1． Refer to the exhibit. A new subnet with 60 hosts has been added to the network. Which subnet address should this network use to provide enough usable addresses while wasting the fewest addresses?
A. 192.168.1.56/26
B. 192.168.1.56/27
C. 192.168.1.64/26
D. 192.168.1.64/27
Answer: C
解释一下：这还是一个地址规划的题目，需要为有60台主机的网络分配IP地址，那么
2n≥60（这里N表示的是网络位），得到N=6，所以主机位为32-6=26，所以可以使用的地址空间有：192.168.1.0/26,192.168.1.64/26,192.168.1.128/64,192.168.1.192/64,这些就是所有可以使用的64位的网段，而前面的几台路由器使用了192.168.1.0/26网段中的一部分的网段，所以可分配给R4的网段就只有192.168.64/26 , 1921.68.1.128/64 , 192.168.1.192/64了。
2． Refer to the exhibit. The routers are running RIPv2. Which addressing scheme would satisfy the needs of this network yet waste the fewest addresses?
A. Network 1: 192.168.10.0/26
Network 2: 192.168.10.64/26
Network 3: 192.168.10.128/26
Serial link 1: 192.168.20.0/24
Serial link 2: 192.168.30.0/24
B. Network 1: 192.168.10.0/26
Network 2: 192.168.10.64/28
Network 3: 192.168.10.80/29
Serial link 1: 192.168.10.88/30
Serial link 2: 192.168.10.96/30
C. Network 1: 192.168.10.0/26
Network 2: 192.168.10.64/27
Network 3: 192.168.10.96/28
Serial link 1: 192.168.10.112/30
Serial link 2: 192.168.10.116/30
D. Network 1: 192.168.10.0/27
Network 2: 192.168.10.64/28
Network 3: 192.168.10.96/29
Serial link 1: 192.168.10.112/30
Serial link 2: 192.168.10.116/30
Answer: C
解释一下：划分IP地址的时候，按从多到少的顺序划分。我们先来安排有50台主机的Network1的网段：2n≥50，得到N=6，所以网段为192.168.10.0/26；然后来分有20台主机的Network2的网段：2n≥≥20，得到N=5，所以网段为192.168.10.64/27；最后计算有10台主机的Network3：2n≥≥10，得到N=4，所以网段为192.168.10.96/28.
串行链路中是只须有两个可用的IP地址的，所以serial link1的网段为192.168.10.112/30,而serial link2的网段为192.168.10.116/30.
3. Refer to the exhibit. S0/0 on R1 is configured as a multipoint interface to communicate with R2 and R3 in this hub-and-spoke Frame Relay topology. While testing this configuration, a technician notes that pings are successful from hosts on the 172.16.1.0/24 network to hosts on both the 172.16.2.0/25 and 172.16.2.128/25 networks. However, pings between hosts on the 172.16.2.0/25 and 172.16.2.128/25 networks are not successful. What could explain this connectivity problem?
A. The ip subnet-zero command has been issued on the R1 router.
B. The RIP v2 dynamic routing protocol cannot be used across a Frame Relay network.
C. Split horizon is preventing R2 from learning about the R3 networks and R3 from learning about the R2 networks.
D. The 172.16.2.0/25 and 172.16.2.128/25 networks are overlapping networks that can be seen by R1, but not between R2 and R3.
E. The 172.16.3.0/29 network used on the Frame Relay links is creating a discontiguous network between the R2 and R3 router subnetworks.
Answer: C
解释一下：在帧中继的环境下，当使用物理接口的时候，水平分割默认是关闭的，而当使用子接口的时候，水平分割默认是开启的。图中的R1上使用的是多点子接口，所以在他的上面水平分割是开启的，所以R2 和R3的下面连接的网段是不能通过R1传递给对方的，所以R2的172.16.2.0/25与R3的172.16.2.128/25不能相互ping通。
4．Refer to the exhibit. What does STATUS=ACTIVE refer to in the output of the show frame-relay pvc command?
A. The PVC is experiencing congestion.
B. The Frame Relay switch is correctly programmed with the DLCI and is operational.
C. The router is actively broadcasting to establish a link to the Frame Relay switch.
D. The router is connected to the local Frame Relay switch, but not to the far end device.
Answer: B
解释一下：PVC的状态为active表示链路是好的可以正常使用的。
5．Refer to the exhibit. What is the meaning of the term dynamic as displayed in the output of the show frame-relay map command shown?
A. The Serial0/0 interface is passing traffic.
B. The DLCI 100 was dynamically allocated by the router.
C. The Serial0/0 interface acquired the IP address of 172.16.3.1 from a DHCP server.
D. The DLCI 100 will be dynamically changed as required to adapt to changes in the Frame Relay cloud.
E. The mapping between DLCI 100 and the end station IP address 172.16.3.1 was learned through Inverse ARP.
Answer: E
解释一下：dynamic表示是通过反向ARP学习到的地址。如果是static则表示为自己手动添加的静态的MAP。
