求MATLAB大神给指出程序错误!

来源：学生作业帮编辑：灵鹊做题网作业帮分类：综合作业时间：2024/04/29 18:08:05

求MATLAB大神给指出程序错误!
function y = objfun(x)
y = 1/x(1) + 2/x(2) + 4/x(3) + 6/x(4)
function [c,ceq]=confun(x)
c = [x(1)+x(2)+3*x(3)+3*x(4)-2;x(1)+2*x(2)+2*x(3)+x(4)-1.8];
ceq = []
x0 = [0,0,0,0]
lb = [0.2,0.2,0.2,0.2]
ub = []
options = optimset('largescale' ,'off');
[x,fval] = fmincon('objfun',x0,[],[],[],[],lb,ub,'confun',options)
[c,ceq] = confun(x)
为什么没法运行,出现如下错误
>> confun
Error using confun (line 2)
Not enough input arguments.

function y = objfun(x)
y = 1/x(1) + 2/x(2) + 4/x(3) + 6/x(4)；
%这个程序独立保存为objfun.m
function [c,ceq]=confun(x)
c = [x(1)+x(2)+3*x(3)+3*x(4)-2;x(1)+2*x(2)+2*x(3)+x(4)-1.8];
ceq = [];
%这个程序独立保存为objfun.m;
clc
clear
x0 = [0,0,0,0]
lb = [0.2,0.2,0.2,0.2]
ub = []
options = optimset('largescale' , 'off','Algorithm','active-set');%Algorithm = 'active-set'
[x,fval] = fmincon('objfun',x0,[],[],[],[],lb,ub,'confun',options)
[c,ceq] = confun(x)
%这个程序独立保存为myfmin.m;运行本主程序就得到下面结果运行结果
x0 =
0 0 0 0

lb =
0.2000 0.2000 0.2000 0.2000

ub =
[]

Local minimum possible. Constraints satisfied.
fmincon stopped because the predicted change in the objective functionis less than the default value of the function tolerance and constraints were satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
Active inequalities (to within options.TolCon = 1e-006): lower upper ineqlin ineqnonlin 1 1
x =
0.2000 0.2791 0.2279 0.2791

fval =
51.2177

c =
0 -0.3070

ceq =
[]

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