Subnetting based on host requirements
There are two considerations for planning a network:
 The number of hosts required for each network.
 The number of subnetwork needed.
The table below in the figure displays the details for subnetting a /24 network. Observe how there is an opposite relationship between the number of hosts and the number of subnets. The more bits borrowed to create subnets, the fewer host bits available. If more host addresses are required, more host bits are necessary, which resulting in fewer subnets.
Here in this lesson, we will discuss the subnetting based on the host requirements. So we will jump right into the examples because that’s the best way to learn.
The examples here we will be talking about will look and feel the same as the ones we did in the previous lesson but there is a most important twist that makes it special.
Let’s suppose you have a network administrator in an organization, which purchased the class C address 200.10.1.0 with the subnet mask 255.255.255.0 and you are required to break that address into 30 hosts per network. The number of host addresses required in the subnet will determine how many bits must be left in the host portion. Remember that two of the addresses cannot be used, so the usable number of addresses can be calculated as 2^{h}2. The process is almost exactly the same with the one based on network requirements.
Convert the number of hosts to binary
Required Host – 30 (So convert 30 into binary)
30 = 11110
Reserve bits in the subnet mask and find the increment
Its mean that we required 5 bits in the host portion of the address. The difference here is that we convert the number of host per network back to binary instead of converting the number of networks.
Remember! We have already learned that 1s represent the network and 0s represent the hosts.
Keep in mind we are still subnetting.But, our focus this time is not how many networks I get but how many hosts I get per network. 30’s binary representation takes up to 5 bits, right?
255.255.255.0=11111111.11111111.11111111.00000000
255.255.255.224=11111111.11111111.11111111.111000000
The new subnet mask is 255.255.255.224 or /27. So, 30 needs 5 bits so we need to save 5 host bits. Notice that instead of going from left to right like we did with the network requirements, I went from right to leave because that’s where my 0s exist. we know that we can get 30 hosts per subnetwork.
Use the increment in order to find the network ranges
Our focus is just on the 0s in Subnetting on hostbased. So let’s now figure out our network ranges. Our increment is 32 because the lowest network bit converted back to a decimal number is 32 in our case.
Net 
Network ID 
Broadcast IP 
Total IP Addresses 
Net0 
200.10.1.0 + 000.00.0.32 
200.10.1.31 + 000.00.0.32 
32

Net1 
200.10.1.32 + 000.00.0.32 
200.10.1.63 + 000.00.0.32 
32

Net2 
200.10.1.64 + 000.00.0.32 
200.10.1.95 + 000.00.0.32 
32

Net3 
200.10.1.96 + 000.00.0.32 
200.10.1.127 + 000.00.0.32 
32

Net4 
200.10.1.128 + 000.00.0.32 
200.10.1.159 + 000.00.0.32 
32

Net5 
200.10.1.160 + 000.00.0.32 
200.10.1.191 + 000.00.0.32 
32

Net6 
200.10.1.192 + 000.00.0.32 
200.10.1.223 + 000.00.0.32 
32

Net7 
200.10.1.224 
200.10.1.255 
32 
We know that first and the last address of each subnetwork aren’t usable therefore we exactly 30 usable hosts per network.