Tutorial: Subnet a network using VLSM.

Introduction

Today I had an exam about CCNA 2 which was made by our teacher. So this was not the standard CISCO multiple choice exam. The first question was to subnet a network using VLSM. Since I do not had a guide for myself I will make this post.

The question

We have a network with the following locations:

• Campus A
• 1000 hosts
• Campus B
• 400 hosts
• Department Administration
• 110 hosts

Subnet this network using VLSM with the required amount of hosts in each subnet. You can choose between the following networks:

• 10.0.0.0/8
• 172.16.0.0/16
• 192.168.1.0/24

Answer

Since we have a network that requires at least 1000 hosts I choose for the 172.16.0.0/16 network. Also the serial interfaces between the routers need to be included in the subnets. So we have a total of 5 subnets in this particular network.

Subnet 1 - 1000 hosts

So the first network is our subnet with at least 1000 hosts. The amount of bits we need is 2^10 (1024) for the hosts.

Subnet mask:

1111 1111 . 1111 1111 . 1111 1100 . 0000 0000 = 255.255.252.0 or /22

Range:

1010 1100 . 0001 0000 . 0000 00 | 00 . 0000 0001 = 172.16.0.1

1010 1100 . 0001 0000 . 0000 00 | 11 . 1111 1110 = 172.16.3.254

Broadcast:

1010 1100 . 0001 0000 . 0000 00 | 11 . 1111 1111 = 172.16.3.255

Subnet 2 - 400 hosts

So for our second network we need at least 400 hosts. The amount of bits we need is 2^9 (512) for the hosts. And we know our first available network is 172.16.4.0.

Subnet mask:

1111 1111 . 1111 1111 . 1111 1110 . 0000 0000 = 255.255.254.0 or /23

Range:

1010 1100 . 0001 0000 . 0000 010 | 0 . 0000 0001 = 172.16.4.1

1010 1100 . 0001 0000 . 0000 010 | 1 . 1111 1110 = 172.16.5.254

Broadcast:

1010 1100 . 0001 0000 . 0000 010 | 1 . 1111 1111 = 172.16.5.255

Subnet 3 - 110 hosts

So for our third network we need at least 110 hosts. The amount of bits we need is 2^7 (128) for the hosts. And we know our first available network is 172.16.6.0.

Subnet mask:

1111 1111 . 1111 1111 . 1111 1111 . 1000 0000 = 255.255.255.128 or /25

Range:

1010 1100 . 0001 0000 . 0000 0110 . 0 | 000 0001 = 172.16.6.1

1010 1100 . 0001 0000 . 0000 0110 . 0 | 111 1110 = 172.16.6.126

Broadcast:

1010 1100 . 0001 0000 . 0000 0110 . 0 | 111 1111 = 172.16.6.127

Subnet 4 - router 1

The subnets for our routers only require 2 usable IP's. The amount of bits we need for this is 2^2 (4) for the hosts. And the first available address is now 172.16.6.128.

Subnet mask:

1111 1111 . 1111 1111 . 1111 1111 . 1111 1100 = 255.255.255.252 or /30

Range:

1010 1100 . 0001 0000 . 0000 0110 . 0000 00 | 01 = 172.16.6.129

1010 1100 . 0001 0000 . 0000 0110 . 0000 00 | 10 = 172.16.6.130

Broadcast:

1010 1100 . 0001 0000 . 0000 0110 . 0000 00 | 11 = 172.16.6.131

Subnet 5 - router 2

The amount of bits we need for this network is also 2^2 (4) for the hosts. And the first available address is now 172.16.6.132.

Subnet mask:

1111 1111 . 1111 1111 . 1111 1111 . 1111 1100 = 255.255.255.252 or /30

Range:

1010 1100 . 0001 0000 . 0000 0110 . 0000 00 | 01 = 172.16.6.133

1010 1100 . 0001 0000 . 0000 0110 . 0000 00 | 10 = 172.16.6.134

Broadcast:

1010 1100 . 0001 0000 . 0000 0110 . 0000 00 | 11 = 172.16.6.135

Summary

Needed Size	Allocated Size	Address	Mask	Dec Mask	Assignable Range	Broadcast
1000	1022	172.16.0.0	/22	255.255.252.0	172.16.0.1 - 172.16.3.254	172.16.3.255
400	510	172.16.4.0	/23	255.255.254.0	172.16.4.1 - 172.16.5.254	172.16.5.255
110	126	172.16.6.0	/25	255.255.255.128	172.16.6.1 - 172.16.6.126	172.16.6.127
2	2	172.16.6.128	/30	255.255.255.252	172.16.6.129 - 172.16.6.130	172.16.6.131
2	2	172.16.6.132	/30	255.255.255.252	172.16.6.133 - 172.16.6.134	172.16.6.135