# Documentation/Calc Functions/DB

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DB

## Category:

Financial Analysis

## Summary:

Calculates the depreciation allowance of an asset.

This form of depreciation is used if you want to get a higher depreciation value at the beginning of the depreciation (as opposed to linear depreciation). The depreciation value is reduced with every depreciation period by the depreciation already deducted from the initial cost.

## Syntax:

DB(Cost; Salvage; Life; Period[; Month])

## Returns:

Returns a positive real number(in Currency units) which is the depreciation of an asset for a specified period using the fixed-declining balance method.

## Arguments:

Cost is a positive real number or a reference to the cell containing that number which is the initial cost of an asset.

Salvage is a non-negative real number or a reference to the cell containing that number which is the value of an asset at the end of the depreciation.

Life is a positive real number or a reference to the cell containing that number which defines the period over which an asset is depreciated.

Period is a non-negative real number or a reference to the cell containing that number which is the length of each period. The length must be entered in the same date unit as the depreciation period.

Month is a positive number from 1 to 12 or a reference to the cell containing that number which denotes the number of months for the first year of depreciation. If an entry is not defined, 12 is used as the default.

• If Month is non-integer then the function uses its truncated value.
• If after truncation Month is not an integer from 1 to 12, then the function returns an error value.
• If Cost is less than or equal to 0 then the function returns an error value.
• If Life is less than Period then the function returns an error value.
• If Period is less than 1 then the function returns an error value.
• If Salvage is less than 0 then the function returns an error value.

## Additional details:

• The formula for DB is:

The rate is calculated as

and is rounded to 3 decimals.

For the first period the residual value is

For all periods, where period <= lifeTime, the residual value is calculated by

If month was specified, the residual value for the period after lifeTime becomes

The depreciation allowance for the first period is

For all other periods the allowance is calculated by

For all periods, where Period > lifeTime + 1 – INT(Month/12), the depreciation allowance is zero.

Here lifetime is Life parameter value.

## Examples:

Formula Description Returns
=DB(25000;1000;36;1;6) A computer system with an initial cost of 25,000 currency units is to be depreciated over a three year period. The salvage value is to be 1,000 currency units. One period is 30 days. The fixed-declining depreciation of the computer system is 1,075.00 currency units. 1,075.00 (Currency units)
=DB(25000;1000;36;1) The function calculates the depreciation of an asset for a specified period using the fixed-declining balance method. The value Month is 12 by default. 2,150.00 (Currency units)
=DB(25000;1000;0;1;6) The function returns an error value since Life cannot be less than Period. Err:502

DB