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
