CompTIA Network+ (N10-009)Network ImplementationMedium
Three switches are interconnected in a triangle topology running Spanning Tree Protocol. All switches have the default priority of 32768. Switch A has MAC address 00:1A:2B:3C:4D:5E, Switch B has 00:1A:2B:3C:4D:5D, and Switch C has 00:1A:2B:3C:4D:5F. Which switch will be elected the root bridge?
- ASwitch A, because it was powered on first
- BSwitch C, because it has the highest MAC address
- CThe switch with the most connected ports
- DSwitch B, because it has the lowest MAC address
Show answer & explanationAnswer & explanation
Correct answer: D. Switch B, because it has the lowest MAC address
When priority values are tied, STP breaks the tie using the lowest MAC address to determine the root bridge. Switch B's MAC address (00:1A:2B:3C:4D:5D) is numerically lower than Switch A's (...5E) and Switch C's (...5F), so Switch B becomes root.
Why the other options are wrong
- A. Power-on order does not factor into STP root bridge election.
- B. The highest MAC address is never favored in STP tie-breaking.
- C. Port count has no bearing on root bridge selection in STP.
STP Root Bridge Election
Spanning Tree Protocol elects the root bridge by comparing Bridge IDs, which consist of a priority value and the switch's MAC address; the lowest Bridge ID wins.
- Default priority is 32768 on most switches
- Tie-breaker is lowest MAC address
- Root bridge becomes the reference point for the loop-free tree
Memory trick: Lowest Bridge ID wins—like golf, low score takes the crown.