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: False (or 0) = straight-line is used when the depreciation is greater than the declining balance calculation (default) True (<> 0) = variable is always used

REMARKS

* For an illustrated example refer to the Depreciation page.
* VDB stands for variable declining balance.
* The "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" = 2, then the double declining balance method is used.
* If "no_switch" is left blank, then False is used.
* You can use the AMORDEGRC function to return the depreciation of an asset in a single period (declining balance method).
* You can use the AMORLINC function to return the depreciation of an asset in a single period (linear method).
* 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-declining balance method).
* You can use the SLN function to return the depreciation of an asset in a single period (straight-line method).
* You can use the SYD function to return the depreciation of an asset in a single period (sum-of-years digits method).
* For the Microsoft documentation refer to support.microsoft.com

	A
1	=VDB(45000, 15000, 4, 0, 1, 3) = $30,000.00_)
2	=VDB(45000, 15000, 4, 0, 1, 2) = $22,500.00_)
3	=VDB(45000, 15000, 4, 0, 1) = $22,500.00_)
4	=VDB(45000, 15000, 4, 0, 1, 1.5) = $16,875.00_)
5	=VDB(45000, 15000, 4, 0, 1, 1) = $11,250.00_)
6	=VDB(45000, 15000, 4, 0, 1, 0.5) = $7,500.00_)
7	=SLN(45000, 15000, 4) = $7,500.00_)
8	=VDB(2400, 300, 10, 0, 0.875, 1.5) = $315.00_)
9	=VDB(2400, 300, 3650, 0, 1) = $1.32_)
10	=VDB(2400, 300, 120, 0, 1) = $40.00_)
11	=VDB(2400, 300, 120, 6, 18) = $396.31_)
12	=VDB(2400, 300, 120, 6, 18, 1.5) = $311.81_)

1 - How much has my car depreciated in the first year if I bought it for £45,000 and its resale value after 4 years is £15,000 with a factor = 3.
2 - How much has my car depreciated in the first year if I bought it for £45,000 and its resale value after 4 years is £15,000 with a factor = 2.
3 - How much has my car depreciated in the first year if I bought it for £45,000 and its resale value after 4 years is £15,000.
4 - How much has my car depreciated in the first year if I bought it for £45,000 and its resale value after 4 years is £15,000 with a factor = 1.5.
5 - How much has my car depreciated in the first year if I bought it for £45,000 and its resale value after 4 years is £15,000 with a factor = 1.
6 - How much has my car depreciated in the first year if I bought it for £45,000 and its resale value after 4 years is £15,000 with a factor = 0.5.