*Pj*, *O t*<sup>0</sup> *i* � � � � . If for *t* 0

0

first position, and *m* from *t*

• *t* 0 <sup>1</sup> � *P*<sup>4</sup> *t* 0 <sup>2</sup> � *P*3.

• *a*<sup>1</sup> ¼ 13 *a*<sup>2</sup> ¼ 15

Now, we will define the following action for *ai*:

8 ><

>:

0

*<sup>i</sup>* ¼ *ai*. For example:

0 *k*.

*<sup>k</sup>* corresponding *b*

0

*j*þ1 *<sup>i</sup>* ¼ *b j*

We denote the new *t<sup>α</sup>* corresponding to *b*

If there is no corresponding position for *t*

*ai* � *n m*

*ai* � *n m* h i

position for which *μ*½ � *x* <sup>1</sup> ¼ *ai*. In this case *n* ¼ *μ*0½ � *P*<sup>1</sup> , and *m* ¼ # *Pj*, *O t*<sup>0</sup>

*b*1

*b*1

We consider all *T* <sup>∗</sup> items from the right to left, starting from *tα*.

*j*

, if ð Þ *ai* � *n* modm ¼ 0

*<sup>i</sup>* transition *P*1, ⋯, *Pk* positions correspond, then we will take the *P*<sup>1</sup>

*a*<sup>1</sup> ¼ ð Þ *a*<sup>1</sup> � 1 *=*1 ¼ ð Þ 13 � 1 *=*1 ¼ 12 *a*<sup>2</sup> ¼ ð Þ *a*<sup>2</sup> � 0 *=*1 ¼ ð Þ 15 � 0 *=*1 ¼ 15*:*

<sup>1</sup> ¼ *a*<sup>1</sup> ¼ 12

<sup>2</sup> ¼ *a*<sup>2</sup> ¼ 15*:*

We will take *T* <sup>∗</sup> last transition or the succession of transition, fix it and mark

*<sup>i</sup>* is marked as *α* which we also fix. The fixed *b*

*<sup>i</sup>* þ *α* � *l*, where *<sup>l</sup>* from *P*<sup>1</sup> position *t<sup>α</sup>* is the number of arrows.

*j*

*<sup>k</sup>* corresponds with *P*1, ⋯, *Pl* positions. If ∃1≤*j*≤*l* in the way

*<sup>i</sup>* þ *α* � *l*, where *l* ¼ # *Pj*,*I t*ð Þ*<sup>α</sup>*

It follows that for each transition or sequence of transitions, there will be a

After that, we fix *t<sup>α</sup>* the previous transition action and denote it as *tα*.

*j i*

of the transition or sequence of transitions. For example:

*<sup>k</sup>* is the considered transition or the transition succession and *P*<sup>1</sup> is the

0

þ 1, *if a*ð Þ *<sup>i</sup>* � *n* modm 6¼ 0

*<sup>i</sup>* or in *σ* corresponding *Pj* position, the number of tokens are in their

*<sup>i</sup>* or *σ* to the number of the arrows in the state:

*i* � � � � .

*j*

*<sup>i</sup>* in the next move will get the following

, which will mark the number of implementation

� �. In the opposite case,

*<sup>i</sup>* as *α* and go to the second step again.

*<sup>i</sup>* does not

*<sup>i</sup>* transition, then we will leave *ai* to

*ai* ¼

*Numerical Modeling and Computer Simulation*

$$\begin{aligned} S(\mathbf{y}) &= \sum\_{j=i\_1}^{i\_k} \mathbf{z}\_j(t), \text{ where } \mathbf{z}\_j(t) = \boldsymbol{\pi}(P\_j, I(t\_k)), \forall t\_k \in T^\*. \\\ \mathbf{t}\_i(\mathbf{y}) &= \sum\_{i=1}^k b'\_i, \text{ where } \boldsymbol{k} = \left| T^\* \left( \mathbf{y}\_1 \right) \right|. \end{aligned}$$

In this number some *b*<sup>0</sup> *i* s are equal to 1. Without breaking the sense we will suppose that the first *d*<sup>0</sup> number of *b*<sup>0</sup> *i* s is equal to 1. We will get

$$t\_s(y\_1) = d' + \sum\_{i=d'+1}^k b'\_i = d' + \sum\_{i=d'+1}^k \left(\mu[\mathbf{x}]\_l + m' \cdot b'\right)^i$$

We have *ts y*<sup>2</sup> � � <sup>¼</sup> <sup>P</sup>*<sup>k</sup>*<sup>0</sup> *<sup>i</sup>*¼<sup>1</sup>*b*<sup>00</sup> *<sup>i</sup>* , where *<sup>k</sup>*<sup>0</sup> <sup>¼</sup> *<sup>T</sup>*<sup>∗</sup> *<sup>y</sup>*<sup>2</sup> � � � � � �. Suppose for *b*<sup>00</sup> *<sup>i</sup>* s, number of *d*<sup>00</sup> is equal to 1. Moreover *d*<sup>00</sup> ≤*d*<sup>0</sup> , as *S y*<sup>1</sup> � �<*S y*<sup>2</sup> � �,

$$\begin{aligned} t\_\circ(\mathsf{y}\_2) &= d'' + \sum\_{\substack{i=d''+1\\ \Rightarrow t\_\circ(\mathsf{y}\_1) \leq t\_\circ(\mathsf{y}\_2)}}^{k'} \left( \mu[\mathsf{x}]\_l + n\_i'' \cdot b'' \right) \geq d' + \sum\_{i=d'+1}^k \left( \mu[\mathsf{x}]\_l + n\_i' \cdot b' \right) = t\_\circ(\mathsf{y}\_1),\\ &\Rightarrow t\_\circ(\mathsf{y}\_1) \leq t\_\circ(\mathsf{y}\_2) \end{aligned}$$

The lemma is proven.

**Lemma 2.** Suppose *y*<sup>1</sup> and *y*<sup>2</sup> are covering peaks. There is *S y*<sup>1</sup> � � <sup>¼</sup> *S y*<sup>2</sup> � �& *T* <sup>∗</sup> 1 � � � � < *T* <sup>∗</sup> 2 � � � � : in this case *ts y*<sup>1</sup> � �<*ts <sup>y</sup>*<sup>2</sup> � �.

Proof:

$$\begin{aligned} t\_\mathbf{t}(\mathbf{y}\_1) &= d' + \sum\_{i=d'+1}^k b'\_i = d + \sum\_{i=d'+1}^k \left( \mu[\mathbf{x}]\_l + \mathbf{n}'\_i \cdot \mathbf{b}' \right)\_{d'$$

The lemma is proven.

**Theorem.** Through the abovementioned number of covering state, transition algorithm is in its minimal state.

Proof: Let *y* be the covering peak in our algorithm and *t* 0 1, ⋯, *t* 0 *<sup>k</sup>* the succession of transitions. It must be shown that the number of *t* 0 1, ⋯, *t* 0 *<sup>k</sup>* move is in minimal state. For this we need to show that ∃ �*y*<sup>0</sup> covering peak has less number of transitions than the number of *t* 0 1, ⋯, *t* 0 *k*.

Let us consider two cases:

1.*y* 6¼ *y*<sup>0</sup>

Suppose the transition number of *y*<sup>0</sup> is less than *t* 0 1, ⋯, *t* 0 *<sup>k</sup>* implementation number. According to the algorithm: *S y*ð Þ<*S y*<sup>0</sup> ð Þ or

$$\mathcal{S}(\mathcal{y}) = \mathcal{S}(\mathcal{y'}) \mathcal{S} \left| T\_1^\* \right| < \left| T\_2^\* \right|.$$

If *S y*ð Þ< *S y*<sup>0</sup> ð Þ) according to Lemma 1: *ts*ð Þ*y* ≤*ts y*<sup>0</sup> ð Þ. We've come into a controversy.

If *S y*ð Þ¼ *S y*<sup>0</sup> ð Þ& *<sup>T</sup>*<sup>∗</sup> 1 � � � �<sup>&</sup>lt; *<sup>T</sup>* <sup>∗</sup> 2 � � � � ) according to Lemma 2: *ts*ð Þ*y* ≤ *ts y*<sup>0</sup> ð Þ. We've come into a controversy.

2.*y* ¼ *y*<sup>0</sup>

Suppose succession transitions of *y*<sup>0</sup> is *s*1, ⋯, *sr*. As the tree does not contain any cycle, *y* ¼ *y*<sup>0</sup> ) *t* 0 1, ⋯, *t* 0 *<sup>k</sup>* and *s*1, ⋯, *sr* are the same ) *ts*ð Þ*y* ≤*ts y*<sup>0</sup> ð Þ. The theorem is proven.

> The distribution of tokens, called marking, determines the state of the simulated system. The dynamic behavior of CPN is due to the triggering of a transition that transfers the system from one state to another. A transition is enabled if the associated arc expressions of all input arches can be evaluated as a multi-set, which is compatible with the current tokens in corresponding input places, and its guard is satisfied. After the transition is triggered, tokens are removed from the input places, respectively, by the specified expression of the arches of all incoming arches, and tokens are placed in the output places, respectively, by the specified expressions of

Let us suppose that there are two processes of producers and consumers [1, 9].

There is a distribution problem in the described system. To use the channel, the pairð Þ *P*1, C1 must have priority toward ð Þ *P*2, C2 in the sense of using the channel. This is described as follows: while the buffer is not empty, the channel cannot report data from the buffer to the consumer. It is impossible to solve this problem with the help of classical Petri nets, since it is permissible in nature. The proof of this fact is

To solve that problem, it is needed to extend Petri net's several properties in such

a manner that the proposed properties are headed toward the opportunity of

The CPN (**Figure 5**) is the model of the solved problem of priority usage

**3.1 The example of modeling consumers' process with CPN**

*The consumers' process with the common usage and buffer is an action.*

*The Possibilities of Modeling Petri Nets and Their Extensions*

*DOI: http://dx.doi.org/10.5772/intechopen.90275*

The following picture shows the process diagram (**Figure 4**).

the outgoing arches [3–6, 17].

**Figure 4.**

described in the literature [1].

**3.2 Declaration**

[17, 19].

**123**

Color E = {e};

Var ct:Control;

Color Control = {0;1}; Color S = product E\*Control;

checking the zero in Petri nets [13].

#### **2.3 Conclusion**

The proven theorem and research reveal some important features of Petri nets from the point of view of optimization, that is, if the idea of Petri nets is used in technical devices, then the idea of sequential transitions save resources and time.
