Exponential reliability and system reliability

(a) Exponential law

$R(t) = e^{-\lambda t}$, MTBF $= 1/\lambda$, with $\lambda = 2.5\times10^{-4}\ \text{h}^{-1}$.

Reliability at 400 h

$$R(400) = e^{-2.5\times10^{-4}\times 400} = e^{-0.1} = 0.9048$$

MTBF

$$\text{MTBF} = \frac{1}{2.5\times10^{-4}} = 4000\ \text{h}$$

Time for $R = 0.95$

$$t = -\frac{\ln 0.95}{\lambda} = \frac{0.05129}{2.5\times10^{-4}} = 205.2\ \text{h}$$

Failures among 500 components in 1000 h

$$R(1000) = e^{-0.25} = 0.7788,\quad \text{failures} = 500(1-0.7788) = 110.6 \approx 111$$

(b) System reliability

        +--- B (0.90) ---+
 --A----|                |----D----
 (0.95) +--- C (0.90) ---+   (0.98)

Parallel pair: $R_{BC} = 1 - (1-0.90)(1-0.90) = 1 - 0.01 = 0.99$

Series system: $R_s = R_A \times R_{BC} \times R_D$

$$R_s = 0.95 \times 0.99 \times 0.98 = 0.9217$$

Note that without the redundancy in B and C the reliability would be $0.95\times0.90\times0.98 = 0.838$, so the parallel pair improves the system by about 8 percentage points. The weakest series element (A) now limits reliability.

Answer: (a) $R(400)=0.905$, MTBF $=4000$ h, $t_{0.95}=205$ h, about 111 failures; (b) $R_s = 0.922$.