Graphics Programs Reference
In-Depth Information
Throughput — The comparison of throughputs of transitions represent-
ing a rendez-vous provides an indication of how often different processes
need to synchronize. Assuming that all transition rates are identically equal
to 1, and n = k = 1, then all transitions of the net have the same throughput
(equal to 0.165468): this is due to the symmetry of P
1
and P
2
and to the
fact that there are only two T-invariants: one for P
1
and one for P
2
, and
that all transitions in the invariants have multiplicity 1. We can destroy the
symmetry by having P
2
requesting Spooler to print more than one packet
(e.g. n = 10). As a consequence the throughputs of the transitions of P
1
are
all the same (0.43274) and they differ from those of P
2
. In particular the
throughput for the transitions of P
2
inside the for loop is 10 times higher
than the throughput of those outside the loop (0.413923 vs 0.041392). This
is not surprising since process P
1
is now covered by a T-invariant where
transitions belonging to the for loop have weight 10.
Mutual exclusion — If we define a process as “active” when it is not
in a state in which it is waiting for a rendez-vous, then the probability
that the two processes P
1
and P
2
are active at the same time is given by
P
{
P
1
active
∧
P
2
active
}
which in terms of distribution of tokens in places
is given by:
P
{
M(while
1
) + M(c
1
) + M(for
1
) + M(c
3
) + M(ec
3
) + M(while
2
)+
+M(c
2
) + M(for
2
) + M(c
4
) + M(ec
4
) > 1
}
The value of this probability is a function of the transition rates, but it is
always greater than zero, meaning that the two processes are not mutually
exclusive, i.e. they can be doing useful work at the same time, and therefore
it may be convenient to allocate them on two different processors.
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