Question

Use sum of power of 2, AND method to find the first address in a block, find the number of address in a block and find the final address in the block using calculation in base 256, Mark remaider methods. Show calculation.

13. 100.8.16.0/19 14. 168.84.64.0/18 15. 4.98.64.0/20

Answer 1

To answer all these 3 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.

13. The address 100.8.16.0 in binary is

0110 0100. 0000 1000. 0001 0000. 0000 0000

Number of unmasked bits = 32-19 = 13

Number of address in this block = 2¹³ = 8192 out of which only 8190 will be available for use.

First address I.e. network address can be obtained by performing AND operation with masked bit as 1 and unmasked bit as 0. So applying AND operation of above address with

1111 1111. 1111 1111. 1110 0000. 0000 0000

The result of AND operation is

0110 0100. 0000 1000. 0000 0000. 0000 0000

And hence converting to decimal the first address is

100.8.0.0

Last address can be obtained by performing OR operation with masked bit as 0 and unmasked bit as 1. So performing OR operation of above address with

0000 0000. 0000 0000. 0001 1111. 1111 1111

We get

0110 0100. 0000 0100. 1111 1111. 1111 1111

Which in decimal is 100.8. 255.255.

14. Doing the same thing here, the address 168.84.64.0 in binary is

1010 1000. 0101 0100. 0100 0000. 0000 0000

To get first address with have to perform AND operation with

1111 1111. 1111 1111. 1100 0000. 0000 0000

We get

1010 1000. 0101 0100. 0100 0000. 0000 0000

So the first address in decimal representation is 168.84.64.0

And to get last address we have to perform OR with

0000 0000. 0000 0000. 0011 1111. 1111 1111

This gives

1010 1000. 0101 0100. 0111 1111. 1111 1111

Which in decimal is

168.84. 127. 255

And number of addresses is 2^32-18 = 2¹⁴ = 16384

15. 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

Please comment for anu clarifcation.