使用遗传算法求解多元函数最值(实例)

public static final int M=22; //每一个变量编码位数 public static String[]pop=new String[POP_SIZE];//种群编码 public static double[][]result=new double[varnum][POP_SIZE];//种群代表的结果 public static final int LENGTH=M*varnum;//编码长度，因为要精确到小数点后六位，所以编为22位长，22*i,i 为变量个数 public static final int MJ2=4194304;//2^22 public static double[]fitness=new double[POP_SIZE];//存放种群适应度 public static final double PC=0.35;//交叉率 public static final double PM=0.08;//变异率 public static double[]p=new double[POP_SIZE];//轮盘赌方法个体适应度概率(按比例的适应度分配) public static double[]q=new double[POP_SIZE];//q[i]是前 n 项 p 之和（累积概率） public static Random random=new Random();//用于产生随机数的工具 public static Best best=new Best();//记录最佳答案的对象
if(a>=0&&a<=92){ if(b>=90&&b<=110){ if(c>=20&&c<=25){ fitness[i]=1000000-(5.3578547*Math.pow(result[2][i], 2)+0.8356891*result[0][i]*result[4][i]+37.293239*result[0][i]-40792.141); } } } }
best.str=pop[i]; //System.out.println(best.fitness); } } rFra Baidu bibliotekturn max; }
public static void main(String[] args) { Random random=new Random();
//d 为初试种群 lower[0]=78; uper[0]=102; lower[1]=33; uper[1]=45; for(int i=2;i<5;i++){ lower[i]=27; uper[i]=45; } for(int i=0;i<varnum;i++){ for(int j=0;j<POP_SIZE;j++){ initpop[i][j]=lower[i]+random.nextDouble()*(uper[i]-lower[i]); } }
}
public void roulettewheel() {
decoding(); fitness();
double sum=0; for (int i = 0; i <POP_SIZE; i++) { sum=fitness[i]+sum; } for (int i = 0; i < POP_SIZE; i++) { p[i]=fitness[i]/sum; q[i]=0; } for (int i = 0; i < POP_SIZE; i++) { for (int j = 0; j < i+1; j++) {
public void mutation() {
for (int i = 0; i < pop.length; i++) { for (int j = 0; j < LENGTH; j++) {
double k=random.nextDouble();
if(PM>k) {
mutation(i,j); } } } } public void mutation(int i,int j) { String s=pop[i]; StringBuffer sb=new StringBuffer(s); if(sb.charAt(j)=='0') sb.setCharAt(j, '1'); else sb.setCharAt(j, '0'); pop[i]=sb.toString();
public double findResult() {
if(best==null) best=new Best(); double max=best.fitness; for (int i = 0; i < fitness.length; i++) { if(fitness[i]>max) {
best.fitness=fitness[i]; for(int m=0;m<varnum;m++){ best.x[m]=result[m][i]; }
}
}
public void decoding()//将2进制编码转换为10进制 {
for (int i = 0; i < pop.length; i++) { for(int j=0;j<varnum;j++){ int k=Integer.parseInt((pop[i].substring(j*22, (j+1)*22-1)), 2); result[j][i]=lower[j]+k*(uper[j]-lower[j])/(MJ2-1); }
}
}
public void fitness() {
for (int i = 0; i < POP_SIZE; i++) { fitness[i]=0; double a=85.334407+0.0056858*result[1][i]*result[4][i]+0.0006262*result[0][i]*result[3][i]-0.0022053*result[2] [i]*result[4][i]; double
使用遗传算法求解多元函数最值
这是我们的待求解问题模型，下面是我们的实现代码： package test;
import java.util.Random; public class GA { public static final int varnum=5;//变量的个数 public static final double []lower=new double[varnum]; public static final double []uper=new double[varnum]; public static final int POP_SIZE=80;//种群数目 public static final double[][]initpop=new double[varnum][POP_SIZE];
//初始化其它参数 GA ga=new GA(initpop);
System.out.println("种群进化中...."); //进化，这里进化10000次 long starttime=System.currentTimeMillis(); ga.dispose(10000); long endtime=System.currentTimeMillis(); System.out.println("进化耗时："+(endtime-starttime)+"ms"); System.out.println("+++++++++++++++++++++++++++结果为："); for(int i=0;i<varnum;i++){ System.out.print("x["+(i+1)+"]="+best.x[i]+"\t"); } System.out.println(); System.out.println("约束条件1的值： "+(85.334407+0.0056858*best.x[1]*best.x[4]+0.0006262*best.x[0]*best.x[3]-0.0022053*best.x[2]*best.x[4])) ; System.out.println("约束条件2的值： "+(80.51249+0.0071317*best.x[1]*best.x[4]+0.0029955*best.x[0]*best.x[1]+0.0021813*best.x[2]*best.x[2])) ; System.out.println("约束条件3的值： "+(9.300961+0.0047026*best.x[2]*best.x[4]+0.0012547*best.x[0]*best.x[2]+0.0019085*best.x[2]*best.x[3])) ; System.out.println("目标函数值："+(5.3578547*Math.pow(best.x[2], 2)+0.8356891*best.x[0]*best.x[4]+37.293239*best.x[0]-40792.141)); System.out.println("Function="+(1000000-best.fitness));
public GA(double initpop[][]) {
for (int i = 0; i < initpop.length; i++) { for(int j=0;j<initpop[0].length;j++){
result[i][j]=initpop[i][j]; }
} }
public void encoding() {
for (int i = 0; i < POP_SIZE; i++) { pop[i]=""; for(int j=0;j<varnum;j++){ double d1=((initpop[j][i]-lower[j])/(uper[j]-lower[j]))*(MJ2-1); String GeneCode=Integer.toBinaryString((int)d1); if(GeneCode.length()<M){ int k=M-GeneCode.length(); for(int l=0;l<k;l++){ GeneCode="0"+GeneCode; } } pop[i]+=GeneCode; }
encoding(); crossover(); mutation(); decoding(); fitness(); roulettewheel(); findResult();
}
public void dispose(int n) {
for (int i = 0; i < n; i++) { evolution(); } }
q[i]+=p[j]; } } double []ran=new double[POP_SIZE]; String[] tempPop=new String[POP_SIZE];
for (int i = 0; i < ran.length; i++) { ran[i]=random.nextDouble(); } for (int i = 0; i < ran.length; i++) { int k = 0; for (int j = 0; j < q.length; j++) {
//System.out.println(fitness[i]); }
public void crossover(){//单点交叉 for(int i=0;i<POP_SIZE;i++){ double d=random.nextDouble(); if(d<PC){ int k1=random.nextInt(POP_SIZE); int k2=random.nextInt(POP_SIZE); int position=random.nextInt(LENGTH); String s11="",s12="",s21="",s22=""; s11=pop[k1].substring(0,position); s12=pop[k1].substring(position,LENGTH); s21=pop[k2].substring(0, position); s22=pop[k2].substring(position, LENGTH); pop[k1]=s11+s22; pop[k2]=s21+s12; } } }
b=80.51249+0.0071317*result[1][i]*result[4][i]+0.0029955*result[0][i]*result[1][i]+0.0021813*result[2][ i]*result[2][i];
double c=9.300961+0.0047026*result[2][i]*result[4][i]+0.0012547*result[0][i]*result[2][i]+0.0019085*result[2][ i]*result[3][i];
if(ran[i]<q[j]) {
k=j; break; } else continue; } tempPop[i]=pop[k]; } for (int i = 0; i < tempPop.length; i++) { pop[i]=tempPop[i]; } }
public void evolution() {