Reliability specification clarified

[E Ellis]

I am working on a new product development project and have been given a reliability specification of 5-year life with 98% reliability. How do I relate this to a MTBF with a confidence level or a B-life.

[tarik]

Specifying a reliability goal in terms of a reliability value at a specified time with a given confidence interval (for example, the 90% lower reliability goal at 5yrs should be 98% ) is a better way to specify a reliability goal then to use an MTBF goal. After collecting test data you can fit it to a life model (such as Weibull, lognormal, exponential …) then use the model to determine what your reliability is at a given time at a certain confidence level.

If you still want to calculate the MTBF or B-life, then that would depend on the life model that you assume or that best fits your data. For example for a 2-parameter Weibull distribution,
MTBF = Eta x GammaFunction(1/Beta+1).

For other models, such as the exponential distribution, the MTBF calculation is simpler to calculate by hand. For the exponential distribution (which assumes a constant failure rate, ie no early failures and no wareout):
R(t)=Exp(-Lambda x t)
MTBF = 1/lambda
So MTBF=1/(Ln(R)/t)
Or in your case:
MTBF=1/Ln(.98)/5yr = 9.899663.
So your MTBF goal assuming an exponential
distribution is 9.89yrs

However the confidence bounds calculation would depend on the variations in your data.

These metrics along with confidence bounds can be obtained easily using Weibull++.

For more info about life data analysis visit: http://www.weibull.com/lifedatawebcontents.htm#Weibull%20Distribution

For more info about Weibull++ visit: http://www.reliasoft.com/Weibull/index.htm

[Pantelis]

As per Tarik, why relate to an MTBF - you already have a better metric. Now as to a B-Life it is B2 of 5 years, which is the same as 98% reliability at 5 years.