引用:

原帖由 非常菜 于 2007-6-22 13:05 发表


传统的子网划分，网络地址要去掉头和尾，正如计算公式：子网数 = 2的N次方-2 ，这个没有问题。
因此，如果是C类地址划分/26子网，可用子网数应该是：2的2次方-2 = 2个。

那27题的答案应该不对。27题找不到正确答案。

高版本的IOS中,默认起用了 subnet zero

引用:

原帖由 赵高 于 2007-6-25 11:43 发表
高版本的IOS中,默认起用了 subnet zero

那就对了，谢谢楼主。

觉得NA就是纯考英文

QUESTION NO: 36
On the topic of VLSM, which one of the following statements best describes the
concept of the route aggregation?
对于VLSM，下列哪种说法确切地描述了路由聚合的概念？

A. Deleting unusable addresses through the creation of many subnets.
B. Combining routes to multiple networks into one supernet.
C. Reclaiming unused space by means of changing the subnet size.
D. Calculating the available host addresses in the AS.

解析：路由聚合(也叫路由归纳或者超网)可以在路由器里面用一条路由代表多个网段。
而VLSM允许在编址计划中有更多的体系分层，因此可以在路由表内进行更好的路由归纳，减少内存占用。

VLSM(Variable Length Subnet Mask 可变长子网掩码)是一种产生不同大小子网的网络分配机制，指一个网络可以配置不同的掩码。开发可变

长度子网掩码的想法就是在每个子网上保留足够的主机数的同时，把一个网分成多个子网时有更大的灵活性。如果没有VLSM，一个子网掩码只

能提供给一个网络。这样就限制了要求的子网数上的主机数。

VLSM和CIDR区别
CIDR是把几个标准网络合成一个大的网络
VLSM是把一个标准网络分成几个小型网络（子网）
CIDR是子网掩码往左边移，VLSM是子网掩码往右边移

Answer: B

QUESTION NO: 37
You have a single Class C IP address and a point-to-point serial link that you want
to implement VLSM on. Which subnet mask is the most efficient?
你有一个单独的C类地址和一条点对点串行链路，如果要在该链路上配置VLSM，哪个掩码是最有效率的?

A. 255.255.255.0
B. 255.255.255.240
C. 255.255.255.248
D. 255.255.255.252
E. 255.255.255.254

解析：VLSM允许在编址计划中有更多的体系分层，如果没有VLSM,一个子网掩码只能提供给一个网络。这样就限制了要求的子网数上的主机数。
串行链路2个IP就可以了，所以掩码采用/30便可，所以/30也叫“串行掩码”

Answer: D

QUESTION NO: 38
You have a network that supports VLSM and you need to reduce IP address waste
in your point to point WAN links. Which of the masks below would you use?
你有一个支持VLSM的网络，为了在点对点广域网链路上减少地址浪费，应该使用哪个子网掩码

A. /38
B. /30
C. /27
D. /23
E. /18
F. /32

解析：思路同上题，VLSM允许在编址计划中有更多的体系分层，所以可以掩码分的很灵活.

Answer: B

QUESTION NO: 39
How would you express the binary number: 10101010 in its decimal and
hexadecimal forms?
如何用十进制表达二进制数:10101010

A. Decimal=160, hexadecimal=00
B. Decimal=170, hexadecimal=AA
C. Decimal=180, hexadecimal=BB
D. Decimal=190, hexadecimal=CC

解析：

二进制->十进制 按权展开法 10101010 = 128+32+8+2 = 170
二进制->十六进制 每4位作变化 1010= 10(A)

太棒了,收益非浅~~~

QUESTION NO: 40
Which of the following IP hosts would be valid for PC users, assuming that a /27
network mask was used for all of the networks? (Choose all that apply.)

所有网络的掩码均为/27，下列哪些IP地址是可用的主机地址(多选)

A. 15.234.118.63
B. 83.121.178.93
C. 134.178.18.56
D. 192.168.19.37
E. 201.45.116.159
F. 217.63.12.192

解析:掩码为255.255.255.224，最后一个八位组子网占了3位，子网应该是256-224=32的倍数，那么合法主机地址条件自然是在
"32的倍数+1" 和"32的下一个倍数－2"之间(拿掉一个广播地址)
A.63=64-1 广播
B.满足条件
C.满足条件
D.满足条件
E.159=160-1 广播
F.192=16*12

答案：B.C.D

QUESTION NO: 41
You are the network administrator at TestKing. TestKing has been provided with
the network address 165.100.27.0/24. The TestKing CEO wants to know how many
subnetworks this address provides, and how many hosts can be supported on each
subnet.What would your reply be? (Choose all that apply)
你是TK的管理员,提供给公司的IP网段是65.100.27.0/24，公司领导想知道这个地址能够提供多少个子网，每个子网提供多少个主机？

A. One network with 254 hosts.
B. 254 networks with 254 hosts per network.
C. 65,534 networks with 255 hosts per network.
D. 30 networks with 64 hosts per network.
E. 254 networks with 65,534 per network.

解析：作为B类地址，我们看到子网掩码占用了主机地址8位，可以说分了256个子网，但是发现第三个八位组是27，可见这是个子网地址
换句话说，我们分到的IP网段是已经别人分的256个子网中的一个。答案显而易见。
此题很容易打错，如果给的IP是165.100.0.0/24，那么答案就是B了。

答案：A

UESTION NO: 42 DRAG DROP
TestKing has three locations and has plans to redesign the network accordingly. The
networking team received 192.168.126.0 to use as the addressing for entire network
from the administrator. After subnetting the address, the team is ready to assign the
address.
The administrator plans to configure ip subnet-zero and use RIP v2 as the routing
protocol. As a member of the networking team, you must address the network and
at the same time converse unused addresses for future growth.
Being mindful of these goals, drag the host addresses on the left to the correct router
interface. One of the routers is partially configured. Move the mouse over a router
to view its configuration (** This information is missing**). Not all of the host
addresses on the left will be used.
拖拽题



TK公司有三个分部，准备计划重新设计网络，分配的地址是92.168.126.0
管理员计划配置启用全0子网，并且使用RIP v2作为路由协议，你分配地址的同时必须保留一些地址给将来备用
图中看到：一个路由器做了一些配置，将合适的主机地址放入正确的位置(不是所有地址都用上)

解析：RIPv2支持VLSM，可以使用不同的掩码
3个路由器连接3个子网，需要的主机数目分别是：3,20,13
为了最大限度的解决IP地址，我们看左边的选项
192.168.126.67/29，可以有6个主机地址，分给TestKing1 Fa0/0
192.168.126.35/27，可以有30个主机地址
192.168.126.2/27，可以有30个主机地址
这两个随意分给TestKing 2 Fa0/0 和TestKing 3 Fa0/0 都可以
(如果有个/28的就可以分给TestKing 3 Fa0/0)
至于两个路由器之间的串行口，自然用"串行掩码",分配192.168.126.49/30

答案：

QUESTION NO: 43
The TestKing network has been divided into 5 separate departments as displayed
below:
Using a Class C IP network, which subnet mask will provide one usable subnet per
department while allowing enough usable host addresses for each department
specified in the graphic?



如图，TK公司的网络分成了5个独立的部门，使用一个C类网络，哪个掩码可以为每部分提供一个可用子网以允许每个部门有足够用的主机数目?

A. 255.255.255.0
B. 255.255.255.192
C. 255.255.255.224
D. 255.255.255.240
E. 255.255.255.248
F. 255.255.255.252

解析：看图可知：需要5个子网，最后主机数是16个，主机位应该有5位。

答案：C

QUESTION NO: 44 DRAG DROP
TestKing has three locations and has plans to redesign the network accordingly. The
networking team received 192.168.55.0 to use as the addressing for entire network
from the administrator. After subnetting the address, the team is ready to assign the
address.
The administrator plans to configure ip subnet-zero and use RIP v2 as the routing
protocol. As a member of the networking team, you must address the network and
at the same time converse unused addresses for future growth.
Being mindful of these goals, drag the host addresses on the left to the correct router
interface. One of the routers is partially configured. Not all of the host addresses on
the left will be used.
公司有三个分部，准备重新设置网络划分子网，管理员给网络分配的地址是192.168.55.0
管理员计划配置启动全0子网和RIPv2作为路由协议，作为网络管理团队的一员，你必须分配地址的同时必须保留一些地址给将来备用
图中看到：一个路由器做了一些配置，将合适的主机地址放入正确的位置(不是所有地址都用上)



解析：RIPv2支持VLSM，可以使用不同的掩码，思路同42题
3个路由器连接3个子网，需要的主机数目分别是：7,90,23
2的4次方-2=14 可见/28适合
2的7次方－2＝126 可见/25适合
2的5次方－2＝30 可见/27适合
两个路由器之间串行链路用/30正好

答案：

QUESTION NO: 45
You are the network administrator at TestKing. TestKing has been assigned the
class C IP address 189.66.1.0 by its Internet Service Provider. If you divide the
network range by using the 255.255.255.224 subnet mask, how many hosts can be
supported on each network?
你是TK公司的管理员，ISP给公司重新分配了IP地址189.66.1.0，如果你决定使用255.255.255.224作为子网掩码，那么每一个网络支持多少个

主机？

A. 14
B. 16
C. 30
D. 32
E. 62
F. 64

解析：作为一个C类地址，默认掩码应该是/24，这里变成了/27,可见借用了3位，那么留下5位给主机.

Answer: C

这个帖子好啊
我每天一看

经典的方法...

QUESTION NO: 46
Which of the following statements are true regarding a network using a subnet mask
of 255.255.248.0? (Choose three)
对于使用子网掩码255.255.248.0的网络，下列说法哪三个正确？

A. It corresponds to a Class A address with 13 bits borrowed.
B. It corresponds to a Class B address with 4 bits borrowed.
C. The network address of the last subnet will have 248 in the 3rd octet.
D. The first 21 bits make the host portion of the address.
E. This subnet mask allows for 16 total subnets to be created.
F. The subnetwork numbers will be in multiples of 8.

解析:这个掩码是/21
如果配给一个A类地址，那么从主机位借用了13比特划分子网
如果配给一个B类地址，那么从主机位借用了5比特划分子网
256-248=8 可见子网号是8的倍数
因为题目中没有说明主网IP，所以不能说具体分了几个子网

答案：A，C, F

QUESTION NO: 47

Which of the following IP addresses is a private IP address? Select all that apply.
下面哪些地址是私有IP地址?

A. 12.0.0.1
B. 168.172.19.39
C. 172.20.14.36
D. 172.33.194.30
E. 192.168.42.34

解析：本题考察RFC1918定义的私有IP概念

继续学习