Computer Networks Lecture 11

Transcription

1 Computer Networks Lecture 11 Wei Liu ( 刘威 ) Dept. of Electronics and Information Eng. Huazhong University of Science and Technology Nov

2 Problem in Chapter4: There Is More Than One Network -2-

3 Lecture 11 Chapter 4. Internetworking Problem: There Is More Than One Network 4.1 Simple Internetworking (IP) 4.2 Routing 4.3 Global Internet -3-

4 Review: IP Service model 1) what kind of communication to support among networks? 支持怎样的网络间通信? Datagram switched communication Merits learned from ARPAnet 2) what kind of service to provide for upper layers? 向上层提供怎样的服务? Best effort service in network layer Reliable service in higher layer protocol, such as TCP 3) how to realize the service in protocol stack? 协议栈的实现 Unique IP protocol Everything over IP IP IP over everything 4) how to realize the service in router? 路由器的实现 Datagram forwarding Routing protocol (Distance Vector based, Link State based) -4-

5 Review: IP service model Implementation IP 服务模型的整体设计与实现 Datagram delivery(subsection 4.1.2) Packet format 分组的格式 Fragmentation and resemble 分组的分段与组装 Hierarchical addressing (subsection 4.1.3) IP 服务模型在路由器的实现 Datagram forwarding (subsection 4.1.4) Routing algorithms and protocols (subsection ) Router Design (subsection 4.2.6) -5-

6 Review: IP Addressing Hierarchical addressing Addressing problem of internetworking Each host should have its unique address Addressing should support scalability IP s solution each address is globally unique 全球唯一地址 hierarchical addressing 层次化编址 two address spaces, {Network: Host} 网络地址与主机地址 -6-

7 Hierarchical Address Space Each IP address has 2 parts a network part and a host part IP address classes (a) 0 Network 7 24 Host (b) 1 0 Network Host (c) Network Host -7-

8 Network address 网络地址 The network part of an IP address uniquely identifies a network All hosts and router interfaces connected to the same network have the same network number Given any IP address, a router has a way of finding out to which network this IP address belongs The focus is to deliver packets to their destination networks -8-

9 Classful addressing 有类别地址 Classful addressing It assumes that the Internet would consist of a small number of wide area networks (class A networks) a modest number of site- (campus-) sized networks (class B networks a large number of LANs (class C networks). However, there are some unfair and un-efficient allocation Some classless approach is proposed in Internet -9-

10 IP 地址的使用范围 网络 最大 第一个 最后一个 每个网络 类别网络数 可用的 可用的 中最大的 网络号 网络号 主机数 A 126 (2 7 2) ,777,214 B 16,384 (2 14 ) ,534 C 2,097,152 (2 21 )

11 Expression of IP address `` four decimal integers expression is proposed to represent IP address for manual usage and memory 四段十进制的整数表达的 IP 地址 IP addresses are written as four decimal integers separated by dots. Each integer represents the decimal value contained in 1 byte of the address, starting at the most significant. For example, the address of the computer on which this sentence was typed is Another naming approach is introduced for Internet domain names, which is different to the hierarchical IP address 由主机域名 URL(Uniform Resource Locator) 表达的地址 Domain names tend to be ASCII strings separated by dots, such as itec.hust.edu.cn. DNS (domain name service) in application layer is responsible to mapping IP address to its domain name. -11-

12 4.3 Global Internet 全球互联网 However, simple internetworking is not enough for scalability 扩展性问题 The simple hierarchy of the IP address can only achieve somewhat scalable Routers need to know about all the networks connected to the internet, which is impossible in real cases There are a variety of techniques that greatly improve scalability of global Internet -12-

13 Real case of Global Internet The tree structure of the Internet in 1990 Internet Services Provider (ISP) provides the area network end users are also autonomous systems -13-

14 Real case of Global Internet The tree structure of the Internet in routing cost = link utili. routing cost = delay The The universities universities can can allocate allocate the the IP IP addresses addresses for for internal internal users users freely freely => => Utilization Utilization problem problem of of address address space space The The ISPs ISPs can can have have different different optimal optimal view view of of routing routing => => Autonomous Autonomous System System -14-

15 Lecture 11 Chapter 4. Internetworking Problem: There Is More Than One Network 4.1 Simple Internetworking (IP) 4.2 Routing 4.3 Global Internet Subnetting Classless Routing (CIDR) Interdomain Routing (BGP) Routing Areas IP Version 6 (IPv6) -15-

16 4.3.1 Subnetting Problem IP Address Depletion Problem IP 地址耗尽的问题 Motivation Introduce another hierarchy 引入新的层次结构 Solution Addressing: Subnet mask IP Forwarding -16-

17 IP Address Depletion Problem IPv4 has a 32-bit address space: the total number is about 4 billion not all addresses are used for hosts or interfaces some are used for multicast or reserved Why are there not enough IP addresses? fast growth of the Internet Inefficient address assignment -17-

18 Inefficiency of Address Assignment (a) 0 Network 7 24 Host (b) 1 0 Network Host (c) Network Host Example 1: a network would like to have 10 hosts requires a class C network efficiency: 10/256 is about 4% Example 2: a network would like to have 300 hosts requires a class B network efficiency: 300/65536 is about 0.5% -18-

19 4.3.1 Subnetting Problem IP Address Depletion Problem Motivation Introduce another hierarchy Solution Addressing: Subnet mask IP Forwarding -19-

20 Motivation of subnetting Subnet introduce another level of hierarchy into the IP address 引入新的层次结构 Subnetting allocate the IP addresses with single network number to several physical networks (subnets) 子网 subnets are close to each other, which have only one network number among them a router will only be able to select one route to reach any of the subnets subnet reduce the required amount of network numbers -20-

21 Subnet Note the address with all 0 in host part represents this network the address with all 1 in host part represents all hosts in this network, i.e., broadcast address -21-

22 Efficiency of Subnetting Example: 3 networks, 2 with 50 hosts, 1 with 100 hosts original assignment requires 3 networks of class C efficiency: 200/(3x256) is about 26% with subnetting requires 1 class C network divide it into 3 subnets: 2 with 64 addresses, 1 with 128 put 50 hosts each in the first two, 100 in the last efficiency: 200/256 is about 78% -22-

23 4.3.1 Subnetting Problem IP Address Depletion Problem Motivation Introduce another hierarchy Solution Addressing: Subnet mask IP Forwarding with subnet -23-

24 子网划分 IP 地址的各字段和子网掩码 两级 IP 地址 因特网部分 网络号 net-id 本地部分 主机号 host-id 因特网部分 本地部分 三级 IP 地址 net-id subnet-id host-id 网络号 子网号 主机号 子网掩码 划分子网时的网络地址 net-id subnet-id host-id 为全 0

25 子网划分 (IP 地址 ) AND ( 子网掩码 ) = 网络地址 两级 IP 地址 因特网部分 网络号 net-id 本地部分 主机号 host-id 因特网部分 本地部分 三级 IP 地址 net-id subnet-id host-id 子网掩码 划分子网时的网络地址 网络号 AND 子网号 主机号 net-id subnet-id host-id 为全 0

26 How to Subnet a Network Example: given a class C network with number , divide it into 3 subnets, 2 with 64 addresses and 1 with 128 for the subnet with 128 hosts, the length for host-id in address is 7 bits subnet mask is for the subnet with 64 hosts, the length for host-id in address is 6 bits subnet mask is

27 How to Subnet a Network (contd.) 128 = = Subnet Number Subnet Mask Addresses

28 How to Subnet a Network (contd.) Do the largest subnet first!! -28-

29 Sample of IP addressing network address: the full 0 in host part broadcast address: the full 1 in host part if one host IP is , subnet mask is , give the network address and broadcast address 137 = = last segment of network address is = 128 so the network address is last segment of broadcast address is = 159 so the broadcast address is

30 Sample of IP addressing in the subnet, network address (full 0) and broadcast address (full 1) will take two additional addresses in real Ethernet, we always leave another address for the Gateway if we want a subnet have 10 addresses for host, give the mask total addresses are = 12, 2 3 < 12 < 2 4 so 4 bits for host part, the last segment of mask is = 240 so the mask is

31 4.3.1 Subnetting Problem IP Address Depletion Problem Motivation Introduce another hierarchy Solution Addressing: Subnet mask IP Forwarding -31-

32 Changes to Forwarding With the introduction of subnet, packet delivery is made to a subnet rather than a network (A, B, or C), and a forwarding table holds a number of entries in the form of <SubNetNum, SubNetMask, NextHop> Forwarding algorithm D = dst. IP addr. for each entry in forwarding table D1 = SubNetMask & D if D1 = SubNetNum if NextHop is an interface else deliver datagram directly to dst. deliver datagram to NextHop (router) -32-

33 Subnet Example -33-

34 Summary: Merits of Subnetting Subnetting helps solve our scalability problems in two ways First, it improves our address assignment efficiency 子网的引入提升了地址分配的有效性 We do not have to use up an entire class C or class B address every time we add a new physical network. Second, it helps us aggregate information. 子网的引入提升了信息的聚合程度 The amount of information that routers need to store to deliver datagrams to those networks can be reduced. -34-

35 Lecture 11 Chapter 4. Internetworking Problem: There Is More Than One Network 4.1 Simple Internetworking (IP) 4.2 Routing 4.3 Global Internet Subnetting Classless Routing (CIDR) Interdomain Routing (BGP) Routing Areas IP Version 6 (IPv6) -35-

36 4.3.2 Classless Routing (CIDR) Problem low efficiency of the fixed IP address structure 有类别固定地址划分的低效率 Motivation classless structure of IP address 无类别的 IP 地址结构 Solution CIDR aggregate routing -36-

37 Problem with fixed address classes Example: an ISP requires a network with 1000 hosts if use address classes a class B address will be given efficiency: 1000/65536 is about 0.15% Real problem the granularity of network size is too coarse network size could only be 256, 65k, 17M Even if we use subnet, we can not avoid this problem -37-

38 Problem with fixed address classes One possible solution We would be handing out address space in chunks of 256 addresses at a time However, raises a problem that is at least as serious: excessive storage requirements at the routers. We have to balance them minimize the number of routes vs. hand out addresses efficiently 地址分配细化与路由表存储的矛盾 aggregate routes: to use a single entry in a forwarding table to tell us how to reach a lot of different networks. 聚合路由, 一条路由记录可以包括到多个网络地址的信息 -38-

39 4.3.2 Classless Routing (CIDR) Problem low efficiency of the fixed IP address structure Motivation classless structure of IP address Solution CIDR aggregate routing -39-

40 Motivation Addressing make the granularity of network size finer network size can be 256, 512, 1024, 2048, Routing find out blocks of class C addresses, sharing a common prefix contain a number of class C networks that is a power of two -40-

41 4.3.2 Classless Routing (CIDR) Problem low efficiency of the fixed IP address structure Motivation classless structure of IP address Solution CIDR aggregate routing -41-

42 Solution CIDR: Classless InterDomain Routing 无差别域间路由 RFC 1518 and RFC 1519, define a new concept of allocation of IP address blocks and new methods of routing IPv4 packets based on variable-length subnet masking (VLSM) to allow allocation on arbitrary-length prefixes 可变尺寸的子网掩码 -42-

43 超网划分 (x.x.x.x/y) IP 地址的各字段和掩码 掩码 网络地址 net-id host-id 为全 0 y bit

44 Example Example: An ISP requires a network with 1000 hosts It is given 4 contiguous blocks of class C addresses assume the class C addresses are the first 22 bits of all these Class C network numbers are the same With CIDR, this becomes a network with a 22-bit network number, which can be written as /22 this notation means the network number is given by the first 22 bits of

45 CIDR and Forwarding Table A forwarding table now looks like Net/Mask Length / / / /24 Next Hop Router A Router B Router C Router D When router receives an IP packet it looks up the destination IP address go through the table entries one-by-one until a match a match: the length of the mask is n, the first n bits of the destination IP = the first n bits of the network number -45-

46 Longest Prefix Match Example: destination IP address second last entry: /20 last entry: /24 both match the destination IP address Principle: use the network that match the longest prefix in this example: /24-46-

47 Longest Prefix Match: Illustration -47-

48 Summary Scalability with Basic IP problem 1: inefficient use of address space problem 2: size of routing and forwarding table Two approaches to basic IP Subnetting: solves problem 1 Classless InterDomain Routing (CIDR): solves problem 1 and 2-48-

49 Example An organization has a class C network and wants to form subnets for four departments 1~4, with hosts of 20, 31, 33, 75. Continuous IP address blocks are preferred, please provide the network address, broadcasting address and network mask for each department. Suppose the 4 departments are equipped with router A~D as the gateway to the inner network of this organization. Router E locates in the edge of this network, and connects to A~D directly. Provide the CIDR based routing table for router E with the destination to A~D. If one IP packet with destination IP of , how can router E find the next hop to forward it, provide the matching process.

50 (a) Solution Since the number of hosts is 20, 31, 33, 75, and each department requires at least 2 more IP addresses ( one for network, one for broadcast, possible additional one more for the gateway in LAN ), then the requirements are 22, 33, 35, 77. The nominal subnet sizes are 128=2 7, 64=2 6, 32=2 5, 32=2 5 respectively. Considering the continuous IP requirement, the IP allocation result is as following #host in Dept No. Subnet Max # IP Subnet bit Subnet Mask The final allocation is shown as following #Hosts required Network add Subnet Mask Address range Network bridge is required to combine the two subnets for this big department -50-

51 (b) Solution E A B C D 20 hosts 31 hosts 33 hosts 75 hosts Routing table for router E: Network add Next Hop /26 Router C (to 33hosts) /26 Router B (to 31hosts) /26 Router D (to 75hosts) /27 Router D (to 75hosts) /27 Router A (to 20hosts) -51-

52 (C) Solution Entry ID Network add Next Hop /26 Router C (to 33hosts) /26 Router B (to 31hosts) /26 Router D (to 75hosts) /27 Router D (to 75hosts) /27 Router A (to 20hosts) Following the ``Longest Prefix Match principle Router E will map with the longest prefix mask``*/27, which are corresponding to the entry 4 and 5 in route table = , network id = Thus it does not match with entry 4 and 5. Router E will map with the longest prefix mask``*/26, which are corresponding to the entry 1, 2 and 3 in route table = ,network id = Thus it does match with entry 3. Thus, Router E will forward this IP packet to Router D -52-

53 谢谢! 刘威副教授 互联网技术与工程研究中心华中科技大学电子与信息工程系 网址 :