GATE 2026 CS (CS2) – Question 65
It is necessary to design a link-layer protocol between two hosts directly connected over a lossless link of length 3000 kilometers. Assume the link bandwidth is $10^8$ bits per second and the propagation delay is 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 so that (i) the sequence numbers do not wrap around before 60 seconds, and (ii) maximum link utilization is achieved. The value of $N$ is __________. (answer in integer)
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Correct answer: 30
Explanation
The one-way propagation delay is $$3000\times10^3\text{ m}\times5\text{ ns/m}=15\text{ ms},$$ so the RTT is 30 ms. At $10^8$ bps, the link carries $12.5\times10^6$ bytes per second. To fully utilize the link, the sender must be able to have one bandwidth-delay product worth of data in flight: $$12.5\times10^6\times0.03=375000\text{ bytes}.$$ Requirement (ii) is therefore satisfied by a window that can cover at least 375000 sequence numbers. Requirement (i) says sequence numbers must not wrap within 60 seconds, so the total number of distinct byte sequence numbers must exceed $$12.5\times10^6\times60=7.5\times10^8.$$ Since $2^{29}\approx5.37\times10^8$ is too small but $2^{30}\approx1.07\times10^9$ is sufficient, the minimum number of bits is 30. Hence the answer is 30.