Edit - I should have said, "counting DOWN down in the second chart gives you a subnet mask of 240".
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Subnetwork id
Network 192.168.17.0 is to be subnetted into 12.Find subnetwork id of second subnetwork.Calculate maximum no of hosts per subnet
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by: pseudocyberPosted on 2004-03-16 at 06:44:33ID: 10606527
From another post of mine:
First learn the powers of 2
1 2
2 4
3 8
4 16
5 32
6 64
7 128
8 256
You can keep going up, but you at least need to learn the first 8.
Then, learn counting in binary, backwards from 10000000
8 10000000 128
7 11000000 192
6 11100000 224
5 11110000 240
4 11111000 248
3 11111100 252
2 11111110 254
1 11111111 255
Your problem:
192.168.17.0 is to be subnetted into 12.Find subnetwork id of second subnetwork.Calculate maximum no of hosts per subnet
A 192.168.17.0 network is naturally a class C with a mask of 255.255.255.0. You want 12 subnetworks, so you count down with the first chart to find the first number > 12. Counting down to 4 gives you 2^4-2 which is 14. If you went with less than that, 8, you wouldn't have enough subnetworks. So, that tells you you have to have 4 bits in the subnetwork portion, leaving you 4 bits in the host portion. Counting up in the second chart gives you a subnet mask of 240.
Now, a neat trick is to take the subnet mask octet (240) and subtract it from 256. 256-240=16. This means your networks will start at 0 (if you're using subnet 0) and continue every 16 decimal numbers (0, 16, 32, 48, etc) So, you want the subnetwork ID of the second subnetwork - the first would be 0, and the second would be .16. Therefore, it's 192.168.17.16 with a mask of 255.255.255.240.
As I said there are 4 host bits available - 2^4-2=14 hosts.
HTH!