GATE CS 2006 Q45 — Fragmentation and IP Addressing, IPv4, CIDR

To determine if a destination IP address is on the same local network, a host performs a bitwise AND operation between the destination IP address and its own subnet mask, and compares the result with its own network address.

1. Analysis for $C1$:

• $C1$'s own network address:
  $203.197.2.53 \text{ AND } 255.255.128.0 = 203.197.0.0$
• $C1$ checks $C2$'s IP:
  $203.197.75.201 \text{ AND } 255.255.128.0 = 203.197.0.0$

Since the results match, $C1$ assumes $C2$ is on the same network.

2. Analysis for $C2$:

• $C2$'s own network address:
  $203.197.75.201 \text{ AND } 255.255.192.0 = 203.197.64.0$ (since $75 \text{ AND } 192 = 01001011_2 \text{ AND } 11000000_2 = 01000000_2 = 64$)
• $C2$ checks $C1$'s IP:
  $203.197.2.53 \text{ AND } 255.255.192.0 = 203.197.0.0$

Since $203.197.0.0 \neq 203.197.64.0$, $C2$ assumes $C1$ is on a different network.

Thus, $C1$ assumes $C2$ is on the same network, but $C2$ assumes $C1$ is on a different network.