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Posted to user@cassandra.apache.org by 魏金仙 <se...@126.com> on 2010/12/05 14:45:23 UTC
probability of node receiving (not be responsible for) the request
If a particular client send 5 requests to a 6-node cluster, then the probability of each node receiving(not be responsible for) the first request is 1/6. Assume that node1 received the 1st request, will node1 receive the 2nd request, the 3rd one, the 4th one and the 5th one with high probability or 1/6?
thanks for your time.
Re: Re: probability of node receiving (not be responsible for) the request
Posted by Brandon Williams <dr...@gmail.com>.
2010/12/6 魏金仙 <se...@126.com>
> thank you.
> but I mean the probability of a node to receive the request not process it
> eventually .
I see. That depends on how the client-side load balancing is written.
-Brandon
Re:Re: probability of node receiving (not be responsible for) the
request
Posted by 魏金仙 <se...@126.com>.
thank you.
but I mean the probability of a node to receive the request not process it eventually .
At 2010-12-06 00:56:58，"Brandon Williams" <dr...@gmail.com> wrote:
2010/12/5 魏金仙<se...@126.com>
If a particular client send 5 requests to a 6-node cluster, then the probability of each node receiving(not be responsible for) the first request is 1/6.
Assuming RF=1 and RandomPartitioner.
Assume that node1 received the 1st request, will node1 receive the 2nd request, the 3rd one, the 4th one and the 5th one with high probability or 1/6?
thanks for your time.
1/6th with RandomPartitioner, something much higher with OrderPreservingPartitioner.
-Brandon
Re: probability of node receiving (not be responsible for) the request
Posted by Brandon Williams <dr...@gmail.com>.
2010/12/5 魏金仙 <se...@126.com>
> If a particular client send 5 requests to a 6-node cluster, then the
> probability of each node receiving(not be responsible for) the first request
> is 1/6.
Assuming RF=1 and RandomPartitioner.
> Assume that node1 received the 1st request, will node1 receive the 2nd
> request, the 3rd one, the 4th one and the 5th one with high probability or
> 1/6?
> thanks for your time.
>
1/6th with RandomPartitioner, something much higher with
OrderPreservingPartitioner.
-Brandon