Root Bridge Election on a Priority Tie: Lowest MAC Wins (SW2)

Interpret basic operations of Rapid PVST+ Spanning Tree Protocol
Answer Correct answer: B — SW2 and SW3 tie at priority 4096, so the lowest MAC decides; SW2 (0C:0E:16:11:05:97) is lower than SW3, making SW2 root.

Refer to the exhibit. Which switch in this configuration will be elected as the root bridge? SW1: 0C:E4:82:33:62:23 - SW2: 0C:0E:16:11:05:97 - SW3: 0C:E0:16:1A:3C:9D - SW4: 0C:00:18:A1:B3:19 - image

  1. SW1
  2. SW2 Correct Answer
  3. SW3
  4. SW4

Community Votes

B
100%

100% of anonymous learners picked answer B. Votes are pick records left by other test-takers — they are not the verified answer.

Community Insight

Priority is compared before MAC. Only when two switches share the lowest priority does the lowest MAC become the decider.

STP elects the root by lowest bridge priority first. SW2 and SW3 are tied at priority 4096, so the tiebreaker is the lowest MAC address, and SW2's MAC (0C:0E:16:11:05:97) is lower than SW3's, so SW2 becomes root.

Ignoring the priority tie and picking the globally lowest MAC (SW4 0C:00:18...) — SW4's priority is higher than the 4096 tie, so it cannot win.

Community Discussion (6 comments)

f2bdbb0 👍 6
the correct answer is sw 4..
Simrankoor 👍 2
B is the correct answer becasue of priority tie between s2 and s3, the other swithes priority is way high the lowest mac address wont have any effect
bezkin 👍 1 Selected: B
stp uses lowest values always (this is the reason we always have examples of old switches with older (lower) mac addresses being connected to the network and taking over as root bridge)
3040636 👍 1
answer is correct B. First two lowest priority tally and next lowest MAC
exiledwl 👍 2
B) SW2 is correct, don't forget the stp priority is checked first then the lowest mac address if there's a tie. In this case there is a tie between Sw2 and Sw3 with both having priority 4096, between these two we check who has the lowest mac add which is SW2 so B is correct
379c9f6 👍 2
correct

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Expert Analysis

Why the Answer Is Correct

B (SW2) is correct. STP compares bridge priority first; SW2 and SW3 are both at the lowest priority (4096), so the election falls to the lowest MAC address. SW2's MAC 0C:0E:16:11:05:97 is lower than SW3's 0C:E0:16:1A:3C:9D, so SW2 wins.

Why the Other Options Are Wrong

A (SW1) and D (SW4) have higher bridge priorities than the tied 4096 pair and are eliminated before the MAC check. C (SW3) loses the MAC tiebreaker to SW2.

Community Comment Notes

The vote is B (100). Commenters explicitly note the SW2/SW3 priority tie at 4096 and that the lowest MAC then elects SW2.

Official Reference

https://learningnetwork.cisco.com/s/ccna-exam-topics

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