Real-Time Scheduling
Scheduling of Reactive Systems
Priority-Driven Scheduling
[Some parts of this lecture are based on a real-time systems course
of Colin Perkins
http://csperkins.org/teaching/rtes/index.html]
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Current Assumptions
Single processor
Fixed number, n, of independent periodic tasks
i.e. there is no dependency relation among jobs
Jobs can be preempted at any time and never suspend
themselves
No aperiodic and sporadic jobs
No resource contentions
Moreover, unless otherwise stated, we assume that
Scheduling decisions take place precisely at
release of a job
completion of a job
(and nowhere else)
Context switch overhead is negligibly small
i.e. assumed to be zero
There is an unlimited number of priority levels
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Fixed-Priority vs Dynamic-Priority Algorithms
A priority-driven scheduler is on-line
i.e. it does not precompute a schedule of the tasks
It assigns priorities to jobs after they are released and places the
jobs in a ready job queue in the priority order
with the highest priority jobs at the head of the queue
At each scheduling decision time, the scheduler updates the
ready job queue and then schedules and executes the job at the
head of the queue
i.e. one of the jobs with the highest priority
Fixed-priority = all jobs in a task are assigned the same priority
Dynamic-priority = jobs in a task may be assigned different priorities
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Priority-Driven Algorithms
Dynamic-priority:
EDF = at the time of a scheduling decision, the job queue is
ordered by the earliest deadline
Fixed-priority:
RM = assigns priorities to tasks based on their periods
DM = assigns priorities to tasks based on their relative deadlines
(In all cases, ties are broken arbitrarily.)
We consider the following questions:
Are the algorithms optimal?
How to efficiently (or even online) test for schedulability?
To measure abilities of scheduling algorithms and to get fast online
tests of schedulability we use a notion of utilization
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Utilization
Utilization ui of a periodic task Ti with period pi and
execution time ei is defined by ui := ei/pi
ui is the fraction of time a periodic task with period pi and execution time
ei keeps a processor busy
Total utilization UT of a set of tasks T = {T1, . . . , Tn} is
defined as the sum of utilizations of all tasks of T , i.e. by
UT
:=
n
i=1
ui
U is a schedulable utilization of an algorithm ALG if all sets
of tasks T satisfying UT ≤ U are schedulable by ALG.
Maximum schedulable utilization UALG of an algorithm ALG
is the supremum of schedulable utilizations of ALG.
If UT
< UALG, then T is schedulable by ALG.
If U > UALG, then there is T with UT
≤ U that is not
schedulable by ALG.
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Utilization – Example
T1 = (2, 1) then u1 = 1
2
T1 = (11, 5, 2, 4) then u1 = 2
5
(i.e., the phase and deadline do not play any role)
T = {T1, T2, T3} where T1 = (2, 1), T2 = (6, 1), T3 = (8, 3)
then
UT
=
1
2
+
1
6
+
3
8
=
25
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Real-Time Scheduling
Priority-Driven Scheduling
Dynamic-Priority
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Optimality of EDF
Theorem 1
Let T = {T1, . . . , Tn} be a set of independent, preemptable
periodic tasks with Di ≥ pi for i = 1, . . . , n. The following
statements are equivalent:
1. T can be feasibly scheduled on one processor
2. UT ≤ 1
3. T is schedulable using EDF
(i.e., in particular, UEDF = 1)
Proof.
1.⇒2. We prove that UT
> 1 implies that T is not schedulable (whiteb.)
2.⇒3. Next slides and whiteboard ...
3.⇒1. Trivial
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Proof of 2.⇒3.
Notation: Given a set of tasks L, we denote by L the set of all
jobs of the tasks in L.
We prove ¬3.⇒ ¬2. assuming that Di = pi for i = 1, . . . , n (note that
the general case immediately follows).
Assume that T is not schedulable by EDF. We show that UT
> 1.
Suppose that a job Ji,k of Ti misses its deadline at time t = ri,k + pi.
Let T be the set of all tasks whose jobs are released in [ri,k , t] (i.e., a
task belongs to T iff at least one job of the task is released in [ri,k , t]).
Let t− be the end of the latest interval before t in which either jobs of
(T T ) are executed, or the processor is idle.
Then ri,k ≥ t− since all jobs of (T T ) waiting for execution during
[ri,k , t] have deadlines later than t (thus have lower priorities than Ji,k ).
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Proof of 2.⇒3. (cont.)
It follows that
no job of (T T ) is executed in [t−, t],
(by definition of t−)
all jobs of T executed in [t−, t] are released in [t−, t] and have
their deadlines in [t−, t],
(since no job of T executes just before t−, and jobs with deadlines
after t have lower priorities than Ji,k )
the processor is fully utilized in [t−, t].
(by definition of t−)
Let G be the set of all jobs that are released in [t−, t] and have their
deadlines in [t−, t].
Note that Ji,k ∈ G since ri,k ≥ t−. Since Ji,k misses its deadline, the
processor should not be able to complete all jobs of G before t.
Denote by EG the sum of all execution times of all jobs in G (the total
execution time of G).
It follows that EG > t − t− because otherwise, all jobs of G (in
particular, Ji,k ) would complete in [t−, t]. 10
Proof of 2.⇒3. (cont.)
How to compute EG?
For T ∈ T , denote by R the earliest release time of a job in T
during the interval [t−, t].
For every T ∈ T , exactly t−R
p jobs of T belong to G. (For every
T ∈ T T , exactly 0 jobs belong to G.)
Thus
EG =
T ∈T
t − R
p
e
As argued above:
t−t− < EG =
T ∈T
t − R
p
e ≤
T ∈T
t − t−
p
e ≤ (t−t−)
T ∈T
u ≤ (t−t−)UT
which implies that UT
> 1.
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Density and EDF
What about tasks with Di < pi ?
Density of a task Ti with period pi, execution time ei and relative
deadline Di is defined by
ei/ min(Di, pi)
Total density ∆T of a set of tasks T is the sum of densities of
tasks in T
Note that if Di < pi for some i, then ∆T
> UT
Theorem 2
A set T of independent, preemptable, periodic tasks can be
feasibly scheduled on one processor if ∆T ≤ 1.
Note that this is NOT a necessary condition! (Example whiteb.)
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Schedulability Test For EDF
The problem: Given a set of independent, preemptable, periodic
tasks T = {T1, . . . , Tn} where each Ti has a period pi, execution time
ei, and relative deadline Di, decide whether T is schedulable by EDF.
Solution using utilization and density:
If pi ≤ Di for each i, then it suffices to decide whether UT ≤ 1.
Otherwise, decide whether ∆T ≤ 1:
If yes, then T is schedulable with EDF
If not, then T does not have to be schedulable
Note that
Phases of tasks do not have to be specified
Parameters may vary: increasing periods or deadlines, or
decreasing execution times does not prevent schedulability
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Schedulability Test for EDF – Example
Consider a digital robot controller
A control-law computation
takes no more than 8 ms
the sampling rate: 100 Hz, i.e. computes every 10 ms
Feasible? Trivially yes ....
Add Built-In Self-Test (BIST)
maximum execution time 50 ms
want a minimal period that is feasible (max one second)
With 250 ms still feasible ....
Add a telemetry task
maximum execution time 15 ms
want to minimize the deadline on telemetry
period may be large
Reducing BIST to once a second, deadline on telemetry
may be set to 100 ms ....
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