Test Series

Question

Comments

Efficiency = $\frac{1}{1+2a}=\frac{1}{2}$

$a=\frac{1}{2}=\frac{Tp}{Tt}$

$Tt=2*Tp$

$2*10ms = \frac{L}{64*10^{3}}*10^{3}$

$L=2*10*64 bits = 2*80= 160Bytes$

Answer 1

link utilization is 50% means efficiency is .5..

1/2=1/(1+2a)  ->  2*T_p/T_t=1 ->T_t=2*10^-2  -> Frame_length=2*10^-2 *64*10³ ->1280bits or 1280/8 ->160bytes