Question

Telecom networks and security

Use sum of powers of 2, AND methods to find the first network address in a block, find the number of addreses in a block and find the last address in the block using calculation in base 256, mask remainder methods. Show calculations.

1) 4.98.64.0/20

2) 205.16.37.32/28

3) 190.16.42.44/20

4)123.45.24.52/3

Answer 1

To answer all these 4 questions we follow below steps :-

1. Represent IP address as 32 bit binary number.

2. Number of unmasked bit will be 32 minus number of masked bit.

3. The first address of Ip Address can be obtained by performing AND operation with masked bits will be 1 and unmasked bit will be zero. The result will be first IP address which is also called as network address.

4. Last IP address can be obtained by performing OR operation with all masked bit as 0 and unmasked bit as 1.

5. Number of IP addresses = 2ⁿ where n is number of unmasked bits. Out of 2ⁿ address the first address is network address and last address is broadcast address and hence these two addresses are not assigned to any host.

1) 4.98.64.0/20

Here number of unmasked bit is 32-20= 12

Number of addresses = 2¹² = 4096

Representing 4.98.64.0 in 32 bit binary is

0000 0100. 0110 0010. 0100 0000. 0000 0000

To get first address, we perform AND with

1111 1111. 1111 1111. 1111 0000. 0000 0000

Which gives 0000 0100. 0110 0010. 0100 0000. 0000 0000

Which in decimal is 4.98.64.0

To get last address we perform OR with

0000 0000. 0000 0000. 0000 1111. 1111 1111

Which gives

0000 0100. 0110 0010. 0100 1111. 1111 1111

Which in decimal representation is

4.68.79.255

2) 205.16.37.32/28

Here number of unmasked bit is 32-28= 4

Number of addresses = 2⁴ = 16

Representing 205.16.37.32 in 32 bit binary is

1100 1101. 0001 0000. 0010 0101. 0010 0000

To get first address, we perform AND with

1111 1111. 1111 1111. 1111 1111. 1111 0000

Which gives 1100 1101. 0001 0000. 0010 0101. 0010 0000

Which in decimal is 205.16.37.32

To get last address we perform OR with

0000 0000. 0000 0000. 0000 0000. 0000 1111

Which gives

1100 1101. 0001 0000. 0010 0101. 0010 1111

Which in decimal representation is

205.16.37.47

3)  190.16.42.44/20

Here number of unmasked bit is 32-20= 12

Number of addresses = 2¹² = 4096

Representing 190.16.42.44 in 32 bit binary is

1011 1110. 0001 0000. 0010 1010 . 0010 1100

To get first address, we perform AND with

1111 1111. 1111 1111. 1111 0000. 0000 0000

Which gives 1011 1110. 0001 0000. 0010 0000. 0000 0000

Which in decimal is 190.16.32.0

To get last address we perform OR with

0000 0000. 0000 0000. 0000 1111. 1111 1111

Which gives

1011 1110. 0001 0000. 0010 1111. 1111 1111

Which in decimal representation is

190.16.47.255

4)123.45.24.52/3

Here number of unmasked bit is 32-3= 29

Number of addresses = 2²⁹  

Representing 123.45.24.52 in 32 bit binary is

0111 1011. 0010 1101. 0001 1000. 0011 0100

To get first address, we perform AND with

1110 0000. 0000 0000. 0000 0000. 0000 0000

Which gives 0110 0000. 0000 0000. 0000 0000. 0000 0000.

Which in decimal is 96.0.0.0

To get last address we perform OR with

0001 1111. 1111 1111. 1111 1111. 1111 1111

Which gives

0111 1111. 1111 1111. 1111 1111. 1111 1111

Which in decimal representation is

127.255.255.255

Please comment for any clarification.