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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 efficiency of channel? (with 2 decimal places)

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Transmission time= $\large \frac{L}{B}$ = 40ms

Propagation time = 400ms

a=$\large \frac{T_p}{T_t}$ = 10

max window size for 100% efficiency = $1+2a$ =  21

efficiency =$\large \frac{10\times 100}{21 } = 47.6$%
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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
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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 

 

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Here, efficiency = N / (1 + 2 a), where N is the sender window size (given as 10).

a = Tp / T                       

Tp = 400 ms (given) and Tt = (length of each frame) / (bandwidth of the satellite link) = (100 * 8) / (20 * 210).

Evaluating the above, we find a = 10.24.

Substituting the values in the above formula, we get efficiency = 0.4655 = 46.55 %.

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