[Gllug] RAID 1+0 vs 0+1

[John Winters]

On Thu, 2006-03-09 at 17:48 +0000, Christian Smith wrote:
[snip]
> Assuming you have 2 disks of the 6 fail, the stripe then mirror will be
> broken if:
>   disk 1 and disk 2 are in different stripes =
>     1/6 + 3/5 = 23 / 30 = 77%
> 
> The mirror plus strip will be broken if:
>    disk 1 and disk 2 are in the same mirror:
>     1/6 + 1/5 = 11 / 30 = 37%
> 
> So, given this analysis, mirror then strip is significantly safer.

Eeeek!  Probability calculations are notoriously tricky, but if I can
stick my maths teacher hat on here for a moment.

P(A & B)  (the probability of events A and B both happening) is
calculated by:

     P(A & B) = P(A) * P(B), provided A and B are independent.

     P(A v B) = P(A) + P(B), provided A and B are mutually exclusive.

You seem to have added to calculate an "and".

I *think* the correct calculation would be:

Stripe then mirror
------------------

	Probability that the first disc is in a stripe = 1
	Probability that the second disc is in the same stripe = 2/5

	P(Safety) = 2/5		P(Disaster) = 3/5

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

	Probability that the first disc is in a mirror pair = 1
	Probability that the second disc in the same pair = 1/5

	P(Safety) = 4/5		P(Disaster) = 1/5

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.


> Someone check my maths!

I'd give it a D1.  (1 for effort, D for achievement).

John

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