GATE | GATE-CS-2016 (Set 1) | Question 65
A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds. Assuming no frame is lost, the sender throughput is __________ bytes/second.
Total time = Transmission-Time + 2* Propagation-Delay + Ack-Time. Trans. time = (1000*8)/80*1000 = 0.1 sec 2*Prop-Delay = 2*100ms = 0.2 sec Ack time = 100*8/8*1000 = 0.1 sec. Total Time = 0.1 + 0.2 + 0.1 = 0.4 sec. Throughput = ((L/B)/Total time) * B, L = data packet to be sent and B = BW of sender. Throughput = L/Total Time = 1000/0.4 = 2500 bytes/sec.
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