I have a Class C Subnet and want to split it in half over a WAN link. I'm using Cisco 2600 routers on both sides.
So I want to take the subnet 10.10.1.0/24
Primary location has the 1st 128 addresses.
Secondary location (over a WAN link) has the rest.
Thanks for any assistance.
Vilmos
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T
Trendkill
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?
C
Chris
Also, the term 'Class C' isn't valid any more. You have a /24.
What exactly is your question?
Chris.
V
vbranyik
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
T
Trendkill
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.
T
Trendkill
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.
V
vbranyik
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
T
Trendkill
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).
V
vbranyik
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?
T
Trendkill
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)
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