Unable to understand RAID PARTITION


 
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# 1  
Old 04-15-2014
Unable to understand RAID PARTITION

Hi

i am studying about raid partion.i am not able to understand RAID level 5.
below is excerpt taken from tutorial.
Code:
 RAID level 5

Quote:

Combines striping and parity. Data is written across all disks as in RAID 0, but parity data is also written to one of the disks. Should a hard drive failure occur, this parity data can be used to recover the data from the failed drive, including while the data is being accessed and the drive is still missing from the array.
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

Regards,
Scriptor

Last edited by Perderabo; 04-15-2014 at 03:21 PM..
# 2  
Old 04-15-2014
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.
# 3  
Old 04-15-2014
Hi jgt

so it means that drive 4 will contain the data of drive 1 drive 2 and drive 3

is this is what you want to say am i right
# 4  
Old 04-15-2014
# 5  
Old 04-16-2014
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
Code:
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

Code:
 
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

Last edited by scriptor; 04-16-2014 at 09:56 AM..
# 6  
Old 04-16-2014
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?

Take a disk:
Say Disk 1
What is written on it?

Reminder: DukeNuke's Link: https://en.wikipedia.org/wiki/RAID_5#RAID_5

Last edited by vbe; 04-16-2014 at 12:04 PM.. Reason: typo
# 7  
Old 04-16-2014
Maybe my explanation of the RAID mechanism is better suited to your understanding. See chapter "RAIDing the storage":

The Most Incomplete Guide to Performance Tuning

The principle is the following. Suppose you write this onto the disks:

Code:
disk1:  1
disk2:  2
disk3:  3
disk4:  4
disk5:  5
disk6: 15

Now, suppose disk6 fails: no problem, because you do not need it - your data is on disk1-5.

Suppose one of the other disks fails, say: disk4. To retrieve what was written on disk 4 compute:

Code:
disk6 - disk1 - disk2 - disk3 - disk5 =
   15 -     1 -     2 -     3 -     5 = 4

Similar for every other disk1-5. It is also obvious, that the system will survive the loss of ONE disk, but not two.

I hope this helps.

bakunin

Last edited by bakunin; 04-16-2014 at 01:08 PM..
This User Gave Thanks to bakunin For This Post:
 
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