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.
_______________________________________________ CentOS mailing list CentOS@xxxxxxxxxx http://lists.centos.org/mailman/listinfo/centos