computer-architecture-a-quantitative-approach-solution-for-5th-edition

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




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
1.1 a.
0.30  3.89
Yield = 1 + -------------------------
–4

 4.0  = 0.36

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
2
b.   30  2 – ------------------------------
  30
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.

1.3 a.
0.75  1.99  2
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

b. $20 = Wafer size
old dpw 
0.28
$20  0.28 = Wafer size/old dpw
Wafer size $20 
x =
0.28 = = $23.33
1/2  old dpw  0.48 1/2  0.48


Copyright © 2012 Elsevier, Inc. All rights
reserved.
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