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 = 104000000000000 reads. But my BER is smaller: 10^14 = 100000000000000 So I am (theoretically) guaranteed to lose another drive on raid rebuild. Then the calculation for _that_ rebuild is: 12*8*10^12 = 96000000000000 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. This message posted from opensolaris.org _______________________________________________ zfs-discuss mailing list zfs-discuss@opensolaris.org http://mail.opensolaris.org/mailman/listinfo/zfs-discuss
