java 线性规划 和lingo 比较 - cndavy - 博客园
model:max=13*A+ 23*B;
5*A + 15*B &480 ;
4*A + 4 *B &160 ;
35* A + 20 *B &1190
Reduced Cost
Slack or Surplus
Dual Price
public static void test0() {
double[][] A = {
double[] c = { 13, 23 };
double[] b = { 480,160,1190};
test(A, b, c);}
value = 800.0x[0] = 11.998x[1] = 28.0y[0] = 1.0y[1] = 2.0y[2] = -0.0
package 线性规划;
* Created by han on .
import java.io.PrintS
/*************************************************************************
Compilation:
javac Simplex.java
Execution:
java Simplex
Given an M-by-N matrix A, an M-length vector b, and an
N-length vector c, solve the
LP { max cx : Ax &= b, x &= 0 }.
Assumes that b &= 0 so that x = 0 is a basic feasible solution.
Creates an (M+1)-by-(N+M+1) simplex tableaux with the
RHS in column M+N, the objective function in row M, and
slack variables in columns M through M+N-1.
*************************************************************************/
public class Simplex {
private static final double EPSILON = 1.0E-10;
private double[][]
// tableaux
private int M;
// number of constraints
private int N;
// number of original variables
private int[]
// basis[i] = basic variable corresponding to row i
private static PrintStream StdOut=System.
// only needed to print out solution, not book
// sets up the simplex tableaux
public Simplex(double[][] A, double[] b, double[] c) {
a = new double[M+1][N+M+1];
for (int i = 0; i & M; i++)
for (int j = 0; j & N; j++)
a[i][j] = A[i][j];
for (int i = 0; i & M; i++) a[i][N+i] = 1.0;
for (int j = 0; j & N; j++) a[M][j]
for (int i = 0; i & M; i++) a[i][M+N] = b[i];
basis = new int[M];
for (int i = 0; i & M; i++) basis[i] = N +
// check optimality conditions
assert check(A, b, c);
// run simplex algorithm starting from initial BFS
private void solve() {
while (true) {
// find entering column q
int q = bland();
if (q == -1) break;
// optimal
// find leaving row p
int p = minRatioRule(q);
if (p == -1) throw new RuntimeException("Linear program is unbounded");
pivot(p, q);
// update basis
basis[p] =
// lowest index of a non-basic column with a positive cost
private int bland() {
for (int j = 0; j & M + N; j++)
if (a[M][j] & 0) return
return -1;
// optimal
// index of a non-basic column with most positive cost
private int dantzig() {
int q = 0;
for (int j = 1; j & M + N; j++)
if (a[M][j] & a[M][q]) q =
if (a[M][q] &= 0) return -1;
// optimal
else return
// find row p using min ratio rule (-1 if no such row)
private int minRatioRule(int q) {
int p = -1;
for (int i = 0; i & M; i++) {
if (a[i][q] &= 0) continue;
else if (p == -1) p =
else if ((a[i][M+N] / a[i][q]) & (a[p][M+N] / a[p][q])) p =
// pivot on entry (p, q) using Gauss-Jordan elimination
private void pivot(int p, int q) {
// everything but row p and column q
for (int i = 0; i &= M; i++)
for (int j = 0; j &= M + N; j++)
if (i != p && j != q) a[i][j] -= a[p][j] * a[i][q] / a[p][q];
// zero out column q
for (int i = 0; i &= M; i++)
if (i != p) a[i][q] = 0.0;
// scale row p
for (int j = 0; j &= M + N; j++)
if (j != q) a[p][j] /= a[p][q];
a[p][q] = 1.0;
// return optimal objective value
public double value() {
return -a[M][M+N];
// return primal solution vector
public double[] primal() {
double[] x = new double[N];
for (int i = 0; i & M; i++)
if (basis[i] & N) x[basis[i]] = a[i][M+N];
// return dual solution vector
public double[] dual() {
double[] y = new double[M];
for (int i = 0; i & M; i++)
y[i] = -a[M][N+i];
// is the solution primal feasible?
private boolean isPrimalFeasible(double[][] A, double[] b) {
double[] x = primal();
// check that x &= 0
for (int j = 0; j & x. j++) {
if (x[j] & 0.0) {
StdOut.println("x[" + j + "] = " + x[j] + " is negative");
return false;
// check that Ax &= b
for (int i = 0; i & M; i++) {
double sum = 0.0;
for (int j = 0; j & N; j++) {
sum += A[i][j] * x[j];
if (sum & b[i] + EPSILON) {
StdOut.println("not primal feasible");
StdOut.println("b[" + i + "] = " + b[i] + ", sum = " + sum);
return false;
return true;
// is the solution dual feasible?
private boolean isDualFeasible(double[][] A, double[] c) {
double[] y = dual();
// check that y &= 0
for (int i = 0; i & y. i++) {
if (y[i] & 0.0) {
StdOut.println("y[" + i + "] = " + y[i] + " is negative");
return false;
// check that yA &= c
for (int j = 0; j & N; j++) {
double sum = 0.0;
for (int i = 0; i & M; i++) {
sum += A[i][j] * y[i];
if (sum & c[j] - EPSILON) {
StdOut.println("not dual feasible");
StdOut.println("c[" + j + "] = " + c[j] + ", sum = " + sum);
return false;
return true;
// check that optimal value = cx = yb
private boolean isOptimal(double[] b, double[] c) {
double[] x = primal();
double[] y = dual();
double value = value();
// check that value = cx = yb
double value1 = 0.0;
for (int j = 0; j & x. j++)
value1 += c[j] * x[j];
double value2 = 0.0;
for (int i = 0; i & y. i++)
value2 += y[i] * b[i];
if (Math.abs(value - value1) & EPSILON || Math.abs(value - value2) & EPSILON) {
StdOut.println("value = " + value + ", cx = " + value1 + ", yb = " + value2);
return false;
return true;
private boolean check(double[][]A, double[] b, double[] c) {
return isPrimalFeasible(A, b) && isDualFeasible(A, c) && isOptimal(b, c);
// print tableaux
public void show() {
StdOut.println("M = " + M);
StdOut.println("N = " + N);
for (int i = 0; i &= M; i++) {
for (int j = 0; j &= M + N; j++) {
StdOut.printf("%7.2f ", a[i][j]);
StdOut.println();
StdOut.println("value = " + value());
for (int i = 0; i & M; i++)
if (basis[i] & N) StdOut.println("x_" + basis[i] + " = " + a[i][M+N]);
StdOut.println();
public static void test(double[][] A, double[] b, double[] c) {
Simplex lp = new Simplex(A, b, c);
StdOut.println("value = " + lp.value());
double[] x = lp.primal();
for (int i = 0; i & x. i++)
StdOut.println("x[" + i + "] = " + x[i]);
double[] y = lp.dual();
for (int j = 0; j & y. j++)
StdOut.println("y[" + j + "] = " + y[j]);
public static void test0() {
double[][] A = {
double[] c = { 13, 23 };
double[] b = { 480,160,1190};
test(A, b, c);
public static void test1() {
double[][] A = {
double[] c = { 1, 1, 1 };
double[] b = { 5, 45, 27, 24, 4 };
test(A, b, c);
// x0 = 12, x1 = 28, opt = 800
public static void test2() {
double[] c = {
double[] b = { 480.0, 160.0, 1190.0 };
double[][] A = {
5.0, 15.0 },
{ 35.0, 20.0 },
test(A, b, c);
// unbounded
public static void test3() {
double[] c = { 2.0, 3.0, -1.0, -12.0 };
double[] b = {
double[][] A = {
{ -2.0, -9.0,
1.0, -1.0, -2.0 },
test(A, b, c);
// degenerate - cycles if you choose most positive objective function coefficient
public static void test4() {
double[] c = { 10.0, -57.0, -9.0, -24.0 };
double[] b = {
double[][] A = {
{ 0.5, -5.5, -2.5, 9.0 },
{ 0.5, -1.5, -0.5, 1.0 },
0.0, 0.0 },
test(A, b, c);
// test client
public static void main(String[] args) {
{ test0();
catch (Exception e) { e.printStackTrace(); }
StdOut.println("--------------------------------");
{ test1();
catch (Exception e) { e.printStackTrace(); }
StdOut.println("--------------------------------");
{ test2();
catch (Exception e) { e.printStackTrace(); }
StdOut.println("--------------------------------");
{ test3();
catch (Exception e) { e.printStackTrace(); }
StdOut.println("--------------------------------");
{ test4();
catch (Exception e) { e.printStackTrace(); }
StdOut.println("--------------------------------");
int M = Integer.parseInt(args[0]);
int N = Integer.parseInt(args[1]);
double[] c = new double[N];
double[] b = new double[M];
double[][] A = new double[M][N];
for (int j = 0; j & N; j++)
c[j] = StdRandom.uniform(1000);
for (int i = 0; i & M; i++)
b[i] = StdRandom.uniform(1000);
for (int i = 0; i & M; i++)
for (int j = 0; j & N; j++)
A[i][j] = StdRandom.uniform(100);
Simplex lp = new Simplex(A, b, c);
StdOut.println(lp.value());
private static class StdRandom {
public static double uniform(int i) {
return Math.random()*i;
阅读(...) 评论()用JAVA求线性规划_百度知道
用JAVA求线性规划
求JAVA解线性规划
∑xy=387.5
要求输出以下结果
b=3.2429, a=4.3999
The regression equation for this data is therefore:
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线性约束规划问题java求解
Linear programming.Linear programming solvers. is a bare-bones versionof the simplex algorithm.is a Java linear program solver created by David Applegate, William Cook,Sanjeeb Dash, and Monika Mevenkamp.It can be used at no cost for research or education purposes. solves a linear programin LP format, such as .contains a linear programming solver in the optimization toolbox.[wayne:tombstone] ~& matlab
& M A T L A B (R) &
The MathWorks, Inc.
Version 7.9.0.529 (R2009b) 64-bit (glnxa64)
August 12, 2009
&& A = [5 15; 4 4; 35 20];
&& b = [480; 160; 1190];
&& c = [13; 23];
&& lb = [0; 0];
&& ub = [ inf];
&& x = linprog(-c, A, b, [], [], lb, ub)
is a high-performance mathematical programming solverfor linear programming, mixed integer programming, and quadratic programming.It supports interfaces to C, C++, Java, Python, Matlab, and Microsoft Excel.It is also accessible via the modeling systems includingAIMMS, AMPL, GAMS, and MPL. is a modeling language for mathematical programming.The files
and specify the model and data for the brewery problem.[wayne:tombstone] ~& ampl
ILOG AMPL 9.100
AMPL Version
(SunOS 5.8)
ampl: model beer.
ampl: data beer.
ILOG CPLEX 9.100
CPLEX 9.1.0: objective 800
2 dual simplex iterations (1 in phase I)
ampl: display constraints.
Microsoft Excel has a primitive solver add-in for Windows, but it isno longer available for Mac.Below is the syntax highlighted version of from ./*************************************************************************
Compilation:
javac Simplex.java
Execution:
java Simplex
Given an M-by-N matrix A, an M-length vector b, and an
N-length vector c, solve the
LP { max cx : Ax &= b, x &= 0 }.
Assumes that b &= 0 so that x = 0 is a basic feasible solution.
Creates an (M+1)-by-(N+M+1) simplex tableaux with the
RHS in column M+N, the objective function in row M, and
slack variables in columns M through M+N-1.
*************************************************************************/
public class Simplex {
private static final double EPSILON = 1.0E-10;
private double[][]
// tableaux
private int M;
// number of constraints
private int N;
// number of original variables
private int[]
// basis[i] = basic variable corresponding to row i
// only needed to print out solution, not book
// sets up the simplex tableaux
public Simplex(double[][] A, double[] b, double[] c) {
a = new double[M+1][N+M+1];
for (int i = 0; i & M; i++)
for (int j = 0; j & N; j++)
a[i][j] = A[i][j];
for (int i = 0; i & M; i++) a[i][N+i] = 1.0;
for (int j = 0; j & N; j++) a[M][j]
for (int i = 0; i & M; i++) a[i][M+N] = b[i];
basis = new int[M];
for (int i = 0; i & M; i++) basis[i] = N +
// check optimality conditions
assert check(A, b, c);
// run simplex algorithm starting from initial BFS
private void solve() {
while (true) {
// find entering column q
int q = bland();
if (q == -1)
// optimal
// find leaving row p
int p = minRatioRule(q);
if (p == -1) throw new RuntimeException("Linear program is unbounded");
pivot(p, q);
// update basis
basis[p] =
// lowest index of a non-basic column with a positive cost
private int bland() {
for (int j = 0; j & M + N; j++)
if (a[M][j] & 0)
return -1;
// optimal
// index of a non-basic column with most positive cost
private int dantzig() {
int q = 0;
for (int j = 1; j & M + N; j++)
if (a[M][j] & a[M][q]) q =
if (a[M][q] &= 0) return -1;
// optimal
// find row p using min ratio rule (-1 if no such row)
private int minRatioRule(int q) {
int p = -1;
for (int i = 0; i & M; i++) {
if (a[i][q] &= 0)
else if (p == -1) p =
else if ((a[i][M+N] / a[i][q]) & (a[p][M+N] / a[p][q])) p =
// pivot on entry (p, q) using Gauss-Jordan elimination
private void pivot(int p, int q) {
// everything but row p and column q
for (int i = 0; i &= M; i++)
for (int j = 0; j &= M + N; j++)
if (i != p && j != q) a[i][j] -= a[p][j] * a[i][q] / a[p][q];
// zero out column q
for (int i = 0; i &= M; i++)
if (i != p) a[i][q] = 0.0;
// scale row p
for (int j = 0; j &= M + N; j++)
if (j != q) a[p][j] /= a[p][q];
a[p][q] = 1.0;
// return optimal objective value
public double value() {
return -a[M][M+N];
// return primal solution vector
public double[] primal() {
double[] x = new double[N];
for (int i = 0; i & M; i++)
if (basis[i] & N) x[basis[i]] = a[i][M+N];
// return dual solution vector
public double[] dual() {
double[] y = new double[M];
for (int i = 0; i & M; i++)
y[i] = -a[M][N+i];
// is the solution primal feasible?
private boolean isPrimalFeasible(double[][] A, double[] b) {
double[] x = primal();
// check that x &= 0
for (int j = 0; j & x. j++) {
if (x[j] & 0.0) {
StdOut.println("x[" + j + "] = " + x[j] + " is negative");
// check that Ax &= b
for (int i = 0; i & M; i++) {
double sum = 0.0;
for (int j = 0; j & N; j++) {
sum += A[i][j] * x[j];
if (sum & b[i] + EPSILON) {
StdOut.println("not primal feasible");
StdOut.println("b[" + i + "] = " + b[i] + ", sum = " + sum);
// is the solution dual feasible?
private boolean isDualFeasible(double[][] A, double[] c) {
double[] y = dual();
// check that y &= 0
for (int i = 0; i & y. i++) {
if (y[i] & 0.0) {
StdOut.println("y[" + i + "] = " + y[i] + " is negative");
// check that yA &= c
for (int j = 0; j & N; j++) {
double sum = 0.0;
for (int i = 0; i & M; i++) {
sum += A[i][j] * y[i];
if (sum & c[j] - EPSILON) {
StdOut.println("not dual feasible");
StdOut.println("c[" + j + "] = " + c[j] + ", sum = " + sum);
// check that optimal value = cx = yb
private boolean isOptimal(double[] b, double[] c) {
double[] x = primal();
double[] y = dual();
double value = value();
// check that value = cx = yb
double value1 = 0.0;
for (int j = 0; j & x. j++)
value1 += c[j] * x[j];
double value2 = 0.0;
for (int i = 0; i & y. i++)
value2 += y[i] * b[i];
if (Math.abs(value - value1) & EPSILON || Math.abs(value - value2) & EPSILON) {
StdOut.println("value = " + value + ", cx = " + value1 + ", yb = " + value2);
private boolean check(double[][]A, double[] b, double[] c) {
return isPrimalFeasible(A, b) && isDualFeasible(A, c) && isOptimal(b, c);
// print tableaux
public void show() {
StdOut.println("M = " + M);
StdOut.println("N = " + N);
for (int i = 0; i &= M; i++) {
for (int j = 0; j &= M + N; j++) {
StdOut.printf("%7.2f ", a[i][j]);
StdOut.println();
StdOut.println("value = " + value());
for (int i = 0; i & M; i++)
if (basis[i] & N) StdOut.println("x_" + basis[i] + " = " + a[i][M+N]);
StdOut.println();
public static void test(double[][] A, double[] b, double[] c) {
Simplex lp = new Simplex(A, b, c);
StdOut.println("value = " + lp.value());
double[] x = lp.primal();
for (int i = 0; i & x. i++)
StdOut.println("x[" + i + "] = " + x[i]);
double[] y = lp.dual();
for (int j = 0; j & y. j++)
StdOut.println("y[" + j + "] = " + y[j]);
public static void test1() {
double[][] A = {
double[] c = { 1, 1, 1 };
double[] b = { 5, 45, 27, 24, 4 };
test(A, b, c);
// x0 = 12, x1 = 28, opt = 800
public static void test2() {
double[] c = {
double[] b = { 480.0, 160.0, 1190.0 };
double[][] A = {
5.0, 15.0 },
{ 35.0, 20.0 },
test(A, b, c);
// unbounded
public static void test3() {
double[] c = { 2.0, 3.0, -1.0, -12.0 };
double[] b = {
double[][] A = {
{ -2.0, -9.0,
1.0, -1.0, -2.0 },
test(A, b, c);
// degenerate - cycles if you choose most positive objective function coefficient
public static void test4() {
double[] c = { 10.0, -57.0, -9.0, -24.0 };
double[] b = {
double[][] A = {
{ 0.5, -5.5, -2.5, 9.0 },
{ 0.5, -1.5, -0.5, 1.0 },
0.0, 0.0 },
test(A, b, c);
// test client
public static void main(String[] args) {
{ test1();
catch (Exception e) { e.printStackTrace(); }
StdOut.println("--------------------------------");
{ test2();
catch (Exception e) { e.printStackTrace(); }
StdOut.println("--------------------------------");
{ test3();
catch (Exception e) { e.printStackTrace(); }
StdOut.println("--------------------------------");
{ test4();
catch (Exception e) { e.printStackTrace(); }
StdOut.println("--------------------------------");
int M = Integer.parseInt(args[0]);
int N = Integer.parseInt(args[1]);
double[] c = new double[N];
double[] b = new double[M];
double[][] A = new double[M][N];
for (int j = 0; j & N; j++)
c[j] = StdRandom.uniform(1000);
for (int i = 0; i & M; i++)
b[i] = StdRandom.uniform(1000);
for (int i = 0; i & M; i++)
for (int j = 0; j & N; j++)
A[i][j] = StdRandom.uniform(100);
Simplex lp = new Simplex(A, b, c);
StdOut.println(lp.value());
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写出下列线性规划问题的对偶问题：
（1)max z=2x1+x2+x3-x4
（2)max z=4x1+5x2+3x3+6x4
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提问人：匿名网友
发布时间：
写出下列线性规划问题的对偶问题：&&(1)max z=2x1+x2+x3-x4&&&&(2)max z=4x1+5x2+3x3+6x4&&&&(3)&&&&(4)&&
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确认密码：java如何调用cplex
java如何调用cplex
　　Cplex是IBM公司的一个优化软件，可以用来求解线性规划(Linear Programming，LP)，二次规划(Quadratic Programming，QP)、混合整数规划(Mixed Integer Programming，MIP)问题。下面就让学习啦小编给大家说说java如何调用cplex吧。
　　java调用cplex的方法
　　首先需要安装Cplex软件，我安装的版本是cplex_studio122.win-x86-32.exe
　　下图安装后打开的Cplex自带的IDE，看上去跟Eclipse差不多。
　　在Cplex的安装目录下有许多值得我们学习的东西，还有一些examples，可供我们参考。
　　我是在Eclipse中使用Java调用Cplex，所以先把一些Cplex依赖加上。
　　运行依赖：cplex.jar(在..\cplex\lib目录下找到)和cplex122.dll(在..\cplex\bin目录下找到)。将cplex.jar加到工程的Build Path中。
　　cplex122.dll可以设置到运行时的环境中(VM arguments)，或者添加到项目的Native library location。
　　接下来我们求解一个具体的线性规划问题。
　　例如，我们求解下面这样一个线性规划问题：
　　Maximize x1 + 2x2 + 3x3
　　subject to
　　-x1 + x2 + x3 ≦20
　　x1 - 3x2 + x3 ≦30
　　with these bounds
　　0 ≦x1 ≦40
　　0 ≦x2 ≦+&
　　0≦ x3≦ +&
　　先创建一个IloCplex对象，它是用来创建所有建模对象所需要的模型。此时会抛出一个异常：IloException，需要try\catch。
　　代码如下：static public class Application {
　　static public main(String[] args) {
　　IloCplex cplex = new IloCplex();
　　// create model and solve it
　　} catch (IloException e) {
　　System.err.println(&Concert exception caught: & + e);
　　定义决策变量：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);
　　定义目标函数：
　　IloNumExpr expr = cplex.sum(x[0], cplex.prod(2.0, x[1]),cplex.prod(3.0, x[2]));
　　cplex.addMaximize(expr);
　　其中这个地方有许多写法，大家在使用的时候可以注意一下。
　　定义决策的约束条件：cplex.addLe(cplex.sum(cplex.negative(x[0]), x[1], x[2]), 20);cplex.addLe(cplex.sum(cplex.prod(1, x[0]), cplex.prod(-3, x[1]),cplex.prod(1, x[2])), 30);
　　最后解决模型问题：if(cplex.solve()){....}
　　如果solve()返回true的话，我们可以获取一些信息，例如问题的解决状态、获取方案的目标值、获取数组中的所有决策变量的解值。
　　cplex.getStatus()返回值类型：Error、Unknown、Feasible、Bounded、Optimal、Infeasible、Unbouded、InfeasibleorUnbounded。
　　获取方案的目标值：double objval = cplex.getObjValue();获取数组中的所有决策变量的解值：double[] xval = cplex.getValues(x);
　　运行程序最后控制台的输出结果如下所示：
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