Why?

T

Tennin

why c = 5.99999999999998E-02
Note: Excel XP SP

Sub test(
Dim a As Doubl
Dim b As Doubl
Dim c As Doubl
a = 1.4
b = 1.3
c = a -
Debug.Print
End Su
 
K

Kenny

Because of this:

Any value less than the result from this program is read
as zero on your machine. Machine Epsilon.

Double precision used in this program
Sub MachineEpsilon()

Dim g, ex, eps As Double
Dim i As Long
g = 1
i = 0
Do
i = i + 1
g = g / 2
ex = g * 0.98 + 1
ex = ex - 1
If ex > 0 Then eps = ex
Loop While ex > 0
MsgBox ("No. of Iterations " & i)
MsgBox ("Machine Epsilon is " & eps)
End Sub
 
D

Dana DeLouis

Thanks Kenny...I gave a program like this an attempt in the past, but had no
luck. Thanks for the code. :>)

Looks like one could change this line of code and get the same results:
ex = g * 0.999 + 1

Even this worked ok.
ex = g * 1 + 1

The result I get is:
Machine Epsilon is 2.22044604925031E-16

Which I think is the value of the last bit (Excel works with 15)

=2^(-52)
2.22044604925031E-16

Which agrees with mma on my system: :>)
$MachineEpsilon
2.220446049250313*^-16

Any value less than the result from this program is read
as zero on your machine. Machine Epsilon.
Click to expand...

I am not an expert, but do you think it would be more correct to say that
Excel can store and work with numbers less than this. For example =3*1.1E-20
is ok. It is only during certain math operations that such a small number
is treated as zero. It is so confusing! :>)

Just to share an idea, here is mma's definition
?$MachineEpsilon
$MachineEpsilon gives the smallest machine-precision number which can be
added to 1.0 to give a result that is distinguishable from 1.0

(They are careful to say 1.0, as 1.0 is a machine number.)

Fun subject, but so confusing. :>)
 
K

Kenny

THe programme is only for Double Precision numbers. Using
numbers outside double precision range using something
different. (I'm not an expert either, so I'm nt exactly
sure what). But for VBA programming, you just have to be
careful when using iterations with small numbers. You may
possibly get unexpected errors when the code seems fine.

see for examples

http://www5.in.tum.de/~huckle/bugse.html
 
N

NickHK

tennin,
Floating point calculations are not guarenteed to give answer you may
expect.

Depending on your needs, either Format the answer or

c=(CLng(a*100)-CLng(b*100)) /100

NickHK
 
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