From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: From: Russell Coker Subject: Re: [linux-lvm] LVM and fault tolerance Date: Wed, 28 Mar 2001 16:43:40 +1000 References: <20010327235600.H20535@vestdata.no> <3AC1142D.C47A9C96@tnonline.net> In-Reply-To: <3AC1142D.C47A9C96@tnonline.net> MIME-Version: 1.0 Message-Id: <0103281643400K.27902@lyta> Content-Transfer-Encoding: quoted-printable Sender: linux-lvm-admin@sistina.com Errors-To: linux-lvm-admin@sistina.com Reply-To: linux-lvm@sistina.com List-Help: List-Post: List-Subscribe: , List-Unsubscribe: , List-Archive: List-Id: Content-Type: text/plain; charset="us-ascii" To: linux-lvm@sistina.com, Anders Widman On Wednesday 28 March 2001 08:29, Anders Widman wrote: > ok. great... so if I make say 20GB partitions on all disks and put them > in a RAID-5 array. Can I add extra disks and rebuild the array? I have > seen some expensive RAID cards that does that. RAID-4 involves having N data disks (N >=3D 2) and 1 parity disk. The pa= rity=20 disk contains the XOR of the blocks on the N data disks. If one of the N= =20 disks dies then it's contents can easily be regenerated by the XOR of the= =20 surviving N-1 disks and the parity disk. RAID-5 is the same but has the parity data spread across all the disks fo= r=20 best performance. Thus RAID-5 has 3 or more disks. In RAID-4 or RAID-5 if you lose two disks at the same time then the XOR w= on't=20 get your data back and you are comprehensively stuffed. If you create a RAID-5 with two partitions on the same disk then you may = as=20 well use a bulk eraser. The scheme I mentioned in my previous message is the simplest way of doin= g=20 this with such disks. But really if your time is worth more than about $= 10=20 per hour you should just buy some more 80G disks. --=20 http://www.coker.com.au/bonnie++/ Bonnie++ hard drive benchmark http://www.coker.com.au/postal/ Postal SMTP/POP benchmark http://www.coker.com.au/projects.html Projects I am working on http://www.coker.com.au/~russell/ My home page