2
$\begingroup$

A system consists of four components that are similar and work independently, say $A,B,C,D$. To operate properly, it is necessary that $(A$ and $B)$ or $(C$ and $D)$ are functioning properly.

Define $T$ as the failure time of the complete system, and $T_{k}$ as the failure time of component $k \in \{A,B,C,D\}$, and suppose that $T_{k}$ ~ $\exp{(\alpha)}$.

Show that $P(T < t) =(1-e^{-2\alpha t})^{2}$

My ideas:

Since we have been given the distribution of $T_{k}'$s, we will have to write $T$ in terms of $T_{A},T_{B},T_{C},T_{D}$

From @antkam's idea, I've got to the following:

Assuming there are only $T_{A},T_{B}$, it follows $T=\min{T_{A},T_{B}}$

$P(T< t)=P(\min{T_{A},T_{B}}<t)=1-P(\min{T_{A},T_{B}}\geq t)=1-P(T_{A}\geq t,T_{B}\geq t)=1-P(T_{A}\geq t)^{2}=1-(\exp(-\alpha t))^{2}=1-\exp(-2\alpha t)$

$\endgroup$
  • 1
    $\begingroup$ Hint: consider just the (A and B) part of the system. This fails at time $\min(T_A, T_B)$, i.e. the minimum of two exponetial random variables. Do you know how to calculate that? If not, see: en.wikipedia.org/wiki/… $\endgroup$ – antkam Feb 11 at 18:11
  • 1
    $\begingroup$ oh, you have multiple misunderstandings then... first, when a component fails, it does NOT recover. it remains failed for all time afterwards. so e.g. if A fails at $t=3$, it remains failed for all $t \ge 3$. the exponential $exp(\alpha)$ describes that one moment when A starts to fail. second, if A fails $t \ge 3$ and B fails $t \ge 5$, then the (A and B) system has failed at time $t \ge 3$ because it requires both A and B to function. $\endgroup$ – antkam Feb 12 at 16:20
  • 1
    $\begingroup$ so lets solve the subsystem E = (A and B) first. we have $T_E = \min (T_A, T_B)$ because whichever of A/B fails earlier, that's when E fails. (and remember, nobody ever recovers.) now $T_A, T_B$ are both exponential $exp(\alpha)$. look into my earlier wiki link to find the distribution of $T_E$. can you answer the question: what is $P(T_E < t)$? $\endgroup$ – antkam Feb 12 at 16:22
  • 1
    $\begingroup$ so you're almost done... F = (C and D) has the same $P(T_F < t) = 1 - exp(-2\alpha t)$. Now back to the whole system of (E or F), what is $P(T < t)$? $\endgroup$ – antkam Feb 12 at 16:52
  • 1
    $\begingroup$ Thanks for your help, I think my confusion simply stemmed from the fact that $T$ shows time at which the system fails rather than time duration of the system failure $\endgroup$ – SABOY Feb 12 at 17:02

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.