Hi Avi,

Actually, in Thomas' example you would need an additional 100G of free disk
space to complete the compaction, in the worst-case situation (the
worst-case would be that neither input SSTable contains any overlapping
data or tombstones, therefore the output SSTable would also be roughly
100G).

STCS progressively compacts SSTables of similar size together, with the
output being a single SSTable containing the data of the input SSTables.

Eventually you may end up with some very large SSTables that combined will
take up 50% of your total disk space. In order to compact those SSTables
together, STCS requires an equal amount of free disk space, which would be
the other (unused) 50% of your total disk space.

Cheers,
Justin

On Mon, 2 Oct 2017 at 12:42 Avi Levi <a...@indeni.com> wrote:

> Hi Thomas ,
> So IIUC in this case you should leave at least 50G for compaction  (half
> of the sstables size). Is that makes sense?
> Cheers
> Avi
>
>
> On Oct 1, 2017 11:39 AM, "Steinmaurer, Thomas" <
> thomas.steinmau...@dynatrace.com> wrote:
>
> Hi,
>
>
>
> half of free space does not make sense. Imagine your SSTables need 100G
> space and you have 20G free disk. Compaction won’t be able to do its job
> with 10G.
>
>
>
> Half free of total disk makes more sense and is what you need for a major
> compaction worst case.
>
>
>
> Thomas
>
>
>
> *From:* Peng Xiao [mailto:2535...@qq.com]
> *Sent:* Samstag, 30. September 2017 10:21
> *To:* user <user@cassandra.apache.org>
> *Subject:* space left for compaction
>
>
>
> Dear All,
>
>
>
> As for STCS,datastax suggest us to keep half of the free space for
> compaction,this is not strict,could anyone advise how many space should we
> left for one node?
>
>
>
> Thanks,
>
> Peng Xiao
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