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+1 vote
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asked in Computer Networks by Boss (6.4k points)   | 181 views

2 Answers

+6 votes
Best answer

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

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

answered by Veteran (47.9k points)  
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For this Question You can refer RavindraBabu Ravula video on CSMA-CD which is free on youtube.

I think Answer will be (A).
answered by Veteran (14.5k points)  


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