Một số câu hỏi về máy tính

Một số câu hỏi về máy tính

Computer system

Computer System

Book I: computer system fundamentals.

Chapter 1: INTRODUCTION TO COMPUTER.

Question 1 What is a computer?

A computer may be defined as a machine which accepts data from an input device, processes it

by performing arithmetical and logic operations in accordance with a program of instructions andreturns the results through an output unit

A computer is basically an electronic machine operating on current

Question 2 Components of a Computer system?

A computer system comprises of the following components:

1 Central Processing Unit (CPU)

- CPU is the heart of the whole sys

- CPU consists of the :

 program counter (PC)

 general purpose register

 Control unit:

 control and co_ordinate all hardware functions of the CS

 examine and decode all program instructions to the computer andinitiate their execution by sending the appropriate signals

 performs all arithmetic <addition, subtraction, multiplication,division & exponentiation> and logic comparision two valuesfunctions required by computer

 keeps track of the status of the accumalator

 for general purpose procedures

Please refer to diagram for an illustratin of the basic components of the CPU

CPU Control unit Arithmetic Logic Unit Accumulator

Program Couter to main Instruction Register memory Memory Address Register

Memory Data Register Status Register

General Purpose Register

Basic components of a CPU

Control Unit Input Unit ALU Output Unit

B

U

S

Main Memory

Backing Storage

Control signals Data flow

Components of a CS

2 Input units

- Used to enter data( raw unprocessed facts) and instructions to the computer

3 Output units

- Used for delevering the processed result from the computer in useful form

4 Backing storage units

- Backing storage units need for high capacity data storage devices that can storedata in a more permanent form for later retrieral, updating and referencing

- Backing storage is also called secondary storage external storage and auxiliarystorage

Chapter 2: MICOPROCESSOR.

Question 1 Cache Memory?

the main memory

- Cache memory is used to temporaryty store data instructions that are likely to beretrieved many times, thus speeds up the processing of data

- Sits between main storage and the processor acting as holding area through whichall data and instructions pass

- Old data in the cache memory is over written by new then cache is full

Question 2 Virtual Memory?

- In a virtual memory sys, each user has the illusion that his program is in the mainmemory all the time

program’s code and data on a backing store device which is usually magnetic disk

is transparent to the users

- Please refer to diagram for virtual memory

Virtual Memory

Chapter 3: BATCH/ ONLINE AND REAL TIME PROCESSING SYSTEM.

Question 1 Batch Processing System?

- Def: Computer processing does not begin until all the input data has beencollected and grouped together called Batched Generally data is accumulated for acertain period of time or unitl a certain quantity.

 Response time is not critical

 Need to process large volumn of data

 Computer efficiency is more important than response time

 Time between recording and processing of source document is long

 Rereen normally required if errors are encountered

 Data is not current

 Error correction is more difficult

Question 2 Online Processing System?

- Def: Inputs data enters the computer directly as soon as it is being transacted.There information will be processed immediately and updated into the master file

 Enter availability of information for decision making

 More accurate data capture

 Schedules suits user

 CPU time is used less efficiently

 Random arrival of transactions, terminal operator process eachtransaction separately

 More expensive than batch processing

Question 3 Real Time Processing System?

returning results sufficiently quickly to affect the functioning of the environment

at that time

 Response time is very critical and sufficient quick

 Expensive hardware & software

 Very complex in terms of hardware & software

Chapter 4: PRINTERS AND TERMINALS.

Question 1 Classification of printers?

1 Classifying printers according to speed

a Serial printers

Slow printers that print one character at a time

Eg: Dot matrix printers

2 Calssifying printers according to method of printing

a Impact printers

Use hammers or prints to strike a print rebbon in order to form the character on thepaper

b Non impact printers

Use more silent methods of printing

Eg: Thermal printers

Ink Jet printersLazers printers

3 Classifying printers according to print quality

Kinds of quality printersDraft quality

Near letter quality(NLQ)Letter quality

Graphic quality

Question 2 Describe some types of printer?

1 According to speed:

a Dot matrix printer

- Serial impact printers that can print draft, near letter quality and a limited amount

f Thermal printers

- Uses special heat sensitive paper and a matrix of print wires that become hot whenexposed to an electric current The heated wires come into close contact with thepaper, burning the image of the character onto it

- The more advanced thermal printers are using thermal transfer printing

- They have a special heat sensitive ribbon and a print head with wires that becomehot when a currents is applied

- The heat from the print wires causes the ink from the ribbon to fuse to a piece ofregular paper

g Inl Jet Printers

- The ink jet prints by using a small droplet generator to break special inks into tinydrops, which are then forced towards a paper supply

h Lazer printers

- A lazer is then used to write the image of the character onto the drum

- After exposure to the lazer, the drum rotates through a developing station, picks

up toner and transfers it to the paper

- The character is fused onto the paper by heat

i Ion deposition printers

- Ions are created in a cavity, and directed electrically through an orifice onto thedielectric surface of a rotating cylinder

surface

- Toner is the applied to the charged image and transferred to the paper on which it

is transfixed by pressure(cold fusion)

Question 3 Characteristics of a page printers?

Chapter 5: DATA STORAGE MEDIA.

Question 1 Data storage Requirements Characteristics?

- Interchangeability: can be change easily

- Transfer rate: fast enough

Question 2 Magnetic disks?

- This comprises a drive unit onto which one or perhaps two magnetic diskcartridges are loaded

- The drive consists of a control unit and a spindle housing that rotates continuouslywhen switch on

- The cartridge are loaded by the operator so as to provide the data currently neededfor the job in hand

- Bach tracks is devided up into sectors(often 4 or 8), sectors are read or written ormore at a time as blocks by means of a read

- There are usually one head for each surface, all the heads are moved

- Sunchronously across the tracks

- Once in position all the data on the equiradial tracks can be read or writtenwithout further movement of the heads

- Cylinder is a set of equiradial tracks

- A cartridge comprises several flat disks mounted on a central sprindle Whenmounted it rotates at a high speed enabling data to be read from or written to it.The data is recorded magnetically on both surfaces of each disk in the form ofconcertric tracks

- The heads are very close disk surface

- Curshion of air carried by the rotating disk

Question 3 Winchester disks( hard disks )?

- Comprises a number of platters(disks) permanently into an airtight enclosure

closer to the surfaces and so enabling greater recording densities to be employed

- The disks have greater storage capacity and a higher rate of data transger

- It has the lubricated surfaces allowing the heads “land” when the platters cease torotate, so eliminating head crashes

- Winchester platters are either 14 in, 8 in, 5¼ in or 3½ in diameter

Question 4 Floppy disks?

- Diskettes, generally called floppy disks, are single disks made of flexible plasticand permanently housed is an envelope

- The data on floppy disks is in concentric tracks on the outer part of the surfacesand access to it is via slot in the envelope

- The most common size are 3½ in, 5¼in, and 8 in diameter disks, the 3½ in diskshave the advantages of a shutter.

- Floppy disks may be either single or double sided and of course the drive needs to

be correspondingly equipped

that they have come into extensive used by small business and home computerbuffs

- The range of capacities is from 1/4 to 2 megabytes and transfer rates around 125

to 250 kilobytes per seconds

Question 5 Optical disks?

- Optical disk are comparatively new development for data storage

- Optical disks consist of a single removable glass, plastic or metal disk coated onone side with tellurium and protected by a 1 mm layer or transpacent plastic

- The disk diameters are mostly between 8 in and 14 in they rotate on a spindle in asimilar fashion to magnetic disks

- The data is recorded in the form of minute pits burned into the telliurium coating

by a finely-focused lazer beam

- Optical disks hold between 0.7 and GBs, this is about 20 times greater thanmagnetic dis cartridges

- The data is read by a low power laser beam which moved across the surface and isreflected into a photo cell

- Optical disks rotate mostly at 1500 r.p.m which, allowing for the movement of thelaser unti, given access time of between 16 & 500 ms and data transfer rates of 0.6

to 3 MVs per second

non-rewriteable

Question 6 Mass storage media?

- Mass storage media is a high capacity disk system as when necessary bytransferring data from a number of “data cartridges” house in cells

- Each cartridge consists of a 3 in wide magnetic modium inside a protective cover

- In order to load the disk system, the data cartridges are moved automatically fromthe cells

- A typical system consists of 9440 cartridges giving a storage capacity of 472000million bytes

Question 7 Magnetic drums?

- A magnetic drum consists of a cylinder upon the surface of which data is stored inmagnetic form in tracks running around its circumference, each track has its ownread/write head

- A typical magnetic drum has 800 tracks each capable of holding 5000 bytes

Question 8 Charge_coupled Device Memory (CCD)?

- CCD consists of thousands tiny metal squares each capable of holding an electriccharge, thus representing a bit

- The squares are in the form of an array 64 x 64 holding 4096 bits

- It is very impact

- CCD is volate lity storage

Question 9 Magnetic Bubble Memory?

- A thin wayer of magnetic garnet is capable of containing tiny domains orcylinders of magnetism, called bubbles

- By erasing unwanted bubbles, the resultant presence of a bubbles represent a 1 or

a 0 bit

non-volitility

Question 10 Megnetic tape?

method of backing storage

- It is often used as a depositony for disk dumped from fixed data storage.

- It is in reells of up 3600 feet and is made of Mylar plastic tape, 1/2 in wide andcoated with a magnetic material on one side

- The data is read from one read and written to another

- A reel of tape is loaded on a magnetic tape drive, and so as many drives areneeded as reels during a processing run

- It is used as a backing medium than a primary method of backing storage

- The seconds usually have to be sequence where store in magnetic tape

Chapter 7: COMPUTER FILES.

Question 1 File Processes?

1 Sorting

a The records in logical file are brought into some sequence as determined by key inthe records

b A computer is capable of sorting record into a “nested” sequence

c Sorting is done by a “sorting generator” This is part of the computer’s softwareand comprises several sophisticated sorting techniques that are called into useaccording to the file and the sort requirements

d The need of sorting has dimished in line with the demise of magnetic tape asbacking storage

b Masmatched records are highlighted for subsequent action

Chapter 8: DIRECT ACCESS FILE ORGANIZATION AND STRUCTURES.

Question 1 Storage and Access Modes?

There are 3 principal modes for storing and accessing accords on a disk or drum:

1 Serial mode:

- The record are stored contigously regardless of their keys

starting with the first record

- It is sometimes possible to partition a serial files thus reducing the search time bystarting the search at the beginning of a known partition

- A serial file is normally of a temporary nature awaiting sorting into a usefulsequence.

3 Indexed_sequential/ selective_sequential mode

and accessed selectively

- Groups of unrequired records are skipped past

- Indexed sequential files may also be accessed haphazandly

4 Random modes:

an add generation algorithm

- The only erricient way to find a record is to use the algorithm

 No index is required thus saving storage space

 It is a fast access method because little or no searching is involved

 Transaction do not need storing, thus saving time

 New records are easily insertly into the random file provided theyare not excessive in number

 The main problem with the random mode is in achieving a uniformspread of records over the storage are allocated to the file

Question 2 Direct Access Addressing?

- The key of record is used to identify by record

- The key of record also is used to decide its storage location(or address)

1 Self addressing:

- Self addressing is a straight forwards method because a record’s address is equal

to its key’s value

- The file is inevitably stored in key sequence

 It leads directly to the wanted record

 No indexing or searching is required

 The key itself need not necessarily be held within the storedrecord- although it generally is

 The storage space per second has to be the same

 When records one missing, storage locations related to its must beleft empty

2 Self addressing with key conversion

- This method a basically similar to self addressing except that the key required alittle processing to turn it into the record’s address

- This leads to either a pricise address

3 Matrix addressing

- In somes case, it is necessary to find the add of a record held within a multidimensional matrix of record it’s called matrix addressing

Question 3 Direct Access Searching?

- Where as addressing determines the location of a record by using algorithmicmethods, searching finds the record by scanning groups of records, and index, orboth

- ]The simplest method is to examine every record a file until the required record isfound a shortcut is generally desiable

1 Indexed sequential searching

- A cylinder index is created to hold the highest cylinder’s key

- Associated with each cylinder is a block index holding the highest key in eachblock within that cylinder

 The cylinder index is examined key_by_key until one is found that

is larger than or equal to the wanted key this directs the search tothe appropriate block index

 The block index a similarly examined and the search

 The block is searched record by record until the wanted record isfound

2 Binary searching( binary chopping )

- The key in the index to be binary search must be in sequence and form a completeset

- The search starts at the midpoint of the index and then moves half way to the left

or right(down or up) depending upon whether are wanted key is less than orgreater than the midpoint key

- In pracice, the index is unlikely to as convenient as this example because it is notalways possible to exactly halve each sucessive move(complete exact holvingispossible only when the total number of keys in the index is 20-1)

- The average number of examinations comparisons is (log2k)-1 ( k is the number ofkeys in the index)

4 Balanced binary tree searching

- A binary tree is a relationship of keys such that the examination of any key leads

to one of two other keys

- The binary tree is actually in the form of an index containing all the keys togetherwith a directory showing the braches stemming left and right from each key

- Binary tree searching is suitable for an unsequenced file

- The search is similar to binary searching in that each key examination holves therinaining keys, on average

Chapter 11: INTRODUCTION TO ARTIFICAL INTELLIGENCE.

Question 1 AI?

Atificial Intelligence

It has three braches

1 Expert systems (or knowledge- base system)

- ESs are programs that contain the knowledge of human expert, encoded so acomputer can understand it with encated- knowledge seasoning machinism, EScan tackle problem that are beyond the seach of conventionally programmedcomputers

2 Natural language systems (everyday native language)

- Natural language systems are programs that understand the native language of theuser, such as E

- The most popular natural language systems are those that act as interfaces to databases

3 Simple perception systems (for vision, speed and touch)

and quality control criteria, or move a robot to the proper location ot grasp a partfor manufacturing

Question 2 Who does the updates?

- Updating the knowledge bases is very diffirent when with updating databasesbecause of the difference in the type of information and in the cause and effectrelationship contained in knowledge bases

- A knowledge in the area, when databases may be modified by a normal users

Chapter 12: EXPERT SYSTEMS.

Question 1 What is an ES( Expert system )?

An ES is a knowledge-intersive program that solves a problem that normally requires humanexpertise

 Characteristics of ESs

- They solve problems as well as or better than human experts

- They use knowledge in the form of rules or frames

 Is the leasts expert or lowest level ESs

 It helps a decision maker by doing routine analysis and porting outthose portion of the work where human expertise is required

 The new discusses the problem until a joint decission is reached

 When system is going wrong, the user adds more information toget it back on track

 Is a system that advises the user without question

 There are no practical areas today in which decission

Question 2 A ES Life Cycle (ESLC)?

There are 6 phases life cycle in an ES

1 Phase1 – Selection of an Appropriate Problem

- Phase 1 involves finding an appropriate problem for an ES, indentifying an expert

to contribute the expertise

2 Phase 2 – Development of a prototype system

how to encode the facts, the relationships and the knowledge of experts

and to develop a deeper understanding of the field of expertise

- Other subtasks in this phase:

 Learning about the domain and the task

 Specifying performance criteria

 Selecting an ES building tool

 Developing an implementation plan

 Developing a detailed design for a complete system

3 Phase 3 – Development of a Complete System

- The main work in this phase is the addition of a very large number of rules

real world and the user interface has to be developed

2 Phase 4 – Evaluation of the system

- This phase involves testing the system against the performance establised inearlier stages

5 Phase 5 – Intergration of the system

- The ES has to be intergrated into the data flow and work patterns of theorganization

- In this stage, the expert system has to be interfaced with other databases,instruments and hardware

6 Phase 6 – Maintenance of the system

- The maintenance of the ES involves is updating, charging in the system whenoperating When operating, more problems occur in the system, so it is necessary

to continue take care the system by expert in a fix period of time

- So expert system, are so complex that in a few year the maintenance costs willequal the development costs.

BOOK II: Computer systems architecture.

Chapter 1 – 2: NUMBER BASES.

Question 1 Common number bases used in computer hardware operation?

- The base is ten – there are 10 different symbols, the digits 0, 1, 2, etc upto 9

- To represent value less than ten involves only one digit larger values need two ormore digits

- The base must be two, with only the digits 0 and 1 available

- To show values of two or ever require two or more binary digits

- Octal system has eight as its base, it uses the symbol 0, 1, 2 up to 7 only

- Two or more digits are needed for values of eight and above

- Hexadecimal system has sixteen as its base, it use the symbols 0, 1, 2 ,9 & A, B,

C, D, E, F, to stand for the “digits” ten, eleven, twelve, thirteen, fourteen, fifteen

Question 2 Converting from Bases To Bases?

1 Change the decimal

1.25 + 0.24 + 1.23 + 0.22 + 1.21 + 0.20 = 42

(101010)2 = (42)10

- To octal 100101101  1st step change into denary = 1.28 + 1.25 + 1.23 + 1.22 + 1.20 = 256 + 32 + 8 + 4 + 1

=(301)10  2nd step: convert to octal 301 8

61 37 8

(301)10 = (455)8 (100101101)2 = (455)8 - To hexadecimal 110111011011 1st step = 1.211 + 1.210 + 1.28 + 1.27 + 1.26 + 1.24 + 1.23 + 1.21 + 1.20 = 2048+ 1024 + 256 + 158 + 64 + 16 + 8 + 2 + 1 = (3547)10 2nd step 3547 16

384 221 16

27 61

(3547)10 = (CCA)16 (110111011011)2 = (CCA)16 3 Convert into binary and display the answer in normalized exponential form 247 1

123 1

61 1

30 1

15 1

7 1

3 1

1 1

0 1 (247)10 = (11110111)2

= 0 1111011 x 2 normalized exponential form

Question 3 Integer and Floating – point arithmetic?

1 Floating – point Addition

a (0.1011 x 25 ) + (0.1001 x 25 )

= (0.1011 + 0 1001) x 25

= 1.0100 x 25

= 0.10100 x 26

b (0.1001 x 23 ) + (0.1110 x 25 )

= (0.001001 x 25 ) + (0.1110 x 25 )

= (0.001001 + 0.111000) x 25

= 1.000001 x 25

= 0.1000 x 26 (here have truncation)

2 Floating – point subtraction

1

1 3