Consider an IP packet with a length of 4,500 bytes that includes a 20-byte…
2018
Consider an IP packet with a length of 4,500 bytes that includes a 20-byte IPv4 header and a 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600 bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20 bytes. Assume that the fragmentation offset value stored in the first fragment is 0.
The fragmentation offset value stored in the third fragment is _______.
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Correct answer: 144
Key idea: find the data carried, the usable payload per fragment (must be a multiple of 8 bytes), then compute offsets in 8-byte units.
Total payload (data) = 4500 - 20 (IP header) - 40 (TCP header) = 4440 bytes.
Maximum data per fragment = MTU - IP header = 600 - 20 = 580 bytes. Must be a multiple of 8, so usable data per fragment = floor(580/8) * 8 = 576 bytes.
Fragment offsets are measured in 8-byte units. First fragment offset = 0. Second fragment offset = 576 / 8 = 72. Third fragment offset = 2 * 576 / 8 = 144.
Answer: the fragmentation offset stored in the third fragment is 144.
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