2
$\begingroup$

The system I am trying to model is as follows:

  • A program takes time $T$ to run
  • The computer crashes on average every $\text{MTBF}$ time units
  • The program is repeatedly run until it completes without a crash
  • What is the expected time to completion?

I tried to calculate this by working out how many times the program is expected to fail before it completes:

If failures are modelled by a Poisson process of rate $\frac{1}{\text{MTBF}}$ then the probability of no failure occurring during the program is $e^{-\frac{T}{\text{MTBF}}}$

Therefore the number of failures until a success is modelled by a Geometric distribution and hence the expected number of failures is $e^{\frac{T}{\text{MTBF}}}-1$ (by standard result that the expectation $=\frac{1-p}{p}$)

Therefore the time to completion is given by $(e^{\frac{T}{\text{MTBF}}}-1)*t+T$ where $t$ is the expected length of each failed run. But I'm not sure how to calculate this value $t$? Is this just going to be the expected time between events in the Poisson process conditioned on this time being $<T$?

Is there a more direct way of working out this expected time until a gap $T$ between two Poisson process events?

$\endgroup$
2
$\begingroup$

Let $X$ denote the time of the first crash and $C$ the time to completion, thus $X$ is exponentially distributed with parameter $a=1/\textrm{MTBF}$, and $C=T$ if $X>T$ while $C=X+C'$ if $X<T$, where $C'$ is distributed as $C$ and independent of $X$. Thus, $$E(C)=E(T\mathbf 1_{X>T})+E((X+C')\mathbf 1_{X<T})=TP(X>T)+E(X\mathbf 1_{X<T})+E(C)P(X<T)$$ which implies that the expected time to completion $E(C)$ is $$E(C)=\frac{TP(X>T)+E(X\mathbf 1_{X<T})}{P(X>T)}=T+\frac{E(X\mathbf 1_{X<T})}{P(X>T)}$$ Now, $P(X>T)=e^{-aT}$ and $$E(X\mathbf 1_{X<T})=\int_0^Tx\,ae^{-ax}dx=\left[-(x+a^{-1})e^{-ax}\right]_0^T=a^{-1}-(T+a^{-1})e^{-aT}$$ which yields $$E(C)=\frac{e^{aT}-1}a$$ that is, $$E(C)=\textrm{MTBF}\cdot(e^{T/\textrm{MTBF}}-1)$$

| cite | improve this answer | |
$\endgroup$
-2
$\begingroup$

t is the expected length of each failed run

You already have the value of t, you state it in bullet #2 in your premise.

The computer crashes on average every MTBF time units

The expected length of each failed run equals how long (on average) it runs before it crashes. So, t = MTBF.

Is this just going to be the expected time between events in the Poisson process conditioned on this time being less than T?

No.

In your Poisson distribution, you use λ = T/MTBF (a time rate).

The expected number of crashes during time T therefore is T/MTBF.

I was initially tempted to calculate the "inter crash rate" of 1/λ (i.e. MTBF/T) but this doesn't apply in your case.

Since you start over the process after a crash, you "re-set" the timer on the Poison process.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.