Oops...
And this is precisely why the mystery remains - because you're only
describing half the picture! Describe the rest of the picture - including
what exactly those two zks can and can't do, including resolution of ties
and the concept of "constituting a majority" and a quorum.
I'm not saying the mystery CAN'T be solved or that you haven't resolved it
in your own mind, but simply that it hasn't been resolved and clearly
described in a complete and consistent manner in the narrative here today.
-- Jack Krupansky
-----Original Message-----
From: Jack Krupansky
Sent: Thursday, December 06, 2012 8:39 PM
To: solr-user@lucene.apache.org
Subject: Re: Minimum HA Setup with SolrCloud
And this is precisely why the mystery remains - because you're only
describing half the picture! Describe the rest of the picture - including
what exactly those two zks can and can't do, including resolution of ties
and the concept of "constitu.
-- Jack Krupansky
-----Original Message-----
From: Walter Underwood
Sent: Thursday, December 06, 2012 8:33 PM
To: solr-user@lucene.apache.org
Subject: Re: Minimum HA Setup with SolrCloud
Configure an ensemble of three. When one goes down, you still have an
ensemble of three, but with one down. The ensemble size is not reset after
failures.
wunder
On Dec 6, 2012, at 5:20 PM, Jack Krupansky wrote:
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
--
Walter Underwood
wun...@wunderwood.org
