Sub Netting Made Easy

By Vishal Pawar

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1 Subnet Prefix mask

1 0 0 0 0 0 0 0 128 /25 1 1 0 0 0 0 0 0 192 /26 1 1 1 0 0 0 0 0 224 /27 1 1 1 1 0 0 0 0 240 /28 1 1 1 1 1 0 0 0 248 /29 1 1 1 1 1 1 0 0 252 /30 1 1 1 1 1 1 1 0 254 /31 1 1 1 1 1 1 1 1 255 /32

Ex.- Class C Address N . N . N . H 11111111.11111111.11111111.00000000 No of Hosts = 2 ^ n (no of host bits) 192.168.1.0 to 192.168.1.255 Network Address - Broadcast Address Actual Host address = 192.168.1.1 to 192.168.1.254

Subnetting Subnetting is dividing one network

into many networks.

Formula for subnetting - 2 ^ n (no of bits borrowed from

host portion) 2 ^ 1 = 2 subnets

Calculate ? 1. Class2. Default Subnet3. Classless Subnet4. Number of Networks 5. Block Size6. Number of Usable Hosts per Network7. Network Address8. Broadcast Address9. Next Network Address10. First Host IP Address11. Last Host IP Address

Example - 1 Given IP Address -

10.10.10.40

Given Prefix -

/27

Step 1- Class 1-126 it is a Class A address 128-191 it is a Class B address 192-223 it is a Class C address 224-239 it is a Class D address 240-255 it is a Class E address________________________________ Answer -

10.10.10.40 - Class A Address

Step 2 – Default Subnet Class A - 255.0.0.0 Class B - 255.255.0.0 Class C - 255.255.255.0_______________________________ Answer -

For - 10.10.10.40 Default Subnet is - 255.0.0.0

Step 3 – Classless Subnet

Calculate Binary of given prefix – /2711111111.11111111.11111111.11100000 8 + 8 + 8 + 3 = 27 bits ON Convert it into Decimal- 255 . 255 . 255 . 224_______________________________________ Answer –

Classless Subnet mask - 255.255.255.224

Step 4 - Number of Networks No. of Networks = 2^n

Where n is No. of bits borrowed

No. of bits borrowed = 11100000 = 3

2^3 = 8 _______________________________ Answer –

Number of Networks = 8

Step 5 – Block SizeDefault subnet mask – Classless subnet

mask 255.255.255.256 - 255.255.255.224 (Subnet mask of /27)------------------------ 0.0.0.32_______________________________ Answer –

Block Size = 0.0.0.32

Step 6 – Total Number of Usable Hosts 2 ^ (Total Host bits) - 2

= 2 ^ 5 - 2 = 32 – 2 = 30

_______________________________ Answer –

Total Number of Usable Hosts = 30

Step 7 – Network Address Use ANDing method

1. Convert IP Add to Binary 2. Convert Subnet mask to Binary3. Use AND operation 4. Convert answer into decimal

IP address : 00001010.00001010.00001010.00101000Subnet mask : 11111111.11111111.11111111.1110000AND -----------------------------------------------------------

Network address : 00001010.00001010.00001010.00100000 Decimal Format: 10.10.10.32

_______________________________ Answer –

Network Address = 10.10.10.32

Step 8 - Broadcast Address Network Add + Block size - 1

10.10.10.32 + 0.0.0.32 - 1 = 10.10.10.63

Answer – Broadcast Address = 10.10.10.63

Step 8 - Broadcast Address Use XOR method

1. Convert IP Add to Binary 2. Convert Subnet mask to Binary3. Invert Binary Subnet mask4. Use XOR operation 5. Convert answer into decimal

IP address : 00001010.00001010.00001010.00101000Inverted Subnet mask : 00000000.00000000.00000000.00011111XOR -------------------------------------------------------------------------------- Network address : 00001010.00001010.00001010.00111111 Decimal Format : 10.10.10.63

______________________________________ Answer –

Broadcast Address = 10.10.10.63

Step 10 – First Host IP Address First Host IP Add = Network Address + 1 = 10.10.10.32 + 1 = 10.10.10.33

____________________________________ Answer –

First Host IP Address = 10.10.10.33

Step 10 – Last Host IP Address Last Host IP Add = Broadcast Address -

1 = 10.10.10.63 - 1 = 10.10.10.62

____________________________________ Answer –

Last Host IP Address = 10.10.10.62

Step 9 – Next Network AddressNext Network Add = Network add + Block

Size = 10.10.10.32 + 0.0.0.32 = 10.10.10.64

____________________________________ Answer –

Next Network Address = 10.10.10.64

Thank You