mediumMultiple ChoiceObjective-mapped
300-410 Practice Question: Runs the following command on Router R1: R1# show…
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?
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 detection time is 3000 ms.
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.
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 detection time is 3000 ms.
Why this is correct
It is the holddown timer shown in the output, which is 3000 ms.
- ✗
The detection time is 1000 ms.
Why it's wrong here
This is the transmit interval, not the detection time.
- ✗
The detection time is 1500 ms.
Why it's wrong here
This would be if the multiplier was 1.5, but it is 3.
- ✗
The detection time is 9000 ms.
Why it's wrong here
This would be if the transmit interval was 3000 ms, but it is 1000 ms.
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