The CLP Procedure |

This example illustrates how you can use the element constraint to represent almost any function between two variables in addition to representing nonstandard domains. Consider the following nonlinear optimization problem:

You can use the CLP procedure to solve this problem by introducing four artificial variables – to represent each of the nonlinear terms. Let , , , and . Since the domains of and are not consecutive integers that start from 1, you can use element constraints to represent their domains by using index variables and , respectively. For example, either of the following two ELEMENT constraints specifies that the domain of is the set of odd integers in :

element(z2,(-5,-3,-1,1,3,5,7,9),x2) element(z2,(-5 to 9 by 2),x2)

Any functional dependencies on or can now be defined using or , respectively, as the index variable in an element constraint. Since the domain of is , you can directly use as the index variable in an element constraint to define dependencies on .

For example, the following constraint specifies the function ,

element(z1,(-125,-64,-27,-8,-1,0,1,8,27,64,125),y1)

The nonlinear optimization problem can now be reduced to a set of element and linear constraints. By expressing the objective function as a linear constraint where *obj* is a supplied parameter, you can find a solution with objective value of at least *obj*.

In a manner similar to that of Example 3.3, you can create a SAS macro %CALLCLP with parameter *obj* that can be called iteratively from a search method such as %FINDMAX to determine the optimal value of the objective function, as follows:

%macro callclp(obj); %put The objective value is: &obj..; proc clp out=clpout; var x1=[-5, 5] x2=[-5, 9] x3=[1, 10] (y1-y4) (z1-z2); /* Use element constraint to represent non-contiguous domains */ /* and nonlinear functions. */ element /* Domain of x1 is [-5,5] */ (z1, ( -5 to 5), x1) /* Functional Dependencies on x1 */ /* y1 = x1^3 */ (z1, (-125, -64, -27, -8, -1, 0, 1, 8, 27, 64, 125), y1) /* y4 = mod(x1, 4) */ (z1, ( -1, 0, -3, -2, -1, 0, 1, 2, 3, 0, 1), y4) /* Domain of x2 is the set of odd numbers in [-5, 9] */ (z2, (-5 to 9 by 2), x2) /* Functional Dependencies on x3 */ /* y2 = 2^x3 */ (x3, (2, 4, 8, 16, 32, 64, 128, 256, 512, 1024), y2) /* y3 = x3^2 */ (x3, (1, 4, 9, 16, 25, 36, 49, 64, 81, 100), y3); lincon /* Objective function: x1^3 + 5 * x2 - 2^x3 */ y1 + 5 * x2 - y2 >= &obj, /* x1 - .5 * x2 + x3^2 <=50 */ x1 - .5 * x2 + y3 <= 50, /* mod(x1, 4) + .25 * x2 >= 1.5 */ y4 + .25 * x2 >= 1.5; run; /* when a solution is found, */ /* &_ORCLP_ contains the string SOLUTIONS_FOUND */ %if %index(&_ORCLP_, SOLUTIONS_FOUND) %then %let clpreturn=SUCCESSFUL; %mend; /* Bisection search to determine the optimal objective value. */ %macro findmax(lb, ub); %do %while (&lb<&ub-1); %put Currently lb=&lb, ub=&ub..; %let newobj=%eval((&lb+&ub)/2); %let clpreturn=NOTFOUND; %callclp(&newobj); %if &clpreturn=SUCCESSFUL %then %let lb=&newobj; %else %let ub=&newobj; %end; %callclp(&lb); %put Maximum possible objective value within given range is &lb.; %put Any value greater than &lb makes the problem infeasible.; proc print; run; %mend; /*Find the maximum objective value between -200 and 200. */ %findmax(lb=-200, ub=200);

Output 3.4.1 shows the solution that corresponds to the optimal objective value of 168.

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