1. Real Time Operating Systems Schedulability - Part 3 Course originally developed by Maj Ron Smith 10/24/2015Dr Alain Beaulieu1

2. Outline  Review of Task Model Assumptions  Removing Task Independence  Review Priority Inversion & Priority Inheritance  Time-Demand Analysis & Blocking  Time-Demand Analysis & Priority-Ceiling Protocols 10/24/2015 Dr Alain Beaulieu

3. Review of Task Model Assumptions  single processor  fixed number of tasks  all tasks are periodic  context switch times are negligible  all tasks are preemptable  all tasks are independent  task priorities are fixed     X X X X - Our schedulability analysis can not handle deviation from the assumption.  - Our schedulability analysis (to date) can handle deviation from the assumption. 10/24/2015 Dr Alain Beaulieu

4. Removing Task Independence  The assumption that tasks are independent is unreasonable for any meaningful system  tasks typically share common resources (R k ) through the use of semaphores and monitors  tasks often must synchronize  This implies that a task(s) may suspend pending some future event which is dependent upon one or more other tasks  How will this dependence affect schedulability? 10/24/2015 Dr Alain Beaulieu

5. Recall : Priority Inversion  Priority inversion  can occur when a high and a low priority task share a common resource  the low priority task gets exclusive access to the shared resource  the higher priority task preempts the lower priority task but is blocked pending release of the shared resource  meanwhile a medium priority task preempts the lower priority task, thus further delaying the execution of the high priority task  since the medium task is guaranteed priority service over the blocked higher priority task => priority inversion 10/24/2015 Dr Alain Beaulieu

6. Recall: Priority Inheritance & Scheduling  Simple priority inheritance is a technique for avoiding priority inversion  each task has a dynamic priority, equal to the maximum of its own static priority and any it inherits due to blocking higher priority tasks  this effectively blocks any “medium” priority tasks  Theorem : if priority inheritance is employed, the number of times a task can be blocked by lower priority tasks is limited by the lessor of:  K - the number of blocking critical sections; or  n - the number of lower priority tasks 10/24/2015 Dr Alain Beaulieu

7. Blocking Time Defined  From the previous theorem, the block time of task i is defined to be -> b i =  usage(k, i ) e CS (k)(1) where usage(k, i) = 1 if resource k is used by at least one task with priority < i and at least one task (including i ) with priority ≥ i = 0 otherwise and e CS (k) = execution time of the k th critical section k=1 K 10/24/2015 Dr Alain Beaulieu

8. Blocking Time Example  Given the 4 task, 2 resource system described below, determine the blocking time of each task using equation (1) Tii12213443Tii12213443 k j e cs (k j )used by tasks 10.51,4 20.251,2,3 Tasks Resource Utilization 10/24/2015 Dr Alain Beaulieu

9. Time-Demand Analysis with Blocking  Yielding a more general response time for task i: w i * = e i + b i + I i * Note: this equation is now pessimistic (sufficient but not necessary)  Why?  Although employing simple priority inheritance will bound the number of blocks a task may experience, recall that it does not prevent transitive blocking nor deadlock 10/24/2015 Dr Alain Beaulieu

10. Blocking Exercise TasksResources Taske i p i D i k e CS (k) used by 1 3257 1 21,3,4 2212- 2 42,4 3517- 4624-  assume DM priority scheduling  simple priority inheritance is employed  determine the schedulability of the tasks 10/24/2015 Dr Alain Beaulieu

11. Blocking Exercise - Solution (1) Task  i e i p i D i k e CS (k) used by 1 13257 1 21,3,4 2 2212- 2 42,4 3 3517- 4 4624-  use DMPO  simple priority inheritance is employed  determine the schedulability of the tasks 10/24/2015 Dr Alain Beaulieu

12. Blocking Exercise - Solution (2) w 1 = e 1 + b 1 + I 1 where b 1 = usage(1,1) e CS (1) + usage(2,1) e CS (2) = (1)(2) + (0)(4) = 2 andI 1 = 0 (there are no higher priorities) therefore w 1 = 3 + 2 + 0 = 5 and w 1  D 1 10/24/2015 Dr Alain Beaulieu

13. Blocking Exercise - Solution (3) w 2 = e 2 + b 2 + I 2 where b 2 = usage(1,2) e CS (1) + usage(2,2) e CS (2) = (1)(2) + (1)(4) = 6 and w 2 = 2 + 6 + Σ  w 2 / p j  e j using recursionw 2 0 = 2 w 2 1 = 8 +  w 2 0 /25  3 = 11 w 2 2 = 8 +  w 2 1 /25  3 = 11 therefore w 2 = 11 andw 2  D 2 10/24/2015 Dr Alain Beaulieu

14. Blocking Exercise - Solution (4) w 3 = e 3 + b 3 + I 3 where b 3 = usage(1,3) e CS (1) + usage(2,3) e CS (2) = (1)(2) + (1)(4) = 6 and w 3 = 5 + 6 + Σ  w 3 / p j  e j againw 3 0 = 5 w 3 1 = 11 +  w 3 0 /25  3 +  w 3 0 /12  2 = 16 w 3 2 = 11 +  w 3 1 /25  3 +  w 3 1 /12  2 = 18 w 3 3 = 11 +  w 3 2 /25  3 +  w 3 2 /12  2 = 18 therefore w 3 = 18 (> 17) - the system is not scheduleable! 10/24/2015 Dr Alain Beaulieu

15. Now Recall: Ceiling Priority Protocols  A class of resource sharing algorithms entitled Ceiling Priority Protocols are introduced as an alternative to simple priority inheritance to:  eliminate transitive blocking  for any given task only a single blocking event may occur  eliminate deadlock  a task holding one resource may not claim another resource that could lead to circular lock/requests 10/24/2015 Dr Alain Beaulieu

16. Blocking & Ceiling Priority Protocols  Given that the ceiling priority protocols guarantee that only a single instance of blocking can occur (with respect to task i’s execution),  We can use the modified blocking time equation: b i = max {usage(k, i ) e CS (k) } k=1 K 10/24/2015 Dr Alain Beaulieu

17. CPP Exercise - Same system as before TasksResources Taske i p i D i k e CS (k) used by 1 3257 1 21,3,4 2212- 2 42,4 3517- 4624-  ** This is the exact same system as previously in the lecture, except the system uses a ceiling priority protocol  Determine if the system is schedulable 10/24/2015 Dr Alain Beaulieu

18. CPP Exercise – Solution (1)  The following changes must be accounted for with the use of a ceiling priority protocol first b 2 = max { usage(1,2) e CS (1), usage(2,2) e CS (2) } = 4 andb 3 = max { usage(1,3) e CS (1), usage(2,3) e CS (2) } = 4 then w 1 = 5 (no change) and w 2 = 9 (less response time) but check out w 3 10/24/2015 Dr Alain Beaulieu

19. CPP Exercise – Solution (2) w 3 = e 3 + b 3 + I 3 with new b 3 = 4 and w 3 = 5 + 4 + Σ  w 3 / p j  e j againw 3 0 = 5 w 3 1 = 9 +  w 3 0 /25  3 +  w 3 0 /12  2 = 14 w 3 2 = 9 +  w 3 1 /25  3 +  w 3 1 /12  2 = 16 w 3 3 = 9 +  w 3 2 /25  3 +  w 3 2 /12  2 = 16 therefore w 3 = 16  17 and the system may now be scheduleable! But now you must check w 4 10/24/2015 Dr Alain Beaulieu

20. CPP Exercise – Solution (3) w 4 = e 4 + b 4 + I 4 where b 4 = max{ usage(1,4) e CS (1), usage(2,4) e CS (2)} = max{ (0)(2), (0)(4) } = 0 (lowest priority task) and w 4 = 6 + 0 + Σ  w 4 / p j  e j againw 4 0 = 6 w 4 1 = 6 +  w 4 0 /25  3 +  w 4 0 /12  2 +  w 4 0 /17  5 = 16 w 4 2 = 6 +  w 4 1 /25  3 +  w 4 1 /12  2 +  w 4 1 /17  5 = 18 w 4 3 = 6 +  w 4 2 /25  3 +  w 4 2 /12  2 +  w 4 2 /17  5 = 23 w 4 4 = 6 +  w 4 3 /25  3 +  w 4 3 /12  2 +  w 4 3 /17  5 = 23 therefore w 4 = 23  24 & the system is now scheduleable! 10/24/2015 Dr Alain Beaulieu

21. References [1] Liu, J.W.S., “Real-Time Systems”, Prentice-Hall, 2000. [2] Burns, A. and Wellings, A., “Real-Time Systems and Programming Languages”, Chapter 13, Addison Wesley, 1997 [3] TimeSys Corp, “The Concise Handbook of Real-Time Systems”, Version 1.0, 1999 10/24/2015 Dr Alain Beaulieu