Given numbers: 1.666015625 × 10 0 , 1.9760 × 10 4 and − 1.9744 × 10 4 . Binary Multiplication: Binary multiplication follows the following rules: 0 x 0 = 0 0 x 1 = 0 1 x 0 = 0 1 x 1 = 1 Given Information: A, B and C are stored in 16-bit half precision format.

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Answer 1

Question

Chapter 3, Problem 3.39E

Program Plan Intro

Given numbers: 1.666015625×100, 1.9760×104 and −1.9744×104.

Binary Multiplication:

Binary multiplication follows the following rules:

0 x 0 = 00 x 1 = 01 x 0 = 01 x 1 = 1

Given Information:

A, B and C are stored in 16-bit half precision format.

Expert Solution & Answer

Explanation of Solution

Calculate:

(1.666015625×100×1.9760×104)+(1.666015625×100×−1.9744×104)

Here, 1.666015625×100 is represented as “A”, 1.9760×104 is represented as “B” and −1.9744×104 is represented as “C”.

Converting the above values to binary:

1.666015625×100= 1.1010101010×20

1.9760×104=1.0011010011×214

−1.9744×104=−1.0011010010×214

The exponents match and hence shifting is not necessary.

Exponent=0+14=14

Here, the sign of the result is positive as all the numbers are positive and hence their product is positive. It is assumed that it has “1” guard, “1” round bit and “1” sticky bit and round to the nearest even.

The fraction of A×B is represented below:

(A) 1.1010101010(B) ×1.0011010011 −−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−−−− 10.0000001001100001111 Normalize, add 1 to exponent

Here, the guard is “1”, Round is “1”, Sticky is “1: Round” and the value of A×B is 10.000000100 11 00001111.

Hence, the final value of A×B is 1.0000000101×215.

The fraction “A×B+A×C” is represented below:

(A) 1.1010101010(C) ×1.0011010010 −−−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−− 10.000000011111011010Normalize and add 1 to exponentA×C −1.0000000100×215A×B 1.0000000101 × 215A×C −1.0000000100×215 −−−−−−−−−−−−A×B+A×C .0000000001×215A×B+A×C 1.0000000000×25

Hence, the value of A×B+A×C is 1.0000000000×25.

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Answer 2

Question

Chapter 3, Problem 3.39E

Program Plan Intro

Given numbers: 1.666015625×100, 1.9760×104 and −1.9744×104.

Binary Multiplication:

Binary multiplication follows the following rules:

0 x 0 = 00 x 1 = 01 x 0 = 01 x 1 = 1

Given Information:

A, B and C are stored in 16-bit half precision format.

Expert Solution & Answer

Explanation of Solution

Calculate:

(1.666015625×100×1.9760×104)+(1.666015625×100×−1.9744×104)

Here, 1.666015625×100 is represented as “A”, 1.9760×104 is represented as “B” and −1.9744×104 is represented as “C”.

Converting the above values to binary:

1.666015625×100= 1.1010101010×20

1.9760×104=1.0011010011×214

−1.9744×104=−1.0011010010×214

The exponents match and hence shifting is not necessary.

Exponent=0+14=14

Here, the sign of the result is positive as all the numbers are positive and hence their product is positive. It is assumed that it has “1” guard, “1” round bit and “1” sticky bit and round to the nearest even.

The fraction of A×B is represented below:

(A) 1.1010101010(B) ×1.0011010011 −−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−−−− 10.0000001001100001111 Normalize, add 1 to exponent

Here, the guard is “1”, Round is “1”, Sticky is “1: Round” and the value of A×B is 10.000000100 11 00001111.

Hence, the final value of A×B is 1.0000000101×215.

The fraction “A×B+A×C” is represented below:

(A) 1.1010101010(C) ×1.0011010010 −−−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−− 10.000000011111011010Normalize and add 1 to exponentA×C −1.0000000100×215A×B 1.0000000101 × 215A×C −1.0000000100×215 −−−−−−−−−−−−A×B+A×C .0000000001×215A×B+A×C 1.0000000000×25

Hence, the value of A×B+A×C is 1.0000000000×25.

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Answer 3

Question

Chapter 3, Problem 3.39E

Program Plan Intro

Given numbers: 1.666015625×100, 1.9760×104 and −1.9744×104.

Binary Multiplication:

Binary multiplication follows the following rules:

0 x 0 = 00 x 1 = 01 x 0 = 01 x 1 = 1

Given Information:

A, B and C are stored in 16-bit half precision format.

Expert Solution & Answer

Explanation of Solution

Calculate:

(1.666015625×100×1.9760×104)+(1.666015625×100×−1.9744×104)

Here, 1.666015625×100 is represented as “A”, 1.9760×104 is represented as “B” and −1.9744×104 is represented as “C”.

Converting the above values to binary:

1.666015625×100= 1.1010101010×20

1.9760×104=1.0011010011×214

−1.9744×104=−1.0011010010×214

The exponents match and hence shifting is not necessary.

Exponent=0+14=14

Here, the sign of the result is positive as all the numbers are positive and hence their product is positive. It is assumed that it has “1” guard, “1” round bit and “1” sticky bit and round to the nearest even.

The fraction of A×B is represented below:

(A) 1.1010101010(B) ×1.0011010011 −−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−−−− 10.0000001001100001111 Normalize, add 1 to exponent

Here, the guard is “1”, Round is “1”, Sticky is “1: Round” and the value of A×B is 10.000000100 11 00001111.

Hence, the final value of A×B is 1.0000000101×215.

The fraction “A×B+A×C” is represented below:

(A) 1.1010101010(C) ×1.0011010010 −−−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−− 10.000000011111011010Normalize and add 1 to exponentA×C −1.0000000100×215A×B 1.0000000101 × 215A×C −1.0000000100×215 −−−−−−−−−−−−A×B+A×C .0000000001×215A×B+A×C 1.0000000000×25

Hence, the value of A×B+A×C is 1.0000000000×25.

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Answer 4

Question

Chapter 3, Problem 3.39E

Program Plan Intro

Given numbers: 1.666015625×100, 1.9760×104 and −1.9744×104.

Binary Multiplication:

Binary multiplication follows the following rules:

0 x 0 = 00 x 1 = 01 x 0 = 01 x 1 = 1

Given Information:

A, B and C are stored in 16-bit half precision format.

Expert Solution & Answer

Explanation of Solution

Calculate:

(1.666015625×100×1.9760×104)+(1.666015625×100×−1.9744×104)

Here, 1.666015625×100 is represented as “A”, 1.9760×104 is represented as “B” and −1.9744×104 is represented as “C”.

Converting the above values to binary:

1.666015625×100= 1.1010101010×20

1.9760×104=1.0011010011×214

−1.9744×104=−1.0011010010×214

The exponents match and hence shifting is not necessary.

Exponent=0+14=14

Here, the sign of the result is positive as all the numbers are positive and hence their product is positive. It is assumed that it has “1” guard, “1” round bit and “1” sticky bit and round to the nearest even.

The fraction of A×B is represented below:

(A) 1.1010101010(B) ×1.0011010011 −−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−−−− 10.0000001001100001111 Normalize, add 1 to exponent

Here, the guard is “1”, Round is “1”, Sticky is “1: Round” and the value of A×B is 10.000000100 11 00001111.

Hence, the final value of A×B is 1.0000000101×215.

The fraction “A×B+A×C” is represented below:

(A) 1.1010101010(C) ×1.0011010010 −−−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−− 10.000000011111011010Normalize and add 1 to exponentA×C −1.0000000100×215A×B 1.0000000101 × 215A×C −1.0000000100×215 −−−−−−−−−−−−A×B+A×C .0000000001×215A×B+A×C 1.0000000000×25

Hence, the value of A×B+A×C is 1.0000000000×25.

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Answer 5

Chapter 3, Problem 3.39E

Program Plan Intro

Given numbers: 1.666015625×100, 1.9760×104 and −1.9744×104.

Binary Multiplication:

Binary multiplication follows the following rules:

0 x 0 = 00 x 1 = 01 x 0 = 01 x 1 = 1

Given Information:

A, B and C are stored in 16-bit half precision format.

Expert Solution & Answer

Explanation of Solution

Calculate:

(1.666015625×100×1.9760×104)+(1.666015625×100×−1.9744×104)

Here, 1.666015625×100 is represented as “A”, 1.9760×104 is represented as “B” and −1.9744×104 is represented as “C”.

Converting the above values to binary:

1.666015625×100= 1.1010101010×20

1.9760×104=1.0011010011×214

−1.9744×104=−1.0011010010×214

The exponents match and hence shifting is not necessary.

Exponent=0+14=14

Here, the sign of the result is positive as all the numbers are positive and hence their product is positive. It is assumed that it has “1” guard, “1” round bit and “1” sticky bit and round to the nearest even.

The fraction of A×B is represented below:

(A) 1.1010101010(B) ×1.0011010011 −−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−−−− 10.0000001001100001111 Normalize, add 1 to exponent

Here, the guard is “1”, Round is “1”, Sticky is “1: Round” and the value of A×B is 10.000000100 11 00001111.

Hence, the final value of A×B is 1.0000000101×215.

The fraction “A×B+A×C” is represented below:

(A) 1.1010101010(C) ×1.0011010010 −−−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−− 10.000000011111011010Normalize and add 1 to exponentA×C −1.0000000100×215A×B 1.0000000101 × 215A×C −1.0000000100×215 −−−−−−−−−−−−A×B+A×C .0000000001×215A×B+A×C 1.0000000000×25

Hence, the value of A×B+A×C is 1.0000000000×25.

Answer 6

Chapter 3, Problem 3.39E

Program Plan Intro

Given numbers: 1.666015625×100, 1.9760×104 and −1.9744×104.

Binary Multiplication:

Binary multiplication follows the following rules:

0 x 0 = 00 x 1 = 01 x 0 = 01 x 1 = 1

Given Information:

A, B and C are stored in 16-bit half precision format.

Expert Solution & Answer

Explanation of Solution

Calculate:

(1.666015625×100×1.9760×104)+(1.666015625×100×−1.9744×104)

Here, 1.666015625×100 is represented as “A”, 1.9760×104 is represented as “B” and −1.9744×104 is represented as “C”.

Converting the above values to binary:

1.666015625×100= 1.1010101010×20

1.9760×104=1.0011010011×214

−1.9744×104=−1.0011010010×214

The exponents match and hence shifting is not necessary.

Exponent=0+14=14

Here, the sign of the result is positive as all the numbers are positive and hence their product is positive. It is assumed that it has “1” guard, “1” round bit and “1” sticky bit and round to the nearest even.

The fraction of A×B is represented below:

(A) 1.1010101010(B) ×1.0011010011 −−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−−−− 10.0000001001100001111 Normalize, add 1 to exponent

Here, the guard is “1”, Round is “1”, Sticky is “1: Round” and the value of A×B is 10.000000100 11 00001111.

Hence, the final value of A×B is 1.0000000101×215.

The fraction “A×B+A×C” is represented below:

(A) 1.1010101010(C) ×1.0011010010 −−−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−− 10.000000011111011010Normalize and add 1 to exponentA×C −1.0000000100×215A×B 1.0000000101 × 215A×C −1.0000000100×215 −−−−−−−−−−−−A×B+A×C .0000000001×215A×B+A×C 1.0000000000×25

Hence, the value of A×B+A×C is 1.0000000000×25.

Answer 7

Chapter 3, Problem 3.39E

Program Plan Intro

Given numbers: 1.666015625×100, 1.9760×104 and −1.9744×104.

Binary Multiplication:

Binary multiplication follows the following rules:

0 x 0 = 00 x 1 = 01 x 0 = 01 x 1 = 1

Given Information:

A, B and C are stored in 16-bit half precision format.

Answer 8

Expert Solution & Answer

Explanation of Solution

Calculate:

(1.666015625×100×1.9760×104)+(1.666015625×100×−1.9744×104)

Here, 1.666015625×100 is represented as “A”, 1.9760×104 is represented as “B” and −1.9744×104 is represented as “C”.

Converting the above values to binary:

1.666015625×100= 1.1010101010×20

1.9760×104=1.0011010011×214

−1.9744×104=−1.0011010010×214

The exponents match and hence shifting is not necessary.

Exponent=0+14=14

Here, the sign of the result is positive as all the numbers are positive and hence their product is positive. It is assumed that it has “1” guard, “1” round bit and “1” sticky bit and round to the nearest even.

The fraction of A×B is represented below:

(A) 1.1010101010(B) ×1.0011010011 −−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−−−− 10.0000001001100001111 Normalize, add 1 to exponent

Here, the guard is “1”, Round is “1”, Sticky is “1: Round” and the value of A×B is 10.000000100 11 00001111.

Hence, the final value of A×B is 1.0000000101×215.

The fraction “A×B+A×C” is represented below:

(A) 1.1010101010(C) ×1.0011010010 −−−−−−−−−− 11010101010 11010101010 11010101010 11010101010 11010101010 −−−−−−−−−−−−− 10.000000011111011010Normalize and add 1 to exponentA×C −1.0000000100×215A×B 1.0000000101 × 215A×C −1.0000000100×215 −−−−−−−−−−−−A×B+A×C .0000000001×215A×B+A×C 1.0000000000×25