DEPDBSL Function

Returns the declining balance with conversion to a straight-line depreciation.

Category:

Financial

Syntax

DEPDBSL(p,v,y,r)

Required Arguments

p

is an integer, the period for which the calculation is to be done.

v

is numeric, the depreciable initial value of the asset.

y

is an integer, the lifetime of the asset.

Range

y > 0

r

is numeric, the rate of depreciation that is expressed as a fraction.

Range

r ≥ 0

Details

The DEPDBSL function returns the depreciation by using the declining balance method with conversion to a straight-line depreciation, which is given by the following equation:

\begin{matrix} D E P D B S L (p, v, y, r) = {\begin{matrix} 0 & p \leq 0 \\ v \frac{r}{y} {(1 - \frac{r}{y})}^{p - 1} & 0 < p \leq t \\ \frac{v (1 - \frac{r}{y}) \begin{matrix} t \end{matrix}}{(y - t)} & t < p \leq y \\ 0 & p > y \end{matrix} \end{matrix}

The following relationship applies to the preceding equation:

\begin{matrix} t = i n t (y - \frac{y}{r} + 1) \end{matrix}

and int( ) denotes the integer part of a numeric argument.

The p and y arguments must be expressed by using the same units of time. The declining balance that changes to a straight-line depreciation chooses for each time period the method of depreciation (declining balance or straight-line on the remaining balance) that gives the larger depreciation.

Example

An asset has a depreciable initial value of $1,000 and a ten-year lifetime. Using a declining balance rate of 150%, the depreciation of the value of the asset in the fifth year can be expressed as

   y5=depdbsl(5,1000,10,1.5);

The value 87.001041667 is returned. The first and the third arguments are expressed in years.