GATE CS 2024 Set 1 Q55 — Fragmentation and IP Addressing, IPv4, CIDR

To find the number of fragments delivered at the receiver, we trace the fragmentation process across both links. Each fragment must include a $20$-byte IP header, and its payload size (except for the last fragment) must be a multiple of $8$ bytes.

Step 1: First Link (Sender to Router)

• Original Datagram: Total size = $1420$ bytes, Payload = $1400$ bytes.
• MTU = $542$ bytes $\implies$ Maximum payload size = $542 - 20 = 522$ bytes.
• The largest multiple of $8$ less than or equal to $522$ is $520$ bytes.
• The $1400$ bytes of payload is fragmented into:
• Fragment 1: Payload = $520$ bytes (Total size = $540$ bytes)
• Fragment 2: Payload = $520$ bytes (Total size = $540$ bytes)
• Fragment 3: Payload = $1400 - 520 - 520 = 360$ bytes (Total size = $380$ bytes)

Step 2: Second Link (Router to Receiver)

• MTU = $360$ bytes $\implies$ Maximum payload size = $360 - 20 = 340$ bytes.
• The largest multiple of $8$ less than or equal to $340$ is $336$ bytes.
• The router fragments each of the incoming fragments:
• Fragment 1 ($520$ bytes payload): Split into $336$ bytes and $184$ bytes ($2$ fragments).
• Fragment 2 ($520$ bytes payload): Split into $336$ bytes and $184$ bytes ($2$ fragments).
• Fragment 3 ($360$ bytes payload): Split into $336$ bytes and $24$ bytes ($2$ fragments).

Thus, the total number of fragments delivered at the receiver is $2 + 2 + 2 = 6$.