hardMultiple ChoiceObjective-mapped
BFD Timer Calculations for CCNP ENARSI
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?
Quick Answer
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.
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 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.
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.
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Same concept, more angles
3 more ways 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 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 (1000000 µs = 1000 ms) and the remote MinTxInt. The remote MinTxInt is not directly shown in the output, but the remote MinRxInt is 1000000 µs. Since the remote MinTxInt must be at least its own MinRxInt (by default, MinTxInt ≥ MinRxInt), the remote MinTxInt is at least 1000000 µs. Thus the maximum is 1000000 µs, giving a negotiated receive interval of 1000 ms. Note that the holdown time (3000 ms) is the product of the multiplier (3) and the negotiated transmit interval, not the receive interval.
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.
JA
Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
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