Discussion
Consider a network that uses Ethernet and IPv4. Assume that IPv4 headers do not use any options field. Each Ethernet frame can carry a maximum of 1500 bytes in its data field. A UDP segment is transmitted. The payload (data) in the UDP segment is 7488 bytes.
Which ONE of the following choices has the CORRECT total number of fragments transmitted and the size of the last fragment including IPv4 header?
Answer: D
The datagram goes as 6 fragments, the last of 116 bytes.
The IP payload is the whole UDP segment: 7488 bytes of data plus the 8-byte UDP header, 7496 bytes. Each fragment carries a 20-byte IP header, leaving 1480 bytes of data, a multiple of 8 as fragment offsets require:
\[ \begin{aligned} 5 \times 1480 &= 7400 \\ 7496 - 7400 &= 96 \\ \text{last fragment} &= 96 + 20 = 116\ \text{bytes} \end{aligned} \]
So there are 5 full fragments and a sixth of 116 bytes. 108 bytes forgets the UDP header (88 + 20), 88 bytes forgets both headers, and five fragments cannot carry the datagram at all, since at most 5 × 1480 = 7400 bytes of data fit in them.
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