i am studying about raid partion.i am not able to understand RAID level 5.
below is excerpt taken from tutorial.
RAID level 5
are they trying to say that the will be one extra disk which contain all the data. let says there are 4 disk. out of 4 , 3 disk are used for storing data and 4th one contain all the data (data of disk 1 2 and 3).
so here 4 th disk is acting as a parity disk or extra disk . which can be access whenever any one disk got corrupted.
please let me know if my understanding is correct or not , if not pls explain me what does raid 5 means then
Suppose you have a RAID 5 configuration consisting of 4 disks.
If you store "1" on drive 1, "2" on drive 2, "3" on drive 3, and "6" on drive 4, then you can always determine the value of any one missing disk.
ie
1+2+3=6
6-3-2=1
etc.
my confusion is still same . looks to me as i am unable to undersatnd it
please answer my query . also before putting my question in forum i have already gone throgh this wiki page
let consider jgt's example
as per him
Suppose you have a RAID 5 configuration consisting of 4 disks.
If you store "1" on drive 1, "2" on drive 2, "3" on drive 3, and "6" on drive 4, then you can always determine the value of any one missing disk.
ie
1+2+3=6
6-3-2=1
it means that drive 4 will contain the data of drive 1 drive 2 and drive 3
and if any drive get failed it will serch data in drive 4
is this is what you want to say am i right
or
each 4 drive have a same data and ata time only one drive is active and rest is passive mode. if suppose drive one got failed in that case drive 2 become active and so on
also please explain below line. i know i am asking stupid type question and reflecting my poor IQ too this time
A RAID 5 comprises block-level striping with distributed parity
here pls explain what is striping and block level striping in raid. with example
what is distributed parity in raid with example pls
You dont understand because you are not open minded...
Forget everything you know and start reading the given link. then try to understand what its about ( Dont start interfeering with what you know or think you know!...)
Look at the schema how A is written - where is Ap ? What is Ap?
Then do the same for the next and... Where is Xp?
Once that is clear
What do all disks have?
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
Drive-A
E1
F1
G1
Hp
Drive-B
E2
F2
Gp
H1
Drive-C
E3
Fp
G2
H2
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
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.
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.
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.