LIFE IN HELL:My thirteen days in Exchange Hell

[Jon Wahlmann]

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. ;-)

-Jon