The answer is that only one path to 10.1.1.0/24 is installed in the routing table, because the alternate route via 10.2.2.2 fails the EIGRP feasible successor condition. For a route to qualify as a feasible successor, its reported distance (RD) must be strictly less than the feasible distance (FD) of the successor. In the topology all-links output, the FD is 1310720, and the RD of the alternate path is also 1310720—equal, not less—so the feasibility condition is not met, and that path remains unused in the topology table. On the ENCOR 350-401 exam, this concept tests your ability to interpret the show ip eigrp topology all-links command and distinguish between successors and feasible successors. A common trap is assuming that any path in the topology table is automatically installed in the routing table; in reality, only successors are installed. Remember the mnemonic: “RD must be less than FD to be a feasible successor—if it’s equal, it’s just a backup that won’t go.”
CCNP EIGRP Practice Question
This 350-401 practice question tests your understanding of eigrp. 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 10.1.1.0/24, 1 successors, FD is 1310720, serno 5
via 192.168.1.2 (1310720/1310720), GigabitEthernet0/0
via 10.2.2.2 (1587200/1310720), GigabitEthernet0/1
P 10.2.2.0/24, 1 successors, FD is 1310720, serno 6
via 192.168.1.2 (1310720/1310720), GigabitEthernet0/0
via 10.2.2.2 (1587200/1310720), GigabitEthernet0/1
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
The router has two paths to 10.1.1.0/24, but only one is in the routing table.
The output shows that for 10.1.1.0/24, the feasible distance (FD) is 1310720 and there is only one successor (the route via 192.168.1.2). The alternate path via 10.2.2.2 has a reported distance (RD) of 1310720, which equals the FD, so it does not satisfy the feasibility condition (RD < FD) and therefore is not a feasible successor. Only the successor route is installed in the routing table.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✗
Both routes have a feasible successor via 10.2.2.2.
Why it's wrong here
The RD from 10.2.2.2 equals the FD, so it does not meet the feasibility condition.
✗
The route 10.1.1.0/24 has two successors.
Why it's wrong here
There is only one successor (FD is 1310720, and only one path has that metric).
✗
The alternate path via 10.2.2.2 will be used immediately if the successor fails.
Why it's wrong here
Since it is not a feasible successor, the router must go Active and query before using it.
✓
The router has two paths to 10.1.1.0/24, but only one is in the routing table.
Why this is correct
Only the successor is installed in the routing table; the other path is not used unless the successor fails and queries are sent.
Related concept
Read the scenario before looking for a memorised answer.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Cisco often tests the distinction between a successor and a feasible successor, and the trap here is that candidates assume any alternate path with a lower metric than the FD is a feasible successor, but the feasibility condition requires the reported distance to be strictly less than the feasible distance, not less than or equal.
Detailed technical explanation
How to think about this question
In EIGRP, the feasibility condition requires that the reported distance from a neighbor be strictly less than the current feasible distance for that route to be considered a feasible successor. When the RD equals the FD, the path is an equal-cost path only if the metrics are identical and the FD is the same; here, the FD is 1310720 and the RD via 10.2.2.2 is also 1310720, but the total metric via that path is higher (1587200), so it is not an equal-cost path. This scenario often occurs in frame-relay or non-broadcast multiaccess networks where the same next-hop is reachable via different interfaces, but the feasibility condition still applies.
KKey Concepts to Remember
Read the scenario before looking for a memorised answer.
Find the constraint that changes the correct option.
Eliminate answers that are true in general but not in this case.
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
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
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.
What to study next
Got this wrong? Here's your next step.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
EIGRP — This question tests EIGRP — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: The router has two paths to 10.1.1.0/24, but only one is in the routing table. — The output shows that for 10.1.1.0/24, the feasible distance (FD) is 1310720 and there is only one successor (the route via 192.168.1.2). The alternate path via 10.2.2.2 has a reported distance (RD) of 1310720, which equals the FD, so it does not satisfy the feasibility condition (RD < FD) and therefore is not a feasible successor. Only the successor route is installed in the routing table.
What should I do if I get this 350-401 question wrong?
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
What is the key concept behind this question?
Read the scenario before looking for a memorised answer.
About these practice questions
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Variation 1. A network engineer runs the following command on Router R1:
R1# show ip eigrp topology
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 10.1.1.0/24, 1 successors, FD is 1310720
via 192.168.1.2 (1310720/1310720), GigabitEthernet0/0
P 10.2.2.0/24, 1 successors, FD is 1310720
via 192.168.1.2 (1310720/1310720), GigabitEthernet0/0
P 10.3.3.0/24, 1 successors, FD is 1310720
via 192.168.1.2 (1310720/1310720), GigabitEthernet0/0
Based on this output, what can be concluded?
medium
A.All routes have a feasible successor.
B.The routes are in Active state, meaning the router is querying for alternate paths.
✓ C.Each route has exactly one successor and no feasible successor.
D.The router is using EIGRP stub routing.
Why C: The output shows each route with a code 'P' (Passive) and exactly one successor, with no feasible successor listed. In EIGRP, a feasible successor is only present if there is a backup route that satisfies the feasibility condition (reported distance < feasible distance). Since only one next-hop is shown per route and no additional entries exist, there is no feasible successor. Option C correctly identifies this.
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Question Discussion
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