LIFE IN HELL:My thirteen days in Exchange Hell
jon at wahlmann.com
Wed Oct 13 15:29:14 PDT 2004
Andrew P. Lentvorski, Jr. wrote:
> On Oct 13, 2004, at 1:14 PM, Jon Wahlmann wrote:
>> Andrew P. Lentvorski, Jr. wrote:
>>> While I recognize that this is a joke, you *never* want to do this.
>>> Do not do RAID 0+1 (stripe/concatenate followed by mirror)!
>>> One failed disk in each RAID 0 bank and you are toast.
>>> Probability: (1/n*1/(n/2))
>>> Do RAID 1+0 (mirror disks and then stripe/concatenate the mirrors).
>>> It will take two disks in the same RAID 1 to fail to take it out.
>>> This is
>>> a much lower probability event. Probability (1/n*1/(n-1))
>> Not to nit-pick, but the probabilities are relatively close to each
>> other based on your descriptions.
>> Multiplying out the first results in: 2/(n^2)
>> The second is: 1/(n^2 - n)
>> Both are still essentially O(1/(n^2)), depending on how larger or
>> small n is.
> This is fallacious reasoning. n does not increase to an infinite limit
> with disk drives. Do not apply asymptotic analysis to non-asymptotic
> systems. Constant factors *are* important.
It would have been useful if you would have defined what 'n' was. I
wanted to assume it was refering to the number of disk drives in the
RAID, but give the probabilities you mentioned, I assumed it would be
something else such as a MTBF number. Taking n to be the number of
drives, my case is made given the lowly number of n=2. But, mea culpa. :-)
> However, looks like I got the probability wrong for RAID 0+1:
> It should be: 1/n * ((n/2) / n-1)
> (initial) * (number of drives in other mirror) / (total drives left)
> And nobody corrected me. The curmudgeon nitpicker contingent of
> kplug-list should be ashamed.
Yeah, I was fully expecting Stewart to chime in here. ;-)
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