（超详细）计算机组成与设计（第四版）全部课后答案.pdf-资料库

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1 Solutions Solution 1.1 1.1.1 Computer used to run large problems and usually accessed via a network: 5 supercomputers 1.1.2 1015 or 250 bytes: 7 petabyte 1.1.3 Computer composed of hundreds to thousands of processors and terabytes of memory: 3 servers 1.1.4 Today’s science ﬁ ction application that probably will be available in near future: 1 virtual worlds 1.1.5 A kind of memory called random access memory: 12 RAM 1.1.6 Part of a computer called central processor unit: 13 CPU 1.1.7 Thousands of processors forming a large cluster: 8 datacenters 1.1.8 A microprocessor containing several processors in the same chip: 10 multi- core processors 1.1.9 Desktop computer without screen or keyboard usually accessed via a net- work: 4 low-end servers 1.1.10 Currently the largest class of computer that runs one application or one set of related applications: 9 embedded computers 1.1.11 Special language used to describe hardware components: 11 VHDL 1.1.12 Personal computer delivering good performance to single users at low cost: 2 desktop computers 1.1.13 Program that translates statements in high-level language to assembly language: 15 compiler

S2 Chapter 1 Solutions 1.1.14 Program that translates symbolic instructions to binary instructions: 21 assembler 1.1.15 High-level language for business data processing: 25 cobol 1.1.16 Binary language that the processor can understand: 19 machine language 1.1.17 Commands that the processors understand: 17 instruction 1.1.18 High-level language for scientiﬁ c computation: 26 fortran 1.1.19 Symbolic representation of machine instructions: 18 assembly language 1.1.20 Interface between user’s program and hardware providing a variety of services and supervision functions: 14 operating system 1.1.21 Software/programs developed by the users: 24 application software 1.1.22 Binary digit (value 0 or 1): 16 bit 1.1.23 Software layer between the application software and the hardware that includes the operating system and the compilers: 23 system software 1.1.24 High-level language used to write application and system software: 20 C 1.1.25 Portable language composed of words and algebraic expressions that must be translated into assembly language before run in a computer: 22 high-level language 1.1.26 1012 or 240 bytes: 6 terabyte Solution 1.2 1.2.1 8 bits × 3 colors = 24 bits/pixel = 4 bytes/pixel. 1280 × 800 pixels = 1,024,000 pixels. 1,024,000 pixels × 4 bytes/pixel = 4,096,000 bytes (approx 4 Mbytes). 1.2.2 2 GB = 2000 Mbytes. No. frames = 2000 Mbytes/4 Mbytes = 500 frames. 1.2.3 Network speed: 1 gigabit network ==> 1 gigabit/per second = 125 Mbytes/ second. File size: 256 Kbytes = 0.256 Mbytes. Time for 0.256 Mbytes = 0.256/125 = 2.048 ms.

Chapter 1 Solutions S3 1.2.4 2 microseconds from cache ==> 20 microseconds from DRAM. 20 micro- seconds from DRAM ==> 2 seconds from magnetic disk. 20 microseconds from DRAM ==> 2 ms from ﬂ ash memory. Solution 1.3 1.3.1 P2 has the highest performance performance of P1 (instructions/sec) = 2 × 109/1.5 = 1.33 × 109 performance of P2 (instructions/sec) = 1.5 × 109/1.0 = 1.5 × 109 performance of P3 (instructions/sec) = 3 × 109/2.5 = 1.2 × 109 1.3.2 No. cycles = time × clock rate cycles(P1) = 10 × 2 × 109 = 20 × 109 s cycles(P2) = 10 × 1.5 × 109 = 15 × 109 s cycles(P3) = 10 × 3 × 109 = 30 × 109 s time = (No. instr. × CPI)/clock rate, then No. instructions = No. cycles/CPI instructions(P1) = 20 × 109/1.5 = 13.33 × 109 instructions(P2) = 15 × 109/1 = 15 × 109 instructions(P3) = 30 × 109/2.5 = 12 × 109 = timeold × 0.7 = 7 s 1.3.3 timenew CPI = CPI × 1.2, then CPI(P1) = 1.8, CPI(P2) = 1.2, CPI(P3) = 3 ƒ = No. instr. × CPI/time, then ƒ(P1) = 13.33 × 109 × 1.8/7 = 3.42 GHz ƒ(P2) = 15 × 109 × 1.2/7 = 2.57 GHz ƒ(P3) = 12 × 109 × 3/7 = 5.14 GHz 1.3.4 IPC = 1/CPI = No. instr./(time × clock rate) IPC(P1) = 1.42 IPC(P2) = 2 IPC(P3) = 3.33 1.3.5 Timenew/Timeold 1.3.6 Timenew/Timeold So Instructionsnew = 7/10 = 0.7. So ƒnew = 9/10 = 0.9. = ƒold/0.7 = 1.5 GHz/0.7 = 2.14 GHz. = Instructionsold × 0.9 = 30 × 109 × 0.9 = 27 × 109.

S4 Chapter 1 Solutions Solution 1.4 1.4.1 P2 Class A: 105 instr. Class B: 2 × 105 instr. Class C: 5 × 105 instr. Class D: 2 × 105 instr. Time = No. instr. × CPI/clock rate −4 P1: Time class A = 0.66 × 10 −4 Time class B = 2.66 × 10 −4 Time class C = 10 × 10 −4 Time class D = 5.33 × 10 −4 Total time P1 = 18.65 × 10 P2: Time class A = 10 −4 −4 Time class B = 2 × 10 −4 Time class C = 5 × 10 −4 Time class D = 3 × 10 −4 Total time P2 = 11 × 10 1.4.2 CPI = time × clock rate/No. instr. CPI(P1) = 18.65 × 10 CPI(P2) = 11 × 10 −4 × 2 × 109/106 = 2.2 −4 × 1.5 × 109/106 = 2.79 1.4.3 clock cycles(P1) = 105 × 1 + 2 × 105 × 2 + 5 × 105 × 3 + 2 × 105 × 4 = 28 × 105 clock cycles(P2) = 105 × 2 + 2 × 105 × 2 + 5 × 105 × 2 + 2 × 105 × 3 = 22 × 105 1.4.4 (500 × 1 + 50 × 5 + 100 × 5 + 50 × 2) × 0.5 × 10–9 = 675 ns 1.4.5 CPI = time × clock rate/No. instr. CPI = 675 × 10–9 × 2 × 109/700 = 1.92 1.4.6 Time = (500 × 1 + 50 × 5 + 50 × 5 + 50 × 2) × 0.5 × 10–9 = 550 ns Speed-up = 675 ns/550 ns = 1.22 CPI = 550 × 10–9 × 2 × 109/700 = 1.57

Chapter 1 Solutions S5 Solution 1.5 1.5.1 a. 1G, 0.75G inst/s b. 1G, 1.5G inst/s 1.5.2 a. b. P2 is 1.33 times faster than P1 P1 is 1.03 times faster than P2 1.5.3 a. b. P2 is 1.31 times faster than P1 P1 is 1.00 times faster than P2 1.5.4 a. 2.05 µs b. 1.93 µs 1.5.5 a. 0.71 µs b. 0.86 µs 1.5.6 a. 1.30 times faster b. 1.40 times faster Solution 1.6 1.6.1 Compiler A CPI Compiler B CPI a. b. 1.00 0.80 1.17 0.58

S6 Chapter 1 Solutions 1.6.2 a. 0.86 b. 1.37 1.6.3 a. b. 1.6.4 a. b. Compiler A speed-up Compiler B speed-up 1.52 1.21 P1 peak 4G Inst/s 4G Inst/s 1.77 0.88 P2 peak 3G Inst/s 3G Inst/s 1.6.5 Speed-up, P1 versus P2: a. 0.967105263 b. 0.730263158 1.6.6 a. 6.204081633 b. 8.216216216 Solution 1.7 1.7.1 Geometric mean clock rate ratio = (1.28 × 1.56 × 2.64 × 3.03 × 10.00 × 1.80 × 0.74)1/7 = 2.15 Geometric mean power ratio = (1.24 × 1.20 × 2.06 × 2.88 × 2.59 × 1.37 × 0.92)1/7 = 1.62 1.7.2 Largest clock rate ratio = 2000 MHz/200 MHz = 10 (Pentium Pro to Pentium 4 Willamette) Largest power ratio = 29.1 W/10.1 W = 2.88 (Pentium to Pentium Pro)

Chapter 1 Solutions S7 1.7.3 Clock rate: 2.667 × 109/12.5 × 106 = 212.8 Power: 95 W/3.3 W = 28.78 1.7.4 C = P/V2 × clockrate −6 80286: C = 0.0105 × 10 −6 80386: C = 0.01025 × 10 −6 80486: C = 0.00784 × 10 −6 Pentium: C = 0.00612 × 10 Pentium Pro: C = 0.0133 × 10 Pentium 4 Willamette: C = 0.0122 × 10 −6 Pentium 4 Prescott: C = 0.00183 × 10 Core 2: C = 0.0294 × 10 1.7.5 3.3/1.75 = 1.78 (Pentium Pro to Pentium 4 Willamette) −6 −6 −6 1.7.6 Pentium to Pentium Pro: 3.3/5 = 0.66 Pentium Pro to Pentium 4 Willamette: 1.75/3.3 = 0.53 Pentium 4 Willamette to Pentium 4 Prescott: 1.25/1.75 = 0.71 Pentium 4 Prescott to Core 2: 1.1/1.25 = 0.88 Geometric mean = 0.68 Solution 1.8 = V2 × clock rate × C. Power2 1.8.1 Power1 C2/C1 = 0.9 × 52 × 0.5 × 109/3.32 × 1 × 109 = 1.03 = 0.9 Power1 1.8.2 Power2/Power1 = V2 2 × clock rate2/V1 2 × clock rate1 Power2/Power1 = 0.87 => Reduction of 13% 1.8.3 2 × 1 × 109 × 0.8 × C1 = 0.6 × Power1 Power2 = V2 Power1 = 52 × 0.5 × 109 × C1 V2 V2 = ( (0.6 × 52 × 0.5 × 109)/(1 × 109 × 0.8) )1/2 = 3.06 V 2 × 1 × 109 × 0.8 × C1 = 0.6 × 52 × 0.5 × 109 × C1

S8 Chapter 1 Solutions 1.8.4 Powernew power scales by 1. = 1 × Cold × V2 old/(2 −1/4)2 × clock rate × 21/2 = Powerold. Thus, 1.8.5 1/2 −1/2 = 21/2 1.8.6 Voltage = 1.1 × 1/2 −1/4 = 0.92 V. Clock rate = 2.667 × 21/2 = 3.771 GHz Solution 1.9 1.9.1 a. 1/49 × 100 = 2% b. 45/120 × 100 = 37.5% 1.9.2 a. b. Ileak = 1/3.3 = 0.3 Ileak = 45/1.1 = 40.9 1.9.3 a. b. Powerst/Powerdyn = 1/49 = 0.02 Powerst/Powerdyn = 45/57 = 0.6 = 0.6 => Powerst = 0.6 × Powerdyn 1.9.4 Powerst/Powerdyn a. Powerst = 0.6 × 40 W = 24 W Powerst = 0.6 × 30 W = 18 W b. 1.9.5 a. b. Ilk = 24/0.8 = 30 A Ilk = 18/0.8 = 22.5 A

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（超详细）计算机组成与设计（第四版）全部课后答案.pdf

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