I am using CBC solver in Pulp.
I have an MILP where I know a feasible solution exists.
Yet, I received a "Solution is not feasible" error.
I added a constraint "var_1 == var_2", which is part of the feasible solution I know exists and ta-dam, optimal solution is found.
How can this be?
Shouldn't the solver be able to always find a solution if one exists
I will note that all variables are either integers bounded between [1,50] or binary variables.
Thank you
Related
Given a matrix's LDLT decomposition, I would like to modify the diagonal - for example floor all the values. Is there a way to do this with eigen?
To be clear, I can do:
auto ldlt_ = matrix.ldlt();
and I would like to follow up with:
ldlt_.vectorD().cwiseMax(Vector::Constant(n,epsilon))
before solving a problem:
ldlt_.solve(a)
I don't see any non const accessors to the vectorD member - what am I missing?
No, you cannot do that, and I don't think that's a good idea to increase small (or negative) diagonal entries this way. If there are too small entries, the usual approach is either to ignore them (default behavior of LDLT::solve), or to redo the factorization with matrix+eps*I. Anyway, if you really want to tweak D yourself, then you have to implement your own solve function.
Hello all :) I'm pretty new to Optimization and barely understand it (was about ready to slit my wrist after figuring out how to write Objective Functions without any formal learning on the matter), and need a little help on a work project.
How would I go about setting a logical constraint when using the Optimization Toolbox, fmincon specifically (using Trust Region Reflective algorithm)?
I am optimizing 5 values (lets call it matrix OptMat), and I want to optimize with the constraint such that
max(OptMat)/min(OptMat) > 10
I assume this will optimize the 5 values of OptMat as low as possible, while keeping the above constraint in mind so that if a set of values for OptMat is found with a lower OF in which it breaks the constraint it will NOT report those values and instead report the next lowest OF where OptMat values meet the above constraint?
For the record, my lower bounds are [0,0,0,0,0]. I'm not sure how to enter it into upper bounds as it only accepts doubles and that would be logical. I tried the Active Set Algorithm and that enabled the Nonlinear Constraint Function box and I think I'm on the right track with that. If so, I'm not sure what the syntax for entering my desired constraint. Another method^that ^may ^or ^may ^not ^work I could think of is using this as an Upper Boundary.
[min(OptMat)*10, min(OptMat)*10, min(OptMat)*10, min(OptMat)*10, min(OptMat)*10]
Again, I'm using the GUI Optimization Toolbox. I haven't looked too much into command line optimization (though I will need to write it command line eventually) and I think I read somewhere that you can set the Upper Boundary and it does not have to be double?
Thank you so very much for the help, if someone is able. I apologize if this is a really nooby question.
What you are looking for are nonlinear constraints, fmincon can handle it (I only know the command, not the GUI) with the argument nonlcon. For more information look at this guide http://de.mathworks.com/help/optim/ug/fmincon.html
How would you implement this? First create a function
function [c, ceq] = mycondition(x)
c = -max(x)/min(x)/10;
ceq = 0;
I had to change the equation to match the correct formalism, i.e. c(x)<=0 is needed.
Maybe you could also create an anonymous function, I'm not sure (http://de.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html).
Then use this function to feed the fmincon function using the # sign, i.e. at the specific location write
fmincon(...., #mycondition, ...)
I am working on solving inequality problems using prolog.I have found a code and it solves ax+b>=0 type equations.
The code I have used is as follows.
:-use_module(library(clpr)).
dec_inc(left,right):-
copy_term(left-right,Copyleft-Copyright).
tell_cs(Copyleft).
max(Copyright,right,Leq).
tell_cs(Leq).
max([],[],[]).
max([E=<_|Ps],[_=<P1|P1s],[K=<P1|Ls]):-
sup(E,K),
max(Ps,P1s,Ls).
tell_cs([]).
tell_cs([C|Cs]):-
{C},
tell_cs(Cs).
for example
when we give {2*X+2>=5}. it gives the correct answer. {X>=1.5}.
2.But if I enter fraction like {(X+3)/(3*X+1)>=1}. it gives {1- (3+X)/ (1+3.0*X)=<0.0}.
How can I solve this type of inequality questions to find the final answer.(questions which include fractions).
Please help me.
If there is any learning material I can refer please let me know.
library(clpr) documentation advises that it deals with non linear constraints only passively, so you're out of luck. You need a more sophisticated algebra system.
I am studing the algorithm given here, and
somewhere it is claimed that it is efficent and always give correct result.
But, I try to run the algorithm and it is not giving me correct or efficent output for the following patterns.
For 5 x 5 grid, where (n) is light number and 0/1 state whethere the light is on/off, 1 ON and 0 OFF.
(1)1 (2)0 (3)0 (4)0 (5)0 the output should be 1,7,13,19,25(Pressing this light will make the full grid OFF. But what I am getting is this
(6)0 (7)1 (8)0 (9)0 (10)0 3,5,6,7,8,10,13,16,18,19,20,21,23.
(11)0 (12)0 (13)1 (14)0 (15)0
(16)0 (17)0 (18)0 (19)1 (20)0
(21)0 (22)0 (23)0 (24)0 (25)1
While for some pattern it is giving me correct output as below.
(1)0 (2)0 (3)0 (4)0 (5)1 the output should be 5,9,13,17,21, and the algorithm is giving me correct result.
(6)0 (7)0 (8)0 (9)1 (10)0
(11)0 (12)0 (13)1 (14)0 (15)0
(16)0 (17)1 (18)0 (19)0 (20)0
(21)1 (22)0 (23)0 (24)0 (25)0
If somebody need a code let me know I can post it.
Can please somebody let me know if this methods will always give correct as well as efficient result or not ?
(I'm the author of the code you linked to.) To the best of my knowledge, the code is correct (and I'm sure that the high-level algorithm of using Gaussian elimination over GF(2) is correct). The solution it produces is guaranteed to solve the puzzle, though it's not necessarily the minimal number of button presses. The "efficiency" I was referring to in the writeup refers to the time complexity of solving the puzzle overall (it can solve a Lights Out grid in polynomial time, as opposed to the exponential-time brute-force solution of trying all possible combinations) rather than to the "efficiency" of the generated solution.
I actually don't know any efficient algorithms for finding a solution requiring the minimum number of button presses. Let me know if you find one!
Hope this helps!
I want to use a least squares problem with the use of Eigen library.
My options are 2,
sysAAA.jacobiSvd( Eigen::ComputeThinU | Eigen::ComputeThinV ).solve( sysBBB )
sysAAA.colPivHouseholderQr().solve( sysBBB );
I was using the first in the beginning, but it proved to be very slow (1)(2).
So I went to the second solution (other methods aren't appropriate for my case, because they require special matrices (2) )
Does colPivHouseholderQr().solve give a least squares solution?
My impression is that it doesn't (3), but I want to be sure before looking for a "workaround".
http://forum.kde.org/viewtopic.php?f=74&t=102088
http://eigen.tuxfamily.org/dox/TopicLinearAlgebraDecompositions.html
http://eigen.tuxfamily.org/dox/TutorialLinearAlgebra.html#TutorialLinAlgLeastsquares
Yes, ColPivHouseholderQr::solve() computes a least-square solution.