Host A is sending data to host B over a full duplex link. A and B are using…
2003
Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 µs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 us. What is the maximum achievable throughput in this communication?
- A.
A
- B.
B
- C.
C
- D.
D
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Correct answer: B
Solution summary: compute the maximum throughput using the sliding-window limit.
Step 1 — Compute RTT:
RTT = 2 × propagation delay + data transmission time (ACK transmission time is negligible).
RTT = 2 × 200 µs + 50 µs = 450 µs.
Step 2 — Compute maximum in-flight data allowed by the window:
In-flight bytes = window size × packet size = 5 × 1000 bytes = 5000 bytes.
Step 3 — Compute window-limited throughput:
Throughput = in-flight bytes / RTT = 5000 bytes / 450 µs ≈ 11.111… × 10^6 bytes/s.
Step 4 — Check link capacity (not limiting here):
Link raw capacity = packet size / packet transmission time = 1000 bytes / 50 µs = 20 × 10^6 bytes/s, which is greater than the window-limited throughput.
Final answer: the maximum achievable throughput is approximately 11.11 × 10^6 bytes per second.
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