<s>[INST] Here's a formula for Chernoff throughput estimation:
\[Y = \log \left( \frac{\overline{f}(1)}{\alpha TM}\right)\]
where $Y$ is Chernoff throughput, $\overline{f}(1)$ is average departure process, $\alpha$ is a safety factor, $T$ is frame length, and $M$ is the number of available mini-slots.
We can calibrate the safety factor $\alpha$ with these measurements:
Online and Offline Estimates $\begin{array}{c|c|c} \alpha & \overline{f}(1) & T \\ \hline 1/3 & 93040 & 5000 \\ 1/2 & 136280 & 5000 \\ 2/3 & 196270 & 5000 \\ 1/3 & 173600 & 2000 \\ \end{array}$
where $\overline{f}(1)$ can take three values, 93040, 136280, and 196270.
We also know that Chernoff throughput is given as 250 when $\overline{f}(1)$ = 196270 and $\alpha = \frac{1}{3}$.
What is the value of $M$?
We have four cases based on values of $\overline{f}(1)$:
(1)  $\overline{f}(1)$ = 93040, $\alpha$ = $\frac{1}{3}$.
(2)  $\overline{f}(1)$ = 93040, $\alpha$ = $\frac{1}{2}$.
(3)  $\overline{f}(1)$ = 196270, $\alpha$ = $\frac{2}{3}$.
(4)  $\overline{f}(1)$ = 173600, $\alpha$ = $\frac{1}{3}$.
However, we only get the Chernoff throughput for case (4): when $\overline{f}(1)$ = 196270 and $\alpha$ = $\frac{1}{3}$.[/INST]