Essentials of Computer Organization and Architecture
Essentials of Computer Organization and Architecture
5th Edition
ISBN: 9781284123036
Author: Linda Null
Publisher: Jones & Bartlett Learning
Question
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Chapter 2, Problem 43E

a)

Program Plan Intro

Booth’s algorithm:

  • The main goal of booths algorithm is to multiply two signed binary numbers by involving two’s complement notation.
  • There are multiple methods available for fast multiplication but those methods will not be applicable for signed multiplication.
  • The advantage of booth’s algorithm is not only performs multination faster but also it is efficient to do multiplication on the signed numbers.

Booth’s algorithm reads the value as follows:

  • Subtraction of multiplicand takes pace when the multiplier’s current and the preceding bits are 1 and 0 respectively.
  • Addition of multiplicand takes pace when the multiplier’s current and the preceding bits are 0 and 1 respectively.
  • Left shifting of multiplicand takes pace when the multiplier’s current and the preceding bits are 0 and 0 or 1 and 1 respectively

a)

Expert Solution
Check Mark

Explanation of Solution

Multiplying two numbers:

1011(-5)×0101(5)---------------

Since both the numbers are given in 4-bit two’s complement the resultant value should 8-bit which is also in two’s complement.

The multiplicand is “-5”, so the 2’s complement is 0101. The 4-bit value should be extended to 8-bit 00000101

00000101(10=subtract 1101=add 00000011)+11111011(01=add 11111011 to product)+00000101(10=subtract 1101=add 00000101)+11111011(01=add 11111011 to product)--------------------100111100111

Discard the last 4 bits because only 8 rightmost bits should be considered.

Hence, the resultant will be “11100111”.

b)

Program Plan Intro

Booth’s algorithm:

  • The main goal of booths algorithm is to multiply two signed binary numbers by involving two’s complement notation.
  • There are multiple method available for fast multiplication but those methods will not be applicable for signed multiplication.
  • The advantage of booth’s algorithm is not only performs multination faster but also it is efficient to do multiplication on the signed numbers.

Booth’s algorithm reads the value as follows:

  • Subtraction of multiplicand takes pace when the multiplier’s current and the preceding bits are 1 and 0 respectively.
  • Addition of multiplicand takes pace when the multiplier’s current and the preceding bits are 0 and 1 respectively.
  • Left shifting of multiplicand takes pace when the multiplier’s current and the preceding bits are 0 and 0 or 1 and 1 respectively

b)

Expert Solution
Check Mark

Explanation of Solution

Multiplying two numbers

0011(3)×1011(-5)---------------

Since both the numbers are given in 4-bit two’s complement the resultant value should 8-bit which is also in two’s complement.

The multiplicand is “-5”, so the 2’s complement is 0101. The 4-bit value should be extended to 8-bit 00000101

11111101(10=subtract 0011=add 11111101)+00000000(11 means simple shift)+00000011(01 means 0011)+11111101(10=subtract 0111= add 11111011)--------------------100011110001

Discard the last 4 bits because only 8 rightmost bits should be considered.

Hence, the resultant will be “11110001”.

c)

Program Plan Intro

Booth’s algorithm:

  • The main goal of booths algorithm is to multiply two signed binary numbers by involving two’s complement notation.
  • There are multiple method available for fast multiplication but those methods will not be applicable for signed multiplication.
  • The advantage of booth’s algorithm is not only performs multination faster but also it is efficient to do multiplication on the signed numbers.

Booth’s algorithm reads the value as follows:

  • Subtraction of multiplicand takes pace when the multiplier’s current and the preceding bits are 1 and 0 respectively.
  • Addition of multiplicand takes pace when the multiplier’s current and the preceding bits are 0 and 1 respectively.
  • Left shifting of multiplicand takes pace when the multiplier’s current and the preceding bits are 0 and 0 or 1 and 1 respectively

c)

Expert Solution
Check Mark

Explanation of Solution

Multiplying two numbers

1011(-5)×1100(-4)---------------

Since both the numbers are given in 4-bit two’s complement the resultant value should 8-bit which is also in two’s complement.

The multiplicand is “-5”, so the 2’s complement is 1101. The 4-bit value should be extended to 8-bit 11111101

00000000(00 means simple shift)+00000000(00 means simple shift)+00000101(10=subtract 1011= add 00000101)+00000000(11 means simple shift)--------------------000000010100

Discard the last 4 bits because only 8 rightmost bits should be considered.

Hence, the resultant will be “00010100”.

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Chapter 2 Solutions

Essentials of Computer Organization and Architecture

Ch. 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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