Classless Interdomain Routing (CIDR)
IP Address Problems
- IP Address Exhaustion
- Class A, B, and C address structure inefficient
- Class B too large for most organizations, but future proof
- Class C too small
- IP routing table size
- Growth in # of networks in Internet reflected in # of table entries
- Stress on router processing power and memory allocation
- Short-term solution:
- Classless Interdomain Routing (CIDR), RFC 1518
- New allocation policy (RFC 2050)
- Private IP Addresses set aside for intranets(企业内部网) (NAT)
- Long-term: IPv6 with much bigger address space
Classless Interdomain Routing Scheme
- CIDR uses an arbitrary(任意的) prefix length to indicate the network number
- 205.100.0.0/22
- /22 means mask: 11111111 11111111 11111100 00000000 - 255.255.252.0,means the mask is 22 bits long
- Packets are routed according to the prefix w/o address classes
- An entry in CIDR routing table contains 32-bit IP address and 32-bit mask
- Enables supernetting to allow a single routing entry to cover a block of classful addresses
CIDR Aggregation(聚合)
- A company is allocated the following four contiguous(相邻的) /24 networks. At some router, it is often true that all of the four networks use the same outgoing line. CIDR aggregation can be done to reduce the number of entry at the router.
- 128.56.24.0/24; 10000000 00111000 00011000 00000000
- 128.56.25.0/24; 10000000 00111000 00011001 00000000
- 128.56.26.0/24; 10000000 00111000 00011010 00000000
- 128.56.27.0/24. 10000000 00111000 00011011 00000000
By Per-bit AND - 128.56.24.0/22 10000000 00111000 00011000 00000000
(Instead of 4 entries in routing table, one entry is sufficient by CIDR)
CIDR Scheme and Range
- CIDR deals with Routing Table Explosion Problem
- Networks represented by prefix and mask
- Summarize a contiguous group of class C addresses using variable-length mask, if all of them use the same outgoing line
- Solution: Route according to prefix of address, not class
- Routing table entry has <IP address, network mask>
- Example: 192.32.136.0/21
- 11000000 00100000 10001000 00000001 min address
- 11111111 11111111 11111--- -------- mask
- 11000000 00100000 10001--- -------- IP prefix
- 11000000 00100000 10001111 11111110 max address
- Eight C networks: 192.32.136.0/24 to 192.32.143.0/24
CIDR Supernetting Example (1)
- Summarize a contiguous group of class C addresses using variable-length mask
- Example: 150.158.16.0/20
- IP Address (150.158.16.0) & mask length (20)
- IP address = 10010110 10011110 00010000 00000000
- Mask = 11111111 11111111 11110000 00000000
- Contains 16 Class C blocks:
- From 10010110 10011110 00010000 00000000
i.e. 150.158.16.0/24 - Up to 10010110 10011110 00011111 00000000
i.e. 150.158.31.0/24
CIDR Supernetting Example (2)
A router has the following CIDR entries in its routing table:
- Address/mask - - - - -Next hop
- 128.56.24.0/22 - - - - - Interface 0
- 128.56.60.0/22 - - - - -Interface 1
- default - - - - -Router 2
A packet comes with IP address of 128.56.63.10. What does the router do?
- 128.56.63.10 and mask 22 bits
- First, the address will be transform into binary, then use MASK by AND operation to get the prefix, finally it will match the table, if match a interface, go to it.
- IP address = 10000000 00111000 01111111 00001010
- Mask = 11111111 11111111 11111100 00000000
By Per-bit AND - Prefix = 10000000 00111000 01111100 00001010
i.e. 128.56.60 - Router table lookup and match, should go to interface 1
New Address Allocation Policy
- Class A & B assigned only for clearly demonstrated need
- Consecutive(连续的) blocks of class C assigned (up to 64 blocks)
- All IP addresses in the range have a common prefix, and every address with that prefix is within the range
- Arbitrary prefix length for network ID improves efficiency
- Address assignment should reflect the physical topology of the network
- Facilitates the aggregation of logical packet flows into physical flows
Longest Prefix Match
- By CIDR, multiple entries may match a given IP destination address ,but different prefix
- Example: perform CIDR on the following three /24 IP addresses (but 128.56.24.0/24 to a different port)
- 128.56.25.0/24;
- 128.56.26.0/24;
- 128.56.27.0/24;
- By CIDR aggregation(Per-bits AND): 128.56.24.0/22
- What if a packet with destination IP address 128.56.24.1 comes? It belongs to another port
Example of Longest Prefix Match
Classless Interdomain Routing (CIDR)的更多相关文章
- Netmask, 子网与 CIDR (Classless Interdomain Routing)
Netmask, 子网与 CIDR (Classless Interdomain Routing) 我们前面谈到 IP 是有等级的,而设定在一般计算机系统上面的则是 Class A, B, C.现在我 ...
- [转]CIDR简介
IP Subnetting and Variable Length Subnet Masks Subnetting Basics 子网划分(subnetting)的优点: 1.减少网络流量 2.提高网 ...
- IP协议
因特网协议(Internet Protocol, IP)是Internet的核心协议之一, 工作在网络层.IP协议提供节点间的寻址,路由以及顺序控制, 流量控制等服务. IP协议分为IPv4和IPv6 ...
- (转) IPv6相关RFC
转自http://blog.csdn.net/lucien_cc/article/details/12688477 IPv6 Spec RFC 2460 : Internet Protocol, Ve ...
- IPv6 tutorial 4 IPv6 address syntax
https://4sysops.com/archives/ipv6-tutorial-part-4-ipv6-address-syntax/ Now that you know about the n ...
- How Network Load Balancing Technology Works--reference
http://technet.microsoft.com/en-us/library/cc756878(v=ws.10).aspx In this section Network Load Balan ...
- wireshark filter manualpage
NAME wireshark-filter - Wireshark filter syntax and reference SYNOPSIS wireshark [other options] [ - ...
- 【计算机网络】 一个小白的网络层学习笔记:总结下IP,NAT和DHCP
前言:这篇文章是学习网络层协议时候总结的笔记,前面的主要部分介绍的都是IP协议, 后半部分介绍NAT协议和DHCP协议 参考书籍 <计算机网络-自顶向下> 作者 James F ...
- OSPF(Open Shortest Path First)
1.概述 路由协议OSPF全称为Open Shortest Path First,也就开放的最短路径优先协议,因为OSPF是由IETF开发的,所以所有厂商都可以用. OSPF的流量使用IP协议号. O ...
随机推荐
- LVS(Linux Virtual Server)
LVS的英文全称是Linux Virtual Server,即Linux虚拟服务器.它是我国的章文嵩博士的一个开源项目.在linux内存2.6中,它已经成为内核的一部分,在此之前的内核版本则需 ...
- Redis 常见命令
0. 5种数据类型 String(字符串) List(列表) Hash(字典) Set(集合) Sorted Set(有序集合) 1. String 字符串 set key value 设置key=v ...
- css移动端:acitve效果的实现
做移动前端也有一些日子了,一直有个问题没有解决,就是与pc端那样的一个:hover的效果,:hover是鼠标指针浮动在其上的元素的一个选择器,但因为在移动端是没有鼠标的,代替的是触摸屏,用户也不是有“ ...
- COGS2216 你猜是不是KMP
第一道自己写的FFT...... 不知为啥这题在网上找不到题解......真是麻烦,害得我推了半天...... 还是写个简要题解吧...... 首先把S和T拆成序列,a~z分别对应成1~26,?是0, ...
- WebClient用法小结(转载)
如果只想从特定的URI请求文件,则使用WebClient,它是最简单的.NET类,它只用一两条命令执行基本操作,.NET FRAMEWORK目前支持以http:.https:.ftp:.和 file: ...
- odps编写UDF的实现
问题 尝试写一个UDF,参数支持输入x,y与一个Polygon,返回结果是(x,y)是否在输入的Geometry之内? 环境 eclipse odps 插件 jts包:jts-1.8.jar 解法 i ...
- 部署node.js的开发环境
1.进入Node.js的官方网站下载安装包: http:nodejs.org 2.安装后打开cmd的dos窗口(在path环境变量中查看到有nodejs说明安装成功): 3.运行node.
- qwewq
- 用Eclipse创建第一个Spring项目(最最入门级)
http://blog.csdn.net/shymi1991/article/details/48085955 网上关于Spring的介绍资料已经数不胜数,但大多篇幅冗长,初学者不易理解记忆.这里先作 ...
- 在win7 windows 上编译 beego 上传到 linux 去执行
在beego的项目目录下,执行: GOOS=linux GOARCH=amd64 go build So easy!但是却搞了好久! 参考连接:http://blog.csdn.net/changji ...