GATE 2016 CS – 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 (1 Kbps = 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
Explanation
Sending a frame takes $\frac{8000 \text{ bits}}{80000 \text{ bits/s}} = 0.1$ s. The acknowledgement takes $\frac{800 \text{ bits}}{8000 \text{ bits/s}} = 0.1$ s to transmit. The frame and the acknowledgement each travel for 0.1 s, so propagation adds 0.2 s. One full cycle takes $0.1 + 0.1 + 0.2 = 0.4$ s, during which 1000 bytes are delivered. The throughput is $\frac{1000}{0.4} = 2500$ bytes per second.