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.
Solutions to Case Studies
and Exercises
1
Copyright © 2012 Elsevier, Inc. All rights reserved.
2 ■ Solutions to Case Studies and Exercises
Case Study 1: Chip Fabrication Cost
1
1.1 a.
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.
1.2 a.
Profit = 416 × 0.65 × $20 = $5408
b.
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.
No defects = 0.28
2
= 0.08
One defect = 0.28 × 0.72 × 2 = 0.40
No more than one defect = 0.08 +
0.40 = 0.48
b.
$20 × 0.28 = Wafer size/old dpw
= $23.33
Chapter 1 Solutions
Yield 1
0.30 3.89×
4.0
---------------------------+
⎝⎠
⎛⎞
4–
0.36==
Dies per wafer
π 30 2⁄()×
2
1.5
-----------------------------=
π 30×
sqrt 2 1.5×()
-------------------------------– 471 54.4 416=–=
Yield 1
0.30 1.5×
4.0
------------------------+
⎝⎠
⎛⎞
4–
0.65==
Dies per wafer
π 30 2⁄()×
2
2.5
-----------------------------=
π 30×
sqrt 2 2.5×()
-------------------------------– 283 42.1 240=–=
Yield 1
0.30 2.5×
4.0
--------------------------+
⎝⎠
⎛⎞
4–
0.50==
Defect – Free single core 1
0.75 1.99 2⁄×
4.0
----------------------------------+
⎝⎠
⎛⎞
4–
0.28==
$20
Wafer size
old dpw 0.28×
------------------------------------=
x
Wafer size
1/2 old dpw× 0.48×
--------------------------------------------------
$20 0.28×
1/2 0.48×
-------------------------==
Copyright © 2012 Elsevier, Inc. All rights reserved.
Chapter 1 Solutions ■ 3
Case Study 2: Power Consumption in Computer Systems
1.4 a. .80x = 66 + 2 × 2.3 + 7.9; x = 99
b. .6 × 4 W + .4 × 7.9 = 5.56
c. Solve the following four equations:
seek7200 = .75 × seek5400
seek7200 + idle7200 = 100
seek5400 + idle5400 = 100
seek7200 × 7.9 + idle7200 × 4 = seek5400 × 7 + idle5400 × 2.9
idle7200 = 29.8%
1.5 a.
b.
c. 200 W × 11 = 2200 W
2200/(76.2) = 28 racks
Only 1 cooling door is required.
1.6 a. The IBM x346 could take less space, which would save money in real estate.
The racks might be better laid out. It could also be much cheaper. In addition,
if we were running applications that did not match the characteristics of these
benchmarks, the IBM x346 might be faster. Finally, there are no reliability
numbers shown. Although we do not know that the IBM x346 is better in any
of these areas, we do not know it is worse, either.
1.7 a. (1 – 8) + .8/2 = .2 + .4 = .6
b.
c. ;
x = 50%
d.
Exercises
1.8 a. (1.35)
10
= approximately 20
b. 3200 × (1.4)
12
= approximately 181,420
c. 3200 × (1.01)
12
= approximately 3605
d. Power density, which is the power consumed over the increasingly small area,
has created too much heat for heat sinks to dissipate. This has limited the
activity of the transistors on the chip. Instead of increasing the clock rate,
manufacturers are placing multiple cores on the chip.
14 KW
66 W2.3 W7.9 W++()
------------------------------------------------------------183=
14 KW
66 W 2.3 W 2+ + 7.9 W×()
--------------------------------------------------------------------- 1 6 6=
Power new
Power old
--------------------------
V 0.60×()
2
F 0.60×()×
V
2
F×
-----------------------------------------------------------0.6
3
0.216===
1
.75
1 x–()x 2⁄+
--------------------------------=
Power new
Power old
--------------------------
V 0.75×()
2
F 0.60×()×
V
2
F×
-----------------------------------------------------------0.75
2
0.6× 0.338===
Copyright © 2012 Elsevier, Inc. All rights reserved.
4 ■ Solutions to Case Studies and Exercises
e. Anything in the 15–25% range would be a reasonable conclusion based on
the decline in the rate over history. As the sudden stop in clock rate shows,
though, even the declines do not always follow predictions.
1.9 a. 50%
b. Energy = ½ load × V
2
. Changing the frequency does not affect energy–only
power. So the new energy is ½ load × ( ½ V)
2
, reducing it to about ¼ the old
energy.
1.10 a. 60%
b. 0.4 + 0.6 × 0.2 = 0.58, which reduces the energy to 58% of the original
energy.
c. newPower/oldPower = ½ Capacitance × (Voltage × .8)
2
× (Frequency × .6)/ ½
Capacitance × Voltage × Frequency = 0.8
2
× 0.6 = 0.256 of the original power.
d. 0.4 + 0 .3 × 2 = 0.46, which reduce the energy to 46% of the original energy.
1.11 a. 10
9
/100 = 10
7
b. 10
7
/10
7
+ 24 = 1
c. [need solution]
1.12 a. 35/10000 × 3333 = 11.67 days
b. There are several correct answers. One would be that, with the current sys-
tem, one computer fails approximately every 5 minutes. 5 minutes is unlikely
to be enough time to isolate the computer, swap it out, and get the computer
back on line again. 10 minutes, 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. $90,000 = (x + x + x + 2x)/4
$360,000 = 5x
$72,000 = x
4th quarter = $144,000/hr
1.13 a. Itanium, because it has a lower overall execution time.
b. Opteron: 0.6 × 0.92 + 0.2 × 1.03 + 0.2 × 0.65 = 0.888
c. 1/0.888 = 1.126
1.14 a. See Figure S.1.
b. 2 = 1/((1 – x) + x
/10)
5/9 = x = 0.56 or 56%
c. 0.056/0.5 = 0.11 or 11%
d. Maximum speedup = 1/(1/10) = 10
5 = 1/((1 – x) + x
/10)
8/9 = x = 0.89 or 89%
Copyright © 2012 Elsevier, Inc. All rights reserved.
