To be honest with 3:8 we could protect the cluster more from osd flapping. Let's say you have less chance to have 8 down pgs on 8 separate nodes then with 8:3 only 3pgs on 3 nodes. Of course this comes with the cost on storage used. Is there any disadvantage performance wise on this? Istvan ________________________________ From: Alexander Patrakov <patrakov@xxxxxxxxx> Sent: Tuesday, December 17, 2024 4:48:25 PM To: Szabo, Istvan (Agoda) <Istvan.Szabo@xxxxxxxxx> Cc: Ceph Users <ceph-users@xxxxxxx> Subject: Re: Erasure coding best practice Email received from the internet. If in doubt, don't click any link nor open any attachment ! ________________________________ Hello Szabo, Some of these "weird" erasure coding setups come from old-style stretch clusters where the cluster was designed to withstand the loss of one datacenter out of two. For example, a 2+4 EC setup could be used together with a rule that selects three hosts from one datacenter and three from the other. On paper, this offers a space advantage over the official stretch cluster setup, which requires four replicas. Please note that this practice itself is questionable, as it is possible to satisfy the requirement of having min_size up-to-date shards total without having enough shards in each datacenter, which is what matters for data safety. In other words, there are scenarios involving partial recovery followed by another incident where such old-style stretch clusters might lose data. Don't do that. On Tue, Dec 17, 2024 at 1:03 PM Szabo, Istvan (Agoda) <Istvan.Szabo@xxxxxxxxx> wrote: > > Hi, > > Couple of time I saw some use-case in the mail list where people using erasure code on a "weird" way. > Normally we use more data chunks than coding chunks, like k=4 m=2 or k=8 m=3. > What I mean "weird" I saw someone using k=3 and m=8 (I might remember wrong). > > I'm trying to understand what is the benefit of the higher coding chunks? You can use smaller object size? > Let's say 4:2 the minimum object size should be 24K at least or with 8:3 it would be 44K because nothing will be stored on smaller space. > In case of k=3 m=8, smallest object can be 12K, but you can lose 8 nodes (pgs) and data still there? > > Ty > _______________________________________________ > ceph-users mailing list -- ceph-users@xxxxxxx > To unsubscribe send an email to ceph-users-leave@xxxxxxx -- Alexander Patrakov _______________________________________________ ceph-users mailing list -- ceph-users@xxxxxxx To unsubscribe send an email to ceph-users-leave@xxxxxxx
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