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1、 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 Todays science fi ction application tha
2、t 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 proce
3、ssors 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 lan
4、guage 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 tran
5、slates 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
6、for scientifi c computation: 26 fortran 1.1.19 Symbolic representation of machine instructions: 18 assembly language 1.1.20 Interface between users program and hardware providing a variety of services and supervision functions: 14 operating system 1.1.21 Software/programs developed by the users: 24
7、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 Porta
8、ble 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,
9、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
10、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
11、= 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 rat
12、e, 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 1.3.3 timenew = timeold 0.7 = 7 s CPI = CPI 1.2, then CPI(P1) = 1.8, CPI(P2) = 1.2, CPI(P3) = 3 = No. instr. CPI/time, then (P1) = 13.33 109
13、 1.8/7 = 3.42 GHz (P2) = 15 109 1.2/7 = 2.57 GHz (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 = 7/10 = 0.7. So new = old/0.7 = 1.5 GHz/0.7 = 2.14 GHz. 1.3.6 Timenew/Timeold = 9/10 = 0.9. So Instructionsn
14、ew = 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 P1: Time class A = 0.66 104 Time class B = 2.66 104 Time class C = 10 104 Time class D =
15、 5.33 104 Total time P1 = 18.65 104 P2: Time class A = 104 Time class B = 2 104 Time class C = 5 104 Time class D = 3 104 Total time P2 = 11 104 1.4.2 CPI = time clock rate/No. instr. CPI(P1) = 18.65 104 1.5 109/106 = 2.79 CPI(P2) = 11 104 2 109/106 = 2.2 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) 0.5 109 = 675 ns 1.4.5 CPI = time clock rate/No. instr. CPI = 675 109 2 109/700 = 1.92 1.4.6 Time = (500 1 + 50 5 + 50 5 + 50 2) 0.5 109 = 550 ns Speed-up = 675 n
