# Subnet Assignment

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CMIT 265 Subnetting Assignment

Subnetting Assignment:
215.251.145.0/24 subnetted to 46 subnets and provide information for subnets #1, #4, #5, and #46

Step 1: Convert the network address to binary
215.251.145.0/24 is a class C address with a subnet mask of 24 bits of ones
First Octet
128.64.32.16.8.4.2.1, 128=1, 64=1, 32=0, 16=1, 8=0, 4=1, 2=1, 1=1
215 = 11010111.
Second Octet = 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=0, 2=1, 1=1
251 = 11111011
Third Octet = 128.64.32.16.8.4.2.1, 128=1, 64=0, 32=0, 16=1, 8=0, 4=0, 2=0, 1=1
145 = 10010001
Fourth Octet = 0

Step 2: Convert Subnet mask into binary
First Octet
128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=1, 2=1, 1=1
255 = 11111111
Second Octet
128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=1, 2=1, 1=1
255 = 11111111
Third Octet
128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=1, 2=1, 1=1
255 = 11111111
Forth Octet = 0
Step 3: Calculate the subnets required
46 subnets required per instructions: Use the formula 2 to the power of X to find out how many bits to borrow from the subnet mask.
2^5= 32 and is insufficient to cover the subnets required so, 2^6 = 64 x is 6 we need 6 bits from last octet of the subnet mask to cover our subnets. I need 46 subnets but I have total 64 available so that leaves me with; 64-46 = 18 extra subnets to use at a later date.
11010111.11111011.10010001.00000000
Replace subnet bits with the subnet number
1, in binary is 1
Make sure it is 6 bits: 000001 = new subnet bits
11010111.11111011.10010001.00000100
New Network Address for subnet #1 is: 215.251.145.4
Change the subnet mask to reflect the bits taken:

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