计算机系统组成与体系结构_答案.pdf

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1、SOLUTIONS MANUALComputer Systems Organizationand ArchitectureJohn D.CarpinelliCopyright 2001,Addison Wesley Longman-All Rights ReservedComputer Systems Organization and Architecture-Solutions ManualTable of ContentsChapter 11Chapter 28Chapter 318Chapter 421Chapter 533Chapter 645Chapter 759Chapter 88

2、0Chapter 992Chapter 10100Chapter 11106Chapter 12.116Copyright 2001 Addison Wesley-All Rights ReservedPage iiComputer Systems Organization and Architecture-Solutions ManualChapter 12.a)wXyzwxxzVwx+xz+yb)wXyzw+x+y+z000000110000000010011000110010000000101001100000011101000011010010101011101011011000000

3、1101011101010111110000011100011001001110011101000001010110110000101111100101111001110111111101111101001111011111110111111c)wXyzw xyz wxyzw yyz w 3 3z+w xyz+卬 *$z+w yyz0000000000001000000010001010011100010100000000101000000110000110111010011000000001001000001010000001011000001100000001101000001110000

4、00111100000Copyright 2001 Addison Wesley-All Rights ReservedPage 1Computer Systems Organization and Architecture-Solutions Manual工abab(abyab a +6.0001111010110110010111110000aba+b(a+ab ab0001111011010010100101110000a)b)c)4wxyz 00011110001|001011id1VlJ.000oo)000加,D00010k.i)011 00J1R10m0(E1a)5卬二夕+卬 +w

5、xy+wx 2 or卬夕 2+xyz+wyz，V+w 旷+xz+xyorx2+w 旷+xz+y26.a)vv zz+xywxyz 00011110()000n011 JX1,i11 0,1X010 0XX0b)wxyz 0001111000显1i,x j010XX011 00X010修XXX17.a)wxyW 00 01-0 011 1001111000001n0(L u71)0000wx+xz+w y zy z+孙，+wzwxyz,00 010001111011 100101101001011010Already tninitnal8.9.wx+xyCopyright 2001 Addiso

6、n Wesley-All Rights ReservedPage 2Computer Systems Organization and Architecture-Solutions Manual10.a)11.Change the AND gates to NAND gates.The rest of the circuit is unchanged.12.Remove the tri-state buffers and do one of the following:a)Change each 2-input AND gate to a 3-input AND gate.Each gates

7、1 inputs should be its two originalinputs and E,orb)Have each AND gates output serve as an input to another 2-input AND gate,one gate fbr each originalAND gate.The second input to the new 2-input AND gates is E.13.14.Copyright 2001 Addison Wesley-All Rights ReservedPage 3Computer Systems Organizatio

8、n and Architecture-Solutions Manual15.Set up Karnaugh maps fbr each output,then develop minimal logic expressions and design the appropriatelogic circuits.(X Y)=XJYI,+X。匕 +XM1(X=Y)=X,Yi,Yo+X/，X()YIrYo+Yo 1+X/X ohYo=(XI 匕)。%)(X +XllX o,e=X2Xl,+X0X3X X,X0 00 01 11 1000 0 I 0 I 0|Uo i-6 6 6 o-“XXX 国io|

9、o I o I x|X3XK,X 00 01 11 10C =X2%X。，f=XlX0+X3fX2X0+X2fXICopyright 2001 Addison Wesley-All Rights ReservedPage 5Computer Systems Organization and Architecture-Solutions Manual20.The four inputs can be in one of 24(=4!)possible orders.Since each sorter has two possible states(MAX=X MIN=X,or MAX =Y MI

10、N=X),sorters can have up to T states.Four sorters can have only 24=16states,not enough to sort all 24 possible input orders.Five sorters have 25=32 states,which could besufficient.(This argument establishes a lower bound;it does not guarantee the existence of a 5-sorternetwork that can sort four inp

11、uts.Since the sorting network of Figure 1.24(b)matches this bound,it is aminimal network.)21.a)22.A flip-flop is clocked if the increment signal and clock are asserted,and all flip-flops to its right are 1.23.Each clock is driven by Q of the flip-flop to its right instead of Q.The clock of the right

12、most flip-flop isunchanged.All other signals are unchanged.XzXX/X。00 01 11 10Copyright 2001 Addison Wesley-All Rights ReservedPage 6Computer Systems Organization and Architecture-Solutions ManualCopyright 2001 Addison Wesley-All Rights ReservedPage 7Computer Systems Organization and Architecture-Sol

13、utions ManualChapter 22.Present State SgKTTTR Next Stateuu _UTT2TT2T3g2TuoTV五yTggigKTU 1 1 u3.Add the following states to the state table.Since all additions are self-loops,it is not necessary to changethe state diagram.Present StatechIoNext StateRGASNOCAR001s NOCAR100SNOCAR010SNOCAR100s NOCAR011sNO

14、CAR100SPMD101SP.M D010SPA II110SPAII,010SPA ID111SPA ID010s CHEAT001s CH EAT101S CHEAT010S CH EAT101S CHEAT011S CH EAT101Copyright 2001 Addison Wesley-All Rights ReservedPage 8Computer Systems Organization and Architecture-Solutions ManualCopyright 2001 Addison Wesley-All Rights ReservedPage 9Comput

15、er Systems Organization and Architecture-Solutions Manual5.6.Address Data(Mealy)Data(Moore)000000000000000100100010001001000100001101100110010010001000010110101010011011011100011111101110100000000000100100100010101001000100101101100110110010001001110110101011111011011100n il111011107.Present StateIN

16、ext StateM000000001010010000011100100110101100110001111011N=P/Pol+PF。NO=P/PO7+PFOT+P R JM=P RCopyright 2001 Addison Wesley-All Rights ReservedPage 10Computer Systems Organization and Architecture-Solutions Manual8.9.Address Data(Mealy)Data(Moore)000000000001010010010000000Oil100100100111110101100100

17、110000001111010Oil10.State value assignments(P3 P.S()=0000 S5=0001 S/o=0010 S15=0011 S20=010025=0101 S30=00 SPAID=0 SN0CAR=1000 SCHEAT=1001M=c，M=P/Chlo+尸 3(匕 +尸/)C/o+P/(P2 +PFo)C“Io+P2CII,IO,N尸 P3P2+匕+Po)ChIo+匕(尸2+P/K W +PNPiP。+P/PJ+P2PlPo)CIl,Io+P.PoChIoN=PAP2+匕 +PQ C。+PNPo+P2Pi)CIlI0,+P式 Po+P2PDCW

18、0+P3 尸。C/4/+P3Poe+2 3(尸2+匕 +尸0)。R=SPAIDG=SPAID4 =ScHEAT1 1.State value assignments(Pj-Po)：S()=0000 S5=0001 Sl0=0010=0011 520=010025=0101 S30=0110 SPAID=0111 SNOCAR=1000 SCHEAT=1001N3=CM =P H+P/(P2+尸3(尸2 +P/P o)C/0 +P2Ch，Io,N】=P3P2+匕 +P0)ChI0+PNP2+尸+PAP1P0+PFo+P2PiPo)CIl,Io+PiPoChVN0=匕(P2+匕 +PQCWo+PN

19、Po+P2PI)C1II(),+尸3 W+P2PI)CII 7。+尸3尸。0+P3Poe+尸3(尸2+匕+尸。)。R=GG=P3P2+PChlo+P；P2Poeh+P；P2PAh+Io)+P3T2PF0CA =匕(尸2+匕+尸。)仁Copyright 2001 Addison Wesley-All Rights ReservedPage 11Computer Systems Organization and Architecture-Solutions Manual12.AddressDataOOOOX X X 1001101 1001101 1001101 1001101 0000100 0

20、001100 0010100 01011000001X X X1001101 1001101 1001101 1001101 0001100 0010100 0011100 01101000010X X X1001101 1001101 1001101 1001101 0010100 0011100 0100100 01110100011X X X1001101 1001101 1001101 1001101 0011100 0100100 0101100 01110100100X X X1001101 1001101 1001101 1001101 0100100 0101100 01101

21、00 01110100101X X X1001101 1001101 1001101 1001101 0101100 0110100011101001110100110X X X1001101 1001101 1001101 1001101 01101000111010011101001110100111X X X1000100 1000100 1000100 1000100 0111010 0111010 0111010 01110101000X X X1000100 1000100 100010()1000100 0000100 000()10()0000100 00001001001X

22、X X1001101 1001101 1001101 1001101 000010()000010()0000100 00001001010X X X 1000100 1000100 1000100 1000100 1000100 1000100 1000100 100010010HX X X1000100 1000100 1000100 1000100 1000100 1000100 100010()1000100uooxxx1000100 1000100 1000100 1000100 1000100 1000100 1000100 10001001101X X X1000100 1000

23、100 1000100 1000100 1000100 1000100 1000100 10001001110X X X 1000100 1000100 1000100 1000100 1000100 1000100 1000100 10001001111X X X1000100 1000100 1000100 1000100 1000100 1000100 1000100 1000100N2=P2Po，+P2U,+PlPoUN1=PIPO，+PIU,+P2,P1,POUNo=PoU+PoUC=P2,PI,POV,+P2PI,POUv2=P2P1P0U+尸2匕了。+PzPfPMVy=PPiPo

24、U+PzHPo+PzPjPoU%=(尸2+PMoU+(P2+Pj)PoU15.All possible next state values are already used.Copyright 2001 Addison Wesley-All Rights ReservedPage 12Computer Systems Organization and Architecture-Solutions Manual16.State value assignments(P3-Po)：So=0000 55=0001 5川=0010 5/5=0011$20=0100S25=0101$30=0110 Spi

25、D=0111 SOCAR=1000 SCHEAT=1001 54=1010ion sc=iioo sD=noi sE=mo sF=miAdd to state diagram:Add to state table:Present StateChIoNext StateRGA1010XXX10000001011XXX10000001100XXX10000001101XXX10000001110XXX10000001111XXX1000000N3=C,P3(P2+PI)M =P H+匕(尸2+匕)C/J +PAP2 +匕 P o)C/o +P2Ch，IoNi=P/(P2+P/+Po)CIlIo+尸

26、 3(P 2+P j)W +PPiPo+P】Po+P2PjPo)CI“o+PiPoChIo1No=PNP?+匕+W h lo +P3P0+P2PoeW +P 3 W+P2PDCW0+P；PoCh7o，+P3Poe+尸3(尸2 +匕+尸。)。R=SPAIDG=SpAIDA =S C H E A T17.N 3 =P 2 P/P 0 U +尸3(尸2,+P/+P o+N2=P3P2(Po+/2匕+PfPFoUN=P/(尸2+尸/)+P z P/P o U +PjUNO=(P3+P2)PIV +P3P2U+P0U1C=P2Pl,Pof匕=尸3尸/尸。+尸3尸2尸0匕=-3巧尸。+尸2尸/尸。V0=P

27、3lP2rP0+P2PP0P3PlfP0Copyright 2001 Addison Wesley-All Rights ReservedPage 13Computer Systems Organization and Architecture-Solutions Manual18.raCopyright 2001 Addison Wesley-All Rights ReservedPage 14Computer Systems Organization and Architecture-Solutions Manual20.States are of the form ABCYZ,where

28、 ABC=0 if a player may signal,or 1 if the player may not signal.YZrepresents the player answering the question(01=player 1,10=player 2,11=player 3,00=no player).Although not shown in the diagram,there is an arc from every state back to state 00000 with condition R.AddressDataX X X X X XXIX 00000 000

29、00000 ooox00000 00000000()1 ox00001 00000000 100X00010 00000000 110X00011 00000001 xxoo00001 10000001 X X 0110000 10000010 xxoo00010 01000010 X X 0101000 01000011 xxoo00011 00100011 X X 0100100 00100100 ooox00100 00000100 010X00101 00000100 100X00110 00000100 110X00100 00000101 xxoo00101 10000101 X

30、X 0110100 100AddressData00110 xxoo0011001000110X X 010110001001000 xoox01000 00001000 010X01001 00001000 110X01011 00001001 xxoo()1001 10001001 X X 0111000 10001011 xxoo01011 00101011 X X 0101100 00101100 ooox()1100 00()01100 010X01101 00001100 1X 0X01100 00001101 xxoo01101 10001101 X X 0111100 1001

31、0000 oxox10000 00010000 100X10010 00010000 110X10011 000AddressData10010 xxoo1001001010010 X X 0111000 01010011 xxoo10011 00110011 X X 0110100 00110100 oxox10100 00010100 100X10110 00010100 110X10100 00010110 xxoo1011001010110X X 011110001011000 oxox11000 00011000 100X11000 00011000 110X11011 000110

32、11 xxoo11011 00111011 X X 0111100 00111100 xxox11100 000AH others()000()000Copyright 2001 Addison Wesley-All Rights ReservedPage 15Computer Systems Organization and Architecture-Solutions Manual21.States are of the form ABCYZ,where ABC=0 if a player may signal,or 1 if the player may not signal.YZrep

33、resents the player answering the question(01=player 1,10=player 2,11=player 3,00=no player).Although not shown in the diagram,there is an arc from every state back to state 00000 with condition R.AddressDataAddressDataX X X X X X X IX 00000 00000110 xxoo0011001000000 ooox00000 00000110X X 0101100 00

34、000000 01 ox00001 1000100()xoox01000 00000000 100X00010 01001000 010X01001 10000000 110X00011 00101000 110X01011 00100001 xxoo00001 10001001 xxoo01001 10000001 X X 0110000 00001001 X X 0111000 00000010 xxoo00010 01001011 xxoo01011 00100010 X X 0101000 00001011 X X 0101100 00000011 xxoo00011 00101100

35、 ooox01100 00000011 X X 0100100 00001100 010X01101 1000010()ooox00100 000()1100 1X 0X01100 00000100 010X00101 10001101 xxoo01101 10000100 100X0011001001101 X X 0111100 00000100 110X00100 00()10000 oxox10000 00000101 xxoo00101 10010000 100X1001001000101 X X 0110100 00010000 110X10011 001AddressData10

36、010 xxoo10010 01010010 X X 0111000 00010011 xxoo10011 00110011 X X 0110100 00010100 oxox10100 00010100 100X1011001010100 110X10100 00010110 xxoo1011001010110X X 0111100 00011000 oxox11000 00011000 100X11000 00011000 110X11011 001n o n xxoo11011 001n o n xxoi11100 00011100 xxox11100 000All others0000

37、0 000Copyright 2001 Addison Wesley-All Rights ReservedPage 16Computer Systems Organization and Architecture-Solutions Manual22.23.24.25.Po：PoX Y should be P()X TB:Po should be P。26.CLR:Counter input D():A:OX K should be X YOX Y should be OX Y1 should be 027.Address Correct Data0011011100100010H10110

38、0111Copyright 2001 Addison Wesley-All Rights ReservedPage 17Computer Systems Organization and Architecture-Solutions ManualChapter 31.a)Data movementb)Data operationC)Program controld)Data operatione)Data operation2.a)Data operationb)Program controlC)Data movementd)Data movemente)Data operation3.a)D

39、irectb)Impliedc)Implicit4.a)Implicitb)Directc)Implicit5.a)Implicitb)DirectC)Implicit6.a)Register Directb)ImmediateC)Implicitd)Immediatee)Direct7.a)Implicitb)DirectC)Indirectd)Register Indirecte)Register Direct8.a)Register Directb)Register IndirectC)Implicitd)Implicite)ImmediateMULADDPOPX9.a)AC=11 b)

40、AC=12 c)AC=10 d)AC=11 e)AC=10f)AC=33 g)AC=4110.a)AC=11 b)AC=12 C)AC=30 d)AC=31e)AC=10f)AC=23 g)4C=3111.a)AC=Il b)AC=12 C)AC=20 d)AC=2e)AC=10f)AC=43 g)AC=2112.a)MULX,B,Cb)MOV X,B C)LOADBd)PUSH AADD X,X,AMUL X,CMULCPUSHBADD X,X,DADD X,AADDAPUSH CADD X,DADDDSTORE XMULPUSH DADDADDPOPX13.a)MUL TABb)MOV T

41、,A c)LOAD Ad)PUSH AMUL T,T,CMUL T,BMULBPUSHBADD X,E,FMUL T,CMULCMULMUL X,X,DMOV X,ESTORETPUSHCADD X,X,TADD X,FLOADEMULMUL X,DADDFPUSHDADD X,TMULTDADDTSTORE XPUSHEPUSHFADDCopyright 2001 Addison Wesley-All Rights ReservedPage 18Computer Systems Organization and Architecture-Solutions ManualPOPX14.a)MU

42、L X,B,CSUB X,A,Xb)MOV T,BMUL T,CC)LOADBMULCd)PUSH APUSH BMUL T,E,FMOV X,ASTORETPUSHCADD T,T,DSUB X,TLOAD AMULMUL X,X,TMOV T,ESUBTSUBMUL T,FSTORE XPUSH DADD T,DLOADEPUSHEMUL X,TMULFPUSHFADDDMULMULXADDSTORE XMUL15.Processor Time per instruction#Instructions Total time035 ns4140 ns150 ns3150 ns270 ns21

43、40 ns3100 ns1100 ns fastest16.Processor Time per instruction#Instructions Total time035 ns8280 ns150 ns5250 ns 0,20 fbr example.13.14.This is the same as the previous problem,except/o/M is not included.15.16.This is the same as the previous problem,except IO/M is not included.Copyright 2001 Addison

44、Wesley-All Rights ReservedPage 26Computer Systems Organization and Architecture-Solutions Manual17.18.This is the same as the previous problem,except/o/M is not included.Copyright 2001 Addison Wesley-All Rights ReservedPage 27Computer Systems Organization and Architecture-Solutions Manual19.Copyrigh

45、t 2001 Addison Wesley-All Rights ReservedPage 28Computer Systems Organization and Architecture-Solutions Manual20.Copyright 2001 Addison Wesley-All Rights ReservedPage 29Computer Systems Organization and Architecture-Solutions Manual21.Copyright 2001 Addison Wesley-All Rights ReservedPage 30Computer

46、 Systems Organization and Architecture-Solutions Manual22.Memory subsystem:A J d r e S s 5Copyright 2001 Addison Wesley-All Rights ReservedPage 31Computer Systems Organization and Architecture-Solutions Manual22(continued).I/O subsystem:Copyright 2001 Addison Wesley-All Rights ReservedPage 32Compute

47、r Systems Organization and Architecture-Solutions ManualChapter 51.a)a：w x,y zb)a:W Xa:r-zC)a:W X.l/普XFCopyright 2001 Addison Wesley-All Rights ReservedPage 33Computer Systems Organization and Architecture-Solutions ManualCopyright 2001 Addison Wesley-All Rights ReservedPage 34Computer Systems Organ

48、ization and Architecture-Solutions ManualA7x7)/)z7l7abcdecghx)zn/)zzlabedengh2)z!/J/!/)zlz!/abcdecghs0011 0010 0000 01000100 1100 1000 00010011()010 0000 01010100 1100 1000 00011011()010 0000 01001100 1100 1000 00011001 0000 0010 00000000 1001 1001 00000000 0111 0010 10100100 0001 110()10100000 OH 1

49、 0010 10111100 0001 110()10101000 0111 0010 10101100 0001 1100 10100011 1001 0101 00000000 1000 0011 1001io n ooio m i oooo0010 1100 1011 1100io n ooio m i oooo0010 1100 1011 11000011 0010 1111 oooo0010 1100 1011 11001001 0111 1000 oooo0000 0101 1001 0111Copyright 2001 Addison Wesley-All Rights Rese

50、rvedPage 35Computer Systems Organization and Architecture-Solutions Manual11.a)Copyright 2001 Addison Wesley-All Rights ReservedPage 36Computer Systems Organization and Architecture-Solutions Manuala)zX,X)/xzA r7X.5 AX?E AX。9 八)xzuzXI/XJ7)/!/abcdeng工X-XO,(n-l)-lX(n-2)-0-X(n-3)-0,0X(n-2)-0-X(n-1)-1X