Suppose one has two machines (machine A and machine B) in sequence with time to machine break down exponentially distributed with rate parameters $\lambda_A$ and $\lambda_B$. Machine A and B have a machine repair time exponentially distributed with respectively rate parameters $\mu_A$ and $\mu_B$. I would like to write the machine repair time as one cumulative distribution function G(s).

I know the probability of machine A breaking down first is the minimum of two exponentially distributed variables with parameters $\lambda_A$ and $\lambda_B$, which gives another exponentially distributed variable with parameter $\lambda_A+\lambda_B$:

$\frac{\lambda_A}{\lambda_A+\lambda_B}$, therefore the probability of machine B breaking down first: $\frac{\lambda_B}{\lambda_A+\lambda_B}$.

Can i now say that the $G(s) = \frac{\lambda_A}{\lambda_A+\lambda_B}(1-\exp(-\mu_At))+ \frac{\lambda_B}{\lambda_A+\lambda_B}(1-\exp(-\mu_Bt))$

thank you for your answer in advance


You are seeking the CDF of the unconditional first machine repair time?

If so, your solution is correct if only one machine can be repaired at the same time. Otherwise there are possible events of the form "machine A breaks down first and starts repairing, but machine B breaks down and is repaired before the repairing of machine A has been completed".

*Your solution should have $s$ on the right hand side instead of $t$.

I have a feeling you are looking for something else here and need to state the question with a little more information.

  • $\begingroup$ Only one machine can be repaired at the same time and if one machine breaks down, the other machine will be shut down until the broken machine is repaired. So indeed i am seeking for the CDF of the unconditional first machine repair time. When machine A breaks down its CDF will be 1−exp(−μAs), when machine B breaks down its CDF will be 1−exp(−μBs). Then I thought of stating the CDF of the machine repair time as the G(s) formula above. ps. thank you for pointing out that the t on the RH side had to be s! $\endgroup$
    – Paul
    Feb 1 '15 at 13:37
  • $\begingroup$ In that case your solution is correct. As a side note, this distribution is called Hyper-exponential: en.wikipedia.org/wiki/Hyperexponential_distribution. $\endgroup$
    – QQQ
    Feb 1 '15 at 13:39
  • $\begingroup$ thank you for your time liron. $\endgroup$
    – Paul
    Feb 1 '15 at 15:58

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