On Tue, 13 Mar 2007 08:14:28 -0000 "Nelson, David \(ED, PAR&D\)" <[EMAIL PROTECTED]> wrote:
> > -----Original Message----- > > From: mwq [mailto:[EMAIL PROTECTED] > > Sent: 10 March 2007 21:00 > > To: [email protected] > > Subject: [gentoo-user] RAID > > > > > > I have one laic question which may not be directly connected > > to Gentoo but I think you'll forgive me that. > > Imagine such a situation: I have two hard drives but drive A > > is twice faster when reading and writing then drive B. I want > > to make RAID 0 using A and B. Why are the stripes sizes on > > both drives excacly the same? (I've googled and didn't find > > any information about different spripes sizes) I think that > > using twice greater stripe on A gives more speed then using > > equal stripes. > > And my question is: where am I doing a mistake? > > PS > > Sorry for my poor English > > > > -- > > [email protected] mailing list > > > > > > Forgive me the maths approach here, but if drive A reads/writes at > speed a, and drive B at speed b. > > a = 2*b from what you have said. > In raid 0 drive speed is limited by the slower drive (if I recall > correctly) so the speed your raid array would be limited to is .... b > + b = 2b = a (from above). > > Hence you would be as well not bothering with RAID if drive A is 2x > as fast as drive B. > > Naturally it's not as clear cut but should be pretty close. RAID 0ing > these drives would gain you little in terms of speed while any 2 > drive RAID 0 setup increases the chance of failure by 2. > > If you just want them to appear as one drive look at something like > LVM which can create one volume from both drives, although parts of > the volume on drive A would be faster than those parts on drive B. > > Just my £0.02, feel free to poke holes in my reasoning ;) well, I'll try a bit perhaps. Given my scheme, we should really be talking about 3 drives -- well, actualy you generally raid _partitions_. A = fast disk, or speed thereof B = slow disk. or speed thereof A1 = first partition, size R/3, fast disk A2 = second partition, size R/3, fast disk B1 = first partition, size R/3, slow disk raid-0 is A1 + A2 + B1 now let's say you have a file N to pull from the disk. Naturally, one third of it will come from each partition. that is, two-thirds of it comes from A, and A being twice as fast as drive B, the one-third that's on B gets done in about the same amount of time. so N mb is copied in max( (2/3*N)/A , (1/3*N)/B ) seconds which should be roughly the same. if N was on A alone obviously the speed would be N/A seconds. (N/A) > (2/3*N)/A. in reality, though, I think the best performance would probaby involve just using the fast drive. RAID introduces too much overhead to make up for itself in this situation I think. -- [email protected] mailing list
