传输层作业

问题1(习题5-23) 主机A向主机B连续发送了两个TCP报文段,其序号分别是70和100.试问:

(1)第一个报文段携带了多少字节的数据?

(2)主机B收到第一个报文段后,发回的确认中的确认号应当是多少?

(3)如果B收到第二个报文段后,发回的确认中的确认号是180,试问A发送的第二个报文段中的数据有多少字节?

(4)如果A发送的第一个报文段丢失了,但第二个报文段到达了B,B在第二个报文段到达后向A发送确认。试问这个确认号应为多少?

 

 

问题2(习题5-24) 一个TCP连接下面使用256 kb/s的链路,其端到端时延为128 ms。经测试,发现吞吐量只有120kb/s。试问发送窗口W是多少?(需要根据接收端发出确认的时机,写出两种情况)

 

 

网络层作业

问题1 如图1所示,网络145.13.0.0/18划分为四个子网N1,N2,N3和N4。这四个子网与路由器R连接的接口分别是m0,m1,m2和m3。路由器R的第五个接口m4连接到因特网。

    (1)试给出路由器R的路由表

    (2)路由器R收到一个分组。其目的地址是145.13.160.78。试给出这个分组是怎样被转发的。

 

              图1  网络145.13.0.0/18划分为四个子网N1,N2,N3和N4

 

问题2、设某路由器建立了如下路由表:

目的网络

子网掩码

下一跳

128.96.39.0

255.255.255.128

接口m0

128.96.39.128

255.255.255.128

接口m1

128.96.40.0

255.255.255.128

R2

192.4.153.0

255.255.255.192

R3

*(默任)

255.255.255.192

R4

现接收到5个分组,其目的地址分别为:

(1)128.96.39.10

(2)128.96.40.12

(3)128.96.40.151

(4)192.4.153.17

(5)192.4.153.90

试分别计算其下一跳。

 

问题3:习题4-20、4-31、4-33.

 

数据链路层作业

P1

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 is used:

(a) Byte count.

(b) Flag bytes with byte stuffing.

(c) Starting and ending flag bytes with bit stuffing.

问题1

数据链路协议使用了下面的字符编码:

A: 01000111 B: 11100011 FLAG: 01111110 ESC: 11100000

为了传输一个包含4个字符的帧: A B ESC FLAG,试问使用下面的成帧方法时所发送的比特序列(用二进制表达)是什么?

(a)字节计数

(b)字节填充的标志字节

(c)比特填充的首尾标志字节

 

 

p2  Two CSMA/CD stations are each trying to transmit long (multiframe) files. After each frame is sent, they contend for the channel, using the binary exponential backoff algorithm. What is the probability that the contention ends on round k, and what is the mean number of rounds per contention period?

问题2 两个CSMA/CD 站都企图传送大文件(多个帧)。每发出一帧,它们就使用二进制指数算法竞争信道。试问在第k轮结束竞争的概率是多少?每个竞争周期的平均次数是多少?

 

P3 Briefly describe the difference between store-and-forward and cut-through switches.

问题3 请简述存储-转发型交换机和直通型交换机之间的区别?

 

问题4 是什么原因使以太网有一个最小帧长和最大帧长?

 

问题5 以太网使用载波监听多点接入碰撞检测协议CSMA/CD。频分复用FDM才使用载波。以太网有没有使用频分复用?

 

问题6 课后习题3-12、3-22

 

 

 

 

 

第2章 

P-1

A noiseless 4-kHz channel is sampled every 1 msec. What is the maximum data rate? How does the maximum data rate change if the channel is noisy, with a signal-to-noise ratio of 30 dB?(每1ms对一条无噪声4kHz信道采样一次。试问最大数据传输率是多少?如果信道上有噪声,信噪比为30dB,试问最大数据速率将如何变化?)

 

P-2

What are the advantages of fiber optics over copper as a transmission medium? Is there any downside of using fiber optics over copper?(试问光纤作为传输介质,相比铜芯有什么优势?是否存在不足?)

P-3 课后习题 2-13

为什么要使用信道复用技术?常用的信道复用技术有哪些?

P-3 课后习题 2-16

共有4 个站进行码分多址CDMA 通信。4 个站的码片序列为:

A:( -1 –1 –1 +1 +1 –1 +1 +1) B:( -1 –1 +1 -1 +1 +1 +1 -1)

C:( -1 +1 –1 +1 +1 +1 -1 -1) D:( -1 +1 –1 –1 -1 –1 +1 -1)

现收到这样的码片序列:(-1 +1 –3 +1 -1 –3 +1 +1)。问哪个站发送数据了?发送数据的站发送的1 还是0?

答案如下:

P-1

A noiseless 4-kHz channel is sampled every 1 msec. What is the maximum data rate? How does the maximum data rate change if the channel is noisy, with a signal-to-noise ratio of 30 dB?(每1ms对一条无噪声4kHz信道采样一次。试问最大数据传输率是多少?如果信道上有噪声,信噪比为30dB,试问最大数据速率将如何变化?)

The answer may be:

A noiseless channel can carry an arbitrarily large amount of information, no matter how often it is sampled. Just send a lot of data per sample. For the 4-kHz channel, make 8000 samples/sec. If each sample is 16 bits, the channel can send 128 kbps. If each sample is 1024 bits, the channel can send 8.2 Mbps. The key word here is “noiseless”.With a normal 4 kHz channel, the Shannon limit would not allow this. A signal-to-noise ratio of 30 dB means S/N = 1000. So, the Shannon limit is about 39.86 kbps.

P-2

What are the advantages of fiber optics over copper as a transmission medium? Is there any downside of using fiber optics over copper?(试问光纤作为传输介质,相比铜芯有什么优势?是否存在不足?)

The answer may be:

Fiber has many advantages over copper. It can handle much higher bandwidth than copper. It is not affected by power surges, electromagnetic interference, power failures, or corrosive chemicals in the air. It does not leak light and is quite difficult to tap. Finally, it is thin and lightweight, resulting in much lower installation costs. There are some downsides of using fiber over copper. First, it can be damaged easily by being bent too much. Second, optical communication is unidirectional, thus requiring either two fibers or two frequency bands on one fiber for two-way communication. Finally, fiber interfaces cost more than electrical interfaces.

2-13 为什么要使用信道复用技术?常用的信道复用技术有哪些?

答:信道复用的目的是让不同的计算机连接到相同的信道上,以共享信道资源。在一条传输介质上传输多个信号,提高线路的利用率,降低网络的成本。这种共享技术就是多路复用技术。

频分复用(FDM,Frequency Division Multiplexing)就是将用于传输信道的总带宽划分成若干个子频带(或称子信道),每一个子信道传输1 路信号。频分复用要求总频率宽度大于各个子信道频率之和,同时为了保证各子信道中所传输的信号互不干扰,应在各子信道之间设立隔离带,这样就保证了各路信号互不干扰(条件之一)。频分复用技术的特点是所有子信道传输的信号以并行的方式工作,每一路信号传输时可不考虑传输时延,因而频分复用技术取得了非常广泛的应用。

时分复用(TDM,Time Division Multiplexing)就是将提供给整个信道传输信息的时间划分成若干时间片(简称时隙),并将这些时隙分配给每一个信号源使用,每一路信号在自己的时隙内独占信道进行数据传 .。时分复用技术的特点是时隙事先规划分配好且固定不变,所以有时也叫同步时分复用。其优点是时隙分配固定,便于调节控制,适于数字信息的传输;缺点是当某信号源没有数据传输时,它所对应的信道会出现空闲,而其他繁忙的信道无法占用这个空闲的信道,因此会降低线路的利用率。时分复用技术与频分复用技术一样,有着非常广泛的应用,电话就是其中最经典的例子,此外时分复用技术在广电也同样取得了广泛地应用,如SDH,ATM,IP 和HFC 网络中CM 与CMTS 的通信都是利用了时分复用的技术。

2-16 共有4 个站进行码分多址CDMA 通信。4 个站的码片序列为:

A:( -1 –1 –1 +1 +1 –1 +1 +1) B:( -1 –1 +1 -1 +1 +1 +1 -1)

C:( -1 +1 –1 +1 +1 +1 -1 -1) D:( -1 +1 –1 –1 -1 –1 +1 -1)

现收到这样的码片序列:(-1 +1 –3 +1 -1 –3 +1 +1)。问哪个站发送数据了?发送数据

的站发送的1 还是0?

答:S·A=(+1-1+3+1-1+3+1+1)/8=1, A 发送1

S·B=(+1-1-3-1-1-3+1-1)/8=-1, B 发送0

S·C=(+1+1+3+1-1-3-1-1)/8=0, C 无发送

S·D=(+1+1+3-1+1+3+1-1)/8=1, D 发送1

第1章 题 目

问题1-1

五层协议的网络体系结构的要点,各层的主要功能是什么?

问题1-2

协议和服务有何区别?有何关系?网络协议的三要素是什么?

P1-3

A client-server system uses a satellite network, with the satellite at a height of 40,000 km. What is the best-case delay in response to a request?

P1-4

An image is 1600 × 1200 pixels with 3 bytes/pixel. Assume the image isuncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet? Over gigabit Ethernet?

答案如下:

问题1-1

五层协议的网络体系结构的要点,各层的主要功能是什么?

P28页

问题1-2

协议和服务有何区别?有何关系?网络协议的三要素是什么?

P31页

P1-3

A client-server system uses a satellite network, with the satellite at a height of 40,000 km. What is the best-case delay in response to a request?

The answer may be:

The request has to go up and down, and the response has to go up and down. The total path length traversed is thus 160,000 km. The speed of light in air and vacuum is 300,000 km/sec, so the propagation delay alone is 160,000/300,000 sec or about 533 msec.

 

P1-4

An image is 1600 × 1200 pixels with 3 bytes/pixel. Assume the image isuncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet? Over gigabit Ethernet?

 

The answer may be:

The image is 1600 × 1200 × 3 bytes or 5,760,000 bytes. This is 46,080,000 bits. At 56,000 bits/sec, it takes about 822.857 sec. At 1,000,000 bits/sec, it takes 46.080 sec. At 10,000,000 bits/sec, it takes 4.608 sec. At 100,000,000 bits/sec, it takes about 0.461 sec. At 1,000,000,000 bits/sec it takes about 46 msec.




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