Reliability and cost optimization of series–parallel system with metaheuristic algorithm

Abstract

This study looks into the numerous reliability aspects of a series–parallel system. The designed system is made up of the three subsystems A, B, and C that interconnected in a series–parallel manner. Subsystem B has two units, whereas subsystems A and C have a single unit. Here, the system reliability measures such as reliability, availability, and mean time to failure using Laplace transformation and Markov’s process are evaluated. This study deals with minimize cost while system having the maximum reliability as constraining from one of the metaheuristic algorithms i.e., Particle Swarm Optimization (PSO). Lastly, a numerical example and graphical representation has been shown that the proposed methods are effective and efficient for solving reliability measures and cost problems.

Appendix

$$\begin{gathered} p_{0} (t + \Delta t) = (1 - 2\delta_{B} \Delta t)\,(1 - \delta_{A} \Delta t)\,(1 - \delta_{C} \Delta t)p_{0} (t + \Delta t) \hfill \\ \quad + \int_{0}^{\infty } {p_{11} (q,t)} \,\gamma (q)\,dq\,\Delta t + \int_{0}^{\infty } {p_{10} (q,t)} \,\gamma (q)\,dq\,\Delta t \hfill \\ \quad + \int_{0}^{\infty } {p_{9} (q,t)} \,\gamma (q)\,dq\,\Delta t + \int_{0}^{\infty } {p_{8} (q,t)} \,\gamma (q)\,dq\,\Delta t \hfill \\ \end{gathered}$$

(34)

$$p_{2} (t + \Delta t) = (1 - \delta_{B} \Delta t)\,(1 - \delta_{C} \Delta t)\,p_{2} (t) + \,p_{0} (t)2\delta_{B} \,\Delta t$$

(35)

$$p_{1} (t + \Delta t) = (1 - \delta_{C} \Delta t)\,(1 - \delta_{A} \Delta t)\,p_{1} (t) + \,p_{2} (t)\delta_{B} \,\Delta t$$

(36)

$$p_{3} (t + \Delta t) = (1 - \delta_{C} \Delta t)\,(1 - 2\delta_{B} \Delta t)\,p_{3} (t) + \,p_{0} (t)\delta_{A} \,\Delta t$$

(37)

$$p_{4} (t + \Delta t) = (1 - \delta_{A} \Delta t)\,(1 - 2\delta_{B} \Delta t)\,p_{4} (t) + \,p_{0} (t)\delta_{C} \,\Delta t$$

(38)

$$p_{5} (t + \Delta t) = (1 - \delta_{B} \Delta t)\,\,p_{5} (t) + \,p_{4} (t)2\delta_{B} \,\Delta t + p_{2} \delta_{C} \Delta t$$

(39)

$$p_{6} (t + \Delta t) = (1 - \delta_{C} \Delta t)\,(1 - \delta_{B} \Delta t)\,p_{6} (t) + \,p_{3} (t)2\delta_{B} \,\Delta t + p_{2} \delta_{A} \Delta t$$

(40)

$$p_{7} (t + \Delta t) = (1 - \delta_{C} \Delta t)\,p_{7} (t) + \,p_{6} (t)\delta_{B} \,\Delta t + p_{1} \delta_{A} \Delta t$$

(41)

$$p_{9} (q + \Delta q,t + \Delta t) = (1 - \gamma \Delta t)\,\,p_{9} (t)$$

(42)

$$p_{8} (q + \Delta q,t + \Delta t) = (1 - \gamma \Delta t)\,\,p_{8} (t)$$

(43)

$$p_{10} (q + \Delta q,t + \Delta t) = (1 - \gamma \Delta t)\,\,p_{10} (t)$$

(44)

$$p_{11} (q + \Delta q,t + \Delta t) = (1 - \gamma \Delta t)\,\,p_{11} (t)$$

(45)

Boundary Conditions

$$p_{9} (0,t) = \delta_{A} \,p_{4} (t) + \delta_{C} \,p_{3} (t)$$

(46)

$$p_{8} (0,t) = \delta_{C} \,p_{2} (t) + \delta_{B} \,p_{5} (t)$$

(47)

$$p_{10} (0,t) = \delta_{A} \,p_{5} (t) + \delta_{C} \,p_{6} (t)$$

(48)

$$p_{11} (0,t) = \delta_{C} \,p_{7} (t)$$

(49)

Initial Condition.

Good Condition $p_{0} = 1$ and all are zero (0).

From Eq. (34)

$$\begin{gathered} p_{0} (t + \Delta t) = (1 - \delta_{C} \Delta t - \delta_{A} \Delta t + \delta_{A} \,\delta_{C} (\Delta t)^{2} - 2\delta_{B} \Delta t + 2\delta_{B} \,\delta_{C} (\Delta t)^{2} \hfill \\ + 2\delta_{B} \,\delta_{A} (\Delta t)^{2} - 2\delta_{B} \,\delta_{C} \,\delta_{A} (\Delta t)^{3} )p_{0} \hfill \\ + \int_{0}^{\infty } {p_{11} (q,t)} \,\gamma (q)\,dq\,\Delta t + \int_{0}^{\infty } {p_{10} (q,t)} \,\gamma (q)\,dq\,\Delta t \hfill \\ + \int_{0}^{\infty } {p_{9} (q,t)} \,\gamma (q)\,dq\,\Delta t + \int_{0}^{\infty } {p_{8} (q,t)} \,\gamma (q)\,dq\,\Delta t \hfill \\ \end{gathered}$$

Neglecting higher order terms of $\Delta t$ since $\Delta t$ is very small. So, authors get-

$$\begin{aligned} p_{0} (t + \Delta t) = & (1 - \delta _{C} \Delta t - \delta _{A} \Delta t - 2\delta _{B} \Delta t)p_{0} (t) \\ & + \int_{0}^{\infty } {p_{{11}} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t + \int_{0}^{\infty } {p_{{10}} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t \\ & + \int_{0}^{\infty } {p_{9} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t + \int_{0}^{\infty } {p_{8} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t \\ \end{aligned}$$

$$\begin{aligned} p_{0} (t + \Delta t) = & p_{0} (t) + ( - \delta _{C} \Delta t - \delta _{A} \Delta t - 2\delta _{B} \Delta t)p_{0} (t) \\ & + \int_{0}^{\infty } {p_{{11}} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t + \int_{0}^{\infty } {p_{{10}} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t \\ & + \int_{0}^{\infty } {p_{9} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t + \int_{0}^{\infty } {p_{8} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t \\ \end{aligned}$$

$$\begin{aligned} p_{0} (t + \Delta t) - p_{0} = & ( - \delta _{C} \Delta t - \delta _{A} \Delta t - 2\delta _{B} \Delta t)p_{0} \\ & + \int_{0}^{\infty } {p_{{11}} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t + \int_{0}^{\infty } {p_{{10}} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t \\ & + \int_{0}^{\infty } {p_{9} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t + \int_{0}^{\infty } {p_{8} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \Delta t \\ \end{aligned}$$

Divide by $\Delta t$ both side

$$\begin{aligned} \frac{{p_{0} (t + \Delta t) - p_{0} (t)}}{{\Delta t}} = & ( - \delta _{C} - \delta _{A} - 2\delta _{B} )p_{0} (t) \\ & + \int_{0}^{\infty } {p_{{11}} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} + \int_{0}^{\infty } {p_{{10}} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \\ & + \int_{0}^{\infty } {p_{9} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} + \int_{0}^{\infty } {p_{8} (q,t)} {\mkern 1mu} \gamma (q){\mkern 1mu} dq{\mkern 1mu} \\ \end{aligned}$$

$$\begin{gathered} \left( {\frac{\partial }{\partial t} + \delta_{C} + \delta_{A} + 2\delta_{B} } \right)\,p_{0} (t) = \int_{0}^{\infty } {p_{11} (q,t)} \,\gamma (q)\,dq\, + \int_{0}^{\infty } {p_{10} (q,t)} \,\gamma (q)\,dq\, \hfill \\ \quad + \int_{0}^{\infty } {p_{9} (q,t)} \,\gamma (q)\,dq\, + \int_{0}^{\infty } {p_{8} (q,t)} \,\gamma (q)\,dq\, \hfill \\ \end{gathered}$$

(50)

Similarly, from Eq. (35) to (41)

$$\left( {\frac{\partial }{\partial t} + \delta_{C} + \delta_{A} + \delta_{B} } \right)\,p_{2} = \,2\delta_{B} \,p_{0}$$

(51)

$$\left( {\frac{\partial }{\partial t} + \delta_{C} + \delta_{A} } \right)\,p_{2} = \,\delta_{B} \,p_{1}$$

(52)

$$\left( {\frac{\partial }{\partial t} + \delta_{C} + 2\delta_{B} } \right)\,p_{3} = \,\delta_{A} \,p_{0}$$

(53)

$$\left( {\frac{\partial }{\partial t} + \delta_{A} + 2\delta_{B} } \right)\,p_{4} = \,\delta_{C} \,p_{0}$$

(54)

$$\left( {\frac{\partial }{\partial t} + \delta_{B} + \delta_{A} } \right)\,p_{5} \, = \,\delta_{C} \,p_{1} + 2\delta_{B} \,p_{4}$$

(55)

$$\left( {\frac{\partial }{\partial t} + \delta_{B} + \delta_{C} } \right)\,p_{6} \, = \,\delta_{A} \,p_{1} + 2\delta_{B} \,p_{3}$$

(56)

$$\left( {\frac{\partial }{\partial t} + \delta_{C} } \right)\,p_{7} \, = \,\delta_{A} \,p_{2} + \delta_{B} \,p_{6}$$

(57)

From Eq. (42), by a Taylor expansion

$$\begin{gathered} p_{9} (q,t) + \Delta q\frac{\partial }{\partial q}p_{9} (q,t) + \Delta t\frac{\partial }{\partial q}p_{9} (q,t) + - - - - - = p_{9} (q,t) \hfill \\ \quad + ( - \gamma \,\Delta t)\,p_{9} (q,t) + \delta_{C} \,\Delta t\,p_{9} (q,t) + \delta_{A} \,\Delta t\,p_{9} (q,t) \hfill \\ \end{gathered}$$

As, $\Delta q \approx \Delta t$, then we get

$$\left( {\frac{\partial }{\partial q} + \frac{\partial }{\partial t} + \gamma } \right)p_{9} (t) = \,0$$

(58)

Similarly, from Eq. (43) to (45)

$$\left( {\frac{\partial }{\partial q} + \frac{\partial }{\partial t} + \gamma } \right)p_{8} (t) = \,0$$

(59)

$$\left( {\frac{\partial }{\partial q} + \frac{\partial }{\partial t} + \gamma } \right)p_{10} (t) = \,0$$

(60)

$$\left( {\frac{\partial }{\partial q} + \frac{\partial }{\partial t} + \gamma } \right)p_{11} (t) = \,0$$

(61)