The Linear Programming Solver |
The following example illustrates how you can use the OPTMODEL procedure to solve linear programs. Suppose you want to solve the following problem:
You can use the following statement to call the OPTMODEL procedure for solving linear programs:
proc optmodel; var x{i in 1..3} >= 0; max f = x[1] + x[2] + x[3] ; con c1: 3*x[1] + 2*x[2] - x[3] <= 1; con c2: -2*x[1] - 3*x[2] + 2*x[3] <= 1; solve with lp / solver = ps presolver = none printfreq = 1; print x; quit;
The optimal solution and the optimal objective value are displayed in Figure 10.1.
Problem Summary | |
---|---|
Objective Sense | Maximization |
Objective Function | f |
Objective Type | Linear |
Number of Variables | 3 |
Bounded Above | 0 |
Bounded Below | 3 |
Bounded Below and Above | 0 |
Free | 0 |
Fixed | 0 |
Number of Constraints | 2 |
Linear LE (<=) | 2 |
Linear EQ (=) | 0 |
Linear GE (>=) | 0 |
Linear Range | 0 |
The iteration log displaying problem statistics, progress of the solution, and the optimal objective value is shown in Figure 10.2.
line |
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NOTE: The problem has 3 variables (0 free, 0 fixed). |
NOTE: The problem has 2 linear constraints (2 LE, 0 EQ, 0 GE, 0 range). |
NOTE: The problem has 6 linear constraint coefficients. |
NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range). |
NOTE: The OPTMODEL presolver removed 0 variables, 0 linear constraints, and 0 |
nonlinear constraints. |
WARNING: No output destinations active. |
NOTE: The OPTLP presolver value NONE is applied. |
NOTE: The PRIMAL SIMPLEX solver is called. |
Objective Entering Leaving |
Phase Iteration Value Variable Variable |
2 1 0.500000 x[3] c2 (S) |
2 2 8.000000 x[2] c1 (S) |
NOTE: Optimal. |
NOTE: Objective = 8. |
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