On Mon, Nov 27, 2000 at 04:58:12PM -0200, Carlos Carvalho wrote:
> 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...
The submect line said 50%, and I missed that, but it was early in the morning ;)
>
> >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?
Parity is XOR of the bits. For disks A, B, C and D we pick a bit a, b, c, d.
a xor b xor c = d (parity)
for writing a new c (let's call it c1) we can either:
read a and b
calculate a xor b xor c1
write this result as d, and write c1
or we can
read c
if c1 != c then d should be !d
write the new d and write c1
You still have full parity. At least as far as I can see...
--
................................................................
: [EMAIL PROTECTED] : And I see the elder races, :
:.........................: putrid forms of man :
: Jakob �stergaard : See him rise and claim the earth, :
: OZ9ABN : his downfall is at hand. :
:.........................:............{Konkhra}...............:
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