操作系统习题解析

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操作系统练习题_及答案解析

操作系统练习题_及答案解析

操作系统练习题第一章引言(一单项选择题1 操作系统是计算机系统的一种( 。

A.应用软件B. 系统软件c. 通用软件D.工具软件2.操作系统目的是提供一个供其他程序执行的良好环境,因此它必须使计算机( A.使用方便B. 高效工作C.合理使用资源D.使用方便并高效工作3.允许多个用户以交互方式使用计算机的操作系统是( 。

A.分时操作系统B. 批处理单道系统C.实时操作系统D.批处理多道系统4.下列系统中( 是实时系统。

A.计算机激光照排系统B. 办公自动化系统C.化学反应堆控制系统D.计算机辅助设计系统5.操作系统是一种系统软件,它( 。

A.控制程序的执行B. 管理计算机系统的资源C.方便用户使用计算机D.管理计算机系统的资源和控制程序的执行6.计算机系统把进行( 和控制程序执行的功能集中组成一种软件,称为操作系统A.CPU 管理B.作业管理C.资源管理D.设备管理7.批处理操作系统提高了计算机系统的工作效率,但( 。

A.不能自动选择作业执行B. 无法协调资源分配c.不能缩短作业执行时间D 在作业执行时用户不能直接干预8.分时操作系统适用于( 。

A.控制生产流水线B.调试运行程序c.大量的数据处理D.多个计算机资源共享9.在混合型操作系统中,“前台”作业往往是指( 。

A.由批量单道系统控制的作业B.由批量多道系统控制的作业c.由分时系统控制的作业D.由实时系统控制的作业10.在批处理兼分时的系统中,对( 应该及时响应,使用户满意。

A.批量作业B.前台作业c.后台作业D.网络通信11.实时操作系统对可靠性和安全性要求极高,它( 。

A.十分注重系统资源的利用率B.不强调响应速度c.不强求系统资源的利用率D.不必向用户反馈信息12.分布式操作系统与网络操作系统本质上的不同之处在于( 。

A.实现各台计算机之间的通信B.共享网络个的资源c.满足较大规模的应用D.系统中若干台计算机相互协作完成同一任务13.SPOOL技术用于( 。

(完整版)操作系统第五版费祥林_课后习题答案解析参考

(完整版)操作系统第五版费祥林_课后习题答案解析参考

第一章操作系统概论1、有一台计算机,具有IMB 内存,操作系统占用200KB ,每个用户进程各占200KB 。

如果用户进程等待I/O 的时间为80 % ,若增加1MB 内存,则CPU 的利用率提高多少?答:设每个进程等待I/O 的百分比为P ,则n 个进程同时等待刀O 的概率是Pn ,当n 个进程同时等待I/O 期间CPU 是空闲的,故CPU 的利用率为1-Pn。

由题意可知,除去操作系统,内存还能容纳4 个用户进程,由于每个用户进程等待I/O的时间为80 % , 故:CPU利用率=l-(80%)4 = 0.59若再增加1MB 内存,系统中可同时运行9 个用户进程,此时:cPu 利用率=l-(1-80%)9 = 0.87故增加IMB 内存使CPU 的利用率提高了47 % :87 %/59 %=147 %147 %-100 % = 47 %2 一个计算机系统,有一台输入机和一台打印机,现有两道程序投入运行,且程序A 先开始做,程序B 后开始运行。

程序A 的运行轨迹为:计算50ms 、打印100ms 、再计算50ms 、打印100ms ,结束。

程序B 的运行轨迹为:计算50ms 、输入80ms 、再计算100ms ,结束。

试说明(1 )两道程序运行时,CPU有无空闲等待?若有,在哪段时间内等待?为什么会等待?( 2 )程序A 、B 有无等待CPU 的情况?若有,指出发生等待的时刻。

答:画出两道程序并发执行图如下:(1)两道程序运行期间,CPU存在空闲等待,时间为100 至150ms 之间(见图中有色部分)(2)程序A 无等待现象,但程序B 有等待。

程序B 有等待时间段为180rns 至200ms 间(见图中有色部分)3 设有三道程序,按A 、B 、C优先次序运行,其内部计算和UO操作时间由图给出。

试画出按多道运行的时间关系图(忽略调度执行时间)。

完成三道程序共花多少时间?比单道运行节省了多少时间?若处理器调度程序每次进行程序转换化时lms , 试画出各程序状态转换的时间关系图。

操作系统第九版部分课后作业习题答案解析

操作系统第九版部分课后作业习题答案解析

CHAPTER 9 Virtual Memory Practice Exercises9.1 Under what circumstances do page faults occur? Describe the actions taken by the operating system when a page fault occurs.Answer:A page fault occurs when an access to a page that has not beenbrought into main memory takes place. The operating system verifiesthe memory access, aborting the program if it is invalid. If it is valid, a free frame is located and I/O is requested to read the needed page into the free frame. Upon completion of I/O, the process table and page table are updated and the instruction is restarted.9.2 Assume that you have a page-reference string for a process with m frames (initially all empty). The page-reference string has length p;n distinct page numbers occur in it. Answer these questions 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?Answer:a. nb. p9.3 Consider the page table shown in Figure 9.30 for a system with 12-bit virtual and physical addresses and with 256-byte pages. The list of freepage frames is D, E, F (that is, D is at the head of the list, E is second, and F is last).Convert the following virtual addresses to their equivalent physical addresses in hexadecimal. All numbers are given in hexadecimal. (A dash for a page frame indicates that the page is not in memory.)• 9EF• 1112930 Chapter 9 Virtual Memory• 700• 0FFAnswer:• 9E F - 0E F• 111 - 211• 700 - D00• 0F F - EFF9.4 Consider the following page-replacement algorithms. Rank these algorithms on a five-point scale from “bad” to “perfect” according to their page-fault rate. Separate those algorithms that suffer from Belady’s anomaly from those that do not.a. LRU replacementb. FIFO replacementc. Optimal replacementd. Second-chance replacementAnswer:Rank Algorithm Suffer from Belady’s anomaly1 Optimal no2 LRU no3 Second-chance yes4 FIFO yes9.5 Discuss the hardware support required to support demand paging. Answer:For every memory-access operation, the page table needs to be consulted to check whether the corresponding page is resident or not and whether the program has read or write privileges for accessing the page. These checks have to be performed in hardware. A TLB could serve as a cache and improve the performance of the lookup operation.9.6 An operating system supports a paged virtual memory, using a central processor with a cycle time of 1 microsecond. It costs an additional 1 microsecond to access a page other than the current one. Pages have 1000 words, and the paging device is a drum that rotates at 3000 revolutions per minute and transfers 1 million words per second. The following statistical measurements were obtained from the system:• 1 percent of all instructions executed accessed a page other than the current page.•Of the instructions that accessed another page, 80 percent accesseda page already in memory.Practice Exercises 31•When a new page was required, the replaced page was modified 50 percent of the time.Calculate the effective instruction time on this system, assuming that the system is running one process only and that the processor is idle during drum transfers.Answer:effective access time = 0.99 × (1 sec + 0.008 × (2 sec)+ 0.002 × (10,000 sec + 1,000 sec)+ 0.001 × (10,000 sec + 1,000 sec)= (0.99 + 0.016 + 22.0 + 11.0) sec= 34.0 sec9.7 Consider the two-dimensional array A:int A[][] = new int[100][100];where A[0][0] is at location 200 in a paged memory system with pages of size 200. A small process that manipulates the matrix resides in page 0 (locations 0 to 199). Thus, every instruction fetch will be from page 0. For three page frames, how many page faults are generated bythe following array-initialization loops, using LRU replacement andassuming that page frame 1 contains the process and the other twoare initially empty?a. for (int j = 0; j < 100; j++)for (int i = 0; i < 100; i++)A[i][j] = 0;b. for (int i = 0; i < 100; i++)for (int j = 0; j < 100; j++)A[i][j] = 0;Answer:a. 5,000b. 509.8 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 replacement• FIFO replacement•Optimal replacement32 Chapter 9 Virtual MemoryAnswer:Number of frames LRU FIFO Optimal1 20 20 202 18 18 153 15 16 114 10 14 85 8 10 76 7 10 77 77 79.9 Suppose that you want to use a paging algorithm that requires a referencebit (such as second-chance replacement or working-set model), butthe hardware does not provide one. Sketch how you could simulate a reference bit even if one were not provided by the hardware, or explain why it is not possible to do so. If it is possible, calculate what the cost would be.Answer:You can use the valid/invalid bit supported in hardware to simulate the reference bit. Initially set the bit to invalid. O n first reference a trap to the operating system is generated. The operating system will set a software bit to 1 and reset the valid/invalid bit to valid.9.10 You have devised a new page-replacement algorithm that you thinkmaybe optimal. In some contorte d test cases, Belady’s anomaly occurs. Is the new algorithm optimal? Explain your answer.Answer:No. An optimal algorithm will not suffer from Belady’s anomaly because —by definition—an optimal algorithm replaces the page that will notbe used for the long est time. Belady’s anomaly occurs when a pagereplacement algorithm evicts a page that will be needed in the immediatefuture. An optimal algorithm would not have selected such a page.9.11 Segmentation is similar to paging but uses variable-sized“pages.”Definetwo segment-replacement algorithms based on FIFO and LRU pagereplacement schemes. Remember that since segments are not the samesize, the segment that is chosen to be replaced may not be big enoughto leave enough consecutive locations for the needed segment. Consider strategies for systems where segments cannot be relocated, and thosefor systems where they can.Answer:a. FIFO. Find the first segment large enough to accommodate the incoming segment. If relocation is not possible and no one segmentis large enough, select a combination of segments whose memoriesare contiguous, which are “closest to the first of the list” andwhich can accommodate the new segment. If relocation is possible, rearrange the memory so that the firstNsegments large enough forthe incoming segment are contiguous in memory. Add any leftover space to the free-space list in both cases.Practice Exercises 33b. LRU. Select the segment that has not been used for the longestperiod of time and that is large enough, adding any leftover spaceto the free space list. If no one segment is large enough, selecta combination of the “oldest” segments that are contiguous inmemory (if relocation is not available) and that are large enough.If relocation is available, rearrange the oldest N segments to be contiguous in memory and replace those with the new segment.9.12 Consider a demand-paged computer system where the degree of multiprogramming is currently fixed at four. The system was recently measured to determine utilization of CPU and the paging disk. The results are one of the following alternatives. For each case, what is happening? Can the degree of multiprogramming be increased to increase the CPU utilization? Is the paging helping?a. CPU utilization 13 percent; disk utilization 97 percentb. CPU utilization 87 percent; disk utilization 3 percentc. CPU utilization 13 percent; disk utilization 3 percentAnswer:a. Thrashing is occurring.b. CPU utilization is sufficiently high to leave things alone, and increase degree of multiprogramming.c. Increase the degree of multiprogramming.9.13 We have an operating system for a machine that uses base and limit registers, but we have modified the ma chine to provide a page table.Can the page tables be set up to simulate base and limit registers? How can they be, or why can they not be?Answer:The page table can be set up to simulate base and limit registers provided that the memory is allocated in fixed-size segments. In this way, the base of a segment can be entered into the page table and the valid/invalid bit used to indicate that portion of the segment as resident in the memory. There will be some problem with internal fragmentation.9.27.Consider a demand-paging system with the following time-measured utilizations:CPU utilization 20%Paging disk 97.7%Other I/O devices 5%Which (if any) of the following will (probably) improve CPU utilization? Explain your answer.a. Install a faster CPU.b. Install a bigger paging disk.c. Increase the degree of multiprogramming.d. Decrease the degree of multiprogramming.e. Install more main memory.f. Install a faster hard disk or multiple controllers with multiple hard disks.g. Add prepaging to the page fetch algorithms.h. Increase the page size.Answer: The system obviously is spending most of its time paging, indicating over-allocationof memory. If the level of multiprogramming is reduced resident processeswould page fault less frequently and the CPU utilization would improve. Another way toimprove performance would be to get more physical memory or a faster paging drum.a. Get a faster CPU—No.b. Get a bigger paging drum—No.c. Increase the degree of multiprogramming—No.d. Decrease the degree of multiprogramming—Yes.e. Install more main memory—Likely to improve CPU utilization as more pages canremain resident and not require paging to or from the disks.f. Install a faster hard disk, or multiple controllers with multiple hard disks—Also animprovement, for as the disk bottleneck is removed by faster response and morethroughput to the disks, the CPU will get more data more quickly.g. Add prepaging to the page fetch algorithms—Again, the CPU will get more datafaster, so it will be more in use. This is only the case if the paging action is amenableto prefetching (i.e., some of the access is sequential).h. Increase the page size—Increasing the page size will result in fewer page faults ifdata is being accessed sequentially. If data access is more or less random, morepaging action could ensue because fewer pages can be kept in memory and moredata is transferred per page fault. So this change is as likely to decrease utilizationas it is to increase it.10.1、Is disk scheduling, other than FCFS scheduling, useful in asingle-userenvironment? Explain your answer.Answer: In a single-user environment, the I/O queue usually is empty. Requests generally arrive from a single process for one block or for a sequence of consecutive blocks. In these cases, FCFS is an economical method of disk scheduling. But LOOK is nearly as easy to program and will give much better performance when multiple processes are performing concurrent I/O, such as when aWeb browser retrieves data in the background while the operating system is paging and another application is active in the foreground.10.2.Explain why SSTF scheduling tends to favor middle cylindersover theinnermost and outermost cylinders.The center of the disk is the location having the smallest average distance to all other tracks.Thus the disk head tends to move away from the edges of the disk.Here is another way to think of it.The current location of the head divides the cylinders into two groups.If the head is not in the center of the disk and a new request arrives,the new request is more likely to be in the group that includes the center of the disk;thus,the head is more likely to move in that direction.10.11、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 disk-scheduling algorithms?a. FCFSb. SSTFc. SCANd. LOOKe. C-SCANAnswer:a. The FCFS schedule is 143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130. The total seek distance is 7081.b. The SSTF schedule is 143, 130, 86, 913, 948, 1022, 1470, 1509, 1750, 1774. The total seek distance is 1745.c. The SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 130, 86. The total seek distance is 9769.d. The LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 130, 86. The total seek distance is 3319.e. The C-SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 86, 130. The total seek distance is 9813.f. (Bonus.) The C-LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 86, 130. The total seek distance is 3363.12CHAPTERFile-SystemImplementationPractice Exercises12.1 Consider a file currently consisting of 100 blocks. Assume that the filecontrol block (and the 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 contiguous-allocation case, assume that there is no room to grow atthe beginning but there is room to grow at the end. Also assume thatthe 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.Answer:The results are:Contiguous Linked Indexeda. 201 1 1b. 101 52 1c. 1 3 1d. 198 1 0e. 98 52 0f. 0 100 012.2 What problems could occur if a system allowed a file system to be mounted simultaneously at more than one location?Answer:4344 Chapter 12 File-System ImplementationThere would be multiple paths to the same file, which could confuse users or encourage mistakes (deleting a file with one path deletes thefile in all the other paths).12.3 Why must the bit map for file allocation be kept on mass storage, ratherthan in main memory?Answer:In case of system crash (memory failure) the free-space list would notbe lost as it would be if the bit map had been stored in main memory.12.4 Consider a system that supports the strategies of contiguous, linked, and indexed allocation. What criteria should be used in deciding which strategy is best utilized for a particular file?Answer:•Contiguous—if file is usually accessed sequentially, if file isrelatively small.•Linked—if file is large and usually accessed sequentially.• Indexed—if file is large and usually accessed randomly.12.5 One problem with contiguous allocation is that the user must preallocate enough space for each file. If the file grows to be larger than thespace allocated for it, special actions must be taken. One solution to this problem is to define a file structure consisting of an initial contiguous area (of a specified size). If this area is filled, the operating system automatically defines an overflow area that is linked to the initialc ontiguous area. If the overflow area is filled, another overflow areais allocated. Compare this implementation of a file with the standard contiguous and linked implementations.Answer:This method requires more overhead then the standard contiguousallocation. It requires less overheadthan the standard linked allocation. 12.6 How do caches help improve performance? Why do systems not use more or larger caches if they are so useful?Answer:Caches allow components of differing speeds to communicate moreefficie ntly by storing data from the slower device, temporarily, ina faster device (the cache). Caches are, almost by definition, more expensive than the device they are caching for, so increasing the number or size of caches would increase system cost.12.7 Why is it advantageous for the user for an operating system to dynamically allocate its internal tables? What are the penalties to the operating system for doing so?Answer:Dynamic tables allow more flexibility in system use growth — tablesare never exceeded, avoiding artificial use limits. Unfortunately, kernel structures and code are more complicated, so there is more potentialfor bugs. The use of one resource can take away more system resources (by growing to accommodate the requests) than with static tables.Practice Exercises 4512.8 Explain how the VFS layer allows an operating system to support multiple types of file systems easily.Answer:VFS introduces a layer of indirection in the file system implementation. In many ways, it is similar to object-oriented programming techniques. System calls can be made generically (independent of file system type). Each file system type provides its function calls and data structuresto the VFS layer. A system call is translated into the proper specific functions for the ta rget file system at the VFS layer. The calling program has no file-system-specific code, and the upper levels of the system call structures likewise are file system-independent. The translation at the VFS layer turns these generic calls into file-system-specific operations.。

操作系统习题解析

操作系统习题解析
试从交互性、及时性以及可靠性三个方面,比较分时系统与实时系统。P11
已知一个求值公式(A2+3B)/(B+5A),若A,B已赋值,试画 出该公式求值过程的前趋图,并用信号量解决公式的求解过程。 开始 S1:x1=A*A S2:x2=3*B S3:x3=5*A S1 S4:x4=x1+x2 S5:x5=B+x3 S4 S6:x6=x4/x5 结束 S5 S2 S3
A:1
B:2
C:4 D:2
7.下面8个系统中,必须是实时操作系统的有( ) A计算机辅助设计系统;B 航空定票系统;C 过程控制系统;D 机器翻译系统;E 办公自动化 系统;F 计算机激光照排系统;G情报检索系统;H导弹的制导系统
ABCGH
二、 简答题 在操作系统中实现虚拟的关键技术是什么?并加以说明。
操作系统中所谓的”虚拟“,是指通过某种技术把一个物理实体变为若干个逻辑上 的对应物,相应的用于实现虚拟的技术称为虚拟技术。在操作系统中利用了多种虚拟技 术分别用来实现虚拟处理机、虚拟内存、虚拟外设和虚拟信道等。虚拟的实现主要是通 过分时技术,例如,多道程序系统中,通过分时技术来实现虚拟处理机;将一台物理处 理机虚拟为多台逻辑上的处理机,是靠多道程序分时地使用同一台物理处理机来实现的。 微观上,该处理机在每一时刻只运行一道程序,它们分时地运行;然而在宏观上,系统 中确有几道程序在同时运行,从而给用户的感觉是系统中同时同时有多台处理机在为其 中的每一道程序服务,显然用户所感觉到的处理机并不实际存在。
A:4 B:5
5.采用(A)结构时,将OS分成用于实现OS最基本功能的内核和提供各种服务的服 务器两个部分。通常,下列模块中必须包含在操作系统内核中的是(B)模块。 A:(1)整体式;(2)模块化;(3)层次式;(4)微内核。 B:(1)内存分配;(2)中断处理;(3)文件处理;(4)命令处理。 A:4 B:2

操作系统习题与解答(含答案)

操作系统习题与解答(含答案)

操作系统习题与解答(含答案)第⼀章操作系统引论⼀、选择题1.在计算机系统中配置操作系统的主要⽬的是(),操作系统的主要功能是管理计算机系统中的(),其中包括()管理和()管理,以及设备管理和⽂件管理。

这⾥的()管理主要是对进程进⾏管理。

(1)A.增强计算机系统的功能;B.为了使⽤系统的资源;C.提⾼系统的运⾏速度;D.提⾼系统使⽤效率,合理地组织系统的⼯作流程,以提⾼系统吞吐量。

(2)A.程序和数据;B.进程;C.资源;D.作业;E.任务。

(3)(4)A.存储器;B.虚拟存储器;C.运算器;D.处理机;E.控制器。

2.操作系统有多种类型:(1)允许多个⽤户以交互⽅式使⽤计算机的操作系统,称为();(2)允许多⽤户将若⼲个作业提交给计算机系统集中处理的操作系统称为();(3)在()的控制下,计算机系统能及时处理由过程控制反馈的数据,并做出响应。

A.批处理操作系统;B.分时操作系统;C.实时操作系统;D.微机操作系统;E.多处理机操作系统。

3.在下列性质中,哪⼀个不是分时系统的特征。

()A.交互性B.多路性C.成批性D.独占性4.实时操作系统追求的⽬标是()。

A.⾼吞吐率B.充分利⽤内存C.快速响应D.减少系统开销5. 现代操作系统的两个基本特征是()和资源共享A.多道程序设计B.中断处理C.程序的并发执⾏D.实现分时与实时处理6.引⼊多道程序的⽬的在于()。

A.有利于代码共享,减少主、辅存信息交换量。

B.提⾼实时响应速度。

C.充分利⽤CPU,减少CPU等待时间D.充分利⽤存储器7.操作系统是⼀组().A.⽂件管理程序B.中断处理程序C.资源管理程序D.设备管理程序8.()不是操作系统关⼼的主要问题.A.管理计算机裸机B.设计、提供⽤户程序与计算机硬件系统的界⾯C.管理计算机系统资源D.⾼级程序设计语⾔的编译器9.⽤户在⼀次计算机过程中,或者⼀次事物处理中,要求计算机完成所做的⼯作的集合,这是指().A.进程B.程序C.作业D.系统调⽤10.订购机票系统处理各⾃各个终端的服务请求,处理后通过终端回答⽤户,所以它是⼀个()。

操作系统习题及参考答案

操作系统习题及参考答案

操作系统习题及参考答案一、选择题1. 操作系统的主要功能是什么?A. 提供用户接口B. 管理和分配计算机的资源C. 控制和管理硬件设备D. 执行用户程序参考答案:B. 管理和分配计算机的资源2. 下列哪项不属于操作系统的资源管理功能?A. 内存管理B. 磁盘管理C. 进程管理D. 数据库管理参考答案:D. 数据库管理3. 在进程调度算法中,哪种调度算法能够保证所有进程公平获得CPU时间片?A. 先来先服务(FCFS)B. 短作业优先(SJF)C. 轮转调度(RR)D. 优先级调度参考答案:C. 轮转调度(RR)4. 下列哪个不是进程状态之一?A. 运行态B. 就绪态C. 阻塞态D. 结束态参考答案:D. 结束态5. 现代操作系统采用的存储管理方式是?A. 分页式存储管理B. 段式存储管理C. 段页式存储管理D. 段和页混合存储管理参考答案:C. 段页式存储管理二、填空题1. 在分页式存储管理中,操作系统将内存和磁盘划分为固定大小的______________。

参考答案:页(page)2. 进程在执行过程中如果发生了某些事件而无法继续执行,则进程会进入阻塞态,也被称为______________。

参考答案:等待态3. 进程管理的主要任务之一是实现进程的______________,即从一个进程的执行转到另一个进程的执行。

参考答案:切换(调度)4. 虚拟内存的实现需要使用到的硬件机制是______________。

参考答案:页表5. 一般而言,操作系统的中断处理程序都是使用______________的方式实现的。

参考答案:中断向量表三、简答题1. 请简要说明操作系统的作用及其主要功能。

操作系统是计算机系统中的一个重要组成部分,起着控制和协调计算机硬件与软件资源,为用户提供接口的作用。

操作系统主要功能包括:- 资源管理:负责管理和分配计算机的各种资源,如 CPU、内存、磁盘等,以满足不同进程的需求。

操作系统-第3章习题解析

操作系统-第3章习题解析

操作系统-第3章习题解析第三章习题解析1.⾼级调度不低级调度的主要任务是什么?为什么要引⼊中级调度?答:⾼级调度的主要任务是根据某种算法,把外存上处于后备队列中的那些作业调⼊内存。

低级调度是保存处理机的现场信息,按某种算法先取进程,再把处理器分配给进程。

引⼊中级调度的主要⽬的是为了提⾼内存利⽤率和系统吞吐量。

使那些暂时不能运⾏的进程不再占⽤内存资源,将它们调⾄外存等待,把进程状态改为就绪驻外存状态或挂起状态。

2.处理机调度算法的共同⽬标是什么?批处理系统的调度⽬标⼜是什么?答:共同⽬标:资源利⽤率,公平性,平衡性,策略强制执⾏。

批处理系统的调度⽬标:平均周转时间短,系统吞吐量⾼,处理机利⽤率⾼。

3.何谓作业、作业步和作业流?答:作业包含通常的程序和数据,还配有作业说明书。

系统根据该说明书对程序的运⾏进⾏控制。

批处理系统中是以作业为基本单位从外存调⼊内存。

作业步是指每个作业运⾏期间都必须经过若⼲个相对独⽴相互关联的顺序加⼯的步骤。

作业流是指若⼲个作业进⼊系统后依次存放在外存上形成的输⼊作业流;在操作系统的控制下,逐个作业进程处理,于是形成了处理作业流。

4.在什么情况下需要使⽤作业控制块JCB?其中包含了哪些内容?答:每当作业进⼊系统时,系统便为每个作业建⽴⼀个作业控制块JCB,根据作业类型将它插⼊到相应的后备队列中。

JCB包含的内容通常有:1)作业标识 2)⽤户名称 3)⽤户账户 4)作业类型(CPU繁忙型、I/0芳名型、批量型、终端型) 5)作业状态 6)调度信息(优先级、作业已运⾏) 7)资源要求 8)进⼊系统时间 9)、开始处理时间 10)作业完成时间 11)作业退出时间 12)资源使⽤情况等5.在作业调度中应如何确定接纳多少个作业和接纳哪些作业?答:作业调度每次接纳进⼊内存的作业数,取决于多道程序度。

应将哪些作业从外存调⼊内存,取决于采⽤的调度算法。

最简单的是先来服务调度算法,较常⽤的是短作业优先调度算法和基于作业优先级的调度算法。

计算机操作系统课后习题答案解析张尧学

计算机操作系统课后习题答案解析张尧学

第一章绪论1.什么是操作系统的基本功能?答:操作系统的职能是管理和控制汁算机系统中的所有硬、软件资源,合理地组织计算机工作流程,并为用户提供一个良好的工作环境和友好的接口。

操作系统的基本功能包括:处理机管理、存储管理、设备管理、信息管理(文件系统管理)和用户接口等。

2.什么是批处理、分时和实时系统?各有什么特征?答:批处理系统(batchprocessingsystem):操作员把用户提交的作业分类,把一批作业编成一个作业执行序列,由专门编制的监督程序(monitor)自动依次处理。

其主要特征是:用户脱机使用计算机、成批处理、多道程序运行。

分时系统(timesharingoperationsystem):把处理机的运行时间分成很短的时间片,按时间片轮转的方式,把处理机分配给各进程使用。

其主要特征是:交互性、多用户同时性、独立性。

实时系统(realtimesystem):在被控对象允许时间范围内作出响应。

其主要特征是:对实时信息分析处理速度要比进入系统快、要求安全可靠、资源利用率低。

3.多道程序(multiprogramming)和多重处理(multiprocessing)有何区别?答;多道程序(multiprogramming)是作业之间自动调度执行、共享系统资源,并不是真正地同时值行多个作业;而多重处理(multiprocessing)系统配置多个CPU,能真正同时执行多道程序。

要有效使用多重处理,必须采用多道程序设计技术,而多道程序设计原则上不一定要求多重处理系统的支持。

6.设计计算机操作系统时与那些硬件器件有关运算器、控制器、存储器、输入设备、输出设备第二章作业管理和用户接口2.作业由哪几部分组成?各有什么功能?答:作业由三部分组成:程序、数据和作业说明书。

程序和数据完成用户所要求的业务处理工作,作业说明书则体现用户的控制意图。

3.作业的输入方式有哪几种?各有何特点答:作业的输入方式有5种:联机输入方式、脱机输入方式、直接耦合方式、SPOOLING(Simultaneous Peripheral OperationsOnline)系统和网络输入方式,各有如下特点:(1)联机输入方式:用户和系统通过交互式会话来输入作业。

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吃水果问题
问题描述:桌上有一只盘子,每次只能放一个水果,爸爸专向盘中放苹果,妈妈专向盘 中放桔子,儿子专等吃盘里的桔子,女儿专等吃盘里的苹果。只要盘子空,则爸爸或妈 妈可向盘中放水果,仅当盘中有自己需要的水果时,儿子或女儿可从中取出,请给出四 人之间的同步关系,并用P、V操作实现四人正确活动的程序。
习题选讲与解析
1
一、选择题 1.在计算机系统中配置操作系统的主要目的是(A)。操作系统的主要功能是管理
计算机系统中的(B),其中包括(C)、(D),以及文件和设备。这里的(C)管 理主要是对进程进行管理。 A:(1)增强计算机系统的功能;(2)提高系统资源的利用率;(3)提高系统的运
行速度;(4)合理组织系统的工作流程,以提高吞吐量。 B:(1)程序和数据;(2)进程;(3)资源;(4)作业;(5)软件;(6)硬件。 C,D:(1)存储器;(2)虚拟存储器;(3)运算器;(4)处理机;(5)控制器。
微机操作系统;(5)多处理机操作系统。
A:2 B:1 C:3 D:4
2
3.在设计分时操作系统时,首先要考虑的是(A);在设计批处理操作系统时,先 要考虑的是(B);在设计实时操作系统时,首先要考虑的是(C)。 A,B,C:(1)灵活性和可适应性;(2)交互性和响应时间;(3)周转时间和 系统吞吐量;(4)实时性和可靠性。
操作系统中所谓的”虚拟“,是指通过某种技术把一个物理实体变为若干个逻辑上的对应 物,相应的用于实现虚拟的技术称为虚拟技术。在操作系统中利用了多种虚拟技术分别用 来实现虚拟处理机、虚拟内存、虚拟外设和虚拟信道等。虚拟的实现主要是通过分时技术, 例如,多道程序系统中,通过分时技术来实现虚拟处理机;将一台物理处理机虚拟为多台 逻辑上的处理机,是靠多道程序分时地使用同一台物理处理机来实现的。微观上,该处理 机在每一时刻只运行一道程序,它们分时地运行;然而在宏观上,系统中确有几道程序在 同时运行,从而给用户的感觉是系统中同时同时有多台处理机在为其中的每一道程序服务, 显然用户所感觉到的处理机并不实际存在。
A:2 B:3 C:4
4.分时系统的响应时间(及时性)主要是根据(A)确定的,而试试系统的响应时间 则是由(B)确定的。 A,B:(1)时间片大小;(2)用户数目;(3)计算机运行速度;(4)用户所能 接受的等待时间;(5)控制对象所能接受的时延;(6)实时调度。
A:4 B:5
5.采用(A)结构时,将OS分成用于实现OS最基本功能的内核和提供各种服务的服 务器两个部分。通常,下列模块中必须包含在操作系统内核中的是(B)模块。 A:(1)整体式;(2)模块化;(3)层次式;(4)微内核。 B:(1)内存分配;(2)中断处理;(3)文件处理;(4)命令处理。
void son (void) { while(TRUE){ P(so); get an orange; V(s); eat an orange;} }
void daught (void) { while(TRUE){ P(sp); get an apple; V(s); eat an apple;} }
A:1 B:2 C:4 D:2
7.下面8个系统中,必须是实时操作系统的有( ) A计算机辅助设计系统;B 航空定票系统;C 过程控制系统;D 机器翻译系统;E 办公自动化 系统;F 计算机激光照排系统;G情报检索系统;H导弹的制导系统
ABCGH
二、 简答题 在操作系统中实现虚拟的关键技术是什么?并加以说明。
4
试从交互性、及时性以及可靠性三个方面,比较分时系统与实时系统。P11
已知一个求值公式(A2+3B)/(B+5A),若A,B已赋值,试画 出该公式求值过程的前趋图,并用信号量解决公式的求解过程。
开始
S1:x1=A*A S2:x2=3*B S3:x3=5*A
S4:x4=x1+x2
S5:x5=B+x3
struct semaphore s,sp,so=1,0,0; cobegin void father (void)
{ while(TRUE){
have an apple; P(s); put an apple; V(sp); } }
void mother (void) { while(TRUE){ have an orange; P(s); put an orange; V(so); } }
S6:x6=x4/x5
S1
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结束
S6
S3 S5
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struct semaphore a,b,c,d,e,=0,0,0,0,0;
cobegin
{S1;V(a);}
{S2;V(b);}
{S3;V(c) ;}
S1
{P(a);P(b);S4;V(d)} a
S2
b
S3
c
{P(c);S5;V(e);} {P(d);P(e);S6);} coend
coend
7
Plate 1 爸私 applempty 盘中无苹果 1 女私applefull盘中有苹果0 妈私orangempty无桔子1 儿私orangefull有桔子0


{
{
p(plate);
P(applempty); p(applefull);
放苹果;
P(plate);
V(applefull); 取苹果;
A:2 B:3 C:4 D:1
2.操作系统有多种类型:允许多个用户以交互方式使用计算机的操作系统称为(A); 允许多个用户将若干个作业提交给计算机系统集中处理的操作系统称为(B);在 (C)的控制下,计算机系统能及时处理由过程控制反馈的数据,并做出响应;在IBM -PC机上的操作系统称为(D)。 A,B,C,D:(1)批处理操作系统;(2)分时操作系统;(3)实时操作系统;(4)
A:4S是(A)操作系统,Windows95是(B)操作系统,WindowsXP是(C) 操作系统,它们都是由(D)开发的。
A,B,C:(1)单用户单任务;(2)单用户多任务;(3)多用户单任务;(4)多用户多任务。 D:(1)IBM公司;(2)Microsoft公司;(3)Microsoft和IBM联合;(4)Bell实验室;
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