From: k c (jwongccie@yahoo.com.hk)
Date: Sat May 14 2005 - 05:22:20 GMT-3
Hi Ed Lui,
My question is:
one side is p2p, the other side is physical interface. By default, physical interface cannot connect to p2p sub-interface, am I correct? Assuming "no fram inverse-arp" is used. So how to make connection between a p2p sub-interface and a physical interface.
Thanks.
Ed Lui <edwlui@gmail.com> <6<g:
k c,
It will look like this. Your question is ?
--------------------------------(s0)(phy int)R4
/ 172.16.14.X
(s0.14)ptp
R1(s0.123)multipoint=========================(S0)(phy int)R2
\ 172.16.123.X
========================(s0)(phy int)R3
HTH,
On 5/13/05, k c wrote:
> Hi Group,
>
> R1 (172.16.123.1/172.16.14.1), R2 (172.16.123.2), R3 (172.16.123.3), R4 (172.16.14.4).
> R1 is Hub, R2 & R3 are spokes within 172.16.123.0 subnet.
> "One and only one router must configured with two and only two logical FR interfaces. The three remaining routers with FR interfaces must use physical interfaces only. The logical interfaces must be of a different type"
>
> Obviously, R1 will be configured with one mp (connect to R2 and R3) and one p2p connect to R4, however, R2-R4 should be physical interfaces. Any hints?
>
> Thanks.
>
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