Simple networking setup

You'll need to pick a separate network for the WAN link itself if you haven't already (as it won't divide up nicely if you carve out a /30 of this range), and simply configure 10.10.1.0/25 (255.255.255.128) on the ethernet interface on one side and 10.10.1.128/25 on the ethernet of the other. Static routes or a routing protocol will do fine, I'd prefer the routing protocol, and advertise as above with no summary if you use eigrp or ospf. What specifically are you asking?

Also, the term 'Class C' isn't valid any more. You have a /24.

What exactly is your question?

Chris.

Simply put I want to route every address above 128 on this /24 to a router at a remote site.

In other words, if the primary router is in New York. I want to send all the traffic from .129 - .254 to a router in Los Angeles.

Vilmos

So configure 1.128/25 on the LA router's ethernet, and turn up a routing protocol. Ensure the NY router is also configured with the same routing protocol, and that NY's network is not /24, but is 1.0/25.

Or are you saying you want to route traffic based on source address? Meaning both 0-127 and 128-255 are on the same router, but you want them to be routed differently based on source? If that is the case, google 'policy-based routing' and that will provide all the stuff you need.

Not by source.

I need 0 -127 on Router A and 128 - 254 on Router B. Router A is connect directly to the Internet. Router B is behind it on a WAN link.

Maybe this will help clear it up (not using real addresses.

|----------------------------| | Router A | | 208.3.102.1 | |----------------------------| | | |----------------------------| | Network | |----------------------------| | | |----------------------------| | Router C | | connected to | | remote location | |----------------------------| | WAN | |----------------------------| | Router B | |----------------------------|

All traffic to 208.3.102.0/24 has to come through 208.3.102.1. Router A is the first hop but redirects all traffic for addresses 0 - 128 on the local network. Address 129 - 254 are routed through Router C over a WAN link to Router B. I need all the addresses on Router B to be publicly routeable.

Thanks for the help.

Vilmos

I have already said how to do this. Turn up 1.128/25 on Router B, and advertise it via a routing protocol. Use two /30s, different from the

1.0 network for the WAN links between B and C, and C and A. Turn up C in the same routing protocol with only the WAN links in the network statements of the routing protocol. Turn up 1.0/25 on Router A, and turn up the same routing protocol but with the 1.0/25 as a network statement rather than 1.128/25 which is already being advertised by router B.

At this point, both 1.0/25 and 1.128/25 will be known by Router A, one locally and one remote.

However, 10.X addresses are not publicly routable. So all of this will need to be nat'ed on router A towards the internet. If you do this, and NAT to whatever public addresses you own (perhaps just one, but doesn't matter), then this will work exactly as I have outlined above. If you have provided 10.X as an example, and you really do own a public /24, then all you need is to either summarize the two /25s into one /24 when advertising to the internet, or run a diff protocol as outlined by your ISP and summarize there (probably BGP if this is the case).

I do own a public IP and was just using 10. as an example.

Sorry to continue this but I want to make sure I understand (using fake addresses again) I have never had to break apart a subnet before.

Router A - Gets an address of 208.3.102.1 and is connected to the Internet Router C - Connected on the same physical network as Router A has an address of 208.3.102.128 Router B - Connected to Router C by Serial Interface has an address of

208.3.102.129

Router A has an IP ROUTE command routing all traffic to

208.3.102.129/25 through 208.3.102.128 (Router C)

My question is how to summarize the /25 block created. Specifically I'm wondering about subnet masks I already have a router connected with the address 208.3.102.1 and a subnet mask of 255.255.255.0. I understand how to use the IP route command to point the 102.129/25 subnet. I don't understand is what subnet masks to use.

Is this right?

Router A - 208.3.102.1 255.255.255.0 Router C - 208.3.102.128 255.255.255.0 Router B - 208.3.102.129 255.255.255.128

Or is it

Router A - 208.3.102.1 255.255.255.128 Router C - 208.3.102.128 255.255.255.128 Router B - 208.3.102.129 255.255.255.128

Or is it something else?

I don't think so. A will have 102.1 as its interface, and router C can have 208.3.102.2 255.255.255.128 as its interface in that same network. You need to pick a private address for the WAN links, say

192.168.1.0/30 which means 192.168.1.1 255.255.255.252 on one side (router C) and 192.168.1.2 255.255.255.252 on the other side (router b). This means that router C has one interface in the 102.0 network, and the other on the WAN. Router B will then have one interface on the WAN, and one interface in the other /25, which is 208.3.102.129 255.255.255.128.

You then want router A to have a routing protocol and network statements for 208.3.102.0 255.255.255.128. You want router C to have the same protocol and network statement as router A, but also add the network statement for the wan link (192.168.1.0 255.255.255.252). Router B will then have the same protocol, and network statements for the WAN (192.168.1.0 255.255.255.252) and his local ethernet (208.3.102.128 255.255.255.128).

Thats my interpretation of what you want to do.

Router A (208.3.102.1) (208.3.102.2) Router C (192.168.1.1) (192.168.1.2) Router B (208.3.102.129)