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).

costThe original cost of the asset.
salvageThe value of the asset at the end of its life.
lifeThe number of periods over which the asset is being depreciated.
start_periodThe first period in the calculation.
end_periodThe last period in the calculation.
factor(Optional) The rate at which the balance declines.
no_switch(Optional) The type of declining balance to use:
True = variable is always used
False = straight-line is used when the depreciation is greater than the declining balance calculation.

Remarks

* VDB stands for variable declining balance.
* "start_period", "end_period" and "life" must all use the same units and be positive.
* 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 "factor" is left blank, then 2 is used.
* 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 the Microsoft documentation refer to support.office.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


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