hardMultiple ChoiceObjective-mapped
EIGRP Feasible Successor Condition
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
Based on this output, what can be concluded?
Quick Answer
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.”
⚠ Common exam trap
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.
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
✓
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.
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.
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.
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Same concept, more angles
1 more way this is tested on 350-401
These questions test the same concept from different angles. Work through them to make sure you can recognise it however the exam phrases it.
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.
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.