Suppose you have a network which uses time division multiplexing (i.e. Slotted Aloha). There are $N$ nodes each transmitting in a given slot with probability $p$.

I understand that the probability a node is successful in a slot is $p(1 - p)^{N-1}$. This is just a geometric distribution. But our book then says that the probability there is success in a slot (i.e. no collisions) is $Np(1-p)^{N-1}$.

I don't understand:

  1. What the difference between the two probabilities (how is the probability that a node is successful in a slot different from the probability of success in a slot).

  2. Why you need to multiply the probability by the number of nodes $N$ since $p(1 -p)^{N-1}$ gives you the probability that one node transmits and all other nodes don't transmit. Could it be because this is the probability that node might transmit given that all other nodes don't transmit, which isn't guaranteed. But we want to find the probability that any of the nodes transmit when all other nodes don't transmit - hence the multiplication by $N$?


1 Answer 1


There is a clear explanation given in these lecture notes MIT course Digital communication systems (read by H. Balakrishnan and G. Verghese).

The specific part that you require is this>

If each node sends with probability $p$, then the probability that exactly one node sends in any given slot is $Np(1 − p)^{N−1}$.

The reason is that the probability that a specific node sends in the time slot is $p$, and for its transmission to be successful, all the other nodes should not send. That combined probability is $p(1 − p)^{N−1}$. Now, we can pick the successfully transmitting node in $N$ ways, so the probability of exactly one node sending in a slot is $Np(1 − p)^{N−1}$.


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