# DB

Returns the depreciation of an asset for a specified period using the fixed-declining balance method.

**Syntax**

**DB**(**cost**,**salvage**,**life**,**period**,month)

**Cost** is the initial cost of the asset.

**Salvage** is the value at the end of the depreciation (sometimes called the salvage value of the asset).

**Life** is the number of periods over which the asset is being depreciated (sometimes called the useful life of the asset).

**Period** is the period for which you want to calculate the depreciation. Period must use the same units as life.

Month is the number of months in the first year. If month is omitted, it is assumed to be 12.

**Remarks**

The fixed-declining balance method computes depreciation at a fixed rate. DB uses the following formulas to calculate depreciation for a period:

(cost - total depreciation from prior periods) * rate

where:

rate = 1 - ((salvage / cost) ^ (1 / life)), rounded to three decimal places

Depreciation for the first and last periods is a special case. For the first period, DB uses this formula:

cost * rate * month / 12

For the last period, DB uses this formula:

((cost - total depreciation from prior periods) * rate * (12 - month)) / 12

**Examples**

Suppose a factory purchases a new machine. The machine costs $1,000,000 and has a lifetime of six years. The salvage value of the machine is $100,000. The following examples show depreciation over the life of the machine. The results are rounded to whole numbers.

`DB(1000000,100000,6,1,7)`

equals $186,083`DB(1000000,100000,6,2,7)`

equals $259,639`DB(1000000,100000,6,3,7)`

equals $176,814`DB(1000000,100000,6,4,7)`

equals $120,411`DB(1000000,100000,6,5,7)`

equals $82,000`DB(1000000,100000,6,6,7)`

equals $55,842`DB(1000000,100000,6,7,7)`

equals $15,845