Tuesday, March 16, 2021

Syd function

Returns a Double specifying the sum-of-years' digits depreciation of an asset for a specified period.

Syntax

SYD( cost, salvage, life, period )

The SYD function syntax has these arguments:

Argument

Description

cost

Required. Double specifying initial cost of the asset.

salvage

Required. Double specifying value of the asset at the end of its useful life.

life

Required. Double specifying length of the useful life of the asset.

period

Required. Double specifying period for which asset depreciation is calculated.

Remarks

The life and period arguments must be expressed in the same units. For example, if life is given in months, period must also be given in months. All arguments must be positive numbers.

Query examples

Expression

Results

SELECT SYD([LoanAmount],[LoanAmount]*.1,20,2) AS Expr1 FROM FinancialSample;

Calculates the depreciation for an asset valued as "LoanAmount", with a salvage value of 10% ("LoanAmount" multiplied by 0.1), considering the useful life of the asset to be 20 years. The depreciation is calculated for the second year.

SELECT SYD([LoanAmount],0,20,3) AS SLDepreciation FROM FinancialSample;

Returns the depreciation for an asset valued as "LoanAmount", with a salvage value of $0, considering the useful life of the asset to be 20 years. The results are displayed in the column SLDepreciation. The depreciation is calculated for the third year.

VBA example

Note: Examples that follow demonstrate the use of this function in a Visual Basic for Applications (VBA) module. For more information about working with VBA, select Developer Reference in the drop-down list next to Search and enter one or more terms in the search box.

This example uses the SYD function to return the depreciation of an asset for a specified 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). The period in years for which the depreciation is calculated is PDepr.

Dim Fmt, InitCost, SalvageVal, MonthLife, LifeTime, DepYear, 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 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
DepYear = CInt(InputBox("For which year do you want depreciation?"))
Do While DepYear < 1 Or DepYear > LifeTime
MsgBox "You must enter at least 1 but not more than " & LifeTime
DepYear = CInt(InputBox("For what year do you want depreciation?"))
Loop
PDepr = SYD(InitCost, SalvageVal, LifeTime, DepYear)
MsgBox "The depreciation for year " & DepYear & " is " & Format(PDepr, Fmt) & "."

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