ceph erasure code: calculate size

cola16

Member
Feb 2, 2024
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I noticed something strange.
I thought that no matter how many OSDs, the case with K=3 M=3 and the case with K=6 M=6 would be the same except for durability.
I predicted that k=6 m=6 would be smaller, especially for capacity.
The reason is that K=6 M=6 requires 12 OSDs to be aggregated.

But surprisingly, with 12 of 1-terabyte disks, only 3 terabytes are available for K=3 M=3.
On the other hand, for K=6 M=6, 6 terabytes are available.

It looks like we are only utilizing K+M OSDs, is that correct?