A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of…
2016
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.
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Correct answer: 2500
Key idea: in Stop-and-Wait the sender must finish transmitting a frame, wait for it to propagate to the receiver, wait for the acknowledgement to be transmitted, and wait for the acknowledgement to propagate back before sending the next frame.
Frame transmission time = (1000 bytes × 8 bits/byte) / 80,000 bits/s = 8,000 / 80,000 = 0.1 s.
Acknowledgement transmission time = (100 bytes × 8) / 8,000 = 800 / 8,000 = 0.1 s.
Two one-way propagation delays = 2 × 100 ms = 0.2 s.
Cycle time = 0.1 + 0.1 + 0.2 = 0.4 s.
Throughput = frame size / cycle time = 1000 bytes / 0.4 s = 2500 bytes/s.
Answer: 2500 bytes/second
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