Real-time Systems Lab, Computer Science and Engineering, ASU Scheduling Algorithm and Analysis (ESP...

Real-time Systems Lab, Computer Science and Engineering, ASU

Scheduling Algorithm and Analysis(ESP – Fall 2014)

Computer Science & Engineering DepartmentArizona State University

Tempe, AZ 85287

Dr. Yann-Hang [email protected](480) 727-7507


Task Scheduling

Schedule: to determine which task is assigned to a processor at any time order of execution meet deadlines, fast response time, utilize resource effectively

Need an algorithm to generate a schedule optimal scheduling algorithm: always find a feasible schedule if and

only if a feasible schedule exists Scheduler or dispatcher: the mechanism to implement a schedule Misconcept:

RTOS will schedule tasks to meet task deadlines A good schedule will reduce CPU load

2


Task Functional Parameters

Preemptivity: suspend the executing job and switch to the other one should a job (or a portion of job) be preemptable context switch: save the current process status (PC, registers, etc.) and

initiate a ready job transmit a UDP package, write a block of data to disk, a busy waiting

loop Preemptivity of resources: concurrent use of resources or

critical section lock, semaphore, disable interrupts

How can a context switch be triggered?

Assume you want to preempt an executing job -- why a higher priority job arrives run out the time quantum

waiting

executing

ready

blocked suspendeddispatched

wake-up

3


Event- and Time-Triggered Systems

Time-triggered control system All activities are carried out at certain points in time know a priori

at design time (based on a globally synchronized time base) Transmission of messages Task execution Monitoring of external states

All nodes have a common notion of time Event-triggered control system

All activities are carried out in response to events external to the system Reception of a message Termination of a task External interrup

4


0 10 20 30 40 50 60

Major and Minor Cycle Model

Time is divided into equal-sized frame minor cycle = length of frame Major cycle = length of schedule = k * minor_cycle

An example: A=(10,4) B=(20,6) C=(30,5) major cycle=60, minor cycle=10 scheduling string AB_AC_AB_AC_AB_A_

Jobs must be done within a minor cycle limit timing error to one frame suspend and resume as background, continue, or abort if overrun

5


An Example

A1 must be done at least every 10ms, and takes 1ms A2 must be completed with 5ms when E occurs and takes 2 ms E must be detected by polling and is detectable for at least 0.5

ms

E would not occur twice within 50 ms polling of E takes 0 overhead

0.5ms

6


Major/Minor Cyclic Scheduling There should be a periodic polling action for E

Assume a timer of 0.5ms to activate polling operation and no polling overhead

Should be an interval of 2ms to execute A2 for an arbitrary 5ms interval May detect E in the first frame and execute A2 in the second frame

period=2.5ms A2 takes 2ms if E, otherwise is 0 WCET=2ms

Should be an interval of 1ms to execute A1 for an arbitrary 10ms interval Period= 10ms, WCET= 1ms Since 2ms + 1ms > 2.5ms, we will divide A1 into two parts of 0.5ms

A2 A1_1 A2 A1_2 A2 A2 A2 A1_1 A2 A1_2 A2 A2

7


Time Triggered Architecture (TTA)

Multiple CPU nodes connected by a TDMA bus. Bus schedule is divided into fixed-length time slots, Each CPU node is assigned a time slot to transmit messages. Messages are delivered to their destination CPU nodes at the

end of the time slot, regardless of their exact arrival times.

8


Summary of Cyclic Schedule

Pros simple, table-driven, easy to validate (knows what is doing at any

moment) fit well for harmonic periods and small system variations static schedule deterministic, static resource allocation, no

preemption small jitter, no scheduling anomalies

Cons difficult to change (need to re-schedule all tasks) fixed released times for the set of tasks difficult to deal with dependencies schedule algorithm may get complex (NP-hard) doesn’t support aperiodic tasks efficiently Processes with sizeable computation times.

9


Priority-Driven Scheduling of Periodic Tasks

Why priority-driven scheduling use priority to represent urgency easy implementation of scheduler (compare priorities and dispatch

tasks) tasks can be added or removed easily no direct control of execution instance

How can we analyze the schedulability if we don’t know when a task is to be executed

Let’s begin a deterministic case in one processor independent periodic tasks deadline = period preemptable no overhead for context switch

10


Priority-Driven Schedules

Assign priority when jobs arrive static -- all jobs of a task have a fixed priority dynamic -- different priorities to individual tasks of a task relative priorities don’t change while jobs are waiting

Static priority schedules Rate-monotonic -- the smaller a task period, the higher its priority Deadline-monotonic – if deadline period, the smaller a task’s deadline

(relative), the higher its priority Dynamic priority schedules

EDF -- earliest deadline (absolute) first Schedulable utilization:

a scheduling algorithm can feasibly schedule any sets of priority tasks if the total utilization is equal to or less than the schedulable utilization of the algorithm

11


Earliest-deadline First (EDF) Schedule

Priority preemptive scheduling a job with earliest (absolute) deadline has the highest priority does not require the knowledge of execution time

Optimal if single processor, no resource contention, preemption

Scheduling anomaly if non-preemptive --- the schedule fails after we reduce job execution times

T1

T2

T3

D1 D2 D3

Missed deadline

( all jobs meet their deadline under EDF after increasing e1 )

12


dk di

JkJi

rk ( rk )

dk di

JkJk Ji

(non-EDF)

(EDF)

Earliest-deadline First (EDF) Schedule

Priority preemptive scheduling a job with earliest (absolute) deadline has the highest priority does not require the knowledge of execution time

Optimal if single processor, no resource contention, preemption why it is optimal: assume a feasible schedule

13


EDF Schedule

A optimal algorithm under single processor and preemptable tasks

How do we know a set of periodic tasks are schedulable under EDF ?

If we know the schedulable utilization SU of EDF, then any sets of tasks are schedulable when U SU

Theorem: A set of n periodic tasks can be scheduled by EDF iff

This schedulability utilization is no long true if deadline < period.

11

n

i i

i

p

eU

14


Example of EDF Schedule

A digital robot with EDF schedule control loop: ec 8ms at 100Hz BIST: eb 50ms given

BIST can be done every 250ms

Add a telemetry task to send and receive messages with et 15ms if BIST is done every 1000ms given

the telemetry task can have a relative deadline of 100ms sending or receiving must be separated at least 100ms

150

10

8

bbc p

uu

115

1000

50

10

8

tbc D

uu

15


Rate-Monotonic Scheduling Algorithm

A base case: no additional overhead, simple periodic tasks with pi =Di

Assign priorities according their periods Ti has a higher priority that Tk if i < k ( pi < pk ) Is RM optimal? if there is a feasible fixed-priority schedule, then RM is

feasible How do we know RM is feasible schedulability test

Results: RM is optimal if pi Di

sufficient condition utilization test

a complete test what is the worst response time given all possible arrivals and preemptions

)12( /1

1

nn

i i

i np

eU

16


Critical Instant

Critical instant of Ti: a job of Ti arriving at the instant has a maximum response time

If we can find the critical instant of Ti, then check whether all jobs of Ti meet their deadlines

let’s increase ei until the maximum response time = Di schedulable utilization

In-phase instant is critical: all higher priority tasks are released at the same instant (assume all jobs are completed before the next job in the tasme task is released.) which T2 has the maximum response time

T1 T2 T1, T2 T2 T1

17


Schedulability: UB Test

Utilization bound (UB) test: a set of n independent periodic tasks scheduled by the rate monotonic algorithm will always meet its deadlines, for all task phasings, if

For harmonic task sets, the utilization bound is U(n)=1.00 for all n.

U(1) = 1.0 U(4) = 0.756 U(7) = 0.728

U(2) = 0.828 U(5) = 0.743 U(8) = 0.724

U(3) = 0.779 U(6) = 0.734 U(9) = 0.720

1)n(2U(n)p

e.......

p

e /n1

n

n

1

1

18


Schedulability Test: Time-Demand Analysis

Consider in-phase instant only If Ji is done at t, then the total work must be done in [0,t] is

(from Ji and all higher priority tasks)

Can we find a t Di such that

wi(t) t cannot check all t [0, Di]

check all arrival instants and Di

The completion time of Ji satisfies

k

i

k kii e

p

tetw )(

1

1

k

i

k ki e

p

tet

1

1

t

w(t)

time Di

19


Schedulability: RT Test

Theorem: for a set of independent, periodic tasks, if each task meets its first deadline, with worst-case task phasing, the deadline will always be met.

Response Time (RT) test: let an = response time of task i. an may be computed by the following

iterative formula:

Test terminates when an+1 = an

Task i is schedulable if its response time is before its deadline: an ≤ Ti

i

1jj0j

1i

1j j

ni1n ea e where

p

aea

20


RMA for Periodic Tasks

A Sample Problem - Periodics

100 msec

40 msec

20 msec

Periodics Servers Aperiodics

20 msec

2 msec

10 msec

10 msec

50 msec

5 msec

Data Server

Comm Server

Emergency100 msec

150 msec

350 msec

Deadline 6 msec after arrival

1

2

3

21


Sample Problem: UB Test

Total utilization is

.200 + .267 + .286 = .753 < U(3) =.779 The periodic tasks in the sample problem are

schedulable according to the UB test.

e p U

Task 1 20 100 0.200

Task 2 40 150 0.267

Task 3 100 350 0.286

22


scheduling points

Timeline for The Sample Problem

0 100 200 300 400

3

2

1

23


Example: Applying RT Test (1)

Taking the sample problem, we increase the compute time of 1ז from 20 to 40; is the task set still schedulable?

Utilization for the first task : 40/100=0.4 < U(1) Utilization of first two tasks: 0.667 < U(2) = 0.828

First two tasks are schedulable by UB test

Utilization of all three tasks: 0.953 > U(3) = 0.779 UB test is inconclusive Need to apply RT test

24



Use RT test to determine if 3ז meets its first deadline:

i = 3

2608080100)40(150

180)40(

100

180100

pp

aee

p

aea

1801004040eeeea

j

2

1j j

03j

1i

1j j

0i1

321

3

1jj0

25



a3 = a2 = 300 Done!

Task is schedulable using RT test. a3 = 300 < p3 = 350

300)40(150

300)40(

100

300100e

p

aea

300)40(150

260)40(

100

260100e

p

aea

j

1i

1j j

2i3

j

1i

1j j

1i2

26


Timeline for The Example

0 100 200300

1

3

2

3 completes its work at t=300

27


Summary

UB test is simple but conservative RT test is more exact but also more complicated To this point, UB and RT tests share the same

limitations: All tasks run on a single processor All tasks are periodic and noninteracting Deadlines are always at the end of the period There are no interrupts Rate monotonic priorities are assigned There is zero context switch overhead Tasks do not suspend themselves

28


Modeling Task Switching

C1

S

e1

e2

e1

e2

1

2

3

e1

40 40

0 100 200

e1C1 C1

e1 S

e2C1 C1

C1

S e2 S S

time

e1

p1 p2 2p1

i

ii p

S2eU

Two scheduling actions per task(start of period and end of period)

29


Schedulability with Interrupts

Interrupt processing can be inconsistent with rate monotonic priority assignment. interrupt handler executes with high priority despite its period interrupt processing may delay execution of tasks with shorter

periods

Effects of interrupt processing must be taken into account in schedulability model.

Question is: how to do that?

30


Determining Schedulability with Interrupts

Task(i) Period(p) Execution Time(e)

Priority Deadline(D)

3200 60 HW 200

1100 20 High 100

2150 40 Medium 150

4350 40 Low 350

3 is an interrupt handler

31


UB Test with Interrupt Priority

Test is applied to each task. Determine effective utilization (fi) of each task i

using

q Compare effective utilization against bound, U(n).

n = num(Hn) + 1

num(Hn) = the number of tasks in the set Hn

1n Hk

kii

i

Hj j

ji e

p

1

p

e

p

ef

Preemption form the tasks

that can hit more than once

(with period less that Di)

Execution of a task

under test

Preemption from tasks

That can hit only once

(with period greater than Di )

32


UB Test with Interrupt Priority: 3

For 3, no tasks have a higher priority:

H = Hn =H1 = { }.

Note that utilization bound is U(1): num(Hn) = 0.

Plugging in numbers:

)1(U0p

e0f

3

33

0.13.0200

60

p

ef

3

33

33



For 1, 3 has a higher priority: H = {3}; Hn = {};

H1 = {3}.

)1(Uep

1

p

e0f

3kk

11

11


Plugging in the numbers:

0.1800.0100

60

100

20

p

e

p

ef

1

3

1

11

34



For 2 : H={1, 3}; Hn={1 }; H1={3 };



)2(Uep

1

p

e

p

ef

3kk

22

2

1j j

j2

828.0867.0150

60

150

40

100

20

p

e

p

e

p

ef

3

3

2

2

1

12

35


UB Test with Interrupt Priority:

For 4 : H={1, 2, 3}; Hn={1 , 2, 3 }; H1={};



)4(U0p

e

p

ef

4

4

3,2,1j j

j4

756.0882.0350

60

200

60

150

40

100

20

p

e

p

e

p

e

p

ef

4

4

3

3

2

2

1

14

36


Priority Inversion in Synchronization

1(H)B

2(M)

3(L)

Time

1 :{…P(S1)…V(S1)…}

3 :{…P(S1)…V(S1)…} S1 unlocked

Attempt to lock S1 (blocked) S1 locked

S1 unlockedS1 locked

37


Priority Inversion

Delay to a task’s execution caused by interference from lower priority tasks is known as priority inversion

Preemption is normal, but priority inversion should be avoided

Priority inversion is modeled by blocking time Identifying and evaluating the effect of sources of priority

inversion is important in schedulability analysis Sources of priority Inversion

Synchronization and mutual exclusion Non-preemtable regions of code FIFO (first-in-first-out) queues

38


Accounting for Priority Inversion

Recall that task schedulability is affected by preemption: two types of preemption

can occur several times per period can occur once per period

execution: once per period blocking: at most once per period for each source

The schedulability formulas are modified to add a “blocking” or “priority inversion” term to account for inversion effects

39


UB Test with Blocking

Include blocking while calculating effective utilization for each tasks:

Hn Preemption(can hit n times)

Execution

Blocking H1 Preemption(can hit once)

1n Hk

kii

i

i

i

Hj j

ji e

p

1

p

B

p

e

p

ef

40


RT Test with Blocking

Blocking is also included in the RT test

Perform test as before, including blocking effect

i

1jji0

j

1i

1j j

nii1n

eBa ere wh

ep

aeBa

41


Consider the following examplePeriodic tasks

What is the worst case blocking effect (priority inversion) experienced by each task ?

Example: Considering Blocking

25 msec

100 msec

50 msec

200 msec

100 msec

300 msec

10 msec

30 msec

Data Structure

1

2

3

42


Example: Adding Blocking

Task 2 does not use the data structure. Task 2 does experiences no priority inversion

Task 1 shares the data structure with 3 . Task 1 could have to wait for 3 to complete its critical section. But worse, if 2 preempts while 1 is waiting for the data structure, 1 could have to wait for 2’s entire computation.

This is the resulting tabletask Period Execution

TimePriority Blocking

delayDeadline

1100 25 High 30+50 100

2200 50 Medium 0 200

3300 100 Low 0 300

43


UB Test for Example

Recall UB test with blocking:

1n Hk

kii

i

i

i

Hj j

ji e

p

1

p

B

p

e

p

ef

00.105.1100

80

100

25

p

B

p

ef

1

1

1

11

)2(U5.0200

50

100

25

p

e

p

ef

2

2

1

12

)3(U84.0300

100

200

50

100

25

p

e

p

e

p

ef

3

3

2

2

1

13

Not schedulable

with additional RT test, 3 is shown to be schedulable

44


Nonpreemption Protocol

2:{…P(S1)…V(S1)…}

4:{…P(S1)…V(S1)…}

readyblocked

ready

ready

S1 locked S1 unlocked

Time

1(H)

4(L)

2

3

45


Basic Inheritance Protocol (BIP)

2:{…P(S1)…V(S1)…}

4:{…P(S1)…V(S1)…}

ready

ready


1(H)

4(L)

2

3

blocked

attempts to lock S1

S1 unlockedS1 locked

ready

inherits the priority of 2

after 2 is blocked46


Highest Locker’s Priority Protocol

2:{…P(S1)…V(S1)…}

4:{…P(S1)…V(S1)…}

ready

ready


1(H)

4(L)

2

3

BB

run with the priority of 2

47


Summary of Synchronization Protocols

1 Only if tasks do not suspend within critical sections2 PCP is not affected if tasks suspend within critical sections.

ProtocolBounded Priority

InversionBlocked at Most Once

Deadlock Avoidance

Nonpreemptible critical sections

Yes Yes1 Yes1

Highest locker's priority

Yes Yes1 Yes1

Basic inheritance Yes No No

48


Case Study

The target system responds to 6 events each event is processed by an ISR and an application routine ISRs are nonpreemption and set up event ready flags Main program checks ready flags in round-robin manner

if flag is set, calls the application routing

Main program

E1 E6E5E4E3E2

RTOS and ISRs

49


Scheduling Discipline

wait for signals

50


Task Data

The total utilization is only 55% in the worst-case

ei ea e p U

event 1 2.0 0.5 2.5 40 0.063

event 2 7.5 8.5 16 75 0.213

event 3 6.0 0.6 6.6 125 0.053

event 4 21.0 27.0 48.0 250 0.192

event 5 5.0 24.0 29.0 1050 0.028

event 6 3.0 1.0 4.0 4000 0.001

total 0.550

51


A Possible Scenario

Examine a possible scenario of event 1, 3 and 4 The main program just checked the flag for event 3

and then three interrupts arrives

0 40 80 120 160 200 240

0 125 250

0 250

event 1

event 4

event 3

miss deadline

52


Applying RMA in the Case Study

Event 4 application is schedulable (f4<69%)

Event Period Preempt{Hn} Execute Preempt

{H1}total(fi)

E1a 40 0.013

E2a 75 0.113

E3a 125 0.005

E4a 250 0.198 0.108 0.254 0.56

E5a 1050 0.023

E6a 4000 0.0003

56.0

250

35211240.68.50.5

250

27

125

6

75

7.5

40

2.0 a4E

53


Improving Response Times

Process events in RM order go back to the main loop after completing an application routine

Streamlined ISR move the work done in ISR to application routines

Preemptable services

Main program

E1 E6E3

RTOS and ISRs

E2 E5E4

54


Analysis After Improvements

Is it scheduable?

ei ea e p U

event 1 2.0 0.5 2.5 40 0.063

event 2 1.5 14.5(1.7) 16 75 0.213

event 3 6.0 0.6 6.6 125 0.053

event 4 6.5 41.5(4.5) 48 250 0.192

event 5 5.0 24(3.9) 29 1050 0.028

event 6 3.0 1.0 4.0 4000 0.001

total 0.550

609.075

0.355.60.65.15.4

75

5.14

40

5.2 f a2E 2a

55

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