Subnet mask 255.255.255.224
your subnet : 255.255.255.224 is binary
11111111.11111111.11111111.11100000
so you have 5 host bits available to distribute for IP adresses. Then you do 2^5(squared) because a bit can be 1 or 0, so you get 30 IP's on this subnet. (actually 32, but you can't use all 0's and all 1's)
Because you have 3 subnet bits, you have 2^3 (squared) subnets so 8 subnets
Subnets :
192.20.2.0
192.20.2.32
192.20.2.64
192.20.2.96
192.20.2.128 -> the range you are in, but because you can't have all 1's and all 0's, this is 130 for the ip adress
192.20.2.160
192.20.2.192
192.20.2.224
(note that the last number is the same as the subnet masks last number)
Host Id ranges :
192.20.2.1 - 192.20.2.30
192.20.2.33 - 192.20.2.62
192.20.2.65 - 192.20.2.94
192.20.2.97 - 192.20.2.126
192.20.2.130 - 192.20.2.158
192.20.2.160 - 192.20.2.190
192.20.2.192 - 192.20.2.224
With thanks to the next user for pointing out m mistake ;-)
To previous answerer: very well explained. I love doing bits & bytes!!
There was just one mistake - it's 2^5 , not 5^2.
So the real number of valid hosts would be:
2^5 - 2 = 30
30
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http://www.subnettingquestions.com/custom/bren/
The second one is completely free.
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