How many fragments are created and what is the last fragment size in GATE CS 2025 Set-2 Q23?

6 fragments are created and the last fragment is 116 bytes (including the IPv4 header of 20 bytes). The UDP segment total size is 7488 (payload) + 8 (UDP header) + 20 (IP header) = 7516 bytes. Each fragment can carry 1480 bytes of data (1500 - 20 IP header). First 5 fragments carry 1480 bytes each = 7400 bytes. Remaining data = 7496 - 7400 = 96 bytes, so last fragment = 96 + 20 = 116 bytes.

How does IPv4 fragmentation work with Ethernet MTU?

Ethernet has an MTU of 1500 bytes. Each IP fragment must fit within this limit including the 20-byte IPv4 header (assuming no options). So each fragment can carry at most 1500 - 20 = 1480 bytes of data. The original IP datagram is split into fragments of this size, and each fragment gets its own IP header with appropriate offset and more-fragments flags.

What is included in the total UDP segment size for fragmentation?

The total size of the IP datagram to be fragmented includes: UDP payload (7488 bytes) + UDP header (8 bytes) + IPv4 header (20 bytes) = 7516 bytes. The UDP header and payload together form the data that gets fragmented at the IP layer, with each fragment carrying a fresh 20-byte IP header.

IPv4 UDP Fragmentation Ethernet MTU | GATE CS 2025 Set-2 Q23

Computer Sciences > GATE 2025 SET-2 > IP Fragmentation

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?

5 fragments, 1488 bytes

6 fragments, 88 bytes

6 fragments, 108 bytes

6 fragments, 116 bytes

Correct : d

The correct answer is Option D - 6 fragments, 116 bytes.
Let''s work through the calculation step by step.
Total IP datagram size:
UDP payload = 7488 bytes
UDP header = 8 bytes
IPv4 header = 20 bytes (no options)
Total = 7488 + 8 + 20 = 7516 bytes
Data per fragment:
Ethernet MTU = 1500 bytes. Each fragment carries a 20-byte IPv4 header, leaving 1500 − 20 = 1480 bytes of data per fragment.
Note: Fragment offsets must be multiples of 8, and 1480 is divisible by 8, so this is valid.
Total data to fragment (UDP header + payload) = 7488 + 8 = 7496 bytes
Number of fragments:
⌈7496 / 1480⌉ = ⌈5.065⌉ = 6 fragments
Last fragment size:
First 5 fragments carry 5 × 1480 = 7400 bytes.
Remaining data = 7496 − 7400 = 96 bytes
Last fragment size (including IPv4 header) = 96 + 20 = 116 bytes
So the answer is 6 fragments and the last fragment is 116 bytes - Option D.

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