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
