Last Time. Priority-based scheduling. Schedulable utilization Rate monotonic rule: Keep utilization below 69% Static priorities Dynamic priorities

Transcription

1 Last Tme Prorty-based schedulng Statc prortes Dynamc prortes Schedulable utlzaton Rate monotonc rule: Keep utlzaton below 69%

2 Today Response tme analyss Blockng terms Prorty nverson And solutons Release jtter Other extensons

3 Response Tme Analyss Rate monotonc result Tells us that a broad class of embedded systems meet ther tme constrants: Scheduled usng fxed prortes wth RM or DM prorty assgnment Total utlzaton not above 69% However, doesn t gve very good feedback about what s gong on wth a specfc system Response tme analyss Tells us for each task, what s ts worst-case response tme Then these can be compared wth deadlnes Gves nsght nto how close the system s to meetng / not meetng ts deadlne Is more precse (rejects fewer systems)

4 Computng Response Tme WC response tme of hghest prorty task R 1 R 1 = C 1 Obvous? WC response tme of second-prorty task R 2 Case 1: R 2 T 1 R 2 = C 2 + C 1 R 1 R 2 T 1 T

5 More Second-Prorty Case 2: T 1 < R 2 2T 1 R 2 = C 2 + 2C 1 T 1 R 2 R 1 2T 1 T Case 3: 2T 1 < R 2 3T 1 R 2 = C 2 + 3C 1 General case of the second-prorty task: R 2 = C 2 + celng ( R 2 / T 1 ) C 1

6 Task Response Tme General case: R = C + j hp() R T j C j hp() s the set of tasks wth prorty hgher than I Only hgher-prorty tasks can delay a task Problem wth usng ths equaton n practce?

7 Computng Response Tmes Rewrte as a recurrence relaton and solve by teratng: R n+ 1 = C Fnshed when R n+1 = R n Or when R n > D + j hp ( ) R T n j C Choose R 0 = 0 or R 0 = C There may be many solutons to the recurrence These startng ponts guarantee convergence to the smallest soluton (unless there s dvergence) Result s nvald f R > T Why? j

8 Response Tme Example Task 1: T = 30, D = 30, C = 10 Task 2: T = 40, D = 40, C = 10 Task 3: T = 52, D = 52, C = 12 Utlzaton = 81% Rejected by the rate monotonc test! R 1 = 10 R 2 = 20 R 3 = 52 R n+ 1 = C + j hp ( ) R T n j C j

9 Sharng Resources So far tasks are assumed to be ndependent Not allowed to block (e.g. on a network devce) Not allowed to contend for shared resources Bg problem n practce! Soluton: Compute worst-case blockng tme for each task Longest tme that task s delayed by a lower-prorty task Why just lower prorty? Now we can analyze the system agan: R n+ 1 = C + B + j hp ( ) R T n j C j

10 Computng Blockng Terms How do we compute blockng terms? Depends on the synchronzaton protocol Tasks synchronze by dsablng nterrupts Best answer: Each task gets blockng term wth length of the longest crtcal secton n a lower-prorty task Smpler answer: Each task gets blockng term wth length of the longest crtcal secton n any task Why do these work? Tasks synchronze usng mutexes Blockng term generally mpossble to bound oops! Standard thread locks are unfrendly to real-tme systems Lock wat queue s FIFO Possble soluton: Prorty queues for mutexes

11 Prorty Inverson Prorty nverson: Low-prorty task delays a hgh prorty task Mutexes (even wth prorty queung) provde unbounded prorty nverson preempton P(s) blocks T P(s) succeeds

12 Prorty Inverson Case Study Mars Pathfnder Lands on Mars July Msson s successful Behnd the scenes Sporadc total system resets on the rover Caused by prorty nverson Debugged on the ground, software patch uploaded to fx thngs Detals Rover controlled by a sngle RS6000 runnng vxworks Rover devces polled over 1553 bus At 8 Hz bc_sched task sets up bus transactons bc_dst task runs (also at 8 Hz) to read back data

13 More Pathfnder Symptom: bc_sched sometmes was not fnshed by the tme bc_dst ran Ths trggered a system reset Should never happen snce these tasks are hgh prorty Problem: bc_sched shared a mutex wth ASI/MET task, whch does meteorologcal scence at low prorty Occasonally the classc prorty nverson happened when there were long-runnng medum prorty tasks Soluton: vxworks supports prorty nhertance wth a global flag They turned t on

14 Prorty Inverson Solutons 1. Avod blockng dsable nterrupts nstead Pros: Effcent Smple Con: Also delays unrelated, hgh prorty tasks 2. Immedate prorty celng protocol before lockng, rase prorty to hghest prorty of any thread that can touch that semaphore Pros: Farly smple Less blockng of unrelated tasks Cons: Requres ahead-of-tme system analyss Stll has some pessmstc blockng

15 Prorty Inverson Solutons 3. Prorty nhertance protocol When a task s blockng other tasks (by holdng a mutex) t executes at the prorty of the hghest-prorty blocked task Pros No pessmstc blockng Cons Complcated n presence of nested lockng Not that effcent Blockng terms larger than IPCP Other solutons exst, such as lock-free synchronzaton

16 IPCP Bonus In IPCP, rasng prorty prevents anyone else who mght access a resource from runnng So why take a lock at all? Turns out that lockng s not necessary rasng prorty s enough HOWEVER: Task must not voluntarly block (e.g. on dsk or HOWEVER: Task must not voluntarly block (e.g. on dsk or network) whle n a crtcal secton

17 Overheads A real RTOS requres tme to: Block a task Make a schedulng decson Dspatch a new task Handle tmer nterrupts For a well-desgned RTOS these tmes can be bounded Worst-case blockng tme of the RTOS needs to be added to each task s blockng term 2x worst-case context swtch tme needs to be added to each task s WCET We always charge the cost of a context swtch to the hgher-prorty task

18 Release Jtter Release jtter J Tme between nvocaton of task and tme at whch t can actually run E.g. task becomes conceptually runnable at the start of ts perod But must wat for the next tmer nterrupt before the scheduler sees t and dspatches t Or, task would lke to run but must wat for network data to arrve before t actually runs R = C + B + j hp() R + Tj J C j

19 Other Extensons Sporadcally perodc tasks Task has an outer perod and smaller nner perod Models bursty processng lke network nterrupts Sporadc servers Provde rate-lmtng for truly aperodc processng E.g. nterrupts from an untrusted devce Arbtrary deadlnes When D > T prevous equatons do not apply Can rewrte Precedence constrants Task A cannot run untl Task B has completed Models scenaro where tasks feed data to each other Makes t harder to schedule a system

20 Summary Prorty based schedulng It s what RTOSs support A strong body of theory can be used to analyze these systems Theory s practcal: Many real-world factors can be modeled Response tme analyss supports worst-case response tme for each prorty-based task Blockng terms Release jtter Prorty nverson can be a major problem Solutons have nterestng tradeoffs