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.
Search WWH ::




Custom Search