Question

Given the IP address and subnet mask 214.20.1.15/255.255.255.192 (/26)

Whats the subnet ip range, network ID, broadcast address, and number of available IPs in the subnet

Answer 1

subnet ip range:            214.20.1.1 to 214.20.1.62
network ID:                 214.20.1.0
broadcast address:          214.20.1.63
number of available IPs:    62

Explanation:
-------------
IP Address: 214.20.1.15
----------------------------------------
Let's first convert this into binary format
214.20.1.15
Let's convert all octets to binary separately
Converting 214 to binary
Divide 214 successively by 2 until the quotient is 0
   > 214/2 = 107, remainder is 0
   > 107/2 = 53, remainder is 1
   > 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 11010110
So, 214 of decimal is 11010110 in binary
214 in binary is 11010110

Converting 20 to binary
Divide 20 successively by 2 until the quotient is 0
   > 20/2 = 10, remainder is 0
   > 10/2 = 5, remainder is 0
   > 5/2 = 2, remainder is 1
   > 2/2 = 1, remainder is 0
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 10100
So, 20 of decimal is 10100 in binary
20 in binary is 00010100

Converting 1 to binary
Divide 1 successively by 2 until the quotient is 0
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 1
So, 1 of decimal is 1 in binary
1 in binary is 00000001

Converting 15 to binary
Divide 15 successively by 2 until the quotient is 0
   > 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 1111
So, 15 of decimal is 1111 in binary
15 in binary is 00001111

==================================================================================
||    214.20.1.15 in binary notation is 11010110.00010100.00000001.00001111    ||
==================================================================================

Subnet address: 255.255.255.192
----------------------------------------
Let's convert this into binary format
255.255.255.192
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 192 to binary
Divide 192 successively by 2 until the quotient is 0
   > 192/2 = 96, remainder is 0
   > 96/2 = 48, remainder is 0
   > 48/2 = 24, remainder is 0
   > 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 11000000
So, 192 of decimal is 11000000 in binary
192 in binary is 11000000

======================================================================================
||    255.255.255.192 in binary notation is 11111111.11111111.11111111.11000000    ||
======================================================================================

Remove all dots to form 11111111111111111111111111000000
Remove all 0's from the right side to form 11111111111111111111111111
Number of 1's in this is 26
so, Subnet mask is /26

For Calculating network ID, keep first 26 bits of 11010110.00010100.00000001.00001111 and set all remaining bits to 0.
so, network ID in binary is 11010110.00010100.00000001.00000000
11010110.00010100.00000001.00000000:
----------------------------------------
11010110.00010100.00000001.00000000
Let's convert all octets to decimal separately
Converting 11010110 to decimal
Converting 11010110 to decimal
11010110
=> 1x2^7+1x2^6+0x2^5+1x2^4+0x2^3+1x2^2+1x2^1+0x2^0
=> 1x128+1x64+0x32+1x16+0x8+1x4+1x2+0x1
=> 128+64+0+16+0+4+2+0
=> 214
11010110 in decimal is 214

Converting 00010100 to decimal
Converting 00010100 to decimal
00010100
=> 0x2^7+0x2^6+0x2^5+1x2^4+0x2^3+1x2^2+0x2^1+0x2^0
=> 0x128+0x64+0x32+1x16+0x8+1x4+0x2+0x1
=> 0+0+0+16+0+4+0+0
=> 20
00010100 in decimal is 20

Converting 00000001 to decimal
Converting 00000001 to decimal
00000001
=> 0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+1x2^0
=> 0x128+0x64+0x32+0x16+0x8+0x4+0x2+1x1
=> 0+0+0+0+0+0+0+1
=> 1
00000001 in decimal is 1

Converting 00000000 to decimal
Converting 00000000 to decimal
00000000
=> 0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+0x2^0
=> 0x128+0x64+0x32+0x16+0x8+0x4+0x2+0x1
=> 0+0+0+0+0+0+0+0
=> 0
00000000 in decimal is 0

==================================================================================
||    11010110.00010100.00000001.00000000 in decimal notation is 214.20.1.0    ||
==================================================================================
=========================================
||    So, Network ID is 214.20.1.0    ||
=========================================

For Calculating broadcast ID, keep first 26 bits of 11010110.00010100.00000001.00001111 and set all remaining bits to 1.
so, broadcast ID in binary is 11010110.00010100.00000001.00111111
11010110.00010100.00000001.00111111:
----------------------------------------
11010110.00010100.00000001.00111111
Let's convert all octets to decimal separately
Converting 11010110 to decimal
Converting 11010110 to decimal
11010110
=> 1x2^7+1x2^6+0x2^5+1x2^4+0x2^3+1x2^2+1x2^1+0x2^0
=> 1x128+1x64+0x32+1x16+0x8+1x4+1x2+0x1
=> 128+64+0+16+0+4+2+0
=> 214
11010110 in decimal is 214

Converting 00010100 to decimal
Converting 00010100 to decimal
00010100
=> 0x2^7+0x2^6+0x2^5+1x2^4+0x2^3+1x2^2+0x2^1+0x2^0
=> 0x128+0x64+0x32+1x16+0x8+1x4+0x2+0x1
=> 0+0+0+16+0+4+0+0
=> 20
00010100 in decimal is 20

Converting 00000001 to decimal
Converting 00000001 to decimal
00000001
=> 0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+1x2^0
=> 0x128+0x64+0x32+0x16+0x8+0x4+0x2+1x1
=> 0+0+0+0+0+0+0+1
=> 1
00000001 in decimal is 1

Converting 00111111 to decimal
Converting 00111111 to decimal
00111111
=> 0x2^7+0x2^6+1x2^5+1x2^4+1x2^3+1x2^2+1x2^1+1x2^0
=> 0x128+0x64+1x32+1x16+1x8+1x4+1x2+1x1
=> 0+0+32+16+8+4+2+1
=> 63
00111111 in decimal is 63

===================================================================================
||    11010110.00010100.00000001.00111111 in decimal notation is 214.20.1.63    ||
===================================================================================
============================================
||    So, broadcast ID is 214.20.1.63    ||
============================================

For Calculating first valid address in subnet, add 1 to the network address.
Network Address is 214.20.1.0
============================================================
||    So, first valid address in subnet is 214.20.1.1    ||
============================================================

For Calculating last valid address in subnet, remove 1 from the broadcast address.
Broadcast Address is 214.20.1.63
============================================================
||    So, last valid address in subnet is 214.20.1.62    ||
============================================================

Number of valid hosts in the network
   > 2^(32 - number of 1's in subnet) - 2
   > 2^(32 - 26) - 2
   > 2^(6) - 2
   > 64 - 2
   > 62
==========================================================
||    So, number of valid hosts in the subnet is 62    ||
==========================================================