[ILUG] Why RAID

[Paul Jakma]

On Tue, 20 Jul 2004, Paul Jakma wrote:

> 1 over 0, probability of second failure taking out array is (n-2)/(n-1), (ie 
> 0.6.. for 4 disk, 0.85 for 8 disk, 0.88 for 10, and gets worse as n is 
> increased), but 0+1 probabilty of second failure killing the whole array is 
> 1/(n-1), and hence actually improves the more disks you add. (but the more 
> disks, the lower the interval between failures of individual disks possibly).

You know, it takes a lot of practice and skill to be as consistently 
wrong as me.. please dont try this at home kids, i'm an expert.

1 over 0, assuming n disks, where n >= 2 and n%2 = 0, divided into 2 
RAID0 arrays which are then mirrored, the probability, P, of a two 
disk failure taking out the complete array is:

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

(not '(n-2)/(n-1)' ). ie, for n disks, the probability P is:

n   P
2   1/1
4   2/3
6   3/5
8   4/7
10  5/9
etc..

So, it starts at 1, descends immediately to 2/3 and decreases slower 
there after, asymptotically bounded by P = 0.5, (as n -> infinity, 
(n-1) tends towards n, so the probability tends towards (n/2)/n which 
= 1/2), but you need a huge number of of disks to get P anywhere near 
0.5 ;) 16 or more, eg even with 6 disks, probability is still 0.6.

0 over 1, for the same scenario (2 RAID1 arrays of n/2 disks each, 
the arrays striped), the probability is 1 for n=2, 1/3 for n=4 and 0 
for all, (i cant think of a way to relate P to n for a fixed number 
of failures, though relating n to F (F -> # of failures that can be 
tolerated) for a given P is possible. should be able to express it as 
a series though.). However, increasing n does not increase capacity, 
only redundancy, capacity is fixed at n*2.

However, you meant n/2 RAID1 arrays, each RAID1 array composed of 2 
disks (obviously), with the the n/2 RAID1 arrays striped with RAID0, 
right? For this the probability is 1/(n-1), ie n=2 -> 1, n=4 -> 1/3, 
and so on, tending towards 0, which isnt bad at all, though risks of 
multi-n failures increase as n increases. And I think you were 
comparing it to the 1 over 0 case i described above (2 RAID0 arrays 
of n/2 disks each, mirrored), which *isnt* what I was thinking of 
when I answered John.

I was thinking of n/2 RAID0 arrays of 2 disks per array (obviously), 
Well actually, I was thinking of 2 RAID0 disks mirrored, 4 disks in 
total and not thinking past that at all, but lets consider a bunch of 
2-disk RAID0 arrays all mirrored. In this case P is:

 	n P
 	2 1
 	4 2/3
 	6 0

and 0 thereafter.

However, the array does not gain space as n increases, only 
redundancy. Size will always be n*2. So 0 over pairs of 1 wins, yes.

> Apologies, I was dangerously wrong. (not new for me).

And even better, in pointing out where I was wrong, I was wrong.. 3rd 
time lucky maybe, but unlikely - though you never know, I could be 
wrong.

regards,
-- 
Paul Jakma	paul at clubi.ie	paul at jakma.org	Key ID: 64A2FF6A
 	warning: do not ever send email to spam at dishone.st
Fortune:
In Pocataligo, Georgia, it is a violation for a woman over 200 pounds
and attired in shorts to pilot or ride in an airplane.