For example if we need to find the broadcast address of the subnet 192.168.116.32/27 we will first list the subnet address in binary, (we have previously discuses the mathematical conversion process in a previous article). To calculate the broadcast address of a subnet change all the host bits values in the subnet address to binary 1s. We will also learn another way to perform these calculations. The methods shown above to calculate the range of valid host IP addresses and the broadcast address are very easy and quick. The host address must exactly be 30 in number since we have reserved 5 bits for the host portion. The host range will start from the next address after the subnet address for example for the subnet 192.168.116.32/27 the host range will start at 192.168.116.33 and end at one less the broadcast address which will be 192.168.116.62 since the broadcast address is 192.168.1.63. For example for the subnet 192.168.116.0/27 the broadcast address will be 192.168.1.31 which is last address of this subnet and in binary it will be all 1s in the host portion, this address is preceded by the next subnet address. The broadcast address will be the last address of the subnet and one less the preceding subnet address. Shown in the table below are the subnet addresses, their respective host ranges and the broadcast addresses.ДЄs shown in the table, once we have listed the subnet addresses, calculating the host range and broadcast address is relatively simple. The first subnet address will be 192.168.116.0/27 and the following subnets will be with increments of 32, the subnet multiplier we calculated in the previous step. Our final step will be to list the subnet address, host range and the broadcast address. We will use the subnet multiplier in the next step to list the subnets. So in this case our subnet multiplier will be 256-224 =32. All we have to do is subtract the last nonzero octet of the subnet mask from 256. Our third step will be to determine the subnet multiplier which is fairly simple. We will set the first 3 bits of the last octet to 1 and last octet will now be 11100000 which from the table above will be 224 in decimal. We can quickly convert 255.255.255.0 to binary by looking at the table above. So our original subnet mask was 255.255.255.0 and we allocated 3 bits from the host portion which allowed us to have 8 subnets and 30 hosts within each subnet. When performing IP subnetting we will refer to the picture shown below which is very handy in this process. We have discussed in detail the conversion process of binary to decimal and vice versa. Since we have borrowed 3 bits from the host portion our new subnet mask will be 11111111.11111111.11111111. Thus in this case we can create 2 3 =8 subnets Where the exponent n is bits borrowed from the host portion. The network 192.168.116.0/24 has 8 bits for host portion and we will reserve 5 bits for the new host portion, the 3 bits remaining can now be used for creating subnets. The -2 in the formula accounts for two addresses the subnetwork address and the broadcast address which cannot be assigned to hosts. 2 5 -2 =30, so 5 bits atleast must be available for host addressing and the remaining can be borrowed to create subnet addresses. We can calculate how many bits will be required so that each subnet has 30 host addresses. Where the exponent n is equal to the number of bits left after subnet bits are borrowed. Our First step will be to determine how many bits do we need to borrow from the host portion such that the requirement of minimum 30 hosts per subnet is fulfilled. The requirement is to perform subnetting such that we create as many subnets as we can with 30 hosts in each subnet. For example we are given a network address 192.168.116.0 with the default subnet mask of 255.255.255.0. The best way to demonstrate subnetting method is by using an example.

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