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From Brandon Williams <dri...@gmail.com>
Subject Re: probability of node receiving (not be responsible for) the request
Date Sun, 05 Dec 2010 16:56:58 GMT
2010/12/5 ้ญ้‡‘ไป™ <sei_wjx@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

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