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This problem looks incorrect; the two links for Certkiller1 switch have a tie-breaker for the Root Port, and so the lowest port id is selected, Fa0/1 on CertKiller1. Also Fa0/2 on Certkiller2 is the DP since it's root, and therefore Fa0/2 on Certkiller1 is placed in blocking state. Who Agrees? (question below)

Two Certkiller switches are shown below:

This network is stable and operating properly. Assuming that default STP configurations are running on both switches, which port will be in blocking mode?

A. Port Fa0/1 on Certkiller 1
B. Port Fa0/2 on Certkiller 1
C. Port Fa0/1 on Certkiller 2
D. Port Fa0/2 on Certkiller 2
Answer: A

Explanation:
Spanning-Tree Protocol (STP) is a Layer 2 protocol that utilizes a special-purpose algorithm to discover physical loops in a network and effect a logical loop-free topology. STP creates a loop-free tree structure consisting of leaves and branches that span the entire Layer 2 network. The actual mechanics of how bridges communicate and how the STP algorithm works will be discussed at length in the following topics. Note that the terms bridge and switch are used interchangeably when discussing STP. In addition, unless otherwise indicated, connections between switches are assumed to be trunks. The switches move on to selecting Root Ports. The Root Port of a bridge is the port that is closest to the Root Bridge in terms of Path Cost. Every non-Root Bridge must select one Root Port. Again, bridges use the concept of cost to measure closeness. As with some routing metrics, the measure of closeness using STP is not necessarily reflected by hop count. Specifically, bridges track what is referred to as Root Path Cost, which is the cumulative cost of all links to the Root Bridge. So, Answer A is correct.