Question 17.6: A B7XX plane has two similar computers onboard for flight co......
A B7XX plane has two similar computers onboard for flight control functions: one that is operational and the second as an active standby. The time to failure for each computer follows an exponential distribution with an MTBF of 4000 hours.
(a) Find the reliability of the computer system (consisting of both computers) for 800 hours when the switching is perfect and the second computer is instantaneously switched on when the first computer fails. Also find the MTBF of the computer system.
(b) Find the MTBF of the computer system when the switching and sensing unit is not perfect and the switching mechanism has a reliability of 0.98 when it is required to function.
(c) Find the reliability of the computer system for 800 hours when the switching mechanism is not perfect and is dynamic. The time to failure for the switching mechanism also has exponential distribution with MTBF of 12,000 hours.
(a) We have, using Equation 17.28,
\theta=4000 \\ \lambda=\frac{1}{4000}=0.00025 \\ R_S(t)=e^{-\lambda t}(1+\lambda t)=e^{-0.00025 \times 800}(1+0.00025 \times 800)=0.982477 \\ \operatorname{MTBF}=\int_0^{\infty} e^{-\lambda t}(1+\lambda t) d t=\int_0^{\infty} e^{-\lambda t} d t+\int_0^{\infty} \lambda t e^{-\lambda t} d t \\ =\frac{1}{\lambda}+\frac{1}{\lambda}=\frac{2}{\lambda}=8000 \text { hours. }R_S(t)=e^{-\lambda t}(1+\lambda t) (equation 17.28)
(b) We have
R_S(t)=e^{-\lambda t}+p_{S W}\left(\lambda t e^{-\lambda t}\right) \\ \mathrm{MTBF}=\int_0^{\infty} R_S(t) d t=\frac{1}{\lambda}+p_{S W} \frac{1}{\lambda}=4000+0.98 * 4000=7920 \text { hours. }(c) We have
\lambda=\frac{1}{4000} \quad \lambda_{S W}=\frac{1}{12,000} \\ R_S(t)=e^{-\lambda t}\left[1+\frac{\lambda}{\lambda_{S W}}\left(1-e^{-\lambda_{S W} t}\right)\right] \\ R_S(800)=e^{-0.00025 \times 800}\left[1+3\left(1-e^{-800 / 12000}\right)\right]=0.977138.