H12-831_V1.0 Exam Dumps : HCIP-Datacom-Advanced Routing & Switching Technology V1.0

Comprehensive and Detailed In-Depth Explanation:

Understanding the OSPF Network Structure in the Question:

OSPF Areas & Routers:

Area 3: Contains R5 (which has loopback 10.0.5.5/32).

Area 1: Contains R6.

Area 2: Contains R4.

Area 0 (Backbone Area): Contains R1, R2, and R3 (which interconnect the other areas).

ACL 2000 Filtering Rule:

The ACL rule 2000 explicitly denies 10.0.5.5/32 from being imported.

The filter 2000 import command applies this filter to Area 0, preventing routes from Area 3 (including R5's loopback) from being learned by Area 0 routers (R1, R2, and R3).

Route Distribution Analysis:

R5 (Area 3):

R5 has 10.0.5.5/32 configured on its Loopback0.

Since OSPF is enabled, R5 advertises the route to R6 in Area 1.

R6 (Area 1) → R2 (Area 0, Backbone Router):

Without filtering, R6 would normally propagate 10.0.5.5/32 to R2 (since R2 is a backbone router in Area 0).

However, since ACL 2000 is applied in Area 0, R2 blocks this route from being imported.

Therefore, R2 does not learn 10.0.5.5/32.

R2 → R1 and R3 (Both in Area 0):

R2 would typically share any learned routes with R1 and R3 in Area 0.

But since R2 itself does not have 10.0.5.5/32, neither R1 nor R3 can receive it.

Thus, R1 and R3 also do not have 10.0.5.5/32 in their routing tables.

R4 (Area 2) is Unaffected:

Since R4 is in Area 2, it is not directly impacted by the filter applied in Area 0.

It can still receive the route via another path if redistribution is configured properly.

Final Conclusion:

R1, R2, and R3 do not have the 10.0.5.5/32 route due to the filter 2000 import command applied in Area 0.

R4 may still have the route since it is not affected by the filtering in Area 0.

Thus, the correct answer is:✅ R1, R3, R2 do not have the route 10.0.5.5/32 in their routing tables.

Comprehensive and Detailed In-Depth Explanation:

To verify the correct answer, let's analyze the LSA types in OSPFv3 and how they work in the given topology.

1. Understanding OSPFv3 LSAs in the Figure

From the command output shown in the image:

LS Type: Intra-Area-Prefix-LSA (Type 9 in OSPFv3).

Originating Router: 10.0.2.2 (R2).

Referenced LS Type: 0x2002, which corresponds to a Network-LSA (Type 2).

Prefix Advertised: 2000:23::/64, which is an IPv6 prefix associated with a network segment.

2. What is an Intra-Area-Prefix LSA (Type 9) in OSPFv3?

In OSPFv3, Type 9 LSAs (Intra-Area-Prefix LSAs) describe IPv6 prefixes associated with either Router-LSAs (Type 1) or Network-LSAs (Type 2).

In multi-access networks (e.g., Ethernet), the DR (Designated Router) generates a Network-LSA (Type 2), which represents the shared network segment.

R2 is likely the DR in this topology, meaning it generates a Network-LSA (Type 2) and an associated Intra-Area-Prefix LSA (Type 9) for the subnet.

3. Analyzing Why the Statement is TRUE

R2 (10.0.2.2) is the originator of the Intra-Area-Prefix LSA (Type 9).

The LSA references a Network-LSA (Type 2), confirming that R2 is the DR for the subnet.

This means the LSA is correctly describing the IPv6 prefix associated with the network.

Since the LSA is visible in R3's LSDB, R3 is learning this information from R2.

All these points confirm that the statement is TRUE.

Comprehensive and Detailed In-Depth Explanation:

Understanding the BGP Network Topology in the Question:

EBGP Peering & Route Distribution

Each router forms EBGP peer relationships with directly connected neighbors.

BGP learns and propagates routes via EBGP neighbors.

R1 and R4 advertise the static route 192.168.1.0/24 into BGP.

Effect of the aggregate 192.168.1.0/16 detail-suppressed Command on R2

R2 performs BGP route aggregation, summarizing 192.168.1.0/24 into a larger 192.168.1.0/16 route.

The detail-suppressed option hides the more specific 192.168.1.0/24 prefix when advertising routes.

As a result, R2 does not advertise the specific 192.168.1.0/24 route to its neighbors.

Route Selection and Traffic Flow Analysis:

R1 and R4 originate 192.168.1.0/24 and advertise it via BGP.

R2 does not advertise the 192.168.1.0/24 route because of the detail-suppressed command.

R3 cannot learn the specific route from R2, so it must choose an alternative path via R5.

R5 learns 192.168.1.0/24 from R1 and advertises it to R3.

Thus, the best available path for R3 to reach 192.168.1.0/24 is R3 → R5 → R1.

✅ Correct Path: R3 → R5 → R1

Checking Each Option:

✅ Option A: R3 → R5 → R1 (Correct)

R3 learns 192.168.1.0/24 from R5, and R5 forwards traffic to R1, which has the static route.

❌ Option B: R3 → R2 → R1 (Incorrect)

R2 does not advertise 192.168.1.0/24 due to route summarization (detail-suppressed), so R3 cannot use R2 as the next hop.

❌ Option C: R3 → R6 → R5 → R1 (Incorrect)

This is an unnecessary detour.

R3 has a direct EBGP connection with R5, so it will use the shortest path (R3 → R5 → R1) instead of going through R6.

❌ Option D: R3 → R5 → R4 (Incorrect)

R3 prefers the shortest AS path to reach 192.168.1.0/24.

Since R1 and R4 both advertise the route, R3 will prefer the route via R5 to R1 rather than R4.

Final Conclusion:

✅ A. The correct path for traffic from R3 to 192.168.1.0/24 is R3 → R5 → R1.

Thus, the correct answer is: A.