Return the depreciation for a specified period using the fixed-declining balance method. Excel: DB

Contents

Interface Function Documentation Page Comments

Interface

#include <codecogs/finance/accounting/deprecfixeddec.h>

using namespace Finance::Accounting;

	double	deprecFixedDec (double valCost, double valSalv, int life, int per, int month=12) Return the depreciation for a specified period using the fixed-declining balance method. Excel: DB
	Real	cc_deprecFixedDec (Real valCost, Real valSalv, Integer life, Integer per, Integer month) This function is available as a Microsoft Excel add-in.

Function Documentation

This function calculates the depreciation of an asset for a specified period. The method used is a fixed-declining method. This function is equivalent to the Microsoft Excel function DB. It works on the following set of equations:

With two special cases, when period=1:

When period=life:

Where:

• salv is the salvage value of the asset,
• cost is the initial cost of the asset,
• life is the number of periods over which the asset is being depreciated.
• prevTot is the total depreciation from previous periods.

References:

http://www.vni.com/products/imsl/jmsl/v30/api/

Example 1:

#include <stdio.h>
#include <codecogs/finance/accounting/deprecfixeddec.h>
 
int main(void)
{
  double costVal=1000000;
  double salvVal=100000;

  printf("Consider an asset with the following terms:\n");
  printf("Cost\t\tSalvage\t\tLife\tPeriod\tMonth\tDepreciation\n");
 
  for (int per=1; per<=7; ++per) { 
    printf("%f\t%f\t6\t%i\t7\t%f\n",
           costVal,
           salvVal,
           per,
           Finance::Accounting::deprecFixedDec(costVal, salvVal, 6, per, 7));
  }

  printf("\n");
  costVal=2500;
  salvVal=500;
  for (int per=1; per<=4; ++per) { 
    printf("%f\t%f\t3\t%i\t6\t%f\n",
           costVal,
           salvVal,
           per,
           Finance::Accounting::deprecFixedDec(costVal, salvVal, 3, per, 6));
   }

   return 0;
}

Output:

Consider an asset with the following terms:
Cost            Salvage         Life    Period  Month   Depreciation
1000000.000000  100000.000000   6       1       7       186083.333333
1000000.000000  100000.000000   6       2       7       259639.416667
1000000.000000  100000.000000   6       3       7       176814.442750
1000000.000000  100000.000000   6       4       7       120410.635513
1000000.000000  100000.000000   6       5       7       81999.642784
1000000.000000  100000.000000   6       6       7       55841.756736
1000000.000000  100000.000000   6       7       7       15845.098474
 
2500.000000     500.000000      3       1       6       518.750000
2500.000000     500.000000      3       2       6       822.218750
2500.000000     500.000000      3       3       6       480.997969
2500.000000     500.000000      3       4       6       140.691906

being depreciated (sometimes called the useful life of the asset).

depreciation. per must have the same units as life.

Parameters:

valCost	The initial cost of the asset.
valSalv	The salvage value of the asset at the end of its life.
life	The is the number of periods over which the asset is
per	The period for which you want to calculate the
month	The number of months in the first year. This value defaults to 12.

Returns:

The depreciation for each accounting period.

Authors:

James Warren (August 2005)