## How to maximize the variable in the given scenario ?

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. Btw can someone let me know if my derived equations (1 and 2) are all right or not!