g) The threshold is set to half the value of the congestion window when packet loss is

detected. When timer out event occurred during transmission round 23, the congestion windows size is 26. Hence the threshold is 13 during the 24th transmission round.

h) During the 1st transmission round, packet 1 is sent; packet 2-3 are sent in the 2nd

transmission round; packets 4-7 are sent in the 3rd transmission round; packets 8-15 are sent in the 4th transmission round; packets 15-31 are sent in the 5th transmission round; packets 32-63 are sent in the 6th transmission round; packets 64 – 96 are sent in the 7th transmission round. Thus packet 70 is sent in the 7th transmission round.

i) The congestion window and threshold will be set to half the current value of the

congestion window (8) when the loss occurred. Thus the new values of the threshold and window will be 4.

17. （教材P38）Host A is sending an enormous file to Host B over a TCP connection. Over this

connection there is never any packet loss and the timers never expire. Denote the transmission rate of the link connecting Host A to the Internet by R bps. Suppose that the process in Host A is capable of sending data into its TCP socket at a rate S bps, where S = 10*R. Further suppose that the TCP receive buffer is large enough to hold the entire file, and the send buffer can hold only one percent of the file. What would prevent the process in Host A from continuously passing data to its TCP socket at rate S bps? TCP flow control? TCP congestion control? Or something else? Elaborate.

Answer: In this problem, there is no danger in overflowing the receiver since the receiver’s receive buffer can hold the entire file. Also, because there is no loss and acknowledgements are returned before timers expire, TCP congestion control does not throttle the sender. However, the process in host A will not continuously pass data to the socket because the send buffer will quickly fill up. Once the send buffer becomes full, the process will pass data at an average rate or R << S.

18. 在TCP协议中,为了使通信不致发生混乱,引入了所谓套接字的概念,这里,套接字由( )

和IP地址两部分组成。

A。端口号 B.域名 C。接口 D。物理地址

解析:端口号和IP地址合起来,称为套接字,套接字可以在全网范围内唯一标识一个端口。在TCP协议中,一条连接两端的套接字就可以唯一标识该连接了。所以选项A为正确答案。 答案:A

19. 面向连接的传输有三个过程:连接建立、( )和连接释放。 A．连接请求 B。连接应答 C。数据传输 D。数据共享

解析:面向连接服务具有连接建立、数据传输和连接释放这三个阶段。所以选项C为正确答案。 答案:C

20. 试述UDP和TCP协议的主要特点及它们的使用场合。(华中科技大学2003年试题)

解析:用户数据报协议是对IP协议组的扩充,它增加了一种机制,发送方使用这种机制可以区分一台计算机上的多个接收者。每个UDP报文除了包含某用户进程发送数据外,还有报文目的端口的编号和报文源端口的编号,从而使UDP的这种扩充,在两个用户进程之间的递送数据报成为可能。

TCP提供的是一种可靠的数据流服务。当传送受差错干扰的数据,或基础网络故障,或网络负荷太重而使网络基本传输系统(无连接报文递交系统)不能正常工作时,就需要通过TCP这样的协议来保证通信的可靠。

答案:UDP是一个简单的面向数据报的传输层协议。应用进程的每个输出操作都产生一个UDP数据报,并组装成一份待发送的IP数据报中发送。UDP提供不可靠、无连接的数据报服务,它把应用程序传给IP层的数据发送出去,但是并不保证它们能到达目的地。因此,uDP通常用于不要求可靠传输的场合,另外也常用于客户机/服务器模式中。

TCP协议被用来在一个不可靠的互联网中为应用程序提供可靠的端点间的字节流服务。所有TCP连接都是全双工和点对点的,因而TCP不支持广播和组播的功能。TCP实体间以“段”为单位进行数据交换。为实现可靠的数据传输服务,TCP提供了对段的检错、应答、重传和排序的功能,提供了可靠地建立连接和拆除连接的方法,还提供了流量控制和阻塞控制的机制。TCP适用于传输大量重要数据的场合。

21. 在使用TCP协议传送数据时,如果有一个确认报文段丢失了,也不一定会引起对方数据的

重传。试说明为什么?

解析:本题考查的是TCP重传机制。

TCP连接的一个重要的特性就是为上层服务提供了一个可靠的数据流。由于TCP是建立在不可靠的IP层的基础之上的,因此就必然涉及报文丢失的问题,这样,报文的重传就成了保证数据可靠到达的一个重要机制。这方面TCP采取了超时重传的策略,对每个TCP连接都维护一个计时器,每发送一个报文:就设置一次计时器,只要计时器设置的重传时间已到但仍然没有收到相应的确认信息,就重传这一报文。

答案:对方还未来得及重传,就收到了对更高序号的确认,相当于对连同被丢失确认的报文段一并确认。

22. 在连续ARQ协议中,设编号用3位,而发送窗口WT=8,试找出一种情况,使得在此情况下协议不能正常工作。

解答:发送端: 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0

接收端: 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0

当发送方发送0~7数据帧,因发送窗口已满,发送暂停,接收方收到所有数据帧,并对每一个帧都发送确认帧,若所有的确认帧都没有到达发送方,经过发送方计时器控制的超时后,发送方会再次发送之前的8个数据帧,而接收方收到这8个帧却无法判断是新的数据帧或是重传的数据帧。

23. 在选择重传ARQ协议中,设编号用3位。再设发送窗口WT=6,而接收窗口WR=3。试找出

一种情况,使得在此情况下协议不能正常工作。 解答:发送端:0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0

接收端:0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0

Chapter 4

1. （R2，答案R3）What is the difference between routing and forwarding?

2. （R3，答案R2）What are the two most important network-layer functions in a datagram

network? What are the three most important network-layer functions in a virtual-circuit network?

3. （R4，答案R4）Do the routers in both datagram networks and virtual-circuit networks use

forwarding tables? If so, describe the forwarding tables for both classes of networks. 4. （R6，答案R9）Describe how packet loss can occur at input ports. Describe how packet loss

at input ports can be eliminated (without using infinite buffers). 5. 6. 7. 8. 9.

（R9，答案R10）Describe how packet loss can occur at output ports.

（R11，答案R11）What is HOL blocking? Does it occur in input ports or output ports? （R12，答案R13）What is the 32-bit binary equivalent of the IP address 223.1.3.27? （R13，答案R？）Do routers have IP addresses? If so, how many?

（R22，答案R23）Is it necessary that every autonomous system use the same intra-AS routing algorithm? Why or why not?

10. （R24，答案R24）Consider Figure 4.35. Starting with the original table in D, suppose that D

receives from A the following advertisement:

Will the table in D change? If so how?

11. （R27，答案R22）Why are different inter-AS and intra-AS protocols used in the Internet? 12. （P8，答案P8，但具体值不对）Consider a datagram network using 32-bit host addresses.

Suppose a router has five links, numbered 0 through 4, and packets are to be forwarded to the link interfaces as follows:

Destination Address Range Link Interface

11100000 00000000 00000000 00000000

Through 0 11100000 00000000 11111111 11111111

11100000 00000001 00000000 00000000

Through 1

11100000 00000001 11111111 11111111

11100000 00000010 00000000 00000000

Through 2

11100000 11111111 11111111 11111111

11100001 00000000 00000000 00000000

Through 3

11100001 11111111 11111111 11111111

Otherwise 4

a. Provide a forwarding table that has five entries, uses longest prefix matching, and forwards packets to the correct link interfaces.

Prefix Match Link Interface 11100000 00000000 0 11100000 00000001 1 11100000 2 11100001 3 Otherwise 4

b. Describe how your forwarding table determines the appropriate link interface for datagrams with destination addresses:

11001000 10010001 01010001 01010101 11100000 10101101 11000011 00111100 11100001 10000000 00010001 01110111

13. （P9，答案P9）Consider a datagram network using 8 bit host addresses. Suppose a router

uses longest prefix matching and has the following forwarding table: Prefix Match 00 01 10 11 Interface 0 1 2 3 For each of the four interfaces, give the associated range of destination host addresses and the number of addresses in the range.

（14、15、16可选讲）

14. （P11，答案P11，但值不对）Consider a router that interconnects three subnets: Subnet 1,

Subnet 2, and Subnet 3. Suppose all of the interfaces in each of these three subnets are required to have the prefix 220.2.240/20. Also suppose that Subnet 1 is required to support up to 2000 interfaces, and Subnets 2 and 3 are each required to support up to 1000 interfaces. Provide three network addresses (of the form a.b.c.d/x) that satisfy these constraints.

总共的地址空间包括232-20=212=4096

220.2.240.0/20 = 11011100 00000010 11110000 00000000 Subnet 1需要2000个IP，因此分配

220.2.240.0/21= 11011100 00000010 11110000 00000000

Subnet 2和3需要1000个IP，因此将剩下的/21均分为两个/22 Subnet 2：220.2.248.0/22 = 11011100 00000010 11111000 00000000 Subnet 3：220.2.252.0/22 = 11011100 00000010 11111100 00000000