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# Availability calculation for a series model

## Availability calculation for a series model

(OP)
Dear All,

Imagine a system consisting of 2 components in series with the same following parameters:
mtbf = 1000 h
mttr = 5 h

Knowing that A = mtbf / ( mtbf + mttr ), we have :
A1 = A2 = 1000/ (1000+5) = 0,995024876

Because they are in series A system = A1 * A2
and then A system = 0,990074503

However I decided to calculate the equivalent mtbf and mttr of the system :

lambda system = lambda 1 + lambda 2 (in series)
mtbf system = 1 / lambda system,
and then mtbf system = 500 h

mttr system = [mttr1 * lambda 1 + mttr2 * lambda 2] / [lambda 1 +lambda 2]
and then mttr system = 5 which looks quite consistent.

The problem appears when I tried to calculate Asystem with the previous system parameters :

A system = mtbf sys / (mtbf sys+ mttr sys)
A system = 500/505 = 0,99009901

This result is not consistent with the previous one computed with A1 and A2 ....

Did I miss something ?

Any help would be really appreciated!!!

Mathieu

### RE: Availability calculation for a series model

Are you in need of 8 to 9th place precision?  Your results start deviating after the 5th place.  Depending upon what you used for calculation, this could be rounding or truncation rules within the processor or program acting on subtotal results.

Regards,

### RE: Availability calculation for a series model

It's the 5th decimal place, and it's definitely not a rounding problem.

You can show that the second calculation is effectively:

A = (MTBF/2)/[(MTBF/2)+MTTR] = MTBF/(MTBF+2MTTR)

likewise, you can reformat the first calcuation to be:

A = MTBF/[MTBF+2MTTR+(MTTR2/MTBF)]

Mathematically, the two calculations are not identical, so they should not result in the same value.

My take on it is that the series availability approach is a shortcut approximation, given that the extra term is usually too small to be of consequence.  The "correct" approach is to redo the failure rate and calculate the new availability from scratch.

Obviously, in a complicated system, there is great desire to simply multiply availabitilies that have already been calculated, rather than cranking the calculation from scratch.

### RE: Availability calculation for a series model

(OP)

First, even if the difference between both of the results is not so big, I will have thousands of blocks to integrate in order for me to model the system. So I need to know which one is the exact result....

From a probabilistic approach, saying that A sys = product of Availabilities doesn't appear to be wrong, since event are independents. we have P(A.B)=P(A) x P(B) when A and B independent.

Besides that, I began to develop tool in vb like a RBD tooling able to compute different kind of configuration (series , k ouf n, active standby, with repair or without, etc) based on markov chain.

For the previous series configuration if I solve the markov system for a steady state solution, it gives the result of the Asystem = mtbf system / (mtbf sys + mttr sys) and not A1 x A2... That is puzzling me...

1 - Maybe, it is because with the markov solution, we only assume one possible failure at a time?

2 - Anyway let's assume that I have computed for each block or configuration the equivalent mtbf and mttr for the steady state. I don't think I can use them as if they were normal mtbf or mttr since associates rates may not be constant...

Which means if I want to calculte A(t) for the system I could't use 1/(mtbf equiv) and 1/(mttr equiv) in the availabilty formula for one compotent A(t)=(m/l+m)+(l/l+m)exp(-t(l+m))?
Am I right ?

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