Question

1. Given the following IP address of 212.134.99.112 with a subnet mask of 255.255.255.224

1. Class of the IP address:
2. Net ID for this IP address:
3. Total number of subnets:
4. First address (subnet ID) and the Last address (broadcast address) that 21.134.99.112 belongs to:

Answer 1

a)  Class D
b)  212.134.99.96
c)  number of subnets = 8
d)  
first address is 212.134.99.97
last address is 212.134.99.126

IP Address: 212.134.99.112
----------------------------------------
Let's first convert this into binary format
212.134.99.112
Let's convert all octets to binary separately
Converting 212 to binary
Divide 212 successively by 2 until the quotient is 0
   > 212/2 = 106, remainder is 0
   > 106/2 = 53, remainder is 0
   > 53/2 = 26, remainder is 1
   > 26/2 = 13, remainder is 0
   > 13/2 = 6, remainder is 1
   > 6/2 = 3, remainder is 0
   > 3/2 = 1, remainder is 1
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 11010100
So, 212 of decimal is 11010100 in binary
212 in binary is 11010100

Converting 134 to binary
Divide 134 successively by 2 until the quotient is 0
   > 134/2 = 67, remainder is 0
   > 67/2 = 33, remainder is 1
   > 33/2 = 16, remainder is 1
   > 16/2 = 8, remainder is 0
   > 8/2 = 4, remainder is 0
   > 4/2 = 2, remainder is 0
   > 2/2 = 1, remainder is 0
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 10000110
So, 134 of decimal is 10000110 in binary
134 in binary is 10000110

Converting 99 to binary
Divide 99 successively by 2 until the quotient is 0
   > 99/2 = 49, remainder is 1
   > 49/2 = 24, remainder is 1
   > 24/2 = 12, remainder is 0
   > 12/2 = 6, remainder is 0
   > 6/2 = 3, remainder is 0
   > 3/2 = 1, remainder is 1
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 1100011
So, 99 of decimal is 1100011 in binary
99 in binary is 01100011

Converting 112 to binary
Divide 112 successively by 2 until the quotient is 0
   > 112/2 = 56, remainder is 0
   > 56/2 = 28, remainder is 0
   > 28/2 = 14, remainder is 0
   > 14/2 = 7, remainder is 0
   > 7/2 = 3, remainder is 1
   > 3/2 = 1, remainder is 1
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 1110000
So, 112 of decimal is 1110000 in binary
112 in binary is 01110000

=====================================================================================
||    212.134.99.112 in binary notation is 11010100.10000110.01100011.01110000    ||
=====================================================================================

Subnet address: 255.255.255.224
----------------------------------------
Let's convert this into binary format
255.255.255.224
Let's convert all octets to binary separately
Converting 255 to binary
Divide 255 successively by 2 until the quotient is 0
   > 255/2 = 127, remainder is 1
   > 127/2 = 63, remainder is 1
   > 63/2 = 31, remainder is 1
   > 31/2 = 15, remainder is 1
   > 15/2 = 7, remainder is 1
   > 7/2 = 3, remainder is 1
   > 3/2 = 1, remainder is 1
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 11111111
So, 255 of decimal is 11111111 in binary
255 in binary is 11111111

Converting 255 to binary
Divide 255 successively by 2 until the quotient is 0
   > 255/2 = 127, remainder is 1
   > 127/2 = 63, remainder is 1
   > 63/2 = 31, remainder is 1
   > 31/2 = 15, remainder is 1
   > 15/2 = 7, remainder is 1
   > 7/2 = 3, remainder is 1
   > 3/2 = 1, remainder is 1
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 11111111
So, 255 of decimal is 11111111 in binary
255 in binary is 11111111

Converting 255 to binary
Divide 255 successively by 2 until the quotient is 0
   > 255/2 = 127, remainder is 1
   > 127/2 = 63, remainder is 1
   > 63/2 = 31, remainder is 1
   > 31/2 = 15, remainder is 1
   > 15/2 = 7, remainder is 1
   > 7/2 = 3, remainder is 1
   > 3/2 = 1, remainder is 1
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 11111111
So, 255 of decimal is 11111111 in binary
255 in binary is 11111111

Converting 224 to binary
Divide 224 successively by 2 until the quotient is 0
   > 224/2 = 112, remainder is 0
   > 112/2 = 56, remainder is 0
   > 56/2 = 28, remainder is 0
   > 28/2 = 14, remainder is 0
   > 14/2 = 7, remainder is 0
   > 7/2 = 3, remainder is 1
   > 3/2 = 1, remainder is 1
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 11100000
So, 224 of decimal is 11100000 in binary
224 in binary is 11100000

======================================================================================
||    255.255.255.224 in binary notation is 11111111.11111111.11111111.11100000    ||
======================================================================================

Remove all dots to form 11111111111111111111111111100000
Remove all 0's from the right side to form 111111111111111111111111111
Number of 1's in this is 27
so, Subnet mask is /27

For Calculating network ID, keep first 27 bits of 11010100.10000110.01100011.01110000 and set all remaining bits to 0.
so, network ID in binary is 11010100.10000110.01100011.01100000
11010100.10000110.01100011.01100000:
----------------------------------------
11010100.10000110.01100011.01100000
Let's convert all octets to decimal separately
Converting 11010100 to decimal
=> 11010100
=> 1x2^7+1x2^6+0x2^5+1x2^4+0x2^3+1x2^2+0x2^1+0x2^0
=> 1x128+1x64+0x32+1x16+0x8+1x4+0x2+0x1
=> 128+64+0+16+0+4+0+0
=> 212
11010100 in decimal is 212

Converting 10000110 to decimal
=> 10000110
=> 1x2^7+0x2^6+0x2^5+0x2^4+0x2^3+1x2^2+1x2^1+0x2^0
=> 1x128+0x64+0x32+0x16+0x8+1x4+1x2+0x1
=> 128+0+0+0+0+4+2+0
=> 134
10000110 in decimal is 134

Converting 01100011 to decimal
=> 01100011
=> 0x2^7+1x2^6+1x2^5+0x2^4+0x2^3+0x2^2+1x2^1+1x2^0
=> 0x128+1x64+1x32+0x16+0x8+0x4+1x2+1x1
=> 0+64+32+0+0+0+2+1
=> 99
01100011 in decimal is 99

Converting 01100000 to decimal
=> 01100000
=> 0x2^7+1x2^6+1x2^5+0x2^4+0x2^3+0x2^2+0x2^1+0x2^0
=> 0x128+1x64+1x32+0x16+0x8+0x4+0x2+0x1
=> 0+64+32+0+0+0+0+0
=> 96
01100000 in decimal is 96

=====================================================================================
||    11010100.10000110.01100011.01100000 in decimal notation is 212.134.99.96    ||
=====================================================================================
============================================
||    So, Network ID is 212.134.99.96    ||
============================================

For Calculating first valid address in subnet, add 1 to the network address.
Network Address is 212.134.99.96
===============================================================
||    So, first valid address in subnet is 212.134.99.97    ||
===============================================================

For Calculating last valid address in subnet, remove 1 from the broadcast address.
Broadcast Address is 212.134.99.127
===============================================================
||    So, last valid address in subnet is 212.134.99.126    ||
===============================================================