5 Aggregate the following set of 64 /24 IP network addresses to the

highest degree possible.

202.1.96.0/24

202.1.97.0/24

202.1.98.0/24

:

202.1.126.0/24

202.1.127.0/24

202.1.128.0/24

202.1.129.0/24

:

202.1.158.0/24

202.1.159.0/24 

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

fix for all of the addresses:

202.1.96.0/24 11001010.00000001.01100000.00000000

202.1.97.0/24 11001010.00000001.01100001.00000000

202.1.98.0/24 11001010.00000001.01100010.00000000

:

202.1.126.0/24 11001010.00000001.01111110.00000000

202.1.127.0/24 11001010.00000001.01111111.00000000

202.1.128.0/24 11001010.00000001.10000000.00000000

202.1.129.0/24 11001010.00000001.10000001.00000000

:

202.1.158.0/24 11001010.00000001.10011110.00000000

202.1.159.0/24 11001010.00000001.10011111.00000000 

b. Note that this set of 64 /24s cannot be summarized as a single /19. 

202.1.96.0/19 11001010.00000001.01100000.00000000

202.1.128.0/19 11001010.00000001.10000000.00000000

c. The CIDR aggregation is:

202.1.96.0/19

202.1.128.0/19

Similar to the previous example, if two /19s are to be aggregated into

a /18, the /19s must fall within a single /18 block. Since each of these

two /19s is a member of a different /18 block, they cannot be aggre 

gated into a single /18. They could be aggregated into 202.1/16, but

this aggregation would include 192 network numbers that were not

part of the original allocation. Thus, the smallest possible aggregate is

two /19s. 

6 How would you express the entire Class A address space as a single

CIDR advertisement?

Since the leading bit of all Class A addresses is  0 , the entire Class A

address space can be expressed as 0/1. 

U N D E R S TA N D I N G   I P   A D D R E S S I N G

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