Suppose an ISP owns the block of addresses of the form 157.192.0.0/12 Suppose th

Question

Suppose an ISP owns the block of addresses of the form 157.192.0.0/12 Suppose that seven organizations, A,B,C,D E,F and G, request 4096, 1024, 4096, 8192, 2048, 256 and 8192 addresses, respectively, and in that order. For organizations A,B,C,D,E,F,G, what is the first IP address assigned, the last IP address assigned, and the subnet prefix in the CIDR w.x.y.z/s notation.? Write the addresses in the table Justify your answer. Important: Each organization must get just one block of addresses. At any time, assign the smallest IP addresses that are available and possible! OrganizFirst address ation How Subnet CIDR man 4096 Last address 57.192.0.0 1024 4096 8192 2048 256 8192

Explanation / Answer

Solution:

For the first organization A, total number of addresses required is 4096 which is power of 2 (212)

Therefore, the subnet mask for A = 32 – log2 4096 = 32 – 12 = 20.

The first address of A can be achieved by doing AND operation between 157.192.0.0 and 11111111 11111111 11110000 00000000 (total number of 1’s is 20 as subnet mask is 20) which results 10011101.11000000.00000000.00000000 or 157.192.0.0/20.

The last address of A can be obtained by doing AND operation between 157.192.0.0 and 00000000 00000000 00001111 11111111 (32 – 20 = 12 rightmost bits set to 1) which results 157.192.15.255/20.

The subnet CIDR (11111111.11111111.11110000.00000000)2 or (255.255.240.0)10.

For the first organization B, total number of addresses required is 1024 which is power of 2 (210)

Therefore, the subnet mask for B = 32 – log2 1024 = 32 – 10 = 22.

The first address of B is 157.192.16.0/22

The last address of B is 157.192.19.255/22

The subnet CIDR (11111111.11111111.11111100.00000000)2 or (255.255.252.0)10.

For the first organization C, total number of addresses required is 4096 which is power of 2 (212)

Therefore, the subnet mask for C = 32 – log2 4096 = 32 – 12 = 20.

The first address of C is 157.192.20.0/20

The last address of C is 157.192.25.255/20

The subnet CIDR (11111111.11111111.11110000.00000000)2 or (255.255.240.0)10.

For the first organization D, total number of addresses required is 8192 which is power of 2 (213)

Therefore, the subnet mask for D = 32 – log2 8192 = 32 – 13 = 19.

The first address of D is 157.192.26.0/19

The last address of D is 157.192.57.255/19

The subnet CIDR (11111111.11111111.11100000.00000000)2 or (255.255.224.0)10.

For the first organization E, total number of addresses required is 2048 which is power of 2 (211)

Therefore, the subnet mask for E = 32 – log2 2048 = 32 – 11 = 21.

The first address of E is 157.192.58.0/21

The last address of E is 157.192.65.255/21

The subnet CIDR (11111111.11111111.11111000.00000000)2 or (255.255.248.0)10.

For the first organization F, total number of addresses required is 256 which is power of 2 (28)

Therefore, the subnet mask for F = 32 – log2 256 = 32 – 8 = 24.

The first address of F is 157.192.66.0/24

The last address of F is 157.192.66.255/24

The subnet CIDR (11111111.11111111.11111111.00000000)2 or (255.255.255.0)10.

For the first organization G, total number of addresses required is 8192 which is power of 2 (213)

Therefore, the subnet mask for G = 32 – log2 8192 = 32 – 13 = 19.

The first address of G is 157.192.67.0/19

The last address of G is 157.192.98.255/19

The subnet CIDR (11111111.11111111.11100000.00000000)2 or (255.255.224.0)10.

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