It would be 4/(4+2) = 4/6 =2/3 or k/(k+m)? -----Original Message----- From: David Orman <orma...@corenode.com> Sent: Tuesday, July 28, 2020 9:32 PM To: ceph-users <ceph-users@ceph.io> Subject: [ceph-users] Usable space vs. Overhead
[CAUTION: External Mail] I'm having a hard time understanding the EC usable space vs. raw. https://urldefense.com/v3/__https://ceph.io/geen-categorie/ceph-erasure-coding-overhead-in-a-nutshell/__;!!B4Ndrdkg3tRaKVT9!79nn4ZG7ADJCY7JEhJwbPvHUn8dvmzAYz9_z-BUG_7Pe0uUETMW_AwDPmgiU4dc$ indicates "nOSD * k / (k+m) * OSD Size" is how you calculate usable space, but that's not lining up with what i'd expect just from k data chunks + m parity chunks. So, for example, k=4, m=2. you'd expect every 4 byte object written would consume 6 bytes, so 50% overhead. however, the prior formula in a 7 server cluster, using 4+2 encoding, would indicate 66.67% usable capacity vs. raw storage. What am I missing here? _______________________________________________ ceph-users mailing list -- ceph-users@ceph.io To unsubscribe send an email to ceph-users-le...@ceph.io _______________________________________________ ceph-users mailing list -- ceph-users@ceph.io To unsubscribe send an email to ceph-users-le...@ceph.io
