How to: IPv4 Subnetting
using the desired new netmask

Let's say you are given IPv4 address 172.16.68.230, the original subnet mask is 255.255.0.0 and the new subnetmask is 255.255.240.0.
Using these parameters, you need to determine the following:

1. Number of subnet bits
2. Number of subnets that will be created
3. Number of host bits
4. Number of hosts per subnet

5. Networkaddress of current IP address
6. First host in network of current IP address
7. Last host in network of current IP address
8. Broadcast adress in network of current IP address

To solve this question, we begin by determining how many subnet bits we need to borrow.

The original subnetmask is 255.255.0.0 is written in binary as follows:

Original subnet mask:       255       . 255       . 0         . 0
Original mask in binary:    1111 1111 . 1111 1111 . 0000 0000 . 0000 0000
Mask length:                /16

The new subnetmask is 255.255.240.0 is written in binary as follows:

Original subnet mask:       255       . 255       . 240       . 0
Original mask in binary:    1111 1111 . 1111 1111 . 1111 0000 . 0000 0000
Mask length:                /20

We can now answer questions a and b by counting the number of 1's in each subnetmask.

The number of bits borrowed is 20 - 16 = 4 bits

We can determine the number of subnets with the formula 2^N where N = the number of the subnet bits.

Finding the number of host bits is fairly easy. A IPv4 address consists of 32 bits. Our new netmask is 20 bits.

So how many hosts can be contained in each subnet? Use the formula 2^B - 2 where B = the number of the host bits.
Note that we subtract 2 IP numbers, because the first IP number in the subnet is the network address and the last last IP number is the broadcast address.

Halfway there! Now we need to determine the network address of the subnet in which the IP address 172.16.68.230 resides.

To accomplish this, we need to perform the binary AND-function on the IP and its subnet mask:

IP address in binary:       1010 1100 . 0001 0000 . 0100 0100 . 1110 0110
Subnetmask in binary:       1111 1111 . 1111 1111 . 1111 0000 . 0000 0000
                        AND _____________________________________________
                            1010 1100 . 0001 0000 . 0100 0000 . 0000 0000

The resulting binary is the network address for this IP and its subnetmask. Now convert the bytes back to decimal:

To determine the first valid host in this subnet, we increment the host portion of the network address by 1

Mask:                       nnnn nnnn . nnnn nnnn . nnnn hhhh . hhhh hhhh
First host in binary:       1010 1100 . 0001 0000 . 0100 0000 . 0000 0001
converted to decimal:       172       . 16        . 64        . 1                        

The broadcast address of a IPv4 network is always the last possible number in the address range.
We find this number by filling up all the bits of the networkaddress with ones, and then converting the binary values back to decimals.

The last valid host IP is the broadcast number decremented by 1. We can use the same binary method to determine the decimal numbers:

Mask:                       nnnn nnnn . nnnn nnnn . nnnn hhhh . hhhh hhhh
Broadcast in binary:        1010 1100 . 0001 0000 . 0100 1111 . 1111 1111
converted to decimal:       172       . 16        . 79        . 255                  
Last host in binary:        1010 1100 . 0001 0000 . 0100 1111 . 1111 1110
converted to decimal:       172       . 16        . 79        . 254                  

Using the methods described above, you should be able to solve any similar subnetting problem. There's an endless supply of automatically generated subnet exercises waiting for you on the exercise page. Go ahead, give it a try!

Try exercise »