SYD Function

Applies To
Access for Microsoft 365 Access 2021 Access 2019 Access 2016

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) & "."