On Tuesday 30 Aug 2011 21:02:54 Arne Babenhauserheide wrote:
> Am Dienstag, 30. August 2011, 01:08:16 schrieb Arne Babenhauserheide:
> > 5) solution: count each SSK as only
> >     average_SSK_success_rate * data_to_transfer_on_success.
> 
> Some more data: 
> 
> chances of having at least this many successful transfers for 40 SSKs with a 
> mean success rate of 16%: 
> 
> for i in {0..16}; do echo $i $(./spielfaehig.py 0.16 40 $i); done
> 
>     0 1.0
>     1 0.999064224991
>     2 0.99193451064
>     3 0.965452714478
>     4 0.901560126912
>     5 0.788987472629
>     6 0.634602118184
>     7 0.463062835467
>     8 0.304359825607
>     9 0.179664603573
>     10 0.0952149293922
>     11 0.0453494074947
>     12 0.0194452402752
>     13 0.00752109980912
>     14 0.0026291447461
>     15 0.000832100029072
>     16 0.00023879002726
> 
> what this means: if a SSK has a mean success rate of 0.16, then using 0.25 as 
> value makes sure that 95% of the possible cases don’t exhaust the bandwidth. 
> We then use only 64% of the bandwidth on average, though. With 0.2, we’d get 
> 68% of the possible distributions safe and use 80% of bandwidth on average.

Only if there are no clusters - a single requestor fetches a bunch of stuff 
that is all already there, rather than polling keys that usually aren't. IMHO 
there will be. For instance, when a new chat client is started up, a lot of 
what it does will be fetching existing messages rather than polling for new 
ones.

Attachment: signature.asc
Description: This is a digitally signed message part.

_______________________________________________
Devl mailing list
[email protected]
http://freenetproject.org/cgi-bin/mailman/listinfo/devl

Reply via email to