto account shortly. But once s ready to run, no lower prorty nterrupt can run, so only one such low prorty nterrupt need be consdered, and the longest executon te one s the worst case. 9 sec:,, Fgure. Executes before,, and are trggered. 7, Phlp Koopan Interrupt esponse Te

3 Now we have a stuaton where ght get to run before by just beatng t to the CPU. Beyond that, t s possble that every hgher prorty nterrupt than starts just after starts, but before ends, so they could also get to run before as well. Fgure shows ths stuaton, wth startng to run, followed quckly by,, and beng trggered. The tes at whch and can be retrggered are also shown n Fgure, snce as we wll fnd out they ght have to be servced one or ore tes before fnally gets a chance to run. When fnshes executng at 9 sec, all three of,, and are pendng. Snce s the hghest prorty task, t goes frst, followed by. Whle s executng, s trggered a second te at 5 sec, and s trggered agan at te sec. Nether of these events dsturbs the executon of snce t s non-preeptable (nterrupts are asked whle executng ISs). Ths leads to a stuaton at te where all three of,, and are stll pendng (Fgure ). sec:,, Fgure. By the te and run once, they have been retrggered. At sec, s the hghest pendng task, so t executes agan, and s agan followed by, at 5 sec. then retrggers at 3 sec, but has not yet trggered agan. So at 3 sec when ends, only and are pendng (Fgure 3). 3 sec:, Fgure 3. At 3 sec, has retrggered and s stll pendng, but t s not te for yet. At 3 sec s stll the hghest prorty nterrupt pendng, so t runs untl 36 sec (Fgure 4). 7, Phlp Koopan 3 Interrupt esponse Te

4 36 sec: Fgure 4. At 3 sec, and are pendng, so runs agan. At 36 sec, has copleted executon and s no longer pendng (t won t be trggered agan untl 45 sec). Moreover, s not due to run untl 4 sec. Ths leaves as the only task pendng, so t starts executon and runs untl 43 sec. By 43 sec has retrggered, so t starts runnng (Fgure 5), but does not nterfere wth the copleton of because nterrupts are non-preeptable once started. Thus, n the worst case, copletes at 43 sec after t s trggered. 43 sec: Fgure 5. At 36 sec, s the only task stll pendng, so t fnally gets to execute. 4. Analytc Approach Now that we have seen the types of coplcatons that can arse when ultple tasks copete for CPU te, we can take a ore rgorous, atheatcal, approach to the analyss. In ths secton we ll create a set of equatons that coputes the worst case latency for any task n a set of tasks. These equatons can be used to deterne f each task n the set wll eet ts own partcular deadlne. The followng notaton s used n the equatons below: T : Task : esponse te of Task, whch s the worst-case te between when T s ready to start executng and the te t actually starts executon. W : Copleton te of Task C : Coputaton te for one executon of T (worst case largest possble C ) 7, Phlp Koopan 4 Interrupt esponse Te

5 P : Perod for executon T (worst case fastest possble P ). If the task s aperodc, then assue a P correspondng to shortest possble te between any two executons of T (.e., recprocal of worst case shortest nter-arrval te of task executons). D : Deadlne for T B : Blockng te caused by background tasks that ask nterrupts, or other dependences. x : floor functon; rounds x down to next lowest nteger The followng assuptons are used n the equatons below as a startng pont: There are prortzed N tasks, nubered through N-, wth Task called T. In the prevous exaple, each IS handler was a task. Any other tasks runnng on the coputer are referred to as background tasks. Background tasks nterfere wth Tasks through N- only va dsablng nterrupts or task swtchng for soe axu blockng te B. (Blockng te was not shown n the precedng graphcal exaple.) Tasks are statcally prortzed, wth Task beng the hghest prorty and Task N- beng the lowest Each task T executes only when no other task wth hgher prorty s ready to execute, then runs to copleton wthout stoppng (.e., tasks are non-preeptable). If no task T s ready to execute, then background tasks are run untl soe task T s trggered to run. Any task T can and wll preept any background tasks, possbly wth a delay caused by blockng te. (For exaple, ISs preept any non-is code.) Each task s trggered for executon no faster than once per stated perod. Moreover, the perod represents the worst-case nu nter-arrval te between trggers for that task to execute. The perod of each task ay be dfferent. The worst-case longest copute te for each task s known and used n the calculatons. The deadlne for each task s known, and s less than or equal to that task s perod. The cost of changng tasks (e.g., processng an nterrupt and correspondng TI nstructon) s accounted for n the worst-case copute te. The values we are nterested n fndng are the copleton tes of all tasks. For a syste to perfor properly, all tasks ust coplete ther work W at or before the applcable deadlne D. So, the ultate goal s to ensure that: () ( W D ) Whch states: for all values of, the copleton te of Task s less than or equal to the deadlne of Task (.e., all tasks coplete before ther deadlne). The copleton te of a task has two coponents: the te spent watng to start executon (the response te ) and the te spent actually dong the work of the task (the coputaton te C ). () W = C In the systes we re lookng at, tasks are non-preeptable, so once the coputaton of a task starts, that task runs to copleton. Thus, C s a known constant value. But, s trcker, because t ust take nto account the fact that Task has to wat for all hgher prorty tasks to execute and also wat for any blockng te. For exaple, f Task 4 s an IS, that IS can t execute untl any nterrupt askng n the an progra s copleted (.e., blockng te B) and all hgher prorty nterrupts,,, and 3 execute at least one te (because n the worst case all four of those 7, Phlp Koopan 5 Interrupt esponse Te

6 nterrupts were trggered just after the begnnng of the blockng te such a sall delay that we wll just be conservatve and consder t to be zero elapsed te n our equatons). Accountng for blockng te starts the buld-up of equatons to obtan : (3) B By ths we ean that s at least as long as the blockng te, but possbly longer. Now let us consder Task. Is B the only factor that could delay the start of executon? Even though ths s the hghest prorty task n the syste, there s soethng else that can delay t. The other factor s soe other task wth a lower prorty that has already begun executon, because tasks are non-preeptable (once started, they run to copleton). The worst case s that the task wth the longest possble coputaton te has just started executon, and ust coplete before Task can run. In other words, a lower prorty task can delay executon of a hgher prorty task because t s allowed to run to copleton. Ths, n effect, s a dfferent for of blockng. In general, for Task, t s possble for soe task wth a hgher task nuber (.e., lower prorty) to be executng, delayng the start of Task. Thus, (4) ax < j< N N ( C ) j Ths eans that the response te for Task ust be at least as bad as the worst case wat caused by the longest coputaton te of any Task j wth a lower prorty than Task. Task tself s not consdered, because we assue that the deadlne for each task s longer than ts perod, so Task ust have copleted executon before t attepts to execute agan. For task N-, whch s the lowest prorty task that sn t a background task, ths delay s zero, snce there s no lower prorty task to get n the way (but, even ths task s subject to blockng te fro the background tasks). Next, we cobne the two startng factors of B and axu C j to get an ntal lower bound on response te. But rather than addng the, we can sply take the axu of the two, because both stuatons can t happen at the sae te. Consder the two possble stuatons. Stuaton (): If a task has to wat for blockng te B, then that eans a task T sn t already runnng (because blockng can only occur due to a task other than Tasks..N- executng). If that s the case, as soon as blockng has fnshed, the hghest prorty task wll begn executng as soon as blockng s over. Ths akes t possble for a task wth lower prorty than Task to delay the start of task after blockng. If Task sn t ready to execute when blockng s copleted, then the blockng te hasn t delayed ts response te, snce t wasn t ready to run. Stuaton (): If Task j, wth lower prorty than Task, s already executng, then blockng can t occur, because when Task j copletes, Task (or soe task wth hgher prorty) wll edately start executng rather than the background tasks. The background tasks that can cause blockng won t resue executon untl all prortzed tasks, ncludng Task, coplete executon. So, let s defne the effectve blockng te B as: ( ) (5) B' = ax ax C j, B ; < N < j< N B' = B ; otherwse 7, Phlp Koopan 6 Interrupt esponse Te

7 Ths eans that the response te for Task s bounded by an effectve blockng te B, whch s the longest lower prorty task that ght execute, or the blockng te B. Because there s no lower prorty task than Task N-, then blockng te B s the only ssue for that partcular task. For Task, our response te calculaton s done. Because there are no hgher prorty tasks, Task wll run to copleton once the effectve blockng effect has passed (ether watng for background task blockng B or the longest lower prorty task to coplete). (6) = B' The next factor n response te calculatons s that hgher prorty tasks can execute before lower prorty tasks. In the worst case, Task wll have to wat for every possble Task wth hgher prorty to execute at least once. Fro ths pont on, the response te wll have to be coputed teratvely to account for the fact that enough te ay pass for hgh prorty tasks to re-trgger. We ll use the notaton,k to represent the kth teraton of the coputaton for, wth the coputaton terated by ncreasng k untl the answer converges to a fnal value. To keep thngs sple, and confor to the graphc approached used prevously, we start the teraton wth the effectve blockng value B : (7), = B' Next, we have to account for the executon te of all tasks wth hgher prorty than Task, because t s possble all of the trggered just as Task was trggerng. The nuber of tes a partcular Task executes n te T s one ore than the rounded-down (nteger floor functon) nuber of tes the response te,k can be dvded by the perod of Task : (8) executons T = P For exaple, wth a perod of 7 and an elapsed te of, a task could have been trggered not /7 = 3.4 tes, but rather that nuber rounded down, whch s 3, plus to account for the fact the task ust assued to have been trggered at te zero, whch gves a total of 4 tes (.e., at tes, 7, 4, and sec). (Note that a celng functon ght see attractve nstead of the floor functon. But, the celng functon doesn t qute work f a response te s an exact ultple of a perod.) Once the nuber of executons s known, the aount of delay that hgher prorty Task causes to the watng Task by the te Task s ready to run s the nuber of executons of Task that have taken place by tes the coputaton te of Task : (9) delay executons C = = P C 7, Phlp Koopan 7 Interrupt esponse Te

8 The aount of te that s taken by each executon s the task s coputaton te C. Therefore, the total aount of watng te for Task caused by watng for hgher prorty tasks s the su across all those tasks: () = = P C ; > We stll need to account for the ntal effectve blockng te before any of those hgh prorty tasks can execute, so the coplete equaton s: () = ' = B = P C ; > But, here s the trcky part. The aount of te durng whch other tasks can execute depends on the te spent watng t s a recursve equaton wth appearng on both sdes. In ths case, we can break the recurson by sply usng an teratve evaluaton, where we keep re-evaluatng for longer and longer tes untl the result converges to a fnal value. (If the result doesn t converge, that eans Task wll never execute n the worst case.) (),, = ' = k ; k B C > = P The response te s the result of teratng the above untl t converges, whch s obtaned by takng the lt of,k as k approaches nfnty. As a practcal atter the process only needs to be repeated untl the sae answer s obtaned on two successve teratons. (3) ( ) = l, k k The worst case copleton te W s then the worst case response te plus the executon te of Task : (4) k (, k ) C W = C = l As a render, we are assue the tasks are non-preeptable, so t s not possble for another task to nterrupt the executon of Task once t has started. 7, Phlp Koopan 8 Interrupt esponse Te

9 Ths copletes all the peces we need. To recap, below s the fnal set of workng equatons: B' D, =, k = B' = W k ( C ), = ax ax j B < j< N = B' ; > = l P, k (, k ) C ; < N C ; > Wth the followng equatons applyng nstead of the above for soe specal cases: B' N = B = B' Fgure 6. Suary of equatons. 5. Exaples After all ths, we can get the answer to whether tasks wll eet ther deadlnes by coputng W for all tasks. Let s do ths usng the exaple fro the prevous graphcal analyss and see how the equatons work. Let us revst the prevous exaple and see f the analytc approach yelds the sae result as the graphcal approach. The exaple we used was: N=5 B= takes 5 sec and occurs at ost once every 5 sec; C = 5 ; P = 5 takes 6 sec and occurs at ost once every sec; C = 6 ; P = takes 7 sec and occurs at ost once every sec; C = 7 ; P = takes 9 sec and occurs at ost once every 5 sec; C 3 = 9 ; P 3 = 5 IS4 takes 3 sec and occurs at ost once every 6 sec; C 4 = 3 ; P 4 = B= exaple For B=, let s fnd the worst case copleton te of, whch s W. B' = ax ax ( C ), ax ax ( 3, 4 ), j B = C C B = ax[ 9,3,] = 9 < j < 5, = B' = 9 Gven ths startng pont (whch corresponds to Fgure ), we us, =9 to terate,k : 7, Phlp Koopan 9 Interrupt esponse Te

10 ,,,, = ' = k B C = B' C C = = P P P = = = Note that P, s sec, whch s the sae result as Fgure. Fro ths, t becoes evdent that the teratve equaton s dong the sae thng atheatcally that we dd graphcally n the prevous approach.,,, = B' C C = = 9 = 3 5 P P Ths teraton brngs us to 3 sec, correspondng to the stuaton shown n Fgure 3.,, 3 3, = B' C C = = 9 5 = 36 5 P P Ths teraton brngs us to 36 sec, correspondng to the stuaton shown n Fgure 4 n whch all ISs wth hgher prorty than have just fnshed executon.,, 36 36,3 = B' C C = = 9 5 = 36 5 P P Because teraton,3 hasn t changed copared to,, we can ternate the coputaton and know that addtonal teratons won t change the answer fro 36. Ths gves us a te to copleton of: whch s 43 sec, the sae answer shown n Fgure 5 and s the worst case copleton te. If the deadlne were 5 sec, ths task would always be able to eet ts deadlne under the gven assuptons. 5.. B=3 Exaple As an exaple of what happens when the effectve blockng te s donated by background task blockng te rather than lower prorty tasks, consder what happens when B=3 sec nstead of sec: B' = ax ax ( C ), ax ax ( 3, 4 ), j B = C C B = ax[ 9,3,3] = 3 < j < 5 = B' = 3, ( ) C = W = l, k = k 7, Phlp Koopan Interrupt esponse Te

11 Blockng te B 3 sec: ; ; Fgure 7. B=3 at te 3 sec.,,,, = ' = k B C = B' C C = = P P P = = 4 Blockng te B 4 sec: ; ; Fgure 8. B=3 at te 4 sec.,, 4 4, = B' C C = = 3 = 35 5 P P Blockng te B 35 sec: ; Fgure 9. B=3 at te 35 sec. 7, Phlp Koopan Interrupt esponse Te

12 ,, 35 35,3 = B' C C = = 3 5 = 4 5 P P Blockng te B 4 sec: ; Fgure. B=3 at te 4 sec.,3,3 4 4,4 = B' C C = = = 46 5 P P Blockng te B 46 sec: ; Fgure. B=3 at te 46 sec.,4, ,5 = B' C C = = P P = 5 Blockng te B 5 sec: Fgure. B=3 at te 5 sec. 7, Phlp Koopan Interrupt esponse Te

13 ,5,5 5 5,5 = B' C C = = 3 8 = 5 5 P P At ths pont the coputaton has converged, so we know that wll start executon at te 5 sec. Blockng te B Fgure 3. executes startng at 5 sec and endng at 58 sec for B=3. ( ) C = W = l, k = k Thus the graphcal and analytc technques both arrve at the sae answer n the sae way, and has a worst-case executon te of 58 sec for ths partcular case Other Exaples As further exercses, the reader should confr the followng results both graphcally and analytcally for ths exaple task set wth varous values of B: B= B= B=4 B= B=3 W = 4 W = 4 W = 4 W = 7 W = 8 W = W = W = W = 8 W = 9 W = 43 W = 43 W = 43 W = 46 W = 58 W 3 = 46 W 3 = 46 W 3 = 47 W 3 = 66 W 3 = 67 W 4 = 57 W 4 = 59 W 4 = 6 W 4 = 9 W 4 = 9 As an addtonal exaple, the graphcal results showng the tng for the B= case are below: 7, Phlp Koopan 3 Interrupt esponse Te

14 IS Fgure 3. worst case latency for B=. causes the longest effectve blockng te. IS Fgure 4. worst case latency for B=. causes the longest effectve blockng te. IS Fgure 5. worst case latency for B=. causes the longest effectve blockng te. 7, Phlp Koopan 4 Interrupt esponse Te

15 IS4 IS Fgure 6. worst case latency for B=. IS4 causes the longest effectve blockng te. Background Tasks Only IS4 B IS Fgure 7. IS4 worst case latency for B=. B s the longest effectve blockng te because there are no lower prorty nterrupts to process. 6. More Inforaton The ore generalzed proble ncludes coputng executon tes for both preeptve tasks runnng under an operatng syste and non-preeptve ISs. A descrpton of the ath for that ore general case can be found n: Y. Wang, M. Saksena, Schedulng fxed-prorty tasks wth preepton threshhold, IEEE Internatonal Conference on eal-te Coputng Systes and Applcatons, Deceber 999. Thanks to Jen Morrs Black for her research work n ths area. 7, Phlp Koopan 5 Interrupt esponse Te