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Subnetting & Supernetting
Marc Khayat, Technical Advocacy Team
makhayat@cisco.com
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
© 2011 Cisco and/or its affiliates. All rights reserved. 3 All webinars are available at http://lms.netacad.net/course/view.php?id=1201
Part I - Subnetting
© 2011 Cisco and/or its affiliates. All rights reserved. 4 All webinars are available at http://lms.netacad.net/course/view.php?id=1201
• Number of required subnets
• Number of required hosts per subnet
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© 2011 Cisco and/or its affiliates. All rights reserved. 6 All webinars are available at http://lms.netacad.net/course/view.php?id=1201
• Using the numerical calculations
• Using the graphical representation
© 2011 Cisco and/or its affiliates. All rights reserved. 7 All webinars are available at http://lms.netacad.net/course/view.php?id=1201
• 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
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• 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!
Numerical Calculations
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• 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
First non-255 number is 0
256 – 0 = 256
256 / 8 = 32 (that’s the magic number)
• Mask: (/20) 255.255.240.0
First non-255 number is 240
256 – 240 = 16
16 / 8 = 2 (that’s the magic number)
Numerical Calculations
On the 4th octet
On the 3rd octet
On the 3rd octet
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• Required number of subnets 12 (should be really looking at 16)
• Mask: (/25) 255.255.255.128
First non-255 number is 128
256 – 128 = 128
128 / 16 = 8 (that’s the magic number)
• Mask: (/20) 255.255.240.0
First non-255 number is 240
256 – 240 = 16
16 / 16 = 1 (that’s the magic number)
• Mask: (/21) 255.255.248.0
First non-255 number is 248
256 – 248 = 8
8 / 16 = 0.5 * 256 = 128 (that’s the magic number)
Numerical Calculations
On the 4th octet
On the 3rd octet
Started on the 3rd but moved to the 4th octet
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• 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:
Numerical Calculations
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
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• Subtract the magic number from 256. So to continue with the previous example of subnetting 192.168.252.0/22:
Numerical Calculations
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)
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• 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
Numerical Calculations
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• 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
Numerical Calculations
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• 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.
Numerical Calculations
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• 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
Numerical Calculations
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• 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.
Numerical Calculations
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• 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
Numerical Calculations
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• 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!
Numerical Calculations
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)
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• 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)
Numerical Calculations
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• 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.
Numerical Calculations
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)
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Numerical Calculations – section end
Questions?
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• 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
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• 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
Graphical Representation
128
64
32
16
8
4
0
256
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• Subnet 192.168.10.0/24 into:
A (100 hosts), B (20 hosts), C (10 hosts)
All on the 4th octet.
Graphical Representation
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
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Graphical Representation – section end
Questions?
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Part II - Supernetting
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© 2011 Cisco and/or its affiliates. All rights reserved. 29 All webinars are available at http://lms.netacad.net/course/view.php?id=1201
• Using the numerical calculations
• Using the graphical representation
• Actually, these are the reverse methods of those of subnetting.
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• 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.
Numerical Calculations
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• 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
• Number is 16, starts from 16, at 3rd octet. Supernet address is 192.168.16.0 255.255.240.0
• Networks: 192.168.96.0/20, 192.168.112.0/21, 192.168.120.0/22, 192.168.124.0/23
• Number is 32, starts from 96, at 3rd octet. Supernet address is 192.168.96.0 255.255.224.0
Numerical Calculations
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Numerical Calculations – section end
Questions?
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• 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.
Graphical Representation
128
64
32
16
8
4
0
256
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Graphical Representation
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
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Graphical Representation
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
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Graphical Representation – section end
Questions?
Thank you.
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