GATE CS 2014 Set 3 Q28 — Fragmentation and IP Addressing, IPv4, CIDR

To find the fields of the third IP fragment, we break down the original IP packet based on the Maximum Transmission Unit (MTU) and the requirement that fragment offsets must be multiples of 8 bytes.

1. Identify Packet Parameters

• Total Packet Size: $4404 \text{ bytes}$
• IP Header Length: $20 \text{ bytes}$
• Total Payload Size: $4404 - 20 = 4384 \text{ bytes}$
• MTU: $1500 \text{ bytes}$

2. Determine Maximum Payload per Fragment

The maximum payload size in each fragment is limited by the MTU minus the header size:

Since the fragment offset field in the IP header is measured in units of 8 bytes, the payload size of intermediate fragments must be a multiple of 8. Since $1480$ is divisible by $8$ ($1480 / 8 = 185$), the maximum payload size for each fragment is exactly $1480 \text{ bytes}$.

3. Fragment Breakdown

• Fragment 1:
  • Payload: $1480 \text{ bytes}$ (bytes $0$ to $1479$)
  • Total Length (Datagram Length): $1480 + 20 = 1500 \text{ bytes}$
  • Offset: $0 / 8 = 0$
  • More Fragments (MF) bit: $1$
• Fragment 2:
  • Payload: $1480 \text{ bytes}$ (bytes $1480$ to $2959$)
  • Total Length: $1480 + 20 = 1500 \text{ bytes}$
  • Offset: $1480 / 8 = 185$
  • MF bit: $1$
• Fragment 3 (Last Fragment):
  • Remaining Payload: $4384 - (1480 + 1480) = 1424 \text{ bytes}$ (bytes $2960$ to $4383$)
  • Total Length (Datagram Length): $1424 + 20 = 1444 \text{ bytes}$
  • Offset: $2960 / 8 = 370$
  • MF bit: $0$ (since no more fragments follow)

Thus, the values for the third fragment are: MF bit: 0, Datagram Length: 1444, Offset: 370.