"Dividing a network into subnets" cheat sheet

An example of dividing a network into subnets

The process of dividing the network into subnets is easier to explain with a specific example. Let’s imagine that we have a network of three routers, three switches and several computers, which needs to be divided into 6 subnets. The network diagram is shown in the figure below.

We were allocated one IP address 192.168.0.0/24 for the entire network, which we need to divide into 6 subnets. In binary form, this IP address looks like this (I’ll bold the network portion): 11000000.10101000.00000000. 00000000.

The required number of nodes (hosts, in this case computers) in each subnet:

Subnet A – 100 nodes
Subnet B – 50 nodes
Subnet C – 20 nodes
Subnets D – 2 nodes
Subnet E – 2 nodes
Subnet F – 2 nodes

Remember that you need to start dividing the network from the network with the maximum number of nodes. “/24” is the subnet mask prefix (shorthand for mask). The full entry of the subnet mask is 255.255.255.0. In binary representation, the subnet mask looks like this: 11111111.11111111.11111111.00000000 which means we have 8 bits available to divide the network.

Use the cheat sheet above (namely the “BIN to DEC” table). The first subnet A needs to be allocated IP addresses for 100 nodes. In the “BIN to DEC” table, we see that by taking one bit out of eight in the mask, we get 1 bit to the network portion (which is 2 subnets) and 7 bits to the address portion (01111111 = 127). 127 together with zero in number equals 128, this is the full number of addresses that satisfies the requirements (and there are even a few addresses to spare). And yes, we change the mask from “/24” to “25” (in binary format it will be 11111111.11111111.11111111.10000000). Let’s apply a new mask to our network and get 2 subnets (I will highlight the portion of the network in bold): 1 – 11000000.10101000.00000000.0 0000000 (network 192.100.000) hedgehog 192.168.0.128/25)

In the two new networks, the network portion is 7 bits. Using the formula (which is in the cheat sheet), we will check whether 7 bits are enough for a network with 100 nodes. 2^7-2=128-2=126, which means 7 bits gives us 126 node addresses. (Let me remind you of the formula: 2^X-2=number of addresses for nodes, where X is equal to the number of zeros, and “-2” is for special addresses that cannot be assigned to nodes.)

We have one network left, 192.168.0.128/25, and we need an address for nodes for subnet B 50. As before, we can see in the table “BIN to DEC” 00111111 = 63, this is more than 50, which means it meets the requirements. We occupy one more bit near the portion of the address, which remains 6 (2 6-2 = 62). The mask becomes one more than /26, we apply it to our network and get two new subnets (I will highlight the portion of the network in bold): 1 – 11000000.10101000.00000000.10 000000 (network 1.0.1.0.1.0.1.0.1.0.1.0.1.0.1.0 .1.0.1.0.1.0.1.0.1.0.1.0.1.0.1.0.1.0.1.0.1.0.1. 0000000.11 000000 (network 192.168.0.192/26) We also separate 1 bit from the node address portion (00011111 = 31, which more than 20, and therefore suits us), the mask is already /27.

Again two networks:

1- 11000000.10101000.00000000.110 00000 (network 192.168.0.192/27)
2 – 11000000.10101000.00000002 27)

It remains for us to allocate 3 subnets with 2 addresses for the nodes. The table shows that only two bits (00000011 = 3) are enough for the node address portion, 2^2-2=2 addresses for two nodes. In turn, two bits (2^2=4, the formula from the cheat sheet) are enough for three subnets of the same size. There are only 32 bits in the IP address, we subtract the 2 we need and get 30, so we use the /30 mask. For our network, it looks like this (I will highlight the portion of the network in bold): 11000000.10101000.00000000.111000 00 (network 192.168.0.224/30).