Một số câu hỏi dành cho máy tính

Chia sẻ: Đỗ Hàn Thuyên | Ngày: | Loại File: DOC | Số trang:40

0
310
lượt xem
93
download

Một số câu hỏi dành cho máy tính

Mô tả tài liệu
  Download Vui lòng tải xuống để xem tài liệu đầy đủ

Kinh nghiệm, mẹo nhỏ giúp bạn giải đàp những thắc mắc về tin học, máy tính.

Chủ đề:
Lưu

Nội dung Text: Một số câu hỏi dành cho máy tính

  1. 1 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  and returns the results through an output unit. A   computer   is   basically   an   electronic   machine   operating   on  current.
  2. 2 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 :   • control unit (CU) • arithmetic logic unit (ALU) • accumulator (ACC) • program counter                   (PC) • instruction register (IR) • memory address register (MAR) • memory data register (MDR) • status register (SR) • general purpose register - The function of each components of CPU: • Control unit:  control   and   co_ordinate   all   hardware  functions of the CS.  examine   and   decode   all   program   instructions  to the computer and initiate their execution  by sending the appropriate signals. • ALU:  performs   all   arithmetic     and   logic   comparision   two  values functions required by computer. • ACC:  holds   the   first   operand   of   the   temporary  result of the ALU. • PC:  contains the add of the next instruction to  be excuted. • IR:  contains   the   current   instruction   to   be  executed. Main memory   • MAR:  holds the address location to or from which  data is to be transferred • MDR:  contains   the   data   to   be   written   to   or   read  out of the addressed location. • SR:  keeps   track   of   the   status   of   the  accumalator.
  3. 3 • G eneral Purpose Regi ste r:  for genera l purpose procedures.  Please   refe r   to   di ram   for   an   illu stra tin   of   the   basi   ag c com ponents of the  CPU .      CPU I           N T                                   Contro l unit E                                   Arithm eti  Log i  Uni c c t R N                                   Accum ul ator A                                                                    Program  Couter   L to  m ai n                                     Instru cti  Regi on ste r   B U m em ory S                                   M em ory Address Registe r                                   M em ory D ata Registe r                                   Status  Registe r                                    G eneral Purpose Registe r Basi  com ponents of a CPU . c                  Contro l Unit               Inpu t Unit        ALU  O utput Unit                                                    M ai n M em ory            Backi  Storage ng        Contro l si gna ls        D ata flow Com ponents of a CS. 2. Input uni ts - U sed   to   enter   data(   raw   unprocessed   facts)   and  i stru cti s  to  the  com put n on er. 3. O utput units
  4. 4 - U sed   f   del or everi ng   the   processed   resu lt   from   the   com put er i  usef l for . n u m 4. Backi  storage  uni ng ts - Backi   storage   uni   need   for   hi   capaci   data   ng ts gh ty storage   devi ces   that   can   store   data   i   a   m or n e  per anent   f m   for   l te r   retri ra l,   updati m or a e ng   and  re fe renci . ng - Backi   storage   i   al   call ng s so ed   secondary   storage   externa l storage  and auxilia ry  storage. Chapter 2: MICOPROCESSOR. Question 1. Cache Memory? - Cache memory is a small amount of very fast store with  faster access time than the main memory. - Cache   memory   is   used   to   temporaryty   store   data  instructions   that   are   likely   to   be   retrieved   many  times, thus speeds up the processing of data. - Sits between main storage and the processor acting as  holding   area   through   which   all   data   and   instructions  pass. - Old   data   in   the   cache   memory   is   over   written   by   new  then cache is full. Question 2. Virtual Memory? - Virtual memory makes use of both the main memory and  backing store. - In   a   virtual   memory   sys,   each   user   has   the   illusion  that his program is in the main memory all the time. - The sys maintains this illusion by keeping some of the  “unused” portion  of the program’s  code and data on a  backing store device which is usually magnetic disk  - The   movement   of   the   unused   portion   from   the   backing  store to the mian memory is transparent to the users.  - Please refer to diagram for virtual memory.         Backing Store   Main Memory A3 A2 A1 A3 A2 A1 Virtual Memory   
  5. 5 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   been   collected   and   grouped   together  called   Batched   Generally   data   is   accumulated   for   a  certain period of time or unitl a certain quantity. - Ads:  Response  time is not critical.  Need to process large volumn of data.  Computer   efficiency   is   more   important   than  response time. - Dis:  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. - Ads:   Enter   availability   of   information   for  decision making.  More accurate data capture.  Schedules suits user. - Dis:   CPU time is used less efficiently.  Random   arrival   of   transactions,   terminal  operator   process   each   transaction  separately.  More expensive than batch processing. Question 3. Real Time Processing System? - Def:   One   which   controls   the   environment   by   receiving  data   processing   them   and   returning   results  sufficiently quickly to affect the functioning of the  environment at that time.  - Ads:  Response   time   is   very   critical   and  sufficient quick. - Dis:  Expensive hardware & software.  Very   complex   in   terms   of   hardware   &  software.
  6. 6 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  Daisywheel printers b. Line printers Medium to high speed printers that can print in excess of  2000 lines per minute. Eg: Chain Printers Band Printers Drum 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 the paper. b. Non impact printers Use more silent methods of printing. Eg: Thermal printers Ink Jet printers Lazers printers 3. Classifying printers according to print quality Kinds of quality printers Draft 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 of graphics. - The   print   resolution   is   generally   lower   than   lazer  printers. b. Daisywheel printers - Are serial impact printers, the speed of a daisywheel  printer is slow(20­55 characters per second), noisy in  operation. - The print head has the letters arranged at the end of  spokes round a central hub. c. Chain printers - The   chains   printers   has   its   characters   set   rapidly  rotating on a print chain. d. Band printers - The band printer has rotating scalloped steel band. e. Drum printers
  7. 7 - Are l i ne prin te rs , the pr in t character are ra i sed in bands around a heavy metal drum which rotates at very high speed. - The pr in t hammers str i ke the paper and a prin t r ibbon against an apropr ia te character on the l i ne . An ent i re l i ne of the same character is pr in ted on one rotat i on of the drum. f. Thermal pr in te rs - Uses specia l heat sensi t i v e paper and a matr ix of pr in t wires that become hot when exposed to an elect r i c current . The heated wires come in to close contact with the paper, burning the image of the character onto i t . - The more advanced thermal pr in te rs are using thermal trans fer prin t i ng . - They have a specia l heat sensi t i ve r ibbon and a pr in t head with wires that become hot when a currents is appl ied . - The heat from the pr in t wires causes the ink from the r ibbon to fuse to a piece of regular paper. g. In l Je t Pr in te rs - The ink je t pr in ts by using a smal l droplet generator to break specia l inks into t iny drops, which are then forced towards a paper supply . h. Lazer pr in te rs - Using a photoconduct ive drum. - A lazer i s then used to wri te the image of the character onto the drum. - After exposure to the lazer , the drum rotates through a develop ing stat i on , picks up toner and transfe rs i t to the paper. - The character is fused onto the paper by heat . i. Ion deposi t i on pr in te rs - Ions are created in a cavi ty , and di rec ted elect r i c a l l y through an or i f i c e onto the die lec t r i c surface of a rotat i ng cyl i nder . - The requi red characters are formed as an elect r i c charge image on the cyl i nders sur face . - Toner i s the appl i ed to the charged image and trans fer red to the paper on which i t i s trans f i xed by pressure(co ld fus ion) . j. Elect ros ta t i c pr in te rs - Letterheads and logos are created elect ros ta t i c a l l y f rom a changeable metal cyl i nder . k. Magnetic prin te rs - A drum in the pr in te r has a surface that can be coated with sows of t iny spots of magnetion by means of thousands of minute record ing heads.
  8. 8 - As the drum rotates it becomes covered with these magnetic spots so as to from a latent image of the page to be printed. - Dry ink partic les are brought into contact with the drum’s surface and these adthere to the magnetised spots. The ink was then pressunal on to the surface and subsequently transferred onto the paper. Question 3. Characteristics of a page printers? - Speed - Characters sets - Copies - Intel l i gence - Output Chapter 5: DATA STORAGE MEDIA. Question 1. Data storage Requirements Characteristics? - Low access time: fast speed - Storage capacity: much enough - Interchangeabil i ty : can be change easily - Security: safe enough - Transfer rate: fast enough - Cost: economic Question 2. Magnetic disks? - This comprises a drive unit onto which one or perhaps two magnetic disk cartr idges are loaded. - The drive consists of a control unit and a spindle housing that rotates continuously when switch on. - The cartr idge are loaded by the operator so as to provide the data currently needed for the job in hand. - Bach tracks is devided up into sectors(often 4 or 8), sectors are read or written or more 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 posit ion all the data on the equiradial tracks can be read or written without further movement of the heads. - Cylinder is a set of equiradial tracks. - A cartr idge comprises several f lat disks mounted on a central sprindle. When mounted it rotates at a high speed enabling data to be read from or written to it . The data is recorded magnetical ly on both surfaces of each disk in the form of concertr ic tracks. •  Certain models of disk units also have a number of f ixed read/write heads in addit ion to the movable heads.
  9. 9  The f ixed head are posit ioned permanently over certain of the outer tracks, there being one head per track, so cl imimating the need for head movement. - The heads are very close disk surface. - Curshion of air carried by the rotat ing disk. Question 3. Winchester disks( hard disks )? - Comprises a number of platters(disks) permanently into an airt ight enclosure. - All dust is excluded thus perimit ing the read/write heads to be posit ioned even closer to the surfaces and so enabling greater recording densit ies to be employed. - The disks have greater storage capacity and a higher rate of data transger. - I t has the lubricated surfaces allowing the heads “land” when the platters cease to rotate, so eliminating head crashes. - Winchester platters are either 14 in, 8 in, 5¼ in or 3½ in diameter. Question 4. Floppy disks? - Diskettes, generally cal led f loppy disks, are single disks made of f lexible plast ic and permanently housed is an envelope. - The data on f loppy disks is in concentric tracks on the outer part of the surfaces and access to it is via slot in the envelope. - The most com mon size are 3½ in, 5¼in, and 8 in diameter disks, the 3½ in disks have the advantages of a shutter. - Floppy disks may be either single or double sided and of course the drive needs to be correspondingly equipped. - Both the drives and the f loppy disks themselves are inexpensive with the result that they have come into extensive used by small business and home computer buffs. - The range of capacit ies is from 1/4 to 2 megabytes and transfer rates around 125 to 250 ki lobytes per seconds. Question 5. Optical disks? - Optical disk are comparatively new development for data storage. - Optical disks consist of a single removable glass, plast ic or metal disk coated on one side with tel lur ium and protected by a 1 m layer or transpacent m plast ic .
  10. 10 - The disk diameters are mostly between 8 in and 14 in they rotate on a spindle in a similar fashion to magnetic disks. - The data is recorded in the form of minute pits burned into the tel l iur ium coating by a f inely- focused lazer beam. - Optical disks hold between 0.7 and GBs, this is about 20 times greater than magnetic dis cartr idges. - The data is read by a low power laser beam which moved across the surface and is ref lected into a photo cell . - Optical disks rotate mostly at 1500 r.p.m which, al lowing for the movement of the laser unti , given access time of between 16 & 500 ms and data transfer rates of 0.6 to 3 MVs per second. - The draw back of optical disks is that the data cannot be erased so making them non-rewriteable. Question 6. Mass storage media? - Mass storage media is a high capacity disk system as when necessary by transferr ing data from a number of “data cartr idges” house in cel ls . - Each cartr idge consists of a 3 in wide magnetic modium inside a protective cover - In order to load the disk system, the data cartr idges are moved automatical ly from the cells . - A typical system consists of 9440 cartr idges giving a storage capacity of 472000 mill ion bytes. Question 7. Magnetic drums? - A magnetic drum consists of a cyl inder upon the surface of which data is stored in magnetic form in tracks running around its circumference, each track has its own read/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 electr ic charge, thus representing a bit . - The squares are in the form of an array 64 x 64 holding 4096 bits. - I t is very impact. - CCD is volate l i ty storage. Question 9. Magnetic Bubble Memory? - A thin wayer of magnetic garnet is capable of containing tiny domains or cyl inders of magnetism, called bubbles. - By erasing unwanted bubbles, the resultant presence of a bubbles represent a 1 or a 0 bit .
  11. 11 - The main ads are low power consumption, compactness, robustness rel iabi l i t y and non-voli t i l i t y . Question 10. Megnetic tape? - The magnetic tape usage is now more as a backup medium rather than a primary method of backing storage. - I t is often used as a depositony for disk dumped from f ixed data storage. - I t is in reel ls of up 3600 feet and is made of Mylar plast ic tape, 1/2 in wide and coated 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 are needed as reels during a processing run. - I t 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 f i le are brought into some sequence as determined by key in the records. b. A computer is capable of sort ing record into a “nested” sequence. c. Sorting is done by a “sort ing generator”. This is part of the computer’s software and comprises several sophist icated sort ing techniques that are called into use according to the f i le and the sort requirements. d. The need of sort ing has dimished in l ine with the demise of magnetic tape as backing storage. 2. Merging - Merging implies that two or more f i les in the same sequence are combined into one f i le . a. Fi le merging  Two or more separate f i les of similar seconds and in the same sequence are marged together so as to form one f i le . b. Record merging  The records from two or more “input” f i les , usually in the same sequence, are combined one record in the output f i le . 3. Matching a. Two or more input f i les (general ly in the same sequence) are compared records against record in order to ensure that there is a complete set of records for each key.
  12. 12 b. Masmatched records are highl ighted for subsequent action 4. Sum manizing a. Records with the same key in one f i le are accumulated together to form one record in the output f i le . b. Sum manizing usually applies to a f i le presorted into a certain sequence and the resultant f i le is in the same sequence. c. Records to be sum marized are generally of a similar type. 5. Searching a. Searching is looking for records with certain keys or holding certain data and in some way making a note of these. b. An instance is a search for and count of all records with a debt balance of above a certain amount. 6. Information retr ieval a. Information retr ieval is the process that involves the bringing together of data from several f i les . b. Data may also be extracted from several f i les and combined before being presented as information. 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 - The sole way of accessing serial seconds is to search through the complete f i le start ing with the f i rst record. - I t is sometimes possible to parti t ion a serial f i les thus reducing the search time by start ing the search at the beginning of a known parti t ion. - A serial f i le is normally of a temporary nature awaiting sort ing into a useful sequence. 2. Sequential mode: - direct access sequential mode normally involves accessing sequential a f i le that is stored sequential ly . - sequential mode is often associated with a master f i le held in a certain sequence and updated by a transaction f i le sorted into the same sequence. 3. Indexed_sequential / select ive_sequential mode - Indexed_sequential is a mode of storage where by records are held sequential ly and accessed select ively. - Groups of unrequired records are skipped past.
  13. 13 - Indexed sequential f i lesmay also be accessed haphazandly. 4. Random modes: - Each record is stored in a location determind from the second’s key by means of an add generation algorithm. - The only erric ient way to f ind a record is to use the algorithm - Random mode is applicable to master f i les • Ads of random modes  No index is required thus saving storage space  I t is a fast access method because l i t t l e or no searching is involved  Transaction do not need storing, thus saving time  New records are easi ly insert ly into the random f i le provided they are not excessive in number • Dis  The main problem with the random mode is in achieving a uniform spread of records over the storage are al located to the f i le Question 2. Direct Access Addressing? - The key of record is used to identi fy 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 f i le is inevitably stored in key sequence • Ads of self addressing  I t leads direct ly to the wanted record  No indexing or searching is required  The key itsel f need not necessari ly be held within the stored record- although it generally is • Dis  The storage space per second has to be the same  When records one missing, storage locations related to its must be left empty 2. Self addressing with key conversion - This method a basical ly similar to self addressing except that the key required a l i t t l e processing to turn it into the record’s address - This leads to either a pricise address
  14. 14 3. Matrix addressing - In somes case, it is necessary to f ind the add of a record held within a multi dimensional matrix of record it ’ s called matrix addressing. Question 3. Direct Access Searching? - Where as addressing determines the location of a record by using algori thmic methods, searching f inds the record by scanning groups of records, and index, or both. - ]The simplest method is to examine every record a f i le unti l the required record is found a shortcut is generally desiable. 1. Indexed sequential searching - A cyl inder index is created to hold the highest cyl inder’s key - Associated with each cyl inder is a block index holding the highest key in each block within that cyl inder - When searching for a record’s key in the index  The cyl inder index is examined key_by_key unti l one is found that is larger than or equal to the wanted key this directs the search to the appropriate block index  The block index a similar ly examined and the search  The block is searched record by record unti l the wanted record is found 2. Binary searching( binary chopping ) - The key in the index to be binary search must be in sequence and form a complete set - 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 or greater than the midpoint key - In pracice, the index is unlikely to as convenient as this example because it is not always possible to exactly halve each sucessive move(complete exact holvingis possible only when the total number of keys in the index is 20- 1) - The average number of examinations comparisons is (log 2k)­1 ( k is the number of keys in the index) 3. Block searching - A block is a subdivis ion of an index. A block is devised to contain, roughly the square root of the number of keys in the whole index - The search is f i rst through the block index to f ind the appropriate block and then through this to f ind the wanted key
  15. 15 - The average number of examinations is square – root – k  (k is the total number of keys) 4. Balanced binary tree searching - A binary tree is a relat ionship of keys such that the examination of any key leads to one of two other keys - The binary tree is actual ly in the form of an index containing all the keys together with a directory showing the braches stemming left and right from each key - Binary tree searching is suitable for an unsequenced f i le - The search is similar to binary searching in that each key examination holves the rinaining keys, on average Chapter 11: INTRODUCTION TO ARTIFICAL INTELLIGENCE. Question 1. AI? Atif ic ia l Intel l igence I t has three braches 1. Expert systems (or knowledge- base system) - ESs are programs that contain the knowledge of human expert, encoded so a computer can understand it with encated- knowledge seasoning machinism, ES can tackle problem that are beyond the seach of conventional ly program med computers. 2. Natural language systems (everyday native language) - Natural language systems are programs that understand the native language of the user, such as E - The most popular natural language systems are those that act as interfaces to data bases 3. Simple perception systems (for vision, speed and touch) - They can interpret visual scenes and decide i f object meet inspection standards and quali ty control cri ter ia , or move a robot to the proper location ot grasp a part for manufacturing Question 2. Who does the updates? - Updating the knowledge bases is very dif f i rent when with updating databases because of the dif ference in the type of information and in the cause and effect relat ionship 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 human expertise • Characterist ics of ESs - They solve problems as well as or better than human experts - They use knowledge in the form of rules or frames
  16. 16 - They can consider multiple hypotheses simultaneouly • Types of ES - An assistant  Is the leasts expert or lowest level ESs  I t helps a decision maker by doing routine analysis and porting out those portion of the work where human expertise is required - A col leage  The new discusses the problem unti l a joint decission is reached  When system is going wrong, the user adds more information to get it back on track - True ES  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)? - An accepted SDLC for expert systems has yet to be developed There are 6 phases l i fe cycle in an ES 1. Phase1 – Selection of an Appropriate Problem - Phase 1 involves f inding an appropriate problem for an ES, indenti fy ing an expert to contribute the expertise - Establ ishing a prel iminary approach - Analysing the cost and benefitsPreparing a development plan 2. Phase 2 – Development of a prototype system - A prototype sys is a small version of an ES designed to test assumptions about how to encode the facts, the relat ionships and the knowledge of experts - The prototype permits the knowledge engineer to gain the expert’s com mitment and to develop a deeper understanding of the f ie ld of expertise - Other subtasks in this phase:  Learning about the domain and the task  Specifying performance cri ter ia  Selecting an ES building tool  Developing an implementation plan  Developing a detai led design for a complete system 3. Phase 3 – Development of a Complete System - The main work in this phase is the addit ion of a very large number of rules
  17. 17 - The knowledge base has to be expanded to ful l knowledge base appropriate to the real world and the user interface has to be developed 2. Phase 4 – Evaluation of the system - This phase involves test ing the system against the performance establ ised in earl ier stages 5. Phase 5 – Intergrat ion of the system - The ES has to be intergrated into the data f low and work patterns of the organization - 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 when operating. When operating, more problems occur in the system, so it is necessary to continue take care the system by expert in a f ix period of time - So expert system, are so complex that in a few year the maintenance costs wil l equal the development costs. BOOK II: Computer systems architecture. Chapter 1 – 2: NUMBER BASES. Question   1. Common   number   bases   used   in   computer   hardware   operation? • Decimal(denary) system: - The base is ten – there are 10 dif ferent symbols, the digits 0, 1, 2, etc. . .upto 9 - To represent value less than ten involves only one digit larger values need two or more digits • Binary system - 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 - 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(hex) - 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, thir teen, fourteen, f i f teen. Question 2. Converting from Bases To Bases? 1. Change the decimal - Binary:
  18. 18 Eg. (2559) 10 2559 1 1279 1 639 1 319 1 159 1 79 1 (2559)10 = (10111111111)2 39 1 19 1 9 1 4 0 2 1 0 0 - Octal: 7690 8 49 96,1 8 10 16 120 8 2 1 40 15 8 0 7 1                                                                            (7690)10  =  (17012)8 - Hexadecimal: 6396 16 159 399 16 156 79 24 16 1 1 8 1 C F (6369)10 = (CF81)16 2. Convert to others from binary - To decimal (101010)2                   (?)10  1.25 + 0.24 + 1.23 + 0.22 + 1.21 + 0.20 = 42 (101010)2 = (42)10  - To octal 100101101  1 step change into denary st
  19. 19 = 1.2 + 1.25 + 1.23 + 1.22 + 1.20 8 = 256 + 32 + 8 + 4 +1 =(301)10  2 step: convert to octal nd 301 8 61 37 8   5 5 4 (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 1 1 1 (3547)10  =   (CCA)16  2 16 (110111011011)  = (CCA) 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) = (11110111)   10 2 = 0. 1111011 x 2 normalized exponential form Question 3. Integer and Floating – point arithmetic?
  20. 20 1. Floating – point Addition a. (0.1011 x 2 ) + (0.1001 x 2 ) 5 5 = (0.1011 + 0. 1001) x 2 5 = 1.0100 x 25 = 0.10100 x 26 b. (0.1001 x 2 ) + (0.1110 x 2 ) 3 5 = (0.001001 x 2 ) + (0.1110 x 25 ) 5 = (0.001001 + 0.111000) x 25 = 1.000001 x 25 = 0.1000 x 26 (here have truncation) (0.1000001 x 26 ) 2. Floating – point subtraction a. (0.1110 x 2 ) – (0.1100 x 2 ) 7 7 = 0.0010 x 2 7 = 0. 10 x 25 b. (0.1001 x 2 ) – ( 0.1000 x 2 ) 8 5 = (0.1001 x 2 ) – ( 0.0001 x 28 ) 8 = 0.1000 x 28 3. Floating – point multipl icat ion a. (0.1010 x 2 ) x (0.1100 x 2 ) 3 3 = (0.1010 x 0.1100) x 2 6 = 0.01111 x 26 = 0.1111 x 25 b. (0.11110 x 2 ) x ((0.01011) x 24 ) 3 = (0.11110 x 0. 01011) x 2 7 = 0.001111 x 27 = 0.1111 x 25 4. Floating – point divis ion. a. (0.11010 x 2 ) : (0.001 x 2 ) 6 6 = (0.11010 x 26) : (1 x 23) = 0.1101 x 26 : 1x 23 = 0.1101 x 23 b. (0.110111 x 2 ) : (0.1001 x 2 ) 6 4 = (0.110111 : 0.1001) x 22 = (1101.11 : 1001) x 22 = 1.100001 x 22 = 0.1100001 x 23 Chapter 3: TYPES OF INSTRUCTION AND ADDRESSING. Question 1. Types of instructions used in CS? 1. Arithmetic instruct ions. Arithmetic instruct ions include direct ives to the computers to perform addit ions, subtraction, multipl i cat ions, divis ions and exponentiat ions. 2. Input/ output instruct ions. They direct the computer to read data values from the specif ied input devices into the main store for processing.

CÓ THỂ BẠN MUỐN DOWNLOAD

Đồng bộ tài khoản