GATE CS 2025 Set 2 — Question 23
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Computer Networks → Medium Access & LANs → Ethernet & MAC
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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?
Correct answer
(D) 6 fragments, 116 bytes
Solution
1.Identify Protocol Overheads and Sizes:
- Ethernet MTU: The maximum data field in an Ethernet frame is 1500 bytes. This is the Maximum Transmission Unit (MTU) for the IP packet.
- IPv4 Header: The problem states no options are used, so the standard header size is bytes.
- Max IP Payload: The maximum amount of data an IP packet can carry per fragment is bytes.
- UDP Header: A standard UDP header is bytes.
- UDP Payload (Data): Given as bytes.
- The IP layer encapsulates the entire UDP segment (Header + Data).
- Total IP Payload = UDP Header + UDP Data = bytes.
- Total data to be fragmented = bytes.
- Max data per fragment = bytes.
- Number of fragments = fragments.
- Data carried in the first 5 fragments = bytes.
- Remaining data for the 6th fragment = bytes.
- Total size of the last fragment = IP Header + Remaining Data = bytes.
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