Pass the Huawei HCIA-Datacom H12-811_V2.0 Questions and answers with ExamsMirror

On Huawei switches, a physical Ethernet interface usually works as a Layer 2 switch port by default. A Layer 2 interface cannot be assigned an IP address directly. If an administrator wants to configure an IP address on such an interface, the interface must first be converted into a Layer 3 interface . The command used for this is undo portswitch , so option B is correct.

After this command is executed, the interface no longer behaves as a Layer 2 switching port and can then support Layer 3 features such as direct IP address configuration. Options A , C , and D are all Layer 2 port-type commands used to configure the interface as a trunk, access, or hybrid port. These commands do not solve the problem because they keep the interface operating at Layer 2. HCIA-Datacom emphasizes the difference between Layer 2 switching interfaces and Layer 3 routed interfaces on switches. Understanding when to use undo portswitch is important in campus network deployment, especially when configuring routed uplinks, management interfaces, or inter-device Layer 3 links on switches.

A /28 subnet mask corresponds to 255.255.255.240, which means each subnet contains 16 IP addresses . The subnet boundaries in the last octet increase in steps of 16: 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, and so on. The host address 192.168.1.147 falls within the subnet range 192.168.1.144 to 192.168.1.159 .

In that subnet, the network address is 192.168.1.144, the usable host range is 192.168.1.145 through 192.168.1.158, and the broadcast address is 192.168.1.159. Therefore, option C is correct. Option D is the first usable host address, not the broadcast address. Option A is just another usable host address. Option B would be the broadcast address only for a /24 network, not for a /28. HCIA-Datacom requires learners to master subnetting because it is essential for IP planning, gateway deployment, route summarization, and troubleshooting Layer 3 communication issues in campus and enterprise networks.

ospf 1 → [R1]

area 0 → [R1-ospf-1]

network 10.0.12.1 0.0.0.0 → [R1-ospf-1-area-0.0.0.0]

ospf enable 1 area 0 → [R1-GE0/0/0]

Huawei devices support two common ways to enable OSPF on an interface.

The first method is the process-based method . In this method, you first enter system view, then enter the OSPF process view using ospf 1, so this command belongs to [R1] . After that, you enter the area view using area 0, so this command belongs to [R1-ospf-1] . Inside the area view, the command network 10.0.12.1 0.0.0.0 is used to match the interface with IP address 10.0.12.1, so it belongs to [R1-ospf-1-area-0.0.0.0] .

The second method is the interface-based method . In this method, OSPF is enabled directly under the interface using ospf enable 1 area 0, so this command belongs to [R1-GE0/0/0] .

This question tests the difference between global OSPF process configuration and direct interface OSPF enabling , both of which are standard Huawei OSPF configuration methods.

A /27 subnet mask is 255.255.255.224, which means each subnet contains 32 IP addresses. The subnet increments in blocks of 32 in the last octet: 0, 32, 64, 96, 128, 160, 192, 224. The address 192.168.1.200/27 belongs to the subnet 192.168.1.192/27.

For this subnet, the network address is 192.168.1.192, the broadcast address is 192.168.1.223, and the valid host range is from 192.168.1.193 to 192.168.1.222. Therefore, 192.168.1.221/27, 192.168.1.193/27, and 192.168.1.222/27 are all valid host addresses in the same subnet and are reachable at Layer 2 without routing. That makes A, C, and D correct.

Option B is not correct because 192.168.1.192/27 is the network address of the subnet and cannot be assigned to a host by default. This question checks IP subnetting and the ability to identify valid hosts within the same broadcast domain.

The configuration first defines a named time range, working-time, covering 08:00 through 18:00 on working days. It then creates a named basic ACL and applies a deny rule for source network 192.168.1.0/24 using wildcard mask 0.0.0.255, with the rule tied to that time range. A packet from any address in 192.168.1.0 through 192.168.1.255 satisfies the source-address portion of the rule. The time-range condition means the rule is active only during the defined working-time period on the device. Outside that period, this particular rule is not considered a match. A basic ACL is appropriate here because the match depends on source IPv4 address and time, both of which are supported basic-ACL conditions. Whether traffic is ultimately permitted or denied also depends on where the ACL is applied and on any other ACL rules, but the question asks only whether the specified source addresses match this rule during the stated working hours. They do. Therefore, the statement is TRUE and option A is correct.

An Ethernet switch learns MAC address entries dynamically by examining the source MAC address of incoming frames. When SW1 first receives frames from PC1 on GE1/0/1, it creates a MAC address table entry mapping PC1’s MAC address to that port. If PC1 later sends frames that arrive on GE1/0/2, SW1 updates the learned entry to reflect the new incoming interface. Therefore, option B is correct: SW1 saves the mapping between PC1’s MAC address and GE1/0/2 and removes or overwrites the old mapping associated with GE1/0/1.

Option A is incorrect because a single dynamic MAC address is not normally maintained on two different ports within the same VLAN at the same time; that would cause forwarding ambiguity. Option C is incorrect because a normal move of one MAC address does not require clearing the entire MAC table. Option D is also incorrect because switches update MAC learning dynamically upon receiving new source traffic and do not have to wait for the previous entry to age out. HCIA-Datacom uses this principle to explain MAC address learning, migration, and troubleshooting of loops or host movement in switched Ethernet networks.

In TCP, the acknowledgment number indicates the next byte that the receiver expects to receive. In the figure, PC1 sends a TCP segment to PC2 with sequence number = a+22 and payload length = 22 bytes . Since TCP sequence numbers count bytes, the last byte carried in this segment corresponds to sequence number a+43 . Therefore, after PC2 successfully receives this data, it sends back an ACK indicating that the next expected byte is a+44 .

That is why option B is correct. This is a standard TCP reliability mechanism. TCP does not acknowledge “packets” as units; it acknowledges the byte stream. The ACK number always points to the next byte expected from the peer. The source and destination port numbers identify the session, but the sequence and acknowledgment numbers track ordered delivery. HCIA-Datacom uses this mechanism to explain connection-oriented transport, reliable delivery, retransmission, and traffic control. Understanding how the payload length affects sequence progression is essential when analyzing packet captures and troubleshooting TCP communication problems in enterprise networks.

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.

RADIUS and HWTACACS are AAA protocols that support accounting functions in addition to authentication and authorization. Accounting records can be used to track login sessions, service usage, command activity, or other information required by the deployment. LDAP and Microsoft Active Directory, by contrast, are directory-based identity technologies. Huawei documentation describes LDAP/AD integration primarily for authentication and authorization by querying directory information and validating user credentials. They are not used as the native AAA accounting protocol between a Huawei AAA client and an AAA server in the same way as RADIUS or HWTACACS. Therefore, when the question asks which listed protocols cannot be used for accounting between the AAA client and AAA server, AD and LDAP are the appropriate selections. RADIUS explicitly carries accounting messages, and HWTACACS also separates and supports authentication, authorization, and accounting functions. The source file leaves the answer blank, but the Datacom AAA model supports a verified answer of B and D.

This question tests the difference between Layer 3 addressing and Layer 2 encapsulation during routing. The IP packet originated by PC1 keeps PC1 ' s IP address as its source IP and PC2 ' s IP address as its destination IP as it traverses the routers. Ethernet MAC addresses, however, are rewritten at every routed hop. When R1 forwards the packet across the link toward R2, R1 builds a new Ethernet frame on its outgoing interface. According to the exhibit, R1 ' s outgoing interface on the R1-R2 link uses MAC_C, so MAC_C is the source MAC address. The next-hop device on that link is R2, whose receiving interface uses MAC_D, so MAC_D is the destination MAC address. MAC_A belongs to PC1 and was relevant only on the first local Ethernet segment. MAC_F belongs to PC2 and is used on the final Ethernet segment from R2 to PC2, not on the R1-R2 link. Therefore the correct statements are B and D. The uploaded key B, C is incorrect.