5

$a^n + C_n^{1}a^{n-1}b + ... C_n^{n-1}a^{1}b^{n-1}+b^n = (a+b)^n$

But how to calculate (maybe approximately)

$a^n + C_n^{1}a^{n-1}b + ... C_n^{i}a^{n-i}b^{i} = ?$

For info, the underlying problem is "How many $n$ packets should I send to deliver at least $i$ packets successfully if packet loss probability is $a$".

Vi0
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1 Answers1

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If one of $a,b$ is much larger than the other you can ignore the terms with high powers of the small one. Say $a \ll b$, corresponding to most packets getting through. Then your expression is $(a+b)^n=b^n(\frac ab +1)^n= b^n\left(1+n\frac ab+\frac {n(n-1)}2(\frac ab)^2\ldots +(\frac ab)^n\right)$. Keep as many terms as you need to get the accuracy you want. If $\frac{na}b \ll 1$ they will decrease rapidly.

Ross Millikan
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  • I expect $a$ to be from about 2% to 80% (i.e. from improving slight problems to hammering thought serious networking issues). – Vi0 Feb 21 '13 at 23:12