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Official PYQ · 2026Set 2 · Q55Data Link Layer: Framing, Error Detection, MAC, Ethernet BridgingNAT2 marks

GATE CS 2026 Set 2 Q55 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

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It is necessary to design a link-layer protocol between two hosts that are directly connected over a lossless link of length 30003000 kilometers. Assume that the link bandwidth is 10810^{8} bits per second and that the propagation delay in the link is 55 nanoseconds per meter. Every transmitted data byte is assigned a unique sequence number.

Let NN be the minimum number of bits needed for the sequence number field in the protocol header such that

  1. the sequence numbers do not wrap around before 6060 seconds, and
  2. the maximum utilization of the link is achieved.

The value of NN is ____\_\_\_\_ . (answer in integer)

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Answer: 30:30

To find the minimum number of bits NN for the sequence number field, we must satisfy both given conditions.

First, calculate the transmission rate in bytes per second since sequence numbers are assigned per byte:

Bandwidth (B)=108 bits/sec=1088 bytes/sec=1.25×107 bytes/sec\text{Bandwidth } (B) = 10^8 \text{ bits/sec} = \frac{10^8}{8} \text{ bytes/sec} = 1.25 \times 10^7 \text{ bytes/sec}

Condition (i): No wrap-around before 60 seconds

The total number of bytes transmitted in 6060 seconds is:

Bytes transmitted=1.25×107 bytes/sec×60 seconds=7.5×108 bytes\text{Bytes transmitted} = 1.25 \times 10^7 \text{ bytes/sec} \times 60 \text{ seconds} = 7.5 \times 10^8 \text{ bytes}

To ensure unique sequence numbers for all these bytes without wrapping around:

2N≥7.5×1082^N \ge 7.5 \times 10^8

Since 229=536,870,9122^{29} = 536,870,912 and 230=1,073,741,8242^{30} = 1,073,741,824, we need at least:

N≥30N \ge 30

Condition (ii): Maximum link utilization

The total propagation delay (TpT_p) is:

Tp=3000 km×5 ns/m=3×106 m×5×10−9 s/m=15 msT_p = 3000 \text{ km} \times 5 \text{ ns/m} = 3 \times 10^6 \text{ m} \times 5 \times 10^{-9} \text{ s/m} = 15 \text{ ms} Round Trip Time (RTT)=2×Tp=30 ms=0.03 s\text{Round Trip Time } (RTT) = 2 \times T_p = 30 \text{ ms} = 0.03 \text{ s}

The bandwidth-delay product (in bytes) is:

RTT Capacity=1.25×107 bytes/sec×0.03 s=3.75×105 bytes\text{RTT Capacity} = 1.25 \times 10^7 \text{ bytes/sec} \times 0.03 \text{ s} = 3.75 \times 10^5 \text{ bytes}

This requires a sequence number space of at least 3.75×1053.75 \times 10^5 (for Go-Back-N) or 7.5×1057.5 \times 10^5 (for Selective Repeat), which requires at most 2020 bits.

To satisfy both conditions, we take the maximum of the two requirements:

N=max⁡(30,20)=30N = \max(30, 20) = 30

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