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SLN Function 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."