Raid 5 performance

[Ralph Doncaster]

On Wed, 22 Sep 2004, John Macdonald wrote:

> On Wed, Sep 22, 2004 at 10:40:24AM -0400, Ralph Doncaster wrote:
> > > If a write
> > > spans some, but not all, of the data disks in a stripe, then
> > > data must be read from the disk to compute the new parity info.
> > This is a correct statement, and this is what happens when you write 16KB
> > chunks to a stripe of 4 disks (3 data + 1 parity).  The Linux RAID code
> > doesn't coalesce writes of multiple chunks (i.e. 3x 16KB chunks) to
> > optimize away the theoretically unecessary read for parity computation.
>
> So, the driver does not compute the parity directly from the
> data for a 3+1 setup when you write exactly one entire data
> stripe of 12k, but it does for a 4+1 setup when you write
> exactly one data stripe of 16k?

No, there you go trying to put words in my mouth again.  I've been polite
up to this point, but you seem intent on being rude because you don't like
being proven wrong.

RAID doesn't write exactly one entire data stripe of 12KB, since that
is not a valid chunk size.  When writing 16KB to a 3+1 drive array, the
first 12KB written should not require a read for parity computation.  It's
the last 4KB that will require the read.

Despite all your efforts to complicate the issue, it's only grade-5 math.
If the number of disk sectors written modulus the number of data drives !=
0, a read will be required for parity computation.  The number of disk
sectors written for a RAID chunk has to be a power of 2.  Therefore with 2
data disks (+1 for parity), a read will NEVER be required for parity
computation when writing a chunk.

If you still can't comprehend that, you'll have to ask someone else for
help as I've sent in my unsubscribe request to majordomo-jmbJ75VLJBo at public.gmane.org

-Ralph
--
The Toronto Linux Users Group.      Meetings: http://tlug.ss.org
TLUG requests: Linux topics, No HTML, wrap text below 80 columns
How to UNSUBSCRIBE: http://tlug.ss.org/subscribe.shtml