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iwill1k

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[求助] 多元函数求最值

刚接触matlab,我想用matlab求下面函数的最大、最小值以及取最值的时候自变量取值,是一个有4个自变量的函数,自变量都有一定的变化范围。具体函数是:
f=[d*(s+b+d)+0.858*r*r-0.858*(r+d/2)*(r+d/2)]/[s*b-0.858*r*r];其中,2.79≤s≤2.83;7.97≤b≤8.07;0.13≤d≤0.18;0.6≤r≤1.有木有大神帮帮忙知道一下啊
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iwill1k

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引用回帖:
3楼: Originally posted by dbb627 at 2012-04-09 12:35:37:
>> f=@(x)(x(3).*(x(1)+x(2)+x(3))+0.858*x(4).*x(4)-0.858*(x(4)+x(3)/2).*(x(4)+x(3)/2))./(x(1).*x(2)-0.858*x(4).*x(4));
  opt=optimset('Algorithm','interior-point');
= fmincon(f,,[],[],[],[], ...

弱弱的问一下,[X,FVAL,EXITFLAG]= fmincon(f,[2.8 8 0.15 0.8],[],[],[],[],[2.79 7.97 0.13 0.6],[2.83 8.07 0.18 1],[],opt)里面的[2.8 8 0.15 0.8]是算的定义域里面的平均值吗?
9楼2012-04-10 12:28:24
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lijie169

铜虫 (著名写手)

【答案】应助回帖

感谢参与,应助指数 +1
百度你会不?http://www.google.com.hk/#hl=zh- ... rgbld.&bav=on.2,or.r_gc.r_pw.,cf.osb&fp=b7e86341b10bb823&biw=1366&bih=667
2楼2012-04-09 11:43:42
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dbb627

荣誉版主 (著名写手)

【答案】应助回帖

感谢参与,应助指数 +1
>> f=@(x)(x(3).*(x(1)+x(2)+x(3))+0.858*x(4).*x(4)-0.858*(x(4)+x(3)/2).*(x(4)+x(3)/2))./(x(1).*x(2)-0.858*x(4).*x(4));
  opt=optimset('Algorithm','interior-point');
[X,FVAL,EXITFLAG]= fmincon(f,[2.8 8 0.15 0.8],[],[],[],[],[2.79 7.97 0.13 0.6],[2.83 8.07 0.18 1],[],opt)

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 were satisfied to within the default value of the constraint tolerance.




X =

    2.8300    8.0700    0.1300    0.9998


FVAL =

    0.0600


EXITFLAG =

     1

>> f1=@(x)-(x(3).*(x(1)+x(2)+x(3))+0.858*x(4).*x(4)-0.858*(x(4)+x(3)/2).*(x(4)+x(3)/2))./(x(1).*x(2)-0.858*x(4).*x(4));
>> [X,FVAL,EXITFLAG]= fmincon(f1,[2.8 8 0.15 0.8],[],[],[],[],[2.79 7.97 0.13 0.6],[2.83 8.07 0.18 1],[],opt)

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 were satisfied to within the default value of the constraint tolerance.




X =

    2.7900    7.9700    0.1800    0.6000


FVAL =

   -0.0853


EXITFLAG =

     1
The more you learn, the more you know, the more you know, and the more you forget. The more you forget, the less you know. So why bother to learn.
3楼2012-04-09 12:35:37
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yqx1985

木虫 (著名写手)

云中仙

楼上正解
给你贴个1stOpt的
Parameters s[2.79,2.83], b[7.97,8.07],d[0.13,0.18],r[0.6,1];
Minimum =f;
Function f=(d*(s+b+d)+0.858*r*r-0.858*(r+d/2)*(r+d/2))/(s*b-0.858*r*r);
====== 结果 ======

迭代数: 18
计算用时(时:分:秒:毫秒): 00:00:00:94
计算中止原因: 达到收敛判定标准
优化算法: 共扼梯度法 + 通用全局优化法
函数表达式: (d*(s+b+d)+0.858*r*r-0.858*(r+d/2)*(r+d/2))/(s*b-0.858*r*r)
目标函数值(最小): 0
s: 2.82999999999992
b: 8.06833196340274
d: 0
r: 0.830131999171392

====== 计算结束 ======
人面不知何时去,桃花依旧笑春风
4楼2012-04-09 13:43:37
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