**6.2 Tangible states and reachability graph creation**

Tangible states are those for timed transitions [21], since we used SPN so there are e tangible states with markings as shown in **Table 3**.

Based on the tangible states of the PN a reachability graph of the PN (**Figure 4**) can be obtained as shown in **Figure 5**.


#### **Table 2.**

*ISO throughput values.*

**Figure 4.** *PN model of ISO.*

*Reliability Analysis of Instrumentation and Control System: A Case Study of Nuclear Power… DOI: http://dx.doi.org/10.5772/intechopen.101099*


#### **Table 3.**

*ISO tangible states with markings of PN.*

#### **6.3 Markov chain model creation**

The MC model shown in **Figure 6** is obtained from the reachability graph of the PN shown in **Figure 4**.

With the help of *Q* which is transition probability matrix, the transition probability *Pij* of MC can be computed from SPN. For the transition matrix *Q*, transitionrate *qij* is the transition of one state to another states unit/per time, therefore we take the ratio of the transition *qij* and the transition rate of the states sum must be zero. The diagonal elements can be defined as:

*Nuclear Reactors - Spacecraft Propulsion, Research Reactors, and Reactor Analysis Topics*

$$q\_{ii} = -\sum\_{j \neq i} q \ddot{y} \tag{3}$$

It is clear that the system is no ergodic, therefore, *Pij* will be zero and defined as:

$$P\_{\vec{\eta}} = \begin{cases} \frac{q\_{\vec{\eta}}}{\sum\_{k \neq i} q\_{ik}}, \text{if } k \neq i \\\\ 0, \text{ otherwise} \end{cases} \tag{4}$$

*<sup>P</sup>* <sup>¼</sup> *<sup>I</sup>* � *<sup>d</sup>*�<sup>1</sup> *<sup>Q</sup> Q*, where *dQ* ¼ *dia Q*ð Þ *diagonal matrix of Q:* The transition matrix is given in Eq. (5) as follows:


**Figure 6.** *Markov chain.*

*Reliability Analysis of Instrumentation and Control System: A Case Study of Nuclear Power… DOI: http://dx.doi.org/10.5772/intechopen.101099*

Now we solve Eq. (5) to get the design metrics and it seriousness of the NPP as defined in Eq. (6). We solve the Eq. (6) then we get the following linear equations.


$$\sum\_{i=0}^{c} \mathcal{M}\_i = \mathbf{1} \tag{17}$$

## **6.4 Reliability analysis of proposed framework**

Let *pi* ð Þ*t* be the probability which component in state at time *t* is *i*. When components execute for *t* ! ∞ then probability leads to the stationary distribution. Then probability is defined as:

$$\Gamma\_p^-[p(M\_0), p(M\_1), p(M\_2), p(M\_3), p(M\_4), p(M\_5), p(M\_6), p(M\_7), p(M\_8), ]\tag{18}$$

$$\sum\_{i \in \mathcal{M}} p(i) = \mathbf{1} \tag{19}$$

$$Rel\_{ISO}^{est} = 1 - \sum\_{i \in \mathcal{G}} \mathcal{M}\_i \tag{20}$$

There is only one failure state *M*<sup>6</sup> in MC. Now we solve the linear equation Eqs. (7)-(16) and Eq. (17) using the standard method, we get steady-state probability of each state as follows:

*M*<sup>0</sup> ¼ 0*:*1282051, *M*<sup>1</sup> ¼ 0*:*1282051, *M*<sup>2</sup> ¼ 0*:*1282051, *M*<sup>3</sup> ¼ 0*:*1282051, *M*<sup>4</sup> ¼ 0*:*1282051, *M*<sup>5</sup> ¼ 0*:*1282051, *M*<sup>6</sup> ¼ 0.025641, *M*<sup>7</sup> ¼ 0.1025641, and *M*<sup>8</sup> ¼ 0*:*1025641

Hence the reliability of ISO is:

$$\text{Reliability} = \mathbf{1} - 0.02564\mathbf{1} = 0.974359.\tag{21}$$
