Virtual Gate Test Series: Computer Networks - Transmission Delay

Question

Comments

$B$?

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I got it it as 61.44Mbps

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is it C??

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@ExamBazaar

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Answer 1

According to the question :

         Transmission time   =   Round trip delay

                                      =    2 * Propogation delay (as per the normal convention)

  ==>  Data size / Data rate  =   2 * Distance between station / Speed of medium

  ==>  2¹⁰ * 2³ bits / x         =    2 *  20 * 10³ /  (3 * 10⁸)

  ==>  x                              =     2¹⁰ * 2³ * 3 * 10⁸ /  (2 * 20 * 10³)

                                          =    61.44 Mbps

 Actually taking 1 KB = 1000 B = 8000 bits , we get x  =   60 Mbps but the thing is that for system oriented things like packet size , memory size etc we take 1 K = 1024 whereas for channel and external to system things like velocity of medium , clock rate etc we take 1 K = 1000 .       

Comments

I got the same answer and everything looks smooth. But one doubt.

It is mentioned in the question that speed of light should be used. But practically, in OFC data travels at 70% of the speed of light, I think that will be 2.8*10^6 m/s.

So, I see that we all took the speed of light, but shouldn't we actually take the practical speed of data in Optic Fibre Cable ?

Answer 2

according to question:2*tp=tx

2*Distance/velocity= packetsize/band width

2*(20*10^3)/3*10^8 =1024 *8/band width

band width=1024*8*3*10^5/40

=614.4*10^5

=61.44Mbps