Re: What is the point of RAID?
On Wed, Nov 12, 2008 at 09:44:47AM -0500, Henning Follmann <hfollmann@itcfollmann.com> was heard to say:
> On Wed, Nov 12, 2008 at 08:53:46AM -0500, Jeff Soules wrote:
> > On Wed, Nov 12, 2008 at 3:44 AM, lee <lee@yun.yagibdah.de> wrote:
> > > 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?
> >
> > It's purely a statistical property, not related to being in a RAID
> > array. But if there's (say) a 5% chance for a given drive to fail on
> > a given day, there's a 95% chance it won't fail.
> > If you have two drives, the chance *both* won't fail is the chance of
> > one not failing, times the chance of the other not failing -- 95%
> > times 95%, or 90.25%.
> >
> > With 24, the chance of all the drives not failing is .95^24 or 29.2%.
> >
> > Of course I just made the rates up, the survival chances of individual
> > drives are higher. But logic holds; the more drives you're watching,
> > the more lucky you'd have to be for none of them to be a dud.
> >
> 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)
Yes, but he was calculating
P(drive 1 fails OR drive 2 fails) =
1 - P(neither drive 1 NOR drive 2 fails) =
1 - P(drive 1 doesn't fail) * P(drive 2 doesn't fail)
and he already said he made the rates up for illustrative purposes.
Besides, the failure rate is meaningless unless we know the interval --
over a sufficiently long time, I expect that drives will have a failure
rate of 5%.
Daniel
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