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Community Member

## VLSM Help

I used to be pretty good with VLSM addressing, but it's been a while since I've looked at it and I'm confused. I don't have a problem subnetting an existing subnet into equal subnets, but I need some help in subnetting with variable subnets. For example, if I need five subnets with 2, 7, 29, 34, and 100 hosts, what is the best way to figure that by hand? I know I can use the calculator, but I want to understand how to do it by hand.

5 REPLIES
Silver

## Re: VLSM Help

Hi William,

Its gud that you know subnetting, though we all tend to forget things if we dont come across it on regular basis.

The concept is, u always start subnetting from high to low. In your case, 1st u subnet a network for 100 hosts, the remaining network will then be subnetted for 34 hosts, then 29 & so on.......

Community Member

## Re: VLSM Help

Thanks. Can you point me to a good example where this is done?

Bronze

## Re: VLSM Help

Hall of Fame Super Blue

## Re: VLSM Help

Hi

You need to look at in binary really. Using 192.168.5.0 255.255.255.0 as the example network

one octet in binary =

128 64 32 16 8 4 2 1

Because we are dealing with a Class C network the first 3 octets in the subnet mask will be 255 ie. 255.255.255.0 so we will concentrate only on the last octet.

So first you need 100 hosts. Well the only number larger than 100 is 128 so your first subnet is

128 64 32 16 8 4 2 1

1 0 0 0 0 0 0 0 = .128

So the first network is

192.168.5.0 255.255.255.128

which includes host 192.168.5.1 -> 192.168.5.126 with 192.168.5.127 as the broadcast address.

Next you need a network with 34 hosts. Unluckily for you this is just above 32 in the binary above so you need to select 64 ie.

128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 0 = .192

(remember subnet masks must be continguous)

So carrying on from last network

192.168.5.128 255.255.255.192

hosts 192.168.5.129 -> 190

Next you need a network with 29 hosts

128 64 32 16 8 4 2 1

1 1 1 0 0 0 0 0 = 224

192.168.5.192 255.255.255.224

hosts = 192.168.5.193 -> 222

7 hosts next. You need to be careful with this one. Seven hosts would suggest you just need to mark out the "8" in the octet but

128 64 32 16 8 4 2 1

1 1 1 1 1 0 0 0 = 248

192.168.5.224 255.255.255.248

hosts 192.168.5.225 -> 230

You only get 6 hosts because of the broadcast and the the subnet mask so you need a "16" in the octet ie.

128 64 32 16 8 4 2 1

1 1 1 1 0 0 0 0 = 240

192.168.5.224 255.255.255.240

host 192.168.5.225 -> 192.168.5 238

and finally 2 hosts

128 64 32 16 8 4 2 1

1 1 1 1 1 1 1 0 = 252

192.168.5.240 255.255.255.252

hosts 192.168.5.241 & 242

As one of the other posters said you should start with the largest requirement first.

** Edit - apologies, the octet values don't quite line up in the examples but hopefully it still makes sense **

HTH

Jon

Silver

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