> Now if a row key hash is mapped to a range owned by a node in DC3, > will the Node in DC3 still store the key as determined by the > partitioner and then walk the ring and store 2 replicas each in DC1 > and DC2 ? No, only nodes in the DC's specified in the NTS configuration will be replicas.
> Or will the co-ordinator node be aware of the > replica placement strategy, > and override the partitioner's decision and walk the ring until it > first encounters a node in DC1 or DC2 ? and then place the remaining > replicas ? The NTS considers each DC to have it's own ring. This can make token selection in a multi DC environment confusing at times. There is something in the DS docs about it. Cheers ----------------- Aaron Morton Freelance Developer @aaronmorton http://www.thelastpickle.com On 23/05/2012, at 3:16 PM, java jalwa wrote: > Hi all, > I am a bit confused regarding the terms "replica" and > "replication factor". Assume that I am using RandomPartitioner and > NetworkTopologyStrategy for replica placement. > From what I understand, with a RandomPartitioner, a row key will > always be hashed and be stored on the node that owns the range to > which the key is mapped. > http://www.datastax.com/docs/1.0/cluster_architecture/replication#networktopologystrategy. > The example here, talks about having 2 data centers and a replication > factor of 4 with 2 replicas in each datacenter, so the strategy is > configured as DC1:2 and DC2:2. Now suppose I add another datacenter > DC3, and do not change the NetworkTopologyStrategy. > Now if a row key hash is mapped to a range owned by a node in DC3, > will the Node in DC3 still store the key as determined by the > partitioner and then walk the ring and store 2 replicas each in DC1 > and DC2 ? Will that mean that I will then have 5 replicas in the > cluster and not 4 ? Or will the co-ordinator node be aware of the > replica placement strategy, > and override the partitioner's decision and walk the ring until it > first encounters a node in DC1 or DC2 ? and then place the remaining > replicas ? > > Thanks.
