In all the examples of ALOHA I've seen, the Poisson distribution is used. Theoretically, how could the throughput be calculated if a Binomial distribution was used instead?

For example, in the case that the offered load follows a Binomial distribution with probability $p$, I know that the probability of $k$ attempts within $t$ time slots is as follows:

$$\Pr\big[k \text{ attempts within $t$ time slots}\big] = \binom{t}{k}p^k(1-p)^{t-k}\,.$$

I know that the throughput, $S$ is equivalent to $G$ (average offered load) multiplied by the probability of success:

$$S = G\cdot\Pr\big[\text{successful}\big]\,,$$

but what would this be in the case that a binomial distribution is used?

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    $\begingroup$ You can include LaTeX directly in your posts here, without needing to import images -- just use $...$ and $$...$$ as you would if you were writing a LaTeX document. $\endgroup$ – David Richerby Oct 28 '16 at 11:27

I don't think it makes sense to use the binomial distribution.

The point of using the Poisson distribution is that network nodes are assumed to want to transmit with an exponential distribution at some rate. This is a reasonable model for objects that randomly emit stuff – for example radioactive decays follow this distribution. Given a bunch of things emitting stuff at intervals sampled from the exponential distribution, the number of emissions in some time interval is given by a Poisson distribution. In other words, the Poisson distribution arises naturally from a model of the individual nodes. A particular property of the Poisson distribution is that it has infinite support: it assigns a probability to the event that you have a million transmission attempts within ten time slots. Another feature of using exponential distributions is that it allows one to model the back-off time after a collision.

In contrast, using a binomial distribution doesn't come from any natural model of the individual nodes' behaviour. It only allows you to model "were there at least $n$ transmissions attempts during this time slot?" for some $n$ of your choosing. It doesn't seem able to deal with back-off and it doesn't seem able to deal with multiple collisions within a single time slot.

  • $\begingroup$ That makes sense, but entirely theoretically, would it be possible to calculate the throughput in this case? $\endgroup$ – tibsar Oct 28 '16 at 12:44

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