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Savatey [412]
3 years ago
13

Given a floating-point formal with a k-bit exponent and an n-bit (fraction, write formulas for the exponent E, significant M, th

e fraction f, and the value V for the quantities that follow. In addition, describe the bit representation.
A. The number 7.0
B. The largest odd integer that can be represented exactly
C. The reciprocal of the smallest positive normalized value
Computers and Technology
1 answer:
ANEK [815]3 years ago
8 0

Answer:

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112  ≈ 7/4

fractional value ( F )  = 0.112

And, numeric value of the quantity ( V )  = 7

The exponent bits will be represented  as :  100----01.

while The fraction bits will be represented  as : 1100---0.

<u>B) The largest odd integer that can be represented exactly </u>

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.11------12 = 2 - 2-n

also the fractional value ( F ) =  

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits  will be represented  as follows:  n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

<u>C) The reciprocal of the smallest positive normalized value </u>

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E )  = 2k-1 – 2

While the significand value ( M )  = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

Explanation:

E = integer value of exponent

M = significand value

F = fractional value

V = numeric value of quantity

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112  ≈ 7/4

fractional value ( F )  = 0.112

And, numeric value of the quantity ( V )  = 7

The exponent bits will be represented  as :  100----01.

while The fraction bits will be represented  as : 1100---0.

<u>B) The largest odd integer that can be represented exactly </u>

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.11------12 = 2 - 2-n

also the fractional value ( F ) =  

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits  will be represented  as follows:  n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

<u>C) The reciprocal of the smallest positive normalized value </u>

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E )  = 2k-1 – 2

While the significand value ( M )  = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

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// here is code in c++.

// include header

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

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Explanation:

Declare two variables "userNum" and "x". Read the value of these. Run a for loop 4 time and divide the "userNum" with "x" and print  the value of "userNum".

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1000 500 250 125  

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Answer:

Following are the program in the C++ programming Language.

//header files

#include <iostream>

#include <vector>

using namespace std;//name space

// define main method

int main()

{

//set integer variables

int size, num;

//get input the size of the list

cout<<"Enter size of list: ";

cin >> size;

//set integer type vector variable

vector<int> vecs;

//Set the for loop

for (int index = 0; index < size; ++index)

{//get elements of the list from the user

cout<<" :";

cin >> num;

//push back the elements of the list

vecs.push_back(num);

}

//storing the first two elements in variable

int n = vecs[0], n1 = vecs[1], current , temp;

//set if conditional statement

if (n > n1)

{

//perform swapping

temp = n;

n = n1;

n1 = temp;

}

//Set for loop

for (int index = 2; index < size; ++index)

{

//store the value of the vector in the variable

current = vecs[index];

//set if conditional statement

if (current < n)

{

//interchange the elements of the variable

n1 = n;

n = current;

}

else if(current < n1)

{

n1 = current;

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}

//print the value of first two smallest number.

cout <<"\n" <<n << " " << n1 << endl;

return 0;

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<u>Output:</u>

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:3

:21

:2

2 3

Explanation:

Here, we define the required header files and namespace then, we define "main()" function inside the main function.

  • Set two integer data type variable "size", "num".
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  • Then, we push back the elements of the list and store first two elements of the list in the variable "n", "n1".
  • Set the conditional statement to perform swapping.
  • Define the for loop to store the list of the vector in the integer type variable "current".
  • Finally, we print the value of the two smallest numbers of the list.
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