GATE CS 2010 Q47 — Fragmentation and IP Addressing, IPv4, CIDR

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

The first three octets of both IP addresses ($10.105.1.113$ and $10.105.1.91$) are identical ($10.105.1$). Thus, we only need to compare the fourth octet of the IP addresses with the fourth octet of the netmask ($N_4$).

Convert the fourth octets of the IP addresses to binary:

• $113 = 01110001_2$
• $91 = 01011011_2$

Now, let's test the given netmask options for the fourth octet ($N_4$):

• Option A ($N_4 = 0$):
  $113 \text{ AND } 0 = 0$
  $91 \text{ AND } 0 = 0$ (Equal, same network)
• Option B ($N_4 = 128 = 10000000_2$):
  $113 \text{ AND } 128 = 0$
  $91 \text{ AND } 128 = 0$ (Equal, same network)
• Option C ($N_4 = 192 = 11000000_2$):
  $113 \text{ AND } 192 = 64$
  $91 \text{ AND } 192 = 64$ (Equal, same network)
• Option D ($N_4 = 224 = 11100000_2$):
  $113 \text{ AND } 224 = 01100000_2 = 96$
  $91 \text{ AND } 224 = 01000000_2 = 64$ (Not equal, different networks)

Therefore, the netmask $255.255.255.224$ should not be used.