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?
Official answer and solution
- Official answer
- D
- From
- Official GATE 2025 answer key
Solution
Add the 8-byte UDP header to the UDP payload to get the IPv4 payload, divide by the maximum fragment payload of 1480 bytes, and add the 20-byte IPv4 header to the last fragment.
- Total IPv4 payload: The UDP segment consists of bytes of UDP data plus an -byte UDP header, giving bytes.
- Maximum data per IPv4 fragment: Ethernet MTU bytes, and the basic IPv4 header is bytes, so each fragment can carry at most bytes of payload (note that is divisible by ).
- Number of fragments and last fragment size:
- The first fragments carry bytes of IPv4 payload.
- The th (last) fragment carries the remaining bytes of payload.
- Including its -byte IPv4 header, the size of the th fragment is bytes.
Option (D)
Forgetting to include the 8-byte UDP header in the IPv4 payload (which would give 88 + 20 = 108 bytes in Option C).
The answer is the one in the official GATE answer key. The solution is this site's own explanation.
Source
- Paper
- GATE 2025 Computer Science and Information Technology - Session 2 (CS2)
- Question
- 23 of 65
- Section
- Computer Science and Information Technology
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