Simple networking setup

Discussion in 'Cisco' started by vbranyik@gmail.com, Nov 1, 2007.

  1. Guest

    Sorry if this is a simple question.....

    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
    , Nov 1, 2007
    #1
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  2. Trendkill Guest

    On Nov 1, 3:20 pm, wrote:
    > Sorry if this is a simple question.....
    >
    > 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


    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?
    Trendkill, Nov 1, 2007
    #2
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  3. Chris Guest

    On Thu, 01 Nov 2007 12:20:29 -0700, wrote:

    > Sorry if this is a simple question.....
    >
    > 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


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

    What exactly is your question?

    Chris.
    Chris, Nov 1, 2007
    #3
  4. Guest


    > 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
    , Nov 1, 2007
    #4
  5. Trendkill Guest

    On Nov 1, 7:03 pm, wrote:
    > > 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.
    Trendkill, Nov 1, 2007
    #5
  6. Trendkill Guest

    On Nov 1, 7:03 pm, wrote:
    > > 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


    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.
    Trendkill, Nov 1, 2007
    #6
  7. Guest

    On Nov 1, 5:09 pm, Trendkill <> wrote:
    > On Nov 1, 7:03 pm, wrote:
    >
    > > > 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

    >
    > 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
    , Nov 2, 2007
    #7
  8. Trendkill Guest

    On Nov 1, 8:16 pm, wrote:
    > On Nov 1, 5:09 pm, Trendkill <> wrote:
    >
    >
    >
    > > On Nov 1, 7:03 pm, wrote:

    >
    > > > > 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

    >
    > > 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).
    Trendkill, Nov 2, 2007
    #8
  9. Guest

    On Nov 2, 4:06 am, Trendkill <> wrote:
    > On Nov 1, 8:16 pm, wrote:


    > > 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?
    , Nov 2, 2007
    #9
  10. Trendkill Guest

    On Nov 2, 8:39 am, wrote:
    > On Nov 2, 4:06 am, Trendkill <> wrote:
    >
    >
    >
    > > On Nov 1, 8:16 pm, wrote:
    > > > 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)
    Trendkill, Nov 2, 2007
    #10
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