We consider either the host requirement or the network requirement for subnetting. The table below in the figure displays the details for subnetting a /16 network. You can examine how there is an opposite relationship between the number of host and the number of subnets. The more borrowed bits result in more subnets, but with fewer host bits available resulting in fewer hosts in the network. If host requirement are more addresses then more host bits are necessary, which resulting in fewer subnets.
So, here in this lesson, we will discuss the subnetting based on the host requirement. So we will jump right into the examples for better understanding. The examples here we will be talking about will seem to be the same as the ones we did in the earlier lesson but there is a most important twist that makes it special.
Let’s suppose we have purchased the address 192.168.100.0 with the subnet mask 255.255.255.0 and we required to break that address into 62 hosts per network. The number of host requirement 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 2h-2. The process is almost exactly the same as the one based on network requirements.
Convert the number of hosts to binary
Required Host – 62 (So convert 64 into binary)
62 = 111110
Reserve bits in the subnet mask and find the increment
S, we need 6 bits in the host portion of the address. The difference is that we convert the number of host per network back to binary instead of converting the number of networks. We already learned that “1s” represent the network and 0s represent the hosts in the subnet mask. Also, remain in mind we are still subnetting with focusing the host requirement instead of networks requirement. So, the binary of 62 takes up to 6 bits, right? Now examine the default subnet mask before subnetting and subnet mask after adding ones in the place of borrowed bits position places.
So the new subnet mask is 255.255.255.192 or /26. So, 62 hosts’ needs 6 bits in the host portion. Also note that instead of going from left to right as 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 sub-network.
Use the increment in order to find the network ranges
Our focus is just on the 0s in subnetting on host requirement based. So let’s now fined the network ranges. Our increment is 64 because the lowest network bit converted back to a decimal number is 64 in our case.
|Net||Network ID||Broadcast IP||Total IP Addresses|
|Net-0||192.168.100.0 + |
|192.168.100.63 + |
|Net-1||192.168.100.64 + |
|192.168.100.127 + |
|Net-2||192.168.100.128 + |
|192.168.100.191 + |
|Net-3||192.168.100.192 +||192.168.100.255 +||64|
We know that first and the last address of each sub-network aren’t usable therefore we exactly 62 usable hosts per network.