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The following snippet of code:

IloNum i = 2*0.6 + 3*0.4 + 0.6 - 0.4 + 1 + 0.6 - 0.4 + 1;
IloNum j = 4.8;
cout << "i=" << i << ", j=" << j << ", (i>j)=" << (i > j) << endl;

has the following output:

i=4.8, j=4.8, (i>j)=1

Have I overlooked some details in the documentation or is this a bug?

Philippe Olivier
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1 Answers1

1

You should remember that as soon as you do some float operations you have some tiny errors.

If I rewrite your model in OPL

float i = 2*0.6 + 3*0.4 + 0.6 - 0.4 + 1 + 0.6 - 0.4 + 1;
float j = 4.8;

execute
{
writeln("i=" , i , ", j=" ,j ,", (i>j)=" , (i > j) );;
}

float i2=i-4.8;
float j2=j-4.8;

execute
{
writeln("i2=",i2);
writeln("j2=",j2);
}

then I get

i=4.8, j=4.8, (i>j)=true
i2=8.881784197e-16
j2=0

which shows the tiny imprecision that leads to what looks like a paradox

And let me underline that

    float i = 2*6 + 3*4 + 6 - 4 + 10 + 6 - 4 + 10;
float j = 48;

execute
{
writeln("i=" , i , ", j=" ,j ,", (i>j)=" , (i > j) );;
}

float i2=i-48;
float j2=j-48;

execute
{
writeln("i2=",i2);
writeln("j2=",j2);
}

gives

i=48, j=48, (i>j)=false
i2=0
j2=0
halfer
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Alex Fleischer
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  • I am aware of the inherent imprecision associated with floating-point operations, but I would have expected to see this imprecision when printing `i` and `j`. I did not think to substract 4.8 from `i` to see its deviation from zero. Thanks. – Philippe Olivier Feb 21 '19 at 21:46