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A 20 Kbps Satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible ?

a) 5 Kbps

b)10 Kbps

c) 15 Kbps

d) 20 Kbps

Do explain your Answer.
asked in Computer Networks by (255 points) | 249 views
0
is answer b)?
0
yes.

please give your solution

3 Answers

+3 votes
Given Data
Bandwidth = 20 Kbps
Propogation dealy = 400 ms, so RTT = 2 * PD = 800 ms
window size = 10
frame size = 100 bytes or 800 bits

Even though bandwidth is 20kbps, we are restricted with window size here

In 1 RTT we are sending 10 * 800 bits
In 800 ms we are sending 10 * 800 bits
In 1 ms we are sending 10 bits
In 1 sec we are sending 10 * 1000 bits
In 1 sec we are sending 10 Kb

So effective bandwidth = 10 Kb/sec, which is option B
answered by Loyal (7.5k points)
edited by
0 votes
10 kbps
answered by Active (2.9k points)
0 votes

We use the formula Throughput =Data Size/(Transmission time +2Propagation Delay)

or data size =Windows size* packet Size (here windows n=10)

Bandwidth =20 Kbps =20*10^bps

Propagation Delay=400ms =400*10^-3 seconds 

transmission time (Data )=100/20*10^-3 seconds =0.4 seconds 

And using  formula  of Throughput = (10*100*8)/(0.4+2*400*10^-3)= 10000 bps =10Kbps 

 

answered by Junior (553 points)
edited by


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