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Subnetting: Network with six subnets

Posted on 2016-09-04
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Last Modified: 2016-09-07
I'm reading "Windows Networking Essential" and my question is about subnetting.
I understand that in order to create subnets, I need to borrow from the host bit of the address.
In the book, it says that

...you can use the following formula to determine how many subnets you can create based on how many bits you borrow: 2n, where n is the number of bits you borrow.

So if I I want to create four subnets, I borrow two bits, if I borrow three, I get eight subnets etc.
But how do I create, for example, six subnets?

For example, with the IP address 192.168.50.0 how is it possible to create six subnets?

To create four subnets, I borrow two bits:
11000000 10101000  00110010  00000000

What would you do to create six subnets?

Another important question is: If no network mask is given, can I assume that it's 255.255.255.0 ?
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Question by:mscola
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Qlemo earned 250 total points
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You need to round up the available subnets to 2^n. 6 subnets require 3 bits = 8 subnets. with two subnets unused. With an original netmask of /24 each of those 6 subnets will only be able to hold 2^5(-2) hosts, as if you had 8 subnets. There is no way to change that.

If the netmask is not provided, you'll have to assume the traditional classes A (10./8), B (172.16..0.0/16-172.31.0.0/16) or C (192.168.x.0/24).
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by:Bing CISM / CISSP
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>> To create four subnets, I borrow two bits

this is for 4 same-size subnets, each has 32 IPs (30 workable IPs available).

>> with the IP address 192.168.50.0 how is it possible to create six subnets

you may simply "borrow" one more bit for 2 of the 4 above subnrts, then 6 subnrts will be available in the original class C space, 32 IPs for two subnets (mask 224), and 16 IPs for four subnets (mask 240).

> 6 subnets require 3 bits = 8 subnets. with two subnets unused.

leaving two unused subnets is not necessary, the unused subnets can be merged into two of existing eight subnets.
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by:mscola
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Then the netmask changes from
11111111 11111111 11111111 00000000 to
11111111 11111111 11111111 11100000 =  255.255.255.224

When I apply it to the IP address (in order to determine the network ID)

IP Address: 11000000  10101000  00110010  00000000
Netmask:    11111111  11111111  11111111  11100000
ID:                11000000  10101000  00110010  00000000

I get 192.168.50.0 as the network ID.  Is that correct?
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by:Bing CISM / CISSP
Bing CISM / CISSP earned 250 total points
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> I get 192.168.50.0 as the network ID.  Is that correct?

correct, for the 1ST subnet having 32 IPs (30 workable).
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by:Qlemo
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and the next one is 192.168.50.32, .64, .96 aso.
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by:mscola
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For the sake of completeness (if following my book) .. so it looks right.


1  000            192.168.50.0                    192.168.50.1 – 192.168.50.30
2  001            192.168.50.32            192.168.50.33 – 192.168.50.62
3  010            192.168.50.64            192.168.50.65 – 192.168.50.94
4  011            192.168.50.96            192.168.50.97 – 192.168.50.126
5  100            192.168.50.128            192.168.50.129 – 192.168.50.158
6  101            192.168.50.160            192.168.50.161 – 192.168.50.190
7  110            192.168.50.192            192.168.50.193 – 192.168.50.222
8  111            192.168.50.224            192.168.50.225 – 192.168.50.254

Bing, you wrote that ,

the unused subnets can be merged into two of existing eight subnets.

I assume this is something more advanced?

Is it like setting the default gateway of network 8 to the subnet of network 7 for example?
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by:Qlemo
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Bing's suggestion using exactly 6 subnets with an uneven split works, but is not recommended. It increases the complexity, and is prone to confusion and misconfiguration. You usually only do that if it is absolutely necessary.

If you want to do uneven splitting, you should go exactly as Bing stated: Use 4 subnets, then split two of them further. This makes sure you use correct subnets and masks, If you go the other way round you have to pay attention to combine the correct ones. E.g. you cannot combine .32 and .64, but .64 and .96.

Whatever, the result would be two /26 networks and four /27. Each one needs a gateway in the same network as the own.
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by:mscola
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Thanks a lot for your help!
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