Unable to understand RAID PARTITION


 
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# 8  
Old 04-17-2014
thx bakunin , I think my understanding is much better now . but still I have one query regarding RAID 5 .
looking forward for your help.
Let say there is a 4 Drive , Drive-A, Drive-B, Drive-C, and Drive-D.
Now assuming there is E , F, G and H is the info which is going to be stored in a Drive , like below

Code:
Drive-A   
E1
F1
G1
Hp

Code:
Drive-B
E2
F2
Gp
H1

Code:
Drive-C
E3
Fp
G2
H2

Code:
Drive-D
Ep
F3
G3
H3





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.
# 9  
Old 04-17-2014
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
# 10  
Old 04-17-2014
Hi Bakunin,

thx once again for above info.

i just want to know ,do we have a seperate parity disk in RAID-5 like we have in RAID-4 ?

my understanding here is as, in RAID-5 we store parity in each HD so, there should not be a sepearate parity disk in RAID-5

I am sorry if it looks like argument . just want to clear my doubt.
# 11  
Old 04-18-2014
On RAID3 the parity drive is fixed. So, for example, given the minimum of 3 drives, you get the usable capacity of 2 drives, and lose the capacity of 1 drive for parity. Similarly, if you had 50 drives in the array, you would get a usable capacity of 49 drives.

The issue with RAID3 is that all files, even very tiny files which have a length of, say, less than 512 bytes (one disk sector) would still involve writing the parity data so the parity disk gets hammered and can cause a bottleneck. (RAID3 is high performance for the likes of big databases where the vast majority of operations are read and, if the array is healthy, the parity doesn't need to be read at all.)

Then, some bright spark thought that if the parity storage was rotated ('rotating parity' being the technical term) then the load would be evenly spread. This is RAID5. So, for example, for one stripe of the array, data could be written on drives 1 and 2 with the data written on drive 3. The very next stripe would have the data written on drives 2 and 3 with the parity on drive 1, and so on.

There's also now a RAID6 which involves a more advanced parity algorithm and uses 2 parity drives (therefore losing the capacity of 2 drives per array) but fault tolerant to 2 drive failures and still remain operational.

Hope that helps.

Last edited by hicksd8; 04-18-2014 at 05:50 PM..
This User Gave Thanks to hicksd8 For This Post:
# 12  
Old 04-22-2014
hi hicksd8,
thx a lot for such a explanation.

here is one more question
then what is the difference between RAID-3 and RAID-4 ? both use one dist for parity ?

Regards
Scriptor
# 13  
Old 04-22-2014
Now you are getting technical!!! I can tell you that in 4 years as the technical director of a storage distributor to corporates, I never once was asked for a RAID4 configuration. The main RAID levels in the market are 0, 1, 3, 5, 6 (recently), 10 (which is 0 +1; many mirrored pairs).

RAID-4 uses large stripes. It can read records from any single drive. This allows you to take advantage of overlapped I/O for read operations. Since all write operations have to update the parity drive, no I/O overlapping is possible.

RAID-5 includes a rotating parity array, thus addressing the write limitation in RAID-4. All read and write operations can be overlapped. RAID-5 stores parity information but not redundant data (however, parity information can be used to reconstruct the data). RAID-5 requires at least three and usually five disks for the array. It's best for multi-user systems in which performance is not critical or which do few write operations.

RAID-4 is faster than RAID-5 because the parity is written to a dedicated disk, rather than scattered around with the data.
If you have N disks, you'd have N-1 data disks, and 1 parity disk.

What happens when the parity disk is lost? You have lost all protection! There is no parity, and hence no way to calculate the data.

For this reason RAID-4 is considered dangerous and is not often used.

So, the advantage of RAID-4 is that it has the highest read data transaction rate, with a medium write data transaction rate.

Data is stripped on disk creating high efficiency along with a good aggregate transfer rate. Parity is stored on a separate disk.

RAID-4 offers no other advantages over RAID-5.

Hope that helps.
# 14  
Old 04-23-2014
hi hicksd8

my question is
what is the difference between RAID-3 and RAID-4 ? both use one disk for parity ?

i just want to the basic difference.

Regards,
Scriptor
 
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