Hello

I have a problem with the round function, if a value = 1.2345 the round figure is 1.23 however if a figure is 1.235 the round figure is 1.23 (should be 1.24 as 3rd figure is 5, i assume this would round up) is there a way i can set the round to round up if 3rd figure is =>5. I will post the expression i am using as well.

Round([Amount]*0.015,2)

Amount = a $ Value ie ($234.00) which i then get 1.5% of this figure and round to 2 decimal places.

Any Ideas

Regards

DB

The round function works correctly for me with everything I've tried. What is the data type of [Amount]? Do you have the full code of the section where the rounding is going wrong? Perhaps it's being truncated before it's being rounded.

Try also to use:
  Math.Round([Amount] * 0.015, 2)
  Round(cDbl([Amount]) * 0.015, 2)

If all else fails you can do your own rounding:
--------------------------------------------------------------
Dim vTmp As Variant, iMakeInt As Long, fRightNumber As Double
vTmp = [Amount] * 0.015 'Check this value, make sure there are several decimal places, not truncated
vTmp = vTmp + .005 'Add the extra '5' to the appropriate decimal place
iMakeInt = Int(vTmp * 100) 'Check and see that the value is correct minus the decimal
fRightNumber = cDbl(iMakeInt / 100) 'fRightNumber should now contain right number
--------------------------------------------------------------
If any of the values above as you step through the equation returns a bad value then you may consider a detect and repair. You may want to anyway if you're using proper data types and Round isn't working right.

Actually it seems like there was some problems with rounding depending on install and some inconsistencies even then. Found a really nice site on vb/vba rounding issues. It's quite extensive and you may find your answer there:
http://ewbi.blogs.com/develops/2003/...und_yet_a.html

Thanks for you suggestion

I went with the solution whick added .005 to the total ie [Amount]*0.015 then i add 0.005 which has resolved my problem.
After further investigation i found that the figures that were being supplied to me via excel had incorrect rounding figures as well. The 0.005 solution got us the closest to the figure required.

Thanks Again
Darren