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计算机体系结构量化分析第六版课后答案

计算机体系结构

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Chapter 1 Solutions

Case Study 1: Chip Fabrication Cost

1.1 a. Yield ¼1/(1 + (0.04 2))

¼0.34

b. It is fabricated in a larger technology, which is an older plant. As plants age,

their process gets tuned, and the defect rate decreases.

1.2 a. Phoenix:

Dies per wafer ¼ π  45=2ðÞ



=2  π  45ðÞ=sqrt 22ðÞ¼795  70:7 ¼ 724:5 ¼ 724

Yield ¼ 1= 1+ 0:04  2ðÞðÞ

¼ 0:340

Profit ¼ 724  0:34  30 ¼ $7384:80

b. Red Dragon:

Dies per wafer ¼ π  45=2ðÞ



=2  π  45ðÞ=sqrt 2  1:2ðÞ¼1325  91:25 ¼ 1234

Yield ¼ 1= 1+ 0:04  1:2ðÞðÞ

¼ 0:519

Profit ¼ 1234  0:519  15 ¼ $9601:71

c. Phoenix chips: 25,000/724 ¼ 34.5 wafers needed

Red Dragon chips: 50,000/1234 ¼40.5 wafers needed

Therefore, the most lucrative split is 40 Red Dragon wafers, 30 Phoenix wafers.

1.3 a. Defect-free single core ¼ Yield ¼ 1/(1 + (0.04  0.25))

¼0.87

Equation for the probability that N are defect free on a chip:

#combinations  (0.87)

 (1  0.87)

8N

Yield for Phoenix

: (0.39 + 0.21 + 0.06 + 0.01) ¼0.57

Yield for Phoenix

: (0.001 + 0.0001) ¼ 0.0011

Yield for Phoenix

: 0.000004

b. It would be worthwhile to sell Phoenix

. However, the other two have such a

low probability of occurring that it is not worth selling them.

# defect-free # combinations Probability

0.32821167

0.39234499

0.20519192

0.06132172

0.01145377

0.00136919

0.0001023

4.3673E-06

8.1573E-08

$20 ¼

Wafer size

odd dpw  0:28

Step 1: Determine how many Phoenix4 chips are produced for every

Phoenix8 chip.

There are 57/33 Phoenix4 chips for every Phoenix8 chip ¼1.73

$30 + 1:73  $25 ¼ $73:25

Case Study 2: Power Consumption in Computer Systems

1.4 a. Energy: 1/8. Power: Unchang ed.

b. Energy: Energy

new

/Energy

old

¼(Voltage  1/8)

/Voltage

¼0.156

Power: Power

new

/Power

old

¼0.156  (Frequency  1/8)/Frequency ¼ 0.00195

c. Energy: Energy

new

/Energy

old

¼(Voltage  0.5)

/Voltage

¼0.25

Power: Power

new

/Power

old

¼0.25  (Frequency  1/8)/Frequency ¼0.0313

d. 1 core ¼ 25% of the original power, running for 25% of the time.

0:25  0:25 + 0:25  0:2ðÞ0:75 ¼ 0:0625 + 0:0375 ¼ 0:1

1.5 a. Amdahl’s law: 1/(0.8/4 + 0.2) ¼ 1/(0.2 + 0.2) ¼ 1/0.4 ¼2.5

b. 4 cores, each at 1/(2.5) the frequency and voltage

Energy: Energy

quad

/Energy

single

¼4  (Voltage  1/(2.5))

/Voltage

¼0.64

Power: Power

new

/Power

old

¼0.64  (Frequency  1/(2.5))/Frequency¼ 0.256

c. 2 cores + 2 ASICs vs. 4 cores

2+ 0:2  2ðÞðÞ=4 ¼ 2:4ðÞ=4 ¼ 0:6

1.6 a. Workload A speedup: 225,000/13,461 ¼16.7

Workload B speedup: 280,000/36,465 ¼7.7

1/(0.7/16.7 + 0.3/7.7)

b. General-purpose: 0.70  0.42 + 0.30 ¼ 0.594

GPU: 0.70  0.37 + 0.30 ¼0.559

TPU: 0.70  0.80 + 0.30 ¼ 0.886

c. General-purpose: 159 W + (455 W  159 W)  0.594 ¼ 335 W

GPU: 357 W + (991 W  357 W)  0.559 ¼ 711 W

TPU: 290 W + (384 W  290 W)  0.86 ¼ 371 W

Speedup A B C

GPU 2.46 2.76 1.25

TPU 41.0 21.2 0.167

% Time 0.4 0.1 0.5

2 ■ Solutions to Case Studies and Exercises

GPU: 1/(0.4/2.46 + 0.1/2.76 + 0.5/1.25) ¼ 1.67

TPU: 1/(0.4/41 + 0.1/21.2 + 0.5/0.17) ¼ 0.33

e. General-purpose: 14,000/504 ¼ 27.8  28

GPU: 14,000/1838 ¼7.62  8

TPU: 14,000/861 ¼16.3  17

d. General-purpose: 2200/504 ¼ 4.37 4, 14,000/(4  504) ¼ 6.74  7

GPU: 2200/1838 ¼1.2  1, 14,000/(1  1838) ¼ 7.62  8

TPU: 2200/861 ¼2.56  2, 14,000/(2  861) ¼ 8.13  9

Exercises

1.7 a. Somewhere between 1.4

and 1.55

, or 28.9  80x

b. 6043 in 2003, 52% growth rate per year for 12 years is 60,500,000 (rounded)

c. 24,129 in 2010, 22% growth rate per year for 15 years is 1,920,000 (rounded)

d. Multiple cores on a chip rather than faster single-core performance

e. 2 ¼x

, x ¼ 1.032, 3.2% growth

1.8 a. 50%

b. Energy: Energy

new

/Energy

old

¼(Voltage  1/2)

/Voltage

¼0.25

1.9 a. 60%

b. 0.4 + 0.6  0.2 ¼ 0.58, which reduces the ene rgy to 58% of the original energy

c. newPower/oldPower¼½Capacitance (Voltage0.8)

(Frequency 0.6)/½

Capacitance Voltage Frequency¼0.8

0.6¼ 0.256 of the original power.

d. 0.4 + 0.3  2 ¼ 0.46, which reduces the energy to 46% of the original energy

1.10 a. 10

/100 ¼10

b. 10

/10

+24¼1

c. [need solution]

1.11 a. 35/10,000  3333 ¼11.67 days

b. There are several correct answ ers. One would be that, with the current system,

one computer fails approximately every 5 min. 5 min is unlikely to be enough

time to isolate the computer, swap it out, and get the computer back on line

again. 10 min, however, is much more likely. In any case, it would greatly

extend the amount of time before 1/3 of the computers have failed at once.

Because the cost of downtime is so huge, being able to extend this is very

valuable.

c. $90,000 ¼(x + x + x +2x)/4

$360,000 ¼5x

$72,000 ¼x

4th quarter ¼$144,000/h

Chapter 1 Solutions

■ 3

1.12 a. See Figure S.1.

b. 2 ¼1/((1 x)+x/20)

10/19 ¼x ¼ 52.6%

c. (0.526/20)/(0.474 + 0.526/20) ¼ 5.3%

d. Extra speedup with 2 units: 1/(0.1 + 0.9/2) ¼ 1.82. 1.82  20  36.4.

Total speedup: 1.95. Extra speedup with 4 units: 1/(0.1 + 0.9/4) ¼3.08.

3.08  20  61.5. Total speedup: 1.97

1.13 a. old execution time ¼ 0.5 new + 0.5 10 new ¼ 5.5 new

b. In the original code, the unenhanced part is equal in time to the enhanced part

(sped up by 10), therefore:

(1 x) ¼ x/10

10 10x ¼ x

10 ¼11x

10/11 ¼x ¼ 0.91

1.14 a. 1/(0.8 + 0.20/2) ¼1.11

b. 1/(0.7 + 0.20/2 + 0.10  3/2) ¼ 1.05

c. fp ops: 0.1/0.95 ¼ 10.5%, cache: 0.15/0.95 ¼15.8%

1.15 a. 1/(0.5 + 0.5/22) ¼1.91

b. 1/(0.1 + 0.90/22) ¼7.10

c. 41%  22¼ 9. A runs on 9 cores. Speedup of A on 9 cores: 1/(0.5 + 0.5/9) ¼

1.8 Overall speedup if 9 cores have 1.8 speedup, others none: 1/(0.6+0.4/1.8)

¼1.22

d. Calculate values for all processors like in c. Obtain: 1.8, 3, 1.82, 2.5,

respectively.

e. 1/(0.41/1.8 + 0.27/3 + 0.18/1.82 + 0.14/2.5) ¼2.12

0 1020304050

Percent vectorization

Net speedup

60 70 80 90 100

Figure S.1 Plot of the equation: y

100/((100 2 x)+x/10).

4 ■ Solutions to Case Studies and Exercises

1.16 a. 1/(0.2 + 0.8/N)

b. 1/(0.2 + 8  0.005 + 0.8/8) ¼ 2.94

c. 1/(0.2 + 3  0.005 + 0.8/8) ¼ 3.17

d. 1/(.2 + logN  0.005 + 0.8/N)

e. d/dN (1/((1  P) + logN  0.005 + P/N) ¼0)

Chapter 1 Solutions

■ 5

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