In terms of the nines of reliability, calculate how much improvement we get by improving a system’s percentage reliability from 62% to 63%.
k = -log(1-p)
For 62% reliability, p = 0.62, k = 0.420.
For 63% reliability, p = 0.63, k = 0.432.
So we get an improvement of 0.432 – 0.420 = 0.012 nines of reliability.
In terms of the nines of reliability, calculate how much improvement we get by improving a system’s percentage reliability from 98% to 99%?
For 98%, p = 0.98, k = 1.699.
For 99%, p = 0.99, k = 2.
So we get an improvement of 2 – 1.699 = 0.301 nines of reliability.
In percentage terms, both situations above represent an improvement of 1%. In terms of the nines of reliability, which situation shows a more meaningful improvement, and how much more meaningful is it?
The second situation shows a more meaningful improvement. It is about 25 times more meaningful than the first situation. (0.301/0.012)