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OS考研模拟试题8及答案

计算机操作系统模拟题八

一．选择题（在下列各小题的备选答案中，请把你认为正确答案的题号，填入题干后的括号内。多选、少选及选错不给分。每题3分，共15分）

1．分时操作系统需要使用下面哪些成份。（）

①多道程序设计技术②作业说明书

③终端命令解释程序④中断处理

⑤优先级调度⑥系统调用

2．进程具有哪些特性。（）

①动态性②共享性③并发性④相互制约性⑤独立性⑥静态性

3. 在页式虚存管理系统中，若常发生抖动影响CPU的利用率，从系统管理员的角度，则下面哪些方法可改善CPU的利用率。（）

①用一个更快的CPU ②用一个更大的辅存③减少多道程序的道数

④增加多道程序的道数⑤增大主存⑥采用更快的I/O设备

4．在文件系统中，为实现文件保护一般应采用下面哪些方法。（）

①口令②密码③访问控制④复制⑤在读写文件之前使用OPEN系统调用

⑥在读写文件之后使用CLOSE系统服务

5. 从资源分配角度，操作系统把外部设备分为( )

①独占型设备②共享型设备③快速型设备④慢速性设备

⑤块设备⑥字符型设备⑦虚拟设备

二、（9分）对访问串:1,2,3,4,1,2,5,1,2,3,4,5, 指出在驻留集大小分别为3,4时，使用FIFO和LRU替换算法的页故障数。结果说明了什么？

三．（8分）简述文件的二级目录组织形式。欲实现文件共享如何处理？

四．（8分）假设有5道作业，它们的提交时间及运行时间由下表给出：

作业提交时间（时）运行时间（小时）

1 10 2

2 10．05 1

3 10．25 0．75

4 12．2

5 0．5

5 12．5 0．25

若采用FCFS和SJF两种调度算法，指出作业以单道串行方式运行时的被调度顺序及平均周转时间。

五．（10分）设有如下图所示的工作模型。

四个进程P0，P1,P2,P3和四个信箱M0,M1,M2,M3进程间借助相邻的信箱传递消息：每次从中取出一条消息，经加工送入中。其中M0,M1,M2,M3分别设有3,3,2,2个格子，每个格子放一条消息，初始时，M0装满了三条消息，其余为空。写出使用信号量实现进程（i=0,1,2,3）同步及互斥的流程。

六．（10分）设系统中仅有一类数量为M的独占型资源，系统中N个进程竞争该类资源，其中各进程对该类资源的最大需求量为W。当M、N、W分别取下列值时，试判断哪些情况会发生死锁？为什么？

① M=2，N=2，W=1 ②M=3，N=2，W=2 ③M=3，N=2，W=3

④M=5，N=3，W=2 ⑤M=6，N=3，W=3

参考答案

一．选择题（每题3分，共15分）

1．（①②④⑥）

2．（①③④⑤）

3.（③）

4．（①②③④）

5.(①②⑦)

二、当驻留集为3时，采用FIFO替换算法，页面故障数为9次；采用LRU替换算法时，页面故障数为10次。

当驻留集为4时，采用FIFO替换算法，页面故障数为10次；采用LRU替换算法时，页面故障数为8次。

结果表明，FIFO替换算法的故障数不随驻留集增大而减少；而LRU算法的故障数随驻留集增大而减少。

三．把记录文件的目录分成主文件目录和由其主管的若干个子目录，各子目录的位置由主目录中的一项指出。应用中常设一个主文件目录，而为系统中每一个用户设立一张主文件目录MFD，每个用户的所有文件均设立一个用户文件目录UFD，作为MFD中的一项。用以描述UFD

的文件名和物理位置，即UFD是用户全部文件的文件控制块的全体。

在二级文件目录中，欲共享文件需给出一个文件的全路径名。由系统从根目录开始检索；或者用户将其当前目录指向另一用户的子目录上，以实现共享访问。

四．采用FCFS调度算法的被调度顺序为1à2à3à4à5

平均周转时间为T ＝（T1+T2+T3+T4+T5）/ 5 = (2+2.95+3.5+2+2) / 5 =2.49 (小时 )

采用SJF调度算法的被调度顺序为1à3à5à4à2

平均周转时间为T=T1+T2+T3+T4+T5）/ 5 = (2 +2.5 +0.5 +1.25 + 4.45 ) / 5 =2.14（小时）

五．定义如下公共信号量：

mutex0 ~ mutex3 : 分别用于控制互斥访问M0 ~ M 3，初值为1。

full0 ~ full3 : 分别用于控制同步访问M0 ~ M3 ,其中full0 初值为3，full1 ~ full3 初值为0，表示信箱中消息条数。

empty0 ~ empty3 : 分别用于同步控制对M0 ~ M3的访问。Empty0初值为0，empty2~ empty3初值为2，empty1初值为3，分别用于表示信箱中空格子个数。

另用send ( Mi , message )表示将消息送到（Mi mod 4）号信箱中；而用receive ( Mi，message )表示接收已存在于( Mi mod 4 )中的消息。

则使用信号量实现进程Pi (i = 0 , 1 ,2 ,3 )同步及互斥的流程如下：

mutex0 , m utex 1, m utex2 , m utex3 : semaphore ;

full0 , ful l1 , ful l2 , ful l3 : semaphore ;

empty0 , em pty1 , em pty2 , em pty3 : semaphore ;

begin

mutex0 : = 1 ; mutex1 : = 1 ; mutex2 : = 1 ; mutex : = 1 ;

full0 : = 3 ; full1 : = 0 ; full2 : = 0 ; full3 : = 0 ;

empty0 : = 0 ; empty1 : = 3 ; empty2 : = 2 ; empty3 : = 2 ;

Parbegin

P0：begin

repeat

P ( mutex0 ) ;

P ( full0 ) ;

Receive ( M0,message);

V (empty0 ) ;

Processing the message until finished;

P ( mutex1 ) ;

P ( empty1 ) ;

Send ( M1,message ) ;

V ( full1 ) ;

V ( mutex1 ) ;

Until false ;

…

end ;

P1：｛可类似于P0实现之｝;

P2：｛可类似于P0实现之｝;

P3：｛可类似于P0实现之｝;

Parend ;

End;

六．③可能会发生死锁。只要一个进程占用了少于3个独占型资源而另一个进程占用了其余的独占型资源，两个进程都会相互处于等待对方进程释放资源的状态。

⑤也可能会发生死锁。当每个进程都分配了两个资源时，3个进程都会彼此等待。

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