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Calculating subnet mask
Hello,
I would like to link my network to the one next door. Currently, he has the address range 192.168.30.1-254 and I have the address range 192.168.0.1-254.
What subnet mask would we need to use to allow these two networks to "talk" to each other? I have tried reading up on subnets but I am struggling to understand it.
Thanks,
Jonathan
I would like to link my network to the one next door. Currently, he has the address range 192.168.30.1-254 and I have the address range 192.168.0.1-254.
What subnet mask would we need to use to allow these two networks to "talk" to each other? I have tried reading up on subnets but I am struggling to understand it.
Thanks,
Jonathan
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Don't worry about classes - we now have CIDR (and, if you insist. it's still class C anyway, as the above mask constructs 32 class C networks containing 8192 addresses each).
Please read this -
http://en.wikipedia.org/wi ki/Classle ss_Inter-D omain_Rout ing
You'll have to adjust the netmask of your server, of course.
Please read this -
http://en.wikipedia.org/wi
You'll have to adjust the netmask of your server, of course.
Oops, made a typo myself - forget the 'each' above! The phrase was planned to look quite different when I began to write it down.
ASKER
Thanks for that information. I will change all of the subnets then.
Thank you all for your help.
Jonathan
Thank you all for your help.
Jonathan
ASKER
Put a router in-between the networks to route the traffic - I would not know how to configure this though.
Add a second IP address
Change them all to use 192.168.0.x or 1092.168.0.30
Change them to use 255.255.224.0 as a subnet mask.
From what I understand, 255.255.224.0 is a class B subnet mask. If I changed the subnet to this but kept the same class C IP addresses, would there be any downside?
(I don't know if it makes a difference but we actually have a server in-between the two networks with a wireless connection to my network and a wired connection to my friends network, these two connections have then been bridged on the server.)