Skip to content
GATEiit

GATE CSE 2026 Session 2 Q65: Sequence number bits for no wrap-around in 60 s at full utilisation

Question

It is necessary to design a link-layer protocol between two hosts that are directly connected over a lossless link of length 3000 kilometers. Assume that the link bandwidth is 10810^8 bits per second and that the propagation delay in the link is 5 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

i. the sequence numbers do not wrap around before 60 seconds, and

ii. the maximum utilization of the link is achieved.

The value of NN is ______. (answer in integer)

Try it: numerical answer: type the value

Numerical answer. No negative marking.
Show answer and explanation

Answer: 30 (official GATE 2026 final answer key)

Concept used: Byte sequence numbers must cover all bytes sent in the wrap-around time and at least the bytes in flight per round trip.

First step: In 60 s the link sends 60×108/8=7.5×10860 \times 10^8 / 8 = 7.5 \times 10^8 bytes; find the smallest NN with 2N2^N at least this, then check the window condition.

Full step-by-step solution

Detailed working, common mistakes and linked mock tests will be available with a free GATEiit account. Log-in is coming soon.

Share this question

WhatsAppDownload image

Source: official GATE 2026 Session 2 CS question paper and final answer key, organised by IIT Guwahati. Question paper · Answer key

More Computer Networks PYQs

PracticePractise Computer Networks questions →Full paperAll GATE CSE 2026 Session 2 questions →