P(two failures within MTTR for first node) = P(one failure)P(second failure within MTTR of first node|one failure)
independence simplifies things
= P(one failure)P(second failure within MTTR of first node)
= P(one failure) * (1 - e^-λx)
where x = MTTR for first node
λ = 1/MTBF
plugging in the numbers from your blog post
P(one failure within 30 days) = 0.01 not sure if this part is correct.
MTTR = 5 minutes + 5 hours =~ 5.083 hours
MTBF = 30 days / 0.01 = 3000 days = 72000 hours
0.01 * (1 - e^(-5.083 / 72000)) = 0.0000007 ~= 0.00007 %
I must be doing something wrong cuz I'm not getting the 0.000001% you have in the blog post. If there's some existing work on this I'd be stoked to read it, I can't quite find a source.
Also there's two nodes that have the potential to fail while the first is down but that would make my answer larger not smaller.
independence simplifies things
= P(one failure)P(second failure within MTTR of first node)
= P(one failure) * (1 - e^-λx)
where x = MTTR for first node
λ = 1/MTBF
plugging in the numbers from your blog post
P(one failure within 30 days) = 0.01 not sure if this part is correct.
MTTR = 5 minutes + 5 hours =~ 5.083 hours
MTBF = 30 days / 0.01 = 3000 days = 72000 hours
0.01 * (1 - e^(-5.083 / 72000)) = 0.0000007 ~= 0.00007 %
I must be doing something wrong cuz I'm not getting the 0.000001% you have in the blog post. If there's some existing work on this I'd be stoked to read it, I can't quite find a source.
Also there's two nodes that have the potential to fail while the first is down but that would make my answer larger not smaller.