Given the reliability at 2100 hours =0.992, then lambda(2100 hours) = -Ln(0.992)=0.008. I need to convert the hazard rate (lambda) for 2100 hours to the normalized value in terms of failure rate per million hours (fpmh, 10^6 hours).
I am foggy on the process from this point; should I use the pdf function to convert such as:
Then take the derivative f(t)=0.008 (1+0.008*10^6)^-2
Then h(t): f(t)/1-F(t)
Can someone point me in the right direction?
Thank you for your comment. The equations for F(t) and f(t) were taken from "Applied Reliability", Tobias and Trindade, pg. 35, Example 2.3 Failure Rate Calculations as method to determine the cummulative hazard rate function h(t). The 0.992 value for reliability is taken from an exponential graph of reliability vs. time resulting from product life testing.
For the notation issue, I think you can also refer to
which has some fundamental relationship between
Here you need to be more specific to say whether you are interested in the instantaneous hazard rate or the cumulative hazard. (supposed to be different)
The second thing is that we are not sure about the parametric form of any one of . If you can type the exact parametric formula for any one of them before doing any numerical substitution, that will be much better.
The most important thing is that we are not sure, say if you are concerning changing from 1 unit/scale to another unit/scale, or you are given some information and want to know how to make use of the information and the relationship I quoted above to find the desired (cumulative) hazard.
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