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From Benjamin>
Subject Re: Worst case #iops to read a row
Date Tue, 13 Apr 2010 00:59:47 GMT
On Mon, Apr 12, 2010 at 4:27 PM, Time Less <> wrote:
> With this formula, we can already begin to formulate more useful answers to
> the question. If I have 10B rows in my CF, and I can fit 10k rows per
> SStable, and the SStables are spread across 5 nodes, and I have 1 bloom
> filter false positive and 1 tombstone and ask the wrong node for the key,
> then:
> Mv = (((2B/10k)+1+1)*3)+1 == ((200,000)+2)*3+1 == 300,007 iops to read a
> key.

This is a nonsensical arrangement.  Assuming each SSTable is the size
of the default Memtable threshold (128MB), then each row is (128MB /
10k) == 12.8k and 10B rows == 128TB of raw data.  A typical RF of 3
takes us to 384TB.  The need for enough space for compactions takes us
to 768TB.  That's not 5 nodes, it's more like 100+, and almost 2
orders of magnitude off your estimate, without addressing shortcomings
in the rest of it (which I leave to more capable folks on this list).


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