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.
If only 6 bits are reserved for sequence number field, then the efficiency of the system is
Answer: C. 0.578 — Given: Link length = 3000 km Data rate = 1.536 Mbps Frame size = 64 bytes = 512 bits Propagation speed = 6 μs/km → one-way propagation delay = 6 μs/km × 3000…
- A.
0.587
- B.
0.875
- C.
0.578
- D.
0.50
Attempted by 172 students.
Show answer & explanation
Correct answer: C
Given:
Link length = 3000 km
Data rate = 1.536 Mbps
Frame size = 64 bytes = 512 bits
Propagation speed = 6 μs/km → one-way propagation delay = 6 μs/km × 3000 km = 18000 μs = 0.018 s
Sequence number bits = 6 → window size = 2^6 − 1 = 63 (for Go-Back-N sliding window)
Step-by-step calculation:
Frame transmission time t_frame = 512 bits / 1.536×10^6 bps = 1/3000 s ≈ 0.0003333 s (333.33 μs).
One-way propagation delay t_prop = 0.018 s, so round-trip propagation contribution = 2 × t_prop = 0.036 s.
Maximum useful transmission time with window = window × t_frame = 63 × (1/3000) = 63/3000 = 0.021 s.
Efficiency η = (window × t_frame) / (t_frame + 2 × t_prop) = 0.021 / (0.0003333 + 0.036) = 0.021 / 0.0363333 ≈ 0.57895.
Final answer: 0.578 (≈0.579).
Note: The existing keyed value 0.587 appears inconsistent with the calculation above; the correct computed efficiency is ≈0.57895, which matches the option 0.578.
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