A system has an n-layer protocol hierarchy. Applications generate messages of length M bytes. At each of the layers, an h-byte header is added. What fraction of the network bandwidth is filled with headers?

中文第一章 17 题。 解：在除顶层之外的其它 n-1 层各加上

n 字节报头。报头总长： h*(n-1) 字节 答案： hn /(M + hn).

21 List two ways in which the OSI reference model and

the TCP/IP reference model are the same. Now list two ways in which they differ. 中文第一章 12 题。

答：相同点： 1. 都是基于独立的协议栈概念。 2. 两者都有功能相似的应用层、传输层、

网络层。 不同点：

1. 在 OSI 模型中，严格地定义了服务、接口、协议；在 TCP/IP 模型中，并没有严格区分服务、接口与协议。

2.OSI 模型支持非连接和面向连接的网络层通信，但在传输层只支持面向连接的通信； TCP/IP 模型只支持非连接的网络层通信，但在传输层有支持非连接和面向连接的两种协议可供用户选择。

3.TCP/IP 模型中不区分、甚至不提起物理层和数据链路层。

Chap 2

3 . Television channels are 6MHz wide. How many bits/sec can be sent if four-level digital signals are used? Assume a noiseless channel.

解：使用 nyquist 定理： 6M 的带宽可以传输每秒 12 兆次的采样值。 4 级的信号电平， V=4

最大数据传输率 ==24M (bit/s) 最大数据传输率 =2H log2V （ b/s ）

4 If a binary signal is sent over a 3-kHz channel whose signal-to noise ration is 20 dB, what is the maximum achievable data rate?

解：利用 shannon 定理： S/N=100, 对任何带宽为 H Hz ，信噪比为 S/N 的信道，

最大数据传输速率 =H log2 （ 1+S/N ） （ b/s ）

最大传输速率＝ (k bit/s) 然而根据 nyquist 定理，最大传输速率＝

(k bit/s) 取两个上限中较小的。答案是 6k bit/s 。

shannon 定理和 nyquist 定理分别是针对信道传输中的两个方面来考虑的。

Shannon 定理考虑的只是信噪比对信道传输速率的影响，如果想达到这个上限的话，需要信道编码无穷的长，打比方说就是 4 个有用比特加上 1 万比特的纠错码。这样便能够接近 shannon 定理算出的极限。

nyquist 定理是针对于带限信道的本质来说的。每一个信号编码后，时域都是有限的，频域都是无限的。经过一般的低通信道的截取，必然会造成信号的失真，不同的编码之间必然会有码间串扰（ SI ）。 nyquist 定理考虑的就是因为低通的信道特性所造成的码间串扰，使得信道的波形传输速率为 2H 波特。如果采用 16 进制的信号也就是说一个波形可以传输 4 比特数据的话， V ＝ 16 ，算出来nyquist 定理的上限值就会大于 shannon 定理的上限值了。

Chap 1 13 What is the principal difference

between connectionless communication and connection-oriented communication? 中文第一章第 7 题。

答：面向连接的服务：首先要在信源与信宿之间建立连接，然后在此连接上通信，最后拆除连接。报文的发送按顺序进行。

非连接服务：将每个数据单元打包，在包头添加地址信息，每个数据包独自寻路，同一数据流的包可能经由不同的路径到达目的地，到达的顺序也可能颠倒。

9 Is Nyquist theorem true for optical fiber of only for copper wire?

答：适用。 Nyquist 定理是数学特性，与实现技术无关。适合任何传输媒介。

就是说一个函数的傅立叶频谱中不包含频率大于 f 的正弦或者余弦信号，那么这个函数就可以用 2f 频率的抽样样值承载此函数全部的信息。

29 Why has the PCM sampling time been set at 125 μsec?

答： 125μs 的采样间隔对应于每秒采样 8000次。根据 Nyquist 定理，这可以捕捉 4K Hz信道中的所有信息。而普通的话音信道就是4K Hz 。（实际上的话音信道带宽小于 4K Hz ，但是截止频率线不是很垂直的）

30 What is the percent overhead on a T1 carrier; that is, what percent of the 1.544Mbps are not delivered to the end user?

答：每帧总比特数为： 24×8 ＋ 1 ＝ 193 bit 其中有用的为： 24×7 ＝ 168 bit ，开销

（ overhead ）为 25bit 25/193×100％＝ 13％。 当 T1被完全用于数据传输时，仅有 23 个信道被用于数据传输，第 24 个信道完全用于同步模式。以便在失去帧同步时快速恢复。

Chap 3

5 A bit string, 0111101111101111110, needs to be transmitted at the data link layer. What is the string actually transmitted after bit stuffing?

答：首尾加上 FLAG (01111110) ，中间每 5个 1 后面加上一个 0 。

结果： 011110111110011111010. （如果首尾加上 flag ，也算对）

14 What is the remainder obtained by dividing by he generator polynomial ?

答：结果是 。 写成 ， 111 ，都算对，

7 5 1x x 3 1x

2 1x x 2 1x x

17 A channel has a bit rate of 4 kbps and a propagation delay of 20 msec. For what range of frame sizes does stop-and-wait give an efficiency of at least 50 percent?

答：当传输数据帧所需的时间等同于往返传播延时时，效率便达到 50% 。传输速率4bit/ms ， 160bit 需要 40秒。所以帧长度为大于 160bit 的时候，停等协议达到 50％的效率。

20 Imagine a sliding window protocol using so many bits for sequence numbers that wraparound never occurs. What relations must hold among the four window edges and the window size, which is constant and the same for both the sender and the receiver?

答：设发送窗口： (Sl , Su) ，接受窗口： (Rl , Ru) ，窗口大小 W.

0 ≤ Su - Sl + 1 ≤ W Ru - Rl + 1 = W Sl ≤ Rl ≤ Su + 1

PPT 上面指示的题目 2 The following character encoding is used in a data link

protocol: A: 01000111; B: 11100011; FLAG: 01111110; ESC:

11100000 Show the bit sequence transmitted (in binary) for the four-

character frame: A B ESC FLAG. When each of the following framing methods are used:

(a) Character count. (b) Flag bytes with byte stuffing (c) Starting and ending flag bytes, with bit stuffing. 答： (a) 字符计数法： 00000100 01000111 11100011 11100000 01111110 即： 4 A B ESC FLAG (b) 带字符填充的首尾界符法 01111110 01000111 11100011 11100000 11100000 11100000 01111110

01111110 即： FLAG A B ESC ESC ESC FLAG FLAG (c) 带位填充的首尾标志符法 01111110 01000111 110100011 111000000 011111010

01111110 即：首尾加上 FLAG ，中间每 5 个 1 后面加上一个 0 。

3 The following data fragment occurs in the middle of a data stream for which the byte-stuffing algorithm described in the text is used: A B ESC C ESC FLAG FLAG D. What is the output after stuffing?

答：题问采用字符填充后的结果。 A B ESC ESC C ESC ESC ESC FLAG

ESC FLAG D

7 Can you think of any circumstances under which an open-loop protocol, (e.g., a Hamming code) might be preferable to the feedback-type protocols discussed throughout this chapter?

题意：在什么情况下，开环协议（如海明编码）比本章中提出的反馈类型的协议更加适用？

在传输延时很长的情况下，比如地球和火星、金星之间的通信，那么前向的误差纠正就比较合适了。还有在军事应用中，如果接收方不希望过多的数据传输以暴露自己。也更适合使用开环协议。如果出错概率较低，纠错码足够好的话，开环协议比反馈类型的协议还更简单。最后，实时系统也不能容忍传输确认帧所造成的延时。

9 Sixteen-bit messages are transmitted using a Hamming code. How many check bits are needed to ensure that the receiver can detect and correct single bit errors? Show the bit pattern transmitted for the message 1101001100110101, Assume that even parity is used in the Hamming code.

题为：需要多少位校验位才能保证纠 1 位差错？写出位模式。假设使用偶校验：

答：校验位插于第 1 、 2 、 4 、 8 、 16 位。也就是 5 个校验位。（ 5个位足够校验总长 31 位的信息（包括校验位自身））。位模式为： 0110,1011,0011,0011,1010,1.

计算方法： 3=1+2; 5=1+4; 6=2+4; 7=1+2+4; 9=1+8; 10=2+8;

11=1+2+8; 12=4+8; 13=1+4+8; 14=2+4+8;

15=1+2+4+8; 17=1+16; 18=2+16; 19=1+2+16; 20=4+16;

21=1+4+16; 1 (3, 5, 7, 9, 11, 13, 15, 17, 19, 21)∈ 2 (3, 6, 7, 9, 11, 14, 15, 18, 19)∈ 4 (5, 6, 7, 12, 13, 14, 15, 20, 21)∈ 8 (9, 10, 11, 12, 13, 14, 15)∈ 16 (17, 18, 19, 20, 21)∈

30 Compute the fraction of the bandwidth that is wasted on overhead (headers and retransmissions) for protocol 6 on a heavily-loaded 50-kbps satellite channel with data frames consisting of 40 header and 3960 data bits. Assume that the signal propagation from the earth to the satellite is 270 msec. ACK frames never occur. NAK frames are 40 bits. The error rate for data frames is 1 percent, and the error rate for NAK frames is negligible. The sequence numbers are 8 bits.

题为：使用协议 6 ，在有严重负荷的 50kbps 的卫星信道上，传输包括 40位头部和 3960位数据位的数据帧，计算浪费在附加头（ overhead ）上的带宽比重。假定：地球和卫星之间的信号传输速度为 270 msec ， ACK帧从不出现， NAK帧 40 比特，数据帧的出错率为 1% ， NAK 的出错率忽略不计。序列数据为 8位。

解： 40 + 4000*1% + 40*1% = 80.4 80.4/(3960+80.4) * 100% = 1.99%

Chap 4 3 Consider the delay of pure ALOHA

versus slotted ALOHA at low load. Which one is less? Explain your answer.

答：在低负载的网络中，纯 ALOHA 协议的数据帧立刻发出，而且出现碰撞的可能性非常小，所以发出去的帧基本上能够成功发送。而时隙 ALOHA ，则需要等待下一个时隙的到来，这样便引入了平均半个时隙长度的等待时间。

15 A seven-story office building has 15 adjacent offices per floor. Each office contains a wall socket for a terminal in the front wall, so the sockets form a rectangular grid in the vertical plane, with a separation of 4 m between sockets, both horizontally and vertically. Assuming that it is feasible to run a straight cable between any pair of sockets, horizontally, vertically, or diagonally, how many meters of cable are needed to connect all sockets using

(a) a star configuration with a single router in the middle? (b) an 802.3 LAN? 答： (a) ： (m)

(b) ： 4×(15-1)×7 + 4×(7-1) ＝ 416 (m)

7 152 2

1 1

4 ( 4) ( 8) 1832i j

i j

21 Consider building a CSMA/CD network running at 1G bps over a 1-km cable with no repeaters. The signal speed in the cable is 200,000 km/sec. What is the minimum frame size?

答： 1km 除以 200,000km/sec = 5 us, 5us×2 = 10 us,

10us × 1G bps = 10,000 (bit) = 1250 (Byte).

23 Ethernet frames must be at least 64 bytes long to ensure that the transmitter is still going in the event of a collision at the far end of the cable. Fast Ethernet has the same 64-byte minimum frame size but can get the bits out ten times faster. How is it possible to maintain the same minimum frame size?

答：在快速以太网 802.3u 中，网络的最大长度变成了 250米，是 10M 以太网的 10 分之一。

24 Some books quote the maximum size of an Ethernet frame as 1518 bytes instead of 1500 bytes. Are they wrong? Explain your answer.

答：净荷 payload 长度最大为 1500 ，如果算上目的地址 6 ，源地址 6 ，长度字段 2 ，校验字段 4 ，就是 1518 。

40 A switch designed for use with fast Ethernet has a backplane that can move 10 Gbps. How many frames/sec can it handle in the worst case?

答：快速以太网中最小帧长度依然是 64Byte ，512bit 。所以：

1×10^10 / 512 = 19,531,250 (frames/sec)

42 Briefly describe the difference between store-and-forward and cut-through switches.

答：存储转发式交换机： 将数据帧从头到尾全部接收后先存放到内存，检查帧的

FCS 后再进行转发。抛弃错误的数据帧，这对要求高可靠性的客户 - 服务器系统十分有效。另外，它还能与不同传输速率的链路和不同类型的链路进行互连。

缺点是延时相对要大一些。 切换方式交换机： 切换方式也可称为在线（ on the line ）方式，它只参照

MAC帧头的目的地址，接收到数据帧的头 6 个字节后就可查找内部MAC 地址表，找到输出端口后立即输出。因此从帧的接收到转发它的内部时延很小，对延时敏感的多媒体业务十分有效。

缺点是不检查帧的正确与否，并且不能匹配不同速率的传输端口，也不能连接异种类型的链路，如以太网与FDDI 、以太网与 ATM等。

Chap 5 6: Assuming that all routers and hosts are

working properly and that all software in both is free of all errors, is there any chance, however small, that a packet will be delivered to the wrong destination?

答：有。大的噪声可能严重影响数据包。在具有 k比特校验和的情况下，错误有 2^-k 的可能性不能被检测出来。如果地址域、或者虚电路的 virtual circuit number改变了，数据包就会被发送到错误的目的地，并且被看成是正确的发送。换句话说，一个突发噪声可以将一个完全合乎规定的数据包改变成另一个合乎规定的数据包。

9: Consider the subnet of Fig.5-13(a).Distance vector routing is used, and the following vectors have just come in to the router C: from B:(5,0,8,12,6,2); from D:(16,12,6,0,9,10); from E:(7,6,3,9,0,4). The measured delays to B,D and E are 6,3,and 5,respectively. What is C’s new routing table? Give both the outgoing line to use and the expected delay.

答： Going via B gives (11, 6, 14, 18, 12, 8). Going via D gives (19, 15, 9, 3, 9, 10). Going via E gives (12, 11, 8, 14, 5, 9). Taking the minimum for each destination except C gives

(11, 6, 0, 3, 5, 8). The outgoing lines are (B, B, –, D, E, B).

10: If delays are recorded as 8-bit numbers in a 50-router network, and delay vectors are exchanged twice a second, how much bandwidth per(full-duplex)line is chewed up by the distributed routing algorithm? Assume that each router has three lines to other routers.

答：一个路由表是 400bit ，每条线路上每秒（单向）交换两次路由表，所以每条链路上每个方向需要 800bit/s 的带宽。

34. Suppose that host A is connected to a router R1, R1 is connected to another router R2,and R2 is connected to host B. Suppose that a TCP message that contains 900bytes of data and 20 bytes of TCP header is passed to the IP code at host A for delivery to B. Show the Total length, Identification, DF, MF, and Fragment offset fields of the IP header in each packet transmitted over the three links. Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frame header, link R1-R2 can support a maximum frame size of 512 bytes, including an 8-byte frame header, and link R2-B can support a maximum frame size of 512 bytes including a 12-byte frame header.

答：三段的 ID 都一样就可以了。这里认为 IP 头标是 20 个字节，（无选项）

Link A-R1: Length = 940; ID = x; DF = 0; MF = 0; Offset = 0 Link R1-R2: (1) Length = 500; ID = x; DF = 0; MF = 1; Offset = 0 (2) Length = 460; ID = x; DF = 0; MF = 0; Offset = 60 Link R2-B: (1) Length = 500; ID = x; DF = 0; MF = 1; Offset = 0 (2) Length = 460; ID = x; DF = 0; MF = 0; Offset = 60

A network on the Internet has a subnet mask of 255.255.240.0. What is the maximum number of hosts it can handle?

答： 2^12 -2 =4094 4094 – 2 = 4092

41. A router has just received the following new IP addresses: 57.6.96.0/21, 57.6.104.0/21, 57.6.112.0/21, and 57.6.120.0/21. If all of them use the same outgoing line, can they be aggregated? If so, to what? If not, why not?

答：如果需要聚类的话，就聚类入这样的子网：57.6.96.0/19

几个地址分别为：00111001.00000110.01100000.00000000

00111001.00000110.01101000.00000000 00111001.00000110.01110000.00000000 00111001.00000110.01111000.00000000

A router has the following(CIDR) entries in its routing table:

Address/mask Next hop 135.46.56.0/22 Interface 0 135.46.60.0/22 Interface 1 192.53.40.0/23 Router 1 default Router 2 For each of the following IP addresses, what does the

router do if a packet with that address arrives? 答： (a) 135.46.63.10 Interface 1 (b) 135.46.57.14 Interface 0 (c) 135.46.52.2 Router 2 (d) 192.53.40.7 Router 1 (e) 192.53.56.7 Router 2

补充

IP 头标中的 TTL 域有什么作用？ 答： 8bits ，单位秒，表示分组的生存时间。

实际操作时，分组每经过一个路由器， TTL值减一，当 TTL 值为 0 时，该分组被丢弃。作用：仿制数据报在网络中无限制的漫游，当路由表崩溃时就会发生这种情况。

ICMP 协议的主要功能是什么？ 答： IP 在网络层提供尽力服务（ best effort service) ，当分组由于各

种原因无法投递而遭丢弃时，就用互连网控制报文协议（ ICMP ）发送差错报告。

ICMP 定义了两类报文：差错报文和信息报文 差错报文： 源抑制（ Source quench ）：抑制发送过多分组的主

机。 超时（ Time exceeded ） ：分组的 TTL 为 0 。 信宿不可达（ Destination unreachable ） ：报告子网、主机不能定

位的信宿。 重定向（ Redirect ）：路由重定向 参数问题（ Parameter problem) ：分组头参数出错。 信息报文： 回音请求 / 响应（ Echo request/reply ） 地址掩码请求 / 响应（ Address mask request/reply ） 路由器发现（ Router discovery ）

3 ）某实验室有若干台 PC （设 20台左右），一台 24 口二层交换机和一个全局 IP 地址（如 202.38.75.11 ），要建立一个内部局域网，请画出网络拓扑结构，内部 IP 地址分配，网关基本功能。

网关启用 NAT 地址转换协议

19: What’s the total size of the minimum TCP MTU, including TCP and IP overhead but not including data link layer overhead?

答：缺省的段长度是 536 字节，加上 TCP 20 字节头标和 IP 20 字节头标，总共是 576字节。

20: Datagram fragmentation and reassembly are handled by IP and are invisible to TCP. Does this mean that TCP does not have to worry about data arriving in the wrong order?

答： IP 层的重组功能是对一个长度超过最小路径MTU 的数据包分片后，使这些分片按序到达的。即使整个的数据包完整的到达了，那么不同的数据包之间的按序到达是 IP 层做不到的。所以 TCP 必须将整个消息中的数据包的到达顺序进行排序重组。

30: If the TCP round-trip time, RTT, is currently 30msec and the following acknowledgements come in after 26, 32 and 24msec, respectively, what is the new RTT estimate using the Jacobson algorithm? Use α ＝ 0.9

答： RTT=RTT+(1- )M

答案： 29.6 29.84 29.256

31: A TCP machine is sending full windows of 65,535 bytes over a 1-Gbps channel that has a 10-msec one-way delay. What is the maximum throughput achievable? What is the line efficiency?

答： 20msec 发送一个窗口，每秒钟发送 50 个窗口。大概比特速率是 3.3M Bytes/sec 。

线路利用率： 3.3×8/1000 = 2.6% TCP 的“窗口比例选项”