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

2 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 = d i – (t + c i ’) where di: deadline; t : current time; c i ’ : remaining computation time. t C’ i Laxity didi

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) c i is the execution time and p i is the period, The term n(2 1/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: T i = (c i, p i ) [computation time, period] T1 = (2,4) and T2 = (1,8) Schedulability check: 2/4 + 1/8 = = ≤ 2(√2 -1) = T11T11 T21T21 T12T Active Tasks : {T1, T2} Active Tasks : {T2} Active Tasks : {T1}

6 RMS scheduler – Example 2 Task set: T i = (c i, p i ) T1 = (2,4) and T2 = (4,8) Schedulability check: 2/4 + 4/8 = = 1.0 > 2(√2 -1) = T11T11 T21T21 T12T Active Tasks : {T1, T2} Active Tasks : {T2} Active Tasks : {T2, T1} T21T21 Active Tasks : {T2} Some task sets that FAIL the utilization-based schedulability test are also schedulable under RMS

RMS is not optimal T 1 =(1,2) and T 2 =(2.5,5) 7

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: T i = (c i, p i, d i ) T1 = (1,3,3) and T2 = (4,6,6) Schedulability check: 1/3 + 4/6 = = 1.0 T11T11 T21T21 T12T Active Tasks : {T1, T2} Active Tasks : {T2} Active Tasks : {T2, T1} Active Tasks : {T1} Unlike RMS, Only those task sets which pass the schedulability test are schedulable under EDF 3 T21T21

11 Comparison of RMS and EDF T1T1 T2T2 RMS schedule T1T1 T2T2 EDF schedule Deadline miss Process Period,T WCET,C T T 2 7 4

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 READYRUN WAITING Activate scheduling Preemption Termination Wait on busy resource Signal free resource 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 T 1, T 2, and T 3 be the three periodic tasks with decreasing order of priorities. Let T 1 and T 3 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 T3 and 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 T2 got executed in between.

17 Priority Inversion example T1 T2 T3T3 T3 0 T3 is the only active task Preempted by higher priority task T1 T1T1 Makes a request for resource S and gets blocked T3T3 Preempted by higher priority task T2 T2T2 T3T3 T3 completes T1T1 Resource S is available and T1 is scheduled here K1 K2 K3 T2 completes L1 Total blocking time for task T1 = (K2+K3) + (L1) Highest priority Least priority Medium priority T1 and T3 share resource S A higher priority task waits for a lower priority task

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

19 Resource access control - example Task Ticici pipi cixcix ciyciy cizciz T T T T2 and T3 have access to a shared resource R c i x : Task duration before entering the critical section c i y : Critical section duration c i z : Task duration after the critical section ci = c i x + c i y + c i z By RMS, T3 > T1 > T2

20 Schedules Locks R Preempted by T3 T3T1T2T3T2T1T2T3 T2 T3T1T2T3T2T3T1T3T2 Direct blocking of T3 by T2 Priority inversion of T3 by T RMS Schedule RMS Schedule with Priority Inheritance Protocol 0 Direct blocking of T3 by T2 Inheritance blocking of T1 by T2 Release R Task Ticici pipi cixcix ciyciy cizciz T T T326110

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 T i 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 T i. If task T i 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 T i and hence inherits the priority of T i.

23 Priority Ceiling Protocol - properties This protocol is the same as the priority inheritance protocol, except that a task T i 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 T i.

24 Priority Celiling Protocol - Example For the previous example, the priority ceiling for both CS 1 and CS 2 is the priority of T 2. From time t 0 to t 2, the operations are the same as before. At time t 3, T 2 attempts to lock CS 1, but is blocked since CS 2 (which has been locked by T 1 ) has a priority ceiling equal to the priority of T 2. Thus T 1 inherits the priority of T 2 and proceeds to completion, thereby preventing deadlock situation.

25 Scheduling tasks with precedence relations Scheduler {T1, T2} Conventional task set T1T2 Modify task parameters in order to respect precedence constraints Scheduler task set with precedence constraints

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 R j* ≥ Max (R j, R i* ) where R i* is the modified ready time of the task T i Priority Prio i ≥ Prio j TiTj

27 Modifying ready times for RMS: example T1 1 T2 2 T3 2 T4 1 T5 3 TaskRiRi CiCi DiDi T1015 T2527 T3025 T40110 T50312 Initial Task Parameters

28 Modifying the Ready times for RMS T1 1 T2 2 T3 2 T4 1 T5 3 TaskRiRi CiCi DiDi T1015 T2527 T3025 T40110 T50312 Initial Task Parameters R1 = 0R2 = 5 R3 = 0 R4 = 0 R5 = 0 R3’ = max(R1, R3) R3’ = 0 R4’ = max(R1, R2,R4) R4’ = 5 R5’ = max(R3’, R4’,R5) R5’ = 5

29 Modified Ready times for RMS T1 1 T2 2 T3 2 T4 1 T5 3 TaskRiRi CiCi DiDi T1015 T2527 T3025 T45110 T55312 Modified Task Parameters R1 = 0R2 = 5 R3’ = 0 R4’ = 5 R5’ = 5

30 Assigning task priorities for RMS T1 1 T2 2 T3 2 T4 1 T5 3 TaskRiRi CiCi DiDi T1015 T2527 T3025 T45110 T55312 Modified Task Parameters R1 = 0R2 = 5 R3’ = 0 R4’ = 5 R5’ = 5 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. Priority

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 R j* ≥ Max (R j, (R i* + C i )) D i* ≥ Min (D i, (D j* – C j )) TiTj

32 Modifying the Ready times for EDF T1 1 T2 2 T3 2 T4 1 T5 3 TaskRiRi CiCi DiDi T1015 T2527 T3025 T40110 T50312 Initial Task Parameters R1 = 0R2 = 5 R3 = 0 R4 = 0 R5 = 0 R3’ = max(R1 + C1, R3) R3’ = 1 R4’ = max(R1+C1, R2+C2,R4) R4’ = 7 R5’ = max(R3’+C3, R4’+C4,R5) R5’ = 8

33 Modifying the Ready times for EDF T1 1 T2 2 T3 2 T4 1 T5 3 TaskRiRi CiCi DiDi T1015 T2527 T3125 T47110 T58312 Modified Task Parameters R1 = 0R2 = 5 R3’ = 1 R4’ = 7 R5’ = 8

34 Modifying the Deadlines for EDF T1 1 T2 2 T3 2 T4 1 T5 3 TaskRiRi CiCi DiDi T1015 T2527 T3125 T47110 T58312 Modified Task Parameters D1 = 5D2 = 7 D3 = 5 D4 = 10 D5 = 12 D3’ = Min( (D5 – C5), D3) D4’ = Min( (D5 – C5), D4) D3’ = 5 D4’ = 9 D2’ = Min( (D4’ – C4), D2) D2’ = Min( (D4’ – C4), (D3’ – C3), D1) D2’ = 7D1’ = 3

35 Modifying the Deadlines for EDF T1 1 T2 2 T3 2 T4 1 T5 3 TaskRiRi CiCi DiDi T1013 T2527 T3125 T4719 T58312 Modified Task Parameters D5 = 12 D3’ = 5 D4’ = 9 D2’ = 7D1’ = 3