Hmm, option-testing leads to more confusion:
With this 922GB-sdb1 I do
mkfs.lustre --reformat --mgs --mdt ... /dev/sdb1
The output of the command says
Permanent disk data:
Target: test0:MDT0000
...
device size = 944137MB
formatting backing filesystem ldiskfs on /dev/sdb1
target name test0:MDT0000
4k blocks 241699072
options -J size=4096 -I 1024 -i 2560 -q -O
dirdata,uninit_bg,^extents,mmp,dir_nlink,quota,huge_file,flex_bg -E lazy_journal_init -F
mkfs_cmd = mke2fs -j -b 4096 -L test0:MDT0000 -J size=4096 -I 1024 -i 2560 -q -O
dirdata,uninit_bg,^extents,mmp,dir_nlink,quota,huge_file,flex_bg -E lazy_journal_init -F /dev/sdb1
241699072
Mount this as ldiskfs, gives 369 M inodes.
One would assume that specifying one / some of the mke2fs-options here in the mkfs.lustre-command will
change nothing.
However,
mkfs.lustre --reformat --mgs --mdt ... --mkfsoptions="-I 1024" /dev/sdb1
says
device size = 944137MB
formatting backing filesystem ldiskfs on /dev/sdb1
target name test0:MDT0000
4k blocks 241699072
options -I 1024 -J size=4096 -i 1536 -q -O
dirdata,uninit_bg,^extents,mmp,dir_nlink,quota,huge_file,flex_bg -E lazy_journal_init -F
mkfs_cmd = mke2fs -j -b 4096 -L test0:MDT0000 -I 1024 -J size=4096 -i 1536 -q -O
dirdata,uninit_bg,^extents,mmp,dir_nlink,quota,huge_file,flex_bg -E lazy_journal_init -F /dev/sdb1
241699072
and the mounted devices now has 615 M inodes.
So, whatever makes the calculation for the "-i / bytes-per-inode" value becomes ineffective if I
specify the inode size by hand?
How many bytes-per-inode do I need?
This ratio, is it what the manual specifies as "one
inode created for each 2kB of LUN" ?
Perhaps the raw size of an MDT device should better be such that it leads to "-I
1024 -i 2048"?
Regards,
Thomas
On 01/26/2018 03:10 PM, Thomas Roth wrote:
Hi all,
what is the relation between raw device size and size of a formatted MDT? Size of inodes + free space
= raw size?
The example:
MDT device has 922 GB in /proc/partions.
Formatted under Lustre 2.5.3 with default values for mkfs.lustre resulted in a 'df -h' MDT of 692G and
more importantly 462M inodes.
So, the space used for inodes + the 'df -h' output add up to the raw size:
462M inodes * 0.5kB/inode + 692 GB = 922 GB
On that system there are now 330M files, more than 70% of the available inodes.
'df -h' says '692G 191G 456G 30% /srv/mds0'
What do I need the remaining 450G for? (Or the ~400G left once all the inodes
are eaten?)
Should the format command not be tuned towards more inodes?
Btw, on a Lustre 2.10.2 MDT I get 369M inodes and 550 G space (with a 922G raw device): inode size is
now 1024.
However, according to the manual and various Jira/Ludocs the size should be 2k
nowadays?
Actually, the command within mkfs.lustre reads
mke2fs -j -b 4096 -L test0:MDT0000 -J size=4096 -I 1024 -i 2560 -F /dev/sdb
241699072
-i 2560 ?
Cheers,
Thomas
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