Consider sending an IP datagram of size 1420 bytes (including 20 bytes of IP…
2024
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 ________
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Correct answer: 6
Key steps: compute how the 1400-byte payload is split on each link.
Total IP payload = 1420 − 20 (IP header) = 1400 bytes.
First link MTU 542 → max data per fragment = 542 − 20 = 522 bytes. Fragment offsets must be a multiple of 8 bytes, so each fragment carries at most 520 bytes of payload.
1400 bytes split by 520 → 520 + 520 + 360. So the router receives three fragments with payloads 520, 520, and 360 bytes.
Second link MTU 360 → max data per fragment = 360 − 20 = 340 bytes. Rounded down to a multiple of 8 gives 336 bytes of payload per fragment.
Fragment each incoming piece for the second link:
520 → 336 + 184
520 → 336 + 184
360 → 336 + 24
Count fragments delivered to receiver: 2 + 2 + 2 = 6.
Answer: 6 fragments.
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