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Debora [2.8K]
2 years ago
6

Develop an sec (single error correction) code for a 16-bit data word. generate the code for the data word 0101000000111001. show

that the code will correctly identify an error in data bit 4.
Computers and Technology
2 answers:
natita [175]2 years ago
7 0

Answer:

010100000001101000101

Explanation:

When an error occurs in data bits, the SEC code is used to determine where the error took place. 5 check-bits are needed to generate SEC code for 16-bits data word. The check bits are:

C16=0, C8=0, C4=0,C2=0,C1=1

Therefore the SEC code is 010100000001101000101

jarptica [38.1K]2 years ago
4 0

Answer:

<em>The code is given as = 010100000001101000101</em>

<em>Explanation:</em>

<em>The steps take is shown below, </em>

<em>The SEC code is used to ascertain where errors had occurred. Such as the occurrence of errors in data  bits. </em>

<em>The inequality given is: 2^k - 1 >= M + K </em>

<em>Where M is =16 </em>

<em>For the arrangement of the data bits and checking of the bits, the following steps is taken below </em>

<em>   Bit position              Number             Check bits             Data Bits </em>

<em> 21                                   10101 </em>

<em> 20                                  10100 </em>

<em> </em>

<em>The bits is therefore checked in a way up to the bit position 1 </em>

<em>Therefore, the code is then written as follows: 010100000001101000101 </em>

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Answer:

Following are the code to this question:

CarCounter::~CarCounter()//Defining destructor CarCounter

{

cout << "Destroying CarCounter\n";//print message Destroying CarCounter

}

Explanation:

Following are the full program to this question:

#include <iostream>//Defining header file

using namespace std;

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{

public:

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{

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CarCounter* parkingLot = new CarCounter();//Defining class object parkingLot

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Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and
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Explanation:

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INT -110 \times 10^6,

CPI - 1

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Time(old) =\frac{50 x 10^6 + 110 x 10^6 + 4 x ( 80 x 10^6) + 2 x (16 x 10^ 6)}{2 x 10^9}

Time(old) = 256 \times 10^ {-3}

Time(new) =  \frac{256 \times 10^{-3}}{2}

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                =\frac{CPI(new) x [50 x 10^6 + 110 x 10^6 + 4 x ( 80 x 10^6) + 2 x (16 x 10^ 6)]}{2 x 10^9}

                =  128 \times 10^{-3}

CPI(new) = \frac{-206}{50}

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iogann1982 [59]

Answer:

<em>The programming language is not stated;</em>

<em>However, I'll answer this question using C++ programming language</em>

<em>The program uses few comments; See explanation section for more detail</em>

<em>Also, the program assumes that all input will always be an integer</em>

#include<iostream>

#include<sstream>

#include<string>

using namespace std;

int main()

{

string input;

cout<<"Enter the first 9 digits of an ISBN as a string: ";

cin>>input;

//Check length

if(input.length() != 9)

{

 cout<<"Invalid input\nLength must be exact 9";

}

else

{

 int num = 0;

//Calculate sum of products

for(int i=0;i<9;i++)

{

 num += (input[i]-'0') * (i+1);    

}

//Determine checksum

if(num%11==10)

{

 input += "X";

 cout<<"The ISBN-10 number is "<<input;

}

else

{

 ostringstream ss;

 ss<<num%11;

 string dig = ss.str();

 cout<<"The ISBN-10 number is "<<input+dig;

}

}  

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}

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cout<<"Enter the first 9 digits of an ISBN as a string: ";  -> This line prompts the user for input

cin>>input;  -> The user input is stored here

if(input.length() != 9)  { cout<<"Invalid input\nLength must be exact 9";  }  -> Here, the length of input string is checked; If it's not equal to then, a message will be displayed to the screen

If otherwise, the following code segment is executed

else  {

int num = 0; -> The sum of products  of individual digit is initialized to 0

The sum of products  of individual digit is calculated as follows

for(int i=0;i<9;i++)

{

num += (input[i]-'0') * (i+1);

}

The next lines of code checks if the remainder of the above calculations divided by 11 is 10;

If Yes, X is added as a suffix to the user input

Otherwise, the remainder number is added as a suffix to the user input

if(num%11==10)  {  input += "X";  cout<<"The ISBN-10 number is "<<input; }

else

{

ostringstream ss;

ss<<num%11;

string dig = ss.str();

cout<<"The ISBN-10 number is "<<input+dig;

}

}  

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