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[学习交流] CCNA640-802 V13题库试题分析23(艾迪飞吴老师讲解) [复制链接]

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发表于 2007-12-6 12:10:39 |显示全部楼层

1．The network security policy requires that only one host be permitted to attach dynamically to each switch interface. If that policy is violated, the interface should shut down. Which two commands must the network administrator configure on the 2950 Catalyst switch to meet this policy? (Choose two.)<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /><o:p></o:p>
A. Switch1(config-if)# switchport port-security maximum 1<o:p></o:p>
B. Switch1(config)# mac-address-table secure<o:p></o:p>
C. Switch1(config)# access-list 10 permit ip host<o:p></o:p>
D. Switch1(config-if)# switchport port-security violation shutdown <o:p></o:p>
<?xml:namespace prefix = st1 ns = "urn:schemas-microsoft-com:office:smarttags" /><st1:place w:st="on">E. Switch1</st1:place>(config-if)# ip access-group 10 <o:p></o:p>
Answer: AD <o:p></o:p>
解释一下：switchport port-security maximum 1这个命令是配置这个端口为安全模式且只允许有学习一个mac地址。<o:p></o:p>
Switchport port-security violation shutdown:这句命令的意思是如果借口违反了安全策略就shutdown该接口。<o:p></o:p>
<o:p> </o:p>
<o:p> </o:p>
2．Which statement is true about full-duplex Ethernet in comparison to half-duplex Ethernet? <o:p></o:p>
A. Full-duplex Ethernet consists of a shared cable segment. Half-duplex Ethernet provides a point-to-point link. <o:p></o:p>
B. Full-duplex Ethernet uses a loopback circuit to detect collisions. Half-duplex Ethernet uses a jam signal. <o:p></o:p>
C. Full-duplex Ethernet can provide higher throughput than can half-duplex Ethernet of the same bandwidth. <o:p></o:p>
D. Full-duplex Ethernet uses two wires to send and receive. Half-duplex Ethernet uses one wire to send and receive. <o:p></o:p>
Answer: C<o:p></o:p>
解释一下：全双工和半双工的区别：全双工的是既能收也能发，而半双工是在收的时候是不能发数据的。所以他能更有效得利用带宽。所以对于相同的带宽，全双工的比半双工的能提供更高的吞吐量。<o:p></o:p>
<o:p> </o:p>
3．Refer to the exhibit.<o:p></o:p>
<?xml:namespace prefix = v ns = "urn:schemas-microsoft-com:vml" /><v:shapetype id=_x0000_t75 stroked="f" filled="f" path="m@4@5l@4@11@9@11@9@5xe" o:preferrelative="t" o:spt="75" coordsize="21600,21600"><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path o:connecttype="rect" gradientshapeok="t" o:extrusionok="f"></v:path><o:lock aspectratio="t" v:ext="edit"></o:lock></v:shapetype><o:p></o:p>
A network associate has configured OSPF with the command:<o:p></o:p>
City(config-router)# network 192.168.12.64  0.0.0.63 area 0<o:p></o:p>
After completing the configuration, the associate discovers that not all the interfaces are participating in OSPF.<o:p></o:p>
Which three of the interfaces shown in the exhibit will participate in OSPF according to this configuration statement? (Choose three.)
<o:p>

3.jpg (32.17 KB, 下载次数: 16) </o:p>
A. FastEthernet0 /0 <o:p></o:p>
B. FastEthernet0 /1<o:p></o:p>
C. Serial0/0<o:p></o:p>
D. Serial0/1.102 <o:p></o:p>
E. Serial0/1.103<o:p></o:p>
F. Serial0/1.104 <o:p></o:p>
Answer: BCD <o:p></o:p>
解释一下：192.168.12.64 0.0.0.63匹配的地址范围是 192.168.12.64-192.168.12.127 ，所以被激活的接口有FastEthernet0 /1, Serial0/0, Serial0/1.102.<o:p></o:p>
<o:p> </o:p>
<o:p> </o:p>
4．Which statements are true about EIGRP successor routes? (Choose two.)
A. A successor route is used by EIGRP to forward traffic to a destination.<o:p></o:p>
B. Successor routes are saved in the topology table to be used if the primary route fails. <o:p></o:p>
C. Successor routes are flagged as "active" in the routing table. <o:p></o:p>
D. A successor route may be backed up by a feasible successor route. <o:p></o:p>
E. Successor routes are stored in the neighbor table following the discovery process. <o:p></o:p>
Answer: AD<o:p></o:p>
解释一下：EIGRP的successor路径是他选出的最优的路径，路由器将选择这条路径到达目的地。而feasible successor 则是successor的备份的路径，如果successor路径出了问题，就立即将feasible successor路径转为successor路径转发数据。<o:p></o:p>
<o:p> </o:p>
<o:p> </o:p>
5．Which statement is correct about the internetwork shown in the diagram?
<o:p>

5.jpg (38.33 KB, 下载次数: 13) </o:p>
<o:p></o:p>
A. Switch 2 is the root bridge. <o:p></o:p>
B. Spanning Tree is not running.<o:p></o:p>
C. Host D and Server 1 are in the same network.<o:p></o:p>
D. No collisions can occur in traffic between Host B and Host C.<o:p></o:p>
E. If Fa0/0 is down on Router 1, Host A cannot access Server 1.<o:p></o:p>
F. If Fa0/1 is down on Switch 3, Host C cannot access Server 2.<o:p></o:p>
Answer: E<o:p></o:p>
解释一下：这个涉及到多个交换机的多个VLAN间的通讯<o:p></o:p>
。而在交换机之间有一个环路，因此生成树阻断了其中的一个接口，根据生成树的原理，block的端口是在非根桥上的，因此switch2不会是根桥。<o:p></o:p>
而host D和server一看就不是在相同的VLAN之间，因此是不在相同的网络中的。<o:p></o:p>
我们说整个hub是一个大的冲突域和广播域，此host B和host C 之间是有冲突存在的。<o:p></o:p>
Host A和server在不同的VLAN之间，他们之间通讯是要借助于Router 1的接口Fa0/0的子接口来实现的，因此如果Router 1的接口Fa0/0 down了，那么Host A和server 1之间就无法通讯了。<o:p></o:p>
同理host C与server 2之间的通讯也是借助于Router 1的Fa0/0来实现的，如果switch 3的Fa0/1 down了，那么生成树的就会重新选举，以前block的接口就会起来转发数据，host C就通过switch 2将他的数据发送出去，他与server 2之间的连通性还是不能失去的。<o:p></o:p>

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[学习交流] CCNA640-802 V13题库试题分析23(艾迪飞吴老师讲解) [复制链接]

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