It is necessary to design a link-layer protocol between two hosts directly connected over a lossless

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GATE-CS is a competitive examination conducted in India. This question is from the 2026 GATE-CS Computer Networks paper.

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This GATE-CS 2026 Computer Networks question has a hard difficulty level and is marked as an important question.

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Question

It is necessary to design a link-layer protocol between two hosts directly connected over a lossless link of length 3000 kilometers. Link bandwidth = 10^8 bits per second. Propagation delay = 5 nanoseconds per meter. Every transmitted data byte is assigned a unique sequence number.Let N be the minimum number of bits needed for the sequence number field such that:(i) The sequence numbers do not wrap around before 60 seconds.(ii) The maximum utilization of the link is achieved.The value of N is ______. (answer in integer)

Accepted Answer

Answer: 30Concept -- Sequence Number Field Size for Link Protocols: Link length = 3000 km = 3×10^6 m. Propagation delay = 5 ns/m → total propagation delay = 3×10^6 × 5×10^-9 = 0.015 s. Bandwidth = 10^8 bps. Condition (i): No wrap-around before 60 seconds. Total bits transmitted in 60 s = 10^8 × 60 = 6×10^9 bits = 7.5×10^8 bytes (since 1 byte = 8 bits). Number of unique sequence numbers needed = 7.5×10^8. Bits needed = ceil(log2(7.5×10^8)) = ceil(29.48) = 30 bits.Condition (ii): Maximum utilization of the link for a stop-and-wait or sliding window protocol. For maximum utilization, we need window size W ≥ 1 + 2a, where a = propagation delay / transmission time per byte. Transmission time per byte = 8 bits / 10^8 bps = 8×10^-8 s. a = 0.015 / (8×10^-8) = 187,500. So W ≥ 1 + 2×187,500 = 375,001. Bits needed for W = ceil(log2(375,001)) = ceil(18.5) = 19 bits. Since condition (i) requires 30 bits, the minimum N = 30 (as it satisfies both).Note: Sequence number bits must be enough to avoid wrap-around for the maximum time or to support the required window size. The larger of the two determines N.

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