- A
Only one path is available; the second path is a backup that is not used.
Correct. Only the first path is a successor and is installed. The second path is not a feasible successor (RD == FD), so it is not immediately usable; it is only a backup that requires a query to become active.
- B
Both paths are feasible successors, but only the first is installed in the routing table.
Why wrong: Incorrect. The second path's reported distance (131072) equals the feasible distance (131072), so it does not satisfy the feasibility condition (RD < FD). Hence, it is not a feasible successor. Only the first path is a feasible successor.
- C
Both paths are installed in the routing table for load balancing.
Why wrong: Incorrect. Only the successor is installed. EIGRP does not load balance over this second path because it is not a feasible successor, and variance is not configured.
- D
The route is in active state, indicating a query is in progress.
Why wrong: Incorrect. The route is in Passive state, as indicated by 'State: Passive'. An active state would show 'State: Active'.
300-410 Feasible Successor Practice Question
This 300-410 practice question tests your understanding of eigrp troubleshooting. The scenario asks you to isolate a root cause — eliminate options that address a different problem before choosing. A key principle to apply: feasible Successor. 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 to troubleshoot an EIGRP issue:
R1# show ip eigrp topology 10.1.1.0/24
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
10.1.3.3 (GigabitEthernet0/1), from 10.1.3.3, Send flag: 0x0
Composite metric: (131328/131072), Route is Internal Vector metric: Minimum bandwidth is 10000 Kbit Total delay is 200 microseconds Reliability is 255/255 Load is 1/255 Minimum MTU is 1500 Hop count is 2
What does this output indicate?
Clue words in this question
Noticing these words before you look at the options changes how you read each choice.
Clue:
"minimum / minimize"Why it matters: Asks for the least resource use — fewest addresses, smallest subnet, lowest overhead. Eliminate over-provisioned options even if they would technically work.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
Only one path is available; the second path is a backup that is not used.
The output shows two paths for 10.1.1.0/24. The first path has a composite metric of (131072/130816), making the feasible distance (FD) 131072. The second path has a composite metric of (131328/131072), where the reported distance (131072) equals the FD. For a path to be a feasible successor, the reported distance must be strictly less than the FD. Since the second path's reported distance equals the FD, it does not satisfy the feasibility condition and is not a feasible successor. Therefore, only the first path is a successor and is installed in the routing table. The second path is stored in the topology table but is not used unless a query process occurs. Thus, option A is correct.
Key principle: Feasible Successor
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
Only one path is available; the second path is a backup that is not used.
Why this is correct
Correct. Only the first path is a successor and is installed. The second path is not a feasible successor (RD == FD), so it is not immediately usable; it is only a backup that requires a query to become active.
Clue confirmation
The clue word "minimum / minimize" in the question point toward this answer.
Related concept
Feasible Successor
- ✗
Both paths are feasible successors, but only the first is installed in the routing table.
Why it's wrong here
Incorrect. The second path's reported distance (131072) equals the feasible distance (131072), so it does not satisfy the feasibility condition (RD < FD). Hence, it is not a feasible successor. Only the first path is a feasible successor.
- ✗
Both paths are installed in the routing table for load balancing.
Why it's wrong here
Incorrect. Only the successor is installed. EIGRP does not load balance over this second path because it is not a feasible successor, and variance is not configured.
- ✗
The route is in active state, indicating a query is in progress.
Why it's wrong here
Incorrect. The route is in Passive state, as indicated by 'State: Passive'. An active state would show 'State: Active'.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Candidates often misread the topology table and assume that any path with a reported distance less than or equal to the FD is a feasible successor. However, the feasibility condition requires a strict inequality (RD < FD). Here, the second path's RD equals the FD, so it is not a feasible successor.
Trap categories for this question
Command / output trap
Incorrect. The route is in Passive state, as indicated by 'State: Passive'. An active state would show 'State: Active'.
Detailed technical explanation
How to think about this question
EIGRP uses the Diffusing Update Algorithm (DUAL) to maintain loop-free paths. A feasible successor must satisfy the feasibility condition: its reported distance (RD) must be less than the feasible distance (FD) of the successor. In this output, the second path has an RD of 130816, which is less than the FD of 131072, so it qualifies as a feasible successor. However, without the 'variance' command, only the successor (lowest metric path) is installed in the routing table. The 'variance' multiplier (default 1) allows unequal-cost load balancing by including routes whose metric is less than or equal to variance times the FD.
KKey Concepts to Remember
- Feasible Successor
- Successor
- EIGRP Topology Table
TExam Day Tips
- Watch for words such as best, first, most likely and least administrative effort.
- Review why wrong options are wrong, not only why the correct option is correct.
Key takeaway
Feasible Successor
Real-world example
How this comes up in practice
A practitioner preparing for the 300-410 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. Feasible Successor Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.
Visual reference
Quick reference
Routing Protocol Comparison
| Protocol | Metric | Max Hops | Algorithm | Type |
|---|---|---|---|---|
| RIP v2 | Hop count | 15 | Bellman-Ford | Distance vector |
| OSPF | Cost (bandwidth) | Unlimited | Dijkstra (SPF) | Link state |
| EIGRP | Composite metric | Unlimited | DUAL | Hybrid |
| IS-IS | Cost | Unlimited | Dijkstra | Link state |
| BGP | Policy / attributes | Unlimited | Path vector | Path vector |
RIP's 15-hop limit makes it unsuitable for large networks. OSPF and EIGRP dominate modern enterprise deployments.
What to study next
Got this wrong? Here's your next step.
Review feasible Successor, then practise related 300-410 questions on the same topic to reinforce the concept.
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EIGRP Troubleshooting — study guide chapter
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FAQ
Questions learners often ask
What does this 300-410 question test?
EIGRP Troubleshooting — This question tests EIGRP Troubleshooting — Feasible Successor.
What is the correct answer to this question?
The correct answer is: Only one path is available; the second path is a backup that is not used. — The output shows two paths for 10.1.1.0/24. The first path has a composite metric of (131072/130816), making the feasible distance (FD) 131072. The second path has a composite metric of (131328/131072), where the reported distance (131072) equals the FD. For a path to be a feasible successor, the reported distance must be strictly less than the FD. Since the second path's reported distance equals the FD, it does not satisfy the feasibility condition and is not a feasible successor. Therefore, only the first path is a successor and is installed in the routing table. The second path is stored in the topology table but is not used unless a query process occurs. Thus, option A is correct.
What should I do if I get this 300-410 question wrong?
Review feasible Successor, then practise related 300-410 questions on the same topic to reinforce the concept.
Are there clue words in this question I should notice?
Yes — watch for: "minimum / minimize". Asks for the least resource use — fewest addresses, smallest subnet, lowest overhead. Eliminate over-provisioned options even if they would technically work.
What is the key concept behind this question?
Feasible Successor
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Last reviewed: Jul 4, 2026
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