RAID 5 Equation

[Richard Meeker]

It's not linear.  If you have 3 disks, then you have 2/3 the total capacity
of all disks for RAID.  If you have 4 disks, then you have 3/4.  5 disks =
4/5.  So, for 5 18gig disks, you would have 72 gig of data storage and 18gig
would be striped across all of the 5 disks in the array. So your total data
storage would be (n-1)/n where n = the number of disks in the array (not
including hot spares), and your total storage availability (bytes) would be
(n-1) x (the size of the smallest disk in the array).

-----Original Message-----
From: owner-kclug at marauder.illiana.net
[mailto:owner-kclug at marauder.illiana.net]On Behalf Of Hanasaki JiJi
Sent: Tuesday, October 01, 2002 8:08 PM
To: Kclug at Kclug. Org
Subject: Re: RAID 5 Equation

Well 5 or 8 or 10 disks would require more storage for partity...

at 3 disks you are saying 1/3 is used for parity.  is this linear?  so:

	60 gig usable =  90 gig total = 5 x 18gig?

Bill Clark wrote:
> RAID 5 parity is not on anyone particular disk.  That would be RAID 3 that
> Mick is describing.  In RAID 5 the parity is spread out and shared across
> all disks.  That is why RAID is so popular.  The odds are you won't loose
> more than one disk at a time.  In most hardware RAID implementations if a
> disks dies.  Then yanking the disk and replacing the disk will start the
> rebuild of the parity redundancy and within minutes you can be fully
> resillilent again.  Or if you have a designated hot spare it can
> automatically take the place of a bad disk.
>
> If you have 3 disks and they are all 9GB disks then your effective
capacity
> for that set is 18 GB.  Three disks is the minimum number of disks allowed
> in RAID 5.
>
> Bill
> ----- Original Message -----
> From: Hanasaki JiJi <hanasaki at hanaden.com>
> Cc: Kclug at Kclug. Org <kclug at kclug.org>
> Sent: Tuesday, October 01, 2002 7:50 PM
> Subject: Re: RAID 5 Equation
>
>
>
>>So in Raid 5 the parity recovers data as well as identifying if data has
>>been corrupted?  There was some equation, years ago that started:
>>- 1 bit parity identifies an odd number of bit errors
>>- 1 bit parity misses an even number of bit errors
>>- 2 bits parity ....  BUT ITS NOT LINEAR, ITS EXP
>>
>>So what if the partity disk bites it?
>>
>>Mick Ohrberg wrote:
>>
>>>Theoretically, providing all the disks are the same size, it's the total
>
> sum
>
>>>of all disks, minus one disk (the parity disk).
>>>
>>>/Mick
>>>
>>>| -----Original Message-----
>>>| From: owner-kclug at marauder.illiana.net
>>>| [mailto:owner-kclug at marauder.illiana.net]On Behalf Of Jeremy Fowler
>>>| Sent: Sunday, September 01, 2002 4:27 PM
>>>| To: Kclug at Kclug. Org
>>>| Subject: RAID 5 Equation
>>>|
>>>|
>>>| Anyone know off hand the equation used to find the usable storage
>
> space
>
>>>| available with RAID 5?
>>>|
>>>|
>>
>>
>>
>>
>
>
>
>