Tài liệu một số câu hỏi dành cho máy tính

Chia sẻ: Hoang The Vinh | Ngày: | Loại File: DOC | Số trang:40

0
95
lượt xem
20
download

Tài liệu một số câu hỏi dành cho máy tính

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

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.

Chủ đề:
Lưu

Nội dung Text: Tài liệu 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. 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   coordinate   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:
  2. 2  hol   the   first   operand   of   the   tem porary   ds resu lt  of the  ALU . • PC :  conta i s  the  add of the  next i n nstru ction  to   be excuted . • I : R  conta i s   the   curren t   i n nstru ction   to   be  executed . M ai n m em ory   •M AR:  hol  the  address  l ds ocati  to  or from  w hi   on ch data  i  to  be transfe rred s • M DR:  conta i s   the   data   to   be  w ri n tten   to   or  read   out of the  addressed  l ocati .on • SR :  keeps   track   of   the   status   of   the   accum al ator. • 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  
  3. 3                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 fl ow 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 uni ts - 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.
  4. 4 - The sys m ai nta i s  th i  illu sion  by keepi  som e of the   n s ng “unused” porti  of the  program ’  code  and data  on  a  on s backi  store  devi  w hi  i  usuall  m agneti  di   ng ce ch s y c sk - The   m ovem ent  of  the   unused   porti   from   the   backi   on ng store  to  the  m ai n m em or s  transparen t to  the  users.  y i - Please  refe r  to  di agram  for vi rtua l m em ory.         Backi  Store ng   M ai n M em ory A3 A2 A1 A3 A2 A1 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   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.
  5. 5  M ore accurate  data  capture .  Schedul  su i  user. es ts - Dis:   CPU ti e i  used l m s ess  effic ien tly .  Random   arri l   of   transacti s,   ter i l  va on m na operator   process   each   transacti   on separate l . y  M ore expensi  than batch  processi . ve ng 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. 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
  6. 6 G raphi  quality    c 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 - Are  line   printers,   the   print   character   are   raised   in  bands around a heavy metal drum which rotates at very  high speed. - The print hammers strike the paper and a print ribbon  against an apropriate character on the line. An entire  line of the same character is printed on one rotation  of the drum. f. Thermal printers - Uses   special   heat   sensitive   paper   and   a   matrix   of  print   wires   that   become   hot   when   exposed   to   an  electric   current.   The   heated   wires   come   into   close  contact   with   the   paper,   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   become   hot   when   a   currents   is  applied. - The heat from the print wires causes the ink from the  ribbon to fuse to a piece of regular paper. g. Inl Jet Printers - The ink jet prints by using a small droplet generator  to break special inks into tiny drops, which are then  forced towards a paper supply. h. Lazer printers - Using a photoconductive drum. - A   lazer   is   then   used   to   write   the   image   of   the  character onto the drum.
  7. 7 - A fter  exposure  to  the  l azer,  the  drum  rota tes  th rough  a devel i  stati ,  pi op ng on cks  up toner and transfe rs  it   to  the  paper. - The character  i  f s used  onto  the  paper by heat. i. I  deposi on tion  pri te rs n - I ons   are   created   i   a   cavi ,   and   di n ty rected   el ectrica lly   th rough   an   orifice   onto   the   di l e ectric   surface  of a rota ti g  cyli n nder. - The   requ i red   characters   are   for ed   as   an   el m ectric   charge i age on the  cyli m nders  surface . - Toner   i   the   appli s ed   to   the   charged   i age   and  m transfe rred  to  the  paper on w hi  it  i  transfi ch s xed  by  pressure(co l  f d usi ). on j. El ectrosta tic  pri te rs n - Lette rheads   and   l ogos   are   created   el ectrosta tica lly   from  a changeabl  m et e al cyli nder. k. M agneti  pri te rs c n - A drum  i  the  pri te r  has a surface  that can be coated   n n w i   sow s   of   ti th ny   spots   of   m agneti   by   m eans  of  on thousands of m i nute  record i  heads. ng - As   the   drum   rota tes   it   becom es  covered   w i   these   th m agneti   spots   so   as   to   from   a   l ten t   i age   of  the   c a m page to  be pri ted . n - D ry   i   parti le s   are   brought  i to   contact   w i   the   nk c n th drum ’   surface   and   these   adthere   to   the   m agneti s sed   spots.   The   i   w as  then   pressunal  on   to   the   surface   nk and subsequentl  transfe rred  onto  the  paper. y Question 3. Characteristics of a page printers? - Speed - Characters sets - Copies - Intelligence - Output Chapter 5: DATA STORAGE MEDIA. Question 1. Data storage Requirements Characteristics? - Low access time: fast speed - Storage capacity: much enough - Interchangeability: 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 cartridges are loaded. - The   drive   consists   of   a   control   unit   and   a   spindle  housing that rotates continuously when switch on. 
  8. 8 -The   cartri ge   are   l d oaded   by   the   operator   so   as   to   provi  the  data  curren tl  needed f  the  j  i  hand. de y or ob n - Bach tracks  i  devi s ded  up i to  sectors(o ften  4 or 8),  n sectors   are   read   or   w ri tten   or   m or   at   a   ti e   as  e m bl ocks by m eans of a read . - There  are  usuall  one head f  each  surface ,  all  the   y or heads are  m oved. - Sunchronousl  across  the  tracks. y - O nce i  posi n tion  all  the  data  on the  equi rad i l  tracks   a can be read  or w ri tten  w i thout f rther  m ovem ent of the   u heads. - Cyli nder i  a set of equi s rad i l  tracks. a - A cartri ge  com pri d ses  severa l  fl t  di a sks  m ount ed on a  centra l   spri l .   W hen  m ount   it   rota tes   at  a   hi   nd e ed gh speed enabli  data  to  be read  from  or w ri ng tten  to  it.   The data  i  recorded  m agneti lly  on both  surfaces  of  s ca each di  i  the  f m  of concertric  tracks. sk n or •  Certa i   m odel   of   di   uni   al   have   a  n s sk ts so num ber of fi xed  read / rite  heads i  addi on   w n ti to  the  m ovabl  heads. e  The   fixed   head   are   posi tioned   per anentl   m y over   certa i   of   the   outer   tracks,   there   n bei  one head per track,  so  cli i ati  the   ng m m ng need f or head m ovem ent. - The heads are  very  cl ose  di  surface . sk - Cursh i  of ai  carri  by the  rotati  di on r ed ng sk. Question 3. Winchester disks( hard disks )? - Comprises a number of platters(disks) permanently into  an airtight enclosure. - All   dust   is   excluded   thus   perimiting   the   read/write  heads to be positioned even 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   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   called   floppy   disks,   are   single  disks made of flexible plastic and permanently housed  is an envelope.
  9. 9 -The   data   on   floppy   disks   i   i   concentric   tracks   on  s n the  outer part of the  surfaces  and access to  it  i  vi   s a sl t  i  the  envel o n ope. - The   m ost  com m on  si   are   3½  i ,   5¼i ,   and   8   i   ze n n n di eter di am sks,  the  3½ i  di n sks  have the  advantages of  a shutte r. - Floppy  di sks  m ay be ei ther  si l  or doubl  si ng e e ded  and  of   course   the   dri ve   needs   to   be   correspond i l  ng y equi pped . - Both   the   drives   and   the   floppy   di sks   them selves   are   inexpensi  w i   the   resu lt  that   they   have   com e i to   ve th n extensi   used   by   sm al   busi ve l ness   and   hom e  com put er  buffs. - The range of capaci s  i  from  1/  to  2 m egabytes and  tie s 4 transfe r   rates   around   125   to   250   kil obytes   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   on   one   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   a   similar   fashion   to  magnetic disks. - The data is recorded in the form of minute pits burned  into the telliurium coating by a finely­focused lazer  beam. - Optical disks hold between 0.7 and GBs, this is about  20 times greater than magnetic dis cartridges. - The data is read by a low power laser beam which moved  across the surface and is reflected into a photo cell. - Optical   disks   rotate   mostly   at   1500   r.p.m   which,  allowing   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 non­rewriteable. Question 6. Mass storage media? - Mass storage  media is a high capacity  disk system  as  when   necessary   by   transferring   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 from the cells.
  10. 10 - A typ i l  system  consi ca sts  of 9440 cartri ges  gi i  a  d v ng storage  capaci  of 472000 m i ty llion  bytes. Question 7. Magnetic drums? - A   magnetic   drum   consists   of   a   cylinder   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   electric   charge,   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   or   cylinders   of   magnetism,  called bubbles. - By erasing unwanted bubbles, the resultant presence of  a bubbles represent a 1 or a 0 bit. - The   main   ads   are   low   power   consumption,   compactness,  robustness reliability and non­volitility. Question 10. Megnetic tape? - The magnetic tape usage is now more as a backup medium  rather than a primary 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 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. - 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 in the records.
  11. 11 b. A   com put   i   capabl   of   sorti er s e ng   record   i to   a  n “nested” sequence. c. Sorti  i  done by a “sorti g  generator”.  Th i  i  part  ng s n s s of   the   com put er’s   softw are   and   com pri ses   severa l  soph i stica ted  sorti  techn i ng ques  that  are  call  i to   ed n use accord i  to  the  file  and the  sort  requ i ng rem ents. d. The   need   of   sorti g   has   di i n m shed   i   li e   w i   the   n n th dem i se of m agneti  tape as backi  storage. c ng 2. M er i   g ng - M er i   i pli   that   t o   or   m or   file s   i   the   sam e  g ng m es w e n sequence are  com bi ned i to  one file . n a. Fil  m er ng e gi  Tw o  or  m or   separate   file s   of  si il r   e m a seconds   and   i   the   sam e  sequence   are   n m arged together so as to  f m  one file . or b. Record  m er ng gi  The   records   from   t o   or   m or   “i w e npu t”   file s,   usuall   i   the   sam e  sequence,  y n are   com bi ned   one   record   i   the   output  n file . 3. M at i ch ng a. Tw o   or   m or   i e nput   file s   (genera lly   i   the   sam e  n sequence) are  com pared records  agai nst  record  i  order  n to  ensure  that there  i  a com pl s ete  set of records  f   or each key. b. M asm at ched   records   are   hi li ted   for   subsequent  gh gh acti on 4. Sum m ani ng zi a. Records  w i  the  sam e key i  one file  are  accum ul th n ated   together to  for  one record  i  the  output file . m n b. Sum m ani ng  usuall  appli  to  a file  presorted  i to  a  zi y es n certa i  sequence and the  resu ltan t  file  i  i  the  sam e  n s n sequence. c. Records   to   be   sum m ari zed   are   genera ll   of  a   si il r   y m a type. 5. Search i ng a. Search i  i  l ng s ooki  f  records  w i  certa i  keys or  ng or th n hol i  certa i  data  and i  som e w ay m aki  a note  of  d ng n n ng these. b. An i stance  i  a search  for  and count of all  records   n s w i  a debt bal th ance  of above a certa i  am ount. n 6. I fo r ati  retrieva l n m on a. I fo r ati  retrieva l  i  the  process  that i n m on s nvo lves  the   bri i  together of data  from  severa l file s. ng ng b. D ata   m ay  al   be   extracted   from   severa l   file s   and  so com bi ned bef ore  bei  presented  as i fo r ati . ng n m on Chapter 8: DIRECT ACCESS FILE ORGANIZATION AND STRUCTURES. Question 1. Storage and Access Modes?
  12. 12 There are 3 pr inc ipa l modes for stor i ng and access ing accords on a disk or drum: 1. Ser ia l mode: - The record are stored cont igous ly regard less of thei r keys - The sole way of access ing ser ia l seconds i s to search through the complete f i l e star t i ng with the f i r s t record . - I t i s sometimes possib l e to part i t i on a ser ia l f i l e s thus reducing the search t ime by star t i ng the search at the beginning of a known part i t i on . - A ser ia l fi le i s normal ly of a temporary nature await i ng sort i ng into a usefu l sequence. 2. Sequentia l mode: - di rec t access sequent ia l mode normal ly invo lves access ing sequent ia l a fi le that is stored sequent ia l l y . - sequent ia l mode i s often assoc ia ted with a master f i l e held in a certa in sequence and updated by a transact i on f i l e sorted in to the same sequence. 3. Indexed_sequent ia l / selec t i ve_sequent ia l mode - Indexed_sequent ia l i s a mode of storage where by records are held sequent ia l l y and accessed selec t i ve l y . - Groups of unrequi red records are skipped past . - Indexed sequent ia l fi les may also be accessed haphazandly . 4. Random modes: - Each record i s stored in a locat i on determind from the second’s key by means of an add generat i on algor i t hm. - The only err i c i en t way to f ind a record i s to use the algor i t hm - Random mode i s appl i cab le to master f i l e s • Ads of random modes  No index is requi red thus saving storage space  I t is a fast access method because l i t t l e or no searching is invo lved  Transact i on do not need stor i ng , thus saving t ime  New records are easi l y inser t l y in to the random f i l e provided they are not excess ive in number • Dis  The main problem with the random mode i s in achiev ing a uni fo rm spread of records over the storage are al l ocated to the f i l e
  13. 13 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 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
  14. 14  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 - 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.
  15. 15 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 - 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
  16. 16 - 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 - 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:
  17. 17 - 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: 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
  18. 18                                                                            (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 = 1.28 + 1.25 + 1.23 + 1.22 + 1.20 = 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
  19. 19 27 61 1 1 1 (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) = (11110111)   10 2 = 0. 1111011 x 2 normalized exponential form Question 3. Integer and Floating – point arithmetic? 1. Floating – point Addition a. (0.1011 x 2 ) + (0.1001 x 2 ) 5 5 = (0.1011 + 0. 1001) x 25 = 1.0100 x 25 = 0.10100 x 26 b. (0.1001 x 2 ) + (0.1110 x 2 ) 3 5 = (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) (0.1000001 x 26 ) 2. Floating – point subtraction a. (0.1110 x 2 ) – (0.1100 x 2 ) 7 7 = 0.0010 x 27 = 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
  20. 20 b. (0.11110 x 2 ) x ((0.01011) x 24 ) 3 = (0.11110 x 0. 01011) x 27 = 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 2 ) : (1 x 2 ) 6 3 = 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 2 2 = (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. They also include instruct ions to write the contents of me mory locations holding the result of processing to a specif ied output device. 3. Decision or control instruct ions. Most data processing applicat ion wil l contain situat ions where alternative calculat ions or procedures wil l have to be executed based on the result of condit ion tests carried out. 4. Data handling instruct ions They include the copying of the content of one me mory location to another or sett ing a me mory locations to an ini t ia l value. Also include the management or insert ion of characters into data items Examples of such instruct ions include branch instruct ions, jump instruct ion & stop instruct ion. Question 2. Types of addressing? 1. Direct addressing The operands of each machine instruct ions is used to retr ieve the data 2. Indirect addressing The operands is used to specify the me mory address which contains the address of the data to be processed Op – code OP – CODE OPERAND OP – CODE OPERAND
Đồng bộ tài khoản