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failure rate probability 2

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Ciaci

Mechanical
Jul 11, 2015
58
I am reading a paper about a Failure Tree Analysis. When I was at university I did a course where the FTA was presented and we worked directly on the probability using the logic gates AND/OR.

In this paper they work on the failure rates. At the beginning I thought that the failure rate and the probability were the same thing, because the units where the same [1/T] but I read here
"Although the failure rate, $\lambda (t)$, is often thought of as the probability that a failure occurs in a specified interval given no failure before time $ t$, it is not actually a probability because it can exceed 1. "
After that I have read here that the probability P is

$P=1-exp(-\lambda t)$

So now my question here is, given the failure rate of the system calculated in the paper, can I obtain the probability with the previous formula?
For example if the failure rate is 9.5 FPMH, the probability per year is:

P=1-exp(-9.5*numberofhourperyears/millionofhours)=8%

 
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"Probability per year" is a horrible term - you're drifting back towards a failure rate.

Units of probability are not 1/T. Probability is dimensionless. This isn't just a bit of foolish pedantry - it's also a reminder that you can't do things like multiplying two failure rates together to get the "rate at which both things happen".

I think what you're trying to calculate there is the "Probability of experiencing one or more failures during any particular interval of one year" and yes, 8% is a good answer given those parameters.

Lambda is only truly constant for a particular type of failure mode (what we'd call a random failure). Once you start dealing with failures that result from ageing, or those that only occur at startup, or those that only occur when the equipment is switched on, then you need to consider carefully whether a constant lambda assumption is good enough for your circumstances.

A.
 
It's been a long time, but doesn't the 'Bathtub curve" come into this somewhere?


Politicians like to panic, they need activity. It is their substitute for achievement.
 
Classical failure rate is defined as -time_duration * ln(reliability), where reliability is the probability of surviving until time_duration. Failure rate is often quoted as 1/MTBF (mean time between failures). By that definition, we can see that the probability of surviving until time=MTBF is only 1/e=36.85%

Net result is that the equation you've quoted is correct, and it makes sense, no? 9.5 failures per million hours applied to one year (8766 hr). However, note again that MTBF comes out to 1/9.5Mhr = 12 yr MTBF. Plugging in 12 yr, the probability of survival is only 36.8% and probability of having failed by then is 63.2%

TTFN (ta ta for now)
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zeusfaber said:
"Probability per year" is a horrible term - you're drifting back towards a failure rate.
I completely agree, you pointed out something that I didnt' noticed reading my reference book of fire risk assessment.
It refers to probability placing in parenthesis "per year" and than it did not place any unit. Suqsequently it identifies f (which I assume is the frequency) as a value with unit yr^-1. Sometime I saw using indifferently probability and frequency, but they have the same value only if the Probability considers a year and the frequency is diveded by a year but they also have different units. @zeusfaber Am I understanding correctly?

 
Ciaci said:
they have the same value only if the Probability considers a year and the frequency is diveded by a year but they also have different units

I think that's quite a good summary.

A.
 
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