Re: RAID question

From: Ivan Marsh (annoyed_at_you.now)
Date: 11/24/04 

Date: Wed, 24 Nov 2004 15:03:53 -0600

On Wed, 24 Nov 2004 21:25:08 +0100, Mark wrote:

> On Wed, 24 Nov 2004 14:06:15 -0600, Ivan Marsh wrote:
>
>>> Your last paragraph shows that you really haven't grasped just what
>>> parity
>>> is. The formula is indeed correct, you lose the capacity of exactly
>>> one drive, regardless of how many drives are in the array
>>
>> Which is physically impossible as the math you snipped from my post
>> using your formula proved.
>>
>>> - which you can
>>> verify for yourself by reading any of the abundant RAID tutorials or
>>> HOWTOs, or even by trying it out for yourself.
>>
>> Yea, lets do that... and find that there isn't one opinion about it out
>> there either. Not only is there support for the formula in question
>> there is also the (1/N)*S formula and many sites that say things like:
>> "The loss of disk space is basically 100 divided by the number of disk
>> drives [which is basically the same at (1/N)*S]. With 3 drives, there is
>> a 33% loss of
>
> 33% * 3 disks = 1 disk..
>
>> disk space. With 5 drives, there is a 20% loss of disk space." Which
>
> 20% * 5 disks = 1 disk...
>
>> supports what I said about it.
>
> No, It seems to contradict what you said about it....

Okay, let's extrapolate a bit:

3 x 10g disks @ 33% : 30g total: 9.9g overhead
5 x 10g disks @ 20% : 50g total: 10g overhead

Sure seems like the amount of overhead isn't static to me?

How 'bout another example:

Two 30g raid arrays,
One made out of 3 disks has 10g of overhead and a 20g capacity,
One made out of 10 disks has 1g of overhead and a 29g capacity.

So... only an idiot would buy anything but the 1g drives. Why waste all
that space buying the 10g ones?

That formula doesn't work.

The overhead on a RAID 5 array is a percentage of the entire disk space
and has nothing to do with how many drives are envolved. 

-- 
We used to have a president that worked with a chimp...
   I guess it was matter of time before we elected a chimp president.   
i.m.

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