Wrong Answer! Why?

[littlefitzer]

Hi,
Why does the following piece of code five an answer of 0.999999999 rather than an answer of 1?

OnePenceCoins = (0.76 - (0.5 * 1 + 0.2 * 1 + 0.1 * 0 + 0.05 * 1 + 0.02 * 0)) / 0.01

what this equate to is 0.01 / 0.01, so why is it giving me the wrong answer?
I have tried it with the divide sign the other way around i.e. \ but it complains about dividing by zero. And I havetried using Fix() but that gives an anser of 0.

How can I get it to give the correct answer of 1?
Any help would be appreciated,
littlefitzer

[steve65]

This is an excerpt from Microsoft's Knowledge Base

INFO: Visual Basic and Arithmetic Precision

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The information in this article applies to:

Microsoft Visual Basic Professional Edition for Windows, versions 5.0 , 6.0
Microsoft Visual Basic Enterprise Edition for Windows, versions 5.0 , 6.0

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SUMMARY
When a Visual Basic program runs a statement such as 1.26 - 1.25, the answer is not 0.01, but 0.00999999 because of the complex way in which numbers are represented in a computer. This article describes how Visual Basic handles arithmetic precision.



MORE INFORMATION
When people count, they use the decimal numbering system, or base 10. Computers count by using the binary numbering system, or base 2. The binary system is perfect for a computer because it uses thousands of tiny electrical switches to store information internally. Each switch can have two states; a high electrical voltage or a low electrical voltage: essentially, on/off, or one/zero. Each switch is called a bit. By stringing these bits together, people can store numbers that are represented by a binary numbering system. Because the computer is always counting in binary and people are always counting in decimal, the computer is always converting from one numbering system to the other. A few simple decimal to binary number conversions are:

1 Decimal = 1 Binary
2 Decimal = 10 Binary
3 Decimal = 11 Binary
4 Decimal = 100 Binary
10 Decimal = 1010 Binary
So far, computer switches have only been representing the simplest of numbers like 1, 2, 3, and so forth. These numbers are often called "natural numbers" or "whole numbers." When you add, subtract, or multiply a whole number, the answer is a whole number. However, if you divide two whole numbers, you do not always get a whole number; you can get a fraction, also known as a "floating-point number." Sometimes these fractions do strange things. For example: 2317/990 = 2.34040404040.... with the 40 repeating over and over. These positive and negative integers and fractions, and the number 0, comprise the "rational number" system. A more formal definition would be: A rational number is any number that can be written as a fraction where the numerator and denominator are both integers.

Because there is no fractional part to an integer, its binary representation is much simpler than it is for a floating-point value. Every base 10 integer can be exactly converted to a base 2 integer. All that the computer needs is enough bits. However, this is not true for fractional numbers. For example, the base 10 fraction 1/100 can be represented as the decimal number 0.01. However, when the same number is converted to binary, it becomes a repeating binary number. The reason that 1/100 cannot be represented exactly in base 2 is similar to the way that 1/3 cannot be represented exactly in base 10; only fractional base 10 numbers that can be represented in the form p/q, where q is an integer power of 2, can be expressed exactly with a finite number of ones (1) and zeros (0).

To add to the issue, a computer has a finite amount of space available to hold a binary number. Computers frequently use a set of eight switches to represent a number, called a byte. If all eight of these switches in one byte are on or set to one (11111111), the binary number converts to a decimal number of 255. So 255 would be the largest number a byte could hold. The number 256 requires another switch, or nine bits. A Visual Basic integer is two bytes long. That is 16 bits or 16 ones and zeros to represent a binary number. A single in Visual Basic is four bytes long.

How Computers Represent Fractions

There is an Institute of Electrical and Electronics Engineers, Inc. (IEEE) standard that dictates the computer representation of fractions, also known as floating-point values. The standard uses 32 bits that are numbered 0 to 31 from left to right. The first bit is the sign bit to represent positive or negative numbers. The next eight bits represent the exponent, and the remaining 23 bits are the fraction. Therefore, not only does a floating-point value have a finite numerical value that it can hold, it has a limit to the accuracy of a number that it can hold. Taking the example of 0.01, this number cannot be represented in a finite amount of space. Therefore, this number is rounded down by approximately -2.78E-17 when it is stored in Visual Basic as a single data type.

When the mathematical operation 1.26 - 1.25 is performed, Visual Basic actually returns 0.009999 instead of 0.01. The value of 0.01 cannot be represented as a non-repeating binary decimal. Microsoft recommends that you always use format statement when numbers are displayed to avoid the confusion of 9.99999E-03 rather than 0.01. Also, if accuracy is important to your calculation, use the Decimal data type. A Double gives you more precision but does not completely resolve the problem.

A simple test to run in Visual Basic is a For loop as follows:

Dim i As Integer
Dim Sum As Single

For i = 0 To 100
  Sum = Sum + 0.01
Next i
Text1.Text = Sum

The sum is equal to 1.01, but the Text1.Text displays 1.009999.

For the same reason, you should always be very cautious when you make comparisons on numbers. The following example illustrates a common programming error:

Dim i As Integer
Dim x As Single
Dim y As Single

For i = 0 To 1000
  x = x + 0.01
Next i
y = 10.082 - 0.072

If (x = y) Then Text3.Text = "Equal!"
Text4.Text = x
Text5.Text = y

This code does not display "Equal!" in the Textbox, because 10.01 cannot be represented exactly in binary, which causes the value that results from the assignment to be slightly different in binary than the value that is generated from the arithmetic expression. In practice, you should always code such comparisons in such a way as to allow for some tolerance, such as:

Dim i As Integer
Dim x As Single
Dim y As Single

For i = 0 To 1000
  x = x + 0.01
Next i
y = 10.082 - 0.072

If ((x - y) < 0.01 And (x - y) > -0.01) Then
  Text3.Text = "Equal!"
End If
Text4.Text = x
Text5.Text = y

In addition, you need to be aware that in long mathematical expressions, Visual Basic stores intermediate values. For example, in the calculation CInt(4.555 * 100), the value 455.5 is temporarily stored and used in the CInt() function. If you break the calculation into the two steps, x = 4.555 * 100 and CInt(x), you avoid this internal storage and thus avoid the floating-point rounding error.

Following are other common floating-point errors:
Round-off error - This error results when all of the bits in a binary number cannot be used in a calculation.


Overflow and underflow - This error occurs when the result is too large or too small to be represented by the data type.


Division by a very small number - This can trigger a "divide by zero" error or can produce bad results.


NOTE : This is not specifically a Visual Basic problem. You have the same problem with C, FORTRAN, or any other language that stores floating-point numbers according to IEEE standards.

[littlefitzer]

Hmmmmm....thanks for that, very interesing. Now I know why it happens but at least it gives me something to think over.
Thanks again,
littlefitzer

[steve65]

No Problem

[RAEsquivelC]

Way back, since I started using BASIC, the fact that computers normally have this kind of "wrong" result, the standard procedure was to use a simple, rounding up procedure, based on the precision required for the specific task. For instance, for a precision of 3 decimals, one would write, for x = 0.1/0.1:

x = INT(0.1/0.1*1000+.5)/1000, and get, as the answer, x = 1

In general, I developed, and used, the simple generalization:

x = INT(blablabla*10^p + .5)/10^p, where p was the precision required.

[DiGiTaIErRoR]

Originally posted by RAEsquivelC
Way back, since I started using BASIC, the fact that computers normally have this kind of "wrong" result, the standard procedure was to use a simple, rounding up procedure, based on the precision required for the specific task. For instance, for a precision of 3 decimals, one would write, for x = 0.1/0.1:

x = INT(0.1/0.1*1000+.5)/1000, and get, as the answer, x = 1

In general, I developed, and used, the simple generalization:

x = INT(blablabla*10^p + .5)/10^p, where p was the precision required.

Good thing we have the Round function to do it for us!