Thanks svante for clearing the doubt. With Regards Aman Tandon
On Thu, Mar 5, 2015 at 2:15 PM, svante karlsson <s...@csi.se> wrote: > The network will "only" split if you get errors on your network hardware. > (or fiddle with iptables) Let's say you placed your zookeepers in separate > racks and someone pulls network cable between them - that will leave you > with 5 working servers but they can't reach each other. This is "split > brain scenario". > > >Are they guaranteed to split 4/0 > Yes. A node failure will not partition the network. > > > any odd number - it could be 21 even > Since all write a synchronous you don't want to use a too large number of > zookeepers since that would slow down the cluster. Use a reasonable number > to reach your SLA. (3 or 5 are common choices) > > >and from a single failure you drop to an even number - then there is the > danger of NOT getting quorum. > No, se above. > > BUT, if you first lose most of your nodes due to a network partition and > then lose another due to node failure - then you are out of quorum. > > > /svante > > > > 2015-03-05 9:29 GMT+01:00 Julian Perry <ju...@limitless.co.uk>: > > > > > 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. > >>>>>> > >>>>> >
