Computing Essentials Answer(Chapter1-10) (1)

前二章作业1.计算机的四个基本功能（Functions）是什么？2.在计算机的top-level structure view中，四个structural components 是什么？3.谁提出了store-program concept ?你能用汉语简单地描述这个存储程序的概念吗？4.CPU的英文全称是什么？汉语意义是什么？5.ALU的英文全称是什么？汉语意义是什么？6.V on Neumann 的IAS机的五大部件都是什么？7.在第一章中我们认识到的四个结构性部件（第2题）与V on Neumann的IAS机（第6题）中部件有本质差别吗？8.Fundamental Computer Elements 有哪几个？它们与计算机的四个基本功能的关系是什么？9.Moore’s Law在中文翻译为什么？它描述了什么事物的一般规律？10.本书的次标题和第二章第二节标题均为“Designing forPerformance”,Performance 主要指什么？Performance Balance的（balance）平衡要平衡什么？11.本书作者将他要研究的范围局限在“desktop, workstation , server“中，它们的中文名称是什么？各自的工作范围是什么？Chapter 3Homework1.PC means _________.A. personal computerB. programming controllerC. program counterD. portable computer2. PC holds _______________ .A. address of next instructionB. next instructionC. address of operandD. operand3. At the end of fetch cycle, MAR holds _____.A. address of instructionB. instructionC. address of operandD. operand4. Interrupt process steps are __________.A. suspending , resuming , branching & processingB. branching , suspending , processing & resumingC. suspending , branching , processing & resumingD. processing , branching , resuming & suspending5. A unsigned binary number is n bits, so it is can represent a value in the range between _________ .A. 0 to n-1B. 1 to nC. 0 to 2n-1D. 1 to 2n6.The length of the address code is 32 bits, so addressing range (or the range of address) is ________________.A. 4GB.from –2G to 2GC.4G-1D. from 1 to 4G7.There are three kinds of BUSes. Which is not belong to them?A. address busB. system busC. data busD. control busQuestions1.Translate the following terms (Note: function)PC, MAR, MBR, IR, AC, bus, system bus, data bus , address bus , control bus , handler*, opcode, Bus arbitrate* , multiplexed bus* , interrupt, ISR, Instruction cycle , fetch cycle , execute cycle(带“*”为选做题)2.Page90 problems3.1What general categories of functions are specified by computer instruction?3. Describe simply the operations of PC and IR in an instruction cycle.4.Suppose the length of word is n-bit, describe simply operand(操作数) format and instruction format.5. Describe simply the procedure of the interruption6. Describe simply the types and functions of the BUS.一：选择题1.The computer memory system refers to _________A.RAMB.ROMC.Main memoryD.Register , main memory, cache, external memory2.If the word of memory is 16 bits, which the following answer is right ?A.The address width is 16 bitsB.The address width is related with 16 bitsC.The address width is not related with 16 bitsD.The address width is not less than 16 bits3.The characteristics of internal memory compared to external memoryA.Big capacity, high speed, low costB.Big capacity, low speed, high costC.small capacity, high speed, high costD.small capacity, high speed, low cost4．On address mapping of cache, any block of main memory can be mapped toany line of cache, it is ___________ .A) Associative Mapping B) Direct MappingC) Set Associative Mapping D) Random Mapping5. Cache’s write-through polity means write operation to main memory _______.A)as well as to cacheB)only when the cache is replacedC)when the difference between cache and main memory is foundD)only when direct mapping is used6.Cache’s write-back polity means write operation to main memory ______________.a)as well as to cacheb)only when the relative cache is replacedc)when the difference between cache and main memory is foundd)only when using direct mapping7. On address mapping of cache, the data in any block of main memory can be mapped to fixed line of cache, it is _________________.associative mapping B) direct mappingC)set associative mapping D) random mapping8.On address mapping of cache, the data in any block of main memory can be mapped to fixed set any line(way) of cache, it is _________________.associative mapping B) direct mappingD)set associative mapping D) random mapping二：计算题(from page 126)Problem 4.1 , Problem 4.3 , Problem 4.4 , Problem 4.5 , , Problem 4.7, Problem 4.10第五章作业1.which type of memory is volatile?A.ROMB. E2PROMC. RAMD. flash memory2.which type of memory has 6-transistor structure?A. DRAMB. SRAMC. ROMD. EPROMing hamming code, its purpose is of one-bit error.A. detecting and correctingB. detectingC. correctingD. none of all4.Flash memory is .A. read-only memoryB. read-mostly memoryC. read-write memoryD. volatile5.Which answer about internal memory is not true?A. RAM can be accessed at any time, but data would be lost when power down..B. When accessing RAM, access time is non-relation with storage location.C. In internal memory, data can’t be modified.D. Each addressable location has a unique address.Page161 Problems: 5.4 5.5 5.6 5.7 5.8第六章作业一、选择题1. RAID levels_________make use of an independent access technique.A. 2B. 3C. 4D. all2. In RAID 4, to calculate the new parity, involves _________reads.A. oneB. twoC. threeD.four3. During a read/write operation, the head is ___________A. movingB. stationaryC. rotatingD. above all4. On a movable head system, the time it takes to position the head at the track is know as______.A. seek timeB. rotational delayC. access timeD. transfer time5. RAID makes use of stored______information that enable the recovery of data lost due to a disk failure.A. parityB. user dataC. OSD. anyone6. Recording and retrieval via _________called a headA. conductive coilB. aluminiumC. glassD. Magnetic field7.In Winchester disk track format, _________is a unique identifier or address used to locate a particular sector.A. SYNCHB. GapC. ID fieldD. Data field8. Data are transferred to and from the disk in ________.A. trackB. sectorC. gapD. cylinder9. In _________, each logical strip is mapped to two separate physical disk.A. RAID 1B. RAID 2C. RAID 3D. RAID 410. With _________, the bits of an error correcting code are stored in the corresponding bit position on multiple parity disk.A. RAID 1B. RAID 2C. RAID 3D. RAID 411. The write-once read-many CD, known as ________.A. CD-ROMB. CD-RC. CD-R/WD. DVD二、How are data written onto a magnetic disk?三、In the context of RAID, what is the distinction between parallel access and independent access?Homework in Chapter 71.“When the CPU issues a command to the I/O module, it must wait until the I/Ooperation is complete”. It is programmed I/O , the word “wait”means ___________________.a. the CPU stops and does nothingb. the CPU does something elsec. the CPU periodically reads & checks the status of I/O moduled. the CPU wait the Interrupt Request Signal2.See Figure 7.7. To save (PSW & PC) and remainder onto stack, why theoperations of restore them is reversed? Because the operations of stack are ________________.a. first in first outb. randomc. last in first outd. sequenceding stack to save PC and remainder, the reason is ____________________ .a.some information needed for resuming the current program at the point ofinterruptb.when interrupt occurs, the instruction is not executed over, so the instructionat the point of interrupt must be executed once againc.the stack must get some information for LIFOd.the start address of ISR must transfer by stack4.The signals of interrupt request and acknowledgement exchange between CPUand requesting I/O module. The reason of CPU’s acknowledgement is ________________a.to let the I/O module remove request signalb. to let CPU get the vectorfrom data busc.both a & bd. other aims5.In DMA , the DMA module takes over the operations of data transferring fromCPU, it means _________________________a.the DMA module can fetch and execute instructions like CPU doesb.the DMA module can control the bus to transfer data to or from memoryusing stealing cycle techniquec.the DMA module and CPU work together(co-operate) to transfer data into orfrom memoryd.when DMA module get ready, it issues interrupt request signal to CPU forgetting interrupt service6.Transfer data with I/O modules, 3 types of techniques can be used. Which one isnot belong them?a. Interrupt-driven I/Ob. programmed I/Oc. direct I/O accessd. DMA7.Think 2 types of different data transferring, to input a word from keyboard and tooutput a data block of some sectors to harddisk. The best choice is to use ___________.a. interrupt-driven I/O and DMAb. DMA and programmed I/O C. both interrupt-driven I/Os d. both DMAsparing with interrupt-driven I/O, DMA further raises the usage rate of CPUoperations, because __________a. it isn’t necessary for CPU to save & restore sceneb. it isn’t necessary for CPU to intervene the dada transferc. it isn’t necessary for CPU to read & check status repeatedlyd. both a and b9.Simply script the all actions when using Interrupt-driven I/O technique totransferring data with I/O module.(please insert the “vector “at step3 & step5)10.See Figure 7.7 & 7.8. Redraw figure 7.8, and mark the sequence numberaccording to Figure 7.7, to indicate the sequence of the information flowing.11. According to DMA technique, write all information of CPU sending to DMA module, and write at which time the DMA module issues interrupt request signal to CPU and why the INTR is issued ?.Chapter9 homework1.Suppose bit long of two’s complement is 5 bits, which arithmetic operation brings OVERFLOW?A. 5+8B. (-8)+(-8)C. 4-(-12)D.15-72.Overflow occurs sometime in ______arithmetic operation.A. addB. subtractC. add and subtractD. multiply3. In twos complement, two positive integers are added, when does overflow occurs?A. There is a carryB. Sign bit is 1C. There is a carry, and sign bit is 0D. Can’t determine4. An 8-bit twos complement 1001 0011 is changed to a 16-bit that equal to____.A.1000 0000 1001 0011B. 0000 0000 1001 0011C.1111 1111 1001 0011D.1111 1111 0110 1101 115. An 8-bit twos complement 0001 0011 is changed to a 16-bit that equal to____.A. 1000 0000 1001 0011B. 0000 0000 0001 0011C. 1111 1111 0001 0011D. 1111 1111 1110 11016.Booth’s algorithm is used for Twos complement ______.A. additionB. subtractionC. multiplicationD. division7. In floating-point arithmetic, addition can divide to 4 steps: ______.A. load first operand, add second operand, check overflow and store resultB. compare exponent, shift significand, add significands and normalizeC. fetch instruction, indirectly address operand, execute instruction and interruptD. process scheduling states: create, get ready, is running and is blocked8. In floating-point arithmetic, multiplication can divide to 4 steps: ______.A. load first operand, add second operand, check overflow and store resultB. fetch instruction, indirectly address operand, execute instruction and interruptC. process scheduling states: create, get ready, is running and is blockedD. check for zero, add exponents, multiply significands, normalize, and round.9.The main functions of ALU are?A. LogicB. ArithmeticC. Logic and arithmeticD. Only addition10. Which is true?A. Subtraction can not be finished by adder and complement circuits in ALUB. Carry and overflow are not sameC.In twos complement, the negation of an integer can be formed with thefollowing rules: bitwise not (excluding the sign bit), and add 1.D. In twos complement, addition is normal binary addition, but monitor sign bit foroverflowPage326：9.4, 9.5 and 9.7(其中9.4选作)To prove: in twos complement, sign-extension rule (converting between different bit length) and negation rule ( (-X)补= X补+ 1).Chapter 10 and Chapter 111: In instruction, the number of addresses is 0, the operand(s)’address is implied, which is(are) in_______.A. accumulatorB. program counterC. top of stackD. any register2: Which the following addressing mode can achieve the target of branch in program?A.Direct addressing modeB.Register addressing modeC.Base-register addressing modeD.Relative addressing mode （有问题）3: In index-register addressing mode , the address of operand is equal toA.The content of base-register plus displacementB.The content of index-register plus displacementC.The content of program counter plus displacementD.The content of AC plus displacement4: The address of operand is in the instruction, it is_________ ?A.Direct addressing modeB.Register indirect addressing modeC.Stack addressing modeD.Displacement addressing mode5: Which the following is not the area that the source and result operands can be stored in ?A.Main or virtual memoryB.CPU registerC.I/O deviceD.Instruction6: Compared with indirect addressing mode , the advantage of register indirect addressing mode isrge address spaceB.Multiple memory referenceC.Limit address spaceD.Less memory access7:With base-register ADDRESSING , the ______________ register can be used.A. BASEB. INDEXC. PCD. ANY8:The disadvantage of INDIRECT ADDRESSING is ____________.A. large addressing rangeB. no memory accessC. more memory accessD. large value range9:Which is not an advantage with REGISTER INDIRECT?A. just one times of operand’s accessB. large memory spaceC. large value rangeD. no memory reference10:The REGISTER ADDRESSING is very fast, but it has _________________.A. very less value rangeB. very less address spaceC. more memory accessD. very complex address’ calculating11:The disadvantage of IMMEDIATE ADDRESSING is ___________.A. limited address rangeB. more memory accessC. limit value rangeD. less memory access12:In instruction, the number of addresses is 2, one address does double duty both _______________.A. a result and the address of next instructionB.an operand and a resultC.an operand and the address of next instructionD.two closed operands13.In instruction, the number of addresses is 3, which are _______________.A. two operands and one resultB. two operands and an address of next instructionC. one operand, one result and an address of next instructionD. two operands and an address of next instruction14.The address is known as a type of data, because it is represented by __________.A. a number of floating pointB. a signed integerC. an unsigned integerD. a number of hexadecimal15.Which is not a feature of Pentium .A. complex and flex addressingB. abundant instruction setC. simple format and fixed instruction lengthD. strong support to high language16. Which is not a feature of Power PC .A. less and simple addressing modeB. basic and simple instruction setC. variable instruction length and complex formatD. strong support to high languageChapter 12 and Chapter 181. After the information flow of fetch subcycle, the content of MBR is_____________.A.oprandB.address of instructionC. instructionD. address of operand2. After the information flow of instruction subcycle, the content of MBR is_____________.A.oprandB.address of instructionC. instructionD. address of operand3. The worse factor that limits the performance of instruction pipeline is _________________.A.conditional branch delaying the operation of target addressB. the stage number of pipeline c an’t exceed 6C. two’s complement arithmetic too complexD. general purpose registers too few4.The most factor to affect instruction pipeline effectiveness is __________.A. The number of stagesB. the number of instructionC. the conditional branch instructionD. the number of pipelines5. RISC rejects ______.A. few, simple addressing modesB. a limited and simple instruction setC. few, simple instruction formatsD. a few number of general purpose registers6. RISC rejects ______.A.a large number of general-purpose registersB. indirect addressingC. a single instruction sizeD. a small number of addressing mode7. Which is NOT a characteristic of RISC processor.A. a highly optimized pipeline.B. Register to registeroperationsC. a large number of general-purpose registersD. a complexed instructionformat8.Control unit use some input signals to produce control signals that open the gatesof information paths and let the micro-operations implement. Which is NOT the input signals of control unit/A.clock and flagsB.instruction registerC.interrupt request signalD.memory read or write9.Control unit use some output signals to cause some operations. Which is notincluded in the output signals?A.signals that cause data movementB.signals that a ctivate specific functions(e.g. add/sub/…)C.flagsD.read or write or acknowledgement10. Symmetric Multi-Processor (SMP) system is tightly coupled by _______.A. high-speed data-link and distributed memoryB. shared RAIDs and high-speed data-linkC. distributed caches and shared memoryD. interconnect network and distributed memory11. The SMP means __________.A.Sharing Memory ProcessesB.Split Memory to PartsD.Stack and Memory Pointer D.Symmetric Multi-Processo r12.The “MESI” means states of ____________ .A.Modified, Exclusive, Stored and InclusiveB.Modified, Expected, Shared and InterruptedC.Modified, Exclusive, Shared and InvalidD.Moved, Exchanged, Shared and Invalid13.The protocol “MESI” is also called __________.A. write back policyB. write-update protocolC. write-invalidate protocolD. write through policyChapter 121.Which register is user –visible but is not directly operated in 8086 ?A. DSB. SPC. IPD. BP2.The indirect sub-cycle is occurred _____________ ?A. before fetch sub-cycleB. after execute sub-cycleC. after interrupt sub-cycleD. after fetch sub-cycle and before execute sub-cycle3.Within indirect sub-cycle , the thing the CPU must do is ______________?A. fetch operand or store resultB. fetch operand’s address from memoryC. fetch next instruction from memoryD. nothing4.In general, which register is used for relative addressing ---- the content inthis register plus the A supplied by instruction to make a target address in branch or loop instructions.A. SPB. IRC. BRD. PC5.The Memory Address Register connects to ____________ BUS .A. systemB. addressC. dataD. control6.The Memory Buffer Register links to ________ BUS.A. systemB. addressC. dataD. control7.After Indirect cycle , there is a ______________ cycle .A. FetchB. IndirectC. ExecuteD. Interrupt8.The Interrupt cycle is __________ ______ Execute cycle .A. always afterB. never afterC. sometime afterD. maybe before9.The correct cycle sequence is _________________ .A. Fetch , Indirect , Execute and InterruptB. Fetch , Execute , Indirect and InterruptC. Fetch , Indirect , Interrupt and ExecuteD. Indirect , Fetch , Execute and Interrup10.The aim of the indirect cycle is to get __________________.A. an operandB. an instructionC. an address of an instructionD. an address of an operand11.Which is not in the ALU ?A. shifterB. adderC. complementerD. accumulator12.The registers in the CPU is divided _____registers and ________registers .A. general purpose , user-visibleB. user-visible , control and statusC. data , addressD. general purpose , control and status13.The Base register is a(n) __________ register in 8086.A. general purposeB. dataC. addressD. control14.The Instruction Pointer is a(n) __________ register in 8086.A. general purposeB. dataC. addressD. control15.The Index register is a(n) __________ register in 8086.A. general purposeB. dataC. addressD. control16.The Stack Pointer is a(n) __________ register in 8086.A. general purposeB. dataC. addressD. control17.The Accumulator is a(n) __________ register in 8086.A. general purposeB. dataC. addressD. control18.The Programming Status W ord is a(n) __________ register .A. general purposeB. dataC. addressD. controlShow all the micro-operations and control signals for the following instruction:1. ADD AX, X; —The contents of AC adds the contents of location X, result is stored to AC.2. MOV AX, [X];—Operand pointed by the content of location X is moved to AX, that means ((X))->AX—[ ] means indirect addressing.3. ADD AX, [BX];—Operand pointed by the content of Register BX is added to AX, that means (AX)+((BX))->AX—[ ] means register indirect addressing.4. JZ NEXT1; —If (ZF)=0,then jump to (PC)+ NEXT1.5. CALL X; —Call x function, save return address on the top of stack.6. RETURN; —From top of stack return to PC.。

版本：冶金工业出版社Keys to Exercises and Examinations第1单元Text11.（1）B（2）A（3）A（4）A（5）A（6）B（7）D（8）A（9）C（10）C2.（1）registering,predicting（2）Logs（3）boolean logic（4）graphical interface（5）integrated circuit3.（1）战争期间，冯·诺依曼在流体力学、弹道学、气象学、博弈论以及统计学等方面的专业性的意见，被很好地运用在了一些工程中。

（2）战后的冯·诺伊曼在高等研究中心专注于计算机和它的拷贝的发展。

4.（1）It was a massive steam-powered mechanical calculator designed to print astronomical tables.（2）John Louis von Neumann’s famous stored program concept says that the program is stored as data in the computer’s memory and the computer is able to manipulate it as data—for example,to load it from disk,store it back on disk, and move it in memory.This concept became a fundamental of modern computing.Text21.（1）A（2）B（3）B（4）B（5）D（6）C（7）B（8）A（9）A（10）C2.（1）parallel-operation（2）capacity（3）peripherals（4）sophisticated（5）microprocessors3.它们（液晶显示器）不像CRT显示那样，需要考虑几何学的、收敛的或是焦点问题，而且它们的清晰度使得人们可以在更小的屏幕上看到更高的分辨率。

计算机英语第二版课后习题答案练习答案PART ONE Computer BasicsUnit 1 My ComputerSection A I(Fill in the blanks with the information given inthe text:1( Charles Babbage; Augusta Ada Byron 2( input; output3( VLSI4( workstations; mainframes 5( vacuum; transistors6( instructions; software7( digit; eight; byte8( microminiaturization; chipII(Translate the following terms or phrases from English intoChinese and vice versa:1( artificial intelligence 人工智能2( paper-tape reader 纸空阅读机3( optical computer 光学计算机4( neural network 神经网络5( instruction set 指令集6( parallel processing 平行处理7( difference engine 差分机air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house thefirst element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materials require higher thermal resistance, good thermal insulation properties, and thicken the North wall must also be taken of thickness and measures such as roof andceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters:ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materialsrequire higher thermal resistance, good thermal insulation properties, and thicken the North wall must also be taken of thickness and measures such as roof and ceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,8( versatile logical element 通用逻辑器件9( silicon substrate 硅基10( vacuum tube 真空管(电子管)11( the storage and handling of data 数据的存储与处理12( very large-scale integrated circuit 超大规模集成电路13( central processing unit 中央处理器14( personal computer 个人计算机15( analogue computer 模拟计算机16( digital computer 数字计算机17( general-purpose computer 通用计算机18( processor chip 处理器芯片19( operating instructions 操作指令20( input device 输入设备III(Fill in each of the blanks with one of the words given in the following list, making changes if necessary:We can define a computer as a device that accepts input, processes data, stores data, and produces output. According to the mode of processing, computers are either analog or digital. They can beclassified as mainframes, minicomputers, workstations, or microcomputers. All else (for example, the age of the machine) being equal, this categorization provides some indication of the computer’s speed, size, cost, and abilities.air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia,hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materials require higher thermal resistance, good thermal insulation properties, and thicken the North wall must also be taken of thickness and measures such as roof andceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilationnatural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materialsrequire higher thermal resistance, good thermal insulation properties, and thicken the North wall must also be taken of thickness and measures such as roof and ceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,Ever since the advent of computers, there have been constant changes. First-generation computers of historic significance, such as UNIVAC, introduced in the early 1950s, were based on vacuum tubes. Second-generation computers, appearing in the early 1960s, were those in which transistors replaced vacuum tubes. In third-generation computers, dating from the 1960s, integrated circuits replaced transistors. In fourth-generation computers such as microcomputers, which first appeared in the mid-1970s, large-scale integration enabled thousands of circuits to be incorporated on one chip. Fifth-generation computers are expected to combine very-large-scale integration with sophisticated approaches to computing, including artificial intelligence and true distributed processing.IV(Translate the following passage from English into Chinese.A computer system includes a computer, peripheral(外围的)devices, and software. The electric, electronic, and mechanical devices used for processing data are referred to as hardware. Inaddition to the computer itself, the term “hardware” refers to components called peripheral devices air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materials require higher thermal resistance, good thermal insulationproperties, and thicken the North wall must also be taken of thickness and measures such as roof and ceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normallighting,air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materials require higher thermal resistance, good thermal insulation properties, and thicken the Northwall must also be taken of thickness and measures such as roof and ceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,that expand the computer’s input, output, and storage capabilities. Computer hardware in and of itself does not provide a particularly useful mind tool. To be useful, a computer requires a set of instructions, called software or a computer program, which tells the computer how to perform a particular task. Computers become even more effective when connected to other computers in a network so users can share information.计算机系统包括计算机、外围设备和软件。

第一章1.(Q1) What is the difference between a host and an end system? List the types ofendsystems. Is a Web server an end system?Answer: There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, Internet-connected PDAs, WebTVs, etc.2.(Q2) The word protocol is often used to describe diplomatic relations. Give an example of adiplomatic protocol.Answer: Suppose Alice, an ambassador of country A wants to invite Bob, an ambassador of country B, over for dinner. Alice doesn’t simply just call Bob on the phone and say, come to our dinner table now”. Instead, she calls Bob and sugges ts a date and time. Bob may respond by saying he’s not available that particular date, but he is available another date. Alice and Bob continue to send “messages” back and forth until they agree on a date and time. Bob then shows up at the embassy on the agreed date, hopefully not more than 15 minutes before or after the agreed time. Diplomatic protocols also allow for either Alice or Bob to politely cancel the engagement if they have reasonable excuses.3.(Q3) What is a client program? What is a server program? Does a server programrequestand receive services from a client program?Answer: A networking program usually has two programs, each running on a different host, communicating with each other. The program that initiates the communication is the client.Typically, the client program requests and receives services from the server program.4.(Q4) List six access technologies. Classify each one as residential access, company access, ormobile access.Answer:1. Dial-up modem over telephone line: residential; 2. DSL over telephone line: residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, 3G/4G): mobile5.(Q5) List the available residential access technologies in your city. For each type of access,provide the advertised downstream rate, upstream rate, and monthly price.Answer: Current possibilities include: dial-up (up to 56kbps); DSL (up to 1 Mbps upstream, up to 8 Mbps downstream); cable modem (up to 30Mbps downstream, 2 Mbps upstream.6.(Q7) What are some of the physical media that Ethernet can run over?Answer: Ethernet most commonly runs over twisted-pair copper wire and “thin” coaxial cable.It also can run over fibers optic links and thick coaxial cable.7.(Q8)Dial-up modems, HFC, and DSL are all used for residential access. For each of theseaccess technologies, provide a range of transmission rates and comment on whether the transmission rate is shared or dedicated.Answer:Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to1 Mbps, bandwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstreamchannel is usually less than a few Mbps, bandwidth is shared.8.(Q13)Why is it said that packet switching employs statistical multiplexing? Contraststatistical multiplexing with the multiplexing that takes place in TDM.Answer:In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.9.(Q14) Suppose users share a 2Mbps link. Also suppose each user requires 1Mbps whentransmitting, but each user transmits only 20 percent of the time. (See the discussion of statistical multiplexing in Section 1.3.)a.When circuit switching is used, how many users can be supported?b.For the remainder of this problem, suppose packet switching is used. Why will there beessentially no queuing delay before the link if two or fewer users transmit at the same time? Why will there be a queuing delay if three users transmit at the same time?c.Find the probability that a given user is transmitting.d.Suppose now there are three users. Find the probability that at any given time, allthree users are transmitting simultaneously. Find the fraction of time during which the queue grows.Answer:a. 2 users can be supported because each user requires half of the link bandwidth.b.Since each user requires 1Mbps when transmitting, if two or fewer users transmitsimultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three users transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this case, there will be queuing delay before the link.c.Probability that a given user is transmitting = 0.2d.Probability that all three users are transmitting simultaneously=33p3(1−p)0=0.23=0.008. Since the queue grows when all the users are transmitting, the fraction oftime during which the queue grows (which is equal to the probability that all three users are transmitting simultaneously) is 0.008.10.(Q16)Consider sending a packet from a source host to a destination host over a fixed route.List the delay components in the end-to-end delay. Which of these delays are constant and which are variable?Answer:The delay components are processing delays, transmission delays, propagation delays, and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable.11.(Q19) Suppose Host A wants to send a large file to Host B. The path from Host A to Host Bhas three links, of rates R1 = 250 kbps, R2 = 500 kbps, and R3 = 1 Mbps.a.Assuming no other traffic in the network, what is the throughput for the file transfer.b.Suppose the file is 2 million bytes. Roughly, how long will it take to transfer the file toHost B?c.Repeat (a) and (b), but now with R2 reduced to 200 kbps.Answer:a.250 kbpsb.64 secondsc.200 kbps; 80 seconds12.(P2)Consider the circuit-switched network in Figure 1.8. Recall that there are n circuits oneach link.a.What is the maximum number of simultaneous connections that can be in progress atany one time in this network?b.Suppose that all connections are between the switch in the upper-left-hand cornerand the switch in the lower-right-hand corner. What is the maximum number ofsimultaneous connections that can be in progress?Answer:a.We can n connections between each of the four pairs of adjacent switches. This gives amaximum of 4n connections.b.We can n connections passing through the switch in the upper-right-hand cornerandanother n connections passing through the switch in the lower-left-hand corner, giving a total of 2n connections.13.(P4) Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 50km/hour.a.Suppose the caravan travels 150 km, beginning in front of one tollbooth, passingthrough a second tollbooth, and finishing just before a third tollbooth. What is theend-to-end delay?b.Repeat (a), now assuming that there are five cars in the caravan instead of ten.Answer: Tollbooths are 150 km apart, and the cars propagate at 50 km/hr, A tollbooth services a car at a rate of one car every 12 seconds.a.There are ten cars. It takes 120 seconds, or two minutes, for the first tollbooth to servicethe 10 cars. Each of these cars has a propagation delay of 180 minutes before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 182 minutes. The whole process repeats itself for traveling between the second and third tollbooths. Thus the total delay is 364 minutes.b.Delay between tollbooths is 5*12 seconds plus 180 minutes, i.e., 181minutes. The totaldelay is twice this amount, i.e., 362 minutes.14.(P5) This elementary problem begins to explore propagation delay and transmission delay,two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B.a.Express the propagation delay, d prop , in terms of m and s.b.Determine the transmission time of the packet, d trans, in terms of L and R.c.Ignoring processing and queuing delays, obtain an expression for the end-to-enddelay.d.Suppose Host A begins to transmit the packet at time t = 0. At time t = d trans, where isthe last bit of the packet?e.Suppose d prop is greater than d trans. At time t = d trans, where is the first bit of thepacket?f.Suppose d prop is less than d trans.At time t = d trans,where is the first bit of the packet?g.Suppose s = 2.5*108, L = 100bits, and R = 28kbps. Find the distance m so that dprop equals d trans .Answer:a. d prop = m/s seconds.b. d trans = L/R seconds.c. d end-to-end = (m/s + L/R) seconds.d.The bit is just leaving Host A.e.The first bit is in the link and has not reached Host B.f.The first bit has reached Host B.g.Wantm=LRS=10028∗1032.5∗108=893 km.15.(P6) In this problem we consider sending real-time voice from Host A to Host B over apacket-switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-Byte packets. There is one link between Host A and B; its transmission rate is 500 kbps and its propagation delay is 2 msec.As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packet’s bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded(as part of the analog signal at Host B)?Answer: Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in thepacket must be generated. This requires56∗8sec=7 msec64∗103The time required to transmit the packet is56∗8sec=896 μsec500∗103Propagation delay = 2 msec.The delay until decoding is7msec +896μsec + 2msec = 9.896msecA similar analysis shows that all bits experience a delay of 9.896 msec.16.(P9) Consider a packet of length L which begins at end system A, travels over one link to apacket switch, and travels from the packet switch over a second link to a destination end system. Let d i, s i, and R i denote the length, propagation speed, and the transmission rate of link i, for i= 1, 2. The packet switch delays each packet by d proc. Assuming no queuing delays, in terms of d i, s i, R i, (i= 1, 2), and L, what is the total end-to-end delay for the packet? Suppose now the packet Length is 1,000 bytes, the propagation speed on both links is 2.5 * 108m/s, the transmission rates of both links is 1 Mbps, the packet switch processing delay is 2 msec, the length of the first link is 6,000 km, and the length of the last link is 3,000 km. For these values, what is the end-to-end delay?Answer: The first end system requires L/R1to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay ofd proc; after receiving the entire packet, the packet switch requires L/R2to transmit the packetonto the second link; the packet propagates over the second link in d2/s2. Adding these five delays givesd end-end = L/R1 + L/R2 + d1/s1 + d2/s2 + d procTo answer the second question, we simply plug the values into the equation to get 8 + 8 +24 + 12 + 2= 54 msec.17.(P10) In the above problem, suppose R1 = R2 = R and d proc= 0. Further suppose the packetswitch does not store-and-forward packets but instead immediately transmits each bit it receivers before waiting for the packet to arrive. What is the end-to-end delay?Answer: Because bits are immediately transmitted, the packet switch does not introduce any delay;in particular, it does not introduce a transmission delay. Thus,d end-end = L/R + d1/s1 + d2/s2For the values in Problem 9, we get 8 + 24 + 12 = 44 msec.18.(P11) Suppose N packets arrive simultaneously to a link at which no packets are currentlybeing transmitted or queued. Each packet is of length L and the link has transmission rate R.What is the average queuing delay for the N packets?Answer:The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)L/R for the nth transmitted packet. Thus, the average delay for the N packets is(L/R + 2L/R + ....... + (N-1)L/R)/N = L/RN(1 + 2 + ..... + (N-1)) = LN(N-1)/(2RN) = (N-1)L/(2R) Note that here we used the well-known fact that1 +2 + ....... + N = N(N+1)/219.(P14) Consider the queuing delay in a router buffer. Let I denote traffic intensity; that is, I =La/R. Suppose that the queuing delay takes the form IL/R (1-I) for I<1.a.Provide a formula for the total delay, that is, the queuing delay plus the transmissiondelay.b.Plot the total delay as a function of L/R.Answer:a.The transmission delay is L / R . The total delay isILR(1−I)+LR=L/R1−Ib.Let x = L / R.Total delay=x 1−αx20.(P16) Perform a Traceroute between source and destination on the same continent at threedifferent hours of the day.a.Find the average and standard deviation of the round-trip delays at each of the threehours.b.Find the number of routers in the path at each of the three hours. Did the pathschange during any of the hours?c.Try to identify the number of ISP networks that the Traceroute packets pass throughfrom source to destination. Routers with similar names and/or similar IP addresses should be considered as part of the same ISP. In your experiments, do the largest delays occur at the peering interfaces between adjacent ISPs?d.Repeat the above for a source and destination on different continents. Compare theintra-continent and inter-continent results.Answer: Experiments.21.(P18) Suppose two hosts, A and B, are separated by 10,000 kilometers and are conn ectedby a direct link of R =2 Mbps. Suppose the propagation speed over the link is 2.5•108 meters/sec.a.Calculate the bandwidth-delay product, R•d prop.b.Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sentcontinuously as one large message. What is the maximum number of bits that will be in the link at any given time?c.Provide an interpretation of the bandwidth-delay product.d.What is the width (in meters) of a bit in the link? Is it longer than a football field?e.Derive a general expression for the width of a bit in terms of the propagation speed s,the transmission rate R, and the length of the link m.Answer:a.d prop= 107 / 2.5•108= 0.04 sec; so R •d prop= 80,000bitsb.80,000bitsc.The bandwidth-delay product of a link is the maximum number of bits that can be inthelink.d. 1 bit is 125 meters long, which is longer than a football fielde.m / (R •d prop ) = m / (R * m / s) = s/R22.(P20) Consider problem P18 but now with a link of R = 1 Gbps.a.Calculate the bandwidth-delay product,R·d prop .b.Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sentcontinuously as one big message. What is the maximum number of bits that will be inthe link at any given time?c.What is the width (in meters) of a bit in the link?Answer:a.40,000,000 bits.b.400,000 bits.c.0.25 meters.23.(P21) Refer again to problem P18.a.How long does it take to send the file, assuming it is sent continuously?b.Suppose now the file is broken up into 10 packet is acknowledged by the receiver andthe transmission time of an acknowledgment packet is negligible. Finally, assumethat the sender cannot send a packet until the preceding one is acknowledged. Howlong does it take to send the file?pare the results from (a) and (b).Answer:a. d trans + d prop = 200 msec + 40 msec = 240 msecb.10 * (t trans + 2 t prop ) = 10 * (20 msec + 80 msec) = 1.0sec。

Chapter 1 Computer Networks and the Internet networks use VC. numbers to forward packets toward their1．The ( ) is a worldwide computer network, that is, a network that destination.interconnects millions of computing devices throughout the world. D datagram networks use VC. numbers and virtual-circuit networks ppt3 use destination addresses to forward packets toward their destination.A public Internet 9．In the following options, which one is not a guided media? ( )B Intranet A twisted-pair wireC switch net B fiber opticsD television net C coaxial cable2．Which kind of media is not a guided media? ( ) D satelliteA twisted-pair copper wire 10．Processing delay does not include the time to ( ).B a coaxial cable A examine the packet’s headerC fiber optics B wait to transmit the packet onto the linkD digital satellite channel C determine where to direct the packetD check bit-error in the packet3．Which kind of media is a guided media? ( )11．In the following four descriptions, which one is correct? ( )A geostationary satelliteB low-altitude satelliteC fiber opticsD wireless LAN A The traffic intensity must be greater than 1.B The fraction of lost packets increases as the traffic intensity decreases.C If the traffic intensity is close to zero, the average queuing delay 4．The units of data exchanged by a link-layer protocol are called ( ).will be close to zero. A FramesD If the traffic intensity is close to one, the average queuing delay B Segmentswill be close to one. C Datagrams12．The Internet’s network layer is responsible for moving network-layer D bit streamspackets known as ( ) from one host to another. 5．Which of the following option belongs to the circuit-switched networks?A frame ( )B datagram A FDMC segment B TDMD message C VC networks13．The protocols of various layers are called ( ). D both A and BA the protocol stack 6．( )makes sure that neither side of a connection overwhelms the otherB TCP/IPside by sending too many packets too fast.C ISPA Reliable data transferD network protocolB Flow control14．There are two classes of packet-switched networks: ( ) networks andC Congestion controlvirtual-circuit networks.D Handshaking procedureA datagram7．( ) means that the switch must receive the entire packet before it canB circuit-switchedbegin to transmit the first bit of the packet onto the outbound link.C television A Store-and-forward transmissionB FDMD telephoneC End-to-end connection 15．Access networks can be loosely classified into three categories:D TDM residential access, company access and ( ) access.8．Datagram networks and virtual-circuit networks differ in that ( ).A cabledA datagram networks are circuit-switched networks, andB wirelessvirtual-circuit networks are packet-switched networks.B datagram networks are packet-switched networks, andC campusD city areavirtual-circuit networks are circuit-switched networks.Question 16～17 C datagram networks use destination addresses and virtual-circuit1D distributed applicationsSuppose, a is the average rate at which packets arrive at the queue, R is24．The Internet provides two services to its distributed applications: a the transmission rate, and all packets consist of L bits, then the trafficintensity is ( 16 ), and it should no greater than ( 17 ). connectionless unreliable service and () service.A flow control16． A LR／aB connection-oriented reliableB La／RC congestion controlC Ra／LD TCP25．It defines the format and the order of messages exchanged between twoD LR／aor more communicating entities, as well as the actions taken on the 17．A 2B 1 transmission and/or receipt of a message or other event. The sentence describes ( ).C 0A InternetD -1B protocol 18．In the Internet, the equivalent concept to end systems is ( ).C intranet A hostsD network B servers26．In the following options, which does not define in protocol? ( )C clientsA the format of messages exchanged between two or more D routerscommunicating entities 19．In the Internet, end systems are connected together by ( ).B the order of messages exchanged between two or more A copper wirecommunicating entities B coaxial cableC the actions taken on the transmission of a message or other eventC communication linksD the transmission signals are digital signals or analog signals 模拟D fiber optics信号20．End systems access to the Internet through its ( ).27．In the following options, which is defined in protocol? ( ) A modemsA the actions taken on the transmission and/or receipt of a message orB protocolsother eventC ISPB the objects exchanged between communicating entities D socketsC the content in the exchanged messages21．End systems, packet switches, and other pieces of the Internet, run ( )D the location of the hosts that control the sending and receiving of information within the28．In the following options, which does not belong to the network edge?Internet.( )A programsA end systemsB processesB routersC applicationsC clientsD protocolsD servers22．There are many private networks, such as many corporate and29．In the following options, which belongs to the network core? ( ) government networks, whose hosts cannot exchange messages withA end systemshosts outside of the private network. These private networks are oftenB routersreferred to as ( ).C clientsA internetsD serversB LAN30．In the following options, which is not the bundled with the Internet ’s C intranetsconnection-oriented service? ( ) D WANA reliable data transfer23．The internet allows ( ) running on its end systems to exchange dataB guarantee of the transmission timewith each other.C flow control***A clients applicationsD congestion-controlB server applications31．An application can rely on the connection to deliver all of its data C P2P applicationswithout error and in the proper order. The sentence describes ( ).2***A flow control C data switchingB congestion-control D message switchingC reliable data transfer 40．In ( ) networks, the resources needed along a path to provide forD connection-oriented service communication between the end system are reserved for the duration 持续of the communication session. 32．It makes sure that neither side of a connection overwhelms 淹没压倒the other side by sending too many packets too fast. The sentence A packet-switcheddescribes ( ). B data-switchedA flow control C circuit-switchedB congestion-control D message-switchedC connection-oriented service 41．In ( ) networks, the resources are not reserved; a session’s messagesD reliable data transfer use the resources on demand, and as a consequence, may have to wait 33．It helps prevent the Internet from entering a state of gridlock. When a for access to communication link.packet switch becomes congested, its buffers can overflow and packet A packet-switchedloss can occur. The sentence describes ( ). B data-switchedA flow control C circuit-switchedB congestion-control D message-switchedC connection-oriented service 42．In a circuit-switched network, if each link has n circuits, for each linkD reliable data transfer used by the end-to-end connection, the connection gets ( ) of the 34．The Internet’s connection-oriented service has a name, it is ( ). link ’s bandwidth for the duration of the connection.A TCP A a fraction 1/nB UDP B allC TCP/IP C 1/2D IP D n times35．In the following options, which service does not be provided to an 43．For ( ), the transmission rate of a circuit is equal to the frame rate application by TCP?( ) multiplied by the number of bits in a slot.A reliable transport A CDMAB flow control B packet-switched networkC video conferencing C TDMD congestion control D FDM36．The Internet’s connectionless service is called ( ). 44．( ) means that the switch must receive the entire packet before it canA TCP begin to transmit the first bit of the packet onto the outbound link.B UDP A Queuing delayC TCP/IP B Store-and-forward transmissionD IP C Packet loss37．In the following options, which does not use TCP?( ) D PropagationA SMTP 45．The network that forwards packets according to host destinationB internet telephone addresses is called ( ) network.C FTP A circuit-switchedD HTTP B packet-switched38．In the following options, which does not use UDP?( ) C virtual-circuitA Internet phone D datagramB video conferencing 46．The network that forwards packets according to virtual-circuit numbersC streaming multimedia is called ( ) network.D telnet A circuit-switched39．There are two fundamental approaches to building a network core, ( ) B packet-switchedand packet switching. C virtual-circuitA electrical current switching D datagramB circuit switching3。

第一章 微型计算机基础题1-1 计算机发展至今，经历了哪几代？答：电子管计算机、晶体管计算机、集成电路计算机、超大规模集成电路计算机、非冯诺伊曼计算机和神经计算机。

题1-2 微机系统由哪几部分组成？微处理器、微机、微机系统的关系是什么？ 答：1、微机系统分硬件和软件，硬件包括CPU、存储器、输入输出设备和输入输出接口，软件包括系统软件和应用软件。

2、微处理器是指微机的核心芯片CPU；微处理器、存储器和输入输出设备组成微机；微机、外部设备和计算机软件组成微机系统。

题1-3 微机的分类方法包括哪几种？各用在什么应用领域中？答：按微处理器的位数，可分为1位、4位、8位、32位和64位机等。

按功能和机构可分为单片机和多片机。

按组装方式可分为单板机和多板机。

单片机在工业过程控制、智能化仪器仪表和家用电器中得到了广泛的应用。

单板机可用于过程控制、各种仪器仪表、机器的单机控制、数据处理等。

题1-4 微处理器有哪几部分组成？各部分的功能是什么？答：微处理器包括运算器、控制器和寄存器三个主要部分。

运算器的功能是完成数据的算术和逻辑运算；控制器的功能是根据指令的要求，对微型计算机各部分发出相应的控制信息，使它们协调工作，从而完成对整个系统的控制；寄存器用来存放经常使用的数据。

题1-5 微处理器的发展经历了哪几代？Pentium系列微处理器采用了哪些先进的技术？答：第一代4位或低档8位微处理器、第二代中高档8位微处理器、第三代16位微处理器、第四代32位微处理器、第五代64位微处理器、第六代64位高档微处理器。

Pentium系列微处理器采用了多项先进的技术，如：RISC技术、超级流水线技术、超标量结构技术、MMX技术、动态分支预测技术、超顺序执行技术、双独立总线DIB技术、一级高速缓冲存储器采用双cache结构、二级高速缓冲存储器达256KB或512KB、支持多微处理器等。

题1-6 何为微处理器的系统总线？有几种？功能是什么？答: 系统总线是传送信息的公共导线，微型计算机各部分之间是用系统总线连接的。

《计算机专业英语》参考答案Chapter 1 Computer ScienceText AExercises2.(a)out (b)with (c)in (d)in (e)in (f)with (g)for (h)aboutText BExercises3.(a)to (b)now (c)in (d)with (e)out (f)uponText CExercises1.(1)analyze, analytic (2)complicate, complex (3) collaborate, collaborative (4)vary, various (5)introduce, introductory (6)base, basic (7)create, creative (8)differ, different (9)free, freeChapter 2 Discrete Mathematics for Computer ScienceText AExercises1.C48 =70，C38 =562.6*25=1923.if a=0then i f b=0then return(0)else return(1+Add(0,b-1))else if b=0then return(1+Add(a-1,0))else return(1+1+Add(a-1,b-1))4.if Rest(S)=Øthen return(First(S))else if (First(S)<Largest(Rest(S)))return(Largest(Rest(S)))else Return(First(S))5.Now we can define function Concat(S1,S2) as:if(Length(S1)=0)then return(S2)else return(Cons(First(S1), Concat(Rest(S1),S2)))Text BExercises1.[Proof]:According to given conditions, we knowa n = a n-1 + 2na n-1 = a n-2 + 2(n-1)……a2 = a1 + 2*2a1 = 3sum all items in left side, and delete same items in the right side of equations, we can result thata n = 3 + 2（2+3+……+n-1+n）=1+n(n+1)=n2+n+1that is what we conclude.Text CExercises1.（1）depend, dependent (2) correspond, correspondent (3)grow, grown (4)solve, solvent (5) relate, relational (6)recur, recursive (7) validate, valid (8) conclude, conclusive (9) justify, justificative2.(1)connect-disconnect (2)possible-impossible (3)regular-irregular (4)measure-countermeasureChapter 3 Algorithms in the Real WorldText AExercises2. finite, solving, processing, effective, eventually, next, randomly3. by, in, on, in, on, from4. the algorithm can terminate.It is correct for sorting.If the length of array A is n, the time for computation is O(n2)Its memory cost n units.As n increase, its computational cost will become large.Text BExercises3.(1)—(e), (2)—(c), (3)—(d), (4)—(a), (5)—(b)4. inconvenience, incapacity, independence5.We couldn’t have lived without water.Chapter 4 Dictation SystemText AExercises2. forefront, institution, turnaround, boost, embrace, portfolio, handle, portable, issue, stringent3. off, on, from, into, in, to, over, to, on, toText BExercises2.(1)—(h), (2)—(g), (3)—(a), (4)—(j), (5)—(e), (6)—(f), (7)—(b), (8)—(i), (9)—(c), (10)—(d) Text CExercises3. organize,organizationaldictate,dictativeproduce,productiveadministrate,administrativetranscribe, transcriptivesimplify,simplicialimplement,implementativeprotect,protectiveChapter 10 Introduction to ComputersText AⅠ.1. hardware, software2. a Central Processing Unit (CPU), memory, an input device, an output device3. Input devices, Output devices4. An input device5. application software, system softwareⅡ.1. hardware 6. integrated circuit2. software 7. secondary storage3. a Central Processing Unit (CPU) 8. computer system4. application software 9. main memory5. operating system 10. scannerText BⅠ.1. The WYSIWYG2. cell3. finding, fixing4. Formulas5. headings across the top6. character, word, phraseⅡ.1. true 6. false2. true 7. false3. true 8. true4. true 9. true5. false 10. falseText C当使用计算机的时候，你必须知道与它“交流”。

大学计算机基础第1至4章习题答案大学计算机基础习题答案第一章一、填空题1.计算机科学是主要研究（计算理论）、（计算机）、和（信息处理）的学科。

2.在模型建立的前提下，利用计算机求解问题的核心工作就是（算法和程序）设计。

3.算法是一组规则，它的主要特性是（有限性）、（可执行性）、（机械性）、（确定性）和（终止性）。

4.要使一个问题能够用计算机解决，其必要条件是（有确定算法）。

5.在计算机内，一切信息都是以（二进制码）形式表示的。

6.如果说图灵机A能够完全模拟图灵机B，则意味着（给定相同的输入，A的输出与B的输出相同）。

如果A和B能够相互模拟，则表示（A与B计算等价）。

7.图灵机中的纸带可以相当于计算机中的（硬盘）。

8.第一代计算机的主要部件是由（电子管）构成的。

9.未来全新的计算机技术主要指（光子计算机）、（量子计算机）和（生物计算机）。

10.未来电子计算机的发展方向是（巨型化）、（微型化）、（网络化）和（智能化）。

11.目前，国际上广泛采用的西文字符编码是标准（ASCII码），它是用（８（７？））位二进制码表示一个字符。

12.采用16位编码的一个汉字储存时要占用的字节数为（2B）。

13.位图文件的存储格式为（BMP），用数码相机拍摄的照片的文件格式一般为（JPG）。

14.若处理的信息包括文字、图片、声音和电影，则其信息量相对最小的是（文字）。

15.模拟信号是指（在时间和幅值上）都连续变化的信号。

16.计算机中对信息的组织和管理方式有两种，即（文件）和（数据库）。

17.软件的测试方法包括（白盒测试）和（黑盒测试）。

18.普适计算的主要特点是（无处不在）。

二、简答题1.P14；2.P5；3.P7；4.输入信息、输出信息、程序、内部状态；5.P6；6.P23；7.P19；8.软硬件技术、通信技术、纳米技术；大规模并行处理体系结构、高性能算法、可重构计算、功耗。

第二章1.计算机系统主要由（硬件系统）和（软件系统）组成。