Huawei Updated H12-811_V2.0 Exam Questions and Answers by macy

This statement is true . ARP is a Layer 2 broadcast-based protocol used to resolve an IPv4 address into a MAC address within the same broadcast domain. If two PCs are connected to the same switch but cannot learn each other’s ARP entries, one likely reason is an incorrect VLAN configuration .

For example, if PC1 and PC2 are placed in different VLANs, or if one interface is configured with the wrong access VLAN, ARP broadcast frames from one PC will not reach the other PC because VLANs separate Layer 2 broadcast domains. As a result, the ARP request will not be received by the peer, and the MAC address cannot be resolved. Other causes may also exist, such as incorrect IP addressing, port isolation, or security policies, but VLAN misconfiguration is a very common and valid cause in campus switching scenarios. HCIA-Datacom teaches that ARP communication depends on correct Layer 2 domain membership, and VLAN planning or interface configuration errors often directly affect host-to-host communication even when both hosts are physically connected to the same switch.

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.

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.

PC1 and PC3 are members of VLAN 2, while PC2 and PC4 are members of VLAN 3. SW1 receives the frame from PC1 on access interface GE1/0/1 and associates it with VLAN 2. Because all switch MAC address tables are initially empty, PC3 ' s destination MAC address is unknown to SW1. The switch therefore floods this unknown unicast frame only through other interfaces that can carry VLAN 2. GE1/0/3 connects to PC3 in VLAN 2, so it is a valid flood port. GE1/0/4 is a trunk that permits VLAN 2 and VLAN 3, so it is also eligible to carry the VLAN 2 frame. SW1 does not send the frame back through GE1/0/1, the ingress interface, and it does not send it through GE1/0/2 because that access port belongs to VLAN 3. Therefore, the interfaces through which SW1 does not send the frame are GE1/0/2 and GE1/0/1, corresponding to A and D.