Exam (elaborations) Computer Architecture (CSY1017) Computer Architecture, ISBN: 9780128119051

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Computer architecture, A Quantitative
Approach (solution for 5th edition)
Computer Communications (University of
Northampton)

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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

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Solutions to Case Studies and Exercises

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2 ■ Solutions to Case Studies and Exercises


Chapter 1 Solutions

Case Study 1: Chip Fabrication Cost
1.1 0.30  3.89
a. 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
  30  2   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.

1.3 a. 0.75  1.99  2 –
4
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
x = 1/2  old dpw $20 
 0.48 = 0.28 1/2 = $23.33
 0.48