[Gluster-users] EC clarification
Jeff Darcy
jdarcy at redhat.com
Wed Sep 21 19:36:19 UTC 2016
> 2016-09-21 20:56 GMT+02:00 Serkan Çoban <cobanserkan at gmail.com>:
> > Then you can use 8+3 with 11 servers.
>
> Stripe size won't be good: 512*(8-3) = 2560 and not 2048 (or multiple)
It's not really 512*(8+3) though. Even though there are 11 fragments,
they only contain 8 fragments' worth of data. They just encode it with
enough redundancy that *any* 8 contains the whole.
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