Mathematical analysis
Table of contents
Pure ALOHA assumptions and analysis
Assumptions:
- All frames have the same length.
- Stations only generates a new frame after a successful transmission ACK.
- The (aggregate) population of stations transmits according to a Poisson distribution.
- Stations transmit a frame when data is ready, regardless of whether the channel is busy or not.
- If a station does not receive an acknoledgement, the frame is retransmitted after a random delay.
Definitions:
- \(T\)
- Time required to send one frame
- \(G\)
- Poisson distribution mean number of frame arrivals in a certain interval of length \(T\).
The Poisson distribution is discrete – the probability of k events occurring in a fixed time interval if the events have a constant mean rate and are independent of the time since the last event.
\( f_{\text{Poisson}}(k; G)= \Pr(X = k)= \dfrac{G^k e^{-G}}{k!} \)
Exponential distribution has mean of \(\mu = 1/\lambda\)
\( f_{\text{exp}}(x;\lambda) = \begin{cases} \lambda e^{-\lambda x} & x \ge 0,
0 & x < 0. \end{cases} \)
The Erlang distribution \((k, \lambda)\) is the sum of \(k\) independent exponential variables with mean \(1/\lambda\) each. It is a continuous distribution with a mean of \(k/\lambda\)
\( f_{\text{Erlang}}(x; k,\mu)=\dfrac{ x^{k-1} e^{-\frac{x}{\mu}} }{\mu^k (k-1)!}\quad\mbox{for }x, \mu \geq 0 \)
PROBLEM: what is probability density function of delays for an individual client that yields a Poisson distribution for number of packets arriving in a time interval \(T\)?