> On 20 Mar 2001 at 8:28am, Martin K. Petersen wrote
> > >>>>> "Joshua" == Joshua Baker-LePain <jlb17@xxxxxxxx> writes:
> > Joshua> My RAID is an IDE-SCSI system -- IDE disks, looks like SCSI to
> > Joshua> the host. It has 8 80GB drives in RAID 5, for about 560GB of
> > Joshua> usable space. I'd like it to be just one partition. My
> > Joshua> question is, are there any special parameters I should pass to
> > Joshua> mkfs for such a big (well, for me at least) partition? What
> > Joshua> about to tailor it to the hardware RAID?
> > Well. You could try and hint to the allocator how your RAID device is
> > set up. Try and look up which chunk size it uses. You can pass that
> > information on to mkfs using the sunit and swidth options. See the
> > mkfs.xfs man page for details.
> The unit's default is a stripe size of 64 512 byte blocks. The stripe
> size can be set between 4 and 128 (incrementally, of course). For the
> defaults, would this be correct:
> mkfs.xfs -d sunit=1,swidth=64
> Or should I multiply both of those numbers by 512? Also, should I make
> the logfile any bigger than the default due to the size of the partition?
Actually this is incorrect, you need
The sunit is the amount of data on each disk in 512 byte chunks, the swidth
is the total amount of data before we go back to the first disk again. You
have 8 disks in raid5 so 7 data disks and a parity disk. You need 7 * 64 as
the stripe width.
You probably want a bigger log file, you have lots of room to play with,
doing 16384b or 32768b is not going to hurt you. You should also mount
with -o logbufs=8 if you expect heavy traffic.
p.s. Let us know if these numbers make things go faster!