Subnetting with host requirements.
Sample 1:
Required Host per subnet - 30
address - 192.168.1.0
Step 1. Create the bin table
| 128 | 192 | 224 | 240 | 248 | 252 | 254 | 255 |
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
Step 2. Convert the host requirement to binary and count how many bits which means how many 1's and 0's
30 = 11100 = 5 bits(or three 1's plus two 0's)
Step 3. Identify the class of address that is given in this case its class c (192.168.1.0)
Step 4. Use this formula to get your actual netmask [24 + (8 - Number of bits)] = netmask note: for class c only
24 + (8-5) = 27
From the original netmask of /24 you have now /27 which is 255.255.255.224
Step 5. Find your increment
To get the increment simply refer your actual netmask to the bin table.
Remember you only need to refer the last octet which .224 because you are subnetting using class c private address. The increment is just below the 224 of the bin table which is 32.
Step 6. Get your network addresses ranges using the increment.
Address is 192.168.1.0
Increment is 32
1st address range 192.168.1.0 - 31
2nd address range 192.168.1.32 - 63
3rd address range 192.168.1.64 - 95
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Last address range 192.168.1.224 - 255
Next will be using class b address... check it out!
Thanks for droping by
