In the post, https://nepcodex.com/2020/05/what-is-subnetting/, we talked about what subnetting is. Similarly, in the previous post, we understood about subnet mask, IP address, Network ID and Host ID. Now, we are going look at some examples to divide a network.
Example 1 – Subnet a network with equal number of hosts
Firstly, we are going to divide a network such that each network has an equal number of hosts. Let’s assume, we have a network 103.5.150.0/24 to divide. This is the address of Pulchowk Campus, IOE, Nepal. This means we get the space for 256 addresses. Also, the requirement is that each network can have at most 26 hosts.
Now, in the previous post, we know that each network has an address to identify it and another address for broadcasting. Hence, the total number of IP addresses per network is 26 + 2 = 28. Furthermore, the nearest number with a raised exponent of 2 is 25 = 32. So, we can use the last 5 bits for the hosts, and first 3 bits for the network. Hence, there will be 23 = 8 networks with capacity of 32 – 2 = 30 hosts. So, let’s convert the last octet of addresses into the binary system.
103.5.150.0 = 103.5.150.00000000
Also, the subnet length prefix is 24, which means the subnet mask is 255.255.255.0 = 255.255.255.00000000
As we have said we are going to borrow the first 3 bits of the last octet for each subnet. Therefore, the range of subnets will be as follows.
103.5.150.00000000 – 103.5.150.00011111 having subnet mask 255.255.255.11100000.
i.e. 103.5.150.0 – 103.5.150.31 having subnet mask 255.255.255.224
i.e. 103.5.150.0/27, Net-ID = 103.5.150.0, Broadcast Address = 103.5.150.31
103.5.150.00100000 – 103.5.150.00111111 having subnet mask 255.255.255.11100000.
i.e. 103.5.150.32 – 103.5.150.63 having subnet mask 255.255.255.224
i.e. 103.5.150.32/27, Net-ID = 103.5.150.32, Broadcast Address = 103.5.150.63
Similarly,
103.5.150.01000000 – 103.5.150.01011111 having subnet mask 255.255.255.11100000.
i.e. 103.5.150.64 – 103.5.150.95 having subnet mask 255.255.255.224
i.e. 103.5.150.64/27, Net-ID = 103.5.150.64, Broadcast Address = 103.5.150.95
And, so on. To summarize, the networks will be:
103.5.150.0/27, 103.5.150.32/27, 103.5.150.64/27, 103.5.150.96/27, 103.5.150.128/27, 103.5.150.160/27, 103.5.150.192/27, 103.5.150.224/27
Example 1 – Subnet a network with different number of hosts
The above example is the simplest to understand the process. Now, suppose we have a requirement that each network can have different hosts. For example, there are 5 departments in a college, say A, B, C, D, E. There are 4 routers to connect the departments with each other. Again, the network is 103.5.150.0/24. The number of hosts for the networks are as follows:
A: 12, B: 60, C: 31, D: 28, E: 15, A-B: 2, B-C: 2, C-D: 2, D-E: 2
Since we have to take account of the network ID and broadcast address, we have to add 2 addresses in the requirement. i.e.
A: 14, B: 62, C: 33, D: 30, E: 17, A-B: 4, B-C: 4, C-D: 4, D-E: 4
Now, we have to arrange the required number of hosts in descending order. i.e.
B: 62, C: 33, D: 30, E: 17, A: 14, A-B: 4, B-C: 4, C-D: 4, D-E: 4
So, the actual total number of hosts for each network are:
B: 64 = 26, C: 64 = 26, D: 32 = 25, E: 32 = 25, A: 16 = 24, A-B: 4 = 22, B-C: 4 = 22, C-D: 4 = 22, D-E: 4 = 22
This means that we have to borrow 8 – 6 = 2 bits for networks B and C, 8 – 5 = 3 bits for D and E, 8 – 4 = 4 bits for A and 8 – 2 = 6 bits for the router links.
So, converting the last octet of the addresses to the binary system, we get.
Network IP address: 103.5.150.0 = 103.5.150.00000000
Subnet Mask: 255.255.255.0 = 255.255.255.00000000
Departments
For network B:
103.5.150.00000000 – 103.5.150.00111111 with subnet mask 255.255.255.11000000
i.e. 103.5.150.0 – 103.5.150.63 with subnet mask 255.255.255.192
i.e. 103.5.150.0/26, Net-ID = 103.5.150.0, Broadcast Address = 103.5.150.63
Similarly, for network C:
103.5.150.01000000 – 103.5.150.01111111 with subnet mask 255.255.255.11000000
i.e. 103.5.150.64 – 103.5.150.127 with subnet mask 255.255.255.192
i.e. 103.5.150.64/26, Net-ID = 103.5.150.64, Broadcast Address = 103.5.150.127
For network D:
103.5.150.10000000 – 103.5.150.10011111 with subnet mask 255.255.255.11100000
i.e. 103.5.150.128 – 103.5.150.159 with subnet mask 255.255.255.224
i.e. 103.5.150.128/27, Net-ID = 103.5.150.128, Broadcast Address = 103.5.150.159
Similarly, for network E:
103.5.150.10100000 – 103.5.150.10111111 with subnet mask 255.255.255.11100000
i.e. 103.5.150.160 – 103.5.150.191 with subnet mask 255.255.255.224
i.e. 103.5.150.160/27, Net-ID = 103.5.150.160, Broadcast Address = 103.5.150.191
For network A:
103.5.150.11000000 – 103.5.150.11001111 with subnet mask 255.255.255.11110000
i.e. 103.5.150.192 – 103.5.150.207 with subnet mask 255.255.255.240
i.e. 103.5.150.192/28, Net-ID = 103.5.150.192, Broadcast Address = 103.5.150.207
Router Links
For router links, A-B:
103.5.150.11010000 – 103.5.150.11010011 with subnet mask 255.255.255.11111100
i.e. 103.5.150.208 – 103.5.150.211 with subnet mask 255.255.255.252
i.e. 103.5.150.208/30, Net-ID = 103.5.150.208, Broadcast Address = 103.5.150.211
For router links, B-C:
103.5.150.11010100 – 103.5.150.11010111 with subnet mask 255.255.255.11111100
i.e. 103.5.150.212 – 103.5.150.215 with subnet mask 255.255.255.252
i.e. 103.5.150.212/30, Net-ID = 103.5.150.212, Broadcast Address = 103.5.150.215
For router links, C-D:
103.5.150.11011000 – 103.5.150.11011011 with subnet mask 255.255.255.11111100
i.e. 103.5.150.216 – 103.5.150.219 with subnet mask 255.255.255.252
i.e. 103.5.150.216/30, Net-ID = 103.5.150.216, Broadcast Address = 103.5.150.219
For router links, D-E:
103.5.150.11011100 – 103.5.150.11011111 with subnet mask 255.255.255.11111100
i.e. 103.5.150.220 – 103.5.150.223 with subnet mask 255.255.255.252
i.e. 103.5.150.220/30, Net-ID = 103.5.150.220, Broadcast Address = 103.5.150.223
Hence, in this way, we can divide a network according to the requirements. As we can see that, we didn’t oversupply the departments with the IP addresses. Therefore, we have some addresses available and ready to be allocated in future.
Conclusion
In this way, we can subnet a network. Check out the others posts for more contents.
