Topic
• 5 replies
• Latest Post - ‏2013-02-08T10:20:32Z by TobiasUni-Hohenheim
TobiasUni-Hohenheim
7 Posts

# Pinned topic Find maximum of function with 2 variables

‏2013-02-06T22:05:42Z |
Hi there,

I am new to CPLEX. Is there a way for cplex to find the maximum of a function with 2 linear variables? For example, find maximum of

f(x,y) = 3x - 2y

I was hoping cplex will give me (x,y) where f(x,y) is maximal. Of course, x and y may be in some predefined bounds.

How can I implement such a model?

Thanks!
Updated on 2013-02-08T10:20:32Z at 2013-02-08T10:20:32Z by TobiasUni-Hohenheim
• T_O
479 Posts

#### Re: Find maximum of function with 2 variables

‏2013-02-06T22:52:18Z
Hi Tobias,

this sounds trivial. Do you really have a linear objective function and only constraints that are in fact bounds of variables? Then you can trivially calculate the optimal value. Just take the upper or lower bound of every variable depending on the sign of the corresponding objective coefficient.

In your case, take the upper bound of x and the lower bound of y, because you are maximizing.

Of course, you can do this with CPLEX but I think you can do this faster with a trivial code.
Or are your "bounds" more difficult (e.g. linear functions)?

Best regards,
Thomas
• TobiasUni-Hohenheim
7 Posts

#### Re: Find maximum of function with 1 variable

‏2013-02-07T18:09:42Z
• T_O
• ‏2013-02-06T22:52:18Z
Hi Tobias,

this sounds trivial. Do you really have a linear objective function and only constraints that are in fact bounds of variables? Then you can trivially calculate the optimal value. Just take the upper or lower bound of every variable depending on the sign of the corresponding objective coefficient.

In your case, take the upper bound of x and the lower bound of y, because you are maximizing.

Of course, you can do this with CPLEX but I think you can do this faster with a trivial code.
Or are your "bounds" more difficult (e.g. linear functions)?

Best regards,
Thomas
Thanks! But sorry, it was my fault. I am looking to find the (local) maximum of a function like this:

f(x) = 4 * sqrt(x) - 2 * x

This is not possible with cplex, or?
Thanks!
7929 Posts

#### Re: Find maximum of function with 1 variable

‏2013-02-07T22:16:09Z
Thanks! But sorry, it was my fault. I am looking to find the (local) maximum of a function like this:

f(x) = 4 * sqrt(x) - 2 * x

This is not possible with cplex, or?
Thanks!
You can't do it directly with CPLEX. You could implement a cutting plane method (which is very old-school) using CPLEX, but it would be easier to use a program designed to solve nonlinear optimization problems.

Paul

Mathematicians are like Frenchmen: whenever you say something to them, they translate it into their own language, and at once it is something entirely different. (Goethe)
7929 Posts

#### Re: Find maximum of function with 1 variable

‏2013-02-08T00:58:29Z
Thanks! But sorry, it was my fault. I am looking to find the (local) maximum of a function like this:

f(x) = 4 * sqrt(x) - 2 * x

This is not possible with cplex, or?
Thanks!
Your function f(x) is concave, so any local maximum will be globally maximal. In general you cannot solve nonlinear problems using CPLEX, but in this case your problem can be made into a conic quadratic that CPLEX can handle: maximize 4*y - 2*x, subject to y^2 <= x and x >= 0. Indeed I wrote it in AMPL as
```var x >= 0; var y;
maximize obj: 4*y - 2*x;
subj to ydef: y^2 <= x;
```

and CPLEX happily solved it:
```ampl: option solver cplex;
ampl: solve;

CPLEX 12.5.0.0: optimal solution; objective 1.999999999
7 barrier iterations
No basis.

ampl: display x, y;
x = 0.999999
y = 0.999999
```

(Probably the reported objective would be exactly 2 if you set some tolerances a little tighter.) In the documentation of the CPLEX quadratic solvers, there's some discussion of the kinds of quadratic problems that can be solved.
Updated on 2014-03-24T22:40:25Z at 2014-03-24T22:40:25Z by iron-man
• TobiasUni-Hohenheim
7 Posts

#### Re: Find maximum of function with 1 variable

‏2013-02-08T10:20:32Z
Your function f(x) is concave, so any local maximum will be globally maximal. In general you cannot solve nonlinear problems using CPLEX, but in this case your problem can be made into a conic quadratic that CPLEX can handle: maximize 4*y - 2*x, subject to y^2 <= x and x >= 0. Indeed I wrote it in AMPL as
<pre class="java dw" data-editor-lang="java" data-pbcklang="java" dir="ltr">var x >= 0; var y; maximize obj: 4*y - 2*x; subj to ydef: y^2 <= x; </pre>
and CPLEX happily solved it:
<pre class="java dw" data-editor-lang="java" data-pbcklang="java" dir="ltr">ampl: option solver cplex; ampl: solve; CPLEX 12.5.0.0: optimal solution; objective 1.999999999 7 barrier iterations No basis. ampl: display x, y; x = 0.999999 y = 0.999999 </pre>
(Probably the reported objective would be exactly 2 if you set some tolerances a little tighter.) In the documentation of the CPLEX quadratic solvers, there's some discussion of the kinds of quadratic problems that can be solved.
Excellent, 4er! That answered my question!