Question 17.12: Consider a 4-out-of-5 system with the reliabilities of the f......
Consider a 4-out-of-5 system with the reliabilities of the five components as given below:
R_{1} = 0.90
R_{2} = 0.95
R_{3} = 0.85
R_{4} = 0.80
R_{5} = 0.75.
(a) Develop the structure function \phi(x) for this 4-out-of-5 system.
(b) Find the system reliability with the reliability values of the components given above
(a) There are five minimum paths for this system. Hence, we can develop the structure function using minimum paths (Eq. 17.54):
\phi(x)=1-\left(1-x_1 x_2 x_3 x_4\right)\left(1-x_1 x_2 x_3 x_5\right)\left(1-x_1 x_2 x_4 x_5\right) \times\left(1-x_1 x_3 x_4 x_5\right)\left(1-x_2 x_3 x_4 x_5\right),
\phi(x)=1-\prod_{j=1}^p\left[1-\alpha_j(x)\right] (17.54)
which can be simplified as
\begin{aligned} = & 1-\left(1-x_1 x_2 x_3 x_4-x_1 x_2 x_3 x_5+x_1 x_2 x_3 x_4 x_5\right) \\ & \left(1-x_1 x_2 x_4 x_5-x_1 x_3 x_4 x_5+x_1 x_2 x_3 x_4 x_5\right)\left(1-x_2 x_3 x_4 x_5\right) \\ = & 1-\left(1-x_1 x_2 x_3 x_4-x_1 x_2 x_3 x_5+x_1 x_2 x_3 x_4 x_5-x_1 x_2 x_4 x_5+x_1 x_2 x_3 x_4 x_5+x_1 x_2 x_3 x_4 x_5\right. \\ & -x_1 x_2 x_3 x_4 x_5-x_1 x_3 x_4 x_5+x_1 x_2 x_3 x_4 x_5+x_1 x_2 x_3 x_4 x_5-x_1 x_2 x_3 x_4 x_5 \\ & \left.+x_1 x_2 x_3 x_4 x_5-x_1 x_2 x_3 x_4 x_5-x_1 x_2 x_3 x_4 x_5+x_1 x_2 x_3 x_4 x_5\right)\left(1-x_2 x_3 x_4 x_5\right) \\ = & 1-\left(1-x_1 x_2 x_3 x_4-x_1 x_2 x_3 x_5-x_1 x_2 x_4 x_5-x_1 x_3 x_4 x_5+3 x_1 x_2 x_3 x_4 x_5\right) \times\left(1-x_2 x_3 x_4 x_5\right) \\ = & 1-\left(1-x_1 x_2 x_3 x_4-x_1 x_2 x_3 x_5-x_1 x_2 x_4 x_5-x_1 x_3 x_4 x_5\right. \\ & \left.+3 x_1 x_2 x_3 x_4 x_5-x_2 x_3 x_4 x_5+4 x_1 x_2 x_3 x_4 x_5-3 x_1 x_2 x_3 x_4 x_5\right) \\ = & x_1 x_2 x_3 x_4+x_1 x_2 x_3 x_5+x_1 x_3 x_4 x_5+x_2 x_3 x_4 x_5+x_1 x_2 x_4 x_5-4 x_1 x_2 x_3 x_4 x_5 \end{aligned}(b) Taking the expected value of the structure function, we can calculate the system reliability as
\begin{aligned} R_S= & R_1 R_2 R_3 R_4+R_1 R_2 R_3 R_5+R_1 R_3 R_4 R_5+R_2 R_3 R_4 R_5+R_1 R_2 R_4 R_5-4 R_1 R_2 R_3 R_4 R_5 \\ = & (0.90)(0.95)(0.85)(0.80)+(0.90)(0.95)(0.85)(0.75)+(0.90)(0.85)(0.80)(0.75) \\ & +(0.90)(0.85)(0.80)(0.75)+(0.90)(0.95)(0.85)(0.75) \\ & -4(0.90)(0.95)(0.85)(0.85)(0.75) \\ = & 0.5814+0.5451+0.4590+0.4845+0.5130-1.744 \\ = & 0.8388 \end{aligned}