Related Function:

The DB function calculates the depreciation of an asset for a given time period based on the fixed-declining balance method. The Excel DB Function uses the Fixed Declining Balance Method (see below) to calculate depreciation.

### Syntax

=DB(cost,salvage,life,period,[num_months])

#### Arguments

Argument Description
cost The original cost of the asset
salvage The salvage value after the asset has been fully depreciated
life The useful life of the asset or the number of periods that you will be depreciating the asset
period The period that you wish to calculate the depreciation for. Use the same units as for the life
[num_month] Optional. The number of months in the first year of depreciation

 • If supplied, the [num_month] argument is used to specify a partial year for the first period of depreciation

#### Examples

##### Example 1

The following spreadsheet shows the DB function used to calculate the yearly depreciation of an asset that cost $10,000 at the start of year 1, and has a salvage value of$1,000 after 5 years.

A B C D E
1 Data
2 $10,000 Initial cost 3$1,000 Salvage value
4 5 Life
5 12 Months in 1st year
6
7 Year Formula Depreciation EOP Value Accumulated Depreciation
8 1 =DB(10000,1000,5,1,12) $3,690.00$6,310.00 $3,690.00 9 2 =DB(A2,A3,5,2,12)$2,328.39 $3,981.61$6,018.39
10 3 =DB(A2,A3,A4,3,A5) $1,469.21$2,512.40 $7,487.60 11 4 =DB(A$2,A$3,A$4,A11,A$5)$927.07 $1,585.32$8,414.68
12 5 =DB(A$2,A$3,A$4,A12,A$5) $584.98$1,000.34 $8,999.66 ##### Example 2 The following spreadsheet shows the DB function used with the same cost, salvage and life argument values as in Example 1 above. However, in the following example, the depreciation calculation starts 7 months into year 1. A B C D E 1 Data 2$10,000 Initial cost
3 $1,000 Salvage value 4 5 Life 5 7 Months in 1st year 6 7 Year Formula Depreciation EOP Value Accumulated Depreciation 8 1 =DB(10000,1000,5,1,7)$2,152.50 $7,847.50$2,152.50
9 2 =DB(A2,A3,5,2,7) $2,895.73$4,951.77 $5,048.23 10 3 =DB(A2,A3,A4,3,A5)$1,827.20 $3,124.57$6,875.43
11 4 =DB(A2,A3,4,A11,A5) $1,152.97$1,971.60 $8,028.40 12 5 =DB(A$2,A$3,A$4,A12,A$5)$727.52 $1,244.08$8,755.92
13 6 =DB(A$2,A$3,A$4,A13,A$5) $191.28$1,052.80 \$98,947.20

Note: The yearly rate of depreciation in both examples, calculated from the equation 1-(Salvage/Cost)^(1/Life), is calculated to be 36.9%

#### Common Function Error(s)

Problem What went wrong
#VALUE! Occurs if any of the supplied arguments are not numeric values
#NUM! Occurs if:

 • the cost or the salvage argument is < 0 • the life or the period argument is ≤ 0 • the [num_month] argument is ≤ 0 or is > 12 • period > life (and the [num_month] argument is omitted) • period > life + 1 (and the [num_month] argument is supplied and is < 12)

When calculating the depreciation of an asset, several different calculations can be used.

One of the most popular methods of calculating depreciation is the Fixed Declining-Balance Method, in which, for each period of an asset’s useful lifetime, the calculated value is reduced by a fixed percentage of the asset’s value at the start of the period.

The Excel DB Function uses the Fixed Declining Balance Method, using the following constant percentage reduction per period (rounded to 3 decimal places):

where:

• Salvage is the final value of the asset at the end of its lifetime
• Cost is the initial cost of the asset
• Life is the number of periods over which the depreciation occurs