[Gllug] RAID 1+0 vs 0+1

[Rich Walker]

John Winters <john at sinodun.org.uk> writes:

>
> I *think* the correct calculation would be:
>
Stripe n then mirror
------------------

 	Probability that the first disc is in a stripe = 1
 	Probability that the second disc is in the same stripe = (n-1)/(2n-1)

 	P(Safety) = (n-1)/(2n-1)		P(Disaster) = (n+1)/(2n-1)

Mirror then stripe
------------------

 	Probability that the first disc is in a mirror pair = 1
 	Probability that the second disc in the same pair = 1/(2n-1)

 	P(Safety) = (2n-2)/(2n-1)		P(Disaster) = 1/(2n-1)

For n=2, this is  1/3, 2/3
                  2/3, 1/3  i.e. mirror then strip better
   n=3 as previous email
   n=4: 3/7, 4/7
        6/7, 1/7

as n increases, the overall likelihood of 2-disks-failing increases as
well, so "mirror-then-stripe" is *always* better.



>
> Of course all this assumes that all discs are equally likely to fail.
> In the second case, it might be that the reason the first disc failed
> was because it was the most heavily used, in which case its partner will
> be equally heavily used and so more likely to fail than the other four.

mirror-then-stripe-then-mirror-the-whole-lot-over-drbd?

cheers, Rich.


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