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

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.

Comprehensive and Detailed In-Depth Explanation:

Step 1: Identify the Primary Path

The shortest OSPF cost to 10.0.3.3/32 from R1 is through R5:

R1 → R5 (Cost 10) → R3 (Cost 10) = Total Cost 20

The primary outbound interface from R1 to R5 is GE0/0/1.

Step 2: Identify the Backup Path (OSPF IP FRR & ECMP)

OSPF IP Fast Reroute (FRR) provides a backup next-hop in case the primary link (R1 → R5) fails.

A feasible backup path must have a cost equal to or close to the primary path, but it must avoid the same failure point.

Alternative Path Calculation:

Path 1 (R1 → R2 → R3)

R1 → R2 (Cost 10), R2 → R3 (Cost 10) → Total Cost = 20 ✅

Uses interface GE0/0/2 on R1.

Path 2 (R1 → R4 → R3)

R1 → R4 (Cost 10), R4 → R3 (Cost 10) → Total Cost = 20 ✅

Uses interface GE0/0/3 on R1.

Final Decision on the Backup Interface

Since both paths R1 → R2 → R3 and R1 → R4 → R3 have equal costs, either can be used for backup.

However, R1 → R2 → R3 shares a direct cost path with R5, meaning it may not be preferred as a full FRR backup.

The best backup outbound interface in this scenario is GE0/0/3 (R1 → R4 → R3).

Final Answer: ✅ C (GE0/0/3)

References:

HCIP-Datacom-Advanced Routing & Switching Technology V1.0 – OSPF IP FRR (Fast Reroute) Mechanism

OSPF Equal-Cost Multi-Path (ECMP) and Backup Route Selection

OSPF Route Calculation and Best-Path Selection