# What is the actual formula for the efficiency in CSMA/CD?

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?