On Fri, 2003-04-11 at 12:14, Derek Glidden wrote:
> (in the meanwhile, I've gone and made a Linux software-RAID5 box in the
> lab to play with)
> On Fri, 2003-04-11 at 12:26, Christian Guggenberger wrote:
> > > "block units" and then in "bytes." So if I have made a RAID volume with
> > > 64K stripes, should I specify a su value of 128? (128*512 = 64K)
> > >
> > no, su=64k _or_ sunit=128 !
> > > data disks in a RAID device." So I take this to mean that if I have 8
> > > disks in a RAID5, my "sw" parameter is "8"?
> > >
> > having an 8 disk RAID5, you'll need swidth=(8-1)*sunit (=896 for your
> > example) - if I remember some
> > posts about that topic correctly.( a 8 disk RAID5 has 7 "data" disks in that
> > term)
> It gets even more confusing when I do:
> # mkfs -d sunit=128,swidth=384 /dev/md4
> on software RAID5 with 4 disks and 64k stripes, and the status screen
> data = bsize=4096 blocks=84286944,
> = sunit=16 swidth=48 blks,
> since it apparently displays the "sunit" and "swidth" values in terms of
> XFS blocks, which are 4096 bytes each....
> Thankfully, "mkfs.xfs" with no options Just Does The Right Thing on this
> machine, and sets sunit and swidth correctly when it makes the
> filesystem. :) (Though it might be different with hardware RAID...)
All the output units are in fsblocks, on the mkfs command line it does
not know what you blocksize is all the time.
Steve Lord voice: +1-651-683-3511
Principal Engineer, Filesystem Software email: lord@xxxxxxx