Question 17.2: Two subsystems of a system functionally operate in series an......
Two subsystems of a system functionally operate in series and have the time to failure random variable with the pdfs given by
f_i(t)=\left(\frac{t}{\eta_i}\right) \exp \left[-\frac{t^2}{2 \eta_i}\right] \quad t \geq 0, \quad i=1,2 .where η_{i} is the parameter for the pdf for the ith subsystem. Time is measured in hours. We want to answer the following five parts.
(a) Find the system hazard function, h_{S}(t).
(b) Find the system reliability function, R_{S}(t).
(c) Find the pdf, f_{S}(t), for the time to failure for the system.
(d) If η_{1} = 300 hours and η_{2} = 400 hours, find R_{S}(20 hours).
(e) For the values in (d) find t* such that R_{S}(t*) = 0.90.
We can easily notice that f_i(t) is a Weibull distribution with
\beta=2, \eta=\left(2 \eta_i\right)^{1 / \beta}.
So, the reliability function and the hazard function for each subsystem are
R_i(t)=\exp \left[-\frac{t^2}{2 \eta_i}\right] \\ h_i(t)=\frac{f_i(t)}{R_i(t)}=\frac{t}{\eta_i}.
(a) Find the system hazard function, h_{S}(t). Using Equation 17.14, we have
h_S(t)=\sum_{i=1}^2 h_i(t)=\frac{t}{\eta_1}+\frac{t}{\eta_2} \equiv \frac{t}{\eta}h_S(t)=\sum_{i=1}^n h_i(t) (17.14)
where 1/η = 1/η_{1} + 1/η_{2} .
(b) Find the system reliability function, R_{s} (t).
From part (a), and using Equation 17.5, we have
R_S(t)=\prod_{i=1}^2 R_i(t) \\ =\exp \left[-\frac{t^2}{2 \eta_1}\right] \times \exp \left[-\frac{t^2}{2 \eta_2}\right] \\ =\exp \left[-\frac{t^2}{2}\left(\frac{1}{\eta_1}+\frac{1}{\eta_2}\right)\right]=\exp \left[-\frac{t^2}{2 \eta}\right].
R_S(t)=R_1(t) \cdot R_2(t) \cdots R_n(t)=\prod_{i=1}^n R_i(t) . (17.5)
(c) Find the system pdf, f_{S}(t).
f_S(t)=h_S(t) \cdot R_S(t)=\frac{t}{\eta} \exp \left[-\frac{t^2}{2 \eta}\right](d) If η_{1} = 300 hours and η_{1} = 400 hours, find R_{S} (20 hours). First, 1/η = 1/η_{1} + 1/η_{2} = 1/300 + 1/400, so η = 171.4 hours
Now,
R_S(20)=\exp \left[-\frac{t^2}{2 \eta}\right] \\ =\exp \left[-\frac{20}{2 \times 171.4}\right] \\ =0.3113 \text {. }(e) For the values in (d), find t* such that R_{s}(t*) = 0.90:
R_S\left(t^*\right)=\exp \left[-\frac{\left(t^*\right)^2}{2 \eta}\right]=0.9 \\ \frac{\left(t^*\right)^2}{2 \eta}=-\ln (0.9) \\ t^*=\sqrt{2 \eta(-\ln (0.9))}=\sqrt{2 * 1714 *(-\ln (0.9))} \\ t^*=6.01 \text { hours. }