Chapter III IP address planning and design technology
IP addresses concepts and research course is divided address of the new technology is divided into four stages
Basics
IP address of the standard classification of
IPv4 address length of 32 bits, in dotted decimal notation, is divided into four parts, each one is 8bit, between each number separated by a dot (.). A 32-bit value for each possibility is 0 or 1.
Each number in the range expressed as a decimal? ?
Each range of 0 to 255.
For example: 202.113.29.119 IPv4 address is
for example: 256.113.29.119 IPv4 address is not a
class 5 IP address
IP address classification, initial aim is to each IP
address can be uniquely determined identifying a network and a host
so that the first IP address of the standard two-layer structure is: the network number (net) + host ID (host)
a class a addresses: the first one and the last one reserved for special use (1.0.0.0 and 127.255.255.255) and a network number 10 reserved for special use, only 125 (net) may be used;
B and class C addresses: host ID are all 0 and 1 are all reserved for special use, so there is class B IP addresses (2 16) are available -2 host number (host), class C IP addresses with a (2 8) -2 available host number (host).
Special address form
Direct broadcast address
In the ABC three IP addresses, host number are all 1's IP address for the direct broadcast address.
Effect: the host sends a packet to all hosts on a particular network (201.161.20.0) in a broadcast manner, the need to use a direct broadcast address (201.161.20.255).
Limited broadcast address
Limited broadcast address: 255.255.255.255
effect: a packet is broadcast to all hosts in the same network, and the packet is blocked by the router, the broadcast packet only present inside the network, which may not host other than the broadcast network according to the present .
"Particular host on this network" address
When a host or a router to send a packet to a specific host of the present network, it requires the use of "a particular host on that network" network address to the whole number 0, host number corresponding to the host to accept the host number. To host such as a host of this network (201.161.20.18) to send a packet, only the host to accept this grouping, that "a particular host on this network" address (0.0.0.18).
Loopback address (loopback address)
A class of IP address 127.0.0.0 loopback address, it is a role reserved address: for inter-network software testing and local interprocess communication.
Subnetting
The basic idea put forward the concept of subnets: Allows network is divided into multiple parts for internal use and for external network but still like a network.
Subnetting art elements:
① the IP address of the three-layer structure: net ID (network ID) ----- sbunet ID (subnet number) ----- host ID (host ID);
the same subnet in ② All hosts must use the same subnet number subnet ID;
concept may be applied to the subnet ③ class a, B, C any IP address in a class;
the distance between subnets ④ must close;
⑤ distribution a subnet is a matter of internal organization and units, both do not apply to ICANN, nor any need to change the external database;
⑥ in the literature in the Internet, also known as a subnet of an IP network or a network.
How subnetting?
1 0 网络号(14位) 主机号(16位)
I want to separate the class B network 64 subnets out how to divide?
Since 2 ^ 64 = 6, so I just need to borrow bits from the host six out.
1 0 网络号(14位) 子网号 主机号(10位)
Processing Standard IP address, the first two bits only need to determine the value of the IP address, if 10 is Classes B, if compared to the class C 11, if compared to the class A is 00 or 01 at the beginning.
Represents a method of treating subnetting IP address, subnet mask, it is necessary to use (subnet mask) subnet mask: below
Two notations Subnet Mask:
four equal dotted decimal (+ subnet network number of the whole number. 1)
/ + (+ length subnet network ID number)
so that the IP address subnetting process, first the first two categories by determining what kind of address, and then determines the number of bits in the subnet subnet Mask.
Classless Inter-Domain Routing
Using CIDR <network prefix> <host number> + host number network number instead of the standard classification of IP address, subnet concept is no longer used. Tertiary structure of the IP address back to the secondary structure, but does not use the classification of secondary structure. Become two address structure without classification.
CIDR address representation (inclined notation), such as 200.16.23.0/20, which is a network prefix 20, the number 12 for the host
IP address planning method
Address planning of the basic method
Selected subnet ID field length value X, Nnet≤2x requirements
such as: the number of subnets in claim 10 open, 10≤24, only four sub-length to meet the requirements.
Selecting host ID field length value Y, requirements Nhost≤2Y
example: the number of subnets host in claim 12, 12≤ 24, so the host No. 4 length to meet the requirements.
Length can be determined subnet and host IP address based on what kind of application
X + Y≤8 can apply Class C IP addresses
X + Y> 8 may apply two Class C or Class B addresses
Computer subnet mask
没有划分子网的地址中网络号全置1，主机号全置0；
比如200.168.1.5 其子网掩码为255.255.255.0；
划分子网的地址中其网络号和子网号全置1，主机号全置0；
比如192.168.1.0/28，其子网掩码为255.255.255.240
计算网络地址
比如一个网络号：192.168.1.0，子网号4个长度，主机号4个长度
192.16.1.00000000
4位子网号有24种子网网络号(主机号全置0)，如下图
计算网络广播地址
子网的网络号中，主机号位全置1
计算机网络的主机地址
主机号除去全0和全1的ip地址
子网地址规划方法
子网划分实例：
用户需求：
①一个校园网获得一个B类IP地址(156.26.0.0),要进行子网划分
②该校园网将由近210个网络组成
③为了便于管理，要求根据目前情况进行子网划分
分析划分步骤：
①确定子网长度，210≤2x，那这个x最小值应该等于多少？
②子网掩码应该为255.255.255.0
③由于子网号和主机号位段全0和1都不能用，所以每一个子网的主机号 由0-255(256个)变成1-254(254个)
所以子网划分方案：
子网1:156.26.0.1~156.26.0.254 子网2:156.26.1.1~156.26.1.254
……
子网255:156.26.254.1~156.26.254.254 子网256:156.26.255.1~156.26.255.254
可变长度子网掩码(VLSM)地址规划方法
用户需求：
①某个公司申请了一个C类202.60.31.0的地址
②该公司有三个部门要为员工主机分配IP地址，销售部门100名员工，财务部门
50员工，设计部门50名员工
③要求网络管理员为这三个部门分别组建子网
分析和划分步骤：
①子网划分子网数量按照前面方法肯定是偶数，x为子网长度，子网数2x肯定是偶数。
②按照可变长度子网掩码划分子网，100≤27,所以先划分两个子网，子网号1长度，主机长度7。
子网1：202.60.31.00000000 202.60.31.1~202.60.31.126(126个主机)
子网掩码：255.255.255.128
子网2：202.60.31.10000000
将子网2再划分为两个更小的子网
子网2-1：202.60.31.10000000 202.60.31.129~202.60.31.190(62个主机)
子网2-2：202.60.31.11000000 202.60.31.193~202.60.31.254(62个主机)
子网掩码：255.255.255.192
CIDR地址规划方法用户需求：
一个校园管理中心获得了200.24.16.0/20的地址块，希望将这个网络分成8个等长的地址块。
分析和划分步骤
①200.24.16.0/20 该IP地址无类域间路由地址，网络前缀20位， 主机12位，这个地址块：200.24.00010000.0~ 200.24.00011111.255。
②将这个大地址块均等分成8小份，23=8，只需要向主机号中前3 位不作为主机号，作为网络前缀，就可以均等分成8个小地址块。网络前缀变为23位。
200.24.00010000.00000000~ 200.24.00010001.11111111 计算机系地址
200.24.00010010.00000000~ 200.24.00010011.11111111 自动化系地址
200.24.00010100.00000000~ 200.24.00010101.11111111 电子系地址
200.24.00011000.00000000~ 200.24.00011001.11111111 物理系地址
200.24.00010110.00000000~ 200.24.00010111.11111111 生物系地址
200.24.00011010.00000000~ 200.24.00011011.11111111 中文系地址
200.24.00011100.00000000~ 200.24.00011101.11111111 化学系地址
200.24.00011110.00000000~ 200.24.00011111.11111111 数学系地址
内部网络的专用IP地址规划与网络地址转换NAT方法
内部网络的专用IP地址
专用IP地址有哪些？
A类地址专用IP地址块：10.0.0.0~10.255.255.255
B类地址专用IP地址块：172.16.0.0~172.31.255.255
C类地址专用IP地址块：192.168.0.0~192.168.255.255
网络地址转换NAT的基本原理
IPv6地址规划基本方法
IPv6地址表示方法
IPv6地址是将128位按每16位划分一位段，每一位段转换为一个4位的十六进制数，并用:号隔开，称为冒号十六进制表示法。
比如：21DA:0000:0000:0000:02AA:000F:FE08:9C5A
可以简化：
①每一位段前导0可以省略，比如02AA和000F可以写成2AA和F，但是中间和末尾0不可以省略，比如FE08不可以省略；
②每一个位段必须保留一个数字比如0000，可以写成0，不可以全部省略；根据上面两步可以写成21DA:0:0:0:2AA:F:FE08:9C5A
③几个连续位段都是0，可以将这些连续位段0写成:: 所以还可以写成21DA::2AA:F:FE08:9C5A
④IPv6不存在子网掩码，只有IPv4存在，所以如果这样书写21DA::2AA:F:FE08:9C5A/24就是错误的。
课后真题（4）
1、下列IPv6地址FE80:0:0:0801:FE:0:0:04A1的简化错误的是(A )
A. FE8::801:FE:0:0:04A1
B. FE80::801:FE:0:0:04A1
C. FE80:0:0:801:FE::04A1
D. FE80:0:0:801:FE::4A1
2、下列对IPv6地址FF60:0:0:601:BC:0:0:05D7简化表示中，错误的是( B )
A. FF60:0:0:601:BC:0:0:5D7
B. FF6::601:BC::05D7
C. FF60:0:0:601:BC::05D7
D. FF60:0:0:601:BC::5D7
3、下列对IPv6地址表示中，错误的是( B )
A. ::601:BC:0:05D7
B. 21DA:0:0:0:0:2A:F:FE08:3
C. 21BC::0:0:1/48
D. EF60::2A90:FE:0:4CA2:9C5A
4、将内部专用IP地址转换为外部公用IP地址的技术是(B )
A. RAPR
B. NAT
C. DHCP
D. ARP
5、下图是网络地址转换NAT的一个示例，图中①和②是地址 转换之后与转换之前的一对地址（含端口号），它们依次应为(B )。
A) 10.0.1.1,1234和59.67.0.6,2341
B) 59.67.0.6,2341和10.0.1.1,1234
C) 10.0.1.1,1234和202.2.1.1,8080
D) 202.2.1.1,8080和10.0.1.1,1234
6、下图是网络地址转换NAT的一个示例,图中①和②是地址转换的一对地址，①应为( D )。
A) 10.0.1.1,3342
B) 58.4.2.1,80
C) 202.112.15.1,3342
D) 202.112.15.1,1524
7、某个IP地址的子网掩码为255.255.255.192,该掩码又可以写为( C )
A. /22
B. /24
C. /26
D. /28
8、IP地址192.168.15.136/29子网掩码可写为( D )
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.255.240
D. 255.255.255.248
9、若某大学分配给计算机系的IP地址块为202.113.16.128/26，分配给自动化系的IP地址块为202.113.16.192/26，那么这两个地址块经过聚合后的地址为©。
计算机系:202.113.16.10000000
自动化系:202.113.16.11000000
划分之前:202.113.16.10000000
202.113.16.128/25
A. 202.113.16.0/24
B. 202.113.16.0/25
C. 202.113.16.128/25
D. 202.113.16.128/24
10、某企业分配给产品部的IP地址块为192.168.31.192/26,分配给市场部的IP地址块为192.168.31.160/27，分配给财务部的IP地址块是192.168.31.128/27，那么这三个地址经过聚合后的地址为(C )。
A. 192.168.31.0/25
B. 192.168.31.0/26
C. 192.168.31.128/25
D. 192.168.31.128/26
产品部:192.168.31.11000000 市场部:192.168.31.10100000
财务部:192.168.31.10000000 子网1:192.168.31.11000000
子网2:192.168.31.10000000 划分之前：192.168.31.100000000
11、某公司拥有IP地址块202.113.77.0/24其中202.111.77.16/28和202.113.77.32 /28已经分配给人事部和财务部，现在技术部需要100个IP地址，可分配的IP地址块是( D )
划分前:202.113.77.0
人事部:202.113.77.00010000
财务部:202.113.77.00100000
202.113.77.10000000
202.113.77.00000000 这两个子网都满足
A. 202.113.77.0/25
B. 202.113.77.48/25
C. 202.113.77.64/25
D. 202.113.77.128/25
12, the IP address, the private IP address does not belong to the (B)
A. 10.1.8.1
B. 172.12.8.1
C. 172.30.8.1
D. 192.168.8.1