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.
Solution: Given Answer is 1425.
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
