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From Roland Hänel <rol...@haenel.me>
Subject Re: Ring management and load balance
Date Sat, 27 Mar 2010 09:26:03 GMT
Mike,

If you have the assumption that your rows are roughly equal in size (at
least statistcally), then you could also just take a node's total load (this
is exposed via Jmx) and divide by the amount of keys/rows on that node. Not
sure how to get the latter, but shouldn't be such a big deal to integrate in
JMX if not already there.

Roland

26.03.2010 22:36 schrieb am "Mike Malone" <mike@simplegeo.com>:

2010/3/26 Roland Hänel <roland@haenel.me>

>
> Jonathan,
>
> I agree with your idea about a tool that could 'propose' good token
choices for op...
With the random partitioner there's no need to suggest a token. The key
space is statistically random so you should be able to just split 2^128 into
equal sized segments and get fairly equal storage load. Your read / write
load could get out of whack if you have hot spots and stuff, I guess. But
for a large distributed data set I think that's unlikely.

For order preserving partitioners it's harder. We've been thinking about
this issue at SimpleGeo and were planning on implementing an algorithm that
could determine the median row key statistically without having to inspect
every key. Basically, it would pull a random sample of row keys (maybe from
the Index file?) and then determine the median of that sample. Thoughts?

Mike

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