Hardware Reference
In-Depth Information
Given that DSS behaves like a periodic task, the following theorem states that a full
processor utilization is still achieved.
Theorem 6.2 (Spuri, Buttazzo)
Given a set of
n
periodic tasks with processor uti-
lization
U
p
and a Dynamic Sporadic Server with processor utilization
U
s
, the whole
set is schedulable
if
and
only if
U
p
+
U
s
≤
1
.
Proof.
1 and suppose there is an overflow at time
t
. The
overflow is preceded by a period of continuous utilization of the processor. Further-
more, from a certain point
t
on (
t
<t
), only instances of tasks ready at
t
or later
and having deadlines less than or equal to
t
are run (the server may be one of these
tasks). Let
C
be the total execution time demanded by these instances. Since there is
an overflow at time
t
, we must have
t
If
. Assume
U
p
+
U
s
≤
t
<C
. We also know that
−
t
C
i
+
C
ape
n
t
−
C
≤
T
i
i
=1
t
C
i
+
t
C
s
n
t
t
−
−
≤
T
i
T
s
i
=1
n
t
t
t
−
C
i
+
t
−
≤
C
s
T
i
T
s
i
=1
t
)(
U
p
+
U
s
)
.
≤
(
t
−
Thus, it follows that
U
p
+
U
s
>
1
,
a contradiction.
Only If
. Since DSS behaves as a periodic task with period
T
s
and execution time
C
s
, the server utilization factor is
U
s
=
C
s
/T
s
and the total utilization factor of the
processor is
U
p
+
U
s
. Hence, if the whole task set is schedulable, from the EDF
schedulability bound [LL73] we can conclude that
U
p
+
U
s
≤
1.
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