信息通信网络概论课件：Chapter 3 Switch Technology Fundamentals

1、Chapter 3 Switch Technology Fundamentals,Required reading: Garcia 1.1 4.1 4.4 7.1 a host can send data as soon as it is ready. Source host has no way of knowing if the network is capable of delivering a packet or if the destination host is even up. Since packets are treated independently , it is pos

2、sible to route around link and node failures. Since every packet must carry the full address of the destination, the overhead per packet is higher than for the connection-oriented model.,Packet 1,Packet 2,Packet 3,Packet 1,Packet 2,Packet 3,Datagram Delay,Packet 1,Packet 2,Packet 3,processing delay

3、of Packet 1 at Node 2,Host 1,Host 2,Node 1,Node 2,propagation delay between Host 1 and Node 2,transmission time of Packet 1 at Host 1,Datagram or VC network: why?,Internet * data exchange among computers “elastic” service, no strict timing req. * “smart” end systems (computers) can adapt, perform co

4、ntrol, error recovery simple inside network, complexity at “edge” * many link types different characteristics uniform service difficult,ATM evolved from telephony human conversation: strict timing, reliability requirements need for guaranteed service “dumb” end systems telephones complexity inside n

5、etwork,Virtual Circuit PK. Datagram (1),分组头： DG方式的每个分组头中要包含详细的目的地址 VC方式由于预先已建立逻辑连接，分组头中只要含有对应于所建立的VC的逻辑信道标识 选路： VC方式预先有建立过程，但一旦虚电路建立，在端到端之间所选定的路由上的各个交换节点都具有映象表，存放出入逻辑信道的对应关系，每个分组到来时只要查找映象表，而不需要进行复杂的选路。 DG方式则不需要有建立过程，但对每个分组都要独立地进行选路。 分组顺序： VC方式中，属于同一呼叫的各个分组在同一条虚电路上传送，分组会按原有顺序到达终点，不会产生失序现象。 DG方式中，各个分组

6、由于是独立选路，可以从不同的路由转送，会引起失序,Virtual Circuit PK. Datagram (2),故障敏感性 VC方式对故障较为敏感，当传输链路或交换节点发生故障时可能引起虚电路的中断，需要重新建立。 （有些分组网具有再连接功能，出现故障时可自动建立新的虚电路，并做到不丢失用户数据） DG方式中各个分组可选择不同路由，对故障的防卫能力较强，从而可靠性较高。 应用 VC方式适用于较连续的数据流传送，其持续时间应显著地大于呼叫建立的时间，如文件传送、传真业务等。 DG方式则适用于面向事务的询问响应型数据业务。,Delay Comparison,采用存储转发方式的分组交换兼有电路交

7、换和报文交换的优点，它与报文交换的不同在于：分组交换将用户要传送的信息分割为若干个分组（packet），每个分组中有一个分组头，含有可供选路的信息和其他控制信息。,3.4 The delay analysis in Networks,Delay in Circuit Switching,Assume: Number of hops = M Per-hop processing delay = P Link propagation delay = L Transmission speed = W bit/s Message size = B bits Total Delay = total pr

8、opagation + total transmission + total processing = 4ML + B/W + (M-1)P,P,L,B/W,Total Delay,Delay g in Datagram Packet Switching,Assume: Number of hops = M Per-hop processing delay = P Link propagation delay = L Packet transmission delay = T Message size = N packets Total Delay = total propagation +

9、total transmission + total store&forward + total processing = ML + NT + (M-1)T + (M-1)P,P,T,L,Total Delay,P,T,Delay Virt. Circ. Packet Switching,Assume: Number of hops = M Per-hop processing delay = P Link propagation delay = L Packet transmission delay = T Message size = N packets Total Delay = tot

10、al propagation + total transmission + total store&forward + total processing =4ML + NT + (M-1)T + 4(M-1)P,P,T,L,Total Delay,P,T,P,Remark,We are often interested only in the delay elapsed from the time the first bit was sent by the sender to the time the last bit was received by the receiver(i.e., we

11、 exclude the time involved in acknowledging connection termination). If this is the case, the delay will be given as follows: Circuit Switching: Delay = 3ML + B/W + (M-1)P Datagram packet switching: Delay = ML + NT + (M-1)T + (M-1)P (same as before) Virtual circuit packet switching: Delay = 3ML + NT

12、 + (M-1)T + 3(M-1)P,Exercises,Q1: A and B are 4 hops apart on a datagram packet-switched network where each link is 100 mile long. Per-hop processing delay is 5ms. Packets are 1500 bytes long. All links have a transmission speed of 56kbit/s . The speed of light in the wire is approximately 125,000 m

13、iles/s. If B sends a 10-packet message to A, how long will it take A to receive the message up to the last bit (measured from the time B starts sending)?,Exercises,Q2: All is the same, except that link transmission speed now is 1Gbit/s. How long will it take A to receive the message up to the last bit (measured f