GATE CS 2003 Q82 — Fragmentation and IP Addressing, IPv4, CIDR

For two IP addresses to belong to the same network, their network addresses (obtained by performing a bitwise AND operation between the IP address and the subnet mask) must be identical.

The given subnet mask is $255.255.31.0$. Let us analyze the options:

• Option A: The second octets are $57$ and $56$. Since the mask for the second octet is $255$, the resulting network prefixes will differ ($57 \neq 56$).
• Option B: The third octets are $28$ and $29$. Bitwise AND with $31$ ($00011111_2$):
  $\bullet\; 28 \text{ AND } 31 = 28$
  $\bullet\; 29 \text{ AND } 31 = 29$
  Since $28 \neq 29$, they belong to different networks.
• Option C: The second octets are $203$ and $234$. Since the mask is $255$, the network prefixes will differ.
• Option D: The third octets are $129$ and $161$. Let's perform the bitwise AND with $31$ ($00011111_2$):
  $\bullet\; 129 \text{ AND } 31 = 10000001_2 \text{ AND } 00011111_2 = 00000001_2 = 1$
  $\bullet\; 161 \text{ AND } 31 = 10100001_2 \text{ AND } 00011111_2 = 00000001_2 = 1$
  Since the first two octets ($128.8$) are identical and the fourth octet is masked to $0$, both IP addresses yield the same network address: $128.8.1.0$.