The part I still find confusing is that if you start with 3 and lose 1, your
have 2, which means you can't always break a tie, right? How is this
explained? As opposed to saying that 4 is the minimum if you need to
tolerate a loss of 1.
-- Jack Krupansky
-----Original Message-----
From: Walter Underwood
Sent: Thursday, December 06, 2012 7:51 PM
To: solr-user@lucene.apache.org
Subject: Re: Minimum HA Setup with SolrCloud
What is the mystery? Two is not more than half of four. Therefore, two
machines is not a quorum for a four machine Zookeeper ensemble.
wunder
On Dec 6, 2012, at 4:50 PM, Jack Krupansky wrote:
It's still an unresolved mystery, for now.
-- Jack Krupansky
-----Original Message----- From: Walter Underwood
Sent: Thursday, December 06, 2012 7:30 PM
To: solr-user@lucene.apache.org
Subject: Re: Minimum HA Setup with SolrCloud
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
--
Walter Underwood
wun...@wunderwood.org
