The Zookeeper ensemble knows the total size. It does not adjust it each time that a machine is partitioned or down.
Two machines is not a quorum for a four machine ensemble. Why do you think that the documentation would get this wrong? wunder On Dec 6, 2012, at 4:14 PM, Jack Krupansky wrote: > But that is the context I was originally referring to - that with 4 zk you > can lose only one, that you can't lose two. So, if you want to tolerate a > loss on one, 4 zk would be the minimum... but then it was claimed that you > COULD start with 3 zk and loss of one would be fine. I mean whether you start > with 4 and lose 2 or start with 3 and lose 1 is the same, right? > > -- Jack Krupansky > > -----Original Message----- From: Yonik Seeley > Sent: Thursday, December 06, 2012 6:34 PM > To: solr-user@lucene.apache.org > Subject: Re: Minimum HA Setup with SolrCloud > > On Thu, Dec 6, 2012 at 5:55 PM, Jack Krupansky <j...@basetechnology.com> > wrote: >> I trust that you have the right answer, Mark, but maybe I'm just struggling >> to parse this statement: "the remaining two machines do not constitute a >> majority." >> >> If you start with 3 zk and lose one, you have an ensemble that does not >> "constitute a majority". > > I think you took that out of context. They were talking about losing > 2 nodes in a 4 node cluster. > > "For example, with four machines ZooKeeper can only handle the failure > of a single machine; if two machines fail, the remaining two machines > do not constitute a majority." > > -Yonik > http://lucidworks.com -- Walter Underwood wun...@wunderwood.org
