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

Explanation of Solution

Multiplying two numbers:

$\begin{array}{l}\text{1011(-5)}\\ \text{×0101(5)}\\ \text{---------------}\end{array}$

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 $00000101$

$\begin{array}{l}00000101\text{(10=subtract 1101=add 00000011)}\\ +11111011\text{(01=add 11111011 to product)}\\ +\text{00000101}\text{(10=subtract 1101=add 00000101)}\\ +\text{11111011}\text{(01=add 11111011 to product)}\\ \text{--------------------}\\ \boxed{1001}11100111\end{array}$

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

Explanation of Solution

Multiplying two numbers

$\begin{array}{l}0011\text{(3)}\\ \text{×1011(-5)}\\ \text{---------------}\end{array}$

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 $00000101$

$\begin{array}{l}\text{11111101}\text{(10=subtract 0011=add 11111101)}\\ +\text{00000000}\text{(11 means simple shift)}\\ +\text{00000011}\text{(01 means 0011)}\\ +\text{11111101}\text{(10=subtract 0111= add 11111011)}\\ \text{--------------------}\\ \boxed{1000}11110001\end{array}$

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

Explanation of Solution

Multiplying two numbers

$\begin{array}{l}1011\text{(-5)}\\ \text{×1100(-4)}\\ \text{---------------}\end{array}$

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 $11111101$

$\begin{array}{l}00000000\text{(00 means simple shift)}\\ +\text{00000000}\text{(00 means simple shift)}\\ +\text{00000101}(\text{10=subtract 1011= add 00000101})\\ +\text{00000000}\text{(11 means simple shift)}\\ \text{--------------------}\\ \boxed{0000}00010100\end{array}$

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

Hence, the resultant will be “$00010100$”.