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## Forum Replies

1. Wonderful explanation. thanks

2. Hi Rene
When creating this blocks, do they have to be of equal size?

For example, you have an address of 10.10.10.0/23, and I am working to create a network/mask for the below requirements

I. 24 hosts
II. 111 hosts
III. 47 hosts
IV. 200 hosts

I came up with this , can you let me know if this is correct

I. 10.10.10.0/27
II. 10.10.10.128/25
III. 10.10.10.64/26
IV. 10.10.11.0/24

If I did this way, is this correct?

I. 10.10.10.192/27
II. 10.10.11.64/25
III. 10.10.11.0/26
IV. 10.10.10.0/24

Thanks
Palani

3. Palani,
Short answer: Everything you have is correct. For any given subnet mask, the block size is always the same.

People do subnetting many different ways. I haven’t heard of others doing it my way, but it works for me. I thought I might share with you how I can do subnetting very fast in my head–no calculators or even pencil and paper required.

I use the formula 2^X >= Y, which reads as 2 to the Xth power is greater than or equal to Y. Y is the number of hosts (or subnets) you are trying to figure out, and X is the variable you are trying to solve for. Let’s walk through one of your examples: 47 hosts. In this case we know that Y = 47. So now we have to figure out what power of 2 is as close to 47 as possible, but cannot be less. Two to the fifth power is 32, so that’s too small. How about 6? Two to the sixth power is 64–so that’s our answer: X = 6.

So what does X do for you? A lot! Once you solve for X, you can easily figure out two things:

1. The network block size (the range of each subnet): This is simply 2^X, which in our case is 64. This means that your subnet ranges would be…
0 - 63
64 - 127
128 - 191
192 - 255
Notice how on the left column above, I just increase the number by the block size? Easy!

2. The subnet mask. You will have to know which octet you are trying to figure out (in this case it is the last octet). To figure out the mask, simply subtract 2^X from 256–so in this case it would be: 256 - (2^6) = 256 - 64 = 192. Now you know the subnet mask will be 255.255.255.192. Easy!

4. Use the formula (2^X) - 2 >= Y

5. Guys, I need help. Every time I try getting into different topics, be it NAT, Routing Policy, etc - I always get back to Subnetting. I finally realized there’s no way to progress without mastering this subject. I google everywhere, but I need your help and hopefully I’ll finally figure this out. Please, show me how would you answer this question;

Write a standard ACL that will cover the host range from 192.168.100.128 through 192.168.100.131.

Let’s not worry about the ACL part, but how would you go about writing an IP address with a subnet (wildcard) mask that would cover the need for 3 usable hosts?

You’d really help me out a lot.