I am building a 14 disk raid 6 array with 1 TB seagate AS (non-enterprise)
drives.
So there will be 14 disks total, 2 of them will be parity, 12 TB space
available.
My drives have a BER of 10^14
I am quite scared by my calculations - it appears that if one drive fails, and
I do a rebuild, I will perform:
13*8*10^12 = 104
reads. But my BER is smaller:
10^14 = 100
So I am (theoretically) guaranteed to lose another drive on raid rebuild. Then
the calculation for _that_ rebuild is:
12*8*10^12 = 96
So no longer guaranteed, but 96% isn't good.
I have looked all over, and these seem to be the accepted calculations - which
means if I ever have to rebuild, I'm toast.
But here is the question - the part I am having trouble understanding:
The 13*8*10^12 operations required for the first rebuild isn't that the
number for _the entire array_ ? Any given 1 TB disk only has 10^12 bits on it
_total_. So why would I ever do more than 10^12 operations on the disk ?
It seems very odd to me that a raid controller would have to access any given
bit more than once to do a rebuild ... and the total number of bits on a drive
is 10^12, which is far below the 10^14 BER number.
So I guess my question is - why are we all doing this calculation, wherein we
apply the total operations across an entire array rebuild to a single drives
BER number ?
Thanks.
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