I start out with 5 zk's. All good.
One zk fails - I'm left with four. Are they guaranteed
to split 4/0 or 3/1 - because if they split 2/2 I'm screwed,
right?
Surely to start with 5 zk's (or in fact any odd number - it
could be 21 even), and from a single failure you drop to an
even number - then there is the danger of NOT getting quorum.
So ... I can only assume that there is a mechanism in place
inside zk to guarantee this cannot happen, right?
--
Cheers
Jules.
On 05/03/2015 06:47, svante karlsson wrote:
Yes, as long as it is three (the majority of 5) or more.
This is why there is no point of having a 4 node cluster. This would also
require 3 nodes for majority thus giving it the fault tolerance of a 3 node
cluster but slower and more expensive.
2015-03-05 7:41 GMT+01:00 Aman Tandon <amantandon...@gmail.com>:
Thanks svante.
What if in the cluster of 5 zookeeper only 1 zookeeper goes down, will
zookeeper election can occur with 4 / even number of zookeepers alive?
With Regards
Aman Tandon
On Tue, Mar 3, 2015 at 6:35 PM, svante karlsson <s...@csi.se> wrote:
synchronous update of state and a requirement of more than half the
zookeepers alive (and in sync) this makes it impossible to have a "split
brain" situation ie when you partition a network and get let's say 3
alive
on one side and 2 on the other.
In this case the 2 node networks stops serving request since it's not in
majority.
2015-03-03 13:15 GMT+01:00 Aman Tandon <amantandon...@gmail.com>:
But how they handle the failure?
With Regards
Aman Tandon
On Tue, Mar 3, 2015 at 5:17 PM, O. Klein <kl...@octoweb.nl> wrote:
Zookeeper requires a majority of servers to be available. For
example:
Five
machines ZooKeeper can handle the failure of two machines. That's why
odd
numbers are recommended.
