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From Walter Underwood <wun...@wunderwood.org>
Subject Re: Minimum HA Setup with SolrCloud
Date Fri, 07 Dec 2012 00:30:09 GMT
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 <jack@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
wunder@wunderwood.org




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