SHREYANS JAIN: January 2010

Welcome to the Web Pages supporting Operating System Concepts

STUDENT’S MANUAL

TO ACCOMPANY

OPERATING

SYSTEM

CONCEPTS

SEVENTH EDITION

ABRAHAM SILBERSCHATZ

Yale University

PETER BAER GALVIN

Corporate Technologies

GREG GAGNE

Westminster College

Preface

This volume is a student’s manual for the Seventh Edition of Operating System

Concepts, by Abraham Silberschatz, Peter Baer Galvin, and Greg Gagne. It

consists of answers to the exercises in the parent text.

Although we have tried to produce a student’s manual that will aid all of

the users of our book as much as possible, there can always be improvements

(improved answers, additional questions, sample test questions, programming

projects, alternative orders of presentation of the material, additional references,

and so on). We invite you, both instructors and students, to help us

improve this manual. If you have better solutions to the exercises or other

items that would be of use with Operating System Concepts, we invite you to

send them to us for consideration in later editions of this manual. All contributions

will, of course, be properly credited to their contributor.

Internet electronic mail should be addressed to os-book@cs.yale.edu.

Physical mail may be sent to Avi Silberschatz, Yale University, Department of

Computer Science, 51 Prospect Street, P.O. box 208285, New Haven, CT 06520,

USA.

A. S.

P. B. G

G. G.

iii

Contents

Chapter 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Chapter 2 Operating-System Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Chapter 3 Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Chapter 4 Threads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Chapter 5 CPU Scheduling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Chapter 6 Process Synchronization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Chapter 7 Deadlocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Chapter 8 Memory Management . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Chapter 9 VirtualMemory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Chapter 10 File-Systems Interface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Chapter 11 File-Systems Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Chapter 12 Mass Storage Structure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

Chapter 13 I/O Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

Chapter 14 Protection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

Chapter 15 Security . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Chapter 16 Distributed System Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Chapter 17 Distributed File Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Chapter 18 Distributed Coordination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Chapter 19 Real-Time Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

Chapter 20 Multimedia Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Chapter 21 The Linux System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Chapter 22 Windows XP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Chapter 23 Influential Operating Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

CH1APTER

Introduction

Exercises

1.1 What are the three main purposes of an operating system?

Answer:

• To provide an environment for a computer user to execute programs

on computer hardware in a convenient and efficient manner.

• To allocate the separate resources of the computer as needed to

solve the problem given. The allocation process should be as fair

and efficient as possible.

• As a control program it serves two major functions: (1) supervision

of the execution of user programs to prevent errors and improper

use of the computer, and (2) management of the operation and

control of I/O devices.

1.2 What are the main differences between operating systems for mainframe

computers and personal computers?

Answer: Generally, operating systems for batch systems have simpler

requirements than for personal computers. Batch systems do not have

to be concerned with interacting with a user as much as a personal

computer. As a result, an operating system for a PC must be concerned

with response time for an interactive user. Batch systems do not have

such requirements. A pure batch system also may have not to handle

time sharing, whereas an operating system must switch rapidly between

different jobs.

1.3 List the four steps that are necessary to run a program on a completely

dedicated machine.

Answer:

2 Chapter 1 Introduction

a. Reserve machine time.

b. Manually load program into memory.

c. Load starting address and begin execution.

d. Monitor and control execution of program from console.

1.4 We have stressed the need for an operating system to make efficient use

of the computing hardware. When is it appropriate for the operating

system to forsake this principle and to “waste” resources? Why is such

a system not really wasteful?

Answer: Single-user systems should maximize use of the system for

the user. A GUI might “waste” CPU cycles, but it optimizes the user’s

interaction with the system.

1.5 What is themain difficulty that a programmermust overcome in writing

an operating system for a real-time environment?

Answer: The main difficulty is keeping the operating system within

the fixed time constraints of a real-time system. If the system does not

complete a task in a certain time frame, it may cause a breakdown

of the entire system it is running. Therefore when writing an operating

system for a real-time system, the writer must be sure that his scheduling

schemes don’t allow response time to exceed the time constraint.

1.6 Consider the various definitions of operating system. Consider whether

the operating system should include applications such asWeb browsers

andmail programs. Argue both that it should and that it should not, and

support your answer.

Answer: Point. Applications such as web browsers and email tools are

performing an increasingly important role inmodern desktop computer

systems. To fulfill this role, they should be incorporated as part of the

operating system. By doing so, they can provide better performance

and better integration with the rest of the system. In addition, these

important applications can have the same look-and-feel as the operating

system software.

Counterpoint. The fundamental role of the operating system is to manage

system resources such as the CPU, memory, I/O devices, etc. In addition,

it’s role is to run software applications such as web browsers and

email applications. By incorporating such applications into the operating

system, we burden the operating system with additional functionality.

Such a burdenmay result in the operating system performing a less-thansatisfactory

job at managing system resources. In addition, we increase

the size of the operating system thereby increasing the likelihood of

system crashes and security violations.

1.7 How does the distinction between kernel mode and usermode function

as a rudimentary form of protection (security) system?

Answer: The distinction between kernel mode and user mode provides

a rudimentary form of protection in the following manner. Certain

instructions could be executed only when the CPU is in kernel mode.

Similarly, hardware devices could be accessed only when the program

is executing in kernel mode. Control over when interrupts could be enExercises

abled or disabled is also possible only when the CPU is in kernel mode.

Consequently, the CPU has very limited capability when executing in

user mode, thereby enforcing protection of critical resources.

1.8 Which of the following instructions should be privileged?

a. Set value of timer.

b. Read the clock.

c. Clear memory.

d. Issue a trap instruction.

e. Turn off interrupts.

f. Modify entries in device-status table.

g. Switch from user to kernel mode.

h. Access I/O device.

Answer: The following operations need to be privileged: Set value of

timer, clear memory, turn off interrupts, modify entries in device-status

table, access I/O device. The rest can be performed in user mode.

1.9 Some early computers protected the operating system by placing it in

a memory partition that could not be modified by either the user job

or the operating system itself. Describe two difficulties that you think

could arise with such a scheme.

Answer: The data required by the operating system (passwords, access

controls, accounting information, and so on) would have to be stored

in or passed through unprotected memory and thus be accessible to

unauthorized users.

1.10 Some CPUs provide for more than two modes of operation. What are

two possible uses of these multiple modes?

Answer: Although most systems only distinguish between user and

kernel modes, some CPUs have supported multiple modes. Multiple

modes could be used to provide a finer-grained security policy. For

example, rather than distinguishing between just user and kernelmode,

you could distinguish between different types of user mode. Perhaps

users belonging to the same group could execute each other’s code. The

machine would go into a specified mode when one of these users was

running code. When the machine was in this mode, a member of the

group could run code belonging to anyone else in the group.

Another possibility would be to provide different distinctions within

kernel code. For example, a specific mode could allow USB device drivers

to run. This would mean that USB devices could be serviced without

having to switch to kernel mode, thereby essentially allowing USB device

drivers to run in a quasi-user/kernel mode.

1.11 Timers could be used to compute the current time. Provide a short description

of how this could be accomplished.

Answer: A program could use the following approach to compute the

current time using timer interrupts. The program could set a timer for

4 Chapter 1 Introduction

some time in the future and go to sleep. When it is awakened by the

interrupt, it could update its local state, which it is using to keep track

of the number of interrupts it has received thus far. It could then repeat

this process of continually setting timer interrupts and updating its local

state when the interrupts are actually raised.

1.12 Is the Internet a LAN or a WAN?

Answer: The Internet is a WAN as the various computers are located

at geographically different places and are connected by long-distance

network links.

CH2APTER Operating-

System

Structures

Exercises

2.1 What is the purpose of system calls?

Answer: System calls allow user-level processes to request services of

the operating system.

2.2 What are the five major activities of an operating system in regard to

process management?

Answer:

a. The creation and deletion of both user and system processes

b. The suspension and resumption of processes

c. The provision of mechanisms for process synchronization

d. The provision of mechanisms for process communication

e. The provision of mechanisms for deadlock handling

2.3 What are the three major activities of an operating system in regard to

memory management?

Answer:

a. Keep track of which parts of memory are currently being used

and by whom.

b. Decide which processes are to be loaded into memory when memory

space becomes available.

c. Allocate and deallocate memory space as needed.

2.4 What are the three major activities of an operating system in regard to

secondary-storage management?

Answer:

6 Chapter 2 Operating-System Structures

• Free-space management.

• Storage allocation.

• Disk scheduling.

2.5 What is the purpose of the command interpreter? Why is it usually

separate from the kernel?

Answer: It reads commands from the user or from a file of commands

and executes them, usually by turning them into one or more system

calls. It is usually not part of the kernel since the command interpreter

is subject to changes.

2.6 What system calls have to be executed by a command interpreter or shell

in order to start a new process?

Answer: In Unix systems, a fork system call followed by an exec system

call need to be performed to start a new process. The fork call clones the

currently executing process, while the exec call overlays a new process

based on a different executable over the calling process.

2.7 What is the purpose of system programs?

Answer: System programs can be thought of as bundles of useful

system calls. They provide basic functionality to users so that users do

not need to write their own programs to solve common problems.

2.8 What is the main advantage of the layered approach to system design?

What are the disadvantages of using the layered approach?

Answer: As in all cases of modular design, designing an operating

system in a modular way has several advantages. The system is easier

to debug and modify because changes affect only limited sections of

the system rather than touching all sections of the operating system.

Information is kept only where it is needed and is accessible only within

a defined and restricted area, so any bugs affecting that data must be

limited to a specific module or layer.

2.9 List five services provided by an operating system. Explain how each

provides convenience to the users. Explain also in which cases it would

be impossible for user-level programs to provide these services.

Answer:

a. Program execution. The operating system loads the contents (or

sections) of a file into memory and begins its execution. A userlevel

program could not be trusted to properly allocate CPU time.

b. I/O operations. Disks, tapes, serial lines, and other devices must

be communicated with at a very low level. The user need only

specify the device and the operation to perform on it, while the

system converts that request into device- or controller-specific

commands. User-level programs cannot be trusted to access only

devices they should have access to and to access them only when

they are otherwise unused.

c. File-systemmanipulation. There aremany details in file creation,

deletion, allocation, and naming that users should not have to perform.

Blocks of disk space are used by files and must be tracked.

Exercises 7

Deleting a file requires removing the name file information and

freeing the allocated blocks. Protections must also be checked to

assure proper file access. User programs could neither ensure adherence

to protection methods nor be trusted to allocate only free

blocks and deallocate blocks on file deletion.

d. Communications. Message passing between systems requires

messages to be turned into packets of information, sent to the network

controller, transmitted across a communications medium,

and reassembled by the destination system. Packet ordering and

data correction must take place. Again, user programs might not

coordinate access to the network device, or they might receive

packets destined for other processes.

e. Error detection. Error detection occurs at both the hardware and

software levels. At the hardware level, all data transfers must be

inspected to ensure that data have not been corrupted in transit.

All data on media must be checked to be sure they have not

changed since they were written to the media. At the software

level, media must be checked for data consistency; for instance,

whether the number of allocated and unallocated blocks of storage

match the total number on the device. There, errors are frequently

process-independent (for instance, the corruption of data on a

disk), so there must be a global program (the operating system)

that handles all types of errors. Also, by having errors processed

by the operating system, processes need not contain code to catch

and correct all the errors possible on a system.

2.10 What is the purpose of system calls?

Answer: System calls allow user-level processes to request services of

the operating system.

2.11 What are the main advantages of the microkernel approach to system

design?

Answer: Benefits typically include the following (a) adding a new

service does not require modifying the kernel, (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.

2.12 Whydo some systems store the operating system in firmware, and others

on disk?

Answer: For certain devices, such as handheld PDAs and cellular telephones,

a disk with a file system may be not be available for the device.

In this situation, the operating system must be stored in firmware.

2.13 How could a system be designed to allow a choice of operating systems

to boot from? What would the bootstrap program need to do?

Answer: Consider a system that would like to run both Windows

XP and three different distributions of Linux (e.g., 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

8 Chapter 2 Operating-System Structures

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 userwith 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.

CH3APTER

Processes

Exercises

3.1 Palm OS provides no means of concurrent processing. Discuss three

major complications that concurrent processing adds to an operating

system.

Answer:

a. A method of time sharing must be implemented to allow each

of several processes to have access to the system. This method

involves the preemption of processes that do not voluntarily give

up the CPU (by using a system call, for instance) and the kernel

being reentrant (so more than one processmaybe executing kernel

code concurrently).

b. Processes and system resources must have protections and must

be protected from each other. Any given process must be limited

in the amount of memory it can use and the operations it can

perform on devices like disks.

c. Care must be taken in the kernel to prevent deadlocks between

processes, so processes aren’t waiting for each other’s allocated

resources.

3.2 The Sun UltraSPARC processor has multiple register sets. Describe the

actions of a context switch if the new context is already loaded into

one of the register sets. What else must happen if the new context is in

memory rather than in a register set and all the register sets are in use?

Answer: The CPU current-register-set pointer is changed to point to the

set containing the new context,which takes very little time. If the context

is in memory, one of the contexts in a register set must be chosen and be

moved to memory, and the new context must be loaded from memory

10 Chapter 3 Processes

into the set. This process takes a little more time than on systems with

one set of registers, depending on how a replacement victim is selected.

3.3 When 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?

a. Stack

b. Heap

c. Shared memory segments

Answer: Only the shared memory segments are shared between the

parent process and the newly forked child process. Copies of the stack

and the heap are made for the newly created process.

3.4 Again considering the RPC mechanism, consider the “exactly once” semantic.

Does the algorithm for implementing this semantic execute correctly

even if the “ACK” message back to the client is lost due to a network

problem? Describe the sequence ofmessages andwhether "exactly once"

is still preserved.

Answer: The “exactly once” semantics ensure that a remore procedure

will be executed exactly once and only once. The general algorithm for

ensuring this combines an acknowledgment (ACK) scheme combined

with timestamps (or some other incremental counter that allows the

server to distinguish between duplicate messages).

The general strategy is for the client to send the RPC to the server along

with a timestamp. The client will also start a timeout clock. The client

will then wait for one of two occurrences: (1) it will receive an ACK from

the server indicating that the remote procedure was performed, or (2) it

will time out. If the client times out, it assumes the server was unable

to perform the remote procedure so the client invokes the RPC a second

time, sending a later timestamp. The client may not receive the ACK for

one of two reasons: (1) the original RPCwas never received by the server,

or (2) the RPC was correctly received—and performed—by the server

but the ACK was lost. In situation (1), the use of ACKs allows the server

ultimately to receive and perform the RPC. In situation (2), the serverwill

receive a duplicate RPC and it will use the timestamp to identify it as a

duplicate so as not to perform the RPC a second time. It is important to

note that the server must send a second ACK back to the client to inform

the client the RPC has been performed.

3.5 Assume that a distributed system is susceptible to server failure. What

mechanisms would be required to guarantee the “exactly once” semantics

for execution of RPCs?

Answer: The server should keep track in stable storage (such as a

disk log) information regarding what RPC operations were received,

whether they were successfully performed, and the results associated

with the operations. When a server crash takes place and a RPC message

is received, the server can check whether the RPC had been previously

performed and therefore guarantee “exactly once” semanctics for the

execution of RPCs.

CH4APTER

Threads

Exercises

4.1 Provide two programming examples in which multithreading provides

better performance than a single-threaded solution.

Answer: (1)AWeb server that services each request in a separate thread.

2) (A parallelized application such as matrix multiplication where (different

parts of the matrix may be worked on in parallel. (3) An (interactive

GUI program such as a debugger where a thread is used (to monitor

user input, another thread represents the running (application, and a

third thread monitors performance.

4.2 What are two differences between user-level threads and kernel-level

threads? Under what circumstances is one type better than the other?

Answer: (1) User-level threads are unknown by the kernel,whereas the

kernel is aware of kernel threads. (2) On systems using either M:1 orM:N

mapping, user threads are scheduled by the thread library and the kernel

schedules kernel threads. (3) Kernel threads need not be associated with

a processwhereas every user thread belongs to a process. Kernel threads

are generally more expensive tomaintain than user threads as they must

be represented with a kernel data structure.

4.3 Describe the actions taken by a kernel to context switch between kernellevel

threads.

Answer: Context switching between kernel threads typically requires

saving the value of the CPU registers fromthe thread being switched out

and restoring the CPU registers of the new thread being scheduled.

4.4 What resources are used when a thread is created? How do they differ

from those used when a process is created?

Answer: Because a thread is smaller than a process, thread creation

typically uses fewer resources than process creation. Creating a process

12 Chapter 4 Threads

requires allocating a process control block (PCB), a rather large data structure.

The PCB includes a memory map, list of open files, and environment

variables. Allocating and managing the memory map is typically

themost time-consuming activity. Creating either a user or kernel thread

involves allocating a small data structure to hold a register set, stack, and

priority.

4.5 Assume an operating system maps user-level threads to the kernel using

the many-to-many model and the mapping is done through LWPs.

Furthermore, the system allows developers to create real-time threads.

Is it necessary to bind a real-time thread to an LWP? Explain.

Answer: Yes. Timing is crucial to real-time applications. If a thread is

marked as real-time but is not bound to an LWP, the thread may have

to wait to be attached to an LWP before running. Consider if a real-time

thread is running (is attached to an LWP) and then proceeds to block (i.e.

must perform I/O, has been preempted by a higher-priority real-time

thread, is waiting for a mutual exclusion lock, etc.) While the real-time

thread is blocked, the LWP itwas attached to has been assigned to another

thread. When the real-time thread has been scheduled to run again, it

must first wait to be attached to an LWP. By binding an LWP to a realtime

thread you are ensuring the threadwill be able to runwithminimal

delay once it is scheduled.

4.6 APthread program that performs the summation function was provided

in Section 4.3.1. Rewrite this program in Java.

Answer: Please refer to the supportingWeb site for source code solution.

CH5APTER

CPU Scheduling

Exercises

5.1 A CPU scheduling algorithm determines an order for the execution of its

scheduled processes. Given n processes to be scheduled on one processor,

how many possible different schedules are there? Give a formula in

terms of n.

Answer: n! (n factorial = n × n – 1 × n – 2 × ... × 2 × 1).

5.2 Define the difference between preemptive and nonpreemptive scheduling.

Answer: Preemptive scheduling allows a process to be interrupted

in the midst of its execution, taking the CPU away and allocating it

to another process. Nonpreemptive scheduling ensures that a process

relinquishes control of the CPU only when it finishes with its current

CPU burst.

5.3 Suppose that the following processes arrive for execution at the times

indicated. Each process will run the listed amount of time. In answering

the questions, use nonpreemptive scheduling and base all decisions on

the information you have at the time the decision must be made.

Process Arrival Time Burst Time

P1 0.0 8

P2 0.4 4

P3 1.0 1

a. What is the average turnaround time for these processes with the

FCFS scheduling algorithm?

14 Chapter 5 CPU Scheduling

b. What is the average turnaround time for these processeswith the

SJF scheduling algorithm?

c. The SJF algorithm is supposed to improve performance, but notice

that we chose to run process P1 at time 0 because we did not know

that two shorter processes would arrive soon. Compute what the

average turnaround timewill be if the CPU is left idle for the first 1

unit and then SJF scheduling is used. Remember that processes P1

and P2 arewaiting during this idle time, so their waiting timemay

increase. This algorithm could be known as future-knowledge

scheduling.

Answer:

a. 10.53

b. 9.53

c. 6.86

Remember that turnaround time is finishing time minus arrival time, so

you have to subtract the arrival times to compute the turnaround times.

FCFS is 11 if you forget to subtract arrival time.

5.4 What advantage is there in having different time-quantum sizes on different

levels of a multilevel queueing system?

Answer: Processes that need more frequent servicing, for instance,

interactive processes such as editors, can be in a queue with a small time

quantum. Processeswith no need for frequent servicing can be in a queue

with a larger quantum, requiring fewer context switches to complete the

processing, and thus making more efficient use of the computer.

5.5 Many CPU-scheduling algorithms are parameterized. For example, the

RR algorithm requires a parameter to indicate the time slice. Multilevel

feedback queues require parameters to define the number of queues,

the scheduling algorithms for each queue, the criteria used to move

processes between queues, and so on.

These algorithms are thus really sets of algorithms (for example, the

set of RR algorithms for all time slices, and so on). One set of algorithms

may include another (for example, the FCFS algorithm is the RR algorithm

with an infinite time quantum).What (if any) relation holds between the

following pairs of sets of algorithms?

a. Priority and SJF

b. Multilevel feedback queues and FCFS

c. Priority and FCFS

d. RR and SJF

Answer:

a. The shortest job has the highest priority.

b. The lowest level of MLFQ is FCFS.

Exercises 15

c. FCFS gives the highest priority to the job having been in existence

the longest.

d. None.

5.6 Suppose that a scheduling algorithm (at the level of short-term CPU

scheduling) favors those processes that have used the least processor

time in the recent past. Why will this algorithm favor I/O-bound programs

and yet not permanently starve CPU-bound programs?

Answer: It will favor the I/O-bound programs because of the relatively

short CPU burst request by them; however, the CPU-bound programs

will not starve because the I/O-bound programs will relinquish the CPU

relatively often to do their I/O.

5.7 Distinguish between PCS and SCS scheduling.

Answer: PCS scheduling is done local to the process. It is how the

thread library schedules threads onto available LWPs. SCS scheduling is

the situation where the operating system schedules kernel threads. On

systems using either many-to-one ormany-to-many, the two scheduling

models are fundamentally different. On systems using one-to-one, PCS

and SCS are the same.

5.8 Assume an operating system maps user-level threads to the kernel using

the many-to-many model where the mapping is done through the use

of LWPs. Furthermore, the system allows program developers to create

real-time threads. Is it necessary to bind a real-time thread to an LWP?

Answer: Yes, otherwise a user thread may have to compete for an

available LWP prior to being actually scheduled. By binding the user

thread to an LWP, there is no latency while waiting for an available LWP;

the real-time user thread can be scheduled immediately.

CH6APTER

Process

Synchronization

Exercises

6.1 In Section 6.4 we mentioned that disabling interrupts frequently could

affect the system’s clock. Explain why it could and how such effects

could be minimized.

Answer: The system clock is updated at every clock interrupt. If interrupts

were disabled—particularly for a long period of time—it is

possible the system clock could easily lose the correct time. The system

clock is also used for scheduling purposes. For example, the time

quantum for a process is expressed as a number of clock ticks. At every

clock interrupt, the scheduler determines if the time quantum for the

currently running process has expired. If clock interruptswere disabled,

the scheduler could not accurately assign time quantums. This effect can

be minimized by disabling clock interrupts for only very short periods.

6.2 The Cigarette-Smokers Problem. Consider a system with three smoker processes

and one agent process. Each smoker continuously rolls a cigarette

and then smokes it. But to roll and smoke a cigarette, the smoker needs

three ingredients: tobacco, paper, and matches. One of the smoker processes

has paper, another has tobacco, and the third has matches. The

agent has an infinite supply of all three materials. The agent places

two of the ingredients on the table. The smoker who has the remaining

ingredient then makes and smokes a cigarette, signaling the agent on

completion. The agent then puts out another two of the three ingredients,

and the cycle repeats. Write a program to synchronize the agent

and the smokers using Java synchronization.

Answer: Please refer to the supportingWeb site for source code solution.

6.3 Give the reasons why Solaris, Windows XP, and Linux implement multiple

locking mechanisms. Describe the circumstances under which they

18 Chapter 6 Process Synchronization

use spinlocks, mutexes, semaphores, adaptive mutexes, and condition

variables. In each case, explain why the mechanism is needed.

Answer: These operating systems provide different locking mechanisms

depending on the application developers’ needs. Spinlocks are

useful formultiprocessor systemswhere a thread can run in a busy-loop

(for a short period of time) rather than incurring the overhead of being

put in a sleep queue. Mutexes are useful for locking resources. Solaris 2

uses adaptive mutexes, meaning that the mutex is implemented with a

spin lock on multiprocessor machines. Semaphores and condition variables

are more appropriate tools for synchronization when a resource

must be held for a long period of time, since spinning is inefficient for a

long duration.

6.4 Explain the differences, in terms of cost, among the three storage types

volatile, nonvolatile, and stable.

Answer: Volatile storage refers to main and cache memory and is very

fast. However, volatile storage cannot survive system crashes or powering

down the system. Nonvolatile storage survives system crashes and

powered-down systems. Disks and tapes are examples of nonvolatile

storage. Recently, USB devices using erasable program read-only memory

(EPROM) have appeared providing nonvolatile storage. Stable storage

refers to storage that technically can never be lost as there are redundant

backup copies of the data (usually on disk).

6.5 Explain the purpose of the checkpoint mechanism. How often should

checkpoints be performed? Describe how the frequency of checkpoints

affects:

• System performance when no failure occurs

• The time it takes to recover from a system crash

• The time it takes to recover from a disk crash

Answer: A checkpoint log record indicates that a log record and its

modified data has been written to stable storage and that the transaction

need not to be redone in case of a system crash. Obviously, themore often

checkpoints are performed, the less likely it is that redundant updates

will have to be performed during the recovery process.

• System performance when no failure occurs—If no failures occur,

the system must incur the cost of performing checkpoints that are

essentially unnecessary. In this situation, performing checkpoints

less often will lead to better system performance.

• The time it takes to recover from a system crash—The existence of a

checkpoint record means that an operation will not have to be

redone during system recovery. In this situation, the more often

checkpoints were performed, the faster the recovery time is from a

system crash.

• The time it takes to recover from a disk crash—The existence of a

checkpoint record means that an operation will not have to be

redone during system recovery. In this situation, the more often

Exercises 19

checkpoints were performed, the faster the recovery time is from a

disk crash.

6.6 Explain the concept of transaction atomicity.

Answer: A transaction is a series of read and write operations upon

some data followed by a commit operation. If the series of operations in

a transaction cannot be completed, the transaction must be aborted and

the operations that did take place must be rolled back. It is important

that the series of operations in a transaction appear as one indivisible

operation to ensure the integrity of the data being updated. Otherwise,

data could be compromised if operations from two (or more) different

transactions were intermixed.

6.7 Show that some schedules are possible under the two-phase locking

protocol but not possible under the timestamp protocol, and vice versa.

Answer: A schedule that is allowed in the two-phase locking protocol

but not in the timestamp protocol is:

step T0 T1 Precedence

1 lock-S(A)

2 read(A)

3 lock-X(B)

4 write(B)

5 unlock(B)

6 lock-S(B)

7 read(B) T1 → T0

8 unlock(A)

9 unlock(B)

This schedule is not allowed in the timestamp protocol because at step

7, the W-timestamp of B is 1.

A schedule that is allowed in the timestamp protocol but not in the

two-phase locking protocol is:

step T0 T1 T2

1 write(A)

2 write(A)

3 write(A)

4 write(B)

5 write(B)

This schedule cannot have lock instructions added tomake it legal under

two-phase locking protocol because T1 must unlock (A) between steps 2

and 3, and must lock (B) between steps 4 and 5.

6.8 The wait() statement in all Java program examples was part of a while

loop. Explain why you would always need to use a while statement

when using wait() and why you would never use an if statement.

Answer: This is an important issue to emphasize! Java only provides

anonymous notification—you cannot notify a certain thread that a cer20

Chapter 6 Process Synchronization

tain condition is true. When a thread is notified, it is its responsibility

to re-check the condition that it is waiting for. If a thread did not recheck

the condition, it might have received the notification without the

condition having been met.

CH7APTER

Deadlocks

Exercises

7.1 List three examples of deadlocks that are not related to a computersystem

environment.

Answer:

• Two cars crossing a single-lane bridge from opposite directions.

• A person going down a ladder while another person is climbing up

the ladder.

• Two trains traveling toward each other on the same track.

• Two carpenters who must pound nails. There is a single hammer

and a single bucket of nails. Deadlock occurs if one carpenter has

the hammer and the other carpenter has the nails.

7.2 Suppose that a system is in an unsafe state. Show that it is possible for

the processes to complete their execution without entering a deadlock

state.

Answer: An unsafe state may not necessarily lead to deadlock, it just

means that we cannot guarantee that deadlock will not occur. Thus, it

is possible that a system in an unsafe state may still allow all processes

to complete without deadlock occurring. Consider the situation where

a system has 12 resources allocated among processes P0, P1, and P2. The

resources are allocated according to the following policy:

Max Current Need

P0 10 5 5

P1 4 2 2

P2 9 3 6

22 Chapter 7 Deadlocks

for (int i = 0; i <>

// first find a thread that can finish

for (int j = 0; j <>

if (!finish[j]) {

boolean temp = true;

for (int k = 0; k <>

if (need[j][k] > work[k])

temp = false;

}

if (temp) { // if this thread can finish

finish[j] = true;

for (int x = 0; x <>

work[x] += work[j][x];

}

} Figure 7.1 Banker’s algorithm safety algorithm.

Currently there are two resources available. This system is in an unsafe

state as process P1 could complete, thereby freeing a total of four

resources. But we cannot guarantee that processes P0 and P2 can complete.

However, it is possible that a processmay release resources before

requesting any further. For example, process P2 could release a resource,

thereby increasing the total number of resources to five. This allows process

P0 to complete, which would free a total of nine resources, thereby

allowing process P2 to complete as well.

7.3 Prove that the safety algorithm presented in Section 7.5.3 requires an

order of m × n2 operations.

Answer:

Figure 7.1 provides Java code that implement the safety algorithm of

the banker’s algorithm (the complete implementation of the banker’s

algorithm is available with the source code download).

As can be seen, the nested outer loops—both of which loop through n

times—provide the n2 performance. Within these outer loops are two

sequential inner loops which loop m times. The big-oh of this algorithm

is therefore O(m × n2).

7.4 Consider a computer system that runs 5,000 jobs per month with no

deadlock-prevention or deadlock-avoidance scheme. Deadlocks occur

about twice permonth, and the operatormust terminate and rerun about

10 jobs per deadlock. Each job is worth about $2 (in CPU time), and the

jobs terminated tend to be about half-done when they are aborted.

A systems programmer has estimated that a deadlock-avoidance

algorithm (like the banker’s algorithm) could be installed in the system

with an increase in the average execution time per job of about 10 percent.

Since the machine currently has 30-percent idle time, all 5,000 jobs per

month could still be run, although turnaround time would increase by

about 20 percent on average.

Exercises 23

a. What are the arguments for installing the deadlock-avoidance

algorithm?

b. What are the arguments against installing the deadlock-avoidance

algorithm?

Answer: An argument for installing deadlock avoidance in the system

is thatwe could ensure deadlockwould never occur. In addition, despite

the increase in turnaround time, all 5,000 jobs could still run.

An argument against installing deadlock avoidance software is that

deadlocks occur infrequently and they cost little when they do occur.

7.5 Can a system detect that some of its processes are starving? If you answer

“yes,” explain how it can. If you answer “no,” explain how the system

can deal with the starvation problem.

Answer: Starvation is a difficult topic to define as it maymean different

things for different systems. For the purposes of this question, we will

define starvation as the situation whereby a process must wait beyond

a reasonable period of time—perhaps indefinitely—before receiving a

requested resource. One way of detecting starvation would be to first

identify a period of time—T—that is considered unreasonable. When a

process requests a resource, a timer is started. If the elapsed time exceeds

T, then the process is considered to be starved.

One strategy for dealing with starvation would be to adopt a policy

where resources are assigned only to the process that has been waiting

the longest. For example, if process Pa has been waiting longer for resource

X than process Pb , the request fromprocess Pb would be deferred

until process Pa ’s request has been satisfied.

Another strategy would be less strict than what was just mentioned. In

this scenario, a resource might be granted to a process that haswaited less

than another process, providing that the other process is not starving.

However, if another process is considered to be starving, its request

would be satisfied first.

7.6 Consider the following resource-allocation policy. Requests and releases

for resources are allowed at any time. If a request for resources cannot

be satisfied because the resources are not available, then we check any

processes that are blocked, waiting for resources. If they have the desired

resources, then these resources are taken away from them and are given

to the requesting process. The vector of resources for which the process

is waiting is increased to include the resources that were taken away.

For example, consider a system with three resource types and the

vector Available initialized to (4,2,2). If process P0 asks for (2,2,1), it gets

them. If P1 asks for (1,0,1), it gets them. Then, if P0 asks for (0,0,1), it

is blocked (resource not available). If P2 now asks for (2,0,0), it gets the

available one (1,0,0) and one thatwas allocated to P0 (since P0 is blocked).

P0’s Allocation vector goes down to (1,2,1) and its Need vector goes up to

(1,0,1).

a. Can deadlock occur? If you answer “yes”, give an example. If you

answer “no,” specify which necessary condition cannot occur.

b. Can indefinite blocking occur? Explain your answer.

24 Chapter 7 Deadlocks

Answer:

a. Deadlock cannot occur because preemption exists.

b. Yes. A process may never acquire all the resources it needs if they

are continuously preempted by a series of requests such as those

of process C.

7.7 Suppose that you have coded the deadlock-avoidance safety algorithm

and now have been asked to implement the deadlock-detection algorithm.

Can you do so by simply using the safety algorithm code and

redefining Maxi = Waitingi + Alloca tioni, where Waitingi is a vector

specifying the resources process i is waiting for, and Alloca tioni is as

defined in Section 7.5? Explain your answer.

Answer:

Yes. The Max vector represents the maximum request a process may

make. When calculating the safety algorithm we use the Need matrix,

which representsMax— Allocation. Another way to think of this is Max

= Need + Allocation. According to the question, theWaiting matrix fulfills

a role similar to the Need matrix, therefore Max = Waiting + Allocation.

7.8 Is it possible to have a deadlock involving only one single process?

Explain your answer.

Answer: No. This follows directly from the hold-and-wait condition.

CH8APTER

Memory

Management

Exercises

8.1 Name two differences between logical and physical addresses.

Answer: A logical address does not refer to an actual existing address;

rather, it refers to an abstract address in an abstract address space. Contrast

thiswith a physical address that refers to an actual physical address

in memory. A logical address is generated by the CPU and is translated

into a physical address by the memory management unit(MMU). Therefore,

physical addresses are generated by the MMU.

8.2 Consider a system in which a program can be separated into two parts:

code and data. The CPU knows whether it wants an instruction (instruction

fetch) or data (data fetch or store). Therefore, two base–limit

instruction base–limit register pair is automatically read-only, so programs

can be shared among different users. Discuss the advantages and

disadvantages of this scheme.

Answer: The major advantage of this scheme is that it is an effective

mechanism for code and data sharing. For example, only one copy of an

editor or a compiler needs to be kept in memory, and this code can be

shared by all processes needing access to the editor or compiler code.

Another advantage is protection of code against erroneous modification.

The only disadvantage is that the code and data must be separated,

which is usually adhered to in a compiler-generated code.

8.3 Why are page sizes always powers of 2?

Answer: Recall that paging is implemented by breaking up an address

into a page and offset number. It is most efficient to break the address

into X page bits and Y offset bits, rather than perform arithmetic on

the address to calculate the page number and offset. Because each bit

26 Chapter 8 Memory Management

position represents a power of 2, splitting an address between bits results

in a page size that is a power of 2.

8.4 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?

Answer:

a. Logical address: 13 bits

b. Physical address: 15 bits

8.5 What is the effect of allowing two entries in a page table to point to the

same page frame in memory? Explain how this effect could be used to

decrease the amount of time needed to copy a large amount of memory

from one place to another. What effect would updating some byte on the

one page have on the other page?

Answer: By allowing two entries in a page table to point to the same

page frame inmemory, users can share code and data. If the code is reentrant,

much memory space can be saved through the shared use of large

programs such as text editors, compilers, and database systems. “Copying”

large amounts of memory could be effected by having different

page tables point to the same memory location.

However, sharing of nonreentrant code or data means that any user

having access to the code can modify it and these modifications would

be reflected in the other user’s “copy.”

8.6 Describe amechanism by which one segment could belong to the address

space of two different processes.

Answer: Since segment tables are a collection of base–limit registers,

segments can be sharedwhen entries in the segment table of two different

jobs point to the same physical location. The two segment tables must

have identical base pointers, and the shared segment number must be

the same in the two processes.

8.7 Sharing segments among processeswithout requiring the same segment

number is possible in a dynamically linked segmentation system.

a. Define a system that allows static linking and sharing of segments

without requiring that the segment numbers be the same.

b. Describe a paging scheme that allows pages to be shared without

requiring that the page numbers be the same.

Answer: Both of these problems reduce to a program being able to

reference both its own code and its datawithout knowing the segment or

page number associated with the address. MULTICS solved this problem

by associating four registers with each process. One register had the

address of the current program segment, another had a base address for

the stack, another had a base address for the global data, and so on. The

idea is that all references have to be indirect through a register thatmaps

to the current segment or page number. By changing these registers, the

Exercises 27

same code can execute for different processes without the same page or

segment numbers.

8.8 In the IBM/370, memory protection is provided through the use of keys.

A key is a 4-bit quantity. Each 2K block of memory has a key (the storage

key) associated with it. The CPU also has a key (the protection key)

associated with it. A store operation is allowed only if both keys are

equal, or if either is zero.Which of the following memory-management

schemes could be used successfully with this hardware?

a. Bare machine

b. Single-user system

c. Multiprogramming with a fixed number of processes

d. Multiprogramming with a variable number of processes

e. Paging

f. Segmentation

Answer:

a. Protection not necessary, set system key to 0.

b. Set system key to 0 when in supervisor mode.

c. Region sizes must be fixed in increments of 2k bytes, allocate key

with memory blocks.

d. Same as above.

e. Frame sizes must be in increments of 2k bytes, allocate key with

pages.

f. Segment sizes must be in increments of 2k bytes, allocate keywith

segments.

CH9APTER

Virtual

Memory

Exercises

9.1 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 been

brought into main memory takes place. The operating system verifies

the 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 pagereplacement

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. n

b. p

9.3 Which of the following programming techniques and structures are

“good” for a demand-paged environment ? Which are “not good”? Explain

your answers.

a. Stack

b. Hashed symbol table

30 Chapter 9 Virtual Memory

c. Sequential search

d. Binary search

e. Pure code

f. Vector operations

g. Indirection

Answer:

a. Stack—good.

b. Hashed symbol table—not good.

c. Sequential search—good.

d. Binary search—not good.

e. Pure code—good.

f. Vector operations—good.

g. Indirection—not good.

9.4 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 fromBelady’s

anomaly from those that do not.

a. LRU replacement

b. FIFO replacement

c. Optimal replacement

d. Second-chance replacement

Answer:

Rank Algorithm Suffer from Belady’s anomaly

1 Optimal no

2 LRU no

3 Second-chance yes

4 FIFO yes

9.5 When virtual memory is implemented in a computing system, there are

certain costs associated with the technique and certain benefits. List the

costs and the benefits. Is it possible for the costs to exceed the benefits?

If it is, what measures can be taken to ensure that this does not happen?

Answer: The costs are additional hardware and slower access time. The

benefits are good utilization ofmemory and larger logical address space

than physical address space.

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

Exercises 31

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 accessed a

page already in memory.

• 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 sec

9.7 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 is in page 0 (locations 0 to 199) for manipulating

the matrix; thus, every instruction fetch will be from page

For three page frames, how many page faults are generated by the following

array-initialization loops, using LRU replacement, and assuming

page frame 1 has the process in it, and the other two are initially empty?

a. for (int j = 0; j <>

for (int i = 0; i <>

A[i][j] = 0;

b. for (int i = 0; i <>

for (int j = 0; j <>

A[i][j] = 0;

Answer:

a. 50

b. 5,000

9.8 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.

32 Chapter 9 Virtual Memory

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 replacement

Answer:

Number of frames LRU FIFO Optimal

1 20 20 20

2 18 18 15

3 15 16 11

4 10 14 8

5 8 10 7

6 7 10 7

7 7 7 7

9.9 Suppose that you want to use a paging algorithm that requires a reference

bit (such as second-chance replacement or working-setmodel), but

the 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.On 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 newpage-replacement algorithmthat you think may

be optimal. In some contorted 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 not be used for the longest time. Belady’s anomaly occurs

when a page-replacement algorithm evicts a page that will be needed

in the immediate future. An optimal algorithm would not have selected

such a page.

9.11 Segmentation is similar to paging but uses variable-sized “pages.” Define

two segment-replacement algorithms based on FIFO and LRU pagereplacement

schemes. Remember that since segments are not the same

size, the segment that is chosen to be replaced may not be big enough

to leave enough consecutive locations for the needed segment. Consider

strategies for systems where segments cannot be relocated, and those

for systems where they can.

Answer:

Exercises 33

a. FIFO. Find the first segment large enough to accommodate the incoming

segment. If relocation is not possible and no one segment

is large enough, select a combination of segments whose memories

are contiguous, which are “closest to the first of the list”

and which can accommodate the new segment. If relocation is

possible, rearrange the memory so that the first N segments large

enough for the incoming segment are contiguous in memory.Add

any leftover space to the free-space list in both cases.

b. LRU. Select the segment that has not been used for the longest

period of time and that is large enough, adding any leftover space

to the free space list. If no one segment is large enough, select

a combination of the “oldest” segments that are contiguous in

memory (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 percent

b. CPU utilization 87 percent; disk utilization 3 percent

c. CPU utilization 13 percent; disk utilization 3 percent

Answer:

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 machine 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 thememory. Therewill be some problemwith internal fragmentation.

C1HA0PTER

File-System

Interface

Exercises

10.1 Some systems automatically delete all user files when a user logs off or

a job terminates, unless the user explicitly requests that they be kept;

other systems keep all files unless the user explicitly deletes them.

Discuss the relative merits of each approach.

Answer: Deleting all files not specifically saved by the user has the

advantage of minimizing the file space needed for each user by not

saving unwanted or unnecessary files. Saving all files unless specifically

deleted is more secure for the user in that it is not possible to lose files

inadvertently by forgetting to save them.

10.2 Why do some systems keep track of the type of a file,while others leave

it to the user or simply do not implement multiple file types? Which

system is “better?”

Answer: Some systems allow different file operations based on the

type of the file (for instance, an ascii file can be read as a stream while a

database file can be read via an index to a block). Other systems leave

such interpretation of a file’s data to the process and provide no help in

accessing the data. The method that is “better” depends on the needs

of the processes on the system, and the demands the users place on the

operating system. If a system runs mostly database applications, itmay

be more efficient for the operating system to implement a databasetype

file and provide operations, rather than making each program

implement the same thing (possibly in different ways). For generalpurpose

systems it may be better to only implement basic file types to

keep the operating system size smaller and allow maximum freedom

to the processes on the system.

36 Chapter 10 File-System Interface

10.3 Similarly, some systems support many types of structures for a file’s

data, while others simply support a stream of bytes. What are the

advantages and disadvantages?

Answer: An advantage of having the system support different file

structures is that the support comes from the system; individual applications

are not required to provide the support. In addition, if the

system provides the support for different file structures, it can implement

the support presumably more efficiently than an application.

The disadvantage of having the system provide support for defined file

types is that it increases the size of the system. In addition, applications

that may require different file types other than what is provided by the

system may not be able to run on such systems.

An alternative strategy is for the operating system to define no support

for file structures and instead treat all files as a series of bytes. This is

the approach taken by UNIX systems. The advantage of this approach

is that it simplifies the operating system support for file systems, as the

system no longer has to provide the structure for different file types.

Furthermore, it allows applications to define file structures, thereby alleviating

the situation where a system may not provide a file definition

required for a specific application.

10.4 Could you simulate a multilevel directory structure with a single-level

directory structure in which arbitrarily long names can be used? If your

answer is yes, explain how you can do so, and contrast this scheme with

the multilevel directory scheme. If your answer is no, explain what

prevents your simulation’s success. How would your answer change

if file names were limited to seven characters?

Answer: If arbitrarily long names can be used then it is possible to

simulate amultilevel directory structure. This can be done, for example,

by using the character “.” to indicate the end of a subdirectory. Thus, for

example, the name jim.java.F1 specifies that F1 is a file in subdirectory

javawhich in turn is in the root directory jim.

If file names were limited to seven characters, then the above scheme

could not be utilized and thus, in general, the answer is no. The next best

approach in this situation would be to use a specific file as a symbol

table (directory) to map arbitrarily long names (such as jim.java.F1)

into shorter arbitrary names (such as XX00743), which are then used

for actual file access.

10.5 Explain the purpose of the open() and close() operations.

Answer:

• The open() operation informs the system that the named file is

about to become active.

• The close() operation informs the system that the named file is

no longer in active use by the user who issued the close operation.

10.6 Give an example of an application in which data in a file should be

accessed in the following order:

a. Sequentially

Exercises 37

b. Randomly

Answer:

a. Print the content of the file.

b. Print the content of record i. This record can be found using

hashing or index techniques.

10.7 In some systems, a subdirectory can be read and written by an authorized

user, just as ordinary files can be.

a. Describe the protection problems that could arise.

b. Suggest a scheme for dealing with each of the protection problems

you named in part a.

Answer:

a. One piece of information kept in a directory entry is file location.

If a user could modify this location, then he could access other

files defeating the access-protection scheme.

b. Do not allow the user to directly write onto the subdirectory.

Rather, provide system operations to do so.

10.8 Consider a system that supports 5000 users. Suppose that you want to

allow 4990 of these users to be able to access one file.

a. How would you specify this protection scheme in UNIX?

b. Could you suggest another protection scheme that can be used

more effectively for this purpose than the scheme provided by

UNIX?

Answer:

a. There are two methods for achieving this:

1. Create an access control list with the names of all 4990

users.

2. Put these 4990 users in one group and set the group access

accordingly. This scheme cannot always be implemented

since user groups are restricted by the system.

b. The universal access to files applies to all users unless their name

appears in the access-control list with different access permission.

With this scheme you simply put the names of the remaining

ten users in the access control list but with no access

privileges allowed.

10.9 Researchers have suggested that, instead of having an access list associated

with each file (specifying which users can access the file, and how),

we should have a user control list associated with each user (specifying

which files a user can access, and how). Discuss the relative merits of

these two schemes.

Answer:

38 Chapter 10 File-System Interface

• File control list. Since the access control information is concentrated

in one single place, it is easier to change access control information

and this requires less space.

• User control list. This requires less overhead when opening a file.

CH1A1PTER

File-System

Implementation

Exercises

11.1 Consider a file currently consisting of 100 blocks. Assume that the

file control 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 in the

beginning, but there is room to grow in the end. Assume that the 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:

Contiguous Linked Indexed

a. 201 1 1

b. 101 52 1

c. 1 3 1

d. 198 1 0

e. 98 52 0

f. 0 100 0

40 Chapter 11 File-System Implementation

11.2 What problems could occur if a system allowed a file system to be

mounted simultaneously at more than one location?

Answer: There would be multiple paths to the same file, which could

confuse users or encourage mistakes (deleting a file with one path

deletes the file in all the other paths).

11.3 Whymust the bitmap for file allocation be kept on mass storage, rather

than in main memory?

Answer: In case of system crash (memory failure) the free-space list

would not be lost as it would be if the bit map had been stored in main

memory.

11.4 Consider a system that supports the strategies of contiguous, linked,

and indexed allocation.What criteria should be used in decidingwhich

strategy is best utilized for a particular file?

Answer:

• Contiguous—if file is usually accessed sequentially, if file is

relatively small.

• Linked—if file is large and usually accessed sequentially.

• Indexed—if file is large and usually accessed randomly.

11.5 One problem with contiguous allocation is that the user must preallocate

enough space for each file. If the file grows to be larger than the

space 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 initial

contiguous area. If the overflow area is filled, another overflow area

is allocated. Compare this implementation of a file with the standard

contiguous and linked implementations.

Answer: This method requires more overhead then the standard contiguous

allocation. It requires less overhead than the standard linked

allocation.

11.6 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

more efficiently by storing data from the slower device, temporarily,

in a 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.

11.7 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 allowmore flexibility in system use growth—

tables are never exceeded, avoiding artificial use limits. Unfortunately,

kernel structures and code are more complicated, so there is more

potential for bugs. The use of one resource can take away more system

resources (by growing to accommodate the requests) than with static

tables.

Exercises 41

11.8 Explain how the VFS layer allows an operating system easily to support

multiple types of file systems.

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

structures to the VFS layer. A system call is translated into the proper

specific functions for the target 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.

Mass C1HA2PTER

Storage

Structure

Exercises

12.1 The accelerating seek described in Exercise 12.3 is typical of hard-disk

drives. By contrast, floppy disks (and many hard disks manufactured

before the mid-1980s) typically seek at a fixed rate. Suppose that the

disk in Exercise 12.3 has a constant-rate seek rather than a constantacceleration

seek, so the seek time is of the form t = x + yL, where t

is the time in milliseconds and L is the seek distance. Suppose that the

time to seek to an adjacent cylinder is 1 millisecond, as before, and is

0.5 milliseconds for each additional cylinder.

a. Write an equation for this seek time as a function of the seek

distance.

b. Using the seek-time function from part a, calculate the total seek

time for each of the schedules in Exercise 12.2. Is your answer

the same as it was for Exercise 12.3(c)?

c. What is the percentage speedup of the fastest schedule over FCFS

in this case?

Answer:

a. t = 0.95 + 0.05L

b. FCFS 362.60; SSTF 95.80; SCAN 497.95; LOOK 174.50; C-SCAN 500.15;

(and C-LOOK 176.70). SSTF is still the winner, and LOOK is the

runner-up.

c. (362.60 − 95.80)/362.60 = 0.74 The percentage speedup of SSTF

over FCFS is 74%, with respect to the seek time. If we include the

overhead of rotational latency and data transfer, the percentage

speedup will be less.

44 Chapter 12 Mass-Storage Structure

12.2 Is disk scheduling, other than FCFS scheduling, useful in a single-user

environment? 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.

12.3 Explain why SSTF scheduling tends to favor middle cylinders over the

innermost and outermost cylinders.

Answer: 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.

12.4 Why is rotational latency usually not considered in disk scheduling?

How would you modify SSTF, SCAN, and C-SCAN to include latency

optimization?

Answer: Most disks do not export their rotational position information

to the host. Even if they did, the time for this information to reach the

scheduler would be subject to imprecision and the time consumed by

the scheduler is variable, so the rotational position information would

become incorrect. Further, the disk requests are usually given in terms

of logical block numbers, and the mapping between logical blocks and

physical locations is very complex.

12.5 Howwould use of a RAM disk affect your selection of a disk-scheduling

algorithm? What factors would you need to consider? Do the same

considerations apply to hard-disk scheduling, given that the file system

stores recently used blocks in a buffer cache in main memory?

Answer: Disk scheduling attempts to reduce the overhead time of

disk head positioning. Since a RAM disk has uniform access times,

scheduling is largely unnecessary. The comparison between RAM disk

and the main memory disk-cache has no implications for hard-disk

scheduling because we schedule only the buffer cache misses, not the

requests that find their data in main memory.

12.6 Why is it important to balance file system I/O among the disks and

controllers on a system in a multitasking environment?

Answer: A system can perform only at the speed of its slowest bottleneck.

Disks or disk controllers are frequently the bottleneck in modern

systems as their individual performance cannot keep up with that of

the CPU and system bus. By balancing I/O among disks and controllers,

neither an individual disk nor a controller is overwhelmed, so that

bottleneck is avoided.

Exercises 45

12.7 What are the tradeoffs involved in rereading code pages from the file

system versus using swap space to store them?

Answer: If code pages are stored in swap space, they can be transferred

more quickly to main memory (because swap space allocation is tuned

for faster performance than general file system allocation). Using swap

space can require startup time if the pages are copied there at process

invocation rather than just being paged out to swap space on demand.

Also, more swap space must be allocated if it is used for both code and

data pages.

12.8 Is there any way to implement truly stable storage? Explain your answer.

Answer: Truly stable storage would never lose data. The fundamental

technique for stable storage is to maintain multiple copies of the data,

so that if one copy is destroyed, some other copy is still available for

use. But for any scheme, we can imagine a large enough disaster that

all copies are destroyed.

12.9 The term “fast wide SCSI-II” denotes a SCSI bus that operates at a data

rate of 20 megabytes per second when it moves a packet of bytes between

the host and a device. Suppose that a fast wide SCSI-II disk drive

spins at 7200 RPM, has a sector size of 512 bytes, and holds 160 sectors

per track.

a. Estimate the sustained transfer rate of this drive in megabytes

per second.

b. Suppose that the drive has 7000 cylinders, 20 tracks per cylinder,

a head switch time (from one platter to another) of 0.5 millisecond,

and an adjacent cylinder seek time of 2 milliseconds. Use

this additional information to give an accurate estimate of the

sustained transfer rate for a huge transfer.

c. Suppose that the average seek time for the drive is 8 milliseconds.

Estimate the I/Os per second and the effective transfer rate for

a random-access workload that reads individual sectors that are

scattered across the disk.

d. Calculate the random-access I/Os per second and transfer rate

for I/O sizes of 4 kilobytes, 8 kilobytes, and 64 kilobytes.

e. If multiple requests are in the queue, a scheduling algorithm such

as SCAN should be able to reduce the average seek distance. Suppose

that a random-access workload is reading 8-kilobyte pages,

the average queue length is 10, and the scheduling algorithm

reduces the average seek time to 3 milliseconds. Now calculate

the I/Os per second and the effective transfer rate of the drive.

Answer:

a. The disk spins 120 times per second, and each spin transfers a

track of 80 KB. Thus, the sustained transfer rate can be approximated

as 9600 KB/s.

46 Chapter 12 Mass-Storage Structure

b. Suppose that 100 cylinders is a huge transfer. The transfer rate is

total bytes divided by total time. Bytes: 100 cyl * 20 trk/cyl * 80

KB/trk, i.e., 160,000 KB. Time: rotation time + track switch time +

cylinder switch time. Rotation time is 2000 trks/120 trks per sec,

i.e., 16.667 s. Track switch time is 19 switch per cyl * 100 cyl * 0.5

ms, i.e., 950 ms. Cylinder switch time is 99 * 2 ms, i.e., 198 ms.

Thus, the total time is 16.667 + 0.950 + 0.198, i.e., 17.815 s. (We

are ignoring any initial seek and rotational latency, which might

add about 12 ms to the schedule, i.e. 0.1%.) Thus the transfer rate

is 8981.2 KB/s. The overhead of track and cylinder switching is

about 6.5%.

c. The time per transfer is 8ms to seek + 4.167 ms average rotational

latency + 0.052 ms (calculated from 1/(120 trk per second * 160

sector per trk)) to rotate one sector past the disk head during

reading. We calculate the transfers per second as 1/(0.012219),

i.e., 81.8. Since each transfer is 0.5 KB, the transfer rate is 40.9

KB/s.

d. We ignore track and cylinder crossings for simplicity. For reads of

size 4 KB, 8 KB, and 64 KB, the corresponding I/Os per second are

calculated from the seek, rotational latency, and rotational transfer

time as in the previous item, giving (respectively) 1/(0.0126),

1/(0.013), and 1/(0.019). Thus we get 79.4, 76.9, and 52.6 transfers

per second, respectively. Transfer rates are obtained from 4,

8, and 64 times these I/O rates, giving 318 KB/s, 615 KB/s, and

3366 KB/s, respectively.

e. From 1/(3+4.167+0.83) we obtain 125 I/Os per second. From 8

KB per I/O we obtain 1000 KB/s.

12.10 More than one disk drive can be attached to a SCSI bus. In particular,

a fast wide SCSI-II bus (see Exercise 12.9) can be connected to at most

15 disk drives. Recall that this bus has a bandwidth of 20 megabytes

per second. At any time, only one packet can be transferred on the bus

between some disk’s internal cache and the host. However, a disk can

be moving its disk arm while some other disk is transferring a packet

on the bus. Also, a disk can be transferring data between its magnetic

platters and its internal cache while some other disk is transferring a

packet on the bus. Considering the transfer rates that you calculated

for the variousworkloads in Exercise 12.9, discuss how many disks can

be used effectively by one fast wide SCSI-II bus.

Answer:

For 8 KB random I/Os on a lightly loaded disk, where the random access

time is calculated to be about 13 ms (see Exercise 12.9), the effective

transfer rate is about 615 MB/s. In this case, 15 disks would have an

aggregate transfer rate of less than 10 MB/s, which should not saturate

the bus. For 64 KB random reads to a lightly loaded disk, the transfer

rate is about 3.4 MB/s, so five or fewer disk drives would saturate the

bus. For 8 KB reads with a large enough queue to reduce the average

seek to 3 ms, the transfer rate is about 1 MB/s, so the bus bandwidth

may be adequate to accommodate 15 disks.

Exercises 47

12.11 Remapping of bad blocks by sector sparing or sector slipping could

influence performance. Suppose that the drive in Exercise 12.9 has a

total of 100 bad sectors at random locations and that each bad sector is

mapped to a spare that is located on a different track, but within the

same cylinder. Estimate the number of I/Os per second and the effective

transfer rate for a random-access workload consisting of 8-kilobyte

reads, with a queue length of 1 (that is, the choice of scheduling algorithm

is not a factor).What is the effect of a bad sector on performance?

Answer: Since the disk holds 22,400,000 sectors, the probability of

requesting one of the 100 remapped sectors is very small. An example

of a worst-case event is that we attempt to read, say, an 8 KB page, but

one sector from the middle is defective and has been remapped to the

worst possible location on another track in that cylinder. In this case,

the time for the retrieval could be 8 ms to seek, plus two track switches

and two full rotational latencies. It is likely that a modern controller

would read all the requested good sectors fromthe original track before

switching to the spare track to retrieve the remapped sector and thus

would incur only one track switch and rotational latency. So the time

would be 8 ms seek + 4.17 ms average rotational latency + 0.05 ms

track switch + 8.3 ms rotational latency + 0.83 ms read time (8 KBis 16

sectors, 1/10 of a track rotation). Thus, the time to service this request

would be 21.8 ms, giving an I/O rate of 45.9 requests per second and

an effective bandwidth of 367 KB/s. For a severely time-constrained

application this might matter, but the overall impact in the weighted

average of 100 remapped sectors and 22.4 million good sectors is nil.

12.12 In a disk jukebox, what would be the effect of having more open files

than the number of drives in the jukebox?

Answer: Two bad outcomes could result. One possibility is starvation

of the applications that issue blocking I/Os to tapes that are not mounted

in drives.Another possibility is thrashing, as the jukebox is commanded

to switch tapes after every I/O operation.

12.13 If magnetic hard disks eventually have the same cost per gigabyte as do

tapes, will tapes become obsolete, or will they still be needed? Explain

your answer.

Answer: Tapes are easily removable, so they are useful for off-site

backups and for bulk transfer of data (by sending cartridges). As a

rule, a magnetic hard disk is not a removable medium.

12.14 It is sometimes said that tape is a sequential-access medium, whereas

magnetic disk is a random-access medium. In fact, the suitability of

a storage device for random access depends on the transfer size. The

term streaming transfer rate denotes the data rate for a transfer that

is underway, excluding the effect of access latency. By contrast, the

effective transfer rate is the ratio of total bytes per total seconds, including

overhead time such as the access latency.

Suppose that, in a computer, the level-2 cache has an access latency

of 8 nanoseconds and a streaming transfer rate of 800 megabytes per

second, the main memory has an access latency of 60 nanoseconds and

a streaming transfer rate of 80 megabytes per second, the magnetic disk

48 Chapter 12 Mass-Storage Structure

has an access latency of 15 millisecond and a streaming transfer rate of

5 megabytes per second, and a tape drive has an access latency of 60

seconds and a streaming transfer rate of 2 megabytes per seconds.

a. Random access causes the effective transfer rate of a device to

decrease, because no data are transferred during the access time.

For the disk described, what is the effective transfer rate if an

average access is followed by a streaming transfer of 512 bytes,

8 kilobytes, 1 megabyte, and 16 megabytes?

b. The utilization of a device is the the ratio of effective transfer

rate to streaming transfer rate. Calculate the utilization of the

disk drive for random access that performs transfers in each of

the four sizes given in part a.

c. Suppose that a utilization of 25 percent (or higher) is considered

acceptable. Using the performance figures given, compute the

smallest transfer size for disk that gives acceptable utilization.

d. Complete the following sentence: A disk is a random-access device

for transfers larger than bytes, and is a sequentialaccess

device for smaller transfers.

e. Compute the minimum transfer sizes that give acceptable utilization

for cache, memory, and tape.

f. Whenis a tape a random-access device, andwhenis it a sequentialaccess

device?

Answer:

a. For 512 bytes, the effective transfer rate is calculated as follows.

ETR = transfer size/transfer time.

If X is transfer size, then transfer time is ((X/STR) + latency).

Transfer time is 15ms + (512B/5MB per second) = 15.0097ms.

Effective transfer rate is therefore 512B/15.0097ms = 33.12 KB/sec.

ETR for 8KB = .47MB/sec.

ETR for 1MB = 4.65MB/sec.

ETR for 16MB = 4.98MB/sec.

b. Utilization of the device for 512B = 33.12 KB/sec / 5MB/sec =

.0064 = .64

For 8KB = 9.4%.

For 1MB = 93%.

For 16MB = 99.6%.

c. Calculate .25 = ETR/STR, solving for transfer size X.

STR = 5MB, so 1.25MB/S = ETR.

1.25MB/S * ((X/5) + .015) = X.

.25X + .01875 = X.

X = .025MB.

d. Adisk is a random-access device for transfers larger than Kbytes

(where K > disk block size), and is a sequential-access device for

smaller transfers.

Exercises 49

e. Calculate minimum transfer size for acceptable utilization of

cache memory:

STR = 800MB, ETR = 200, latency = 8 * 10−9.

200 (XMB/800 + 8 X 10−9) = XMB.

.25XMB + 1600 * 10−9 = XMB.

X = 2.24 bytes.

Calculate for memory:

STR = 80MB, ETR = 20, L = 60 * 10−9.

20 (XMB/80 + 60 * 10−9) = XMB.

.25XMB + 1200 * 10−9 = XMB.

X = 1.68 bytes.

Calculate for tape:

STR = 2MB, ETR = .5, L = 60s.

.5 (XMB/2 + 60) = XMB.

.25XMB + 30 = XMB.

X = 40MB.

f. It depends upon how it is being used. Assume we are using

the tape to restore a backup. In this instance, the tape acts as a

sequential-access device where we are sequentially reading the

contents of the tape. As another example, assume we are using

the tape to access a variety of records stored on the tape. In this

instance, access to the tape is arbitrary and hence considered

random.

12.15 Suppose that we agree that 1 kilobyte is 1,024 bytes, 1 megabyte is

1,0242 bytes, and 1 gigabyte is 1,0243 bytes. This progression continues

through terabytes, petabytes, and exabytes (1,0246). There are currently

several new proposed scientific projects that plan to record and store a

few exabytes of data during the next decade. To answer the following

questions, you will need to make a few reasonable assumptions; state

the assumptions that you make.

a. How many disk drives would be required to hold 4 exabytes of

data?

b. Howmanymagnetic tapes would be required to hold 4 exabytes

of data?

c. How many optical tapes would be required to hold 4 exabytes

of data (see Exercise 12.21)?

d. How many holographic storage cartridges would be required to

hold 4 exabytes of data (see Exercise 12.20)?

e. How many cubic feet of storage space would each option require?

Answer:

a. Assume that a disk drive holds 10 GB. Then 100 disks hold 1

TB, 100,000 disks hold 1 PB, and 100,000,000 disks hold 1 EB. To

store 4 EBwould require about 400 million disks. If a magnetic

tape holds 40 GB, only 100 million tapes would be required. If

50 Chapter 12 Mass-Storage Structure

an optical tape holds 50 times more data than a magnetic tape, 2

million optical tapeswould suffice. If a holographic cartridge can

store 180 GB, about 22.2 million cartridges would be required.

b. A 3.5" disk drive is about 1" high, 4" wide, and 6" deep. In feet,

this is 1/12 by 1/3 by 1/2, or 1/72 cubic feet. Packed densely,

the 400 million disks would occupy 5.6 million cubic feet. If we

allow a factor of two for air space and space for power supplies,

the required capacity is about 11 million cubic feet.

c. A 1/2" tape cartridge is about 1" high and 4.5" square. The volume

is about 1/85 cubic feet. For 100 million magnetic tapes

packed densely, the volume is about 1.2 million cubic feet. For 2

million optical tapes, the volume is 23,400 cubic feet.

d. A CD-ROM is 4.75" in diameter and about 1/16" thick. If we

assume that a holostore disk is stored in a library slot that is 5"

square and 1/8" wide, we calculate the volume of 22.2 million

disks to be about 40,000 cubic feet.

C1HA3PTER

I/O Systems

Exercises

13.1 State three advantages of placing functionality in a device controller,

rather than in the kernel. State three disadvantages.

Answer: Three advantages: Bugs are less likely to cause an operating

system crash

Performance can be improved by utilizing dedicated hardware and

hard-coded algorithms

The kernel is simplified by moving algorithms out of it

Three disadvantages: Bugs are harder to fix—a new firmware version

or new hardware is needed

Improving algorithms likewise require a hardware update rather than

just a kernel or device-driver update

Embedded algorithms could conflict with application’s use of the device,

causing decreased performance.

13.2 The example of handshaking in Section 13.2 used 2 bits: a busy bit and a

command-ready bit. Is it possible to implement this handshaking with

only 1 bit? If it is, describe the protocol. If it is not, explain why 1 bit is

insufficient.

Answer: It is possible, using the following algorithm. Let’s assume

we simply use the busy-bit (or the command-ready bit; this answer

is the same regardless). When the bit is off, the controller is idle. The

host writes to data-out and sets the bit to signal that an operation is

ready (the equivalent of setting the command-ready bit). When the

controller is finished, it clears the busy bit. The host then initiates the

next operation.

This solution requires that both the host and the controller have read

and write access to the same bit, which can complicate circuitry and

increase the cost of the controller.

52 Chapter 13 I/O Systems

13.3 Why might a system use interrupt-driven I/O to manage a single serial

port, but polling I/O to manage a front-end processor, such as a terminal

concentrator?

Answer: Polling can be more efficient than interrupt-driven I/O. This

is the case when the I/O is frequent and of short duration. Even though

a single serial port will perform I/O relatively infrequently and should

thus use interrupts, a collection of serial ports such as those in a terminal

concentrator can produce a lot of short I/O operations, and interrupting

for each one could create a heavy load on the system. A well-timed

polling loop could alleviate that load without wasting many resources

through looping with no I/O needed.

13.4 Polling for an I/O completion can waste a large number of CPU cycles

if the processor iterates a busy-waiting loop many times before the I/O

completes. But if the I/O device is ready for service, polling can be much

more efficient than is catching and dispatching an interrupt. Describe

a hybrid strategy that combines polling, sleeping, and interrupts for

I/O device service. For each of these three strategies (pure polling, pure

interrupts, hybrid), describe a computing environment in which that

strategy is more efficient than is either of the others.

Answer: A hybrid approach could switch between polling and interrupts

depending on the length of the I/O operation wait. For example,

we could poll and loop N times, and if the device is still busy at N+1,

we could set an interrupt and sleep. This approach would avoid long

busy-waiting cycles. This method would be best for very long or very

short busy times. It would be inefficient it the I/O completes at N+T

(where T is a small number of cycles) due to the overhead of polling

plus setting up and catching interrupts.

Pure polling is best with very short wait times. Interrupts are best with

known long wait times.

13.5 How does DMA increase system concurrency? How does it complicate

hardware design?

Answer: DMA increases system concurrency by allowing the CPU

to perform tasks while the DMA system transfers data via the system

and memory buses. Hardware design is complicated because the DMA

controller must be integrated into the system, and the system must

allow the DMA controller to be a bus master. Cycle stealing may also

be necessary to allow the CPU and DMA controller to share use of the

memory bus.

13.6 Why is it important to scale up system bus and device speeds as the

CPU speed increases?

Answer: Consider a system which performs 50% I/O and 50% computes.

Doubling the CPU performance on this system would increase

total system performance by only 50%. Doubling both system aspects

would increase performance by 100%. Generally, it is important to remove

the current system bottleneck, and to increase overall system

performance, rather than blindly increasing the performance of individual

system components.

13.7 Distinguish between a STREAMS driver and a STREAMS module.

Exercises 53

Answer: The STREAMS driver controls a physical device that could be

involved in a STREAMS operation. The STREAMS module modifies the

flow of data between the head (the user interface) and the driver.

C1HA4PTER

Protection

Exercises

14.1 What are the main differences between capability lists and access lists?

Answer: An access list is a list for each object consisting of the domains

with a nonempty set of access rights for that object. A capability list is

a list of objects and the operations allowed on those objects for each

domain.

14.2 A Burroughs B7000/B6000 MCP file can be tagged as sensitive data.

When such a file is deleted, its storage area is overwritten by some

random bits. For what purpose would such a scheme be useful?

Answer: This would be useful as an extra security measure so that the

old content of memory cannot be accessed, either intentionally or by

accident, by another program. This is especially useful for any highly

classified information.

14.3 In a ring-protection system, level 0 has the greatest access to objects,

and level n (greater than zero) has fewer access rights. The access rights

of a program at a particular level in the ring structure are considered as

a set of capabilities. What is the relationship between the capabilities

of a domain at level j and a domain at level i to an object (for j > i)?

Answer: Dj is a subset of Di .

14.4 The RC 4000 system (and other systems) have defined a tree of processes

(called a process tree) such that all the descendants of a process are given

resources (objects) and access rights by their ancestors only. Thus, a

descendant can never have the ability to do anything that its ancestors

cannot do. The root of the tree is the operating system, which has the

ability to do anything. Assume the set of access rights was represented

by an access matrix, A. A(x,y) defines the access rights of process x to

56 Chapter 14 Protection

object y. If x is a descendant of z,what is the relationship between A(x,y)

and A(z,y) for an arbitrary object y?

Answer: A(x,y) is a subset of A(z,y).

14.5 What protection problems may arise if a shared stack is used for parameter

passing?

Answer: The contents of the stack could be compromised by other

process(es) sharing the stack.

14.6 Consider a computing environment where a unique number is associated

with each process and each object in the system. Suppose that we

allow a process with number n to access an object with number m only

if n > m.What type of protection structure do we have?

Answer: Hierarchical structure.

14.7 Consider a computing environment where a process is given the privilege

of accessing an object only n times. Suggest a scheme for implementing

this policy.

Answer: Add an integer counter with the capability.

14.8 If all the access rights to an object are deleted, the object can no longer

be accessed. At this point, the object should also be deleted, and the

space it occupies should be returned to the system. Suggest an efficient

implementation of this scheme.

Answer: Reference counts.

14.9 Why is it difficult to protect a system in which users are allowed to do

their own I/O?

Answer: In earlier chapters we identified a distinction between kernel

and user mode where kernel mode is used for carrying out privileged

operations such as I/O. One reason why I/O must be performed in

kernel mode is that I/O requires accessing the hardware and proper

access to the hardware is necessary for system integrity. If we allow

users to perform their own I/O, we cannot guarantee system integrity.

14.10 Capability lists are usually kept within the address space of the user.

How does the system ensure that the user cannot modify the contents

of the list?

Answer: A capability list is considered a “protected object” and is

accessed only indirectly by the user. The operating system ensures the

user cannot access the capability list directly.

C1HA5PTER

Security

No Exercises

Distributed C1HA6PTER

System

Structures

Exercises

16.1 Most WANs employ only a partially connected topology. Why is this

so?

Answer: Cost. A fully connected network requires a link between

every node in the network. For a WAN, this may be too costly as communication

links between physically distant hosts may be expensive.

16.2 Under what circumstances is a token-passing network more effective

than an Ethernet network?

Answer: A token ring is very effective under high sustained load, as

no collisions can occur and each slot may be used to carry a message,

providing high throughput. A token ring is less effective when the

load is light (token processing takes longer than bus access, so any one

packet can take longer to reach its destination) or sporadic.

16.3 Why would it be a bad idea for gateways to pass broadcast packets

between networks? What would be the advantages of doing so?

Answer: All broadcasts would be propagated to all networks, causing

a lot of network traffic. If broadcast traffic were limited to important

data (and very little of it), then broadcast propagation would save

gateways from having to run special software to watch for this data

(such as network routing information) and rebroadcast it.

16.4 Discuss the advantages and disadvantages of caching name translations

for computers located in remote domains.

Answer: There is a performance advantage to caching name translations

for computers located in remote domains: repeated resolution of

the same name from different computers located in the local domain

could be performed locally without requiring a remote name lookup

operation. The disadvantage is that there could be inconsistencies in the

60 Chapter 16 Distributed System Structures

name translations when updates are made in the mapping of names to

IP addresses. These consistency problems could be solved by invalidating

translations, which would require state to be managed regarding

which computers are caching a certain translation and also would require

a number of invalidation messages, or by using leases whereby

the caching entity invalidates a translation after a certain period of time.

The latter approach requires less state and no invalidationmessages but

might suffer from temporary inconsistencies.

16.5 What are the advantages and disadvantages of using circuit switching?

For what kinds of applications is circuit switching a viable strategy?

Answer: Circuit switching guarantees that the network resources required

for a transfer are reserved before the transmission takes place.

This ensures that packets will not be dropped and their deliverywould

satisfy quality of service requirements. The disadvantage of circuit

switching is that it requires a round-trip message to set-up the reservations

and it also might overprovision resources, thereby resulting

in suboptimal use of the resources. Circuit switching is a viable strategy

for applications that have constant demands regarding network

resources and would require the resources for long periods of time,

thereby amortizing the initial overheads.

16.6 What are two formidable problems that designersmust solve to implement

a network-transparent system?

Answer: One such issue is making all the processors and storage

devices seem transparent across the network. In other words, the distributed

system should appear as a centralized system to users. The

Andrew file system and NFS provide this feature: the distributed file

system appears to the user as a single file system but in reality it may

be distributed across a network.

Another issue concerns the mobility of users. We want to allow users

to connect to the “system” rather than to a specific machine (although

in reality they may be logging in to a specific machine somewhere in

the distributed system).

16.7 Process migration within a heterogeneous network is usually impossible,

given the differences in architectures and operating systems. Describe

a method for process migration across different architectures

running:

a. The same operating system

b. Different operating systems

Answer: For the same operating system, process migration is relatively

straightforward, as the state of the process needs tomigrate from

one processor to another. This involves moving the address space, state

of the CPU registers, and open files from the source system to the destination.

However, it is important that identical copies of the operating

system are running on the different systems to ensure compatibility.

If the operating system are the same, but perhaps different versions

are running on the separate systems, then migrating processesmust be

Exercises 61

sure to follow programming guidelines that are consistent between the

different versions of the operating system.

Java applets provide a nice example of process migration between different

operating systems. To hide differences in the underlying system,

the migrated process (i.e., a Java applet) runs on a virtual machine

rather than a specific operating system. All that is required is for the

virtual machine to be running on the system the process migrates to.

16.8 To build a robust distributed system, you must know what kinds of

failures can occur.

a. List three possible types of failure in a distributed system.

b. Specify which of the entries in your list also are applicable to a

centralized system.

Answer: Three common failures in a distributed system include:

(1) network link failure, (2) host failure, (3) storage medium failure.

Both (2) and (3) are failures that could also occur in a centralized system,

whereas a network link failure can occur only in a networkeddistributed

system.

16.9 Is it always crucial to know that the message you have sent has arrived

at its destination safely? If your answer is yes, explain why. If your

answer is no, give appropriate examples.

Answer: No.Many status-gathering programswork fromthe assumption

that packets may not be received by the destination system. These

programs generally broadcast a packet and assume that at least some

other systems on their network will receive the information. For instance,

a daemon on each system might broadcast the system’s load

average and number of users. This information might be used for process

migration target selection. Another example is a program that

determines if a remote site is both running and accessible over the network.

If it sends a query and gets no reply it knows the system cannot

currently be reached.

16.10 Present an algorithm for reconstructing a logical ring after a process in

the ring fails.

Answer: Typically distributed systems employ a coordinator process

that performs functions needed by other processes in the system. This

would include enforcing mutual exclusion and—in this case of a ring

—replacing a lost token.

A scheme similar to the ring algorithm presented in Section 18.6.2 can

be used. The algorithm is as follows:

The ring algorithm assumes that the links are unidirectional and that

processes send their messages to the neighbor on their right. The main

data structure used by the algorithm is the active list, a list that contains

the priority numbers of all active processes in the system when the

algorithm ends; each process maintains its own active list.

a. If process Pi detects a coordinator failure, it creates a new active

list that is initially empty. It then sends a message elect(i) to its

right neighbor, and adds the number i to its active list.

62 Chapter 16 Distributed System Structures

b. If Pi receives a message elect(j) from the process on the left, it

must respond in one of three ways:

1. If this is the first elect message it has seen or sent, Pi creates

a new active listwith the numbers i and j. It then sends the

message elect(i), followed by the message elect(j).

2. If i = j, that is, the message received does not contain Pi ’s

number, then Pi adds j to its active list and forwards the

message to its right neighbor.

3. If i = j, that is, Pi receives the message elect(i), then the

active list for Pi now contains the numbers of all the active

processes in the system. Process Pi can now determine

the largest number in the active list to identify the new

coordinator process.

16.11 Consider adistributed system with two sites,AandB.Considerwhether

site A can distinguish among the following:

a. B goes down.

b. The link between A and B goes down.

c. B is extremely overloaded and its response time is 100 times

longer than normal.

What implications does your answer have for recovery in distributed

systems?

Answer: One technique would be for B to periodically send a I-am-up

message to A indicating it is still alive. If A does not receive an I-amup

message, it can assume either B—or the network link—is down.

Note that an I-am-up message does not allow A to distinguish between

each type of failure. One technique that allows A better to determine

if the network is down is to send an Are-you-up message to B using an

alternate route. If it receives a reply, it can determine that indeed the

network link is down and that B is up.

If we assume that A knows B is up and is reachable (via the I-amup

mechanism) and that A has some value N that indicates a normal

response time, A could monitor the response time fromB and compare

values to N, allowing A to determine if B is overloaded or not.

The implications of both of these techniques are that A could choose

another host—say C—in the system if B is either down, unreachable,

or overloaded.

C1HA7PTER

Distributed File

Systems

No Exercises

C1HA8PTER

Distributed

Coordination

No Exercises

C1HA9PTER

Real-time

Systems

No Exercises

2CHAP0TER

Multimedia

Systems

No Exercises

C2HAP1TER

The Linux

System

Exercises

21.1 Dynamically loadable kernel modules give flexibility when drivers are

added to a system, but do they have disadvantages too? Under what

circumstances would a kernel be compiled into a single binary file, and

when would it be better to keep it split into modules? Explain your

answer.

Answer: There are two principal drawbacks with the use of modules.

The first is size: module management consumes unpageable kernel

memory, and a basic kernel with a number of modules loaded will

consume more memory than an equivalent kernel with the drivers

compiled into the kernel image itself. This can be a very significant

issue on machines with limited physical memory.

The second drawback is that modules can increase the complexity

of the kernel bootstrap process. It is hard to load up a set of modules

from disk if the driver needed to access that disk itself a module that

needs to be loaded. As a result, managing the kernel bootstrap with

modules can require extra work on the part of the administrator: the

modules required to bootstrap need to be placed into a ramdisk image

that is loaded alongside the initial kernel image when the system is

initialized.

In certain cases it is better to use a modular kernel, and in other

cases it is better to use a kernelwith its device drivers prelinked.Where

minimizing the size of the kernel is important, the choice will depend

on how often the various device drivers are used. If they are in constant

use, then modules are unsuitable. This is especially true where

drivers are needed for the boot process itself. On the other hand, if

some drivers are not always needed, then the module mechanism al-

72 Chapter 21 The Linux System

lows those drivers to be loaded and unloaded on demand, potentially

offering a net saving in physical memory.

Where a kernel is to be built that must be usable on a large variety

of very different machines, then building it with modules is clearly

preferable to using a single kernel with dozens of unnecessary drivers

consuming memory. This is particularly the case for commercially distributed

kernels, where supporting the widest variety of hardware in

the simplestmanner possible is a priority.

However, if a kernel is being built for a single machine whose

configuration is known in advance, then compiling and usingmodules

may simply be an unnecessary complexity. In cases like this, the use of

modules may well be a matter of taste.

21.2 Multithreading is a commonly used programming technique. Describe

three different ways that threads could be implemented. Explain how

these ways compare to the Linux clone mechanism. When might each

alternative mechanism be better or worse than using clones?

Answer: Thread implementations can be broadly classified into two

groups: kernel-based threads and user-modethreads. User-mode thread

packages rely on some kernel support—they may require timer interrupt

facilities, for example—but the scheduling between threads is not

performed by the kernel but by some library of user-mode code. Multiple

threads in such an implementation appear to the operating system

as a single execution context. When the multithreaded process is running,

it decides for itselfwhich of its threads to execute, using non-local

jumps to switch between threads according to its own preemptive or

non-preemptive scheduling rules.

Alternatively, the operating system kernel may provide support for

threads itself. In this case, the threadsmay be implemented as separate

processes that happen to share a complete or partial common address

space, or they may be implemented as separate execution contexts

within a single process.Whicheverway the threads are organized, they

appear as fully independent execution contexts to the application.

Hybrid implementations are also possible, where a large number of

threads are made available to the application using a smaller number

of kernel threads. Runnable user threads are run by the first available

kernel thread.

In Linux, threads are implemented within the kernel by a clone

mechanism that creates a new process within the same virtual address

space as the parent process. Unlike some kernel-based thread packages,

the Linux kernel does not make any distinction between threads and

processes: a thread is simply a process that did not create a new virtual

address space when it was initialized.

The main advantage of implementing threads in the kernel rather

than in a user-mode library are that:

• kernel-threaded systems can take advantage of multiple

processors if they are available; and

• if one thread blocks in a kernel service routine (for example, a

system call or page fault), other threads are still able to run.

Exercises 73

A lesser advantage is the ability to assign different security attributes

to each thread.

User-mode implementations do not have these advantages. Because

such implementations run entirely within a single kernel execution context,

only one thread can ever be running at once, even if multiple CPUs

are available. For the same reason, if one thread enters a system call,

no other threads can run until that system call completes. As a result,

one thread doing a blocking disk read will hold up every thread in the

application. However, user-mode implementations do have their own

advantages. The most obvious is performance: invoking the kernel’s

own scheduler to switch between threads involves entering a new protection

domain as the CPU switches to kernelmode, whereas switching

between threads in user mode can be achieved simply by saving and

restoring the main CPU registers. User-mode threads may also consume

less system memory: most UNIX systems will reserve at least a

full page for a kernel stack for each kernel thread, and this stack may

not be pageable.

The hybrid approach, implementing multiple user threads over a

smaller number of kernel threads, allows a balance between these tradeoffs

to be achieved. The kernel threads will allow multiple threads to

be in blocking kernel calls at once and will permit running on multiple

CPUs, and user-mode thread switching can occur within each

kernel thread to perform lightweight threading without the overheads

of having too many kernel threads. The downside of this approach

is complexity: giving control over the tradeoff complicates the thread

library’s user interface.

21.3 The Linux kernel does not allow paging out of kernel memory. What

effect does this restriction have on the kernel’s design? What are two

advantages and two disadvantages of this design decision?

Answer: The primary impact of disallowing paging of kernel memory

in Linux is that the non-preemptability of the kernel is preserved. Any

process taking a page fault, whether in kernel or in user mode, risks

being rescheduled while the required data is paged in from disk. Because

the kernel can rely on not being rescheduled during access to its

primary data structures, locking requirements to protect the integrity

of those data structures are very greatly simplified. Although design

simplicity is a benefit in itself, it also provides an important performance

advantage on uniprocessor machines due to the fact that it is

not necessary to do additional locking onmost internal data structures.

There are a number of disadvantages to the lack of pageable kernel

memory, however. First of all, it imposes constraints on the amount of

memory that the kernel can use. It is unreasonable to keep very large

data structures in non-pageable memory, since that represents physical

memory that absolutely cannot be used for anything else. This has two

impacts: first of all, the kernelmust prune back many of its internal data

structures manually, instead of being able to rely on a single virtualmemory

mechanism to keep physical memory usage under control.

Second, itmakes it infeasible to implement certain features that require

large amounts of virtual memory in the kernel, such as the /tmp74

Chapter 21 The Linux System

filesystem (a fast virtual-memory-based file system found on some

UNIX systems).

Note that the complexity of managing page faults while running

kernel code is not an issue here. The Linux kernel code is already able

to deal with page faults: it needs to be able to deal with system calls

whose arguments reference user memory that may be paged out to

disk.

21.4 What are three advantages of dynamic (shared) linkage of libraries

compared to static linkage? What are two cases where static linkage is

preferable?

Answer: The primary advantages of shared libraries are that they

reduce thememory and disk space used by a system, and they enhance

maintainability.

When shared libraries are being used by all running programs,

there is only one instance of each system library routine on disk, and

at most one instance in physical memory. When the library in question

is one used by many applications and programs, then the disk and

memory savings can be quite substantial. In addition, the startup time

for running new programs can be reduced, since many of the common

functions needed by that program are likely to be already loaded into

physical memory.

Maintainability is also a major advantage of dynamic linkage over

static. If all running programs use a shared library to access their system

library routines, then upgrading those routines, either to add new

functionality or to fix bugs, can be done simply by replacing that shared

library. There is no need to recompile or relink any applications; any

programs loaded after the upgrade is complete will automatically pick

up the new versions of the libraries.

There are other advantages too. A program that uses shared libraries

can often be adapted for specific purposes simply by replacing one or

more of its libraries, or even (if the system allows it, and most UNIXs

including Linux do) adding a new one at run time. For example, a

debugging library can be substituted for a normal one to trace a problem

in an application. Shared libraries also allow program binaries to be

linked against commercial, proprietary library code without actually

including any of that code in the program’s final executable file. This is

important because on most UNIX systems, many of the standard shared

libraries are proprietary, and licensing issues may prevent including

that code in executable files to be distributed to third parties.

In some places, however, static linkage is appropriate.One example is

in rescue environments for system administrators. If a system administrator

makes a mistakewhile installing any new libraries, or if hardware

develops problems, it is quite possible for the existing shared libraries

to become corrupt. As a result, often a basic set of rescue utilities are

linked statically, so that there is an opportunity to correct the fault

without having to rely on the shared libraries functioning correctly.

There are also performance advantages that sometimes make static

linkage preferable in special cases. For a start, dynamic linkage does

increase the startup time for a program, as the linking must now be

Exercises 75

done at run time rather than at compile time. Dynamic linkage can also

sometimes increase the maximum working set size of a program (the

total number of physical pages ofmemoryrequired to run the program).

In a shared library, the user has no control over where in the library

binary file the various functions reside. Since most functions do not

precisely fill a full page or pages of the library, loading a function will

usually result in loading in parts of the surrounding functions, too.With

static linkage, absolutely no functions that are not referenced (directly

or indirectly) by the application need to be loaded into memory.

Other issues surrounding static linkage include ease of distribution:

it is easier to distribute an executable file with static linkage than with

dynamic linkage if the distributor is not certain whether the recipient

will have the correct libraries installed in advance. There may also be

commercial restrictions against redistributing some binaries as shared

libraries. For example, the license for the UNIX “Motif” graphical environment

allows binaries using Motif to be distributed freely as long

as they are statically linked, but the shared libraries may not be used

without a license.

21.5 Compare the use of networking sockets with the use of shared memory

as amechanism for communicating data between processes on a single

computer.What are the advantages of each method? When might each

be preferred?

Answer: Using network sockets rather than shared memory for local

communication has a number of advantages. The main advantage is

that the socket programming interface features a rich set of synchronization

features. A process can easily determine when new data has

arrived on a socket connection, how much data is present, and who

sent it. Processes can block until new data arrives on a socket, or they

can request that a signal be delivered when data arrives. A socket also

manages separate connections. A process with a socket open for receive

can accept multiple connections to that socket and will be told

when new processes try to connect or when old processes drop their

connections.

Shared memory offers none of these features. There is no way

for a process to determine whether another process has delivered or

changed data in shared memory other than by going to look at the

contents of that memory. It is impossible for a process to block and

request a wakeup when shared memory is delivered, and there is no

standard mechanism for other processes to establish a shared memory

link to an existing process.

However, shared memory has the advantage that it is very much

faster than socket communications in many cases. When data is sent

over a socket, it is typically copied from memory to memory multiple

times. Shared memory updates require no data copies: if one process

updates a data structure in sharedmemory, that update is immediately

visible to all other processes sharing that memory. Sending or receiving

data over a socket requires that a kernel system service call be made

to initiate the transfer, but shared memory communication can be performed

entirely in user mode with no transfer of control required.

76 Chapter 21 The Linux System

Socket communication is typically preferred when connection management

is important or when there is a requirement to synchronize

the sender and receiver. For example, server processes will usually establish

a listening socket to which clients can connect when they want

to use that service. Once the socket is established, individual requests

are also sent using the socket, so that the server can easily determine

when a new request arrives and who it arrived from.

In some cases, however, sharedmemory is preferred. Sharedmemory

is often a better solution when either large amounts of data are to

be transferred or when two processes need random access to a large

common data set. In this case, however, the communicating processes

may still need an extra mechanism in addition to shared memory to

achieve synchronization between themselves. The XWindow System, a

graphical display environment for UNIX, is a good example of this: most

graphic requests are sent over sockets, but shared memory is offered

as an additional transport in special cases where large bitmaps are to

be displayed on the screen. In this case, a request to display the bitmap

will still be sent over the socket, but the bulk data of the bitmap itself

will be sent via shared memory.

21.6 UNIX systems used to use disk-layout optimizations based on the rotation

position of disk data, but modern implementations, including

Linux, simply optimize for sequential data access. Why do they do so?

Of what hardware characteristics does sequential access take advantage?

Why is rotational optimization no longer so useful?

Answer: The performance characteristics of disk hardware have

changed substantially in recent years. In particular, many enhancements

have been introduced to increase the maximum bandwidth that

can be achieved on a disk. In a modern system, there can be a long

pipeline between the operating system and the disk’s read-write head.

A disk I/O request has to pass through the computer’s local disk controller,

over bus logic to the disk drive itself, and then internally to the

disk, where there is likely to be a complex controller that can cache data

accesses and potentially optimize the order of I/O requests.

Because of this complexity, the time taken for one I/O request to be

acknowledged and for the next request to be generated and received

by the disk can far exceed the amount of time between one disk sector

passing under the read-write head and the next sector header arriving.

In order to be able efficiently to read multiple sectors at once, disks

will employ a readahead cache. While one sector is being passed back

to the host computer, the disk will be busy reading the next sectors in

anticipation of a request to read them. If read requests start arriving in

an order that breaks this readahead pipeline, performance will drop.

As a result, performance benefits substantially if the operating system

tries to keep I/O requests in strict sequential order.

A second feature of modern disks is that their geometry can be very

complex. The number of sectors per cylinder can vary according to the

position of the cylinder: more data can be squeezed into the longer

tracks nearer the edge of the disk than at the center of the disk. For an

operating system to optimize the rotational position of data on such

Exercises 77

disks, it would have to have complete understanding of this geometry,

as well as the timing characteristics of the disk and its controller. In general,

only the disk’s internal logic can determine the optimal scheduling

of I/Os, and the disk’s geometry is likely to defeat any attempt by the

operating system to perform rotational optimizations.

2CHA2PTER

Windows XP

Exercises

22.1 What type of operating system is Windows XP? Describe two of its

major features.

Answer: A 32/64 bit preemptive multitasking operating system supporting

multiple users. (1) The ability automatically to repair application

and operating system problems. (2) Better networking and device

experience (including digital photography and video).

22.2 List the design goals ofWindows XP. Describe two in detail.

Answer: Design goals include security, reliability,Windows and POSIX

application compatibility, high performance, extensibility, portability

and international support. (1) Reliability was perceived as a stringent

requirement and included extensive driver verification, facilities for

catching programming errors in user-level code, and a rigorous certification

process for third-party drivers, applications, and devices. (2)

Achieving high performance required examination of past problem areas

such as I/O performance, server CPU bottlenecks, and the scalability

of multithreaded and multiprocessor environments.

22.3 Describe the booting process for aWindows XP system.

Answer: (1) As the hardware powers on, the BIOS begins executing

fromROMand loads and executes the bootstrap loader fromthe disk. (2)

The NTLDR program is loaded from the root directory of the identified

system device and determineswhich boot device contains the operating

system. (3) NTLDR loads the HAL library, kernel, and system hive. The

system hive indicates the required boot drivers and loads them. (4)

Kernel execution begins by initializing the system and creating two

processes: the system process containing all internal worker threads,

and the first user-mode initialization process: SMSS. (5) SMSS further

80 Chapter 22 Windows XP

initializes the system by establishing paging files and loading device

drivers. (6) SMSS creates two processes: WINLOGON, which brings up

the rest of the system, and CSRSS (theWin32 subsystem process).

22.4 Describe the three main architectural layers of Windows XP.

Answer: (1) The HAL (Hardware Abstraction Layer) creates operating

system portability by hiding hardware differences from the upper

layers of the operating system. Administrative details of low-level facilities

are provided by HAL interfaces. HAL presents a virtual-machine

interface that is used by the kernel dispatcher, the executive and device

drivers. (2) The kernel layer provides a foundation for the executive

functions and user-mode subsystems. The kernel remains in memory

and is never preempted. Its responsibilities are thread scheduling, interrupt

and exception handling, low-level processor synchronization,

and power failure recovery. (3) The executive layer provides a set of services

used by all subsystems: object manager, virtualmemorymanager,

process manager, local procedure call facility, I/O manager, security

monitor, plug-and-play manager, registry, and booting.

22.5 What is the job of the object manager?

Answer: Objects present a generic set of kernelmode interfaces to usermode

programs. Objects are manipulated by the executive-layer object

manager. The job of the object manager is to supervise the allocation

and use of all managed objects.

22.6 What types of services does the process manager provide? What is a

local procedure call?

Answer: The process manager provides services for creating, deleting,

and using processes, threads and jobs. The process manager also

implements queuing and delivery of asynchronous procedure calls to

threads. The local procedure call (LPC) is a message-passing system.

The operating system uses the LPC to pass requests and results between

client and server processeswithin a single machine, in particular

betweenWindows XP subsystems.

22.7 What are the responsibilities of the I/O manager?

Answer: The I/O manager is responsible for file systems, device

drivers, and network drivers. The I/O manager keeps track of which

device drivers, filter drivers, and file systems are loaded and manages

buffers for I/O requests. It furthermore assists in providing memorymapped

file I/O and controls the cache manager for the whole I/O

system.

22.8 Does Windows XP offer any user-mode processes that enable it to run

programs developed for other operating systems? Describe two of these

subsystems.

Answer: Environmental subsystems are user-mode processes layered

over the native executable services to enable Windows XP to run programs

developed for other operating systems. (1) AWin32 application

called the virtual DOS machine (VDM) is provided as a user-mode process

to run MS-DOS applications. The VDM can execute or emulate Intel

486 instructions and also provides routines to emulate MS-DOS BIOS

Exercises 81

services and provides virtual drivers for screen, keyboard, and communication

ports. (2)Windows-on-windows (WOW32) provides kernel

and stub routines forWindows 3.1 functions. The stub routines call the

appropriate Win32 subroutines, converting the 16-bit addresses into

32-bit addresses.

22.9 What types of networking doesWindows XP support? How doesWindows

XP implement transport protocols? Describe two networking protocols.

Answer: Support is provided for both peer-to-peer and client-server

networking. Transport protocols are implemented as drivers. (1) The

TCP/IP package includes SNMP, DHCP, WINS, and NetBIOS support. (2)

Point-to-point tunneling protocol is provided to communicate between

remote-accessmodules running onWindowsXP servers and other client

systems connected over the internet. Using this scheme, multi-protocol

virtual private networks (VPNs) are supported over the internet.

22.10 How is the NTFS namespace organized? Describe.

Answer: The NTFS namespace is organized as a hierarchy of directories

where each directory uses a B+ tree data structure to store an index of

the file names in that directory. The index root of a directory contains

the top level of the B+ tree. Each entry in the directory contains the

name and file reference of the file as well as the update timestamp and

file size.

22.11 How does NTFS handle data structures? How does NTFS recover from

a system crash? What is guaranteed after a recovery takes place?

Answer: In NTFS, all file-system data structure updates are performed

inside transactions. Before a data structure is altered, the transaction

writes a log record containing redo and undo information. A commit

record is written to the log after a transaction has succeeded. After a

crash the file system can be restored to a consistent state by processing

the log records, first redoing operations for committed transactions and

undoing operations for transactions that did not successfully commit.

This scheme does not guarantee that user file contents are correct after

a recovery, but rather that the file-system data structures (file metadata)

are undamaged and reflect some consistent state that existed before the

crash.

22.12 How doesWindows XP allocate user memory?

Answer: User memory can be allocated according to several schemes:

virtual memory, memory-mapped files, heaps, and thread-local storage.

22.13 Describe some of the ways an application can use memory via the

Win32 API.

Answer: (1) Virtual memory provides several functions that allow

an application to reserve and release memory, specifying the virtual

address at which the memory is allocated. (2) A file may be memorymapped

into address space, providing a means for two processes to

share memory. (3) When a Win32 process is initialized, it is created

with a default heap. Private heaps can be created that provide regions of

82 Chapter 22 Windows XP

reserved address space for applications. Threadmanagement functions

are provided to allocate and control thread access to private heaps. (4)

A thread-local storage mechanism provides a way for global and static

data to work properly in a multithreaded environment. Thread-lock

storage allocates global storage on a per-thread basis.

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