操作系统基础基本理论、基本方法复习

Chapter 1 Introduction

1.Abstract View of System Components

2.Multiprogrammed Batch Systems

Several jobs are kept in main memory at the same time, and the

CPU is multiplexed among them.

Chapter 2: Computer-System Structures

https://www.360docs.net/doc/6117908801.html,puter-System Architecture

2. Interrupt Time Line For a Single Process Doing Output

3.Moving-Head Disk Mechanism

4.Direct Memory Access Structure

Six Step Process to Perform DMA Transfer 5.Storage-Device Hierarchy

https://www.360docs.net/doc/6117908801.html,e of A Base and Limit Register

7.Hardware Address Protection

Chapter 3: Operating-System Structures 1.MS-DOS Execution

2.MS-DOS Layer Structure

Chapter 4: Processes

1.Diagram of Process State

2.Process Control Block (PCB)

3.CPU Switch From Process to Process

4.Representation of Process Scheduling

5.Addition of Medium Term Scheduling

Chapter 5: Threads 1.Single and Multithreaded Processes

2.Many-to-One Model

3.One-to-one Model

4.Many-to-Many Model

Chapter 6: CPU Scheduling

1.Alternating Sequence of CPU And I/O Bursts

2.First-Come, First-Served (FCFS) Scheduling

Process Burst Time

P124

P2 3

P3 3

●Suppose that the processes arrive in the order: P1 , P2 , P3

The Gantt Chart for the schedule is:

●Waiting time for P1= 0; P2= 24; P3 = 27

●Average waiting time: (0 + 24 + 27)/3 = 17

3.Example of Non-Preemptive SJF

Process Arrival Time Burst Time

P10.0 7

P2 2.0 4

P3 4.0 1

P4 5.0 4

●SJF (non-preemptive)

●Average waiting time = (0 + 6 + 3 + 7)/4 - 4

4.Example of Preemptive SJF

Process Arrival Time Burst Time

P10.0 7

P 2 2.0 4 P 3 4.0 1 P 4 5.0 4

● SJF (preemptive)

● Average waiting time = (9 + 1 + 0 +2)/4 - 3 5.

Example of RR with Time Quantum = 20 Process Burst Time P 1 53 P 2 17 P 3 68 P 4 24

● The Gantt chart is:

● Typically, higher average turnaround than SJF, but better response . 6.

Dispatch Latency

Chapter 7: Process Synchronization

1.

Implementation

? Semaphore operations now defined as wait (S ): S.value--; if (S.value < 0) { add this process to S.L; block; }

signal (S ):

S.value++;

if (S.value <= 0) {

remove a process P from S.L;

wakeup(P);

}

2.Deadlock

?– two or more processes are waiting indefinitely for an event that can be caused by only one of the waiting processes.

?Let S and Q be two semaphores initialized to 1

P0P1

wait(S); wait(Q);

wait(Q); wait(S);

signal(S); signal(Q);

signal(Q) signal(S);

3.Bounded-Buffer Problem

?Shared data

semaphore full, empty, mutex;

Initially:

full = 0, empty = n, mutex = 1

Problem Producer Process

do {

…

produce an item in nextp

…

wait(empty);

wait(mutex);

…

add nextp to buffer

…

signal(mutex);

signal(full);

} while (1);

Problem Consumer Process

do {

wait(full)

wait(mutex);

…

remove an item from buffer to nextc

…

signal(mutex);

signal(empty);

…

consume the item in nextc

…

} while (1);

4.Readers-Writers Problem

?Shared data

semaphore mutex, wrt;

Initially

mutex = 1, wrt = 1, readcount = 0 Writer Process

wait(wrt);

…

writing is performed

…

signal(wrt);

Reader Process

wait(mutex);

readcount++;

if (readcount == 1)

wait(rt);

signal(mutex);

…

reading is performed

…

wait(mutex);

readcount--;

if (readcount == 0)

signal(wrt);

signal(mutex):

5.Dining-Philosophers Problem

?Shared data

semaphore chopstick[5];

Initially all values are 1

?Philosopher i:

do {

wait(chopstick[i])

wait(chopstick[(i+1) % 5])

…

eat

…

signal(chopstick[i]);

signal(chopstick[(i+1) % 5]);

…

think

…

} while (1);

Chapter 8: Deadlocks

1.Example of a Resource Allocation Graph

2.Basic Facts

?If graph contains no cycles ? no deadlock.

?If graph contains a cycle ?

–if only one instance per resource type, then deadlock.

–if several instances per resource type, possibility of deadlock.

3.Safe State

?When a process requests an available resource, system must decide if immediate allocation leaves the system in a safe state.

?System is in safe state if there exists a safe sequence of all processes.

?Sequence is safe if for each P i, the resources that Pi can still request can be satisfied by currently available resources + resources held by all the P j, with j

–If P i resource needs are not immediately available, then P i can wait until all P j have finished.

–When P j is finished, P i can obtain needed resources, execute, return allocated resources, and terminate.

–When P i terminates, P i+1 can obtain its needed resources, and so on.

4.Banker’s Algorithm

?Multiple instances.

?Each process must a priori claim maximum use.

?When a process requests a resource it may have to wait.

?When a process gets all its resources it must return them in a finite amount of time. Data Structures for the Banker’s Algorithm

Let n = number of processes, and m = number of resources types.

?Available:Vector of length m. If available [j] = k, there are k instances of resource type R j available.

?Max: n x m matrix. If Max [i,j] = k, then process P i may request at most k instances of resource type R j.

?Allocation: n x m matrix. If Allocation[i,j] = k then P i is currently allocated k instances of R j.

?Need: n x m matrix. If Need[i,j] = k, then P i may need k more instances of R j to complete

its task.

Need [i,j] = Max[i,j] –Allocation [i,j].

Safety Algorithm

1) Let Work and Finish be vectors of length m and n, respectively. Initialize:

Work = Available

Finish [i] = false for i - 1,3, …, n.

2) Find and i such that both:

(a) Finish [i] = false

(b) Need i≤Work

If no such i exists, go to step 4.

3) Work = Work + Allocation i

Finish[i] = true

go to step 2.

4) If Finish [i] == true for all i, then the system is in a safe state.

Resource-Request Algorithm for Process P i

Request = request vector for process P i. If Request i[j] = k then process P i wants k instances of resource type R j.

1) If Request i≤Need i go to step 2. Otherwise, raise error condition, since process has exceeded its maximum claim.

2) If Request i≤Available, go to step 3. Otherwise P i must wait, since resources are not available.

3) Pretend to allocate requested resources to P i by modifying the state as follows:

Available = Available = Request i;

Allocation i= Allocation i + Request i;

Need i= Need i–Request i;;

?If safe ? the resources are allocated to P i.

?If unsafe ?P i must wait, and the old resource-allocation state is

restored

5.Example of Banker’s Algorithm

? 5 processes P0 through P4; 3 resource types A

(10 instances),

B (5instances, and

C (7 instances).

?Snapshot at time T0:

?Allocation Max Available

? A B C A B C A B C

?P0 0 1 0 7 5 3 3 3 2

?P1 2 0 0 3 2 2

?P2 3 0 2 9 0 2

?P3 2 1 1 2 2 2

?P40 0 2 4 3 3

?T he content of the matrix. Need is defined to be Max – Allocation.

?Need

? A B C

?P0 7 4 3

?P1 1 2 2

?P2 6 0 0

?P30 1 1

?P4 4 3 1

?T he system is in a safe state since the sequence < P1, P3, P4, P2, P0> satisfies safety criteria.

Chapter 9: Memory Management

1.Multistep Processing of a User

2.Dynamic relocation using a relocation register

3.Overlays for a Two-Pass

4.

Schematic View of Swapping

5.Hardware Support for Relocation and Limit Registers

6.Address Translation Architecture

7.Paging Hardware With TLB

8.Valid (v) or Invalid (i) Bit In A Page Table

Chapter 10: Virtual Memory 1.Transfer of a Paged Memory to Contiguous Disk Space

2.Steps in Handling a Page Fault

3.First-In-First-Out (FIFO) Algorithm

?Reference string: 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5

? 3 frames (3 pages can be in memory at a time per process)

? 4 frames

FIFO Replacement –Belady’s Anomaly

–more frames ? less page faults

Optimal Page Replacement

Least Recently Used (LRU) Algorithm

Chapter 11: File-System Interface

1.Sequential-access File

2.Access Lists and Groups

●Mode of access: read, write, execute

●Three classes of users

RWX

a) owner access7 ? 1 1 1

RWX

b) group access 6 ?1 1 0

RWX

c) public access 1 ?0 0 1

●Ask manager to create a group (unique name), say G, and add some users to the group.

For a particular file (say game) or subdirectory, define an appropriate access.

Chapter 12: File System Implementation

https://www.360docs.net/doc/6117908801.html,yered File System

2. A Typical File Control Block

3.In-Memory File System Structures

Chapter 13: I/O Systems

1.Interrupt-Driven I/O Cycle

Chapter 14: Mass-Storage Systems

Disk Scheduling

●Several algorithms exist to schedule the servicing of disk I/O requests.

●We illustrate them with a request queue (0-199).

98, 183, 37, 122, 14, 124, 65, 67

Head pointer 53

1.FCFS

Illustration shows total head movement of 640 cylinders.

2.SSTF

Illustration shows total head movement of 236 cylinders.

3.SCAN

Illustration shows total head movement of 208 cylinders.

4.C-SCAN

5.C-LOOK

6.MS-DOS Disk Layout

§1．True-False Questions: (Please answer with ╳or√)

( )1.Only one process can be in the state of running on any processor at any time.√( )2.Context-switch time is not overhead; the system still can do useful work while switching. ╳

( )3.There is at least one thread belongs to a running process. √

( )4.Many-to-one model provides more concurrency than the one-to-one threading model. √

( )5.All processes, either I/O-bound or CPU-bound, must perform I/O for

meaningful execution. √

( )6.CPU scheduling deals with the problem of deciding which of the processes located in the hard disk is to be allocated the CPU. ╳

( )7.FCFS is a kind of preemptive scheduling algorithm. ╳

( )8.SJF is optimal – gives minimum average waiting time for a given set of processes √

( )9.With multilevel queue scheduling, the RoundRobin should be used in the foreground for the interactive processes. √

( )10.It is better to determine where the program will reside in memory before loading than making this decision until program execution.╳

( )11.The logical and physical addresses are the same in compile-time and

load-time address binding but different in execution-time address binding.

√

( )12.With memory management unit (MMU), the user program never sees the real physical addresses.√

( )13.With overlay, the system is able to keep all the necessary instructions and data in memory at all time.╳

( )14.Multiprogramming is impossible in fixed-sized partition because each process may require different sized memory space.√

( )15.It is possible for paging scheme to generate external fragmentation, while the internal fragmentation is inevitable.√

( )16.Both associative registers and page tables can be searched simultaneously.

╳

( )17. A bad page replacement choice increases the page-fault rate and slows process execution, but this will bot affect the correct execution of the

processes.╳

( )18.The working-set window should be fixed within the execution of a particular program.╳

( )19.SSTF scheduling is a form of SJF scheduling; may cause starvation of some requests.√

( )20.Since OS is only software, the changes in the design of the hardware has no influence on i╳t.

( )21.When a process executes, it typically executes for a long time before it either finishes or needs to perform I/O.√

( )22.Random access memory (RAM) can only be accessed by CPU, no other device can directly communicate with it.╳

( )23.Being another kind of processor, device controllers cannot execute

concurrently with the CPU.╳

( )24.The concept of cache memory can be applied only between CPU and

memory in the storage device hierarchy.√

( )25. A program is a passive entity and a process is an active entity with a program counter specifying the next instruction to execute.√

( )26.The execution of a process is not necessarily to be sequential. √

( )27.With layered approach, an operating system can be debugged and verified without any concern for the lower layer. ╳

( )28.During system design, although mechanisms are strongly affecting policies, it is better to separate the policy from the mechanism. √

( )29.Two processes associated with same program are considered the same separate execution sequences. ╳

( )30.There are five states a process can be in, and only one process can be in running state at any instance.√

( )31.The long-term scheduler selects which processes should be brought in to the ready queue. √

( )32.Context-switch time is an overhead because the system cannot do useful work while switching. √

( )33.The parent processes and the children processes created by the parent can be executed concurrently. √

( )34.There will be no race condition when the instructions involved can be executed atomically. ╳

( )35.When one thread is executing inside its critical section, all other threads should stop running and wait. ╳

( )36.Without proper control, it is possible for two processes to be in their critical sections at the same time. √

( )37.When solving critical section problems, we could make possible assumption concerning the relative speed of the threads. ╳

( )38.Semaphores can be implemented as synchronization tool that does not require busy waiting. √

( )39. A system built with semaphores cannot guarantee mutual exclusion. ╳

( )40. A semaphore is always as powerful as a monitor and it always has value 0 or 1. ( )41.Deadlock prevention differs from avoidance because deadlock avoidance make deadlock impossible. √

( )42.If the processes in the system has no hold-and-wait condition, no deadlock can happen. √

( )43.In order to prevent deadlock, there is no other ways except ensuring no cycle can happen in the resource allocation graph associated with this system. ╳( )44.It is impossible for a system with deadlock prevention scheme to become deadlocked. ╳