Frequently Asked Question

We have a 4oo4 final element. We realised that when we increase the beta factor the probability of failure on demand decreases. How is it technically possible? Theoretically, when the the beta (common failure mode) is increased, the PFD should decrease?‚Äč
Last Updated a year ago

When common cause is considered in a MooM voting you will notice that the PFDavg will decrease with increasing beta. This seems counterintuitive, common cause is expected to have a negative effect on a redundant configuration. Consider a simplified PFDavg equation for a 1oo2 voting arrangement:
PFDavg = ((LambdaDN * TI)^2)/3 + LambdaDC * TI/2
In this equation we see a portion that represents the redundant architecture and a portion that represents a single architecture (i.e. the common cause portion). A similar equation can be defined for a 2oo2 voting arrangement:
PFDavg = LambdaDN * TI + LambdaDC * TI/2
This equation has a similar structure, however the redundant architecture portion of the equation has been replaced by the PFDavg formula for a 2oo2 configuration. Now substitute Lambda and Beta in this equation. LambdaDN = (1-beta) * LambdaD and LambdaDC = beta * LambdaD. This leads to:
PFDavg = (1-beta) * LambdaD * TI + beta * LambdaD * TI/2 = (1-beta/2) * LambdaD * TISo from this equation it can be seen that if beta increases PFDavg will actually decrease. Though this is counterintuitive it makes sense when you consider that a 2oo2 configuration is typically used to reduce spurious trips. Beta will have a negative impact on the spurious trip rate for a 2oo2 configuration. Similarly beta has a negative impact on PFDavg for a 1oo2 configuration (a configuration to improve safety integrity) but will have a positive effect on the spurious trip rate.

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