Network Design and Operations
Each of the following three sets consists of four consecutive /24 networks. For which set can the four routes be summarized into exactly one /22 route without including any addresses outside the set?
- A10.0.2.0/24 to 10.0.5.0/24
- B10.0.4.0/24 to 10.0.7.0/24
- C10.0.6.0/24 to 10.0.9.0/24
Show answer and explanation
Answer: B. 10.0.4.0/24 to 10.0.7.0/24
In a /22, the lowest two bits of the third octet fall in the host part, so one /22 consists of four consecutive /24s whose third octet starts at a multiple of 4. For 10.0.4.0/24 to 10.0.7.0/24, the third octet runs from 4 to 7 (00000100 to 00000111 in binary) and the upper 22 bits are common, so they summarize to 10.0.4.0/22. 10.0.2.0/24 to 10.0.5.0/24 straddles 10.0.0.0/22 and 10.0.4.0/22, and 10.0.6.0/24 to 10.0.9.0/24 straddles 10.0.4.0/22 and 10.0.8.0/22, so neither forms a single /22. Summarizing routes (route summarization) keeps routing tables small, but including networks that do not exist pulls in packets that have nowhere to go.
- A. The smallest route that covers all four is 10.0.0.0/21, not a /22, because 2 is not a multiple of 4.
- B. Correct. They summarize exactly to 10.0.4.0/22.
- C. The smallest route that covers all four is 10.0.0.0/20, not a /22, because 6 is not a multiple of 4.