The Sequential Quadratic Programming Solver

Example 13.1: Solving a Highly Nonlinear Problem

Consider the following example of minimizing a highly nonlinear problem:

\min & \sum_{j = 1}^{10} e^{x_j} (c_j + x_j - \log {\sum_{k=1}^{10} e^{x_k} } )&...   ...e^{x_9} + e^{x_{10} } & {=} & 1 \    & -100 \le x_i \le 100,  i = 1, ... ,10 &

In this instance, the constants c_j are (\,-6.089, -17.164, -34.054, -5.914, -24.721, \linebreak   -14.986, -24.100, -10.708, -26.662, -22.179\,). Assume the starting point \mathbf{x}^0 = (\,-2.3, ...,-2.3,..., -2.3\,). 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 13.1.1.

Output 13.1.1: 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



The OPTMODEL Procedure

Solution Summary
Solver SQP
Objective Function obj
Solution Status Optimal
Objective Value -102.086323
Iterations 67
   
Infeasibility 8.7541812E-7
Optimality Error 1.3610787E-6
Complementarity 0




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