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?

1. Determine the total IP payload size:

   • UDP payload $= 7488\text{ bytes}$

   • UDP header size $= 8\text{ bytes}$

   • Total UDP segment size (which forms the IPv4 payload) $= 7488 + 8 = 7496\text{ bytes}$

2. Determine maximum data payload per fragment:

   • Ethernet Maximum Transmission Unit (MTU) $= 1500\text{ bytes}$

   • Standard IPv4 header size (without options) $= 20\text{ bytes}$

   • Maximum IPv4 payload per fragment $= 1500 - 20 = 1480\text{ bytes}$

   • Since $1480$ is a multiple of $8$ ($1480 / 8 = 185$), each full fragment can carry $1480\text{ bytes}$ of data.

3. Calculate number of fragments and remaining bytes:

   • Fragment 1 to 5: Each carries $1480\text{ bytes}$ of payload $\implies 5 \times 1480 = 7400\text{ bytes}$.

   • Fragment 6 (the last fragment): Carries the remaining $7496 - 7400 = 96\text{ bytes}$ of payload.

   • Total number of fragments $= 6$.

4. Size of the last fragment including IPv4 header:

   • Size $= 96\text{ bytes (payload)} + 20\text{ bytes (IPv4 header)} = 116\text{ bytes}$.

Thus, the total number of fragments is 6 and the size of the last fragment is 116 bytes, corresponding to Option (D).