Real-Time Scheduling
Priority-Driven Scheduling
Fixed-Priority
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Fixed-Priority Algorithms
We consider a set of tasks T = {T1, . . . , Tn}
Any fixed-priority algorithm schedules tasks of T according to
fixed (distinct) priorities assigned to tasks
We write Ti Tj whenever Ti has a higher priority than Tj
We denote by Ti↑ the set of tasks that have the same or higher
priority than Ti.
Recall that Fixed-Priority Algorithms might not be optimal
Consider T = {T1, T2} where T1 = (2, 1) and T2 = (5, 2.5)
UT = 1 and thus T is schedulable by EDF
If T1 T2, then J2,1 misses its deadline
If T2 T1, then J1,1 misses its deadline
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Critical Instant – Informally
To be able to further analyze fixed-priority algorithms we need
to consider a notion of critical instant
Intuitively, a critical instant is the time instant in which the
system is most loaded, and has its worst response time
Schedulability of a set of tasks is determined by response times
of jobs released at critical instants
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Critical Instant – Formally
Definition 1
A critical instant tcrit of a task Ti is a time instant in which a job
Ji,k in Ti is released so that Ji,k either does not meet its
deadline, or has the maximum response time of all jobs in Ti
Denote by Wi the response time of such Ji,k
Theorem 2
In a fixed-priority system where every job completes before the
next job in the same task is released, a critical instant of a task
Ti occurs when one of its jobs Ji,k is released at the same time
with a job from every higher-priority task.
Note that the situation described in the theorem does not have to occur if
tasks are not in phase. So we use critical instants either to study tasks in
phase, or to get upper bounds on schedulability as follows:
Set phases of all tasks to zero, which gives a new set of tasks
T = {T1
, . . . , Tn}
Determine the response time w of the first job Ji,1 in Ti
Then w ≥ Wi, the response time of a job in T released at the critical instant 4
RMa and DM Algorithms (reminder)
RM = assigns priorities to tasks based on their periods
the priority is inversely proportional to the period pi
DM = assigns priorities to tasks based on their relative deadlines
the priority is inversely proportional to the relative deadline Di
(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?
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Optimality of RM for Simply Periodic Tasks
Definition 3
A set {T1, . . . , Tn} is simply periodic if for every pair Ti, Tk
satisfying pi < pk we have that pk is an integer multiple of pi
Example 4
The helicopter control system from the first lecture
Theorem 5
A set T of n simply periodic, independent, preemptable tasks
with Di = pi is schedulable on one processor according to RM
iff UT = n
i=1
ei
pi
≤ 1.
i.e. on simply periodic tasks RM is as good as EDF
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Optimality of DM (RM) among Fixed-Priority Algs.
Theorem 6
A set of independent, preemptable periodic tasks with Di ≤ pi
that are in phase (i.e., ϕi = 0 for all i = 1, . . . , n) can be feasibly
scheduled on one processor according to DM if it can be
feasibly scheduled by some fixed-priority algorithm.
Proof.
Assume a fixed-priority feasible schedule with T1 · · · Tn.
Consider the least i such that the relative deadline Di of Ti is
larger than the relative deadline Di+1 of Ti+1.
Swap the priorities of Ti and Ti+1.
The resulting schedule is still feasible.
DM is obtained by using finitely many swaps.
Note: If the assumptions of the above theorem hold and all relative deadlines
are equal to periods, then RM is optimal among all fixed-priority algorithms.
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Fixed-Priority Algorithms: Schedulability
We consider two schedulability tests:
Schedulable utilization URM of the RM algorithm.
Time-demand analysis based on response times of jobs
released at critical instants
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Schedulable Utilization for RM
Theorem 7
Let us fix n ∈ N and consider only independent, preemptable
periodic tasks with Di = pi.
If T is a set of n tasks satisfying UT ≤ n(21/n − 1), then UT
is schedulable by the RM algorithm.
For every U > n(21/n − 1) there is a set T of n tasks
satisfying UT ≤ U that is not schedulable by RM.
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Schedulable Utilization for RM
It follows that the maximum schedulable utilization URM over
independent, preemptable periodic tasks satisfies
URM = inf
n
n(21/n
− 1) = lim
n→∞
n(21/n
− 1) = ln 2 ≈ 0.693
Note that UT
≤ n(21/n
− 1) is a sufficient but not necessary condition for
schedulability of T using the RM algorithm (an example will be given later)
In what follows we assume that p1 < p2 < . . . < pn which implies
that T1 T2 · · · Tn
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Proof Sketch of Theorem 7
A set of tasks T fully utilizes the processor if it is schedulable by RM
but any increase in execution time makes the set unschedulable.
Given n ∈ N, we denote by bn the greatest lower bound on utilization
over all sets of n tasks that fully utilize the processor.
We prove that
bn = n(2
1
n − 1)
Proof:
Given p1, . . . , pn, denote by U[p1, . . . , pn] the minimum utilization over
sets of n tasks with periods p1, . . . , pn that fully utilize the processor.
(A) Show that for every set that fully utilizes the processor there is
another one whose utilization cannot be larger and
1. is in phase,
2. satisfies pn ≤ 2p1.
In the rest of the proof we assume 1. and 2.
(B) For a fixed set of periods p1, . . . , pn, find a set T with periods
p1, . . . , pn and utilization U[p1, . . . , pn].
(C) Show that minp1,...,pn
U[p1, . . . pn] = n(21/n
− 1) 11
Proof Sketch of Theorem 7
A set of tasks T fully utilizes the processor if it is schedulable by RM
but any increase in execution time makes the set unschedulable.
Given n ∈ N, we denote by bn the greatest lower bound on utilization
over all sets of n tasks that fully utilize the processor.
We prove that
bn = n(2
1
n − 1)
It immediately follows that for every U > bn there is a set of n tasks T
that is not schedulable by RM but satisfies UT
≤ U
We prove that if UT
≤ bn for a given set T of n tasks, then T is
schedulable using the RM algorithm by induction on n
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Proof Sketch of Theorem 7 – Step (B)
In general, for pn ≤ 2p1, the following instance gives bn:
0 p1 2p1
0 p2
0 p3
0 pn−1
0 pn
...
T3
T2
T1
Tn
Tn−1
ek = pk+1 − pk for k = 1, . . . , n − 1
en = pn − 2
n−1
k=1
ek = 2p1 − pn
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Time-Demand Analysis
Assume that Di ≤ pi for every i.
Compute the total demand for processor time by a job released
at a critical instant of a task, and by all the higher-priority tasks,
as a function of time from the critical instant
Check if this demand can be met before the deadline of the job:
Consider one task Ti at a time, starting with highest priority
and working to lowest priority
Focus on a job Ji,c in Ti, where the release time, t0, of that
job is a critical instant of Ti
At time t0 + t for t ≥ 0, the processor time demand wi(t) for
this job and all higher-priority jobs released in [t0, t] is
bounded by
wi(t) = ei +
i−1
k=1
t
pk
ek for 0 < t ≤ pi
(Recall that by the critical instant theorem, the longest response time
occurs when jobs of all tasks with higher priority are released at t0)
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Time-Demand Analysis
Compare the time demand, wi(t), with the available time, t:
If wi(t) ≤ t for some t ≤ Di, the job Ji,c released at critical
instant of Ti meets its deadline, t0 + Di
If wi(t) > t for all 0 < t ≤ Di, then the task probably cannot
complete by its deadline; and the system likely cannot be
scheduled using a fixed priority algorithm
(Note that this condition is only sufficient as the expression for
wi(t) relies on the fact that jobs of all higher priority tasks are
released at the critical instant t0)
Use this method to check that all tasks are schedulable if
released at their critical instants; if so conclude the entire
system can be scheduled
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Time-Demand Analysis – Example
Example: T1 = (3, 1), T2 = (5, 1.5), T3 = (7, 1.25), T4 = (9, 0.5)
This is schedulable by RM even though
U{T1,...,T4}
= 0.85 > 0.757 = URM(4) 16
Time-Demand Analysis
The time-demand function wi(t) is a staircase function
Steps in the time-demand for a task occur at multiples of
the period for higher-priority tasks
The value of wi(t) − t linearly decreases from a step until
the next step
If our interest is the schedulability of a task, it suffices to
check if wi(t) ≤ t at the time instants when a higher-priority
job is released and at Di
Our schedulability test becomes:
Compute wi(t)
Check whether wi(t) ≤ t for some t equal either to Di, or to
j · pk where k = 1, 2, . . . , i and j = 1, 2, . . . , Di/pk
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Time-Demand Analysis
Time-demand analysis schedulability test is more complex
than the schedulable utilization test but more general:
Works for any fixed-priority scheduling algorithm, provided
the tasks have short response time (Di ≤ pi)
Can be extended to tasks with arbitrary deadlines
Still more efficient than exhaustive simulation
Only a sufficient test (as well as the utilization test for fixedpriority
systems)
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Dynamic vs Fixed Priority
EDF
pros:
optimal
very simple and complete test for schedulability
cons:
difficult to predict which job misses its deadline
strictly following EDF in case of overloads assigns higher
priority to jobs that missed their deadlines
larger scheduling overhead
DM (RM)
pros:
easier to predict which job misses its deadline (in particular,
tasks are not blocked by lower priority tasks)
easy implementation with little scheduling overhead
(optimal in some cases often occurring in practice)
cons:
not optimal
incomplete and more involved tests for schedulability
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