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Program to find the Break Even Point

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Given the list of monthly expenditure 

\sum exp
 

of an organization, selling price 


S
 

and the overhead maintenance 


M
 

of each item, the task is to calculate the Break Even Point. 
Break Even Point refers to the number of items sold in order to neutralize the total expenditure i.e. Overall, neither profit nor loss.
Examples:
 


 

Input: Expenditure = 18000, S = 600, M = 100 
Output: 36 
We need to sell 36 items to cover expenditure and maintenance overhead
Input: Expenditure = 3550, S = 90, M = 65 
Output: 142


 


 


 

Approach:

  1. Calculate the sum of all the expenditures.
  2. Subtract the maintenance (Cost price) from the selling price.
  3. Divide the expenditure sum by the above-obtained amount to get the minimum number of items to be sold (Break Even Point).

Below is the implementation of the above approach:  

C++

// C++ program to find the break-even point.
 
#include <iostream>
#include <math.h>
using namespace std;
 
// Function to calculate Break Even Point
int breakEvenPoint(int exp, int S, int M)
{
    float earn = S - M;
 
    // Calculating number of articles to be sold
    int res = ceil(exp / earn);
 
    return res;
}
 
// Main Function
int main()
{
    int exp = 3550, S = 90, M = 65;
    cout << breakEvenPoint(exp, S, M);
    return 0;
}

                    

Java

// Java program to find Break Even Point
import java.io.*;
import java.lang.*;
 
class GFG
{
// Function to calculate
// Break Even Point
public static int breakEvenPoint(int exp1,
                                 int S, int M)
{
    double earn = S - M;
     
    double exp = exp1;
 
    // Calculating number of
    // articles to be sold
    double res = Math.ceil(exp / earn);
 
    int res1 = (int) res;
     
    return res1;
}
 
// Driver Code
public static void main (String[] args)
{
    int exp = 3550, S = 90, M = 65;
    System.out.println(breakEvenPoint(exp, S, M));
}
}
 
// This code is contributed
// by Naman_Garg

                    

Python 3

# Python 3 program to find
# Break Even Point
import math
 
# Function to calculate
# Break Even Point
def breakEvenPoint(exp, S, M):
     
    earn = S - M
 
    # Calculating number of
    # articles to be sold
    if res != 0:
      res = math.ceil(exp / earn)
    # if profit is 0, it will never make ends meet
    else:
      res = float('inf')
         
    return res
     
# Driver Code
if __name__ == "__main__" :
         
    exp = 3550
    S = 90
    M = 65
     
    print (int(breakEvenPoint(exp, S, M)))
 
# This code is contributed
# by Naman_Garg

                    

C#

// C# program to find Break Even Point
using System;
 
class GFG
{
// Function to calculate
// Break Even Point
public static int breakEvenPoint(int exp1,
                                int S, int M)
{
    double earn = S - M;
     
    double exp = exp1;
 
    // Calculating number of
    // articles to be sold
    double res = Math.Ceiling(exp / earn);
 
    int res1 = (int) res;
     
    return res1;
}
 
// Driver Code
public static void Main ()
{
    int exp = 3550, S = 90, M = 65;
    Console.WriteLine(breakEvenPoint(exp, S, M));
}
}
 
// This code is contributed
// by inder_verma..

                    

PHP

<?php
// PHP program to find the break-even point.
 
// Function to calculate Break Even Point
function breakEvenPoint($exp, $S, $M)
{
    $earn = $S - $M;
 
    // Calculating number of articles
    // to be sold
    $res = ceil($exp / $earn);
 
    return $res;
}
 
// Driver Code
$exp = 3550; $S = 90; $M = 65;
echo breakEvenPoint($exp, $S, $M);
 
// This code is contributed
// by inder_verma..
?>

                    

Javascript

<script>
 
// Javascript program to find the break-even point.
 
// Function to calculate Break Even Point
function breakEvenPoint(exp, S, M)
{
    var earn = S - M;
 
    // Calculating number of articles to be sold
    var res = Math.ceil(exp / earn);
 
    return res;
}
 
// Main Function
var exp = 3550, S = 90, M = 65;
document.write( breakEvenPoint(exp, S, M));
 
</script>

                    

Output: 
142

 

Time Complexity: O(1)

Auxiliary Space: O(1)



Last Updated : 11 Jul, 2022
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