I need to solve the following SOCP in Matlab:
argmin_x ||R*x||_2 s.t. s^H * x = 1 and ||x||_2 < d,
where x is an Nx1 vector and R is an MxN matrix.
CVX can solve this type of problem. However, CVX requires me to give R and does not allow me to instead give a function handle that will return R*x. This is a problem for me since once R becomes large, computing R*x directly takes too long. There exists an efficient algorithm for computing R*x that I would like to take advantage of, so I am hoping that there is another SOCP solver that I could use.
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.
Here I've a transfer function matrix with size of (3*7):
G = [G11,G12,G13,G14,G15,G16,G17;
G21,G22,G23,G24,G25,G26,G27;
G31,G32,G33,G34,G35,G36,G37]
Is it possible to get A = [G*(G^(-1))T] symbolically in Matlab:
Where :
G^(-1) = inv(G) and (G^(-1))T = transpose of (inv(G))
Yeah it's possible but it may take a lot of time, and also it's possible that your computer runs out of memory. Matlab's symbolic operations are not very good, but here's the solution. First define the elements of your matrix as symbolic variables. syms G11 defines G11 as symbolic. than define your G matrix and go on. Than you can find the A matrix.
I should also mention that since your matrix is 3*7, I don't know if matlab have inverse command for non-square matrix, but you can have pseudo inverse. And if you want to do symbolic computations, Maple and mathematica are so much better. But matlab is better in Numerical computations .
I have the following differential equation which I'm not able to solve.
We know the following about the equation:
D(r) is a third grade polynom
D'(1)=D'(2)=0
D(2)=2D(1)
u(1)=450
u'(2)=-K * (u(2)-Te)
Where K and Te are constants.
I want to approximate the problem using a matrix and I managed to solve
the similiar equation: with the same limit conditions for u(1) and u'(2).
On this equation I approximated u' and u'' with central differences and used a finite difference method between r=1 to r=2. I then placed the results in a matrix A in matlab and the limit conditions in the vector Y in matlab and ran u=A\Y to get how the u value changes. Heres my matlab code for the equation I managed to solve:
clear
a=1;
b=2;
N=100;
h = (b-a)/N;
K=3.20;
Ti=450;
Te=20;
A = zeros(N+2);
A(1,1)=1;
A(end,end)=1/(2*h*K);
A(end,end-1)=1;
A(end,end-2)=-1/(2*h*K);
r=a+h:h:b;
%y(i)
for i=1:1:length(r)
yi(i)=-r(i)*(2/(h^2));
end
A(2:end-1,2:end-1)=A(2:end-1,2:end-1)+diag(yi);
%y(i-1)
for i=1:1:length(r)-1
ymin(i)=r(i+1)*(1/(h^2))-1/(2*h);
end
A(3:end-1,2:end-2) = A(3:end-1,2:end-2)+diag(ymin);
%y(i+1)
for i=1:1:length(r)
ymax(i)=r(i)*(1/(h^2))+1/(2*h);
end
A(2:end-1,3:end)=A(2:end-1,3:end)+diag(ymax);
Y=zeros(N+2,1);
Y(1) =Ti;
Y(2)=-(Ti*(r(1)/(h^2)-(1/(2*h))));
Y(end) = Te;
r=[1,r];
u=A\Y;
plot(r,u(1:end-1));
My question is, how do I solve the first differential equation?
As TroyHaskin pointed out in comments, one can determine D up to a constant factor, and that constant factor cancels out in D'/D anyway. Put another way: we can assume that D(1)=1 (a convenient number), since D can be multiplied by any constant. Now it's easy to find the coefficients (done with Wolfram Alpha), and the polynomial turns out to be
D(r) = -2r^3+9r^2-12r+6
with derivative D'(r) = -6r^2+18r-12. (There is also a smarter way to find the polynomial by starting with D', which is quadratic with known roots.)
I would probably use this information right away, computing the coefficient k of the first derivative:
r = a+h:h:b;
k = 1+r.*(-6*r.^2+18*r-12)./(-2*r.^3+9*r.^2-12*r+6);
It seems that k is always positive on the interval [1,2], so if you want to minimize the changes to existing code, just replace r(i) by r(i)/k(i) in it.
By the way, instead of loops like
for i=1:1:length(r)
yi(i)=-r(i)*(2/(h^2));
end
one usually does simply
yi=-r*(2/(h^2));
This vectorization makes the code more compact and can benefit the performance too (not so much in your example, where solving the linear system is the bottleneck). Another benefit is that yi is properly initialized, while with your loop construction, if yi happened to already exist and have length greater than length(r), the resulting array would have extraneous entries. (This is a potential source of hard-to-track bugs.)
I want to solve, in MatLab, a linear system (corresponding to a PDE system of two equations written in finite difference scheme). The action of the system matrix (corresponding to one of the diffusive terms of the PDE system) reads, symbolically (u is one of the unknown fields, n is the time step, j is the grid point):
and fully:
The above matrix has to be intended as A, where A*U^n+1 = B is the system. U contains the 'u' and the 'v' (second unknown field of the PDE system) alternatively: U = [u_1,v_1,u_2,v_2,...,u_J,v_J].
So far I have been filling this matrix using spdiags and diag in the following expensive way:
E=zeros(2*J,1);
E(1:2:2*J) = 1;
E(2:2:2*J) = 0;
Dvec=zeros(2*J,1);
for i=3:2:2*J-3
Dvec(i)=D_11((i+1)/2);
end
for i=4:2:2*J-2
Dvec(i)=D_21(i/2);
end
A = diag(Dvec)*spdiags([-E,-E,2*E,2*E,-E,-E],[-3,-2,-1,0,1,2],2*J,2*J)/(dx^2);`
and for the solution
[L,U]=lu(A);
y = L\B;
U(:) =U\y;
where B is the right hand side vector.
This is obviously unreasonably expensive because it needs to build a JxJ matrix, do a JxJ matrix multiplication, etc.
Then comes my question: is there a way to solve the system without passing MatLab a matrix, e.g., by passing the vector Dvec or alternatively directly D_11 and D_22?
This would spare me a lot of memory and processing time!
Matlab doesn't store sparse matrices as JxJ arrays but as lists of size O(J). See
http://au.mathworks.com/help/matlab/math/constructing-sparse-matrices.html
Since you are using the spdiags function to construct A, Matlab should already recognize A as sparse and you should indeed see such a list if you display A in console view.
For a tridiagonal matrix like yours, the L and U matrices should already be sparse.
So you just need to ensure that the \ operator uses the appropriate sparse algorithm according to the rules in http://au.mathworks.com/help/matlab/ref/mldivide.html. It's not clear whether the vector B will already be considered sparse, but you could recast it as a diagonal matrix which should certainly be considered sparse.
I'm new to matlab and try to do some energy minimization work with it. The energy function takes a 3-channel image as input. For every channel, there's a energy term looks like this:
E = x'Ax + ||Bx||^2 + w*||x-c||^2,
where x,c are vectors of length N, A is a matrix of size N*N. A is sparse and positive semi-definite and has 25 non-zero elements per row, giving constraints to all elements of x. B is of size M*N. B is sparse too and has 2 non-zero elements per row. N is about 850,000. M is about 1,000,000. Although B gives more than N constraints, some elements of x have nothing to do with ||Bx||^2 term. The weight w of term ||x-c||^ is quite small, say 1e-3.
I've searched matlab documentation. It looks like I should use lsqnonlin for this problem. Is there a special designed function or option for quadratic form minimization in matlab?
For those who are familiar with computer vision literature, I'm actually trying to implement the algorithm in "Coherent Intrinsic Images from Photo Collections". The authors said they use matlab backslash operator to minimize the energy, but I can't see how a backslash operator can be used in quadratic form problem.
Yes, there is a function specifically for optimizing quadratic cost functions: quadprog. However, if you don't have any linear constraints, then you should be able to write your cost function as
E = x'Mx/2 + vx + k
Finding the point of zero gradient (hopefully a minimum) can then be achieved by taking first derivatives:
dE/dx = Mx + v
setting them to zero giving the solution:
x = -M\v