DDB |
DDB(cost, salvage, life, period [,factor]) |
Returns the depreciation of an asset in a single period (double or higher declining balance method) (Double). |
cost | The initial cost of the asset (Double). |
salvage | The value at the end of its useful life (Double). |
life | The length of useful life of the asset (Double). |
period | The period you want to calculate the depreciation over (Double). |
factor | (Optional) The number indicating the type of declining balance to use: 2 - double declining balance depreciation (default) 3 - triple declining balance depreciation 4 - four times declining balance, etc |
REMARKS |
* The "cost", "salvage", "life, "period" and "factor" must all be positive numbers. * The "life" and "period" arguments must be expressed in the same units (days, months, years). * If "factor" is left blank, then 2 is used. * If "factor" = 2, then the double declining balance method is used. * If "factor" = 3, then the triple declining balance method is used. * If "factor" >= "life", then all the depreciation is taken off in the first year. * You can use the SLN function to return the straight-line depreciation of an asset over a single period of time. * You can use the SYD function to return the sum-of-years' digits depreciation of an asset. * The equivalent Excel function is Application.WorksheetFunction.DDB * The equivalent .NET function is Microsoft.VisualBasic.Financial.DDB * For the Microsoft documentation refer to learn.microsoft.com |
'the depreciation of a £500 asset over 1 year
Debug.Print Ddb(500, 0, 1, 1) '= 500
Debug.Print Ddb(500, 0, 1, 1, 2) '= 500
Debug.Print Ddb(500, 0, 1, 1, 4) '= 500
'over 2 years, factor = 2
Debug.Print Ddb(500, 0, 2, 1) '= 500, the amount of depreciation in the first year
Debug.Print Ddb(500, 0, 2, 2) '= 0, the amount of depreciation in the second year
'over 3 years, factor = 2
Debug.Print Ddb(500, 0, 3, 1) '= 500, the anount of depreciation in the first year
Debug.Print Ddb(500, 0, 3, 2) '= 0, the amount of depreciation in the second year
Debug.Print Ddb(500, 0, 3, 3) '= 0, the amount of depreciation in the third year
'over 4 years, factor = 2
Debug.Print Ddb(500, 0, 4, 1, 2) '= 250
Debug.Print Ddb(500, 0, 4, 2, 2) '= 125
Debug.Print Ddb(500, 0, 4, 3, 2) '= 62.5
Debug.Print Ddb(500, 0, 4, 4, 2) '= 31.25
'over 4 years, factor = 3
Debug.Print Ddb(500, 0, 4, 1, 3) '= 375
Debug.Print Ddb(500, 0, 4, 2, 3) '= 93.75
Debug.Print Ddb(500, 0, 4, 3, 3) '= 23.44
Debug.Print Ddb(500, 0, 4, 4, 3) '= 5.86
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