Re: [PATCH] md: Use new topology calls to indicate alignment and I/O sizes

On Wednesday June 10, martin.petersen@oracle.com wrote:

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"Neil" == NeilBrown  [off-list ref] writes:

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+ blk_queue_io_opt(mddev->queue, chunk_size * (mddev->raid_disks >>
 	blk_queue_io_opt(mddev->1));

Neil>  ">> 1" is wrong.  A raid10 may have more than 2 copies for each
Neil>  block.

I recall pondering when I wrote that.  Never got around to revisiting
the issue.  Your RAID10 implementation is a bit hipper than most :)

And I know where my towel is, too :-)

Neil>  This calculation needs to be a bit more subtle and take about of
Neil>  the particular layout (near, offset, or far).

Why would layout interfere with preferred I/O size?

For a 'far' or 'offset' layout, you probably want
    chunksize * raid_disks
just like raid0. 

For a 'near' layout, what you suggest seems good.

The driver actually allows both 'near' and 'far'.  I doubt if anyone
would use it, but it would be nice to get it 'right'.

How about:

    if (conf->raid_disks % conf->copies)
	blk_queue_io_opt(mddev->queue, chunk_size * conf->raid_disks);
    else
	blk_queue_io_opt(mddev->queue, chunk_size *
			 (conf->raid_disks / conf->copies));

I think that if you replace "conf->copies" by 'nc' (which is
 (mddev->layout & 255)
) then that will be perfect.  For a 'far' layout, nc is normally '1',
and either branch will give the correct value.

Thanks,
NeilBrown

I.e.:

2 drives, 2 copies: a1 a1       optimal I/O = 1 * chunk_size

3 drives, 2 copies: a1 a1 a2
                    a2 a3 a3    optimal I/O = 3 * chunk_size

4 drives, 2 copies: a1 a1 a2 a2	optimal I/O = 2 * chunk_size

4 drives, 3 copies: a1 a1 a1 a2
                    a2 a2 a3 a3
                    a3 a4 a4 a4 optimal I/O = 4 * chunk_size

And so on.  Does that make sense?  I.e. we try to keep all drives
equally busy.


Neil> You seem to be treating raid4 like raid6 :-)

Whoops.  I always forget about raid4.  Fixed.

-- 
Martin K. Petersen	Oracle Linux Engineering
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