September 16, 2019 Sam Siewert

CEC 450Real-Time Systems

Lecture 4 – Rate Monotonic TheoryPart 2

Quiz Results86% Average, 73 low, 100 high

Goal is to learn what you’re not learning yet– Motivation to keep up with reading– Practice and prepare for future Exams– 4 Total Quizzes + Take-away Quiz (replaces lowest)

2 on Fundamental concepts

One before each Mid-term exam as well

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Assignment #1 – Scheduling LinuxLook over candidate solution for scheduling - lab1.cNote Method for Enabling all CPU cores on Jetson– Set processor affinity for an AMP RT core– Use binary semaphore to BLOCK and to release Services – Set SCHED_FIFO RT priorities– Timed Execution for C, based on benchmark synthetic load– Cache warmup (first few iterations are invalid)– Use of SW nanosleep 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 threadsLinux provides predictable responseRTOS more deterministicHW state machine most deterministic

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http://mercury.pr.erau.edu/%7Esiewerts/cec450/code/sequencer/lab1.c

Scheduling Linuxlab1.c

Modify the Sequencer loop for specific RMA Scenario over Major Period

Scale Time Units by 10

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Example A#1 T1 2 C1 1 U1 0.5 LCM = 10T2 5 C2 2 U2 0.4

Utot = 0.9

RM Schedule 1 2 3 4 5 6 7 8 9 10F10F20S3

System has 4 processors configured and 4 available.Using CPUS=1 from total available.Pthread Policy is SCHED_FIFOPTHREAD SCOPE SYSTEMrt_max_prio=99rt_min_prio=1Service threads will run on 1 CPU coresF10 runtime calibration 0.101834 msec per 100 test cycles, so 98 requiredF20 runtime calibration 0.101417 msec per 100 test cycles, so 197 requiredStart sequencerStarting Sequencer: [S1, T1=20, C1=10], [S2, T2=50, C2=20], U=0.9, LCM=100

**** CI t=0.004333F10 start 1 @ 0.241167 on core 3F10 complete 1 @ 10.001917, 98 loopsF20 start 1 @ 10.184167 on core 3t=20.259917F10 start 2 @ 20.353500 on core 3F10 complete 2 @ 30.146167, 98 loopsF20 complete 1 @ 39.858917, 197 loopst=40.539083F10 start 3 @ 40.659917 on core 3F10 complete 3 @ 50.289333, 98 loopst=50.713667F20 start 2 @ 50.799833 on core 3t=60.833167F10 start 4 @ 60.917250 on core 3F10 complete 4 @ 70.678500, 98 loopsF20 complete 2 @ 80.251417, 197 loopst=80.985000F10 start 5 @ 81.078167 on core 3F10 complete 5 @ 90.812750, 98 loops

http://mercury.pr.erau.edu/%7Esiewerts/cec450/code/sequencer/lab1.c

General Rate Observations

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Rate Clock Measure Human Interaction

Thread Context

ISR Instructions FPGA / ASIC

1 Hz sec seconds X X

10 Hz 1/10thsec

millisec X X

100 Hz 1/100thsec

millisec X

1000 Hz (KHz)

millisec microsec X X

MHz microsec nannosec X X X

GHz nanosec picosec X X

THz picosec femtosec sub-nanosecond

CPU cores run at 100’s MHz to several GHz1000 Instructions in one microsecond on GHz CPU coreThread scheduled processing down to 1 millisecondISR processing down to 1 microsecondCo-processor, Instruction below 1 microsecond to sub-nanosecond

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A Service Release and ResponseCi WCETInput/Output LatencyInterference Time

EventSensed Interrupt Dispatch Preemption Dispatch

Interference

Completion(IO Queued)

Actuation(IO Completion)

Input-Latency

Dispatch-Latency

Execution Execution

Output-Latency

Time

Response Time = TimeActuation – TimeSensed(From Release to Response)

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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

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Derivation of RM LUB for 2 TasksClaim - General RM Least Upper Bound:(Guarantees that all Service Releases Can meet Deadlines)

Goal - Derive RM LUB For 2 Tasks:

For a System, Can All C’s fit in largest T over LCM time?Given: Services S1, S2 with periods T1 and T2 and C1 and C2, Assume T2 > T1E.g. T1=2, T2=5, C1=1, C2= 1, then if prio(S1) > prio(S2), we can see that ...

)12()/(1

1

−≤=∑=

m

i

mmTiCiU

83.0)12(2// 21

2211 ≤−≤+= TCTCU

1/ 2 1/ 5 0.7U = + = 22112 / TCCTT ≤+

T2T1

C.I.

S1S2

22112 / TCCTT ≤+

83.0)12(27.0 21

≤−≤=U

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Can You Safely Exceed LUB?YES!In some cases, but RM LUB will never pass an infeasible systemRM LUB is a SUFFICIENT Feasibility TestRM 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 S1, S2 with periods T1 and T2 and C1 and C2, Assume T2 > T1E.g. T1=2, T2=5, C1=1, C2= 2, then if prio(S1) > prio(S2), note that:

9.05/22/1 =+=U

83.09.0)12(29.0 2

1

>=−>=

UU

22112 / TCCTT ≤+

T2T1

C.I.

S1S2

22112 / TCCTT ≤+

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RM Priority Assignment PolicyRM Policy:Given: services S1, S2 with periods T1 and T2 and C1 and C2 with T2 > T1E.g. T1=2, T2=5, C1=1, C2=2, then if prio(S1) > prio(S2), note that:

S1S2T2T1

S1 Makes Deadline if prio(S1) > prio(S2)

D1 D1 D2

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RM Priority Assignment PolicyConsider Alternative Policy:Given: services S1, S2 with periods T1 and T2 and C1 and C2 with T2 > T1E.g. T1=2, T2=5, C1=1, C2=2, then if prio(S2) > prio(S1), note that:

S1S2T2T1

S1 Misses Deadline if prio(S2) > prio(S1)

Conclusion:If {Sn} feasible with prio(S2) > prio(S1), then {Sn} is also always feasible given prio(S1) > prio(S2), but converse is not necessarily TRUE!!Therefore, prio(S1) > prio(S2) is OPTIMAL

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RM LUB Derivation

LUB Derivation Strategy:Case 1: C1 short enough to fit all 3 releases in T2 (fits S2 critical time zone)Case 2: C1 too large to fit last release in T2 (doesn’t fit S2 critical time zone)Examine U in both cases to find common U bound.

12 /TT

T2T1

C.I.S1

S2

#1 #2 #3

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 releases of S1 during T2!

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RM LUB Derivation

LUB Derivation Strategy:Case 1: C1 short enough to fit all 3 releases in T2 (fits S2 critical time zone)Case 2: C1 too large to fit last release in T2 (doesn’t fit S2 critical time zone)Examine U in both cases to find common U bound.

12 /TT

T2T1

C.I.S1

S2

#1 #2 #3

Finding RM Safe Upper Bound (LUB):If system feasible over LCM with RM policy, then by inspection, can we say that it is safe? Certainly feasible!Note that there can be up to releases of S1 during T2!

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RM LUB DerivationCase 1 (All 3 C1 releases fit in T2):

2

2

1

1

12122

12121

//

TC

TCU

TTCTCTTTTC

+=

−=−≤

[ ]

( )

−+=

−+=

2

1211

2

1212

1

1

//11

/

TTTTCU

TTTCT

TCUPlugging C2 Expression into U:

U monotonically decreaseswith increasing C1 when (T2>T1)

T2T1

C.I.S1

S2

#1 #2 #3

(I.e, C2 = T2 - Interference from C1 releases)

(I.e, C1 is small enough to fit into fractional 3rd T1 shown below as )

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Proof that U Monotonically Decreases in Case-1

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RM LUB DerivationCase 2 (Last C1 release does not fit in T2):

2

2

1

1

1211212

12121

///

TC

TCU

TTCTTTCTTTTC

+=

−=−≥

T2T1

C.I.#1 #2 #3S1

S2

( ) ( ) ( )

1 2 1 1 2 11

1 2

1 2 2 1 1 1 2 2 1

/ /

/ / 1/ 1/ /

T T T C T TCUT T

U T T T T C T T T T

− = +

= + −

Plugging C2 Expression into U:

U monotonically increaseswith increasing C1 when (T2>T1)

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Proof that U Monotonically Increases in Case-2

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RM LUB DerivationGiven Cases 1 and 2 (T1=2, T2=5)

2

2

1

1

1211212

12121

///

TC

TCU

TTCTTTCTTTTC

+=

−=−=

Case 1 Case 2

U

C1

2

2

1

1

1211212

12121

///

TC

TCU

TTCTTTCTTTTC

+=

−=−≥

2

2

1

1

12122

12121

//

TC

TCU

TTCTCTTTTC

+=

−=−≤

=?

1.0

LUB

Intersection

Note: we want the LUB for any given T2 and T1, 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 T1 and T2 only for general LUB.

T1= 2 T2= 5Case 1 Case 2C1 U C1 U

0 1 0 0.80.1 0.99 0.1 0.810.2 0.98 0.2 0.820.3 0.97 0.3 0.830.4 0.96 0.4 0.840.5 0.95 0.5 0.850.6 0.94 0.6 0.860.7 0.93 0.7 0.870.8 0.92 0.8 0.880.9 0.91 0.9 0.89

1 0.9 1 0.9

( ) ( ) ( ) [ ]122111221 //1/1// TTTTCTTTTU −+=

( )

−+=

2

1211

//11T

TTTCUCase 1:

Case 2:

Sheet1

T1=2T2=5

Case 1Case 2

C1UC1U

0100.8

0.10.990.10.81

0.20.980.20.82

0.30.970.30.83

0.40.960.40.84

0.50.950.50.85

0.60.940.60.86

0.70.930.70.87

0.80.920.80.88

0.90.910.90.89

10.910.9

Sheet2

Sheet3

CEC 450�Real-Time SystemsQuiz ResultsAssignment #1 – Scheduling LinuxScheduling LinuxGeneral Rate ObservationsA Service Release and ResponseRM Assumptions and ConstraintsDerivation of RM LUB for 2 TasksCan You Safely Exceed LUB?RM Priority Assignment PolicyRM Priority Assignment PolicyRM LUB DerivationRM LUB DerivationRM LUB DerivationProof that U Monotonically Decreases in Case-1RM LUB DerivationProof that U Monotonically Increases in Case-2RM LUB Derivation