The forwarding table of a router is shown below. A packet addressed to a destination address $200.150.68.118$ arrives at the router. It will be forwarded to the interface with ID _________.

To determine the forwarding interface, we apply the Longest Prefix Match rule by performing a bitwise AND between the destination IP address 200.150.68.118 and each subnet mask in the routing table:

1. Interface 1:

   • Subnet Mask: $255.255.0.0$ (/16)

   • Bitwise AND: 200.150.68.118$\text{AND}$255.255.0.0 = 200.150.0.0

   • Matches the Subnet Number 200.150.0.0 (Prefix length: $16$).

2. Interface 2:

   • Subnet Mask: $255.255.224.0$ (/19)

   • 3rd octet: $68 = 01000100_2$, $224 = 11100000_2$

   • $68 \text{ AND } 224 = 01000000_2 = 64$

   • Bitwise AND: 200.150.68.118$\text{AND}$255.255.224.0 = 200.150.64.0

   • Matches the Subnet Number 200.150.64.0 (Prefix length: $19$).

3. Interface 3:

   • Subnet Mask: $255.255.255.0$ (/24)

   • Bitwise AND: 200.150.68.118$\text{AND}$255.255.255.0 = 200.150.68.0

   • Matches the Subnet Number 200.150.68.0 (Prefix length: $24$).

4. Interface 4:

   • Subnet Mask: $255.255.255.224$ (/27)

   • 4th octet: $118 = 01110110_2$, $224 = 11100000_2$

   • $118 \text{ AND } 224 = 01100000_2 = 96 \neq 64$

   • Bitwise AND: 200.150.68.96$\neq$200.150.68.64 (Does not match).

Among the matching interfaces (1, 2, and 3), Interface 3 has the longest prefix match (/24).

Therefore, the packet is forwarded to interface ID 3.