# VDB

VDB(cost, salvage, life, start_period, end_period [,factor] [,no_switch])

Returns the depreciation of an asset in a single period (variable declining balance method).

 cost The original cost of the asset. salvage The value of the asset at the end of its life. life The number of periods over which the asset is being depreciated. start_period The first period in the calculation. end_period The last period in the calculation. factor (Optional) The rate at which the balance declines (2). no_switch (Optional) A logical value indicating the type of declining balance to use:True = variable is always usedFalse = straight-line is used when the depreciation is greater than the declining balance calculation (default).

 REMARKS
 * VDB stands for variable declining balance.* "start_period", "end_period" and "life" must all use the same units and be positive.* If "factor" is left blank, then 2 is used.* If "factor" = 1, then ??* If "factor" = 2, then the double declining balance method is used.* If "factor" = 3, then* If "factor" = 1.5, then* If "no_switch" is left blank, then ?? Is used.* You can use the DB function to return the depreciation of an asset in a single period (declining balance method).* You can use the DDB function to return the depreciation of an asset in a single period (double or triple declining balance method).* You can use the SLN function to return the straight line depreciation of an asset for one period.* You can use the SYD function to return the depreciation of an asset using the sum of years method.* For a working example refer to the Quant > Accounting > Depreciation page.* For the Microsoft documentation refer to support.microsoft.com

 A 1 =VDB(2400,300,10,0,0.875,1.5) = 315 2 =VDB(2400,300,3650,0,1) = 1.315 3 =VDB(2400,300,120,0,1) = 40 4 =VDB(2400,300,10,0,1) = 480 5 =VDB(2400,300,120,6,18) = 396.306 6 =VDB(2400,300,120,6,18,1.5) = 311.809