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H3C Constructing H3C High-Performance Campus Networks Sample Questions:

1. 由于网络规模扩大可能导致管理工作越来越繁重,因此在网络建设初期,网络管理员都希望采取各种各样的技术来降低网络管理工作量其中最为常用的方法是简化网络结构,统一网络设备等方法下列哪些技术属于简化网络结构的技术

A) 万兆以太网
B) 集群技术
C) SNMP 协议
D) 堆叠技术

2. 三台路由器 RTA 、 RTB 、 RTC 通过局域网连接在一起,组成 VRRP 备份组,各路由器接口配置如下:
RTA 的 GigabitEthernet ip address 192.168.0.252 255.255.255.0 vrrp vrid 1 virtual-ip 192.168.0.254 vrrp vrid 1 priority 120 RTB 的 GigabitEthernet ip address 192.168.0.253 255.255.255.0 vrrp vrid 1 virtual-ip 192.168.0.254 vrrp vrid 1 priority 90 RTC 的 GigabitEthernet ip address 192.168.0.254 255.255.255.0 vrrp vrid 1 virtual-ip 192.168.0.254
从上述信息可以得知

A) RTC 为备份组 Master 路由器
B) RTA 为备份组 Master 路由器
C) RTB 为备份组 Master 路由器
D) RTA 发后, RTB 为备份组 Master 路由器

3. 客户的二台交换机 SWA 、 SWB 通过七根以太网线连接在一起,并配置了动态链路聚合,交换机 SWA 上对应端口为 Ethernet 1/0 Ethernet 1/0/2 、 Ethernet 1/0/3 、 Ethernet 1/0/4 、 Ethernet 1/0/5 、 Ethernet 1/0/6 、 Ethernet1/0/7 如果交换机 SWA 、 SWB 每个聚合组只支持 6 个端口,且 SWA 交换机有如下配置:
[SWA]interface Ethernet 1/0/7
[SWA-Ethernet1/0/7]lacp port-priority 4096
设备上査看 LACP 状态输出如下:
[SWA]display lacp system-id
Actor System ID: 0x8000, 00e1-fc00-5000
[SWB]display lacp system-id
Actor System ID: 0x8000, 00e0-fc43-7384
从上述信息可以得知 _

A) 在各端口配置一致的情况下, Ethernet1/0/6 端口为聚合组 Unselected 端口
B) 在各端口配置一致的情况下, Ethernet1/0/1 端口为聚合组 Unselected 端口
C) 在各端口配置一致的情况下, Ethernet1/0/7 端口为聚合组 Unselected 端口
D) 在各端口配置一致的情况下, Ethernet1/0/2 端口为聚合组 Unselected 端口

4. 在如图所示的交换网络中,交换机 SWA 上设置 VLAN10 为 Isolate-user-vlan VLAN2 和 VLAN3 为 VLAN 10 的 Secondary VLAN ；交换机 SWB 上创建 VLAN2?VLAN20 ,并且设置 VLAN20 为 Isolate-user-vlan , VLAN4 为 VLAN20 的 Secondary VLAN 设置好各设备的 IP 地址后,在 SWA 上査看 MAC 地址表,将显示 PCB 属于（选择一项或多项)

A) VLAN3
B) VLAN20
C) VLAN4
D) VLAN2
E) VLAN 10
F) VLAN 1

5. 如图所示组网中,路由器的单播路由表如下:
<H3C>display ip routing-table
Routing Tables: Public
Destinations : 9Routes : 9
Destination/MaskProtoPreCost NextHop Interface
10.1.1.0/30 Direct 0010.1.1.2S0/2/0
10.1.1.1/32 Direct 0010.1.1.1 S0/2/0
10.1.1.2/32 Direct 00127.0.0.1 InLoopO
11.1.1.0/30 Direct 0011.1.1.1 Eth0/110
11.1.1.1/32 Direct 00127.0.0.1 InLoopO
12.1.1.0/30 Direct 0012.1.1.1Eth0/1/1
12.1.1.1/32 Direct 00127.0.0.1 InLoopO
13.1.1.0/30 Direct 0013.1.1.1EthO/1/2
13.1.1.1/32 Direct 00127.0.0.1 InLoopO
100.1.1.0/240SPF 10 1563 10.1.1.1S0/2/0
127.0.0.0/8 Direct 00127.0.0.1 InLoopO
127.0.0.1/32Direct 00127.0.0.1 InLoopO
200.1.1.0/240SPF 10 212.1.1.2Eth0/1/1
那么,接收者（ Receiver )会收到的组播数据为

A) （ 100.1.1.1 , 226.3.3.3 )
B) （ 100.1.1.1 , 226.1.1.1 )
C) （ 200.1.1.1 , 226.2.2.2 )

Solutions: