March 24, 2011 at 8:06 pm
Comments posted to this topic are about the item RAID 5 Parity
March 24, 2011 at 8:44 pm
Mmm, terminology is getting in my way for this one. My definition for parity (rightly or wrongly) applies at the byte level (Parity Bit).
For RAID 5, I have normally used the term "checksum" which is the value stored on 1 of the 3 disks required to store a single value,
March 24, 2011 at 8:51 pm
A Better reference: http://technet.microsoft.com/en-us/library/cc938485.aspx
Checksum isn't a term I've seen associate with RAID. Almost always things are listed as parity calculations.
March 24, 2011 at 8:56 pm
Cat, it's for both, and same purpose. In a RAID 5, for each section of disk (I forget the break out) in say a 4 disk array, 3 have data, one has a parity to make sure the other three stay intact. Because of that parity, you can assume the data on the 'lost' disk while it's replaced, and rebuild it, due to reverse construction of the bit/bytes. Same concept, expanded further.
Lose two drives and it's game over though, as most folks are well aware. 🙂
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March 24, 2011 at 11:40 pm
Thanks for the question.
Here is another explanation that makes some sense.
http://riceball.com/d/content/raid-5-parity-what-it-and-how-does-it-work
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March 25, 2011 at 12:42 am
Thanks for the question Steve!
March 25, 2011 at 2:05 am
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March 25, 2011 at 2:46 am
Nice question, thanks.
Luckily I had the Wikipedia page for RAID bookmarked, in case new QotD featuring RAID would arrise 🙂
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March 25, 2011 at 3:46 am
Craig Farrell (3/24/2011)
Cat, it's for both, and same purpose. In a RAID 5, for each section of disk (I forget the break out) in say a 4 disk array, 3 have data, one has a parity to make sure the other three stay intact. Because of that parity, you can assume the data on the 'lost' disk while it's replaced, and rebuild it, due to reverse construction of the bit/bytes. Same concept, expanded further.Lose two drives and it's game over though, as most folks are well aware. 🙂
I believe that the data is actually stored on 3 disks...2 have the actual data and the 3rd has the exclusive OR of the other 2 disks. Using this, if you loose any one of the 3 disks that store the data you are after, the other 2 have enough information to give to retrieve the data. This basic pattern is used regardless of the number of disks in the array. When there are more than 3 disks, each chunk of data is still stored on 3 of the disks. The controller manages the allocation of disk space to ensure that all disks on the array are utilised.
Steve - whilst you may not have heard of checksum (Vs parity) in relation to RAID 5 , try searching google for "RAID 5 checksum". I do agree that you can view this as a "parity" but I am one of them dinosaurs who have been around since the days of punch cards, paper tape and magnetic tape. Parity back then was more to do with detecting error.
March 25, 2011 at 7:42 am
Nice question. Love these easy harware quesitons, used to make money and impress friends with good knowledge about RAID and how the differant levels performed and why.
These days I am more interested in the default block size used when writing and reading data to the Disk Volume.
March 25, 2011 at 12:15 pm
happycat59 (3/25/2011)
Steve - whilst you may not have heard of checksum (Vs parity) in relation to RAID 5 , try searching google for "RAID 5 checksum". I do agree that you can view this as a "parity" but I am one of them dinosaurs who have been around since the days of punch cards, paper tape and magnetic tape. Parity back then was more to do with detecting error.
Isn't that still what the parity/checksum in RAID5 is for to detect and correct errors?
March 25, 2011 at 10:29 pm
Good one 🙂
M&M
March 26, 2011 at 2:04 am
I learned some thing form this question.
I referred : http://download.oracle.com/docs/cd/E19494-01/820-1260-15/appendixf.html#50548797_98008
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Gobikannan
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