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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Question
Chapter 5, Problem 5.9.3E
Program Plan Intro
Error detection code:
It enables the detection of an error in given data, instead of changing the correct position. Detecting single bit errors is referred as single bit “error detection code”. For this, Hamming uses a “parity code” for error detection.
Method of selecting bits:
The position of parity bit finds sequence of data bits, which checks the bits according to the following rules:
- For bit “1”, check the bits “1”, “3”, “5”, “7”, “9”, and so on. Because these bits contains the value as “1” in rightmost bit address.
- Consider an example, for bit 1 the binary representation is “0001” and for bit 3 the binary value is “0011” and so on.
- For bit “2”, check the bits “2”, “3”, “6”, “7”, “10”, and so on. Because these bits contains the value as “1” in second bit to the right of the bit address.
- Consider an example, for bit 2 the binary representation is “0010” and for bit 3 the binary value is “0011” and so on.
- For bit “4”, check the bits “4-7”, “12-15”, “20-23”, and so on. Because these bits contains the value as “1” in third bit to the right of the bit address.
- Consider an example, for bit 4 the binary representation is “0100” and for bit 5 the binary value is “0101” and so on.
- For bit “8”, check the bits “8-15”, “24-31”, “40-47”, and so on. Because these bits contains the value as “1” in fourth bit to the right of the bit address.
- Consider an example, for bit 8 the binary representation is “1000” and for bit 9 the binary value is “1001” and so on.
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Chapter 5 Solutions
Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design)
Ch. 5 - Prob. 5.1.1ECh. 5 - Prob. 5.1.2ECh. 5 - Prob. 5.1.3ECh. 5 - Prob. 5.1.4ECh. 5 - Prob. 5.1.5ECh. 5 - Prob. 5.1.6ECh. 5 - Prob. 5.2.1ECh. 5 - Prob. 5.2.2ECh. 5 - Prob. 5.2.3ECh. 5 - Prob. 5.2.4E
Ch. 5 - Prob. 5.2.5ECh. 5 - Prob. 5.2.6ECh. 5 - Prob. 5.3.1ECh. 5 - Prob. 5.3.2ECh. 5 - Prob. 5.3.3ECh. 5 - Prob. 5.3.4ECh. 5 - Prob. 5.3.5ECh. 5 - Prob. 5.3.6ECh. 5 - Prob. 5.4.1ECh. 5 - Prob. 5.4.2ECh. 5 - Prob. 5.4.3ECh. 5 - Prob. 5.4.5ECh. 5 - Prob. 5.4.6ECh. 5 - Prob. 5.5.1ECh. 5 - Prob. 5.5.2ECh. 5 - Prob. 5.5.3ECh. 5 - Prob. 5.5.4ECh. 5 - Prob. 5.5.5ECh. 5 - Prob. 5.5.6ECh. 5 - Prob. 5.6.1ECh. 5 - Prob. 5.6.2ECh. 5 - Prob. 5.6.3ECh. 5 - Prob. 5.6.4ECh. 5 - Prob. 5.6.5ECh. 5 - Prob. 5.6.6ECh. 5 - Prob. 5.7.1ECh. 5 - Prob. 5.7.2ECh. 5 - Prob. 5.7.3ECh. 5 - Prob. 5.7.4ECh. 5 - Prob. 5.7.5ECh. 5 - Prob. 5.7.6ECh. 5 - Prob. 5.8.1ECh. 5 - Prob. 5.8.2ECh. 5 - Prob. 5.8.3ECh. 5 - Prob. 5.8.4ECh. 5 - Prob. 5.9.1ECh. 5 - Prob. 5.9.2ECh. 5 - Prob. 5.9.3ECh. 5 - Prob. 5.10.1ECh. 5 - Prob. 5.10.2ECh. 5 - Prob. 5.10.3ECh. 5 - Prob. 5.10.4ECh. 5 - Prob. 5.10.5ECh. 5 - Prob. 5.10.6ECh. 5 - Prob. 5.11.1ECh. 5 - Prob. 5.11.2ECh. 5 - Prob. 5.11.3ECh. 5 - Prob. 5.11.4ECh. 5 - Prob. 5.11.5ECh. 5 - Prob. 5.11.6ECh. 5 - Prob. 5.12.1ECh. 5 - Prob. 5.12.2ECh. 5 - Prob. 5.12.3ECh. 5 - Prob. 5.12.4ECh. 5 - Prob. 5.12.5ECh. 5 - Prob. 5.12.6ECh. 5 - Prob. 5.13.1ECh. 5 - Prob. 5.13.2ECh. 5 - Prob. 5.13.3ECh. 5 - Prob. 5.13.4ECh. 5 - Prob. 5.13.5ECh. 5 - Prob. 5.13.6ECh. 5 - Prob. 5.14.1ECh. 5 - Prob. 5.14.2ECh. 5 - Prob. 5.14.3ECh. 5 - Prob. 5.14.4ECh. 5 - Prob. 5.14.5ECh. 5 - Prob. 5.14.6ECh. 5 - Prob. 5.15.1ECh. 5 - Prob. 5.15.2ECh. 5 - Prob. 5.15.3ECh. 5 - Prob. 5.15.4ECh. 5 - Prob. 5.16.1ECh. 5 - Prob. 5.16.2ECh. 5 - Prob. 5.17.1ECh. 5 - Prob. 5.17.2ECh. 5 - Prob. 5.17.3ECh. 5 - Prob. 5.17.4ECh. 5 - Prob. 5.17.5ECh. 5 - Prob. 5.17.6ECh. 5 - Prob. 5.18.1ECh. 5 - Prob. 5.18.2ECh. 5 - Prob. 5.18.3ECh. 5 - Prob. 5.18.4ECh. 5 - Prob. 5.18.5ECh. 5 - Prob. 5.18.6ECh. 5 - Prob. 5.19.1ECh. 5 - Prob. 5.19.2ECh. 5 - Prob. 5.19.3ECh. 5 - Prob. 5.19.4ECh. 5 - Prob. 5.19.6E
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