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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.

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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  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 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 - 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  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 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 - 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 -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 -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 - 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 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 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 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  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 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 - 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 - 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 (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 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 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

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