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#1
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Raid 6 in use?
Does anyone actually use RAID 6??
I've looked at several items recently from 3ware stuff, Dell stuff, and some of IBM's big stuff, and there is never any mention of Raid 6. |
#2
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Jerry Cloe wrote:
Does anyone actually use RAID 6?? I've looked at several items recently from 3ware stuff, Dell stuff, and some of IBM's big stuff, and there is never any mention of Raid 6. That's probably because no one has ever actually defined precisely what it is beyond marketing :-) Different vendors have used the term for different technologies in the past. -- Nik Simpson |
#3
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On Sun, 11 Jul 2004 17:54:31 -0400, "Nik Simpson"
wrote: Jerry Cloe wrote: Does anyone actually use RAID 6?? I've looked at several items recently from 3ware stuff, Dell stuff, and some of IBM's big stuff, and there is never any mention of Raid 6. That's probably because no one has ever actually defined precisely what it is beyond marketing :-) Different vendors have used the term for different technologies in the past. Ummm... while certain notorious individuals have claimed their implementations of some other scheme are actually a new scheme, I think you'll find that RAID 6 is usually defined as a mechanism using two different parity/ECC schemes such that a set can survive the loss of any two disks. Malc. |
#4
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Malcolm Weir wrote:
On Sun, 11 Jul 2004 17:54:31 -0400, "Nik Simpson" wrote: Jerry Cloe wrote: Does anyone actually use RAID 6?? I've looked at several items recently from 3ware stuff, Dell stuff, and some of IBM's big stuff, and there is never any mention of Raid 6. That's probably because no one has ever actually defined precisely what it is beyond marketing :-) Different vendors have used the term for different technologies in the past. Ummm... while certain notorious individuals have claimed their implementations of some other scheme are actually a new scheme, Yes, that was precisely what/who I was thinking of :-) I think you'll find that RAID 6 is usually defined as a mechanism using two different parity/ECC schemes such that a set can survive the loss of any two disks. I stand corrected, I realised that HP had defined RAID-6 this way, but I din't realize that is a generally accepted definition. -- Nik Simpson |
#5
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On Mon, 12 Jul 2004 15:05:29 -0700, Malcolm Weir
wrote: On Sun, 11 Jul 2004 17:54:31 -0400, "Nik Simpson" wrote: Jerry Cloe wrote: Does anyone actually use RAID 6?? I've looked at several items recently from 3ware stuff, Dell stuff, and some of IBM's big stuff, and there is never any mention of Raid 6. That's probably because no one has ever actually defined precisely what it is beyond marketing :-) Different vendors have used the term for different technologies in the past. Ummm... while certain notorious individuals have claimed their implementations of some other scheme are actually a new scheme, I think you'll find that RAID 6 is usually defined as a mechanism using two different parity/ECC schemes such that a set can survive the loss of any two disks. Malc. Prior to this post I had never heard of raid 6 and have (still) no idea what it's structure is. Going on Malcolm's allusion I assume Double Disk Parity or Diagonal Parity or whatever is raid 6 nowadays? ~F |
#6
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"Faeandar" wrote in message ... .... Prior to this post I had never heard of raid 6 and have (still) no idea what it's structure is. Going on Malcolm's allusion I assume Double Disk Parity or Diagonal Parity or whatever is raid 6 nowadays? Well, whatever allows you to survive the loss of any two disks without loss of data (and isn't doubly-mirrored - i.e., is a parity-like mechanism). Michael Rabin did some work a while ago on generalized mechanisms that allow data to be spread across m + n disks such that up to n disks can be lost without loss of data. Special solutions exist that allow the data on the m disks to be usable directly (just as you can read directly from a member of a RAID-5 set rather than having to read m disks to get anything at all). Similar algorithms are IIRC used in the Berkeley 'Ocean Store' project, which allows reconstruction of data from any m of m + n sources (unless I'm thinking of Freenet - they are similar in some respects). - bill |
#7
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"Bill Todd" writes:
Prior to this post I had never heard of raid 6 and have (still) no idea what it's structure is. Going on Malcolm's allusion I assume Double Disk Parity or Diagonal Parity or whatever is raid 6 nowadays? Well, whatever allows you to survive the loss of any two disks without loss of data (and isn't doubly-mirrored - i.e., is a parity-like mechanism). Michael Rabin did some work a while ago on generalized mechanisms that allow data to be spread across m + n disks such that up to n disks can be lost without loss of data. Special solutions exist that allow the data on the m disks to be usable directly (just as you can read directly from a member of a RAID-5 set rather than having to read m disks to get anything at all). There are pretty straightforward ways to do that. For example, choose k so that 2**k is larger than m+n. k=8 is convenient and lets you have up to 256 drives, which is more than you'll find in most RAID's, and it means you do your operations on 8-bit data bytes. Let b(x,y) denote the contents of byte #y on drive #x where the drives go from 0,1,...,m+n-1, and we'll assume k=8 below. Now for the first m disks, just write the data directly, i.e. b(x,y) for x m is just whatever is in the file system at that spot. So now you can use those disks directly. For the other n disks, set b(x,y)=L(x) where L(x) is the value of the degree m-1 Lagrange interpolation polynomial over GF(2**8) running through (0, b(0,y)), (1, b(1,y)), ..., (m-1, b(m-1, y)) and evaluated at x. That L(x) calculation requires some arithmetic in GF(2**8) that amount to a few 8-bit xor's and 256-element table lookups for each byte written to each of the spare drives, which can be done fast in either hardware or software. It's more work than simply computing parity (m xor's per byte) but it's still not too bad. Is that similar to what Rabin did? In general, error correcting codes designed to recover data where the error locations are known are called "erasure codes" and there's a body of literature about them. The example above is something I figured out a while ago that I've never seen published, but I haven't studied the field much so I don't know what else is out there. Then again, I haven't implemented this scheme so maybe I made some unfixable error and it just plain doesn't work ;-). BTW, the above is loosely inspired by the Blakeley-Shamir secret sharing scheme from cryptography, that lets you split up a secret number into m+n pieces so that any m are enough to recover the number. |
#8
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"Paul Rubin" wrote in message ... "Bill Todd" writes: Prior to this post I had never heard of raid 6 and have (still) no idea what it's structure is. Going on Malcolm's allusion I assume Double Disk Parity or Diagonal Parity or whatever is raid 6 nowadays? Well, whatever allows you to survive the loss of any two disks without loss of data (and isn't doubly-mirrored - i.e., is a parity-like mechanism). Michael Rabin did some work a while ago on generalized mechanisms that allow data to be spread across m + n disks such that up to n disks can be lost without loss of data. Special solutions exist that allow the data on the m disks to be usable directly (just as you can read directly from a member of a RAID-5 set rather than having to read m disks to get anything at all). There are pretty straightforward ways to do that. For example, choose k so that 2**k is larger than m+n. k=8 is convenient and lets you have up to 256 drives, which is more than you'll find in most RAID's, and it means you do your operations on 8-bit data bytes. Let b(x,y) denote the contents of byte #y on drive #x where the drives go from 0,1,...,m+n-1, and we'll assume k=8 below. Now for the first m disks, just write the data directly, i.e. b(x,y) for x m is just whatever is in the file system at that spot. So now you can use those disks directly. For the other n disks, set b(x,y)=L(x) where L(x) is the value of the degree m-1 Lagrange interpolation polynomial over GF(2**8) running through (0, b(0,y)), (1, b(1,y)), ..., (m-1, b(m-1, y)) and evaluated at x. That L(x) calculation requires some arithmetic in GF(2**8) that amount to a few 8-bit xor's and 256-element table lookups for each byte written to each of the spare drives, which can be done fast in either hardware or software. It's more work than simply computing parity (m xor's per byte) but it's still not too bad. Is that similar to what Rabin did? My recollection (somewhat vague at this point) is yes. I meant to add in the previous response that as the size of the disk group grows surviving the loss of more than a single disk becomes more important. This may not be a significant issue for static configurations (where you can just gang up multiple small RAID-5 groups instead of one large group with minimal increase in disk count), but for more innovative use of large disk groups (e.g., accommodating growth/shrinkage without requiring a complete data shuffle) it can start to become one. - bill |
#9
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Bill Todd wrote:
Well, whatever allows you to survive the loss of any two disks without loss of data (and isn't doubly-mirrored - i.e., is a parity-like mechanism). Michael Rabin did some work a while ago on generalized mechanisms that allow data to be spread across m + n disks such that up to n disks can be lost without loss of data. Special solutions exist that allow the data on the m disks to be usable directly (just as you can read directly from a member of a RAID-5 set rather than having to read m disks to get anything at all). Similar algorithms are IIRC used in the Berkeley 'Ocean Store' project, which allows reconstruction of data from any m of m + n sources (unless I'm thinking of Freenet - they are similar in some respects). - bill [I agree] I found in John May's 2001 book "Parallel IO for high performance computing" that RAID 6 is a collection of techniques that allow a system to lose 2 disks, for example: -P+Q parity is like RAID5 with 2 algorithms -MxN is like having the drives in a 2D array with M+N spare drives. I think the max number of faulty drives with no data loss is M+N-1, with adequate location; but it can not handle 4 drives failing if they are in square. [my question] Inostor launched RAIDn last year, with attractive features. http://www.inostor.com/products/prod...AIDn_index.htm None is said about algorithms, data placement, etc... Is that marketing (some known RAID6 issues) or something different? Stephane |
#10
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Nik Simpson wrote:
I think you'll find that RAID 6 is usually defined as a mechanism using two different parity/ECC schemes such that a set can survive the loss of any two disks. I stand corrected, I realised that HP had defined RAID-6 this way, but I din't realize that is a generally accepted definition. I think HP/Compaq calls this Advanced Data Guard (ADG). Isn't the RAID-6 name copyrighted to someone? B. |
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