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subneting 10.12.0.0/14

Posted on 2013-12-10
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Last Modified: 2014-04-08
Hello experts,

I have a 10.12.0.0/14 network and I'd like to know the following:
- How many subnet I can have within the /14 subnet mask?
- I plan to have 10 VLANs and some /30 links. What the best practice to subnet 10.12.0.0

Thanks
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Question by:leblanc
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tmoore1962 earned 100 total points
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should be able to have 1024 subnets w/16382 host per sn
10.12.0.1 - 10.12.63.254 for ex of 1 host range
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by:leblanc
ID: 39709628
10.12.0.0/14 gives me 262142 host IP addresses. Now is there a tool that I can use to help me to see how many subnets I can have? How did you come up with 1024 subnets w/ 16382 hosts? Thanks
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by:gt2847c
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You could have 16 subnets with 16,382 hosts, but that's a lot of addresses in a single broadcast domain.  Best practice would be to size the network appropriate to what it is you are doing.  Also consider the hardware limitations for your network equipment (some switches have limits on the number of MAC entries per VLAN or per switch).

Best would be to size or segment the network based on usage and need (per floor, per department, etc) as makes sense.  Then size the network with some growth available.
Networks larger than a /22 (1024) can start to get unwieldy.  Reserve the unused space for other things you may eventually need.  No reason to chunk it into huge pieces and not use the space...

For the /30 links, chunk out a piece of the beginning or the end of the range and reserve it for your links.
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by:leblanc
ID: 39710002
I understand that but let say I have given 10.12.0.0/14 to use for my headquarter. I need 10 VLANs. So is there a tool or techniques I can use to assist me in breaking up my block of 10.12.0.0/14? I would like to limit my host to 500 as Cisco recommends. Thanks
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by:Fred Marshall
Fred Marshall earned 200 total points
ID: 39710013
I have a 10.12.0.0/14 network and I'd like to know the following:
- How many subnet I can have within the /14 subnet mask?
To answer your question literally, you can have subnets with CIDR of /30.
/14 has 262,142 possible host addresses.
/30 has 2 possible host addresses
The difference between /14 and /30 is 17 if you count the end points.
2^(32-14) = 262,144 less the end points for network and broadcast is 262,142
2^(32-15) = 131,072 less the end points for network and broadcast is 131,070.
..
..
2^(32-30) = 4 less the end points for network and broadcast is 2.

So, I would divide 262,144 (the size of the one big subnet) by 4 (the size of the smallest possible subnet) to get 65,536 subnets of 4 with 2 host addresses each.  That's how many subnets you *could* have .. MAX.

- I plan to have 10 VLANs and some /30 links. What the best practice to subnet 10.12.0.0
Well, 10.20.0.0 as the initial network address has 16,382 addresses in total.
I would divide by 10 to get 1638 and then go down to the nearest power of 2 which is 1,024 or CIDR /22.  Ten of these yields 10,240 out of the 16,382  I would then add at least a couple for margin (new VLANs) for a total of12,288.  This leaves 4094 which, again, is enough for some /30 subnets and another couple of VLANs beyond the margin above.
At this point it's a matter of what somebody calls "good practice".  Working it out this way so you understand what you've done and the implications of it would be the ultimate "good practice".  After that one can quibble about numbers.
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by:Infamus
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So is there a tool or techniques I can use to assist me in breaking up my block of 10.12.0.0/14?
http://www.subnet-calculator.com/
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by:leblanc
ID: 39713190
Yes I tried that but this is what I am looking for:
- How many /23 do I have in 10.0.0.0/14?
- How many /24 do I have in 10.0.0./14?

I use the subnet-calculator. But it just listed all subnets of 10.0.0.0/14. It does not have an option where I can see the number if subnet within another subnets. When I get a block of /14 for example, I'd like to be able to see, before I subnet it, the number of future subnets.

Thanks
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by:Fred Marshall
Fred Marshall earned 200 total points
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- How many /23 do I have in 10.0.0.0/14?
- How many /24 do I have in 10.0.0./14?
I tried to show how to do that:

/23 means there are 23 bits in the network address and 32-23=9 bits in the subnet address space.  2^9 = 512.
/24 means there are 24 bis in the network address and 32-24=8 bits in the subnet address space.  2^8=256.

That's a starting point.
/14 means there are 14 bits in the network address and 32-14=18 bits in the subnet address space. 2^14=262,144 addresses.

So, you ask how many /23 in /14?
262,144  / 512 = 512 subnets

how many /24 in /14?
262,144 / 256= 1,024 subnets.

If you are comfortable working with powers of 2 then you might notice this:

262,144 / 512 = 2^18 / 2^9 = 2^(18-9) = 2^9 = 512 subnets
So, the number of subnets of some CIDR value B in a larger subnet of CIDR value A would be:

2^[(32-A)-(32-B)] = 2^(B-A)  So, in this case 2^(24-14) = 2^10 = 1,024 subnets - as above.

That answers "how many subnets of  /B in a subnet of /A?"
It is 2^(B-A) subnets.
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by:leblanc
ID: 39715418
Got it. I just realized that you have to look at the number of bits and find it that way. The powers of 2 solution looks easier. Thank you for the solution.
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