CEC 450 Real-Time Systems

Transcription

1 CEC 450 Real-Time Systems Lecture 3 Real-Time Services Part 2 (Rate Monotonic Theory - Policy and Feasibility for RT Services) September 7, 2018 Sam Siewert

2 Quick Review Service Utility RM Policy, Feasibility, Safety Timing Diagrams over LCM Recall Cyclic Executive Design from Microcontrollers and Read about CE s used on Space Shuttle (MFE) and General Theory (Assignment #2) Use Cheddar to Check Work Overview Today - Timing Diagrams with Mathematical Observations Sam Siewert 2

3 A Service Release and Response C i WCET Input/Output Latency Interference Time Response Time = Time Actuation Time Sensed (From Release to Response) Event Sensed Interrupt Dispatch Preemption Dispatch Completion (IO Queued) Actuation (IO Completion) Interference Time Input-Latency Dispatch-Latency Execution Execution Output-Latency Sam Siewert 3

4 Real-Time Services Theoretical Timing Diagrams Lehoczky, Shah, and Ding Theorem Look at Timing Over LCM of All Periods Simple for Small Number of Services Necessary and Sufficient Proof of Feasibility Trace Tools for Real Timing Software Trace Tools WindView Linux Trace Toolkit C.I. Given: Services S1, S2 with periods T1 and T2 and C1 and C2, Assume T2 > T1 E.g. T1=2, T2=5, C1=1, C2= 1, then if prio(s1) > prio(s2), we can see that the service set is feasible by diagramming. T / T 2 1 C + C T T / T 2 1 C + C T S 1 S 2 T 2 T 1 LCM Sam Siewert 4

5 RM Assumptions and Constraints A1: All Services Requested on Periodic Basis, the Period is Constant A2: Completion-Time < Period A3: Service Requests are Independent (No Known Phasing) A4: Run-time is Known and Deterministic (WCET may be Used) C1: Deadline = Period by Definition C2: Fixed Priority, Preemptive, Run-to-Completion Scheduling Critical Instant: longest response time for a service occurs when all system services are requested simultaneously (maximum interference case for lowest priority service) No Other Shared Resources Not in Paper, but key assumption e.g. shared memory Sam Siewert 5

6 Derivation of RM LUB for 2 Tasks Claim - General RM Least Upper Bound: (Guarantees that all Service Releases Can meet Deadlines) Goal - Derive RM LUB For 2 Tasks: U m = ( Ci / Ti) m(2 i= m 1) U = C / / 2(22 1 T1 + C2 T2 1) 0.83 For a System, Can All C s fit in largest T over LCM time? Given: Services S 1, S 2 with periods T 1 and T 2 and C 1 and C 2, Assume T 2 > T 1 E.g. T 1 =2, T 2 =5, C 1 =1, C 2 = 1, then if prio(s 1 ) > prio(s 2 ), we can see that... U = 1 / 2 + 1/ 5 = U = 0.7 2(2 1 1) 0.83 S 1 C.I. T / T 2 1 C + C T T / T 2 1 C + C T S 2 T 2 T 1 Sam Siewert 6

7 Can You Safely Exceed LUB? YES! In some cases, but RM LUB will never pass an infeasible system RM LUB is a SUFFICIENT Feasibility Test RM LUB is NOT a NECESSARY and SUFFICIENT Feasibility Test (Systems with Harmonic Periods can often have U=1.0 and be feasible!) Example Where RM LUB is Safely Exceeded: Given: Services S 1, S 2 with periods T 1 and T 2 and C 1 and C 2, Assume T 2 > T 1 E.g. T 1 =2, T 2 =5, C 1 =1, C 2 = 2, then if prio(s 1 ) > prio(s 2 ), note that: U = 1 / / 5 = 0.9 C.I. T 2 / T1 C1 + C2 T2 T 2 / T1 C1 + C2 T2 U U = = > 2(2 0.9 > ) S 1 S 2 T 2 T 1 Sam Siewert 7

8 RM Priority Assignment Policy RM Policy: Given: services S 1, S 2 with periods T 1 and T 2 and C 1 and C 2 with T 2 > T 1 E.g. T 1 =2, T 2 =5, C 1 =1, C 2 =2, then if prio(s 1 ) > prio(s 2 ), note that: S 1 Makes Deadline if prio(s 1 ) > prio(s 2 ) S 1 S 2 T 2 T 1 D 1 D 1 D 2 Sam Siewert 8

9 Harmonic Services YES NO Example 9 harmonic f0 multiple LCM/T f T1 6 C1 1 U LCM = 24 4 f T2 8 C2 2 U f T3 12 C3 4 U f T4 24 C4 6 U Utot = 1 1 Example 9 harmonic f0 multiple LCM/T f T1 7.5 C1 1 U LCM = f T2 10 C2 2 U f T3 15 C3 4 U f T4 30 C4 6 U4 0.2 Utot = YES Example 9 harmonic f0 multiple LCM/T f T1 9 C1 1 U LCM = 36 4 f T2 12 C2 2 U f T3 18 C3 4 U f T4 36 C4 6 U Utot = Sam Siewert 9

10 Smallest f is largest T All other f s (T s) must be such that they are a multiple of the smallest f, f 0 f 0 = fundamental Harmonics 1 st harmonic is 1xf 0 2 nd harmonic is 2xf 0 Nth harmonic is Nxf 0 Add f 0 to get any next harmonic Sam Siewert 10

11 So, Two Simple Examples Harmonic Services S 1, S 2, S 3 with T 1 =3, f 1 =1/3 = 5 x f 3 (f 3 is the fundamental a.k.a f 0 ) T 2 =5, f 2 =1/5 = 3 x f 3 T 3 =15, f 3 =1/15 = f 3 is the smallest (lowest frequency) NOT Harmonic T 1 =4, f 1 =1/4 = 5 x f 3 (f 3 is the fundamental here) T 2 =8, f 2 =1/8 = 2/3 x f 3 T 3 =12, f 3 =1/12 = f 3 is the smallest (lowest frequency) Check with S are harmonic by comparing all f Multiples of smallest T, not a good check [common mistake] LCM / T should reveal harmonics, factors of largest T Or T n /2, T n /3, T n /4, Sam Siewert 11

12 RM Priority Assignment Policy Consider Alternative Policy: Given: services S 1, S 2 with periods T 1 and T 2 and C 1 and C 2 with T 2 > T 1 E.g. T 1 =2, T 2 =5, C 1 =1, C 2 =2, then if prio(s 2 ) > prio(s 1 ), note that: S 1 Misses Deadline if prio(s 2 ) > prio(s 1 ) Conclusion: S 1 S 2 T 2 T 1 If {S n } feasible with prio(s 2 ) > prio(s 1 ), then {S n } is also always feasible given prio(s 1 ) > prio(s 2 ), but converse is not necessarily TRUE!! Therefore, prio(s 1 ) > prio(s 2 ) is OPTIMAL Sam Siewert 12

13 RM LUB Derivation Finding RM Safe Upper Bound (LUB): If system feasible over LCM with RM policy, then by Worst-Case Analysis, it is safe! Note that there can be up to T /T 2 1 releases of S 1 during T 2! S 1 T 2 T 1 C.I. S 2 #1 #2 #3 LUB Derivation Strategy: Case 1: C 1 short enough to fit all 3 releases in T 2 (fits S 2 critical time zone) Case 2: C 1 too large to fit last release in T 2 (doesn t fit S 2 critical time zone) Examine U in both cases to find common U bound. Sam Siewert 13

14 Next Time Complete RM LUB Derivation Discuss Pitfalls Introduce Extensions to overcome Pitfalls Sam Siewert 14