On 03/16/2016 10:57 PM, Oleksandr Natalenko wrote:
OK, I've repeated the test with the following hierarchy:
* 10 top-level folders with 10 second-level folders each;
* 10 000 files in each second-level folder.
So, this composes 10×10×10000=1M files and 100 folders
Initial brick used space: 33 M
Initial inodes count: 24
After test:
* each brick in replica took 18G, and the arbiter brick took 836M;
* inodes count: 1066036
So:
(836 - 33) / (1066036 - 24) == 790 bytes per inode.
So, yes, it is slightly bigger value than with previous test due to, I guess,
lots of files in one folder, but it is still too far from 4k. Given a good
engineer should consider 30% reserve, the ratio is about 1k per stored inode.
Correct me if I'm missing something (regarding average workload and not corner
cases).
Looks okay to me Oleksandr. You might want to make a github gist of your
tests+results as a reference for others.
Regards,
Ravi
Test script is here: [1]
Regards,
Oleksandr.
[1] http://termbin.com/qlvz
On вівторок, 8 березня 2016 р. 19:13:05 EET Ravishankar N wrote:
On 03/05/2016 03:45 PM, Oleksandr Natalenko wrote:
In order to estimate GlusterFS arbiter brick size, I've deployed test
setup
with replica 3 arbiter 1 volume within one node. Each brick is located on
separate HDD (XFS with inode size == 512). Using GlusterFS v3.7.6 +
memleak
patches. Volume options are kept default.
Here is the script that creates files and folders in mounted volume: [1]
The script creates 1M of files of random size (between 1 and 32768 bytes)
and some amount of folders. After running it I've got 1036637 folders.
So, in total it is 2036637 files and folders.
The initial used space on each brick is 42M . After running script I've
got:
replica brick 1 and 2: 19867168 kbytes == 19G
arbiter brick: 1872308 kbytes == 1.8G
The amount of inodes on each brick is 3139091. So here goes estimation.
Dividing arbiter used space by files+folders we get:
(1872308 - 42000)/2036637 == 899 bytes per file or folder
Dividing arbiter used space by inodes we get:
(1872308 - 42000)/3139091 == 583 bytes per inode
Not sure about what calculation is correct.
I think the first one is right because you still haven't used up all the
inodes.(2036637 used vs. the max. permissible 3139091). But again this
is an approximation because not all files would be 899 bytes. For
example if there are a thousand files present in a directory, then du
<dirname> would be more than du <file> because the directory will take
some disk space to store the dentries.
I guess we should consider the one
that accounts inodes because of .glusterfs/ folder data.
Nevertheless, in contrast, documentation [2] says it should be 4096 bytes
per file. Am I wrong with my calculations?
The 4KB is a conservative estimate considering the fact that though the
arbiter brick does not store data, it still keeps a copy of both user
and gluster xattrs. For example, if the application sets a lot of
xattrs, it can consume a data block if they cannot be accommodated on
the inode itself. Also there is the .glusterfs folder like you said
which would take up some space. Here is what I tried on an XFS brick:
[root@ravi4 brick]# touch file
[root@ravi4 brick]# ls -l file
-rw-r--r-- 1 root root 0 Mar 8 12:54 file
[root@ravi4 brick]# du file
*0 file**
*
[root@ravi4 brick]# for i in {1..100}
> do
> setfattr -n user.value$i -v value$i file
> done
[root@ravi4 brick]# ll -l file
-rw-r--r-- 1 root root 0 Mar 8 12:54 file
[root@ravi4 brick]# du -h file
*4.0K file**
*
Hope this helps,
Ravi
Pranith?
[1] http://termbin.com/ka9x
[2]
http://gluster.readthedocs.org/en/latest/Administrator%20Guide/arbiter-vo
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