TEST BANK FOR Computer Architecture A Quantitative Approach 4th Edition By John L. Hennessy

Chapter 1 Solutions
Yield 1 0.7 × 1.99
4.0
+ -------------------------
 
 
–4
= = 0.28
Yield 1 0.75 × 3.80
4.0
+ ----------------------------
 
 
–4
= = 0.12
Yield 1 0.30 × 3.89
4.0
+ ---------------------------
 
 
–4
= = 0.36
Dies per wafer π × (30 ⁄ 2)2
3.89
= ------------------------------ π × 30
sqrt(2 × 3.89)
– ---------------------------------- = 182 – 33.8 = 148
Cost per die $500
148 × 0.36
= -------------------------- = $9.38
Yield 1 .7 × 1.86
4.0
+ ---------------------
 
 
–4
= = 0.32
Dies per wafer π × (30 ⁄ 2)2
1.86
= ------------------------------ π × 30
sqrt(2 × 1.86)
– ---------------------------------- = 380 – 48.9 = 331
Cost per die $500
331 × .32
= ----------------------- = $4.72
Yield 1 .75 × 3.80 ⁄ 8
4.0
+ -------------------------------
 
 
–4
= = 0.71
Prob of error = 1 – 0.71 = 0.29
L.1 Chapter 1 Solutions

L
-
3
d.
0.51 ⁄ 0.06 = 8.5
e.
x
×
$150 + 8.5
x
×
$100 – (9.5
x
×
$80) – 9.5
x
×
$1.50 = $200,000,000
x
= 885,938 8-core chips, 8,416,390 chips total
Case Study 2: Power Consumption in Computer Systems
1.4 a.
.70
x
= 79 + 2
×
3.7 + 2
×
7.9
x
= 146
b.
4.0 W
×
.4 + 7.9 W
×
.6 = 6.34 W
c.
The 7200 rpm drive takes 60 s to read/seek and 40 s idle for a particular job.
The 5400 rpm disk requires 4/3
×
60 s, or 80 s to do the same thing. Therefore,
it is idle 20% of the time.
1.5 a.
b.
c.
1.6 a.
See Figure L.1.
b.
Sun Fire T2000
c.
More expensive servers can be more compact, allowing more computers to be
stored in the same amount of space. Because real estate is so expensive, this
is a huge concern. Also, power may not be the same for both systems. It can
cost more to purchase a chip that is optimized for lower power consumption.
1.7 a.
50%
b.
c.
Sun Fire T2000 IBM x346
SPECjbb 213 91.2
SPECweb 42.4 9.93
Figure L.1
Power/performance ratios.
14 KW
(79 W + 2.3 W + 7.0 W)
----------------------------------------------------------- = 158
14 KW
(79 W + 2.3 W + 2 × 7.0 W)
---------------------------------------------------------------------- = 146
MTTF
1
9 × 106 ------------------ + 8 × 1
4500
----------- 1
3 × 104 + ------------------ 8 × 2000 + 300
9 × 106 ------------------------------------- 16301
9 × 106 = = ------------------
1
Failure rate
--------------------------- 9 × 106
16301
= ------------------ = 522 hours
=
Power new
Power old
-------------------------- (V × 0.50)2 × (F × 0.50)
V2 × F
------------------------------------------------------------- 0.53 = = = 0.125
.70
(1 – x) + x ⁄ 2