QID:NA354 - Incorrect answer & explaination in H2P?

[vtnl241]

Hi all,
This question has already been posted but it's not discussed thoroughly, so I re-post it
It seems to me that the correct answer will be "RT2 will obtain a full routing table, including a default route, from RT1".

I've already simulated the question on the lab and the correct one is as above.

Extract configuration on RT1:

router ospf 1
log-adjacency-changes
network 172.16.100.0 0.0.0.3 area 0
network 172.16.100.64 0.0.0.63 area 0
default-information originate
!
ip route 0.0.0.0 0.0.0.0 Serial0/0
!

------------

routing table on RT2
RT2#sh ip route

    172.16.0.0/16 is variably subnetted, 3 subnets, 3 masks
O      172.16.100.64/26 [110/65] via 172.16.100.1, 00:02:13, Serial0/0
C      172.16.32.0/22 is directly connected, FastEthernet1/0
C      172.16.100.0/30 is directly connected, Serial0/0
O*E2 0.0.0.0/0 [110/1] via 172.16.100.1, 00:02:13, Serial0/0
RT2#

-------------------
Routing table in RT2 includes default route advertised from RT1 through OSPF.
Looking further into the OSPF database table on RT2 for default route:

RT2#sh ip ospf database external

            OSPF Router with ID (3.3.3.3) (Process ID 1)

                Type-5 AS External Link States

  Routing Bit Set on this LSA
  LS age: 497
  Options: (No TOS-capability, DC)
  LS Type: AS External Link
  Link State ID: 0.0.0.0 (External Network Number )
  Advertising Router: 2.2.2.2
  LS Seq Number: 80000001
  Checksum: 0xFEAB
  Length: 36
  Network Mask: /0
        Metric Type: 2 (Larger than any link state path)
        TOS: 0
        Metric: 1
        Forward Address: 0.0.0.
        External Route Tag: 1


Which means the default route from RT1 didn't include next hop in its LSA, thus RT2 has no idea
if there will be a loop if it put default route in its routing table.

Would anybody advice?

Thanks,
Peter

[jumbobob]

I don't know the answer, but this question is very much like (identical to?) NA12 except that NA12 has a very different answer.  I'm not sure which one to believe.