I meant sync not fsync(O_SYNC flag).
My main question is why there is better throughput when I make the agsize smaller?
On Wed, Oct 30, 2013 at 5:59 AM, Matthias Schniedermeyer <ms@xxxxxxx> wrote:
On 29.10.2013 18:10, K T wrote:I get the same bad performance with your dd statement.
> Hi,
>
> I have a 1 TB SATA disk(WD1003FBYX) with XFS. In my tests, I preallocate a
> bunch of 10GB files and write data to the files one at a time. I have
> observed that the default mkfs setting(4 AGs) gives very low throughput.
> When I reformat the disk with a agsize of 256mb(agcount=3726), I see better
> throughput. I thought with a bigger agsize, the files will be made of fewer
> extents and hence perform better(due to lesser entries in the extent map
> getting updated). But, according to my tests, the opposite seems to be
> true. Can you please explain why this the case? Am I missing something?
>
> My test parameters:
>
> mkfs.xfs -f /dev/sdbf1
> mount -o inode64 /dev/sdbf1 /mnt/test
> fallocate -l 10G fname
> dd if=/dev/zero of=fname bs=2M count=64 oflag=direct,sync conv=notrunc seek=0
fallocate -l 10G fname
time dd if=/dev/zero of=fname bs=2M count=64 oflag=direct,sync conv=notrunc seek=0
64+0 records in134217728 bytes (134 MB) copied, 4,24088 s, 31,6 MB/s
64+0 records out
After pondering the really hard to read dd-man-page.
Sync is for 'synchronized' I/O. aka REALLY BAD PERFORMANCE. And i assume
you don't really that.
I think what you meant is fsync. (a.k.a. File (and Metadata) has hit
stable-storage before dd exits).
That is: conv=fsync
So:
time dd if=/dev/zero of=fname bs=2M count=64 oflag=direct conv=notrunc,fsync seek=0
64+0 records in134217728 bytes (134 MB) copied, 1,44088 s, 93,2 MB/s
64+0 records out
That gets much better performance, and in my case it can't get any
better because the HDD (and encryption) just can't go any faster.
--
Matthias
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