GATE2025_CS1_Q57

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)

✅ Final Answer:

The correct answer is 7.

🔹 Short Answer:

The initial 15000-byte message is fragmented by Router-1 (MTU=5000) into 4 packets. When these 4 packets reach Router-2 (MTU=3000), the three large packets are too big and must be re-fragmented into 2 packets each (3x2=6). The last small packet passes through without change. The total fragments arriving at the destination are 6 + 1 = 7.

🔸 Long Answer:

This problem requires understanding how IP packets are fragmented when they encounter a network link with a smaller Maximum Transmission Unit (MTU). We will track the message as it passes through each router.

Key Assumptions

• IPv4 Header: We assume a standard IPv4 header of 20 bytes (as no options field is mentioned).
• MTU Includes Header: The problem states the MTU includes the IP header. This means the total size of a packet (Header + Data) cannot exceed the MTU.
• Data Payload: The maximum data a single fragment can carry is MTU - Header Size. While technically the data size must be a multiple of 8, for most GATE problems this can be ignored unless it's critical. We will proceed with the simpler calculation, which yields the correct answer.

Fragmentation at Router-1 (MTU = 5000 bytes)

The first router, R1, receives a large datagram carrying 15000 bytes of data.

Max Data per Fragment at R1 = MTU - Header = 5000 - 20 = 4980 bytes

Number of fragments needed to carry 15000 bytes:

Number of Fragments = ceil(Total Data / Max Data per Fragment)
= ceil(15000 / 4980) = ceil(3.012...) = 4 Fragments

The fragments created by R1 are:

• Fragment 1: 4980 bytes data + 20 byte header = 5000 bytes
• Fragment 2: 4980 bytes data + 20 byte header = 5000 bytes
• Fragment 3: 4980 bytes data + 20 byte header = 5000 bytes
• Fragment 4: 60 bytes data (15000 - 3*4980) + 20 byte header = 80 bytes

So, 4 packets leave R1 and head towards R2.

Re-fragmentation at Router-2 (MTU = 3000 bytes)

R2 receives these 4 packets and inspects their size against its own MTU of 3000 bytes.

Max Data per Fragment at R2 = MTU - Header = 3000 - 20 = 2980 bytes

Now, we check each incoming fragment:

• Process Fragment 1 (5000 bytes): Too large. It must be re-fragmented.
  Data to re-fragment: 4980 bytes.
  Number of new fragments = ceil(4980 / 2980) = ceil(1.67...) = 2 fragments.
• Process Fragment 2 (5000 bytes): Same as above. Re-fragmented into 2 fragments.
• Process Fragment 3 (5000 bytes): Same as above. Re-fragmented into 2 fragments.
• Process Fragment 4 (80 bytes): Size (80) is less than MTU (3000). It passes through unchanged as 1 fragment.

Final Count

The total number of fragments that finally arrive at the destination is the sum of all fragments leaving R2.

Total Fragments = 2 (from F1) + 2 (from F2) + 2 (from F3) + 1 (from F4) = 7

Bibliography:
- Kurose, J. F., & Ross, K. W. (2016). Computer Networking: A Top-Down Approach. Pearson. (Chapter 4: The Network Layer: Data Plane).
- Tanenbaum, A. S., & Wetherall, D. J. (2011). Computer Networks. Pearson. (Chapter 5: The Network Layer).

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