2011年2009级《操作系统》-Test1-试题及答案

北京邮电大学2011--2012学年第一学期计算机科学与技术学院(09405-08408)“Operating Systems” Test（1）1. (10 points) Describe the principle of signalsAnswers:Unix、Linux等操作系统提供的一种基于事件驱动的进程间通信机制。

当系统内发生某种特定事件时，OS将产生1个信号，并以信号方式通知特定进程；当进程收到信号后，将转入内核态下的信号处理程序，以响应所发生的事件，对其进行处理。

2. (10 points) Describe the principle of pipe-based inter-process communications. Answer:是一种OS提供的基于共享文件的进程间通信方式。

2个需要进行通信的进程共享1个称为管道的文件，发送方通过执行“写文件”系统调用向管道中写入数据，接收方通过“读文件”系统调用从管道中读取数据，传递的数据为无格式字节流，由此实现2个进程间数据量通信。

OS提供了管道的创建/打开/关闭/读写等原语。

或（书上）：相互通信进程由1个共享文件——管道连接。

管道是用于连接读进程和写进程、实现它们之间通信的软件，向管道提供输入的发送进程以字符流形式将数据批量写入管道，接收进程从管道中读取数据。

3. (25 points) In a computer system, the users submit to the system their computational tasks as jobs, and all these jobs are then stored as the standby jobs on the disk.The job scheduler (also known as long-term scheduler) selects the standby jobs on the disk, creates new processes in memory for them, and then starts executing these processes. Each job’s ID is the same as that of the process created for it, for example, J i and P i.When the number of concurrent processes in memory is lower than three, the job scheduler takes the FCFS algorithm to select a standby job on the disk to create a new process. Otherwise, the processes should wait in the disk.For the processes in memory, the process scheduler (also known as short-term schedule r) takes the non-preemptive priority-based algorithm to select a process and allocates the CPU to it.It is assumed the system costs resulting from job and process scheduling are omitted.Consider the following set of Jobs J1, J2, J3 , J4 and J5. For 1≤ i≤5, the arrival time of each J i, the length of the CPU burst time of each process P i, and the priority number for each J i/P i are given as below, and a smaller priority number implies a higher priority.Job Arrival Time Burst Time Priority Number(minute)J114:00 40 4J214:20 30.01 2J314:30 50.01 3J414:50 20.01 5J515:05 10.01 5(1)Illustrate the execution of each job/process by charts.(2)What is the turnaround time of each job?(3)What is the waiting time of each job?Note: The waiting time of a job includes the time it waits on the disk and that it waits in memory.Answer:答案：（1）注：图中J i部分表示作业被调入内存，P i表示进程被调度执行。

J1到达时，并发进程数<3，作业被直接调入内存，创建进程P1；P1被调度程序选中，开始执行。

J2到达时，并发进程数<3，作业被直接调入内存，创建进程P2；此时P1在执行，因此P2处于就绪等待状态。

P1结束后，系统内有2个作业J1、J2对应的进程P1、P2，优先级高的进程P2开始执行。

J3到达时，并发进程数=2<3，作业被直接调入内存，创建进程P3，等待P2执行完毕。

P2执行完毕后，系统内有2个作业J3、J4对应的进程P3、P4，优先级高的进程P3被调度程序选中，开始执行。

J4到达时，并发进程数=2<3，作业被直接调入内存，创建进程P4，等待P3执行完毕。

P3执行完毕后，系统内有2个作业J4、J5对应的进程P4、P5，2者的优先级相同，按照FCFS原则，P4先执行。

J5到达时，并发进程数=3，必须等到P2结束后，系统内并发进程数<3，方能创建进程P5。

等到P4执行结束后，P5开始执行。

（2）J1：T1 = 40 （min）J2：T2 = 20 + 30.01 = 50.01 （min）J3：T3 = 40.01 + 50.01 = 90.02 （min）J4：T4 = 70.02 + 20.01 = 90.03 （min）J5：T5 = 5.01 + 70.02 + 10.01 = 85.04（min）（3）J1：W1 = 0 （min）J2：W2 = 20（min）J3：W3 = 40.01（min）J4：W4 = 70.02（min）J5：W5 =75.03（min）4-1. (25 points) There is a coop (笼子) that can hold only one tiger or two pigs. If one pig is already kept in it, another pig is allowed to be in it, but a tiger is not; if one tiger is in the coop, other tigers or pigs are not permitted to be in it.A tiger-hunter will hunt a tiger at a time and need to put the tiger into the coop; a pig-hunter will hunt a pig at a time and need to put the pig into the coop.Then the feeder (饲养员) will take out the tiger from the coop and send it to a park. The kitchener (厨师) will take out one pig from the coop at a time, and then send it to a restaurant.The process structures for the tiger-hunter, the pig-hunter, the feeder, and the kitchener are shown as follows.tiger-hunter:while(true){Catch a tiger;(1)Put the tiger into the coop;(2)} feeder:while(true){(3)Take the tiger from the coop;Send the tiger to the park;(4)}pig-hunter:while(true){Catch a pig;(5)Put the pig into the coop;(6)} kitchener:while(true){(7)Take a pig from the coop;Send the pig to restaurant;(8)}Please design semaphores and wait and signal operations on the semaphores to synchronize the above-mentioned processes.It is required that:1) definitions and initial values of the semaphores should be given, and2) appropriate code sections should be described for the blanks marked by numbersfrom (1) to (8).Answers：Binary SemaphoresSCOOP=1; used for mutual exclusion use of the coop.STIGER=0; used for synchronization between the process tiger-hunter andfeeder.SPIG=0; used for synchronization between the process pig-hunter and kitchener.MUTEX=1; used for mutual exclusion operation on variable pigcount. SemaphorePIGROOM=2; used to record the rooms left for keeping pigs.V ARIABLEPigcount=0; used to record the number of pigs kept in the coop.CODE SECTIONS:(1) wait(SCOOP);(2) signal(STIGER);(3) wait(STIGER);(4) signal(SCOOP);(5) wait(PIGROOM);wait(MUTEX);pigcount++;if (pigcount==1) wait(SCOOP);signal(MUTEX);(6) signal(SPIG);(7) wait(SPIG);(8) wait(MUTEX);pigcount--;if (pigcount==0) signal(SCOOP);signal(MUTEX);signal(PIGROOM);4-2 (25 points) As illustrated in the figure, on the two sides of a one-plank bridge(独木桥), there are two groups of soldiers that are composed of m and n people respectively and need to cross the bridge, but the narrow bridge allows only one group of the soldiers in the same direction to cross at the same time. One group of the soldiers is permitted to cross as long as there are no people on the bridge. Once one group of the soldiers begins walking on the bridge, the other group should be waiting to start crossing until all members of the first group have passed the bridge.Please design two semaphore-based processes to describe the crossing actions of the soldiers in the two groups. It is required(1) to define the semaphores and variables needed, explain their roles?, and give theirinitial values; and(2) to illustrate the structures of processes for the soldiers in each group.Answers:要求：完整的答案包括3部分1)信号量的定义和含义，定义中需区分二元、多元信号量2)信号量的初始化3)基于信号量的过程该问题可归结为读写者问题，但2队士兵均为读者。