Trying to intuitively understand this:
If I have $N$ packets flowing through a network, and the probability of a packet being successfully transmitted is $p$, how is the total number of packets that successfully went through $S$ equal $Np$ ?
The reason I have trouble understanding this is, assume $p=10\%$. If I send $10$ packets through a network, it doesn't necessarily mean I get only $Np=10×10\%=1$ packet to transmit successfully right? Couldn't I potentially get more than one packet to successfully transmit?
I came across this reasoning from Computer Networks 5th edition.
Here's an excerpt from the book:
Under all loads, the throughput, $S$, is just the offered
load, $G$, times the probability, $P_0$, of a transmission succeeding—that is, $S=GP_0$, where $P_0$ is the probability that a frame does not suffer a collision.
Best Answer
If you're carrying out $N$ transmissions, each of which is a bernoulli trial with success probability of $p$, the distribution of number of successful transmissions is the binomial $(N, p)$ distribution.
The "expected value" (i.e. the average) of this distribution is $Np$, which is what the textbook claims.