Adrian,
Apart from the question of whether you feel it improves or degrades
readability of code, With block speed code up if they reduce the levels of
the hierarchy that the compiler needs to resolve. You can tell this by
counting the periods (.).

e.g.
ThisWorkbook.Worksheets(1).Range("A1").Borders(xlE dgeTop).LineStyle =
xlContinuous

'4 periods

With ThisWorkbook.Worksheets(1).Range("A1").Borders(xlE dgeTop)
.LineStyle = xlContinuous
End With

'Still 4 periods, so no gain

But:
ThisWorkbook.Worksheets(1).Range("A1").Borders(xlE dgeTop).LineStyle =
xlContinuous
ThisWorkbook.Worksheets(1).Range("A1").Borders(xlE dgeTop).Weight = xlThin
ThisWorkbook.Worksheets(1).Range("A1").Borders(xlE dgeTop).ColorIndex =
xlAutomatic
'Total 12 periods

With ThisWorkbook.Worksheets(1).Range("A1").Borders(xlE dgeTop)
.LineStyle = xlContinuous
.Weight = xlThin
.ColorIndex = xlAutomatic
End With
'Total 6 periods

You can also define your own variables to help the compiler and cut down the
number of levels
Dim brder As Border

Set brder = ThisWorkbook.Worksheets(1).Range("B2").Borders(xlE dgeTop)

With brder
.LineStyle = xlContinuous
.Weight = xlThin
.ColorIndex = xlAutomatic
End With
'Although in this case you gain nothing from the With, only from the
variable.

Not aware of a downside to either method, so at a loss to explain your
slower execution.
In a related comment, avoid using the generic "Object", unless you need to,
e.g. because of late binding.

NickHK

"Adrian D. Bailey" wrote in message
...
I have a large set of books and loads of macros - they all work OK, but
I've
been trying to tweak some of them to make them go a little faster.
I've discovered that sometimes adding the With - End With construct to a
macro can make it go faster (as I expected) but othertimes it makes it go
slower.
What are the rules?

Here is an example of a change that made things slower (part of a
function)...
Original:
Set rng = Workbooks("data.xls").Worksheets("students").Range ("1:1")
lkcol = Application.WorksheetFunction.Match(lookupkey, rng, 0)
rtncol = Application.WorksheetFunction.Match(rtnkey, rng, 0)
With Workbooks("data.xls").Worksheets("students")
Set rng2 = .Range(.Cells(1, lkcol), .Cells(65535, lkcol))
End With
rtnrow = Application.WorksheetFunction.Match(lookupval, rng2, 0)
Sinfo = Workbooks("data.xls").Worksheets("students").Cells (rtnrow, rtncol)

Revised, and markedly slower:
With Workbooks("data.xls").Worksheets("students")
Set rng = .Range("1:1")
lkcol = Application.WorksheetFunction.Match(lookupkey, rng, 0)
rtncol = Application.WorksheetFunction.Match(rtnkey, rng, 0)
Set rng2 = .Range(.Cells(1, lkcol), .Cells(65535, lkcol))
rtnrow = Application.WorksheetFunction.Match(lookupval, rng2, 0)
Sinfo = .Cells(rtnrow, rtncol)
End With
--
Adrian D.Bailey, Information and Systems Manager, Dept.Human Sciences
Loughborough University, Loughborough Leics, LE11 3TU, UK.
Tel: 01509 223007 Fax: 01509 223940

Community Warden, Storer and Burleigh Areas. Out-of-hours Tel: 01509
563263
--