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Subnet - Subnetting - classless address block

Posted on 2011-02-16
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Last Modified: 2012-08-14
Hello!
I have an address block of 192.168.0.0/24
I have to get a smaller block so I can get 6 subnets out of it.

I think 192.168.0.0/28 can be that block? so what will be the first subnet ip (network) address?

Is there an easy way to subnet the subnet, to use the address block more efficiently?

What is a gateway, is it the router address?

Last how do I create a routing table for a router with more than one address to get to a host?
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Question by:dev2030
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Sommerblink earned 500 total points
ID: 34912372
Well. We have a problem

You specified how many subnets you need, but not how many addresses per subnet are required.

So, to answer your actual question, I'm taking a liberty here.


Subnet      Mask      Inverse Mask      Subnet Size      Host Range      Broadcast
192.168.0.0      255.255.255.192      0.0.0.63      62      192.168.0.1  to  192.168.0.62      192.168.0.63
192.168.0.64      255.255.255.192      0.0.0.63      62      192.168.0.65  to  192.168.0.126      192.168.0.127
192.168.0.128      255.255.255.192      0.0.0.63      62      192.168.0.129  to  192.168.0.190      192.168.0.191
192.168.0.192      255.255.255.192      0.0.0.63      62      192.168.0.193  to  192.168.0.254      192.168.0.255



Or you can do 8 subnets:

Subnet      Mask      Inverse Mask      Subnet Size      Host Range      Broadcast
192.168.0.0      255.255.255.224      0.0.0.31      30      192.168.0.1  to  192.168.0.30      192.168.0.31
192.168.0.32      255.255.255.224      0.0.0.31      30      192.168.0.33  to  192.168.0.62      192.168.0.63
192.168.0.64      255.255.255.224      0.0.0.31      30      192.168.0.65  to  192.168.0.94      192.168.0.95
192.168.0.96      255.255.255.224      0.0.0.31      30      192.168.0.97  to  192.168.0.126      192.168.0.127
192.168.0.128      255.255.255.224      0.0.0.31      30      192.168.0.129  to  192.168.0.158      192.168.0.159
192.168.0.160      255.255.255.224      0.0.0.31      30      192.168.0.161  to  192.168.0.190      192.168.0.191
192.168.0.192      255.255.255.224      0.0.0.31      30      192.168.0.193  to  192.168.0.222      192.168.0.223
192.168.0.224      255.255.255.224      0.0.0.31      30      192.168.0.225  to  192.168.0.254      192.168.0.255

Your router, can be any IP address within each range. It really doesn't matter.
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Expert Comment

by:Sommerblink
ID: 34912394
As far as routing between this goes:

You will need one more subnet. This subnet will contain all the 'WAN' interfaces for your 6 routers. (Perhaps take one of those two spare subnets from the /27)

The 6 routers will need to know the subnets for your remaining 5 routers.


This can be done either via static entries or you can use RIP.
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Author Comment

by:dev2030
ID: 34912578
Thank you for your answer. Can you subnet the first subnet and create 6 subnets. Each subnet have not more than 4 hosts. Sorry for not mentioning it, but thank you for complete reply, really appreciate it:)
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Expert Comment

by:Sommerblink
ID: 34912819
Unfortunately no.

With your /24, you can either subnet that to a /25, which will have two subnets. Or a /26 which will contain 4 subnets or a /27 which will have 8 subnets. (a /28 will have 16 subnets, within a /24, and so on and so forth).

This is just an byproduct of binary math..

But as I said originally, there are two and only two things which confine you with subnetting:

1) The number of subnets you need and ;
2) The number of host / subnet.

You must factor in the largest of BOTH, in order to successfully divide your network.

In this case, you have just 3 options:
The largest subnet you can go with is a /27 (provides 8 subnets with 32 address [30 usable])
/28 provides 16 subnets with 16 addresses (14 usable)
The most subnets you can go with is a /29 (provides 32 subnets with 8 addresses [6 usable])

Of course you can go with non-symetric subnetting and get many more combinations...
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Author Comment

by:dev2030
ID: 34913010
Thank u for explaining this in such detail and making it easier to understand.

So when it comes down to it, which is better use of address block, getting lower number of subnets, or higher. I think lower number will give us lower Masks bit but at the same time it will be better use of address space. AM I RIGHT HERE:)?

For example, 32 subnets will give the right amount of hosts. But we need only 6 subnet. So if we use mask 27, is it better than using 32, I mean it saves address space right?
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Assisted Solution

by:Sommerblink
Sommerblink earned 500 total points
ID: 34913157
Heh... as they say, the devil is in the details.

The question here is really, what is your growth: Will your growth be in the number of hosts per subnets or in the number of subnets you will need.

And when I speak of growth, think longer-term... the next 5 years or so. And its PERFECTLY OK to over-estimate. In fact, it's encouraged. :)

Now, to be perfectly honest with you.. you are using non-routable (RFC 1918 http://www.faqs.org/rfcs/rfc1918.html) IP space.

Within the 192.168.0.0/16 (basically from 192.168.0.0 - 192.168.255.255), you do have a rather large chunk of IP space available to you [of course, you may have other constraints that I am unaware of].

This means you don't necessarly have to be stingy with your subnets within this space. What is important is you want to make sure that the subnet which binds all your routers 'WAN' addresses within the same subnet. While this isn't a necessity, it will help.

Also, as I said, you have this wide expanse of addresses available. There is no rule that says they need to be next to each other or that they need to be the same size.

I do hope that this helps you
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Author Comment

by:dev2030
ID: 34913360
Thanks a lot, great help:) REALLY AWESOME:)

Thank you!!!!
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