Computer architecture, a
quantitative approach (solution for
5th edition)
Advanced Computer Architecture
Karunya University
91 pag.
Document shared on https://www.docsity.com/en/computer-architecture-a-quantitative-approach-solution-for-5th-edition/581185/
, Chapter 1 Solutions 2
Chapter 2 Solutions 6
Chapter 3 Solutions 13
Chapter 4 Solutions 33
Chapter 5 Solutions 44
Chapter 6 Solutions 50
Appendix A Solutions 63
Appendix B Solutions 83
Appendix C Solutions 92
Copyright © 2012 Elsevier, Inc. All rights reserved.
Document shared on https://www.docsity.com/en/computer-architecture-a-quantitative-approach-solution-for-5th-edition/581185/
, Solutions to Case Studies
and Exercises
Copyright © 2012 Elsevier, Inc. All rights reserved.
Document shared on https://www.docsity.com/en/computer-architecture-a-quantitative-approach-solution-for-5th-edition/581185/
, 2 ■ Solutions to Case Studies and Exercises
Chapter 1 Solutions
Case Study 1: Chip Fabrication Cost
1.1 0.30 3.89 –4
a. Yield = 1 + ------------------------- = 0.36
4.0
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.
(30 2)2 30
1.2 a. Dies per wafer = ----------------------------- – ------------------------------ = 471 – 54.4 = 416
1.5 sqrt(2 1.5)
0.30 1.5 –4
Yield = 1 + ----------------------- = 0.65
4.0
Profit = 416 0.65 $20 = $5408
(30 2) 30
2
b. Dies per wafer = ----------------------------- – ------------------------------ = 283 – 42.1 = 240
2.5 sqrt(2 2.5)
0.30 2.5 –4
Yield = 1 + -------------------------
4.0 = 0.50
Profit = 240 0.50 $25 = $3000
c. The Woods chip
d. Woods chips: 50,000/416 = 120.2 wafers needed
Markon chips: 25,000/240 = 104.2 wafers needed
Therefore, the most lucrative split is 120 Woods wafers, 30 Markon wafers.
1.3 0.75 1.99 2
a. Defect – Free single core = 1 + ---------------------------------
–4
= 0.28
4.0
No defects = 0.282 = 0.08
One defect = 0.28 0.72 2 = 0.40
No more than one defect = 0.08 + 0.40 = 0.48
Wafer size
b. $20 =
old dpw 0.28
$20 0.28 = Wafer size/old dpw
Wafer size $20 0.28 = $23.33
x = =
1/2 old dpw 0.48 1/2 0.48
Copyright © 2012 Elsevier, Inc. All rights reserved.
Document shared on https://www.docsity.com/en/computer-architecture-a-quantitative-approach-solution-for-5th-edition/581185/
quantitative approach (solution for
5th edition)
Advanced Computer Architecture
Karunya University
91 pag.
Document shared on https://www.docsity.com/en/computer-architecture-a-quantitative-approach-solution-for-5th-edition/581185/
, Chapter 1 Solutions 2
Chapter 2 Solutions 6
Chapter 3 Solutions 13
Chapter 4 Solutions 33
Chapter 5 Solutions 44
Chapter 6 Solutions 50
Appendix A Solutions 63
Appendix B Solutions 83
Appendix C Solutions 92
Copyright © 2012 Elsevier, Inc. All rights reserved.
Document shared on https://www.docsity.com/en/computer-architecture-a-quantitative-approach-solution-for-5th-edition/581185/
, Solutions to Case Studies
and Exercises
Copyright © 2012 Elsevier, Inc. All rights reserved.
Document shared on https://www.docsity.com/en/computer-architecture-a-quantitative-approach-solution-for-5th-edition/581185/
, 2 ■ Solutions to Case Studies and Exercises
Chapter 1 Solutions
Case Study 1: Chip Fabrication Cost
1.1 0.30 3.89 –4
a. Yield = 1 + ------------------------- = 0.36
4.0
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.
(30 2)2 30
1.2 a. Dies per wafer = ----------------------------- – ------------------------------ = 471 – 54.4 = 416
1.5 sqrt(2 1.5)
0.30 1.5 –4
Yield = 1 + ----------------------- = 0.65
4.0
Profit = 416 0.65 $20 = $5408
(30 2) 30
2
b. Dies per wafer = ----------------------------- – ------------------------------ = 283 – 42.1 = 240
2.5 sqrt(2 2.5)
0.30 2.5 –4
Yield = 1 + -------------------------
4.0 = 0.50
Profit = 240 0.50 $25 = $3000
c. The Woods chip
d. Woods chips: 50,000/416 = 120.2 wafers needed
Markon chips: 25,000/240 = 104.2 wafers needed
Therefore, the most lucrative split is 120 Woods wafers, 30 Markon wafers.
1.3 0.75 1.99 2
a. Defect – Free single core = 1 + ---------------------------------
–4
= 0.28
4.0
No defects = 0.282 = 0.08
One defect = 0.28 0.72 2 = 0.40
No more than one defect = 0.08 + 0.40 = 0.48
Wafer size
b. $20 =
old dpw 0.28
$20 0.28 = Wafer size/old dpw
Wafer size $20 0.28 = $23.33
x = =
1/2 old dpw 0.48 1/2 0.48
Copyright © 2012 Elsevier, Inc. All rights reserved.
Document shared on https://www.docsity.com/en/computer-architecture-a-quantitative-approach-solution-for-5th-edition/581185/
