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A sender uses the stop and wait ARQ protocol for reliable transmission of frames. Frames are of size 100 bytes and the transmission rate at the sender is 20 Kbps. Size of an acknowledgement is 10 bytes and the transmission rate at the receiver is 8 Kbps. The one way propagation delay is 10 msec.

Assuming no frame is lost, the sender throughput is ________ bytes/sec.

However, I use following approach and my answer is incorrect, Can you please tell me whats I am doing wrong.

Throughput → Useful data sent in 1sec.

Useful data → 100B

Total time taken = Transmission time (sender) + Propagation delay (sender to receiver) + Transmission time (receiver for ack) + propagation delay(receiver to sender)

Transmission Time(Sender) = $\frac{100X8}{20X10^{3}}$ = 40msec
Propagation Delay(Sender to Receiver) = 10ms

Transmission Time(Reciver for ack) =  $\frac{10X8}{8X10^{3}}$ = 10ms

Propagation Delay( Receiver to Sender) = 10ms

So total 70msec took to transfer 100Bytes of data meaning

1msec →  $\frac{100}{70}$ Bytes of data

hence 1sec → $\frac{10}{7}X1000$ = 1428.57

Your solution seems right to me. Could you please post their solution too? @sharmanikhil610

The solution you have provided is correct

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Length of frame = 100 bytes

Bandwidth (sender) = 20 Kbps = 20*1000 bits per second = 2500 bytes per second

Length of acknowledgement = 10 bytes

Bandwidth (receiver) = 8 Kbps = 8*1000 bits per second = 1000 bytes per second

Transmission time (sender), Td (frame) = 100/2500 = 0.04 sec

Transmission time (receiver), Td (ack) = 10/1000 = 0.01 sec

Propagation time, Pd = 10 msec = 0.01sec

Efficiency,ŋ = Useful time / total time

= Td (frame) / (Td (frame) + Td (ack) + 2Pd)

= 0.04/ (0.04+0.02+0.01) = 0.5714 ≈ 0.57

Throughput = Efficiency * Bandwidth (sender) = 0.57 *2500 = 1425 bytes/sec
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