I am working on pure LP problem and using Multi-Objective to solve it. When I am using objective with weights:
solver1.minimize( A* 1000
+ B* 10000
+ C* 100
+ D* 1
)
Solver is able to get the optimal result successfully
But when I am using ObjectiveSense package with objective:
solver1.set_multi_objective(ObjectiveSense.Minimize, [A, B, C,D]
, priorities=[1, 2, 0, 1], weights=[1000,1,1,1])
Solver is getting infeasible after first hierarchy of solve and give: "Non-optimal status: infeasibleCPLEX Error 1300: Failure to solve multi-objective subproblem."
I am trying to figure out how to use parameters reltol and abstol to get a feasible solution. Any ideas or examples?
Also trying to understand why Multi-Objective is going infeasible though we have solution through weighted objective? Any input about same will help.
Thanks!
This page
https://www.ibm.com/docs/en/icos/12.10.0?topic=optimization-solving-multiple-objective-problems
explains in detail how tolerances are used to solve multi objective problems.
To summarize, two different algorithms are used for LP and MIP.
The LP algorithm uses LP bases, for which absolute tolerance is interpreted as a limit for reduced costs, and relative costs are not used.
The MIP algorithm uses an iterative solve approach, for which tolerances are interpreted as tolerances on each intermediate sub-objective.
We have seen cases where the LP algorithm may fail, when the model has difficult numerical properties. To check this case, run the following line on your model:
The range of coefficients should not exceed 1e+6.
print(mdl.cplex_matrix_stats)
As a workaround, you can try the Model.solve_with_goals method in Docplex, which runs the iterative multi objective algorithm for any kind of model.
Related
I am working on Linear Programming Problem with 800K Constraints and the problem takes 20 mins to solve but if I solve the problem for half horizon it just takes 1 min. Is there a way in DoCPLEX where I can solve for partial horizon and then use the solution to solve for other half of the problem without using a for-loop
Three suggestions:
load your problem as LP or SAV into cplex interactive optimizer and run display problem stats. This might show (or rule out) precision issues (ill-conditioned problem). Also it will output number of nonzeros
set datacheck parameters to 2, this might detect numerical issues in data
have you tried different LP algorithms? Using the lpmethod parameter you could try primal, dual or barrier algorithm to see whether one runs faster on your problem.
Reference:
https://www.ibm.com/support/knowledgecenter/SSSA5P_12.10.0/ilog.odms.cplex.help/CPLEX/Parameters/topics/LPMETHOD.html
In DOcplex:
model.parameters.datacheck = 2
model.parameters.lpmethod = 4 # for barrier
From your answers, I can think of the following:
if you are in pure LP (is this true?) I see no point in rounding numbers (but yes, that would help in a MIP, try rounding coefficients whose fractional part is say less than 1e-7: 4.0000001 -> 4)
1e+14 conditioning denotes serious modeling issue: a common source is mixing different objectives with coefficients. Have you tried multi-objective to avoid that?
Another source is big_M formulations, to which you should prefer indicator constraints. If you are not in these two cases, then try to renormalize the data to keep in a smaller condition range...
Finally, you might try setting markowitz tolerance to 0.99, to add extra cautiouness in simplex factorizations, but behavior may vary from one dataset to the other...
Hello stackoverflow community,
I have troubles in understanding a least-square-error-problem in the c++ armadillo package.
I have a matrix A with many more rows than columns (5000 to 100 for example) so it is overdetermined.
I want to find x so that A*x=b gives me the least square error.
If i use the solve function of armadillo on my data like "x = Solve(A,b)" the error of "(A*x-b)^2" is sometimes way to high.
If on the other hand I solve for x with the analytical form by "x = (A^T * A)^-1 *A^T * b" the results are always right.
The results for x in both cases can differ by 10 magnitudes.
I had thought that armadillo would use this analytical form in the background if the system is overdetermined.
Now I would like to understand why these two methods give such different results.
I wanted to give a short example program, but i can't reproduce this behavior with a short program.
I thought about giving the Matrix here, but with 5000 times 100 it's also very big. I can deliver the values for which this happens though if needed.
So as a short background.
The matrix I get from my program is a numerically solved reaction of a nonlinear oscillator in which I put information inside by wiggeling a parameter of this system.
Because the influence of this parameter on the system is small, the values of my different rows are very similar but never the same, otherwise armadillo should throw an error.
I'm still thinking that this is the problem, but the solve function never threw any error.
Another thing that confuses me is that in a short example program with a random matrix, the analytical form is waaay slower than the solve function.
But on my program, both are nearly identically fast.
I guess this has something to do with the numerical convergence of the pseudo inverse and the special case of my matrix, but for that i don't know enough about how armadillo works.
I hope someone can help me with that problem and thanks a lot in advance.
Thanks for the replies. I think i figured the problem out and wanted to give some feedback for everybody who runs into the same problem.
The Armadillo solve function gives me the x that minimizes (A*x-b)^2.
I looked at the values of x and they are sometimes in the magnitude of 10^13.
This comes from the fact that the rows of my matrix only slightly change. (So nearly linear dependent but not exactly).
Because of that i was in the numerical precision of my doubles and as a result my error sometimes jumped around.
If i use the rearranged analytical formular (A^T * A)*x = *A^T * b with the solve function this problem doesn't occur anymore because the fitted values of x are in the magnitude of 10^4. The least square error is a little bit higher but that is okay, as i want to avoid overfitting.
I now additionally added Tikhonov regularization by solving (A^T * A + lambda*Identity_Matrix)*x = *A^T * b with the solve function of armadillo.
Now the weight vectors are in the order of around 1 and the error nearly doesn't change compared to the formular without regularization.
I have to find all basic solutions of some tiny linear-programming problems.
Here's an example (in lp_solve format):
max: x1 + x2;
x1 + x2 <= 1;
x1 <= 0.8;
x2 <= 0.8;
All 2 basic solutions:
x1 = 0.2, x2 = 0.8
x1 = 0.8, x2 = 0.2
Of course there is a way of finding alternative solutions, but I really prefer using existing libraries instead of crafting my own simplex code.
I'm using Python as my programming language, and hoping there's some method in lp_solve or GLPK's C API can do this.
Thanks.
There is no routine to do that with glpk; and IMHO it is very unlikely that any real-world solver implements something like that, since it is not very useful in practise and it is certainly not a simple problem.
What is indeed easy to find one other basic solution once you reached optimality with the simplex algorithm, which does not mean that it is easy to list them all.
Consider a LP whose domain has dimension n; the set S of the optimal solutions is a convex polyhedron whose dimension m can be anything from 0 to n-1.
You want a method to list all the basic solutions of the problem, that is all the vertices of S: as soon as m is greater than 2, you will need to carefully avoid cycling when you move from one basic solution to another.
However, there is (luckily!) no need to write your own simplex code: you can access the internals of the current basis with the glpk library, and probably with lpsolve too.
Edit: two possible solutions
The better way would be to use another library such as PPL for this.
Assume that you have a problem of the form:
min cx; subject to: Ax <= b
First solve your problem with glpk, this will give you the optimal value V of the problem. From this point, you can use PPL to get the description of the polyedron of optimal values:
cx = V and Ax <= b
as the convex hull of its extreme points, which correspond to the BFSs you are looking for.
You can (probably) use the glpk simplex routines. Once you get an optimal BFS, you can get the reduced cost associated with all non-basic columns using the routine glp_get_row_dual (the basis status of the variable can be obtained with glp_get_row_stat), so you can find a non-basic variables with a null reduced cost. Then, I think that you can use function glp_set_row_stat to change the basis status of this column in order to have it enter the basis.
(And then, you can iterate this process as long as you avoid cycling.)
Note that I did not try any of those solutions myself; I think that the first one is by far the best, although it will require that you learn the PPL API. If you wish to go for the second one, I strongly suggest that you send an e-mail to the glpk maintainer (or look at the source code), because I am really not sure it will work as is.
Here's the problem:
I am currently trying to create a control system which is required to find a solution to a series of complex linear equations without a unique solution.
My problem arises because there will ever only be six equations, while there may be upwards of 20 unknowns (usually way more than six unknowns). Of course, this will not yield an exact solution through the standard Gaussian elimination or by changing them in a matrix to reduced row echelon form.
However, I think that I may be able to optimize things further and get a more accurate solution because I know that each of the unknowns cannot have a value smaller than zero or greater than one, but it is free to take on any value in between them.
Of course, I am trying to create code that would find a correct solution, but in the case that there are multiple combinations that yield satisfactory results, I would want to minimize Sum of (value of unknown * efficiency constant) over all unknowns, i.e. Sigma[xI*eI] from I=0 to n, but finding an accurate solution is of a greater priority.
Performance is also important, due to the fact that this algorithm may need to be run several times per second.
So, does anyone have any ideas to help me on implementing this?
Edit: You might just want to stick to linear programming with equality and inequality constraints, but here's an interesting exact solution that does not incorporate the constraint that your unknowns are between 0 and 1.
Here's a powerpoint discussing your problem: http://see.stanford.edu/materials/lsoeldsee263/08-min-norm.pdf
I'll translate your problem into math to make things a bit easier to figure out:
you have a 6x20 matrix A and a vector x with 20 elements. You want to minimize (x^T)e subject to Ax=y. According to the slides, if you were just minimizing the sum of x, then the answer is A^T(AA^T)^(-1)y. I'll take another look at this as soon as I get the chance and see what the solution is to minimizing (x^T)e (ie your specific problem).
Edit: I looked in the powerpoint some more and near the end there's a slide entitled "General norm minimization with equality constraints". I am going to switch the notation to match the slide's:
Your problem is that you want to minimize ||Ax-b||, where b = 0 and A is your e vector and x is the 20 unknowns. This is subject to Cx=d. Apparently the answer is:
x=(A^T A)^-1 (A^T b -C^T(C(A^T A)^-1 C^T)^-1 (C(A^T A)^-1 A^Tb - d))
it's not pretty, but it's not as bad as you might think. There's really aren't that many calculations. For example (A^TA)^-1 only needs to be calculated once and then you can reuse the answer. And your matrices aren't that big.
Note that I didn't incorporate the constraint that the elements of x are within [0,1].
It looks like the solution for what I am doing is with Linear Programming. It is starting to come back to me, but if I have other problems I will post them in their own dedicated questions instead of turning this into an encyclopedia.
I am trying to solve a minimisation problem and I want to minimise an expression
a/b
Where both a & b are variables. Hence this is not a linear problem...
How can I transform this function into an other one (being a linear one).
There is a detailed section on how to handle ratios in Linear Programming on the lpsolve site. It should be general enough to apply to AMPL and CPLEX as well.
There are several ways to do this, but the simplest to explain requires that you solve a series of linear programs. First, remove the objective and add a constraint
a <= c * b
Where c is a known upper bound on the solution. Then do a binary search on c you can a range where c_l, c_u where the problem is infeasible for
a <= c_l * b
but feasible for
a <= c_u * b
The general form of the obj should be a linear fractional function, something like f_{0}(x)=(c^Tx+d)/(e^Tx+f). For your case, X=(a,b),c=(1,0),(e=0,1),d=f=0.
To solve this kind of opt, something called linear fractional programming can be used. it's like linear constrainted version of linear fractional function and Charnes-Cooper transformation is applied to transform into a LP. You can find the main idea from wiki. Many OR books talk more about this such as pp53, pp165 in the Boyd's "convex optimization" (free to download).