On Sun, 28 Apr 2019 at 21:45, Igor Podlesny <ceph-u...@poige.ru> wrote:
> On Sun, 28 Apr 2019 at 16:14, Paul Emmerich <paul.emmer...@croit.io> > wrote: > > Use k+m for PG calculation, that value also shows up as "erasure size" > > in ceph osd pool ls detail > > So does it mean that for PG calculation those 2 pools are equivalent: > > 1) EC(4, 2) > 2) replicated, size 6 > Correct > > ? Sounds weird to be honest. Replicated with size 6 means each logical > data is stored 6 times, what needed single PG now requires 6 PGs. > And with EC(4, 2) there's still only 1.5 overhead in terms of raw > occupied space -- how come PG calculation distribution needs adjusting > to 6 instead of 1.5 then? > A single logical data unit (an object in ceph terms) will be allocated to a single PG. For a replicated pool of size n this PG will simply be stored on n OSDs. For an EC(k+m) pool this PG will get stored on k+m OSDs with the difference that this single PG will contain different parts of the data on the different OSDs. http://docs.ceph.com/docs/master/architecture/#erasure-coding provides a good overview on how this is actually achieved. > Also, why does CEPH documentation say "It is equivalent to a > replicated pool of size __two__" when describing EC(2, 1) example? > This relates to fault tolerance. A replicated pool of size 2 can loose one OSD without data loss and so can a EC(2+1) pool > > -- > End of message. Next message? > _______________________________________________ > ceph-users mailing list > ceph-users@lists.ceph.com > http://lists.ceph.com/listinfo.cgi/ceph-users-ceph.com >
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