Jakob �stergaard ([EMAIL PROTECTED]) wrote on 27 November 2000 18:18:
>On Mon, Nov 27, 2000 at 02:24:03PM -0700, Hans Reiser wrote:
>> using reiserfs over raid5 with 5 disks. This is unnecessarily
>> suboptimal, it should be that parity writes are 20% of the disk
>> bandwidth. Comments?
>
>It would be great to have a number describing ``unnecessarily suboptimal''
>more accurately ;)
He just said 20%, 1 in 5...
>Last I looked, the RAID-5 code would, upon performing a write:
> 1) Read a block from all disks but the parity and the destination disk
> 2) Calculate parity from those reads along with the new block
> 3) Write the new block and the new parity
>
>However, I cannot see a reason why (1) cannot be replaced with reading the
>old parity, and the old data block, then changing (2) to calculating parity
>based on that information instead.
You mean read just the old parity and the old data block? If so that's
not enough, because to calculate the parity you need all blocks so
that if any one fails you can reconstruct it. You need all the others
to reconstruct, that's why there's redundancy of only one disk.
Did I miss anything obvious?
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