操作系统名词解释整理

==================================名词解释======================================

Operating system: operating system is a program that manages the computer hardware. The operating system is the one program running at all times on the computer (usually called the kernel), with all else being systems programs and application programs.

操作系统：操作系统一个管理计算机硬件的程序，他一直运行着，管理着各种系统资源Multiprogramming: Multiprogramming is one of the most important aspects of operating systems. Multiprogramming increases CPU utilization by organizing jobs (code and data) so that the CPU always has one to execute.

多程序设计：是操作系统中最重要的部分之一，通过组织工作提高CPU利用率，保证了CPU始终在运行中。batch system: A batch system is one in which jobs are bundled together with the instructions necessary to allow them to be processed without intervention.

批处理系统：将许多工作和指令捆绑在一起运行，使得它们不必等待插入，以此提高系统效率。

time sharing is a logical extension of by switching among them, but the

switches occur so frequently that the users can interact with each program while it is running.

分时系统：是对多道程序设计的逻辑扩展。通过在多个作业间转换CPU，可以同时执行多个作业，但是这种转换发生的如此频繁以至于用户在程序运行的同时可以与计算机交互。

A real-time system has well-defined, fixed time constraints. Processing must be done within the defined constraints, or the system will fail.

实时系统：实时系统具有明确定义的、不变的时间约束。处理过程必须在规定的时间内完成，否则系统就失效了。

it allows devices to transfer data without subjecting the CPU to a heavy overhead.

直接内存访问：一种不经过CPU而直接从内存存取数据的数据交换模式。

System call provide the interface between a running program and the operating system.

系统调用：系统调用为运行中的程序和操作系统提供了一个接口。

nonessential components from the kernel and implementing them as system and user-level

programs.

微内核：微内核将所有不必要的组件从内核中去掉，并将它们作为系统和用户层程序来实现。

It is program in execution。

进程：一个正在运行的程序

PCB: each process is represented in the operating system by a process control block (PCB). A PCB contains many pieces of information associated with a specific process.

PCB:操作系统通过进程控制块（PCB）表示进程，一个PCB存储了进程的许多信息。

Job scheduler: the long-term scheduler, the job scheduler, selects processes which are spooled to a mass-storage device and loads them into memory for execution.

作业调度：又称长程调度，从存储器中选择进程并将其载入内存。

CPU scheduler: the short-term scheduler, selects from among the processes that are ready to execute and allocates the CPU to one of them.

CPU。

A thread is basic unit of CPU utilization; it comprises a thread ID, a program counter, a register set, and a stack.

线程：线程是一个基本的CPU执行单元；它包含了一个线程ID，一个程序计数器，一个寄存器组合一个堆栈。

one measure of work is the number of processes that are completed per time unit. 吞吐量：对工作量的一种测量单位，单位时间内运行完的进程数量。

Under nonpreemptive scheduling, once the CPU has been allocated to a process, the process keeps the CPU until it releases the CPU either by terminating or by switching to the wait state. A form of scheduling designed so that jobs can be interrupted and their resources transferred to more urgent jobs. An interrupted job can be either terminated completely or resumed tater.

非抢占式/抢占式调度：在非抢占式调度下，一旦CPU分配给一个进程，那么该进程就会保持CPU直到终止或转换到等待状态。抢占式调度：允许进程被中断，使其资源用在更重要的地方。

the dispatcher is the module that gives the control of the CPU to the process selected by short-term scheduler.

调度程序：调度程序是一个模块，他将CPU控制提交给短程调度程序选择的进程。

Turnaround time: the interval from the time of submission of a process to the time of completion is the turnaround time.

周转时间：从进程提交到进程完成的时间间隔。

Waiting time: waiting time is the sum of the periods spent waiting in the ready queue.

等待时间：进程在就绪队列中耗费时间的总和。

Response time: response time is the time from the submission of a request until the first response is produced.

响应时间：从进程提交到产生首次响应的时间。

critical section is a segment of code in which the process may be changing common variables, updating a table, writing a file, and so on.

临界区：每个进程有一个代码段，被称之为临界区，进程在临界区内可能会修改公有变量、更新一个表、写一个文件等。

Semaphore: A semaphore S is integer variable that, apart from initialization, is accessed only through two standard atomic operations: wait() and signal().

信号量：信号量s是一个整形数，除初始化之外，对它的访问只能通过两个标准原子操作：wait和signal。

Atomic operation: Atomic operation means an operation that completes in its entirety without interruption.

原子操作：原子操作指不会被中断，一次运行完的操作。

Deadlock: a set of blocked processes each holding a resource and waiting to acquire a resource held by another process in the set.

死锁：一些进程各自拥有一个资源，同时等待着另一个已被占有的资源，造成互相锁定的情况。Starvation: starvation is a situation in which processes wait indefinitely within the semaphore.

饥饿：饥饿是指一个进程在信号量中无休止的等待。

Deadlock prevention: Restrain the ways request can be made, attacking four conditions.

死锁预防：只要确保造成死锁的四个条件中任意一个不成立，就可以预防死锁。

Deadlock avoidance: A decision is made dynamically whether the current resource allocation request will, if granted, potentially lead to a deadlock. Requires that the system has some additional a priori information available.

死锁避免：就是要求操作系统预先提供进程在其生命周期里要请求和释放资源的信息。

Logical address: An address generated by the CPU is commonly referred to as a logical address. 逻辑地址：CPU产生的地址通常称之为逻辑地址

Physical address: an address seen by the memory unit – that is, the one loaded into the

memory-address register of the memory – is commonly referred to as a physical address.

物理地址：物理地址是内存单元的地址，也就是被装入内存的内存地址寄存器的地址。

Dynamic linking: linking is postponed until execution time.

动态链接：动态链接直到进行时才进行链接。

Contiguous allocation: In contiguous memory allocation, each process is contained in a single contiguous section of memory.

连续分配：又称连续内存分配，每个进程都包含在一个单独的连续内存区中。

External fragmentation: External fragmentation exists when there is enough total memory space to satisfy a request, but the available spaces are not contiguous; storage is fragmented into a large number of small holes.

外部碎片：伴随着进程的装入和移除，空闲内存空间被分成了众多的小块。当这些存在的内存空间的总大小可以满足一个请求时，外部碎片存在，但它却是不连续的（这样就不能满足请求）；存储器被分成了众多的碎片（小的分区）

memory that is internal to a partition but is not being used.

内部碎片：内存在一个块内部，但不被使用。

Page: logical memory broken into fixed-sized blocks of the same size as frames called pages

页：把逻辑内存分成同样大小的块，称为页。

Frame: physical memory broken into fixed-sized blocks called frames

帧：把物理内存分为固定大小的块，称为帧。

Segmentation: segmentation is a memory-management scheme that supports this user view of memory . A logical address space is a collection of segments. Each segment has a name and a length.

分段：分段是一种内存管理机制，他支持内存的用户视角。逻辑地址是段的集合。每个段都有其名称和长度。

Thrashing: This high paging activity is called thrashing. A process is thrashing if it is spending more time paging than executing.

抖动：也叫颠簸，高频率的换页行为，如果一个进程在换页上用的时间要多于运行时间，那么这个进程就在抖动。

Working set: The set of pages in the most recent △ page references is the working set.

工作集合：最近X个被引用的页集合。

File: the operating system abstracts from the physical properties of its storage devices to define a logical storage unit, the file. A file is an abstract data type.

文件：文件是记录在外存上相关信息的具体名称的集合。从用户角度而言，文件是逻辑外存的最小分配单元。

FCB: A File Control Block (FCB) is a file system structure in which the state of an open file is maintained.

FCB:文件控制块FCB是系统为管理文件而设置的一个数据结构.FCB是文件存在的标志,它记录了系统管理文件所需要的全部信息.

文件逻辑结构：文件逻辑结构是从用户观点出发所看到的文件组织形式，是用户可以直接处理的数据及其结构

文件物理结构：文件物理结构是指文件在外存上的存储组织形式，它直接关系到存储器空间的利用率。

Sequential access: sequential access is the simplest access method which information in the file is processed in order, one record after the other.

顺序访问：最为简单的访问方式，文件信息按顺序，一个接一个得处理

A file is made up of fixed-length logical records that allow programs to read and write records rapidly in no particular order.

直接访问：文件由固定长度的逻辑记录组成，以允许程序按任意顺序进行读写。

the directory records information—such as name, location, size, and type—for all files on that volume.

目录：目录是文件说明的集合，用于文件的描述与控制，包含名字，位置，大小和类型等信息Contiguous allocation: contiguous allocation requires that each file occupy a set of contiguous

blocks on the disk.

连续分配：连续分配方法要求每个文件在磁盘上占有一组连续的块。

Linked allocation: with linked allocation, each file is in a linked list of disk blocks; the disk blocks may be scattered anywhere on the disk.

链接分配：采用链接分配，每个文件都是磁盘块的链表，磁盘块分部在磁盘的任何地方。

Indexed allocation: each file has its own index block, which is an array of disk-block addresses. the i-th entry in the index block points to the i-th block of the file.

索引分配：每个文件都有其空间的索引块，这是一个磁盘块地址的数组。索引块的第i个条目指向文件的第i个块。

Seek time: Seek time is the time for the disk to move the heads to the cylinder containing the desired sector.

寻道时间：寻道时间是磁臂将磁头移动到包含目标扇区的柱面的时间。

Rotational latency: Rotational latency is the additional time waiting for the disk to rotate the desired sector to the disk head.

旋转延迟：是磁盘需要将目标扇区转动到磁头下的时间。

==================================上课提问题目================================== Chapter 1

1. What is the three main purpose of OS? 操作系统的目的？

，control program and a interface。

资源分配程序控制和接口。

2. What are the three major activities of an operating system in regard to secondary-storage management?

Free-space management. 空间管理

Storage allocation. 存储分配

Disk scheduling. 磁盘调度

3. What is the purpose of the command interpreter（命令解释程序）? Why is it usually separate from the kernel?

them into one or more system calls. It is usually not part of the kernel since the command interpreter is subject to changes.

它从用户或者命令文件中读取命令，并常常把他们转化为系统调用的形式。和核心分离的原因是命令解释器容易受各种变化的影响。

4. What is the main advantage for an operating-system designer of using a virtual-machine architecture? What is the main advantage for a user?

对操作系统设计者而言，虚拟机技术的好处是什么？对用户的好处是什么？

a good platform for operating system research since many different operating systems may run on one

physical system.

系统容易调试，安全问题也容易解决。此外，可以将多种操作系统运行在一个物理系统上，为研究操作系统提供了较好的平台。

Chapter 2

What are the main advantages of the microkernel approach to system design?微核模式的好处？

(b) it is more secure as more operations are done in user mode than in kernel mode, and (c) a simpler kernel design and functionality typically results in a more reliable operating system.

好处有：a.当增加新的服务、功能时不用改变核心。b.由于大部分的操作运行在用户模式下，因此变得更为安全。

c.微核模式使得系统更为可靠。

2. Why do some systems store the operating system in firmware, and others on disk?为什么有些系统把操作系统固化在固件上，而有些则写在磁盘中？

be not be available for the device. In this situation, the operating system must be stored in firmware.

对一些掌上电脑或者手机而言，无法做到提供一个内置的磁盘和操作系统，因此只能存在固件内。

3. How could a system be designed to allow a choice of operating systems to boot from? What would the bootstrap program need to do?如何设计系统实现选择引导的功能？引导程序需要做什么？

RedHat, Debian, and Mandrake). Each operating system will be stored on disk. During system boot-up, a special program (which we will call the boot manager) will determine which operating system to boot into. This means that rather initially booting to an operating system, the boot manager will first run during system startup. It is this boot manager that is responsible for determining which system to boot into. Typically boot managers must be stored at certain locations of the hard disk to be recognized during system startup. Boot managers often provide the user with a selection of systems to boot into; boot managers are also typically designed to boot into a default operating system if no choice is selected by the user

考虑到我们可能同时安装XP和其他的一些LINUX系统。每个操作系统都存放在磁盘上。在系统启动时，一个特殊的程序（我们称之为引导管理程序）会决定启动哪个操作系统。这意味着，在操作系统运行前，引导程序应该随着系统启动而运行。一般而言，该程序存放在硬盘上一个确定的位置以便于系统启动时读取。引导程序还会为用户提供一个操作系统的界面，另外，需要设置一个默认启动的操作系统以防止用户未做任何选择。

Chapter 3

1. Describe the differences among short-term, medium-term, and longterm scheduling. (课后题目)写出短调度，中调度和长调度的区别

Short-term (CPU scheduler)—selects from jobs in memory those jobs that are ready to execute and allocates the CPU to them.

短程调度（CPU调度）——从内存中选取就绪进程，并为之分配CPU

Medium-term—used especially with time-sharing systems as an intermediate scheduling level. A swapping scheme is implemented to remove partially run programs from memory and reinstate them later to continue where they left off.

中程调度——有些操作系统（如分时系统）可能会引入另外一种中间级的调度。它将进程从内存中移除，稍后再把进程重新载入内存并从停止的地方继续运行。

Long-term (job scheduler)—determines which jobs are brought into memory for processing.

长程调度（工作调度）——选择进程并调入内存。

The primary difference is in the frequency of their execution. The short-term must select a new process quite often. Long-term is used much less often since it handles placing jobs in the system and may wait a while for a job to finish before it admits another one.

他们主要的区别在于运行的频繁程度。。短程调度程序必须要频繁的为CPU选择一个新进程。对长程调度来说，在它为系统选择进程，直到该进程结束并设置新进程的期间，它并不会频繁执行。

2. Describe the actions taken by a kernel to context-switch between processes. (课后题目)简述当进程间进行上下文切换时内核所采取的措施。

state of the process scheduled to be run next. Saving the state of a process typically includes the values of all the CPU registers in addition to memory allocation. Context switches must also perform many architecture-specific operations, including flushing data and instruction caches.

a process creates a new process using the fork() operation, which of the following state is shared between the parent process and the child process?当一个进程里有FORK指令创建一个子进程时，以下的哪些选项父进程和子进程时共享的？

a. Stack

b. Heap

c. Shared memory segment

child process. Copies of the stack and the heap are made for the newly created process.

Chapter 6

1. Prove Peterson’s solution is right

Two process solution

Assume that the LOAD and STORE instructions are atomic; that is, cannot be interrupted.

The two processes share two variables:

int turn;

Boolean flag[2]

The variable turn indicates whose turn it is to enter the critical section. Turn表示轮到谁进入临界区

The flag array is used to indicate if a process is ready to enter the critical section. flag[i] = true implies that process Pi is ready! Flag[i]表示进程i已经准备好

Algorithm for Process Pi

do {

flag[i] = TRUE; Prove Dekker’s solution is right

var boolean flag[2] ; Show that, if the wait() and signal() semaphore operations are not executed atomically, then mutual exclusion may be violated.

Semaphore operations now defined as

wait(S) :

;

if < 0) {

add this process to ;

block;

} signal(S) :

++;

if <= 0) {

remove a process P from ;

wakeup(P);

}

4. If we interchange the order of two wait() semaphore operations in Consumer Process or Producer Process, what will happen?

The structure of Producer Process

do {

…

produce an item in nextp

…

wait(empty);

wait(mutex);

…

add nextp to buffer

…

signal(mutex);

signal(full);

} while (1); structure of the Consumer Process

do {

wait(full);

wait(mutex);

…

remove an item from buffer to nextc

…

signal(mutex);

signal(empty);

…

consume the item in nextc

…

} while (1);

如果将两个wait 操作，即wait(full)和wait(mutex)互换位置，或者wait(empty)和wait(mutex)互换位置，都可能引起死锁。

考虑系统中缓冲区全满时，若一生产者进程先执行了wait(mutex)操作并获得成功，当再执行wait(empty)操作时，它将因失败而进入阻塞状态，它期待消费者执行signal(empty)来唤醒自己，在此之前，它不可能执行signal(mutex)操作，从而使企图通过wait(mutex)进入自己的临界区的其他生产者和所有的消费者进程全部进入阻塞状态，从而引起系统死锁。类似地，消费者进程若先执行wait(mutex)，后执行wait(full)，同样可能造成死锁。

signal(full)和signal(mutex)互换位置，或者signal(empty)和signal(mutex)互换位置，则不会引起死锁，其影响只是使临界资源的释放略为推迟一些。

Chapter 8

1. Consider a logical address space of eight pages 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 physical address?

2. Why are segmentation and paging sometimes combined into one scheme?(

课后习题)

and paging are often combined in order to improve upon each other. Segmented paging is helpful when the page table becomes very large. A large contiguous section of the page table that is unused can be collapsed into a single segment table entry with a page-table address of zero. Paged segmentation handles the case of having very long segments that require a lot of time for allocation. By

paging the segments, we reduce wasted memory due

to external fragmentation as well as simplify the

allocation.

3. What is the purpose of paging the page tables? (课后习题)

课堂练习

1. 页表某系统采用页式存储管理策略，拥有逻辑空间32 页，每页2K，拥有物理空间1M。(1) 写出逻辑

地址的格式。(2) 若不考虑访问权限等，进程的页表有多少项？每项至少有多少位？(3) 如果物理空间减少一半，结构应相应作怎样的改变？

答案：（1）该系统拥有逻辑空间32 页，故逻辑地址中页号必须用 5 位来描述；而每页为2K，因此，页内

地址必须用11 位来描述，这样可得到它的逻辑地址格式如下：

（2）每个进程最多有32 个页面，因此，进程的页表项最多为32 项；若不考虑访问权限等，则页表项中只需给出页所对应的物理块块号，1M/2K=2^9,1M 的物理空间可分成2^9个内存块，故每个页表项至少有9 位。（3）如果物理空间减少一半，则页表中页表项数仍不变，但每项的长度可减少 1 位。

2. 已知某分页系统，主存容量为64K，页面大小为1K，对一个4 页大的作业，其0、1、2、3 页分别被分配到主存的2、4、6、7 块中。(1) 将十进制的逻辑地址1023、2500、3500、4500 转换成物理地址。(2) 以十进制的逻辑地址1023 为例画出地址变换过程图。

答案：(1) 对上述逻辑地址，可先计算出它们的页号和页内地址(逻辑地址除以页面大小，得到的商为页号，余数为页内地址)，然后通过页表转换成对应的物理地址。①逻辑地址1023：1023/1K，得到页号为0，页内地址为1023，查页表找到对应的物理块号为2，故物理地址为2×1K+1023=3071。②逻辑地址2500：2500/1K，得到页号为2，页内地址为452，查页表找到对应的物理块号为6，故物理地址为6×1K+452=6596。

③逻辑地址3500：3500/1K，得到页号为3，页内地址为428，查页表找到对应的物理块号为7，故物理地址为7×1K+428=7596。④逻辑地址4500：4500/1K，得到页号为4，页内地址为404，因页号不小于页表长度，故产生越界中断。

(2) 逻辑地址1023 的地址变换过程如图所示，其中的页表项中没考虑每页的访问权限。

===================================课本习题=================================== Chapter 1

Under what circumstances would a user be better off using a time-sharing system, rather than a personal computer or single-user workstation?什么情况下会使用分时系统，而不是个人电脑或者工作站Answer: When there are few other users, the task is large, and the hardware is fast, timesharing makes sense. The full power of the system can be brought to bear on the user’s problem. The problem can be solved faster than on a personal computer. Another case occurs when lots of other users need resources at the same time.

A personal computer is best when the job is small enough to be executed reasonably on it and when performance is suf ficient to execute the program to the user’s satisfaction.

Describe the differences between symmetric and asymmetric multiprocessing. What are three advantages and one disadvantage of multiprocessor systems?

Answer: Symmetric multiprocessing treats all processors as equals, and I/O can be processed on any CPU.

Asymmetric multiprocessing has one master CPU and the remainder CPUs are slaves. The master distributes tasks among the slaves, and I/O is usually done by the master only. Multiprocessors can save money by not duplicating power supplies, housings, and peripherals. They can execute programs more quickly and can have increased reliability. They are also more complex in both hardware and software than uniprocessor systems.

How do clustered systems differ from multiprocessor systems?What is

required for twomachines belonging to a cluster to cooperate to provide

a highly available service?

perform a computational task distributed across the cluster. Multiprocessor systems on the other hand could be a single physical entity comprising of multiple CPUs. A clustered system is less tightly coupled than a multiprocessor system. Clustered systems communicate using messages, while processors in a multiprocessor system could communicate using shared memory. In order for twomachines to provide a highly available service, the state on the two machines should be replicated and should be consistently one of the machines fail, the other could then take-over

the functionality of the failed machine.

What is the purpose of interrupts? What are the differences between a trap and an interrupt? Can traps be generated intentionally by a user program? If so, for what purpose?

An interrupt is a hardware-generated change-of-flow within the system. An interrupt handler is summoned to deal with the cause of the interrupt; control is then returned to the interrupted context and instruction. A trap is a software-generated interrupt. An interrupt can be used to signal the completion of an I/O to obviate the need for device polling. A trap can be used to call operating system routines or to catch arithmetic errors.

Chapter 2

What are the advantages and disadvantages of using the same system-

call interface for manipulating both files and devices?

Answer: Each device can be accessed as though it was a file in the file system. Since most of the kernel deals with devices through this file in- terface, it is relatively easy to add a new device driver by implementing the hardware-specific code to support this abstract file interface. There- fore, this benefits the development of both user program code, which can bewritten to access devices and files in the samemanner, and device driver code, which can be written to support a well-defined API. The disadvantage with using the same interface is that it might be difficult to capture the functionality of certain devices within the context of the file access API, thereby either resulting in a loss of functionality or a loss of performance. Some of this could be overcome by the use of ioctl oper- ation that provides a general purpose interface for processes to invoke operations on devices.

What is the purpose of the command interpreter? Why is it usually separate from the kernel? Would it be possible for the user to develop a new command interpreter using the system-call interface provided by the operating system?

them into one or more system calls. It is usually not part of the kernel since the command interpreter is subject to changes.

Why is the separation of mechanism and policy desirable?

Answer: Mechanismand policymust be separate to ensure that systems are easy to modify. No two system installations are the same, so each installation may want to tune the operating system to suit its needs. With mechanism and policy separate, the policy may be changed at will while the mechanism stays unchanged.

This arrangement provides a more flexible system.

What are the main advantages of the microkernel approach to system design? How do user programs and system services interact in a microkernel architecture? What are the disadvantages of using the microkernel approach?

(b) it is more secure as more operations are done in user mode than in kernel mode, and (c) a simpler kernel design and functionality typically results in a more reliable operating system.

What is the main advantage for an operating-system designer of using a virtual-machine architecture? What is the main advantage for a user?

Answer:The system is easy to debug, and security problems are easy to solve. Virtual machines also provide a good platform for operating systemresearch sincemany different operating systemsmay run on one physical system.

Chapter 3

Describe the differences among short-term, medium-term, and long- term scheduling.

Answer:

? Short-term (CPU scheduler)—selects from jobs in memory those jobs that are ready to execute and allocates the CPU to them.

? Medium-term—used especially with time-sharing systems as an intermediate scheduling level. A swapping scheme is implemented to remove partially run programs from memory and reinstate them later to continue where they left off.

? Long-term (job scheduler)—determines which jobs are brought into

memory for processing. The primary difference is in the frequency of their execution. The short- termmust select a new process quite often. Long-termis usedmuch less often since it handles placing jobs in the system and may wait a while for a job to finish before it admits another one.

Describe the actions taken by a kernel to context-switch between pro-

cesses.

Answer: In general, the operating system must save the state of the currently running process and restore the state of the process sched- uled to be run next. Saving the state of a process typically includes the values of all the CPU registers in addition to memory allocation. Con- text switches must also perform many architecture-specific operations, including flushing data and instruction caches.

P117

Chapter 4

Under what circumstances does a multithreaded solution using multi-

ple kernel threads provide better performance than a single-threaded

solution on a single-processor system?

Answer: When a kernel thread suffers a page fault, another kernel thread can be switched in to use the interleaving time in a usefulmanner. A single-threaded process, on the other hand, will not be capable of performing useful work when a page fault takes place. Therefore, in scenarios where a program might suffer from frequent page faults or has to wait for other system events, a multi-threaded solution would perform better even on a single-processor system.

Which of the following components of program state are shared across

threads in a multithreaded process?

a. Register values

b. Heap memory

c. Global variables

d. Stack memory

Answer: The threads of a multithreaded process share heap memory and global variables. Each thread has its separate set of register values and a separate stack.

Can a multithreaded solution using multiple user-level threads achieve betterperformance on amultiprocessor systemthan on a single-processor system?

Answer: A multithreaded system comprising of multiple user-level threads cannot make use of the different processors in a multiprocessor system simultaneously. The operating system sees only a single process and will not schedule the different threads of the process on separate processors. Consequently, there is no performance benefit associated with executing multiple user-level threads on a multiprocessor system. Chapter 5

Why is it important for the scheduler to distinguish I/O-bound programs

from CPU-bound programs?

Answer: I/O-bound programs have the property of performing only a small amount of computation before performing IO. Such programs typically do not use up their programs, on the other hand, use their entire quantum without performing any blocking IO operations. Consequently, one could make better use of the computer’s resouces by giving higher priority to I/O-bound programs and allow them to execute ahead of the CPU-bound programs.

Discuss how the following pairs of scheduling criteria conflict in certain

settings.

a. CPU utilization and response time

b. Average turnaround time and maximum waiting time

c. I/O device utilization and CPU utilization

Answer:

? CPU utilization and response time: CPU utilization is increased if the overheads associated with context switching is minimized. The context switching overheads could be lowered by performing context switches infrequently. This could however result in increasing the response time for processes.

? Average turnaround time and maximum waiting time: Average turnaround time is minimized by executing the shortest tasks first. Such a scheduling policy could however starve long-running tasks and thereby increase their waiting time.

? I/O device utilization and CPU utilization: CPU utilization is maximized by running long-running CPU-bound tasks without performing context switches. I/O device utilization is maximized by scheduling I/O-bound jobs as soon as they become ready to run, thereby incurring the overheads of context switches.

Consider the following set of processes, with the length of the CPU-burst time given in milliseconds:

The 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(先到先服务), SJF （最短优先）, a nonpreemptive priority（非抢占式的优先级）(a smaller priority number implies a higher priority), and RR (quantum = 1)（时间片为1的分时分配）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:

Chapter 6

What is the meaning of the term busy waiting? What other kinds of waiting are there in an operating system? Can busy waiting be avoided altogether? Explain your answer.

Answer: Busy waiting means that a process is waiting for a condition to be satisfied in a tight loopwithout relinquish the , a process could wait by relinquishing the processor, and block on a condition and wait to be awakened at some appropriate time in the future. Busy waiting can be avoided but incurs the overhead associated with putting a process to sleep and having to wake it up when the appropriate program state is reached.

Show that, if the wait() and signal() semaphore operations are not

executed atomically, then mutual exclusion may be violated.

Answer: A wait operation atomically decrements the value associated with a semaphore. If two wait operations are executed on a semaphore when its value is 1, if the two operations are not performed atomically, then it is possible that both operations might proceed to decrement the semaphore value thereby violating mutual exclusion.

The Sleeping-Barber Problem. A barbershop consists of a waiting room with n chairs and a barber room with one 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.

Chapter 7

Consider a system consisting of four resources of the same type that are shared by three

processes, each of which needs at most two resources. Show that the system is deadlock free. Answer: Suppose the system is deadlocked. This implies that each process is holding one resource and is waiting for one more. Since there are three processes and four resources, one process must be able to obtain two resources. This process requires no more resources and, therefore it will return its resources when done.

Consider a system consisting of m resources of the same type, being shared by n processes. Resources can be requested and released by processes only one at a time. Show that the system is deadlock free if the following two conditions hold:

a. The maximum need of each process is between 1 and m resources

b. The sum of all maximum needs is less than m + n

Answer: Using the terminology of Section 7.6.2, we have:

This implies that there exists a process Pi such that Needi = 0. Since Max i≥1 it follows that Pi has at least one resource that it can release. Hence the system cannot be in a deadlock state.

Consider the following snapshot of a system:

Allocation Max Available

ABCD ABCD ABCD

P0 0012 0012 1520

P1 1000 1750

P2 1354 2356

P3 0632 0652

P4 0014 0656Exercises 53

Answer the following questions using the 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? Answer:

a. What is the content of the matrix Need? The values of Need for processes P0 through P4 respectively are (0, 0, 0, 0), (0, 7, 5, 0), (1, 0, 0, 2), (0, 0, 2, 0), and (0, 6, 4, 2).

b. Is the system in a safe state? Available being equal to (1, 5, 2, 0), either process P0 or P3 could run. Once process P3 runs, it releases its resources which allow all other existing processes to run.

c. If a request from process P1 arrives for (0,4,2,0), can the request be granted immediately? Yes it can. This results in the value of Available being (1, 1, 0, 0).One ordering of processes that can finish is P0, P2, P3, P1,and P4.

Chapter 8

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?

Answer:

First-fit:

212K is put in 500K partition

417K is put in 600K partition

112K is put in 288K partition (new partition 288K =

500K - 212K)

426K must wait

Best-fit:

212K is put in 300K partition

417K is put in 500K partition

112K is put in 200K partition

426K is put in 600K partition

Worst-fit:

212K is put in 600K partition

417K is put in 500K partition

112K is put in 388K partition

426K must wait

In this example, Best-fit turns out to be the best.

Consider a paging system with the page table stored in memory.

a. If a memory reference takes 200 nanoseconds, how long does a paged memory reference take?

b. If we add associative registers, and 75 percent of all page-table references are found in the associative registers, what is the effective memory reference time? (Assume that finding a page-table entry in the associative registers takes zero time, if the entry is there.)

Answer:

a. 400 nanoseconds; 200 nanoseconds to access the page table and 200 nanoseconds to access the word in memory.

b. Effective access time = ×(200 nanoseconds) + ×(400 nanoseconds) = 250 nanoseconds.

Explain why it is easier to share a reentrant module using segmentation than it is to do so when pure paging is used.

Answer: Since segmentation is based on a logical division of memory rather than a physical one, segments of any size can be shared with only one entry in the segment tables of each paging there must be a common entry in the page tables for each page that is shared.

Consider the following segment table:

What are the physical addresses for the following logical addresses?

a. 0,430

b. 1,10

c. 2,500

d. 3,400

e. 4,112

Answer: a. 219 + 430 = 649

b. 2300 + 10 = 2310

c. illegal reference, trap to operating system

d. 1327 + 400 = 1727

e. illegal reference, trap to operating system

Chapter 9

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 would have to be performed in hardware. A TLB could serve as a cache and improve the performance of the lookup operation.

9.4 A certain computer provides its users with a virtual-memory space of 232 bytes. The computer has

218 bytes of physical memory. The virtual memory is implemented by paging, and the page size is 4096 bytes. A user process generates the virtual address . Explain how the system establishes the corresponding physical location. Distinguish between software and hardware operations.

0001 0001 0001 0010 0011 0100 0101 0110

Since the page size is 212, the page table size is 220. Therefore the low-order 12 b its “01000101 0110” are used as the displacement into the page, while the remaining 20 bits “00010001 0001 0010 0011” are used as the displacement in the page table.

9.13 A page-replacement algorithm should minimize the number of page faults. We can do this minimization by distributing heavily used pages evenly over all of memory, rather than having them compete for a small number of page frames. We can associate with each page frame a counter of the number of pages that are associated with that frame. Then, to replace a page, we search for the page frame with the smallest counter.

a. Define a page-replacement algorithm using this basic idea. Specifically address the problems of (1) what the initial value of the counters is, (2) when counters are increased, (3) when counters are decreased, and (4) how the page to be replaced is selected.

b. How many page faults occur for your algorithm for the following reference string, for four page frames? 1, 2, 3, 4, 5, 3, 4, 1, 6, 7, 8, 7, 8, 9, 7, 8, 9, 5, 4, 5, 4, 2.

c. What is the minimum number of page faults for an optimal page-replacement strategy for the reference string in part b with four page frames?

a. Define a page-replacement algorithm addressing the problems of:

i. Initial value of the counters—0.

ii. Counters are increased—whenever a new page is associated with that frame.

iii. Counters are decreased—whenever one of the pages associated with that frame is no longer required. iv. How the page to be replaced is selected—find a frame with the smallest counter. Use FIFO for breaking ties.

b. 14 page faults

c. 11 page faults

What is the cause of thrashing? How does the system detect发现thrashing? Once it detects thrashing, what can the system do to eliminate this problem?

forcing it to continuously page fault. The system can detect thrashing by evaluating the level of CPU utilization as compared to the level of multiprogramming. It can be eliminated by reducing the level of multiprogramming.

Chapter 10

The open-file table is used to maintain information about files that are currently open. Should the operating system maintain a separate table for each user or just maintain one table that contains references to files that are being accessed by all users at the current time? If the same file is being accessed by two different programs or users, should there be separate entries in the open file table? Answer: By keeping a central open-file table, the operating system can perform the following operation that would be infeasible other- wise. Consider a file that is currently being accessed by one or more processes. If the file is deleted, then it should not be removed from the disk until all processes accessing the file have closed it. This check could be performed only if there is centralized accounting of number of processes

accessing the file. On the other hand, if two processes are accessing the file, then separate state needs to be maintained to keep track of the current location ofwhich parts of the file are being accessed by the two processes. This requires the operating system to maintain separate entries for the two processes.

Some systems automatically open a file when it is referenced for the first time, and close the file when the job terminates. Discuss the advantages and disadvantages of this scheme as compared to the more traditional one, where the user has to open and close the file explicitly.

thus makes it more convenient to the user; however, it requires more overhead than the case when explicit opening and closing is required.

Chapter 11

What are the advantages of the variation of linked allocation that uses a FAT to chain together the blocks of a file?

Answer: The advantage is that while accessing a block that is stored at the middle of a file, its location can be determined by chasing the pointers stored in the FAT as opposed to accessing all of the individual blocks of the file in a sequentialmanner to find the pointer to the target block. Typically,most of the FAT can be cached inmemory and therefore the pointers can be determined with just memory accesses instead of having to access the disk blocks.

Consider a file system on a disk that has both logical and physical block sizes of 512 bytes. Assume that the information about each file is already in memory. For each of the three allocation strategies (contiguous, linked, and indexed), answer these questions:

a. How is the logical-to-physical address mapping accomplished in this system? (For the indexed allocation, assume that a file is always less than 512 blocks long.)

b. If we are currently at logical block 10 (the last block accessed was block 10) and want to access logical block 4, how many physical blocks must be read from the disk?

a. Contiguous. Divide the logical address by 512 with X and Y the resulting quotient and remainder respectively.

i. Add X to Z to obtain the physical block number. Y is the displacement into that block.

ii. 1

b. Linked. Divide the logical physical address by 511 with X and Y the resulting quotient and remainder respectively.

i. Chase down the linked list (getting X + 1 blocks). Y + 1 is the displacement into the last physical block.

ii. 4

c. Indexe

d. Divide the logical address by 512, with X and Y the resulting quotient and remainder, respectively.

i. Get the index block into memory. Physical block address is contained in the index block at location X. Y is the displacement into the desired physical block.

ii. 2

Chapter 12

Suppose that a disk drive has 5000 cylinders柱面, 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, is 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130 Starting 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 diskscheduling algorithms?

a. FCFS先来先服务

b. SSTF最短寻道时间有限

c. SCAN来回扫描到0或者4999返回

d. LOOK 到

最远距离就返回，即本题的1774返回 e. C-SCAN扫描至底时返回磁盘开始处 f. C-LOOK

a. The FCFS schedule is 143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130. The total seek distance is

7081.

143-86+1470-86+1470-913+1774-913+1774-948+1509-948+1509-1022+1750-1022+1750-130=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.