My MInverse function will not work in VBA

Question

EDIT: I fixed it, the ReDim and all starts at 0 and not 1 so I had a cell that was empty which wasn't supposed to be there! It now works, thanks for the help!

I'm trying to take a matrix and invert it, but for some reason I get this error:

Unable to get the MInverse property of the WorksheetFunction class.

My (relevant) code is as following:

Dim covar() As Variant
ReDim covar(UBound(assetNames), UBound(assetNames))
Dim corr() As Double
ReDim corr(UBound(assetNames), UBound(assetNames))

Dim covarTmp As Double
For i = 0 To UBound(assetNames) - 1
    For j = 0 To UBound(assetNames) - 1
        covarTmp = 0
        For t = 1 To wantedT
            covarTmp = covarTmp + (Log((prices(histAmount + 1 - t, i + 1)) / (prices(histAmount - t, i + 1))) - mu(i) * dt) * (Log((prices(histAmount + 1 - t, j + 1)) / (prices(histAmount - t, j + 1))) - mu(j) * dt)
        Next t
        covar(i, j) = covarTmp * (1 / ((wantedT - 1) * dt))
        corr(i, j) = covar(i, j) / (sigma(i) * sigma(j))
    Next j
Next i


Dim covarInv() As Variant
ReDim covarInv(UBound(assetNames), UBound(assetNames))
'ReDim covar(1 To UBound(assetNames), 1 To UBound(assetNames))

covarInv = Application.WorksheetFunction.MInverse(covar)

This last row is where the error occurs.

I've tried many things, having covar and covarInv dim as double, variant etc. Different ReDims on covar and covarInv.

Unable to get the MInverse property of the WorksheetFunction class. — Sam

Charles Williams Charles Williams · Accepted Answer · 2013-11-12T14:06:00

You don't say what version of Excel you are using, but with Excel 2010 there seems to be a Minverse maximum limit of 200 * 200 (for Excel 2003 its probably around 80 * 80): How many asset names do you have?

My MInverse function will not work in VBA

1 Answers