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.san...@heinlein-support.de> 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 > > _______________________________________________ > ceph-users mailing list > ceph-users@lists.ceph.com > http://lists.ceph.com/listinfo.cgi/ceph-users-ceph.com >
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