# Reliability and MTTF, what's going on?

1. Dec 7, 2013

### oneamp

I am reading a reference that says, with $f(x)$ being the failure probability density function, reliability of a system with that PDF can be expressed as $1 - \int_0^t f(x) dx$. For a given t, this gives a number. Next, the same reference says that the mean time to failure of the system, $MTTF = \int_0^\infty R(t) dt$. How is this possible? R(t) returns a number, integrating to infinity over a constant returns infinity...

I guess I am missing some concept.

Thanks

2. Dec 7, 2013

### AlephZero

$R(t)$ is not "a number", it is a function of $t$. $R(t)$ will decrease to $0$ as $t$ goes to infinity, so $\int_0^\infty R(t)\,dt$ can be finite.

For example if $f(x) = ae^{-ax}$ for some constant $a$

$R(t) = 1 - (1/a)(ae^{-at} - a) = e^{-at}$

and the MTTF = $1/a$.

3. Dec 7, 2013

Thanks

4. Dec 7, 2013

### oneamp

Also I notice that a lot of sites online show examples with the mean=0. Since it's a PDF, we want only zero onward. So why are there so many examples with mean=0?

5. Dec 9, 2013

### MrAnchovy

MTTF = 0 implies that the component is already broken: you must be misinterpreting the data - perhaps 0 means that there is no data available for the component? Can you link to an example?