saeed darvish pazoki – mcse, ccna ip subnetting 1

Saeed Darvish Pazoki – MCSE, CCNA

IP Subnetting

1

Calculate NetworksCalculate Networks

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Subnetting ExamplesNumber of networks

Process:1. Determine The number of Networks and Convert

to Binary2. Reserve Bits in Subnet Mask and Find the

INCREMENT3. Use Increments to find the networks

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Class C Example #1: 198.200.100.0 Mask: 255.255.255.0 Required networks: 5 Binary: 0 0 0 0 0 1 0 1 3 bits Required Mask: 11111111.1111111.11111111.11100000

255.255.255.224 Increments Number: 32 (128 64 32 16 8 4 2 1) Networks:

198.200.100.0 198.200.100.32 198.200.100.64 198.200.100.96 198.200.100.128 198.200.100.160 198.200.100.192 198.200.100.224

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Class C Example #2 : 198.200.100.0 Mask: 255.255.255.0 Required networks: 12 Binary: 0 0 0 0 1 1 0 0 4 bits Required Mask: 11111111.1111111.11111111.11110000


198.200.100.0 198.200.100.16 198.200.100.32 198.200.100.48 . . 198.200.100.208 198.200.100.224 198.200.100.240

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Class C Example #3 : 198.200.100.0 Mask: 255.255.255.0 Required networks: 60 Binary: 0 0 1 1 1 1 1 0 6 bits Required Mask: 11111111.1111111.11111111.11111100


198.200.100.0 198.200.100.4 198.200.100.8 198.200.100.12 . . 198.200.100.244 198.200.100.248 198.200.100.252

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Class B Example #1 : 180.64.0.0 Mask: 255.255.0.0 Required networks: 30 Binary: 0 0 0 1 1 1 1 0 5 bits Required Mask: 11111111.1111111.11111000.00000000


180.64.0.0 180.64.8.0 180.64.16.0 180.64.24.0 . . 180.64.232.0 180.64.240.0 180.64.248.0

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Class B Example #2 : 180.64.0.0 Mask: 255.255.0.0 Required networks: 100 Binary: 0 1 1 0 0 1 0 0 7 bits Required Mask: 11111111.1111111.11111110.00000000


180.64.0.0 180.64.2.0 180.64.4.0 180.64.6.0 . . 180.64.250.0 180.64.252.0 180.64.254.0

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Subnetting Examples Number of networks

Class B Example #3 : 180.64.0.0 Mask: 255.255.0.0 Required networks: 280 Binary: 1 0 0 1 0 1 1 0 0 9 bits Required Mask: 11111111.1111111.11111111.10000000


180.64.0.0 180.64.0.128 180.64.1.0 180.64.1.128 180.64.2.0 180.64.2.128 . 180.64.254.0 180.64.254.128 180.64.255.0 180.64.255.128

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Class A Example #1 : 60.0.0.0 Mask: 255.0.0.0 Required networks: 500 Binary: 1 0 0 1 0 1 1 0 0 9 bits Required Mask: 11111111.1111111.10000000.00000000 255.255.128.0 Increments Number: 128 (128 64 32 16 8 4 2 1) Networks:

60.0.0.0 60.0.128.0 60.1.0.0 60.1.128.0 60.2.0.0 60.2.128.0 . 60.254.0.0 60.254.128.0 60.255.0.0 60.255.128.0

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Class A Example #1 : 60.0.0.0 Mask: 255.0.0.0 Required networks: 1000 Binary: 1 1 1 1 1 0 1 0 0 0 10 bits Required Mask: 11111111.1111111.11000000.00000000 255.255.192.0 Increments Number: 64 (128 64 32 16 8 4 2 1) Networks:

60.0.0.0 60.0.64.0 60.0.128.0 60.0.192.0 60.1.0.0 60.1.64.0 60.1.128.0 60.1.192.0 . 60.255.0.0 60.255.64.0 60.255.128.0 60.255.192.0

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Calculate HostsCalculate Hosts

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Subnetting ExamplesNumber of Hosts

Process:1. Determine The number of Hosts and Convert to

Binary2. Reserve Bits in Subnet Mask and Find the

INCREMENT3. Use Increments to find the networks

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Class C Example #1:198.200.100.0 Mask: 255.255.255.0Required hosts: 50Binary: 0 0 1 1 0 0 1 0 6 bits RequiredMask: 11111111.1111111.11111111.11000000

255.255.255.192Increments Number: 64 (128 64 32 16 8 4 2 1)Networks:

198.200.100.0 (198.200.100.1 – 198.200.100.62)198.200.100.64 (198.200.100.65 – 198.200.100.126)198.200.100.128 (198.200.100.129 – 198.200.100.190)198.200.100.192 (198.200.100.193 – 198.200.100-254)

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Class C Example #1:198.200.100.0 Mask: 255.255.255.0Required hosts: 100Binary: 0 1 1 0 0 1 0 0 7 bits RequiredMask: 11111111.1111111.11111111.10000000

255.255.255.128Increments Number: 128 (128 64 32 16 8 4 2

1)Networks:

198.200.100.0 (198.200.100.1 – 198.200.100.126)198.200.100.128 (198.200.100.129 –

198.200.100.254)

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Class B Example #1: 180.200.0.0 Mask: 255.255.0.0 Required hosts: 100 Binary: 0 1 1 0 0 1 0 0 7 bits Required Mask: 11111111.1111111.11111111.10000000


180.200.0.0 (198.200.0.1 – 198.200.0.126) 180.200.0.128 (198.200.0.129 – 198.200.0.254) 180.200.1.0 (198.200.1.1 – 198.200.1.126) 180.200.1.128 (198.200.1.129 – 198.200.1.254) . . 180.200.253.0 (198.200.253.1 – 198.200.253.126) 180.200.253.128 (198.200.253.129 – 198.200.253.254)

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Class B Example #2:180.200.0.0 Mask: 255.255.0.0Required hosts: 300Binary: 1 0 0 1 0 1 1 0 0 9 bits RequiredMask: 11111111.1111111.11111110.00000000

255.255.254.0 Increments Number: 2 (128 64 32 16 8 4 2 1)Networks:

180.200.0.0 (180.200.0.1 – 180.200.1.254)180.200.2.128 (180.200.0.129 – 180.200.0.254)180.200.4.0 (180.200.1.1 – 180.200.1.126) . .180.200.253.0 (180.200.253.1 – 180.200.253.126)180.200.253.128 (180.200.253.129 –

180.200.253.254)

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Class B Example #2:180.200.0.0 Mask: 255.255.0.0Required hosts: 1500Binary: 1 0 1 1 1 0 1 1 1 0 0 11 bits RequiredMask: 11111111.1111111.11111000.00000000

255.255.248.0 Increments Number: 8 (128 64 32 16 8 4 2 1)Networks:

180.200.0.0 (180.200.0.1 – 180.200.7.254)180.200.8.128 (180.200.8.1 – 180.200.15.254)180.200.16.0 (198.200.16.1 – 198.200.23.254) .180.200.240.0 (180.200.240.1 – 180.200.247.254)180.200.248.0 (180.200.248.1 – 180.200.255.254)

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Reverse Calculations!Reverse Calculations!

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Subnetting ExamplesFinding the network of a given IP Address

and Mask:192.168.30.179/28

179 = 101100114bit = 11110000And Result = 10110000 = 176 192.168.30.176/28

192.168.30.115/29115 = 011100115bit = 11111000And Result = 01110000 = 112 192.168.30.112/29

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saeed darvish pazoki – mcse, ccna ip subnetting 1

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