# Evaluation of Machinery Readiness Using Semi-Markov Processes
## 4 Case Study  
### Estimation of the Semi-Markov Model Parameters  
First of all, based on the actual relationship between the states defined in Figure 1, the transition probability matrix was calculated. If \(n_{i}\) denotes the number of instants of the system waiting in state \(s_{i}\), while \(n_{ij}\) denotes the number of state transitions from state \(s_{i}\) to state \(s_{j}\), then the transition estimator from state \(s_{i}\) to state \(s_{j}\) shall be determined from the formula:  
\[\hat{p}_{ij}=\frac{n_{ij}}{n_{i}}. \tag{17}\]  
The distribution of probability of changes of the distinguished operating states (in one step), assuming that each graph arch of the exploitation process representation (Equations (2) and (10)) corresponds to the value of probability \(p_{ij}\), is presented in Table 4.  
For the process studied, some limits exist:  
\[\underset{n\rightarrow\infty}{\lim}p_{ij}(n)=\pi_{j}\;i,\;j=1,2,\ldots,6\;\;i \
eq j, \tag{18}\]  
where \(p_{ij}(n)\)--probability of transition from state \(S_{i}\) to state \(S_{j}\) in \(n\) steps. **Definition 2**. : _A probability distribution [25]_:  
\[\pi=\Big{[}\pi_{j}:\;j\in S\Big{]}, \tag{19}\]  
Satisfying a system of linear equations:  
\[\sum\
olimits_{i\in S}\pi_{i}\cdot p_{ij}=\pi_{j},\quad j\in S, \tag{20}\]  
and  
\[\sum\
olimits_{i\in S}\pi_{i}=1, \tag{21}\]  
is said to be a stationary probability distribution of the Markov chain witch transition matrix \(P=\Big{[}p_{ij}:i,\;j\in S\Big{]}\). In matrix form, the Equation (20) takes the following form:  
\[\Pi^{T}P=\Pi^{T}\leftrightarrow\Big{(}\mathbb{P}^{T}-I\Big{)}\cdot\Pi=0, \tag{22}\] \begin{table}
\begin{tabular}{c c c c c c c} \hline \(p_{ij}\) & \(\mathbf{S_{1}}\) & \(\mathbf{S_{2}}\) & \(\mathbf{S_{3}}\) & \(\mathbf{S_{4}}\) & \(\mathbf{S_{5}}\) & \(\mathbf{S_{6}}\) \\ \hline \(\mathrm{S_{1}}\) & 0 & 0.141 & 0.143 & 0.716 & 0 & 0 \\ \(\mathrm{S_{2}}\) & 0.029 & 0 & 0.117 & 0.854 & 0 & 0 \\ \(\mathrm{S_{3}}\) & 0.043 & 0 & 0 & 0.957 & 0 & 0 \\ \(\mathrm{S_{4}}\) & 0.003 & 0 & 0 & 0.997 & 0 \\ \(\mathrm{S_{5}}\) & 0.676 & 0.004 & 0.001 & 0 & 0 & 0.319 \\ \(\mathrm{S_{6}}\) & 0.982 & 0 & 0.018 & 0 & 0 & 0 \\ \hline \end{tabular}
\end{table} 
Table 4: The \(p_{ij}\) inter-states transition probability matrix. Stationary probabilities \(\pi_{j}\) were calculated in accordance with Equation (22). For the process studied, for the 6-state model, the determination of the stationary probabilities \(\pi_{j}\) required solving the following matrix equation:  
\[\left[ \begin{array}{c}\pi_{1}\\ \pi_{2}\\ \pi_{3}\\ \pi_{4}\\ \pi_{5}\\ \pi_{6}\end{array} \right]^{\mathrm{T}}\left[ \begin{array}{cccccc}0&p_{12}&p_ {13}&p_{14}&0&0\\ p_{21}&0&p_{23}&p_{24}&0&0\\ p_{31}&0&0&p_{34}&0&0\\ p_{41}&0&0&0&p_{45}&0\\ p_{51}&p_{52}&p_{53}&0&0&p_{56}\\ p_{61}&0&p_{63}&0&0&0\end{array} \right]=\left[ \begin{array}{c}\pi_{1}\\ \pi_{2}\\ \pi_{3}\\ \pi_{4}\\ \pi_{5}\\ \pi_{6}\end{array} \right]^{\mathrm{T}}, \tag{23}\]  
with the normalization condition:  
\[\pi_{1}+\pi_{2}+\pi_{3}+\pi_{4}+\pi_{5}+\pi_{6}=1, \tag{24}\]  
which is equivalent to the following system of equations:  
\[\left\{ \begin{array}{l}\pi_{2}\text{:}p_{21}+\pi_{3}\text{:}p_{31}+\pi_{4} \text{:}p_{41}+\pi_{5}\text{:}p_{51}+\pi_{6}\text{:}p_{61}=\pi_{1}\\ \pi_{1}\text{:}p_{12}+\pi_{5}\text{:}p_{52}=\pi_{2}\\ \pi_{1}\text{:}p_{13}+\pi_{2}\text{:}p_{23}+\pi_{5}\text{:}p_{53}+\pi_{6}\text {:}p_{63}=\pi_{3}\\ \pi_{1}\text{:}p_{14}+\pi_{2}\text{:}p_{24}+\pi_{3}\text{:}p_{34}=\pi_{4}\\ \pi_{4}\text{:}p_{45}=\pi_{5}\\ \pi_{5}\text{:}p_{56}=\pi_{6}\\ \pi_{1}+\pi_{2}+\pi_{3}+\pi_{4}+\pi_{5}+\pi_{6}=1\end{array} \right.\text{.} \tag{25}\]  
After substituting the figures we get:  
\[\left\{