Question 1,752 of 2,011
hardMultiple ChoiceObjective-mapped
Troubleshooting EIGRP Active Routes and Stuck-in-Active Conditions
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 10.10.10.0/24, 1 successors, FD is 28160 via 10.1.1.2 (28160/28160), GigabitEthernet0/0 P 10.20.20.0/24, 1 successors, FD is 28160 via 10.2.2.2 (28160/28160), GigabitEthernet0/1 P 10.30.30.0/24, 1 successors, FD is 28160 via 10.3.3.2 (28160/28160), GigabitEthernet0/2 A 10.40.40.0/24, 0 successors, FD is Infinity via 10.4.4.2 (Infinity/Infinity), GigabitEthernet0/3
Based on this output, what is the problem?
Quick Answer
The answer is that the route to 10.40.40.0/24 is stuck-in-active, which indicates a potential network instability. This is correct because the topology table shows the route in the Active state (coded as "A") with an infinite feasible distance (FD is Infinity) and zero successors, meaning the router has sent out queries for a successor but has not received all replies, leaving the route in an unresolved active state. On the Cisco CCNP ENARSI 300-410 exam, this scenario tests your ability to troubleshoot EIGRP active state and stuck-in-active conditions, a common topic where routers fail to receive replies from neighbors due to unidirectional links, packet loss, or a neighbor going down. A common trap is confusing the "A" code with a normal query process; remember that a route stuck in Active for more than three minutes triggers the SIA condition, which can destabilize the entire EIGRP domain. Memory tip: "Active with Infinity means SIA—no successor, just a mess."
⚠ Common exam trap
Cisco often tests the distinction between the 'P' (Passive) and 'A' (Active) codes in the 'show ip eigrp topology' output, where candidates may overlook the 'A' code and assume all routes are stable, missing the SIA condition indicated by 0 successors and Infinity FD.
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 route to 10.40.40.0/24 is stuck-in-active and may cause a network instability.
The route to 10.40.40.0/24 is in the Active (A) state with 0 successors and an FD of Infinity, meaning the router has lost its only feasible successor and is actively querying neighbors for a new path. This is a stuck-in-active (SIA) condition because the neighbor 10.4.4.2 is not replying, causing the route to remain active indefinitely and potentially destabilizing the EIGRP domain.
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 route to 10.40.40.0/24 is stuck-in-active and may cause a network instability.
Why this is correct
A route in Active state with no successor indicates that the router is querying neighbors and has not received a reply, which can lead to SIA.
- ✗
The route to 10.40.40.0/24 is passive and stable.
Why it's wrong here
The code 'A' indicates Active, not Passive.
- ✗
The route to 10.30.30.0/24 has a feasible successor.
Why it's wrong here
The output shows only one successor for 10.30.30.0/24; no feasible successor is listed.
- ✗
All routes are in a stable passive state.
Why it's wrong here
The route to 10.40.40.0/24 is Active, so not all routes are passive.
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 300-410
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 10.50.50.0/24 EIGRP-IPv4 Topology Entry for AS(100)/ID(192.168.1.1) for 10.50.50.0/24 State: Active, Reply status: 0, Originating router: 192.168.1.1 Routing Descriptor Blocks: 10.1.1.2 (GigabitEthernet0/0), from 10.1.1.2, Send flag: 0x0 Composite metric: (4294967295/4294967295), Route is Internal Vector metric: Minimum bandwidth: 100000 Kbit Total delay: 100 microseconds Reliability: 255/255 Load: 1/255 Minimum MTU: 1500 Hop count: 1 Based on this output, what is the problem?
hard- ✓ A.The route is in Active state with an infinite metric, indicating that the router has lost the route and is querying for a new path.
- B.The route is passive and stable.
- C.The metric of 4294967295 is normal for a summary route.
- D.The hop count of 1 indicates the route is one hop away and reachable.
Why A: The route is in Active state with a composite metric of 4294967295 (the maximum 32-bit value, effectively infinite), which indicates that the router has lost the feasible successor and is actively sending queries to neighbors to find an alternative path. This is a classic sign of an EIGRP query process in progress, meaning the route is not stable or reachable.
Last reviewed: Jul 4, 2026
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