This example demonstrates the use of the NLP solver to solve the following highly nonlinear optimization problem, which appears in Hock and Schittkowski (1981):

The initial point used is . You can call the NLP solver within PROC OPTMODEL to solve the problem by writing the following SAS statements:
proc optmodel; var x{1..8} >= 0.1 <= 10; min f = 0.4*(x[1]/x[7])^0.67 + 0.4*(x[2]/x[8])^0.67 + 10  x[1]  x[2]; con c1: 1  0.0588*x[5]*x[7]  0.1*x[1] >= 0; con c2: 1  0.0588*x[6]*x[8]  0.1*x[1]  0.1*x[2] >= 0; con c3: 1  4*x[3]/x[5]  2/(x[3]^0.71*x[5])  0.0588*x[7]/x[3]^1.3 >= 0; con c4: 1  4*x[4]/x[6]  2/(x[4]^0.71*x[6])  0.0588*x[8]/x[4]^1.3 >= 0; con c5: 0.1 <= f <= 4.2; /* starting point */ x[1] = 6; x[2] = 3; x[3] = 0.4; x[4] = 0.2; x[5] = 6; x[6] = 6; x[7] = 1; x[8] = 0.5; solve with nlp / algorithm=activeset; print x; quit;
The summaries and the solution are shown in Output 8.1.1.
Output 8.1.1: Summaries and the Optimal Solution
Problem Summary  

Objective Sense  Minimization 
Objective Function  f 
Objective Type  Nonlinear 
Number of Variables  8 
Bounded Above  0 
Bounded Below  0 
Bounded Below and Above  8 
Free  0 
Fixed  0 
Number of Constraints  5 
Linear LE (<=)  0 
Linear EQ (=)  0 
Linear GE (>=)  0 
Linear Range  0 
Nonlinear LE (<=)  0 
Nonlinear EQ (=)  0 
Nonlinear GE (>=)  4 
Nonlinear Range  1 
Performance Information  

Execution Mode  On Client 
Number of Threads  2 
Solution Summary  

Solver  NLP 
Algorithm  Active Set 
Objective Function  f 
Solution Status  Optimal 
Objective Value  3.951163829 
Iterations  24 
Optimality Error  1.8631421E8 
Infeasibility  0 
[1]  x 

1  6.46511 
2  2.23271 
3  0.66740 
4  0.59576 
5  5.93268 
6  5.52724 
7  1.01332 
8  0.40067 