### 计算机组成与设计第五版1-6章答案 英文版

《计算机组成与设计：软硬件接口》英文第五版 1-6章答案
s=4 Chapter 1 Solutions b. cycles(P1)=10×3×109=30×109s cycles(P2)=10×2.5×109=25×109s cycles(P3)=10×4×109=40×109s C.No. instructions(P1)=30×10/1.5=20×109 No. instructions(P2)=25×101=25×109 No. instructions(P3)=40×1092.2=18.18×109 CPI= CPI- X 1. 2, then CPI(P1)=1.8, CPI(P2)=1.2, CPI(P3)=2.6 f= No instr. X CPI/time, then f(P1)=20×109×1.8/7=5.14GHz f(P2)=25×109×1.2/7=4.28GHz f(P1)=1818×10×2.6/7=6.75GHz 1.6 a. Class A: 105 instr. Class B: 2 x 105 instr. Class C: 5 x 105 instr Class i:2×105 instr Time= No instr. X CPI/clock rate Total time p1=(105+2×105×2+5×105×3+2×105×3)/(2.5× 10)=104×10-4s otal time p2=(105×2+2×105×2+5×105×2+2×105×2) (3×10)=6.66×10-4s CPI(P1)=10.4×10-4×2.5×10/10°=2.6 CPI(P2)=6.66×104×3×10°/10°=2.0 b. clock cycles(P1)=105×1+2×105×2+5×105×3+2×105×3 26×10 clock cycles(P2)=105×2+2×105×2+5×105×2+2×105×2 20×105 a.CPI=T×f/No. instr Compiler a cpi= 1.1 Compiler b cpi= 1.25 b. f, /f,=(No instr. (B)X CPI(B)/(No instr (A)X CPI(A))=1.37 T / T T/T 2.27 Chapter 1 Solutions 18 1.8.1C=2×DP/(V2F Pentium 4: C=3.2E-8F Core i5 Ivy Bridge: C=2.9E-8F 1.8.2 Pentium4:10/100=10% Core i5 Ivy Bridge: 30/70= 42.9% 18.3(S+D)(S1,+D 0.90 new C×V2×F V,×I V×I Therefore [D/(C×F)]2 D=0.90×(S+D S=V×(S/V Pentium 4 S=V×(10/1.25)=V×8 0.90×100V×8=90-V×8 new (90-V×8)/(3.2E8×36E9) 0.85V Core 15 S=V×(30/0.9)=V×33.3 D=0.90×70-V×33.3=63V×33.3 new V=[(63-Vm×333)/(29E8×34E9)]2 V=0.64V 19 l9.1 p# arith inst.I# L/S inst.# branch inst. cycles ex, time speedup 2.56E9 1.28E9 2.56E8 7.94E10 39.7 1.83E9 9.14E8 2.56E8 5.67E10 28.3 1.4 4 9.12E8 4.57E8 2.56E8 2.83E10 14.2 2.8 84.57E8229E8256E8142E1071056 Chapter 1 Solutions 19.2 p ex, time 41.0 2 29.3 4 14.6 8 7.33 19.33 110 1.10.1 die area,m wafer area/dies per wafer= pi*7.52/84=2.10 cm yield=1/(1+(0.020*210/2)2=0.9593 d le area wafer area/dies per wafer= pi*102/1( vle ld2n=1/(1+(0.0313.14/2)2=0.9093 110.2cost| diesem=12/(840.9593)=0.1489 cost/die20mn=15/(100-0.9093)=0.1650 1. 10.3 die area, cm wafer area/dies per wafer= pit75/(84*1.1)=1.91cm 1/(1+(0.020*1.15191/2)2=0.9575 die area2ocm wafer area/dies per wafer= pi*102/(100*1. 1)=2.86 cm yield=1/(1+(00311542.86/2)2=0.9082 1. 10.4 defects per area. 92 (1-y^.5)/(y^.5- die area/2)=(1-0.92^.5)/ (0.92^.5*2/2)=0.043 defects/cm defects per areco 1-yA.5)/(yA.5*die_area/2 (1-0.95^.5) (0.95^.5*2/2)=0.026 defects/cm 1.11 1.11.1 CPI= clock rate X CPU time/instr. count clock rate 1/ cycle time 3 GHz CPI(bzip2)=3×10×750/(2389×10°)=0.94 1.11.2 SPEC ratio ref time/execution time SPEC ratio(bzip 2)=9650/750= 12.86 1.11.3. CPU time= No instr. X CPI/clock rate If CPi and clock rate do not change, the Cpu time increase is equal to the increase in the of number of instructions that is 10% Chapter 1 Solutions S-7 1. 11.4 CPU time(before)= No instr. X CPI/clock rate CPU time( after)=1.1×No. instr.×1.05×CPI/ clock rate CPU time(after)/CPU time(before)=1. 1 X 1.05=1.155. Thus, CPU time is increased by 15.5% 1.11.5 SPECratio reference time/CPU time SPECratio(after)/SPECratio(before)= CPU time(before)/CPU time(after 1/1. 1555=0.86. The SPeCratio is decreased by 14% 1.11.6 CPI=(CPU time X clock rate)/No instr CPI=700×4×10(0.85×2389×10°)=1.37 1.11. 7 Clock rate ratio 4 GHZ/3 GHz=1.33 CPI@ 4 GHz= 1.37, CPI@ 3 GHz =0.94, ratio = 1.45 They are different because, although the number of instructions has been reduced by 15%, the Cpu time has been reduced by a lower percentage 1.11.8700/750=0.933. CPU time reduction:6.7% 1.11.9 No instr.= CPU time x clock rate/CPI No.inst.=960×0.9×4×10%/1.61=2146×10 1.11.10 Clock rate= No instr. X CPI/CPU time Clock rate No instr. X CPI/0.9 x CPU time =1/0.9 clock rate 3.33 GHZ 1.11.11 Clock rate= No instr. X CPI/CPU time Clock rate= No instr.x085X CPI/0.80 CPU time =0.85/0.80, clock rate 3. 1 8 GHZ 1.12 1.121T(P1)=5×109×0.9/(4×109=1.125s T(P2)=109×0.75/(3×109)=0.25s clock rate(P1)> clock rate(P2), performance(P1)< performance(P2) 1.12.2 T(P1)=No instr. X CPI/clock rate T(P1)=2.2531021s T(P2)5N×0.75/(3×109), then n=9×108 1. 12. 3 MIPS= Clock rate X 10-6/CPI MIPS(P1)=4×109×10-6/0.9=44×103 Chapter 1 Solutions MIPS(P2)=3×109×1060.75=40×10 MIPS(P1)> MIPS(P2), performance(P1)< performance(P2)(from 11a) 1. 12.4 MFLOPS= No FP operations X 10 /T MFLOPS(P1)=4×5E9×1E-6/1.125=178E3 MFLOPS(P2)=4×1E9×1E-6/.25=1.60E3 MFLOPS(P1)> MFLOPS(P2), performance(P1)<performance(P2 (from 11a) 113 11311=70×0.8=56 56+85+55+40=236s. Reduction:5.6% 1.13.2T=250×0.8=200s,T+T+T,=165s,T=35s. Reduction time int: 58.8% 113.3T=250×0.8=200s,①十工+T,=210s.NO 114 1. 14.1 Clock cycles= CPI X No FP instr. CPI.X No INT instr. CPI X No. L/S instr.+ cr)branch ×No. branch instr Tp clock cycles/clock rate= clock cycles/2 105 clock cycles=512 X 106, TD =0.256 S To have the number of clock cycles by improving the CPI of FP instructions CPI improved X NO FP instr. CPI. X NO INT instr. CPI X No L/S instr.+ C X No branch instr. clock cycles/2 ran CPlimproved f=(clock cycles/2-(CPI X No INT instr. CPL X No L/S instr.+ CPIbranch X No branch instr ))/No FP instr CPI (256-462)/50<0 not possible 1. 14.2 USing the clock cycle data from a To have the number of clock cycles improving the cpi of l/s instructions CPIp X No. FP instr. t CPIint X No INT instr. CPIimproved Is X No. L/S instr. + CPlbranch No branch instr clock cycles/2 CPIimproved s=(clock cycles/2(CPI X No FP instr. CPIint X No INT instr.+ CPIranch X No branch instr. ))/No L/S instr CPI =(256-198)/80=0.725 1. 14.3 Clock cycles= CPI X No FP instr. CPI.X No. INT instr.+ CPL X int No L/S instr.+ CPl X No branch instr Chapter 1 Solutions S-9 clock cycles/clock rate clock cycles/2X 10 CPI=0.6×1=0.6;CPI=0.6×1=0.6;CPI1=0.7×4=28; CPI,=0.7×2=14 Top (before improv. )=0.256 S; Tcp (after improv. )=0.171s 1.15 exec, time/ time actual speedup/ideal processors processor w/overhead speedup speedup 1 100 50 54 100/54=185 185/2=.93 4 25 29 100/29=3.44 3.44/4=0.86 8 12.5 16.5 100/16.5=6.06 606/8=0.75 16 6.25 10.25 100/10.25=9.76 9.76/16=0.61 Solutions Patterson-1610874978-0-12-407726-3 PIL Chapter 2 Solutions S3 21 addi f, h-5 (note, no subi) add f. f 22f 2.3sub\$t0,\$S3,\$S4 add sto. ss6, sto 1W\$t1,16(\$t0) SW\$t1,32(\$s7) 24B[g]=A[f]+A[1+f] 2.5 add sto \$s6 \$s0 add st1, \$s7. \$s1 lw\$s0,0(\$t0) lW\$七0,4(\$t0) add sto, sto. \$ sO W\$t0,0(\$t1) 2.6 2.6.1 temp Array[O] temp2 Array[l]; Array[o]= Array[4]; Array temp; Array[4]= Array[3]; Array[3]=temp2 2.621W\$t0,0(\$s6) lw\$t1,4(\$S6) lW\$t2,16(\$S6) sW\$t2,0(\$S6) sW\$t0,4(\$S6) lW\$七0,12(\$S6) SW\$t0,16(\$s6) W\$t1,12(\$S6)

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