CSMA/CD

Question

Answer 1

$A = KP(1-P)^{K-1}$

For the maximum value of $A$, $\frac{dA}{dP}=0$

Solving, we have

$P(K-1)(1-P)^{K-2} = (1-P)^{K-1}$

$P=\frac{1}{K}$

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Hence, $A =\lim_{K\rightarrow \infty } (1-\frac{1}{K})^{K-1} = \frac{1}{e}$