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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)? From: Caspar Smit <casparsmit@xxxxxxxxxxx> 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, |
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