3 Aggregate the following set of four /24 IP network addresses to the highest

degree possible. 

212.56.132.0/24

212.56.133.0/24

212.56.134.0/24

212.56.135.0/24

a. List each address in binary format and determine the common pre 

fix for all of the addresses: 

212.56.132.0/24 11010100.00111000.10000100.00000000

212.56.133.0/24 11010100.00111000.10000101.00000000

212.56.134.0/24 11010100.00111000.10000110.00000000

212.56.135.0/24 11010100.00111000.10000111.00000000

Common Prefix: 

11010100.00111000.10000100.00000000 

b. The CIDR aggregation is:

212.56.132.0/22 

4 Aggregate the following set of four /24 IP network addresses to the

highest degree possible.

212.56.146.0/24

212.56.147.0/24

212.56.148.0/24

212.56.149.0/24

a. List each address in binary format and determine the common pre 

fix for all of the addresses:

212.56.146.0/24 11010100.00111000.10010010.00000000

212.56.147.0/24 11010100.00111000.10010011.00000000

212.56.148.0/24 11010100.00111000.10010100.00000000

212.56.148.0/24 11010100.00111000.10010101.00000000

b. Note that this set of four /24s cannot be summarized as a single /23.

212.56.146.0/23 11010100.00111000.10010010.00000000

212.56.148.0/23 11010100.00111000.10010100.00000000

c. The CIDR aggregation is:

212.56.146.0/23

212.56.148.0/23

Note that if two /23s are to be aggregated into a /22, then both /23s

must fall within a single /22 block. Since each of the two /23s is a

member of a different /22 block, they cannot be aggregated into a sin 

gle /22 (even though they are consecutive). They could be aggregated

into 222.56.144/21, but this aggregation would include four network

numbers that were not part of the original allocation. Hence, the

smallest possible aggregate is two /23s. 

7 1