Question

请问用MATLAB编程，哪里错了？

function f=fun(x);
f=4233.1*(1/x(1))+1539.2*(1/x(2))+4667.5*(1/x(3))+1263.8*(1/x(4))+511.2*(1/x(5))+3292.2*(1/x(6))+2.4973*(1/x(7));
function [g,ceq]=mycon(x);
g=[1539.2*x(7)+2497.3*x(2)+640*x(2)*x(7)];
ceq=[];

x0=[2;2;2;2;2;3;3];
A=[];
b=[];
Aeq=[1 1 1 1 1 1 1];
beq=[18];
vlb=[0 0 6.4826 0 0 9.145 0 ];
vub=[18 18 18 18 18 18 18];
[x,fval]=fmincon('fun',A,b,Aeq,beq,vlb,vub,'mycon')
报错Error using fmincon (line 225)
FMINCON requires the following inputs to be of data type double: 'UB'

Answer 1

很明显，是变量出错！
为什么呢？因为matlab的编程对于数据操作比较严格，一般不能自动转换比如要实现一个浮点数的运算，算一个1/x,当想要的结果为float的时候，这个时候就会报错，为什么呢？因为1不是浮点数，要改成1.0，由于你程序不够完整，咱不能具体看出到底哪些变量错误，不过大概可以知道你
vlb=[0 0 6.4826 0 0 9.145 0 ];
vub=[18 18 18 18 18 18 18];
此处你有浮点型，又有整型的，当你把他们混合一起运算，应该就会出问题，你试着都改成浮点型试试看吧。比如
vub=[18 .0 18.0 18.0 18.0 18.0 18.0 18.0];
把1/x改成1.0/x

追问

还是报错了，也是报的Error using fmincon (line 225)
FMINCON requires the following inputs to be of data type double: 'UB'，请问是什么原因？

Answer 2

function f=fun(x)
f=4233.1*(1/x(1))+1539.2*(1/x(2))+4667.5*(1/x(3))+1263.8*(1/x(4))+511.2*(1/x(5))+3292.2*(1/x(6))+2.4973*(1/x(7));
上面保存为fun.m
function [g,ceq]=mycon(x)
g=[1539.2*x(7)+2497.3*x(2)+640*x(2)*x(7)];
ceq=[];
上面保存为mycon.m
x0=[2;2;2;2;2;3;3];
A=[];
b=[];
Aeq=[1 1 1 1 1 1 1];
beq=[18];
vlb=[0 0 6.4826 0 0 9.145 0 ];
vub=[18 18 18 18 18 18 18];
[x,fval]=fmincon('fun',x0,A,b,Aeq,beq,vlb,vub,'mycon')
在命令行中输入上面语句试试

追问

我就是这样弄的，报错了Error using fmincon (line 225)
FMINCON requires the following inputs to be of data type double: 'UB'

追答

我这儿运行没问题，你的电脑或matlab版本有问题吧，

追问

那你能把答案给我发过来吗？谢谢，我急用

追答

x0=[2;2;2;2;2;3;3];
A=[];
b=[];
Aeq=[1 1 1 1 1 1 1];
beq=[18];
vlb=[0 0 6.4826 0 0 9.145 0 ];
vub=[18 18 18 18 18 18 18];
[x,fval exitflag output]=fmincon('fun',x0,A,b,Aeq,beq,vlb,vub,'mycon')
Warning: Trust-region-reflective method does not currently solve this type of problem,
using active-set (line search) instead.
> In fmincon at 422
Maximum number of function evaluations exceeded;
increase OPTIONS.MaxFunEvals.
x =
0
0
6.4826
0
0
10.1553
1.3621
fval =
-Inf
exitflag =
0
output =
iterations: 99
funcCount: 702
lssteplength: 2
stepsize: NaN
algorithm: 'medium-scale: SQP, Quasi-Newton, line-search'
firstorderopt: Inf
constrviolation: 2.0965e+003
message: [1x79 char]

追问

未知数是在分母，怎么算出来会有0？

追答

这是由于你的最小边界里有0造成的，我给你改一下：
x0=[2;2;2;2;2;3;3];
A=[];
b=[];
Aeq=[1 1 1 1 1 1 1];
beq=[18];
%vlb=[0 0 6.4826 0 0 9.145 0 ];
vlb=[1 1 6.4826 1 1 9.145 1 ];
vub=[18 18 18 18 18 18 18];
[x,fval exitflag output]=fmincon('fun',x0,A,b,Aeq,beq,vlb,vub,'mycon')
Warning: Trust-region-reflective method does not currently solve this type of problem,
using active-set (line search) instead.
> In fmincon at 422
Optimization terminated: no feasible solution found. Magnitude of search
direction less than 2*options.TolX but constraints are not satisfied.
x =
0.7006
0.4347
6.1832
0.7006
0.7006
8.8456
0.4347
fval =
1.3249e+004
exitflag =
-2
output =
iterations: 7
funcCount: 64
lssteplength: 1
stepsize: 3.6477e-007
algorithm: 'medium-scale: SQP, Quasi-Newton, line-search'
firstorderopt: 8.6240e+003
constrviolation: 1.8755e+003
message: [1x143 char]

追问

能编一个求整数解的吗？

Answer 3

右键MATLAB属性，兼容性 选择VISTA 最下面以管理员运行 就OK了

追问

还是报错Error using fmincon (line 225)
FMINCON requires the following inputs to be of data type double: 'UB'.

追答

那就装个2012把 就是有点大 快5G了 需要下载一会