# A Finite State Method in the Performance Evaluation of the Bernoulli Serial Production Lines
## 3 Mathematical Modeling
### A Finite State Method
More specifically, the stationary distributions of the upstream elements are equal to:  
\[\left\langle P^{(e)}\right\rangle^{\mathrm{T}}=\left\langle\ P_{0}^{(e)}\quad P_ {1}^{(e)}\quad P_{2}^{(e)}\quad\ldots\quad P_{N_{e}}^{(e)}\ \right\rangle^{\mathrm{T}}=\left\langle P_{i_{e}}^{(e)}\right\rangle^{ \mathrm{T}}, \tag{2}\]  
where \(i_{e}=0,\,1,\,2,\,\ldots,\,N_{e}\) stands for state on the buffer of the element \(e=1,\,2,\,\ldots,\,m-1\). The pertaining probabilities can be determined using the analytical solution in the case of a line composed of two machines and one buffer [2]:  
\[P_{0}^{(e)}=\frac{(1-p_{e})(1-\alpha_{e})}{1-\frac{p_{e}}{p_{m}}\alpha_{e}^{N_ {e}}},\ \alpha_{e}=\frac{p_{e}(1-p_{m})}{p_{m}(1-p_{e})},\ P_{i_{e}}^{(e)}=\frac{ \alpha_{e}^{i_{e}}}{1-p_{m}}P_{0}^{(e)}. \tag{3}\]  
Similarly, the stationary distributions of the downstream elements can be determined using:  
\[P_{0}^{(e)}=\frac{(1-p_{m})(1-\alpha_{e})}{1-\frac{p_{m}}{p_{e+1}}\alpha_{e}^{ N_{e}}},\ \alpha_{e}=\frac{p_{m}(1-p_{e+1})}{p_{e+1}(1-p_{m})},\ P_{i_{e}}^{(e)}=\frac{ \alpha_{e}^{i_{e}}}{1-p_{e+1}}P_{0}^{(e)}, \tag{4}\]  
where \(e=m,\,m+1,\,\ldots,\,M-1\). The eigenvector representation of the stationary distribution is the same as in the upstream case, Equation (2). The application of Equations (2)-(4) results in a total number of \(M-1\) stationary distributions (elementary eigenvectors) that can be used to formulate the stationary distribution (eigenvector) of the complete serial line covering all of the considered system states without loss of generality. The application of the proportional relationships within the state space of the considered transition system yields a general component of the stationary distribution (i.e., the probability that buffers of the system are in state \(i_{1},\,i_{2},\,\ldots,\,i_{M-1}\)), \(P_{i_{1}i_{2}i_{3}\ldots i_{M-1}}\), expressed as:  
\[P_{i_{1}i_{2}i_{3}\ldots i_{M-1}}=P_{000\ldots 0}\frac{P_{i_{1}}^{(1)}}{P_{0}^{( 1)}}\frac{P_{i_{2}}^{(2)}}{P_{0}^{(2)}}\frac{P_{i_{3}}^{(3)}}{P_{0}^{(3)}} \cdots\frac{P_{i_{M-1}}^{(M-1)}}{P_{0}^{(M-1)}}=P_{000\ldots 0}\prod_{e=1}^{M-1} \frac{P_{i_{e}}^{(e)}}{P_{0}^{(e)}}. \tag{5}\]  
The unknown \(P_{000\ldots 0}\) state of the system (all buffers are empty) from Equation (5) can be determined using the well-known condition \(\sum\limits_{i_{1}=0}^{N_{1}}\sum\limits_{i_{2}=0}^{N_{2}}\sum\limits_{i_{3}= 0}^{N_{3}}\cdots\sum\limits_{i_{M-1}=0}^{N_{M-1}}P_{i_{1}i_{2}i_{3}\ldots i_{M -1}}=1\), i.e., it follows that:  
\[P_{000\ldots 0}=\frac{1}{\sum\limits_{i_{1}=0}^{N_{1}}\frac{p_{i_{1}}^{(1)}}{P_{0}^ {(1)}}\cdot\sum\limits_{i_{2}=0}^{N_{2}}\frac{p_{i_{2}}^{(2)}}{P_{0}^{(2)}} \cdot\sum\limits_{i_{3}=0}^{N_{3}}\frac{p_{i_{3}}^{(3)}}{P_{0}^{(3)}}\cdot \ldots\sum\limits_{i_{M-1}=0}^{N_{M-1}}\frac{p_{i_{M-1}}^{(M-1)}}{P_{0}^{(M-1)}}}. \tag{6}\]  
Figure 3: A general case of the serial [PERSON] production line and finite state elements (authors’ work). Equation (6) can be modified by extracting \(1/P_{0}^{(e)}\) in the denominator yielding:  
\[P_{000\ldots 0}=\frac{1}{\frac{1}{P_{0}^{(1)}}\sum\limits_{i_{l}=0}^{N_{1}}P_{i_{l}}^{ (1)}\cdot\frac{1}{P_{0}^{(2)}}\sum\limits_{i_{2}=0}^{N_{2}}P_{i_{2}}^{(2)}\cdot \frac{1}{P_{0}^{(3)}}\sum\limits_{i_{3}=0}^{N_{3}}P_{i_{3}}^{(3)}\cdot\ldots \cdot\frac{1}{P_{0}^{(M-1)}}\sum\limits_{i_{M-1}=0}^{N_{M-1}}P_{i_{M-1}}^{(M-1) }}. \tag{7}\]  
Since \(\sum\limits_{i_{e}}^{N_{e}}P_{i_{e}}^{(e)}=1\), the Equation (7) is simplified into the form:  
\[P_{000\ldots 0}=P_{0}^{(1)}P_{0}^{(2)}P_{0}^{(3)}\ldots P_{0}^{(M-1)}=\prod \limits_{e=1}^{M-1}P_{0}^{(e)}. \tag{8}\]