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σ
(t) = 4
σ
(t ) = 2
J
1
E
t
J
2
t
J
3
t
J
4
t
0
1
2
3
456789
10
11
12
3
14
15
t = 4
t = 6
(a)
E
σ
(t) = 2
(t ) = 4
E
J
σ
1
t
J
2
t
J
3
t
J
4
t
0
123456
7
8
9
10
11
12
13
14
15
t = 4
t = 6
(b)
E
Figure 3.4
Proof of the optimality of the EDF algorithm.
a.
schedule
σ
at time
t
=4
.
b.
new schedule obtained after a transposition.
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