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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 to the receiver is

1. 6
2. 7
3. 8
4. 9

Correct answer: 6

Solution

Data = 1420-20 = 1400 bytes. Link1 MTU 542 gives payload floor(522/8)*8 = 520, yielding data fragments 520,520,360. Link2 MTU 360 gives payload 336, so each splits: 520->336+184, 520->336+184, 360->336+24, for 6 fragments delivered. Stored '9' is wrong; correct is 6.

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