ISRO2014-45

Question

Consider a $50$ kbps satellite channel with a $500$ milliseconds round trip propagation delay. If the sender wants to transmit $1000$ bit frames, how much time will it take for the receiver to receive the frame?

• A. $250$ milliseconds
• B. $20$ milliseconds
• C. $520$ milliseconds
• D. $270$ milliseconds

Answer 1

Round trip time = 2 * Propagation time ===>Propagation time =500/2= 250 ms

Transmission time = Message size / bandwidth ===>1000 bits / 50 kbps = 20 ms

Time to receive the frame by the receiver ===> 250 + 20 = 270 ms

Answer 2

receiver to receive frame=transmission time + propagation time from sender to receiver

transmission time=L/B=1000b/50Kb

                                 =20ms

RTT=2*PD

=>PD=RTT/2=250 ms

so total time for receiver to receive frame=250+20 ms=270ms

ans=D

Answer 3

Option D
RTT = 500 ms ==> one way propagation time is 250 ms
Bandwidth is 50 kbps ==> per sec it transmits 50k bits
So for 1k bits  it will take 20ms
So the total time it taken is 250 + 20 ms= 270 ms

Answer 4

=> RTT = 4 * Tp

=> 500 * 10^-3 = 4 * Tp

=> Tp = 125 * 10^-3

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=> Tt = 10³ / 50 * 10³

=> Tt = 20 * 10^-2

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Time for a frame to reach receiver from sender = Tt + 2 * Tp

=> 2 * 10^-2 + 2 * 125 * 10^-3

=> 20 * 10^-3 + 250 * 10^-3

=> 270 * 10^-3

 

Cross Check and point out mistakes if any ?

Comments

what will be the efficiency of the channel??  

efficiency=20/500  OR

efficiency=20/520 ???

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efficiency = transmission time / (transmission time + propagation time)

so 20 / 520