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Finding Solution to IP/Subnetting
19 years 6 months ago #8218
by decolnz
#4ever n 4always#
Finding Solution to IP/Subnetting was created by decolnz
I have been given an IP address 192.168.2.0 and as a networking admin they want me to assign the IP addresses to the following. There is a diagram being given, but I have failed to attach it.
On the diagram there are 3 routers A,B, and connected to each other B being in the middle and on each router there are LAN 1 on router A, LAN 2 on router B and LAN 3 on router C. So please may you help me out with the answer
On the diagram there are 3 routers A,B, and connected to each other B being in the middle and on each router there are LAN 1 on router A, LAN 2 on router B and LAN 3 on router C. So please may you help me out with the answer
#4ever n 4always#
19 years 6 months ago #8240
by MaXiMuS
Replied by MaXiMuS on topic Re: Finding Solution to IP/Subnetting
Hi decolnz,
I think it would be better if you could attach the diagram of your network, after replacing your live ip's of course.
192.168.2.0 is a network add. and not an ip address.
If you are going to connect 3 routers , with a lan to each , you would require a minimum of 4 network addresses.
I think it would be better if you could attach the diagram of your network, after replacing your live ip's of course.
192.168.2.0 is a network add. and not an ip address.
If you are going to connect 3 routers , with a lan to each , you would require a minimum of 4 network addresses.
19 years 6 months ago #8269
by mew
Replied by mew on topic Re: Finding Solution to IP/Subnetting
If you meant Routers A,B, and C then there are 5 networks. Two WAN links and three LAN's. It also looks a lot like a Cisco Lab problem.
Ask yourself how many binary bits does it take to count to 5. Then borrow that many host bits to be used for subnet bits.
If you still have more questions I have more answers but I won't do the work for you.
Ask yourself how many binary bits does it take to count to 5. Then borrow that many host bits to be used for subnet bits.
If you still have more questions I have more answers but I won't do the work for you.
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