Quote:
Originally Posted by
scriptor
Now here my understanding is each drive itself acting as Parity Drive But in some tutorial I have seen that they mentioned Drive-D or last drive (3+1) is used as a parity drive
Please let me know which one is correct.
First off, things in reality are more complicated than i showed you. Just the principle remains the same:
You have several disks. You create a formula, by which you can "add" (this is to be understood loosely - in my example above it was a true addition, but in reality other procedures are used - Hamming codes, CRC, ...) the content of all the datadisks and write the "sum" (again - this word is loosely used here) onto the parity disk.
If any of the data disks fails you can take the checksum from the parity disk, "subtract" all the data disks and get the missing data. If the parity disk itself fails you still have all the data intact without any calculation.
In theoretical terms: the information stored on the N datadisks is now stored on the N+1 disks in the RAID set. If any disk fails you either do not need its information (in case of the parity disk) or you can compute the missing data by using the data from the other data disks plus the parity disk. It is pointless to argue where the
information is stored, because there is no difference
in information between "1+2" and "3" - just a difference in the
data representing this information.
I hope this helps.
bakunin