Subnetting is one of those infuriating subjects which are somehow confusing to learn and so difficult to retain. Practice with textbook Q&A is only a partial solution, as within days the knowledge disappears. There is, however, a simpler way to retain and consolidate the knowledge and that is through practice. Practice makes perfect, or so they say.

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When practicing subnetting it is helpful to take a varied approach, which utilizes both theoretical questions and also some real world examples. Using a mix of questions and scenarios increases the ability to calculate subnets quickly, as it becomes more intuitive with practice. However, using real world scenarios whereby the techniques learned are applied to network models is where knowledge is consolidated and understood and becomes second nature.

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**Quick & Easy Subnetting**

*Consider the IP address 172.16.1.28 with a subnet mask of 255.255.255.240*

- What subnet does this address belong too?
- What is its Broadcast Address?
- How many hosts are in the subnet?
- What is the next subnet address?

In this question, the subnet mask is presented in decimal notation, 255.255.255.240, so there is an easy technique to find the subnet block. Simply take the first number in the subnet mask, working from the right, which is not 255. In this case, it is 240. Then subtract 256 – 240 which equals 16. We now have our subnet range or block. By then simply counting from zero we can find the corresponding subnet, which hosts 172.16.1.28

1^{st} Subnet 172.16.1.0 – 172.16.1.15

2^{nd} Subnet 172.16.1.16 – 172.16.1.31 (172.16.1.28 belongs here)

3^{rd} Subnet 172.16.1.32 – 172.16.1.47

Applying that simple technique determined the subnet it also provided the broadcast address 172.16.1.31, which is the final address in the subnet. Subnets always start with an even number, the network address, and end with an odd number, which is the broadcast address. These two addresses are not available for use by hosts, so our subnet block of 16 actually has 16 -2 = 14 valid hosts. Finally if the subnets broadcast is 172.16.1.31, and is the subnets final address then the next address 172.16.1.32 will be the following subnets address.

*Consider the IP address 10.34.23.22/12*

- What subnet does this address belong too?
- What is its Broadcast Address?
- How many hosts are in the subnet?
- What is the next subnet address?

With the subnet in this format, there is another technique to find our subnet range. First, find the nearest network boundary, in this case the mask is /12 so the closest boundary is 16 (16 is 2^4). (16 -12) = 4, then 2^4 = 16 … The subnet range or block is 16.

Again count from zero

1^{ST} subnet 10.0.0.0 – 10.15.255.255

2^{nd} subnet 10.16.0.0 – 10.31.255.255

3^{rd} subnet 10.32.0.0 – 10.47.255.255 (10.34.23.22/12 belongs to this subnet)

4^{th} subnet 10.48.0.0 – 10.63.255.255

The subnet address is 10.32.0.0 the broadcast is 10.47.255.255 and the number of hosts is 16 -2= 14 finally the next subnet address is 10.48.0.0

*What is the valid host range for the subnet 172.16.0.0/17?*

When calculating number of hosts per subnet the closest boundary is 24, therefore 24 – 17 (the mask) gives a block size of 7, 2^7= 128

The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254.

**Practice Real World Scenarios**

Practicing with subnets using the techniques above certainly improves speed and mental agility however, using the skills learned in a real world scenario is very fulfilling. Consider the table below,

Department | No of Hosts |

Engineering | 27 |

Customer Care | 18 |

IT | 8 |

Finance | 12 |

HR | 4 |

The requirement is for 5 subnets with the largest number of hosts being 27.

In order to create the subnets, bits will be borrowed from the fourth octet by extending the subnet mask from 24 to 27. The three extra borrowed bits are required as 2^3 will allow for eight subnets. By borrowing three bits however there are now only 5 bits remaining for hosts. That will be sufficient as 2^5 = 32 and our biggest requirement is for 27.

The network plan will be

Department | No of Hosts | Subnet | Spare IP |

Engineering | 27 | 192.168.0.0/27 | 3 |

Customer Care | 18 | 192.168.0.32/27 | 12 |

IT | 8 | 192.168.0.64/27 | 22 |

Finance | 12 | 192.168.0.96/27 | 18 |

HR | 4 | 192.168.0.128/27 | 26 |

The network requirement have been satisfied in a theoretical sense, by providing 5 subnets and sufficient IP host addresses to meet the requirement. However, it is not a good real world design. HR and IT subnets are very wasteful. By splitting the address range into eight equal subnets, three subnets went unused, it was less than optimal. Is there another way to subnet?

**Design the Network using VLSM**

Using the same network scenario redesign the IP Plan using subnetting with Variable Length Subnet Masking (VLSM).

With this task it is necessary to consider what masks we can apply for each subnet that provides a more efficient use of the address space.

Hint: Start with the largest departments first, then size the subnet mask to fit by using the minimum number of host bits. Remember, the binary bits to decimal mapping.

**Solution**

Department | No of Hosts | Subnet /Mask |

Engineering | 27 | 192.168.0.0/27 (30 Hosts) |

Customer Care | 18 | 192.168.0.32/27 (30 Hosts) |

IT | 8 | 192.168.0.64/28 (14 Hosts) |

Finance | 12 | 192.168.0.80/28 (14 Hosts) |

HR | 4 | 192.168.0.96/29 ( 6 Hosts) |

By addressing the number of hosts required per department, a more efficient IP address plan can be constructed, which minimizes IP wasted addresses albeit that is not a concern with private addresses. However, practice on conserving address space is necessary for when real valuable IP addresses are used.

Having a practice strategy that focuses on both textbook style subnet questions and network design scenarios, provides a deeper long lasting understanding of subnetting. If you are keen to learn more about IP network design then these resources will help you on the journey to mastering the mysteries of IP networks.