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由于我后台是springcloud，所以我调查到两种实现方式。
第一种是运用python的scipy开源库写一个python脚本，然后java后台调用python脚本，求最优解，然后再将结果返回。
第二种是运用java中ibm组件Cplex直接求解，但分为收费和免费版，免费版决策变量限制为1000个，但也够了。找jar包和dll费劲点。
发现他们俩算出来的最优解是相同的，但各个决策变量不太相同。
这里记录一下运用java中ibm组件Cplex求解的方法。
首先看一下官方说明，免费版本的限制。
申请下载免费版本的时候，要先注册一下填一些信息，然后会把下载地址发送到申请时的邮箱中，可以下载win、linux、macos版本。
这个安装完之后是一个ide开发工具CPLEX_Studio，由于我是用idea开发的，所以还要配置如下的步骤：
CPLEX_Studio安装完以后，安装目录下会有一个jar包，需要导入到我们的idea中，我是将jar包导入到我本地maven仓库中，用pom引入的：
CPLEX_Studio中jar包的位置：
D:\CPLEX_Studio_Community201\cplex\lib\cplex.jar
mvn install:install-file -Dfile=D:\CPLEX_Studio_Community201\cplex\lib\cplex.jar -DgroupId=cplex -DartifactId=cplex -Dversion=20.1.0.0.R1 -Dpackaging=jar
cplex
cplex
20.1.0.0.R1
然后在我们的启动项目时还要加入java.library.path参数，指定CPLEX运行库：
-Djava.library.path="D:\CPLEX_Studio_Community201\cplex\bin\x64_win64"
我在开发中是直接idea配置的：
之后就可以愉快的使用了。
我项目中的不等式方程组是∑求和形式的，这里就手动打码先假设决策变量的数量为 5 * 3 个，经过化简后得到多元一次不等式组如下：
求解最大值 = AX + AY + AZ + BX + BY + BZ + CX + CY + CZ + DX + DY + DZ + EX + EY + EZ
AX + AY + AZ <= 25400
BX + BY + BZ <= 18600
CX + CY + CZ <= 39800
DX + DY + DZ <= 53200
EX + EY + EZ <= 5900
AX + AY + AZ <= 10000
BX + BY + BZ <= 10000
CX + CY + CZ <= 10000
DX + DY + DZ <= 10000
EX + EY + EZ <= 10000
AX + BX + CX + DX + EX >= 15000
AY + BY + CY + DY + DY >= 5000
AZ + BZ + CZ + DZ + DZ >= 10000
(50.25-50)*AX + (49.86-50)*BX + (68.80-50)*CX + (49.79-50)*DX + (48.77-50)*EX >= 0
(50.25-60)*AY + (49.86-60)*BY + (68.80-60)*CY + (49.79-60)*DY + (48.77-60)*EY >= 0
(50.25-55)*AZ + (49.86-55)*BZ + (68.80-55)*CZ + (49.79-55)*DZ + (48.77-55)*DZ >= 0
(30.95*(1-2/100)-30)*AX + (31.52*(1-3/100)-30)*BX + (30.58*(1-1/100)-30)*CX + (30.17*(1-1/100)-30)*DX + (27.83*(1-1/100)-30)*EX >= 0
(30.95*(1-2/100)-30)*AY + (31.52*(1-3/100)-30)*BY + (30.58*(1-1/100)-30)*CY + (30.17*(1-1/100)-30)*DY + (27.83*(1-1/100)-30)*EY >= 0
(30.95*(1-2/100)-30)*AZ + (31.52*(1-3/100)-30)*BZ + (30.58*(1-1/100)-30)*CZ + (30.17*(1-1/100)-30)*DZ + (27.83*(1-1/100)-30)*EZ >= 0
(11.32*(1-2/100)-10)*AX + (12.83*(1-3/100)-10)*BX + (16.06*(1-1/100)-10)*CX + (5.68*(1-1/100)-10)*DX + (8.54*(1-1/100)-10)*EX >= 0
(11.32*(1-2/100)-10)*AY + (12.83*(1-3/100)-10)*BY + (16.06*(1-1/100)-10)*CY + (5.68*(1-1/100)-10)*DY + (8.54*(1-1/100)-10)*EY >= 0
(11.32*(1-2/100)-10)*AZ + (12.83*(1-3/100)-10)*BZ + (16.06*(1-1/100)-10)*CZ + (5.68*(1-1/100)-10)*DZ + (8.54*(1-1/100)-10)*EZ >= 0
(6*(1-2/100)-5)*AX + (4*(1-3/100)-5)*BX + (5*(1-1/100)-5)*CX + (2*(1-1/100)-5)*DX + (5*(1-1/100)-5)*EX <= 0
(6*(1-2/100)-5)*AY + (4*(1-3/100)-5)*BY + (5*(1-1/100)-5)*CY + (2*(1-1/100)-5)*DY + (5*(1-1/100)-5)*EY <= 0
(6*(1-2/100)-5)*AZ + (4*(1-3/100)-5)*BZ + (5*(1-1/100)-5)*CZ + (2*(1-1/100)-5)*DZ + (5*(1-1/100)-5)*EZ <= 0
非负约束：
AX,AY,AZ,BX,BY,BZ,CX,CY,CZ,DX,DY,DZ,EX,EY,EZ >= 0
上面这些是根据我自己的项目得到的不等式组，可以根据自己的项目来做相应改动。
try {
// creat a model
IloCplex cplex = new IloCplex();
// 变量的取值范围
double[] lb = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
double[] ub = {10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000};
IloNumVar[] x = cplex.numVarArray(15, lb, ub);
// 求解目标
double[] objvals = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
cplex.addMaximize(cplex.scalProd(x, objvals));
// 不等式约束
cplex.addLe(cplex.scalProd(x, new double[]{1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}), 25400);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0}), 18600);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0}), 39800);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0}), 53200);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1}), 5900);
cplex.addLe(cplex.scalProd(x, new double[]{1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}), 10000);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0}), 10000);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0}), 10000);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0}), 10000);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1}), 10000);
cplex.addLe(cplex.scalProd(x, new double[]{1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}), 20000);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0}), 20000);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0}), 20000);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0}), 20000);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1}), 20000);
cplex.addLe(cplex.scalProd(x, new double[]{-1, 0, 0, -1, 0, 0, -1, 0, 0, -1, 0, 0, -1, 0, 0}), -15000);
cplex.addLe(cplex.scalProd(x, new double[]{0, -1, 0, 0, -1, 0, 0, -1, 0, 0, -1, 0, 0, -1, 0}), -5000);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, -1, 0, 0, -1, 0, 0, -1, 0, 0, -1, 0, 0, -1}), -10000);
cplex.addLe(cplex.scalProd(x, new double[]{50 - 50.25, 0, 0, 50 - 49.86, 0, 0, 50 - 68.80, 0, 0, 50 - 49.79, 0, 0, 50 - 48.77, 0, 0}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{0, 60 - 50.25, 0, 0, 60 - 49.86, 0, 0, 60 - 68.80, 0, 0, 60 - 49.79, 0, 0, 60 - 48.77, 0}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 55 - 50.25, 0, 0, 55 - 49.86, 0, 0, 55 - 68.80, 0, 0, 55 - 49.79, 0, 0, 55 - 48.77}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{30 - 30.95 * 0.98, 0, 0, 30 - 31.52 * 0.97, 0, 0, 30 - 30.58 * 0.99, 0, 0, 30 - 30.17 * 0.99, 0, 0, 30 - 27.83 * 0.99, 0, 0}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{0, 30 - 30.95 * 0.98, 0, 0, 30 - 31.52 * 0.97, 0, 0, 30 - 30.58 * 0.99, 0, 0, 30 - 30.17 * 0.99, 0, 0, 30 - 27.83 * 0.99, 0}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 30 - 30.95 * 0.98, 0, 0, 30 - 31.52 * 0.97, 0, 0, 30 - 30.58 * 0.99, 0, 0, 30 - 30.17 * 0.99, 0, 0, 30 - 27.83 * 0.99}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{10 - 11.32 * 0.98, 0, 0, 10 - 12.83 * 0.97, 0, 0, 10 - 16.06 * 0.99, 0, 0, 10 - 5.68 * 0.99, 0, 0, 10 - 8.54 * 0.99, 0, 0}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{0, 10 - 11.32 * 0.98, 0, 0, 10 - 12.83 * 0.97, 0, 0, 10 - 16.06 * 0.99, 0, 0, 10 - 5.68 * 0.99, 0, 0, 10 - 8.54 * 0.99, 0}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 10 - 11.32 * 0.98, 0, 0, 10 - 12.83 * 0.97, 0, 0, 10 - 16.06 * 0.99, 0, 0, 10 - 5.68 * 0.99, 0, 0, 10 - 8.54 * 0.99}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{6 * 0.98 - 5, 0, 0, 4 * 0.97 - 5, 0, 0, 5 * 0.99 - 5, 0, 0, 2 * 0.99 - 5, 0, 0, 5 * 0.99 - 5, 0, 0}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{0, 6 * 0.98 - 5, 0, 0, 4 * 0.97 - 5, 0, 0, 5 * 0.99 - 5, 0, 0, 2 * 0.99 - 5, 0, 0, 5 * 0.99 - 5, 0}), 0);
cplex.addLe(cplex.scalProd(x, new double[]{0, 0, 6 * 0.98 - 5, 0, 0, 4 * 0.97 - 5, 0, 0, 5 * 0.99 - 5, 0, 0, 2 * 0.99 - 5, 0, 0, 5 * 0.99 - 5}), 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 (Exception e) {
System.err.println("Concert exception caught: " + e);
}
// Iteration log . . .
// Iteration: 1 Dual objective = 99025.641026
// Solution status = Optimal
// Solution value = 44280.112731282905
// x1 = 4022.4510446323093
// x2 = 0.0
// x3 = 5977.548955367691
// x4 = 10000.0
// x5 = 0.0
// x6 = 0.0
// x7 = 211.60418010554542
// x8 = 3814.5789532616586
// x9 = 5973.816866632794
// x10 = 0.0
// x11 = 801.3074331974203
// x12 = 9198.69256680258
// x13 = 2913.635241579134
// x14 = 384.1136135409201
// x15 = 982.3638761628501
// [INFO] ------------------------------------------------------------------------
// [INFO] BUILD SUCCESS
// [INFO] ------------------------------------------------------------------------
// [INFO] Total time: 2.757 s
// [INFO] Finished at: 2021-04-12T09:58:14+08:00
// [INFO] ------------------------------------------------------------------------感谢各位的阅读，以上就是“java线性规划问题举例分析”的内容了，经过本文的学习后，相信大家对java线性规划问题举例分析这一问题有了更深刻的体会，具体使用情况还需要大家实践验证。这里是创新互联，小编将为大家推送更多相关知识点的文章，欢迎关注！
文章题目：java线性规划问题举例分析
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