Consider an aloha like wireless communication algorithm with $n$ nodes, with each node transmitting with an exponential distribution with rate $\lambda_o$, with $\tau$ be the packet transmission time.

Then $P(\Delta T> 2\tau) = e^{-2\tau n\lambda_o}$ is the probability that the packet is successfully received from transmitting node $B$ to receiving node $A$.

In the presence of $n-1$ neighbors of node A, the rate of successful packets from $B$ to $A$ is

\begin{align} \lambda_a = n \lambda_oP(\Delta T > 2\tau) \ \ \ \ \ \ \ \ (1) \end{align}

Similarly, since the successful packet from $B$ to $A$ form a Poisson process with rate $\lambda_a$, it is only possible if $A$ receives at least one or more packets i.e.

\begin{align} P_{1,a}(t) = 1 - \frac{e^{-\lambda_a t}(-\lambda_a t)^0}{0 !} = 1 - e^{-\lambda_a t} \ \ \ \ \ \ \ \ (2) \end{align}

How am I supposed to find the optimal value of $\tau$? If I find it through eq(1), I get $\tau = -\infty$ (Should I differentiate eq.1 or eq.2 btw.)

\begin{align} \frac{\partial \lambda_a}{\partial \tau} = \frac{\partial \ n\lambda_oP(\Delta T > 2\tau)}{\partial \tau} = \frac{\partial \ n \lambda_o e^{-2\tau n \lambda_o}}{\partial \tau} = 0 \end{align}

\begin{align} \lambda_a' = -2(n\lambda_o)^2 e^{-2\tau n\lambda_o} = 0 \end{align}

\begin{align} \ln(e^{-2\tau n\lambda_o}) => \tau = -\infty \end{align}

PS. By the way can someone let me know if my derived equations (1 and 2) are all right or not?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.