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A $20$ kbps satellite link has propagation delay of $300$ ms , transmitter use Go Back $10$ ARQ scheme, if each frame is $50$ bytes long then the maximum data rate is ________  kbps
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Go back N used where n value is 10

Transmission time  = 50 * 8 bits /  20 kbps =  20 ms

Propagation Time = 300 ms
Efficiency = Window Size*Transmission Time/(Transmission Time + 2*Propagation Time)

= (10 * 20 ) / ( 20 + 2 * 300 )

= 200 / 620

= 0.322580

Maximum Data Rate = 0.322580 * 20 kbps = 6.45
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Given Bandwidth = 20 kbps, Tp = 300 ms, Frame Size = 50 Bytes = 50 * 8 bits,

N = 10

Now, Tt = $\frac{Frame Size}{Bandwidth}$ = $\frac{50 * 8 b}{20 * 10^{3} bps}$ = 20 ms

We know Throughput = Efficiency * Bandwidth = $\frac{N}{1+2*\frac{Tp}{Tt}}* Bandwidth$

                                =$\frac{10}{1+2*\frac{300}{20}} * 20 kbps$ = 6.45 kbps

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