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