The function integrations over a ratiometric scale from the low limit to the high limit. I reckon it could integrate from a smaller value, but not zero. I don't think it would change the answers out to several decimal places, but it's a valid point, thanks.
I'd be very interested if you...
Right; it's a 95% two-sided confidence interval -- 2.5% above and below. The function ConfidenceLimit computes one- or two-sided limits for time- or failure-truncated data.
Thanks for looking.
The true MTBF is unknowable; that was the point. The code integrates over the 95% confidence interval (the constant MTBFConfInt in the first function).
Here's the code:
Function NumSpares(HrsSoFar As Double, nFail As Long, _
HrsToGo As Double, ConfMin As Double, _...
Last year I asked a question in https://www.eng-tips.com/viewthread.cfm?qid=437516 about a function to give the number of spares required for a lifetime purchase given only the number of operating hours and failures to date, the operating hours remaining, and the confidence level. -- something I...
Not intuitive, perhaps. The probability of having 0 events (failures) when the expected number is 2.3 is
=POISSON.DIST(0, 2.3, FALSE)
... which returns 10%.
More generally, it's easy enough to show that if mean = -ln(r), where r is between 0 and 1, then Poisson(0, mean) = r.
I assume you're talking about the -LN(RAND()) * MTBF formula. It looks that way at a glance, but that's not the way it works:
As rand() approaches 1, failure time approaches 0
As rand() approaches 0, failure time increases without limit.
Sure. Comments and suggestions welcome.
In the prior post, I said the random MTBFs have a log-normal distribution; that's wrong, it's a power law distribution (their logs are uniformly...
I think I've resolved this in a way that makes sense.
I made a Monte Carlo model in Excel. In each of 10,000 rows, I calculate a random (log-normal distributed) MTBF between 5,000 and 500,000 hours, and then a bunch of random (exponentially-distributed) failures for that MTBF. Each failure time...
#Mike: >> alternative is industrial motherboards
We buy enterprise everything, which is why we spend $4K for graphics cards instead of $500.
#ramseng >> Where H(t) is the number of failures expected in the interval (0,t)
Without having read the book, knowing the expected number of failures...
@MikeHalloran: I think most car parts (other than electronics) have wearout mechanisms rather than random failures. I'm not saying that's easier to sort out, but it's different.
@IRStuff: The Crow-AMSAA math is a little daunting, but at a glance, it appears to solve for the shape parameter of...
I've asked this question at https://math.stackexchange.com/q/2717852/ without response.
We sell systems comprising line-replaceable units (LRUs). We know how many operating hours have accumulated on each LRU type (because we know when they went into service, and we ask our customers what their...
