Maximum amount of uncertain data tolerated in a reliability analysis

[manuelfr99]

Dear all,

I am dealing with a problem within my company which is as follows: I have a series of lifetime data from a specific part (or family of parts) which are labelled as failures (F), suspensions (S, the part has not failed when we stop observing it: normally it is replaced by a newer part to avoid lots of failures in wear out stages) and uncertain (X, we don't know whether the lifetime is a failure or not). My superiors are asking me to indicate what amount of uncertain data we can tolerate so that the impact on the lifetime distribution is small. I can have single failure modes or multiple failure modes. My idea (at least for single failure modes) is as follows: compute point estimates and their confidence intervals (using Weibull analysis). Anyone has an idea of what level of variability can we tolerate on the parameters the Weibull so that the lifetime distribution is not really impacted?

I am open to discuss more ideas for multiple failure modes.

Cheers and thanks for reading!

 

[rb1957]

ok, I pity you !

You know N (the total number of parts), Nf (the number of known failures), Ns (the number of known replacements), and Nu (the number of parts with an unknown fate, = N-Nf-Ns)
Are you trying to assess the fate of the part ? ... the number of failures being somewhere between Nf and (Nf+Nu) ?

Or are you trying to establish the useful life ... for each failure and substitution you have a life ? This seems more amiable to analysis. I guess you can determine the mean (or 5% or 95%) life of your sample (Nf+Ns), but then you want to say "how reliable is this sample compared with the full population, N ?" I can vaguely remember some statistics, something like if you can determine the standard deviation (and mean) of a sample set, how reliably does this predict the standard deviation of the full population ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.