大學(xué)計(jì)算機(jī)基礎(chǔ)-從數(shù)據(jù)操作到計(jì)算理論課后習(xí)題答案

CHAPTER1

4

CHAPTER1INTRODUCTION

Introduction

ReviewQuestions

Weusethedefinition“issuesrelatedtothecomputer”.

ThevonNeumannModelisthebasisfortoday'scomputers.

Thisdefinitionistoogeneral.Itdoesnotspecifywhattypeofdataisprocessed,howitisprocessed,orifotherkindsofprocessingarepossibleusingthesamedevice.

Itrequiresinputdataandaprogramtoprocesstheinputdata.

Thesubsystemsarethememory,arithmetic/logicunit,controlunit,andinput/out-put.

Memoryisthestorageareausedforprogramsanddata.

TheALUiswherecalculationsandlogicaloperationstakeplace.

Thecontrolunitcontrolstheoperationsofthememory,ALU,andtheinput/outputsubsystem.

Theinputsubsystemacceptsinputdataandtheprogramfromoutsidethecom-puter;theoutputsubsystemsendstheresultoftheprocessingtotheoutside.

Thememorycontentsofearlycomputersheldonlythedatatobeprocessed.ThememorycontentofcomputersbasedonthevonNeumannModelholdsthedatatobeprocessedandtheprogramtoprocessit.

BeforethevonNeumannModel,programmingwastheactofmanipulatingasetofswitchesorchangingthewiringsystemofthecomputeritself.WiththevonNeu-mannModel,programmingbecamethetaskofwritingafinitesetofinstructionstobeexecutedonthedatainsequenceandstoringtheseinstructionsinaformthatcouldbereadintothememoryofthecomputer.

Multiple-ChoiceQuestions

b

c

a

3

b

d

c

d

c

a

d

a

c

d

d

c

Exercises

Answermayvary

Answermayvary

Answermayvary

Theharddiskoftodaymaybeusedaseitheraninputdeviceoranoutputdevice.Whendataisreadfromthedisk,itisconsideredaninputdevice.Whendataiswrittentothedisk,itisconsideredanoutputdevice.

Thereare30,2405-instructionprograms.Thereare604,8007-instructionpro-grams.

Answermayvary

Answermayvary

Awordprocessorprogramseesafileasaseriesofcharacters.Ituseswhitespaceandpunctuationtodividethisseriesintowordsandsentencesanditinterpretsspe-cialcharactersorcharactersequencesasformattinginstructions.

Answermayvary

Answermayvary

Answermayvary

Answermayvary

Answermayvary

CHAPTER2

PAGE

4

CHAPTER2DATAREPRESENTATION

DataRepresentation

ReviewQuestions

Text,numbers,images,audio,andvideo.

Alldatatypesfromoutsidethecomputeraretransformedintoauniformrepresen-tationwhenstoredinthecomputerandthentransformedbackwhenleavingthecomputer

Abitpatternisastringofbits(0sor1's)inaparticularorder.

ASCIIuses7bitstorepresentacharacterandcanrepresentupto128differentsymbols.ExtendedASCIIuses8bitstorepresentacharacterandcanrepresentupto256differentsymbols.Thefirst128symbols(00000000to01111111)arethesameastheASCIIcodewhiletheremainingsymbolstendtobeanon-standard,manufacturer-specificcode.

EBCDICstandsfor'ExtendedBinary-CodedDecimalInterchangeCode'.IBMdevelopedthiscodewhichisusedonlyinitsmainframes.Ituses8bitstorepresentacharacterandcanrepresentupto256differentsymbols.

Abitpatternoflengthncanrepresent2ndifferentcharacters.

Thebitpatternrepresentsthecolorforeachpixel(pictureelement)inamatrixofpixelsthatrepresentstheimage.

Theadvantageofthevectorgraphicmethodisthattheimageisclearerwhenitisrescaled.

Sampling-Themagnitudeofthesignalismeasuredatregularintervals.Quantization-Thesampleisassignedavaluefromaset.

Coding-Thequantizedvaluesarechangedtobinarypatterns.Storing-Thebinarypatternsarestored.

Videodataismadeupofaseriesofimagestobeshowninsequence.Itisusuallystoredinacompressedformat.

Multiple-ChoiceQuestions

d

3

d

c

c

d

b

d

a

c

d

d

b

d

c

a

b

a

d

Exercises

32differentpatternscanberepresentedby5bits.

If0isallowed:100.If0isnotallowed:81.

If0isallowed,1,757,600.;if0isnotallowed,1,423,656.

3bitsmaybeusedtorepresenteachcycle.

3bitsareneededtorepresentthegradewith1patternunused.

Using10bitswouldprovide1024possibleID's.124patternswouldbeunused.Ifthecompanyweretohireanother300people,itwouldhavetoincreasethenumberofbitsunusedinitsIDnumbersto11toprovide2048possibleIDpatterns.

Afour-bitpatterncanrepresentupto16uniquepatterns.6patternswouldbewasted.

20,000bitswouldbeneededtorepresentthepicture.

8bitswouldbeneededtorepresent256differentsoundlevels,so64,000bitswouldberequiredtorepresent1secondofaudio.

x8F0

x20D

x11

xFF

000100100000

0010101000110100

SECTION 5

00000000

11111111

04360

01015

021

0377

001010

010111

100101

010000

Fivehexadecimaldigitsareneededtoconverta19-bitpattern.

Sevenoctaldigitsareneededtoconverta19-bitpattern.

Twelvehexadecimaldigitsareneededtoconverta6-byte(48bit)pattern.

PAGE

6

CHAPTER2DATAREPRESENTATION

CHAPTER3

NumberRepresentation

ReviewQuestions

Repetitivedivisionby2isused,enteringtheremainder(0or1)totheleftofanypreviousremainders.

Multiplyeachbinarydigitbyitsweightandcombinetheresults.

2

10

Sign-and-magnitude,one'scomplement,andtwo'scomplement.

The“maximumunsignedinteger”isthelargestnumberthatcanbestoredinanintegerlocation.Foragivenbitallocation,n,thelargeststorableunsignedintegerwouldbe2n-1.

Bitallocationisthenumberofbitsusedtorepresentaninteger.

Thedecimal256wouldrequireatleast9bitsforstorage.

Unsignedintegersareusedforcounting,addressing,oranywherenegativenum-bersarenotused.

Thelargestpositivenumberstorableinabitallocationof8bitsusingsign-and-magnituderepresentationis127;therefore,overflowwouldoccur.

Therepresentationofpositivenumbersinsign-and-magnitude,one'scomplement,andtwo'scomplementareallidentical.

Negativenumbersinallthreerepresentationswillallstartwith'1'(theleft-mostbitwillbe'1').Therestofthenumberswillvarydependingontherepresentation.

Thereare2valuesforzeroinbothsign-and-magnitude(00000000and10000000)andone'scomplement(00000000and11111111)representations.Two'scomple-mentrepresentationhasonly1valueforzero(00000000).

ForabitallocationofNbits:Therangeofnumbersstorableinsign-and-magni-tudeisfrom-(2N-1-1)to+(2N-1-1);therangeofnumbersstorableinone'scomplementisfrom-(2N-1-1)to+(2N-1-1);therangeofnumbersstorableintwo'scomplementisfrom-(2N-1)to+(2N-1-1).

3

PAGE

4

CHAPTER3NUMBERREPRESENTATION

SECTION

PAGE

5

Ineachoftheserepresentations,a'1'intheleftmostbitpositionindicatesanega-tivenumber.

Excess_Xisusedprimarilytostoretheexponentialvalueofafraction.'X'repre-sentsthemagicnumberbeingused:normally(2N-1)or(2N-1-1)where"N"isthebitallocation.

Normalizationisnecessarytomakecalculationseasier.

Themantissaisthebitsequencetotherightofthedecimalpointafternormaliza-tion.

Thecomputermuststorethesignofthenumber,theexponentusedduringnormal-ization,andthemantissa.

Multiple-ChoiceQuestions

c

a

d

d

b

c

d

d

d

d

d

c

b

d

d

a

c

b

c

b

c

b

d

d

c

c

b

Exercises

a.

00010111

b.

01111001

c.

d.

00100010

101010110(overflow)

48.

a.

0000000000101001

b.

0000000110011011

c.

0000010011010010

49.

d.

0000000101010110

a.

00100000

b.

11100101

c.

00111000

50.

d.

a.

10000001(overflow)

0000000010001110

b.

1000000010110100

c.

0000001000110000

d.

0000100110011000

51.

a.

01111010

b.

10110000

52.

c.

d.

11111111

10000001(overflow)

a.

0000000010100010

b.

1111111110010001

c.

0000101000000000

53.

d.

a.

0010111101011011

11110100

b.

10011011

c.

d.

00111000

10001110(overflow)

54.

a.

0000000001100110

b.

1111111101001101

c.

d.

0000001000010110

1111001001101000(overflow)

55.

a.

107

b.

148

c.

6

56.

d.

80

a.

123

b.

-52

c.

99

57.

d.

a.

-80

99

b.

-123

c.

115

d.

-63

58.

a.

119

b.

-4

c.

116

59.

d.

-50

a.

11110111

b.

01111100

c.

11110111

60.

d.

a.

01001110

10001000

b.

00000011

c.

10001000

d.

00110001

61.

a.

10001001

b.

00000100

c.

10001001

62.

d.

00110010

a.

b.

Originalmethod:10001001Newmethod:10001000+00000001=10001001

Originalmethod:00000100Newmethod:00000011+00000001=00000100

c.

d.

Originalmethod:10001100Newmethod:10001011+00000001=10001100

Originalmethod:00110010Newmethod:00110001+00000001=00110010

63.

a.

01110111=>10001000=>01110111

b.

11111100=>00000011=>11111100

64.

c.

d.

01110100=>10001011=>01110100

11001110=>00110001=>11001110

a.

b.

c.

01110111=>10001001=>01110111

11111100=>00000100=>11111100

01110100=>10001100=>01110100

65.

d.

a.

11001110=>00110010=>11001110

9743

b.

9887

c.

9874

d.

9888

66.Wecanrepresentnumbersfrom-99(900)to+99(099).Wecandeterminethesignofthenumberbytheleftmostdigit.Iftheleftmostdigitisa'9',thenumberisneg-ative;iftheleftmostdigitisa'0',thenumberispositive.Wedohave2zeros:+0

(000)and-1(999).

67.

a.

Alreadynormalized

b.

c.

25x1.111111

20x1.01110011

d.

20x1.0110100000110011000

68.

a.

10111111110001000000000000000000

b.

01000001011111100000000000000000

c.

00111101101110011000000000000000

d.

10111101001101000000000000000000

69.

a.

1011111111111000100000000000000000000000000000000000000000000000

b.

0100000000101111110000000000000000000000000000000000000000000000

c.

0011111110110111001100000000000000000000000000000000000000000000

d.

1011111110100110100000000000000000000000000000000000000000000000

70.

a.

11.101or21x1.1101

b.

c.

1100.00011or23x1.10000011

100.001101or22x1.00001101

d.

1100.0000101or23x1.1000000101

71.

a.

1/8+1/16or3/16(.001+.0001or.0011)

b.

1/2+1/8+1/64or41/64(.1+.001+.000001or.101001)

c.

1/4+1/8+1/32or13/32(.01+.001+.00001or.01101)

d.

1/4+1/8or3/8(.01+.001or.011)

72.

a.

111.0011or22x1.110011

b.

c.

1100.101001or23x1.100101001

1011.01101or23x1.01101101

d.

0.011or2-2x1.1

73.

a.

01000000111001100000000000000000

b.

01000001010010100100000000000000

c.

11000001001101101000000000000000

d.

10111110110000000000000000000000

74.

a.

000000010010101000000000

+000100101010101011111111

000100111101010011111111

=>00000000000100111101010011111111

b.

0000000000000000000000010001

+ 100000100000000000000000

0000100000100000000000010001

=>00000000100000100000000000010001

c. 1001000100010001000100010001

+ 001000010001000100010001

1001001100100010001000100010

=>00001001001100100010001000100010

d.

1111000100010001000100010001

+ 011101110111011101110111

1111100010001000100010001000

=>00001111100010001000100010001000

CHAPTER4

OperationsOnBits

ReviewQuestions

Arithmeticoperationsinvolveinterpretingthebitsasnumbersanddoingarith-meticoperationsonthosenumbers.Logicaloperationsinvolveinterpretingthebitsaslogicalvalues(trueorfalse)anddoingoperationsonthosevalues.

Multiplicationcanbelookedatasrepetitiveaddition.YxZcanbeaccomplishedbyaddingZnumberofYstogether.

Theleftmostcarryisdiscarded.

Thebitallocationcanequal1.Thisdatatypecouldrepresentalogicalvalue.

Overflowhappenswhentheresultofanarithmeticoperationisoutsidetherangeofpossiblevaluesforthebitallocationbeingused.

Thedecimalpointofthenumberwiththesmallerexponentisshiftedtotheleftuntiltheexponentsareequal.

Aunaryoperationtakesasingleoperand.Abinaryoperationtakestwooperands.

Thelogicalbinaryoperationsare:AND,OR,andXOR.

Atruthtablelistsallofthepossibleinputcombinationswiththeresultingoutput.

NOTinvertsvalues:itchangetruetofalseandfalsetotrue.

TheresultofanANDoperatoristruewhenbothoftheoperandsaretrue.

TheresultofanORoperatoristruewheneitheroftheoperandsistrue.

TheresultofanXORoperatoristruewhenbothoftheoperandsaredifferent.

TheinherentruleoftheANDoperatoristhatifoneoftheoperandsisfalse,theresultisfalse.

TheinherentruleoftheORoperatoristhatifoneoftheoperandsistrue,theresultistrue.

TheinherentruleoftheXORoperatoristhatifoneoftheoperandsistrue,theresultwillbetheinverseoftheotheroperand.

TheORoperatorcanbeusedtosetbits.Setthedesiredpositionsinthemaskto1.

TheANDoperatorcanbeusedtoclearbits.Setthedesiredpositionsinthemaskto0.

3

PAGE

4

CHAPTER4OPERATIONSONBITS

SECTION

PAGE

5

TheXORoperatorcanbeusedtoinvertbits.Setthedesiredpositionsinthemaskto1.

Multiple-ChoiceQuestions

c

d

c

c

c

b

d

b

a

c

d

c

d

c

b

a

a

c

a

b

Exercises

00010011+00010111=00101010=42

00010011-00010111=000010011+(-00010111)=00010011+11101001=

11111100=-00000100=-4

(-00010011)+00010111=11101101+00010111=00000100=4

(-00010011)-00010111=(-00010011)+(-00010111)=11101101+11101001

=11010110=-00101010=-42

0000000010100001+0000001111111111=0000010010100000=1184

0000000010100001-0000001111111111=0000000010100001+

(-0000001111111111)=0000000010100001+1111110000000001=11111100

10100010=-0000001101011110=-862

(-0000000010100001)+0000001111111111=1111111101011111+

0000001111111111=0000001101011110=862

(-0000000010100001)-0000001111111111=(-0000000010100001)+

(-0000001111111111)=1111111101011111+1111110000000001=11111011

01100000=-0000010010100000=-1184

Nooverflow

Nooverflow

Nooverflow

Nooverflow

Overflow

Nooverflow

Nooverflow

Overflow

x012A+x0E27=0000000100101010+0000111000100111=00001111

01010001=x0F51

x712A+x9E00=0111000100101010+1001111100000000=00010000

00101010=x102A

x8011+x0001=1000000000010001+0000000000000001=10000000

00010010=x8012

xE12A+x9E27=1110000100101010+1001111000100111=01111111

01010001=x7F51

a.

34.075+23.12=

100010.00010011+10111.0001111=

25x1.0001000010011+24x1.01110001111=

25x(1.0001000010011+0.1011100011110)=

25x1.1100100110001=111001.00110001=57.19140625

b.

-12.00067+451.00=451.00-12.00067=

111000011.0-1100.00000000001010111=

28x1.11000011-23x1.10000000000001010111=

28x1.11000011-28x0.0000110000000000001010111=

28x(1.1100001100000000000000000+28xl.1111001111111111110101001)=

281.1011011011111111110101001=110110110.11111111110101001=

438.99933

c.

33.677-0.00056=

100001.1010110101-.00000000001001001=

25x1.000011010110101-2-12x1.001001=

25x1.000011010110101-25x0.00000000000000001001001=

25x(1.000011010110101+ 1.11111111111111110110111)=

25x1.00001101011010010110111=100001.101011010010110111=

33.676479339599609375

d.

-344.23-123.8902=-(344.23+123.8902)=

-(101011000.0011101+1111011.11100011111001)=

-(28x1.010110000011101+26x1.11101111100011111001)=

-(28x(1.010110000011101+0.0111101111100011111001))=

-(28x(1.1101010000011101111001))=-111010100.00011101111001=

-468.11676025390625

a.

23.125+12.45=

10111.001+1100.0111001=

24x1.0111001+23x1.1000111001=

24x(1.0111001+0.11000111001)=

24x10.00111001001=100011.1001001=34.4703125

b.

0.234-7.192=-(7.192-0.234)=

-(111.0011000100-.0011101111)=

-(22x1.110011000100-2-3x1.1101111)=

-(22x(1.110011000100-0.000011101111))=

-(22x(1.110011000100+1.111100010001))=

-(22x(1.101111010101))=-110.1111010101=-6.9580078125

c.

-0.345+45.123=45.123-0.345=

101101.0001111101-.0101100001=

25x1.011010001111101-2-2x1.01100001=

25x(1.011010001111101-0.000000101100001)=

25x(1.011010001111101+1.111111010011111)=

25x1.011001100011100=101100.1100011100=44.7773437500

d.

-.234-5.345=-(.234+5.345)=

-(.0011101111+101.0101100001)=

-(2-3x1.1101111+22x1.010101100001)=

-(22x(0.000011101111+1.010101100001))=

-(22x1.011001010000)=-101.100101=-5.578125

Overflowcanoccur.

b.

Overflowcanneveroccurbecausetheabsolutevalueoftheresultmustbelessthanorequaltothelargeroftheabsolutevaluesoftheoperands.Toputitanotherway,theresultwillbeclosertozerothantheoperandthatisfarthest

fromzero.

c.

Overflowcanoccur.

d.

Overflowcanneveroccurbecausebysubtractinganegativenumber,youareessentiallyaddingapositivenumberandweagainhavethecasewhereweareaddingapositiveandnegativenumber.

48.

a.

NOTx99=NOT10011001=01100110

b.

NOTxFF=NOT11111111=00000000

c.

NOTx00=NOT00000000=11111111

d.

NOTx01=NOT00000001=11111110

49.

a.

x99ANDx99=10011001AND10011001=10011001

b.

x99ANDx00=10011001AND00000000=00000000

c.

x99ANDxFF=10011001AND11111111=10011001

d.

xFFANDxFF=11111111AND11111111=11111111

50.

a.

x99ORx99=10011001OR10011001=10011001

b.

x99ORx00=10011001OR00000000=10011001

c.

x99ORxFF=10011001OR11111111=11111111

d.

xFFORxFF=11111111OR11111111=11111111

51.

a.

x99XORx99=10011001XOR10011001=00000000

b.

x99XORx00=10011001XOR00000000=10011001

c.

x99XORxFF=10011001XOR11111111=01100110

d.

xFFXORxFF=11111111XOR11111111=00000000

52.

a.

NOT(x99ORx99)=NOT(10011001OR10011001)=NOT10011001=01100110

b.

x99OR(NOTx00)=10011001OR11111111=11111111

c.

(x99ANDx33)OR(x00ANDxFF)=(10011001AND00110011)OR(00000000AND11111111)=00010001OR00000000=00010001

d.

(x99ORx33)AND(x00ORxFF)=(10011001OR00110011)AND(00000000OR11111111)=10111011AND11111111=10111011

Mask:00001111

Operation:00001111ANDXXXXXXXX=0000XXXX

Mask:00001111

Operation:00001111ORXXXXXXXX=XXXX1111

Mask:11000111

Operation:11000111XORXXXXXXXX=YYXXXYYY

Mask1:00011111

Mask2:00000011

Operation:(00011111ANDXXXXXXXX)OR00000011=000XXX11

"abcdefgh"rightshiftedby2becomes'00abcdef',whichdividestheunsignednumberby4.

"abcdefgh"leftshiftedby3becomes'defgh000',whichmultipliesthenumberby

8.Overflowwilloccurifanyofthe3leftmostbitsarenotzero.

Mask:00000001

Number:abcdefghOperations:

Step1:Shifttotherightby3,whichgivesus'000abcde'.

Step2:000abcdeAND00000001=0000000e(iftheresultis1,thebitwasset)Step3:Shiftoncemoretotheright,whichgivesus'0000abcd'.

Step4:0000abcdAND00000001=0000000d(iftheresultis1,thebitwasset)

CHAPTER5

ComputerOrganization

ReviewQuestions

ThethreesubsystemsaretheCPU,themainmemory,andtheinput/outputsub-system.

TheCPUhasanALU,acontrolunit,andvariouskindsofregisters.

TheALUperformsarithmeticandlogicaloperations.

Thedataregistersholddatatemporarily;instructionregistersholdtheinstructionsthathavebeenfetchedfrommainmemory;andtheprogramcounterisaregisterthatholdstheaddressoftheinstructioncurrentlybeingexecuted.

Thecontrolunitisresponsibleforcontrollingtheoperationofalloftheothersub-systems.

Abyteisalwaysagroupof8bits,butthesizeofawordismachine-dependant.Awordcanbe8,16,32,or64bits.

Thefunctionofmainmemoryistostorethedataandprogramscurrentlybeingusedbytheuser.

Theapproximationofamegabyteisbasedonpowersof10(106or1,000,000bytes).Theactualsizeofamegabyteisbasedonpowersof2(220or1,048,576bytes).

Memoryaddressesarerepresentedbyunsignedintegers.

RAMisrandomaccessmemoryandcanbereadfromandwrittentobytheuser.ROMisreadonlymemory.ThecontentsofROMarewrittenbythemanufacturerandcannotbeoverwrittenbytheuser.

SRAMisstaticRAMthatusesflip-flopgatestoholddata.SRAMisfastbutexpensive.DRAMisdynamicRAMthatusescapacitorstoholdthedata.DRAMneedsperiodicrefreshingtokeepfromloosingdata.Itisslow,butcheap.

PROMisprogrammablereadonlymemoryandcanbeprogrammedbytheuserusingspecialequipment.EPROMiserasablePROMandcanbeerasedbytheuserusingaspecialdevicethatusesUVlight.EEPROMiselectronicallyerasablePROMthatcanbereprogrammedwhileitisstillinstalledinthecomputer.

3

PAGE

4

CHAPTER5COMPUTERORGANIZATION

SECTION

PAGE

5

ThepurposeofcachememoryistoprovidetheCPUwithaholdingareaforapor-tionofmainmemory.

HighspeedmemoryisusedfortheregistersinsidetheCPU.Mediumspeedmem-oryisusedforthecache.Theslowestmemoryisusedformainmemorywhereallofthedataandprogramsarestored.

Thekeyboard,monitor,andprinterareexamplesofnonstorageI/Odevices.

Magneticandopticalarethetwomainclassificationsofstoragedevices.

Amagneticdiskconsistsofoneormorediskswithamagneticcoatingandoneread/writeheadforeachdisksurface.

Thesurfaceofamagneticdiskisdividedintocircularringscalledtracks.Eachtrackisdividedintosectionscalledsectors.

Therotationalspeedofthedisk,theseektime,andthetransfertimearethemainfactorsthatdeterminetheperformanceofamagneticdisk.

Magnetictapeconsistsofaplastictapewithamagneticcoating.Thetapedriveconsistsoftworeelsandaread/writehead.

Thewidthofamagnetictapeisdividedinto9tracks.Thelengthofthetapemaybedividedintoblocks.

Thedataonatapeisaccessedsequentially;thatis,togettoaparticularblock,wemustreadallofthepreviousblocks.

FivetypesofopticalstoragedevicesareCD,CD-ROM,CD-R,CD-RW,andDVD.

ACD-ROMiswrittentoonlybythemanufacturer.ACD-RandaCD-RWarewrittentobytheuser.

TheadvantageofaCD-RWisthatitcanbeoverwrittenwithnewdata.

ThepitsandlandsonaCD-ROMareactualpitsandlandsinthepolycarbonateresinusedforthebodyofthedisk.ThepitsandlandsonaCD-Raresimulatedusingadyewithinthebodyofthedisk.ThepitsandlandsofaCD-RWaresimu-latedbychangingthestateofanalloyofsilver,indium,antimony,andtelluriumtoeithercrystalline(transparent)oramorphous(nontransparent).

Allthreedisksusealow-powerlaserbeamtoreadthedisks.Theyallusethedif-ferenceinthereflectionbetweenalandandpit(realorsimulated)todetectdataonadisk.

DataiserasedonaCD-RWwiththeuseofamedium-powerlaserbeam.Itchangesanyspotsintheamorphous(nontransparent)statetothecrystalline(trans-parent)state.

ADVDusesmuchsmallerpitsandlandsthatallowthedisktoholdmuchmoredata(4.7GBto17GB)comparedtoaCD-ROM(650MB).

Thedatabusisusedtotransferdatatoandfrommemory.Theaddressbusallowsaccesstoparticularwordsinmemory.Thecontrolbusisusedtocontroltheopera-tionsoftheCPUandmemory.

TheI/Odevicecontrollersactasanintermediatebetweentheelectronicdevices(CPUandmemory)andtheslowerelectromechanicaldevices(i.e.,diskdrives)toregulatethedevice'saccesstobuses.

ASCSI(smallcomputersysteminterface)controllerisaparallelinterfacethatprovidesadaisychainconnectionbetweendevicesandthebuses.

TheFireWireinterfaceisahighspeedserialinterfacethattransfersdatainpackets.Itcanuseadaisychainortreeconfiguration.

TheUSB(universalserialbus)controllerisaserialcontrollerusedtocommuni-catewithslowerdevices.Italsosuppliespowertothedevices.

IsolatedI/Ousesadifferentsetofinstructionstoread/writetomemorythantoread/writetoI/Odevices.Memory-mappedI/Ousesthesameinstructionsetbutdifferentaddressesfortheseoperations.

Thebasicstepsinamachinecyclearefetch,decode,andexecute.

IntheprogrammedI/Omethod,theCPUwaitsfortheI/Odevice.AlotofCPUtimeiswastedbycheckingforthestatusofanI/Ooperation.Intheinterrupt-drivenI/Omethod,theI/OdeviceinformstheCPUofitsreadystatusviaaninter-upt.Indirectmemoryaccess(DMA),theCPUsendsitsI/OrequeststotheDMAcontrollerwhichmanagestheentiretransaction.

CISC(complexinstructionsetcomputer)hasalargesetofinstructionstoexecutecommandsatthemachinelevel.ThismakesthecircuitryoftheCPUandthecon-trolunitverycomplicated.RISC(reducedinstructionsetcomputer)usesasmallsetofinstructions.Complexoperationsareaccomplishedusingalargernumberofsimplecommands.

Multiple-ChoiceQuestions

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Exercises

64MB/4bytesperword=16million=224;so24bitsareneededtoaccessaword.

24x80=1920bytes

Youneed4bitstodeterminetheinstruction(24=16).Youneed2bitstodeterminearegister(2=4).Youneed10bitstodetermineawordinmemory(210=1024).Youthereforeneed4+2+10or16bitsforaninstruction.

Sincethesmallestwordsizethatcanholdaninstructionis16bits,wemusthave16-bitregisters.

Theinstructionregistermustbeatleast16bits.

Theprogramcountermustbelargeenoughtoholdthenumberofinstructionsinthelargestprogramused.

Thedatabusmustbewideenoughtohold1word.

Theaddressbusmustbe10bitswide.

Theminimumsizeofthecontrolbusis4bitswide.

SincetheaddressspacecanbesharedbetweencontrollerregistersandmemoryonacomputerusingisolatedI/O,64differentcontrollerscanbeaccessed(64control-lersx16registers/controller=1024registerlocations).

SincetheaddressspacemustbesharedbetweenI/Oaddressingandmemoryaddressing,therecanbe64-registercontrollers(6x4=24,24+1000words=

1024or210).

CHAPTER6

ComputerNetworks

ReviewQuestions

Amodelisaspecificationsetbyastandardsorganizationasaguidelinefordesign-ingnetworks,suchastheOSI(OpenSystemsInterconnection)model.Aprotocolisasetofrulesthatcontrolstheinteractionofdifferentdevicesinanetworkoraninternetwork,suchasTCP/IP.

ThesevenlayersoftheOSImodelare:physical,datalink,network,transport,ses-sion,presentation,andapplication.

ThelayersoftheTCP/IPprotocolsuiteare:network,transport,andapplication.

Thephysicallayerisresponsiblefortransmittingabitstreamoveraphysicalmedium.Itincludesthemechanicalandphysicalspecificationsforthephysicaldevices.Thedatalinklayerorganizesthebitstreamintologicalunitscalledframes.Itisresponsiblefornode-to-nodedeliveryoftheseindividualframes.Thenetworklayerisresponsibleforthepacketfromtheoriginalsource