> -----Original Message-----
> From: centos-bounces@xxxxxxxxxx
> [mailto:centos-bounces@xxxxxxxxxx] On Behalf Of Aleksandar Milivojevic
> Sent: Monday, January 08, 2007 1:00 PM
> To: centos@xxxxxxxxxx
> Subject: RE: Disk Elevator
>
> Quoting "Ross S. W. Walker" <rwalker@xxxxxxxxxxxxx>:
>
> > The biggest performance gain you can achieve on a raid
> array is to make
> > sure you format the volume aligned to your raid stripe
> size. For example
> > if you have a 4 drive raid 5 and it is using 64K chunks, your stripe
> > size will be 256K. Given a 4K filesystem block size you
> would then have
> > a stride of 64 (256/4), so when you format your volume:
> >
> > Mke2fs -E stride=64 (other needed options -j for ext3, -N
> <# of inodes>
> > for extended # of i-nodes, -O dir_index speeds up directory
> searches for
> > large # of files) /dev/XXXX
>
> Shouldn't the argument for stride option be how many file system
> blocks there is per stripe? After all, there's no way for OS
> to guess
> what RAID level you are using. For 4 disk RAID5 with 64k chunks and
> 4k file system blocks you have only 48 file system blocks per stripe
> ((4-1)x64k/4k=48). So it should be -E stride=48 in this particular
> case. If it was 4 disk RAID0 array, than it would be 64
> (4x64k/4k=64). If it was 4 disk RAID10 array, than it would be 32
> ((4/2)*64k/4k=32). Or at least that's the way I understood it by
> reading the man page.
You are correct, leave one of the chunks off for the parity, so for 4
disk raid5 stride=48. I had just computed all 4 chunks as part of the
stride.
BTW that parity chunk still needs to be in memory to avoid the read on
it, no? In that case wouldn't a stride of 64 help in that case? And if
the stride leaves out the parity chunk then will not successive
read-aheads cause a continuous wrap of the stripe which will negate the
effect of the stride by not having the complete stripe cached?
-Ross
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