Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design)
5th Edition
ISBN: 9780124077263
Author: David A. Patterson, John L. Hennessy
Publisher: Elsevier Science
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Expert Solution & Answer
Chapter 1, Problem 1.10.4E
Explanation of Solution
The defects per area can be calculated using the following formula
Substitute, “0.92” for “yield” and “2” for “die_area” is given below:
Expert Solution & Answer
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Check out a sample textbook solution
Students have asked these similar questions
1. Show that S₁ = S2 if and only if S₁ U S2 = S₁n S2.
2. Show that if S1 and S2 are finite sets with |S₁ = n and |S₂| = m, then
S₁ U S2 ≤n+m.
3. If language L is regular and is given by L ={awa : w = {a, b} *}
Then show that L² is regular.
Problem 3: Consider the two-link planar elbow manipulator shown below with link information: a₁ = 3,
a₂ = 3.
a2
S
3/0
Link | a | ai | di
ai
| di | 0i |
มา
02
21
02.
12
01
0 0 01
02
0
0
02
01
To
(a) Find the location of the end-effector P° if the joint angles are:
0₁ = 140°, 0₂ = 30°
(b) Find the values of joint variables 01, 02 if the robotic manipulator is commanded to reach the desired
location of Pº = [4.5, 3,0]. Notice there are possibly two sets of solutions.
(c) Also sketch the two poses of the robot for the computed joint variables.
Page 5 of 7
Problem 5: The following shows the schematic of a SCARA robot and its DH parameters. Complete the
following questions based on this information.
20
01
21
02.
22
done by
ai
α;
di
Ꮎ ;
21
d3
1
a1
0
0
0₁**
2
a2
180°
0
0%
3
0
0
d3
0
The Jacobian matrix of this robot is given below:
-a1S1a2S12
-a2S12
J(q) =
a1c1 + a2c12
0
a2C12
0
10°
(a) For a₁ = 1, a2 = 2, is q = 180° a singularity configuration of this robot?
a1
3
[10°1
(b) For a₁ = 1, a2 = 2, is q = | 0° a singularity configuration of this robot?
3
Chapter 1 Solutions
Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design)
Ch. 1 - Prob. 1.1ECh. 1 - Prob. 1.2ECh. 1 - Prob. 1.3ECh. 1 - Prob. 1.4ECh. 1 - Prob. 1.5ECh. 1 - Prob. 1.6ECh. 1 - Prob. 1.7ECh. 1 - Prob. 1.8.1ECh. 1 - Prob. 1.8.2ECh. 1 - Prob. 1.8.3E
Ch. 1 - Prob. 1.9.1ECh. 1 - Prob. 1.9.2ECh. 1 - Prob. 1.9.3ECh. 1 - Prob. 1.10.1ECh. 1 - Prob. 1.10.2ECh. 1 - Prob. 1.10.3ECh. 1 - Prob. 1.10.4ECh. 1 - Prob. 1.11.1ECh. 1 - Prob. 1.11.2ECh. 1 - Prob. 1.11.3ECh. 1 - Prob. 1.11.4ECh. 1 - Prob. 1.11.5ECh. 1 - Prob. 1.11.6ECh. 1 - Prob. 1.11.7ECh. 1 - Prob. 1.11.8ECh. 1 - Prob. 1.11.9ECh. 1 - Prob. 1.11.10ECh. 1 - Prob. 1.11.11ECh. 1 - Prob. 1.12.1ECh. 1 - Prob. 1.12.2ECh. 1 - Prob. 1.12.3ECh. 1 - Prob. 1.12.4ECh. 1 - Prob. 1.13.1ECh. 1 - Prob. 1.13.2ECh. 1 - Prob. 1.13.3ECh. 1 - Prob. 1.14.1ECh. 1 - Prob. 1.14.2ECh. 1 - Prob. 1.14.3ECh. 1 - Prob. 1.15E
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