Sub Netting and Super Netting

Subnetting & Supernetting

Marc Khayat, Technical Advocacy Team

[email protected]

October 2011

© 2011 Cisco and/or its affiliates. All rights reserved. 2 All webinars are available at http://lms.netacad.net/course/view.php?id=1201


Part I - Subnetting


• Number of required subnets

• Number of required hosts per subnet



• Using the numerical calculations

• Using the graphical representation


• Network size is the number of IP addresses in a subnet, including the network ID and the broadcast address

• Magic nb is the same as the network size if prefix > /24

• Magic nb must always be between 1 and 255. If greater than 255, divide by 256. If smaller than 1, multiply by 256.

• Every time you divide by 256, you move 1 octet to the left

• Every time you multiply by 256, you move 1 octet to the right

• Everything is a multiple of the magic nb (therefore, the name )

• Don’t bother googling the term “magic nb”

Numerical Calculations


• Nb of required subnets should be an exponential multiple of 2 (so, 2, 4, 8, 16, 32, etc.). If not, then round up.

• To get the magic nb, few easy steps:

1. Identify the first non-255 number in the mask

2. Subtract that number from 256

3. Divide the outcome by the number of required subnets

4. That’s your magic nb!



• Required number of subnets 5 (should be really looking at 8)

• Mask: (/24) 255.255.255.0

First non-255 number is 0

256 – 0 = 256

256 / 8 = 32 (that’s the magic number)

• Mask: (/16) 255.255.0.0


256 – 0 = 256


• Mask: (/20) 255.255.240.0


256 – 240 = 16



On the 4th octet

On the 3rd octet

On the 3rd octet


• Required number of subnets 12 (should be really looking at 16)

• Mask: (/25) 255.255.255.128


256 – 128 = 128


• Mask: (/20) 255.255.240.0


256 – 240 = 16


• Mask: (/21) 255.255.248.0


256 – 248 = 8

8 / 16 = 0.5 * 256 = 128 (that’s the magic number)


On the 4th octet

On the 3rd octet

Started on the 3rd but moved to the 4th octet


• Simply add the magic nb to the initial network. Example: subnetting 192.168.252.0 255.255.252.0 (/22) into 2, 4 or 8 subnets:


2 Subnets 4 Subnets 8 Subnets

Magic Nb =

(256-252)/2 = 2

Magic Nb =

(256-252)/4 = 1

Magic Nb =

(256-252)/8 = 128

(multiplied by 256)

192.168.252.0 192.168.252.0 192.168.252.0

192.168.254.0 192.168.253.0 192.168.252.128

192.168.254.0 192.168.253.0

192.168.255.0 192.168.253.128

192.168.254.0

192.168.254.128

192.168.255.0

192.168.255.128


• Subtract the magic number from 256. So to continue with the previous example of subnetting 192.168.252.0/22:


2 Subnets 4 Subnets 8 Subnets

Magic Nb = 2 (3rd octet) Magic Nb = 1 (3rd octet) Magic Nb = 128 (4th octet)

256 – 2 = 254 256 – 1 = 255 256 – 128 = 128

255.255.254.0 (/23) 255.255.255.0 (/24) 255.255.255.128 (/25)


• With a subnet mask of 255.255.255.224, all network IDs would be a multiple of (256 – 224 = 32), 4th octet, so x.x.x.0, .32, .64, …, .224

• With a subnet mask of 255.255.128.0, all network IDs would be a multiple of (256 – 128 = 128), 3rd octet, so x.x.0.0, .128.0



• Subnet 192.168.16.0/20 into networks of 4 subnets

• /20 => 255.255.240.0 => 256 – 240 = 16

• 16 / 4 = 4 (magic number, on the 3rd octet)

• Networks are: 192.168.16.0, 192.168.20.0, 192.168.24.0,192.168.28.0

• Mask is 256 – 4 = 252 => 255.255.252.0 or /22



• 1st and foremost: add the 2 IPs of network ID and broadcast address

• Look for the closest exponentional multiple of 2

• That’s your network size.

• Continue as previously explained.



• Subnet 192.168.16.0/20 into networks of 700 hosts per subnet

• 700 => 702 => 1024 (network size)

• 1024 / 256 = 4 (magic number, on the 3rd octet)

• Networks are: 192.168.16.0, 192.168.20.0, 192.168.24.0,192.168.28.0

• Mask is 256 – 4 = 252 => 255.255.252.0 or /22



• 123.45.164.255/22: network, broadcast or valid host? If broadcast or valid host address, what is the network address?

Answer: valid host, network address is 123.45.164.0/22

• 100.198.7.64/18: network, broadcast or valid host? If broadcast or valid host address, what is the network address?

Answer: valid host, network address is 100.198.0.0/18

• Hint: from prefix, get mask, then get magic number, then see the closest multiple of that number and match it to the appropriate value in the address provided above.



• It gets a little bit confusing when it is requested to subnet a network into several subnets, each with a different size.

• So, let’s start with an example!

• VLSM = Variable Length Subnet Mask



• 192.168.1.0/24 to be subnetted into A (100 hosts), B (40 hosts), C (10 hosts), D (2 hosts).

• Remember to always start with the largest network first!


Subnet Nb of

Hosts

Net

Size

Magic

Nb

Mask

(4th octet)

Prefix Subnet address

A 100 128 128 256-128=128 /25 192.168.1.0

B 40 64 64 256-64=192 /26 192.168.1.(0+128=128)

C 10 16 16 256-16=240 /28 192.168.1.(128+64=192)

D 2 4 4 256-4=252 /30 192.168.1.(192+16=208)


• For subnet A, start from the initial network address (192.168.1.0). The next network would be a multiple of the magic nb, as explained before. So, 192.168.1.128. Remember that we’re working on the 4th octet;

• For subnet B, the address is 192.168.1.128. The next network would be a multiple of the magic nb, so we add 64 to 128. So, the next network is: 192.168.1.192.

• For subnet C, address is 192.168.1.192. For the next network, we add the magic nb (16) to the current address, so 192.168.1.208.

• For subnet D, address is 192.168.1.208.

• The next network and all remaining addresses are free

(192.168.1.212 192.168.1.255)



• 192.168.160.0/19 to be subnetted into A (2000 hosts), B (500 hosts), C (100 hosts), D (4 hosts).

• Free addresses: 192.168.170.(128+8) 192.168.191.255.


Subnet Nb of

Hosts

Net

Size

Magic Nb Mask Prefix Subnet address

A 2000 2048 2048/256=8 256-8=248

(3rd octet)

/21 192.168.160.0

B 500 512 512/256=2 256-2=254

(3rd octet)

/23 192.168.(160+8=168).0

C 100 128 128 256-128=128

(4th octet)

/25 192.168.(168+2=170).0

D 4 8 8 256-8=248

(4th octet)

/29 192.168.170.(0+128)


Numerical Calculations – section end

Questions?


• Graphically map addresses of an octet

• Very easy to operate within one octet

• Rule of thumb: always divide in half

• Will ensure that you won’t waste/forget networks

Graphical Representation

• Identify your magic numbers (already know how to do it) Step 1 • Segment and reserve networks on the bar graph starting

with the largest network first. Step 2

• You’re done! Step 3


• The bargraph shows a complete octet (from 0 till 255), so 256 values

• You are only allowed to split a segment in half

• Golden rule still applies: always start with the largest network first

• Choose the subnet locations according to the requirements


128

64

32

16

8

4

0

256


• Subnet 192.168.10.0/24 into:

A (100 hosts), B (20 hosts), C (10 hosts)

All on the 4th octet.


A (128)

B (32)

C (16)

FREE

Addresses

Width of 128

Width of 64

Width of 32

Width of 16

Subnet Magic Nb Subnet ID

A 128 192.168.10.0 /25

B 32 192.168.10.128 /27

C 16 192.168.10.160 /28

FREE 192.168.10.176

192.168.10.255

Subnet Magic Nb

A 128

B 32

C 16

Width of 256

This represents the 4th octet

0

128

160

176

255


Graphical Representation – section end

Questions?


Part II - Supernetting



• Using the numerical calculations

• Using the graphical representation

• Actually, these are the reverse methods of those of subnetting.


• Look for the number that encompasses all subnets

• Must be an exponential multiple of 2 (so, 2, 4, 8, 16, 32, etc.)

• Has to start from a multiple of that number. If not, then round up.

• To get the mask, deduct this number from 256 on the appropriate octet.



• Networks: 192.168.1.0/24, 192.168.2.0/24, 192.168.3.0/24

• Number is 4, starts from 0, at 3rd octet. Supernet address is 192.168.0.0 255.255.252.0

• Networks: 192.168.23.0/24, 192.168.25.0/24, 192.168.27.0/24, 192.168.29.0/24


• Networks: 192.168.96.0/20, 192.168.112.0/21, 192.168.120.0/22, 192.168.124.0/23




Numerical Calculations – section end

Questions?


• Very easy to operate within one octet

• Map subnets to the famous bar graph

• Stack networks onto each other, in order

• Supernet should be a number that:

• Is an exponential multiple of 2

• Has to start from a multiple of that number. If not, then round up.


128

64

32

16

8

4

0

256



A (128)

B (32)

C (16)

FREE

Addresses

0

128

160

176

255

Subnet Net Size Subnet ID

A 128 192.168.10.0 /25

B 32 192.168.10.128 /27

C 16 192.168.10.160 /28

Supernet address: 192.168.10.0/24

192



0

128

255

Subnet Net Size Subnet ID

A 1 192.168.10.0 /24

B 2 192.168.12.0 /23

C 1 192.168.14.0 /24

Supernet address: 192.168.8.0/21

64

32

0

16

8

10

12

16

11 A (1)

B (2)

C (1) 14

15

FREE

FREE

FREE


Graphical Representation – section end

Questions?

Thank you.

Documents

Sub Netting and Super Netting