Which IP host address range is allowable given an IP address of 184.108.40.206 using 12 bits of subnetting?
Answer 1 (Sticks and stones):
170 . 73 . 3 . 00111000
255 . 255.255. 11110000
170 . 73 . 3 . 00110000
Subnet Range ::: 220.127.116.11 à 18.104.22.168
Useable Numbers ::: 22.214.171.124 à 126.96.36.199
Because 170 is a Class-B address range, given to us by ARIN, we have 16 bits for our Network ID. We can’t modify that number at all, but we have 65535 host addresses available to work with. By borrowing 12 of those host bits, we can get 4094 usable subnets with 14 usable hosts each subnet. That gives us the CIDR notation of 188.8.131.52/28 for this ip address. Or we could convert this to Dotted Decimal notation: IP Address: 184.108.40.206, Subnet Mask 255.255.255.240. This is the first step in setting up the problem. If we miss this first step, the rest of our charts and formulas will not work.
If we know that the interesting octet has a 240 subnet mask, then we can assume the first usable subnet begins with 16. How do we know this?
1. After we build Frank and E.J.’s chart, we can cross-reference 240, find our Address Start Block is 16,
2. Or we can determine through sticks and stones that the least significant digit in a 240 subnet mask is the fourth from the left, 111(1)0000, which converts to decimal 16. If we want to make the smallest change possible to our Network ID, we can change the digit worth a decimal 16.
3. Or we could subtract our Subnet Mask from 256 and get 16.(256-240=16)
So our subnets start at:
Our rules tell us that our usable host addresses cannot contain the first number or the last number in the subnet, because the first is reserved for the network number, and the last is our broadcast address.
Our usable hosts in the ranges we mapped out above would be:
Host ranges: 1-14,17-30,33-46,49-62,65-78,81-94
For this problem, 56 will fall into subnet 48 (220.127.116.11), with usable hosts from 49-62, and the broadcast address of 18.104.22.168.