A k=3, m=3 scheme would be 3:3 = 50% , you get to use 4 bytes out of 6 bytes = 
4:6 = 2:3 = 66.6%?

From: David Orman <orma...@corenode.com>
Sent: Wednesday, July 29, 2020 2:17 AM
To: Alan Johnson (System) <al...@supermicro.com>
Cc: ceph-users <ceph-users@ceph.io>
Subject: Re: [ceph-users] Usable space vs. Overhead

[CAUTION: External Mail]
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<mailto: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<mailto:orma...@corenode.com>>
Sent: Tuesday, July 28, 2020 9:32 PM
To: ceph-users <ceph-users@ceph.io<mailto: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$<https://urldefense.com/v3/__https:/ceph.io/geen-categorie/ceph-erasure-coding-overhead-in-a-nutshell/__;!!B4Ndrdkg3tRaKVT9!79nn4ZG7ADJCY7JEhJwbPvHUn8dvmzAYz9_z-BUG_7Pe0uUETMW_AwDPmgiU4dc$>
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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