## Example 15.1 Solving a Highly Nonlinear Problem

Consider the following example of minimizing a highly nonlinear problem:

In this instance, the constants are . Assume the starting point . You can use the following SAS code to solve the problem:

```proc optmodel;
var x {1..10} >= -100 <= 100   /* variable bounds */
init -2.3;   /* starting point */

number c {1..10} = [-6.089 -17.164 -34.054 -5.914 -24.721
-14.986 -24.100 -10.708 -26.662 -22.179];

number a{1..3,1..10}=[1 2 2 0 0 1 0 0 0 1
0 0 0 1 2 1 1 0 0 0
0 0 1 0 0 0 1 1 2 1];

number b{1..3}=[8 1 1];

minimize obj =
sum{j in 1..10}exp(x[j])*(c[j]+x[j]
-log(sum {k in 1..10}exp(x[k])));

con cons{i in 1..3}:
sum{j in 1..10}a[i,j]*exp(x[j])=b[i];

solve with sqp / printfreq = 0;
quit;
```

The output is displayed in Output 15.1.1.

Output 15.1.1 OPTMODEL Output
The OPTMODEL Procedure

Problem Summary
Objective Sense Minimization
Objective Function obj
Objective Type Nonlinear

Number of Variables 10
Bounded Above 0
Bounded Below 0
Bounded Below and Above 10
Free 0
Fixed 0

Number of Constraints 3
Linear LE (<=) 0
Linear EQ (=) 0
Linear GE (>=) 0
Linear Range 0
Nonlinear LE (<=) 0
Nonlinear EQ (=) 3
Nonlinear GE (>=) 0
Nonlinear Range 0

Solution Summary
Solver SQP
Objective Function obj
Solution Status Optimal
Objective Value -102.0863279
Iterations 80

Infeasibility 4.7356948E-7
Optimality Error 3.3847789E-6
Complementarity 0

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