Question

An IP datagram carries 3980 bytes of data (with a separate 20-byte header) and must be transmitted over a network with an MTU of 1500 bytes. Given that the maximum data payload per fragment must be a multiple of 8 bytes, and the usable data space per fragment is 1480 bytes (1500 − 20-byte header), how many fragments are required, and what is the data size of the last fragment?

Q: Answer-An IP datagram carries 3980 bytes of data (with a separate 20-byte header) and must be transmitted o...

Each fragment except possibly the last can carry at most 1480 bytes of data already a multiple of 8, so no further adjustment is needed. Dividing the total data 3980 1480 2.69, meaning 2 full fragments of 1480 bytes each account for 2960 bytes, leaving 3980 2960 1020 bytes for the third final fragment. So 3 fragments total are required, with the last one carrying 1020 bytes of data less than the maximum, as is typical for a final fragment.