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@marauder.illiana.net [mailto:owner-kclug@marauder.illiana.net]On Behalf Of Hanasaki JiJi Sent: Tuesday, October 01, 2002 8:08 PM To: Kclug@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 > Cc: Kclug@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@marauder.illiana.net >>>| [mailto:owner-kclug@marauder.illiana.net]On Behalf Of Jeremy Fowler >>>| Sent: Sunday, September 01, 2002 4:27 PM >>>| To: Kclug@Kclug. Org >>>| Subject: RAID 5 Equation >>>| >>>| >>>| Anyone know off hand the equation used to find the usable storage > > space > >>>| available with RAID 5? >>>| >>>| >> >> >> >> > > > >