SLN function

Returns a Double specifying the straight-line depreciation of an asset for a single period.

Syntax

SLN(cost, salvage, life)

The SLN function has these named arguments:

Remarks

The depreciation period must be expressed in the same unit as the life argument. All arguments must be positive numbers.

Example

This example uses the SLN function to return the straight-line depreciation of an asset for a single period given the asset's initial cost (InitCost), the salvage value at the end of the asset's useful life (SalvageVal), and the total life of the asset in years (LifeTime).

Dim Fmt, InitCost, SalvageVal, MonthLife, LifeTime, PDepr
Const YEARMONTHS = 12    ' Number of months in a year.
Fmt = "###,##0.00"    ' Define money format.
InitCost = InputBox("What's the initial cost of the asset?")
SalvageVal = InputBox("What's the asset's value at the end of its useful life?")
MonthLife = InputBox("What's the asset's useful life in months?")
Do While MonthLife < YEARMONTHS ' Ensure period is >= 1 year.
    MsgBox "Asset life must be a year or more."
    MonthLife = InputBox("What's the asset's useful life in months?")
Loop
LifeTime = MonthLife / YEARMONTHS    ' Convert months to years.
If LifeTime <> Int(MonthLife / YEARMONTHS) Then
    LifeTime = Int(LifeTime + 1)    ' Round up to nearest year.
End If
PDepr = SLN(InitCost, SalvageVal, LifeTime)
MsgBox "The depreciation is " & Format(PDepr, Fmt) & " per year."