Consider a network that uses Ethernet and IPv4. Assume that IPv4 headers do…

Correct answer: 6 fragments, last fragment size 116 bytes (including the 20-byte IPv4 header).

Step-by-step calculation:

• Total UDP segment size (IP payload) = UDP header (8 bytes) + UDP data (7488 bytes) = 7496 bytes.

• Maximum IP payload per fragment = Ethernet data field max (1500 bytes) - IPv4 header (20 bytes) = 1480 bytes.

• Number of full-size fragments = floor(7496 / 1480) = 5, which carry 5 × 1480 = 7400 bytes.

• Remainder for the last fragment = 7496 - 7400 = 96 bytes of IP payload.

• Total number of fragments = 5 full fragments + 1 last fragment = 6 fragments.

• Size of the last fragment including the IPv4 header = IPv4 header (20 bytes) + last fragment IP payload (96 bytes) = 116 bytes.

Common mistakes to avoid:

• Forgetting to add the 8-byte UDP header to the UDP payload when computing the total IP payload to fragment.

• Using the Ethernet MTU (1500) directly without subtracting the IPv4 header to find the usable IP payload per fragment.

• Ignoring that fragment payload sizes (except possibly the last) must align with the 8-byte fragmentation unit; here 1480 is divisible by 8, so it is valid.