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Perform the following hexadecimal additions and subtractions. Assume the numbers are stored in 32-bit 2’s complement binary numbers. Indicate the sign of the answer and whether overflow occurs. c. E9B20F5D – FE605C8D Solution:-         E9B20F5D       – FE605C8D         =  -14AE4D30  (How calculate this value, show the process???) now,   -14AE4D30 = 0001 0100 1010 1110 0100 1101 0011 0000 now the value in 2' complement is: 1110 1011 0101 0001 1011 0010  1100  1111_________________________________+11110 1011 0101 0001 1011 0010  1101 0000 the final result is:- 1110 1011 0101 0001 1011 0010 1101  0000 => EB51B2D0 The final answer is EB51B2D0 with a positive sign and no overflow.

Perform the following hexadecimal additions and subtractions. Assume the numbers are stored in 32-bit 2’s complement binary numbers. Indicate the sign of the answer and whether overflow occurs. c. E9B20F5D – FE605C8D Solution:-         E9B20F5D       – FE605C8D         =  -14AE4D30  (How calculate this value, show the process???) now,   -14AE4D30 = 0001 0100 1010 1110 0100 1101 0011 0000 now the value in 2' complement is: 1110 1011 0101 0001 1011 0010  1100  1111_________________________________+11110 1011 0101 0001 1011 0010  1101 0000 the final result is:- 1110 1011 0101 0001 1011 0010 1101  0000 => EB51B2D0 The final answer is EB51B2D0 with a positive sign and no overflow.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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11. Perform the following hexadecimal additions and subtractions. Assume the
numbers are stored in 32-bit 2’s complement binary numbers. Indicate the sign of
the answer and whether overflow occurs.

c. E9B20F5D – FE605C8D

Solution:-

        E9B20F5D       – FE605C8D         =  -14AE4D30  (How calculate this value, show the process???)

now, 

 -14AE4D30 = 0001 0100 1010 1110 0100 1101 0011 0000

now the value in 2' complement is:

1110 1011 0101 0001 1011 0010  1100  1111_________________________________+11110 1011 0101 0001 1011 0010  1101 0000

the final result is:-

1110 1011 0101 0001 1011 0010 1101  0000 => EB51B2D0

The final answer is EB51B2D0 with a positive sign and no overflow.

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