logical subnet overlaying...

hello. I'm not sure if this is the right mailing list to ask, but I hope
that someone could possiably help..

I have a network with a cisco 2501 router running IOS 10.2(9), and we have
a block of addresses from our internet service provider (60 addresses,, start. with 1 for network, and 1 for brodcast)

we recently requested another block of 32 from our ISP, and they gave us
the following:, start...with 1 for network/1 for brodcast.

I asked a similar question on a mailing list before (not sure if it was
this one) and was told to login to the router, then:


router# conf term
router <conf> int e 0
router <subif> ip address existing_ip existing_subnetmask secondary

i tryed this, but it did not work. I also tryed replacing the exiting
ip/subnet with the new one... but that did not work either...

am i doing something wrong? or do i just need a 11.xx(xx) IOS ?
Any help would be greatly apreciated. thanks.

Hi, Kyle--you'll probably get *lots* of responses to your question,
but I haven't seen any yet, so I apologize if this turns out to be
a duplicate.

don't worry about IOS versions--this has been supported in IOS since
before it was even *called* IOS!
what you'll end up with is TWO "ip address" lines in your configuration,
one for the primary ip address and one for the additional (which cisco
calls "secondary") one. in general, the syntax is:
  int eth 0 0
  ip addr <primary ip address> <primary ip address's mask>
  ip addr <second ip addr> <second ip addr's mask> secondary
in your case, the lines would look like (assuming you're using the first
available address in your block for your router--put the right thing in):
  int eth 0 0
  ip addr ! this is the config you already had
  ip addr secondary ! this is your new one
then user devices can use addresses out of either block on that Ethernet.
Note, however, that if a user on the first subnet wants to talk to a device
in the second subnet range their traffic will go to the router and then back
to the Ethernet to the destination, since the source doesn't know that the
destination is on the same wire as he is. Hope that helps,
  Brent Sweeny
  Indiana University