yep, for odd n that's right.

On Thu, May 30, 2019 at 1:30 PM Joel Mestres <joel.mest...@globant.com>
wrote:

> ok great so the minimum quorum always is F + 1 considering n as odd, right
> ?
>
> On Thu, May 30, 2019 at 3:38 PM Alexander Shraer <shra...@gmail.com>
> wrote:
>
> > If you're using "majority quorums" (the default in ZK),
> F=floor((n-1)/2). A
> > quorum is any set containing a majority (or more) of servers.
> > The basic requirement is that any two quorums must intersect.
> >
> >
> > On Wed, May 29, 2019 at 6:55 PM Patrick Hunt <ph...@apache.org> wrote:
> >
> > > 2n+1 = ensemble size required to survive n failed zkservers (servers
> not
> > in
> > > the quorum)
> > >
> > > iow: 3 nodes means 1 zkserver can fail and the service is still up. 5
> and
> > > you can survive 2 failures.
> > >
> > > Patrick
> > >
> > > On Wed, May 29, 2019 at 4:43 PM Joel Mestres <joel.mest...@globant.com
> >
> > > wrote:
> > >
> > > > Hello which is the relation between the number of F failling machines
> > > that
> > > > the cluster can tolerate and the quorum configuration? F determines
> the
> > > > posibles numbers of quorums? or F can be greater / smaller than
> quorum
> > ?
> > > > thanks in advance for your response!
> > > >
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