The sum of the advertised distance (AD) of the successor and the link cost to the successor is used by EIGRP to calculate the feasible distance (FD) of a route. This is correct because the FD represents the total metric from the local router to the destination network, combining what the neighbor reports as its own distance (the AD) with the cost of reaching that neighbor. On the ENCOR 350-401 exam, this concept tests your understanding of how EIGRP selects the successor route and evaluates feasible successors for loop-free path selection. A common trap is confusing the FD with the AD alone—remember that the FD always includes the local link cost, while the AD is only the neighbor’s reported metric. For a quick memory tip: think of FD as “Full Distance,” meaning you add your own hop cost to what the neighbor advertises.
350-401 EIGRP Practice Question
This 350-401 practice question tests your understanding of eigrp. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. 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.
Which of the following is used by EIGRP to calculate the feasible distance (FD) of a route?
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
The sum of the advertised distance (AD) of the successor and the link cost to the successor.
EIGRP calculates the Feasible Distance (FD) as the sum of the Advertised Distance (AD) from the successor neighbor and the link cost (metric) to that neighbor. This represents the total metric from the local router to the destination network via that path. The FD is used to determine the best route (successor) and to compare against feasible successors.
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.
✓
The sum of the advertised distance (AD) of the successor and the link cost to the successor.
Why this is correct
Correct. The feasible distance is the metric of the successor, which is the AD (reported distance from neighbor) plus the link cost to that neighbor.
Related concept
Read the scenario before looking for a memorised answer.
✗
The lowest hop count among all paths to the destination.
Why it's wrong here
Incorrect. EIGRP does not use hop count in its metric calculation.
✗
The highest bandwidth among all paths to the destination.
Why it's wrong here
Incorrect. While bandwidth is a factor, the FD is not simply the highest bandwidth; it is a composite metric.
✗
The sum of all delays along the path.
Why it's wrong here
Incorrect. Delay is a component, but the FD also includes bandwidth and possibly other K values.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Cisco often tests the distinction between Feasible Distance (FD) and Advertised Distance (AD), and the trap here is that candidates confuse FD with just the link cost or one metric component (like delay or bandwidth), rather than recognizing it as the sum of the AD and the cost to the successor.
Detailed technical explanation
How to think about this question
The EIGRP composite metric is calculated using the formula: metric = [K1 * bandwidth + (K2 * bandwidth) / (256 - load) + K3 * delay] * 256, with default K values (K1=1, K3=1, others=0) simplifying to metric = (bandwidth + delay) * 256. The Feasible Distance (FD) is the total metric from the local router to the destination, while the Advertised Distance (AD) is the metric reported by the neighbor. In real-world scenarios, understanding this distinction is critical for troubleshooting EIGRP path selection and ensuring loop-free routing via the feasibility condition (AD < FD).
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.
Related glossary terms
Concepts from this question explained
These glossary pages explain the core terms tested in this 350-401 question in full detail.
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 sum of the advertised distance (AD) of the successor and the link cost to the successor. — EIGRP calculates the Feasible Distance (FD) as the sum of the Advertised Distance (AD) from the successor neighbor and the link cost (metric) to that neighbor. This represents the total metric from the local router to the destination network via that path. The FD is used to determine the best route (successor) and to compare against feasible successors.
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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