A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter…
2004
A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?
Answer: B. 10Kbps — Key idea: the go-back-N throughput is limited by the transmission time, the round-trip propagation delay, and the window size. Use U = (Window × frame…
- A.
5Kbps
- B.
10Kbps
- C.
15Kbps
- D.
20Kbps
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Correct answer: B
Key idea: the go-back-N throughput is limited by the transmission time, the round-trip propagation delay, and the window size. Use U = (Window × frame transmission time) / (RTT + frame transmission time).
Frame size = 100 bytes = 800 bits. Frame transmission time = 800 bits / 20000 bps = 0.04 s (40 ms).
Round-trip time (RTT) = 2 × propagation delay = 2 × 400 ms = 800 ms = 0.8 s.
Window size W = 10. Utilization U = (10 × 0.04) / (0.8 + 0.04) = 0.4 / 0.84 ≈ 0.47619.
Throughput = U × link capacity = 0.47619 × 20 kbps ≈ 9.52 kbps. The closest provided option is 10 kbps.
Conclusion: The maximum achievable data rate with the given parameters is approximately 9.52 kbps, so the correct choice is 10 kbps.
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