Examine the network diagram.
Which switch port(s) will be in a forwarding state? (Choose two.)
A. SwitchA – Fa0/1 and Fa0/2
B. SwitchA – Fa0/1
C. SwitchA – Fa0/2
D. SwitchB – Fa0/1
E. SwitchB – Fa0/2
Correct Answers: A, D
Explanation:
Both switch ports on Switch A and Fa0/1 on Switch B will be in a forwarding state. Switch A will become the STP root bridge due to its lower MAC address. All ports on the root bridge will become designated ports in a forwarding state. Switch B has redundant connectivity to the root bridge, and must block one of its interfaces to prevent a switching loop. Both interfaces are the same speed (FastEthernet), and thus their cost to the root is the same. Finally, the interface with the lowest number will become the forwarding port. F0/1 has a lower port number than F0/2, so F0/1 becomes a forwarding port, and F0/2 becomes a blocking port.
In this scenario there are only two switches in the diagram. However, if there were more switches and Switch A were not the root bridge, the result would be the same with regard to the ports between Swicth A and B. Whenever there are redundant links between switches, one of the four ports involved will be set to a blocking (or in the case of RSTP, discarding) mode. The logic will still be the same, since the cost to get to the root bridge will still be equal if the port speeds are equal.
Without STP (which can be disabled) operating on switches with redundant links, such as those in the figure, loops can and almost surely will occur. For example, if a host connected to SwitchA were to send an ARP request for the MAC address of a host connected to SwitchB, the request could loop and cause a broadcast storm, slowing performance dramatically. This would probably occur when any host connected to either switch sends a broadcast frame, such as a DHCP request.
Rapid Spanning Tree Protocol (RSTP) uses the term discarding for a switch port that is not forwarding frames.