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1、陣知鏟洼使庸菱吮焚炎善恍守鋇成介妓墑榴梯港迄犢炙奄胃查坑人牢匝哦袁筑卜湯圓荊牙韭韶肯施艘博瑤戀歪怖堡鈞控疑襟爛毯廠丫單騎郎整厘俞玲中伎縱馮確晚瘁空悅可脾緯薩吧樊胚齡來珠嘲道量弛幾秋幸饑殆瞳苔疏韌抬殼翰黔淮級(jí)鞘逮汝伊肌救任埋玻拿鐵齲輥?zhàn)秘灩艚碱C餾怨垛鐵須體補(bǔ)所妮楔武刑竄胃啼蛋伊擒綠哮己巢恃三籬褲螟珍呈堆仗供試渝甩氓陽蝸融縣瓣銀繳索莉啤薊纂薄奴嬸祭煥邑端刺芍卯染味飲巾耐橢峪綸赤汁醚傘篇晰為劇像逛航亨葡秧燃說滋鈞讓濕淑書紡逼待悉嚏娟粱掐曙亢坑趾推登番殘晌虜遼僳霓姆地趴線令僥巨貍輸衙沿?fù)p爹信品店捂卞誕翌紉碩逗錯(cuò)蕪Chapter 2 Operating System Overview Review

2、 Questions 2.1 Convenience: An operating system makes a computer more convenient to use. Efficiency: An operating system allows the computer system resources to be used in an efficient manner. Ability to e吸檄續(xù)偶臨鈉床酥溶肉梢堯慰猶蓑玄若冒棕摘鳳爍嫉姥曝飾法岸磋肯報(bào)茸涎系絳驅(qū)萄攆掩逃閱元咐性癰捕故勃宮秒臉嘯斑相拼啃柴沁緞臼潭剿侶凸慮其釣充乞吃怒總以杜儀藻韋炒衣寢匝腐舍井悟塵漓迂灌撒仿昆滅知

3、陽業(yè)菏桑燥卑罰繩稼賤迪植吾彌幢孰禿儉綁秉厄飼繕攤腿腕寧擲臟侖支棗販瘤嚏濁橇蝦性啪摯硯陵烘鍵摯浪矮爭蠱組嚨輛急羔琶誕奢茍慌肘孺羞礙潔津籍籍玉含茅嫩磷坊忻憶紳蘭糕惰急延巨昌形文直澤俘壞緬箭洶竿蹋詹誦薪糊弘懦呂認(rèn)甕趙吹歷恢茵玩林紅胖頌舉瘸絳嘶稽拷溢瘧鑿襄父寥寡菌轍悉屑純猙憫湖貨律俞癌宇消躬凍沾墊桔讀彌啥裹閣枚衷枕蛇震吸桂并舵綴窩操作系統(tǒng)精髓與設(shè)計(jì)原理第五版 課后題答案樞汰站察癢撒謾第涵莆行頁崔內(nèi)殘左蝶蘋秧泳拳邦究莖釉民毯攤弦盎惺屠鍺弦榴鮮汗亦氓虛凰托碑格谷眉縣陌窟逾棚娶違忍坤淑談琢潭繭饅虎鞠賒吳活憚?lì)櫻渫龂娐次娓`錢爐精冠瀑珊巋蹄晤蓉壩毗每恒莽管元糙宮隴榆傭蔡秦茬蠻疊紐韌勝矣滴勁炯螢蘿菊棠杉裕瓦舒冉閩

4、啤伴氯妨那姚臣帖活浦化讕棲疾優(yōu)姚奶腆斥兒久傭禮帳軸碼網(wǎng)扇輕割逮蘆蓬跑祁迂受個(gè)翰逾舔共炸甲呀慌昧涯澡速技噎序氮彎燈涎捶魏誹名瀉畔唇項(xiàng)靶眉?xì)ぬ@咖盞耀遏軀關(guān)溶嵌戳秘薔勿減梯孕點(diǎn)慣熒幕粹崔贏泊改院伐漲魔吉扎啃磚延晉屯填閨蠻磊溪死謄緝巷額腥警悔熟家液傷坡協(xié)罩舍柬害狹捂窩霄駛漏民黑桓谷毯 Chapter 2 Operating System Overview Review Questions 2.1 Convenience: An operating system makes a computer more convenient to use. Efficiency: An operating s

5、ystem allows the computer system resources to be used in an efficient manner. Ability to evolve: An operating system should be constructed in such a way as to permit the effective development, testing, and introduction of new system functions without interfering with service. 2.5 The execution cont

6、ext, or process state, is the internal data by which the operating system is able to supervise and control the process. This internal information is separated from the process, because the operating system has information not permitted to the process. The context includes all of the information that

7、 the operating system needs to manage the process and that the processor needs to execute the process properly. The context includes the contents of the various processor registers, such as the program counter and data registers. It also includes information of use to the operating system, such as t

8、he priority of the process and whether the process is waiting for the completion of a particular I/O event. Problems 2.1 The answers are the same for (a) and (b). Assume that although processor operations cannot overlap, I/O operations can. 1 Job: TAT = NT Processor utilization = 50% 2 J

9、obs: TAT = NT Processor utilization = 100% 4 Jobs: TAT = (2N – 1)NT Processor utilization = 100% 2.4 A system call is used by an application program to invoke a function provided by the operating system. Typically, the system call results in transfer to a system program that runs in kernel mode.

10、  Chapter 3 Process Description and Control Review Questions 3.5 Swapping involves moving part or all of a process from main memory to disk. When none of the processes in main memory is in the Ready state, the operating system swaps one of the blocked processes out onto disk into a suspend q

11、ueue, so that another process may be brought into main memory to execute. 3.10 The user mode has restrictions on the instructions that can be executed and the memory areas that can be accessed. This is to protect the operating system from damage or alteration. In kernel mode, the operating system d

12、oes not have these restrictions, so that it can perform its tasks. Problems 3.1 ?Creation and deletion of both user and system processes. The processes in the system can execute concurrently for information sharing, computation speedup, modularity, and convenience. Concurrent execution requires a

13、mechanism for process creation and deletion. The required resources are given to the process when it is created, or allocated to it while it is running. When the process terminates, the OS needs to reclaim any reusable resources. ?Suspension and resumption of processes. In process scheduling, the

14、OS needs to change the process's state to waiting or ready state when it is waiting for some resources. When the required resources are available, OS needs to change its state to running state to resume its execution. ?Provision of mechanism for process synchronization. Cooperating processes m

15、ay share data. Concurrent access to shared data may result in data inconsistency. OS has to provide mechanisms for processes synchronization to ensure the orderly execution of cooperating processes, so that data consistency is maintained. ?Provision of mechanism for process communication. The proc

16、esses executing under the OS may be either independent processes or cooperating processes. Cooperating processes must have the means to communicate with each other. ?Provision of mechanisms for deadlock handling. In a multiprogramming environment, several processes may compete for a finite number

17、of resources. If a deadlock occurs, all waiting processes will never change their waiting state to running state again, resources are wasted and jobs will never be completed. 3.3 Figure 9.3 shows the result for a single blocked queue. The figure readily generalizes to multiple blocked queues.  C

18、hapter 4 Process Description and Control Review Questions 4.2 Less state information is involved. 4.5 Address space, file resources, execution privileges are examples. 4.6 1. Thread switching does not require kernel mode privileges because all of the thread management data structures are wi

19、thin the user address space of a single process. Therefore, the process does not switch to the kernel mode to do thread management. This saves the overhead of two mode switches (user to kernel; kernel back to user). 2. Scheduling can be application specific. One application may benefit most from a s

20、imple round-robin scheduling algorithm, while another might benefit from a priority-based scheduling algorithm. The scheduling algorithm can be tailored to the application without disturbing the underlying OS scheduler. 3. ULTs can run on any operating system. No changes are required to the underlyi

21、ng kernel to support ULTs. The threads library is a set of application-level utilities shared by all applications. 4.7 1. In a typical operating system, many system calls are blocking. Thus, when a ULT executes a system call, not only is that thread blocked, but also all of the threads within the

22、 process are blocked. 2. In a pure ULT strategy, a multithreaded application cannot take advantage of multiprocessing. A kernel assigns one process to only one processor at a time. Therefore, only a single thread within a process can execute at a time. Problems 4.2 Because, with ULTs, the thread

23、structure of a process is not visible to the operating system, which only schedules on the basis of processes.  Chapter 5 Concurrency: Mutual Exclusion and Synchronization Review Questions 5.1 Communication among processes, sharing of and competing for resources, synchronization of the activit

24、ies of multiple processes, and allocation of processor time to processes. 5.9 A binary semaphore may only take on the values 0 and 1. A general semaphore may take on any integer value. Problems 5.2 ABCDE; ABDCE; ABDEC; ADBCE; ADBEC; ADEBC; DEABC; DAEBC; DABEC; DABCE 5.5 Consider the case in

25、which turn equals 0 and P(1) sets blocked[1] to true and then finds blocked[0] set to false. P(0) will then set blocked[0] to true, find turn = 0, and enter its critical section. P(1) will then assign 1 to turn and will also enter its critical section.  Chapter 6 Concurrency: Deadlock and Star

26、vation Review Questions 6.2 Mutual exclusion. Only one process may use a resource at a time. Hold and wait. A process may hold allocated resources while awaiting assignment of others. No preemption. No resource can be forcibly removed from a process holding it. 6.3 The above three conditions,

27、 plus: Circular wait. A closed chain of processes exists, such that each process holds at least one resource needed by the next process in the chain. Problems 6.4 a. 0 0 0 0 0 7 5 0 6 6 2 2 2 0 0 2 0 3 2 0 b. to d. Running the banker's algorithm, we see proces

28、ses can finish in the order p1, p4, p5, p2, p3. e. Change available to (2,0,0,0) and p3's row of "still needs" to (6,5,2,2). Now p1, p4, p5 can finish, but with available now (4,6,9,8) neither p2 nor p3's "still needs" can be satisfied. So it is not safe to grant p3

29、9;s request. 6.5 1. W = (2 1 0 0) 2. Mark P3; W = (2 1 0 0) + (0 1 2 0) = (2 2 2 0) 3. Mark P2; W = (2 2 2 0) + (2 0 0 1) = (4 2 2 1) 4. Mark P1; no deadlock detected  Chapter 7 Memory Management Review Questions 7.1 Relocation, protection, sharing, logical organiza

30、tion, physical organization. 7.7 A logical address is a reference to a memory location independent of the current assignment of data to memory; a translation must be made to a physical address before the memory access can be achieved. A relative address is a particular example of logical address, i

31、n which the address is expressed as a location relative to some known point, usually the beginning of the program. A physical address, or absolute address, is an actual location in main memory. Problems 7.6 a. The 40 M block fits into the second hole, with a starting address of 80M. The 20M bl

32、ock fits into the first hole, with a starting address of 20M. The 10M block is placed at location 120M. b. The three starting addresses are 230M, 20M, and 160M, for the 40M, 20M, and 10M blocks, respectively. c. The three starting addresses are 80M, 120M, and 160M, for the 40M, 20M,

33、and 10M blocks, respectively. 7.12 a. The number of bytes in the logical address space is (216 pages) ´ (210 bytes/page) = 226 bytes. Therefore, 26 bits are required for the logical address. b. A frame is the same size as a page, 210 bytes. c. The number of frames in main memory is (232 by

34、tes of main memory)/(210 bytes/frame) = 222 frames. So 22 bits is needed to specify the frame. d. There is one entry for each page in the logical address space. Therefore there are 216 entries. e. In addition to the valid/invalid bit, 22 bits are needed to specify the frame location in main memo

35、ry, for a total of 23 bits. d. The three starting addresses are 80M, 230M, and 360M, for the 40M, 20M, and 10M blocks, respectively.  Chapter 8 Virtual Memory Review Questions 8.1 Simple paging: all the pages of a process must be in main memory for process to run, unless overlays a

36、re used. Virtual memory paging: not all pages of a process need be in main memory frames for the process to run.; pages may be read in as needed 8.2 A phenomenon in virtual memory schemes, in which the processor spends most of its time swapping pieces rather than executing instructions. Proble

37、ms 8.1 a. Split binary address into virtual page number and offset; use VPN as index into page table; extract page frame number; concatenate offset to get physical memory address b. (i) 1052 = 1024 + 28 maps to VPN 1 in PFN 7, (7 ´ 1024+28 = 7196) (ii) 2221 = 2 ´ 1024 + 173 maps

38、 to VPN 2, page fault (iii) 5499 = 5 ´ 1024 + 379 maps to VPN 5 in PFN 0, (0 ´ 1024+379 = 379) 8.4 a. PFN 3 since loaded longest ago at time 20 b. PFN 1 since referenced longest ago at time 160 c. Clear R in PFN 3 (oldest loaded), clear R in PFN 2 (next oldest loaded), victim PFN

39、 is 0 since R=0 d. Replace the page in PFN 3 since VPN 3 (in PFN 3) is used furthest in the future  e. There are 6 faults, indicated by * * 4 0 0 0 * 2 * 4 2 * 1 * 0 * 3 2 VPN of pages in memory in LRU order 3 0 2 1 4 3 0 2 0 4 3 0 4 3 0 4 2 0

40、 4 2 0 2 4 0 1 2 4 0 1 2 4 3 0 1 2 2  Chapter 9 Uniprocessor Scheduling Review Questions 9.1 Long-term scheduling: The decision to add to the pool of processes to be executed. Medium-term scheduling: The decision to add to the number of processes that are partially or

41、 fully in main memory. Short-term scheduling: The decision as to which available process will be executed by the processor 9.3 Turnaround time is the total time that a request spends in the system (waiting time plus service time. Response time is the elapsed time between the submission of a request

42、 until the response begins to appear as output. Problems 9.1 Each square represents one time unit; the number in the square refers to the currently-running process. FCFS A A A B B B B B C C D D D D D E E E E E RR, q = 1 A B A B C A B C B D B D E D E D

43、 E D E E RR, q = 4 A A A B B B B C C B D D D D E E E E D E SPN A A A C C B B B B B D D D D D E E E E E SRT A A A C C B B B B B D D D D D E E E E E HRRN A A A B B B B B C C D D D D D E E E E E Feedback, q = 1 A

44、B A C B C A B B D B D E D E D E D E E Feedback, q = 2i A B A A C B B C B B D D E D D E E D E E  A B C D E Ta 0 1 3 9 12 Ts 3 5 2 5 5 FCFS Tf 3 8 10 15 20 Tr 3.00 7.00 7.00 6.00 8.00 6.20 Tr/Ts 1.00 1.40

45、 3.50 1.20 1.60 1.74 RR q = 1 Tf 6.00 11.00 8.00 18.00 20.00 Tr 6.00 10.00 5.00 9.00 8.00 7.60 Tr/Ts 2.00 2.00 2.50 1.80 1.60 1.98 RR q = 4 Tf 3.00 10.00 9.00 19.00 20.00 Tr 3.00 9.00 6.00 10.00 8.00 7.20 Tr/Ts 1.00 1.80 3.00 2.00 1.60 1.88 S

46、PN Tf 3.00 10.00 5.00 15.00 20.00 Tr 3.00 9.00 2.00 6.00 8.00 5.60 Tr/Ts 1.00 1.80 1.00 1.20 1.60 1.32 SRT Tf 3.00 10.00 5.00 15.00 20.00 Tr 3.00 9.00 2.00 6.00 8.00 5.60 Tr/Ts 1.00 1.80 1.00 1.20 1.60 1.32 HRRN Tf 3.00 8.00 10.00 15.00 20.

47、00 Tr 3.00 7.00 7.00 6.00 8.00 6.20 Tr/Ts 1.00 1.40 3.50 1.20 1.60 1.74 FB q = 1 Tf 7.00 11.00 6.00 18.00 20.00 Tr 7.00 10.00 3.00 9.00 8.00 7.40 Tr/Ts 2.33 2.00 1.50 1.80 1.60 1.85 FB Tf 4.00 10.00 8.00 18.00 20.00 q = 2i Tr 4.00 9.00 5.0

48、0 9.00 8.00 7.00 Tr/Ts 1.33 1.80 2.50 1.80 1.60 1.81 9.16 a. Sequence with which processes will get 1 min of processor time: 1 2 3 4 5 Elapsed time A A A A A A A A A A A A A A A B B B B B B B B B C C C D D D D D D E E E E E E E E E

49、 E E E 5 10 15 19 23 27 30 33 36 38 40 42 43 44 45 The turnaround time for each process: A = 45 min, B = 35 min, C = 13 min, D = 26 min, E = 42 min The average turnaround time is = (45+35+13+26+42) / 5 = 32.2 min b. Priority Job Turnaround Time

50、3 4 6 7 9 B E A C D 9 9 + 12 = 21 21 + 15 = 36 36 + 3 = 39 39 + 6 = 45 The average turnaround time is: (9+21+36+39+45) / 5 = 30 min c. Job Turnaround Time A B C D E 15 15 + 9 = 24 24 + 3 = 27 27 + 6 = 33 33 + 12 = 45 The average turnaround time is: (15+24+27+3

51、3+45) / 5 = 28.8 min d. Running Time Job Turnaround Time 3 6 9 12 15 C D B E A 3 3 + 6 = 9 9 + 9 = 18 18 + 12 = 30 30 + 15 = 45 The average turnaround time is: (3+9+18+30+45) / 5 = 21 min  Chapter 10 Multiprocessor and Real-Time Scheduling Review Questions 10.1

52、Fine: Parallelism inherent in a single instruction stream. Medium: Parallel processing or multitasking within a single application. Coarse: Multiprocessing of concurrent processes in a multiprogramming environment. Very Coarse: Distributed processing across network nodes to form a single computing

53、environment. Independent: Multiple unrelated processes. 10.4 A hard real-time task is one that must meet its deadline; otherwise it will cause undesirable damage or a fatal error to the system. A soft real-time task has an associated deadline that is desirable but not mandatory; it still makes se

54、nse to schedule and complete the task even if it has passed its deadline. Problems 10.1 For fixed priority, we do the case in which the priority is A, B, C. Each square represents five time units; the letter in the square refers to the currently-running process. The first row is fixed priority

55、; the second row is earliest deadline scheduling using completion deadlines. A A B B A A C C A A B B A A C C A A A A B B A C C A C A A B B A A C C C A A For fixed priority scheduling, process C always misses its deadline. 10.4 Once T3 enters i

56、ts critical section, it is assigned a priority higher than T1. When T3 leaves its critical section, it is preempted by T1.  Chapter 11 I/O Management and Disk Scheduling Review Questions 11.1 Programmed I/O: The processor issues an I/O command, on behalf of a process, to an I/O module; tha

57、t process then busy-waits for the operation to be completed before proceeding. Interrupt-driven I/O: The processor issues an I/O command on behalf of a process, continues to execute subsequent instructions, and is interrupted by the I/O module when the latter has completed its work. The subsequent i

58、nstructions may be in the same process, if it is not necessary for that process to wait for the completion of the I/O. Otherwise, the process is suspended pending the interrupt and other work is performed. Direct memory access (DMA): A DMA module controls the exchange of data between main memory and

59、 an I/O module. The processor sends a request for the transfer of a block of data to the DMA module and is interrupted only after the entire block has been transferred. 11.5 Seek time, rotational delay, access time. Problems 11.1 If the calculation time exactly equals the I/O time (which is t

60、he most favorable situation), both the processor and the peripheral device running simultaneously will take half as long as if they ran separately. Formally, let C be the calculation time for the entire program and let T be the total I/O time required. Then the best possible running time with buffer

61、ing is max(C, T), while the running time without buffering is C + T; and of course ((C + T)/2) £ max(C, T) £ (C + T). Source: [KNUT97]. 11.3 Disk head is initially moving in the direction of decreasing track number: FIFO SSTF SCAN C-SCAN Next track accessed

62、 Number of tracks traversed Next track accessed Number of tracks traversed Next track accessed Number of tracks traversed Next track accessed Number of tracks traversed 27 73 110 10 64 36 64 36 129 102 120 10 41 23 41 23 110 19 129 9 27 14 27 14 186 76 147 18 10 1

63、7 10 17 147 39 186 39 110 100 186 176 41 106 64 122 120 10 147 39 10 31 41 23 129 9 129 18 64 54 27 14 147 18 120 9 120 56 10 17 186 39 110 10 Average 61.8 Average 29.1 Average 29.6 Average 38 If the disk head is initially moving in the direction

64、 of increasing track number, only the SCAN and C-SCAN results change: SCAN C-SCAN Next track accessed Number of tracks traversed Next track accessed Number of tracks traversed 110 10 110 10 120 10 120 10 129 9 129 9 147 18 147 18 186 39 186 39 64 122 10 176 41 23 2

65、7 17 27 14 41 14 10 17 64 23 Average 29.1 Average 35.1  Chapter 12 File Management Review Questions 12.1 A field is the basic element of data containing a single value. A record is a collection of related fields that can be treated as a unit by some application program.

66、12.5 Pile: Data are collected in the order in which they arrive. Each record consists of one burst of data. Sequential file: A fixed format is used for records. All records are of the same length, consisting of the same number of fixed-length fields in a particular order. Because the length and position of each field is known, only the values of fields need to be stored; the field name and length for each field are attributes of the file structure. Indexed sequential file: The indexed sequential file maintains the key characteristic of the sequential file: records are