Syntax
DDB(cost, salvage, life, period, [factor])
| Parameter | Description |
|---|---|
| cost | Parameter of the DDB function. |
| salvage | Parameter of the DDB function. |
| life | Parameter of the DDB function. |
| period | Parameter of the DDB function. |
| [factor] | (Optional.) Parameter of the DDB function. |
Examples
Year 1 computer depreciation
Formula
=DDB(8000, 500, 5, 1)
$3,200.00. An $8,000 computer with $500 salvage over 5 years gets $3,200 depreciation in year 1 (40% of $8,000).
Year 2 depreciation
Formula
=DDB(8000, 500, 5, 2)
$1,920.00. In year 2, 40% is applied to the remaining $4,800 book value.
Custom factor (150% declining balance)
Formula
=DDB(8000, 500, 5, 1, 1.5)
$2,400.00. Using a 1.5 factor instead of the default 2 gives a less aggressive depreciation of $2,400 in year 1.
Common Errors
#NUM!
Period exceeds life, salvage is negative, or life is zero. Also occurs if salvage exceeds cost.
#VALUE!
A required argument is not a valid number.
Tips
Won't depreciate below salvage
DDB automatically limits depreciation so the book value never drops below the salvage value, even if the formula rate would push it lower.
Adjust the factor
The default factor is 2 (double-declining). Use 1.5 for 150% declining balance, or 3 for triple-declining. Higher factors = more aggressive early depreciation.
Pair with SLN for switchover
In practice, companies often use DDB until the depreciation amount drops below what SLN would give, then switch to straight-line. You can model this with a MAX(DDB(...), SLN(...)) approach.
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