Graphics Programs Reference
In-Depth Information
Neither the inhibitor arc from p
6
to t
5
nor the test arc from p
5
to t
4
is
represented in the matrix.
The relevance of the incidence matrix is due to the fact that it allows the
net dynamics to be expressed by means of linear algebraic equations. In
particular, we can observe that for any marking M, the firing of a transition
t enabled in M produces the new marking
M
0
= M + C(.,t)
T
(2.11)
where M and M
0
are row vectors, and C(.,t) is the column vector of C
corresponding to transition t.
A similar relation holds for transition sequences. Given a transition sequence
σ = t
(1)
,
···
,t
(k)
, we define the transition count vector V
σ
whose ith entry
indicates how many times transition t
i
appears in the transition sequence.
V
σ
is a
|
T
|
-component column vector. The marking M
00
obtained by firing
the transition sequence σ from marking M (M[σ
i
M
00
) can be obtained using
the following equation:
M
00
= M + [CV
σ
]
T
(2.12)
Observe that only the number of times a transition fires is important: the
order in which transitions appear in σ is irrelevant. The order is important
for the definition of the transition sequence, and for checking whether the
sequence can be fired, but it plays no role in the computation of the marking
reached by that sequence. This remark leads to important consequences
related to the definition of invariant relations for PN models.
In our readers & writers example, starting from the initial marking we can
fire the sequence t
1
,t
2
,t
1
,t
4
; we can thus define a firing count vector V
σ
=
[2, 1, 0, 1, 0, 0, 0]
T
, whose firing leads to the new marking
[K, 0, 0, 0, 1, 0, 0] + [CV
σ
]
T
[K, 0, 0, 0, 1, 0, 0] + 2C(.,t
1
)
T
+ C(.,t
2
)
T
+ C(.,t
4
)
T
=
[K
−
2, 1, 0, 0, 1, 1, 0]
=
where the second equation is obtained by writing the vector V
σ
as the sum
of seven vectors: each vector has all elements equal to zero but the ith one,
which is equal to V
σ
[i].
The same result can be computed by repeatedly applying
(2.11)
:
[K, 0, 0, 0, 1, 0, 0] + C(.,t
1
)
T
+ C(.,t
2
)
T
+ C(.,t
1
)
T
+ C(.,t
4
)
T
[K, 0, 0, 0, 1, 0, 0] + 2C(.,t
1
)
T
+ C(.,t
2
)
T
+ C(.,t
4
)
T
=
[K
−
2, 1, 0, 0, 1, 1, 0]
=
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