Hello Caspar That makes a great deal of sense, thank you for elaborating. Am I correct to assume that if we were to use a k=2, m=2 profile, it would be identical to a replicated pool (since there would be an equal amount of data and parity chunks)? Furthermore, how should the proper erasure profile be determined then? Are we to strive for a as high as possible data chunk value (k) and a low parity/coding value (m)?
Kind regards Ziggy Maes DevOps Engineer CELL +32 478 644 354 SKYPE Ziggy.Maes [http://www.be-mobile.com/mail/bemobile_email.png]<http://www.be-mobile.com/> www.be-mobile.com<http://www.be-mobile.com> From: Caspar Smit <caspars...@supernas.eu> Date: Friday, 20 July 2018 at 14:15 To: Ziggy Maes <ziggy.m...@be-mobile.com> Cc: "ceph-users@lists.ceph.com" <ceph-users@lists.ceph.com> Subject: Re: [ceph-users] Default erasure code profile and sustaining loss of one host containing 4 OSDs Ziggy, For EC pools: min_size = k+1 So in your case (m=1) -> min_size is 3 which is the same as the number of shards. So if ANY shard goes down, IO is freezed. If you choose m=2 min_size will still be 3 but you now have 4 shards (k+m = 4) so you can loose a shard and still remain availability. Of course a failure domain of 'host' is required to do this but since you have 6 hosts that would be ok. Met vriendelijke groet, Caspar Smit Systemengineer SuperNAS Dorsvlegelstraat 13 1445 PA Purmerend t: (+31) 299 410 414 e: caspars...@supernas.eu<mailto:caspars...@supernas.eu> w: www.supernas.eu<http://www.supernas.eu>
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