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199107011523

新虫 (小有名气)

[求助] 优化分析 已有1人参与

已知t1=67;t2=110;t3=82;t4=73;t5=33;t6=0.5;t7=0.54;69≤t4≤75;31≤t5≤35;3%≤t7-t6≤5%;
y1=2t5-5t5^2-3/t5^2+(a-1)t5-at3
y2=2t5-5t5^2-3/t5^2+3t4
y3=3t4+5t5^2-3/t5^3+(a-1)t-at1
z1=y1/(t1+3*t2);
z2=y2/(3*t3+2*t2);
z3=y3/(t1+t5);
a=t7/(t7-t6);
10≤a≤50;
Z=z1+z2+z3;
F=(y1+y2)/y3;
求min f=Z/F   maxF  minZ

以min f=Z/F为例,编写m.文件如下,运行有警告提示,希望得到高手指点:

function f=fun567(t)
a=t(7)/(t(7)-t(6));
y1=2*t(5)-5*t(5)^2-3/t(5)^2+(a-1)*t(2)-a*t(3);
y2=2*t(5)-5*t(5)^2-3/t(5)^2+3*t(4);
y3=3*t(4)+5*t(5)^2-3/t(5)^3+(a-1)*t(2)-a*t(1);
z1=y1/(t(1)+3*t(2));
z2=y2/(3*t(3)+2*t(2));
z3=y3/(t(1)+t(5));
Z=z1+z2+z3;
F=(y1+y2)/y3;
f=Z/F

function [c,ceq]=mycon3_1(t)
c(1)=t(4)-8;
c(2)=5-t(4);
c(3)=t(5)-7;
c(4)=3-t(5);
c(5)=t(7)/(t(7)-t(6))-50;
c(6)=10-t(6)/(t(7)-t(6));
c(7)=t(7)-t(6)-0.06;
c(8)=0.03-t(7)+t(6);
ceq=[];
调用函数如下:
t=[67;110;82;73;33;0.5;0.54];
[t,fval]=fmincon(@(t) fun567(t),t,[],[],Aeq,beq,[],[],@mycon3_1)@laosam280
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fungarwai

新虫 (小有名气)

【答案】应助回帖

★ ★ ★ ★ ★
感谢参与,应助指数 +1
199107011523: 金币+5, ★★★很有帮助 2016-06-23 15:38:20
你没有Aeq,beq,空着

[t,fval]=fmincon(@(t) fun567(t),t,[],[],[],[],[],[],@mycon3_1)
3楼2016-06-22 17:57:54
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199107011523

新虫 (小有名气)

哪位好心人帮帮忙了,急急急

发自小木虫Android客户端
2楼2016-06-22 10:18:08
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199107011523

新虫 (小有名气)

引用回帖:
3楼: Originally posted by fungarwai at 2016-06-22 17:57:54
你没有Aeq,beq,空着

=fmincon(@(t) fun567(t),t,[],[],[],[],[],[],@mycon3_1)

好像不是这个问题的
4楼2016-06-22 18:32:19
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fungarwai

新虫 (小有名气)

引用回帖:
4楼: Originally posted by 199107011523 at 2016-06-22 18:32:19
好像不是这个问题的...

我看到的是这样

Warning: The default trust-region-reflective algorithm does not solve problems with the constraints you
have specified. FMINCON will use the active-set algorithm instead. For information on applicable
algorithms, see Choosing the Algorithm in the documentation.
> In fmincon at 500
  In test at 4
Warning: Your current settings will run a different algorithm (interior-point) in a future release.
> In fmincon at 505
  In test at 4

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in
feasible directions, to within the default value of the function tolerance,
and constraints are satisfied to within the default value of the constraint tolerance.

<stopping criteria details>

Active inequalities (to within options.TolCon = 1e-06):
  lower      upper     ineqlin   ineqnonlin
                                     2
                                     5

t =

   1.0e+08 *

   -0.8936
   -0.2746
    1.2088
    0.0000
    0.0000
    0.0000
    0.0000


fval =

   -3.4171

你的c里面那些值怎么取得跟题目不一样
c(1)=t(4)-75;
c(2)=69-t(4);
c(3)=t(5)-35;
c(4)=31-t(5);
c(7)=t(7)-t(6)-0.05;
5楼2016-06-22 18:45:12
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