failure rate of parallel networks..

[thiravia]

http://src.alionscience.com/pdf/1Q2006.pdf

When I try a simple parallel network with two components , the answers from BlockSim are way off ..from the 2 labda / 3 formulae..

Can anyone explain this..? Also does the system failure rate of a parallel system change with time..? (this seems counter intuitive..). The approximate forumla (2lambda/3) also doesnt seem to be a function of time.

[Arai.M]

The link below does not stress the fact that the equation that leads to the 2 * lambda / 3 formula is just an approximation. Approximations are just that.

Now, the approximation used, assumes that the failure rate of two components in series with a constant failure rate is constant. That is not the case. It changes with time.

Imagine that you have a redundant system with 2 components. As soon as one of these components fails, the failure rate of your system suddenly changes. After that first failure, you can think of it as one component's failure rate (constant). If the time to failure of the first component was a deterministic event, you could think of it as a step function.

However, the time to failure of the first component is not a fixed point in time so you end up with a smooth curve that is time varying.

Unfortunatly, I don't know of an easy way of calculating the failure rate of redundant systems. One way to obtain the failure rate is by taking the ratio FR(t) = f(t)/R(t).
In the example with two identical (and independent) exponential components in parallel then
R(t) = 1 - (1-R)^2 = 2*exp(-lambda*t)-exp(-2*lambda*t)
and
f(t) = -2*lambda*exp(-lambda*t)+2*lambda*exp(-2*lambda*t)

so FR is not pretty.

Hope this helps,
Arai

[Pantelis]

And to add to Arai's comments from my soapbox…

I also just took a look at the referenced publication and I must point out that the approach used to arrive at the constant failure rates is WRONG….

As an example I am looking at what was done on page 2, bottom of first column, under the title “Root Cause of the Problem”. Here they provide an equation for FR.
While the reciprocal of the mean is the failure rate when you have an exponential distribution, this is NOT TRUE when R(t) is not an exponential distribution (as is the case when doing most system reliability solutions with networks).

In other words if that was true then if R(t) was the Weibull Reliability function this would give you a constant failure rate value for the Weibull distribution.

Unfortunately when things like this get published, and not thoroughly reviewed it gets all of us into trouble.