没有合适的资源?快使用搜索试试~ 我知道了~
首页计算机体系结构量化分析第六版课后答案
计算机体系结构量化分析第六版课后答案
需积分: 47 886 浏览量
更新于2023-05-26
评论 1
收藏 3.73MB PDF 举报
computer-architecture-sixth-edition-a-quantitative-approach-solutions-manual-6
资源详情
资源评论
资源推荐

Chapter 1 Solutions
Case Study 1: Chip Fabrication Cost
1.1 a. Yield ¼1/(1 + (0.04 2))
14
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
=2 π 45ðÞ=sqrt 22ðÞ¼795 70:7 ¼ 724:5 ¼ 724
Yield ¼ 1= 1+ 0:04 2ðÞðÞ
14
¼ 0:340
Profit ¼ 724 0:34 30 ¼ $7384:80
b. Red Dragon:
Dies per wafer ¼ π 45=2ðÞ
2
=2 π 45ðÞ=sqrt 2 1:2ðÞ¼1325 91:25 ¼ 1234
Yield ¼ 1= 1+ 0:04 1:2ðÞðÞ
14
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))
14
¼0.87
Equation for the probability that N are defect free on a chip:
#combinations (0.87)
N
(1 0.87)
8N
Yield for Phoenix
4
: (0.39 + 0.21 + 0.06 + 0.01) ¼0.57
Yield for Phoenix
2
: (0.001 + 0.0001) ¼ 0.0011
Yield for Phoenix
1
: 0.000004
b. It would be worthwhile to sell Phoenix
4
. However, the other two have such a
low probability of occurring that it is not worth selling them.
# defect-free # combinations Probability
8
7
6
5
4
3
2
1
0
1
0.32821167
0.39234499
0.20519192
0.06132172
0.01145377
0.00136919
0.0001023
4.3673E-06
8.1573E-08
8
28
56
70
56
28
8
1

c.
$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)
2
/Voltage
2
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)
2
/Voltage
2
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 0:2ðÞ0:75 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))
2
/Voltage
2
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
d.
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
10
and 1.55
10
, 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
4
, x ¼ 1.032, 3.2% growth
1.8 a. 50%
b. Energy: Energy
new
/Energy
old
¼(Voltage 1/2)
2
/Voltage
2
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 (Voltage0.8)
2
(Frequency 0.6)/½
Capacitance Voltage Frequency¼0.8
2
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
9
/100 ¼10
7
b. 10
7
/10
7
+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
12
10
8
6
4
2
0
Figure S.1 Plot of the equation: y
5
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
剩余141页未读,继续阅读


















安全验证
文档复制为VIP权益,开通VIP直接复制

评论0