How to find max and min value of a function ?
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Hi ... I have a function u_g
a=2; b=2; c=1; e=0.75 ; l=0.5;
u_g = @(x, x_e, N)(-0.5*a*x.^2+b*(x-x_e)-c*(N.^l)+e*u_p(x,x_e,N));
x and N are variables, the rest of them are parameters... I want to find the value of x which will bring u_g the maximum value and corresponding max value of u_g
and I want to find the value of N which will bring minimum value to u_g and corresponding value of u_g ..how can I do it ?.. do I have to fix one of the variables?
2 Kommentare
Antworten (3)
Abdolkarim Mohammadi
am 22 Mai 2020
Bearbeitet: Abdolkarim Mohammadi
am 22 Mai 2020
Finding the value of inputs that minimzes or maximizes the objective function value is an optimization problem. If your function is linear, then you run the following code and optimize your function:
[x, fval] = linprog (u_g, [], []);
If your function is unimodal and relatively smooth, then you run the following code and optimize your function:
[x, fval] = fmincon (u_g, x0, [], []);
And if the landscape of your function is unknown, i.e., you don't know whether it is linear, nonlinear, multi-modal, non-smooth, etc, then you run the following code and optimize your function:
nvars = 3;
[x, fval] = ga (u_g, nvars);
You can refer to the documentation of each solver for more information.
2 Kommentare
John D'Errico
am 22 Mai 2020
You seem to be advocating linprog for all problems. (At least those I've seen you answer.) Note that this is NOT a linear objective, so linprog is completely useless here.
Abdolkarim Mohammadi
am 22 Mai 2020
I just wanted to give a general idea of the optimization tools besides the nonlinear ones that are suitale for those problems.
Walter Roberson
am 22 Mai 2020
a=2; b=2; c=1; e=0.75 ; l=0.5;
u_g = @(x, x_e, N)(-0.5*a*x.^2+b*(x-x_e)-c*(N.^l)+e*u_p(x,x_e,N));
funmin = @(xxeN) u_g(xxeN(1), xxeN(2), xxeN(3));
funmax = @(xxeN) -u_g(xxeN(1), xxeN(2), xxeN(3));
lb = [-10 -10 -10]; %adjust as appropriate
ub = [10 10 10]; %adjust as appropriate
xxeN0 = [-.1 .2 .3]; %initial guess
[best4min, fvalmin] = fmincon(funmin, xxeN0, [], [], [], [], lb, ub);
[best4max, fvalmax] = fmincon(funmax, xxeN0, [], [], [], [], lb, ub);
0 Kommentare
Cristian Garcia Milan
am 22 Mai 2020
I think that what you want is the function
fminbnd(fun)
that finds local minimum.
If you use
fminbnd(-fun)
you will get it max.
3 Kommentare
Cristian Garcia Milan
am 22 Mai 2020
How about using symbolic toolbox? Then you can derivate alomg x or N and solve making equal 0
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