From: Connary, Julie Ann (jconnary@xxxxxxxxx)
Date: Thu May 03 2001 - 21:12:11 GMT-3
Hi,
working a lab that has the following:
r2------ethernet ---r5-----
r2's e0 = 137.20.20.2
r5's e0 = 137.20.20.10
r2 has an ip route 0.0.0.0 0.0.0.0 137.20.20.3 (another router on the
ethernet.)
r2 has an ip ospf default information-originate always metric-type 1 statement.
r5 sees this default route as:
O*E1 0.0.0.0/0 [110/21] via 137.20.20.3, 00:01:40, Ethernet0
this is ccbootcamp lab 5. However - in the routing tables provided R5 should
see this as
O*E1 0.0.0.0/0 [110/21] via 137.20.20.2, 00:02:40, Ethernet0
If I look at the database I see:
OSPF Router with ID (172.168.200.1) (Process ID 1)
Type-5 AS External Link States
Routing Bit Set on this LSA
LS age: 143
Options: (No TOS-capability, DC)
LS Type: AS External Link
Link State ID: 0.0.0.0 (External Network Number )
Advertising Router: 200.200.200.1
LS Seq Number: 80000006
Checksum: 0x9C6E
Length: 36
Network Mask: /0
Metric Type: 1 (Comparable directly to link state metric)
TOS: 0
Metric: 20
Forward Address: 137.20.20.3
External Route Tag: 1
So my question is - how in the answers does the forwarding address become
router 2 (137.20.20.2) and not 137.20.20.3?
Is there a trick to this that I am missing? Or should OSPF when it
originates a default route within the same area not change the
forwarding address?
Thanks,
Julie Ann
p.s. I also got the ospf-demand circuit to work in 5b by making sure that
my ISDN interface was not being redistributed
back into ospf from igrp and having LSA type 5's keep the circuit up. The
lab says that ospf demand circuit does not
work for this topology - but mine worked fine.
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Julie Ann Connary
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c i s c o S y s t e m s Email: jconnary@cisco.com
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