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计算机组成与设计（第四版）课后习题答案

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正确详细的课后习题解答，帮助您更好的学好并掌握计算机组成的基本原理以及习题解答过程

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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 10

or 2

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 10

or 2

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 × 10

/1.5 = 1.33 × 10

performance of P2 (instructions/sec) = 1.5 × 10

/1.0 = 1.5 × 10

performance of P3 (instructions/sec) = 3 × 10

/2.5 = 1.2 × 10

1.3.2 No. cycles = time × clock rate

cycles(P1) = 10 × 2 × 10

= 20 × 10

cycles(P2) = 10 × 1.5 × 10

= 15 × 10

cycles(P3) = 10 × 3 × 10

= 30 × 10

time = (No. instr. × CPI)/clock rate, then No. instructions = No. cycles/CPI

instructions(P1) = 20 × 10

/1.5 = 13.33 × 10

instructions(P2) = 15 × 10

/1 = 15 × 10

instructions(P3) = 30 × 10

/2.5 = 12 × 10

1.3.3 time

new

= time

old

× 0.7 = 7 s

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 × 10

× 1.8/7 = 3.42 GHz

ƒ(P2) = 15 × 10

× 1.2/7 = 2.57 GHz

ƒ(P3) = 12 × 10

× 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 Time

new

/Time

old

= 7/10 = 0.7. So ƒ

new

= ƒ

old

/0.7 = 1.5 GHz/0.7 = 2.14 GHz.

1.3.6 Time

new

/Time

old

= 9/10 = 0.9.

So Instructions

new

= Instructions

old

× 0.9 = 30 × 10

× 0.9 = 27 × 10

S4 Chapter 1 Solutions

Solution 1.4

1.4.1 P2

Class A: 10

instr.

Class B: 2 × 10

instr.

Class C: 5 × 10

instr.

Class D: 2 × 10

instr.

Time = No. instr. × CPI/clock rate

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

−4

P2: Time class A = 10

−4

Time class B = 2 × 10

−4

Time class C = 5 × 10

−4

Time class D = 3 × 10

−4

Total time P2 = 11 × 10

−4

1.4.2 CPI = time × clock rate/No. instr.

CPI(P1) = 18.65 × 10

−4

× 1.5 × 10

/10

= 2.79

CPI(P2) = 11 × 10

−4

× 2 × 10

/10

= 2.2

1.4.3

clock cycles(P1) = 10

× 1 + 2 × 10

× 2 + 5 × 10

× 3 + 2 × 10

× 4 = 28 × 10

clock cycles(P2) = 10

× 2 + 2 × 10

× 2 + 5 × 10

× 2 + 2 × 10

× 3 = 22 × 10

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 × 10

/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 × 10

/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. P2 is 1.33 times faster than P1

b. P1 is 1.03 times faster than P2

1.5.3

a. P2 is 1.31 times faster than P1

b. 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

1.00 1.17

0.80 0.58

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