Understanding IP Addressing: Everything You Ever Wanted To Know

Appendix E – CIDR Examples

CIDR Practice Exercises
1. List the individual networks numbers defined by the CIDR block 200.56.168.0/21.
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2. List the individual networks numbers defined by the CIDR block 195.24/13.
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3. Aggregate the following set of (4) IP /24 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
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4. Aggregate the following set of (4) IP /24 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
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5. Aggregate the following set of (64) IP /24 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
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6. How would you express the entire Class A address space as a single CIDR advertisement?
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7. How would you express the entire Class B address space as a single CIDR advertisement?
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8. How would you express the entire Class C address space as a single CIDR advertisement?
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Solutions for CIDR Pracitice Exercises
1. List the individual networks numbers defined by the CIDR block 200.56.168.0/21.
a. Express the CIDR block in binary format:

200.56.168.0/21 11001000.00111000.10101 000.00000000
b. The /21 mask is 3 bits shorter than the natural mask for a traditional /24. This means that the CIDR block identifies a block of 8 (or 2 3 ) consecutive /24 network numbers.

c. The range of /24 network numbers defined by the CIDR block 200.56.168.0/21 includes:

Net #0: 11001000.00111000.10101000 .xxxxxxxx 200.56.168.0
Net #1: 11001000.00111000.10101001 .xxxxxxxx 200.56.169.0
Net #2: 11001000.00111000.10101010 .xxxxxxxx 200.56.170.0
Net #3: 11001000.00111000.10101011 .xxxxxxxx 200.56.171.0
Net #4: 11001000.00111000.10101100 .xxxxxxxx 200.56.172.0
Net #5: 11001000.00111000.10101101 .xxxxxxxx 200.56.173.0
Net #6: 11001000.00111000.10101110 .xxxxxxxx 200.56.174.0
Net #7: 11001000.00111000.10101111 .xxxxxxxx 200.56.175.0

2. List the individual networks numbers defined by the CIDR block 195.24/13.
a. Express the CIDR block in binary format:

195.24.0.0/13 11000011.00011 000.00000000.00000000

b. The /13 mask is 11 bits shorter than the natural mask for a traditional /24. This means that the CIDR block identifies a block of 2,048 (or 2 11 ) consecutive /24 network numbers.

c. The range of /24 network numbers defined by the CIDR block 195.24/13 include:
Net #0: 11000011.00011000.00000000 .xxxxxxxx 195.24.0.0
Net #1: 11000011.00011000.00000001 .xxxxxxxx 195.24.1.0
Net #2: 11000011.00011000.00000010 .xxxxxxxx 195.24.2.0
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Net #2045: 11000011.00011111.11111101 .xxxxxxxx 195.31.253.0
Net #2046: 11000011.00011111.11111110 .xxxxxxxx 195.31.254.0
Net #2047: 11000011.00011111.11111111 .xxxxxxxx 195.31.255.0

3. Aggregate the following set of (4) IP /24 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 prefix 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: 11010100.00111000.100001 00.00000000

b. The CIDR aggregation is:
212.56.132.0/22

4. Aggregate the following set of (4) IP /24 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 prefix 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 11010100.00111000.1001001 0.00000000
212.56.148.0/23 11010100.00111000.1001010 0.00000000

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 single /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.

5. Aggregate the following set of (64) IP /24 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 prefix 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 11001010.00000001.011 00000.00000000
202.1.128.0/19 11001010.00000001.100 00000.00000000

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 aggregated 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 a “0″, the entire Class A address space can be expressed as 0/1.

7. How would you express the entire Class B address space as a single CIDR advertisement?
Since the leading two bits of all Class B addresses are “10″, the entire Class B address space can be expressed as 128/2.

8. How would you express the entire Class C address space as a single CIDR advertisement?
Since the leading three bits of all Class C addresses are “110″, the entire Class C address space can be expressed as 192/3.

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