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and the schedulability condition is given by:
n
C i
T 1 .
(4.12)
i =1
From Equations (4.11), the schedulability condition can also be written as
C n
2 T 1
T n
(4.13)
Now, defining
R i = T i +1
T i
U i = C i
T i
and
.
Equations (4.11) can be written as follows:
U 1 = R 1
1
U 2 = R 2
1
(4.14)
...
U n− 1 = R n− 1
1 .
Now we notice that:
n− 1
R i = T 2
T 1
T 3
T 2 ···
T n
T n− 1
T n
T 1
=
.
i =1
If we divide both sides of the feasibility condition (4.13) by T n , we get:
2 T 1
T n
U n
1 .
Hence, the feasibility condition for a task set that fully utilizes the processor can be
written as
2
n− 1
i =1
U n +1
.
R i
Since R i = U i +1for all i =1 ,...,n
1,wehave
n− 1
( U n +1)
( U i +1)
2
i =1
and finally
n
( U i +1)
2 ,
i =1
which proves the theorem.
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