没有合适的资源?快使用搜索试试~ 我知道了~
1. 概述 1. (网络构成要素)列举计算机网络的构成元素,分别举一个例子,并说明这些构成要素之 2. (网络连通性)说明通过直接连通实现网络连通性的局限 3.
资源详情
资源评论
资源推荐
版本: 1.0
《计算机网络》课程习题集
《计算机网络》课程组
华中科技大学
电信学院
2020.11
《计算机网络》课程习题集
I
目 录
目 录 ................................................................................................................................................ I
1. 概述 ........................................................................................................................................... 1
1.1. 基本概念 ................................................................................................................................. 1
1.2. 网络架构 ................................................................................................................................. 1
1.3. 网络性能 ................................................................................................................................. 2
2. 物理/数据链路层基础 ............................................................................................................... 4
2.1 基础 ......................................................................................................................................... 4
2.2 检错/纠错 ................................................................................................................................ 4
2.3 滑动窗口/ARQ ......................................................................................................................... 4
3. 局域网 ....................................................................................................................................... 8
3.1 基础 ......................................................................................................................................... 8
3.2 以太网 ..................................................................................................................................... 8
3.3 无线局域网 ............................................................................................................................. 9
4. 分组交换 ................................................................................................................................. 10
4.1 数据报交换 ........................................................................................................................... 10
4.2 虚电路交换 ........................................................................................................................... 10
4.3 网桥与自学习 ....................................................................................................................... 11
4.4 局域网交换机 ....................................................................................................................... 13
5. 网络层 ..................................................................................................................................... 15
5.1 基础 ....................................................................................................................................... 15
5.2 路由算法 ............................................................................................................................... 17
5.3 IP 网络路由 ........................................................................................................................... 19
5.4 IP 进阶(多播、IPV6、MPLS) ................................................................................................. 21
6. 传输层 ..................................................................................................................................... 23
6.1 UDP ........................................................................................................................................ 23
6.2 TCP ......................................................................................................................................... 23
6.3 TCP 拥塞控制 ........................................................................................................................ 24
7. 应用层 ..................................................................................................................................... 26
8. 综合 ......................................................................................................................................... 27
参考文献 ......................................................................................................................................... 28
《计算机网络》课程习题集
1
1. 概述
1.1. 基本概念
1. (网络构成要素)列举计算机网络的构成元素,分别举一个例子,并说明这些构成要素之
间的关联。
答:(对于每一类构成元素,举一个正确的例子即可,可能不在下列参考答案之列)
计算机网络的构成元素包括(1)主机、(2)交换节点、(3)链路、(4)网络应用、(5)协议,对应的
例子包括(1)台式机/服务器/智能手机等、(2)交换机/路由器等、(3)双绞线/光纤/无线链路等、
(4)浏览器/QQ/微信等、(5)IP/TCP/HTTP 等。
2. (网络连通性)说明通过直接连通实现网络连通性的局限。
答:直接连通链路有点对点、多路接入两种。(1) 采用点对点链路实现 N 台主机的网络连通,
所需链路条数与 N
2
成正比,开销随着网络规模的增大急剧增加,而且很难用于地理分布较
广的网络,没有可扩展性。(2) 采用多路接入链路实现网络连通,发送冲突的可能性随着主
机数量的增加而增大,因此浪费更多的带宽资源,同时多路接入(MAC)算法的效果和效率也
受限于主机之间的距离,难以有效解决空间分布较广的主机竞争使用共享信道的冲突。
3. (网络资源共享, Ex 1.29, [PD12])假设共享介质 M 以循环方式向主机 A
1
、A
2
、…、A
N
提
供传输一个分组的机会,没有分组要传的主机立即放弃 M。它与 STDM 有何不同?与
STDM 相比,这种方式对网络的利用率如何?
答:In STDM the offered timeslices are always the same length, and are wasted if they are unused
by the assigned station. The round-robin access mechanism would generally give each station only
as much time as it needed to transmit, or none if the station had nothing to send, and so network
utilization would be expected to be much higher.
4. (编址,Ex. 1.7, [PD12]) 邮政地址与网络编址有何相似和不同之处?电话号码与网络编
址又有何相似之处?
Answer: Postal addresses are strongly hierarchical (with a geographical hierarchy, which network
addressing may or may not use). Addresses also provide embedded “routing information”. Unlike
typical network addresses, postal addresses are long and of variable length and contain a certain
amount of redundant information. This last attribute makes them more tolerant of minor errors and
inconsistencies. Telephone numbers, at least those assigned to landlines, are more similar to
network addresses: they are (geographically) hierarchical, fixed-length, administratively assigned,
and in more-or-less one-to-one correspondence with nodes.
1.2. 网络架构
5. (五层网络体系架构) 对于合并 OSI 参考模型和 Internet 架构得到的五层网络体系架构,
图解并说明每一层的功能。
《计算机网络》课程习题集
2
Answer: It is shown below that there are 5 layers in the layered network architecture, as well as
some sample nodes on a network.
The functionalities of 5 layers are respectively explained as follows:
Physical: bit streaming — deal with the transmission of raw bits over a physical link
Data link: framing — collect a stream of bits into a bit aggregate called frame
Network: host-to-host communication — handle packet forwarding along the nodes within a
network
Transport: process-to-process communication — implement the communication between two
processes running on two hosts
Application: application-specific services — concern various types of application services
provided to end users, e.g., ftp, telnet
1.3. 网络性能
6. (文件传输时延与 RTT, Ex. 1.3 rev., [PD12]) 设 RTT 为 50ms、分组长度为 1KB、数据发
送前需要时长为 2RTT 的握手过程,计算下列情况下传输 1000KB 长度文件所需时间:
(a) 速率为 2 Mbps,分组连续发送;
(b) 速率为 2 Mbps,但发送完一个分组后需要等待一个 RTT 才能发送下一个分组;
(c) 速率无限高,但每个 RTT 只能发送 20 个分组;
(d) 速率无限高,但首个 RTT 只能发送 1 个分组,第二个 RTT 可以发送 2 个分组,第
三个 RTT 可以发送 4 个分组,依此类推。
Answer: We will count the transfer as completed when the last data bit arrives at its destination.
An alternative interpretation would be to count until the last ACK arrives back at the sender, in
which case the time would be half an RTT (25ms) longer.
(a) 2 initial RTT’s (100ms) + 1000KB/2Mbps (transmit) + RTT/2 (propagation = 25ms)
≈ 0.125 + 8Mbit/2Mbps = 0.125 + 4 sec = 4.125 sec.
If we pay more careful attention to when a mega is 10
6
versus 2
20
, we get 8,192,000 bits
/2,000,000bps = 4.096 sec, for a total delay of 4.221 sec.
(b) To the above we add the time for 999 RTTs (the number of RTTs between when packet 1
arrives and packet 1000 arrives), for a total of 4.221 + 49.95 = 54.171 sec.
(c) This is 49.5 RTTs, plus the initial 2, for 2.575 seconds.
(d) Right after the handshaking is done we send one packet. One RTT after the handshaking we
send two packets. At n RTTs past the initial handshaking we have sent 1 + 2 + 4 + ... + 2
n
=
2
n
+1
−1 packets. At n = 9 we have thus been able to send all 1,000 packets; the last batch
《计算机网络》课程习题集
3
arrives 0.5 RTT later. Total time is 2+9.5 RTTs, or 0.575 sec.
7. (文件传输时延)设长度为 F 的文件从源主机经过 M 台分组交换机传输至另一主机,分组
长度固定为 L,封装开销可以忽略,每台交换机中分组连续转发,节点处理和排队时延
均为零,每条链路速率、传播时延分别为 C、p。给出下列情况下的文件传输时延:(a) 交
换机均为存储转发模式;(b) 交换机均为直通式,只需收到分组前 N 比特即可转发。
Answer: Considering (1) file transmission time at source, (2) propagation delays over M+1 links,
and (3) packet transmission delays at M switches, end-to-end file transmission latency can be
achieved by:
(a) D = F/C + (M+1)p + ML/C
(b) D = F/C + (M+1)p + MN/C
8. (电路交换与分组交换时延比较)比较在一个电路交换网和在一个(负载轻的)分组交换网
上将 x(bit)报文沿 k 个跳段传输的通路传输的延迟.假定电路建立时间是 s,每跨段上的传
输延迟为 d,分组大小为 p(bit),数据传输速率是 b(b/s).在什么情况下,分组交换网的延迟更
短?(忽略分组头的开销)
答:对于电路交换,t=s 时电路就会建立起来;t=s+x/b 时报文的最后一位发送完毕;t=s+x/b+kd
时报文到达目的地。而对于分组交换,最后一位在 t=x/b 时发送完毕。为到达最终的目的地,
最后 1 个分组必须被中间的路由器重发 k-1 次,每次重发花时间 p/b(一个分组的所有比特
都接收齐了,才能开始重发,因此最后 1 位在每个中间结点的停滞时间为最后一个分组的发
送时间),所以总的延迟为:
x/b+(k-1)p/b+kd
为使分组交换比电路交换快,令:
x/b+(k-1)p/b+kd < s+x/b+kd
得: s > (k-1)p/b
当满足此条件时,分组交换网得延迟更短。
剩余29页未读,继续阅读
茶啊冲的小男孩
- 粉丝: 26
- 资源: 326
上传资源 快速赚钱
- 我的内容管理 展开
- 我的资源 快来上传第一个资源
- 我的收益 登录查看自己的收益
- 我的积分 登录查看自己的积分
- 我的C币 登录后查看C币余额
- 我的收藏
- 我的下载
- 下载帮助
最新资源
- 5.23-Java概述,JDK安装及注释、关键字、标识符、数据类型、变量、常量的介绍
- 《Python基础》实验三指导书(1).doc
- TensorFlow 深度学习、机器学习-任何能够用计算流图形来表达的计算,都可以使用TensorFlow
- 一个基于springboot+sureness的面向REST API资源无状态认证权限管理系统
- 王博外文文献.pdf
- python毕业设计基于社区检测的多任务聚类联邦学习项目源码+使用说明(高分项目).zip
- Javaweb项目源码-编程爱好者博客地带.zip
- java各个技术栈相关知识点
- PYthon代码 pdf合并
- 内容涵盖:Java、MyBatis、ZooKeeper、Dubbo、Elasticsearch、Memcached、 Redis
资源上传下载、课程学习等过程中有任何疑问或建议,欢迎提出宝贵意见哦~我们会及时处理!
点击此处反馈
安全验证
文档复制为VIP权益,开通VIP直接复制
信息提交成功
评论0