Hardware Reference
In-Depth Information
τ
1
0
6
12
18
24
τ
2
0
8
16
24
d
1
d
2
d
1
2
1
3
aperiodic
requests
0
1
2
3
4
56
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Figure 6.6
Total Bandwidth Server example.
task,
τ
2
, with a shorter deadline, equal to 16. Finally, the third aperiodic request
arrives at time
t
=14and gets a deadline
d
3
=max(
r
3
,d
2
)+
C
3
/U
s
=21. It does
not receive immediate service, since at time
t
=14task
τ
1
is active and has an earlier
deadline (18).
6.4.1
SCHEDULABILITY ANALYSIS
In order to derive a schedulability test for a set of periodic tasks scheduled by EDF in
the presence of a TBS, we first show that the aperiodic load executed by TBS cannot
exceed the utilization factor
U
s
defined for the server.
Lemma 6.2
In each interval of time
[
t
1
,t
2
]
,if
C
ape
is the total execution time de-
manded by aperiodic requests arrived at
t
1
or later and served with deadlines less
than or equal to
t
2
, then
C
ape
≤
(
t
2
−
t
1
)
U
s
.
Proof.
By definition
C
ape
=
C
k
.
t
1
≤r
k
,d
k
≤t
2
Given the deadline assignment rule of the TBS, there must exist two aperiodic requests
with indexes
k
1
and
k
2
such that
k
2
C
k
=
C
k
.
t
1
≤r
k
,d
k
≤t
2
k
=
k
1
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