in Computer Networks edited ago by
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7 votes
7 votes

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?

  1. $250$ milliseconds
  2. $20$ milliseconds
  3. $520$ milliseconds
  4. $270$ milliseconds
in Computer Networks edited ago by
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4 Answers

17 votes
17 votes
Best answer

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

edited by
by
4 votes
4 votes

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

by
3 votes
3 votes
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
1 vote
1 vote

=> RTT = 4 * Tp

=> 500 * 10-3 = 4 * Tp

=> Tp = 125 * 10-3


=> Tt = 103 / 50 * 103

=> Tt = 20 * 10-2


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 ?

2 Comments

what will be the efficiency of the channel??  

efficiency=20/500  OR

efficiency=20/520 ???
0
0
efficiency = transmission time / (transmission time + propagation time)

so 20 / 520
0
0
Answer:

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