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  • Writer's pictureClifford McGraw Jr.

Subnetting with VLSM



*I made some changes since I recorded the video.


As you all should know, I'm now the Sr. Network Engineer at a fictitious company named Dunder Mifflin. No, not the one in Scranton, they're opening a branch in Philly and need me to set up their network. Funding is little and we're borrowing internet from next door with a single IP Address to subnet, so VLSM is going to be used to conserve addresses as much as possible.


The IP address given was 192.168.185.0/24


The first thing we need to think about when designing a network is what's going to be using the network. This is important because when using VLSM WE NEED TO SUBNET FROM THE GREATEST TO LEAST HOST ADDRESSES NEEDED. This is key. Here's the layout of the office floor on both floors of the building... Oh yeah, the building has two floors.



It may be hard for you to see here but there are 21 computers and 3 printers in this office. We're going to have to have some phones so for now I figure we'll have one at the receptionist desk and one in the big office. That's how we will subnet the networks.


The phones need there own subnet/VLAN because that's how they're generally configured on Cisco devices. For security reasons you want to segment the printers, so they get there own subnet/VLAN. Computers for each floor will get separate subnets/VLANs as well.


So from biggest to smallest.

  • 1st Floor Computers - 21

  • 2nd Floor Computers - 21

  • 1st Floor Phones - 2

  • 2nd Floor Phones - 2

  • 1st Floor Printers - 3

  • 2nd Floor Printers - 3

These are the end devices but they aren't the only devices that need host addresses. The switch and router interfaces also need host addresses so it's important we come up with a draft of the general network architecture. We'll focus on that in another blog post.


For now, lets subnet using the information we have.


IP Addresses:


192.168.185.0/24 255.255.255.0 is the given address.

IP add and CIDR

Subnet Mask

Host Addr. Avail.

Devices

192.168.185.0/26

255.255.255.192

60

1st Floor Computers

192.168.185.64/26

255.255.255.192

60

2nd Floor Computers

192.168.185.128/27

255.255.255.224

30

1st Floor Phones

192.168.185.160/27

255.255.255.224

30

2nd Floor Phones

192.168.185.192/28

255.255.255.240

14

Printers

Methodology:


Bit Conversion Table

2^7

2^6

2^5

2^4

2^3

2^2

2^1

2^0

128

64

32

16

8

4

2

1

128

192

224

240

248

252

254

255

/25

/26

/27

/28

/29

/30

/31

/32

1st bit

2nd bit

3rd bit

4th bit

5th bit

6th bit

7th bit

8th bit

[2^n where n = the number in the exponent]


1. Find the number of bits used in the least significant octet.

The least significant octet(the octet where the host bits begin) in a /24 isn't using any bits. It looks like this: 11111111.11111111.11111111.00000000 The least significant octet in a /25 looks like this: 11111111.11111111.11111111.10000000 It uses 1 bit, and as a result /25 is 255.255.255.128


Just to drive this home here's what /20 looks like: 11111111.11111111.11110000.00000000 It uses 4 bits in the least significant octet. /20 is 255.255.240.0

2. Count how many hosts are needed and start with the biggest number of hosts.

In this case, the 1st or 2nd floor computers had the most hosts. They added up to 21. We will start with a network that can fit at least 21 host addresses.

3. Find the lowest number in the second row of the bit conversion table that the number of host addresses needed is less than.

We are going to give room for scalability so we are going with 64. If I wanted to be more conservative I could go with 32. Both numbers are higher than the 21 computers.

4. Determine how many bits need to be used in that octet for that number. (2^n - 2 should be > host addresses needed.)

For us to use a subnetwork with 64 addresses (62 host addresses because you have to subtract for the two unusable addresses. 1 for the network address and 1 for their broadcast) we need to use up to the 2nd bit. [2^n - 2 is the formula for finding the number of available host addresses]


2^6 - 2 = 62 > 21


I could use 2^5 - 2 which gives 30 host addresses, but I wanted more room for scaleability.

5. This makes the chosen subnet 255.255.255.192

Which provides 62 host addresses.

6. The next available network address is the network address 192.168.185.0 + 2^6 or 64

Which equals 192.168.185.64 /26

7. Add 64 and the next available address is 192.168.185.128

However, we don't need 62 host addresses so we will use the next bit 2^5. 2^5 - 2 = 30 host addresses. Subnet address for this is 255.255.255.224 or /27

8. Add 2^5 to 192.168.185.128 and you get 192.168.185.160

We're keeping the same number of host addresses

9. 192.168.185.160 + 32 = 192.168.185.192

We don't need as many host addresses so we will use the next bit 2^4. 2^4 - 2 = 14 host addresses. Subnet = 255.255.255.240 or /28


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