计算机组成原理第五章答案

5151CPU(1)IR(2)AR(3)DR25.15STORl(R2)Rl(R2)STOR1,(R2)52(PC)AR(DR)IR(M)DRPCO,G,ARiR/W=RDRO,G,IRi(R2)ARR2O,G,ARi(R1)DRR1O,G,DRi(DR)MR/W=WSTOR1,(R2)~35.15LAD(R3)R0(R3)R2LADR3,(R0)53(PC)AR(DR)IR(M)DRPCO,G,ARiR/W=RDRO,G,IRi(R3)ARR3O,G,ARi(M)DRDRO,G,R0i(DR)R0R/W=RLAD(R3),R0~410MHz5545CPU3Tl200nsT2=400nsT3=200nsTlT2T3200ns5MHz.3T324C43ClC2C3211CCT321CCT313CCT5568043280480(4-1)+1=24132241327ALUMS3S2S1Cy0l(ABHDEFG)MS3S2SlCMS3S2S1CA,BH,DEFG0000101010111111001101yyM=G56S3=H+D+FS2=A+B+D+H+E+F+GS1=A+B+F+GC=H+D+Ey+Fy88I1I8aj108abcdefghijI1I2I3I4I5I6I7I810108(e,f,h)(b,i,j)(d,i,j)(e,f,h)(g,b,j)(i,f,h)2:4(00)efhbijacdgXXXXefhdijabcgXXXXfhibgjacdeXXXX9A8=P1IR6T4A7=P1IR5T4A6=P2CT457A8A6P1P2CIR5IR6T4A5=P3IR5T4A4=P3IR4T4A3=P1IR3T4A2=P1IR2T4A1=P1IR1T4A0=P1IR0T4+P2CT4()51A2A1A010ALUMMDRMARIRR0R3CD(1)(2)ADDR1R2(1)CDALUALUALUIRPCMDRMARMR0R1R2R3DCALU+58(2)ADDR1R2(R1)(PC)MAR(M)MDR(MDR)IRPC+1(R1)C(R2)D(C)+(D)()R1ADDR1,R2~11512*484(1)?(2)(1)512*48944348-9-4=35(48-9-3=36)(2)P.1475.2359124100ns100ns80ns50ns(1)?(2)2?(3)?(1)max(100,100,80,50)=100ns(2)122200ns(3)12100ns13(IF)(ID)(EX)(MEM)(WB)20(1)100ns(2)()(3)(1)100nsSI1I2I15I16I17I18I19I20WBI1I2I15I16I17I18I19I20MEMI1I2I3I16I17I18I19I20OPPIR510EXI1I2I3I4I17I18I19I20IDI1I2I3I4I5I18I19I20IFI1I2I3I4I5I6I19I20123456192021222324T(2)205+119=242400ns/333.8102400209(3)nkn*k*nk(k+n-1)*20517.4195520144IFIDEXWBSI1I2I3I4I5WBI1I2I3I4I5EXI1I2I3I4I5I6IDI1I2I3I4I5I6I7IFI1I2I3I4I5I6I7I812345678TSI1I2WBI1I2EXI1I2IDI1I2IFI1I212345678T85251115kn1nkT11nknHnknTknnH21121nkknnkknHHn=121HHn121HH16?(1)I1LADR1AM(A)R1M(A)I2ADDR2Rl(R2)+(R1)R2(2)I1ADDR3R4(R3)+(R4)R3I2MULR4R5(R4)(R5)R4(3)I1LADR6BM(B)R6M(B)I2MULR6R7(R6)(R7)R6(1)I1R1I2R1RAW()(2)WAR(3)RAWWAW175.396I1LADR1BM(B)R1M(B)I2SUBR2Rl(R2)-(R1)R2I3MULR3R4(R3)*(R4)R3I4ADDR4R5(R4)+(R5)R4I5LADR6AM(A)R6M(A)512I6ADDR6R7(R6)+(R7)R6(1)(2)(1)(FDFDW123/1)I1I2I3I4I5I6I1I2I2I3I4I5I6I6I1I2I2I4I3I5I4I3I6I3I6/I1I2I3I4I5I6(2)1234567891011I1FDEWI2FDEEWI3FDEEEWI4FDEEWI5FDEWI6FDEEW