如何使用fmincon函数求含有积分的函数的最小值
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clc;
clear;
a=0;
alpha=0.01;
beta=0.01;
Caql=1.33;
Crql=1;
b1=3*Caql*(1+a^2)^0.5;
b2=3*Crql*(1+a^2)^0.5;
fun1=@(t)chi2cdf(x(1)*b1^2/(9*x(3)^2)-t^2,x(1)-1)*(normpdf(t+a*x(1)^0.5)+normpdf(t-a*x(1)^0.5));%PN(Caql)
fun2=@(t)chi2cdf(x(1)*b2^2/(9*x(3)^2)-t^2,x(1)-1)*(normpdf(t+a*x(1)^0.5)+normpdf(t-a*x(1)^0.5));%PN(Crql)
fun3=@(t)chi2cdf(x(2)*b1^2/(9*x(3)^2)-t^2,x(2)-1)*(normpdf(t+a*x(2)^0.5)+normpdf(t-a*x(2)^0.5));%PT(Caql)
fun4=@(t)chi2cdf(x(2)*b2^2/(9*x(3)^2)-t^2,x(2)-1)*(normpdf(t+a*x(2)^0.5)+normpdf(t-a*x(2)^0.5));%PT(Crql)
fun=@(x)1/2*((integral(fun3,0,b1*x(2)^0.5/(3*x(3)))*x(1)+(1-integral(fun1,0,b1*x(1)^0.5/(3*x(3))))*x(2))/(1-integral(fun1,0,b1*x(1)^0.5/(3*x(3)))+integral(fun3,0,b1*x(2)^0.5/(3*x(3)))))+...
1/2*((integral(fun4,0,b2*x(2)^0.5/(3*x(3)))*x(1)+(1-integral(fun2,0,b2*x(1)^0.5/(3*x(3))))*x(2))/(1-integral(fun2,0,b2*x(1)^0.5/(3*x(3)))+integral(fun4,0,b2*x(2)^0.5/(3*x(3)))));
x0=[110,130,1.1];
A=[];%AX<=b(线性不等式约束),G(X)<=0(非线性不等式约束)
b=[];
Aeq=[];%AeqX=beq(线性等式约束),Geq(X)=0(非线性等式约束)
beq=[];
vlb=[2,2];%lb<=x<=ub(变量约束)
vub=[];
exitflag=1;
[x,fval,exitflag]=fmincon(fun,x0,A,b,Aeq,beq,vlb,vub,'myfun');
function [g,ceq]=myfun(x)
g=1-alpha-(integral(fun3,0,b1*x(2)^0.5/(3*x(3)))/(1-integral(fun1,0,b1*x(1)^0.5/(3*x(3)))+integral(fun3,0,b1*x(2)^0.5/(3*x(3)))));
ceq=(integral(fun4,0,b2*x(2)^0.5/(3*x(3)))/(1-integral(fun2,0,b2*x(1)^0.5/(3*x(3)))+integral(fun4,0,b2*x(2)^0.5/(3*x(3)))))-beta;
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