How to: IPv4 Subnetting
using the desired number of networks
Task
Let's say you are given the network range 172.16.68.0/24. Now you're asked to create 7 subnets in this range by subnetting.
Using these parameters, you need to determine the following:
 The actual number of subnets to create
 Number of subnet bits to borrow from the hostportion
 The new subnetmask
 Number of hosts per subnet
 Networkaddress of the second subnet
 First host in the second subnet
 Last host in the second subnet
 Broadcast adress of the second subnet
Solution
We are asked to create 7 subnets in the available range. Howevery, we can't optimally divide the network range in 7 equal subnets, because the number of possible subnets for any given range is always a power of 2.
Instead, we must determine the next highest power of 2, which is 8. This means we will not be using the last (eighth) subnet.

Actual networks to createnext highest power of two = 8
Knowing the optimal number of subnets to create, we should now calculate the new subnet mask each subnet should be using
First we need to determine how many extra bits to use in the new subnet mask. There's a simple rule: each bit added will multiply the previously available number subnets by 2. Using 1 extra bit creates 2 subnets, 2 extra bits creates 4 and 3 extra bits creates 8.
Of course, you can also mathematically determine this using the formula S = 2^{N} where N = the number of the subnet bits and S = the number of subnets.
Thus, changing the formula to get number of bits to borrow makes it N = Log_{2}(S). This results in, you guessed it, 3.

Subnet bits to borrowLog_{2}(8) = 3
Time to write this out in binary so we can calculate the new subnet mask. Notice the three added subnet bits.
Original mask length: /24 Original mask in binary: 1111 1111 . 1111 1111 . 1111 1111 . 0000 0000 Original mask in decimal: 255 . 255 . 255 . 0 New mask in binary: 1111 1111 . 1111 1111 . 1111 1111 . 1110 0000 New mask in decimal: 255 . 255 . 255 . 224 New mask length: /27

New subnet mask:1111 1111 . 1111 1111 . 1111 1111 . 1110 0000 = 255.255.255.224
To calculate the number of hosts per subnet we need to check how many host bits are left in the subnetmask. A IPv4 address consists of 32 bits. Our new netmask is 27 bits. This leaves 32  27 = 5 bits for use as host IPs.
Now we can satisfy the formula 2^{B}  2 where B = the number of the host bits.

Number of host bits:32  27 = 5

Number of hosts per subnet:2^{5}  2 = 30
Halfway there!
The final four questions are all about calculating some interesting addresses in the second subnet.
Let's keep the overview by writing out every possible network using the new subnetmask:
Net 1: 1010 1100 . 0001 0000 . 0100 0100 . 0000 0000 = 172. 16. 68. 0/27 Net 2: 1010 1100 . 0001 0000 . 0100 0100 . 0010 0000 = 172. 16. 68. 32/27 (to calculate) Net 3: 1010 1100 . 0001 0000 . 0100 0100 . 0100 0000 = 172. 16. 68. 64/27 Net 4: 1010 1100 . 0001 0000 . 0100 0100 . 0110 0000 = 172. 16. 68. 96/27 Net 5: 1010 1100 . 0001 0000 . 0100 0100 . 1000 0000 = 172. 16. 68. 128/27 Net 6: 1010 1100 . 0001 0000 . 0100 0100 . 1010 0000 = 172. 16. 68. 160/27 Net 7: 1010 1100 . 0001 0000 . 0100 0100 . 1100 0000 = 172. 16. 68. 192/27 Net 8: 1010 1100 . 0001 0000 . 0100 0100 . 1110 0000 = 172. 16. 68. 224/27 (unused)
We now know the network address of the second subnet.

Network address of net 2:1010 1100 . 0001 0000 . 0100 0100 . 0010 0000 = 172.16.68.32
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 nnnn . nnnh hhhh First host in binary: 1010 1100 . 0001 0000 . 0100 0100 . 0010 0001 converted to decimal: 172 . 16 . 68 . 33

First host in net 2:1010 1100 . 0001 0000 . 0100 0100 . 0010 0001 = 172.16.68.33
The broadcast address of a IPv4 network is always the last possible number in the network.
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 nnnn . nnnh hhhh Broadcast in binary: 1010 1100 . 0001 0000 . 0100 0100 . 0011 1111 converted to decimal: 172 . 16 . 68 . 63 Last host in binary: 1010 1100 . 0001 0000 . 0100 0100 . 0011 1110 converted to decimal: 172 . 16 . 68 . 62

Last host in net 2:1010 1100 . 0001 0000 . 0100 0100 . 0011 1110 = 172.16.68.62

Broadcast address of net 2:1010 1100 . 0001 0000 . 0100 0100 . 0011 1111 = 172.16.68.63
Now there we have it. Let summarize the answers again:
Try it yourself!
Using the methods described above, you should be able to calculate the properties of any IPv4 adress. There's an endless supply of automatically generated subnet exercises waiting for you on the exercise page. Go ahead, give it a try!