Failure probability of a valve in one year.

[Terp]

Hi all. Probability definitely isn't my specialty, and I've run across this problem at work.

There's one gate valve that isolates two 2400 psi hydrogen gas banks from each other, and I'm trying to find the probability that it fails in one year. I've done some digging and found failure rates of gate valves are between 1.5 failures and 15 failure / 10^6 hrs. Because I want to be conservative, I used the 15 failures / 10^6 hours and did some simple converting. It comes out to 0.13 failures per year.

Now, to my untrained eye (and brain), it seems that .13 failures / year is the probability of it failing in one year, but at the same time, it's a rate, not a probability. Unfortunately I don't have any more information than I've already given. If more is needed I'll be doing some digging.

Currently consulting Google for this problem, too...

Any guidance would be appreciated. Thank you!

 

[moonman239]

That is correct.

 

[Terp]

That is correct.

What is correct? That the probability of failure is 0.13 per year? I still seem to think that I can get a "probability" of failure out a failure rate.

 

[moonman239]

I still seem to think that I can get a "probability" of failure out a failure rate.

You are correct. In a real-life (non-theoretical) situation, the probability of the event occurring = the rate at which it occurs.

 

[SW VandeCarr]

The range of failures was given as 1.5 to 15 per one million hours. I take these to be confidence intervals (CIs). Typically these are 95% CIs. Without knowing the actual source I can't say for sure but the mean (expectation) would usually be near the midpoint of this interval or about 8.25 failures per one million hours assuming a normal distribution of independent events. The CIs are typically calculated based on a normal (Gaussian) probability distribution.

 

[GeorgeRaetz]

This is classic Poisson process. The probability of the valve lasting a time t with NO failure is:
P₀(t)=exp(-lambda * t)
Here, The rate constant lambda=15/10⁶ hr =.13 yr^-1.
Example:The probability of the valve lasting one year is
exp(-1yr*.13/yr)~.88

 

[SW VandeCarr]

This is classic Poisson process. The probability of the valve lasting a time t with NO failure is:
P₀(t)=exp(-lambda * t)
Here, The rate constant lambda=15/10⁶ hr =.13 yr^-1.
Example:The probability of the valve lasting one year is
exp(-1yr*.13/yr)~.88

If you use \lambda=15, the distribution is essentially normal with a known variance. I took the range of values given to be confidence intervals. Even with \lambda=8.25 the distribution is close enough to normal to construct confidence intervals in the usual way. In any case p=0.13 is the high end of range. The proper way to approach data like this is to use all of it and not discard useful information by taking most extreme value "just to be safe".

By knowing that the high end of a 95% confidence interval is 15, the OP can estimate that there is only about a 5% chance that the failure rate will be higher.

http://www.math.mcmaster.ca/peter/s743/poissonalpha.html

 

[GeorgeRaetz]

... the distribution is essentially normal with a known variance...

OP (Terp) is asking to convert from failures per year to probability that failure will occcur in one year. That can only be answered with knowledge of statistical distribution of failures over time. However, if detailed statistical data is not available, one resorts to a statistical model.
I used the simplest model: the Poisson process, which can only be an approximation since the (failure) rate "constant" should increase as the device ages. However, as long as the time span is less than the MTBF(at least 1/.13yr^-1~7.6 yr) Poisson should be OK for an estimate. A more sophisticated analysis might use the Weibull distribution, for example.
Now taking the model you have in mind, can you answer the question: What is probability of failure when a given time has elapsed?

 

[SW VandeCarr]

Now taking the model you have in mind, can you answer the question: What is probability of failure when a given time has elapsed?

It depends on what value you use for \lambda. You are using the extreme value of a range. Lambda is the mean, not the extreme. Since the OP gave a range, but no other information, I approached this as if the interval were a 95% CI which is standard. If it were simply a range from lowest to highest, then 15 would likely be an outlier and the probability of a value greater than 15 would be even be smaller than .05.

It seems you want to use an extreme value as a mean value. If you and the OP agree that's what you want do, then fine. The OP already knows that the rate associated with 15 failures per year is 0.13.

EDIT: I'd be very interested to see how you would use the Weibull distribution in this particular problem.