DDB(cost, salvage, life, period [,factor])

Returns the depreciation of an asset in a single period (double or triple 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.
periodThe period to calculate the depreciation over.
factor(Optional) The number indicating the type of declining balance to use:
2 - double declining balance depreciation (default)
3 - triple declining balance depreciation

Remarks

* The double-declining balance method computes depreciation at an accelerated rate. Depreciation is highest in the first period and decreases in successive periods.
* All five arguments must be positive numbers.
* The value at the end of the depreciation (sometimes called the salvage value of the asset).
* The number of periods over which the asset is being depreciated (sometimes called the useful life of the asset).
* The "period" and "life" must be expressed in the same units of time: years, months or days.
* If "cost" < "salvage", then ??
* If "factor" is left blank, then 2 is used.
* This is an example of a geometric progression.
* Use the VDB function to switch to straight-line depreciation when the depreciation is greater than the declining balance calculation.
* This function uses the following formula to calculate depreciation for a period: ((cost-salvage) - total depreciation from prior periods) * (factor/life).
* You can use the DB function to return the depreciation of an asset in a single period (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.
* You can use the VDB function to return the depreciation of an asset using a variable declining method.
* The equivalent VBA function is DDB
* For the Microsoft documentation refer to support.office.com

 A
1=DDB(45000,15000,4,1) = 22500
2=DDB(45000,15000,4,2) = 7500
3=DDB(2400,300,120,1,2) = 40
4=DDB(2400,300,10,1,2) = 480
5=DDB(2400,300,10,10) = 22.123
6=DDB(10000,3000,12,1,3) = 2500

1 - What is the depreciation of my car in the first year if it cost me £45,000 and has a resale value of £15,000 after 4 years.
2 - What is the depreciation of my car in the second year.

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