The answer is the feasibility condition, which is the correct mechanism because it enforces a strict mathematical rule to prevent routing loops in EIGRP. This condition requires that a router will only accept an alternate route from a neighbor if that neighbor’s reported distance (RD) to the destination is strictly less than the router’s own feasible distance (FD). By rejecting any route where the RD is equal to or greater than the FD, the router guarantees that the neighbor is not using a path that loops back through itself, effectively blocking routes learned from any neighbor that is not the successor. On the ENCOR 350-401 exam, this concept tests your understanding of EIGRP’s Diffusing Update Algorithm (DUAL) and is often presented in a scenario where a router must choose between multiple feasible successors. A common trap is confusing the feasibility condition with the simpler split horizon or poison reverse; remember that the feasibility condition is unique to EIGRP and relies on comparing RD to FD. A useful memory tip is “RD must be less than FD to be FD-free,” meaning the reported distance must be smaller than the feasible distance to guarantee a loop-free path.
CCNP Architecture Practice Question
This 350-401 practice question tests your understanding of architecture. The scenario asks you to isolate a root cause — eliminate options that address a different problem before choosing. 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 is troubleshooting a routing loop in an EIGRP network. Which mechanism is designed to prevent routing loops by causing a router to reject routes that are learned from a neighbor that is not the successor?
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
Feasibility condition
The feasibility condition is a loop-prevention mechanism unique to EIGRP. It ensures that a router only accepts a route from a neighbor if that neighbor's reported distance (RD) to the destination is less than the router's own feasible distance (FD). This guarantees that the path through that neighbor is loop-free, effectively rejecting routes learned from any neighbor that is not the successor.
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.
✗
Split horizon
Why it's wrong here
Split horizon prevents updates from being sent out the interface they were learned on.
✗
Route poisoning
Why it's wrong here
Route poisoning sets the metric to infinity to indicate a down route.
✗
Hold-down timers
Why it's wrong here
Hold-down timers are used in RIP, not EIGRP, to prevent loop information from being used.
✓
Feasibility condition
Why this is correct
The feasibility condition ensures loop-free paths by verifying that the neighbor's reported distance is lower than the feasible distance.
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 EIGRP's feasibility condition and other distance-vector loop-prevention mechanisms like split horizon or hold-down timers, expecting candidates to confuse these concepts because they all prevent loops but operate at different stages of the routing process.
Detailed technical explanation
How to think about this question
Under the hood, the feasibility condition is part of EIGRP's DUAL algorithm. When a router receives a route update, it compares the neighbor's reported distance to its own feasible distance. If the reported distance is less than the feasible distance, the neighbor is considered a feasible successor and can be used as a backup path without causing a loop. This condition is mathematically proven to guarantee loop-free paths at every instant, even during convergence. In real-world scenarios, misconfigured route summarization or redistribution can cause the feasibility condition to fail, leading to stuck-in-active (SIA) routes.
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 network engineer at a university connects two campus buildings via a fibre link. Both routers run OSPF, but no adjacency forms — even though both routers can ping each other. The engineer finds one router is in area 0 and the other in area 1. OSPF adjacency requires matching area numbers, hello/dead timers, and network type. IP reachability alone is not enough.
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.
Architecture — This question tests Architecture — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: Feasibility condition — The feasibility condition is a loop-prevention mechanism unique to EIGRP. It ensures that a router only accepts a route from a neighbor if that neighbor's reported distance (RD) to the destination is less than the router's own feasible distance (FD). This guarantees that the path through that neighbor is loop-free, effectively rejecting routes learned from any neighbor that is not the successor.
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.
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Question Discussion
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