[求助] 非线性方程组求解

I(1) * (2 * 0.875+ R(1)) - I(2) * R(1) - 36=0 I(2) *(2 * 0.875 + R(1) + R(2))- I(1) * R(1)=0 %灯具功率表达式 I(2)^2 * R (2) - Power=0 (I(2) - I(1))^2 * R(1) - Power=0

%定义遗传算法参数 NIND=50; %个体数目(Numbe of individuals) MAXGEN=1000; %最大遗传代数(Maximum number of generations) NVAR=4; %变量的维数 PRECI=10; %变量的二进制位数(Precision of variables) GGAP=0.9; %代沟(Generation gap) % lbi1=0; %i1取值范围[0.138,0.25] % lbi2=0; %i2取值范围[0.138,0.25] % lbr1=80; %r1取值范围[80,259.2] % lbr2=80; %r2取值范围[80,259.2] % ubi1=10; % ubi2=10; % ubr1=259.2; % ubr2=259.2; lbi1=0; %i1取值范围[0.138,0.25] lbi2=0; %i2取值范围[0.138,0.25] lbr1=50; %r1取值范围[80,259.2] lbr2=50; %r2取值范围[80,259.2] ubi1=10; ubi2 =10; ubr1=300; ubr2=300; trace=zeros(MAXGEN, 2); %建立区域描述器(Build field descriptor) FieldD=[rep([PRECI],[1,NVAR]);lbi1 lbi2 lbr1 lbr2;ubi1 ubi2 ubr1 ubr2;rep([1;0;1;1],[1,NVAR])]; %FieldD Chrom=crtbp(NIND, NVAR*PRECI); %创建初始种群 %Chrom gen=0; %代计数器 X=bs2rv(Chrom,FieldD);%种群二进制到十进制转换 %X % % X*X % X.^2 ObjV=ff(X); %目标函数 % ObjV while gen<MAXGEN %迭代 FitnV=ranking(ObjV);%分配适应度值(Assign fitness values) %FitnV SelCh=select('sus', Chrom, FitnV, GGAP); %选择 %SelCh SelCh=recombin('xovsp', SelCh, 0.7); %重组 %SelCh SelCh=mut(SelCh); %变异 %SelCh X=bs2rv(SelCh,FieldD); %种群二进制到十进制转换 %X ObjVSel=ff(X); %目标函数 [Chrom ObjV]=reins(Chrom, SelCh, 1, 1, ObjV, ObjVSel); %重插入 %X=bs2rv(Chrom,FieldD); gen=gen+1; %代计数器增加 trace(gen, 1)=min(ObjV); %遗传算法性能跟踪 trace(gen, 2)=sum(ObjV)/length(ObjV); end plot(trace(:,1));hold on; plot(trace(:,2),'-.');grid; legend(' 种群均值的变化','解的变化') %输出最优解及其对应的20个自变量的十进制值,Y为最优解,I为种群的序号 [Y, I]=min(ObjV); Y X=bs2rv(Chrom, FieldD); X(I,:)

function [ y ] = ff(x) f1=2*x(:,1)*0.875+(x(:,1)-x(:,2)).*x(:,3)-36; f2=2*x(:,2)*0.875+(x(:,2)-x(:,1)).*x(:,3)+x(:,2).*x(:,4); f3=5-(x(:,1)-x(:,2)).*(x(:,1)-x(:,2)).*x(:,3); f4=5-x(:,2).*x(:,2).*x(:,4); % f1(1,1)^2+f2(1,1)^2+f3(1,1)^2+f4(1,1)^2 y=f1.^2+f2.^2+f3.^2+f4.^2; end

1楼: Originally posted by Honey155752 at 2017-07-10 16:06:39
1.解决目标：方程组如下
I(1) * (2 * 0.875+ R(1)) - I(2) * R(1) - 36=0 I(2) *(2 * 0.875 + R(1) + R(2))- I(1) * R(1)=0 %灯具功率表达式 I(2)^2 * R (2) - Power=0 (I(2) - I(1))^2 * R(1) - Power=0

2.经过 ...