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:
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:
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
Paul Jakma paul at clubi.iepaul at jakma.org Key ID: 64A2FF6A
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