
Re: Failure rate of population of components: Underdamped response to step function
Posted:
Nov 15, 2013 12:03 PM


On Monday, November 11, 2013 5:47:01 PM UTC5, Rich Ulrich wrote: >On Sun, 10 Nov 2013, paul.....@gmail.com wrote: >> I think I should have been clearer about the fact that I'm not >> trying to model second order systems. Rather, I'm trying to find a >> reference for the failure rate with time of an ensemble of parts >> with Poisson failure rates, each of which are replaced upon >> failure. I assumed (perhaps wrongly) that it is wellknown and >> iconic, since it shows up in reliability material that I alluded to >> in my original post. > > I've looked back at your original post. You do mention there, > correctly, that there is a curve that becomes asymptotically > "Poison" in distribution of failures counted in small time > intervals; the failures eventually tend to occur uniformly. > > I think you have a serious misunderstanding of the vocabulary, and > of the point being made in your source. > > EVERY curve with a continuous, increasing failure rate is going to > have some a defined MTTF. This is not peculiar to some single > failure curve. If you replace every failure as it occurs with a > "new" part, the curve is going to be "damped" when you look at later > peaks, and the curve is going to evolve toward a uniform rate of > failures as the starting points become heterogeneous. It is > "uniform rate" that invokes "Poisson" as one of the possible > descriptors.
I could very well have some vocabulary wrong, but I don't see where it is from our discussion so far. The situation of interest in my case is that of an ensemble of (let's say) identical parts with identical MTBF (if they are repaired as good as new upon failure) or MTTF if they are simply replaced upon failure. And specifically, the parts failure follows a Poisson process. Admittedly, the fact that they are repaired as good as new almost makes the Poisson aspect immaterial, since MTTF=MTBF in this scenario (replaced or repaired as good as new). Is that what you are referring to?
In any case, I haven't had a chance to eke out the derivation from scratch.

