On 19/12/2009 01:05, Bernd Schubert wrote:

quoted

On Saturday 19 December 2009, Matt Tehonica wrote:

quoted

I have a 4 disk RAID5 using a 2048K chunk size and using XFS

4 disks is a bad idea. You should have 2^n data disks, but you have 2^1

+ 1 =

quoted

3 data disks. As parity information are calculated in the power of two

and

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blocks are written in the power of two

Sorry, but where did you get that from? p = d1 xor d2 xor d3 has nothing
to do with powers of two, and I'm sure blocks are written whenever they
need to be, not in powers of two.

	Yeah, I was scratching my head over that one, too.  It sounded bogus
to me, but I didn't want to open my mouth, so to speak, when I was unsure of
myself.  Being far from expert in the matter, I can't be certain, but I
surely can think of no reason why writes would occur in powers of two, or
even be more efficient because of it.

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you probably have read operations,
when you only want to write.

That will depend on how much data you're trying to write. With 3 data
discs and a 2M chunk size, writes in multiples of 6M won't need reads.
Writing a 25M file would therefore write 4 stripes and need to read to
do the last 1M. With 4 data discs, it'd be 8M multiples, and you'd write
3 stripes and need a read to do the last 1M. No difference.

	I hadn't really considered this before, and I am curious.  Of course
there is no reason for md to read a stripe marked as being in use if the
data to be written will fill an entire stripe.  However, does it only apply
this logic if the data will completely fill a stripe?  The most efficient
use of disk space of course will be accomplished if the system reads the
potential partially used target stripe whenever the write buffer contains
even 1 chunk less than a full stripe, but the most efficient write speeds
will only check on writing to a partially used stripe if the write buffer
contains less than half a stripe worth of data.  Does anyone know which is
the case?