Priority-Based Scheduling (Periodic Tasks)
• “A preemptive method where the priority of the process
determines whether it continues to run or is disrupted”
(most important process first)
• On-line scheduler (does not precompute schedule)
• Fixed priorities:
– same priority to all jobs in a task
• Dynamic priorities:
– different priorities to individual jobs in each task
– task-level dynamic priorities
– job-level fixed priorities
RMS: Rate Monotonic Scheduling
• On-line
• Preemptive
• Priority-based with static priorities
• Period Ti that is the shortest interval between its arrival
times
• Processes are assigned priorities dependent on length of
Ti, the shorter it is, the higher the priority (or the higher
the rate, the higher the priority)
T i < T j ! Pi > P j
• RM algorithm or RMS
Example Priority Assignment
Process
a
b
c
d
e
Period
25
60
42
105
75
Priority
1
Example
P1
20
10
high
• Period (Ti)
• WCET (ei)
• Priority
arrival time
t=0
t=20
t=30
t=40
t=50
t=60
0
P2
P3
50
30
10
5
low
medium
process
P1,P2,P3
P1
P3
P1
P2
P1,P3
10
20
30
40
50
60
70
time
preemption
Schedulability Test
• For n processes, RMS will guarantee their schedulability
if the total utilization U does not exceed the guarantee
level G = n * (21/n - 1)
• When test fails:
– try with the worst case: assume that all processes are released
simultaneously at time 0, and then arrive according to their
periods
– check whether each process meets its deadline for all releases
before the first deadline for the process with the lowest priority
• Otherwise:
– change U by reducing c i (code optimization, faster processor, …)
– or increase Ti for some process (possible?)
Process Set A
Process Period ComputationTime Priority Utilization
T
C
P
U
a
b
c
50
40
30
12
10
10
1
2
3
0.24
0.25
0.33
• The combined utilization is 0.82 (or 82%)
• This is above the threshold for three processes (0.78)
and, hence, this process set fails the utilization test
2
Timeline for Process Set A
Process
a
Process Release Time
Process Completion Time
Deadline Met
Process Completion Time
Deadline Missed
b
Preempted
c
Executing
0
10
20
30
40
50
60
Time
Process Set B
Process Period ComputationTime Priority Utilization
T
C
P
U
a
b
c
80
40
16
32
5
4
1
2
3
0.400
0.125
0.250
• The combined utilization is 0.775
• This is below the threshold for three processes (0.78)
and, hence, this process set will meet all its deadlines
Process Set C
Process Period ComputationTime Priority Utilization
T
C
P
U
a
b
c
80
40
20
40
10
5
1
2
3
0.50
0.25
0.25
• The combined utilization is 1.0
• This is above the threshold for three processes (0.78) but
the process set will meet all its deadlines
3
Timeline for Process Set C
Process
a
b
c
0
10
20
30
40
50
60
70
80
Time
Response Time Analysis
• Here task i's worst-case response time, R, is calculated
first and then checked (trivially) with its deadline
R i ≤ Di
Ri = Ci + I i
Where I is the interference from higher priority tasks
Calculating R
During R, each higher priority task j will execute a number of
times:
$R "
Number of Releases = # i !
#Tj !
The ceiling function gives the smallest integer greater than the fractional
number on which it acts. So the ceiling of 1/3 is 1, of 6/5 is 2, and of 6/3 is 2.
Total interference is given by:
$ Ri "
# !C j
#Tj !
4
Reponse Time Equation
%R #
Ri = Ci + ! $ i "C j
j&hp ( i ) T
$ j"
Where hp(i) is the set of tasks with priority higher than task i
Solve by forming a recurrence relationship:
win +1 = Ci +
0
% win #
$
"C j
j&hp ( i )
$ Tj "
!
1
2
n
The set of values wi , wi , wi ,..., wi ,.. is monotonically non decreasing
When win = win +1 the solution to the equation has been found, wi0
must not be greater than Ri (e.g. 0 orCi )
Process Set D
Process
Period
T
7
12
20
a
b
c
ComputationTime
C
3
3
5
Priority
P
3
2
1
wb0 = 3
Ra = 3
$3"
wb1 = 3 + # ! 3 = 6
#7!
$6"
2
wb = 3 + # ! 3 = 6
#7!
Rb = 6
wc0 = 5
$5"
$ 5 "
wc1 = 5 + # ! 3 + #
3 = 11
#7!
#12 !
!
11
11
$
"
$
"
wc2 = 5 + #
3+ #
3 = 14
#7 !
!
#12 !
!
$14 "
$14 "
wc3 = 5 + #
3+ #
3 = 17
# 7 !
!
#12 !
!
$17 "
$17 "
wc4 = 5 + #
3+ #
3 = 20
#7 !
!
#12 !
!
$ 20 "
$ 20 "
wc5 = 5 + #
3+ #
3 = 20
# 7 !
!
# 12 !
!
Rc = 20
5
Revisit: Process Set C
Process Period ComputationTime Priority Response
time
T
C
P
R
a
b
c
80
40
20
40
10
5
1
2
3
80
15
5
• The combined utilization is 1.0
• This was above the ulilization threshold for three
processes (0.78), therefore it failed the test
• The response time analysis shows that the process set
will meet all its deadlines
• RTA is necessary and sufficient
Response Time Analysis
• Is sufficient and necessary
• If the process set passes the test they will meet all their
deadlines; if they fail the test then, at run-time, a process
will miss its deadline (unless the computation time
estimations themselves turn out to be pessimistic)
Exact Schedulability Test
for (each task T j) {
I = 0;
do {
R = I + cj
if (R > dj) return UNSCHEDULABLE;
I = ∑R/pkck ; /* k = 1..j-1 */
} while (I + cj > R)
return SCHEDULABLE;
}
6
Deadline-Monotonic Algorithm (DM)
• Fixed-priority
• Uses relative deadlines: the shorter the relative deadline,
the higher the priority
• RM and DM are identical if the relative deadline is
proportional to its period
• Otherwise DM performs better in the sense that it can
sometimes produce a feasible schedule when RM fails,
while RM always fails when DM fails
Example
T1
0
50
100
150
200
250
T2
0
62.5
125
187.5
T3
T1
0
50
100
150
200
250
T2
0
62.5
125
187.5
T3
7