Essentials of Computer Organization and Architecture
Essentials of Computer Organization and Architecture
4th Edition
ISBN: 9781284074482
Author: Linda Null, Julia Lobur
Publisher: Jones & Bartlett Learning
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Chapter 2, Problem 23E

a)

Explanation of Solution

Determining the decimal equivalent for unsigned number 11111100:

Step 1: Multiply the given binary number with the positive integers for base 2.

27×1+26×1+25×1+24×1+23×1+22×1+21×0+20×0

Step 2: Add all the values to get the final result.

128+64+32+16+8+4=252

Therefore, the decimal equivalent of the number represented in unsigned number 11111100 is 252.

Determining the decimal equivalent for unsigned number 01110000:

Step 1: Multiply the given binary number with the positive integers for base 2

b)

Program Plan Intro

Two’s complement:

Two’s complement is another way of representing the binary numbers. To implement the two’s complement to the number, the given binary number should be one’s complemented and then add 1 to the result obtained after one’s complement.

c)

Explanation of Solution

Interpreting 11111100 as a signed magnitude representation:

In the given binary number, the MSB bit is 1. Hence, the binary number is negative.

Step 1: Multiply the given binary number with the positive integers for base 2.

26×1+25×1+24×1+23×1+22×1+21×0+20×0

Step 2: Add all the values to get the final result.

64+32+16+8+4=124

Therefore, the decimal equivalent of the number represented in signed magnitude is 124

Interpreting 01110000 as a signed magnitude representation:

Step 1: Multiply the given binary number with the positive integers for base 2

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Chapter 2 Solutions

Essentials of Computer Organization and Architecture

Ch. 2 - Prob. 7RETCCh. 2 - Prob. 8RETCCh. 2 - Prob. 9RETCCh. 2 - Prob. 10RETCCh. 2 - Prob. 11RETCCh. 2 - Prob. 12RETCCh. 2 - Prob. 13RETCCh. 2 - Prob. 14RETCCh. 2 - Prob. 15RETCCh. 2 - Prob. 16RETCCh. 2 - Prob. 17RETCCh. 2 - Prob. 18RETCCh. 2 - Prob. 19RETCCh. 2 - Prob. 20RETCCh. 2 - Prob. 21RETCCh. 2 - Prob. 22RETCCh. 2 - Prob. 23RETCCh. 2 - Prob. 24RETCCh. 2 - Prob. 25RETCCh. 2 - Prob. 26RETCCh. 2 - Prob. 27RETCCh. 2 - Prob. 28RETCCh. 2 - Prob. 29RETCCh. 2 - Prob. 30RETCCh. 2 - Prob. 31RETCCh. 2 - Prob. 32RETCCh. 2 - Prob. 33RETCCh. 2 - Prob. 34RETCCh. 2 - Prob. 1ECh. 2 - Prob. 2ECh. 2 - Prob. 3ECh. 2 - Prob. 4ECh. 2 - Prob. 5ECh. 2 - Prob. 6ECh. 2 - Prob. 7ECh. 2 - Prob. 8ECh. 2 - Prob. 9ECh. 2 - Prob. 10ECh. 2 - Prob. 11ECh. 2 - Prob. 12ECh. 2 - Prob. 13ECh. 2 - Prob. 14ECh. 2 - Prob. 15ECh. 2 - Prob. 16ECh. 2 - Prob. 17ECh. 2 - Prob. 18ECh. 2 - Prob. 19ECh. 2 - Prob. 20ECh. 2 - Prob. 21ECh. 2 - Prob. 22ECh. 2 - Prob. 23ECh. 2 - Prob. 24ECh. 2 - Prob. 25ECh. 2 - Prob. 26ECh. 2 - Prob. 27ECh. 2 - Prob. 29ECh. 2 - Prob. 30ECh. 2 - Prob. 31ECh. 2 - Prob. 32ECh. 2 - Prob. 33ECh. 2 - Prob. 34ECh. 2 - Prob. 35ECh. 2 - Prob. 36ECh. 2 - Prob. 37ECh. 2 - Prob. 38ECh. 2 - Prob. 39ECh. 2 - Prob. 40ECh. 2 - Prob. 41ECh. 2 - Prob. 42ECh. 2 - Prob. 43ECh. 2 - Prob. 44ECh. 2 - Prob. 45ECh. 2 - Prob. 46ECh. 2 - Prob. 47ECh. 2 - Prob. 48ECh. 2 - Prob. 49ECh. 2 - Prob. 50ECh. 2 - Prob. 51ECh. 2 - Prob. 52ECh. 2 - Prob. 53ECh. 2 - Prob. 54ECh. 2 - Prob. 55ECh. 2 - Prob. 56ECh. 2 - Prob. 57ECh. 2 - Prob. 58ECh. 2 - Prob. 59ECh. 2 - Prob. 60ECh. 2 - Prob. 61ECh. 2 - Prob. 62ECh. 2 - Prob. 63ECh. 2 - Prob. 64ECh. 2 - Prob. 65ECh. 2 - Prob. 66ECh. 2 - Prob. 67ECh. 2 - Prob. 68ECh. 2 - Prob. 69ECh. 2 - Prob. 70ECh. 2 - Prob. 71ECh. 2 - Prob. 72ECh. 2 - Prob. 73ECh. 2 - Prob. 74ECh. 2 - Prob. 75ECh. 2 - Prob. 76ECh. 2 - Prob. 77ECh. 2 - Prob. 78ECh. 2 - Prob. 79ECh. 2 - Prob. 80ECh. 2 - Prob. 81ECh. 2 - Prob. 82E
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