Hardware Reference
In-Depth Information
4.4
Applying the Response Time Analysis, we can easily verify that
R
3
=10(see
the solution of the previous exercise); hence the task set is not schedulable by
RM.
4.5
Since
1
4
+
2
6
+
3
U
=
=0
.
96
<
1
8
the task set is schedulable by EDF, as shown in Figure 13.7.
τ
1
τ
2
τ
3
0
2
4
6
8
10
12
14
16
18
20
22
24
Figure 13.7
Schedule produced by EDF for the task set of Exercise 4.5.
4.6
Applying the processor demand criterion, we have to verify that
L
+
T
i
−
C
i
≤
n
D
i
∀
L
∈D
L.
T
i
i
=1
where
min(
L
∗
,H
)
D
=
{
d
k
|
d
k
≤
}
.
For the specific example, we have
2
6
+
2
4
12
=
11
12
U
=
8
+
i
=1
(
T
i
−
D
i
)
U
i
L
∗
=
=32
−
U
1
H
=
lcm(6
,
8
,
12) = 24
.
Hence, the set of checking points is given by
D
=
{
4
,
5
,
8
,
11
,
12
,
17
,
20
,
23
}
.
Since the demand in these intervals is
we can con-
clude that the task set is schedulable by EDF. The resulting schedule is shown
in Figure 13.8.
{
2
,
4
,
8
,
10
,
12
,
14
,
20
,
22
}
4.7
Applying the Response Time Analysis, we have to start by computing the re-
sponse time of task
τ
2
, which is the one with the shortest relative deadline, and
hence the highest priority:
R
2
=
C
2
=2
.
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