Hardware Reference
In-Depth Information
SOLUTIONS FOR CHAPTER 7
7.1
Applying Equation (7.19), we can verify that
i
C
k
T
k
+
B
i
i
(2
1
/i
∀
i,
1
≤
i
≤
n,
T
i
≤
−
1)
.
k
=1
So we have
C
1
+
B
1
T
1
9
10
<
1
=
C
1
T
1
+
C
2
+
B
2
T
2
4
10
+
6
15
=0
.
8
<
0
.
83
=
C
1
T
1
+
C
2
T
2
+
C
3
T
3
4
10
+
3
15
+
4
20
=0
.
8
>
0
.
78
.
=
Being condition (7.19) only sufficient, we cannot conclude anything about fea-
sibility. By applying the response time analysis we have to verify that
∀
i,
1
≤
i
≤
n,
R
i
≤
D
i
where
R
i
T
k
C
k
.
i−
1
R
i
=
C
i
+
B
i
+
k
=1
So we have
R
1
=
C
1
+
B
1
=9
<
10
R
(0)
2
=
C
1
+
C
2
+
B
2
=10
C
2
+
B
2
+
10
10
4=10
<
15
R
(1)
2
=
R
(0)
3
=
C
1
+
C
2
+
C
3
=11
C
3
+
11
10
4+
11
15
3=15
R
(1)
3
=
C
3
+
15
10
4+
15
15
3=15
<
20
.
R
(2)
3
=
Hence, we can conclude that the task set is schedulable by RM.
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