The answer is that Router 10.1.1.2 is the Designated Router (DR) on the 192.168.12.0/24 segment. This conclusion follows directly from interpreting the OSPF router link states and net link states in the LSDB: the only Net Link State entry shows a Link ID of 192.168.12.2 with an ADV Router of 10.1.1.2, and since Net Link States are only generated by the DR for each broadcast or NBMA segment, the advertising router must be the DR. On the Cisco CCNP ENARSI 300-410 exam, this tests your ability to read the OSPF database output and differentiate between Router LSA (Type 1) and Network LSA (Type 2) responsibilities—a common trap is confusing the Link ID (which is the DR’s interface IP) with the router ID of the DR. Remember: the ADV Router field in a Net Link State always reveals the DR’s router ID, while the Link ID shows the DR’s interface IP on that segment. A helpful memory tip: “Net Link ADV = DR’s RID, Net Link ID = DR’s IP.”
300-410 OSPF Troubleshooting (v2/v3) Practice Question
This 300-410 practice question tests your understanding of ospf troubleshooting (v2/v3). Examine the command output carefully: the correct answer depends on what the output actually shows, not on general recall alone. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
A network engineer runs the following command on Router R1:
Router R1 is the DR on the segment 192.168.12.0/24.
Why wrong: The net link state is advertised by 10.1.1.2, not by R1 (10.1.1.1).
B
Router 10.1.1.2 is the DR on the segment 192.168.12.0/24.
The net link state with Link ID 192.168.12.2 is advertised by 10.1.1.2, meaning it is the DR on that segment.
C
Router 10.1.1.3 has the highest router ID.
Why wrong: Router IDs are 10.1.1.1, 10.1.1.2, and 10.1.1.3; 10.1.1.3 is the highest numerically, but this is not directly relevant to the database output.
D
There is a network LSA for every broadcast segment in area 0.
Why wrong: Only one net link state is present, so either there is only one broadcast segment or other segments have no DR elected.
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
Router 10.1.1.2 is the DR on the segment 192.168.12.0/24.
The database shows router link states for three routers and one net link state. The net link state is advertised by 10.1.1.2, indicating it is the DR on the segment 192.168.12.0/24. The absence of other net link states suggests only one broadcast segment exists.
Key principle: OSPF neighbour adjacency depends on matching area, hello/dead timers, network type, and authentication — IP reachability alone is not enough.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✗
Router R1 is the DR on the segment 192.168.12.0/24.
Why it's wrong here
The net link state is advertised by 10.1.1.2, not by R1 (10.1.1.1).
✓
Router 10.1.1.2 is the DR on the segment 192.168.12.0/24.
Why this is correct
The net link state with Link ID 192.168.12.2 is advertised by 10.1.1.2, meaning it is the DR on that segment.
Router IDs are 10.1.1.1, 10.1.1.2, and 10.1.1.3; 10.1.1.3 is the highest numerically, but this is not directly relevant to the database output.
✗
There is a network LSA for every broadcast segment in area 0.
Why it's wrong here
Only one net link state is present, so either there is only one broadcast segment or other segments have no DR elected.
Common exam traps
Common exam trap: OSPF can fail even when IP connectivity looks correct
OSPF neighbour formation depends on matching areas, timers, network type, authentication and passive-interface behaviour. Do not choose an answer only because the devices can ping.
Trap categories for this question
Command / output trap
Router IDs are 10.1.1.1, 10.1.1.2, and 10.1.1.3; 10.1.1.3 is the highest numerically, but this is not directly relevant to the database output.
Detailed technical explanation
How to think about this question
OSPF questions usually test the details that control adjacency and route selection. Read the neighbour state, area, router ID and interface configuration before deciding what is wrong.
KKey Concepts to Remember
OSPF neighbours must agree on key parameters.
Router ID selection can affect neighbour relationships and LSDB output.
OSPF cost influences the preferred path.
A route can appear in OSPF information but not become the installed route.
TExam Day Tips
→Check area mismatch first when OSPF adjacency fails.
→Review passive interfaces when a network is advertised but no neighbour forms.
→Use show ip ospf neighbor and show ip route clues carefully.
Key takeaway
OSPF neighbour adjacency depends on matching area, hello/dead timers, network type, and authentication — IP reachability alone is not enough.
Real-world example
How this comes up in practice
A network engineer at a university connects two campus buildings via a fibre link. Both routers run OSPF, but no adjacency forms — even though both routers can ping each other. The engineer finds one router is in area 0 and the other in area 1. OSPF adjacency requires matching area numbers, hello/dead timers, and network type. IP reachability alone is not enough.
Related glossary terms
Concepts from this question explained
These glossary pages explain the core terms tested in this 300-410 question in full detail.
Review OSPF neighbour requirements — matching area type, hello and dead timers, network type, stub flags, and authentication. Study show ip ospf neighbor states (INIT, 2-WAY, FULL). Then practise related 300-410 OSPF questions on adjacency and route selection.
OSPF Troubleshooting (v2/v3) — This question tests OSPF Troubleshooting (v2/v3) — OSPF neighbours must agree on key parameters..
What is the correct answer to this question?
The correct answer is: Router 10.1.1.2 is the DR on the segment 192.168.12.0/24. — The database shows router link states for three routers and one net link state. The net link state is advertised by 10.1.1.2, indicating it is the DR on the segment 192.168.12.0/24. The absence of other net link states suggests only one broadcast segment exists.
What should I do if I get this 300-410 question wrong?
Review OSPF neighbour requirements — matching area type, hello and dead timers, network type, stub flags, and authentication. Study show ip ospf neighbor states (INIT, 2-WAY, FULL). Then practise related 300-410 OSPF questions on adjacency and route selection.
What is the key concept behind this question?
OSPF neighbours must agree on key parameters.
About these practice questions
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Variation 1. A network engineer runs the following command on Router R1:
R1# show ip ospf database router 10.1.1.2
OSPF Router with ID (10.1.1.1) (Process ID 1)
Router Link States (Area 0)
LS age: 150
Options: (No TOS-capability, DC)
LS Type: Router Links
Link State ID: 10.1.1.2
Advertising Router: 10.1.1.2
LS Seq Number: 80000002
Checksum: 0x1234
Length: 48
Number of Links: 2
Link connected to: a Transit Network
(Link ID) Designated Router address: 192.168.12.2
(Link Data) Router Interface address: 192.168.12.2
Number of TOS metrics: 0
TOS 0 Metrics: 10
Link connected to: a Stub Network
(Link ID) Network/subnet number: 10.2.2.0
(Link Data) Network Mask: 255.255.255.0
Number of TOS metrics: 0
TOS 0 Metrics: 10
Based on this output, which statement is correct?
hard
✓ A.Router 10.1.1.2 is connected to a transit network and a stub network.
B.Router 10.1.1.2 is the DR on the transit network.
C.Router 10.1.1.2 has a cost of 20 to reach the stub network.
D.Router 10.1.1.2 is not advertising any external routes.
Why A: The router LSA from 10.1.1.2 shows two links: one transit network (to the DR address) and one stub network (10.2.2.0/24). The transit link indicates it is connected to a broadcast segment where it is the DR (since it lists its own interface address as DR address).
Last reviewed: Jun 18, 2026
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