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 B, Problem 9E
Explanation of Solution
A3 A2 A1 A0 F 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0
- In the above table, A0, A1, A2 and A3 represent four different inputs and F represents odd parity function.
- In odd parity, if the function includes odd number of ones then the function has value one and if the function includes even number of ones then the function has zero value.
- If A3=0, A2=0, A1=0 and A0=0 then F=0 because the number of 1 inputs is 0 which is an even number.
- If A3=0, A2=0, A1=0 and A0=1 then F=1 because the number of 1 inputs is 1 which is an odd number.
- If A3=0, A2=0, A1=1 and A0=0 then F=1 because the number of 1 inputs is 1 which is an odd number.
- If A3=0, A2=0, A1=1 and A0=1 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=0, A2=1, A1=0 and A0=0 then F=1 because the number of 1 inputs is 1 which is an odd number.
- If A3=0, A2=1, A1=0 and A0=1 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=0, A2=1, A1=1 and A0=0 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=0, A2=1, A1=1 and A0=1 then F=1 because the number of 1 inputs is 3 which is an odd number.
- If A3=1, A2=0, A1=0 and A0=0 then F=1 because the number of 1 inputs is 1 which is an odd number.
- If A3=1, A2=0, A1=0 and A0=1 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=1, A2=0, A1=1 and A0=0 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=1, A2=0, A1=1 and A0=1 then F=1 because the number of 1 inputs is 3 which is an odd number.
- If A3=1, A2=1, A1=0 and A0=0 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=1, A2=1, A1=0 and A0=1 then F=1 because the number of 1 inputs is 3 which is an odd number.
- If A3=1, A2=1, A1=1 and A0=0 then F=1 because the number of 1 inputs is 3 which is an odd number.
- If A3=1, A2=1, A1=1 and A0=1 then F=0 because the number of 1 inputs is 4 which is an even number.
| A3 | A2 | A1 | A0 | F |
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 0 |
Expert Solution & Answer
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Students have asked these similar questions
• Analyze the logs to determine the consistency model implemented in the system.
• Compare the observed behavior with theoretical models like eventual, strong, or causal
consistency.
• Suggest potential improvements to balance consistency, availability, and partition tolerance (CAP
theorem).
Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qo Hazb9tC440 AZF/view?usp=sharing]
Refer to page 120 for problems on cellular network design.
Instructions:
•
Explain the concept of frequency reuse in cellular networks and its impact on capacity.
•
•
Analyze a given cell layout and calculate the reuse factor.
Discuss challenges of interference management in dense cellular environments.
Link: [https://drive.google.com/file/d/1wKSrun-GlxirS3IZ9qoHazb9tC440AZF/view?usp=sharing]
Refer to page 115 for problems on user interface design.
Instructions:
•
Evaluate a given user interface design for cognitive load using established heuristics.
• Propose improvements to reduce cognitive load and enhance usability.
•
Discuss the relationship between user experience and cognitive psychology principles.
Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qoHazb9tC440AZF/view?usp=sharing]
Chapter B Solutions
Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design)
Ch. B - Prob. 1ECh. B - Prob. 2ECh. B - Prob. 3ECh. B - Prob. 4ECh. B - Prob. 5ECh. B - Prob. 6ECh. B - Prob. 7ECh. B - Prob. 8ECh. B - Prob. 9ECh. B - Prob. 10E
Ch. B - Prob. 11ECh. B - Prob. 14ECh. B - Prob. 15ECh. B - Prob. 16ECh. B - Prob. 17ECh. B - Prob. 19ECh. B - Prob. 20ECh. B - Prob. 25ECh. B - Prob. 26ECh. B - Prob. 27ECh. B - Prob. 28ECh. B - Prob. 29ECh. B - Prob. 30ECh. B - Prob. 31ECh. B - Prob. 32ECh. B - Prob. 33ECh. B - Prob. 34ECh. B - Prob. 35ECh. B - Prob. 36ECh. B - Prob. 37ECh. B - Prob. 38ECh. B - Prob. 39ECh. B - Prob. 40ECh. B - Prob. 41ECh. B - Prob. 42ECh. B - Prob. 43E
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