[tylerdurden]

[ QUOTE ]
uuDevil,

I think your method underestimates the number of overlaps because it implies that failures start exactly on a three hour boundary.

Here is a second method for estimating the number of overlaps. Suppose there are exactly 400 failures. We will say that the ith failure and the jth failure overlap if their start times differ by 3 hours or less. The probability that the ith failure overlaps the jth failure is approximately

6/(24*356) = 0.000684932.

There are 400*399/2 = 79800 possible pairs of i's and j's, so the expected number of overlaps (with "overlap" defined as above) is

6/(24*356) * 400*399/2 = 54.6575.

[/ QUOTE ]


I came up with a similar number in a different manner.

As uuDevil pointed out, The expected number of failures in a three hour period is 400/(24*365/3)=.137

We expect to have 400 failure events (averaging three hours each) in a year. During each one of those, the probability that another machine will fail is 0.137.

Now if we take 400*0.137 = 54.8. However, that means we'd actually have 454.8 failures, not 400 (we're counting duplicates twice.

We just need to solve this for x: (x*0.137)+x=400

That gives us x=351.8. 351.8 single failure events.

351*0.137=48.2

48.2 overlapping events.

351.8+48.2=400.