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.
Note: this is just a binomial spread:
from math import factorial
fac = factorial
def n?k(n, k):
if k > n: return 0
return fac(n) / (fac(k)*fac(n-k))
def binom(p, n, k):
return n?k(n, k) * p** k * (1-p)**(n-k)
def spielf?hig(p, n, min_spieler):
return sum([binom(p, n, k) for k in range(min_spieler, n+1)])
? USK at 6~ZDYdvAgMoUfG6M5Kwi7SQqyS-
gTcyFeaNN1Pf3FvY,OSOT4OEeg4xyYnwcGECZUX6~lnmYrZsz05Km7G7bvOQ,AQACAAE/bab/9/Content-
D426DC7.html
Best wishes,
Arne
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