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.Aggregate the following set of (64) IP /24 network addresses to the highest degreepossible.202.1.96.0/24202.1.97.0/24202.1.98.0/24:202.1.126.0/24202.1.127.0/24202.1.128.0/24202.1.129.0/24:202.1.158.0/24202.1.159.0/24__________________________________________________________________6.How would you express the entire Class A address space as a single CIDRadvertisement?__________________________________________________________________ 7.How would you express the entire Class B address space as a single CIDRadvertisement?__________________________________________________________________8.How would you express the entire Class C address space as a single CIDRadvertisement?__________________________________________________________________Solutions for CIDR Pracitice Exercises1.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 11001000.00111000.10101000.00000000b.The /21 mask is 3 bits shorter than the natural mask for a traditional /24.Thismeans that the CIDR block identifies a block of 8 (or 23) consecutive /24network numbers.c.The range of /24 network numbers defined by the CIDR block 200.56.168.0/21includes:Net #0: 11001000.00111000.10101000.xxxxxxxx 200.56.168.0Net #1: 11001000.00111000.10101001.xxxxxxxx 200.56.169.0Net #2: 11001000.00111000.10101010.xxxxxxxx 200.56.170.0Net #3: 11001000.00111000.10101011.xxxxxxxx 200.56.171.0Net #4: 11001000.00111000.10101100.xxxxxxxx 200.56.172.0Net #5: 11001000.00111000.10101101.xxxxxxxx 200.56.173.0Net #6: 11001000.00111000.10101110.xxxxxxxx 200.56.174.0Net #7: 11001000.00111000.10101111.xxxxxxxx 200.56.175.02.List the individual networks numbers defined by the CIDR block 195.24/13.a.Express the CIDR block in binary format:195.24.0/13 11000011.00011000.00000000.00000000b.The /13 mask is 11 bits shorter than the natural mask for a traditional /24.Thismeans that the CIDR block identifies a block of 2,048 (or 211) consecutive /24network numbers. c.The range of /24 network numbers defined by the CIDR block 195.24/13include:Net #0: 11000011.00011000.00000000.xxxxxxxx 195.24.0Net #1: 11000011.00011000.00000001.xxxxxxxx 195.24.1.0Net #2: 11000011.00011000.00000010.xxxxxxxx 195.24.2.Net #2045: 11000011.00011111.11111101.xxxxxxxx 195.31.253.0Net #2046: 11000011.00011111.11111110.xxxxxxxx 195.31.254.0Net #2047: 11000011.00011111.11111111.xxxxxxxx 195.31.255.03.Aggregate the following set of (4) IP /24 network addresses to the highest degreepossible.212.56.132.0/24212.56.133.0/24212.56.134.0/24212.56.135.0/24a.List each address in binary format and determine the common prefix for all ofthe addresses:212.56.132.0/24 11010100.00111000.10000100.00000000212.56.133.0/24 11010100.00111000.10000101.00000000212.56.134.0/24 11010100.00111000.10000110.00000000212.56.135.0/24 11010100.00111000.10000111.00000000Common Prefix: 11010100.00111000.10000100.00000000b.The CIDR aggregation is:212.56.132.0/224.Aggregate the following set of (4) IP /24 network addresses to the highest degreepossible.212.56.146.0/24212.56.147.0/24212.56.148.0/24212.56.149.0/24a.List each address in binary format and determine the common prefix for all ofthe addresses:212.56.146.0/24 11010100.00111000.10010010.00000000212.56.147.0/24 11010100.00111000.10010011.00000000212.56.148.0/24 11010100.00111000.10010100.00000000212.56.148.0/24 11010100.00111000.10010101.00000000 b.Note that this set of four /24s cannot be summarized as a single /23!212.56.146.0/23 11010100.00111000.10010010.00000000212.56.148.0/23 11010100.00111000.10010100.00000000c.The CIDR aggregation is:212.56.146.0/23212.56.148.0/23Note that if two /23s are to be aggregated into a /22, then both /23s must fall within asingle /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 includefour network numbers that were not part of the original allocation.Hence, thesmallest possible aggregate is two /23s.5.Aggregate the following set of (64) IP /24 network addresses to the highest degreepossible.202.1.96.0/24202.1.97.0/24202.1.98.0/24:202.1.126.0/24202.1.127.0/24202.1.128.0/24202.1.129.0/24:202.1.158.0/24202.1.159.0/24a.List each address in binary format and determine the common prefix for all ofthe addresses:202.1.96.0/24 11001010.00000001.01100000.00000000202.1.97.0/24 11001010.00000001.01100001.00000000202.1.98.0/24 11001010.00000001.01100010.00000000:202.1.126.0/24 11001010.00000001.01111110.00000000202.1.127.0/24 11001010.00000001.01111111.00000000202.1.128.0/24 11001010.00000001.10000000.00000000202.1.129.0/24 11001010.00000001.10000001.00000000:202.1.158.0/24 11001010.00000001.10011110.00000000202.1.159.0/24 11001010.00000001.10011111.00000000b.Note that this set of 64 /24s cannot be summarized as a single /19!202.1.96.0/19 11001010.00000001.01100000.00000000202.1.128.0/19 11001010.00000001.10000000.00000000 c.The CIDR aggregation is:202.1.96.0/19202.1.128.0/19Similar to the previous example, if two /19s are to be aggregated into a /18, the /19smust fall within a single /18 block! Since each of these two /19s is a member of adifferent /18 block, they cannot be aggregated into a single /18.They could beaggregated into 202.1/16, but this aggregation would include 192 network numbersthat were not part of the original allocation.Thus, the smallest possible aggregate istwo /19s.6.How would you express the entire Class A address space as a single CIDRadvertisement?Since the leading bit of all Class A addresses is a "0", the entire Class A addressspace can be expressed as 0/1.7.How would you express the entire Class B address space as a single CIDRadvertisement?Since the leading two bits of all Class B addresses are "10", the entire Class Baddress space can be expressed as 128/2.8.How would you express the entire Class C address space as a single CIDRadvertisement?Since the leading three bits of all Class C addresses are "110", the entire Class Caddress space can be expressed as 192/3
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