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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