+1 vote
86 views
A fiber Optic token Ring used as a MAN is 200 km Long and runs at 100 Mbps . After sending a frame, a Station drains the frame from the ring before regenerating the token. The signal propagation speed in the fibre is 200,000 km/sec and maximum frame size is 1 KB . What is the Maximum Efficiency at N=1 ?
this question is out of syllabus right?

Prop. delay=$\frac{d}{v}$=$\frac{200\times 1000}{200000\times 1000}$=$\frac{1}{1000}$sec. =$1000$microsec.

Trans. delay=$\frac{1\times 1000\times 8}{100\times10 ^{6}}=80$ microsec.

$a=\frac{1000}{80}$=12.5

Efficiency=$\frac{1}{1+\left (\frac{n+1}{n} \right )a}$=$\frac{1}{1+\left (\frac{2}{1} \right )12.5}$

$=0.038=3.8$%
selected
+1 vote
Here in the total cycle time the token is seen by n stations and every station will transmit one packet.n packets are transmitted in 1 cycle time.

efficiency=n*$T_{t}$/($T_{p}+n(T_{t}+T_{p})$=$\frac{1}{1+(\frac{n+1}{n})*a}$  where a=$T_{p}$/$T_{t}$

$T_{p}$=d/v=200/200000=0.001 sec=1ms

$T_{t}$ =Length /bandwidth=1024*8/100*$10^{6}$=0.01msec

So,efficiency=$\frac{1}{1+(\frac{2}{1})*12.5}$=0.038=3.8%