I need an elegant, simple system to find out what is the highest value returned from a deterministic function given one, or more, parameters.
I know that there is a nice implementation of genetic algorithms in MATLAB, but actually, in my case this is an overkill. I need something simpler.
Any idea?
You cannot find a maximum with Matlab directly, but you can minimize something. Multiplying your function by -1 transformes your "find the maximum"-problem into a "find the minimum"-problem, which can be found with fminsearch
f = #(x) 2*x - 3*x.^2; % a simple function to find the maximum from
minusf = #(x) -1*f(x); % minus f, find minimum from this function
x = linspace(-2,2,100);
plot(x, f(x));
xmax = fminsearch(minusf, -1);
hold on
plot(xmax,f(xmax),'ro') % plot the minimum of minusf (maximum of f)
The result looks like this:
A real simple idea is to use a grid search approach, maybe with mesh refinements. A better idea would be to use a more advanced derivative-free optimizer, such as the Nelder-Mead algorithm. This is available in fminsearch.
You could also try algorithms from the global optimization toolbox: for example patternsearch or the infamous simulannealbnd.
Related
I want to minimize a function like below:
Here, n can be 5,10,50 etc. I want to use Matlab and want to use Gradient Descent and Quasi-Newton Method with BFGS update to solve this problem along with backtracking line search. I am a novice in Matlab. Can anyone help, please? I can find a solution for a similar problem in that link: https://www.mathworks.com/help/optim/ug/unconstrained-nonlinear-optimization-algorithms.html .
But, I really don't know how to create a vector-valued function in Matlab (in my case input x can be an n-dimensional vector).
You will have to make quite a leap to get where you want to be -- may I suggest to go through some basic tutorial first in order to digest basic MATLAB syntax and concepts? Another useful read is the very basic example to unconstrained optimization in the documentation. However, the answer to your question touches only basic syntax, so we can go through it quickly nevertheless.
The absolute minimum to invoke the unconstraint nonlinear optimization algorithms of the Optimization Toolbox is the formulation of an objective function. That function is supposed to return the function value f of your function at any given point x, and in your case it reads
function f = objfun(x)
f = sum(100 * (x(2:end) - x(1:end-1).^2).^2 + (1 - x(1:end-1)).^2);
end
Notice that
we select the indiviual components of the x vector by matrix indexing, and that
the .^ notation effects that the operand is to be squared elementwise.
For simplicity, save this function to a file objfun.m in your current working directory, so that you have it available from the command window.
Now all you have to do is to call the appropriate optimization algorithm, say, the quasi Newton method, from the command window:
n = 10; % Use n variables
options = optimoptions(#fminunc,'Algorithm','quasi-newton'); % Use QM method
x0 = rand(n,1); % Random starting guess
[x,fval,exitflag] = fminunc(#objfun, x0, options); % Solve!
fprintf('Final objval=%.2e, exitflag=%d\n', fval, exitflag);
On my machine I see that the algorithm converges:
Local minimum found.
Optimization completed because the size of the gradient is less than
the default value of the optimality tolerance.
Final objval=5.57e-11, exitflag=1
I have a histogram that seems to fit a poisson distribution.
In order to fit it, I declare the function myself as follows
xdata; ydata; % Arrays in which I have stored the data.
%Ydata tell us how many times the xdata is repeated in the set.
fun= #(x,xdata) (exp(-x(1))*(x(1).^(xdata)) )/(factorial(xdata)) %Function I
% want to use in the fit. It is a poisson distribution.
x0=[1]; %Approximated value of the parameter lambda to help the fit
p=lsqcurvefit(fun,x0,xdata,ydata); % Fit in the least square sense
I find an error. It probably has to do with the "factorial". Any ideas?
Factorial outputs a vector from vector xdata. Why are you using .xdata in factorial?
For example:
data = [1 2 3];
factorial(data) is then [1! 2! 3!].
Try ./factorial(xdata) (I cannot recall if the dot is even necessary at this case.)
You need to use gamma(xdata+1) function instead of factorial(xdata) function. Gamma function is a generalized form of factorial function which can be used for real and complex numbers. Thus, your code would be:
fun = #(x,xdata) exp(-x(1))*x(1).^xdata./gamma(xdata+1);
x = lsqcurvefit(fun,1,xdata,ydata);
Alternatively, you can MATLAB fitdist function which is already optimized and you might get better results:
pd = fitdist(xdata,'Poisson','Frequency',ydata);
pd.lambda
I would like to solve the following problem:
x* = argmin f( max_{j=1,...,J} g_j(x)),
where f and g_j are known smooth, convex functions, and x is a scalar. I also have analytical gradients of f and g_j. Is it possible to use fminunc (using Newton method) or another solver in MatLab to find the solution? Obviously max_{j=1,...,J} g_j(x) is not differentiable (unless there is a k such that g_k(x) >= g_j(x) for all j), but it is sub-differentiable. Therefore, I cannot use fminunc directly, as it requests the gradient of the objective function. If I can use fminunc, how would I proceed?
As an example, consider f(x) = x, J=1,2, g_1(x) = (x-1)^2, g_j(x) = (x+1)^2. Then max g_1, g_2 is the upper envelope of g_1 and g_2. The minimum is located at x=0.
Note: The function fminimax is close to what I want. fminimax solves
x* = argmin max_{j=1,...,J} g_j(x).
I have one potential solution to my problem. We can replace the maximum function with a smoothed or soft maximum, see this article by John Cook. In particular, we can approximate max(x,y) with the function g(x,y;k) = log(exp(kx)+exp(ky))/y (k is a tuning parameter). Now the objective function is differentiable, so the Newton method can be applied (i.e., fminunc).
i'm new in matlab. I didn't understand how to derive a dirac delta function and then shift it using symbolic toolbox.
syms t
x = dirac(t)
why can't i see the dirac delta function using ezplot(x,[-10,10]) for example?
As others have noted, the Dirac delta function is not a true function, but a generalized function. The help for dirac indicates this:
dirac(X) is not a function in the strict sense, but rather a
distribution with int(dirac(x-a)*f(x),-inf,inf) = f(a) and
diff(heaviside(x),x) = dirac(x).
Strictly speaking, it's impossible for Matlab to plot the Dirac delta function in the normal way because part of it extends to infinity. However, there are numerous workarounds if you want a visualization. A simple one is to use the stem plot function and the > operator to convert the one Inf value to something finite. This produces a unit impulse function (or Kronecker delta):
t = -10:10;
x = dirac(t) > 0;
stem(t,x)
If t and x already exist as symbolic variables/expressions rather than numeric ones you can use subs:
syms t
x = dirac(t);
t2 = -10:10;
x2 = subs(x,t,t2)>0;
stem(t2, x2)
You can write your own plot routine if you want something that looks different. Using ezplot is not likely to work as it doesn't offer as much control.
First, I've not met ezplot before; I had to read up on it. For things that are functionals like your x, it's handy, but you still have to realize it's exactly giving you what it promises: A plot.
If you had the job of plotting the dirac delta function, how would you go about doing it correctly? You can't. You must find a convention of annotating your plot with the info that there is a single, isolated, infinite point in your plot.
Plotting something with a line plot hence is unsuitable for anything but smooth functions (that's a well-defined term). Dirac Delta definitely isn't amongst the class of functions that are smooth. You would typically use a vertical line or something to denote the point where your functional is not 0.
Is it possible to use vector limits for any matlab function of integration? I have to avoid from for loops because of the speed of my program. Can you please give me a clue on do
k=0:5
f=#(x)x^2
quad(f,k,k+1)
If somebody need, I found the answer of my question:quad with vector limit
I will try giving you an answer, based on my experience with quad function.
Starting from this:
k=0:5;
f=#(x) x.^2;
Notice the difference in your f definition (incorrect) and mine (correct).
If you only mean to integrate f within the range (0,5) you can easily call
quad(f,k(1),k(end))
Without handle function, you may reach the same results in a different way, by making use of trapz:
x = 0:5;
y = x.^2;
trapz(x,y)
If, instead, you mean to perform a step-by-step integration in the small range [k(i),k(i+1)] you may type
arrayfun(#(ii) quad(f,k(ii),k(ii+1)),1:numel(k)-1)
For a sake of convenince, notice that
sum(arrayfun(#(ii) quad(f,k(ii),k(ii+1)),1:numel(k)-1)) == quad(f,k(1),k(end))
I hope this helps.