The answer is that the network has redundant paths with feasible successors for both routes. This is correct because the all-links option in the EIGRP topology table reveals every known path, including those that meet the feasibility condition—where the reported distance from a neighbor is strictly less than the feasible distance (FD) of the successor. For 10.10.10.0/24, the route via 10.2.2.2 has a reported distance of 28160, which equals the FD of 28160, satisfying the condition and making it a feasible successor; similarly, for 10.20.20.0/24, the route via 10.3.3.2 qualifies. On the Cisco CCNP ENARSI 300-410 exam, this command tests your ability to distinguish between successors and feasible successors, a common trap being that the all-links output does not automatically mean all routes are feasible—you must check the reported distance against the FD. A quick memory tip: “Reported must be less than Feasible to be a Feasible successor.”
300-410 EIGRP Troubleshooting Practice Question
This 300-410 practice question tests your understanding of eigrp troubleshooting. 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:
R1# show ip eigrp topology all-links
EIGRP-IPv4 Topology Table for AS(100)/ID(192.168.1.1)
Codes: P - Passive, A - Active, U - Update, Q - Query, R - Reply,
r - reply Status, s - sia Status
P 10.10.10.0/24, 1 successors, FD is 28160, Qos: 0
via 10.1.1.2 (28160/28160), GigabitEthernet0/0
via 10.2.2.2 (28672/28160), GigabitEthernet0/1
P 10.20.20.0/24, 1 successors, FD is 28160, Qos: 0
via 10.2.2.2 (28160/28160), GigabitEthernet0/1
via 10.3.3.2 (28672/28160), GigabitEthernet0/2
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
The network has redundant paths with feasible successors for both routes.
The all-links output shows all known routes, including feasible successors. For 10.10.10.0/24, there is a successor via 10.1.1.2 and a feasible successor via 10.2.2.2 (since its reported distance of 28160 is less than the FD of 28160). For 10.20.20.0/24, there is a successor via 10.2.2.2 and a feasible successor via 10.3.3.2. This indicates good redundancy.
Key principle: NAT direction and interface roles matter as much as the IP address mapping. Inside/outside designation controls which traffic is translated.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✓
The network has redundant paths with feasible successors for both routes.
Why this is correct
Both routes have a feasible successor, providing backup paths.
Related concept
Static NAT maps one inside address to one outside address.
✗
The route to 10.20.20.0/24 has no feasible successor.
Why it's wrong here
It has a feasible successor via 10.3.3.2.
✗
The FD for 10.10.10.0/24 is 28672.
Why it's wrong here
The FD is 28160, as shown in the output.
✗
The route to 10.10.10.0/24 is in Active state.
Why it's wrong here
The code 'P' indicates Passive state.
Common exam traps
Common exam trap: NAT rules depend on direction and matching traffic
NAT is not only about the public address. The inside/outside interface roles and the ACL or rule that matches traffic are just as important.
Trap categories for this question
Command / output trap
The FD is 28160, as shown in the output.
Detailed technical explanation
How to think about this question
NAT questions usually test address translation, overload/PAT behaviour, static mappings and whether the right traffic is being translated. Read the interface direction and address terms carefully.
KKey Concepts to Remember
Static NAT maps one inside address to one outside address.
PAT allows many inside hosts to share one public address using ports.
Inside local and inside global describe the private and translated addresses.
NAT ACLs identify traffic for translation, not always security filtering.
TExam Day Tips
→Identify inside and outside interfaces first.
→Check whether the scenario needs static NAT, dynamic NAT or PAT.
→Do not confuse NAT matching ACLs with normal packet-filtering intent.
Key takeaway
NAT direction and interface roles matter as much as the IP address mapping. Inside/outside designation controls which traffic is translated.
Real-world example
How this comes up in practice
A small business has 20 workstations on the 192.168.1.0/24 network and one public IP from its ISP. The router uses PAT (NAT overload) so all 20 devices share one public address using different source ports. NAT questions test whether you understand the four address terms and which direction each translation applies.
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 the four NAT address types (inside local, inside global, outside local, outside global), PAT port overload, and static vs dynamic NAT use cases. Then practise related 300-410 NAT questions on configuration and troubleshooting.
EIGRP Troubleshooting — This question tests EIGRP Troubleshooting — Static NAT maps one inside address to one outside address..
What is the correct answer to this question?
The correct answer is: The network has redundant paths with feasible successors for both routes. — The all-links output shows all known routes, including feasible successors. For 10.10.10.0/24, there is a successor via 10.1.1.2 and a feasible successor via 10.2.2.2 (since its reported distance of 28160 is less than the FD of 28160). For 10.20.20.0/24, there is a successor via 10.2.2.2 and a feasible successor via 10.3.3.2. This indicates good redundancy.
What should I do if I get this 300-410 question wrong?
Review the four NAT address types (inside local, inside global, outside local, outside global), PAT port overload, and static vs dynamic NAT use cases. Then practise related 300-410 NAT questions on configuration and troubleshooting.
What is the key concept behind this question?
Static NAT maps one inside address to one outside address.
About these practice questions
Courseiva creates original exam-style practice questions with explanations and wrong-answer analysis. It does not publish real exam questions, exam dumps, or protected exam content. Learn why practice questions differ from exam dumps →
These questions test the same concept from different angles. Work through them to make sure you can recognise it however the exam phrases it.
Variation 1. A network engineer runs the following command to troubleshoot an EIGRP issue:
R1# show ip eigrp topology 10.1.1.0/24 detail
IP-EIGRP (AS 100): Topology entry for 10.1.1.0/24
State: Passive, Query origin flag: 1, 1 Successor(s), FD is 131072
Routing Descriptor Blocks:
10.1.2.2 (GigabitEthernet0/0), from 10.1.2.2, Send flag: 0x0
Composite metric: (131072/130816), Route is Internal
Vector metric:
Minimum bandwidth is 10000 Kbit
Total delay is 100 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 1
Originating router: 10.1.2.2
External data: Not advertised
Protocol: EIGRP
Route tag: 0
Extended community: None
What does this output indicate?
medium
A.The route is an external EIGRP route redistributed from another protocol.
✓ B.The route is learned from a single neighbor and is in a stable state.
C.The route has multiple successors and is load-balanced.
D.The route is in Active state, meaning a query is in progress.
Why B: The output shows detailed information about the EIGRP topology entry for 10.1.1.0/24. The route is internal, with a single successor via 10.1.2.2. The FD is 131072, and the RD is 130816. The route is in Passive state.
Last reviewed: Jun 18, 2026
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