Consider routing table of an organization’s router shown below: Which of the…

Answer: 12.20.164.0/22 and 12.20.168.0/22

Given routes (from the table):

• 12.20.164.0, mask 255.255.252.0 (/22) — covers 12.20.164.0 through 12.20.167.255.

• 12.20.170.0, mask 255.255.254.0 (/23) — covers 12.20.170.0 through 12.20.171.255.

• 12.20.168.0, mask 255.255.254.0 (/23) — covers 12.20.168.0 through 12.20.169.255.

• 12.20.166.0, mask 255.255.254.0 (/23) — covers 12.20.166.0 through 12.20.167.255.

Aggregation steps:

1. Combine 12.20.164.0 (/22) and 12.20.166.0 (/23):

   Third octet binaries: 164 = 10100100, 166 = 10100110. They share the first 6 bits (101001), so the common prefix length is 16 (first two octets) + 6 = 22 bits → 12.20.164.0/22.

2. Combine 12.20.168.0 (/23) and 12.20.170.0 (/23):

   Third octet binaries: 168 = 10101000, 170 = 10101010. They share the first 6 bits (101010), so the common prefix length is 16 + 6 = 22 bits → 12.20.168.0/22.

Why the other prefixes are incorrect:

• 12.20.164.0/20 is invalid as written because /20 networks start on third-octet multiples of 16 (for example 12.20.160.0/20). A /20 that covered all four subnets would be 12.20.160.0/20, but that is much larger than necessary and includes addresses not in the routing table.

• 12.20.164.0/21 is also incorrect: a /21 groups third-octet blocks of 8 (e.g., 12.20.160.0/21 covers 160–167) and would not include 12.20.168.0 and 12.20.170.0, so it fails to cover all listed subnets.

Conclusion: The minimal correct aggregation is two prefixes: 12.20.164.0/22 (covering the 164.0 and 166.0 blocks) and 12.20.168.0/22 (covering the 168.0 and 170.0 blocks).