First of all, Stan, thanks for such detailed answer, I greatly appreciate this!
On Sat, Feb 11, 2012 at 8:05 AM, Stan Hoeppner<stan@xxxxxxxxxxxxxxxxx> wrote:
On 2/10/2012 9:17 AM, CoolCold wrote:
I've got server with 7 SATA drives ( Hetzner's XS13 to be precise )
and created mdadm's raid10 with two near copies, then put LVM on it.
Now I'm planning to create xfs filesystem, but a bit confused about
stripe width/stripe unit values.
Why use LVM at all? Snapshots? The XS13 has no option for more drives
so it can't be for expansion flexibility. If you don't 'need' LVM don't
use it. It unnecessarily complicates your setup and can degrade
performance.
There are several reasons for this - 1) I've made decision to use LMV
for all "data" volumes (those are except /, /boot, /home , etc) 2)
there will be mysql database which will need backups with snapshots 3)
I often have several ( 0-3 ) virtual environments (OpenVZ based) which
are living on ext3/ext4 (because of extensive metadata updates on xfs
makes it the whole machine slow) filesystem and different LV because
of this.
As drives count is 7 and copies count is 2, so simple calculation
gives me datadrives count "3.5" which looks ugly. If I understand the
whole idea of sunit/swidth right, it should fill (or buffer) the full
stripe (sunit * data disks) and then do write, so optimization takes
place and all disks will work at once.
Pretty close. Stripe alignment is only applicable to allocation i.e new
file creation, and log journal writes, but not file re-write nor read
ops. Note that stripe alignment will gain you nothing if your
allocation workload doesn't match the stripe alignment. For example
writing a 32KB file every 20 seconds. It'll take too long to fill the
buffer before it's flushed and it's a tiny file, so you'll end up with
many partial stripe width writes.
Okay, got it - I've thinked in similar way.
My read load going be near random read ( sending pictures over http )
and looks like it doesn't matter how it will be set with sunit/swidth.
~13TB of "pictures" to serve eh? Average JPG file size will be
relatively small, correct? Less than 1MB? No, stripe alignment won't
really help this workload at all, unless you upload a million files in
one shot to populate the server. In that case alignment will make the
process complete more quickly.
Basing on current storage, estimations show (df -h / df -i ) average
file size is ~200kb . Inodes count is near 15 millions and it will be
more.
I've just thought that may be I should change chunk size to 256kb,
just to let one file be read from one disk, this may increase latency
and increase throughput too.
root@datastor1:~# cat /proc/mdstat
Personalities : [raid0] [raid1] [raid6] [raid5] [raid4] [raid10]
md3 : active raid10 sdg5[6] sdf5[5] sde5[4] sdd5[3] sdc5[2] sdb5[1] sda5[0]
10106943808 blocks super 1.2 64K chunks 2 near-copies [7/7] [UUUUUUU]
[>....................] resync = 0.8%
(81543680/10106943808) finish=886.0min speed=188570K/sec
bitmap: 76/76 pages [304KB], 65536KB chunk
Almost default mkfs.xfs creating options produced:
root@datastor1:~# mkfs.xfs -l lazy-count=1 /dev/data/db -f
meta-data=/dev/data/db isize=256 agcount=32, agsize=16777216 blks
= sectsz=512 attr=2, projid32bit=0
data = bsize=4096 blocks=536870912, imaxpct=5
= sunit=16 swidth=112 blks
naming =version 2 bsize=4096 ascii-ci=0
log =internal log bsize=4096 blocks=262144, version=2
= sectsz=512 sunit=16 blks, lazy-count=1
realtime =none extsz=4096 blocks=0, rtextents=0
As I can see, it is created 112/16 = 7 chunks swidth, which correlate
with my version b) , and I guess I will leave it this way.
The default mkfs.xfs algorithms don't seem to play well with the
mdraid10 near/far copy layouts. The above configuration is doing a 7
spindle stripe of 64KB, for a 448KB total stripe size. This doesn't
seem correct, as I don't believe a 7 drive RAID10 near is giving you 7
spindles of stripe width. I'm no expert on the near/far layouts, so I
could be wrong here. If a RAID0 stripe would yield a 7 spindle stripe
width, I don't see how a RAID10/near would also be 7. A straight RAID10
with 8 drives would give a 4 spindle stripe width.
I've drawn nice picture from my head in my original post, it was:
A1 A1 A2 A2 A3 A3 A4
A4 A5 A5 A6 A6 A7 A7
A8 A8 A9 A9 A10 A10 A11
A11 ...
So here is A{X} is chunk number on top of 7 disks. As you can see, 7
chunks write (A1 - A7) will fill two rows. And this will made 2 disk
head movements to write full stripe, though that moves may be very
near to each other. Real situation may differ of course, and I'm not
expert to make a bet too.
So, I'll be glad if anyone can review my thoughts and share yours.
To provide you with any kind of concrete real world advice we need more
details about your write workload/pattern. In absence of that, and
given what you've already stated, that the application is "sending
pictures over http", then this seems to be a standard static web server
workload. In that case disk access, especially write throughput, is
mostly irrelevant, as memory capacity becomes the performance limiting
factor. Given that you have 12GB of RAM for Apache/nginx/Lighty and
buffer cache, how you setup the storage probably isn't going to make a
big difference from a performance standpoint.
Yes, this is standard static webserver workload with nginx as frontend
with almost only reads.
That said, for this web server workload, you'll be better off it you
avoid any kind of striping altogether, especially if using XFS. You'll
be dealing with millions of small picture files I assume, in hundreds or
thousands of directories? In that case play to XFS' strengths. Here's
how you do it:
Hundreds directories at least, yes.
After reading you ideas and refinements, I'm making conclusion that I
need to push others [team members] harder to remove mysql instances
from the static files serving boxes at all, to free RAM for least
dcache entries.
About avoiding striping - later in the text.
1. You chose mdraid10/near strictly because you have 7 disks and wanted
to use them all. You must eliminate that mindset. Redo the array with
6 disks leaving the 7th as a spare (smart thing to do anyway). What can
you really to with 10.5TB that you can't with 9TB?
Hetzner's guys were pretty fast on chaning failed disks (one - two
days after claim) so I may try without spares I guess... I just wanna
use more independent spindles here, but I'll think about your
suggestion one more time, thanks.
2. Take your 6 disks and create 3 mdraid1 mirror pairs--don't use
partitions as these are surely Advanced Format drives. Now take those 3
mdraid mirror devices and create a layered mdraid --linear array of the
three. The result will be a ~9TB mdraid device.
3. Using a linear concat of 3 mirrors with XFS will yield some
advantages over a striped array for this picture serving workload.
Format the array with:
/$ mkfs.xfs -d agcount=12 /dev/mdx
That will give you 12 allocation groups of 750GB each, 4 AGs per
effective spindle. Using too many AGs will cause excessive head seeking
under load, especially with a low disk count in the array. The mkfs.xfs
agcount default is 4 for this reason. As a general rule you want a
lower agcount when using low RPM drives (5.9k, 7.2k) and a higher
agcount with fast drives (10k, 15k).
Good to know such details!
Directories drive XFS parallelism, with each directory being created in
a different AG, allowing XFS to write/read 12 files in parallel (far in
excess of the IO capabilities of the 3 drives) without having to worry
about stripe alignment. Since your file layout will have many hundreds
or thousands of directories and millions of files, you'll get maximum
performance from this setup.
So, as I could understand, you are assuming that "internal striping"
by using AGs of XFS will be better than MD/LVM striping here? Never
thought of XFS in this way and it is interesting point.
As I said, if I understand your workload correctly, array/filesystem
layout probably don't make much difference. But if you're after
something optimal and less complicated, for piece of mind, etc, this is
a better solution than the 7 disk RAID10 near layout with XFS.
Oh, and don't forget to mount the XFS filesystem with the inode64 option
in any case, lest performance will be much less than optimal, and you
may run out of directory inodes as the FS fills up.
Okay.
Hope this information was helpful.
Yes, very helpful and refreshing, thanks for you comments!
P.S. As I've got 2nd server of the same config, may be i'll have time
and do fast& dirty tests of stripes vs AGs.
--
Stan
