Subnetting based on host requirement

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 hosts and the number of subnets. The more borrowed bits result in more subnets, but with fewer host bits available, there are fewer hosts in the network. If the host requirement is more addresses, then more host bits are necessary, which results in fewer subnets.

In this lesson, we will discuss the subnetting based on the host requirement. We will jump right into the examples for better understanding. The examples 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 them special.

Let’s suppose we have purchased the address 192.168.100.0 with the subnet mask 255.255.255.0 and are required to break that address into 62 hosts per network. The number of hosts 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 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 hosts per network back to binary instead of the number of networks. We already learned that “1s” represent the network and 0s represent the hosts in the subnet mask. Also, remember we are still subnetting and focusing on the host rather than the network requirement. So, the binary of 62 takes up to 6 bits, right? Now examine the default subnet mask before subnetting and the subnet mask after adding ones in the place of borrowed bits position places.

255.255.255.0=11111111.11111111.11111111.00000000

255.255.255.192=11111111.11111111.11111111.110000000

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 left because that’s where my 0s exist. We know that we can get 30 hosts per sub-network.

Use the increment to find the network ranges.

Our focus is just on the 0s in subnetting on host requirement-based. So, let’s now find the network ranges. Our increment is 64 because the lowest network bit converted back to a decimal number is 64.

We know that the first and last addresses of each sub-network aren’t usable, so we have exactly 62 usable hosts per network.