GATE CS 2025 Set 1 Q47 — Fragmentation and IP Addressing, IPv4, CIDR

To find the number of fragments delivered to the destination, we analyze the fragmentation process at each hop. Let the IP header size be $20$ bytes.

1. First Hop (Source to Router 1): $\text{MTU} = 5000$ bytes

• Maximum payload size per fragment $= 5000 - 20 = 4980$ bytes.
• Since the fragment offset must be a multiple of $8$, the payload size must be a multiple of $8$.
• The largest multiple of $8$ less than or equal to $4980$ is $\lfloor 4980 / 8 \rfloor \times 8 = 4976$ bytes.
• The $15000$ bytes message payload is fragmented into:
  • $3$ fragments with $4976$ bytes payload (total size $= 4996$ bytes each)
  • $1$ fragment with $15000 - (3 \times 4976) = 72$ bytes payload (total size $= 92$ bytes)
• Total fragments sent to Router 1 $= 4$.

2. Second Hop (Router 1 to Router 2): $\text{MTU} = 3000$ bytes

• Maximum payload size per fragment $= 3000 - 20 = 2980$ bytes.
• The largest multiple of $8$ is $\lfloor 2980 / 8 \rfloor \times 8 = 2976$ bytes.
• Router 1 fragments each incoming packet that exceeds the MTU:
  • Each of the $3$ large fragments ($4976$ bytes payload) is split into two sub-fragments: one of size $2976$ bytes payload and another of size $4976 - 2976 = 2000$ bytes payload.
  • The small fragment ($72$ bytes payload) is forwarded as-is.
• Total fragments leaving Router 1 $= (3 \times 2) + 1 = 7$.

3. Third Hop (Router 2 to Destination): $\text{MTU} = 3000$ bytes

• The maximum size among the $7$ incoming fragments is $2996$ bytes (payload $2976$ bytes + header $20$ bytes).
• Since $2996 \le 3000$ bytes, no further fragmentation occurs at Router 2.

Thus, the number of fragments delivered to the destination is 7.