一:cplex的使用:
1.1 导入cplex.jar,包的地址:https://pan.baidu.com/s/1Q0Bv24EQdelV2rY-IrLoZQ 提取码:xn14
1.2 将cplex1290.dll (地址:https://pan.baidu.com/s/1sQXqYTSJWywwaZt5AHVkZQ 提取码:nm0l)添加到VM option中,这里以IDEA为例,
同样eclipse
**二、求解问题。
- 一个简单的线性规划问题:**
maxs.t.x1+2x2+3x3
−x1+x2+x3≤20
x1−3x2+x3≤30
0≤x1≤40
java 中的 cplex 代码如下:
import ilog.concert.IloException;
import ilog.concert.IloNumVar;
import ilog.cplex.IloCplex;
public class LP1 {
public static void main(String[] args) {
try {
IloCplex cplex = new IloCplex(); // creat a model
double[] lb = {0.0, 0.0, 0.0};
double[] ub = {40.0, Double.MAX_VALUE, Double.MAX_VALUE};
IloNumVar[] x = cplex.numVarArray(3, lb, ub);
double[] objvals = {1.0, 2.0, 3.0};
cplex.addMaximize(cplex.scalProd(x, objvals));
double[] coeff1 = {-1.0, 1.0, 1.0};
double[] coeff2 = {1.0, -3.0, 1.0};
cplex.addLe(cplex.scalProd(x, coeff1), 20.0);
cplex.addLe(cplex.scalProd(x, coeff2), 30.0);
if (cplex.solve()) {
cplex.output().println("Solution status = " + cplex.getStatus());
cplex.output().println("Solution value = " + cplex.getObjValue());
double[] val = cplex.getValues(x);
for (int j = 0; j < val.length; j++)
cplex.output().println("x" + (j+1) + " = " + val[j]);
}
cplex.end();
} catch (IloException e) {
System.err.println("Concert exception caught: " + e);
}
}
}
输出结果:
Tried aggregator 1 time.
N
o LP presolve or aggregator reductions.
Presolve time = 0.00 sec. (0.00 ticks)
Iteration log . . .
Iteration: 1 Dual infeasibility = 0.000000
Iteration: 2 Dual objective = 202.500000
Solution status = Optimal
Solution value = 202.5
x1 = 40.0
x2 = 17.5
x3 = 42.5
这里注意一下,求的是线性公式的最大值,而在MATLAB中利用x=linprog(f,[],[],Aeq,beq,lb);
求解的是最小值(我怎么调试都不一致的原因),这里的参数就不在赘述了。
还有就是设置cplex API在https://wenku.baidu.com/view/7095dd76777f5acfa1c7aa00b52acfc789eb9fed.html有介绍(虽然是c#的,但是java可以借鉴。
三 线性最优解的java实现
代码地址:https://pan.baidu.com/s/1UeyQdPLMideXKqUaDBFYZQ 提取码:rte1
里面有详细注解,这里就不再解释,希望对大家有帮助!
借鉴:https://blog.csdn.net/robert_chen1988/article/details/78678289 ,感谢这位博主的文章。