Subnetting is a process of dividing large network into the smaller networks based on layer 3 IP address. Every computer on network has an IP address that represent its location on network. This Presentation explains subnetting process. The Image used in Presentations are CISCO Property.

1. Sub Netting Made Easy
By Vishal Pawar

2. 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

4.  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

6. 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

7. Calculate ?
1. Class
2. Default Subnet
3. Classless Subnet
4. Number of Networks
5. Block Size
6. Number of Usable Hosts per Network
7. Network Address
8. Broadcast Address
9. Next Network Address
10. First Host IP Address
11. Last Host IP Address

8. Example - 1
 Given IP Address -
 10.10.10.40
 Given Prefix -
 /27

9. 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

10. 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

11. Step 3 – Classless Subnet
 Calculate Binary of given prefix – /27
11111111.11111111.11111111.11100000
8 + 8 + 8 + 3 = 27 bits ON
 Convert it into Decimal-
255 . 255 . 255 . 224
_______________________________________
 Answer –
Classless Subnet mask - 255.255.255.224

12. 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

13. Step 5 – Block Size
Default 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

14. 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

15. Step 7 – Network Address
 Use ANDing method
1. Convert IP Add to Binary
2. Convert Subnet mask to Binary
3. Use AND operation
4. Convert answer into decimal
IP address : 00001010.00001010.00001010.00101000
Subnet mask : 11111111.11111111.11111111.1110000
AND -----------------------------------------------------------
Network address : 00001010.00001010.00001010.00100000
Decimal Format: 10.10.10.32
_______________________________
 Answer –
 Network Address = 10.10.10.32

16. 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

17. Step 8 - Broadcast Address
 Use XOR method
1. Convert IP Add to Binary
2. Convert Subnet mask to Binary
3. Invert Binary Subnet mask
4. Use XOR operation
5. Convert answer into decimal
IP address : 00001010.00001010.00001010.00101000
Inverted Subnet mask : 00000000.00000000.00000000.00011111
XOR --------------------------------------------------------------------------------
Network address : 00001010.00001010.00001010.00111111
Decimal Format : 10.10.10.63
______________________________________
 Answer –
 Broadcast Address = 10.10.10.63

18. 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

19. 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

20. Step 9 – Next Network Address
Next 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

21. Thank You