Hardware Reference
In-Depth Information
τ
1
τ
2
τ
3
0
2
4
6
8
10
12
14
16
18
20
22
24
Figure 13.8
Schedule produced by EDF for the task set of Exercise 4.6.
So
τ
2
does not miss its deadline. For
τ
1
we have
2
R
(0)
1
C
j
=
C
1
+
C
2
=4
=
j
=1
C
1
+
R
(0)
C
2
=2+
4
8
2=4
R
(1)
1
1
T
2
=
So
R
1
=4, meaning that
τ
1
does not miss its deadline. For
τ
3
we have
3
R
(0)
3
=
C
j
=
C
1
+
C
2
+
C
3
=8
j
=1
C
3
+
R
(0)
C
2
+
R
(0)
C
1
=4+
8
8
2+
8
6
2=10
R
(1)
3
3
T
2
3
T
1
=
And since
R
(1
3
>D
3
, we can conclude that the task set is not schedulable by
DM. The resulting schedule is shown in Figure 13.9.
τ
1
τ
2
τ
3
0
2
4
6
8
10
12
14
16
18
20
22
24
Figure 13.9
Schedule produced by Deadline Monotonic for the task set of Exercise 4.7.
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