EE5900 Advanced Embedded System For Smart Infrastructure

1. EE5900 Advanced Embedded System For Smart Infrastructure RMS and EDF Scheduling

2. C’i t di Laxity Priority-driven Preemptive Scheduling Assumptions & Definitions • Tasks are periodic • No aperiodic or sporadic tasks • Job (instance) deadline = end of period • No resource constraints • Tasks are preemptable • Laxity (Slack) of a Task • Ti = di – (t + ci’) where di: deadline; t : current time; ci’ : remaining computation time.

3. Rate Monotonic Scheduling (RMS) • Schedulability check (off-line) - A set of n tasks is schedulable on a uniprocessor by the RMS algorithm if the processor utilization (utilization test) ci is the execution time and pi is the period, The term n(21/n -1) approaches ln 2, (0.69 as n). - This condition is sufficient, but not necessary.

4. RMS (cont.) • Schedule construction (online) - Task with the smallest period is assigned the highest priority (static priority). - At any time, the highest priority task is executed.

5. RMS Scheduler - Example 1 Task set: Ti = (ci, pi) [computation time, period] T1 = (2,4) and T2 = (1,8) Schedulability check: 2/4 + 1/8 = 0.5 + 0.125 = 0.625 ≤ 2(√2 -1) = 0. 82 Active Tasks : {T2} Active Tasks : {T1} Active Tasks : {T1, T2} T11 T21 T12 0 2 3 4 6 8

6. RMS scheduler – Example 2 Task set: Ti = (ci, pi) T1 = (2,4) and T2 = (4,8) Schedulability check: 2/4 + 4/8 = 0.5 + 0.5 = 1.0 > 2(√2 -1) = 0. 82 Active Tasks : {T2} Active Tasks : {T2, T1} Active Tasks : {T2} Active Tasks : {T1, T2} T11 T21 T12 T21 6 8 0 2 3 4 Some task sets that FAIL the utilization-based schedulability test are also schedulable under RMS

7. RMS is not optimal • T1=(1,2) and T2=(2.5,5)

8. Earliest Deadline First (EDF) • Schedulability check (off-line) - A set of n tasks is schedulable on a uniprocessor by the EDF algorithm if the processor utilization. • This condition is both necessary and sufficient. • Least Laxity First (LLF) algorithm has the same schedulability check.

9. EDF/LLF (cont.) • Schedule construction (online) • EDF/LLF: Task with the smallest deadline/laxity is assigned the highest priority (dynamic priority). • At any time, the highest priority task is executed. • It is optimal (i.e., whenever there is a feasible schedule, EDF can always compute it) when preemption is allowed and no resource constraint is considered. • Given any two tasks in a feasible schedule, if they are not scheduled in the order of earliest deadline, you can always swap them and still generate a feasible schedule.

10. EDF scheduler - Example Task set: Ti = (ci, pi, di) T1 = (1,3,3) and T2 = (4,6,6) Schedulability check: 1/3 + 4/6 = 0.33 + 0.67 = 1.0 Active Tasks : {T2} Active Tasks : {T2, T1} Active Tasks : {T1} Active Tasks : {T1, T2} T11 T21 T21 T12 0 1 3 5 6 Unlike RMS, Only those task sets which pass the schedulability test are schedulable under EDF

11. Comparison of RMS and EDF Process Period, T WCET, C T 5 2 1 T 7 4 EDF schedule 2 T1 0 5 10 15 20 25 30 35 T2 7 14 21 28 35 0 RMSschedule T1 0 5 10 15 20 25 30 35 Deadline miss T2 7 14 21 28 35 0

12. Resource sharing • Periodic tasks • Task can have resource access • Semaphore is used for mutual exclusion • RMS scheduling

13. Background – Task State diagram • Ready State: waiting in ready queue • Running State: CPU executing the task • Blocked: waiting in the semaphore queue until the shared resource is free • Semaphore types – mutex (binary semaphore), counting semaphore

14. Task State Diagram scheduling Activate Termination READY RUN Preemption Signal free resource Wait on busy resource WAITING Process/Task state diagram with resource constraints

15. Priority Inversion Problem Priority inversion is an undesirable situation in which a higher priority task gets blocked (waits for CPU) for more time than that it is supposed to, by lower priority tasks. Example: • Let T1 , T2, and T3 be the three periodic tasks with decreasing order of priorities. • Let T1 and T3 share a resource S.

16. Priority Inversion - Example • T3 obtains a lock on the semaphore S and enters its critical section to use a shared resource. • T1 becomes ready to run and preempts T3. Then, T1 tries to enter its critical section by first trying to lock S. But, S is already locked by T3and hence T1 is blocked. • T2 becomes ready to run. Since only T2 and T3 are ready to run, T2 preempts T3 while T3 is in its critical section. Ideally, one would prefer that the highest priority task (T1) be blocked no longer than the time for T3 to complete its critical section. However, the duration of blocking is, in fact, unpredictable because task T2got executed in between.

17. A higher priority task waits for a lower priority task Priority Inversion example Resource S is available and T1 is scheduled here Makes a request for resource S and gets blocked T1 T1 T1 Highest priority T2 completes L1 Preempted by higher priority task T1 T3 completes T2 T2 T3 is the only active task Medium priority Preempted by higher priority task T2 K3 K2 K1 T3 T3 T3 T3 Least priority 0 T1 and T3 share resource S Total blocking time for task T1 = (K2+K3) + (L1)

18. Priority Inheritance Protocol Priority inheritance protocol solves the problem of priority inversion. Under this protocol, if a higher priority task TH is blocked by a lower priority task TL, because TL is currently executing critical section needed by TH, TL temporarily inherits the priority of TH. When blocking ceases (i.e., TL exits the critical section), TL resumes its original priority. Unfortunately, priority inheritance may lead to deadlock.

19. Resource access control - example T2 and T3 have access to a shared resource R cix : Task duration before entering the critical section ciy : Critical section duration ciz : Task duration after the critical section ci = cix + ciy + ciz By RMS, T3 > T1 > T2

20. Schedules Locks R Preempted by T3 RMS Schedule Direct blocking of T3 by T2 Release R T3 T1 T2 T3 T2 T1 T2 T3 T3 T2 10 11 12 0 2 4 6 7 8 14 16 Priority inversion of T3 by T1 RMS Schedule with Priority Inheritance Protocol Direct blocking of T3 by T2 T3 T1 T2 T3 T2 T3 T1 T3 T2 0 Inheritance blocking of T1 by T2

21. Priority Inheritance Protocol – Deadlock Assume T2 has higher priority than T1

22. Priority Ceiling Protocol • Priority ceiling protocol solves the priority inversion problem without getting into deadlock. • For each semaphore, a priority ceiling is defined, whose value is the highest priority of all the tasks that may lock it. • When a task Ti attempts to execute one of its critical sections, it will be suspended unless its priority is higher than the priority ceiling of all semaphores currently locked by tasks other than Ti. • If task Ti is unable to enter its critical section for this reason, the task that holds the lock on the semaphore with the highest priority ceiling is said to be blocking Ti and hence inherits the priority of Ti.

23. Priority Ceiling Protocol - properties • This protocol is the same as the priority inheritance protocol, except that a task Ti can also be blocked from entering a critical section if any other task is currently holding a semaphore whose priority ceiling is greater than or equal to the priority of task Ti.

24. Priority Celiling Protocol - Example • For the previous example, the priority ceiling for both CS1 and CS2 is the priority of T2. • From time t0 to t2, the operations are the same as before. • At time t3, T2 attempts to lock CS1, but is blocked since CS2 (which has been locked by T1) has a priority ceiling equal to the priority of T2. • Thus T1 inherits the priority of T2 and proceeds to completion, thereby preventing deadlock situation.

25. Scheduling tasks with precedence relations Conventional task set Scheduler {T1, T2} task set with precedence constraints T1 T2 Modify task parameters in order to respect precedence constraints Scheduler

26. Modifying the task parameters for RMS • While using the RMS scheduler the task parameters (ready time) need to be modified in order to respect the precedence constraints • Rj*≥ Max (Rj, Ri*) where Ri* is the modified ready time of the task Ti • Priority Prioi≥ Prioj Ti Tj

27. Modifying ready times for RMS: example T1 1 T2 2 Initial Task Parameters T3 2 T4 1 T5 3

28. Modifying the Ready times for RMS R1 = 0 R2 = 5 T1 1 T2 2 R4’ = max(R1, R2,R4) R3’ = max(R1, R3) Initial Task Parameters R3 = 0 R3’ = 0 R4 = 0 T3 2 T4 1 R4’ = 5 T5 3 R5 = 0 R5’ = 5 R5’ = max(R3’, R4’,R5)

29. Modified Ready times for RMS R1 = 0 R2 = 5 T1 1 T2 2 Modified Task Parameters R3’ = 0 T3 2 T4 1 R4’ = 5 T5 3 R5’ = 5

30. Assigning task priorities for RMS Assume all tasks of a connected component have the same period. Therefore, as per RMS all tasks will have a tie. We assign priorities to break the ties. R1 = 0 R2 = 5 T1 1 T2 2 R3’ = 0 T3 2 T4 1 R4’ = 5 Modified Task Parameters T5 3 R5’ = 5

31. Modifying task parameters for EDF • While using the EDF scheduler the task parameters need to be modified in order to respect the precedence constraints • Rj*≥ Max (Rj, (Ri* + Ci)) • Di*≥ Min (Di, (Dj* – Cj)) Ti Tj

32. Modifying the Ready times for EDF R1 = 0 R2 = 5 T1 1 T2 2 R4’ = max(R1+C1, R2+C2,R4) R3’ = max(R1 + C1, R3) Initial Task Parameters R3 = 0 R3’ = 1 R4 = 0 T3 2 T4 1 R4’ = 7 T5 3 R5 = 0 R5’ = 8 R5’ = max(R3’+C3, R4’+C4,R5)

33. Modifying the Ready times for EDF R1 = 0 R2 = 5 T1 1 T2 2 Modified Task Parameters R3’ = 1 T3 2 T4 1 R4’ = 7 T5 3 R5’ = 8

34. Modifying the Deadlines for EDF D2’ = Min( (D4’ – C4),(D3’ – C3), D1) D1 = 5 D2 = 7 D1’ = 3 D2’ = 7 T1 1 T2 2 D2’ = Min( (D4’ – C4), D2) Modified Task Parameters D3 = 5 T3 2 D3’ = 5 T4 1 D4 = 10 D4’ = 9 T5 3 D5 = 12 D3’ = Min( (D5 – C5), D3) D4’ = Min( (D5 – C5), D4)

35. Modifying the Deadlines for EDF D1’ = 3 D2’ = 7 T1 1 T2 2 Modified Task Parameters T3 2 D3’ = 5 T4 1 D4’ = 9 T5 3 D5 = 12