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IPv4 Fragmentation

GATE CSE 2021 Set 1 · Q45

MSQ2 marksMedium

Time

Consider two hosts PP and QQ connected through a router RR. The maximum transfer unit (MTU) value of the link between PP and RR is 1500 bytes, and between RR and QQ is 820 bytes. A TCP segment of size 1400 bytes was transferred from PP to QQ through RR, with IP identification value as 0x1234. Assume that the IP header size is 20 bytes. Further, the packet is allowed to be fragmented, i.e., Don’t Fragment (DF) flag in the IP header is not set by PP. Which of the following statements is/are correct?

Select all that applyOne or more correct

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Solution

Step-by-step verified solution

Official answer: A) Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes.; C) If the second fragment is lost, P is required to resend the whole TCP segment.

IPv4 Fragmentation

Let us analyze the fragmentation process:

  1. Packet creation at host PP:

    • TCP segment size (Data payload for IP) =1400 bytes= 1400\text{ bytes}.

    • IP header size =20 bytes= 20\text{ bytes}.

    • Total IP datagram size =1400+20=1420 bytes= 1400 + 20 = 1420\text{ bytes}.

    • Since MTU of link P−RP-R is 1500 bytes≥1420 bytes1500\text{ bytes} \ge 1420\text{ bytes}, no fragmentation occurs between PP and RR.

  2. Fragmentation at Router RR (Link R−QR-Q has MTU=820 bytes\text{MTU} = 820\text{ bytes}):

    • Maximum IP payload size per fragment =820−20=800 bytes= 820 - 20 = 800\text{ bytes}.

    • Since 800800 is a multiple of 88, the maximum allowable payload per fragment is 800 bytes800\text{ bytes}.

    • Fragment 1:

      • IP Payload =800 bytes= 800\text{ bytes}

      • Total IP Datagram size =800+20=820 bytes= 800 + 20 = 820\text{ bytes}

      • Offset =0= 0, More Fragments (MF)=1(\text{MF}) = 1

    • Fragment 2:

      • Remaining IP Payload =1400−800=600 bytes= 1400 - 800 = 600\text{ bytes}

      • Total IP Datagram size =600+20=620 bytes= 600 + 20 = 620\text{ bytes}

      • Offset =800/8=100= 800 / 8 = 100, More Fragments (MF)=0(\text{MF}) = 0

    • Thus, exactly two fragments are created, and the second fragment carries an IP datagram of size 620 bytes620\text{ bytes}. (A is correct).

  3. Reliability and Retransmission:

    • Routers operate at the Network Layer (IP) which is connectionless and unreliable; routers do not buffer or retransmit lost fragments. Hence, (B is incorrect).

    • Fragment reassembly happens only at the destination host QQ. If any fragment is lost, the entire IP datagram is discarded by QQ. Therefore, the source transport layer (TCP at host PP) will time out and retransmit the entire original TCP segment. (C is correct).

  4. TCP Header inspection:

    • The TCP header (which contains the source and destination port numbers) is located at the very beginning of the TCP segment (bytes 00 to 1919). It resides entirely in the first fragment. The second fragment only contains the subsequent payload data and cannot reveal the destination TCP port. Hence, (D is incorrect).

Therefore, the correct options are A and C.

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This IPv4 Fragmentation previous year question is Q45 of the GATE CSE 2021 Set 1 paper. Find more GATE CSE PYQs on IPv4 Fragmentation in Computer Networks, or browse all GATE CSE PYQs.

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IPv4 Fragmentation in GATE CSE

IPv4 Fragmentation has 8 GATE CSE PYQs across 8 papers from 2014 to 2025. This is the only 2-mark MSQ among them, and it is the only one in GATE CSE 2021 Set 1.

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GateAI solution cardGATE CSE 2021 Set 1 · Q45 · MSQ

Consider two hosts P and Q connected through a router R. The maximum transfer unit (MTU) value of the link between P and R is 1500 bytes, and between R and Q is 820 bytes. A TCP segment of size 1400 bytes was transferred from P to Q through R, with IP identification value as 0x1234. Assume that the IP header size is 20 bytes. Further, the packet is allowed to be fragmented, i.e., Don’t Fragment (DF) flag in the IP header is not set by P. Which of the following statements is/are correct?

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.