Raid 5
Shane Hathaway
shane at hathawaymix.org
Fri Dec 2 17:22:40 MST 2005
Nicholas Leippe wrote:
> RAID 10 and RAID 0+1 offer significantly different failure tolerances.
>
> RAID 0+1:
> odds of failure:
> RAID0 1/1000000 * 2 = 1/500000 (the RAID0)
> RAID1 RAID1 (1/1000)^2 = 1/1000000 (per mirror)
> A B C D 1/1000 (per drive)
>
> is better than RAID 10:
>
> RAID1 (1/500)^2 = 1/250000
> RAID0 RAID0 1/1000 * 2 = 1/500
> A B C D 1/1000
While your point stands, the calculation is imprecise. By this logic
(1/1000 * 2 = 1/500), flipping a coin twice guarantees the coin will
land on both sides, which observation disproves:
1/2 * 2 = 1
However, your method yields a close approximation for small fractions.
A simple but correct way to compute the reliability of a RAID 0 array is
to subtract the probability of each drive failing from 1, yielding the
probability of each drive surviving, then multiply those probabilities
together to figure out the probability of the set surviving, then
subtract from 1 again to figure out the probability of the set failing.
IOW:
1 - (1 - 1/1000) ^ 2 = 1999 / 1000000
... which is really close to 1/500. The difference matters when you
talk about a longer period of time than the 1/1000 estimate implies.
BTW, has anyone tried RAID 6? It's in recent Linux kernels. It claims
to survive the loss of any two drives in a set.
If anyone's interested, I've written about storage reliability in my
blog, although I'm only discussing theory, not practice.
http://hathawaymix.org/Weblog/2005-10-26
Shane
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