On Wednesday June 10, martin.petersen@xxxxxxxxxx wrote:
> >>>>> "Neil" == NeilBrown <neilb@xxxxxxx> writes:
>
> >> + 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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