Design of Real-Time Software Part 1: Real-time Scheduling

© O.S. van Roosmalen, 2016

Part 1: Real-time Scheduling

Version July 2016

Onno van RoosmalenO.S.vanRoosmalen@chello.nl

Content

• Introduction• Static priority scheduling

– RM scheduling without resource contention• Periodic tasks• Sporadic tasks• Aperiodic tasks

1

• Aperiodic tasks

– RM scheduling with resource contention – DM scheduling

Content

• Introduction

2

Recording external events

• The controlling software samples the external state with a certain frequency (time-driven) – typically: continuous external variables

• temperature, humidity• water level• pressure

3

• pressure

• The controlling software is informed about the external events (event-driven)– typically: changes in discrete environmental variables

• a packet passes a sensor• a hazardous condition occurs• a frame must be placed on the screen

• The choice between these two approaches is made during design of the controlling system (SW and HW)

Tasks

• Task definition (again): a sequence of actions that must be carried out to generate a response to an external event or a time event

• Task types (note: all cyclic):– Periodic (time driven; released with period exactly T )

4

– Periodic (time driven; released with period exactly T )– Sparse (event driven; released with distance larger than a certain T )– Aperiodic (event driven; released arbitrarily)

• A task has – a name/index (the j th task) j

– a (worst case) execution time (WCET) Cj

– a period (periodic and sporadic tasks) Tj

– a deadline within the period Dj

Tj

Dj

Tj

Cj

j

Dj

Types of tasks

• Software is most often a mix of time-driven and event driven tasks – depends on the choices on measuring environmental state– depends on the choice of HW implementation of sensors

• Both time-driven and event-driven tasks:– Tasks are cyclic

5

– Tasks are cyclic• events recur

– Cycles are started by• a clock (time driven)• an external event occurrence (event driven)

– Each execution of the action-sequence is an instance of a task

instance

Task

instance

Value after deadline

• Soft– A response is still valuable after the deadline, but value decreases

steadily after that.• Example: interaction with human users. People get impatient.

• Firm– A response has no value after the deadline.

• Example: a video frame that cannot be shown in time can be skipped.

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• Example: a video frame that cannot be shown in time can be skipped.

• Hard– Damage is done if a response does not come in time.

value

Time

Soft Firm Hard

deadline

Examples

• Dependable real-time systems– High cost of failure

• Possibly loss of life on failure

– Guaranteed dependability (especially timeliness)– Example: Industrial control

7

• High performance real-time systems– Low probability of failure

• Constant quality of service

– High regularity in performance – Example: Consumer electronics

Example: water vessel

• Requirements:– Vessel may not overflow– Pump may not run dry

• Properties – Water level has a maximum rate of change (up and down)– Sensor positions are chosen

8

• Derived SW requirements– Response deadline can be calculated

• The state “pump is off and the water is high”may not persist longer than a time span td1

– only then risk of overflow

• The state “pump is on and the water is low”may not persist longer than a time span td2

– only then risk of underflow

pump

water high point

water low point h2

h1

Example: water vessel

• Requirements:– Vessel may not overflow– Pump may not run dry

• Properties – Water level has a maximum rate of change (up and down)– Sensor positions are chosen

9

• Derived SW requirements– Response deadline can be calculated

• The state “pump is off and the water is high”may not persist longer than a time span td1

– only then risk of overflow

• The state “pump is on and the water is low”may not persist longer than a time span td2

– only then risk of underflow

pump

water high point

water low point h2

h1

Requirements on tasks

• Detection criterion: situations must be detected in which– the water is above (gets above) the high point– the water is below (gets below) the low point

• Response criterion: after detection the system must respond before

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– the water is above the high point for a time td1 or longer– the water is below the low point for a time td2 or longer

• These requirements can be met by a polling task (!)– Requirements like: “the system must respond when the water

reaches the high point” can not be met by a polling task• Why?

Polling task

– Critical state c should not exist longer than a time span td without response that cancels this critical state

• critical state c: water above/below sensor and pump off/on

– Periodic task is released with period T and satisfies deadline D within this period.

• If water at low sensor: Task stops pump• If water at high sensor: Task starts pump

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• If water at high sensor: Task starts pump

– Schedulability conditions (see diagram): T + D < td• If the task may finish anywhere within the period (D = T ) : 2.T < td

TD

td

c ct

Release of periodic task TD

Predictable computation time

• Make sure that the task can finish within D ( T )– Worst case execution times C (WCET) should be calculable (C D )– The load on the processor resource is not more than C per period T

• The predictability of this is endangered by – anomalies of the hardware and system software

• use of DMA that locks the bus• cache behaviour

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• cache behaviour• interrupts• memory management

– constraints• finding required resources locked by other tasks • precedence constraints

– absence of transparency in high level languages• dynamic variables, garbage collection (memory management)• repetition, recursion

Example: Video device

• Device characteristics– audio/video: perception is main concern– high volume turnover: cost constraints

• Real-time– high quality audio and video pose stringent real-time requirements– regularity is more important than latency

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• Quality of service– “collective effect of service performances that determine the degree

of satisfaction of the user of that service

Average case ( concerns are Quality+Cost) resource allocation is more important than Worst case ( concerns are Latency + Predictability)

Average case: why?

• Average case and worst case are far apart – worst case leads to over-dimensioning of resources

• High volumes: low bill of material is desired• Also low power conflicts with over-dimensioning

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Average case: how?

• Lesser picture quality often better than temporal incorrectness– deadline misses may lead to wrong picture– deadline misses tend to come in bursts

• Reducing quality of manipulating the video stream may reduce load in high load/overload situations

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may reduce load in high load/overload situations – Quality fluctuations are perceived as non-quality– Regularly recurring errors are very visible and annoying

• Only video specialists can make trade-off

Quality of Service

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Example of processing pipeline

• Various periodic tasks in a pipeline• Some tasks can perform processing at various

quality levels (green arrows)• Buffers help to balance the load over time• Periods can be different (compare audio with video

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• Periods can be different (compare audio with video decoding)

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Processing pipeline

PAlgorithm= while true doreceive(i, frame); process(frame);send(o, frame);

od

19

stream inputinterrupt

frameinterrupt

Data in Algorithm Data out

Acceptable WCET’sPAlgorithm= while true do

receive(i, frame); process(frame);send(o, frame);

od

• Average execution time to process frame • Period T is given by frame rate • Assume Gaussian probability distribution of execution times: 2

2)(1

Ct

C

20

• Assume Gaussian probability distribution of execution times:• Probability of exceeding : Favors large

time

Probability density

22

2

1)(

etP

)(Q

CC )(Q

C C

BufferingFailures:• Input buffer overflow• Output buffer underflow

m m

PAlgorithm= while true do do m times

receive(i, frame);process(frame);send(o, frame);

odod

21

stream inputinterrupt

frameinterrupt

Data in Algorithm Data out

• Buffer size ~ 2m• Period : mT• Latency : ~ 2mT

input m frames; (process m frames // input m frames);

Buffer size

• Period of task instance is now m times the original period: mT• Worst case fluctuations in actual start time of periodic process: ~ mT• This requires both filled and empty buffer space of about m • Total buffer size should be about 2m

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mT 2mT0 3mT

m

2m

Processing m frames per task instance

PAlgorithm= while true do do m times

receive(i, frame);process(frame);send(o, frame);

odod

m

23

time

Probability density

mm QQ )()(

m

• Probability of exceeding execution time :• With buffer 1 m : failure probability e.g. 0.1 0.1m

)( Cmm

m QQ )()(

Cm mCm

Correct Assumptions?

• Not statistical– Increased execution times come in bursts – Therefore: execution times are not Gaussian distributed

• Other measures – Different functions (e.g. audio decoding and sharpness enhancing) may be

independent

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independent • Instead of m cycles of a single process combine m cycles of different processes

• We have seen qualitative arguments to illustrate the issues

Guaranteeing Deadlines

• General approach:– Make hypothesis on behavior of environment.

• E.g. water vessel:– The influx of water into the vessel will be less then x liter/second

• E.g. video processing:– The frame processing time will be less than x milliseconds per frame

– The correctness of the hypothesis may have a calculable probability

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• E.g. water vessel:– The influx on average may exceeds this value once in every 100 years.

• E.g. video processing:– The probability distribution for frame processing times is f(t)

– Within the hypothesis meeting deadlines is guaranteed– Failures that fall outside the hypothesis have a certain probability

• E.g. water vessel:– The possibility of overflow is limited to once in every 100 years.

• E.g. video processing:– The probability for missing a frame is 1 in 109

Scheduling theory

• Focus is on how we guarantee meeting deadlines

• Many conclusions are qualitative as well– What scheduling approach is optimal– Optimality depends on the boundary conditions

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– One can be optimal according to many criteria

Content

• Introduction• Static priority scheduling

– Rate Monotonic (RM) scheduling without resource contention

• Periodic tasks

27

• Periodic tasks

Rate-Monotonic Scheduling

• Rate monotonic=task priority proportional to rate (frequency) • Basic assumptions:

– One processor– Tasks are independent and periodically released (polling tasks)– Tasks have fixed individual priorities (j has priority pj )– Preemption allowed

• High priority tasks may preempt low priority tasks.

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• High priority tasks may preempt low priority tasks.

– Tasks have fixed (maximal) execution duration (WCET=Cj)

• For the moment– The processor is the only shared resource– Deadlines within periods are equal to periods (Dj = Tj)

• Terminology:– “There is a correct schedule” or “There is a feasible schedule” means

“There is a schedule that satisfies all timing constraints”

Critical Instant

•• Critical instantCritical instant of a task (fixed priority scheduling only!!) :– Definition: the situation that gives the largest response time

• Lemma:–– The worst case response time of a task occurs when it is released at The worst case response time of a task occurs when it is released at

the same time with all higher priority tasksthe same time with all higher priority tasks

• Example: shift blue task until critical instant is reached

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t’ -T2 t’ +T2 t’ + 2.T2t’

t -T2 t +T2 t + 2.T2t T1

T1

T1

0

0

0

-T2 T2 2.T2

r

r

r

• Consider two tasks 1= < C1 ,T1 , p1 > and 2= < C2 ,T2 , p2 >

• Assume: – T1 < T2

– C1+ C2 T1

• Possible schedules:

Example

C1 + C2C1

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• Conclusion:– If C1+ C2 T1 any priority assignment will lead to a correct schedule

T1 2.T1

T2

C1 + C2

C1

C2

1

2

2

1

Exercise: arbitrary priority assignments

• Task set:• 1 : C1 = 2 T1 = 6• 2 : C2 = 2 T2 = 10• 3 : C3 = 4 T3 = 12

• Draw the six schedules for each of the possible

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• Draw the six schedules for each of the possible priority assignments

• Determine the worst case response time for each task in each case and check if it is smaller than the period (=deadline) for that task.

Optimality of RM Priorities

• Rate monotonic priority assignment – highest rate has highest priority

•• If a priority assignment leads to a feasible schedule If a priority assignment leads to a feasible schedule then the RM assignment leads to a feasible then the RM assignment leads to a feasible schedule as well.schedule as well.

• Example (two tasks): C + C

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• Example (two tasks):

T1

T2

p1 > p2

T1

T2p2 > p1

C1 + C2C2

C1C1 + C2

Schedulability Condition: 2 Tasks• Assume one period the larger of the two: T1 < T2

• We can have two possible priority assignments – non-RM: p2 > p1

– RM: p1 > p2

•• nonnon--RMRM:– At critical instant both tasks should finish in time:

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– At critical instant both tasks should finish in time:

Schedulable iff C1 + C2 T1

T1

T2p2 > p1

C1 + C2C2

•• RM:RM:– More difficult case because 2 can be preempted by 1

– Two situations depending on where period T2 ends– RM: Case 1)

Schedulability Condition: 2 Tasks

0 T1 2.T1 3.T1

T2C1

34

– RM: Case 2) 0 T2

p1> p2

T2 /T1 T1 + C1

T2

T2

0 T1 2.T1

0

3.T1

p1> p2T2 /T1 T1 + C1

> T2

T2 /T1 T1

C1

RM: Case 1

• Assumption: C1 T2 - T2 /T1 T1

• Time occupied by 1 in period T2

is T2 /T1 C1

0 T1 2.T1 3.T1

T2 /T1 T1 + C1

T2

T2C1

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is T2 /T1 C1

• Maximally available for 2 in period T2

is T2 - T2 /T1 C1

• Thus

Schedulable iff C2 T2 - T2 /T1 C1

RM: Case 2

• Assumption: C1 > T2 - T2 /T1 T1

• Time occupied by 1 in period T2 /T1 T1

is T2 /T1 C1

0 T2

0 T1 2.T1 3.T1

T2 /T1 T1 + C1

> T2

T2 /T1 T1

C1

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is T2 /T1 C1

• Maximally available for 2 in period T2 /T1 T1

is T2 /T1 T1 - T2 /T1 C1

• Thus

Schedulable iff C2 T2 /T1 T1 - T2 /T1 C1

Graphically

T2

slope is -T2 /T1

slope is -T2 /T1

Not schedulable, C2 is too large for given C1RM schedulable,

value of C2

is small enough

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C1

C2

T1

0

T2 - T2 /T1 T1

non-RM slope is -1

non-RM schedulable,value of C2

is small enough

• Consider variations around magic pointmagic point C1 = T2 - T2 /T1 T1

– Increasing C1 by can be compensated by decreasing C2 by T2 /T1

Slope Change

T2 /T1 T1 + C1

= T

0 T1 2.T1 3.T1

38

– Decreasing C1 by can be compensated by increasing C2 by T2 /T1

= T2

0 T2

0 T1 2.T1

0

3.T1

T2

Utilization Bound for Schedulability

• Consider two tasks 1=< C1 ,T1 , p1 > and 2=< C2 ,T2 , p2 >• Rate monotonic priorities: pi depends on Ti

• Ci /Ti is worst case time-fraction that i requires processor• Ci /Ti >1 can not be guaranteed (no feasible schedule)

– If Ci ,Ti are all known: a precise schedulability criterion can be derived (see later)

39

derived (see later)

– If only the utilization is known:

We can look for a Umin for which • U Umin the tasks always meet their deadlines• U >Umin the tasks can not be guaranteed to meet their deadlines

m

i i

i

T

CU

1

Determining Umin (two tasks)

• Terminology:– Tasks are fully utilizing the processor if no increase of C1 or C2 possible

• Critical instant• At critical instant: tasks fully occupy the processor until T2

• Schedulable but increasing either C1 or C2 breaks schedulability

– Note: this does not imply Ufull = 1

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– Note: this does not imply Ufull = 1

• Approach– Consider free variables in utilization: C1 ,C2 ,T1 , T2

– Consider largest feasible value for C2 given C1 , T1 , T2

• C2 can be eliminated giving full utilization Ufull(C1 , T1 , T2 )

– Find C1 for which full utilization is minimal, this is Umin

– Then, for any C1 and C2 that yield a utilization below Umin schedule is feasible.

– Minimize Umin with respect to T1 and T2 as well.

Determining Umin

• Consider same two cases as before– Case 1: maximal C2 slopes at - T2 /T1 C1

0 T1 2.T1 3.T1

p > pT2 /T1 T1 + C1

T

T2C1

41

– Case 2: maximal C2 slopes at - T2 /T1 C1

0 T2

p1> p2 T2

T2

0 T1 2.T1

0

3.T1

p1> p2T2 /T1 T1 + C1

> T2

T2 /T1 T1

C1

Case 1

• Assumption: C1 T2 - T2 /T1 T1

• With full utilization: C2 = T2 - T2 /T1 C1

• Thus

0 T1 2.T1 3.T1

T2 /T1 T1 + C1

T2

T2C1

42

1

2

2

1

1

1

2

11

22

1

1

1

1)1( 11T

T

T

T

T

C

T

CTTT

T

C

T

CU full

constant

negativeslope

Case 2

• Assumption: C1 > T2 - T2 /T1 T1

• With full utilization: C2 = T2 /T1 T1 - T2 /T1 C1

• Thus

0 T2

0 T1 2.T1 3.T1

T2 /T1 T1 + C1

> T2

T2 /T1 T1

C1

43

1

2

2

1

1

1

1

2

2

1

2

1

211

1

1

1

1)2( 1)(

T

T

T

T

T

C

T

T

T

T

T

TTCT

T

C

T

CU full

constant smaller 1

positiveslope

Graphically

• Introduce: f = C1/T1 ( 0 f 1)

Given C1, not schedulable. C2 is too large

C1

C2

T10

T2

slope is -T2 /T1

slope is -T2 /T1

44

Umin

1f = C1/T1

0

1

U=C1/T1+C2/T2

Given U, schedulableregardless of value of C1

f0 = T2 /T1 - T2 /T1

fU full)2( fU full

)1( Given U, schedulable

depending on value of C1

Minimal Value of Full Utilization varying T2 /T1

100%

95%

1.0 2.0 3.0T2 /T1

45

95%

90%

85%

80%

Ufull min

T2 /T1 =7/5

Result

• Combine:

• With: f0 = T2 /T1 - T2 /T1

• Introduce:

1

2

2

100

)2(0

)1(min 11

T

T

T

TffUfUU fullfull

2

T

TI I1

46

• Introduce:

• Then:

• And thus:

1T

I

10 0 f

)/()1(1),(min fIffIfU

012 / fITT

I1

Further simplification

• We can even eliminate f and I– Minimal value of Umin when I = 1

– Minimize for values of 0 f 1

12)1/()1(0)(min ffff

dfdU

47

12)1/()1(01

ffffdfdf

I

829.0)12(2min U

T2 /T1 = T2 /T1 + f = 2

Example: Full Utilization

• 1 : C1 = 2 , T1 = 5 • 2 : C2 = 3 , T2 = 7 (Ufull minimal at T2 = 2 . T1 = 1.41 T1 )

0 T1 2.T1 3.T1

C1

4.T1 5.T1 6.T1 7.T1

Full Utilization: C1 nor C2 can be increased

48

0 T2 2.T2 3.T2 4.T2 5.T2

C2

• Worst case response times: 1 : r1 = 2 2 : r2 = 5

• Utilization Ufull = (2/5 +3/7) = (29/35) = 0.828

0 T1 2.T1 3.T1 4.T1 5.T1 6.T1 7.T10 T1 2.T1 3.T1 4.T1 5.T1 6.T1 7.T1

Example: Increase C1 but keep Full Utilization

• 1 : C1 = 2+1 = 3 , T1 = 5 • 2 : C2 = 3–1 = 2 , T2 = 7

49

0 T2 2.T2 3.T2 4.T2 5.T2

0 T2 2.T2 3.T2 4.T2 5.T2

• Worst case response times: 1 : r1 = 3 2 : r2 = 5

• Utilization Ufull = (3/5 +2/7) = (31/35) = 0.89

0 T1 2.T1 3.T1 4.T1 5.T1 6.T1 7.T10 T1 2.T1 3.T1 4.T1 5.T1 6.T1 7.T1

Example: Decrease C1 but keep Full Utilization

• 1 : C1 = 2–1 = 1 , T1 = 5 • 2 : C2 = 3+2 = 5 , T2 = 7

50

0 T2 2.T2 3.T2 4.T2 5.T2

0 T2 2.T2 3.T2 4.T2 5.T2

• Worst case response times: 1 : r1 = 1 2 : r2 = 7

• Utilization Ufull = (1/5 +5/7) = (32/35) = 0.91

Example: Full Utilization

• 1 : C1 = 2 , T1 = 4 • 2 : C2 = 4 , T2 = 10 (Ufull minimal at T2 = 61/2 T1 = 2.45 T1 )

Full Utilization: C1 nor C2 can be increased

0 T1 2.T1 3.T1 4.T1 5.T1

C1 C1

51

• Worst case response times: 1 : r1 = 2 2 : r2 = 8

• Utilization Ufull = (2/4 +4/10) = (9/10) = 0.90

0 T22.T2

C2

Minimal Value of Full Utilization varying T2 /T1

100%

95%

1.0 2.0 3.0T2 /T1

52

95%

90%

85%

80%

Ufull min

T2 /T1 =10/4

Arbitrarily many tasks

• Approach– Consider free variables in utilization: U (C1 ,T1,... ,Cm ,Tm)– Eliminate Cm to obtain full utilization: Ufull (C1 ,T1,... ,Cm-1 ,Tm-1 ,Tm)– Find the Ci’s for which the value of Ufull is minimal.– Minimize also with respect to the Ti’s

53

• Result:

)12( /1min mmU 780.03min U

696.0min U

Example: Full Utilization for 3 Tasks

• 1 : C1 = 1 , T1 = 4 • 2 : C2 = 1 , T2 = 5 (Ufull minimal at T2 = 2 1/3 T1 = 1.26 T1 )• 3 : C3 = 2 , T3 = 6 (Ufull minimal at T3 = 2 2/3 T1 = 1.59 T1 )

54

• Worst case response times: 1 : r1 = 1 ; 2 : r2 = 2 ; 3 : r3 = 4

• Utilization Ufull = (1/4 + 1/5 + 2/6) = (47/60) = 0.7833

Full Utilization: C1 nor C2 nor C3 can be increased

0 5 10 15 20 25 30 35 40 45 50 55 60

Exercises

• Calculate RM schedulability of the following task sets• Use the utilization bound

– Exercise 1. Task set:• 1 : C1 = 1 T1 = 3• 2 : C2 = 1 T2 = 4

– Exercise 2. Task set:• 1 : C1 = 2 T1 = 5

55

• 1 : C1 = 2 T1 = 5• 2 : C2 = 4 T2 = 15• 3 : C3 = 5 T3 = 20

– Exercise 3. Task set:• 1 : C1 = 1 T1 = 4• 2 : C2 = 2 T2 = 6• 3 : C3 = 3 T3 = 10

Calculation of Response Times

• Utilization bound is sufficient but not necessary– Also task sets that do not satisfy the utilization bound may be

schedulable even if U >Umin

• A more precise bound can be given for each specific task set

• This is done on the basis of response times.

56

• Note: we can simply draw a schedule and calculate worst case task response times.

• This calculation by construction can be formalized

Response-Time Approach

•• The worst case response time The worst case response time rrii of a of a ii happens at the happens at the critical instant and satisfiescritical instant and satisfies

i r 1

Execution time of task itself

57

ij

ii

jji C

T

rCr

1

1

Maximum number of times higher priority task j may

be executed in period ri

Worst case waiting time for execution of higher priority tasks

of task itself

Schedulability of Task i

• Consider:

– Task i can be scheduled if a solution for t in Wi(t)/t =1 exists– The response time ri equals the smallest of such solution.

• Take:

j

i

j

ji

T

t

t

C

t

tW

1

)(

58

• Take:

– Then there is a solution for ri :•

• Wi(t)/t is continuous and strictly decreasing

• Except when t is a multiple of any of the Ti.

• Then Wi(t)/t jumps up discontinuously

1)(

min0

t

tWi

Tt i

ttWt

/)(lim0

Wi(t)/t

t

1

Tiri

Single Schedulability Criterion

• A solution ri must exist for each task i

• Therefore all tasks are schedulable iff

1)(

minmax01

t

tWi

Ttmi i

59

– Reason: max1 only if each mini 1

• We only need to consider Wi(t) at integer multiples of the periods Ti (see graph)

01 tTtmi i

Example

• Task set:• 1 : C1 = 2 T1 = 8• 2 : C2 = 2 T2 = 10• 3 : C3 = 4 T3 = 12• 4 : C4 = 1 T4 = 20

– Utilization bound: U = (2/8 +2/10+4/12+1/20) = 0.8333

2T1T1 T2 T3

T4

Here we know schedulability

1

2

3

4

8 10 12 16 20

60

– Utilization bound: U = (2/8 +2/10+4/12+1/20) = 0.8333

• U(4) 0.757; U(3) = 0.780 so last two tasks have no guarantee

– Consider: t = 8, 10, 12, 16, 20

• W4(8) = C1 + C2 + C3 + C4 = 9 not ok (1 short: length of 4 )• W4(10) = 2C1 + C2 + C3 + C4 = 11 not ok (still 1 short: length of 4

• W4(12) = 2C1 + 2C2 + C3 + C4 = 13 not ok (still 1 short: length of 4 )• W4(16) = 2C1 + 2C2 + 2C3 + C4 = 17 not ok (still 1 short: length of 4 )• W4(20) = 3C1 + 2C2 + 2C3 + C4 = 19 ok r4 =19

Exercises

• Calculate RM schedulability of the following task sets• Use the response time formula

– Exercise 1. Task set:• 1 : C1 = 1 T1 = 3• 2 : C2 = 1 T2 = 4

– Exercise 2. Task set:• 1 : C1 = 2 T1 = 5

61

• 1 : C1 = 2 T1 = 5• 2 : C2 = 4 T2 = 15• 3 : C3 = 5 T3 = 20

– Exercise 3. Task set:• 1 : C1 = 1 T1 = 4• 2 : C2 = 2 T2 = 6• 3 : C3 = 3 T3 = 10

Content

• Introduction• Static priority scheduling

– RM scheduling without resource contention• Periodic tasks• Sporadic tasks

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Sporadic tasks

• Sporadic tasks – Released irregularly– Have a maximum rate at which they are released– Thus assuming a minimum inter-arrival time T

•• Sporadic tasks can be treated as periodic tasks Sporadic tasks can be treated as periodic tasks with period with period TT

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with period with period TT

Content

• Introduction• Static priority scheduling

– RM scheduling without resource contention• Periodic tasks• Sporadic tasks• Aperiodic tasks

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• Aperiodic tasks

Aperiodic Tasks

• True aperiodic tasks – Can be arbitrarily released– No guarantee can be given if released too often– Deal with them without spoiling schedulability of periodic

and sporadic tasks

• Several approaches among which:

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• Several approaches among which:– Run as background tasks – Reserving space with a high priority periodic task– Deferred server– Sporadic server – ...

Run at Background

• Run the aperiodic tasks at lowest priority.– Periodic and sporadic tasks experience no interference.– Aperiodic tasks may run when the processor is not

allocated to the periodic/sporadic ones.– Possibly no timeliness of any of the aperiodic task-

instances.

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instances.

Run in High-Priority Task

• Introduce a fictitious high priority server task– Period Ts

– Execution time Cs

– Aperiodic tasks run in slots of this high priority one.– If no aperiodic task are released before or within this slot

the processor is available

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the processor is available– Outside these slots no running of the aperiodic tasks

• Schedulability of the periodic tasks is determined including the fictitious task

• If execution time of an aperiodic task is smaller Csthen such task can finish within time Ts + Cs

– At least, if aperiodic tasks are not released too oftenare not released too often

Deferred Server

• Slight modification– Avoid idle processor if no aperiodic tasks are released– Slot for server process starts every period Td

• If aperiodic processes have been released

– Execute them for period Cd

• If no aperiodic processes are released: deferdefer server execution to moment such tasks are released

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moment such tasks are released– Preempt other tasks

– Execute them for period Cd

– Capacity reset to Cd at start of period Td

– Utilization bound • Aperiodic tasks handled by the deferred server have utilization Ud

• Periodic tasks have utilization Up

652.0 dp UU

Sporadic Server

• Slight modification – To avoid server to execute two periods back-to-back

– Slot for server process starts every period Ts and has capacity Cs

• When aperiodic processes are released

– Execute them for requested period Cc<Cs

– Capacity Cc is added to remaining capacity after Ts time from startfrom start of

C

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cconsumption of Cc

– Sporadic server behaves like a sporadic task with period Ts

– Utilization bound • Aperiodic tasks handled by the deferred server have utilization Us

• Periodic tasks have utilization Up

1

2ln

sp U

U

Content

• Introduction• Static priority scheduling

– RM scheduling without resource contention• Periodic tasks• Sporadic tasks• Aperiodic tasks

70

• Aperiodic tasks

– RM scheduling with resource contention

Shared Resources

• Tasks may need other resources apart from processor– Typical example: mutual exclusive access to data

• Important instances– Reserve the resource: R(a)– Wait for the resource (blocked). Blocking period: B

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– Wait for the resource (blocked). Blocking period: Bi

– Hold the resource– Release (de-reserve) the resource: D(a)

R(a) D(a)

Bi

holdingwaiting

Problem: Priority Inversion

• Consider three tasks– Priorities p1> p2> p3

– Share a single resource a– Events

• 3 holds resource• 1 reserves resource and waits

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• 1 reserves resource and waits • 2 starts, executes and finishes• 3 releases resource• 1 gets resource but executes after 2 !!!

R(a)

1

2

3

R(a) D(a)

D(a)

1 has to wait forlower priority task 2

3 1 2

• Blocking task inherits higher priority of blocked task

Solution: Priority Inheritance

1

2

R(a) D(a)

3 1 2

3 not preempted by 2

2 experiences push-through blocking

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• Disadvantage:– Deadlocks remains possible

• If tasks reserves resources while holding others (hold and block)• Creating a circular wait condition

R(a)

2

3

D(a)

3 inherits priority of 1

1 experiences direct blocking

Utilization bound

• A given task i can be scheduled if

12 /1

1

i

i

ii

j j

j iT

B

T

C

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• For m tasks:

12max /11

11

m

i

im

i

m

j j

j mT

B

T

C

Response-Time Approach

• Assume a single resource shared by tasks i

• Blocking delay Bi for task i

– Direct blocking– Push-through blocking

• The worst case response time ri of a i and should now include these blocking contributions

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i inow include these blocking contributions

iij

ii

jji BC

T

rCr

1

1 Blocking

Blocking Contribution

• Single resource a• Ca,i is execution time of resource a by task i

– Note: execution times Ci include such contributions!

• pa is the highest priority among tasks possibly reserving a• Two cases

– pa pi > pm then m

CB max

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– pa pi > pm then

– pa < pi i = m then

• Computation intensive for more resources due to chained chained blockingblocking.

jaij

i CB ,1

max

0iB

i does not reserve a

m does not experience

blocking

i only blocked by lower priority

tasks

Schedulability of Task i

• Consider:

– Again task i can be scheduled if a solution for t in Wi(t)/t =1 exists– The response time ri equals the smallest of such solution.

• We only need to consider W (t) at integer multiples of the

ij

i

jji B

T

tCtW

1

)(

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• We only need to consider Wi(t) at integer multiples of the periods Ti

Example

• Task set:• 1 : C1 = 40 T1 = 100 B1 = 40• 2 : C2 = 40 T2 = 150 B2 = 30• 3 : C3 = 100 T3 = 350 B3 = 0

– Consider: t = 100, 150, 200, 300, 350

T3

Here we know schedulability

T1 T2 T3

1

2

3

100 200 300 400

2T1

3T1

2T2

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• W1(100) = C1 + B1 = 80 ok

• W2(100) = C1 + C2 + B2 = 110 not ok• W2(150) = 2C1 + C2 + B2 = 150 ok

• W3(100) = C1 + C2 + C3 = 180 not ok • W3(150) = 2C1 + C2 + C3 = 220 not ok• W3(200) = 2C1 + 2C2 + C3 = 260 not ok• W3(300) = 3C1 + 2C2 + C3 = 300 ok• W3(350) = 4C1 + 3C2 + C3 = 380 not ok

Exercises

• Calculate RM schedulability of the following task sets– Exercise 4. Task set:

• 1 : C1 = 4 T1 = 10 B1 = 3• 2 : C2 = 3 T2 = 15 B2 = 2• 3 : C3 = 4 T3 = 20 B3 = 0

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• 3 : C3 = 4 T3 = 20 B3 = 0

• Chained blocking when using multiple resources:

Problem: Chained Blocking

1

2

3

R(a)3 12R(b)D(a) D(b)

80

• Disadvantage:– Not just a single low-priority task can block the higher

one.

R(a) D(a) D(b)R(b)

Solution: Priority Ceiling

• To avoid chained blocking– Ceiling cr of a resource r is highest priority found among the

potentially reserving tasks– System ceiling c is highest ceiling of all currently held resources– A task i may hold a resource r if pi > c or i already holds a resource

with cr = c– Priority inheritance like before when tasks block because of resource

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– Priority inheritance like before when tasks block because of resource usage by other tasks

– ca= cb = p1 ;R(a)

c:= caD(a)

R(a)p1 c3 12

R(b)p1> cD(a) D(b)

D(b)R(b)p2c

Priority Ceiling

• Also avoids deadlock

R(d)p1> cc:= cd

3 12

R(a)holds c

D(d)c:= ca D(a) D(b)

R(b)p2 c

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R(a)c:= ca

D(a)D(b)R(b)

holds c

Blocking Contribution

• Ca,i is execution time of resource a by task i

• BRi are the resources that can block i

– BRi = { r | cr pi }

• Blocking for task i

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jrBRr

m

iji CB

i,

1maxmax

i only blocked by lower priority

tasks

Content

• Introduction• Static priority scheduling

– RM scheduling without resource contention• Periodic tasks• Sporadic tasks• Aperiodic tasks

84

• Aperiodic tasks

– RM scheduling with resource contention – DM scheduling

Deadline Period

• Up to now the assumption with regard to deadline Di= Ti

• If Di< Ti , two possibilities – DM assignment of priorities: Di< Dj pi> pj

– RM assignment of priorities

• In either case, the response time formula is

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• In either case, the response time formula is applicable– Modified use of the response time formula

• Compute response time ri for each task

• Compare ri with Di