Real-Time Scheduling
Priority-Driven Scheduling
Fixed-Priority
[Some parts of this lecture are based on a real-time systems course
of Colin Perkins
http://csperkins.org/teaching/rtes/index.html]
1
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
2
Fixed-Priority Algorithms
Consider a set of tasks T = {T1, . . . , Tn}
Any fixed-priority algorithm schedules tasks according to fixed
priorities assigned to tasks, w.l.o.g. assume
T1 T2 · · · Tn
i.e. T1 has the highest priority, Tn has the lowest priority
Define
T↑i := {Tk | Tk Ti}
3
Utilization
Utilization ui of a periodic task Ti with period pi and
execution time ei is defined by ui := ei/pi
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.
4
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
5
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 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 jobs in T↑i to zero, which gives a new set of tasks T
Determine the response time w of the first job Ji,1 in Ti
Then w ≥ Wi, the response time of a job in the original set T released at the
critical instant 6
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 ≤ 1.
i.e. on simply periodic tasks RM is as good as EDF
7
Optimality of DM (RM) among Fixed-Priority Algs.
Theorem 6
A set of independent, preemptable periodic tasks with Di ≤ pi
that are in phase 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.
8
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
9
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 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.
10
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
11
Proof Sketch of Theorem 7
For the proof we need the following notions:
A set of tasks is difficult-to-schedule if it is schedulable by RM
but any increase in execution time and/or decrease in period
makes the set unschedulable
A set of tasks is most-difficult if its total utilization is smallest
among all difficult to schedule sets
Step 1. Show that for every difficult-to-schedule set there is another one
whose utilization cannot be larger and
1. is in phase
2. satisfies pn ≤ 2p1
from this moment on we may concentrate on sets T satisfying 1. and 2.
Step 2. For a fixed set of periods p1, . . . , pn, find the most-difficult set T
with these periods, which gives U[p1, . . . , pn] := UT
Step 3. Find T with the “most-difficult” values of periods p1, . . . , pn, that is
minimize U[p1, . . . , pn], and show that UT
= n(21/n
− 1)
12
Proof Sketch of Theorem 7 – Step 2.
Most-difficult in phase set with pn ≤ 2p1:
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
13
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
14
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
15
Time-Demand Analysis
Example: T1 = (3, 1), T2 = (5, 1.5), T3 = (7, 1.25), T4 = (9, 0.5)
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
17
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)
18
Real-Time Scheduling
Priority-Driven Scheduling
Aperiodic Tasks
19
Current Assumptions
Single processor
Fixed number, n, of independent periodic tasks
Jobs can be preempted at any time and never suspend
themselves
No resource contentions
Aperiodic jobs exist
They are independent of each other, and of the periodic
tasks
They can be preempted at any time
There are no sporadic jobs (for now)
Jobs are scheduled using a priority driven algorithm
20
Scheduling Aperiodic Jobs
Consider:
A set T = {T1, . . . , Tn} of periodic tasks
An aperiodic task A
Recall that:
A schedule is feasible if all jobs with hard real-time
constraints complete before their deadlines
⇒ This includes all periodic jobs
A scheduling algorithm is optimal if it always produces a
feasible schedule whenever such a schedule exists, and if
a cost function is given, minimizes the cost
⇒ We consider a cost function which to a schedule
assigns the average response time of aperiodic jobs
We assume that the periodic tasks can be scheduled using a
priority-driven algorithm
21
Background Scheduling of Aperiodic Jobs
Aperiodic jobs are scheduled and executed only at times
when there are no periodic jobs ready for execution
Advantages
Clearly produces feasible schedules
Extremely simple to implement
Disadvantages
Not optimal since it is almost guaranteed to delay execution
of aperiodic jobs in favour of periodic ones
Example: T1 = (3, 1), T2 = (10, 4)
22
Polled Execution of Aperiodic Jobs
We may use a polling server
A periodic job (ps, es) scheduled according to the periodic
algorithm, generally as the highest priority job
When executed, it examines the aperiodic job queue
If an aperiodic job is in the queue, it is executed for up to es
time units
If the aperiodic queue is empty, the polling server
self-suspends, giving up its execution slot
The server does not wake-up once it has self-suspended,
aperiodic jobs which become active during a period are not
considered for execution until the next period begins
Simple to prove correctness, performance less than ideal –
executes aperiodic jobs in particular timeslots
23
Polled Execution of Aperiodic Jobs
Example: T1 = (3, 1), T2 = (10, 4), poller = (2.5, 0.5)
Can we do better?
Yes, polling server is a special case of periodic-server for
aperiodic jobs
24
Periodic Severs – Terminology
periodic server = a task that behaves much like a periodic task,
but is created for the purpose of executing aperiodic jobs
A periodic server, TS = (pS, eS)
pS is a period of the server
eS is the (maximal) budget of the server
The budget can be consumed and replenished; the budget
is exhausted when it reaches 0
(Periodic servers differ in how they consume and replenish the budget)
A periodic server is
backlogged whenever the aperiodic job queue is non-empty
idle if the queue is empty
eligible if it is backlogged and the budget is not exhausted
When a periodic server is eligible, it is scheduled as any
other periodic task with parameters (pS, eS)
25
Periodic Severs
Each periodic server is thus specified by
consumption rules: How the budget is consumed
replenishment rules: When and how the budget is
replenished
Example – polling server:
consumption rules:
Whenever the server executes, the budget is consumed at
the rate one per unit time.
Whenever the server becomes idle, the budget gets
immediately exhausted
replenishment rule: At each time instant k · pS replenish
the budget to eS
26
Deferrable Sever
Consumption rule:
The budget is consumed at the rate of one per unit time
whenever the server executes
Unused budget is retained throughout the period, to be
used whenever there are aperiodic jobs to execute
(i.e. instead of discarding the budget if no aperiodic job to execute
at start of period, keep in the hope a job arrives)
Replenishment rule:
The budget is set to eS at multiples of the period
i.e. time instants k · pS for k = 0, 1, 2, . . .
(Note that the server is not able tu cumulate the budget over
periods)
We consider both
Fixed-priority scheduling
Dynamic-priority scheduling (EDF)
27
Deferrable Server – RM
Here the tasks are scheduled using RM.
Is it possible to increase the budget of the server to 1.5 ?
28