运筹学建模与源代码 - 题目是《某公司需要招聘甲乙两种工人其工作可分为1 2 3三大类》.docx

运筹学建模与源代码题目：某公司需要招聘甲，乙两种工人，其工作可分为1 2 3三大类。已知甲，乙两种工人的公子分别为每小时10元和5元，他们每周都工作40小时，但实际做出的等效生产小时数分别为40小时和30小时。为完成生产进度，该公司每周需完成的生产小时数分别为：第1类工作，10000小时；第2类工作，1500小时；第3类工作，20000小时。由于劳动力缺乏，甲种工人不能多于500个，乙种工人不能多于800个。试确定使总劳动力费用最少时工厂应招聘的工人数。并按要求分别完成下列分析：（1）第1类工作每周需完成的生产小时数在何范围内变化时最优基不变？（2）甲种工人的招聘数量在何范围内变化时最优基不变？（3）乙种工人的工资在何范围内变化时最优招聘方案不变？ 需要运筹学课程设计报告的同学可以在我上传的文档中找到一、运筹学建模设x1 x2 x3分别表示第1、2、3种工作类型中甲种工人的数量，x4 x5 x6分别为1、2、3种工作类型中乙种工人的数量x1+x2+x3为三种工作类型中甲种工人的数量，x4+x5+x6为三种工作类型中乙种工人的数量。40 x1+30 x4为第一类工作类型中工人的工作时间，40 x2+30 x5为第二类工作类型中工人的工作时间，40 x3+30 x6为第一类工作类型中工人的工作时间Maxz=-400(x1+x2+x3)-200(x4+x5+x6)x1+x2+x3500x4+x5+x680040 x1+30 x4=1000040 x2+30 x5=150040 x3+30 x6=20000xj0, j=1, 2.6 二、程序源代码 该程序使用JAVA语言实现的二阶段法。Lpframe类用来实现窗体的输出package net.dbet.LP;import javax.swing.Box;import javax.swing.ButtonGroup;import javax.swing.JButton;import javax.swing.JFrame;import javax.swing.JLabel;import javax.swing.JPanel;import javax.swing.JRadioButton;import javax.swing.JScrollPane;import javax.swing.JTextArea;import javax.swing.JTextField;import javax.swing.SwingConstants;public class Lpframe extends JFrameJPanel panel=new JPanel();JRadioButton jr1=new JRadioButton(MAX);JRadioButton jr2=new JRadioButton(MIN);JLabel lbl3=new JLabel(约束条件个数：);JLabel lbl4=new JLabel(变量个数：);JLabel lbl5=new JLabel(=的约束条件个数：);JLabel lbl8=new JLabel(请输入系数矩阵先=的约束条件以“；”结尾：, SwingConstants.LEFT);JTextField txt1=new JTextField(5);JTextField txt2=new JTextField(5);JTextField txt3=new JTextField(5);JTextField txt4=new JTextField(5);JTextField txt5=new JTextField(5);JLabel lbl9=new JLabel(请输入目标函数系数矩阵：);JTextField txt6=new JTextField(20);JTextArea area=new JTextArea(15,15);ButtonGroup bg=new ButtonGroup();JButton btn1=new JButton(确定);JButton btn2=new JButton(取消);JScrollPane jsp=new JScrollPane(area);Mylistener listener=new Mylistener(this);public Lpframe()bg.add(jr1);bg.add(jr2);Box box1=Box.createHorizontalBox();box1.add(jr1);box1.add(Box.createHorizontalStrut(400);box1.add(jr2);Box box2=Box.createHorizontalBox();box2.add(lbl3);box2.add(txt1);box2.add(lbl4);box2.add(txt2);Box box3=Box.createHorizontalBox();box3.add(lbl5);box3.add(txt3);box3.add(lbl6);box3.add(txt4);box3.add(lbl7);box3.add(txt5);Box box4=Box.createHorizontalBox();box4.add(lbl8);Box box5=Box.createHorizontalBox();box5.add(jsp);box5.createVerticalStrut(50);Box box6=Box.createHorizontalBox();box6.add(lbl9);box6.add(txt6);Box box7=Box.createHorizontalBox();box7.add(btn1);box7.add(Box.createHorizontalStrut(100);box7.add(btn2);Box box=Box.createVerticalBox();box.add(box1);box.add(box2);box.add(box3);box.add(box4);box.add(box5);box.add(box6);box.add(box7);panel.add(box);btn1.addActionListener(listener);btn2.addActionListener(listener);this.getContentPane().add(panel);this.setBounds(0,0,600,600);this.setDefaultCloseOperation(3);this.setVisible(true);public static void main(String args)new Lpframe();Mylistener类用来侦听窗体类中的JButton按钮package net.dbet.LP;import java.awt.event.ActionEvent;import java.awt.event.ActionListener;import java.util.regex.Pattern;import javax.swing.JOptionPane;public class Mylistener implements ActionListenerprivate Lpframe lf;public Mylistener(Lpframe lf)this.lf=lf;public void actionPerformed(ActionEvent e) if(e.getSource()=lf.btn1)if(!(lf.txt1.getText().equals()&lf.txt2.getText().equals()&lf.txt3.getText().equals()&lf.txt4.getText().equals()&lf.txt5.getText().equals()&lf.area.getText().equals()double minmax=0; int m; /约束条件的个数 int n; /变量个数 int m1; /=的约束条件个数if(lf.jr1.isSelected()minmax=1;if(lf.jr2.isSelected()minmax=-1;m=Integer.parseInt(lf.txt1.getText();n=Integer.parseInt(lf.txt2.getText();m1=Integer.parseInt(lf.txt3.getText();m2=Integer.parseInt(lf.txt4.getText();m3=Integer.parseInt(lf.txt5.getText();double a=new doublemn+1;String str=lf.area.getText();String s1=str.split(;);for(int j=0;js1.length;j+)System.out.println(s1j);for(int i=0;im;i+)double b=new doublen+1;String regex=,;Pattern p=Ppile(regex);String r=p.split(s1i);for(int k=0;kr.length;k+)if(!(rk.trim().equals()bk=Double.parseDouble(rk);aik=bk; String g=lf.txt6.getText();double goal=new doublen;String regex=,;Pattern p=Ppile(regex);String r=p.split(g);for(int j=0;jr.length;j+)if(!(rj.trim().equals()goalj=Double.parseDouble(rj);System.out.println(goalj);LP lp = new LP(minmax,m,n,m1,m2,m3,a,goal);lf.lbl9.setText(最优值结果是：);String result=lp.solve(); lf.area.setText(result);elseJOptionPane.showMessageDialog(lf.panel,请输入数据);if(e.getSource()=lf.btn2)lf.txt1.setText();lf.txt2.setText();lf.txt3.setText();lf.txt4.setText();lf.txt5.setText();lf.txt6.setText();lf.area.setText();LP类用来实现二阶段法功能的代码package net.dbet.LP;public class LP private int m; /约束条件的个数private int n; /变量个数private int m1; /=的约束条件个数private int error; /判断是否是错误的private int basic;private int nonbasic;private double a; /约束条件的系数矩阵private double minmax; /目标函数的最大值或最小值public LP(double minmax,int m,int n,int m1,int m2,int m3,double a,double x)double value;this.error = 0;this.minmax = minmax;this.m = m;this.n = n;this.m1 = m1;this.m2 = m2;this.m3 = m3;if(m != m1+m2+m3)this.error = 1;/系数矩阵的行数this.a = new doublem+2;for(int i = 0; i m+2; i+)/每一行系数矩阵的列数this.ai = new doublen+m+m3+1;/基变量的个数this.basic = new intm+2;/非基变量的个数this.nonbasic = new intn+m3+1;for(int i = 0; i = m+1; i+)for(int j = 0; j = n+m+m3; j+)this.aij = 0.0;/非基变量for(int j = 0; j = n+m3; j+)nonbasicj = j;/基变量for(int i = 1,j = n+m3+1; i = m; i+,j+)basici=j;/约束条件=的添加变量在系数矩阵的系数for(int i = m-m3+1,j = n+1; i= m; i+,j+)this.aij=-1;this.am+1j=-1;for(int i = 1; i = m; i+)for(int j = 1; j = n; j+)value = ai-1j-1;this.aij=value;value = ai-1n;if(value0)error = 1;this.ai0=value;for(int j = 1; j = n; j+) /价值系数value = xj - 1;this.a0j = value * minmax;for(int j = 1; j = n; j+)value = 0;for(int i = m1+1; i = m; i+)value+=this.aij;this.am+1j=value;/换入变量public int research_in(int objrow)int col = 0;for(int j = 1; j = this.n + this.m3; j+)if(this.nonbasicj 10e-8)col=j;break;return col;/换出变量public int research_out(int col)double temp=-1;int row = 0;for(int i = 1; i 10e-8)val = this.ai0/val;if(val temp | temp = -1)row = i;temp = val;return row;/换入和换出变量交换public void swapbasic(int row,int col)int temp = this.basicrow;this.basicrow = this.nonbasiccol;this.nonbasiccol = temp;/交换规则找出主元并化简public void pivot(int row,int col)for(int j = 0;j = this.n + this.m3; j+)if(j != col)this.arowj = this.arowj / this.arowcol;this.arowcol = 1.0 / this.arowcol;for(int i = 0; i = this.m + 1; i+)if(i != row)for(int j = 0; j = this.n + this.m3; j+)if(j != col)this.aij = this.aij - this.aicol * this.arowj;if(Math.abs(this.aij) 0)row=research_out(col);elsereturn 0;if(row 0)pivot(row,col);elsereturn 2;/判断有无可行解public int phase1()this.error = simplex(this.m + 1);if(this.error 0)return this.error;for(int i = 1; i this.n + this.m1 + this.m3)if(this.ai0 10e-8)return 3;for(int j = 1; j = 10e-8)pivot(i,j);break;return 0;public int phase2()return simplex(0);/判断是无可行解还是无界解public int compute()if(this.error 0)return t