Re: What is the point of RAID?
> Jeff,
> you math is off - way off.
>
> P(one fails) != 5/100
>
> P(two drives fail at the same time) = P(one fails) * P(one fails)
>
> = 25/10000
Henning -- I'm not talking about the chance that the array will fail
-- just that the more drives are under observation, the more the
chance that *one* of them will fail on a given day. The comment I was
responding to:
> Do you mean it is more likely that any one drive in the array fails when
> you have more drives, or do you mean that it is more likely for a drive
> in the array to fail when you have more drives? If drives fail more
> often when being used in an array with more drives, what makes them
> fail more often under those conditions?
seemed to think (mistakenly) that the chances of any one drive failing
would be increased by putting it into an array. Adding drives to an
array doesn't increase the chance that Drive #1, or #2, etc. will
fail, but it does increase the chance that you will see a drive
failure in that array on a given day.
That's a trivial point -- unrelated to whether they're in a RAID, just
simply that you're looking at more drives.
The probability that a RAID-5 *array* will fail in a way that results
in data loss is a separate issue; that's what you're calculating
below.
On Wed, Nov 12, 2008 at 9:44 AM, Henning Follmann
<hfollmann@itcfollmann.com> wrote:
> Jeff,
> you math is off - way off.
>
> P(one fails) != 5/100
>
> P(two drives fail at the same time) = P(one fails) * P(one fails)
>
> = 25/10000
>
> If you have more than 2 drives in the raid you have to make the
> cobinatoric calculations of how many configuration can be there for two
> drives out of n.
> that would be
>
> 2! * (n-2)! / n!
>
> multiply that to P( two drives fail at the same time) where n is the
> number of all drives.
>
> Henning
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