Question 38

Packet and MTU Parameters:

• Total original packet size = $4404$ bytes.
• IP header length = $20$ bytes.
• Total data payload = $4404 - 20 = 4384$ bytes.
• Link MTU = $1500$ bytes.
• Maximum fragment payload must be a multiple of $8$ bytes and satisfy:
  $$\text{Payload} + 20 \le 1500 \implies \text{Payload} \le 1480.$$
  Since $1480$ is divisible by $8$ ($1480 = 185 \times 8$), maximum payload per non-terminal fragment is $1480$ bytes.

Fragment Details:

• Fragment 1:
  • Data bytes: $0$ to $1479$ ($1480$ bytes).
  • Datagram length = $1480 + 20 = 1500$ bytes.
  • Fragment offset = $0 / 8 = 0$.
  • MF (More Fragments) bit = $1$.
• Fragment 2:
  • Data bytes: $1480$ to $2959$ ($1480$ bytes).
  • Datagram length = $1480 + 20 = 1500$ bytes.
  • Fragment offset = $1480 / 8 = 185$.
  • MF bit = $1$.
• Fragment 3:
  • Remaining data payload = $4384 - (1480 \times 2) = 4384 - 2960 = 1424$ bytes.
  • Datagram length = $1424 + 20 = 1444$ bytes.
  • Fragment offset = $2960 / 8 = 370$.
  • MF bit = $0$ (since this is the final fragment carrying the last byte of the original packet).

Thus, for the third fragment:

This matches option A.