CEC 450 Real-Time Systems

Transcription

1 E 450 Real-ime Systems Lecture 4 Rate Monotonic heory Part September 7, 08 Sam Siewert

2 Quiz Results 93% Average, 75 low, 00 high Goal is to learn what you re not learning yet Motivation to keep up with reading Practice and prepare for future Exams 4 otal Quizzes Fundamental concepts One before each Mid-term exam as well Sam Siewert

3 Assignment # Scheduling Linux Look over candidate solution for scheduling - lab.c Enabling all PU cores on Jetson Set processor affinity for an AMP R core Use binary semaphore to BLOK and to release Services Set SHED_FIFO R priorities imed Execution for, based on benchmark synthetic load ache warmup (first few iterations are invalid) Use of SW usleep vs. SW interval timer and signal vs. HW interval timer and ISR Resulting predictability - ISR driven request for service is best Limitations of Linux and user space processes and threads Linux provides predictable response ROS more deterministic HW state machine most deterministic Sam Siewert 3

4 Scheduling Linux System has 4 processors configured and 4 available. Using PUS= from total available. Pthread Policy is SHED_FIFO PHREAD SOPE SYSEM rt_max_prio=99 rt_min_prio= Service threads will run on PU cores F0 runtime calibration msec per 00 test cycles, so 98 required F0 runtime calibration msec per 00 test cycles, so 97 required Start sequencer Starting Sequencer: [S, =0, =0], [S, =50, =0], U=0.9, LM=00 **** I t= F on core 3 F , 98 loops F on core 3 t= F on core 3 F , 98 loops F , 97 loops t= F0 start on core 3 F0 complete , 98 loops t= F on core 3 t= F0 start on core 3 F0 complete , 98 loops F , 97 loops t= F0 start on core 3 F0 complete , 98 loops lab.c Modify the Sequencer loop for specific RMA Scenario over Major Period Scale ime Units by 0 Example A# U 0.5 LM = 0 5 U 0.4 Utot = 0.9 RM Schedule F0 F0 S3 Sam Siewert 4

5 General Rate Observations PU cores run at 00 s MHz to several GHz 000 Instructions in one microsecond on GHz PU core hread scheduled processing down to millisecond ISR processing down to microsecond o-processor, Instruction below microsecond to subnanosecond Rate lock Measure Human Interaction Hz sec seconds X X hread ontext ISR Instructions FPGA / ASI 0 Hz /0 th sec 00 Hz /00 th sec 000 Hz (KHz) millisec X X millisec millisec microsec X X MHz microsec nannosec X X X GHz nanosec picosec X X Hz picosec femtosec subnanosecond X Sam Siewert 5

6 A Service Release and Response i WE Input/Output Latency Interference ime Response ime = ime Actuation ime Sensed (From Release to Response) Event Sensed Interrupt Dispatch Preemption Dispatch ompletion (IO Queued) Actuation (IO ompletion) Interference ime Input-Latency Dispatch-Latency Execution Execution Output-Latency Sam Siewert 6

7 RM Assumptions and onstraints A: All Services Requested on Periodic Basis, the Period is onstant A: ompletion-ime < Period A3: Service Requests are Independent (No Known Phasing) A4: Run-time is Known and Deterministic (WE may be Used) : Deadline = Period by Definition : Fixed Priority, Preemptive, Run-to-ompletion Scheduling ritical 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 7

8 Derivation of RM LUB for asks laim - General RM Least Upper Bound: (Guarantees that all Service Releases an meet Deadlines) Goal - Derive RM LUB For asks: U m = ( i / i) m( i= m ) U = / / ( + ) 0.83 For a System, an All s fit in largest over LM time? Given: Services S, S with periods and and and, Assume > E.g. =, =5, =, =, then if prio(s ) > prio(s ), we can see that... U = / + / 5 = 0.7 U = 0.7 ( ) 0.83 S.I. / + / + S Sam Siewert 8

9 an You Safely Exceed LUB? YES! In some cases, but RM LUB will never pass an infeasible system RM LUB is a SUFFIIEN Feasibility est RM LUB is NO a NEESSARY and SUFFIIEN Feasibility est (Systems with Harmonic Periods can often have U=.0 and be feasible!) Example Where RM LUB is Safely Exceeded: Given: Services S, S with periods and and and, Assume > E.g. =, =5, =, =, then if prio(s ) > prio(s ), note that: U = / + / 5 = 0.9.I. / + / + U U = = 0.9 > ( 0.9 > 0.83 ) S S Sam Siewert 9

10 RM Priority Assignment Policy RM Policy: Given: services S, S with periods and and and with > E.g. =, =5, =, =, then if prio(s ) > prio(s ), note that: S Makes Deadline if prio(s ) > prio(s ) S S D D D Sam Siewert 0

11 RM Priority Assignment Policy onsider Alternative Policy: Given: services S, S with periods and and and with > E.g. =, =5, =, =, then if prio(s ) > prio(s ), note that: S Misses Deadline if prio(s ) > prio(s ) onclusion: S S If {S n } feasible with prio(s ) > prio(s ), then {S n } is also always feasible given prio(s ) > prio(s ), but converse is not necessarily RUE!! herefore, prio(s ) > prio(s ) is OPIMAL Sam Siewert

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

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

14 Sam Siewert 4 RM LUB Derivation ase (All 3 releases fit in ): / / U + = = [ ] ( ) + = + = / / / U U Plugging Expression into U: U monotonically decreases with increasing when ( > ).I. S S # # #3 (I.e, = - Interference from releases) (I.e, is small enough to fit into fractional 3rd shown below as )

15 Proof that U Monotonically Decreases in ase- Sam Siewert 5

16 RM LUB Derivation ase (Last release does not fit in ): = U = / / / + S S.I. # # #3 Plugging Expression into U: U monotonically increases with increasing when ( > ) / / U = + ( ) ( ) ( ) U = / / + / / / Sam Siewert 6

17 Proof that U Monotonically Increases in ase- Sam Siewert 7

18 RM LUB Derivation Given ases and ( =, =5) = U = ase + ase : / ase : /.0 U U = + U = Intersection = = U = / / / + LUB ( ) / / =? ase / / / Sam Siewert 8 = U = + ( / ) / + [ ( / ) ( / ) / ] = = 5 ase ase U U Note: we want the LUB for any given and, not the ones assumed here in particular, so the general LUB could be something other than 0.9, and must be found in terms of and only for general LUB.

19 RM Questions From Ex # Sam Siewert 9

20 RM - Not lear.i. - all services are tasks in ready queue at same time We do proof this week Not necessary as shown later (partial and full interference cases) A key limitation of the RM LUB is that it does not accurately account for partial interference Sam Siewert 0

21 RM Questions. Deadline drive is DYNAMI priority and therefore adjusts to use 00% PU for most urgent deadline service, even in overload. Availability is linear if queue ordering overhead is diminishingly small, otherwise, overhead is an issue! 3. When all are ready at the same time - worst case Sam Siewert

22 RM Questions. heorem 4 is a step they later say is not necessary (partial and full interference). Forget this 3. hat s what we derive this week Sam Siewert