How does the "lowest sender port ID" tiebreaker work when two switches are connected by multiple links?

The port connected to the sender's lowest port ID becomes the root port; the other becomes an alternate port.

This tiebreaker applies when a switch (e.g., S4) is connected to the root bridge (S1) via two parallel links and all other tiebreakers are equal:

Scenario: S4 connects to S1 via F0/6↔F0/1 and F0/5↔F0/2

1. Sender BID (Bridge ID)? Same switch (S1), so it's a tie
2. Sender port priority? Both are 128 (default), so it's a tie
3. Sender port ID? F0/1 on S1 < F0/2 on S1 → F0/1 wins

Result:

• S4's port F0/6 (connected to S1's F0/1) → Root Port (forwarding)
• S4's port F0/5 (connected to S1's F0/2) → Alternate Port (blocking)

Important: If an administrator configures a lower port priority on S1's F0/2, that would override the port ID tiebreaker and make F0/5 the root port instead.

Go deeper:

• Spanning-Tree Port Roles (NetworkAcademy.IO) — the parallel-link case that falls through to the sender's lowest port ID.