I may be wrong, but your correct with your m=6 statement.
Your need atleast K amount of shards available. If you had k=8 and m=2 equally across 2 rooms (5 each), a faidlure in either room would cause an outrage.
With M=6 your atleast getting better disk space availability than 3 replication. But not sure if you may end up with some form of split brain if just was a network issue between both sides and each side was still online and working independently. As both would technically have enough shards to continue to operate.
On Fri, 3 May 2019, 11:46 PM Robert Sander, <r.sander@xxxxxxxxxxxxxxxxxxx> wrote:
Hi,
I would be glad if anybody could give me a tip for an erasure code
profile and an associated crush ruleset.
The cluster spans 2 rooms with each room containing 6 hosts and each
host has 12 to 16 OSDs.
The failure domain would be the room level, i.e. data should survive if
one of the rooms has a power loss.
Is that even possible with erasure coding?
I am only coming up with profiles where m=6, but that seems to be a
little overkill.
Regards
--
Robert Sander
Heinlein Support GmbH
Schwedter Str. 8/9b, 10119 Berlin
https://www.heinlein-support.de
Tel: 030 / 405051-43
Fax: 030 / 405051-19
Amtsgericht Berlin-Charlottenburg - HRB 93818 B
Geschäftsführer: Peer Heinlein - Sitz: Berlin
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