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Old July 17th, 2009, 12:03 AM
pom15595 pom15595 is offline
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Default PFD = - λt/2 vs PFD = 1 – exp (- λt)

I am looking at emergency systems that are normally “inactive” but are required to function on demand (the demand rate on these systems may be ~0.1 times per year on average).
I would like to know which is the “correct” equation to use:
PFD = - λt/2
or
PFD = 1 – exp (- λt)
And why it is the correct equation to use.
Many thanks
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  #2  
Old July 17th, 2009, 07:00 AM
Pantelis Pantelis is offline
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Default Re: PFD = - λt/2 vs PFD = 1 – exp (- λt)

First: What is PFD ~ Prob. of failure on demand? Second you state a usage but not any failure rate characteristics. .

Now with respect to what you put there from a mathematical point of view, the second equation is the probability of failure given a constant failure rate (lambda) after a given time t. Just by looking at it -- I have no idea where you got the first equation or what you are trying to do with it.

If you are looking for a prob of failure given a constant failure rate for a mission duration of t, then the first equation does not give you that. The second does.

Hope it helps.
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Last edited by Pantelis; July 17th, 2009 at 07:17 AM.
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Old July 18th, 2009, 07:30 PM
pom15595 pom15595 is offline
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Default Re: PFD = - λt/2 vs PFD = 1 – exp (- λt)

Yes PFD = probability of failure on demand

PFD = - λt/2 is often used as an approximation (from memory it is "valid" when λt << 1.

What I would like to know is the views of people with experience of analysing similar systems as to the limitations, which should be used when, etc.
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Old July 18th, 2009, 09:45 PM
Pantelis Pantelis is offline
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Default Re: PFD = - λt/2 vs PFD = 1 – exp (- λt)

Let’s go back to basics… The only reason one uses an approximation to an equation is when the original equation is not easily solvable.

Now in this equation (and the approximation) the perceived difficulty lies in computing e^x.

So if you were stranded on a deserted island without calculator (with an exponent function on it) and you didn’t want to go about working this out long hand (say using Taylor series expansion), then the approximation may be useful and of value, as it will give you extra time to work on shelter, as well as creating signals needed for an eventual rescue. If you are however in front of your computer, or have easy access to a pocket calculator, then there is absolutely no reason on earth why you should even be contemplating the use of an approximation.

Hope my explanation (with the associated humor ) makes sense and is useful.
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Last edited by Pantelis; July 18th, 2009 at 10:17 PM.
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Old July 19th, 2009, 07:30 PM
pom15595 pom15595 is offline
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Default Re: PFD = - λt/2 vs PFD = 1 – exp (- λt)

Thanks

I agree, however the approximation is used in certain design standards documents, eg ISA - TR84.00.002-2002-Part 2.

So if it is easy to calculate, why do these organisations (eg ANSI / ISA) use the approximations?
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Old July 20th, 2009, 06:38 AM
Pantelis Pantelis is offline
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Default Re: PFD = - λt/2 vs PFD = 1 – exp (- λt)

Well that’s the $10 dollar question…

The ones you mention are not the only ones. There are many others with approximations for different reliability analysis (and where some are not even correct) that are still out there today, and are getting republished all the time.

The only logical explanation I can think of, is that many of them were written many years ago (when the simplification was of value), and even though revised, when they are revised the team responsible may either not be that well versed on the subject, or they don’t want to rock the boat … (i.e. It’s been like these for the last 50 years why change it now type of mentality).

If you really want to get philosophical, let’s take it a step further, why are most of these standards only considering an exponential distribution! It implies a constant failure rate, something that is not true for most items (and thusly WRONG in these cases), which again is a simplification over the more complex time varying failure rate models (such as the Weibull model, etc.) . See one of my articles related to this at http://www.reliasoft.com/newsletter/...xponential.htm

I can go on ... but I think you get my point.
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Last edited by Pantelis; July 20th, 2009 at 06:44 AM.
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  #7  
Old August 26th, 2009, 02:18 AM
Robbie Robbie is offline
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Default Re: PFD = - λt/2 vs PFD = 1 – exp (- λt)

Its worth looking through IEC standard 61508, part 6 (functional safety of electrical/electronic/programmable electronic safety-related systems)
- if you have access to it. It has quite a bit around PFD calculations for different equipment and system configurations
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Old August 26th, 2009, 02:33 AM
Robbie Robbie is offline
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Default Re: PFD = - λt/2 vs PFD = 1 – exp (- λt)

When I studied Reliability, our main reference text was "Introduction to Reliability Engineering" by E.E. Lewis.T The section on maintained systems and testing for unrevealed failures is excellent, providing the theory and the reasons and circumstances for approximation. In my experience over several years, most of the folks I've come across using λt/2 have no idea about how the equation has been developed. They usually do not understand that they are using a constant rate assumption, and often use the equation incorrectly (by technical/process safety people in particular). We use it in the offshore oil & gas industries, often in relation to safey protective systems which have, for example a pressure or temperature sensor, logic/processing unit, and end elements (e.g. shutdown valves). The users are often trying to demonstrate that a SIL (safety integrity level) target is met in terms of the PFD by adjusting the test frequency. However, when complex redundancy is involved, the simple λt/2 equation start to change. Thats why I've referenced IEC 61508 in my earlier reply. Hope this is of some background help or interest to you.
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