Originally posted by: victortkl

Hi, using opl script, the float precision problem occures
i subtract 0.01 from 0.3 in each iteration, and got 0.199999999 instead of 0.2.
how can i solve this problem

here is my code example

dvar float x;
maximize x*2;
subject to
{
    x<=2;    
}
 
main
{
    for (var i=0.2;i>=0;i=i-0.01)
    {
        writeln("i:\t"+i);    
    }    
}

here is the output:

i: 0.2
i: 0.19
i: 0.18
i: 0.17
i: 0.16
i: 0.15
i: 0.14
i: 0.13
i: 0.12
i: 0.11
i: 0.09999999999999996
i: 0.08999999999999997
i: 0.07999999999999997
i: 0.06999999999999998
i: 0.05999999999999998
i: 0.04999999999999998
i: 0.03999999999999997
i: 0.02999999999999997
i: 0.01999999999999997
i: 0.009999999999999969
#DecisionOptimization
#OPLusingCPLEXOptimizer

Originally posted by: SystemAdmin

There is no function for setting display precision of script data. You can convert it into a string and do appropriate formatting as described in technote:
http://www-01.ibm.com/support/docview.wss?uid=swg21401490

For OPL model, display precision can be set as described in the documentation:
IDE and OPL > Optimization Programming Language (OPL) > Parameters and Settings in OPL > OPL language options

Regards,
Faisal
#DecisionOptimization
#OPLusingCPLEXOptimizer

Originally posted by: victortkl

thank you
this problem also occurs when i write into database and make some data wrong
but it's seems not a good alternative to change the number into string.
regards
victortkl
#DecisionOptimization
#OPLusingCPLEXOptimizer

Originally posted by: SystemAdmin

Have you considered using SQL Function like TRUNCATE when you insert your data?
#DecisionOptimization
#OPLusingCPLEXOptimizer

Originally posted by: victortkl

it's not a wise idea. because in every iteration, the model will use this value as a parameter. this imprecise value will cause some problem
#DecisionOptimization
#OPLusingCPLEXOptimizer

Originally posted by: John Cui

Set your CPLEX algorithm to dual or primal simplex will help you fix the problem.
Please give a try, and tell us the result.

John Cui
#DecisionOptimization
#OPLusingCPLEXOptimizer

Originally posted by: John Cui

Sorry，I misunderstood.

John Cui
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#OPLusingCPLEXOptimizer

Originally posted by: GuangFeng

You cannot convert 0.2 and 0.01 into IEEE 754 float numbers without losing any precision (round off errors). The way you are doing it:

i=i-0.01

accumulates round off errors over time.

The best way is to use integers. For example, if 0.2 represents 0.2 dollar, try to express it in cents: 20. Or at least you should avoid the accumulation of the round off errors. Try to use something as below at least, although it is still not perfect:

main
{
    for (var n=0;n<=20;n++)
    {
        writeln("i:\t"+((20-n)/100));    
    }    
}

#DecisionOptimization
#OPLusingCPLEXOptimizer

Originally posted by: victortkl

thank you
it's a good alternative. but still can solve this difficulty absolutely.
here is the answer.

i: 0.2
i: 0.19
i: 0.18
i: 0.17
i: 0.16
i: 0.15
i: 0.14
i: 0.13
i: 0.12
i: 0.11
i: 0.1
i: 0.09
i: 0.08
i: 0.07000000000000001
i: 0.06
i: 0.05
i: 0.04
i: 0.03
i: 0.02
i: 0.01
i: 0
#DecisionOptimization
#OPLusingCPLEXOptimizer

Originally posted by: GuangFeng

The codes demonstrate the idea of avoiding round off error accumulation. For your simple scenario, converting the numbers to integers is a better and easier alternative.
#DecisionOptimization
#OPLusingCPLEXOptimizer