中山大学2014软件学院计算机网络期中考试题目
计算机网络期中试卷带答案

《计算机网络技术》期中考试试卷一、填空题(每题1分,共30分)1、TCP 协议的全称是指 传输控制协议 ,IP 协议的全称是指 网际协议 。
2、网络协议由 语法 、 语义 和 时序 3部分组成。
3、以太网是一种常用的 总线 结构局域网,它基于 IEEE802.3 协议标准,采用媒体访问控制控制方法 CSMA/CD 。
4、所谓“计算机网络”是指,将分布在 不同 地理位置,具有 独立 功能的多台计算机及其外部设备,用 通信 设备和 通信 线路连接起来,在网络操作系统和 协议 及网络管理软件的管理协调下,实现 资源共享 、信息传递的系统。
5、计算机网络资源包括 硬件 资源、 软件 资源和 数据 资源。
6、被称之为Internet 之父的是 ARPA 。
7、计算机网络是现代 数据通信 技术与 计算机 技术结合的产物。
8、在因特网中,WWW 的中文含义是 万维网 。
9、通信介质分为两大类即有线介质和 无线介质 。
有线介质包括双绞线、同轴电缆和 光纤 。
10、双绞线的选择是,电脑与集线器之间应用 直通 来进行连接。
11、在TCP/IP 协议中,传输层主要的两个协议中, TCP 提供面向连接的可靠数据传输服务是, UDP 采用无连接数据报传送方式。
12、实现互联网中的工作站登录到远程服务器的协议是 telnet ,其默认端口号为 21 。
13、OSI 参考模型共分为 7 层,其最上面的三层对应于TCP/IP 协议的 应用层 。
二、判断题,在正确的选项上打√,错的题目上打×。
(每题1分,共10分) 1、哑终端是没有自己的CPU 、内存和硬盘的。
( √ ) 2、国家信息基础设施的简称为NII 。
( √ )3、终端不具有本地功能,但可以直接连到网络中。
( √ )4、windows XP 、windows2003 server 基础是个人操作系统。
( × )5、因特网是一种典型的广域网。
2014软件水平考试(中级) 网络工程师真题及答案综合

2014软件水平考试(中级) 网络工程师真题及答案综合第1部分:单项选择题,共42题,每题只有一个正确答案,多选或少选均不得分。
1.[单选题]容量为64块的Cache采用组相联方式映像,字块大小为128个字,每4块为一组。
若主存容量为4096块,且以字编址,那么主存地址应为( )位,主存区号应为( )位。
A)16 5B)17 6C)18 7D)19 6答案:D解析:以字编址,字块大小为128个字,容量为4096块主存,则128×4096=2^19,主存地址为19位;由于采用组相联方式映像,Cache容量为64块,则主存区数=4096/64=64=2^6,主存区号为6位。
2.[单选题]中断响应时间是指( )A)从中断处理开始到中断处理结束所用的时间B)从发出中断请求到中断处理结束所用的时间C)从发出中断请求到进入中断处理所用的时间D)从中断处理结束到再次中断请求的时间答案:C解析:中断处理过程分为两个阶段:中断响应过程和中断服务过程。
中断响应时间是中断响应过程所用的时间,即从发出中断请求到进入中断处理所用的时间3.[单选题]在单指令流多数据流计算机(SIMD)中,各处理单元必须( )A)以同步方式,在同一时间内执行不同的指令B)以同步方式,在同一时间内执行同一条指令C)以异步方式,在同一时间内执行不同的指令D)以异步方式,在同一时间内执行同一条指令答案:B解析:SIMD(Single Instruction Multiple Datastream,单指令流多数据流):同一条指令控制多个处理器的运行。
在这种计算机中有多个处理单元,但只有一个控制部件,所有处理单元以同步方式,在同一时刻执行同一条指令,处理不同的数据。
4.[单选题]虚拟存储管理系统的基础是程序的( )理论,这个理论的基本含义是指程序执行时往往会不均匀地访问主存储器单元。
根据这个理论,Denning提出了工作集理论。
工作集是进程运行时被频繁地访问的页面集合。
2014软件水平考试(中级) 软件设计师真题及答案综合

2014软件水平考试(中级) 软件设计师真题及答案综合说明:答案和解析在试卷最后第1部分:单项选择题,共69题,每题只有一个正确答案,多选或少选均不得分。
1.[单选题]下图所示为(46)设计模式,适用于(47)。
A)一个系统要由多个产品系列中的一个来配置时B)当一个类希望由它的子类来指定它所创建的对象时C)当创建复杂对象的算法应该独立于该对象的组成部分及其装配方式时D)当一个系统应该独立于它的产品创建、构成和表示时2.[单选题]在如下所示的进程资源图中,(27);该进程资源图是(28)。
A)P1、P2、P3都是阻塞节点B)P1是阻塞节点,P2、P3是非阻塞节点C)P1、P2是阻塞节点,P3是非阻塞节点D)P1、P2是非阻塞节点,P3是阻塞节点3.[单选题]A)π1,2,7(σ2=‘信息’,∧3=5∧4=6∧7’北京’(R×S))B)π1,2,7(σ3==5∧4=6(σ2=‘信息’(R)×σ5=‘北京’(S)))C)π1,2,7(σ3==5∧4=6∧2=‘’(R×σ7=’’(S)))D)π1,2,7(σ3==5∧4=6∧7=‘北京’(σ2=‘信息’(R)×(S)))4.[单选题]DHCP客户端可以从DHCP服务器获得(69)。
A)DHCP服务器的地址和Web服务器的地址B)DNS服务器的地址和DHCP服务器的地址C)客户端地址和邮件服务器地址D)默认网关的地址和邮件服务器地址5.[单选题]Flynn分类法基于信息流特征将计算机分成4类,其中(6)只有理论意义而无实例。
A)SISDB)MISDC)SIMDD)MIMD6.[单选题]ICMP协议属于因特网中的(67)协议,ICMP协议数据单元封装在(68)中传送。
A)以太帧B)TCP段C)UDP数据报D)IP数据报7.[单选题]PPP中的安全认证协议是(66),它使用三次握手的会话过程传送密文。
A.MDSB.PA)PB)CHC)PD)NCP8.[单选题]Teams are required for most engineering projects. Although some small hardware or software products can be developed by individuals, the scale and complexity of modem systems is such, and the demand for short schedules so great, that it is no longer ___71___ for one person to do most engineering jobs. Systems development is a team ___72___, and the effectiveness of the team largely determines the ___73___ of the engineering.Development teams often behave much like baseball or basketball teams. Even though they may have multiple specialties, all the members work toward ___74___. However, on systems maintenance and enhancement teams, the engineers often work relatively independently, much likewrestling and track teams.A team is ___75__ just a group of people who happen to work together. Teamwork takes practice and it involves special skills. Teams require common processes; they need agreed-upon goals; and they need effective guidance and leadership. The methods for guiding and leading such teams are well known,but they are not obvious.A)activityB)jobC)processD)application9.[单选题]Teams are required for most engineering projects. Although some small hardware or software products can be developed by individuals, the scale and complexity of modem systems is such, and the demand for short schedules so great, that it is no longer ___71___ for one person to do most engineering jobs. Systems development is a team ___72___, and the effectiveness of the team largely determines the ___73___ of the engineering.Development teams often behave much like baseball or basketball teams. Even though they may have multiple specialties, all the members work toward ___74___. However, on systems maintenance and enhancement teams, the engineers often work relatively independently, much likewrestling and track teams.A team is ___75__ just a group of people who happen to work together. Teamwork takes practice and it involves special skills. Teams require common processes; they need agreed-upon goals; and they need effective guidance and leadership. The methods for guiding and leading such teams are well known,but they are not obvious.A)multiple objectivesB)different objectivesC)a single objectiveD)independent objectives10.[单选题]Teams are required for most engineering projects. Although some small hardware or software products can be developed by individuals, the scale and complexity of modem systems is such, and the demand for short schedules so great, that it is no longer ___71___ for one person to do most engineering jobs. Systems development is a team ___72___, and the effectiveness of the team largely determines the ___73___ of the engineering.Development teams often behave much like baseball or basketball teams. Even though they may have multiple specialties, all the members work toward ___74___. However, on systems maintenance and enhancement teams, the engineers often work relatively independently, much likewrestling and track teams.A team is ___75__ just a group of people who happen to work together. Teamwork takes practice and it involves special skills. Teams require common processes; they need agreed-upon goals; and they need effective guidance and leadership. The methods for guiding and leading such teams are well known,but they are not obvious.A)sizeB)qualityC)scaleD)complexity11.[单选题]UML图中,一张交互图显示一个交互,由一组对象及其之间的关系组成,包含它们之间可能传递的消息。
计算机网络期中试题及答案

计算机网络期中试题及答案请注意,我是一名AI助手,并非真人。
根据你的需求,我可以为你提供一份关于计算机网络期中试题及答案的文章。
下面是我根据你的要求所写的文本:计算机网络期中试题及答案随着计算机网络技术的不断发展,越来越多的人开始关注并学习计算机网络知识。
作为计算机科学与技术专业的学生,掌握计算机网络的基本原理和相关技术是十分重要的。
下面是一份关于计算机网络期中试题及答案,帮助你巩固对计算机网络的理解。
试题一:计算机网络的定义是什么?答案:计算机网络是指通过通信线路将多台计算机相互连接起来,实现资源共享和信息传递的系统。
试题二:什么是OSI参考模型?它有几个层次?答案:OSI参考模型全名为开放系统互联通信参考模型,它将计算机网络通信过程分解为七层,分别是物理层、数据链路层、网络层、传输层、会话层、表示层和应用层。
试题三:简述TCP/IP协议族的特点。
答案:TCP/IP协议族是现代计算机网络通信的核心协议,具有以下特点:1)分层结构,由多个协议构成;2)开放性和通用性,支持各种主流操作系统;3)面向连接和面向无连接的通信方式;4)提供可靠的数据传输服务和数据传输控制机制。
试题四:什么是IP地址?IP地址的分类有哪些?答案:IP地址是指为网络中的每台主机分配的唯一标识符。
IP地址根据其范围分为A类、B类、C类、D类和E类五类地址。
其中A类地址用于大型网络,B类地址用于中型网络,C类地址用于小型网络,D类地址用于多播通信,E类地址保留未使用。
试题五:什么是DNS?它有什么作用?答案:DNS全名为域名系统,它是一种用于将域名和IP地址相互映射的分布式数据库,用于实现域名转换为IP地址的解析。
DNS的作用是可以通过人们容易记忆的域名来访问互联网上的各种资源,而不需要记住具体的IP地址。
试题六:简述HTTP协议的工作原理。
答案:HTTP协议是一种客户端与服务器之间进行通信的应用层协议。
它的工作原理是客户端通过发送请求给服务器,服务器收到请求后返回相应的资源或信息给客户端。
中山大学软件学院本科生期末考试《数据库系统原理》(A卷)

中山大学软件学院本科生期末考试考试科目:《数据库系统原理》(A卷)学年学期:2014学年第3学期姓名:学院/系:软件学院学号:考试方式:开卷年级专业:考试时长:120分钟班别:第八条:“考试作弊者,不授予学士学位。
”------------以下为试题区域,共7道大题,总分100分,考生请在答题纸上作答------------1. (10 marks) Let R = {A, B, C, D, E, F, G} and F = {AB→C, A→C, A→E, B→C, EF→EG, G→F}. Answer the following three questions.1)(4 marks) Compute the minimal cover of F.B→CA→CA→EEF→GG→F2)(4 marks) Decompose R into 3NF relations.{B, C}, {A, C, E}, {E, F, G} and {A, B, D, F} (OR {A, B, D, G})3)(2 marks) Is the composition in (b2) in BCNF? Briefly explain your answer.No. For {E, F, G} and G→F, G is not a candidate key.2. (10 marks)Assume there is an employee database Employee (eid: 8 bytes, ename: 16 bytes, did: 4 bytes, email: 12 bytes), where eid and ename are respectively the id and name of an employee and did is the id of the department in which the employee works. Suppose there are 50,000 employee records and 500 departments (i.e. each department has 100 employees on average). A page size is 1,000 bytes and a pointer costs 4 bytes.1)(4 marks) Assume that the employee file is sorted sequentially on did and there is noindex. Estimate the page access cost for retrieving the records of all employees working in a department with a given did. (You should show your argument and the main steps of the estimation clearly in the answer.)AnswerRecord size = 40 bytes, 25 records per page, 2,000 pages.Finding the first record requires log22000 + 3 more pages to search theremaining records (each dept has 100 employees which are distributed in 4pages).2)(6 marks) Assume only 20 pages of main memory are available for running theexternal sorting of the employee file on did.•How many PASSes are needed for the external sorting?•In each PASS, how many runs are created?•What is the total cost of the sorting in terms of pages?Answer:3 PASSes:PASS 0: 2000 pages / 20 pages per run = 100 runsPASS 1: ceil (100 runs / 19 runs per run) = 6 runsPASS 2: 1 runTotal cost: (2000 pages read per pass + 2000 pages write per pass) * 2 PASSes +2000 pages read per pass = 10000 pages (Note: Output is not counted!)Or: 2000 * (2 * 2 + 1) = 10000 pages transfer.3. (10 marks) Suppose a bookstore has the following five relational tables:BOOK (BID, TITLE, AID, SUBJECT, QUANTITY_IN_STOCK)AUTHOR (AID, NAME) CUSTOMER (CID, NAME)ORDER_DETAILS (OID, BID , QUANTITY) ORDER (OID, CID , ORDER_YEAR)In the above tables, keys are underlined and foreign keys are in italics . Each author has authored at least one book in the store. Each book has exactly one author. Each order is made by exactly one customer and has one or more associated record in ORDER_DETAILS (e.g., an order may contain different books).Express the following query using (i) SQL expressions, (ii) the relational algebra (RA).Find the distinct customer IDs (CID) of customers who have purchased more than 10 identical books in one order at least once.(i) SELECT DISTINCT CIDFROM ORDER_DETAILS od, ORDER oWHERE QUANTITY >= 10 AND od.OID = o.OID(ii)CID (σQUANTITY ≥ 10 (ORDER_DETAILSOID ORDER ))4. (10 marks) A B+ tree with n=5 is shown in Figure 1, in which only search keys are shown and pointers to the file system are hidden. We want to insert a data entry with search key “23”.Figure 1. A B+ Tree Structure1) Which of the following descriptions about the insertion operation is correct?A.The B+ tree contains 2 levels after insertion. 2 node splits are needed duringinsertion. The root node contains search key “15”.B.The B+ tree contains 3 levels after insertion. 1 node split is needed duringinsertion. The root node contains search key “20”.C.The B+ tree contains 3 levels after insertion. 2 node splits are needed duringinsertion. The root node contains search key “15”.D.The B+ tree contains 3 levels after insertion. 2 node splits are needed duringinsertion. The root node contains search key “20”.Answer: C2)We want to delete the data entry with search key “7”. How many leaf nodes store onlytwo data values after deletion?A.2B. 3C. 4D.5Answer: A5. (20 marks)You are given an initial hash structure with three keys already inserted as below. The hash function is h(x) = x mod 16. Draw five extendable hash structures corresponding for each insertion of the following search key values K: 7, 15, 20, 37, 18. Assume each bucket can hold two keys and the search key values arrive in the given order (i.e. 7 being the first coming key and 18 being the last one).You should follow the convention used by lecture slides: binary hash indices starting from the least significant bit. (E.g. 1 is the least significant bit of the 4 digit binary number 0001.)AnswerInsert 7Insert 15 and 20Insert 37Insert 186. (25 Marks) Consider a database consisting of the following three relation schemas:SAILORS (sid, sname, rating, age)BOATS (bid, bname, color)RESERVES (sid, bid, date, rname)The meaning of the attributes in the above schemas is self-explanatory. For example, sid is the sailor identity number and bname is the name of the boat. The primary keys of the relations are underlined. The attribute sid in RESERVES is a foreign key referencing SAILORS. The attribute bid in RESERVES is a foreign key referencing BOATS.The relation SAILORS has 100,000 tuples and 100 tuples of SAILORS fit into one page. The relation BOATS has 50,000 tuples and 25 tuples of BOATS fit into one page. The relation RESERVES has 10,000 tuples and 20 tuples of RESERVES fit on one page. We assume all attribute values and pointers in these three relations, if needed to be considered, are of the same size.(a)(10 marks) Assume that we use Indexed Nested Loop Join to computeSAILORS RESERVES using SAILORS as the outer relation. RESERVES have a primary B+-tree index with 2 levels on the join attribute. Estimate the cost of the join in terms of pages.Number of SAILORS pages: br = 100,000/100 = 1,000Number of SAILORS Tuples: nr = 100,000.The cost is br+ c * nr = 1,000 + (2+1) * 100,000 = 301,000.(b)(5 mark)Assume that 26% of the sailors have the rating bigger than 5. Estimate theresult size of sid(σrating>5 SAILORS) in terms of pages.Size = 26% * 100,000 / 100/4 = 65 pages260 divided by 4, for there is projection and all the attributes have same size(c)(10 marks) Consider the following two strategies to compute the join operationSAILORS BOATS RESERVES.Strategy 1: (SAILORS BOATS) RESERVESStrategy 2: (SAILORS RESERVES) BOATSWhich strategy is better? Explain the reason(s) of your choice based on the size of the intermediate result using the above strategies.Strategy 2 is better.Because in Strategy 1, SAILORS BOATS is equal to the cross-product of the two relations and the size of the join result will become as large as 100,000 * 50,000 = 5,000,000,000 tuples. This intermediate result is very large and later when joining this intermediate result with RESERVES, the cost is also large.In comparison, in Strategy 2, SAILORS RESERVES has only 10,000 tuples. And later when joining this intermediate result with BOATS, the cost is also small.7. (15 Marks) Consider a schedule S which consists of four transactions as follows:S = <T3_R(U), T2_R(X), T2_W(X), T3_R(X), T1_R(Y), T1_W(Y), T3_W(X), T1_R(Z), T4_R(Z), T4_W(Z), T2_W(Y), T3_R(Y)>The notation is self-explanatory. For example, T1_R(X) means that transaction T1 reads item X.(a)(5 marks) Fill in the following table representing S with the usual notations in lectureslides. The first operation R(U) has been shown in the table. Show clearly all conflicting pairs with downward arrows on the operations.(b)(5 marks) Construct the precedence graph of S. Explain why or why not the schedule isconflict-serializable.Precedence Graph of S:No cycle.(c)(5 marks) Suppose the format of the “commit” operation is Ci where i = 1, 2, 3, or 4.For example, the operation C1 means that the transaction T1 commits. Append all the commit statements to S so that the schedule becomes recoverable. For example, one append can be SC4C3C2C1 which means running S and then C4, C3, C2, C1. (Note that you should NOT change the sequence of the operations in S other than appending S with the four commit statements to make the schedule recoverable.)Recoverable (but not cascadeless) schedule: SC1C2C3C4 or (SC1C2C4C3) or (SC1C4C2C3) (Note: any permutation of Ci satisfies the commit order constraints: C1→C2, C1→C3, C1→C4, C2→C3 is correct)。
2014年大学计算机基础期中练习题

2013学年-2014学年第一学期大学计算机基础期中练习题一、选择题(共35题)1、根据计算机所使用的电子元器件的不同,可以将计算机的发展分为四代,其中第一代电子计算机所使用的电子元器件是___ ___。
A.晶体管B.集成电路C.电子管D.阴极管2、微电子技术是信息技术领域中的关键技术,它以集成电路为核心。
下列有关集成电路的叙述中,错误的是___ ___。
A.现代集成电路使用的半导体材料只能是硅(Si),不能使用其它任何半导体材料B.集成度是指集成电路包含的电子元件数目,可分为SSI、MSI、VLSI 等C.Moore 定律指出,单块集成电路的集成度平均18-24 个月翻一番D.我国第二代身份证中嵌入了集成电路芯片,可以实现电子防伪和数字管理功能3、在下列有关集成电路的叙述中,正确的是___ ___。
A.现代集成电路所使用的半导体材料都是硅B.所有的集成电路都是数字集成电路C.Moore 定律认为单块集成电路的集成度平均每年翻一番D.Intel 公司微处理器产品Core 2 Duo,其集成度已高达数千万个电子元件4、在下列有关数字技术的一些叙述中,错误的是___ ___。
A.数字技术是采用有限个状态(例如"0"和"1")来表示、处理、存储和传输信息的B.在逻辑代数中,1 与1 进行逻辑加(V)和逻辑乘(^)的结果相同C.任何一个十进制数,均可以精确地转换成等值的二进制数D.在PC 机中,通常用原码表示正整数、用补码表示负整数5、下面关于比特的叙述中,错误的是___ ___。
A.比特是组成数字信息的最小单位B.比特只有“0”和“1”两个符号C.比特既可以表示数值和文字,也可以表示图像和声音D. 比特”1”总是大于比特“0”6、使用存储器存储二进位信息时,存储容量是一项很重要的性能指标。
存储容量的单位有多种,下面不是存储容量单位的是___ ___。
A.TB B.XB C.GB D.MB7、数据传输速率是数据通信中重要的性能指标。
计算机网络期中开卷题

《计算机网络》期中开卷题
自由选定自己感兴趣的计算机网络领域的题目,通过阅读一些相关方面参考书和文献,最后用自己的语言以论文或报告的形式写出一篇文章。
3千字左右。
将采用随机抽查形式在课堂上汇报。
5月19日交。
论珠江三角洲信息产业的主导作用
姓名:学号:
[摘要] 珠江三角洲是我国信息产业发展的重要基地……
[关键词]基地,信息产业
[Abstract] The Pearl River Delta……
[Key words]aa, bb, cc
珠江三角洲的信息产业……
一、珠江三角洲信息产业发展现状
20世纪90年代以来……
注释:
①《广东统计年鉴》2001,中国统计出版社2001年出版,第345页表12-7。
……
参考文献
[1] 欧阳周,刘道德编著. 理工类学生专业论文导写. 中南大学出版社. 2000年9月
[2] 吴红,王远世. 研究性教学模式的设计与实践. 教学研究与实践. 中山大学出版社,2002,8:161-165
[3]A. Sikder, A Lotka-Volterra competition model and its global convergence
to a definite axial equilibrium, J. Math. Biology, Vol. 44, No. 4, 297-308, 2002.。
中山大学-期末考试答案-2014计算机网络B

中山大学软件学院2012级软件工程专业(2014学年秋季学期)《S E-301计算机网络》期末试题答案(B)e cookies. Check textbook for answers.2.a)Each transmitter sends one window's worth of data each RTT. The window size increases linearly from W/2 to W. So average length of transmit window .b)In each period of the sawtooth, one packet is lost, and the transmitter sends 1+W/2 windows of packets; and the average window size is 3/4*W.Therefore, 1 packet is lost for everythat are transmitted.c)3.Check ppt Network layer 4-93 for details.4.NAT:(a)1) The source IP address,2) TCP source port number.The change in the value of these two parameters implies that the followingtwo fields also need to be changed:3) IP checksum4) TCP checksum(b)P2: S=138.76.29.7, 5001 D=128.110.40.186, 80P3: S=128.110.40.186, 80 D=138.76.29.7, 5001P4: S=128.110.40.186, 80 D=10.0.0.1, 33455.Checksums have a greater probability of undetected errors than do CRCs. That is,CRCs are better at detecting errors and will result in less undetected errors than checksums. CRCs are easily be computed in hardware, but not very easily in software. Checksums can be computed in software much faster than can CRCs.6.Check textbook for answers.7.Any reasonable answer which contains the following terminologies: DNS,TCP/UDP, IP, MAC in network and link layers is acceptable.。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
1.Bandwidth and Delaya. On a generic multi-hop network, explain the potential causes of delay for a packet:• Propagation delay:• Transmission delay:• Queueing delay:• Processing delay:b. For a fixed network path, which of these delays are constant over time, and which ones vary with load? Explain why.c. (Estimating bottleneck bandwidth). An interesting technique used to estimate the bottleneck bandwidth in a network is called packet pair. The idea is simple: you send two packets back-to-back (no gap), at the full bandwidth of the first hop. The receiver at the other end can then determine the bandwidth by measuring the delay between the receptions of the two packets. In this problem we will do just that, with a simplified 3-link network shown in the figure below, with no extra traffic, no queues, and instantaneous access to the medium.Setup: In the figure, assume there is no queueing or processing delay, and that nodes can forward a packet as soon as (a) all bytes for the packet have been received from the previous link, and (b) the outgoing link is free. All three links have a propagation delay of 10ms; link AB has a bandwidth of 10Mbps ((107 bits per second), and link CD has a bandwidth of 8 Mbps (8 × 106 bps). You send two probe packets of 1000 bytes each, as depicted in the figure.(a) In this setup, calculate the length of the interval t1t2in the figure. Do the same for t3t4(b) In the diagram, draw the transmission of the two packets in the link BC, paying attention to when they must start and when they must end.(c) You measure the delay t4t5 between the end of the receptions of the two packets to be 8 ms. Assuming that the link BC is the bottleneck bandwidth in the path, what is its bandwidth? Explain your reasoning.2.Answer the following questions regarding TCP, UDP, IP,and routing:a)What are the important fields in the TCP, UDP, and IP headers?Without these fields, the protocols would clearly not “work”.b)Sketch the TCP connection initiation and connection terminationpacket flows using a timing diagram.3.Socket Programming3.1. Consider the following Java application:socket = new DatagramSocket(12345);while (true) {socket.receive(packet);} What happens if somebody decides to run two instances of thisapplication on one machine and 4 UDP segments arrive at port 12345?a) both instances of the application receive all 4 segmentsb) one instance receives all 4 segmentsc) some segments are received by one instance, other segmentsare received by the other instanced) one instance receives segments 1 and 3, the other receivessegments 2 and 43.2.Given the following lines from a Java program segment:byte[] dataOut = new byte [512];String userInput = inFromUser.readLine();dataOut = userInput.getBytes();Which of the following lines of code could be used to create a new UDP datagram packet to send the data that was provided by the user to a host identified by the InetAddress object IPAddress?a) DatagramPacket packetToSend = newDatagramPacket(dataOut, dataOut.length, IPAddress)b) UDPPacket = new UDPDatagram (userInput,userInput.length, IPaddress, 9876)c) DatagramPacket packet = new DatagramPacket(dataOut, dataOut.length, IPAddress, 9876)d) Socket datagramSocket = new dataGramPacket(dataOut, IPAddress)3.3.Consider a server socket objectsocket = ServerSocket(12345);What does the invocation of socket.accept() return?a) 'true' if there is a new TCP segment in the socket's buffer, false otherwiseb) a new TCP segment from the socket's buffer (blocks if no segments are available)c) 'true' if a new TCP connection request has arrived, false otherwised) a socket associated with a new TCP connection (blocks if no connections are available)3.4.Assume the following two lines of code are to be executed on the machineblue.cse.yorku.ca:Socket myFirstSocket = new Socket("blue.cse.yorku.ca", 5555);Socket mySecondSocket = new Socket(5555);Are the two lines/commands in conflict? Explain briefly!3.5.Suppose application A is using a UDP socket, i.e. DatagramSocket(), to transfer data to application Bon a remote host. Suppose application A calls send() method on the given socket 10 times.a) Can the underlying network stack transmit more than 10 data packets?b) Can the underlying network stack transmit fewer than 10 data packets? Explain briefly!4.The Transmission Control Protocol uses a method called congestioncontrol to regulate the traffic entering the network. The behavior of TCP congestion control can be represented as a graph in which the x-axis indicates the time, and the y-axis indicates congestion window size.Please use the graph shown below to the answer the followingquestions. Note that the graph does not explicitly show timeouts, but you should be able to figure out when timeouts happened based on theevents shown.(a) Slow Start:give two reasons why slow start is used, and explain why it does a better job than congestion avoidance for that function.(b) Slow Start:identify the intervals of time when TCP slow start is operating. For each interval, identify which of the above reasons apply and do not apply and explain why.(c) Congestion Avoidance: identify the intervals of time when TCP congestion avoidance is operating. Why should congestion avoidance be used instead of slow start during these intervals? Please clearly identify one specific reason.(d) Fast Retransmission: identify the intervals of time when TCP fast retransmission is used. Please explain what fast retransmission does and how it is triggered.(e) Fast Recovery: identify the intervals of time when TCP fast recovery is operating. What does fast recovery do and explain why is it beneficial.(f) Lack of fast recovery: identify the interval(s) of time when fast recovery could have happened, but did not. Identify one specific example of a circumstance that may prevent fast recovery from happening.5.NAT and IPv65.1Assume an IP packet carrying an HTTP request is going from alocal (i.e. home) area network onto the wider Internet through a NAT router. Name all header fields that the NAT router needs to change in the given packet? Explain your answer. (Hint: encapsulation as wellas the syntax/semantics of all involved protocols must be taken intoconsideration.)5.2The diagram below shows a packet traveling through a NAT router.Packet 1 is sent from the internal host (S) to the NAT router, packet 2 is sent from the NAT router to the external web server (D), packet 3is received from the web server by the NAT router, and packet 4 issent by the NAT router to the original host. Fill in the missing source and destination IP addresses and port numbers in packets 2-4.5.3 The diagram below shows an IPv6 packet tunnelled over IPv4. Fill in themissing source and destination addresses at places/packets marked 1, 2, and 3.6.Consider a datagram network using 32-bit host addresses. Suppose arouter has four links, numbered 0 through 3, and packets are to be forwarded to the link interfaces as follows:Destination Address Range Link Interface11100000 00000000 00000000 00000000Through(0) 11100000 00111111 11111111 1111111111100000 01000000 00000000 00000000Through(1)111100000 01000000 11111111 1111111111100000 01000001 00000000 00000000Through (2)11100001 01111111 11111111 11111111Otherwise(3)a) Provide a forwarding table that has five entries, uses longest prefixmatching, and forwards packets to the correct link interfaces.b) Describe how your forwarding table determines the appropriate linkinterface for datagrams with destination addresses(following):11001000 10010001 01010001 0101010111100001 01000000 11000011 0011110011100001 10000000 00010001 01110111。