The answer is 3000 milliseconds, though the output misleadingly displays 1500 ms. This discrepancy arises because the BFD holddown timer calculation uses the maximum of the local MinRxInt and the received MinRxInt, multiplied by the multiplier. Here, the local MinRxInt is 1,000,000 microseconds and the received MinRxInt is 500,000 microseconds, so the maximum is 1,000,000 microseconds; multiplied by the multiplier of 3 gives 3,000,000 microseconds, or 3000 ms. However, Cisco’s implementation halves this value for display purposes in the show bfd neighbors detail output, showing 1500 ms instead of the actual holddown time. On the CCNP ENARSI 300-410 exam, this is a classic trap: candidates often mistakenly read the displayed holddown as the real value, but the exam tests your understanding that the holddown timer is always max(local MinRxInt, received MinRxInt) × multiplier. A reliable memory tip is “Double the display, trust the math”—always calculate the holddown from the MinRxInt values and multiplier, ignoring the displayed number.
300-410 Bidirectional Forwarding Detection (BFD) Practice Question
This 300-410 practice question tests your understanding of bidirectional forwarding detection (bfd). Examine the command output carefully: the correct answer depends on what the output actually shows, not on general recall alone. 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 runs the following command on Router R1:
R1# show bfd neighbors detail
IPv4 Sessions
NeighborAddr LD/RD Int State Holdown(mult) Intf
10.1.1.2 1/3 Gi0/0 Up 3000(3) Gi0/0
Session state is UP and not using echo function.
OurAddr: 10.1.1.1
Handle: 1
Local Diag: 0, Demand mode: 0, Poll bit: 0
MinTxInt: 1000000, MinRxInt: 1000000, Multiplier: 3
Received MinRxInt: 500000, Received Multiplier: 3
Holddown (hits): 1500(0)
Rx Count: 200, Tx Count: 200
Based on this output, what is the holddown timer value in milliseconds and why?
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
The holddown timer is 3000 ms, but the output shows 1500 ms because the holddown timer displayed is half of the actual holddown time.
The holddown timer is calculated as the maximum of the local MinRxInt and the received MinRxInt, multiplied by the multiplier. Local MinRxInt is 1000000 microseconds, received MinRxInt is 500000 microseconds. The maximum is 1000000 microseconds. Multiplied by 3 gives 3000000 microseconds = 3000 ms. However, the output shows 1500 ms. This is because the holddown timer displayed is actually half of the calculated value due to a Cisco implementation detail where the holddown timer is divided by 2 for display purposes. The actual holddown time is 3000 ms.
Key principle: OSPF neighbour adjacency depends on matching area, hello/dead timers, network type, and authentication — IP reachability alone is not enough.
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 holddown timer is 1500 ms, which is the received MinRxInt (500 ms) multiplied by the multiplier (3).
Why it's wrong here
The holddown timer uses the maximum of local and received MinRxInt, not just the received.
✓
The holddown timer is 3000 ms, but the output shows 1500 ms because the holddown timer displayed is half of the actual holddown time.
Why this is correct
Cisco IOS divides the holddown timer by 2 for display; the actual holddown is 3000 ms.
The holddown timer is 1500 ms, which is the local MinRxInt (1000 ms) multiplied by the multiplier (3) divided by 2.
Why it's wrong here
The calculation uses the maximum of MinRxInt values, not just local.
✗
The holddown timer is 3000 ms, and the output is correct as is.
Why it's wrong here
The output shows 1500 ms, not 3000 ms.
Common exam traps
Common exam trap: OSPF can fail even when IP connectivity looks correct
OSPF neighbour formation depends on matching areas, timers, network type, authentication and passive-interface behaviour. Do not choose an answer only because the devices can ping.
Trap categories for this question
Command / output trap
The output shows 1500 ms, not 3000 ms.
Detailed technical explanation
How to think about this question
OSPF questions usually test the details that control adjacency and route selection. Read the neighbour state, area, router ID and interface configuration before deciding what is wrong.
KKey Concepts to Remember
OSPF neighbours must agree on key parameters.
Router ID selection can affect neighbour relationships and LSDB output.
OSPF cost influences the preferred path.
A route can appear in OSPF information but not become the installed route.
TExam Day Tips
→Check area mismatch first when OSPF adjacency fails.
→Review passive interfaces when a network is advertised but no neighbour forms.
→Use show ip ospf neighbor and show ip route clues carefully.
Key takeaway
OSPF neighbour adjacency depends on matching area, hello/dead timers, network type, and authentication — IP reachability alone is not enough.
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 300-410 question in full detail.
Review OSPF neighbour requirements — matching area type, hello and dead timers, network type, stub flags, and authentication. Study show ip ospf neighbor states (INIT, 2-WAY, FULL). Then practise related 300-410 OSPF questions on adjacency and route selection.
Bidirectional Forwarding Detection (BFD) — This question tests Bidirectional Forwarding Detection (BFD) — OSPF neighbours must agree on key parameters..
What is the correct answer to this question?
The correct answer is: The holddown timer is 3000 ms, but the output shows 1500 ms because the holddown timer displayed is half of the actual holddown time. — The holddown timer is calculated as the maximum of the local MinRxInt and the received MinRxInt, multiplied by the multiplier. Local MinRxInt is 1000000 microseconds, received MinRxInt is 500000 microseconds. The maximum is 1000000 microseconds. Multiplied by 3 gives 3000000 microseconds = 3000 ms. However, the output shows 1500 ms. This is because the holddown timer displayed is actually half of the calculated value due to a Cisco implementation detail where the holddown timer is divided by 2 for display purposes. The actual holddown time is 3000 ms.
What should I do if I get this 300-410 question wrong?
Review OSPF neighbour requirements — matching area type, hello and dead timers, network type, stub flags, and authentication. Study show ip ospf neighbor states (INIT, 2-WAY, FULL). Then practise related 300-410 OSPF questions on adjacency and route selection.
What is the key concept behind this question?
OSPF neighbours must agree on key parameters.
About these practice questions
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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 bfd neighbors detail
IPv4 Sessions
NeighborAddr LD/RD Int State Holdown(mult) Intf
10.1.1.2 1/3 Gi0/0 Up 3000(3) Gi0/0
Session state is UP and not using echo function.
OurAddr: 10.1.1.1
Handle: 1
Local Diag: 0, Demand mode: 0, Poll bit: 0
MinTxInt: 1000000, MinRxInt: 1000000, Multiplier: 3
Received MinRxInt: 1000000, Received Multiplier: 3
Holddown (hits): 3000(0)
Rx Count: 100, Tx Count: 100
Based on this output, what is the BFD session's negotiated transmit interval?
medium
✓ A.The negotiated transmit interval is 1000 ms.
B.The negotiated transmit interval is 500 ms.
C.The negotiated transmit interval is 3000 ms.
D.The negotiated transmit interval is 100 ms.
Why A: The negotiated transmit interval is the maximum of the local MinTxInt and the received MinRxInt. Local MinTxInt is 1000000 microseconds, received MinRxInt is 1000000 microseconds. The maximum is 1000000 microseconds, which is 1000 ms. The BFD session will transmit control packets every 1000 ms.
Variation 2. A network engineer runs the following command on Router R1:
R1# show bfd neighbors detail
IPv4 Sessions
NeighborAddr LD/RD Int State Holdown(mult) Intf
10.1.1.2 1/3 Gi0/0 Up 3000(3) Gi0/0
Session state is UP and not using echo function.
OurAddr: 10.1.1.1
Handle: 1
Local Diag: 0, Demand mode: 0, Poll bit: 0
MinTxInt: 1000000, MinRxInt: 1000000, Multiplier: 3
Received MinRxInt: 1000000, Received Multiplier: 3
Holddown (hits): 3000(0)
Rx Count: 100, Tx Count: 100
Based on this output, what is the BFD session's negotiated receive interval?
medium
✓ A.The negotiated receive interval is 1000 ms.
B.The negotiated receive interval is 500 ms.
C.The negotiated receive interval is 3000 ms.
D.The negotiated receive interval is 100 ms.
Why A: The negotiated receive interval is the maximum of the local MinRxInt and the received MinTxInt. However, the output does not show the received MinTxInt directly. The received MinRxInt is shown, but the receive interval is determined by the local MinRxInt. The local MinRxInt is 1000000 microseconds, so the router can receive BFD packets at intervals of 1000 ms or slower. The negotiated receive interval is typically the local MinRxInt, which is 1000 ms.
Variation 3. A network engineer runs the following command on Router R1:
R1# show bfd neighbors detail
IPv4 Sessions
NeighborAddr LD/RD Int State Holdown(mult) Intf
10.1.1.2 1/3 Gi0/0 Up 3000(3) Gi0/0
Session state is UP and not using echo function.
OurAddr: 10.1.1.1
Handle: 1
Local Diag: 0, Demand mode: 0, Poll bit: 0
MinTxInt: 1000000, MinRxInt: 1000000, Multiplier: 3
Received MinRxInt: 1000000, Received Multiplier: 3
Holddown (hits): 3000(0)
Rx Count: 100, Tx Count: 100
Based on this output, what is the BFD session's detection time?
medium
✓ A.The detection time is 3000 ms.
B.The detection time is 1000 ms.
C.The detection time is 1500 ms.
D.The detection time is 9000 ms.
Why A: The detection time is the holddown timer, which is the negotiated transmit interval multiplied by the multiplier. The negotiated transmit interval is 1000 ms (maximum of local MinTxInt and received MinRxInt), and the multiplier is 3, so the detection time is 3000 ms.
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