Wrong result is coming while adding tow figures

P

pol

Hi all,
I wrote the following function . But in some time the result is showing
wrong format. When VarSum gets value 185.42 and ValueSum gets -185.42 the
result should be 0.

VarSum = VarSum + ValueSum
0= 185 + - 185 . But some cases Excel shows the balance as
8.5265128E-14.

Please help. I given the function below.

Sub CashBookMacrosnew()
Dim VarSum As Double
Dim ValueSum As Double
Dim VarSumOld As Double
Dim totalsum As Double

lastrow = Range("B" & Rows.Count).End(xlUp).Row
currow = ActiveCell.Row
pre_row = ActiveCell.Row
Col_A = ""

For RowCount = 1 To lastrow
If RowCount > 1 Then
pre_row = RowCount - 1
If Range("A" & RowCount) <> "" Then
VarSumOld = VarSum
VarSum = Range("R" & RowCount)
Range("Y" & pre_row) = VarSumOld
Else

ValueSum = (Range("R" & RowCount) + Range("S" & RowCount) + Range("T" &
RowCount) + _
Range("U" & RowCount) + Range("V" & RowCount) + Range("W" & RowCount) + _
Range("X" & RowCount))
VarSum = VarSum + ValueSum
End If
End If


Next RowCount

End Sub
 
J

JLatham

The extremely small difference is caused by the inability of a digital system
to accurately represent all analog values.

If you're only interested in 2 decimal places, replace
VarSum = VarSum + ValueSum
with
VarSum = INT((VarSum + ValueSum) * 100)/100
 
B

Bernard Liengme

The problem has to do with computers way of converting digital numbers
(number using base 10) to binary (base 2) and is a well known problem - see
sites below. I am not sure what JLatham mean by "analog".
Excel (and most other apps) store number with 15 digit precision,

To avoid this: suppose your formula is =A1+A2, replace this by
=ROUND(A1+A2,10)
The 10 could be another value such as 6 or 12, depending on the precision of
you work
best wishes
--
Bernard Liengme
Microsoft Excel MVP
http://people.stfx.ca/bliengme


Chip's clear explanation
http://www.cpearson.com/excel/rounding.htm

Floating-point arithmetic may give inaccurate results in Excel
http://support.microsoft.com/kb/78113/en-us

(Complete) Tutorial to Understand IEEE Floating-Point Errors
http://support.microsoft.com/kb/42980

What Every Computer Scientist Should Know About Floating Point
http://docs.sun.com/source/806-3568/ncg_goldberg.html
http://www.cpearson.com/excel/rounding.htm

Visual Basic and Arithmetic Precision
http://support.microsoft.com/defaul...port/kb/articles/Q279/7/55.ASP&NoWebContent=1

Others:
http://support.microsoft.com/kb/214118

http://docs.sun.com/source/806-3568/ncg_goldberg.html
Go to source web page>>
 
P

pol

Many Thanks for the reply. Your formula is working but still have some prolem
with some figures. For example

When do the calculation for

12165.68
-1118.08
-660.55
-10.51
-59.80
-8.62
-10308.12
----------------

After applying that formula , the result is getting as 0.03 insetad of
0. Other values are correct.
 
P

pol

Now it is working fine


Bernard Liengme said:
The problem has to do with computers way of converting digital numbers
(number using base 10) to binary (base 2) and is a well known problem - see
sites below. I am not sure what JLatham mean by "analog".
Excel (and most other apps) store number with 15 digit precision,

To avoid this: suppose your formula is =A1+A2, replace this by
=ROUND(A1+A2,10)
The 10 could be another value such as 6 or 12, depending on the precision of
you work
best wishes
--
Bernard Liengme
Microsoft Excel MVP
http://people.stfx.ca/bliengme


Chip's clear explanation
http://www.cpearson.com/excel/rounding.htm

Floating-point arithmetic may give inaccurate results in Excel
http://support.microsoft.com/kb/78113/en-us

(Complete) Tutorial to Understand IEEE Floating-Point Errors
http://support.microsoft.com/kb/42980

What Every Computer Scientist Should Know About Floating Point
http://docs.sun.com/source/806-3568/ncg_goldberg.html
http://www.cpearson.com/excel/rounding.htm

Visual Basic and Arithmetic Precision
http://support.microsoft.com/defaul...port/kb/articles/Q279/7/55.ASP&NoWebContent=1

Others:
http://support.microsoft.com/kb/214118

http://docs.sun.com/source/806-3568/ncg_goldberg.html
Go to source web page>>



.
 

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