Approximation of blocking probability

Question

I want to compute the blocking probability \begin{align*} \operatorname{B}\left(\rho,m\right) & = \dfrac{\rho^{m}/m!} {\sum_{i= 0}^{m}\rho^{i}/i!} \\[5mm] \mbox{with}\ \left(\rho,m\right) & = \left(ka, k\left[a - 1\right]\right) \end{align*} ( Erlang $\operatorname{B}$ formula ) for 'billionaire' $k \gg a$.

• My naïve approach is by Stirling approximation: \begin{align*} \frac{\rho^{m}}{m!} & \approx\frac{\rho^{m}}{\,\sqrt{\,2\pi m\,}\,\left(m/{\rm e}\right)^{m}}, \quad\sum_{i= 0}^{m}\frac{\rho^{i}}{i!}\approx e^{\rho} \\[2mm] & \mbox{when}\ m\ \mbox{large and}\ \rho\ \mbox{also large} \end{align*} but this seems not good enough.
• I need your help to get a better approximation.
• However, I googled many times about 'Approximation of blocking probability' but cannot find such an alternative approximated formula, I feel interested in understanding why.

Thanks a real lot !.

Edit: I find a better approximation for the numerator here. And I hope for an approximation of factorial sum $$ \sum_{iv= 0}^{m}\frac{\rho^{i}}{i!}\quad \mbox{as}\quad c\,{\rm e}^{\rho + o\left(1\right)} $$ sharpening the traditional exponential function ${\rm e}^{\rho}$

Edit $\bf 2$: I think Taylor's series with remainder helps here.

Answer 1

To first order $$B\approx \frac{1}{a} \tag1$$

Proof follows (I moved to the end my original proof, much more convoluted and less reliable).

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Consider $$ \frac{1}{B}=\frac{m!}{\rho ^m} \sum_{i= 0}^{m}\frac{\rho^i}{i!}= \sum_{j= 0}^{m} \rho^{-j} \frac{m!}{(m-j)!} \tag 2$$

Now, assuming $m=k(a-1)$ , $\rho = ka$, with $a>1$, $k\gg a$: $$\begin{align}\rho^{-j} \frac{m!}{(m-j)!}&= \left(\frac{m}{\rho} \right)^j \left(1-\frac{1}{m}\right)\left(1-\frac{2}{m}\right)\cdots \left(1-\frac{j-1}{m}\right) \\ &= \left(\frac{a-1}{a}\right)^j \left(1 - \frac{1+2 +\cdots +(j-1)}{m}+o(m^{-1})\right) \\ &= \alpha^j \left(1 - \frac{j (j-1)}{2m}+o(m^{-1})\right) \tag3 \\ \end{align}$$ where $\alpha = (a-1)/a$. Replacing the finite sum in $(2)$ by an infinite sum, and using $\sum_{j=0}^{\infty} \alpha^j j(j-1)=2 \alpha^2/(1-\alpha)^3$ this results in

$$ \begin{align} \frac{1}{B}\approx \frac{1}{1-\alpha} - \frac{1}{m}\frac{\alpha^2}{(1-\alpha)^3}=a - \frac{1}{k(a-1)}\frac{a^3 (a-1)^2}{a^2}=a - \frac{1}{k} a(a-1) \tag4 \end{align} $$

or

$$ B \approx \frac{1}{a - \frac{1}{k} a(a-1) } \tag 5$$

For example, for $k=50$ and $a=3.5$ ($m=125$) we get:

             B          rel error  
exact      0.298436
aprox (1)  0.285714      4.2%
aprox (5)  0.300752     -0.8%

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Alternative original derivation (less reliable).

$$\begin{align} g(m,x)=\sum_{j=0}^{m-1} \frac{x^j}{j!} = e^x \frac{\Gamma(m,x)}{\Gamma(m)}\end{align} $$

Setting $m=k(a-1)$, $x=ka$, with $a>1$ fixed and $k\to \infty$, applying asympotic formula $8.11.7$ from here for the incomplete Gamma function, and Stirling approximation for the Gamma function , we get

$$\log g(m,x) \approx -\frac12\left(\alpha \,k + \log k + \beta\right)$$

with $\alpha = 2 (a-1)\left(\log(\frac{a-1}{a}) -1 \right)$ and $\beta = \log \frac{2\pi}{a-1}$

Then, doing the math...

$$ \log \frac{g(m,x)}{\frac{x^m}{m!}} \approx \log(a-1) + \frac{1}{12 (ak)^2} + o(1/k^2) $$

and finally

$$ B = \frac{1}{1+\frac{g(m,x)}{\frac{x^m}{m!}}}\approx \frac{1}{a} $$