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

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

Mathematical Formula:

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

Arguments:

Name	Type	Description
cost	number	The initial cost of the asset.
salvage	number	The value at the end of the depreciation. (This is sometimes called the salvage value of the asset.)
life	number	The number of periods over which the asset is being depreciated. (This is sometimes called the useful life of the asset.)
period	number	The period for which the depreciation is to be calculated. (period shall use the same units as life.)
month	number	The number of months in the first year. If omitted, a value of 12 is used.

Return Type and Value: number – The depreciation of an asset for a specified period using the fixed-declining balance method.

However, if

• cost, salvage, life, or period < 0, #NUM! is returned.

• month is outside the range 1–12, #NUM! is returned.

[Example:

DB(1000000,100000,6,1,7) results in 186,083.33
DB(1000000,100000,6,2,7) results in 259,639.42
DB(1000000,100000,6,3,7) results in 176,814.44
DB(1000000,100000,6,4,7) results in 120,410.64
DB(1000000,100000,6,5,7) results in 81,999.64
DB(1000000,100000,6,6,7) results in 55,841.76
DB(1000000,100000,6,7,7) results in 15,845.10

end example]