Hello,

How do you calculate PFD/PFH for safety instrumented functions?
The simplified calculations in IEC61508 are very basic and assume for voted sub systems (e.g. 1oo2, 2oo3 etc.) that the failure rates of the components are the same.

Is there an easy method to calculate PFD/PFH without using Markov, Monte Carlo, etc?

Thanks

H

The basic equations are fine for most purposes. If there are different elements in redundant branches, a simple and conservative approach is to assume the least reliable element failure rate applies to all parallel branches.

If there are multiple serial elements in a branch, combine these first.

Any calculation purporting to offer accuracy better than the second significant figure is a bit of a joke anyway given the broad uncertainties in so much of functional safety assessment and design.

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HTDearden CEng
Consulting Engineer
www.tdsl.org.uk

This calculation is found in 61508. You can go simple using lamdda(DU)TI/2 or there are more detailed calculations using the effectiveness of proof test as well. In any case the information is fed from your functional safety assesment.
Regards,
Tom

You can build it as a reliability block diagram and put you failure rates on each branch. You need to be a bit careful with this though, because it's not completely conservative - tends to lead to lambda^2*T^2/4 etc. rather than lambda^2*T^2/3. However if you have a few channels and testing is staggered then there's not much difference.

Alternatively you can expand all the power (^y) terms in the 61508 equations and put in the individual failure rates in each term, using either geometric mean or most pessimistic where a single value is needed.

BUT - HT Dearden's point about supposed accuracy is very important. Far too many people seem to think you can calculate PFD/PFH to any desired level of accuracy. It ain't so. In practice, if you look at the statistical bases of the figures, the first significant figure is often shaky. That's why SIL is based on orders of magnitude.

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Reality is that which, when you stop believing in it, doesn't go away.

For a simple SIS :
PFDavg = 0.5 ×λDU × TI
where
λDU = total dangerous failure rates of all components in the loop (1/hour)
TI = the test interval of the SIS (years)

417303

If lambdaDU is /hr then TI needs to be in hours.

If TI is in years then lambdaDU must be failures/year.

The difference is a factor of about 8760 (3 - 4 SIL levels)

-------------------------
Reality is that which, when you stop believing in it, doesn't go away.