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Subject: Re: Minimum HA Setup with SolrCloud
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From: Walter Underwood <wunder@wunderwood.org>
In-Reply-To: <D980BB0825F54B22A4D3619E5902DDC6@JackKrupansky>
Date: Thu, 6 Dec 2012 17:33:31 -0800
Message-Id: <031C01AE-C52D-44F6-8DBC-12931B323459@wunderwood.org>
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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.
>=20
> -- Jack Krupansky
>=20
> -----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
>=20
> 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.
>=20
> wunder
>=20
> On Dec 6, 2012, at 4:50 PM, Jack Krupansky wrote:
>=20
>> It's still an unresolved mystery, for now.
>>=20
>> -- Jack Krupansky
>>=20
>> -----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
>>=20
>> The Zookeeper ensemble knows the total size. It does not adjust it =
each time that a machine is partitioned or down.
>>=20
>> Two machines is not a quorum for a four machine ensemble.
>>=20
>> Why do you think that the documentation would get this wrong?
>>=20
>> wunder
>>=20
>> On Dec 6, 2012, at 4:14 PM, Jack Krupansky wrote:
>>=20
>>> 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?
>>>=20
>>> -- Jack Krupansky
>>>=20
>>> -----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
>>>=20
>>> 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."
>>>>=20
>>>> If you start with 3 zk and lose one, you have an ensemble that does =
not
>>>> "constitute a majority".
>>>=20
>>> I think you took that out of context.  They were talking about =
losing
>>> 2 nodes in a 4 node cluster.
>>>=20
>>> "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."
>>>=20
>>> -Yonik
>>> http://lucidworks.com
>>=20
>> --
>> Walter Underwood
>> wunder@wunderwood.org
>>=20
>>=20
>>=20
>=20
> --
> Walter Underwood
> wunder@wunderwood.org
>=20
>=20
>=20

--
Walter Underwood
wunder@wunderwood.org


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