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How do I calculate the parity bit? In a RAID 3 odd parity with 5 disks. How do I get the parity bit?

The answer was

But I want to know how is this calculated?

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closed as off topic by Jason Berg, EEAA, gWaldo, Iain, MDMarra Nov 19 '11 at 22:14

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Homework question? RAID 3 is bad and nobody uses it. –  Nic Nov 19 '11 at 4:14
    
Why ever would you need to know this?! The only thing that I can think of is that you're writing a RAID Controller! Also, don't use RAID3; use RAID5 or 6 –  gWaldo Nov 19 '11 at 4:14
    
@Nic, yes homework. Computer Organization to be exact :) Just for knowledge? What is commonly used? RAID0 & 1 for home, 5 for enterprise? I suppose there are higher levels ones that I missed? –  Jiew Meng Nov 19 '11 at 6:24
    
@gWaldo, like Nic mentioned, its homework. I also don't know why the want to teach a CS student just low level stuff actually. Its only useful if you really specialize into maybe hardware? Or hardcore optimization etc I'd say –  Jiew Meng Nov 19 '11 at 6:26
    
Agreed, @jiewmeng - Nic got his comment in 4 seconds before I did... –  gWaldo Nov 19 '11 at 19:33
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2 Answers 2

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"Odd parity" means the sum of all the its is odd. This is the same as saying the number of 1 bits is odd. So if the number of one bits is odd, the parity bit must be 0 to keep it that way. If it's even, the parity bit must be 1 to make it odd.

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The parity is calculated via a logical "exclusive-or" operation, XOR.

It results in true (1) if one or the other of the values that it's calculating on is true, but not if both are true - if neither or both values are true, the result is false (0). Using this logic, the array generates extra data that it is able to use to reconstruct the data on a lost disk.

The catch with odd-parity is that it's reversed - if the XOR operation results in a 1, a 0 is stored instead. I'm not 100% sure, but I don't believe that odd parity gets much use in the real world.

So, in the examples above, the second row:

1(disk0)       XOR 0(disk1) = 1
1(last result) XOR 1(disk2) = 0
0(last result) XOR 1(disk3) = 1 (result of the XOR across the entire row)

In odd parity, flip the result and that's the result bit; 0, in this case.

The reason that this is interesting and useful? Let's say disk 2 is lost, and you need to figure out what was in it.

1(parity bit from disk4) XOR 1(disk0) = 0
0(last result)           XOR 0(disk1) = 0
0(last result)           XOR 1(disk3) = 1 (calculated disk2 data)

Using the parity, data from a lost disk can be reconstructed with the data from the parity disk plus all of the other disks - this is the reason that this type of RAID group can tolerate the loss of any one disk, but will fail if a second disk fails.

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