[lustre-discuss] size of MDT, inode count, inode size

[Thomas Roth]

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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Thomas Roth

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