From: dean gaudet <dean@arctic.org>
To: Peter Rabbitson <rabbit@rabbit.us>
Cc: linux-raid@vger.kernel.org
Subject: Re: XFS sunit/swidth for raid10
Date: Sun, 25 Mar 2007 12:16:20 -0700 (PDT) [thread overview]
Message-ID: <Pine.LNX.4.64.0703251214460.30863@twinlark.arctic.org> (raw)
In-Reply-To: <46038DD0.1060803@rabbit.us>
On Fri, 23 Mar 2007, Peter Rabbitson wrote:
> dean gaudet wrote:
> > On Thu, 22 Mar 2007, Peter Rabbitson wrote:
> >
> > > dean gaudet wrote:
> > > > On Thu, 22 Mar 2007, Peter Rabbitson wrote:
> > > >
> > > > > Hi,
> > > > > How does one determine the XFS sunit and swidth sizes for a software
> > > > > raid10
> > > > > with 3 copies?
> > > > mkfs.xfs uses the GET_ARRAY_INFO ioctl to get the data it needs from
> > > > software raid and select an appropriate sunit/swidth...
> > > >
> > > > although i'm not sure i agree entirely with its choice for raid10:
> > > So do I, especially as it makes no checks for the amount of copies (3 in
> > > my
> > > case, not 2).
> > >
> > > > it probably doesn't matter.
> > > This was essentially my question. For an array -pf3 -c1024 I get swidth =
> > > 4 *
> > > sunit = 4MiB. Is it about right and does it matter at all?
> >
> > how many drives?
> >
>
> Sorry. 4 drives, 3 far copies (so any 2 drives can fail), 1M chunk.
my mind continues to be blown by linux raid10.
so that's like raid1 on 4 disks except the copies are offset by 1/4th of
the disk?
i think swidth = 4*sunit is the right config then -- 'cause a read of 4MiB
will stride all 4 disks...
-dean
prev parent reply other threads:[~2007-03-25 19:16 UTC|newest]
Thread overview: 6+ messages / expand[flat|nested] mbox.gz Atom feed top
2007-03-21 23:23 XFS sunit/swidth for raid10 Peter Rabbitson
2007-03-22 8:23 ` dean gaudet
2007-03-22 12:05 ` Peter Rabbitson
2007-03-23 6:48 ` dean gaudet
2007-03-23 8:20 ` Peter Rabbitson
2007-03-25 19:16 ` dean gaudet [this message]
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