hi! i don't know how to get the increments. Can you please help me how to solve these:?


A ___ B ___ C
( X ) z___ ( X ) Z___ ( X )
| | |
| | |
| | |
| | |
[] [] [_]
[___] [___] [___]

create 16 subnets

create 16 subnets

create 500 subnets

***Thank you very kindly.

Can you elaborate on what you mean by "don't know how to get the increments"? What part of subnetting are you having an issue with? Binary conversion, formulas?

I have a tutorial regarding CIDR and Subnetting that may help you better understand how subnetting works.

I also have an online subnetting calculator that you can use to check your work.

If this is a home work assignment, it is best that you attempt to do the work and ask specific questions about what you need help with.

Hi Loid123,
Please elaborate. I for one cannot make out that drawing of A_B_C. But from what I was able to get from your question, if you want to create 16 subnets in A, you'll need to borrow 5 bits from the host portion of network So the first subnet will be with 6 usable addresses starting at and ending at The b/cast address for this network is For network B, to create 16 subnets you'll do the same except this time, the host part begins at the 26th bit, so you'll borrow bit 26,27,28,29 and 30 to be able to get 16 subnets each with 2 usable host addresses. The procedure is the same for network C. The idea is to find a power of 2 that is close but of course greater than the number of subnets you are asked to create. So for 500, the closest power of 2 but greater than 500 is 2^9 which translates to 512, so you'll need to borrow 9 bits from the host portion of the existing network.

Hi JorgeM,
Thank you very much for the links.. :)
I'm having problem getting the increments of the IP range... :(
I'll try to go over with the link you provided...:) thank you...

Hi c-tech,

I'm sorry for my illustration. I did it using my phone and didn't realize it will look like this after I made the draft.. :(
But thank you very kindly for your help. I find subnetting and ip addressing difficult. Hopefully I can solve problems of these kind as quick as you can someday... :)