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A service provider has given you the Class C network range 209.50.1.0. Your company must break

the network into 20 separate subnets.

Step 1) Determine the number of subnets and convert to binary

- In this example, the binary representation of 20 = 00010100.

Step 2) Reserve required bits in subnet mask and find incremental value

- The binary value of 20 subnets tells us that we need at least 5 network bits to satisfy this requirement

(since you cannot get the number 20 with any less than 5 bits -- 10100)

- Our original subnet mask is 255.255.255.0 (Class C subnet)

- The full binary representation of the subnet mask is as follows:

255.255.255.0 = 11111111.11111111.11111111.00000000

- We must "convert" 5 of the client bits (0) to network bits (1) in order to satisfy the requirements:

New Mask = 11111111.11111111.11111111.11111000

- If we convert the mask back to decimal, we now have the subnet mask that will be used on all the new

networks -- 255.255.255.248

- Our increment bit is the last possible network bit, converted back to a binary number:

New Mask = 11111111.11111111.11111111.1111(1)000 -- bit with the parenthesis is your increment

bit. If you convert this bit to a decimal number, it becomes the number "8‟

Step 3) Use increment to find network ranges

- Start with your given network address and add your increment to the subnetted octet:

209.50.1.0

209.50.1.8

209.50.1.16

...etc

- You can now fill in your end ranges, which is the last possible IP address before you start the next range

209.50.1.0 -- 209.50.1.7

209.50.1.8 -- 209.50.1.15

209.50.1.16 -- 209.50.1.23

...etc

- You can then assign these ranges to your networks! Remember the first and last address from each

range (network / broadcast IP) are unusable

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12y ago

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