Re: btrfs: why default 4M readahead size?

[Shaohua Li <shaohua.li@intel.com>]

On Thu, Mar 18, 2010 at 08:53:13PM +0800, Chris Mason wrote:
> On Thu, Mar 18, 2010 at 09:42:57AM +0800, Shaohua Li wrote:
> > Btrfs uses below equation to calculate ra_pages:
> > 	fs_info->bdi.ra_pages = max(fs_info->bdi.ra_pages,
> >               		4 * 1024 * 1024 / PAGE_CACHE_SIZE);
> > is the max() a typo of min()? This makes the readahead size is 4M by default,
> > which is too big.
> 
> Looks like things have changed since I tuned that number.  Fengguang has
> been busy ;)
> 
> > I have a system with 16 CPU, 6G memory and 12 sata disks. I create a btrfs for
> > each disk, so this isn't a raid setup. The test is fio, which has 12 tasks to
> > access 12 files for each disk. The fio test is mmap sequential read. I measure
> > the performance with different readahead size:
> > ra size		io throughput
> > 4M		268288 k/s
> > 2M		367616 k/s
> > 1M		431104 k/s
> > 512K		474112 k/s
> > 256K		512000 k/s
> > 128K		538624 k/s
> > The 4M default readahead size has poor performance.
> > I also does sync sequential read test, the test difference in't that big. But
> > the 4M case still has about 10% drop compared to the 512k case.
> 
> I'm surprised the 4M is so much slower.  At any rate, the larger size
> was selected because btrfs checksumming means we need a bigger buffer to
> keep the disks saturated.  Were you on a fancy intel box with hardware
> crc32c enabled?
yes, this machine supports sse4.2 instruction. Let me check the result with checksum
disabled.

Thanks,
Shaohua