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Advisor

## High Avaiablity equiations, what is the maths explaination?

In system Admin III H3045S page 2-16 to 2-18
I do not put explaination for the caluculation that I can be conviced?
For people don't have this doc: If your system has 4 failures/year and each failure takes 4 hours to get repaired. Meanwhile NIC had 2 failures/year and each failure needs 6 hours to get repaired. So what is the math explaination between the following equations:

MTTR = ( 4/6 * 4 ) + ( 2/6 * 6 ) = 4.67 ????
Knowing that there are 8760 hours in one year.
P = MTBF / MTBF + MTTR
MTBF= 8760/ (2 + 4) = 1460 ???
P= 1460/(1460+4.67) =0,9968 then R=99.68 %.
Now if there is a redundant LAN?
P1=MTBF/(MTBF+MTTR) P2=MTBF/(MTBF+MTTR)
P1=(8760/2)/((8760/2) + 6) = P2 ???
MTBF = (MTTR1 + MTTR2) /((1-P1)*(1-P2)) ?????
=12/(0.001368 * 0.001368) ??????
MTBF=12/0.0000018714=6412332 Hours
P=6412332/(6412332 + 12) = 0.999998192
R=99.9998192% Available. This is the probability that both at least one of the LAN's is UP!
For the system:
4/4 * 4 = 4 ???
8760/4 = 2190, P=2190/(2190 + 4) = 0.9981
R=99.81 %
1 REPLY
Honored Contributor

## Re: High Avaiablity equiations, what is the maths explaination?

scheduled downtime is not included in the calcn.
ie those 6 hours (you work slooooowww! - 10 minutes more like!) are scheduled!

Later,
Bill
It works for me (tm)