Let there are $n$ devices in the setup. Let each device sends data with a probability $p$.

Successful transmission for a slot (hence of the setup in general) occurs when only one device communicates.

Hence using the binomial theorem, we have $P_{success}= \binom{n}{1}\times p \times (1-p)^{n-1}$.

Differentiating $P{success}$ with respect to $p$ and equating it with $0$, we have $p=\frac{1}{n}$. So,

$P_{success} |max = (1-\frac{1}{n})^{n-1}$

For infinitely large number of devices we have $P_{success} |max = \frac{1}{e}$.

So the probability of success in the system is $\frac{1}{e}$ and let be $n$ trials.

So expected number of successes = $n \times p = \frac{n}{e}.$

Now for 1 success, we have $1 = \frac{n}{e} \implies n=e$.

Or in other words there is success in $1$ out $e$ trials. [$P_{success} |max =\frac{1}{e}$ ] So the number of trials to get one success is $e$.

Here we are finding the average situation. Before a success there are $e-1$ no. of failure in a slot with $e$ trials.$^\dagger$ So these slots with $e$ trials repeat. And hence if $c$ is the no. of failures,

$$\text{Efficiency} = \frac{T_t} {2*c*T_p+T_t+T_p}$$

$$= \frac{1}{2\times c\times a+1+a}= \frac{1}{1+(2c+1)a}= \frac{1}{1+(2(e-1)+1)a}$$

$$ =\frac{1}{1+4.4a}$$

[Here is a material from MIT] (see page 14)

But in many places, I find the formula as $\frac{1}{1+6.44a}$. (Simply because in the $\dagger$ they assume $c=e$ which is not correct. Just as in here.)

Again in the Computer Networking : A Top-Down Approach by Kurose and Ross states the formula to be:

$$\text{Efficiency}=\frac{1}{1+5. \frac{T_p}{T_t}}$$

Which is the correct one and which one to follow?


1 Answer 1


Differentiation of np(1-p)^(n-1) is = n(1-p)^(n-1) + n(n-1)p(1-p)^(n-2) = np^(n-2)(1-p + (n-1)p) = n(1-p)^(n-2)(1 + (n-2)p)= 0 has no solution except p = 1. So the first step of your deduction is wrong.

  • $\begingroup$ please check your calculations once again. $\endgroup$ Jun 23, 2021 at 17:03
  • $\begingroup$ @AbhishekGhosh please mention a particular step where you think that there is an error. $\endgroup$ Jul 8, 2021 at 4:56

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