1. Subnetting leading to VLSM Ed Deacon

2. Splitting up a network you own If you have been given a network address eg 195.112.50.0/24 You require addresses for 4 classrooms. There are 2 ways to approach this. – The number of subnetworks required? – The number of hosts required in each subnetwork?

3. Number of Subnetworks Workout the number of bits required to make the number of subnets In this case 4 128 64 32 16 8 4 2 1 – 1 1 2 bits will make the number 4 but in this case we can’t use the last subnet which has it’s broadcast address which is 255 so we must use 3 bits 128 64 32 16 8 4 2 1 1 1 1 0 0 0 0 0 Look where the last one indicates this case 32 which will give 32 hosts (30 useable the first network id and last broadcast addresses are not useable host addresses)

4. Number of hosts Each classroom has 25 PC in them 128 64 32 16 8 4 2 1 1 1 1 0 0 0 0 0 See where the number of host fits with binary bits again. In this case I require 5 bits to make the number 25 eg 1 1 0 0 1 = 5 bits Host bits are ‘0’s so fill in the ‘0’ and what is left fill with ‘1’s the column with the furthest to the right ‘1’ the number above will indicate the total number of hosts ( in this case 32 but remember 30 useable the first network id and last broadcast addresses are not useable host addresses) 25

5. 255.255.255.224 or /27 This is known as a /27 network because 255 = 1 1 1 1 1 1 1 1 and 224 = 1 1 1 so 1 1 1 1 1 1 1 1. 1 1 1 1 1 1 1 1. 1 1 1 1 1 1 1 1. 1 1 1 =27 So if we look at the possible network ID’s 128 64 32 16 8 4 2 1 0 0 0 = 0 0 0 1 = 32 0 1 0 = 64 0 1 1 = 96 1 0 0 = 128 1 0 1 = 160 1 1 0 = 192 1 1 1 = 224 (because it has 255 as its broadcast address)

6. Design the networks Sub networkNetwork IDFirst UseableLast useableBroadcast Address 1013031 232336263 364659495 49697126127 5128129158159 6160161190191 7192193222223 8224*** 255 So the first sub network will have an ID of 195.112.50.0 the second sub network ID 195.112.50.32 and so on. So the first sub network will have the host addresses of 195.112.50.1 up to 30 the second sub network the host addresses of 195.112.50.33 up to 62 and so on. So the first sub network will have a broadcast address of 195.112.50.31 the second sub network broadcast address of 195.112.50.63 and so on.

7. Then add labels to your design

8. In this example we had spare sub networks that we could use in the future But if we looked at this again we could this differently if the number of host require was different classroom A, B, C has 50 hosts each, classroom D has 10. We need 4 sub networks = 2 bits = 64 host 3 sub networks require /26 255.255.255.192 This give me 64 total hosts Get a blank sheet of A4 to do this exercise.

9. 0 In the top left corner put ‘0’ this is the network ID for this /24 so you need to put ‘255’ in the bottom right this is the broadcast address of this C class network Now fold the paper in half this is to represent subnetting the network to a /25 network now do the same put the sub netowk ID and Broadcast. You will see the 256 addresses have been split into 2 networks with 128 addresses 126 of which can be assigned. We need 50 hosts so fold the paper in half again to represent a /27 network 255

10. 0128 63191 64192 127255 We can’t use this sub network as it has the Broadcast address for the whole C class network. But we could subnet it again to create a smaller network that doesn’t contain ’255’ (subnet a subnet)

11. 0128 63191 64192 127255 223 224

12. Subnetting is not hard It is logical! If you convert the address into binary and do a logical AND with the subnet mask, this will give you the Network ID. In Networking the network ID and subnet mask is the most important information and from that we can work out all other information (that is why it is called Networking not Hosting) like Routers you are only interested in the network information there could be millions of hosts but they belong to “a network”