mediumMultiple Choice
CCNP Practice Question: Runs the following command on Router R1: R1# show…
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?
⚠ Common exam trap
Cisco often tests the distinction between Passive and Active states in EIGRP topology table output, where candidates mistakenly think 'P' stands for 'Primary' or 'Path' instead of 'Passive', leading them to misinterpret the route state and miss the absence of feasible successors.
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
✓
Each route has exactly one successor and no feasible successor.
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.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
All routes have a feasible successor.
Why it's wrong here
A feasible successor in EIGRP must satisfy the feasibility condition: the reported distance (RD) from the next-hop router must be strictly less than the feasible distance (FD) to the destination. In this topology table, each route shows RD equal to FD, so the strict inequality fails, meaning no feasible successor exists for any route. Saying 'all routes have a feasible successor' is therefore the opposite of what the output reveals.
- ✗
The routes are in Active state, meaning the router is querying for alternate paths.
Why it's wrong here
The topology table displayed uses the 'P' code to mark routes, which stands for Passive state in EIGRP's DUAL finite state machine. Passive means the router is not currently performing a route computation or querying neighbors for alternate paths; it is simply using the existing successor. Active state, indicated by 'A', would show that the router lost a successor and is actively sending queries to neighbors, but none of the routes here are in that condition.
- ✓
Each route has exactly one successor and no feasible successor.
Why this is correct
Each destination in the output has exactly one successor as shown by the single 'via' entry on the first line of each topology entry. A feasible successor would require an alternate next-hop route with RD < FD, but here the RD equals the FD for every route, making the feasibility condition false. Therefore, exactly one successor and zero feasible successors is the correct interpretation of the output.
- ✗
The router is using EIGRP stub routing.
Why it's wrong here
EIGRP stub routing is a feature that restricts a router's query propagation and is configured to prevent it from being used as a transit route for EIGRP-injected traffic. This configuration is not directly observable in a standard topology table display; it would be checked via commands like show ip eigrp neighbors detail or show running-config. The topology output here simply shows normal successor/FD relationships, not any indication of stub marking or query behavior, so claiming stub routing based on this output is unsupported.
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.
About these practice questions
One of 1,923 original 350-401 practice questions on Courseiva, each with a full explanation and wrong-answer analysis — not exam dumps or protected exam content. Learn why practice questions differ from exam dumps →
JA
Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
This 350-401 practice question is part of Courseiva's free Cisco certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the 350-401 exam.