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In a token ring network the transmission speed is 10$^7$ bps and the propagation speed is 200 meters/$\mu$s. The 1-bit delay in this network is equivalent to:

1. 500 meters of cable.

2. 200 meters of cable.

3. 20 meters of cable.

4. 50 meters of cable.

As, token ring network

So,        transmission delay = length / bandwidth

transmission delay = 1/10^7 = 0.1micro sec.

as,  propagation speed =200m/micro sec.

In , 1micro sec.  it covers 200m

than in 0.1micro sec. it is 20metres.(C) answer

Can it be solve using, d<= L/2*V/B and by playing with units?
i have the same doubt as @smartmeet. i tried using formula but couldn't approach. Anyone can help?

It works on CSMA/CD and note that there we deal only with one sender and one receiver at distant location (so wait  2Tp  for both collision S-R & R-S direction)

but in ring each station(sender/receiver) is connected with another adjacent station. So we don't require that much time to wait for.

L/B = d/v

So it will work but remove 2 from your from your TRICK

for another example working

https://www.avatto.com/computer-science/test/mcqs/networking/questions/81/10.html

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THIS IS MY OBSERVATION ANY KIND OF CORRECTION IS WELCOMED