Project ForgeA 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.
Explanation
Frame transmission time = 8000 bits / 80000 bps = 100ms. Ack transmission time = 800 bits / 8000 bps = 100ms. Total cycle = frame Tt(100) + propagation(100) + ack Tt(100) + propagation(100) = 400ms. Throughput = 1000 bytes / 0.4s = 2500 bytes/second.
Source: GATE CS 2016 (Session 1 / Set A) (2016)
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