Computer architecture, a
quantitative approach (solution for
5th edition)
Advanced Computer Architecture
Karunya University
91 pag.
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, 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.
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, Solutions to Case Studies
and Exercises 1
Copyright © 2012 Elsevier, Inc. All rights reserved.
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, 2 ■ Solutions to Case Studies and Exercises
Chapter 1 Solutions
Case Study 1: Chip Fabrication Cost 1
0.30 × 3.89 –4
1.1 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.
2
π × ( 30 ⁄ 2 ) π × 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
2
π × ( 30 ⁄ 2 ) π × 30
b. Dies per wafer = ----------------------------- – ------------------------------- = 283 – 42.1 = 240
2.5 sqrt ( 2 × 2.5 )
0.30 × 2.5 – 4
Yield = ⎛ 1 + --------------------------⎞ = 0.50
⎝ 4.0 ⎠
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.
0.75 × 1.99 ⁄ 2 –4
1.3 a. Defect – Free single core = ⎛ 1 + ----------------------------------⎞ = 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
x = -------------------------------------------------- = ------------------------- = $23.33
1/2 × old dpw × 0.48 1/2 × 0.48
Copyright © 2012 Elsevier, Inc. All rights reserved.
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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/
Downloaded by: kareey ()
, 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/
Downloaded by: kareey ()
, Solutions to Case Studies
and Exercises 1
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/
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, 2 ■ Solutions to Case Studies and Exercises
Chapter 1 Solutions
Case Study 1: Chip Fabrication Cost 1
0.30 × 3.89 –4
1.1 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.
2
π × ( 30 ⁄ 2 ) π × 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
2
π × ( 30 ⁄ 2 ) π × 30
b. Dies per wafer = ----------------------------- – ------------------------------- = 283 – 42.1 = 240
2.5 sqrt ( 2 × 2.5 )
0.30 × 2.5 – 4
Yield = ⎛ 1 + --------------------------⎞ = 0.50
⎝ 4.0 ⎠
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.
0.75 × 1.99 ⁄ 2 –4
1.3 a. Defect – Free single core = ⎛ 1 + ----------------------------------⎞ = 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
x = -------------------------------------------------- = ------------------------- = $23.33
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/
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