Fast and Easy Subnetting. Marc Khayat, Technical Advocacy Team makhayat@cisco.com May 2011



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Transcription:

Fast and Easy Subnetting Marc Khayat, Technical Advocacy Team makhayat@cisco.com May 2011

Number of required subnets Number of required hosts per subnet 2011 Cisco and/or its affiliates. All rights reserved. 2

Using the numerical calculations Using the graphical representation 2011 Cisco and/or its affiliates. All rights reserved. 3

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 should 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 2011 Cisco and/or its affiliates. All rights reserved. 4

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! 2011 Cisco and/or its affiliates. All rights reserved. 5

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) On the 4 th octet On the 3 rd octet On the 3 rd octet 2011 Cisco and/or its affiliates. All rights reserved. 6

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) On the 4 th octet On the 3 rd octet Mask: (/21) 255.255.248.0 First non-255 number is 248 Started on the 3 rd but moved to the 4 th octet 256 248 = 8 8 / 16 = 0.5 * 256 = 128 (that s the magic number) 2011 Cisco and/or its affiliates. All rights reserved. 7

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 2011 Cisco and/or its affiliates. All rights reserved. 8

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 (3 rd octet) Magic Nb = 1 (3 rd octet) Magic Nb = 128 (4 th 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) 2011 Cisco and/or its affiliates. All rights reserved. 9

With a subnet mask of 255.255.255.224, all network IDs would be a multiple of (256 224 = 32), 4 th 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), 3 rd octet, so x.x.0.0,.128.0 2011 Cisco and/or its affiliates. All rights reserved. 10

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 3 rd 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 2011 Cisco and/or its affiliates. All rights reserved. 11

1 st 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. 2011 Cisco and/or its affiliates. All rights reserved. 12

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 3 rd 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 2011 Cisco and/or its affiliates. All rights reserved. 13

123.45.164.255/22: network, broadcast or valid host? If broadcast or valid host address, what is the network address? 100.198.7.64/18: network, broadcast or valid host? If broadcast or valid host address, what is the network address? 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. 2011 Cisco and/or its affiliates. All rights reserved. 14

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! 2011 Cisco and/or its affiliates. All rights reserved. 15

192.168.1.0/24 to be subnetted into A (100 hosts), B (40 hosts), C (10 hosts), D (2 hosts). Subnet Nb of Hosts Net Size Magic Nb Mask (4 th 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) Remember to always start with the largest network first! 2011 Cisco and/or its affiliates. All rights reserved. 16

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) 2011 Cisco and/or its affiliates. All rights reserved. 17

192.168.160.0/19 to be subnetted into A (2000 hosts), B (500 hosts), C (100 hosts), D (4 hosts). Subnet Nb of Hosts Net Size Magic Nb Mask Prefix Subnet address A 2000 2048 2048/256=8 256-8=248 (3 rd octet) B 500 512 512/256=2 256-2=254 (3 rd octet) C 100 128 128 256-128=128 (4 th octet) D 4 8 8 256-8=248 (4 th octet) /21 192.168.160.0 /23 192.168.(160+8=168).0 /25 192.168.(168+2=170).0 /29 192.168.170.(0+128) Free addresses: 192.168.170.(128+8) 192.168.191.255. 2011 Cisco and/or its affiliates. All rights reserved. 18

Magic Number section end Questions? 2011 Cisco and/or its affiliates. All rights reserved. 19

Graphical Representation 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 Step 1: identify your magic numbers (already know how to do it) Step 2: Segment and reserve networks on the bargraph starting with the largest network first. Step 3: You re done! 2011 Cisco and/or its affiliates. All rights reserved. 20

Graphical Representation 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 needs 0 128 8 64 256 4 32 16 2011 Cisco and/or its affiliates. All rights reserved. 21

Graphical Representation 0 Subnet 192.168.10.0/24 into: A (100 hosts), B (20 hosts), C (10 hosts) A (128) 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 Width of 256 This represents the 4 th octet Width of 32 Width of 16 Width of 128 Width of 64 128 160 176 B (32) C (16) FREE Addresses 255 2011 Cisco and/or its affiliates. All rights reserved. 22

Graphical representation section end Questions? 2011 Cisco and/or its affiliates. All rights reserved. 23

Thank you.