# Variable Length Subnet Mask (VLSM)

(Rene Molenaar) #1

This topic is to discuss the following lesson:

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(Gabriel S) #2

Rene,

A good way to explain this subject that is a little confuse.
About challenge, I tried to solve it…

A network 10.0.0.0

One subnet for 600 hosts -> It’s need a block 1024
One subnet for 250 hosts. -> It’s need a block 256
One subnet for 120 hosts. -> It’s need a block 128
One subnet for 30 hosts. -> It’s need a block 32
One subnet for 2 hosts. -> It’s need a block 4

A block of 1024 is like 4x256, which is need 10 hosts bits.

So…

Subnet 1: (size 1024)

first host: 10.0.0.1
last host: 10.0.3.254

Subnet 2: (size 250)

first host: 10.0.4.1
last host: 10.0.4.254

Subnet 3: (size 120)

first host: 10.0.5.1
last host: 10.0.5.126

Subnet 4: (size 30)

first host: 10.0.5.129
last host: 10.0.5.158

Subnet 5: (size 2)

first host: 10.0.5.161
last host: 10.0.5.162

I hope I’m right…

When you have a free time, please, fix the details below.
“Subnet 2: 256 – 32 = <>” so the subnet mask is 255.255.255.224"
and
on subnet2 the last host IP should be 172.16.2.254

Hug and more a time, thanks for excellent work!!!

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(Rene Molenaar) #3

Hi Gabriel,

Your example is correct, good job! Also thanks for reporting those errors, just fixed them.

Rene

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(Gabriel S) #4

Rene,

Thanks!!!

Hug

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(RAJAGOPAL K) #5

Hi Rene,

Can you pls elaborate on how you got this subnet mask? I didn’t get it. thanks
<>

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(Rene Molenaar) #6

Sure, which exact subnet mask do you refer to?

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(RAJAGOPAL K) #7

Subnet 1. 255.255.254.0

Subnet 1: this subnet is a block of “512” which is a combination of 2x “256”. A subnet with a size of 256 addresses normally has a subnet mask of 255.255.255.0 (/24). When we combine 2x 256 to get a 512 block, we give back one host bit so that means we have a /23 which is a subnet mask of 255.255.254.0.

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(Rene Molenaar) #8

Let me describe it this way, we know that a subnet mask of 255.255.255.0 has 256 addresses in total:

255.255.255.0 / 24 - 256 addresses

When we create smaller subnets, the list will be like this:

You can see that whenever the subnet becomes smaller, we can only use half of the addresses. I can apply the same logic if I move into the other direction:

Does this help?

Rene

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(Iynkaran T) #9

Hi I’m having hard time figuring out the Subnet Mask, Hosts, Subnets, Network ID and Broadcast ID for the following IP Address. I hope someone can help me answers. Cheers!

150.12.110.10/25

112.10.78.40/22

50.1.112.10/21

28.10.145.10/18

150.50.50.50/23

100.10.185.10/20

172.16.221.10/19

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(Rene Molenaar) #10

Hi Lynkaran,

I’ll explain how to do the first example, see if you can solve them with my technique. Let’s start with 112.10.78.40/22:

First we need to figure out what the subnet mask since /22 doesn’t tell us much. You need to write this down in binary and convert it to decimal:

first 8 bits = 11111111 (255 in decimal)
next 8 bits = 11111111 (255 in decimal)
next 8 bits = 11111100 (252 in decimal)
next 8 bits = 00000000 (0 in decimal)

So now we know the subnet mask is 255.255.252.0

How many hosts do we have per subnet? There are 2 + 8 host bits so 10 host bits in total. You can use this formula for this:

210 - 2 = 1024.

What subnets do we have? There’s a quick way to calculate this. Take number 256 - subnet mask = size of subnet:

256 - 252 = 4

#1 112.10.0.0/22
#2 112.10.4.0/22
#3 112.10.8.0/22
#4 112.10.12.0/22
#5 112.10.16.0/22
#6 112.10.20.0/22
#7 112.10.24.0/22
…and so on

For subnet #1 that will be 112.10.3.255 and for subnet #5 it’s 112.10.19.255.

See if you can do the other examples on your own with these steps. All you need is practice, practice, practice.

Hope this helps.

Rene

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(Iynkaran T) #11

WOW! Awesome explanation! Thank you so much

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(Iynkaran T) #12

Quick question how did you get this answer when the slash value is /22?

first 8 bits = 11111111 (255 in decimal)
next 8 bits = 11111111 (255 in decimal)
next 8 bits = 11110000 (240 in decimal)
next 8 bits = 00000000 (0 in decimal)

So now we know the subnet mask is 255.255.240.0

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(Rene Molenaar) #13

The value behind the slash are the number of bits that we use for the subnet mask:

11111111 11111111 11111100 00000

Those are 22 bits, when you convert it to decimal you get 255.255.252.0

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(Iynkaran T) #14

Wouldnt it be 255.255.252.0?

The nearest classfull mask that is bigger then our /22 is /24. So we subtract 22 from 24 and we get 2. This digit is the power of 2 we need.
So as we get 2 we have to power up 2 by our result - 2.
2^2=4
Now we have to subtract this result (4) from 256(this is maximum quantity of subnet that we can get from 255 in netmask)
256-4=252.
So the result is /22 prefix is 255.255.252.0 netmask it is not the 255.255.240.0

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(Rene Molenaar) #15

Whoops, early morning error You are totally right, I just edited my calculations.

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(Vikas A) #16

Hi Rene,
Why do you always consider the subnet with the largest number of hosts first ?
And then decide the block sizes of the smaller ones.
Why cant we do the other way round ?

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(Rene Molenaar) #17

Hi Vikas,

We do this because you can have issues with overlapping subnets. If you start with the largest subnets first and move on to the smaller ones then you won’t run into issues with overlapping addresses.

Rene

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(Apurva P) #18

Hi,

Thank you for awesome explanation. It is helping me a lot for my ccna studies

I was solving a problem as stated:-

Network ip- 172.16.0.0
No. of host needed-120

CIDR notation will be /25

so ip address range will be from 172.16.0.0 to 172.16.0.128

have I solved problem correctly.

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(Lazaros Agapides) #19

Hello Apurva.

That is exactly correct! Keep in mind as well that the first and last addresses in the range are the network and broadcast addresses respectively so you will have 126 addresses available for hosts.

I hope this has been helpful!

Laz

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(Gerson F) #20

Hi Rene,

I was trying to figure out a way of “knowing” the amount of bits needed for huge hosts without the need of a calculator, like:

• 2^X-2= hosts (too complicated).

This way sorts the problem!

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