My understanding has been that the election algorithm requires [(n/2)+1] (where n is the configured size of the cluster) nodes to reach consensus on who the master is. So [(2/2)+1 = 2] nodes, meaning you have no way to lose a node and still run. Unless I'm mistaken, that's typically why you see an odd number of nodes, since both [(3/2)+1 = 2] and [(4/2)+1 = 3] mean you can only lose one node. Go from 3 to 5 and you get the extra redundancy of being able to lose 2 nodes.
Jim On Thu, Jul 23, 2015 at 3:38 AM khandelwalanuj <khandelwal.anu...@gmail.com> wrote: > Any updates here. > > > > -- > View this message in context: > http://activemq.2283324.n4.nabble.com/ActiveMQ-replicated-leveldb-with-2-brokers-replica-1-tp4699679p4699739.html > Sent from the ActiveMQ - User mailing list archive at Nabble.com. >
