I have found that in (limited) practice that it's fairly hard to estimate
due to compression and compaction behaviour. I think measuring and
extrapolating (with an understanding of the datastructures) is the most
effective.
Tim
Sent from my phone
On 6 Dec 2013 20:54, "John Sanda" wrote:
> I hav
Nice work John. If you learn any more, please share.
S
On Sat, Dec 7, 2013 at 11:50 AM, John Sanda wrote:
> I finally got the math right for the partition index after tracing through
> SSTableWriter.IndexWriter.append(DecoratedKey key, RowIndexEntry
> indexEntry). I should also note that I am
I finally got the math right for the partition index after tracing through
SSTableWriter.IndexWriter.append(DecoratedKey key, RowIndexEntry
indexEntry). I should also note that I am working off of the source for
1.2.9. Here is the break down for what gets written to disk in the append()
call (my ke
I have done that, but it only gets me so far because the cluster and app
that manages it is run by 3rd parties. Ideally, I would like to provide my
end users with a formula or heuristic for establishing some sort of
baselines that at least gives them a general idea for planning. Generating
data as
Not sure what your end setup will be, but I would probably just spin up a
cluster and fill it with typical data to and measure the size on disk.
__
Sent from iPhone
> On 7 Dec 2013, at 6:08 am, John Sanda wrote:
>
> I am trying to do some disk capacity planning. I h
I should have also mentioned that I have tried using the calculations from
the storage sizing post. My lack of success may be due to the post basing
things off of Cassandra 0.8 as well as a lack of understanding in how to do
some of the calculations.
On Fri, Dec 6, 2013 at 3:08 PM, John Sanda wr
I am trying to do some disk capacity planning. I have been referring the
datastax docs[1] and this older blog post[2]. I have a column family with
the following,
row key - 4 bytes
column name - 8 bytes
column value - 8 bytes
max number of non-deleted columns per row - 20160
Is there an effective
