GATE CS 2025 Set 1 Q.57: IPv4/IPv6 Addressing, Subnetting, CIDR, ARP, ICMP and NAT

Suppose a message of size 15000 bytes is transmitted from a source to a destination using IPv4 protocol via two routers as shown in the figure. Each router has a defined maximum transmission unit (MTU) as shown in the figure, including IP header. The number of fragments that will be delivered to the destination is ________ . (Answer in integer)

Network path showing Source connected to Router-1 (MTU=5000 bytes), which connects to Router-2 (MTU=3000 bytes), which connects to Destination.

Your answer

Official answer and solution

Official answer: 7
From: Official GATE 2025 answer key

Solution

Fragment the 15000-byte payload first at Router-1 (MTU = 5000 B, 20 B IPv4 header, payload multiple of 8 B) and then fragment each resulting datagram at Router-2 (MTU = 3000 B).

Assume the standard 20-byte IPv4 header. In IPv4 fragmentation, every non-final fragment's data payload must be a multiple of $8$ bytes:

At Router-1 ( $\text{MTU} = 5000$ bytes):
- Maximum data per fragment $= \lfloor \frac{5000 - 20}{8} \rfloor \times 8 = \lfloor \frac{4980}{8} \rfloor \times 8 = 622 \times 8 = 4976$ bytes.
- Dividing the $15000$ -byte message payload: $15000 = 3 \times 4976 + 72$ .
- Thus Router-1 outputs 4 fragments with data payloads of $4976, 4976, 4976,$ and $72$ bytes.
At Router-2 ( $\text{MTU} = 3000$ bytes):
- Maximum data per fragment $= \lfloor \frac{3000 - 20}{8} \rfloor \times 8 = \lfloor \frac{2980}{8} \rfloor \times 8 = 372 \times 8 = 2976$ bytes.
- Each of the three $4976$ -byte fragments splits into 2 fragments of data sizes $2976$ bytes and $4976 - 2976 = 2000$ bytes.
- The 4th fragment ( $72$ bytes of data + $20$ bytes of header $= 92 \le 3000$ bytes) is not fragmented further (1 fragment).

Total fragments delivered to the destination $= (3 \times 2) + 1 = 7$ .

Final answer

Common trap

Dividing $15000$ directly by Router-2's payload capacity ( $\lceil 15000 / 2976 \rceil = 6$ ) instead of fragmenting sequentially through Router-1 and then Router-2.

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 1 (CS1)
Question: 57 of 65
Section: Computer Science and Information Technology

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