GATE CS 2024 Set 1 — Question 65

NAT+2 / -0HardIPv4 Packet & FragmentationNetwork Layer: AddressingComputer Networks

Computer Networks → Network Layer: Addressing → IPv4 Packet & Fragmentation

Last updated 20 August 2026

Question

Consider sending an IP datagram of size 1420 bytes (including 20 bytes of IP header) from a sender to a receiver over a path of two links with a router between them. The first link (sender to router) has an MTU (Maximum Transmission Unit) size of 542 bytes, while the second link (router to receiver) has an MTU size of 360 bytes. The number of fragments that would be delivered at the receiver is _________

Correct answer

6 to 6

Solution

1.Original Datagram:

Total size = 1420 bytes
IP Header = 20 bytes
Data Payload = $1420 - 20 = 1400$ bytes

2. Fragmentation at Link 1 (MTU = 542 bytes):

Maximum IP packet size = 542 bytes
Maximum data payload per fragment = $542 - 20 = 522$ bytes
Since the Fragment Offset field is in units of 8 bytes, the data length in all but the last fragment must be a multiple of 8.
Largest multiple of 8 $\leq 522$ is $520$ ( $65 \times 8$ ).
Fragments generated:
Fragment 1: Data = 520 bytes, Total size = 540 bytes
Fragment 2: Data = 520 bytes, Total size = 540 bytes
Fragment 3: Data = $1400 - (520 + 520) = 360$ bytes, Total size = 380 bytes

3. Fragmentation at Link 2 (MTU = 360 bytes):

Maximum IP packet size = 360 bytes
Maximum data payload per fragment = $360 - 20 = 340$ bytes
Largest multiple of 8 $\leq 340$ is $336$ ( $42 \times 8$ ).
Each fragment from Link 1 is checked against the new MTU:
Fragment 1 (520 bytes data): Needs fragmentation. $520 = 336 + 184$ . (2 fragments)
Fragment 2 (520 bytes data): Needs fragmentation. $520 = 336 + 184$ . (2 fragments)
Fragment 3 (360 bytes data): Needs fragmentation (since total size 380 > 360). $360 = 336 + 24$ . (2 fragments)

4. Total Fragments at Receiver:

Total = $2 + 2 + 2 = 6$ fragments.

GATE CS 2024 Set 1 — Question 65

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