Supervisory control to maximize mean time to failure in discrete event systems

Abstract

In this paper, we investigate the use of supervisory control to maximize mean time to failure in a discrete event system framework. A complex engineering system is modeled as a discrete event system. Some of the states of the system are essential to the functionality of the system and are called required states. Some other states represent failures in the system and are called failure states. The control objective is to maximize the mean time to failure (MTTF) while allowing the system to visit all required states. The control is achieved by a supervisor that disables some controllable events based on monitoring the observable events as in classical supervisory control. To design such a supervisor, the MTTF of a supervised system is calculated by converting a discrete event system into a Markov chain having the same MTTF. Based on MTTF, two algorithms are developed that together allow us to design an optimal supervisor. The theoretical results are applied to power systems by investigating the maintenance management of equipment such as transformers.

Appendix: Proof of Theorem 3

Start from the derivative of $LP_{i}(s)$ in Eq. (23), let us take the second derivative of $LP_{i}(s)$:

$$\begin{aligned} \displaystyle \frac{d^{2}LP_{i}(s)}{ds^{2}} =&- \sum \limits _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} E[(-\rho _{i})^{2} e^{-\rho _{i} s}] LP_{j}(s) - 2 \sum \limits _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} E[-\rho _{i} e^{-\rho _{i} s}] \frac{dLP_{i}(s)}{ds} \\&- \sum \limits _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} E[e^{-\rho _{i} s}] \frac{d^{2}LP_{j}(s)}{ds^{2}} - \sum \limits _{j =n-l+1}^{n} \frac{\lambda _{ij}}{\lambda _{ii}} E[(-\rho _{i})^{2} e^{-\rho _{i} s}] . \end{aligned}$$

Since

$$\begin{aligned}{} & {} \lim _{s \rightarrow 0} \frac{d^{2}LP_{i}(s)}{ds^{2}} = \lim _{s \rightarrow 0} \int _{0}^{\infty } (-x)^{2}e^{-xs} h_{i}(x)dx = \int _{0}^{\infty } x^{2} h_{i}(x)dx = E[(\pi _{i})^{2}] \\{} & {} \lim _{s \rightarrow 0} \frac{dLP_{i}(s)}{ds} = \lim _{s \rightarrow 0} \int _{0}^{\infty } (-x)e^{-xs} h_{i}(x)dx = - \int _{0}^{\infty } x h_{i}(x)dx = - E[\pi _{i}] = - \eta _{i} \\{} & {} \lim _{s \rightarrow 0} LP_{j}(s) = \lim _{s \rightarrow 0} \int _{0}^{\infty } e^{-xs} h_{i}(x)dx = \int _{0}^{\infty } h_{i}(x)dx = 1 \\{} & {} \lim _{s \rightarrow 0} E[(-\rho _{i})^{2} e^{-\rho _{i} s}] = E[(\rho _{i})^{2}] = 2 (\frac{1}{\lambda _{ii}})^{2} \\{} & {} \lim _{s \rightarrow 0} E[-\rho _{i} e^{-\rho _{i} s}] = \frac{1}{\lambda _{ii}} \\{} & {} \lim _{s \rightarrow 0} E[e^{-\rho _{i} s}] = 1, \end{aligned}$$

we have, by letting $s \rightarrow 0$,

$$\begin{aligned} \displaystyle E[(\pi _{i})^{2}] =&- \sum \limits _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} 2 (\frac{1}{\lambda _{ii}})^{2} - 2 \sum \limits _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} \frac{1}{\lambda _{ii}} (- \eta _{i})\\&- \sum \limits _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} E[(\pi _{j})^{2}] - \sum \limits _{j =n-l+1}^{n} \frac{\lambda _{ij}}{\lambda _{ii}} 2 (\frac{1}{\lambda _{ii}})^{2} . \end{aligned}$$

In other words,

$$\begin{aligned} \sum _{j=1}^{n-l} \lambda _{ij} E\left[ (\pi _{j})^{2}\right]= & {} \lambda _{ii} E\left[ (\pi _{i})^{2}\right] + \sum _{j \not = i, j=1}^{n-l} \lambda _{ij} E\left[ (\pi _{j})^{2}\right] \\= & {} - 2 \sum _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}^{2}} - 2 \sum _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} (- \eta _{i}) - 2 \sum _{j =n-l+1}^{n} \frac{\lambda _{ij}}{\lambda _{ii}^{2}} \\= & {} - 2 \sum _{j \not = i, j=1}^{n} \frac{\lambda _{ij}}{\lambda _{ii}^{2}} + 2 \sum _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} \eta _{i} \\{} & {} \!\!\!\displaystyle (\text {by Eq.~(19)}) \\= & {} 2 \frac{1}{\lambda _{ii}} + 2 \sum _{j \not = i, j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} \eta _{i} \\= & {} 2 \frac{1}{\lambda _{ii}} - 2 \eta _{i} + 2 \sum _{ j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} \eta _{i} . \end{aligned}$$

Since

$$ {\textbf {A}} \left[ \begin{array}{c} \eta _{1} \\ \eta _{2} \\ ... \\ \eta _{n-l} \end{array} \right] = - \left[ \begin{array}{c} 1 \\ 1 \\ ... \\ 1 \end{array} \right] , $$

we have

$$ 2 \sum _{ j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} \eta _{i} = 2 \frac{1}{\lambda _{ii}} \sum _{ j=1}^{n-l} \lambda _{ij} \eta _{i} = -2 \frac{1}{\lambda _{ii}}. $$

Hence,

$$\begin{aligned} \sum _{j=1}^{n-l} \lambda _{ij} E[(\pi _{j})^{2}] = 2 \frac{1}{\lambda _{ii}} - 2 \eta _{i} + 2 \sum _{ j=1}^{n-l} \frac{\lambda _{ij}}{\lambda _{ii}} \eta _{i} = - 2 \eta _{i}. \end{aligned}$$

In the matrix form

$$\begin{aligned} {\textbf {A}} \left[ \begin{array}{c} E[{\pi _{1}^{2}}] \\ E[{\pi _{2}^{2}}] \\ ... \\ E[\pi _{n-1}^{2}] \end{array} \right] = - 2 \left[ \begin{array}{c} \eta _{1} \\ \eta _{2} \\ ... \\ \eta _{n-l} \end{array} \right] = 2 {\textbf {A}}^{-1} \left[ \begin{array}{c} 1 \\ 1 \\ ... \\ 1 \end{array} \right] . \end{aligned}$$

Therefore,

$$\begin{aligned} \left[ \begin{array}{c} E[{\pi _{1}^{2}}] \\ E[{\pi _{2}^{2}}] \\ ... \\ E[\pi _{n-1}^{2}] \end{array} \right] = 2 {\textbf {A}}^{-2} \left[ \begin{array}{c} 1 \\ 1 \\ ... \\ 1 \end{array} \right] . \end{aligned}$$