操作系统(双语)复习资料

操作系统课程英文词汇_1.操作系统Operating System2.计算机Computer3.内核映像Core Image4.超级用户Super-user5.进程Process6.线程Threads7.输入/输出I/O (Input/Output)8.多处理器操作系统Multiprocessor OperatingSystems9.个人计算机操作系统Personal ComputerOperating Systems10.实时操作系统Real-Time Operating Systems11.处理机Processor12.内存Memory13.进程间通信Interprocess Communication14.输入/输出设备I/O Devices15.总线Buses16.死锁Deadlock17.内存管理Memory Management18.输入/输出Input/Output19.文件Files20.文件系统File System21.文件扩展名File Extension22.顺序存取Sequential Access23.随机存取文件Random Access File24.文件属性Attribute25.绝对路径Absolute Path26.相对路径Relative Path27.安全Security28.系统调用System Calls29.操作系统结构Operating System Structure30.层次系统Layered Systems31.虚拟机Virtual Machines32.客户/服务器模式Client/Server Mode33.线程Threads34.调度激活Scheduler Activations35.信号量Semaphores36.二进制信号量Binary Semaphore37.互斥Mutexes38.互斥Mutual Exclusion39.优先级Priority40.监控程序Monitors41.管程Monitor 42.管道Pipe43.临界区Critical Section44.忙等待Busy Waiting45.原子操作Atomic Action46.同步Synchronization47.调度算法Scheduling Algorithm48.剥夺调度Preemptive Scheduling49.非剥夺调度Nonpreemptive Scheduling50.硬实时Hard Real Time51.软实时Soft Real Time52.调度机制Scheduling Mechanism53.调度策略Scheduling Policy54.任务Task55.设备驱动程序Device Driver56.内存管理器Memory Manager57.引导程序Bootstrap58.时间片Quantum59.进程切换Process Switch60.上下文切换Context Switch61.重定位Relocation62.位示图Bitmaps63.链表Linked List64.虚拟存储器Virtual Memory65.页Page66.页面Page Frame67.页面Page Frame68.修改Modify69.访问Reference70.联想存储器Associative Memory71.命中率Hit Ration72.消息传递Message Passing73.目录Directory74.设备文件Special File75.块设备文件Block Special File76.字符设备文件Character Special File77.字符设备Character Device78.块设备Block Device79.纠错码Error-Correcting Code80.直接内存存取Direct Memory Access81.统一命名法Uniform Naming82.可剥夺资源Preemptable Resource83.不可剥夺资源Nonpreemptable Resource84.先来先服务First-Come First-Served85.最短寻道算法Shortest Seek First86.电梯算法Elevator Algorithm87.引导参数Boot Parameter88.时钟滴答Clock Tick89.内核调用Kernel Call90.客户进程Client Process91.服务器进程Server Process92.条件变量Condition Variable93.信箱Mailbox94.应答Acknowledgement95.饥饿Starvation96.空指针Null Pointer97.规范模式Canonical Mode98.非规范模式Uncanonical Mode99.代码页Code Page100.虚拟控制台Virtual Console101.高速缓存Cache102.基地址Base103.界限Limit104.交换Swapping105.内存压缩Memory Compaction 106.最佳适配Best Fit107.最坏适配Worst Fit108.虚地址Virtual Address109.页表Page Table110.缺页故障Page Fault111.最近未使用Not Recently Used 112.最久未使用Least Recently Used 113.工作集Working Set114.请调Demand Paging115.预调Prepaging116.访问的局部性Locality of Reference 117.颠簸Thrashing118.内零头Internal Fragmentation 119.外零头External Fragmentation 120.共享正文Shared Text121.增量转储Incremental Dump 122.权限表Capability List123.访问控制表Access Control List。

计算机复习资料最终定稿版Ji suan ji fu xi zi liao zui zhong ding gao ban1.key words of Definition for computer: instructions,input,processes ,outputA computer is an electronic device which operates under the control of stored instructionsaccepts data inputprocesses the data according to specified rulesproduces and stores output1、A computer’s power is derived from Speed Reliability Accuracy Storage Communications2、Data: The symbols that represent people, events, things, and ideas. Consists of the raw, unprocessed facts, including text, numbers, images, and sounds.3、Information ：Data becomes information when it is presented in a format that people can understand and use .6.分类：Supercomputers (巨型计算机)fastest, most powerful, most expensive Mainframe computers（大型机）provide great processing speeds and data storage. Minicomputers （小型机）Microcomputers（微机）即personal computer or PCleast powerful, but the most widely used.rmation can be represented in one of two waysAnalog data（模拟） A continuous representation，analogous（相似）to the actual informationDigital data A discrete representation8.1 Byte=8 bits，B, KB, MB, GB, TB):9. Binary（Base 2）Octal（8）Decimal（Base 10）Hexadecimal（16）0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F10.distinguish number system(1) Subscript(下标) (101)2 (101) 8（2）Postfix（后缀）101B 101O 101D 101H11. Converting numberNon-Decimal to DecimalDecimal to Non-Decimal❖10—2 integer整数（1）短除法（2）从高位到低位逆用2—10思想❖Fraction小数Non-Decimal to Non-Decimal2—8 (10 111 011.110 1)2 = (273.64)88—2 (6754.32)8 = (110 111 101 100. 011)216—2 （A7B8.C9）16=(1010 0111 1011 1000 .1100 1001)12.Binary Arithmetic12．original code–(2n–1–1) ~ (2n–1–1) Total number is2n–1 n=8 -127 ～+127 Pro：Simple & Direct Cons：Subtraction operation may fail （overflow）13．Original code and Complement code, to do subtraction减法in addition（补码的意义正在于把减法做成加法）▪For positive number：complement = original▪For negative number:•Original: sign bit is 1, the rest is its absolute value•Complement: reverse all digits except the sign, the plus 1▪Complement of complement = original14．complement code –2n–1~ (2n–1–1) Total number is 2n n=8, the range is: -128 ～+12715．Floating point notation to represent very large real numbers and very small real numbers（1）-1.6875=(-1.1011)2=(-0.11011×21)2（2）指数尾数上数=0 0000001 1 11011000000000000000000 （用八位表示指数位，24位表示尾数位）16．In a 32 bits computer, if 8 bits for exponent part and 24 bits for mantissa part, the range of number is 1038 ~ 2-150 0 -1038 ~ -2-15016．Code of charactersASCII美国信息交换标准码character setStandard 7-bits, with 128 charactersExtended 8-bits, with 256 charactersUnicode 16-bits, a superset超集of ASCI17.von Neumannthree main ideas about computer*Five basic components *Use binary information *Stored program18．The principle of stored programA couple of abilities: store program and then automatic execute program .（存储程序、自动执行程序）Computers use “memory” (primary memory主存) to store program which will be processed ❖19．Five basic components▪Control Unit▪Arithmetic/logic Unit (ALU) 运算器▪Input devices▪Output devices▪Storage – primary and secondary storage19．CPUcan fetch each instruction in that program from memory one by onedecode the informationGet data if neededand then execute the information until all instructions arefinished.巧记有一个全控制，有两个互相传数据和信息，有一个只入，有一个只出20．ALU（运算器）（arithmetic operation logical operations）---（register）CPUControl Unit（directs and coordinates使协调operations in computer.）--registerRandom access memory (RAM) temporary or volatile storageMain memory Read-only memory (ROM) ---BIOS Basic Input Output SystemMemory CMOS required every time the computer system is turned onSecondary Storage 外存CPU不能直接访问注释1 CPU定义Interprets（翻译）and carries out basic instructions that operate a computer注释2 ROM cannot be changed 而CMOS can be changed to reflect changes in the computer system注释3Memory used to store the most frequently accessed information stored in RAM –cache,21．two registers in the control unit:The instruction register (IR) contains the instruction that is being executed.The program counter (PC) contains the address of the next instruction to be executed18．#Two steps of a machine cycle: Instruction cycle and execution cycle22.CPU speed Clock rate Word size Cache Instruction set size（指令集大小）23. Computers are electronic devices that accept instructions, process input, and produce information24.术语：Floppy disk 软盘Optical disk/compact disk 光盘Flash disk 闪存盘23．System Unit (主机)❖Motherboard❖Processor – CPU▪ALU▪Control Unit❖Memory▪RAM, ROM, CMOS, Cache▪All data and instructions must be loaded into RAM before they can be executed❖Bus – data highway between devices❖Expansion slots/cards (扩展槽/卡) to add extra devices▪video card, sound card, network card❖Port (端口) – a plug for a cable that leads to a device▪usb ports, RJ-45, VGA port for monitor 《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《《OS》24.系统软件实用软件操作系统外壳，命令解释程序内核注释1Utility software fundamental to computer installations, but not part of the OS.注释2 shell provides an interface through which a human can interact with the computer kernel manages computer resources, such as memory and input/output device强调：System software定义manage a computer system at a fundamental level25. Every operating system performs three basic functions：（different os also have other specific functions ）manages computer resources, such as 1. memory,2. CPU, and3. processesprovides an interface through which a human can interact with the computerallow s an application program to interact with other system resources26. Managing Resources consists of:Memory managementProcess managementCPU scheduling26.1. Memory management optimizes（优化）the use of RAM26.2-1 A process can be defined as a program in execution.26.2-2 A single user operating system allows only one user to run one program at a time.A multiuser operating system enables two or more users to run a program simultaneously.A multitasking operating system allows a single user to work on two or more applications thatreside in memory at the same time.A multiprocessing operating system can support two or more processors running programs atthe same time.26.3 CPU SchedulingThe operating system determines which process in memory is executed by CPU Nonpreemptive scheduling 无优先调度Preemptive scheduling27.OTHER FUNCTIONS27.1System Booting Cold boot&Warm bootThe process of starting or restarting a computer is called booting27.2System Shutting DownSleep mode（睡眠）saves any open documents and programs to RAM, turns off all unneeded functions, and then places the computer in a low-power stateHibernate（休眠）saves any open documents and programs to a hard disk before removing power from the computerUser Interface Device Driver Network Support Performance Monitor System Update 28.Types of Operating Systems28.1A stand-alone operating system（独立操作系统）is a complete operating system that works on a desktop computer, notebook computer, or mobile computing device 28.2An embedded （嵌入式）operating system resides on a ROM chip on a mobile device or consumer electronic device29．File A named collection of related dataFile is stored in secondary storage A file is a set of data that has been given a nameFile system The operating system's logical view of the files it manages Directory A named group of files. Also called a folder30．File typeProgram files（程序文件）- contains instructions to tell the computer what to do.Microsoft Word, Photoshop, AutoCADData Files - not executable and only used with a program.Word document, text file, Excel file, PowerPoint fileText file A file in which the bytes of data are organized ascharacters from the ASCII or Unicode character setsBinary file A file that contains data in a specific format, requiringinterpretation（翻译）31．The root directory is typically identified by the device letter followed by a backslash(反斜线)32．File managementWindows Explorer(资源管理器) Clipboard（剪贴板）Recycle Bin(回收站) 33．File Properties(属性)34．Path NamesAbsolute path D:\Courses\Computer\2011\OS.pptRelative path 2011\OS.ppt 〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉〉Multimedia35．Representing CharacterASCII （America Standard Code for Information Interchange）❖Original ASCII uses 7 bits to code 128 characters❖Extended to 8 bits for 128 extra characters2. Unicodeuses 16 bits, can represent 216 = 6553436．Representing Audio❖Sound is generated by the vibration of air❖Sound wave is analog❖Computers convert sound to digits by sampling❖ A stereo sends an electrical signal to a speaker to reproduce soundA sampling rate of about 40,000 times per second is enough to create a reasonable sound reproductionAudio Formats▪WA V, AU, AIFF, VQF, and MP3❖MP3 is dominant▪MPEG-2, audio layer 3 file >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Graphics 37．Representing Images and Graphics❖Raster(光栅)-graphics▪The storage of image information is on a pixel-by-pixel（像素）basis▪Popular formats•BMP (bitmap), GIF (Graphics Interchange Format), JPEG (JointPhotographic Experts Group)❖Vector-graphics（矢量图）▪Describes an image in terms of lines and geometric shapes▪Include a series of commands that describe a line’s direction, thickness, and colour.▪The file sizes tend to be smaller because not every pixel is described▪38． A graph is treated as a matrix of dots, which is called pixels❖The number of pixels is called resolution（分辨率）39．Color is expressed as an RGB (Red,Green,Blue) value three primary colors40．Color depth▪The amount of data that is used to represent a color41．TrueColor▪ A 24-bit color depth: eight bits used for each number in an (R,G,B) value 42．Each pixel is normally recorded by three bytes, for red-green-blue respectively. Therefore, a pixel is written as (r, g, b), where 0 <= r,g,b <= 25543．Resolution▪The number of pixels used to represent a picture44．In the resolution 800×600, an RGB true-color image needs❖800×600×3×8 bits = 1440KB = 1.4MB。

1.1What are the three main purposes of an operating system?(1) Interface between the hardware and user;(2) manage the resource of hardware and software;(3) abstraction of resource;1.2 List the four steps that are necessary to run a program on a completely dedicated machine. Preprocessing > Processing > Linking > Executing.1.6 Define the essential properties of the following types of operating systems:a. Batchb. Interactivec. Time sharingd. Real timee. Networkf. Distributed1.7 We have stressed the need for an operating system to make efficient use of the computing hardware. When is it appropriate for the operating system to forsake this principle and to“waste” resources? Why is such a system not really wasteful?2.2 How does the distinction between monitor mode and user mode function as a rudimentary form of protection (security) system?2.3 What are the differences between a trap and an interrupt? What is the use of each function?2.5 Which of the following instructions should be privileged?a. Set value of timer.b. Read the clock.c. Clear memory.d. Turn off interrupts.e. Switch from user to monitor mode.2.8 Protecting the operating system is crucial to ensuring that the computer system operates correctly. Provision of this protection is the reason behind dual-mode operation, memory protection, and the timer. To allow maximum flexibility, however, we would also like to place minimal constraints on the user.The following is a list of operations that are normally protected. What is the minimal set of instructions that must be protected?a. Change to user mode.b. Change to monitor mode.c. Read from monitor memory.d. Write into monitor memory.e. Fetch an instruction from monitor memory.f. Turn on timer interrupt.g. Turn off timer interrupt.3.6 List five services provided by an operating system. Explain how each provides convenience to the users. Explain also in which cases it would be impossible for user-level programs to provide these services.3.7 What is the purpose of system calls?3.10 What is the purpose of system programs?4.1 MS-DOS provided no means of concurrent processing. Discuss three major complications that concurrent processing adds to an operating system.4.6 The correct producer–consumer algorithm in Section 4.4 allows only n-1 buffers to be full at any one time. Modify the algorithm to allow all buffers to be utilized fully.5.1 Provide two programming examples of multithreading giving improve performance overa single-threaded solution.5.3 What are two differences between user-level threads and kernel-level threads? Under what circumstances is one type better than the other?6.3 Consider the following set of processes, with the length of the CPU-burst time given inmilliseconds:Process Burst Time PriorityP1 10 3P2 1 1P3 2 3P4 1 4P5 5 2The processes are assumed to have arrived in the order P1, P2, P3, P4, P5, all at time 0.a. Draw four Gantt charts illustrating the execution of these processes using FCFS, SJF, a nonpreemptive priority (a smaller priority number implies a higher priority), and RR (quantum = 1) scheduling.b. What is the turnaround time of each process for each of the scheduling algorithms in part a?c. What is the waiting time of each process for each of the scheduling algorithms in part a?d. Which of the schedules in part a results in the minimal average waiting time (over all processes)?Answer:6.4 Suppose that the following processes arrive for execution at the times indicated. Eachprocess will run the listed amount of time. In answering the questions, use nonpreemptive scheduling and base all decisions on the information you have at the time the decision must be made.a. What is the average turnaround time for these processes with the FCFS scheduling algorithm?b. What is the average turnaround time for these processes with the SJF scheduling algorithm?c. The SJF algorithm is supposed to improve performance, but notice that we chose to run process P1 at time 0 because we did not know that two shorter processes would arrive soon. Compute what the average turnaround time will be if the CPU is leftidle for the first 1 unit and then SJF scheduling is used. Remember that processes P1 and P2 are waiting during this idle time, so their waiting time may increase. This algorithm could be known as future-knowledge scheduling.6.10 Explain the differences in the degree to which the following scheduling algorithms discriminate in favor of short processes:a. FCFSb. RRc. Multilevel feedback queues7.7 Show that, if the wait and signal operations are not executed atomically,then mutual exclusion may be violated.7.8 The Sleeping-Barber Problem. A barbershop consists of a waiting room with n chairs and the barber room containing the barber chair. If there are no customers to be served,the barber goes to sleep. If a customer enters the barbershop and all chairs are occupied, then the customer leaves the shop.If the barber is busy but chairs are available, then the customer sits in one of the free chairs. If the barber is asleep, the customer wakes up the barber. Write a program to coordinate the barber and the customers.8.2 Is it possible to have a deadlock involving only one single process? Explain your answer.8.4 Consider the traffic deadlock depicted in Figure 8.11.a. Show that the four necessary conditions for deadlock indeed hold in this example.b. State a simple rule that will avoid deadlocks in this system.8.13 Consider the following snapshot of a system:Allocation Max AvailableA B C D A B C D A B C DP0 0 0 1 2 0 0 1 2 1 5 2 0P1 1 0 0 0 1 7 5 0P2 1 3 5 4 2 3 5 6P3 0 6 3 2 0 6 5 2P4 0 0 1 4 0 6 5 6Answer the following questions using th e banker’s algorithm:a. What is the content of the matrix Need?b. Is the system in a safe state?c. If a request from process P1 arrives for (0,4,2,0), can the request be granted immediately?9.5 Given memory partitions of 100K, 500K, 200K, 300K, and 600K (in order), how would each of the First-fit, Best-fit, and Worst-fit algorithms place processes of 212K, 417K, 112K, and 426K (in order)? Which algorithm makes the most efficient use of memory?9.8 Consider a logical address space of eight pages of 1024 words each, mapped onto a physicalmemory of 32 frames.a. How many bits are there in the logical address?b. How many bits are there in the physical address?9.16 Consider the following segment table:Segment Base Length0219 60012300 14290 10031327 58041952 96What are the physical addresses for the following logical addresses?a. 0,430b. 1,10c. 2,500d. 3,400e. 4,11210.2 Assume that you have a page reference string for a process with m frames (initially all empty). The page reference string has length p with n distinct page numbers occur in it. For any page-replacement algorithms,a. What is a lower bound on the number of page faults?b. What is an upper bound on the number of page faults?10.11 Consider the following page reference string:1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6.How many page faults would occur for the following replacement algorithms, assuming one, two, three, four, five, six, or seven frames? Remember all frames are initially empty, so your first unique pages will all cost one fault each.LRU replacementFIFO replacementOptimal replacement11.7 Explain the purpose of the open and close operations.11.9 Give an example of an application in which data in a file should be accessed in the following order:a. Sequentiallyb. Randomly11.12 Consider a system that supports 5000 users. Suppose that you want to allow 4990 of these users to be able to access one file.a. How would you specify this protection scheme in UNIX?b. Could you suggest another protection scheme that can be used more effectively for this purpose than the scheme provided by UNIX?12.1 Consider a file currently consisting of 100 blocks. Assume that the file control block (andthe index block, in the case of indexed allocation) is already in memory. Calculate how many disk I/O operations are required for contiguous, linked, and indexed (single-level) allocation strategies, if, for one block, the following conditions hold. In the contiguousallocation case, assume that there is no room to grow in the beginning, but there is room to grow in the end. Assume that the block information to be added is stored in memory.a. The block is added at the beginning.b. The block is added in the middle.c. The block is added at the end.d. The block is removed from the beginning.e. The block is removed from the middle.f. The block is removed from the end.13.2 Consider the following I/O scenarios on a single-user PC.a. A mouse used with a graphical user interfaceb. A tape drive on a multitasking operating system (assume no device preallocation is available)c. A disk drive containing user filesd. A graphics card with direct bus connection, accessible through memory-mappedI/OFor each of these I/O scenarios, would you design the operating system to use buffering, spooling, caching, or a combination? Would you use polled I/O, or interrupt-driven I/O?Give reasons for your choices.14.2 Suppose that a disk drive has 5000 cylinders, numbered 0 to 4999. The drive is currently serving a request at cylinder 143, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130Starting from the current head position, what is the total distance (in cylinders) that the disk arm moves to satisfy all the pending requests, for each of the following diskschedulingalgorithms?a. FCFSb. SSTFc. SCANd. LOOKe. C-SCAN1.1 1.62.3 2.53.7 6.3 6。

1、操作系统：Operating System 13、临界资源：Critical Resource2、串行处理系统：Serial System 14、临界段：Critical Section3、批处理系统：Batch System 15、调度：Scheduler4、分时系统：Time Sharing System 16、响应时间：Response Time5、实时系统：Real Time System 17、最短作业优先：Shortest Process Next6、多道程序：Multi-programming 18、最高响应比优先：Highest Response Ratio Next7、进程：Process 19、时间片轮转：Round-Robin8、进程表：Process T ables 20、死锁：Deadlock9、进程映像：Process Image 21、死锁预防：Deadlock Prevention10、进程控制块：Process Control Block22、死锁避免：Deadlock Abstention11、并发：Concurrence 23、死锁检测：Deadlock Detection12、互斥：Mutual Exclusion 24、死锁恢复：Deadlock Restoration1、试修改下面生产者－消费者问题解法中的错误，改正在错误处：struct semaphore mutex=1,empty=n,full=0;main(){{{producer();consumer();}}}producer(){while(true){生产产品;V(full);V(mutex);把产品送入缓冲;P(mutex);}}consumer(){while(true){V(empty);V(mutex);从缓冲取得产品;P(mutex);消费产品;}}2、已知内存管理采用页式存储管理。

操作系统（Operat ing System ）复习要点操作系统：计算机系统中的一组系统软件，由它统一管理计算机系统的各种资源并合理组织计算机的工作流程，方便用户使用。

具有管理和服务功能操作系统的特征：并发性，共享性，随机性，可重构性，虚拟性。

并发是指计算机系统中同时存在多个程序，宏观上看，这些程序是同时向前推进的。

共享性：批操作系统程序与多个用户程序共用系统中的各种资源虚拟性：物理实体转化为若干逻辑上的对应物。

操作系统的功能：1，进程管理；2,存储管理；3,文件管理；4,作业管理；5,设备管理；6,其他功能（系统安全，网络通信）。

传统OS中，进程是系统调度的最小单位，是程序的一次执行；而现代OS中则是线程，是程序一次相对独立的执行过程。

操作系统的发展历史1,手工操作：穿孔卡片2,监督程序一一早期批处理：计算机高级语言岀现，单道批处理单道批处理：串行执行作业中，由监督程序识别一个作业，进行处理后再取下一个作业的自动定序处理方式第作业的定义：用户要求计算机系统处理的一个计算问题。

（或参考调度性能的衡量一一周转时间、平均周转时间、带权周转时间、平作业的两种控制方式1,批处理：操作系统按各作业的作业控制说明书的要求，分别控制相应的作业按指定步骤执行。

2,交互：在作业执行过程中，操作系统与用户之间不断交互作用。

作业调度：从后备作业队列中选取某个作业投入主存参与多道运行。

调度算法原则：①尽可能运行更多的作业，优先考虑短作业；②使处理机保持繁忙，优先考虑计算量大的作业；③使I/O设备保持繁忙，优先考虑I/O繁忙的作业；④对所有的作业都是公平合理的。

选择原则：①选择的调度算法与系统的整体设计目标一致；②注意系统资源的均衡使用，使I/O作业与CPU作业搭配合理；③作业应该在规定时间内完成，能缩短作业周转时间。

第进程的定义：具有独立功能的并行程序一次执行过程进程和程序的区别与联系：区别：①程序是指令的有序集合，静态；进程是程序的一次运行活动，动态；②进程是一个独立运行单位，共享资源的实体，能并发执行；而程序不能。

XX大学2011——2012学年第一学期《操作系统》期中考试试题（A）一、选择（每题1分，共20分）1.Which function does the operating system can not complete directly of the following four options? ( b )A.Managing computer's hard drivepile the programC.Virtual memoryD.Delete files2.Considering the function of the operating system, ( b ) must give timely response for the external request within the specified time.A.multiuser time sharing systemB.real-time operating systemC.batch operating systemwork operating system3. A process can transform from waiting state to ready state relying on ( d )A.programmer commandB.system serviceC.waiting for the next time sliceD.wake-up of the 'cooperation' process4.As we all know,the process can be thought of as a program in execution.We can deal with the the problem about ( b ) easier after importing the concept of process.A.exclusive resourcesB.shared resourcesC.executing in orderD.easy to execute5.CPU-scheduling decisions may take place under the following circumstances except which one？（D ）A.When a process switches from the running state to the waiting stateB.When a process switches from the running state to the ready stateC.When a process switches from the waiting state to the ready stateD.When a process switches from the ready state to the waiting state6.In the four common CPU scheduling algorithm, Which one is the best choice for the time-sharing system in general？（ C ）A.FCFS scheduling algorithmB.Priority scheduling algorithmC.Round-robin scheduling algorithmD.Shortest-job-first scheduling algorithm7.If the initial value of semaphore S is 2 in a wait( ) and signal( ) operation，its current value is -1,that means there are ( B ) processes are waiting。

操作系统学习指导书操作系统课程组信息工程学院计算机系第1章操作系统引论1.1 知识点总结1、什么是操作系统?操作系统：是控制和管理计算机系统内各种硬件和软件资源、有效地组织多道程序运行的系统软件（或程序集合），是用户与计算机之间的接口。

1) OS是什么：是系统软件（一整套程序组成，如UNIX由上千个模块组成）2) 管什么：控制和管理系统资源（记录和调度）2、操作系统的主要功能?操作系统的功能：存储器管理、处理机管理、设备管理、文件管理和用户接口管理。

1) 存储器管理：内存分配，地址映射，内存保护和内存扩充2) 处理机管理：作业和进程调度，进程控制和进程通信3) 设备管理：缓冲区管理，设备分配，设备驱动和设备无关性4) 文件管理：文件存储空间的管理，文件操作的一般管理，目录管理，文件的读写管理和存取控制5) 用户接口：命令界面/图形界面和系统调用接口3、操作系统的地位操作系统是裸机之上的第一层软件，是建立其他所有软件的基础。

它是整个系统的控制管理中心，既管硬件，又管软件，它为其它软件提供运行环境。

4、操作系统的基本特征？操作系统基本特征：并发，共享和异步性。

1) 并发：并发性是指两个或多个活动在同一给定的时间间隔中进行。

2) 共享：共享是指计算机系统中的资源被多个任务所共用。

3) 异步性：每个程序什么时候执行，向前推进速度快慢，是由执行的现场所决定。

但同一程序在相同的初始数据下，无论何时运行都应获得同样的结果。

5、操作系统的主要类型？多道批处理系统、分时系统、实时系统、个人机系统、网络系统和分布式系统1) 多道批处理系统(1) 批处理系统的特点：多道、成批(2) 批处理系统的优点：资源利用率高、系统吞吐量大(3) 批处理系统的缺点：等待时间长、没有交互能力2) 分时系统(1) 分时：指若干并发程序对CPU时间的共享。

它是通过系统软件实现的。

共享的时间单位称为时间片。

(2) 分时系统的特征：同时性：若干用户可同时上机使用计算机系统交互性：用户能方便地与系统进行人--机对话独立性：系统中各用户可以彼此独立地操作，互不干扰或破坏及时性：用户能在很短时间内得到系统的响应(3) 优点主要是：响应快，界面友好多用户，便于普及便于资源共享3) 实时系统(1) 实时系统：响应时间很快，可以在毫秒甚至微秒级立即处理(2) 典型应用形式：过程控制系统、信息查询系统、事务处理系统4) 个人机系统(1) 单用户操作系统单用户操作系统特征：个人使用：整个系统由一个人操纵，使用方便。

操作系统课程英文词汇Operating System Course English VocabularyIntroductionIn today's rapidly advancing technological world, the study of operating systems holds immense importance. An operating system serves as the backbone of any computer system, managing hardware and software resources efficiently. To fully grasp the concepts and principles associated with operating systems, it is essential to have a strong foundation in the relevant English vocabulary. This article aims to provide a comprehensive list of essential English terms commonly used in the context of operating system courses.1. KernelThe kernel is the core component of an operating system. It acts as an intermediary between hardware and software, managing system resources and providing essential services for the execution of programs.2. ProcessA process refers to a task or program in execution. It includes the program code, associated data, and various execution contexts necessary for its execution. Processes are managed by the operating system and can run concurrently on a computer system.3. ThreadA thread represents a section of a process that can be executed independently. Threads share the same memory space and resources within aprocess, allowing for efficient parallelism and enhanced responsiveness in multitasking environments.4. Memory ManagementMemory management involves the allocation and deallocation of memory resources to different processes and threads. It ensures optimal utilization of memory and prevents conflicts among multiple programs competing for resources.5. File SystemA file system is a method used by the operating system to organize and store data on storage devices such as hard drives and solid-state drives. It provides a structured hierarchy of directories and files, enabling efficient retrieval and storage of information.6. Device DriverA device driver is a software component that allows the operating system to communicate with hardware devices. It provides a standardized interface for controlling and accessing various hardware components, such as printers, keyboards, and network adapters.7. SchedulingScheduling refers to the process of determining the order in which processes or threads are executed on a system's CPU. Various scheduling algorithms, such as round-robin, priority-based, and shortest job first, ensure fair allocation of resources and efficient system performance.8. Virtual MemoryVirtual memory extends the physical memory of a computer system by allowing parts of the operating system or processes to reside in secondary storage, such as hard disks. It provides an illusion of a larger memory space, enabling the execution of more extensive programs.9. DeadlockA deadlock occurs when two or more processes are unable to proceed due to circular dependencies on resources. It can lead to a system-wide stalemate, requiring careful resource allocation and scheduling algorithms to prevent such situations.10. InterruptInterrupts are signals generated by hardware devices or software, causing the operating system to temporarily suspend the execution of a process and handle the event. Interrupt handling ensures timely response to hardware events, such as keyboard input or disk I/O.11. File Allocation MethodsFile allocation methods determine how files are stored on a storage medium. Common methods include contiguous allocation, linked allocation, and indexed allocation. Each method has its advantages and trade-offs concerning storage efficiency and file access time.12. PagingPaging is a memory management technique that divides the physical memory into fixed-size blocks called pages. It allows for efficient memoryallocation and virtual memory usage, reducing fragmentation and enhancing system performance.13. SwappingSwapping involves moving an entire process from main memory to secondary storage and vice versa. It enables the system to free memory resources when necessary, allowing for the execution of more significant processes.14. CachingCaching is a technique used to store frequently accessed data in a cache memory. It helps to improve system performance by reducing the time required to access data from slower storage devices.ConclusionUnderstanding the fundamental concepts and vocabulary associated with operating systems is crucial for students pursuing a course in this field. This comprehensive list of English terms provides a solid foundation for studying operating systems and related topics. Remembering and utilizing these terms will enhance communication, comprehension, and overall learning experience in the exciting world of operating systems.。

Chapter 1 – Computer Systems OverviewTrue / False Questions:1.T / F –The operating system acts as an interface between the computer hardware and the human user.2.T / F –One of the processor’s main functions is to exchange data with memory.3.T / F – User-visible registers are typically accessible to system programs but are not typically available to application programs.4.T / F –Data registers are general purpose in nature, but may be restricted to specific tasks such as performingfloating-point operations.5.T / F –The Program Status Word contains status information in the form of condition codes, which are bits typically set by the programmer as a result of program operation.6.T / F – The processing required for a single instruction on a typical computer system is called the Execute Cycle.7.T / F –A fetched instruction is normally loaded into the Instruction Register (IR).8.T / F –An interrupt is a mechanism used by system modules to signal the processor that normal processing should be temporarily suspended.9.T / F – To accommodate interrupts, an extra fetch cycle is added to the instruction cycle.10. T / F –The minimum information that must be saved before the processor transfers control to the interrupt handler routine is the program status word (PSW) and the location of the current instruction.11. T / F –One approach to dealing with multiple interrupts is to disable all interrupts while an interrupt is being processed.12. T / F –Multiprogramming allows the processor to make use of idle time caused by long-wait interrupt handling.13.T / F –In a two-level memory hierarchy, the Hit Ratio is defined as the fraction of all memory accesses found in the slower memory.14. T / F – Cache memory exploits the principle of locality by providing a small, fast memory between the processor and main memory.15. T / F –In cache memory design, block size refers to the unit of data exchanged between cache and main memory16. T / F – The primary problem with programmed I/O is that the processor must wait for the I/O module to become ready and must repeatedly interrogate the status of the I/O module while waiting.Multiple Choice Questions:1.The general role of an operating system is to:a.Act as an interface between various computersb.Provide a set of services to system usersc.Manage files for application programsd.None of the above2.The four main structural elements of a computer system are:a.Processor, Registers, I/O Modules & Main Memoryb.Processor, Registers, Main Memory & System Busc.Processor, Main Memory, I/O Modules & System Busd.None of the above3.The two basic types of processor registers are:er-visible and Control/Status registersb.Control and Status registerser-visible and user-invisible registersd.None of the above4.Address registers may contain:a.Memory addresses of datab.Memory addresses of instructionsc.Partial memory addressesd.All of the above5. A Control/Status register that contains the address of the next instruction to be fetched is called the:a.Instruction Register (IR)b.Program Counter (PC)c.Program Status Word (PSW)d.All of the above6.The two basic steps used by the processor in instruction processing are:a.Fetch and Instruction cyclesb.Instruction and Execute cyclesc.Fetch and Execute cyclesd.None of the above7. A fetched instruction is normally loaded into the:a.Instruction Register (IR)b.Program Counter (PC)c.Accumulator (AC)d.None of the above8. A common class of interrupts is:a.Programb.Timerc.I/Od.All of the above9.When an external device becomes ready to be serviced by the processor, the device sends this type of signal to the processor:a.Interrupt signalb.Halt signalc.Handler signald.None of the above10. Information that must be saved prior to the processor transferring control to the interrupt handler routine includes:a.Processor Status Word (PSW)b.Processor Status Word (PSW) & Location of nextinstructionc.Processor Status Word (PSW) & Contents of processorregistersd.None of the above11.One accepted method of dealing with multiple interrupts is to:a.Define priorities for the interruptsb.Disable all interrupts except those of highestpriorityc.Service them in round-robin fashiond.None of the above12. In a uniprocessor system, multiprogramming increases processor efficiency by:a.Increasing processor speedb.Taking advantage of time wasted by long waitinterrupt handlingc.Eliminating all idle processor cyclesd.All of the above13. As one proceeds down the memory hierarchy ., from inboard memory to offline storage), the following condition(s) apply:a.Increasing cost per bitb.Decreasing capacityc.Increasing access timed.All of the above14. Small, fast memory located between the processor and main memory is called:a.WORM memoryb.Cache memoryc.CD-RW memoryd. None of the above15.When a new block of data is written into cache memory, the following determines which cache location the block will occupy:a.Block sizeb.Cache sizec.Write policyd.None of the above16. Direct Memory Access (DMA) operations require the following information from the processor:a.Address of I/O deviceb.Starting memory location to read from or write toc.Number of words to be read or writtend.All of the aboveQuestions，，，Problems ，，，，Chapter 2 – Operating System OverviewTrue / False Questions:1.T / F – An operating system controls the execution ofapplications and acts as an interface between applications and the computer hardware.2.T / F – The operating system maintains information thatcan be used for billing purposes on multi-user systems.3.T / F – The operating system typically runs in parallelwith application programs, on it’s own special O/S processor.4.T / F – One of the driving forcesin operating system evolution is advancement in the underlying hardware technology.5.T / F –In the first computers, users interacted directlywith the hardware and operating systems did not exist.6.T / F –In a batch-processing system, the phrase “controlis passed to a job” means that the processor is now fetching and executing instructions in a user program.7.T / F – Uniprogramming typically provides betterutilization of system resources than multiprogramming.8.T / F – In a time sharing system,a user’s program is preempted at regular intervals, but due torelatively slow human reaction time this occurrence is usually transparent to the user.9.T / F – A process can be defined as a unit of activitycharacterized by a single sequential thread of execution, a current state, and an associated set of system resources.10. T / F – A virtual memory address typically consists ofa page number and an offset within the page.11. T / F –Implementing priority levels is a common strategyfor short-term scheduling, which involves assigning each process in the queue to the processor according to its level of importance.12. T / F –Complex operating systems today typically consistof a few thousand lines of instructions.13.T / F – A monolithic kernel architecture assigns only afew essential functions to the kernel, including address spaces, interprocess communication and basic scheduling.14. T / F –The hardware abstraction layer (HAL) maps betweengeneric hardware commands/responses and those unique to a specific platform.Multiple Choice Questions:17.A primary objective of an operating system is:a.Convenienceb.Efficiencyc.Ability to evolved.All of the above18.The operating system provides many types of services to end-users, programmers and system designers, including:a.Built-in user applicationsb.Error detection and responsec.Relational database capabilities with the internalfile systemd.All of the above19.The operating system is unusual in it’s role as a control mechanism, in that:a.It runs on a special processor, completelyseparated from the rest of the systemb.It frequently relinquishes control of the systemprocessor and must depend on the processor to regain control of the systemc.It never relinquishes control of the systemprocessord.None of the above20.Operating systems must evolve over time because:a.Hardware must be replaced when it failsers will only purchase software that has a currentcopyright datec.New hardware is designed and implemented in thecomputer systemd.All of the above21.A major problem with early serial processing systems was:a.Setup timeck of input devicesc.Inability to get hardcopy outputd.All of the above22.An example of a hardware feature that is desirable in a batch-processing system is:a.Privileged instructionsb. A completely accessible memory arearge clock cyclesd.None of the above23.A computer hardware feature that is vital to the effective operation of a multiprogramming operating system is:a.Very large memoryb.Multiple processorsc.I/O interrupts and DMAd.All of the above24.The principle objective of a time sharing, multiprogramming system is to:a.Maximize response timeb.Maximize processor usec.Provide exclusive access to hardwared.None of the above25.Which of the following major line of computer system development created problems in timing and synchronization that contributed to the development of the concept of the processa.Multiprogramming batch operation systemsb.Time sharing systemsc.Real time transaction systemsd.All of the above26. The paging system in a memory management system provides for dynamic mapping between a virtual address used in a program and:a. A virtual address in main memoryb. A real address in main memoryc. A real address in a programd.None of the above27. Relative to information protection and security in computer systems, access control typically refers to:a.Proving that security mechanisms perform accordingto specificationb.The flow of data within the systemc.Regulating user and process access to variousaspects of the systemd.None of the above28. A common problem with full-featured operating systems, due to their size and difficulty of the tasks they address, is:a.Chronically late in deliverytent bugs that show up in the fieldc.Sub-par performanced.All of the above29. A technique in which a process, executing an application, is divided into threads that can run concurrently is called:a.Multithreadingb.Multiprocessingc.Symmetric multiprocessing (SMP)d.None of the aboveQUESTIONS，，，，PROBLEMS，，，。

l.lWhat are the three main purpo ses of an op erat ing system? ⑴ In terface betwee n the hardware and user; (2) man age the resource of hardware and software; (3) abstracti on of resource;Answer:• To provide an environment k>r a computer user to< execute programs on computtr hardware in a ccnvenient and efficient manner• Tb 吕 lk>cat 特 the 艺t?parattz resources of the compuler as nteded to strive the prtjblem given* The allocation p TOCESS should be as fair and efficient as possible.• Asa ct>ntroJ p a>gram it serves two major functions ： (1) sup ervision of the execution of user programs to p re vent errors and improper use of the computer, and (2) manage ment of the «p erat it) JI and control of [/O devices.1.2 List the four steps that are necessary to run a program on a completely dedicated machine. Prep rocess ing > Process ing > Linking > Executi ng.Answer乩 Reserve machine time*b. Manually load program into memory.c. Load starting address and begin execution.d.Monitor and control execution of program fnim ct>nsoie.1.6 Define the esse ntial prop erties of the follow ing types of op erat ing systems: a. Batch b. In teractive c. Time shari ng d. Real time e. Network f. DistributedAnswerBatch. Jobs with similar needs are batched together and run through the computer as a grcup by anoperator or automatic job st^quencer.[咆jrformHrk ：电 is increased by atteiTipting to keep CPU andI/O devices busv 試 all times throughoff-lineoperation, Kptx?ling, and multiprogramming* Batch is good for executing large jobs that netxlinteraction; it can be siubmilted and picked up laterInteractive. Thkin J of many short transactiorifd where the results ofthe next transaction may be unpredictable. Response rirne needs to be short (seconds] since the user submils and w^iih For the result.Time sharing. This systems u&es 匚n scheduling and mLiltiprt)gramming toprtividt* economical interactive of a system. The CP 匚 占w 让 chem rap idly fn>nn one user to another Instead of havinga job defined by spooled card images^ each program re^dsa” b. c.its next control card from the terminal, and output is normally printed immediately to the screen. Real time. Often usvd in a dedicated application, tliis system reads information fram sensors and mustrespond within a fixed amount of time to ensure correct performance- Network.Distributed .This system distributes computation arntmg se¥t?ral physical prtxzesKors, Theprt>cesst>rs do not share memory or a cltKk. Instead, each prixessor has its t>wn kxzal memory. They communicate with each other through various communicatitm lines, such as a high-speed bus or telephone line.1.7 hardware. When is it approp riate for the op erat ing system to forsake this principle and to waste " resources? Why is such a system not really wasteful?Answer Single*user systems shtiuId maximize use of the systenn for the user. A GUI might “waste" CPU cy<les, but it optimizes the user's interaction with the system.2.2 How does the distinction between monitor mode and user mode function as a rudimentary form of p rotecti on (security) system?亠 ■ ■Answer; By establishing a set of privileged instructions that can be executed only when in the mt>niu)r mode, the operating system is assured of ct^ntrolling the entire system at all times.2.3 What are the differe nces betwee n a tra p and an in terru pt? What is the use of each fun cti on?Answer An interrupt is a ha rd w a re-genera ted change-of-flow within the system. An interrupt handler is summoned to deal with the c<iuse oF the interrupt; control is then re*turned to the interrupted context and instruction. A trap is a sof t wa re-genera ted in terru p 匕 An interrupt can bv usvd to signal the coiTipJctk*n of an [/O to obviate the need fnr du\ icy poiling. A trap can be used to call operating system routines or to catch arithmetic errors.2.5 Which of the follow ing in structi ons should be p rivileged? a. Set value of timer. b. Read the clock. c. Clear memory.d. Turn off i nterru pts.e. Switch from user to mon itor mode.d. Wehave stressed the n eed for an op erat ing system to make efficie ntuse of the comp ut ing3OS Exercise BookClass No. NameAnswer: The following instructions should be privileged: a. Set value of timer. b. Clear memory. c. Turn off interruptacL Switch from user to monitor mode.2.8 Protect ing the op erati ng system is crucial to en suri ng that the comp uter system op erates correctly. P rovisi on of this p rotecti on is the reas on beh ind dual-mode op erati on, memory pr otecti on, and the timer. To allow maximum flexibility, however, we would also like to p lace mini mal con stra ints on the user. The followi ngis a list of op erati ons thatof in structi ons that must be p rotected? a. Change to user mode. b. Change to mon itor mode. c. Read from mon itor memory. d. Write into mon itor memory.e. Fetch an in structi on from mon itor memory.f. Tur n on timer in terru pt.g. Turn off timer in terru pt.Answen The minimal set of instructions that must be protected are:Read from monitor memory* Write into monitor me mor v.*Turn off timer interrupt.3.6 List five services p rovided by an op erat ing system. Exp lain how each p rovides convenience to the users. Explainalso in which cases it would be impossible for user-levelpr ograms to p rovide these services.Answer:are no rmally p rotected.What is the minimal seta. Change to monitor mod 匕b. c.Program execution. The operating system loads the contents (or sections) of a file into memory and begins its execution. A user-level pm呂ram could not be trusted to properly allocate CPU time.I/O' op e rat ions. Disks, tapes, serial lines^ and other devices must be communicated with at a \ ery low level. The user need only specify the device and the operation to perform on it, while the system converts that request into device- or controller-specific commands. User-level programs cannot be trusted to only access devices they should have access to nnd to only access them when they art otherwise unused.File-system manipulation. There arv many details in file creation, deletion, alkKation, and naming that users should not have to perform. Blocks of disk spact? are used by files and must be tracked.Deleting a file requires removing the name file information and freeing the 说[located bk>cks.[Protections must also be checked to assure prtiper file access. User prt^grams could neither ensure adherence tct protection methexJs nor be trusted to allocate only free blocks and deallocate blocks on file deletion.Communications. Message passing between systems requires messages be turned into packets of information, sent to the network contnUlei; transmitted across a communications medium, and reassembled by the destination system. Packet ordering and data correction must take place. Again, user programs might not coordinate access to the network device, or they might reevi^T packets destined for other processes. • Error detection. Error detection occurs at both the hardware and sofirware levels. At the hardware le\-el, all data transfers must be inspected to ensure that data have not been c(>rrupted in transit AU data on media must bi checked to be sure they have not changed since they were written to the media. At the software level, media must be checked for data consistency；for instance, do the number of allt>cated and unallocated blocks of storage match the total number on the device. There, errors are frequently prexzess-independent (for instance, the corruption of data on a disk), so there must be a global program (ttiE op erating system) that handles all type 吕 of errors. Also, by having errors processed by the operating system, p rocesses need not contain code to catch and correct all thE errors possible on a syst Em.3.7What is the purpose of system calls?Answer System ualls(J1U>W user-lewl lii request ices uf tl咤uperdting sv;»-tem.3.10 What is the purpose of system programs?jVnswcr: Evstem p rograms can be thought of bundlcz^ uf useful systctu oils. Tbevprovide basic functicrahty to users and so users de not need to write their own programs to sol、忙common problems.4.1MS-DOS pr ovided no means of con curre nt pr ocess ing. Discuss three major comp licati ons that con curre nt p rocess ing adds to an op erat ing system.5OS Exercise BookClass No. NameAnswer:A method of time sharing must be implemented to allow each of several processes to have access to the systen'i. This method inxoJvcs the prompt ion of processes that do not voluntarily giA-e up the CPU (by using A system cal], for instance) <ind the kernel being reentrant (so more than one prtxzess may be executing kernel code concurrently).Processes and system resources must hav E protections and must be prciteck?d frtiTn each other Any given process must be limited in the amount of memory it can use and the operations it can perform on devices like disks.Care must be taken in the kernel to pre\ ent deadkxzks between processe 鬲 so prcicesses aren't waiting for each other's allcxated rest>urces.4.6full at any one time. Modify the algorithm to allow all buffers to be utilized fully.Answer No answer.5.1 P rovide two p rogram ming exa mp les of multithread ing givi ng imp rove p erforma nee over a sin gle-threaded soluti on.Answer (1) A server that services each request in a sep a rate thread. (2) A paral* lelized application such as matrix multi plication where different p arts of the matrix may be worked on in parallel. (3) An interactive GUI program such as a dibugger where a thread is used to monitor user input, another thread represents the running flppHcation, and a third thread monitors performance.5.3 What are two differences between user-level threads and kernel-level threads? Under what circumsta nces is one type better tha n the other?r■Answer Context switching between user tlireads is quite similar tG switching between kernel threads, although it is dependent on the threads library and how it maps user threads to kernel threads. In general, context switching between user threads involves taking a user thread of its LWP and replacing it with another thread. This act typically involves saving and restoring the st^te of the registers.6.3 Con sider the follow ing set of p rocesses, with the len gth of the CPU-burst time give n in millisec on ds: P rocessP1 P2 P3 F4 P5The p rocessesare assumed to have arrived in the orderThe correct p roducer— con sumer algorithm in Secti on 4.4 allows only n-1 buffers to beBurst 10 1 2 1 5Time P riority 3 1 3 4 2P1, P2, P3, P4, P5, all at time 0.a.Draw four Gantt charts illustrati ng the executi on of these p rocesses using FCFS, SJF, a nonpreemp tive p riority (a smaller p riority n umber imp lies a higher p riority), and RR (qua ntum = 1) scheduli ng.b.What is the tur narou nd time of each pr ocess for each of the scheduli ng algorithms in part a?c.What is the wait ing time of each p rocess for each of the scheduli ng algorithms in p art a?d.Which of the schedules in p art a results in the mini mal average wait ing time (over all pr ocesses)?An swer:AnswerThe four Gantt charts area.b- Turnaround timeFCFS円101113卩414 卩519RR■19274SJF191 ^riority18196Waiting time (turnaround time minus burst time)FCFS RR卩210Ih11卩413卩514 SJF214Prit^ritv•J61618d. Shortest Job First6.4 Suppose that the followingpr ocess will run the listed amou nt of time. In an sweri ng the questi ons, use nonpreemp tivescheduli ng and base all decisi ons on the in formati on you have at the time the decisi onp rocesses arrive for execution at the times in dicated. Each70.110.4a. What is the average tur narou nd time for these p rocesses with the FCFS scheduli ng algorithm?b. What is the average turnaround time for these processes with the SJF scheduling algorithm?c. The SJF algorithm is supp osed to impr ove p erforma nee, but no tice that we chose to run p rocessP1 at time 0 because we did not know that two shorter p rocesses wouldarrive soon. Comp ute what the average turn arou nd time will be if the CPU is left idle for the first 1 un it and the n SJF scheduli ng is used. Remember that p rocesses and P2 are wait ing duri ng this idle time, so their wait ing time may in crease. This algorithm could be known as future-k no wledge scheduli ng.Answer; a. 10,53b. 9.53c. 6.86Remember that turnaround time LS finishing time minus arrival time^ so you have h) subtract the arrival times to compute the turnaround times. F 匚FS is 11 if yw forget subtract arrival time.6.10 Explain the differe nces in the degree to which the follow ing scheduli ng algorithms discrim in ate in favor of short p rocesses: a. FCFS b. RRc. Multilevel feedback queuesAnswen孔 FCFS —discriminates against sliort jobs since any short jobs arriving after long jobs will have a kmger waiting time.b. RR 一treats all jobs equally (giving them eqiiaL bursts of CPU time) so short Qbm will be able toleave the system faster since they will finish first.c. Multilevel feedback queues —work similar to the RR algorithm —they discriminate favorably towardshort jt>bs.7.7 Show that, if the wait and sig nal op erati ons are not executed atomically, then mutual exclusi on may be violated.Answer No answer,must be made.ClassOS Exercise BookNo.NameAi rhal rimeBuist 1 i[uvPi7.8 The Slee pin g-Barber Pr oblem. A barbersho p con sists of a wait ing room with n chairsand the barber room containing the barber chair. If there are no customers to be served,the barber goes to slee p. If a customer en ters thebarbersho p and all chairs are occu pi ed,then the customer leaves the sho p.lf the barber is busy but chairs are available, the nthe customer sits in one of the free chairs. If the barber is aslee p, the customer wakes up the barber. Write a p rogram to coord in ate the barber and the customers.Answer: Please refer to the supporting Web site for source code solution.8.2 Is it possible to have a deadlock involving only one single process? Explain your answer.Answer No. This follows directly from the hold-and-wait condition.8.4 Con sider the traffic deadlock dep icted in Figure 8.11.a. Show that the four n ecessary con diti ons for deadlock in deed hold in this exa mple.b. State a sim pie rule that will avoid deadlocks in this system.Answer No answerCon Sider the follow ing snap shot of a system:Allocati onMax AvailableA B C DA B C D A B C D P0 0 0 1 2 0 0 1 2 1 5 2 0P1 1 0 0 0 1 7 5 0 P2 13 54 2 35 6P 3 0 6 3 2 0 6 5 2P40 0 1 40 6 5 6An swer the follow ing questi ons using th e ban kera. What is the content of the matrix Need?b. Is the system in a safe state?c. If a request from p rocess immediately?Answer:A. Deadlock cannot occur because preempticin exists.b. Yes. A process may never acquire all the resources 让 needs if they are continuously preempted by aseries of reqviests such as those of process C.9.5 Given memory partitions of 100K, 500K, 200K, 300K, and 600K (in order), how would each of the First-fit, Best-fit, and Worst-fit algorithms place processes of 212K, 417K, 112K,and 426K (in order)? Which algorithm makes the most efficie nt use of memory?8.13 s algorithm:P1 arrives for (0,4,2,0), can the request be gran ted9OS Exercise BookClassNo. NameAnswer:212K is put in 500K partition 417K is put in BOOK partidon112K is put in 288K partition (new partition 288K = 500K = 212K) e. 426 K must wa 让212K 仏 put in 300K partidon 417K is put in 500K partition1I2K is put in 2nOK parti tin n 426K put in 600K partidon212K is put in 600K partition417K is put in 500K partitio 口 112K is put in 388K partidon 426K must waitIn this example. Best-fit turns out to be the best9.8 Con Sider a logical address sp ace of eight p ages of 1024 words each, mapped onto a physical memory of 32 frames.a. How many bits are there in the logical address?b. How many bits are there in the p hysical address?Answera. Logical addrej^s: 13 bitsb.卩hvsical address: 15 bits9.16 Con Sider the follow ing segme nt table: Segme nt Base Len gtha. First-fit:d ・ f. Best-fit:8- h* k. Worst-fit:m. n.600 14 100 580 96What are the p hysical addresses for the follow ing logical addresses? a. 0,430 b. 1,10 c. 2,500 d. 3,400 e. 4,112Answena. 219 + 43(1-649illegal reference, trap ki Dpvra ting systemillegal reference, trap tu op grating system10.2 Assume that you have a page referenee string for a process with m frames (initiallyall emp ty). The p age refere nee stri ng has len gth p with n disti net p age n umbers occur init. For any p age-re pl aceme nt algorithms,a. What is a lower bou nd on the n umber of p age faults?b. What is an upper bou nd on the n umber of p age faults?Answer a. tr b.卩10.11 Con Sider the follow ing p age refere nee stri ng: 1, Z 3, 4, 2,1,5, 6, Z 1,2, 3, 7, 6, 3, Z 1, Z 3, 6.How many p age faults would occur for the follow ing repl aceme nt algorithms, assum ing one, two, three, four, five, six, or seve n frames? Remember all frames are in itially emp ty, so your first uni que p ages will all cost one fault each.LRU rep laceme nt FIFO rep laceme nt Op timal rep laceme nt0 1 2 3 4219 2300 90 1327 1952 b. 2300 + 10 = 2310c. d. 1327 + 400= 17271111.7 Explain the purp ose of the openAnswer ：■ The open operatit>n informs the system that the named file is about to become active^* Tlie cloiie «peration informs the system that tlie named file is no longer in active use by the user who issued the 匚lose operation.11.9 Give an example of an application in which data in a file should be accessed in the followi ng order: a. Seque ntially b. Ran domlyAnsiver ：Print the content of the file.Print the content of record /. This record can be found using hashing or index techniques.11.12 theseusers to be able to access one file.a. How would you sp ecify this p rotecti on scheme in UNIX?b. Could you suggest ano ther pr otecti on scheme that can be used more effectively for this purpose tha n the scheme p rovided by UNIX?Answera. There are two methods for achieving this:Answer:ClassOS Exercise BookNo.NameNumber of framesLRU FIFO Optimal12 3 48-5 00 8 6 4and close op erati ons.a. b. Con sider a system that supports 5000 users. Suppose that you want to allow 4990 ofi. Create an access list with the names of all 4990ii. Put these 4990 users in one group and set the group access accordingly. This scheme cannot alwaysbe implemented since user groups are restricted by th# system.b. The univErse access information applies to all users unless their name appears in the access-controlHst with different access permission. With this scheme you simply put the names of the remaining ten users in the access control list but no access privileges alh)wcd.■r2Tl _Consider _a filecurrently consisting of 100 blocks.__Assume that _the file control block(ande in dex block, in the case of i ndexed allocati on) is already in memory. Calculate how any disk I/O op erati ons are required for con tiguous, li nked, and in dexed (sin gle-level) locati on strategies, if, for one block, the follow ing con diti ons hold. In thecase, assume that there is no room to grow in the beg inning, but there to be added is stored inAnswer198 9813.2 Con Sider the follow ing I/O see narios on a sin gle-user PC.a. A mouse used with a grap hieal user in terfaeeb. A tape drive on a multitask ing op erat ing system (assume no device p realloeati on is available) e. A disk drive containing user filesd. A gra phics eard with direet bus conn eeti on, aeeessible through memory-ma pped I/OFor each of these I/O see narios, would you desig n the op erat ing system to use bufferi ng, spo oli ng, eaehi ng, or a comb in ati on? Would you use p olled I/O, or in terru pt-drive n I/O? Give reas ons for your choices.teon tiguousalloeati onis room to grow in the end. Assume that the block in formatio nmemory.a. The block is added at the beg inning.b. The block is added in the middle. e. The block is added at the end.d. The block is removed from the begi nning.e. The block is removed from the middle.f. T he block is removed from the end.ContiffljQus LinkedIndexeda. b.201 101 52 3 52 10013OS Exercise BookClass No. NameAnswenA mouse used with a graphical user interfaceBuffering may bt? needed to record mouse movement during times when higher- priority op erations aretaking place. Spooling and caching are inap propriat 巳 Interrupt driven I/O is m 〔＞st appropriate^ A t 日pe drive on a multitasking operating system (assume no device preallocation is availabl £) Buffering may be needed to manage through put difference behveen the tape drive and the souro? or destination of the I/O, C 白匚hing can bv used to hold copies of daU that resides on the tape, for faster access. Spooling could be used to stage data to the device when multiple users desire to read from or write to it. Interrupt driven I/G is likely to allow the best performance. A disk drive containing user filesBuffering can be used to hold data while in transit from user space to the disk, and visa versa. Caching can be used to hold disk-resident data for impmvtd performance. Spociling is not necessary because disks are shared-access devices. Interrupt- driven I/O is best for devices such as disks that transfer data at slow rates-A graphics card with direct bus connection, accessible through memory-rnapped I/O 'Buffering may be needed to control multiple access and for performance (doublebuffering can be used to hold the next screen image while displaying the current one). Caching and 5pooling are not necessary^ due to IH E fast and shared-access natures of the device. Folling and in term pts are only useful for input and for I/O completion detection, neither of which is needed for a memory^mapped device.14.2Suppose that a disk drive has 5000 cylinders,numbered 0 to 4999. The drive is currentlyserving a request at cylinder 143, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130Start ing from the curre nt head p ositi on, what is the total dista nee (in cyli nders) that the disk arm moves to satisfy all the pending requests, for each of the follow ing diskscheduli ng algorithms? a. FCFS b. SSTF c. SCAN d. LOOK e. C-SCANa. b.d.Answer:a. The FCFS schedule bi 143, 86, 1470, 913, 1774, 94S, 1509, 1022, 1750, 130. The tntal seekdistance is 7081.b. The SSTF schedule is 143, 130, 86, 913, 948, 1D22, 1470, 1509, 1750, 1774. The total seekdistance is 1745.c. The SCAN schedule is 143, 913, 94«, 1022,1470, 1509, 1750, 1774, 4999,130, 86. The total seekdistance is 9769.The LCX)K scheduJe is 143, 913, 94H, 1022, 1470, 1509, 1750, 1774, 130, H6. The total seekdistance is 3319.e. The C-SCAN schedule is 143,913,948,1022,1470,1509.1750,1774,4999；130. Thetotal seek distance is 9813.{Bonus.} The C-LOOK schedule is 143” 913,94& 1022,1470,1509,1750,1774,86,130. The tohal seek distance is 3363.1.1 1.62.3 2.53.7 6.3 6。