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# Netmask

What would the netmask be for each of three subnets using a fraction part of a block of ip numbers such as 192.168.10.nnn?
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jude53
Asked:
1 Solution

Commented:
255.255.255.0
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Commented:
Oh well, JBurghardt beat me to answering the question ... what he said is exactly right, 255.255.255.0 is the netmask for the situation you describe.
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Commented:
192.168.10.xxx would be a class C network, which means that the first three octets are fixed and the last one is variable.
Yes, indeed, for a class C subnet, the netmask is 255.255.255.0
A class B network, for example, would have a variable part of two octets, hence a subnet mask of 255.255.0.0

Cheers,
schmox
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Author Commented:
Thanks to all of you for your attention but,
I understand the netmask for a class c network. What I needed to know is, what would be the netmask for three subnets set up as a portion of the original host and what would their ip addresses be given my first example?
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Commented:
You have only 254 addressed to subnet, If you choose 255.255.255.252 as you subnet mask,  that means, the addresses 192.168.10.254,192.168.10.253 and will be the same subnet.

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Commented:
OK, got it now. So you don't want the whole 192.168.10.nnn class C network, but you want to divide it into smaller subnets, right?
First of all, you can divide the last octet to get 2, 6, 14, 30 or 62 subnets. If you need 3 subnets, you'd have to take 6. Your subnet mask would be 255.255.255.224 and you would have 30 hosts per subnet. Specifically, the subnets would be:

1.
subnet address: 192.168.10.0
netmask: 255.255.255.224
hosts: 192.168.10.1 through 192.168.10.30
broadcast address: 192.168.10.31
2.
subnet address: 192.168.10.32
netmask: 255.255.255.224 (same for all subnets)
hosts: 192.168.10.33 through 192.168.10.62
broadcast address: 192.168.10.63
3.
and so on...

For more calculations, you need a Java-capable browser and the following URL:
http://www.ccci.com/subcalc/subcalc.htm

Guess that was the answer...

Cheers,
Schmox
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