Got it. Thanks

On Mon, Oct 2, 2017 at 4:54 AM, Justin Cameron <jus...@instaclustr.com>
wrote:

> 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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>> --
>
>
> *Justin Cameron*Senior Software Engineer
>
>
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