2 votes 2 votes Computer Networks computer-networks csma-cd ethernet + – Arnabi asked Mar 13, 2017 Arnabi 1.7k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 7 votes 7 votes $A = KP(1-P)^{K-1}$ For the maximum value of $A$, $\frac{dA}{dP}=0$ Solving, we have $P(K-1)(1-P)^{K-2} = (1-P)^{K-1}$ $P=\frac{1}{K}$ Hence, $A =\lim_{K\rightarrow \infty } (1-\frac{1}{K})^{K-1} = \frac{1}{e}$ Kapil answered Mar 14, 2017 selected Mar 14, 2017 by 2018 Kapil comment Share Follow See all 0 reply Please log in or register to add a comment.