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Gwar [14]
2 years ago
7

Process redesign and reengineering in 1500 words

Computers and Technology
1 answer:
timama [110]2 years ago
7 0

Answer:

Process re-design is the process in which re-work is done to achieve the improvement in the output. The major aim is to set the goals for achieving the main objective.

The various essential steps in the process of re-design are:

  • Data capturing process are used to identify the process and making the necessary changes.
  • Identify every step of the process to ensure that process is in organized way along with the workflow.

Re-engineering is the process of re-design the system with focuses to improve the efficiency and quality. It basically analysis the whole business re-engineering process. The main step is to identify the need for the changes.

 

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1. The following programs require using arrays. For each, the input comes from standard input and consists of N real numbers bet
Mamont248 [21]

Answer:

import java.util.*;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.Arrays;

class GFG

{

  // Function for calculating mean

  public static double findMean(double a[], int n)

  {

      int sum = 0;

      for (int i = 0; i < n; i++)

          sum += a[i];

 

      return (double)sum / (double)n;

  }

  // Function for calculating median

  public static double findMedian(double a[], int n)

  {

      // First we sort the array

      Arrays.sort(a);

      // check for even case

      if (n % 2 != 0)

      return (double)a[n / 2];

 

      return (double)(a[(n - 1) / 2] + a[n / 2]) / 2.0;

  }

  public static double findMode(double a[], int n)

{

// The output array b[] will

// have sorted array

//int []b = new int[n];

 

// variable to store max of

// input array which will

// to have size of count array

double max = Arrays.stream(a).max().getAsDouble();

 

// auxiliary(count) array to

// store count. Initialize

// count array as 0. Size

// of count array will be

// equal to (max + 1).

double t = max + 1;

double[] count = new double[(int)t];

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

{

count[i] = 0;

}

 

// Store count of each element

// of input array

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

{

count[(int)(10*a[i])]++;

}

 

// mode is the index with maximum count

double mode = 0;

double k = count[0];

for (int i = 1; i < t; i++)

{

if (count[i] > k)

{

k = count[i];

mode = i;

}

}

return mode;

}

public static double findSmallest(double [] A, int total){

Arrays.sort(A);

return A[0];

}

 

public static void printAboveAvg(double arr[], int n)

{

 

// Find average

double avg = 0;

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

avg += arr[i];

avg = avg / n;

 

// Print elements greater than average

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

if (arr[i] > avg)

System.out.print(arr[i] + " ");

System.out.println();

}

 

public static void printrand(double [] A, int n){

Arrays.sort(A);

for(int i=0;i<n;i++){

System.out.print(A[0]+"/t");

}

System.out.println();

}

 

public static void printHist(double [] arr, int n) {

 

for (double i = 1.0; i >= 0; i-=0.1) {

System.out.print(i+" | ");

for (int j = 0; j < n; j++) {

 

// if array of element is greater

// then array it print x

if (arr[j] >= i)

System.out.print("x");

 

// else print blank spaces

else

System.out.print(" ");

}

System.out.println();

}

// print last line denoted by ----

for(int l = 0; l < (n + 3); l++){    

System.out.print("---");

}

 

System.out.println();

System.out.print(" ");

 

for (int k = 0; k < n; k++) {

System.out.print(arr[k]+" ");

}

}

  // Driver program

  public static void main(String args[]) throws IOException

  {

      //Enter data using BufferReader

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

double [] A = new double[100];

int i=0;

System.out.println("Enter the numbers(0.0-1.0) /n Enter 9 if u have entered the numbers. /n");

do

{

A[i++]=Double.parseDouble(br.readLine());

}while(A[i-1]==9);

      i--;

      System.out.println("Average = " + findMean(A,i) );

      System.out.println("Median = " + findMedian(A,i));

      System.out.println("Element that occured most frequently = " + findMode(A,i));

      System.out.println("number closest to 0.0 =" + findSmallest(A,i));

      System.out.println("Numbers that are greater than the average are follows:");

      printAboveAvg(A,i);

      System.out.println("Numbers in random order are as follows:");

      printrand(A,i);

      System.out.println("Histogram is bellow:");

      printHist(A,i);

  }

}

Explanation:

3 0
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DerKrebs [107]
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2 years ago
You need to design a data storage scheme for Hayseed Heaven library data system. There are several hundred thousand large data r
Alekssandra [29.7K]

Answer:

Make use of hash tables

Explanation:

The appropriate thing to use for this should be a hash table.

A Hash Table can be described as a data structure which stores data in an associative manner. In a hash table, data is stored in an array format, where each data value has its own unique index value. Access of data becomes very fast if we know the index of the desired data. So we can perform Hashing on ISBN Number since its unique and based on the Hash Function w ecan store the Information record.

There is no requirement for printing the file in order - HashTables dont store the data in order of insertions, so no problems with that

It becomes a data structure in which insertion and search operations are very fast irrespective of the size of the data. So Querying books details can be fast and searching will take less time.

It can also be pointed out that it wont be too expensive for Hardware implemtation as HashTables stores data based on Hash Functions and memory consumption is also optimal which reduces memory wastages.

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