DB

DB(cost, salvage, life, period [,month])

Returns the depreciation of an asset in a single period (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.
month(Optional) The number of months in the first year between 1 and 12 (12).

REMARKS
* For an illustrated example refer to the [[Depreciation]] page.
* The declining balance method computes depreciation at a fixed rate.
* The depreciation amount is calculated using: (cost - total depreciation from prior periods) * rate
* The value at the end of the depreciation (sometimes called the salvage value of the asset).
* The "period" and "life" must both use the same units.
* If "month" is left blank, then 12 is used.
* The number of periods over which the asset is being depreciated (sometimes called the useful life of the asset).
* If the cell contains the General number format, then the format is automatically changed to "£#,##0.00;".
* 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.
* You can use the VDB function to return the depreciation of an asset using a variable declining method.
* For the Microsoft documentation refer to support.microsoft.com
* For the Google documentation refer to support.google.com

 A
1=DB(45000, 15000, 4, 1, 12) = $10,800.00
2=DB(45000, 15000, 4, 2, 12) = $8,208.00
3=DB(45000, 15000, 4, 3, 12) = $6,238.08
4=DB(45000, 15000, 4, 4, 12) = $4,740.94
5=A1+A2+A3+A4 = $29,987.02
6=DB(45000, 15000, 48, 1, 12) = $1,035.00
7=DB(45000, 15000, 48, 2, 12) = $1,011.20
8=DB(45000, 15000, 48, 12, 12) = $801.27
9=DB(1000000, 100000, 6, 7, 7) = 15,845.10
10=DB(10000, 3000, 12, 1, 12) = 950
11=DB(0, 5000, 12, 1, 12) = 0
12=DB(10000, 3000, 12, "some text", 2) = #VALUE!

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.
2 - How much has my car depreciated in the second year.
3 - How much has my car depreciated in the third year.
4 - How much has my car depreciated in the fourth year.
5 - This is checking the above four results. The total depreciation over 4 years is £30,000
6 - What is the depreciation of my car in the first month if I bought it for £45,000 and its resale value after 4 years is £15,000.
7 - What is the depreciated of my car in the second month.
8 - What is the depreciation of my car depreciated in the twelfth month.

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