I do just about everything in Matlab but I have yet to work out a good way to replicate Mathematica's FindInstance function in Matlab. As an example, with Mathematica, I can enter:
FindInstance[x + y == 1 && x > 0 && y > 0, {x, y}]
And it will give me:
{{x -> 1/2, y -> 1/2}}
When no solution exists, it will give me an empty Out. I use this often in my work to check whether or not a solution to a system of inequalities exists -- I don't really care about a particular solution.
It seems like there should be a way to replicate this in Matlab with Solve. There are sections in the help file on solving a set of inequalities for a parametrized solution with conditions. There's another section on spitting out just one solution using PrincipalValue, but this seems to just select from a finite solution set, rather than coming up with one that meets the parameters.
Can anybody come up with a way to replicate the FindInstance functionality in Matlab?
Building on what jlandercy said, you can certainly use MATLAB's linprog function, which is MATLAB's linear programming solver. A linear program in the MATLAB universe can be formulated like so:
You seek to find a solution x in R^n which minimizes the objective function f^{T}*x subject to a set of inequality constraints, equality constraints, and each component in x is bounded between a lower and upper bound. Because you want to find the minimum possible value that satisfies the above constraint given, what you're really after is:
Because MATLAB only supports inequalities of less than, you'll need to take the negative of the first two constraints. In addition, MATLAB doesn't support strict inequalities, and so what you'll have to do is enforce a constraint so that you are checking to see if each variable is lesser than a small number, perhaps something like setting a threshold epsilon to 1e-4. Therefore, with the above, your formulation is now:
Note that we don't have any upper or lower bounds as those conditions are already satisfied in the equality and inequality constraints. All you have to do now is plug this problem into linprog. linprog accepts syntax in the following way:
x = linprog(f,A,b,Aeq,beq);
f is a vector of coefficients that work with the objective function, A is a matrix of coefficients that work with the inequality, b is a vector of coefficients that are for the right-hand side of each inequality constraint, and Aeq,beq, are the same as the inequality but for the equality constraints. x would be the solution to the linear programming problem formulated. If we reformulate your problem into matrix form for the above, we now get:
With respect to the linear programming formulation, we can now see what each variable in the MATLAB universe needs to be. Therefore, in MATLAB syntax, each variable becomes:
f = [1; 1];
A = [-1 0; 0 -1];
b = [1e-4; 1e-4];
Aeq = [1 1];
beq = 1;
As such:
x = linprog(f, A, b, Aeq, beq);
We get:
Optimization terminated.
x =
0.5000
0.5000
If linear programming is not what you're looking for, consider looking at MATLAB's MuPAD interface: http://www.mathworks.com/help/symbolic/mupad_ug/solve-algebraic-equations-and-inequalities.html - This more or less mimics what you see in Mathematica if you're more comfortable with that.
Good luck!
Matlab is not a symbolic solver as Mathematica is, so you will not get exact solutions but numeric approximations. Anyway if you are about to solve linear programming (simplex) such as in your example, you should use linprog function.
Related
I have the following matrix
R=(A-C)*inv(A+B-C-C')*(A-C');
where A and B are n by n matrices. I want to find n*n matrix C such that the determinant of R is minimized, SO:
C=arg min (det(R));
Is there any function in MATLAB that can handle this problem?
It seems like you are trying to find the minimum of an unconstrained multivariable function. This can probably be achieved with fminunc
fun = #(x)x(1)*exp(-(x(1)^2 + x(2)^2)) + (x(1)^2 + x(2)^2)/20;
x0 = [1,2];
[x,fval] = fminunc(fun,x0)
Note that there are no examples in the documentation where a matrix is used, this is probably because horrendous performance could be expected when trying to solve this problem for a matrix of any nontiny size. (This is not because of matlab, but because of the nature of the problem).
It is also good to realize that this method does not (cannot) guarantee an optimum, only a local optimum.
I am struggling to solve an optimization problem, numerically, of the following (generic) form.
minimize F(x)
such that:
___(1): 0 < x < 1
___(2): M(x) >= 0.
where M(x) is a matrix whose elements are quadratic functions of x. The last constraint means that M(x) must be a positive semidefinite matrix. Furthermore F(x) is a callable function. For the more curious, here is a similar minimum-working-example.
I have tried a few options, but to no success.
PICOS, CVXPY and CVX -- In the first two cases, I cannot find a way of encoding a minimax problem such as mine. In the third one which is implemented in MATLAB, the matrices involved in a semidefinite constraint must be affine. So my problem falls outside this criteria.
fmincon -- How can we encode a matrix positivity constraint? One way is to compute the eigenvalues of the matrix M(x) analytically, and constraint each one to be positive. But the analytic expression for the eigenvalues can be horrendous.
MOSEK -- The objective function must be a expressible in a standard form. I cannot find an example of a user-defined objective function.
scipy.optimize -- Along with the objective functions and the constraints, it is necessary to provide the derivative of these functions as well. Particularly in my case, that is fine for the objective function. But, if I were to express the matrix positivity constraint (as well as it's derivative) with an analytic expression of the eigenvalues, that can be very tedious.
My apologies for not providing a MWE to illustrate my attempts with each of the above packages/softwares.
Can anyone please suggest a package/software which could be useful to me in solving my optimization problem?
Have a look at a nonlinear optimization package with box constraints, where different type of constraints may be coded via penalty or barrier techniques.
Look at the following URL
merlin.cs.uoi.gr
What is the least computational time consuming way to solve in Matlab the equation:
exp(ax)-ax+c=0
where a and c are constants and x is the value I'm trying to find?
Currently I am using the in built solver function, and I know the solution is single valued, but it is just taking longer than I would like.
Just wanting something to run more quickly is insufficient for that to happen.
And, sorry, but if fzero is not fast enough then you won't do much better for a general root finding tool.
If you aren't using fzero, then why not? After all, that IS the built-in solver you did not name. (BE EXPLICIT! Otherwise we must guess.) Perhaps you are using solve, from the symbolic toolbox. It will be more slow, since it is a symbolic tool.
Having said the above, I might point out that you might be able to improve by recognizing that this is really a problem with a single parameter, c. That is, transform the problem to solving
exp(y) - y + c = 0
where
y = ax
Once you know the value of y, divide by a to get x.
Of course, this way of looking at the problem makes it obvious that you have made an incorrect statement, that the solution is single valued. There are TWO solutions for any negative value of c less than -1. When c = -1, the solution is unique, and for c greater than -1, no solutions exist in real numbers. (If you allow complex results, then there will be solutions there too.)
So if you MUST solve the above problem frequently and fzero was inadequate, then I would consider a spline model, where I had precomputed solutions to the problem for a sufficient number of distinct values of c. Interpolate that spline model to get a predicted value of y for any c.
If I needed more accuracy, I might take a single Newton step from that point.
In the event that you can use the Lambert W function, then solve actually does give us a solution for the general problem. (As you see, I am just guessing what you are trying to solve this with, and what are your goals. Explicit questions help the person trying to help you.)
solve('exp(y) - y + c')
ans =
c - lambertw(0, -exp(c))
The zero first argument to lambertw yields the negative solution. In fact, we can use lambertw to give us both the positive and negative real solutions for any c no larger than -1.
X = #(c) c - lambertw([0 -1],-exp(c));
X(-1.1)
ans =
-0.48318 0.41622
X(-2)
ans =
-1.8414 1.1462
Solving your system symbolically
syms a c x;
fx0 = solve(exp(a*x)-a*x+c==0,x)
which results in
fx0 =
(c - lambertw(0, -exp(c)))/a
As #woodchips pointed out, the Lambert W function has two primary branches, W0 and W−1. The solution given is with respect to the upper (or principal) branch, denoted W0, your equation actually has an infinite number of complex solutions for Wk (the W0 and W−1 solutions are real if c is in [−∞, 0]). In Matlab, lambertw is only implemented for symbolic inputs and thus is very slow method of solving your equation if you're interested in numerical (double precision) solutions.
If you wish to solve such equations numerically in an efficient manner, you might look at Corless, et al. 1996. But, as long as your parameter c is in [−∞, 0], i.e., -exp(c) in [−1/e, 0] and you're interested in the W0 branch, you can use the Matlab code that I wrote to answer a similar question at Math.StackExchange. This code should be much much more efficient that using a naïve approach with fzero.
If your values of c are not in [−∞, 0] or you want the solution corresponding to a different branch, then your solution may be complex-valued and you won't be able to use the simple code I linked to above. In that case, you can more fully implement the function by reading the Corless, et al. 1996 paper or you can try converting the Lambert W to a Wright ω function: W0(z) = ω(log(z)), W−1(z) = ω(log(z)−2πi). In your case, using Matlab's wrightOmega, the W0 branch corresponds to:
fx0 =
(c - wrightOmega(log(-exp(c))))/a
and the W−1 branch to:
fxm1 =
(c - wrightOmega(log(-exp(c))-2*sym(pi)*1i))/a
If c is real, then the above reduces to
fx0 =
(c - wrightOmega(c+sym(pi)*1i))/a
and
fxm1 =
(c - wrightOmega(c-sym(pi)*1i))/a
Matlab's wrightOmega function is also symbolic only, but I have written a double precision implementation (based on Lawrence, et al. 2012) that you can find on my GitHub here and that is 3+ orders of magnitude faster than evaluating the function symbolically. As your problem is technically in terms of a Lambert W, it may be more efficient, and possibly more numerically accurate, to implement that more complicated function for the regime of interest (this is due to the log transformation and the extra evaluation of a complex log). But feel free to test.
I'd like to find the principal components of a data matrix X in Matlab by solving the optimization problem min||X-XBB'||, where the norm is the Frobenius norm, and B is an orthonormal matrix. I'm wondering if anyone could tell me how to do that. Ideally, I'd like to be able to do this using the optimization toolbox. I know how to find the principal components using other methods. My goal is to understand how to set up and solve an optimization problem which has a matrix as the answer. I'd very much appreciate any suggestions or comments.
Thanks!
MJ
The thing about Optimization is that there are different methods to solve a problem, some of which can require extensive computation.
Your solution, given the constraints for B, is to use fmincon. Start by creating a file for the non-linear constraints:
function [c,ceq] = nonLinCon(x)
c = 0;
ceq = norm((x'*x - eye (size(x))),'fro'); %this checks to see if B is orthonormal.
then call the routine:
B = fmincon(#(B) norm(X - X*B*B','fro'),B0,[],[],[],[],[],[],#nonLinCon)
with B0 being a good guess on what the answer will be.
Also, you need to understand that this algorithms tries to find a local minimum, which may not be the solution you ultimately want. For instance:
X = randn(1,2)
fmincon(#(B) norm(X - X*B*B','fro'),rand(2),[],[],[],[],[],[],#nonLinCon)
ans =
0.4904 0.8719
0.8708 -0.4909
fmincon(#(B) norm(X - X*B*B','fro'),rand(2),[],[],[],[],[],[],#nonLinCon)
ans =
0.9864 -0.1646
0.1646 0.9864
So be careful, when using these methods, and try to select a good starting point
The Statistics toolbox has a built-in function 'princomp' that does PCA. If you want to learn (in general, without the optimization toolbox) how to create your own code to do PCA, this site is a good resource.
Since you've specifically mentioned wanting to use the Optimization Toolbox and to set this up as an optimization problem, there is a very well-trusted 3rd-party package known as CVX from Stanford University that can solve the optimization problem you are referring to at this site.
Do you have the optimization toolbox? The documentation is really good, just try one of their examples: http://www.mathworks.com/help/toolbox/optim/ug/brg0p3g-1.html.
But in general the optimization function look like this:
[OptimizedMatrix, OptimizedObjectiveFunction] = optimize( (#MatrixToOptimize) MyObjectiveFunction(MatrixToOptimize), InitialConditionsMatrix, ...optional constraints and options... );
You must create MyObjectiveFunction() yourself, it must take the Matrix you want to optimize as an input and output a scalar value indicating the cost of the current input Matrix. Most of the optimizers will try to minimise this cost. Note that the cost must be a scalar.
fmincon() is a good place to start, once you are used to the toolbox you and if you can you should choose a more specific optimization algorithm for your problem.
To optimize a matrix rather than a vector, reshape the matrix to a vector, pass this vector to your objective function, and then reshape it back to the matrix within your objective function.
For example say you are trying to optimize the 3 x 3 matrix M. You have defined objective function MyObjectiveFunction(InputVector). Pass M as a vector:
MyObjectiveFunction(M(:));
And within the MyObjectiveFunction you must reshape M (if necessary) to be a matrix again:
function cost = MyObjectiveFunction(InputVector)
InputMatrix = reshape(InputVector, [3 3]);
%Code that performs matrix operations on InputMatrix to produce a scalar cost
cost = %some scalar value
end
I have a system of 2 equations in 2 unknowns that I want to solve using MATLAB but don't know exactly how to program. I've been given some information about a gamma distribution (mean of 1.86, 90% interval between 1.61 and 2.11) and ultimately want to get the mean and variance. I know that I could use the normal approximation but I'd rather solve for A and B, the shape and scale parameters of the gamma distribution, and find the mean and variance that way. In pseudo-MATLAB code I would want to solve this:
gamcdf(2.11, A, B) - gamcdf(1.61, A, B) = 0.90;
A*B = 1.86;
How would you go about solving this? I have the symbolic math toolbox if that helps.
The mean is A*B. So can you solve for perhaps A in terms of the mean(mu) and B?
A = mu/B
Of course, this does no good unless you knew B. Or does it?
Look at your first expression. Can you substitute?
gamcdf(2.11, mu/B, B) - gamcdf(1.61, mu/B, B) = 0.90
Does this get you any closer? Perhaps. There will be no useful symbolic solution available, except in terms of the incomplete gamma function itself. How do you solve a single equation numerically in one unknown in matlab? Use fzero.
Of course, fzero looks for a zero value. But by subtracting 0.90, that is resolved.
Can we define a function that fzero can use? Use a function handle.
>> mu = 1.86;
>> gamfun = #(B) gamcdf(2.11, mu/B, B) - gamcdf(1.61, mu/B, B) - 0.90;
So try it. Before we do that, I always recommend plotting things.
>> ezplot(gamfun)
Hmm. That plot suggests that it might be difficult to find a zero of your function. If you do try it, you will find that good starting values for fzero are necessary here.
Sorry about my first try. Better starting values for fzero, plus some more plotting does give a gamma distribution that yields the desired shape.
>> B = fzero(gamfun,[.0000001,.1])
B =
0.0124760672290871
>> A = mu/B
A =
149.085442218805
>> ezplot(#(x) gampdf(x,A,B))
In fact this is a very "normal", i.e, Gaussian, looking curve.