real-time scheduling - penn engineeringlee/10cis541/papers/... · 2010. 1. 15. · real-time...
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1
Real-Time Scheduling
Introduction to Real-Time
Review
� Main vocabulary� Definitions of tasks, task invocations, release/arrival time,
absolute deadline, relative deadline, period, start time, finishtime, …
� Preemptive versus non-preemptive scheduling
� Priority-based scheduling
� Static versus dynamic priorities
� Utilization (U) and Schedulability� Main problem: Find Bound for scheduling policy such that
U < Bound � All deadlines met!
� Optimality of EDF scheduling� BoundEDF = 100%
2
Schedulability Analysis of Periodic Tasks
� Main problem:� Given a set of periodic tasks, can they meet their deadlines?
� Depends on scheduling policy
� Solution approaches� Utilization bounds (Simplest)
� Exact analysis (NP-Hard)
� Heuristics
� Two most important scheduling policies� Earliest deadline first (Dynamic)
� Rate monotonic (Static)
Schedulability Analysis of Periodic Tasks
� Main problem:� Given a set of periodic tasks, can they meet their deadlines?
� Depends on scheduling policy
� Solution approaches� Utilization bounds (Simplest)
� Exact analysis (NP-Hard)
� Heuristics
� Two most important scheduling policies� Earliest deadline first (Dynamic)
� Rate monotonic (Static)
3
Utilization Bounds
� Intuitively:
� The lower the processor utilization, U, the easier it is to meet deadlines.
� The higher the processor utilization, U, the more difficult it is to meet deadlines.
� Question: is there a threshold Ubound such that
� When U < Ubound deadlines are met
� When U > Ubound deadlines are missed
Example (Rate-Monotonic Scheduling)
Task 1P1=2
C1=1
Task 2P2=3C2=1.01
%3.833
01.1
2
1
2
2
1
1 ≈+=+=P
C
P
CU
0 1 2 3 4 5 6 time
?
100%
0
U
� Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines are missed
4
Example(Rate-Monotonic Scheduling)
Task 1P1=2
C1=1
Task 2P2=3C2=1.01
%3.833
01.1
2
1
2
2
1
1 ≈+=+=P
C
P
CU
0 1 2 3 4 5 6 time
?
100%
0
U
� Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines are missed
Example(Rate-Monotonic Scheduling)
Task 1P1=2
C1=1
Task 2P2=3C2=1.01
%3.833
01.1
2
1
2
2
1
1 ≈+=+=P
C
P
CU
0 1 2 3 4 5 6 time
Missed deadline!
Unschedulable
?
100%
0
U
83.3%
� Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines are missed
5
Another Example(Rate-Monotonic Scheduling)
Task 1P1=2
C1=1
Task 2P2=6C2=2.4
%906
4.2
2
1
2
2
1
1 =+=+=P
C
P
CU
0 1 2 3 4 5 6 time
?
100%
0
U
83.3%
� Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines are missed
Another Example(Rate-Monotonic Scheduling)
Task 1P1=2
C1=1
Task 2P2=6C2=2.4
%906
4.2
2
1
2
2
1
1 =+=+=P
C
P
CU
0 1 2 3 4 5 6 time
?
100%
0
U
83.3%
� Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines are missed
6
Another Example(Rate-Monotonic Scheduling)
Task 1P1=2
C1=1
Task 2P2=6C2=2.4
%906
4.2
2
1
2
2
1
1 =+=+=P
C
P
CU
0 1 2 3 4 5 6 time
Schedulable!
?
100%
0
U
83.3%90%
� Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines are missed
Another Example(Rate-Monotonic Scheduling)
Task 1P1=2
C1=1
Task 2P2=6C2=2.4
%906
4.2
2
1
2
2
1
1 =+=+=P
C
P
CU
0 1 2 3 4 5 6 time
Schedulable!
?
100%
0
U
83.3%90%
� Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines are missed
Schedulability depends on task set!No clean utilization threshold between schedulable and unschedulable
task sets!
7
A Conceptual View of Schedulability
Utilization
Task Set
SchedulableUnschedulable
� Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines are missed
∑=i i
i
P
C
A Conceptual View of Schedulability
Utilization
Task Set
SchedulableUnschedulable
� Modified Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines may or may not be missed
?
All green area (sche
dulable)
∑=i i
i
P
C
8
A Conceptual View of Schedulability
Utilization
Task Set
SchedulableUnschedulable
� Modified Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines may or may not be missed
?
All green area (sche
dulable)
U < Ubound is a sufficient but not necessaryschedulabilitycondition
∑=i i
i
P
C
A Conceptual View of Schedulability
Utilization
Task Set
SchedulableUnschedulable
� Modified Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines may or may not be missed
?
All green area (sche
dulable)
Equivalent question:What’s the lowest utilization of an unschedulable task set?
∑=i i
i
P
C
9
A Conceptual View of Schedulability
Utilization
Task Set
SchedulableUnschedulable
� Modified Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines may or may not be missed
?
All green area (sche
dulable)
Equivalent question:What’s the lowest utilization of an unschedulable task set?
critically(Called the Utilization Bound, U
bound)
∑=i i
i
P
C
Solution Approach: Look at Critically-Schedulable Task Sets
Utilization
Task Set� Modified Question: is there a threshold Ubound such that� When U < Ubound deadlines are met
� When U > Ubound deadlines may or may not be missed
?
All green area (sche
dulable)
Case (a) Case (b)Utilization increases with x
Utilization decreases with x
Find some task set parameter xsuch thatCase (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
10
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider a simple case: 2 tasks
Find some task set parameter xsuch that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider a simple case: 2 tasks
P1
P2
Task 1
Task 2
Find some task set parameter xsuch that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
11
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider a simple case: 2 tasks
C1
P1
P2
Task 1
Task 2
Find some task set parameter xsuch that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider a simple case: 2 tasks
C1
P1
P2
Task 1
Task 2
Critically schedulable
total = C2
Find some task set parameter xsuch that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
12
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider a simple case: 2 tasks
C1
P1
P2
Task 1
Task 2
?
Find some task set parameter xsuch that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider a simple case: 2 tasks
C1
P1
P2
Task 1
Task 2
1
1
22 P
P
PP
−
Find some task set parameter xsuch that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
13
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider these two sub-cases:
Case (a):
C1
P1
P2
Task 1
Task 2
Case (b):
P1
P2
C1
1
1
22 P
P
PP
− 1
1
22 P
P
PP
−
Find some task set parameter xsuch that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider these two sub-cases:
Case (a):
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−≤
Case (b):
P1
P2
C1
1
1
221 P
P
PPC
−>
Find some task set parameter xsuch that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
14
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider these two sub-cases:
Case (a):
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−≤
Case (b):
P1
P2
C1
1
1
221 P
P
PPC
−>
C1Find some task set parameter x
such that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider these two sub-cases:
Case (a):
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−≤
Case (b):
P1
P2
C1
1
1
221 P
P
PPC
−>
C1Find some task set parameter x
such that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
15
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider these two sub-cases:
Case (a):
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−≤
Case (b):
P1
P2
C1
1
1
221 P
P
PPC
−>
C1Find some task set parameter x
such that
Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x
Thus U(x) is minimum when x=xo
Find U(xo)
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider these two sub-cases:
Case (a):
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−≤
Case (b):
P1
P2
C1
1
1
221 P
P
PPC
−>
+
−= 1
1
2122
P
PCPC
−=
1
2112 )(
P
PCPC
−
−+=+= 11
1
2
1
2
2
1
2
2
1
1
P
P
P
P
P
C
P
C
P
CU
−+
=
1
2
1
2
2
1
1
2
2
1
P
P
P
P
P
C
P
P
P
PU
16
Deriving the Utilization Bound for Rate Monotonic Scheduling
� Consider these two sub-cases:
Case (a):
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−≤
Case (b):
P1
P2
C1
1
1
221 P
P
PPC
−>
+
−= 1
1
2122
P
PCPC
−=
1
2112 )(
P
PCPC
−+
=
1
2
1
2
2
1
1
2
2
1
P
P
P
P
P
C
P
P
P
PU
C1
���� UC
1���� U
−
−+=+= 11
1
2
1
2
2
1
2
2
1
1
P
P
P
P
P
C
P
C
P
CU
Deriving the Utilization Bound for Rate Monotonic Scheduling
� The minimum utilization case:
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−=
−
−+= 11
1
2
1
2
2
1
P
P
P
P
P
CU
17
Deriving the Utilization Bound for Rate Monotonic Scheduling
� The minimum utilization case:
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−=
−
−+= 11
1
2
1
2
2
1
P
P
P
P
P
CU
0
1
2
=
PP
d
dU
83.0≈⇒ U
21
2 =⇒P
P
−
−
−+= 11
1
2
1
2
1
2
1
2
2
1
P
P
P
P
P
P
P
P
P
PU
11
2 =
⇒
P
P
1
2
1
2
1
2 21
1
PP
PP
PP
U
−
−
+=⇒
Note that C1 = P2 – P1
−=
1
2
1
211
P
P
P
PPC
Generalizing to N Tasks
C1 = P2 – P1
C2 = P3 – P2
C3 = P4 – P3
…
...3
3
2
2
1
1 +++=P
C
P
C
P
CU
18
Generalizing to N Tasks
C1 = P2 – P1
C2 = P3 – P2
C3 = P3 – P2
…
0
1
2
=
PP
d
dU0
2
3
=
PP
d
dU0
3
4
=
PP
d
dU …
...3
3
2
2
1
1 +++=P
C
P
C
P
CU
Generalizing to N Tasks
C1 = P2 – P1
C2 = P3 – P2
C3 = P3 – P2
…
0
1
2
=
PP
d
dU0
2
3
=
PP
d
dU0
3
4
=
PP
d
dU …
...3
3
2
2
1
1 +++=P
C
P
C
P
CU
n
i
i
P
P 11 2=⇒ +
−=⇒ 12
1nnU
19
Generalizing to N Tasks
C1 = P2 – P1
C2 = P3 – P2
C3 = P3 – P2
…
0
1
2
=
PP
d
dU0
2
3
=
PP
d
dU0
3
4
=
PP
d
dU …
...3
3
2
2
1
1 +++=P
C
P
C
P
CU
n
i
i
P
P 11 2=⇒ +
−=⇒ 12
1nnU
2ln→∞→ Un
Periodic Tasks
Periodic Task Scheduling
Rate Monotonic EDF
Bound Optimality Bound Optimality
20
Periodic Tasks
Periodic Task Scheduling
Rate Monotonic EDF
Bound Optimality Bound Optimality
Coming Up
Periodic Task Scheduling
Rate Monotonic EDF
Bound Optimality Bound Optimality
21
Rate Monotonic Continued
� Rate monotonic scheduling is the optimal fixed-priority scheduling policy for periodic tasks.
� Optimality (Trial #1):
Rate Monotonic Continued
� Rate monotonic scheduling is the optimal fixed-priority scheduling policy for periodic tasks.
� Optimality (Trial #1): If any other fixed-priority scheduling policy can meet deadlines, so can RM.
22
Rate Monotonic Continued
� Rate monotonic scheduling is the optimal fixed-priority scheduling policy for periodic tasks.
� Optimality (Trial #1): If any other fixed-priority scheduling policy can meet deadlines, so can RM
Rate Monotonic Continued
� Rate monotonic scheduling is the optimal fixed-priority scheduling policy for periodic tasks.
� Optimality (Trial #1): If any other fixed-priority scheduling policy can meet deadlines, so can RM
23
Rate Monotonic Continued
� Rate monotonic scheduling is the optimal fixed-priority scheduling policy for periodic tasks.
� Optimality (Trial #1): If any other fixed-priority scheduling policy can meet deadlines, so can RM
Rate Monotonic Continued
� Rate monotonic scheduling is the optimal fixed-priority scheduling policy for periodic tasks.
� Optimality (Trial #1): If any other fixed-priority scheduling policy can meet deadlines, so can RM
24
Rate Monotonic Continued
� Rate monotonic scheduling is the optimal fixed-priority scheduling policy for periodic tasks.
� Optimality (Trial #2): If any other fixed-priority scheduling policy can meet deadlines in the worst case scenario, so can RM.
� How to prove it?
Rate Monotonic Continued
� Rate monotonic scheduling is the optimal fixed-priority scheduling policy for periodic tasks.
� Optimality (Trial #2): If any other fixed-priority scheduling policy can meet deadlines in the worst case scenario, so can RM.
� How to prove it?
� Consider the worst case scenario
� If someone else can schedule then RM can
25
The Worst-Case Scenario
� Q: When does a periodic task, T, experience the maximum delay?
� A: When it arrives together with all the higher-priority tasks (critical instance)
� Idea of Proof� If some higher-priority task does not arrive together with T, aligning the arrival times can only increase the completion time of T
Proof (Case 1)
Task 1
Task 2
Case 1: higher priority task 1 is running when task 2 arrives
26
Proof
Task 1
Task 2
Case 1: higher priority task 1 is running when task 2 arrives
� shifting task 1 right will increase completion time of 2
Proof
Task 1
Task 2
Case 1: higher priority task 1 is running when task 2 arrives
� shifting task 1 right will increase completion time of 2
Task 1
Task 2
27
Proof (Case 2)
Task 1
Task 2
Case 2: processor is idle when task 2 arrives
Proof (Case 2)
Task 1
Task 2
Case 2: processor is idle when task 2 arrives
� shifting task 1 left cannot decrease completion time of 2
28
Proof (Case 2)
Task 1
Task 2
Task 1
Task 2
Case 2: processor is idle when task 2 arrives
� shifting task 1 left cannot decrease completion time of 2
Optimality of Rate Monotonic
� If any other policy can meet deadlines so can RM
Policy X meets deadlines?
29
Optimality of Rate Monotonic
� If any other policy can meet deadlines so can RM
Policy X meets deadlines?
� RM meets deadlines
YES
Coming Up
Periodic Task Scheduling
Rate Monotonic EDF
Bound Optimality Bound Optimality
30
Coming Up
Periodic Task Scheduling
Rate Monotonic EDF
Bound Optimality Bound Optimality
Utilization Bound of EDF
� Why is it 100%?
� Consider a task set where:
� Imagine a policy that reserves for each task i a fraction fi of each clock tick, where fi = Ci /Pi
1=∑i i
i
P
C
Clock tick
31
Utilization Bound of EDF
� Imagine a policy that reserves for each task i a fraction fi of each time unit, where fi = Ci /Pi
� This policy meets all deadlines, because within each period Pi it reserves for task i a total time
� Time = fi Pi = (Ci / Pi) Pi = Ci (i.e., enough to finish)
Clock tick
Utilization Bound of EDF
� Pick any two execution chunks that are not in EDF order and swap them
32
Utilization Bound of EDF
� Pick any two execution chunks that are not in EDF order and swap them
� Still meets deadlines!
Utilization Bound of EDF
� Pick any two execution chunks that are not in EDF order and swap them
� Still meets deadlines!
� Repeat swap until all in EDF order� EDF meets deadlines
33
Periodic Tasks
Periodic Task Scheduling
Rate Monotonic EDF
Bound Optimality Bound Optimality
Done
Periodic Task Scheduling
Rate Monotonic EDF
Bound Optimality Bound Optimality
34
Exercise:Know Your Worst Case Scenario
� Consider a periodic system of two tasks
� Let Ui = Ci /Pi (for i = 1,2)
� What is the maximum value of:
Πi(1+Ui)
for a schedulable system?
Deriving the Utilization Bound for Rate Monotonic Scheduling
� The minimum utilization case:
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−=
−
−+= 11
1
2
1
2
2
1
P
P
P
P
P
CU
−
−
−+= 11
1
2
1
2
1
2
1
2
2
1
P
P
P
P
P
P
P
P
P
PU
11
2 =
⇒
P
P
−=
1
2
1
211
P
P
P
PPC
( )
−=
1
2112
P
PCPC
35
Deriving the Utilization Bound for Rate Monotonic Scheduling
� The minimum utilization case:
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−=
−
−+= 11
1
2
1
2
2
1
P
P
P
P
P
CU
−
−
−+= 11
1
2
1
2
1
2
1
2
2
1
P
P
P
P
P
P
P
P
P
PU
11
2 =
⇒
P
P
−=
1
2
1
211
P
P
P
PPC
C1
P1
P2
Task 1
Task 2
C2
Deriving the Utilization Bound for Rate Monotonic Scheduling
� The minimum utilization case:
C1
P1
P2
Task 1
Task 2
1
1
221 P
P
PPC
−=
−
−+= 11
1
2
1
2
2
1
P
P
P
P
P
CU
−
−
−+= 11
1
2
1
2
1
2
1
2
2
1
P
P
P
P
P
P
P
P
P
PU
11
2 =
⇒
P
P
−=
1
2
1
211
P
P
P
PPC
C1
P1
P2
Task 1
Task 2
C2
112
121
CPP
PCC
=−
=+
36
Solutions
( )∏ =+
=+
=+=+
=+
=+=+
−=−=
−=
i iU
P
P
P
PC
P
CU
P
P
P
PC
P
CU
PPCPC
PPC
21
211
11
2
2
1
2
22
2
22
1
2
1
11
1
11
21112
121
Critically
Schedulable
Schedulable ( )∏ ≤+i iU 21
Solutions
( )∏ =+
=+
=+=+
=+
=+=+
−=−=
−=
i iU
P
P
P
PC
P
CU
P
P
P
PC
P
CU
PPCPC
PPC
21
211
11
2
2
1
2
22
2
22
1
2
1
11
1
11
21112
121
Critically
Schedulable
Schedulable ( )∏ ≤+i iU 21
37
Solutions
( )∏ =+
=+
=+=+
=+
=+=+
−=−=
−=
i iU
P
P
P
PC
P
CU
P
P
P
PC
P
CU
PPCPC
PPC
21
211
11
2
2
1
2
22
2
22
1
2
1
11
1
11
21112
121
Critically
Schedulable
Schedulable ( )∏ ≤+i iU 21
Solutions
( )∏ =+
=+
=+=+
=+
=+=+
−=−=
−=
i iU
P
P
P
PC
P
CU
P
P
P
PC
P
CU
PPCPC
PPC
21
211
11
2
2
1
2
22
2
22
1
2
1
11
1
11
21112
121
Critically
Schedulable
Schedulable ( )∏ ≤+i iU 21
38
The General Case
( )∏ ==+
=+
=+=+
=+
=+=+
−=
−=
−
+
+
inn
n
i
nn
nn
n
n
n
i
i
i
ii
i
i
i
nn
iii
P
P
P
P
P
P
P
PU
P
P
P
PC
P
CU
P
P
P
PC
P
CU
PPC
PPC
22
...1
211
11
2
1
12
3
1
2
1
1
1
1
Critically
Schedulable
Schedulable ( )∏ ≤+i iU 21
The General Case
( )∏ ==+
=+
=+=+
=+
=+=+
−=
−=
−
+
+
inn
n
i
nn
nn
n
n
n
i
i
i
ii
i
i
i
nn
iii
P
P
P
P
P
P
P
PU
P
P
P
PC
P
CU
P
P
P
PC
P
CU
PPC
PPC
22
...1
211
11
2
1
12
3
1
2
1
1
1
1
Critically
Schedulable
Schedulable ( )∏ ≤+i iU 21
39
The General Case
( )∏ ==+
=+
=+=+
=+
=+=+
−=
−=
−
+
+
inn
n
i
nn
nn
n
n
n
i
i
i
ii
i
i
i
nn
iii
P
P
P
P
P
P
P
PU
P
P
P
PC
P
CU
P
P
P
PC
P
CU
PPC
PPC
22
...1
211
11
2
1
12
3
1
2
1
1
1
1
Critically
Schedulable
Schedulable ( )∏ ≤+i iU 21
The General Case
( )∏ ==+
=+
=+=+
=+
=+=+
−=
−=
−
+
+
inn
n
i
nn
nn
n
n
n
i
i
i
ii
i
i
i
nn
iii
P
P
P
P
P
P
P
PU
P
P
P
PC
P
CU
P
P
P
PC
P
CU
PPC
PPC
22
...1
211
11
2
1
12
3
1
2
1
1
1
1
Critically
Schedulable
Schedulable ( )∏ ≤+i iU 21
40
The Hyperbolic Bound for Rate Monotonic Scheduling
� A set of periodic tasks is schedulable if:
( )∏ ≤+i iU 21
The Hyperbolic Bound for Rate Monotonic Scheduling
� A set of periodic tasks is schedulable if:
� It’s a better bound than
� Example:
� A system of two tasks with U1=0.8, U2=0.1
( )∏ ≤+i iU 21
( )∑ −≤i
n
inU 12 /1
41
The Hyperbolic Bound for Rate Monotonic Scheduling
� A set of periodic tasks is schedulable if:
� It’s a better bound!
� Example:
� A system of two tasks with U1=0.8, U2=0.1
� Liu and Layland bound: U1+U2 = 0.9 > 0.83
( )∏ ≤+i iU 21
The Hyperbolic Bound for Rate Monotonic Scheduling
� A set of periodic tasks is schedulable if:
� It’s a better bound!
� Example:
� A system of two tasks with U1=0.8, U2=0.1
� Liu and Layland bound: U1+U2 = 0.9 > 0.83
� Hyperbolic bound (U1+1)(U2+1) =1.8 x 1.1=1.98 < 2
( )∏ ≤+i iU 21
42
Scheduling Taxonomy
Periodic Task Scheduling
Rate Monotonic EDF
Bound Optimality Bound Optimality
Scheduling Taxonomy
Periodic Task Scheduling
Rate Monotonic EDF
Bound Optimality Bound Optimality
Hyperbolic Bound
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