I have a linear program of the form min(f*x) s.t. A1*x < d1; A2*x < d2. The form with one constraint is implemented in Matlab in command linprog. What command can I use to solve linear program with two constrraints?
I could of course create a block diagonal matrix, and double the size of the variable x, but if there is more efficient way I would like to use it, because the size of the matrix is quite large.
Possibly I don't understand the question right but can't you combine the matrixes A1 und A2 by A = [A1; A2]?
You maybe interested in Dantzig-Wolfe Decomposition algorithm for solving linear programming. It takes advantage of this block diagonal structure. However, I don't think there is an out-of-the box implementation of it in commercial softwares.
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I have a linear system of about 2000 sparse equations in Matlab. For my final result, I only really need the value of one of the variables: the other values are irrelevant. While there is no real problem in simply solving the equations and extracting the correct variable, I was wondering whether there was a faster way or Matlab command. For example, as soon as the required variable is calculated, the program could in principle stop running.
Is there anyone who knows whether this is at all possible, or if it would just be easier to keep solving the entire system?
Most of the computation time is spent inverting the matrix, if we can find a way to avoid completely inverting the matrix then we may be able to improve the computation time. Lets assume I'm only interested in the solution for the last variable x(N). Using the standard method we compute
x = A\b;
res = x(N);
Assuming A is full rank, we can instead use LU decomposition of the augmented matrix [A b] to get x(N) which looks like this
[~,U] = lu([A b]);
res = U(end,end-1)/U(end,end);
This is essentially performing Gaussian elimination and then solving for x(N) using back-substitution.
We can extend this to find any value of x by swapping the columns of A before LU decomposition,
x_index = 123; % the index of the solution we are interested in
A(:,[x_index,end]) = A(:,[end,x_index]);
[~,U] = lu([A b]);
res = U(end,end)/U(end,end-1);
Bench-marking performance in MATLAB2017a with 10,000 random 200 dimensional systems we get a slight speed-up
Total time direct method : 4.5401s
Total time LU method : 3.9149s
Note that you may experience some precision issues if A isn't well conditioned.
Also, this approach doesn't take advantage of the sparsity of A. In my experiments even with 2000x2000 sparse matrices everything significantly slowed down and the LU method is significantly slower. That said full matrix representation only requires about 30MB which shouldn't be a problem on most computers.
If you have access to theory manuals on NASTRAN, I believe (from memory) there is coverage of partial solutions of linear systems. Also try looking for iterative or tri diagonal solvers for A*x = b. On this page, review the pqr solution answer by Shantachhani. Another reference.
I have a linear equation such as
Ax=b
where A is full rank matrix which its size is 512x512. b is a vector of 512x1. x is unknown vector. I want to find x, hence, I have some options for doing that
1.Using the normal way
inv(A)*b
2.Using SVD ( Singular value decomposition)
[U S V]=svd(A);
x = V*(diag(diag(S).^-1)*(U.'*b))
Both methods give the same result. So, what is benefit of using SVD to solve Ax=b, especially in the case A is a 2D matrix?
Welcome to the world of numerical methods, let me be your guide.
You, as a new person in this world wonders, "Why would I do something this difficult with this SVD stuff instead of the so commonly known inverse?! Im going to try it in Matlab!"
And no answer was found. That is, because you are not looking at the problem itself! The problems arise when you have an ill-conditioned matrix. Then the computing of the inverse is not possible numerically.
example:
A=[1 1 -1;
1 -2 3;
2 -1 2];
try to invert this matrix using inv(A). Youll get infinite.
That is, because the condition number of the matrix is very high (cond(A)).
However, if you try to solve it using SVD method (b=[1;-2;3]) you will get a result. This is still a hot research topic. Solving Ax=b systems with ill condition numbers.
As #Stewie Griffin suggested, the best way to go is mldivide, as it does a couple of things behind it.
(yeah, my example is not very good because the only solution of X is INF, but there is a way better example in this youtube video)
inv(A)*b has several negative sides. The main one is that it explicitly calculates the inverse of A, which is both time demanding, and may result in inaccuracies if values vary by many orders of magnitude.
Although it might be better than inv(A)*b, using svd is not the "correct" approach here. The MATLAB-way to do this is using mldivide, \. Using this, MATLAB chooses the best algorithm to solve the linear system based on its properties (Hermation, upper Hessenberg, real and positive diagonal, symmetric, diagonal, sparse etc.). Often, the solution will be a LU-triangulation with partial permutation, but it varies. You'll have a hard time beating MATLABs implementation of mldivide, but using svd might give you some more insight of the properties of the system if you actually investigates U, S, V. If you don't want to do that, do with mldivide.
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 have a linear system of equations AX = B to solve in MATLAB. What I have known is A is sparse, positive-definite and symmetric. I know the command x = A \ b works yet I am not sure MATLAB takes full advantage of A's good properties so as to maximize the efficiency. Is there any way to specify the algorithm to solve it, for example Conjugate Gradient algorithm in MATLAB?
If your matrix is sparse, you can use all these iterative functions, for example bicg for a biconjugate gradients method.
MATLAB's mldivide operator does indeed take advantage of properties of A. See the documentation for details - expand the "Algorithm" section.
from a simulation problem, I want to calculate complex square matrices on the order of 1000x1000 in MATLAB. Since the values refer to those of Bessel functions, the matrices are not at all sparse.
Since I am interested in the change of the determinant with respect to some parameter (the energy of a searched eigenfunction in my case), I overcome the problem at the moment by first searching a rescaling factor for the studied range and then calculate the determinants,
result(k) = det(pre_factor*Matrix{k});
Now this is a very awkward solution and only works for matrix dimensions of, say, maximum 500x500.
Does anybody know a nice solution to the problem? Interfacing to Mathematica might work in principle but I have my doubts concerning feasibility.
Thank you in advance
Robert
Edit: I did not find a convient solution to the calculation problem since this would require changing to a higher precision. Instead, I used that
ln det M = trace ln M
which is, when I derive it with respect to k
A = trace(inv(M(k))*dM/dk)
So I at least had the change of the logarithm of the determinant with respect to k. From the physical background of the problem I could derive constraints on A which in the end gave me a workaround valid for my problem. Unfortunately I do not know if such a workaround could be generalized.
You should realize that when you multiply a matrix by a constant k, then you scale the determinant of the matrix by k^n, where n is the dimension of the matrix. So for n = 1000, and k = 2, you scale the determinant by
>> 2^1000
ans =
1.07150860718627e+301
This is of course a huge number, so you might expect that it should fail, since in double precision, MATLAB will only represent floating point numbers as large as realmax.
>> realmax
ans =
1.79769313486232e+308
There is no need to do all the work of recomputing that determinant, not that computing the determinant of a huge matrix like that is a terribly well-posed problem anyway.
If speed is not a concern, you may want to use det(e^A) = e^(tr A) and take as A some scaling constant times your matrix (so that A - I has spectral radius less than one).
EDIT: In MatLab, the log of a matrix (logm) is calculated via trigonalization. So it is better for you to compute the eigenvalues of your matrix and multiply them (or better, add their logarithm). You did not specify whether your matrix was symmetric or not: if it is, finding eigenvalues are easier than if it is not.
You said the current value of the determinant is about 10^-300.
Are you trying to get the determinant at a certain value, say 1? If so, rescaling is awkward: the matrix you are considering is ill-conditioned, and, considering the precision of the machine, you should consider the output determinant to be zero. It is impossible to get a reliable inverse in other words.
I would suggest to modify the columns or lines of the matrix rather than rescale it.
I used R to make a small test with a random matrix (random normal values), it seems the determinant should be clearly non-zero.
> n=100
> M=matrix(rnorm(n**2),n,n)
> det(M)
[1] -1.977380e+77
> kappa(M)
[1] 2318.188
This is not strictly a matlab solution, but you might want to consider using Mahout. It's specifically designed for large-scale linear algebra. (1000x1000 is no problem for the scales it's used to.)
You would call into java to pass data to/from Mahout.