GATE CSE 2007 | Question: 66, ISRO2016-71

Question

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

• A. $500$ meters of cable.

• B. $200$ meters of cable.

• C. $20$ meters of cable.

• D. $50$ meters of cable.

Comments

Out of syllabus now

Answer 1

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 

Comments

Can it be solve using, d<= L/2*V/B and by playing with units?

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i have the same doubt as @smartmeet. i tried using formula but couldn't approach. Anyone can help?

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@smartmeet & @Regina Phalange

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

navigate to last question 

THIS IS MY OBSERVATION ANY KIND OF CORRECTION IS WELCOMED

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Tranmission delay is Packet size/BW na??
How length got here??