From: Fred Ingham (fingham@xxxxxxx)
Date: Fri Jun 28 2002 - 12:23:28 GMT-3
Emmanuel: To have the host on r1 (Ethernet) connect to the hosts on r2
(Token Ring) you would not configure SR/TLB, DLSW does the conversion.
SR/TLB is only used when you want to connect hosts on EN and Token-ring on
the same router.
Cheers, Fred
----- Original Message -----
From: "Emmanuel Oppong" <e-oppong@attbi.com>
To: <fingham@cox.net>; <ccielab@groupstudy.com>; <fningham@att.net>
Sent: Thursday, June 27, 2002 9:20 PM
Subject: RE: Bitswapping and Mac Filtering
> Thanks, Fred. Actually there is a 3rd router as border peer. I didn't
want
> to make this a long scenario and forgot to convert my config from
> peer-on-demand to prom-peer, in this truncated scenario. Anyway, you
> answered my question: so I don't have to bitswap the mac address in r1
> filter, in this case. But if I wanted the hosts on r1 to communicate with
> hosts on r2, then I would probably config SR/TLB and use the "bridge x
> bitswap-layer-addresses" on r1, I suppose?
>
> This is a scenario from another vendor lab. Looks like they stole your
mac
> or may be a coicidence (:
>
> Thanks
>
> -----Original Message-----
> From: fingham@cox.net [mailto:fingham@cox.net]
> Sent: Thursday, June 27, 2002 5:38 PM
> To: Emmanuel Oppong; ccielab@groupstudy.com; fningham@att.net
> Subject: Re: Bitswapping and Mac Filtering
>
>
> Emmanuel: What you have shouldn't do anything. There are some
fundamental
> errors if I understand your description correctly:
>
> First, if you have r1 and r2 as dlsw peers, i.e., no border peer, then
on
> r1 you want the filter applied to the dlsw prom-peer defaults command not
> the dlsw peer-on-demand-defaults. You would only use the prom-peer
command
> when you have a border and r1 and r2 are not configured as peers. For
your
> configuration to work as shown you would have a remote-peer statement on
r2
> pointing to r1.
>
> Second, you have converted a non-canonical address to a canonical
> address. The token-ring address is already in non-canonical format and
> should not be changed. (As an aside, the MAC address you are using
belongs
> to one of our PC's in the NMC course).
>
> Cheers, Fred
> >
> > From: "Emmanuel Oppong" <e-oppong@attbi.com>
> > Date: 2002/06/27 Thu PM 05:01:01 EDT
> > To: <ccielab@groupstudy.com>
> > Subject: Bitswapping and Mac Filtering
> >
> > Guys/gals
> >
> > I need your help on this one:
> >
> > r1 and r2 are configured as dlsw peers. r1 has a host on e0 network.
r2
> > has host-1 and host-2
> > on its to0 network. I want r1's host to communicate with only host-1 on
> r2.
> > MAC addresses for
> > the hosts area :
> >
> > r1 host = 0000-8616-3F04
> > r2 host-1 = 0000-F669-5EE7
> > r2 host-2 = 0000-F669-5F25
> >
> > My relevant config for r1 are:
> >
> > r1:
> > dlsw local-peer peer-id 172.16.101.1 promiscuous
> > dlsw peer-on-demand-defaults dmac-output-list 700
> > dlsw bridge-group 1
> > !
> > access-list 700 permit 0000.6F96.7AE7 0000.0000.0000
> > !
> > int e0
> > bridge-group 1
> > !
> > bridge 1 protocol ieee
> > !
> >
> > My questions are:
> > I have bitswapped the mac address of host-1 in the access-list. Is that
> the
> > right thing to do,
> > convert r2 host-1 from non-canonical to canonical? Even if it is right,
> how
> > does r1 associate
> > these 2 mac addresses? I am thinking that r1 receives a frame from r2
> with
> > host-1 source mac address
> > in the noncanonical form, right? But how does r1 know that the
canonical
> > form in the access-list
> > matches the host-1 source mac address coming from r2? Can someone
explain
> > this?
> >
> > What is the right way to configure thsi scenario?
> >
> > Thanks.
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