Skip to content
GATEiit

GATE CSE 2025 Session 2 Q23: IPv4 fragments of a 7488-byte UDP payload over Ethernet

Question

Consider a network that uses Ethernet and IPv4. Assume that IPv4 headers do not use any options field. Each Ethernet frame can carry a maximum of 1500 bytes in its data field. A UDP segment is transmitted. The payload (data) in the UDP segment is 7488 bytes.

Which ONE of the following choices has the CORRECT total number of fragments transmitted and the size of the last fragment including IPv4 header?

Try it: multiple choice: exactly one correct option

Select one option
Wrong answer costs 1/3 mark in GATE.
Show answer and explanation

Answer: D (official GATE 2025 final answer key)

Concept used: IP fragments the whole UDP segment (8-byte header plus data); each fragment carries at most MTU − 20 bytes, a multiple of 8.

First step: The IP payload is 7488+8=74967488 + 8 = 7496 bytes and each fragment holds 1500−20=14801500 - 20 = 1480 bytes of it.

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 2025 Session 2 CS question paper and final answer key, organised by IIT Roorkee. Question paper · Answer key

More Computer Networks PYQs

PracticePractise Computer Networks questions →Full paperAll GATE CSE 2025 Session 2 questions →