Hardware Reference
In-Depth Information
SOLUTIONS FOR CHAPTER 9
9.1
Applying the definition of instantaneous load we have
time
ρ
1
(
t
)
ρ
2
(
t
)
ρ
(
t
)
t
=0
0
5/10 = 0.5
0.5
t
=1
0
4/9 = 0.444
0.444
t
=2
0
3/8 = 0.375
0.375
t
=3
3/5 = 0.6
(3+2)/7 = 0.714
0.714
t
=4
2/4 = 0.5
(2+2)/6 = 0.667
0.667
t
=5
1/3 = 0.333
(1+2)/5 = 0.6
0.6
t
=6
0
2/4 = 0.5
0.5
t
=7
0
1/3 = 0.333
0.333
t
=8
0
0
0
9.2
Checking condition (9.24), necessary for the schedulability of the task set, we
have
n
C
i
(
S
i
−
1)
=
2
5
+
2
·
3
4
+
4
·
4
21
20
5
=
>
1
.
T
i
S
i
6
·
8
·
i
=1
Hence, we conclude that the task set is not schedulable by EDF.
9.3
From the service intervals provided by the server, it is clear that the longest
service delay occurs when a task is ready at time
t
=2, since it has to wait
for 3 units of time. Then, the service will be provided according to the supply
function illustrated in Figure 13.22.
From the graph, it is easy to see that the associated bounded delay function has
parameters
α
=0
.
4 and Δ=3
.
5.
9.4
By applying Equation (9.33) to the tasks we have
U
d
)
E
1
E
0
1
.
0)
1
U
1
U
1
0
−
(
U
0
−
=0
.
6
−
(1
.
4
−
4
=0
.
5
=
U
d
)
E
2
E
0
1
.
0)
3
U
2
=
U
2
0
−
(
U
0
−
=0
.
8
−
(1
.
4
−
4
=0
.
5
.
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