Hardware Reference
In-Depth Information
τ
1
τ
2
0
2
4
6
8
10
12
14
16
18
20
22
24
TB*
3
1
1
d1
d2
d3
U = 1/3
0
2
4
6
8
10
12
14
16
18
20
22
24
Figure 13.18
Schedule produced by EDF + TB* for the task set of Exercise 6.6.
τ
1
τ
2
0
2
4
6
8
10
12
14
16
18
20
22
24
TBS1
= 1/10
1
1
d1
d2
U
s
0
2
4
6
8
10
12
14
16
18
20
22
24
TBS2
1
2
d1
d2
U
= 1/6
s
0
2
4
6
8
10
12
14
16
18
20
22
24
Figure 13.19
Schedule produced by EDF+
TB
1
+
TB
2
for the task set of Exercise 6.7.
6.8
First of all, the utilization of the periodic task set is
8
20
+
30
=
3
6
U
p
=
=0
.
6;
5
hence, the largest utilization that can be assigned to a CBS is
U
s
=1
U
p
=
0
.
4. Then, converting all times in microseconds and substituting the values in
Equation (6.15) we obtain
−
C
avg
=
1200
μs
20 +
20
= 550
μs
·
10
4
1200
0
.
6
T
s
=
Q
s
T
s
U
s
= 220
μs.
=
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