Program to find the Depreciation of Value
Last Updated :
28 May, 2022
The value of any article or item subject to wear and tear, decreases with time. This decrease is called its Depreciation. Given three variable V1, R and T where V1 is the initial value, R is the rate of depreciation and T is the time in years. The task is to find the value of the item after T years.
Examples:
Input: V1 = 200, R = 10, T = 2
Output: 162
Input: V1 = 560, R = 5, T = 3
Output: 480.13
Approach: As in Compound Interest, interest is regularly added to the principal at the end of the agreed intervals of time to generate a new and fresh principal. Similarly, Depreciated value is the decreased value from the amount at the end of agreed intervals of time to generate a new Value.
Thus if V1 is the value at a certain time and R% per annum is the rate (the rate can not be more than 100%) of depreciation per year, then the value V2 at the end of T years is:
Below is the implementation of the above approach :
C++
#include <bits/stdc++.h>
using namespace std;
float Depreciation( float v, float r, float t)
{
float D = v * pow ((1 - r / 100), t);
return D;
}
int main()
{
float V1 = 200, R = 10, T = 2;
cout << Depreciation(V1, R, T);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static float Depreciation( float v,
float r, float t)
{
float D = ( float )(v * Math.pow(( 1 - r / 100 ), t));
return D;
}
public static void main(String[] args)
{
float V1 = 200 , R = 10 , T = 2 ;
System.out.print(Depreciation(V1, R, T));
}
}
|
Python3
from math import pow
def Depreciation(v, r, t):
D = v * pow (( 1 - r / 100 ), t)
return D
if __name__ = = '__main__' :
V1 = 200
R = 10
T = 2
print ( int (Depreciation(V1, R, T)))
|
C#
using System;
class GFG
{
static float Depreciation( float v, float r, float t)
{
float D = ( float ) (v * Math.Pow((1 - r / 100), t));
return D;
}
public static void Main()
{
float V1 = 200, R = 10, T = 2;
Console.WriteLine(Depreciation(V1, R, T));
}
}
|
Javascript
function Depreciation( v, r, t)
{
var D = v * Math.pow((1 - r / 100), t)
return D;
}
var V1 = 200, R = 10, T = 2;
document.write(Depreciation(V1, R, T));
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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