Question 17.4: Consider an electronics system consisting of two parts with ......
Consider an electronics system consisting of two parts with constant failure rates as given below:
\lambda_{1} = 6.5 failures/10^{6} hours
\lambda_{2} = 26.0 failures/10^{6} hours.
Assume that failures are governed by a constant failure rate \lambda_{i} for the ith part. Determine:
(a) The system reliability for a 1000-hour mission
(b) The system MTTF
(c) The failure probability density function
(d) The system “failure rate.”
For a constant failure rate, the reliability R_{i} of the ith part has the form:
R_i(t)=e^{-\lambda_{i} t} .
For a parallel system:
R_S(t)=1-\prod_{i=1}^2\left(1-e^{-\lambda_i(t)}\right)=e^{-\lambda_1 t}+e^{-\lambda_2 t}-e^{-\left(\lambda_1+\lambda_2\right) t} .
The failure probability density function is:
f_S(t)=-\frac{d\left[R_S(t)\right]}{d t}=\lambda_1 e^{-\lambda_1 t}+\lambda_2 e^{-\lambda_2 t}-\left(\lambda_1+\lambda_2\right) e^{-\left(\lambda_1+\lambda_2\right) t}.
Substituting numbers in the equation for system reliability, we get the answer for part (a):
R_S(1000)=0.99352+0.97434 – 0.96802=0.99983.
The MTTF (part b) for the parallel system is
\operatorname{MTTF}_S=\int_0^{\infty} R_S(t) d t=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}-\frac{1}{\left(\lambda_1+\lambda_2\right)}=161,538 \text { hours. }The failure probability density function (part c) for the parallel system is
f_S(t)=-\frac{d\left[R_S(t)\right]}{d t} \\ =6.5 \times 10^{-6} e^{-6.5 \times 10^{-6} t}+26.0 \times 10^{-6} e^{-26.0 \times 10^{-6} t}-32.5 \times 10^{-6} e^{-32.5 \times 10^{-6} t}.
The system hazard rate for the parallel system is given by:
h_S(t)=\frac{f_S(t)}{R_S(t)}The system failure rate for the parallel system (part (d)) can be obtained by substituting the results in the equation stated above. We will find that h_S(t) is a function of time and is not constant over time.