Chapter 1 Solutions
Case Study 1: Chip Fabrication Cost
1.1 a. Yield¼1/(1 + (0.042))14¼0.34
b. It is fabricated in a larger technology, which is an
... [Show More] older plant. As plants age,
their process gets tuned, and the defect rate decreases.
1.2 a. Phoenix:
Dies per wafer ¼ πð Þ 45=2 2 =2ð Þ π45 =sqrt 2ð Þ¼ 2 79570:7 ¼ 724:5 ¼ 724
Yield ¼ 1=ð Þ 1+ 0ð Þ :04 2 14 ¼ 0:340
Profit ¼ 724 0:34 30 ¼ $7384:80
b. Red Dragon:
Dies per wafer ¼ πð Þ 45=2 2 =2ð Þ π45 =sqrt 2ð Þ¼ 1:2 132591:25 ¼ 1234
Yield ¼ 1=ð Þ 1+ 0ð Þ :04 1:2 14 ¼ 0:519
Profit ¼ 1234 0:519 15 ¼ $9601:71
c. Phoenix chips: 25,000/724¼34.5 wafers needed
Red Dragon chips: 50,000/1234¼40.5 wafers needed
Therefore, the most lucrative split is 40 Red Dragon wafers, 30 Phoenix wafers.
1.3 a. Defect-free single core¼Yield¼1/(1 + (0.040.25))14¼0.87
Equation for the probability that N are defect free on a chip:
#combinations (0.87)N (10.87)8N
Yield for Phoenix4
: (0.39 + 0.21 + 0.06 + 0.01)¼0.57
Yield for Phoenix2
: (0.001 + 0.0001)¼0.0011
Yield for Phoenix1
: 0.000004
b. It would be worthwhile to sell Phoenix4
. 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 dpw0:28
Step 1: Determine how many Phoenix4 chips are produced for every
Phoenix8 chip.
There are 57/33 Phoenix4 chips for every Phoenix8 chip¼1.73
$30 + 1:73 $25 ¼ $73:25
Case Study 2: Power Consumption in Computer Systems
1.4 a. Energy: 1/8. Power: Unchanged.
b. Energy: Energynew/Energyold¼(Voltage 1/8)2
/Voltage2
¼0.156
Power: Powernew/Powerold¼0.156 (Frequency 1/8)/Frequency¼0.00195
c. Energy: Energynew/Energyold¼(Voltage 0.5)2
/Voltage2
¼0.25
Power: Powernew/Powerold¼0.25 (Frequency 1/8)/Frequency¼0.0313
d. 1 core¼25% of the original power, running for 25% of the time.
0:25 0:25 + 0ð Þ :25 0:2 0:75 ¼ 0:0625 + 0:0375 ¼ 0:1
1.5 a. Amdahl’s law: 1/(0.8/4 + 0.2)¼1/(0.2 + 0.2)¼1/0.4¼2.5
b. 4 cores, each at 1/(2.5) the frequency and voltage
Energy: Energyquad/Energysingle¼4 (Voltage 1/(2.5))2
/Voltage2
¼0.64
Power: Powernew/Powerold¼0.64 (Frequency 1/(2.5))/Frequency¼0.256
c. 2 cores + 2 ASICs vs. 4 cores
ð Þ 2+ 0ð Þ :2 2 =4 ¼ ð Þ 2:4 =4 ¼ 0:6
1.6 a. Workload A speedup: 225,000/13,461¼16.7
Workload B speedup: 280,000/36,465¼7.7
1/(0.7/16.7 + 0.3/7.7)
b. General-purpose: 0.70 0.42 + 0.30¼0.594
GPU: 0.70 0.37 + 0.30¼0.559
TPU: 0.70 0.80 + 0.30¼0.886
c. General-purpose: 159 W + (455 W159 W) 0.594¼335 W
GPU: 357 W + (991 W357 W) 0.559¼711 W
TPU: 290 W + (384 W290 W) 0.86¼371 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)¼1.67
TPU: 1/(0.4/41 + 0.1/21.2 + 0.5/0.17)¼0.33
e. General-purpose: 14,000/504 ¼ 27.8 28
GPU: 14,000/1838¼7.628
TPU: 14,000/861¼16.317
d. General-purpose: 2200/504¼4.374, 14,000/(4 504)¼6.747
GPU: 2200/1838¼1.21, 14,000/(1 1838)¼7.628
TPU: 2200/861¼2.562, 14,000/(2 861)¼8.139
Exercises
1.7 a. Somewhere between 1.410 and 1.5510, or 28.980x
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. 2¼x
4
, x¼1.032, 3.2% growth
1.8 a. 50%
b. Energy: Energynew/Energyold¼(Voltage 1/2)2
/Voltage2
¼0.25
1.9 a. 60%
b. 0.4 + 0.60.2¼0.58, which reduces the energy to 58% of the original energy
c. newPower/oldPower¼½Capacitance(Voltage0.8)2
(Frequency0.6)/½
CapacitanceVoltageFrequency¼0.82
0.6¼0.256 of the original power.
d. 0.4 + 0.32¼0.46, which reduces the energy to 46% of the original energy
1.10 a. 109
/100¼107
b. 107
/107
+ 24¼1
c. [need solution]
1.11 a. 35/10,0003333¼11.67 days
b. There are several correct answers. 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. $90,000¼(x+x+x+ 2x)/4
$360,000¼5x
$72,000¼x
4th quarter¼$144,000/h
Chapter 1 Solutions ■ 3
1.12 a. See Figure S.1.
b. 2¼1/((1x) +x/20)
10/19¼x¼52.6%
c. (0.526/20)/(0.474 + 0.526/20)¼5.3%
d. Extra speedup with 2 units: 1/(0.1 + 0.9/2)¼1.82. 1.82 2036.4.
Total speedup: 1.95. Extra speedup with 4 units: 1/(0.1 + 0.9/4)¼3.08.
3.08 2061.5. Total speedup: 1.97
1.13 a. old execution time¼0.5 new + 0.510 new¼5.5 new
b. In the original code, the unenhanced part is equal in time to the enhanced part
(sped up by 10), therefore:
(1x)¼x/10
1010x¼x
10¼11x
10/11¼x¼0.91
1.14 a. 1/(0.8 + 0.20/2)¼1.11
b. 1/(0.7 + 0.20/2 + 0.103/2)¼1.05
c. fp ops: 0.1/0.95¼10.5%, cache: 0.15/0.95¼15.8%
1.15 a. 1/(0.5 + 0.5/22)¼1.91 [Show Less]