Thanks again Russ, I will try your suggestion.
Lennox.

Russ Jones <[EMAIL PROTECTED]> wrote: Hi Lennox.

Your observation may be correct - but in floating point, you need to  
know that 2.00000... is the same thing as 1.99999... And this applies  
to all numbers. The fraction is not precise, particularly when  
passing through a decimal to binary or binary to decimal conversion.

If you want to study up on it, there is a whole area of interest  
called numerical analysis.

Russ

On Nov 23, 2006, at 10:09 PM, Lennox Jacob wrote:

> Thanks Russ, I will try that but I should point out one observation:-
> With the current code, if the sum is exactly 100,000.00 or less,  
> all calculations are correct but once I go above 100,000.00 then  
> that problem occurs.
> Thanks again Russ.
> Lennox.
>
> Russ Jones  wrote:
> On Nov 23, 2006, at 9:01 PM, Lennox Jacob wrote:
>
>>
>> The problem is that when I use large numbers like 60000.23 +
>> 25000.23 + 25000.23 I get 110000.70 instead of 110000.69
>
> Hi Lennox.
>
> You have come up against a "problem" that has existed in computer
> science for at least the last 45 years - since that is how long I
> have been doing this stuff - and it probably existed before that.
>
> The problem is that you are using "Floating point" (or "double")
> variables to store your numbers, and expect to get fixed-point
> results. Floating point numbers are represented by a fraction,
> multiplied by an exponent. Works pretty well if all is in decimal,
> but computers seem to prefer binary numbers - and there is not a 1-
> to-1 correspondence between decimal fractions and binary fractions.
>
> If you really need the "exact" result, as is often the case in
> financial calculations, you will need to avoid floating point.
>
> You can do all your arithmetic in long integers, if you first
> multiply all incoming numbers by 100, an at the output end, divide
> the results by 100.
>
> HTH
>
> Russ
>
>
>
> Russ Jones
> [EMAIL PROTECTED]
> RB 2006r4 MacOS 10.4.8
>
>
>
>
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