Based on the following passage, answer the Questions : A 3000 km long trunk…
2022
Based on the following passage, answer the Questions :
A 3000 km long trunk operates at 1.536mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 𝜇 sec/km.
The minimum number of bits required in the sequence number field of the packet is
- A.
6 bits
- B.
7 bits
- C.
5 bits
- D.
4 bits
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Correct answer: B
Given: link speed = 1.536 Mbps, distance = 3000 km, frame size = 64 bytes (512 bits), propagation = 6 μs/km.
Step 1 — Propagation delay: 3000 km × 6 μs/km = 18000 μs = 0.018 s (one-way).
Step 2 — Transmission time per frame: 512 bits ÷ 1.536×10^6 bps = 0.000333333 s = 333.33 μs.
Step 3 — Round‑trip time (RTT): 2 × 0.018 s = 0.036 s.
Step 4 — Number of frames that can be in transit (bandwidth–delay product in frames): RTT ÷ transmission time = 0.036 ÷ 0.000333333 ≈ 108 frames.
Step 5 — Sequence number bits for sliding window (assuming Go‑Back‑N): We need a window at least 108 frames. For Go‑Back‑N the maximum window = 2^k − 1, so choose the smallest k with 2^k − 1 ≥ 108. 2^6 − 1 = 63 (too small), 2^7 − 1 = 127 (sufficient).
Answer: 7 bits.
Note: This uses the Go‑Back‑N assumption for sliding window. If Selective Repeat were used, the window must satisfy a different relation (window ≤ 2^(k−1)), which would change the required k.
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