That's what the formula on the ceph link arrives at, a 2/3 or 66.66%
overhead. But if a 4 byte object is split into 4x1 byte chunks data (4
bytes total) + 2x 1 byte chunks parity (2 bytes total), you arrive at 6
bytes, which is 50% more than 4 bytes. So 50% overhead, vs. 33.33% overhead
as the other formula arrives at. I'm curious what I'm missing.

On Tue, Jul 28, 2020 at 3:40 PM Alan Johnson <al...@supermicro.com> wrote:

> 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?
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