Huawei HCIA-Datacom V2.0 H12-811_V2.0 Dumps in PDF

Free Huawei H12-811_V2.0 Real Questions (page: 2)

On the network shown in the figure, the administrator creates Eth-Trunk 1 in manual mode on SW1 and SW2, and adds GE1/0/1 and GE1/0/2 on both switches to the Eth-Trunk. After the administrator runs the display interface brief command on SW1, the administrator finds that GE1/0/1 is Up and GE1/0/2 is Down.
Which of the following statements is true about this scenario?

  1. Eth-Trunk 1 is Up because it is Up as long as one member interface is Up.
  2. Eth-Trunk 1 is Down because the states of member interfaces must be the same in manual mode.
  3. Eth-Trunk 1 is Down because it has only two member interfaces and the minimum number of active links is not reached after one member interface is Down.
  4. Eth-Trunk 1 is Down because all member interfaces of the Eth-Trunk must be Up.

Answer(s): A

Explanation:

For an Eth-Trunk operating in manual load-balancing mode, the logical Eth-Trunk interface can remain Up as long as at least one member link is operational and properly added to the trunk. Therefore, if GE1/0/1 is Up and GE1/0/2 is Down, Eth-Trunk 1 can still stay Up, making option A correct.
Option B is incorrect because member interfaces in a manual Eth-Trunk do not need to be in identical physical states for the trunk itself to remain operational. Option C is also incorrect because there is no inherent rule here that two-member trunks must have both links active unless a separate minimum-links mechanism has been explicitly configured. No such condition is stated in the question. Option D is false because one of the key purposes of Eth-Trunk is redundancy; requiring all member interfaces to be Up would defeat that design advantage. HCIA-Datacom teaches that Eth-Trunk enhances link reliability by allowing continued forwarding when part of the bundle fails, provided the trunk still has an active forwarding member.



The administrator configures an Eth-Trunk in LACP mode between two switches, and sets the maximum number of active links in the Eth-Trunk to 3 and the number of remaining standby links to 1. If one of the active links fails, the two switches automatically adjust the number of active links to 2 through negotiation, and the standby link remains in the standby state.

  1. TRUE
  2. FALSE

Answer(s): B

Explanation:

This statement is false. In an Eth-Trunk operating in LACP mode, if the administrator sets the maximum number of active links to 3 and there is 1 standby link, then when one active member fails, LACP can automatically promote the standby member to active status. The purpose of the standby member is to maintain the configured active-link capacity whenever possible.
Therefore, after one active link fails, the normal expectation is not that the active-link count is reduced from 3 to 2 while the standby link remains idle. Instead, the standby link should participate and replace the failed member, so that the Eth-Trunk continues operating with 3 active links if the standby link and negotiation conditions are normal. This reflects the redundancy and fast recovery capabilities of LACP-based aggregation. HCIA-Datacom highlights that compared with manual mode, LACP provides more intelligent link selection, active/standby management, and fault adaptation. The statement is wrong because it ignores the intended behavior of the standby link in maintaining the number of active forwarding links after a failure.



In the figure, both PC1 and PC2 belong to VLAN 10. Interface GE1/0/1 on SW1 is a trunk interface with the default PVID, and interface GE1/0/2 on SW1 is an access interface in VLAN 10.
Which of the following statements are true about the forwarding of data frames between PC1 and PC2? (Select all that apply)

  1. After receiving the data frame sent from PC1 to PC2, SW1 adds the VLAN 10 tag to the data frame and then accepts the data frame.
  2. SW1 accepts the data frame sent from PC1 to PC2, but discards the data frame returned from PC2 to PC1.
  3. The data frame processed inside SW1 does not carry any VLAN tag.
  4. SW1 removes the VLAN 10 tag from the data frame before forwarding it from GE1/0/2 to PC2.

Answer(s): A,D

Explanation:

When a switch receives an untagged frame from a user host on an access interface, it associates that frame with the VLAN configured on the interface. In this scenario, PC1 and PC2 are in VLAN 10, and GE1/0/2 is an access interface in VLAN 10. After SW1 receives the frame from PC1, it internally associates the frame with VLAN 10, so option A is correct in the practical VLAN-processing sense used in HCIA-Datacom questions.
When the frame is forwarded out an access interface toward PC2, the switch sends it without a VLAN tag, because end hosts normally do not process 802.1Q tags. Therefore, option D is correct. Option B is incorrect because both hosts are in the same VLAN and can communicate normally; SW1 does not discard the return traffic for that reason. Option C is not correct in the standard VLAN switching model used by Huawei training, because the switch performs forwarding based on VLAN membership and internally processes the frame as belonging to VLAN 10. This question mainly checks understanding of access/trunk behavior and tag handling on ingress and egress.



In the STP topology shown in the figure, all links have the same path cost, and SW1 is the root bridge.
Which of the following ports will become designated ports? (Select all that apply)

  1. GE1/0/2 of SW2
  2. GE1/0/1 of SW2
  3. GE1/0/3 of SW3
  4. GE1/0/1 of SW4

Answer(s): A,B,C

Explanation:

In STP, the root bridge has all its active ports elected as designated ports on each connected segment. On non-root switches, one port is selected as the root port, and on each LAN segment, one designated port is elected based on the best BPDU. Since SW1 is the root bridge and all links have equal cost, the election depends first on the path cost to the root, then on bridge ID, and then on port ID if needed.
According to the topology, the ports that become designated are GE1/0/2 of SW2, GE1/0/1 of SW2, and GE1/0/3 of SW3, which correspond to option A, B, and
C. Option D is not a designated port in this topology because that segment has another port with a superior BPDU and therefore SW4’s port does not win the designated-port election. HCIA-Datacom emphasizes that designated ports are elected per segment, not per switch, and that every non-root switch must compare received BPDUs to determine root and designated roles. This question tests the ability to apply STP election logic rather than simply memorizing port-role definitions.



On the network shown in the figure, all switches run STP. All links have a path cost of 20, and SW1 is the root bridge.
What is the root path cost (RPC) of SW4?

  1. 20
  2. 40
  3. 60
  4. 80

Answer(s): B

Explanation:

The root path cost (RPC) of a switch is the total path cost from that switch to the root bridge along the selected spanning-tree path. Since SW1 is given as the root bridge, its RPC is 0. Each link in the topology has a path cost of 20. To determine SW4’s RPC, you identify the shortest active STP path from SW4 to SW1.
Based on the topology used in the question, SW4 reaches SW1 through a path containing two links. Therefore, the total root path cost is 20 + 20 = 40, which makes option B correct. This is a standard STP calculation. HCIA-Datacom uses RPC as one of the main comparison parameters in root-port election on non-root switches. A lower RPC means a better path toward the root bridge. If multiple candidate ports have the same RPC, the switch then compares the sender’s bridge ID and port I
D. This question checks whether the learner understands that RPC is cumulative across the spanning-tree forwarding path and is not simply the cost of a single directly connected link unless the switch is directly attached to the root bridge.



On a switched network where STP is enabled on all devices, when a downstream device detects a topology change, it continuously sends configuration BPDUs to the upstream device until the root bridge is informed of the topology change.

  1. TRUE
  2. FALSE

Answer(s): B

Explanation:

This statement is false. In classic STP, when a non-root switch detects a topology change, it does not continuously send configuration BPDUs upstream to notify the root bridge. Instead, it sends a Topology Change Notification (TCN) BPDU toward the root bridge through its root port. Each upstream switch acknowledges the TCN and forwards it further until it reaches the root bridge.
After the root bridge receives the topology change notification, the root bridge sets the Topology Change (TC) flag in its configuration BPDUs, and those configuration BPDUs are then propagated throughout the network. Therefore, topology-change notification and normal configuration BPDU transmission are different functions. The statement is wrong because it confuses TCN BPDUs with configuration BPDUs, and also inaccurately describes the notification behavior. HCIA-Datacom requires learners to distinguish between these BPDU types and understand how STP reacts to topology changes by accelerating MAC address aging and informing the network through the root bridge. This process helps the Layer 2 topology converge and reduces long-lasting forwarding inconsistencies after a link or port-state change.



After the root bridge is elected on an STP network, which of the following parameters may be compared by ports on non-root bridge nodes to elect the root port? (Select all that apply)

  1. PID of a port on the local device
  2. Root path cost (RPC)
  3. BID of the device that sends BPDUs
  4. PID of a port on the device that sends BPDUs

Answer(s): A,B,C,D

Explanation:

On a non-root bridge, the root port is the port that receives the best BPDU toward the root bridge. STP selects the root port by comparing several parameters in order. The first important parameter is the root path cost (RPC), so option B is correct. If multiple ports have the same RPC, the switch then compares the bridge ID (BID) of the upstream device sending the BPDU, making option C correct. If those are still equal, the switch compares the port ID (PID) of the upstream sending port, so option D is also correct.
If all of those values remain identical from the switch’s perspective, the device can finally compare the local port ID to determine which local interface becomes the root port, so option A is also correct. HCIA-Datacom teaches this comparison logic as part of STP election rules. The process ensures deterministic selection of a single root port on every non-root switch. Understanding the comparison sequence is essential for predicting STP topology behavior and for influencing port roles through path cost tuning or bridge-priority adjustments during campus network design and troubleshooting.



On an STP network, an administrator wants SW2 to be elected as the root bridge.
Which of the following methods can be used to achieve this goal?

  1. Set a lower port priority on SW2.
  2. Set a higher system priority on SW2.
  3. Set a lower system priority on SW2.
  4. Set a lower root path cost on SW2.

Answer(s): C

Explanation:

In STP, the root bridge is elected based on the lowest bridge ID (BID). A BID is composed of the bridge priority and the MAC address of the switch. The device with the smallest BID becomes the root bridge. Therefore, if an administrator wants SW2 to become the root bridge, the most direct and correct method is to configure a lower system priority on SW2, making option C correct.
Option A is incorrect because port priority affects port-role selection in some tie-breaking scenarios, but it does not determine root-bridge election. Option B is the opposite of what is required, because a higher system priority makes the BID larger and reduces the chance of being elected as root. Option D is also incorrect because root path cost is used by non-root bridges to select the best path to the existing root bridge; it does not determine which switch becomes the root bridge in the first place. HCIA-Datacom stresses that bridge priority tuning is the standard administrative method used to control root bridge placement in campus networks for stable and predictable Layer 2 topology design.



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